Pure Mathematics (Volumes 1 & 2)

PURE MATHEMATICS 1 S. L. PARSONSON Senior Mathematics Master Harrow School CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE LONDON...

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PURE MATHEMATICS 1 S. L. PARSONSON Senior Mathematics Master Harrow School

CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE LONDON • NEW YORK • MELBOURNE

Published by the Syndics of the Cambridge University Press The Pitt Building, Trumpington Street, Cambridge CB2 IRP Bentley House, 200 Euston Road, London NW1 2DB 32 East 57th Street, New York, NY 10022, USA 296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia © Cambridge University Press 1970 Library of Congress catalogue card number: 70-100026 ISBN: 0 521 07683 8 First published 1970 Reprinted 1972 1974 1975 1977 PRINTED IN GREAT BRITAIN AT THE UNIVERSITY PRESS, CAMBRIDGE

Contents

Preface 1 Numbers and inequalities

page v 1

2 Vectors and vector geometry

24

3 Coordinates

45

4 Polynomials

68

5 Functions and inequalities

78

6 The trigonometric functions

87

7 Probability in finite outcome spaces

103

8 Finite series and the binomial theorem

130

Revision exercise A 9 Mathematical induction

151 155

10 Expectation

169

11 Further vectors

189

12 Further trigonometry

215

13 Matrices 1

238

14 Matrices 2

258

15 Linear equations

281

16 Discrete probability distributions

301

Revision exercise B

328

Bibliography

339

Answers

343

Index

375

iii

Preface The present book is the first of a two volume course covering those parts of modern ' A' level pure mathematics syllabuses not normally included in standard calculus texts. The reason for omitting calculus is simply to keep the book within manageable proportions and it is felt that there are a number of excellent modern calculus books available. Although no formal calculus is done here, it is expected that a student will be studying calculus concurrently with this text; thus, for example, it is assumed in Chapter 5 that the reader can differentiate simple algebraic expressions and sketch the graphs of rational functions. In Chapter 16, the exponential function is used. It is hoped that the order of presentation of topics in this book will offer an effective teaching programme, but variations can be made at the discretion of the teacher. Thus, for example, the chapters on probability can be deferred while the two chapters on trigonometry could be taken in conjunction. Again certain chapters contain work which might be deemed suitable for a second reading; for example, Sections 4 and 5 of Chapter 10. In some ways Chapter 1 offers the most difficult problem of presentation in the whole book: it is necessary later to be able to refer to rational and irrational numbers, and yet to devote too much time in the early stages to such topics may not be desirable. The author hopes that he has found an acceptable compromise but some might yet find parts of Chapter 1 too formal, in which case they are strongly advised to leave the detailed study of Sections 2, 3 and 4 for a second reading. To learn mathematics, constant practice is necessary, some of it repetitive. The book is therefore liberally supplied with exercises for the student. Most of the questions represent fairly straightforward applications of the bookwork, although in the Miscellaneous Exercises, included at the end of all but two of the chapters, will be found some rather more searching questions. Furthermore, questions marked Ex. occur in the text; it is hoped that most, if not all, of these will be attempted by the pupil as he covers the associated bookwork—certainly those marked with an asterisk should be regarded as obligatory. The syllabuses for M.E.I. ' A' level, the Joint Matriculation Board syllabus in Mathematics (Advanced) and Further Mathematics (Advanced) and the University of London Revised Syllabus in Mathematics (Advanced)

PREFACE

have been particularly kept in mind in the writing of this book and the planning of the next. Other boards are in the process of devising new syllabuses and draft copies published by them indicate that this book will probably cover the necessary work. The School Mathematics Project have produced their own text-books but the present book may be used to supplement these if desired. It is a pleasure to record my thanks to Mr M. J. Rawlinson who read part of the text and made a number of valuable suggestions; to Mr A. J. Moakes who read the entire book and whose detailed criticisms have done much to remove obscurities and improve the presentation; and to my wife, who also read the entire book and who lent invaluable assistance in checking answers. I am grateful for permission to reprint examination questions from the following Boards: Oxford and Cambridge (0 &C, M.E.I., S.M.P.), the Joint Matriculation Board (J.M.B.), the Cambridge Local Examination Syndicate (Cambridge), University of London (London), and the Oxford Delegacy of Local Examinations (Oxford); also to the Clarendon Press for permission to use Oxford Scholarship questions (0.S.) and to the Cambridge University Press and University Registry for the use of Cambridge Scholarship (C.S.) and Mathematical Tripos (M.T.) questions. S.L.P.

vi

1. Numbers and inequalities

1. THE INTEGERS AND THEIR REPRESENTATION The process of counting is deeply rooted in the history of mankind and its development is obscure. Many systems have been evolved but the Arabic (originally Hindu) which we now employ is almost universal. The entities used for counting are the whole numbers, or positive integers, and zero. Ten ciphers are used to represent the whole numbers : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The introduction of the cipher zero was an advance of considerable significance and it gave to the Arabic system its great flexibility and versatility by making possible a system of place value recording; place value calculation (by the abacus) had already been in use for a long time. A base of ten is most frequently used for continued counting; thus, the one hundred and forty second integer is written 142, which stands for

1 x 102 +4x 10+2 x 1. The base, 10, of this method of enumeration has no special significance and any other number would do as well. Indeed, were ten not so deeply rooted in us for physical reasons, other bases would no doubt be preferable; for example, eight (23) or twelve (22x 3). Had primitive man ignored his thumbs we could well have inherited a more efficient system. Other bases have been used. A base which has acquired special significance recently is the number two. Counting from 1, with 2 as base, the first eight integers are written: 1, 10, 11, 100, 101, 110, 111, 1000. The value of this, the binary system, is clear: it requires only two ciphers, 0 and 1. It is thus eminently suitable for recording numbers in two-state systems as, for example, in electronic digital computers. Conversion from one base to another is easily effected. For example 75 (written in the scale of ten, or the denary scale) = 1 x 26 +0x 23+0 x 24+1 x 23+0x22 +1 x21+1 = 1001011 (written in the binary scale). The same result may be arrived at more quickly by continued

NUMBERS AND INEQUALITIES

division by 2, the remainders recorded being the required digits (in reverse order) 2 75 2 37

1

2 18

1

2

9

0

2

4

1

2

2

0

2

1

0

0

1

Ex. 1. Explain why the method of conversion from one scale to another by continued division works. Ex. 2. Evaluate 11011 x 1011 in the binary scale. Perform several other multiplications of this type and check your answers by conversion to the denary scale. Ex. 3. Evaluate 1110011 - 11011 in the binary scale. Perform several other divisions of this type and check your answers by conversion to the denary scale.

Exercise 1(a) 1. Solve the following equation for x, working in the binary scale throughout: (i) 101x+11 = 1101; (ii) 11x-11111 = 1011; ... 1 x 0 + 11 = 111; (iii) 101 (iv) 11(10x — 101) = 10x + 101. 2. Express the denary number 275 in the scales of 2, 3 and 12. (For the last part, you will need to supplement the digits 0, 1, ..., 9 by t = ten and e = eleven. Why ?) 3. With the notation of Question 2, evaluate 4te x 19t in the scale of 12. 4. With the notation of Question 2, evaluate e89t + 2ee in the scale of 12. 5. Explain the following method of conversion from the binary form 1100110111 to denary form 823 1

100

110

111

1

4

6

7

Binary Octal

512 + 256 + 48 + 7 = 823 Denary

6. Show that 1331 is a perfect cube whatever the base b, provided that b is greater than 3. 2

THE INTEGERS

1]

7. The number x lies between 100 and 999 (denary scale) and the number y is formed by writing the digits of x in the reverse order. Prove that x y is divisible by 99, where x y means the difference between x and y. 8. Show that the difference between any number with four digits (in the denary scale) and the number with these digits reversed is divisible by 9, and that, if the two middle digits are the same, the difference is also divisible by 37.

2. THE INTEGERS (CONTINUED) We have already tacitly assumed that the positive integers obey certain laws of combination, which are summarized below. These laws are seen to hold irrespective of the base employed for representing the numbers : the integer seven remains the integer seven, whether it be written 7 or 111 (binary) or 21 (ternary); consistency is all that is demanded. (i) If a = b, then a+ c = b+ c and ac = bc. (ii) The commutative laws: a+b = b+ a; ab = ba. (iii) The associative laws: a+(b+ c) = (a +b)+ c;a(bc) = (ab) c. (iv) The distributive law: a(b+ c) = ab+ac. (v) The additive and multiplicative identity laws: a+0 = a; al = a. Considerably later historically, zero and the positive integers were augmented by the negative integers; the word ìnteger' will in future be taken to mean a positive or negative integer, together with zero. To enable negative integers to be combined, a further rule is required: (vi) The additive inverse law: a+ (— a) = 0.

Example 1. Prove, using only the laws (i)—(vi) above, that (— a) (— b) = ab. First observe that aa+ a0 = a(a+ 0) by (iv) = as

by (v)

= aa+ 0 by (v), a0 = 0

on adding — aa to both sides and using (ii) and (i). Thus, any integer multiplied by zero gives the answer zero. Next, consider ab + [a(— b)+ (— a) (—b)] = a[b + (— b)]+ (— a) (—b) by (iii), (iv) = a0 +(— a) (—b) = 0 + (— a) (—b) =(— a)(—b)

by (vi); by result proved above; by (v). 3


1]

But ab + [a(— b)+ (— a) (—b)] = ab + [a + (— a)] (—b) = ab + 0(— b) = ab +0 = ab ab = (— a) (— b).

Thus

by (ii) and (iv); by (vi); by result proved above; by (v).

Laws (i)—(vi) above supply almost all the apparatus required for the manipulation of integers. However, as Example 2 below shows, they need to be supplemented by one more law: (vii) The cancellation law: ab = a c b = c, provided a 0. (The sign =- is read as ' implies '; -= means ìs implied by' and means ìmplies and is implied by'.) In this section we have given only a brief survey of the logical structure of the integers. The reader interested in acquiring a deeper understanding of this topic and, indeed, of the other topics mentioned in this chapter should consult one of the books mentioned in the Bibliography at the end of the book. Example 2. Arithmetic modulo 12 is defined as follows: any two of the integers 0, 1, 2, ..., 11 are added or multiplied together and the answer is taken to be the remainder on dividing the sum or product by 12. Thus 8+6 -= 2 (mod 12); 4 x 11 = 8 (mod 12); and we also write

4-6 = 10 (mod 12).

Show that arithmetic modulo 12 satisfies laws (i)—(vi) above but that the equation 4x = 4 (mod 12) does not have a unique solution (and so law (vii) does not hold). If a + b = 12c+ r , then b + a = 12c+ r and the commutative rule holds. All the other laws may similarly be verified. But x = 1, 4, 7, 10 all satisfy the equation 4x = 4 (mod 12). Ex. 4. Prove that

—

(a b) = — a+ b. (a— b means a + (— b).) —

Ex. 5. Prove that, if ax = a, then x = 1, provided a * 0. (Notice carefully what you have to prove; it is not sufficient merely to verify that x = 1 satisfies ax = a.) Ex. 6. Show that arithmetic modulo 11 does not suffer from the same defect as that exhibited by arithmetic modulo 12 in Example 2. Suggest why this is so. 4

RATIONAL NUMBERS

3]

3. RATIONAL NUMBERS As soon as numbers began to be applied to problems more complex than the mere counting of objects, the value of subdividing the interval between two integers must have become apparent. For example, when a unit of length was defined, lengths must have been met that were not an exact integral number of units. Thus the concept of a fraction or, as we shall prefer to call it, a rational number, was evolved. In order to emphasize the fact that we are now dealing with a new type of number, we shall avoid the familiar fractional form for rational numbers at first, and instead we shall define a rational number as a pair of integers written in a definite order thus : (p, q), where q + 0. (We may call this an ordered pair of integers: (2, 3) and (3, 2) represent different rational numbers. See Chapter 4.) (In what follows it may help the reader to see what is happening if he bears in mind that our aim is to demonstrate that the rational number (p, q) is what he would call p/q.) Two rational numbers (pi, q1) and (p,, q2) are said to be equal if p2q1 = 0; if pia,— p,q, + 0, they are said to be unequal. Ex. 7. Is

it true that (a, b)

(b, a)?

Ex. 8. Prove that (1, 2), (2, 4), (13, 26) are all equal. Prove more generally that (p, q) = (kp, kq), provided k * 0. Since rational numbers are newly defined objects, rules for adding, subtracting, multiplying and dividing them must be given, for these operations have so far only been applied to integers. The rules are (Pr, q1)+ (P2, q2) = (P1q2+ P2qi, q1q2);

(Pi,

(p1, q2) x (pa, q2) = (P1P2, qiq,);

(Pi, 0

— (P2, q2) = (Piq2 —P2qi, q1q2); (P2, q2) = (P1q2, P20,

provided p, + 0, otherwise the left-hand side is undefined. Example 3. Verify that the distributive law holds for rational numbers. We have to show that (p,, q1) x

[(P2, q2)± (pa, q3)] =

L.H.S. =

q1) x (Pa, q2)± (pi, q1) x (ma),

q1) x (P2q3+P3q2) 9290

= (P1P293+P1Paq2, 91q2q3))

5

NUMBERS AND INEQUALITIES R.H.S. =

[1

(pi, qi) x (p2, q2)+(Pv q1) x (p3, q3)

= (PiPz, qrq2)+ (pip3, 91q3) = (P11929193+ PiP3M2, 912q2q3) = (1,1132q3+ PiP3q2, 91q2q3),

by Ex. 8, since q14 0.

In a similar way, all the rules for combining integers may be verified to hold also for rational numbers. Ex. 9. Verify the laws (i)—(vii) for rational numbers. Ex. 10. Verify that, if the rule for division is ignored, rational numbers of the form (p, 1) have properties identical to those possessed by the integers. Ex. 11. Verify that, if (a, 1) x (x, 1) = (b, 1), then (x, 1) = (b, a), provided a * 0. Ex. 10 shows us that we may identify the integers with rational numbers of the form (p, 1); Ex. 11 then shows us that the rational number (b, a), when multiplied by the integer (a, 1) gives the integer (b, 1). In more familiar language (b, a) has just the properties we associate with the 'number' b/a. We may thus regard a rational number as `the quotient of two integers'. It thus follows that the statement ' x is a rational number' is equivalent to the statement 'integers p and q may be found such that x = p/q'. The reader will be familiar with the process of expressing a rational number as a decimal. For example, 18 1 1 1 = 0.024 = 0 x fo +2+Tiy2 +4 x 625 Conversely, a terminating decimal can always be expressed as a rational number in the form p/q; for example 0.175 =

=

However, the terms 'rational number' and `terminating decimal' are not synonymous. Thus, .* is certainly a rational number, but = 1.142857142857142857... and the process of division cannot be brought to an exact conclusion. However, this expression does recur; the reader will probably be familiar with the notation. 1.142857. In fact, any rational number may be expressed either as a terminating or a recurring decimal, and conversely, any terminating or recurring decimal represents a rational number. The proof of the italicized part of this state6

RATIONAL NUMBERS

3]

ment depends upon the notion of a limit (see Chapter 8; geometric sequences) but good grounds for believing it are given in the following particular example. Example 4. Express 2.378 as a rational number in standard form. Let x = 2.378 ; then 1000x = 2378478 2.i7g x= 999x = 2376 by subtraction -

2376

-

88 37.

Ex. 12. Express the following recurring decimals as fractions: (i) 0.7, (ii) 0.402, (iii) 6.28.

Ex. 13. Using only the digits 0 and 1 we may express any rational number as a bicimal. For example 1 1 11.01101 = 1x 2+1+0x- +1x—+1 x- +0x- +1 x—. 2 22 23 24 24 Express as bicimals the decimals 3.75, 0.703125, 4.6, 0.82. [Hint: see Ex. 1 and substitute repeated multiplication for repeated division.]

Ex. 14. Compare 0.9 (decimal) with 0.1 (bicimal). Ex. 15. Express 0.01 and 1.101 (both bicimals) as decimals.

4. IRRATIONAL NUMBERS It is useful to depict the integers and rational numbers as points on a straight line. In Figure 1.1, an origin 0 is taken, representing the number zero, and equal intervals are measured to the right, the end-points representing 1, 2, 3, ... and to the left representing —1, —2, — 3, .... The position of a point representing a rational number may be defined quite simply ; I I I I I I I —3 —2 —1 0 1 2 3

Fig. 1.1

for example, the point representing 3.28 is obtained by sub-dividing the interval between 3 and 4 equally into 100 divisions and marking the endpoint of the twenty-eighth division. We may thus associate with the rational numbers a definite ordering: xis greater than y (written x > y) if the point representing x lies to the right of that representing y. 7


[1

All rational numbers may thus be represented by points on the line and, furthermore, however close together two points representing rational numbers may be there will always be another point representing a rational number lying between them. For example, between the points representing 3.286 and 3.287 lies the point representing 3.2865. Ex. 16. Show that -1-(x+y) always lies between x and y. Our last remark shows that, however close together we choose two rational numbers, we can always fit another rational number in between them; surprisingly, however, we can never succeed in 'filling the line up' with points representing rational numbers. In Figure 1.2 an isosceles right-angled triangle OPQ has been drawn in which OQ = QP = 1. The theorem of Pythagoras tells us that, if OP represents the number x, then x2= 2; our next example shows that P does not represent a rational number. Q

• 2

Fig. 1.2

Example 5. Prove that J2 is not a rational number. We have to prove that V2 cannot be expressed in the form plq where p, q are integers. Our method of proof will consist in assuming that V2 can be expressed in the form p/q and showing that this leads to a contradiction. Suppose that V2 = p/q, where p, q have no common factors. Then p2 = 2q2 ; but, if the square of an integer is even, the integer itself must be even, and so p = 2r where r is an integer .*. 4r2 = 2q2 and so 2r2 =q2. By the same token q must be even. Thus, p, q must both be even, contradicting our initial assumption.

Numbers such as V2 which are not rational are called irrational. Since rational numbers are represented by terminating or recurring decimals, the decimal representation (if it exists) of an irrational number is nonterminating and non-recurring. Irrational numbers obey the same laws of combination as rational numbers. The proof of this lies beyond the scope of this book, as it requires a systematic definition of irrational numbers in terms of the rationals ; 8

IRRATIONAL NUMBERS

4]

we shall be content to assume this result, which is intuitively fairly obvious if we consider rational (terminating decimal) approximations to irrational numbers. Exercise 1(b) 1. Where does the proof of Example 5 break down if you try to prove ,/4 is irrational ? 2. By assuming that ,;/2 = plq, where p, q are integers cancelled down into their

lowest form, show that ./2 is irrational. 3. Prove that V3 is irrational. 4. Express as (i) a decimal, (ii) a bicimal, the fraction 3(denary notation). If a decimal recurs, does it bicimal representation necessarily recur ?

5. Prove that, if a rational number of the form pl q (cancelled down into its lowest form) is expanded as a recurring decimal, the length of the block of digits that recurs is less than q. 6. Which of the following statements are always true and which are not? If a statement is true, prove it; if it may be false, give a counter-example (that is, an example that illustrates its falsity): (i) rational + rational = rational; (ii) rational + irrational = irrational; (iii) irrational + irrational = irrational; (iv) rational x rational = rational; (v) rational x irrational = irrational; (vi) irrational x irrational = irrational.

7. If a and b are integers and Vb is irrational, prove that (a+ Jb)3is irrational. (The results of Question 6 may be used in this question.) 8. Prove that it is always possible to find an irrational number that lies between two given rational numbers a and b. 9. If pi, p„Paare three unequal prime numbers, prove that 1 1 1 — +— +P1 P2 P3 is a rational number, but not an integer. Can you generalize this result? 10. You are given a ruler with only integral units of length marked on it. Show how to construct, geometrically, the following lengths: (i) A/3 ;

; (iii) \/(2 + A/2).

11. Prove that, if k is an integer and Vk is rational, then Vk is an integer. Show further that, for k > 1, V(k2—1) is irrational and deduce that, for k 1, V(k — 1) + ,/(k+ 1) is irrational.

9


[I

5. SURDS Although the existence of irrational numbers is mathematically significant, from the point of view of practical arithmetic all numbers may be regarded as rational, since any irrational number may be approximated to by a terminating decimal. Indeed, wherever measurements are concerned, the answers obtained must necessarily be in the form of rational numbers, since every measuring device must eventually reach the limits of its possible accuracy. However, just as is simpler to handle than its decimal approximation 0.3333, so J2 is often easier to deal with than its approximation 1-414. An expression involving only rational numbers and their roots (not necessarily square roots) is called a surd; thus surds form a class of irrational numbers, though there are irrational numbers, such as 7r, which cannot be expressed as surds. Examples of surds are: 1 2 4J2; :/;; (V2 + v2 11); ,?,/3 + V2. Note: if x is positive, mix = xion means the positive ?nth root of x; by convention, the 2 is omitted for square roots, e.g. V9 = 3. Again, 27V( — x) has no meaning, while 2m+V(—

x) = (— xy/(2m-Fl) = — (xy/(2m+i).

A great number of complicated surds may be simplified using the three results: ba = V(ab) b ; (i) V(ab) = (ii) (iii) (Va+Vb) (Va — Vb) = a — b. Example 6. Simplify the expression V128 — V32 — V8 +V2. V128 — V32 — V8 + V2 = V(26. 2) — V(24. 2) — V(22. 2) + V2 = 23V2 — 22.2 — 2V2 + V2 = 8V2-4A/2-2V2+ V2 = 3V2. The process of removing surds from the denominator of surd expressions is known as rationalizing the denominator. Conventionally, a surd is usually expressed with its denominator rationalized. 10

SURDS

5]

Example 7. Rationalize the denominators of the following surds:

A/2+1 1 (i) /3.(,..\ . (iii) "' Al 5' "1' V5— V2' ' ' V3+ A/2. 13 V3V5 _1/15 RI /3 ''' 4 5 — V5 V5 V5 — 5 ' 1 1(V5+ V2\ = A/5+V2. (ii) 3 ' V5— V2 = V5— V2 kA/5 + V2) A/2+1 iV2+1 \ (V3 — V2\ _(V2+1) (V3 (iii) 3-2 ' ' A/3 + V2 kA/3 + V2) kA/3 — V2)

/2)

— A

= V6+ A/3 — A/2— 2.

Ex. 17. V45 = 3A/5. Simplify in a similar fashion the surds: (i) A/8; (ii) V18; (iii) V54; (iv) V250; (v) V5292.

Ex. 18. Simplify the following surd expressions. (i) (V3 + 2A/2)2; (ii) A/216— A/150 + V24; (iii) A/75 + A/147 — V300; (iv) (A/32 + V50 — 1/98)2. Ex. 19. Rationalize the denominators of the following expressions: V3 2 /3 1 .. V2+1 ... 1 ; (m) 2— v2; (iv) v3— v2; (v) 2+ v3. (i) -v--3 ; .

(ii)

— A

v2

We conclude this section with some further examples of manipulation

of surd quantities and the use of the V sign. Example 8. Simplify: (i) 1+A/3— V2' (ii) A/(14 —4A/6). 11(A/3 — A/2) (i) 1+ V3 — A/2 = 1—(V3 /2)2 —

— 1

=

(ii) Suppose Then

1 — V3 + V2 1 —(5-2A/6)

= 1+A/2—V3 2(V6— 2) _ (1+1/2-1/3) (A/6+2) 2(6-4) — 1/6+2+1/12+2,/2-1/18-24/3 4 2 + V6 — V2 _ 4 14-41/6 = (Va 21/b)2. 14 41/6 = (a + 4h) 4 V(ab). -

-

—

11


These two expressions will be equal if

a+4b = 14, ab = 6. By inspection (or by solution of the simultaneous quadratic equations) a -= 2, b = 3. Thus V(14 — 4A/6) = 2,/3 — (Notice that we must choose the positive square root.)

Example 9. Solve the equation ,I(3x+ 4)— ,/(x + 2) = 2. (i) Squaring both sides, (3x + 4) + (x+ 2)— 2,/(3x2+ 10x + 8) = 4. (ii) Collecting terms and dividing both sides by 2, 2x+1 = Al(3x2+10x+8). (iii) Squaring both sides and collecting terms, x2 — 6x— 7 = 0. This gives x = 7 or x = —1. However, we must check both of these solutions in the original equation since, after step (iii), we could equally well have been solving the equation /(3x + 4) + ,/(x + 2) = 2. (Why ?)

A

Inspection shows us that only x = 7 is a valid root of the original equation.

Exercise 1(c) 1. Express in the form aV b, where b has no perfect squares as factors (i) V50; (ii) V363; (iii) V2400; (iv) V192; (v) V1452. 2. Simplify the following surds: (i) (V5 1)2; (ii) (20—V2)2; (iii) (V3+1)2; (iv) (V3 V2)3; (v) (✓3 04. -

—

—

3. Simplify the following expressions: (i) V18 — V2; (ii) V80—V5; (iii) V108 V75 + V48; V117. V147— V3 (iv) (v) 3 ' 4 —

12

SURDS

51 4. Rationalize the denominators of the following expressions: 2 . 2V3+1 1+4 v7 _ 2; (iv) 2 v3 _1; (v) ; j(j) ; (i) V3 + V2* 5. Simplify: (i) V(6 —2V5); (ii) V(10 (v) 3/V(7— V40).

(iii) V(30 + 12V6); (iv) V(47 —6V60);

6. Solve for x the following equations: (i) 3Vx — V(x + 5) = 3, (ii) V(2x+ 5)— V(x +2) = 1, (iii) V(x + 6) + V(4— x) = V(1 — 3x). 7. Rationalize the denominators of the expressions: 1 1 ; 2 V2 + V5 — 1 . V3 + V2 + 1 8. The equation

V(x+4.7)+ V(x +4.9) = V33

was solved to give an answer x = 3.1 (to one decimal point). Without using tables, state why this is clearly wrong. 223607, J2 9. Given that .‘15 1 .. 1 (i) —; V (ii)V5+ V2'

F41421, evaluate:

giving your answers to the greatest accuracy that you can guarantee. 10. Given that, to six significant figures, \/6 = 2.44949, V3 = 1.73205, evaluate as accurately as possible 1/(20 —V6), justifying the accuracy you give. 11. Rationalize the denominator of the expression 4/3/(;/3 +1), by using the factorization a3+ b3 = (a + b) (a2— ab + b2). 12. Rationalize the denominators, and simplify as far as possible the following expressions: (i)

V(x + 1) V(x— 1)

(iii)

; (ii) V(x+ 1) +

V(x1+ 1);

1 1 1 : (iv) Al (x — 1)+1 V (x + 1) — 1. V(2x+ a) ,I(2x — a)"

6. SETS OF NUMBERS Suppose we have a collection of numbers, the whole collection being denoted by the letter U. Suppose, too, we ask a question which may be answered unambiguously 'Yes' or 'No' for each number (or element) of 11. Then the collection of those elements of U for which the answer is `Yes' is said to form a set S, which is a subset of the universal set U, 13


(1

written S s U. The collection of those elements of U for which the answer is `No' is called the complement of S, written S'. It may happen that the answer to the question is 'No ' for each element of U; in this case we say S is empty and write this as S = 0. (0 is usually referred to as the null set, or empty set.) As an example, suppose that U is the set of integers from 1 to 10. We may write this as U = {1, 2, 3, ..., 10}. The question may be posed: 'Is the element x of U divisible by 3 ?' Those elements for which the answer is 'Yes' form a set A where A = {3, 6, 9). The complement of A is given by A' = {1, 2, 4, 5, 7, 8, 10). Again, the alternative question 'Is the element x of U irrational ?' may be asked and unambiguously answered for each element of U. The answer is `No' in every case; the set defined is thus 0 and its complement is U. In this example, the universal set U contains only a finite number of elements. It may well happen, however, that U contains an infinite number of elements; for example, U might be the set of all positive integers. The proper subset E defined by the question 'Is the element x of U even ?' also contains an infinite number of elements. (P is a proper subset of Q if P is contained in Q but is not the whole of Q, written P = Q. P s Q means that P is a subset of Q, but may be the whole of Q.) It is convenient to develop a shorthand notation for the somewhat cumbersome method we have used so far in defining our sets. The method adopted is to use braces {:} with a colon (:) or vertical line (I) (read as `such that') separating the two necessary pieces of information: (i) within the bracket and to the left of the colon is stated the universal set from which the elements are drawn; (ii) within the bracket and to the right of the colon is the statement defining which particular elements of the universal set are to be chosen. For instance, for the sets U and A mentioned above we may write A = {x e U:x is divisible by 3).

The sign c means 'is a member of ' or 'is an element of '. Ex. 20. If U is the set of positive integers {1, 2, 3, ..., 9} write down the elements of the following sets: (i) {x e 21 : x + 4 E IT} ; (ii) {x a 1T : Jx is rational). 14

6]

SETS OF NUMBERS

Ex. 21. If U is the set of all positive integers {1, 2, 3, ...} describe in words the following sets: + 1) e 1.4; (ii) {x 11: x-1 e 11}; (i) {x (iii) {x e -Ayx ell}; (iv) {x e : x2 — 5x 6 = 0} ; (v) {x E : x2— 5x 7 = 0). —

-

Ex. 22. Two sets A and B which are both subsets of the same universal set U are said to be equal (A = B) if they contain precisely the same elements. Show that A = B -.44- A B and B S A. We may combine two subsets A g U and B s U according to the two rules of union (U) and intersection (n) defined as follows: A U B is the set of all elements of U that are members of either A or B or both. A n B is the set of all elements of U that are members of both A and B. From these definitions it follows at once that the operations of union and intersection are commutative; that is,

AUB=BUA and

AnB=BnA.

The operations of union, intersection and complementation may be exhibited pictorially using a Venn diagram. In a Venn diagram, the universal set U is represented by a rectangle; any subset A g U is depicted by a closed region lying within the rectangle; see Figure 1.3. In Figures 1.4, 1.5, 1.6, A U B, A n B and A' are shown shaded:

11

Fig. 1.3

B

An B

A'

Fig. 1.5

Fig. 1.6

Ex. 23. Verify, using Venn diagrams, that the operations of union and intersection are associative; that is

A u (B u C) = (A U B)U C; A n (B

= (A n B) n c. 15


Ex. 24. Verify, using Venn diagrams, that the operation of union is distributive over intersection, and that intersection is distributive over union; that is A u (B n C) = (A U B) n (A U C); A n (B U C) = (A n B) U (A n C). Ex. 25. Verify de Morgan's Laws, using a Venn diagram: (i) (A u

= A' n B'; (ii) (A n

= A' U B'.

(The reader may be able to identify these laws with the logic of and NOT 0.)

AND

(n),

Ex. 27. If 11 = {a, b, c, d, e} and A = {a, c, e} write down the subsets: (i) A U A'; (ii) A n A'; (iii) A U 11; (iv) A n 11; (v) A U 0; (vi) A n (vii) 11'; (viii) 0'.

;

OR (U)

Ex. 26. If U is the set of all positive integers and A= {x E

e 111,

B= {x e

e IX}

describe A n B, using the {:} notation.

Ex. 28. If, with the notation of Ex. 27, B = {a, b, e}, write down the subsets (i) A n B; (ii) A' u B; (iii) A n B'. Verify de Morgan's Laws (see Ex. 25) in this particular case.

Three particular sets, the set of all integers, the set of all rational numbers and the set of all real numbers occur throughout mathematics with such unfailing regularity that it is convenient to introduce a notation by which to refer to them. The set of all integers is denoted by Z. The set of all rational numbers is denoted by Q. The set of all real numbers (that is, the union of the set of all rational numbers and the set of all irrational numbers) is denoted by R. With a suitable definition of each of the terms integer, rational and real we have the following relation between them Z Q R.

The sets of all positive integers is denoted by Z+, of all positive rational numbers by Q+ and of all positive real numbers by R+. Ex. 29. Enumerate the elements of the following sets: (i) {x e Z: (x-2) (2x + 1) (x2 -2) = 01; (ii) {x e Q±: (x-2) (2x + 1) (x2 -2) = 01; (iii) {x e Q: (x— 2) (2x + 1) (x2 — 2) = 01; (iv) {x c R+: (x-2) (2x+ 1) (x2 -2) = 0); (v) {x e R:(x — 2) (2x +1) (x2 -2) = 01. 16

7]

INEQUALITIES

7. INEQUALITIES We have already asserted that the integers and the rational numbers may be ordered; that is, given two integers (or rational numbers) x and y, we may answer the question 'Is x greater than y ?' The same is true for the set of all irrational numbers, or, indeed, for the union of the sets of rational and irrational numbers (the set of real numbers). If x, y are two real numbers, x > y (read x is greater than y') means that the point representing the number x lies to the right of the point representing the number y. Similarly for x y and z < 0 xz < yz. Thus, if a set is defined by an inequality x > y, the same set is defined by

xz > yz if z is positive, xz < yz if z is negative.

or by

A similar result holds for division. We now append an alternative solution to Example 10: 3—x>2 — 3+x < — 2, multiplying both sides by —1, x < 1,

adding 3 to both sides,

and we have, as before, the solution set

{xe R:x < 1}. Ex. 31. Solve the inequalities (i) 3 + x < 1; (ii) 2— 3x > —1; (iii) 3 — 4x > 1. The reader should have observed that in, for example, the result x > y and z > 0

xz > yz

the two-way implication was not used: it is not valid to deduce from xz > yz that x > y and z > 0. However, inequalities involving products may be solved by observing that, if the product is positive, then the factors are either both positive or both negative. Another useful observation is that a squared number is always positive or zero.

Example 11. Solve the inequality (x +2) (x 1) > 0 (x real). If (x +2) (x— 1) > 0 then —

either or

(i) (x + 2) > 0 and (x — 1) > 0, (ii) (x +2) < 0 and (x — 1) < 0.

The first pair of inequalities has as solution set {xeR:x > 18

—

2} n {x e R:x > 1} = {x e R:x > 1}.

7]

INEQUALITIES

The second pair of inequalities has as solution set {x e R: x
G unless u2 = v2in which case A = G. Example 13. 200 m of wattle fencing are to be bent to form three sides of a rectangular enclosure, the fourth side being a straight hedge. Find the length of the rectangle if the area to be enclosed is to be a maximum. Let x m be the length, y m the breadth, A m2the area of the enclosure. Then we have (i) x + 2y = 200, (ii) xy = A. Both x and 2y are positive; thus, by the theorem just proved, their is at least as great as their G.M. That is

A.M.

1-(x + 2y) V(2xy)• Using (i) and (ii), this gives

100

V(2A),

i.e.

A 5 5000. Equality occurs only if x = 2y, i.e. x = 100, y = 50. Thus the maximum area occurs when the length is 100 m. Exercise 1(d) 1. Verify, using a Venn diagram, the results: (i) A U (A' U B)' = A n (B' U A), (ii) (A U B) n (A' n = A U (B n C'), (iii) [A' u (B' n C')]' = (A n B) U (A n C). 2. If R is the set of real numbers and A = {x e R: x > 3},

B = {x e R; x < 4}

describe, using the {:} notation (i) A U B; (ii) A n B; (iii) A' U B. 3. Taking the universal set, U as the set of all integers, describe using the {:} notation (i) the set of all negative integers; (ii) the set of all positive integers divisible by 6; (iii) the set of all integers, excluding 0, ± 1, ± 2. 4. A = {x e R: — 1 x < 3},

B = {x e R: 2 -4 x < 4}, C = {x e R:x > 3}.

Find expressions for the following sets, using the {:} notation: (i) A n B;(ii) A' n C; (iii) (A U B) n C'; (iv) A U (B n C); (v) (A' U B') n C; (vi) A' U (B' n C); (vii) A' n B n C.

20

INEQUALITIES

5. A = {x e R: x > 0}, B = {x R: x 1}, C = {x e R: —1 < x < 2}, D = {x e R: — 2 x < 1}. Find expressions for the following sets, using the {:} notation: (i) A nBn C fl D; (ii) A'nBn C 11 D'; (iii) (A U B') n (C n D'); (iv) (A' n B') U (C' n D). 6. Solve the following inequalities : (i) x+2 < —3; (ii) 3x — 1 > 5; (iii) 4x — 3 < 3x + 4; (iv) 2x > x; (v) x2 > x. 7. Solve the following inequalities: (i) (x + 3) (x — 1) > 0; (ii) (2x — 1) (3x +1) > 0; (iii) (x 3) (2x + 3) < 0; (iv) x2 — 6x + 9 < 0; (v) x2 — 4x — 5 < 0. -

8. Solve the inequality (2x + 1) (x — 2) (x + 3) > 0. 9. Prove that the following expressions are positive, for all real values of x, and find their minimum values : (i) x2 +2x + 2; (ii) x2 — 6x + 12; (iii) 2x2 — 2x + 1. 10. Find the signs, for all values of x, of: (i) 2x — x2 — 2; (ii) 2x2 — x +1. 11. For what values of x is the expression (x — 1) (x+ 2) (x 3) (x + 4) -

positive? 12. If xy = 25, find the least possible value for x + y, if both x and y are positive. 13. If xy = 18, find the least possible value for 2x + y, if both x and y are positive. 14. If x + y = 2, find the maximum value of xy.

15. If 2x + 3y = 120, find the maximum value of xy. 16. Some netting is required to make three sides of a rectangular chicken-run, the fourth side being an existing wall. Find the least length of netting needed for an area of 50 m2.

A

17. Figure 1.7 shows part of the framework of a kite. ABCD is a rectangle; P and Q are the mid-points of Q AB, CD respectively. Find the maximum area of Fig. 1.7 ABCD that can be made if 4 m of wood are available. What are the dimensions of the kite necessary to attain this area?

C

Miscellaneous Exercise I 1. Express 120 as a binary number.

Find the least number of stamps required so that any value from p to 60p (in steps of -1-p) may be selected, and give their values. (You may assume stamps of any value are available if required.) 21


[1

2. Show that it is impossible to choose values of a and b so that the number written as ab in the scale of ten is equal to the number written as ba in the scale of twelve. Show, however, that ab (scale of ten) = ba (scale of seven) is possible, and give an example. 3. Prove the well known result that the remainder on dividing a number by 9 is the same as the remainder on dividing the sum of its digits by 9. Show that this result may be generalized as follows: if a number is divided by s then the remainder is the same as the remainder on dividing by s the sum of its digits, when it is expressed to the base s +1. 4. Prove that, if a number is divided by 11, the remainder is either the same as the remainder on dividing the difference between the sums of the digits in the even and odd places, or else the sum of the two remainders is 11. Generalize this result along the lines of Question 3. 5. Prove that, if the digits of an integer (expressed to the base ten) are rearranged in any way to form another integer, the difference between the two integers is divisible by 9. 6. If abc+ abc = cba in the scale of five (when abc here means 52a + 5b + c) find values for a, b, c. Show that this question is always soluble provided a base of 3n — 1 is used, where n is a positive integer. 7. Prove that V3 — 8/2 is irrational. 8. Solve the equation V(4x — 2) + V(x+ 1) — V(7 —5x) = 0.

(0 & C)

9. Verify that the expression az+b2+2_ c 2bc — 2ca— 2ab is equal to (a + b — c)2— 4ab. Hence, or otherwise, prove that the expression is equal to (ce+fi+y)(ce—fl—v)(ce—fi+v) (cc+18-1'), where

a = Va, fi

y = Vc.

Hence, or otherwise, find one solution of each of the equations (i) V(x — 6) + V(x — 1) = V(3x— 5); (ii) V(6 — x) — V(1 — x) = V(5 — 3x).

(0 & C)

10. If a > b and c > d, prove that ac+bd > bc+ da. What happens to the third inequality if a < b, c < d? 11. If the universal set is taken as the set R of all real numbers, the three sets A, B and C are defined as follows: A= {x e R: x > 2}; B= {x e R:1 < x < 4}; C= {x e R: x < 3}. Express, in terms of any or all of A, B, C (and using the notation of union, intersection and complement) the following sets: (ii) (x e R: x < 1); (i) {x E R : 2 < x < 3}; (iv) {x c R: 1 < x < 2 or 3 < x < 4}. (iii) {x e R: 3 < x < 4); 22

7]

MISCELLANEOUS EXERCISE 1

12. Show that the expression x2+ 8xy — 5y2— k(x2 +y2) can be put in the form a(x+by)2when k has either one or other of two values. Find these values and the values of a and b corresponding to each value of k. Prove that when the variables x and y are restricted by the relation x2 + y2 = 1, but are otherwise free, then (0 & C) —7 < x2 + 8xy — 5y2 < 3. 13. a, b, c, d are four unequal positive numbers. By using Theorem 1.1 and considering first the pair of numbers -Ka +b) and +(c+ d), and then, separately, the pairs a, b and c, d, prove that 1(a b + c+ d)> (abcd)1. Deduce, by considering the four unequal numbers a, b, c and +(a+ b + c), that +(a + b + c) > (abc)1. What happens to the last inequality (i) if a = b = c; (ii) a = b c? Suggest a generalization of the results proved above. 14. If x+ 2y+ 3z = 1, where x, y, z are positive numbers, find the maximum

value of xyz. If u+v = 1, where u, v are positive numbers, find the maximum value of u2v. 15. A cylindrical vessel, with one end open, is made from a given piece of material. Show that its volume is greatest if the height and radius are equal.

16. a, b, c are non-zero rational numbers. Show that, if al-Fg+c* = 0, and if none of al, bl, c* is rational, then al, bl, cl are each rational multiples of the same irrational number. 17. Show that, if pig is a good approximation to V2, then (p2 +2q2)/(2pq) is a better one. Starting with p = q = 1, show that V2 Ps,' 577/408, and estimate the accuracy of this approximation.

23

2.

Vectors and vector geometry

1. SCALAR AND VECTOR QUANTITIES Many physical quantities are completely specified by their magnitude alone. For example, if the mass of the box is m kg and we are told that m = 8, we know the mass of the box precisely. Not all physical quantities are so easily described as this; for example, if the position of the point P relative to the fixed point 0 is denoted by r, then the statement `r = 8 m' is not sufficient for us to locate P: it may be anywhere on a sphere, centre 0 and radius 8 m. In order to determine the position of P we need to be given its direction, specified in any suitable way. The position of P relative to 0 is an example of a displacement; that is, a line segment whose length, direction and sense are given. Experimental evidence reveals that many physical quantities obey the same mathematical laws as do displacements; such quantities are called vector quantities. We shall give a precise definition of a vector quantity in the next section, observing here merely that vector quantities require a direction and sense, as well as a magnitude, for their specification. Physical quantities which require only a magnitude (that is, a pure number) to describe them are called scalar quantities. Here are a few examples of each type: Vector quantities: displacement, velocity, acceleration, force, momentum, electric intensity. Scalar quantities: mass, time, temperature, energy, electric charge, electrostatic potential.

2. VECTORS AND THE TRIANGLE RULE Since all vector quantities obey the same mathematical laws as displacements we shall concern ourselves exclusively with displacements in building up the relevant mathematics. It should be borne in mind, however, that the theory being developed is applicable to a wide range of physical quantities. Before embarking upon a study of operations involving displacements we must define what we mean by the equality of two such quantities. A displacement is typified by its magnitude, sense and direction and so it seems reasonable to regard two displacements of the same magnitude which are parallel in the same sense to be equal. Thus, in Figure 2.1, we have 24

2]

VECTORS AND THE TRIANGLE RULE

AB = CD = EF = GH = .... We shall use a single symbol to represent any member of the class of such equal displacements a = AB = CD = EF = Such a representative of a whole class of displacements is called a vector. In other words, a vector is a mathematical entity, to which a whole class of geometrical objects correspond. Vectors are usually printed in bold face type: a, b, c, In manuscript they should be indicated by a wavy line beneath the letters, 4, b, c. The equality of two vectors a = b, means that a and b are both representative of the same class of parallel and equal displacements.

Fig. 2.1

Fig. 2.2

The magnitude of a vector, a, written a I, is the magnitude of any one of the displacements of which it is representative. It is most important to remember that la = does NOT imply that a = b; put another way, two unequal vectors may have the same magnitude. Two displacements AB and BC may be combined together, or added, according to the triangle rule, AB+BC = AC (see Figure 2.2). Two vectors may similarly be added: to find a +b we draw any displacement AB represented by a; we then select the unique displacement BC, represented by b which has as its initial point the point B and the vector a +b will be representative of that class of displacements of which AC is a typical member. We have said that vector quantities are physical quantities that may be completely represented by a displacement. Since displacements obey the triangle rule of combination, so must vector quantities generally. Indeed, we are now in the position to make a formal definition: A physical quantity is a vector quantity if (i) it has magnitude, direction and sense; (ii) it obeys the triangle law of addition.f t The reader should observe that certain vector quantities must have their position specified too. For example, to describe a force we need to know not only its magnitude and direction, but also its line of application.

25

VECTORS AND VECTOR GEOMETRY

[2

The observation that any particular physical quantity is indeed a vector quantity is the result of experimental evidence, which must include a verification of the triangle rule. Ex. 1. Can you suggest any addition to the list of vector and scalar quantities given in Section 1?

Ex. 2. Can three displacements be added together? Does it matter in what order it is done? Ex. 3. A rotation may be given a magnitude (size of angle turned through), direction (axis of rotation) and sense (positive being direction for which rotation is anticlockwise). Show that rotations are NOT vector quantities. (This example shows that physical quantities exist that are neither scalar nor vector quantities.)

3. OPERATIONS WITH VECTORS Since all parallel and equal line segments represent the same vector a we may, when verifying the various rules of vector algebra geometrically, choose as representative those line segments which are most convenient. Rule 1. Vector addition is commutative. a +b = b + a. If AB = DC = a and BC = AD = b, ABCD is a parallelogram and

a+b = AC = b+a.

Fig. 2.3

Rule 2. Vector addition is associative. (a +b)+ c = a+ (b + c). If then and

AB = a, BC = b, CD = c, (a+b)+c = AC+CD = AD, a+(b+c) = AB+BD = AD.

We may thus drop the brackets when adding three vectors together, and write a +b +c without ambiguity. If AB represents the vector a then BA will be said to represent the vector 26

OPERATIONS WITH VECTORS

3]

a. We shall write a+ ( a) = 0 for all a and call 0 the zero vector. (Note that the zero vector, unlike all others, has no direction.) If AB = a and AC = b then, since CA + AB = CB, the line segment CB represents a — b (see Figure 2.5). (Notice that we have used the cornmutativity of vector addition here.) To develop the algebra of vectors further, we need a new definition, the multiplication of a vector by a number (scalar). If k is any positive number, ka is the vector of magnitude k la I in the same direction as a and with the same sense. —

—

b

Fig. 2.6

Fig. 2.5

If k is any negative number, ka is the vector of magnitude —klaj in the same direction as a and with the opposite sense. Thus, multiplying a vector by a number simply stretches or contracts the vector along its length. Notice that a = la. Rule 3. Multiplication by numbers is associative. —

k(la)

—

(kl) a.

This is immediately apparent, since a, la and (kl)a all have the same direction, and the magnitude and senses of k(la) and (kl)a are both (kl) la Rule 4. Multiplication by numbers is distributive over vector addition.

k(a+b) = ka + kb. Representing a by AB and b by BC, ka by AB' and kb by B'C', ACC' is a straight line by similar triangles and the result follows, since the corresponding sides are in proportion (Figure 2.6). Rule 5. Multiplication by numbers is also distributive in the following sense. (k +1) a = ka + /a. The vectors on either side of the above identity have the same magnitude, sense and direction, as can be seen immediately from a diagram. *Ex. 4. Prove that 0.a = 0. What does this tell us about the magnitude of the zero vector? 2

PPM

27


[2

Ex. 5. If a is a vector of magnitude 4 units due north and b is a vector of magnitude 1 unit due east, describe the vectors: (i) 2a; (ii) -3b; (iii) a -4b; (iv) 2(a +b); (v) 2a - 6b. Ex. 6. If a = 3x + y, b = x - 2y, find x, y in terms of a, b, justifying your argument in terms of Rules 1-5. Ex. 7. If a is a vector of magnitude 1 unit due north and b is a vector of magnitude 1 unit in the direction N 60° E, describe the following vectors (using a scale drawing or trigonometry, if necessary): (i) 2a + b; (ii) - (2a - b); (iii) b -a; (iv) - a - 2b. Ex. 8. If a is a vector of 2 units due north in a horizontal plane, b is a vector of 3 units due east in a horizontal plane and c is a vector of 1 unit vertically up out of the plane, describe the vectors: (i) a - 2c; (ii) - b- 3c; (iii) 3a + 2b + 6c; (iv) 3a - 2b- 6c. Ex. 9. ABCD is a parallelogram with AB 2 units and BC 1 unit. If a is a vector of 1 unit in the direction AB and b is a vector of 1 unit in the direction of BC, and if E is the mid-point of CD, find, in terms of a and b, the vectors AC, DB, AE, EB. 4. COMPONENTS OF A VECTOR It is often useful to express a given vector r as the sum of a number of vectors in specified directions r If this is done, a1, a2, anare called components of r. Given a vector r and specified directions two questions immediately arise : (i) Can r be split into components in these directions? (ii) If the answer to (i) is ' Yes', in how many alternative ways can this be done ? We shall answer these questions for a vector r and three non-parallel, non-coplanar directions in three dimensions by proving two theorems of types that constantly recur in mathematics. The first theorem is an example of an Existence Theorem: it will answer question (i) above, telling us that a solution does exist; the second theorem is a Uniqueness Theorem: it will answer question (ii) above, telling us that the solution which we know exists is the only possible one, that is, that it is unique. The significance of these two theorems will become apparent as we proceed. Theorem 2.1 (The Existence Theorem). Given a non-zero vector r and three non-coplanar t, non-parallel vectors a, b, c, there exist numbers A, it, v, such that r = Aa-Ficb+vc.

f That is, displacements represented by a, b, c, cannot all be chosen to lie in one plane. 28

COMPONENTS OF A VECTOR

4]

Proof. Represent the vectors a, b, c, r, by line segments OA, OB, OC, OP respectively. Through P draw a plane parallel to the plane OBC to cut the line OA at A' (this may always be done, by the data). Let P' be the point in the plane OBC such that OP' = A'P. Through P' draw a line parallel to CO to cut OB at B' (this again may be done using the data). The constructions are illustrated in Figure 2.7. r = OA' +A'P = Aa + OP' = Aa + OB' +B'P' Aa+,ub+ vc.

Fig. 2.7

We have thus proved our Existence Theorem by showing that, with the given data, a set of components may be constructed. To demonstrate the truth of the Uniqueness Theorem we employ a common device in mathematical arguments: we assume that it is false and show that this leads to a contradiction (see Chapter 9). Theorem 2.2 (The Uniqueness Theorem). The solution shown to exist in Theorem 2.1 is unique. Proof By Theorem 2.1 a solution exists, say r = Aia+,u,b+vic. Assume that a different solution also exists r = A2a±,u2b±v2c where, say, Al

A2. Then

Aia -kitib+ Pi c = A2a+,u2b+v2 c. and so But Cui — 2-2

— AD a =

— /22) b + (v1 — v2) c.

b +(v, — v2) c is a vector in the plane determined by b and c 29


[2

and any non-zero multiple of a cannot lie in this plane, by the data. Thus = A2, which contradicts our initial assumption. Theorems 2.1 and 2.2 have been proved for three dimensions; they hold equally well in two dimensions, for a vector r lying in the plane determined by a and b, where a kb. An expression such as Aa +#b + vc is called a linear combination of the vectors a, b, c. The theorems we have just proved may be restated in the form 'Given three non-parallel, non-coplanar vectors a, b, c any non-zero vector r may be expressed uniquely as a linear combination of a, b, c'. If numbers A, #, v (not all zero) exist such that Aa+,ub+vc = 0 the vectors a, b, c are said to be linearly dependent; if no such numbers exist, a, b, c are said to be linearly independent. *Ex. 10. Prove that, if 0, A, B, C are coplanar, then the position vectors a, b, c are linearly dependent and, conversely, if a, b, c are linearly dependent, then 0, A, B, C are coplanar.

*Ex. 11. Prove the Existence and Uniqueness Theorems for components in two dimensions. Ex. 12. ABCD is a parallelogram, and E is the mid-point of CD. AB = a, AD = b, AE = x, BE = y. Express x and y in terms of components in the directions a and b and also express a and b in terms of components in the directions x and y.

Ex. 13. Do either the Existence or the Uniqueness Theorems hold in three dimensions if four directions a, b, c, d are given? Exercise 2 a 1. If v = a +2b, w = 2a—b, express the following vectors in the form A.a (i) v+w; (ii) 2v +3w; (iii) v — 3w; (iv) 2(v—w). 2. If u = a +3b, v = 2a— b, express a and b in terms of u and v. 3. If u = a + b + c, v = a +2b +c, w = a— b— 2c, find, in terms of a, b, c: (i) u+ v+w; (ii) u—v—w; (iii) 2u+ v—w; (iv) u+ 2v + 3w. 4. If u = a+b—c, v = 2a— b+c, w = 3a +2b+c, find a, b, c in terms of u, v, w. 5. If i and j are perpendicular vectors of unit magnitude, i pointing due east and j due north, find the magnitudes and directions of the following vectors: (i) 2i; (ii) —3j; (iii) 3i + 4j; (iv) i— j; (v) —3i +4j. 30

4]

COMPONENTS OF A VECTOR

6. If p is a vector of magnitude 2 pointing due east and q is a vector of magnitude 5 pointing due south, find the magnitudes and directions of the following vectors: (i) 6p +q; (ii) 6p—q; (iii) — p— q. 7. If p is a vector of magnitude 1 pointing due east and q is a vector of magnitude 2 pointing north-west find (by accurate drawing, if you wish) the vectors (i) 2p + q; (ii) p — q; (iii) — p— q; (iv) — 4p + 2q. 8. x is a vector of magnitude 1 pointing due east; y is a vector of magnitude 2 pointing south-west; z is a vector of magnitude 20 pointing due north. Express z in the form Ax +gy, giving the explicit numerical values of A and #. 9. If a is a vector of magnitude 1 pointing due north, b is a vector of magnitude 1 pointing N 20° E and c is a vector of magnitude 2 pointing N 70° W, express c in the form c = Aa +/A. If d is a vector of magnitude 2 pointing N 10° E, find approximate values for A', where d = A'a +,u'b. 10. If i, j, k are vectors of unit magnitude pointing respectively due east, due north and vertically upwards out of the plane containing i, j, find the magnitude of the vector u = 4i+ j + 8k. If v = 4i — j + 8k and w = 4i — j — 8k, find the angles between the vectors (i) u, v; (ii) v, w. 11. Figure 2.8 represents a lattice of congruent parallelograms; OA = a, OB = b, AB = c. Express OP in terms of (i) a, b; (ii) a, c; (iii) b, c;

Fig. 2.8 and do the same for the vectors OQ and OR. Express PQ in terms of a, b, and RQ in terms of b, c. Find two different expressions of the form aa + /3b +7c for the vector OQ, with none of a, fl, y zero. Show that, if one of these expressions is subtracted from the other, the known relation b = c+ a is obtained. 12. In Figure 2.8 OC = u, OD = v, QP = w. Obtain expressions for OP, OQ, OR in terms of u, v and hence obtain w in terms of u and v. Why is it not possible to express CD in terms of the two vectors PQ, AR? 13. ABCDEF is a regular hexagon, and AB = a, BC = b. Find CD, DE, EF, FA in terms of a and b. 14. ABCDA'B'C'D' is a cuboid whose base ABCD is a square of side 2 units. The sides AA', etc., are vertical and of magnitude 4 units. E is the mid-point 31


[2

of AB, G is the mid-point of B'C' and F is the mid-point of CC'. a, b, c are three vectors, each of magnitude 1 unit in the directions AB, AD, AA' respectively. Find, in terms of a, b, c the vectors DE, AF, EF, GF, GE. 15. With the data of Question 14, express ED', EG and EF in terms of components in the directions a, b and c and also express a, b, c in terms of components in the directions ED', EG and EF. 16. OABCO' A'B'C' is a rectangular box with square ends OABC and O'A'B'C'. 00', AA', BB', CC' are parallel edges. If AO = OC =1 unit, 00' = 2 units, and if i, j, k are unit vectors (that is, vectors of unit magnitude) along OA, OC, 00' respectively, find, in terms of i, j, k:

(i) OC'; (ii) OB'; (iii) C'A; (iv) OD, where D is the point on C'B' produced such that C'B' = B'D; (vi) MD. (v) OM, where M is the mid-point of AA'; 17. OABC is a regular tetrahedron of side a; p, q, r are unit vectors (see Question 16) along OA, OB, OC respectively. Find in terms of p, q, r: (i) AB; (ii) OD, where ABCD is a rhombus. 18. ABCD is a square and P, Q are the mid-points of BC, CD respectively. Find, in terms of u = AP and v = AQ, (i) AB; (ii) AD; (iii) BD.

19. With the notation of Question 16, OB' = u, AC' = v, OB = w; find, in terms of u, v, w: (i) OA'; (ii) BC'; (iii) OB'; (iv) BC. 20. With the notation of Question 19, if CA' = x, find two distinct expressions for CB' in terms of u, v, w, x. 21. ABCDA'B'C'D' is a cube, with faces ABCD, A'B'C'D' and edges AA', etc. Find, in terms of AD', AC', AC, the vectors D'B and A'C. 22. ABCDEFGH is a regular octagon. If AB = a, BC = b, find, in terms of a, b,

the displacements CD, DE, EF, FG, GH, HA.

5. APPLICATIONS TO GEOMETRY So far we have developed the algebra of vectors using geometrical arguments; we now reverse the process and show that the algebra of vectors may usefully be employed to deduce geometrical results. The vector treatment of geometrical problems has the great advantage that it is equally applicable to two or three dimensions. To describe a geometrical configuration consisting of a number of points A, B, C, ... we must be able to locate each point. This may be done by taking a fixed point 0 (called the origin) and referring to a point A by the line segment OA. If a is the vector representative of all the line segments equal and parallel to OA, then a is called the position vector of the point A with respect to the origin 0. Thus, given an origin 0, we may 32

5]

APPLICATIONS TO GEOMETRY

refer to all points in the plane by their position vectors relative to this origin 0. First, we establish a result known as the Section Formula. This theorem enables us to write down the position vector of any point on a given line and so is of importance in setting up a vectorial description of a geometrical configuration. Theorem 2.3 (The Section Formula). If APB is a straight line, with APIPB = it, and if the position vectors of A, P, B, relative to any origin 0, are a, p, b respectively, then Ab ua P — A d_itt Proof. We have (Figure 2.9) AB = b — a and

AP = p — a.

But

A AB AP = , A+ p,

and so

(A +,u) (p — a) = A(b — a),

or p = Ab+,ua (A and the result follows. Notice that the proof holds equally well for positive or negative values of the ratio Al it. Notice also the important special case in which A = it: the mid-point of AB has position vector +(a + b).

Fig. 2.9

Fig. 2.10

Example 1. Show that the medians of a triangle ABC are concurrent and find the position vector of the centroid (meet of the medians) in terms of the position vectors a, b, c, referred to some origin 0. Let the mid-points of BC, CA, AB be L, M, N respectively and call their position vectors 1, m, n (Figure 2.10). 33


[2

Then, by the Section Formula

1=

e), m = +(c+ a); n =(a+b).

Any point on AL has a position vector Al +/la A -FA '

4A(b+c)+ Aa

i.e.

A+ A

For varying values of A and we obtain different points of the line AL; in particular, if we choose A = 2, is = 1, we obtain a point whose position vector is symmetrical in a, b, c and so lies equally well on BM or CN. Thus, the medians of a triangle are concurrent at the centroid, G, whose position vector is given by

g—

a+b+c 3 •

Notice, incidentally, that we have shown that AG/GL = 2, etc. Example 2. Show that the mid-points of the sides of a skew quadrilateral form the vertices of a parallelogram. Let ABCD be the skew quadrilateral and the mid-points of AB, BC, CD, DA be P, Q, R, S respectively. Take any point 0 as origin, let a denote the position vector of A with respect to 0 and similarly for the other points. Then we have, on using the Section Formula,

p=

+b), q = i(b + c), r = 1(c + d), s = (d+ a)

and so

p q = -1(a — c)

and

s— r =

—

c)

Thus QP = RS and so PQRS is a parallelogram (one pair of sides equal and parallel). Ex. 14. Show that the joins of mid-points of opposite edges of the tetrahedron ABCD are concurrent at a point G. Ex. 15. Show that, in a tetrahedron ABCD, the joins of vertices to the centroids of opposite faces are concurrent at the same point G as that obtained in Ex. 14. *Ex. 16. What condition must the numbers a, /3 possess for the points with position vectors a, b and aa +fib to be collinear? 34

APPLICATIONS TO GEOMETRY

The Section Formula may be restated in a form that gives a test for the collinearity of three points whose position vectors are known. Theorem 2.4 (condition for three distinct points to lie in a line). Three points A, B, C have position vectors a, b, c. Then (i) V A, B, C lie on a straight line, there exist numbers ; /3,7, not all zero, such that aa+fib+yc = 0 oc-1-13+7 = 0.

and

Conversely, (ii) if there exist numbers a, y not all zero such that cla+ fib+yc = 0 fld-y = 0 and a+/3+y then A, B, C are collinear. Proof. (i) Given that A, B, C are collinear, suppose AB/BC = Alit. Then by the Section Formula pa + Ac b= 1H-A

pa—(,1+A) b+Ac = 0.

i.e.

Take a = — + A), y = A and the result follows. (ii) Given that numbers a, y exist such that aa+fib+7c = 0 and we have

a+/3+y = 0 as +fib = (a +ft) c, by substitution.

Now we know that a, f3 y are not all zero and we may assume, without any loss of generality, that y 0. It follows that a +13 + 0 and so ,

c=

as + fib +,3 '

i.e. c is the position vector of the point dividing AB in the ratio fl:a, from which it follows, that A, B, C are collinear. Example 3 (Desargues's Theorem). Two triangles ABC, A'B'C' (not necessarily in the same plane) are so positioned that AA', BB', CC' all pass through a point V. BC, B'C' meet at L; CA, C' A' at M; AB, A' B' at N. Prove that L, M, N are collinear. 35


[2

Denoting the position vectors of the various points in the usual way, we have, using Theorem 2.4: fv+aa+a'a' = 0, jl

1+a+a' = 0;

fv+fib+rb' = 0, 1. 1+fl+fi' = 0; _fv+7e+7'e' = 0, 1 1+7+7' = O.

Fig. 2.11

From the second and third pairs of equations fib — ye_ fi'b' —7'e' /6-7 — and, by the Section Formula, these must both represent 1. Thus Similarly, Thus

(3-7)1 = flb— ye. (y — a) m = ye— aa and (a— ,G) n = as—fib. { (i6'

—7)1+

— cc) m+(cc—fl) n = 0,

(8-7)+(7—ct)+(a —fl) = 0 and so L, M, N are collinear. Ex. 17. If ABC, A'B'C' are two triangles not in the same plane and such that AA', BB', CC' all pass through a point V, prove Desargues's Theorem by showing that the meets of corresponding sides of the two triangles must all lie on the line of intersection of the planes ABC, A'B'C'. Ex. 18. Prove the Converse of Desargues's Theorem.

36

6]

EQUATIONS OF LINES AND PLANES

6. THE VECTOR EQUATIONS OF LINES AND PLANES Throughout the present section we shall adhere to the generally accepted custom that P denotes a variable point whose position vector is r. A locus is a set of points (in a plane or in space) subject to some condition; examples are a straight line, a plane, a sphere, the interior of a sphere, etc. In this section we shall confine our attention to lines and planes. The equation of a line is the algebraic condition that is satisfied by the positions vector r of a general point P lying on the line. Similarly for the equation of a plane. Later we shall meet loci that are defined by inequalities; the interior of a sphere would be such a locus. A straight line is completely specified if two points A, B of the line are given. Suppose that an origin 0 is taken and that the position vectors of A and B relative to 0 are a and b respectively. The equation of the line AB is the condition satisfied by the position vector r of a general point on AB. (Of course, if a different origin 0' were chosen, a different equation would be obtained: the equation of any locus depends upon the choice of origin.) Now, since P lies on AB, or AB produced (Fig. 2.12), AP = AAB, where A is a number. For different points on the line, different values of A are taken. This equation may be rewritten in terms of position vectors as

r — a = A(b— a) which represents the equation of the line AB.

Fig. 2.12

A plane is completely specified by three (non-collinear) points A, B, C (see Chapter 3, Section 5). For a general point P of the plane, AP is a vector lying in the plane of AB and AC and so we may split AP into components in these two directions (remember that A, B, C were non-collinear and so two directions are defined). Thus AP = AAB ±,ctAC (Fig. 2.13) and so, in terms of position vectors relative to some origin 0,

r— a = A(b—a)+gc— a) 37


[2

which may be rewritten as r = —(A+ F-1) a+Ab+,uc. Our final theorem of this chapter is essentially a re-phrasing of the result above. It is structurally very similar to Theorem 2.4 and the reader is advised to refer back to that theorem before continuing. Theorem 2.5 (condition for four points to lie on a plane). Four points A, B, C, D have position vectors a, b, c, d. Then (i) If A, B, C, D lie on a plane, there exist numbers a, ft, y, 8, not all zero, such that aad - igb+yc+8d = 0, a+fl+y+8 = 0. Conversely, (ii) If there exist numbers a, fl, y, 8, not all zero, such that aa-Fflb+yc+ad = 0, a+fi-Fy+8 = 0 then A, B, C, D are coplanar. Proof. (i) Given that A, B, C, D are coplanar, d satisfies an equation of the form r = —(A+11-1)a+Ab+ite, i.e.

—(A+,a-1)a+Ab+icc—d = 0

and the first result follows if we set cc = — (A -Fit -1), A' = A, Y = It, a = —1. (ii) Given that numbers a, ft, y, 8 exist such that aa+flb+yc+ad = 0, a-1-13+y+8 = 0. We have

—(13±y±8)a+ Ab+yc+ad = 0,

i.e.

fi(b— a) + y(c — a) = 8(a — d),

or

fiAB+yAC = &DA.

But AB, AC both lie in the plane ABC and so therefore does (MA. It follows that D lies in the plane ABC and the proof is complete. *Ex. 19. Show that the equation of the line through A parallel to the direction defined by the vector u is r = a+ Au. 38

6]


*Ex. 20. Show that the equation of the plane through A parallel to the directions defined by the two vectors m, n is r = a + Am +mi. Ex. 21. Show that the mid-points of two pairs of opposite edges of a tetrahedron ABCD are coplanar. Example 4. ABCD is the base of a cube whose vertical edges are AA', BB', CC', DD'. X is the point of trisection of BB' nearer B, Y is the point of trisection of CC' nearer C' and M is the mid-point of BC. If D'M cuts the plane AXY at Z, find the ratio in which Z divides D'M. Take side of cube as 6 units and call vectors of unit magnitude in the directions AB, AD, AA' respectively i, j, k.

Working with A as origin, and denoting the position vector of X by x, etc., we have x 6i +2k, y = 6i+6j+4k and so the position vector of any point on the plane AXY is given by r = Ax+ity = 6(A+µ) i +6,4 +2(A +2,u) k. Again,

m = 6i + 3j, d' = 6j + 6k

and so

D'M = 6i — 3j— 6k. 39


[2

The position vector of any point on D' M is given by

r = AD' +D'P = d' + vD'M = 6j + 6k + v(6i — 3j— 6k) = 6vi+ (6 — 3v) j + (6 — 6v) k. Thus, the point of intersection is given by 6vi + (6 — 3v) j + (6 — 6v) k = 6(A+ tc) + 6,aj +2(A + 2/t) k. But i, j, k are non-coplanar vectors so, by the Uniqueness Theorem, v = A+,u, 1

= ,u,

3-3v = A+2,u and thus Thus D'Z:ZM = 4:3.

V =4

Exercise 2 (b) 1. ABC is a triangle. U lies on AB produced so that AB = }AU, V lies on AC so that AV = 2VC. Find, in terms of the position vectors a, b, c of A, B, C, the position vectors of U, V and the mid-point of UV. 2. Draw a diagram showing the relative positions of the points whose position vectors are a, b and 3a — 2b. Prove that these three points are collinear. 3. If G is the centroid of the triangle ABC, prove that GA + GB + GC = 0. Suggest a generalization of this result. 4. ABC is an equilateral triangle of side 3 cm. P, Q lie on BC, CA respectively and are such that AQ = CP = 2 cm. R lies on AB produced so that BR = 1 cm. Prove that PQR is a straight line. 5. OABC is a parallelogram and the position vectors of A, B, C relative to 0 are respectively a, b, c. M is the mid-point of BC. Write down, in terms of b and c, and hence in terms of a and c, the position vector of a general point X on the line OM. Deduce that, if X lies on AC, then X is the point of trisection of AC nearer C and also the point of trisection of OM nearer M. 6. OAB is a triangle; M is the mid-point of AB and T is the point of trisection of OB nearer B. TM produced meets OA at X. If OA = a and OB = b, write down the position vectors of T and M, and hence of a general point P of TM. Deduce the position vector of X and find a relation between the points 0, A and X.

40

6]


7. The position vectors of the vertices A, B, C of a triangle are respectively a, b, c. M is the point of trisection of AC nearer A, N is the point of trisection of AB nearer B. Write down the position vectors of M and N and hence of general points of BM, CN. Deduce the position vector of X, the intersection of BM and CN. 8. ABCDA'B'C'D' is a parallelepiped, with parallel edges AA', BB', CC', DD'. U is the point of trisection of AA' nearer A, V is the point of trisection of C'D' nearer C' and W is the mid-point of B'C'. With A as origin, the position vectors of B, D, A' are respectively b, d, a'. Write down the position vectors of U, V, W and a general point of DD'. Locate the point at which the plane UVW cuts DD'. 9. OUVW is a tetrahedron, the position vectors of U, V, W being u, v, w. A is a vertex of the parallelogram OUAV, B is the mid-point of VW, C is the reflection in the origin of the point of trisection of UV nearer U. Find, in terms of u, v, w, the position vectors of A, B, C and hence find where the plane ABC cuts O W. 10. ABCD is a tetrahedron, P is the mid-point of AB, Q is the mid-point of AD and R is the point of trisection of AC nearer C. Taking A as the origin, write down the position vector of the general point X of the plane PQR in terms of the position vectors of B, C and D. Write down also, in terms of the same frame of reference, the position vector of the general point Y of the line DM, where M is the mid-point of BC. The plane PQR cuts the line DM at Z; find the ratio ZMIMD.

11. Four points P, Q, R, S in a plane through the origin 0 have position vectors OP, OQ, OR, OS given by 2i + 3j, 3i + 2j, 4i + 6j, 9i + 6j respectively, where i and j are given non-parallel vectors. Express the vectors PR and QS in terms of i and j. Show that the position vectors OA and OB of the points A and B on PQ and RS respectively, and such that PA/PQ = a and RB/RS = b, are (2 + a) i + (3 — a) j and (4 +56)1+ 6j, respectively. Hence determine the position vector with respect to 0 of the point of intersection of the lines PQ and RS. (J.M.B.) 12. Two vectors are represented by OP, OQ and R divides PQ in the ratio n: m. Show that mOP + n0Q = (m+ n) OR. The points D, E, F divide the sides BC, CA, AB of a triangle in the ratios 1:4, 3:2, 3: 7 respectively. Show that the sum of the vectors AD, BE, CF is parallel to CX, where X divides AB in the ratio 1:3. (J.M.B.) 13. ABCD is a skew quadrilateral and a plane cuts AB, BC, CD, DA at W, X, Y, Z respectively. Prove that AW BX CY DZ WB.XC• YD' ZA = 1.

Suggest a generalization of this result. 41


[2

14. ABCD is a parallelogram and 0 is a point in the same plane. OD cuts AB at P and BC at Q; OB cuts CD at R and DA at S. Prove that PS and QR are

parallel. 15. OABC is a square of side 2a. i, j are unit vectors along 0A,OC. The mid-point of AB is L; the mid-point of BC is M; OL, AM meet at P; BP meets OA at N.

Show that the segment OP can be measured by the vector A(2ai + aj) and also by the vector 2ai +,u(2aj — ai). Hence determine A and ,u. Prove that ON = BOA. (0 & C)

Miscellaneous Exercise 2 1. ABCDE is a regular pentagon. AB = a and AE = b. Show that CD = (b — a)/(1 +2 cos 72°) and express BC and ED in terms of a and b. 2. ABC is a triangle. L divides BC in the ratio 1 : 2, M divides CA in the ratio 1:2 and N divides AB in the ratio 1:2. Prove that the triangles ABC and LMN have the same centroid. Suggest a generalization of this result and prove your assertion. 3. Referred to some origin 0, the position vectors of A, B, C, D are a, b, c, d respectively. Express in terms of a, b, c, d the displacements BD, AC. What can be said about the quadrilateral ABCD: (i) Ifa+c = b+d; (ii) if la—cl = lb—di; (iii) if both (i) and (ii) hold? 4. The triangles ABC, A'B'C' have centroids respectively at G and G'. Prove that AA' +BB' + CC' = 3GG'. 5. P, Q are the mid-points of the sides BC and CD respectively of the parallelogram ABCD. Prove that: AB+AC+AD =4 (AP+AQ). 6. ABCD is a tetrahedron and B', C', D' lie on AB, AC, AD produced. The centroids of the triangles BC'D', B'CD', B'C'D are G1, G2, G3 respectively and the centroids of the triangles B'CD, BC'D, BCD' are H1, H2, H3respectively. If the centroids of the triangles BCD, G1G2G3, H1ll2H3are F, G, H respectively, prove that FGH is a straight line. 7. ABCD is a parallelogram and points X, Y are taken on the diagonal BD such that BX = YD. Prove vectorially that AXCY is a parallelogram. 8. ABC is a triangle and Y, Z are points on AC, AB respectively such that ZY is parallel to BC. BY and CZ meet at P. Prove vectorially that AP produced bisects BC. 9. ABC, A'B'C' are two skew lines (that is, no plane contains both lines). If AB:BC = A'B':B'C' prove that the mid-points of the lines AA', BB', CC' are collinear. Is the converse true? 42


6]

10. Prove that, if ABC is a plane through the origin 0 from which position vectors are measured, and if A, B, C have position vectors a, b, c, then there necessarily exits a relation of the form pa +qb+rc = 0 (p, q, r * 0) where 0 is the zero vector. The lines AO, BO, CO meet BC, CA, AB in L, M, N respectively. Prove that the position vector of L is c. b+ q+r q+r Deduce that

BL.CM . AN

LC .MA .NB — +1,

where the magnitude and sense of each line segment is taken into account. (M.E.I.) (The result proved in Question 10 is known as Ceva's Theorem: it and its converse are very useful for proving concurrency theorems.) 11. If a transversal cuts the sides BC, CA, AB of a triangle ABC at L, M, N respectively, prove that BL. CM. AN _ —1 LC . MA .NB magnitudes and senses of each line segment being taken into account. (This result is known as Menelaus's Theorem: with its converse it is useful for proving collinearity properties.) 12. ABCD is a plane quadrilateral. AB and DC meet at P; BC and AD meet at Q. Prove that the mid-points of AC, BD and PQ are collinear. 13. P, Q are variable points on two skew lines. Find the locus of the mid-point of PQ. Can you generalize this result? 14. ABCDA'B'C'D' is a parallelepiped, with ABCD, A'B'C'D' congruent parallelograms and AA', etc., edges. The tetrahedron ACB'D' is inscribed in the parallelepiped. Prove that AC' passes through the centroid X of B'CD' and deduce that the joins of the vertices of the tetrahedron to the centroids of opposite faces are concurrent at a point G. Determine the ratios AG: GX: XC'. 15. Two vectors a and b, such that b is not a multiple of a, are given in a plane. Two other vectors c and d are defined by the equations c = yia+y2b, d

= ai a+82b.

Prove that any vector ma +fib can be expressed in terms of c and d provided 0. Find the coefficients of c and d. Explain the geometrical significance of the condition y182 — y2Si *0. (0 & C)

Y3.82 — Y2.61. *

43


[2

16. Let 0, A, B, C be four distinct points in three-dimensional space, no three of which are collinear. The position vectors of A, B, C with respect to 0 are a, b, c respectively and X is the point given by x = Aa -FA+ vc. Prove that: (i) If Xis a point of the line AB, then A+ it = 1, v = 0. (ii) If X is a point of the plane ABC, then Al- p + v = 1. (iii) X is in the interior of triangle ABC if and only if A- F p+v = 1 and A > 0, > 0, v > 0 and indicate on a sketch the regions in the plane ABC in which X lies for other combinations of the signs A, tt, v. Obtain an expression for the position vector x of a general point X in the interior of the tetrahedron (M.E.I.) OABC and find the values of A, 1.1,v corresponding to the centroid.

44

3.

Coordinates

1. UNIT VECTORS A unit vector is a vector of unit magnitude. Thus, a unit vector is a sort of ' signpost ' giving a direction and sense; any vector may be split into the product of a number (its magnitude) and a unit vector with the required sense and direction. Thus we may write r = rP, where r = In and t is a unit vector in the direction of r. (The notation employed here is useful and should be adopted by the reader: given any vector x, its magnitude may be written simply as x while a unit vector with the same sense and direction as x is written x.) Unit vectors are useful in that they enable us to deal separately with the magnitude and direction of a given vector. Now suppose we choose two directions in a plane and specify them by unit vectors i and j. By the Existence and Uniqueness Theorem for a plane (see Chapter 2) any displacement AB in the plane, represented by d may be split into components in the i and j directions

d = pi+ qj where p and q are uniquely determined. The vector d, which is representative of a whole class of displacements, of which AB is a typical member, may be written in the alternative form d=

.

1 Thus, ( —1) represents the vector 2i — j, ( ) the vector i and (_ 10) the 0 vector — j. Of course, the representation of a vector in this new notation depends upon our original choice of base vectors i and j. Again, there is no necessity for the base vectors to be unit vectors, though they will almost invariably be so. To illustrate these last two remarks, consider two unit vectors i and j and the vector a where

a = 3i — j. Now and so we have

a = (i + j) + 2(i — j) a =— 3) with i, j as base vectors, 45

[3

COORDINATES

while

a = ( ) with (i + j), (1— j) as base vectors. 2

Notice that (i + j) and (i — j) are not unit vectors. (For example, if i and j are perpendicular, neither (i+j) nor (i—j) is a unit vector.) Similarly, in three dimensions, if non-coplanar base vectors i, j, k are chosen, any vector d may be split into three components in the i, j, k directions d = pi+qj+rk and d may be represented in the alternative notation as

d = (q) 11 relative to the base vectors i, j, k. We shall now adopt the convention that i, j, k represent unit vectors each one of which is perpendicular to the other two. Furthermore, positive senses are determined by the following rule: if the thumb, first and second fingers of the right hand are splayed out at right angles to one another, the thumb points in the positive i direction, the first finger in the positive j direction and the second finger in the positive k direction. i, j, k are then said to form a right-handed orthogonal triple of unit vectors. (Orthogonal means ' at right-angles to one another'.) Similarly, but more simply, one may define a right-handed orthogonal pair of unit vectors i, j for a plane.

Example 1. If a = i +2j+3k, express a as a column vector (i) with i, j, k as base vectors; (ii) with i+ j, i—k, k — 2j as base vectors. (i) With i, j, k as base vectors, 1 a = (2) . 3 (ii) Write u = i+j, v = i—k, w = k-2j. Solving for i, j, k we have i = 1(2u+ v + w); j = Yu— v — w); k = 1(2u — 2v + w). Thus,

a = i+2j+3k = (iu + +1w) + (3u— iv — iw) + (2u — 2v + w) = Vu—iv+iw

46

1]

UNIT VECTORS

and we have, with u, v, w as base vectors, 10) 3

a= ( —i . a3 Ex. 1. Rewrite 3i +4j— k and (i + j) — 2(j — 2k) in the column vector notation, with i, j, k as base vectors. Ex. 2. Express 3i— j—k as a column vector with (j+k), (k +1), (i+ j) as base vectors. Where do you use the Uniqueness Theorem in your solution? Ex. 3. Evaluate a+ b, 2a— 3b and 3(a— 2b) where ( 1

2)

0), b=

a=

—1

1. —1

Ex. 4. If 2 ( 3) x = (— y) 4 —1 —2 what are the values of x and y? Explain how you make your deductions.

2. COORDINATES Suppose we now have some geometrical configuration in a plane which we want to describe algebraically. Let us choose any point 0 of the plane as origin; then all the points of the plane may be specified by their position vectors with respect to 0. To give more detailed information, these position vectors may be split into their components in the direction defined by the right-handed pair of orthogonal unit vectors i, j. Thus, if

r = OP

we may write

r = xi +yj.

x and y are called, respectively, the x and y coordinates of the point P and are usually written as (x, y). The lines through 0 in the i and j directions are called the x and y axes respectively, and may be denoted by Ox and Oy. In Figure 3.1 the following points have been plotted : A(1, 3); B(3, — 1); C( 1, —2); D( 3, 0). —

—

0 C

•B

.

Fig. 3.1

Note that the x coordinate is always written first. 47

COORDINATES

[3

1 The reader should note the distinction between (1, 3) and ( ) : the 3 first gives the x and y coordinates of the unique point A, while the second denotes the vector (referred to i, j as base) representative of the class of displacements of which OA is a typical member. Coordinates may similarly be defined in three dimensions, though now we require a right-handed triplet of orthogonal unit vectors i, j, k to define the directions of the coordinate axes OX, O Y, OZ through the origin 0. Thus, if P is the point (2, - 1, - 3), the position vector of P is 2i - j - 3k. Note again that the coordinates are given in the order x, y, z. Ex. 5. Draw a sketch to denote the approximate positions of the points: A(1, 0), B( -2, - 1), C(1, -3), D( -2, 0), E( -1, -2). Ex. 6. Write down the position vectors of the points A(1, - 2), B(3, 4). Deduce the coordinates of the mid-point of AB. Can you state a general rule for finding the coordinates of the mid-point of a line ? Ex. 7. Write down the position vectors of the points A(1,

-

1),

B(5, -5).

Use the Section Formula to deduce the coordinates of the two points of trisection of AB. Ex. 8. What are the coordinates of the mid-point of the line joining A(0, -1, 2) and B(2, 3, 2)? Ex. 9. If ABCD is a parallelogram, what relation must hold between the position vectors a, b, c, d? Find the coordinates of the vertex D of the parallelogram ABCD whose other vertices are given by A(1, 2), B(4, 3), C(3, 5). Ex. 10. The coordinates of the points A, B, C are as follows: A( -1, 1, 2), B(1, 0, - 3), C(0, 2, 4). Find the coordinates of the fourth vertex D of the parallelogram ABCD and of the fourth vertex E of the parallelogram ACBE. Ex. 11. Find the coordinates of the centroid of the triangle ABC of Ex. 9. Ex. 12. A rectangular box ABCD A'B'C'D' has base ABCD and the edges AA', BB', CC', DD' are all vertical. M, N, P are the mid-points of BC, CC' and A' B' respectively. AB has magnitude 3 units, BC 2 units and CC' 1 unit. If A is taken as origin and unit vectors i, j, k are taken along AB, AD, AA' respectively, find the position vector of the centroid of triangle MNP and deduce its coordinates. 48

3]

DISTANCES IN TERMS OF COORDINATES

3. DISTANCES IN TERMS OF COORDINATES Before proceeding with our study of coordinates we shall develop an economic notation whose value should be readily apparent to the reader. It would be convenient to refer to all points by using, say, the letter P, to all x coordinates by x, to all y coordinates by y and to all z coordinates by z but to do so without further clarification would clearly lead to appalling confusion. We may, however, differentiate between the various points and their coordinates by the use of suffixes : thus we may call two points P1 and P2and take as their coordinates (x1, yi, z1) and (x2, y„ z2) respectively. One advantage of such a notation is obvious: without further explanation we know that Pt, stands for a point, that xnis its x coordinate and so on. (The reader must guard against confusion between suffixes and indices: x2simply means the x coordinate of the point 132, whereas x2represents the result of multiplying the number x by itself.) The choice of a good notation is often more than half the battle in the solution of a mathematical problem; as he gains experience in its use, the reader will come to appreciate the deeper significance of the suffix notation. One evident further advantage may, however, be noted here: the use of suffixes effects a considerable economy in notation if we have to deal with a large (or indeterminate) number of points. (It should be noted that not all sets may be enumerated, that is, counted in the form 1, 2, 3, .... In such cases, the suffix notation as we have presented it breaks down. Consider, for example, the problem of naming all the points on the line segment joining A(0, 0) and B(1, 0).) Now consider a plane and two points P1and P2 lying in it whose coordinates relative to perpendicular axes through some origin 0 are (x1, yi) and (x2, Y2).

>- x Fig. 3.2

From Figure 3.2 we have P2P1 = P2Q + = (x 1 —x2) i + 0,1—Y2) j. Thus, applying Pythagoras's Theorem to the right-angled triangle P2P1Q, P1 P2 = [(Xi X2)2+ (Y1 Y2)21

49

COORDINATES

[3

For three-dimensional coordinates the same argument applies but Pythagoras's Theorem must be applied twice. In Figure 3.3, the feet of the perpendicular from P1, P2to the plane Oxy are Q1, Q2, and RQ1, Q2R are respectively parallel to Oy and Ox. Thus we have

+ =+R(21+ QPI = + (yi y2 - ) j+(z1 -z2) k

P2P1 =

and so

P2 Q

P1P2 = AiRxi - x2)2 + (Yi

+(z1-z2)2]

on applying Pythagoras's Theorem to triangles P1P2 Q and Q1Q2R.

Fig. 3.3 Example 2. (i) The distance between A(3, -1) and B(-1, 2) is

/[(3 - - 1)2 + (- 1 -2)2] = 5.

0

(ii) The distance between A(1, 0, -2) and B(-1, 2, - 3) is 4J[(1 - - 1)2 + (0 - 2)2 + ( - 2 - -3)2] = 3. Ex. 13. A, B, C, D, E, F have coordinates as follows: C( - 1, - 2, 4), D(0, -4, 3); A(-1, - 2), B(- 3,2); E(a, - a, 0), F(-k, -k, k). Find the lengths of AB, CD and EF. Ex. 14. Show that ABCD is a rectangle, A, B, C, D having coordinates: A(1, 2), B(5, -1), C(8, 3),

D(4, 6).

Ex. 15. What is the fourth vertex of the parallelogram ABCD where A is (1, 1, 2), B(2, 0, -1),

C(3, 3, 0)?

Ex. 16. Find the circumcentre of the triangle

A(0, -1, 1), B(1, 0, 1), C(- 1, 1, 0). What is the circumradius? (The circumcentre of a triangle is the centre of the circle which passes through the three vertices.) -

50

3]

DISTANCES IN TERMS OF COORDINATES

Exercise 3(a) 1. Evaluate as single column vectors: 1 1 (i) (_3)— 1 _3 (1);

(ii) 2 ( 12) + 3 ( 10,• —2 ) —1

(i ) (1) (iii) ( ° 1) + 4 ( —1v ; 2 + 3 ( —11) + 2 (1) ; 3 —2 0 1 —1 1 2 1 1 +4 —1 . (v) 2 ( 2 —3 —2 1 —1 1 ( 1 —1 2. If a= ( 1) , b= ( 3) , c= —3 , 3 —3 —2 find, in column vector form, the vectors: (i) 2a—b; (ii) a + b— 3c; (iii) (3a + b)— 2(b + 3c). 3. If a, b, c are defined as in Question 2, solve for x the equations: (i) 2x = a— b; (ii) 3(x + a) = b— 2c giving your answers in column vector form. 4. Again using the notation of Question 2, solve for x, y, z the following systems of linear equations, giving your answers in column vector form: x+y—z = 2a+2b, {3x— y = a+ 2b, (ii) 2x+y+z = a+3b+2c, 0) x+2y = b—c; x— y— 3z = 4a-4c.

5. Two vectors p, q (where p t kq) are given in a plane. Referred to p, q as base, the vectors a, b, c, d are given by: a = (0)) b = 4 ( — 1 1 c = 2 (— 1) d = (— 4) . ' 4 ' 4 Express a, b, c, d as column vectors, taking p-2q, p+2q as base. 6. Three non-coplanar vectors u, v, w are given. Referred to these vectors as base vectors, the vectors, e, f, g, h are given by: 1 1 2 —1 e= (0) , f= (1) , g = (3), h = 4 ( 1) . 0 1 3 0 Express e, f, g, h as column vectors taking v +w, w +u, u + v as base vectors. 7. Write down the coordinates of the mid-points of the lines joining the following pairs of points: (i) (2, 4), (4, 6); (ii) (-2, 4), (4, 2); (iii) (— 1, — 3), (2, —4); (iv) (3, 1, —5), (2, —4, 0); (v) (a + b, a, a — b), (a —b, —a, — a — b).

51

COORDINATES

[3

8. Write down the coordinates of the two points of trisection of the lines joining each of the following pairs of points: (i) (1, 4), (4, 10); (ii) (- 2, 1), (1, 8); (iii) (a +2b, b -2a), (a-b, a+ b); (iv) (1, 2, - 1), (2, 1, -5); (v) (a, 2a, b), (a-b, b -2a, a+ b). 9. Find the lengths of each of the line segments of Question 7. (Leave your answers in the form Vm.) 10. A(3, 0), B(4, - 1), C(6, 2) are vertices of a parallelogram ABCD. Find the coordinates of D. Find also the coordinates of E if ACBE is a parallelogram. 11. Prove that the triangle whose vertices are A(-1, 2), B(3, 5), C( -4, 6) is isosceles. What is its area? 12. A(3, 1, - 1), B(1, 2, -2), C(0, 0, 2) are vertices of a parallelogram ABCD. Find the coordinates of D. What are the coordinates of the meet of the diagonals ? 13. Find the area of the triangle whose vertices are A(1, - 3), B(2, 5), C(- 4, - 3). 14. Show that the line joining A(-1, 4, - 3) and B(5, -8, 6) meets the x axis. If the point of intersection is C, find the ratio AC/CB. 4. COORDINATE GEOMETRY IN A PLANE;

STRAIGHT LINES AND THEIR GRADIENTS Throughout this section we shall confine our attention to a plane, so that only two coordinates are required to specify a point. The reader will recall that the vector equation of a straight line is the algebraic condition that must be satisfied by the position vector r of a point P if P is to lie on the line. In the same way, the Cartesian equation of a straight line is the algebraic condition that must be satisfied by the coordinates x, y, of a point P(x, y) of the line (and, conversely, no point P not on the line has coordinates satisfying the equation). The equations of lines parallel to the coordinate axes may readily be obtained. For example, the reader should have no difficulty in seeing that the line through (- 3, 1) parallel to the x axis is y = 1; similarly, the line through this point parallel to the y axis is x + 3 = 0. In particular, the equations of the x and y axes are respectively y = 0 and x = 0. A straight line is determined completely by two points on it. Suppose now that we wish to find the Cartesian equation of the straight line through P1(x1, yi) and P2(x2, y2), where x1 + x2 and yi + y2 and so the line is not parallel to the axes. Let the position vectors of P1and P2 be r1 and r2respectively. Then (see page 37) the vector equation of the line P1P2 is r = ri+ A(r2 -r1) where r1 = xii +y, j and r2 = x2i +y2 j. 52

4]

COORDINATE GEOMETRY IN A PLANE

This may be rewritten in column vector notation (with i, j as base) Y

= (x) a (x2— x1) + Y2 Y1 •

By the Uniqueness Theorem in two dimensions this gives the equations (x x1= A(x2 — x1), —

{

(Y —Yi = A(Y2 —Y1). Eliminating a between these two equations we obtain the following result. The Cartesian equation of the line joining the points P1(x1, Yi)P2(x2,Y2) is x—x1= y— yl x2 —x1 Y2—Y1 (The reader must note carefully the distinction in this equation between x, y on the one hand and x1, yl, x2, y2on the other: x, y are the coordinates of a general point P on the line; x1, yi, x2, Y2 are the coordinates of two specified fixed points of the line. If we take U as the set of all points (x, y) in the plane, the equation (1) is the defining condition for the set of points comprising the line; that is, the set {(x,

u: x — xl = x2 — xl Y — Yi Y2 —Y1) .

The same remarks hold for the vector equation r = r, + A.(r, —r1), in which r1, r2are the position vectors of two given points.) Equation (1) may be rewritten in the form Y— or where

=

Y2

Y1 (x

-

y—yl= m(x — x1), m—

(2)

Y2 - Y1 .X2 -

The number m is called the gradient of the line. The sign of the gradient tells us which way the line is sloping (see Figures 3.4 and 3.5). In case (i), x1 — x2and y,—y, have the same sign and m > 0; in case (ii) x, and Y2 have opposite signs and m < 0. In both cases, the positive value of Y1 — m is equal to tan a, where a is the acute angle made by the line with the x axis. Equation (2) gives the form for the straight line when a point on the line and its gradient are known.t t The reader who has already met trigonometric ratios of obtuse angles will realize that if the straight line AB meets the x-axis at P, then the gradient of AB is tan LxPA in all cases, where LxP A is measured in the positive (anticlockwise) sense from the x axis (see Chapter 6).

53

COORDINATES

ol

y

[3

>x

y

0 Fig. 3.4

X

Fig. 3.5

Equation (2) may be rewritten as y = tnx+c.

(3) Here the constant c represents the intercept cut off on the y axis, as may easily be seen by setting x = 0. Having derived these various forms for the equation of a straight line, we may easily deduce the result that any equation of degree 1 in the two variables x, y represent a straight line (hence the word linear usually applied to such equations). Consider the general linear equation in the two variables x, y: ax+by+c = 0. If a = 0 or b = 0 this represents a line parallel to one of the axes. If b 0, the equation may be rewritten in the form a c y = — x-b b and comparison with (3) shows that this is the equation of a line with gradient —alb and making an intercept —clb on the y axis. Example 3. Find the equation of the line L, through (2, —1) with gradient —1 and the equation of the line L2joining the points (0, 7) and (-1, 4). What are the coordinates of the intersection of L, and L2? Draw a sketch showing the relative positions of L, and L2and the coordinate axes. L, has equation y+ 1 = — -1-(x— 2), i.e. 54

x+2y = 0,


41

L2has equation

x —0 _ y —7 —1-0— 4 — 7'

i.e.

3x—y+7 = 0.

Since all points lying on L1satisfy the x + 2y = 0 and all points lying on satisfy 3x —y + 7 = 0, the point of intersection of the two lines must L2 satisfy both equations simultaneously. Solving we have

l

x+2y = 0, ax—y = —7

x = —2, y= 1

and so the point of intersection of L, and L2is ( -2, 1). (In this case the sketching of the lines L1 and L2is straightforward: we know two points on L2, and L1clearly passes through the origin. In general, to sketch the position of a straight line whose equation is given, first find the coordinates of the points where it cuts through the coordinate axes.)

Fig. 3.6

Ex. 17. Draw a sketch to show the positions of the straight lines: (i) 3x-2y-6 = 0; (ii) 2x+y + 3 = 0; (iii) 2x-5y = 0. Ex. 18. Find the equations of the lines through (2, —3) with gradients: CO 2; (ii) — i. Ex. 19. Find the equations of the lines joining the pairs of points: (ii) (0, 1), (2, —1); (i) (2, — 3), (3, 2); (iii) (-3, —1), (-1, 2); (iv) (1, 2), (-3, 2). Ex. 20. What are the gradients of the following lines: (i) 2x—y-3 = 0; (ii) x+y+1 = 0; (iii) 3x+4y+2 = 0?

Suppose now we have two lines, L1and L2, the gradients of which are respectively m1and m2. (i) If L1and L2are parallel, the angles that they make with the x axis are equal and

m1= m2.

(ii) If L1and L2 are perpendicular then either they are parallel to the coordinate axes or they have gradients of opposite sign: in the second case, 55

COORDINATES

[3

if m, = T- tan a, and m2 = + tan a2 then m,m2= — tan a, tan a2= —1, i.e. mi.m2 = 1.

Fig. 3.7

*Ex 21. With the notation above, prove that, if mi. = m2, L1and L2 are parallel. *Ex. 22. Again with the notation above, prove that, if m1m2= —1, Lland L2 are perpendicular. Example 4. Find the equations of the lines L1, through (1, —1) parallel to 3x— y +7 = 0 and L2, through (2, —3) perpendicular to 2x+5y +1 = 0. Find also the equation of the line L3joining the origin to the intersection of L1and L2. The gradient of 3x —y + 7 = 0 is +3 and so L1has equations y+ 1 = 3(x-1), i.e.

3x — y — 4 = 0.

The gradient of 2x + 5y + 1 = 0 is --I and so the gradient of a perpendicular line is +I. Thus, the equation of L2 is y +3 = i(x — 2), i.e.

5x-2y-16 = 0.

(With practice, the reader will soon be able to derive such equations more rapidly than is done here. For example, in the second case, any line perpendicular to 2x + 5y +1 = 0 is clearly of the form 5x-2y = k and, since the required line passes through (2, —3), k = 5 (2)— 2 (-3) = 16.) Now consider the equation (3x — y — 4) + A(5x — 2y — 16) = 0, where A is any number. This is a linear equation and so represents a straight line; furthermore, the point common to L1and L2 clearly satisfies this equation. Thus (3x —y — 4) + A(5x —2y —16) = 0 56

4]


represents a straight line through the intersection of L1and L2. If it passes through the origin, (0, 0), — 4 +A(— 16) = 0, A = —1 and the equation simplifies down to 7x — 2y = 0 which is the required equation of L3. Ex. 23. Find the equations of the lines L1, through (4, —10) parallel to 3x — y = 0 and L2, through (1, 2) perpendicular to 2x —y — 1 = 0. What are the coordinates of the intersection of L1and L2? Ex. 24. Find the line joining the point ( — 1, 2) to the meet of the two lines x+3y-1 = 0,x-4y+2 = 0. Ex. 25. Find the equation of the line through the meet of the two lines 3x—y+1 = 0, 4x-3y+2 = 0 perpendicular to the line 3x —y— 1 = 0. Ex. 26. Find the intercept cut off on the transversal 5x-2y+ 3 = 0 by the two parallel lines 3x—y + 1 = 0 and 3x —y + 4 = 0.

Exercise 3 (b) 1. Find the equations of the lines through the stated points with the given gradients: (i) (0, — 2), 2; (ii) (-1, — 1), —2; (iii) (3, —1), — 1; (iv) (a, b), —a/b. 2. Find the equations of the lines joining the following pairs of points: (i) (1, — 2), (2, 1); (ii) ( 2, 3), (— 1, 4); (iii) (2, — 1), (— 1, — 1); (iv) (a, b), (2a, --lb). -

—

3. Write down the gradients of the following lines: (i) 2x— 3y+ 1 = 0; (ii) 3x+ 6y + 2 = 0; (iii) 5x — y + 1 = 0; (iv) (a + b) x+ (a — b) y+ ab = 0. 4. Find the equations of the lines through (3, — 1) (a) parallel and (b) perpendicular to: (i) 2x— y + 3 = 0; (ii) 5x + 4y + 3 = 0; (iii) x + 7 = 0. In case (i), what is the distance between the two parallel lines?

5. Find the points of intersection of the following pairs of lines: 14x +3y— 6 = 0, (i)

(ax+by+a2 = 0,

(ii)

x— y— 5 = 0; bx— ay+ b2 = 0. 6. The two lines 3x— 5y— 7 = 0, 4x + 4y + 5 = 0 meet at the point A. Find the equations of the line through A (i) which passes through the origin; (ii) is parallel to 7x — y + 2 = 0; (iii) is perpendicular to 3x + 5y — 1 = 0.

57

COORDINATES

[3

7. Find the orthocentre of the triangle whose vertices are (0, 1), (1, 2), (4, 3). (The orthocentre is the meet of the altitudes of a triangle.) 8. The mid-points L, M, N of the sides, BC, CA, AB of a triangle ABC have coordinates (2, 1), (3, — 3), (4, — 5). Find the coordinates of A, B, C. 9. The coordinates of the vertices of a triangle are (6, 0), (— 1, 1) and (5, —7). Find the coordinates of the centre of the circumcircle. 10. The coordinates of the vertices of a triangle are (1, —4), (3, —2) and (- 11, 12). Find the coordinates of the centre of the circumcircle and determine the coordinates of the points where this circle cuts the axes. 11. Show that, for all values of A, the line whose equation is Ax +y = 1-2A always passes through the point (-2, 1). What is the equation (in terms of A) of the perpendicular line through (1, 2)? Show that, whatever the value of A, the intersection of these two lines always satisfies the equation xa ±y2 +x-3y = 0. What locus does this last equation represent? 12. OABC is a rectangle in which OA = 30C. M is the mid-point of OC and L is the point of trisection of CB nearer C. OL, AM meet at X. By setting up axes along OA and OC and assigning suitable coordinates to the various points, determine the ratio in which X divides OL.

13. Prove that, for all values of A, the line (1— A) x+ Ay = 3 — 7A

passes through a fixed point. What are the coordinates of the point? 14. Find the coordinates of the centre of the circumcircle of the triangle ABC, where A, B, C have coordinates (— 1, — 3), (-2, —2), (5, 5) respectively. Prove that the point P(6, 4) lies on the circumcircle and prove further that the feet of the perpendiculars from P on to BC, CA, AB are collinear. 15. Find the equation of the line joining A(a, 0) and B(0, b). A', B' are the feet of the perpendiculars from A, B to a variable line through the origin. If A'P, B'P are respectively parallel to the y and x axes, what can be said about the position of P? 16. What are the equations of the reflections of the line 2x— 3y+ 6 = 0 in (i) the x axis; (ii) the y axis; (iii) the line 2x — 3y = 0; (iv) the line 2x— 3y = 3 ?

5. COORDINATE GEOMETRY IN SPACE: THE PLANE AND STRAIGHT LINE The effective use of Cartesian coordinates in three dimensions requires rather more vector technique than we have at present at our disposal and so a more detailed study will be deferred until a later chapter (Chapter 11). 58

COORDINATE GEOMETRY IN SPACE

5]

We shall content ourselves for the moment with demonstrating the general form of the Cartesian equations of lines and planes in space. We may define a plane as a set of points in space which (i) contains at least three non-collinear points; (ii) has the property that, given any two points R, and R2 of the set, all points of the line R1R2 belong to the set. From this definition, three non-collinear points P1(x1, 311., z].),

P2 (x2, Y2, z2),

P3 (x3, Y3, z3)

clearly define a plane. If r1, r2, r3are respectively the position vectors of P1, P2, /33, we have seen that the plane P1P2P3is (see Chapter 2.6)

r = r, + A(r2 —r„)+p(r3 —r1). In column vector form this may be written (X1

Y1)

=

Z X)

Z1

X2 -XI

X3 -

Y3 — Yl

A (Y2 Y1) +

Z3 - Z1

Z1

which, by the Uniqueness Theorem, yields three equations:

x = x1+A(x2 —x1)+Ax3 —x1), Y = + A(Y2 YO+/-03

z = z,+A(z, — z„)+

z1).

Solving the first two equations simultaneously we obtain A and it as linear expressions in x and y. Hence, substituting these values of A and it in the third equation, we see that the Cartesian equation of a plane is linear in x, y, z. Now let us prove the converse result: a linear equation in x, y, z represents the equation of a plane. Consider the equation ax+by+cz+d = 0 where we shall suppose, for simplicity, that a + 0, b 0, c 0, d 0. (The fact that one or several of them may be zero requires a modification of the proof given below; see Ex. 30.) The three points (—d/a, 0, 0), (0, —d/b, 0), (0, 0, —d/c) certainly all satisfy the equation and are non-collinear (since the first two have zero z coordinate, for example). Thus condition (i) for a plane is satisfied. To demonstrate condition (ii), suppose P1and P2 satisfy the given equation. Then axi +byi + cz,+ d = 0,

f

3

PPM

ax, + by, + cz2 + d = 0 59

COORDINATES

[3

and thus, for any values of k and 1

lxi+ kx,\ + b k +1 1+

a(

k+1)

+c tlzi+kz2\ _ 0. k +1 I

But, by the Section Formula,

1lyi+ky2\ . 11z1+ kz2\ k k+1 k k+1 is the position vector of a point on the line joining P1and P2. Thus, the r—

lri+kr,11xi+kx,\ k+1 k k+1

P+ k

coordinates of any point on the line joiningP1P2satisfies the given equation. In three dimensions two intersecting (that is, non-parallel) planes define a straight line and we should therefore expect a line to be represented by two linear equations in x, y, z. The vector equation of the line through P, parallel to the vector

u = li+mj+nk is (see Chapter 2)

r = 1.1+ Au.

In column vector form this reads

1 (Y) = Y1) +A (m) and so, by the Uniqueness Theorem, we have three equations: x = xi + AI, y = + Am, Z = z1+ Art giving the equation of the line in the form — — y —y, — z — z, = a. m n

(4)

We use this form even if one or more of 1, m, n are zero. For example, the line through the point (1, —1, 2) in the direction of the vector 2i — 3k may be written as x-1 y +1 z— 2 = A 2 = 0 = —3 x-1 z — 2 or —3y' + 1 = O. 2 Notice that the Cartesian coordinates of a second general point of the line (4) may be expressed in terms of one variable as (xi + AI, yi+ Am, zi+ An).

A is called a parameter for points of the line. 60

5]


The vector u defines the direction of the line and may be termed a direction vector. Since u = ii+mj +nk, 1, m, n are proportional to the cosines of the angles that u makes with the directions i, j, k, that is, with the directions of the coordinate axes. 1, m, n are usually called direction ratios for the line; if u is a unit vector, the constant of proportionality is one and 4 m, n are called direction cosines (see Figure 3.8, where 1 = cos 01, m = cos 02, n = cos 03).

Fig. 3.8

Example 5. Find the equation of the plane through A (1, 1, 1), B(0, 1, — 2), C (0, 0, — 1) and the point of intersection of this plane with the line x— 3 _ y — 2 _ z— 2 1 — —2 — 3 ' Let the equation of the required plane be ax+by+cz+d = 0. Since the coordinates of A, B, C all satisfy this equation,

a+b+c =—d, b-2c = —d, —c = —d which gives a = —3d, b = d, c = d. The equation of the plane ABC is thus 3x — y — z —1 = 0. Any point of the given line has coordinates (3 +A, 2-2A, 2 + 3A). This lies on the plane if 3(3 + A) — (2— 2A)— (2 + 3A)— 1 = 0 A = —2

giving and so the point of intersection has coordinates

(1, 6, —4). 3-2

61

COORDINATES

[3

Example 6. Two straight lines are given parametrically by L,: x = 1-4A, y = 1+A, z = 1+A and

L2:

x = 2u,

y = 1— ,u, z = 2+#.

Prove that L1and L2intersect, and find the equation of the plane containing the two lines. If we can choose A, # such that the three equations 1— 4A = 2,u, 1+A = 1 — 1+A = 2+11 are simultaneously satisfied, then L, and L2intersect. By inspection, A = 2 u = —1 fulfils the required condition (giving the point of intersection (— 1, 1, Eliminating A (a) between the x and y coordinates of a general point of L1, and (b) between the y and z coordinates, we see that the two planes ,

x+4y = 5, y—z = 0 both contain L1. It follows that, for all values of k, the equation (x + 4y — 5)+k(y—z) = 0

(1)

represents a plane containing L1. By setting # = 0, we see that the point (0, 1, 2) lies on L2(but not on L1). This lies on the plane (1) if (0 +4 — 5) + k(1 — 2) = 0, i.e.

k = —1.

The equation of the plane containing L1and L2 is thus seen to be x + 4y — 5 — 1(y — z) = 0, i.e.

x+3y+z-5 = 0.

Ex. 27. What is the equation of the plane (i) containing Ox and Oy; (ii) through (1, —1, 2) perpendicular to Oz?

Ex. 28. Describe the position of the plane ax+ by = 0. Ex. 29. Why was condition (i) included in the definition of a plane? *Ex. 30. Show that three non-collinear points may be found on ax+by+cz+d = 0 62

5]


in the following cases: (i) d = 0, a * 0, b 0, c * 0; (iii) d = a = 0, b * 0, c * 0;

(ii) d * 0, a = b =- 0, c * 0; (iv) a = 0, b * 0, c * 0, d *O.

Ex. 31. Find the equation of the plane through the line of intersection of the planes x-y+ 2z+ 1 = 0, 2x + y- z +2 = 0 which contains the origin. Ex. 32. What are the coordinates of a general point of the line x-3 - y+2 - z+59 2 -1 -3 Where does this line cut the plane x+y+z+ 2 = 0? Ex. 33. What are the equations of the line joining (-3, 1, 1) and (2, 2, -1)? Ex. 34. Find the equation of the plane containing the origin and the line x-2 y-3 z +4 2 3 -1

Exercise 3 (c) 1. Find (a) direction ratios, (b) the direction cosines, for the lines joining the following pairs of points: (i) (1, 2, 3), (2, 3, 4); (ii) (2, -1, 3), (1, 1, 2); (iii) (3, 1, 2), (5, -1, 1); (iv) (1, 3, 5), (5, 3, 1); (v) (2, -3, - 4), ( -3, 2, 1); (vi) (a+ A, 2a + 2A, 3a + 3A), (a- it, 2a - 211, 3a - 3,a); (vii) (a2, ab, b2), (a2, ac, c2); (viii) (1, a, a2), (a2, a, 1). 2. Find the equations of the planes through the following sets of points:

(i) (1, 0, 0), (0, 1, 0), (0, 0, 1); (ii) (3, 2, 0), (0, 3, -1), (1, 0, -2); (iii) (2, 0, 0), (1, 1, -1), (6, -5, 3); (iv) (2, 2, 0), (-1, 1, -4), (1, 1, -1); (v) (-1, 3, 1), (1, -3, -3), (3, -1, 5); (vi) (1, 2, 3), (2, -2, 8), (-1, 2, -7). 3. Find the equations of the lines joining the following pairs of points: (i) (0, 0, 0), (2, 1, -3); (ii) (1, 2, -1), (2, 3, -3); (iii) (1, 1, -1), ( -1, 4, 2); (iv) (-2, 1, 3), (2, 1, -3); (v) (1, 4, 2), (1, -1, 2); (vi) (a, 2a, 3a), ( -a, a, 2a). 4. Find the equation of the plane through the line of intersection of the planes x+y -3z = 2, 2x-y-z =1

(i) containing the origin; (ii) containing the point (1, 1, -1); (iii) parallel to the x axis.

5. Find the equation of the plane through the line of intersection of the planes 2x+2y+3z-1 = 0, 3x-y-z+2 = 0 (i) containing the origin; (ii) containing the point (1, 1, 1); (iii) parallel to the z axis. 63

COORDINATES

[3

6. Write down the equation of the plane parallel to the x axis which contains the line x-1 y-3 =z+1 = —1 2 —3 • Find the equation of the plane containing this line and the origin. 7. Find the equation of the plane through the point (1, 1, — 5) containing the line x+1 _ y+5 _ z 3 — 4 — —5' 8. Write down, in terms of a parameter A, the coordinates of a general point of the line x-8 = y 1 = z-1 3 1 0• Where does this line cut the plane x -2y — 3z— 2 = 0? —

9. Find the coordinates of the point where the line joining (2, 3, 1) and (4, 7, 3) cuts the plane 2x +y—z— 3 = 0. 10. Find the equations of the line joining (2, 1, 1) and (1, — 1, 2). Where does this line meet the plane x — y +az = 0? Is your result true for all a? Explain. 11. Determine the direction cosines of the line of intersection of the planes 2x—y+z— 9 = 0, 4x+y+2z — 6 = 0. Where does this line meet the plane x+y+z = 0? 12. Prove that the two lines x-1y+1 z+2 = = 1 2 3 ' x+1 y z-1 = = 1 1 1 ' are skew. 13. Prove that the two lines x-2 y+4 z -1 = = 3 1 —1 x+1 y+4 z+2 = = and 1 —2 2 meet. What are the coordinates of their point of intersection? Find the equation of the plane containing them. 14. Prove that the two lines x -3 y— 2 z-4 4 = = 2 1 1 x +1 y -1 z = = and 0 3 2 meet. What are the coordinates of their point of intersection? Find the equation of the plane containing them.

64

5]


15. Find the coordinates of the point common to the three planes: x— 3y+ z— 6 = 0, 2x— y + 2z— 2 = 0, 3x+2y+z+2 = 0. 16. Find the coordinates of the point common to the three planes: x — 2y + z — 7 = 0,

2x + 3y — 4z + 26 = 0, 3x+y+3z+7 = 0.

17. Find the coordinates of the point where the line determined by the two planes x — 3y— z + 8 = 0,

x — y+ z— 2 = 0 cuts the plane containing the x and y axes. Miscellaneous Exercise 3 1. Find the area of the trapezium whose vertices are: (x3, 0).

(x2, Y2), (x3, Y3), (x2, 0),

Deduce that the area of the triangle P1(x1,

P2(x2, Y2), P3(x3, Y3) is

Mx2Y3 —x3Y2+ x3Y1 —x1Y3+ x1Y2 —x2Y1)• What is the condition that the three points P1P2P3should be collinear? 2. ABCD is a rectangle with AB = 2AD. E is the mid-point of AB and F is the mid-point of BC. CE, AF meet at X. By setting up a suitable coordinate system, prove that DXB is a straight line and determine the ratio DX/XB. What is the area of the quadrilateral AXCD? 3. ABCD is a rectangle and points P, Q are taken on AB, AD respectively. The rectangle APRQ is completed. If the lines BQ, DP meet at X, prove that X, R, C lie on a straight line. 4. Find, in terms of a parameter m, representing the gradient, the equation of a variable straight line through the fixed point (a, b). If this line cuts the x and y axes at A and B respectively, and the parallelogram OABP is completed, prove that, whatever the value of m, the point P lies on the curve xy = xb — ay. Sketch this curve for the case a = b = 1 for values of x >

—

1.

5. Show that, by a suitable choice of coordinate system, the equations of two coplanar non-parallel lines may be taken in the form Y = ax, y = — ax.

65

COORDINATES

[3

Two fixed lines OA, OB are drawn and a variable line, passing through the fixed point C, cuts OA, OB at P and Q respectively. If the feet of the perpendiculars from P, Q to OB, OA are U, V, prove that UV passes through a fixed point. 6. Two fixed straight lines, L1and L2, meet at 0. Through a fixed point A two lines AP1P2and AQ1Q2are drawn to cut L1at P1and Q1, and L2 at P2 and Q2. Prove that whatever the position of the two lines drawn through A, the point of intersection of P1Q2 and P&Llies on a fixed straight line through 0. 7. Points P1, P2, P3 are taken on the x axis, and Q1, Q2, Q3 on the y axis. L1 is the point of intersection of P2 Q3 and P3 Q2, L2 of P3 Q1and P1 Q3, L3 of P1 Q2 and P2 Q1. Prove that L1, L2, L3 lie on a straight line. (This is a particular case of Pappus's Theorem, which holds more generally for two sets of three collinear points on any two straight lines.) 8. Prove that the lines x+1 y-1 z-3

and

—2

2

1

x—1 3

y— 3 1

z— 2

6 are skew. By considering the family of planes through one of the lines, prove that there is just one common transversal to the two lines which passes through the origin and find its equation. 9. Prove that no three of the points (1, 1, 2), (-2, — 6, 3), (— 1, 1, 5), (2, 4, 2) are collinear but that all four points lie on a plane. 10. Explain why the three planes x— z +1 = 0, x — y —z+5 = 0, x+2y—z+2 = 0 have no common point. What can you say about the intersections of the planes: x—z+1 =0, x—y—z+5 = 0, x+2y— z-7 = 0? 11. ABCDA'B'C'D' is a cube, with square faces ABCD, A'B'C'D' and vertical edges AA' etc. M is the mid-point of C'D'. The plane AB'M cuts BD' at X and CC' produced at Y. Find the ratios BX: XD' and CC': C' Y. 12. Prove that the three lines . x _ y _ z ——

.x-1 _y _z .

——

+1 L3.

are skew. 66

3

y z 1—2

5]


If general points of L1and L2 have parameters A and urespectively, find the condition that the line joining the point on L1with parameter A to the point on L2 with parameter it should intersect L3. Deduce that a unique common transversal to the three lines may be drawn through a given point on L1. 13. In the tetrahedron ABCD, each of the faces ABC, ABD, ACD has a right angle at A, and AB = AC = AD. X is the point on AD such that AX = 2XD and Yis the point on BC such that CY = 2 YB. The mid-point of AY is Z. Prove that the mid-point of DZ lies on the plane BCX. 14. A is the fixed point (a, 0, 0) and variable points Q (0, Al 0), R (0, 0, ft) (Al, it positive) are taken on the y and z axes such that the plane AQR passes through the fixed point B /3,7). Prove that

(a— a) Alt ctfl

ct7A. = 0.

Find the value of A that makes the triangle 0 QR isosceles. 15. Explain why the equation x2 +y2 +z2 = 1 represents the surface of a sphere, centre the origin. Find the equations of the planes through the point (2, 0, 0) which are tangential to the sphere and parallel to the y axis.

67

4. Polynomials 1. POLYNOMIALS

Expression of the form and

xs - 3x2+ 4x + 2 x5- x + 1

are called polynomials. More generally, a polynomial of degree n is an expression of the form ao xn + ai xn-+ a2 xn-2+

an (ac, + 0)

involving only multiples of positive integral powers of x and a constant term, an. A polynomial of degree n is completely determined if its (n +1) coefficients (including the constant term or coefficient of x°) are given and two polynomials of degree n are said to be identically equal if the coefficient of xr in each polynomial is the same for all values of r, and conversely. For example, writing = to mean 'identically equal to' ax3 -x -2 -E 2x8 -bx2 -x + c -4=> a = 2, b = 0, c = - 2. Ex. 1. If ax2+ bx + c

3X2 +2ax+ b, find a, b and c.

A convenient shorthand is to write a polynomial as P(x); for example, if we are considering the polynomial xs - 3x- 2 we could write ' Let P(x) = xs - 3x-2' and subsequently refer to this polynomial simply as P(x). By P(1) we

mean the value of the polynomial when x = 1; that is, the numerical value attained by the polynomial when 1 is substituted for x. In the case just quoted, P(1) = 1 - 3 -2 = -4 and P( -2) = -8+6-2 = -4. Ex. 2. If P(x) x4 -3x3+ 4x2 - x-1, show that P(1) = 0 and find P(- 1), P(0) and P(2). Ex. 3. Show that, if P(x) Q(x), then P(k) = Q(k) for all values of k. The converse of this proposition is also true: if P(k) = Q(k) for all values of k, then the coefficients of xr in P(x) and Q(x) are the same for all r (see Ex. 7). Either statement may be taken as the definition of 'identically equal to'. 68

POLYNOMIALS

1]

A non-zero number may be regarded as a polynomial of degree zero, or a constant polynomial. The number zero may be regarded as the zero polynomial.t No ambiguity will arise if we write the zero polynomial as 0. A polynomial is identically equal to the zero polynomial if all its coefficients are zero, and conversely. Ex. 4. If (a-1) x2+ (a+ b) x + (a + b+ c)

E

0, find the values of a, b, c.

Example 1. If P(x) ax2+ bx+ c, distinguish between the identity P(x) 0 and the equation P(x) = 0. P(x) 0 means that P(x) is identically equal to the zero polynomial and so we deduce that a = b = c = 0. Thus, whatever value x may have, say x = k, then P(k) = 0. P(x) = 0 is a statement which holds only for certain values of x. In fact, provided b2 > 4ac and a + 0,

•-•

p [ b +,l(b2— 4ac)1 p [b — V(b2-4ac)1 0 2a 2a and P(x)

0 for any other value of x.

Exercise 4(a) 1. If P(x)

x3+5x2+ 3x + 1; find P(1), P(— 1), P(1), P(3).

2. If (x- 2) (x— 3) (x-4) -.7- ax3+ bx2+ cx+ d, find a, b, c, d. 3. Prove that (x — a)3 a - x3— 3ax2+ 3a2x — a3. For what real values of x does

(x — a)3= — a3? 4. If a(x— 2)2 + b(x— 2)+ c

3x2 -8x-1, find a, b, c.

5. Express 7x3 — x2 + 3x-4 in the form a(x — 1)3+ b(x — 1)2+ c(x-1)+ d. 6. Express 4x3 +12x2 + 6x in the form a(x +1)3+ b(x +1)2+ c(x +1)+ d. 7. Express 3x3 +2x2— 11x— 10 in the form a(x +1)3+ b(x +1)2+ c(x + 1)+d and also in the form cc(x-2)3 + fi(x— 2)2+y(x— 2) + a. Find the three roots of the cubic equation 3x3 +2x2 — 11x— 10 = 0. 8. Express x3+ 4kx2+ 3k2x— k3in the form a(x+ k)3+ b(x+ k)2+c(x+ k)+ d. 9. If (ax2+ bx+ c) (x +1) E- 0, prove that ax2+bx+c -m 0.

f We shall not associate a degree with the zero polynomial. The reader will appreciate that, by elementary algebra, the product of a polynomial of degree m by a polynomial of degree n is a polynomial of degree m+n. This result holds good even if one or both of the polynomials is a constant polynomial, but fails if one of the polynomials is the zero polynomial if a finite degree is associated with this polynomial.

69

[4

POLYNOMIALS 10. If (ax+b) (cx+d) a-(ax+ c) (bx+ d) prove that: (i) if a 0, then b = c; (ii) if b c, then a = d = 0. -

11. If (x—c)3 E x3+ax2+ bx 27, find a, b, c. —

12. If (x a) (x— b) (x— c) a (x + a) (x+ fi)(x+y), prove that —

{a, b, c} =

2. THE FACTOR AND REMAINDER THEOREMS If P(x) (x — a) Q(x), where P(x) is a polynomial of degree n and Q(x) is a polynomial of degree n -1, P(x) is said to have a linear factor. Conversely, if P(x) has a linear factor, then P(x) may be expressed in the form P(x) = (x — a) Q(x). For example,

2x3 + 5x2 — 3x — 10 = (x + 2) (2x2 + x— 5)

and so the cubic polynomial 2x3+ 5x2 — 3x — 10 has a linear factor (x + 2). (It also has a quadratic factor (2x2 +x — 5)) It is important to realise at the outset that possession of a factor is not an absolute property of a polynomial but depends on the restrictions we place upon the coefficients. For example, if we restrict our polynomials to have only rational numbers as coefficients, the polynomial x2— 2 has no linear factors; on the other hand, if we allow our coefficients to be real numbers, x2 -2 = (x— V2) (x + V2). (This is often expressed by the statement: 'x2 -2 is irreducible over the rational field but reducible over the real field'. The word 'field' has a technical meaning and will be defined in Volume 2; all that is necessary at the moment is to realize that over the rational field' means that all polynomials under discussion must have rational numbers as coefficients.) The theorems we are about to prove do not depend upon whether we choose the rational field or the real field for our coefficients provided, of course, that we are consistent. We shall not, therefore, allude to the field from which the coefficients are drawn, but the reader may, if he wishes interpret all the coefficients as, say, rational numbers. It is easy to see that, if P(x) has a factor (x a), then P(a) = 0. For P(x) has a factor (x — a) P(x) (x a) Q(x) =- P(a) = 0 Q(a) P(a) = O. A converse of this result is known as the Factor Theorem. 70

2]

FACTOR AND REMAINDER THEOREMS

Theorem 4.1 (The Factor Theorem). If P(x) is a polynomial and if P(a) = 0, then P(x) has a linear factor x —a. Proof. First observe that

xr_e

(x

-i ±axp-2 0,2xr-3

ar-1).

Suppose

P(x) :-=- bo xn+bi xn-1+b2 xn-2+ • • • +bn-ix+bn,

Then and so

P(a) = bo an +bi an--1+bo an-2+

P(x)—P(a) = bo(xn — + bi(xn-1— ocn-l+ +br,_,(x — a) (x a) [bo(xn--1+ . . . + ocn-1) + bi(xn-2+ . . . + —

+ b n _1].

But P(a) = 0 thus we have P(x) (x — oc) Q(x) and therefore P(x) has a linear factor (x —a). Consider now the following division process: x2 -2x-3 x-2) x3-4x2 +x+2 x3-2x2 — 2X2

— 2x2+ 4x —3x+2 —3x +6 —4 In elementary algebra the result of this process would be described by saying that, if x3 — 4x2 +x +2 is divided by x— 2, the quotient is x 2— 2x — 3 and the remainder is —4. There are advantages, however, in restating the process as follows: x3 — 4x2 +x + 2 is identically equal to the product of (x-2) by the quotient (x2— 2x-3), plus the remainder —4. In symbols x3-4x2 + x + 2(x-2)(x2-2x-3)-4. =

(1)

This identity holds for all values of x, whereas the division process given above is true for all values of x other than x = 2. Ex. 5. Check that x = 2 does indeed satisfy (1). We now prove our second result, the Remainder Theorem, which general-

izes the ideas outlined above. 71

[4

POLYNOMIALS

Theorem 4.2 (The Remainder Theorem). If P(x) is any polynomial of degree n 1, and x— a is any linear polynomial, then P(x) may be expressed in the form P(x) —= (x — a) Q(x)+R where Q(x) is a polynomial of degree n-1 and R = P(a). Proof. As in Theorem 1, we have P(x)—P(a) —= (x — a) Q(x) and the result follows immediately. Note (i). The Remainder Theorem is often stated in the following form: if a polynomial P(x) is divided by x — a, then the remainder is P(a). (ii) Strictly speaking, we should prove that the expression given above for P(x) is unique in the sense that, if P(x) (x— a) Q1(x)+R1, then Q1(x) = Q(x) and R1= R. The proof of this result should be supplied by the reader. Example 2. Find the remainder when x3— 5x2— x + 2 is divided by (i) x-1; (ii) x+2; (iii) 2x+ 1. Write P(x) _= x3— 5)0— x + 2. (i) P(x) (x —1) Qi(x)+ R1; put x = 1:

R1= P(1) = 1-5-1+2 —3;

(ii) P(x) (x +2) Q2(x)+ R2; put x = —2:

R2 = P(-2) = —8-20+2+2 = —24;

(iii) P(x) -= (2x +1) Q3(x)+R3 ; put x =

R3 = P(--1) = --g--1+1+2 = g. Exercise 4(b)

1. Find the remainder when: (i) x3— x + 2 is divided by x-4; (ii) 3x3— 5x2+ x + 2 is divided by x — 2; (iii) 4x4—2x2 + x — 3 is divided by x + 1; (iv) x5— x-1 is divided by x+ 3 ; (v) 4x3— 5x2+2 is divided by 2x — 1; (vi) 4x3— 8x + 1 is divided by 2x + 1. 2. Show that x — 2 is a factor of the polynomial x3— 4x2 + x+ 6 and hence factorize the expression completely. 72

FACTOR AND REMAINDER THEOREMS

2]

3. Factorize the following polynomials over the rational field as far as possible: (ii) 12x3 + 5x2— 19x — 12; (i) 2x3 + 7x2 — 5x— 4; (iv) x3 + 3x2— 2x — 6; (iii) 2x3 +7x2 — 17x — 10; (vi) x3 — 7x2 +7x +15; (v) x3 — x2 — x— 2; (viii) 8x3 +12x2 — 2x — 3. (vii) x3 + 2x2 — 7x— 2; 4. Factorize the polynomials given in Question 3 over the real field as far as possible. 5. If x3 — 5x2+7x— a has a factor x — 2, find a. 6. If 2x3 +ax2 — 5x-1 is divisible by 2x + 1, find a. 7. If x3 + 3x2 + ax— 1 leaves a remainder of 3 on division by (2x +1), find a.

8. If 3x3 + ax2+ bx— 2 is divisible by both x+2 and 3x +1, find a and b. 9. If ax3+ 3x2+ bx — 3 is divisible by both x-1 and 2x+ 3, find a and b. 10. By expressing x4 + 1 in the form x4 + 2x2 + 1 — 2x2, show that x4 + 1 is reducible over the real field. 11. Factorize x4 + 3x3 — 1 5x2 +9x+ 2: (i) over the rational field; (ii) over the real field. 12. ax4+ 2x3— 4x2—2x+ b has factors (x— 1) and (x— 2), find a and b and factorize the expression completely.

3. THE FACTOR THEOREM (CONTINUED) We now show how the Factor Theorem may be extended for cubic polynomials and, in particular, how this extended result may be used to derive certain algebraic identities. The results we prove are valid for the general polynomial (substituting n for 3 in their enunciation) but proofs for this require the use of mathematical induction (see Chapter 9).

Theorem 4.3. If P(x) is a cubic polynomial with leading term ao x3, and if P(a) = = P(y) = 0 for 3 unequal numbers a, 13, y, then P(x) = ao(x ct) (x-13) — Proof. Since P(a) = 0, P(x) (x — a) Q1(x), where Q1(x) is a polynomial of degree 2. Thus, since PO = 0, 0 = (fi— a) Q,(13). But 161 — + 0 and so Q103) = 0. Q1(x) therefore has a factor (x—fl). Q1(x) (x — ie) Q2(x), where Q2(x) is a polynomial of degree 1. By the same argument,

Q2(x)

(x—Y) Q3(x), where Q3(x) is a polynomial of degree 0, that is,

Q3(x) = k, where k is some number. Substituting, P(x) = k(x — a) (x— fi)(x— y). 73

[4

POLYNOMIALS

Since these two polynomials are identically equal, the coefficient of x3 must be the same in each, and so k = a, and the proof is complete.

Corollary (The Identity Theorem). If P(x) is a cubic polynomial and if 4 unequal numbers a, y, 8 can be found such that

P(a)

= P(y) = P(8) = 0, then P(x)

0.

Proof By Theorem 4.3, P(x) ao(x — ct) (x — ,8) (x — y). But P(8) = 0 and so 0 --- a0(6 — cc) (8 — (8 — y). Now 8 + a, 8 + p), S + y and it follows that a, = 0. P(x) thus reduces to a quadratic polynomial a1x2+ a,x + a, which vanishes for 4 distinct values of x. Successive repetitions of the argument show that a1 = a2 = a3 = 0 and the corollary is proved. Ex. 6. Prove that, if P(x) is a quadratic polynomial and if 3 unequal numbers a, y can be found such that P(a) = P(fl) = P(y) = 0, then P(x) = 0. Ex. 7. If P(x) and Q(x) are cubic polynomials, and if unequal numbers a, Q y, 8 can be found such that P(x) = Q(c R defined -

by g(x) = x3 3x2 5x + 2. -

-

A useful pictorial representation of certain functions f: R-> R (or subsets of these sets) may be obtained by taking the x axis to represent the domain and the y axis to contain the range. If y is the image of the number x under f, that is, if y =f(x), we represent the element (x, y) off as the point with Cartesian coordinates (x, y). The set of all such points is called the graph of the function. For example, suppose f: R->. R is defined by the equation f(x) = x2 -5x+ 4. The graph of this function for the part of the domain -1 x S 6 is shown in Figure 5.2. This graph may alternatively be referred to as ' the curve y = x2 -5x + 4'. Certain relations which are not functions may also be represented graphically; for example, consider the subset of R x R consisting of all ordered pairs of the form (x, y) where x e R, y e R such that y2 = x. That part of the graph for x < 6 is shown in Figure 5.3. y

(0,4)

Fig. 5.2

Fig. 5.3

Ex. 5. Explain why the relation depicted in Figure 5.3 is not a function.

B; under what circumstances Suppose now we have a function f: (f(a), a), a E R, constitute a function will the set of all ordered pairs F:B-›-A?

80

I]

FUNCTIONS

Viewed pictorially, the question may be rephrased: if all the arrows in a diagram such as Figure 5.4 were reversed, under what circumstances would we still have a diagram depicting a function?

The question is easily answered if we recall what is demanded of such a function F: it must associate each point of B with a unique point of A. Thus, the range off must be the whole of B (since each point of B has an image under F) and f must be (1-1) (since each point of B has a unique image under F). If f satisfies both of these conditions F is defined as a function: it is called the inverse function off: A B and is written B --> A. Ex. 6. Show that no inverse function exists for the function f: R R where f(x) = x2, but that an inverse function does exist for the function g: R+ R+, whereg(x) = x2.

Exercise 5(a) 1. The function f:R—> R is defined by f(x) f(3),f(-1).

—x-1. Find f(0), f(1), f(2),

2. The function g: R+ ---> R is defined by g(x)

x2 +x+1

x.i I

Find g(1), g(2), g(3). 3. The function f: R R is defined by f(x) =

(— x if x < 0, 0 if x = 0, x if x > 0.

(This is normally written f(x) = IxI; read f(x) equals the modulus of x'.) What is the range of f? Sketch the graph of f. A second function g: R --> R is defined by g(x) = x— Ix 1. What is the range of g? Sketch the graph of g. 81

FUNCTIONS AND INEQUALITIES

[5

4. The function f: R -> R is defined by x f(x) = trx/ , x * 0, 0, x = O. Write down the values of f(- 1) and f(1). Sketch the graph of f. 5. A g R and B g R. The set f of all ordered pairs (x, y) is formed, where x e A, y e B and y = 4,/x. Show that f is not a function if (i) A = R+, B = Q; (ii) A = Q, B = R. Suggest sets A and B for which f is a function. 6. A = {x e R: 1 --5 x ... . 2} and f: A -> R is defined by f(x) = 2/x. The function g: A -)- R is defined by g(y) = 1 +y2. A new function h: A -> R is defined by h(x) = g{f(x)] (this is usually written h = g o f). Find (i) h(); (ii) the range of h. 7. Sketch the graph of the function!: R -+ R where f(x) = lx + 11. 8. Sketch the graph of the function g: R -÷ R where g(x) = lx+11+1x+21+1x+31. 9. A = {xe R: 0 < x < 3} and f: A -* R is defined by f(x) = x2 - 3x+2. Find (i) !(1),.f(2); (ii) the range of f; (iii) the subset of A whose elements have the image 1 under f; (iv) the subset of A whose elements have the image 2 under f. 10. A = {x e R: - V6 -. R is defined by f(x) = x3- 6x. Find the range and sketch the graph of f. 11. If f: R R is defined by f(x) = (x- 1)s, show that f-1: R -> R exists and determine its form explicitly. 12. Determine the numbers which are invariantt under the mapping f: R -+ R where f is defined by {(x2 +x+1) (x+3)-1 (x * -3), f(x) = 0 (x = - 3). Show how to illustrate invariance graphically for a general function g: R -+ R. 13. f: R R is a given function, A a Rand the set of images of all the elements in A is denoted by B. If C is the set of numbers whose images belong to B, prove that A c C and that the inclusion may not be strict. 14. f: R -)- R is a given function, X c R and the set of images of all the elements in X is denoted by f(X). If A c R, B c R is it necessarily true that (i) f(A n B) = f(A) n f(B); (ii) f(A U B) = f(A) U f(B)? 15. Sketch the graph of the function f: R+ R defined by f(x) = x2. Show that the function g: R+ -> R, where g(x) = xi-, is the inverse off. Sketch the graph of g. What connection exists between the graphs off and g? t An element a is invariant under the mapping f if f (a) = a.

82

11

FUNCTIONS

16. If x is a real number, [x] denotes the greatest integer less than or equal to x. (For example, [- 2-1-] = - 3, [4k] = 4, [- 1] - 1.) The function f: R -› Z is defined by f(x) = [x]; sketch its graph. A functionfis called periodic if there exists a number k such that f(x+ k) = f(x), for all x in the domain; k is then called the period of f. Show that the function defined by f(x) = x-[x] is periodic and find its period. What feature does the graph of any periodic function possess?

2. INEQUALITIES The function, f, defined by

1 fix) 3 - 2x

has for its domain the set R with the single point x = i deleted; if we call this set D, we may write D = R-{3}. The inequality

1 3 -2x

(1)

x; x-1

8.

x- 3

9.

-.C. x;

10.

2 < 1. x2 - 5x+ 6 '

x4 < 2; 11. x2-1

13.

x-4 < 1. x2 4 '

14.

6(3x + 1) 0, y < 0 and the corresponding acute angle is +.77 (Figure 6.4 (iii)). Thus, tan -4-7T = — tan in = — 1. (iv) For 0 = 4n,, y > 0 and the corresponding acute angle is 17i (Figure 6.4 (iv)). Thus, cosec 47T = cosec In = +V2. x < 0 and the corresponding acute angle is (v) For 0 = (Figure 6.4(v)). Thus, sec (--i-n) = — sec in = — V2. x < 0, y > 0 and the cor(vi) For 0 = responding acute angle is in (Figure 6.4(vi)) Thus, cot (—in) = —cot in = —,j3. Figure 6.5 gives a useful method for seeing which sign to attach according to the quadrant in which the angle falls. A stands for all, S for sine, T for tangent and C for cosine. In the A quadrant, all trigonoFig. 6.5 metric ratios are positive; in the S quadrant, only ,

4

PPM

91

[6

THE TRIGONOMETRIC FUNCTIONS

sines (and cosecants) are positive; in the T quadrant, only tangents (and cotangents) are positive; in the C quadrant, only cosines (and secants) are positive. Notice, as an aid to remembering the diagram, that the letters (read anticlockwise) spell the word 'CAST'.

Example 2. Solve for 0 the equation 2 sin' + sin 0 1 =0 giving all solutions in the interval 0 0 < 27r. —

The left-hand-side factorizes: (2 sin 0 -1) (sin 0+1) = 0 and so is satisfied by values of 0 such that either (i) sin 0 = +4-, or (ii) sin 0 = —1. The relevant solutions of (i) are 0 = bir 0 = -g-rr; and of (ii) B = Thus, the required set of solutions of the given equation is ,

{-1 6Tr,

fir,

3

Tm} •

Since OP is a unit vector, x2 +y2 = 1 for all 0. Thus, sin2 +cost 0 E- 1, for all 0. Division by cost 0 gives 1 + tan2 0 E sec2 0, for all 0 (2k+1) 77/2. Division by sin2 0 gives 1 + cote 0 E cosec2 0, for all 0 * These three identities are true for all values of 0, provided the functions mentioned are defined.

Fig. 6.6

Suppose OP is the unit vector defined by the angle 0 and OQ the unit vector defined by the angle -17r — 0 (see Figure 6.6). If OP = xi +yj, then OQ = yi+xj. It follows that sin (17r — 0) a- cos 0, cos (-1-7r — 0) E sin 0, cosec 92

— 0)

sec 0, sec (-3:77-- 19)

tan (-177.— 0)

cot 0,

cosec 0, cot (Pr — 0)

tan 0


2]

(provided, of course, that 0 lies in domains for which both sides are defined). *Ex. 5. Find similar simplifications for the values of the trigonometric functions for the angles (i) -1-7/- + 0; (ii) Tr 0; (iii) n +0. N.B. Be careful about the signs! —

Example 3. Solve the equation sin 0 + cos 20 = 0, giving all values of 0 between 0 and 2n. Since cos (-Pr + 0) = — sin 0, the equation may be rewritten cos 20 = cos (171- +0). Now cos 0 = cos cb if 0 = + 0 or 0 = 2n± cb or 0 =

47T ± q5 ....

Thus the original equation is satisfied for values of 0 given by

20 = 27-r ± (zn- + 0); 20 = zirr ± (-1-7r +0); 20 = 671- ± (Pr+ 0); .... 20 = ± an + 0);

Trial of these solutions shows that the values of 0 lying in the given range are fig, in, -16117*}.

Example 4. Prove that the identity tan 0+cot 0 1 — sin 0+ cos 0 — sec 0+ cosec 0

holds, provided 0 + Val., 0 * mr— Pr.. The restrictions on 0 ensure (i) (0 + +Jar) that tan 0, cot 0, sec 0, cosec 0 are defined; (ii) (0 + rut —177.) that sin 0+ cos 0 * 0 and sec 0+ cosec 0 + 0. With these restrictions in 0 cos0) it 1 1 \ R.H.S. = + ± (cos k 0 sin 0 / \cos 0 sin 0) (sine 0 + cos2 0)/(sin 0 + cos 0) 1/(sin 0+ cos 0) = L.H.S.

Example 5. Eliminate 0 between the equation

4-2

x = a cos 0,

(1)

y = b sec 0 + c tan 0,

(2) 93


[6

x = a cos 0 and y = b sec 0 c tan 0 xy = a cos 0 (b sec 0 c tan 0), xy = ab+ac sin 0 ac sin 0 = xy—ab. But ac cos 0 = cx. Squaring and adding a2c2 = c2x2 (xy — ab)2.

Exercise 6(b) 1. Write down the values of cos 47r, tan lir, cosec lir, cot PT, sin P ' T, sec iv, cos fa, tan in, cosec (—in), sec (-177).

2. Write down the values of cos 315°, tan 135°, cosec 330°, sin (— 135°), cot (-120°), sec 240°, sin 480°, sec (— 210°), cot (— 60°), sin 1020°. 3. Use your tables to find the values of sin 215°, cos 128°, tan (— 40°), cosec 161°, sin (-200°). 4. Solve the following equations for 0, giving all values lying in the interval 0 27r: (i) sin 0 = —1; (ii) tan 0 +1 = 0; (iii) sec 0 +2 = 0; (iv) cot 0 = V3; (v) 4 sin2 0 = 3; (vi) 2 cos (0 — 170+ ,/3 = 0; (ix) sec (0 + Pr) = 1; (vii) sin 0+ cos 0 = 0; (viii) 2 sin 30 = 1; (x) 3 sect + = 4; (xi) cosec2(0 + Pr) = 1; (xii) tang 20 = 3.

5. Solve the following equations for 0, giving all values lying in the interval — 7T < 0 < (i) 2 sin2 0+sin 0 = 0; (iii) cos 20 = sin 0; (v) sin (iir— 0)+ cos 0 = 0; (vii) tan 0 + cot air— = 0; (ix) cote 0+ cosec 0 +1 = 0;

(ii) 2 cos2 0+3 sin 0 = 0; (iv) 2 tan 0 + sin 0 = 0; (vi) cos 0 = 2 cot 0; (viii) tan 0+ cot 0+sec 0 = 0; (x) sec 30+ cosec 0 = 0.

6. Find the maximum and minimum values of: (i) 3/(2 + sin 0); (ii) (1 + cos 20)2 ; (iii) sin2 0+ 2 sin 0 + 2.

7. No tables to be used in this question. (i) If lir < x < it and sin x = 4, find cos x and tan x. (ii) If lir < x < it and tan x = —1, find sin x and cos x. (iii) If it < x < irr and sec x = —3, find cos x and tan x. (iv) If 4ir < x < it and sin x = I, find tan x and sec x; (v) If zit < x < 2ir and sec x = 4, find cosec x and cot x. 8. Prove that, provided sin 0 1

—

0. cos 0 _

sin 0 94

1

cosec 0+ cot 0'

2]


9. Prove that, provided sin 0 * 0, cos 0 t 0, tan 0 + cot 0

sec 0 cosec 0.

10. Prove that, provided all the values of the functions are defined, cosecs 0 = 1 + cos 0 cot 0 cosec 0. 11. If sin 0 = s find, without using tables, the possible values of sec 0 +cosec 0. 12. If tan = find, without using tables, the possible values of 2 cos 0 + cot O. 13. Use your tables to find all values of 0, lying in the interval 0° < -4 360°, which satisfy the following equations: (i) 2 sec2 0° = 5—tan 00; (ii) 3 cost 0° = 7 cos 0°-2; (iii) 2 sine 0° — sin 0° cos 0° — cos20° = 0; (iv) cote 8° = cosec 0'; (v) 16 tang 8° = 9; (vi) 3 sec2 0° = 2 cosec 0° ; (vii) tan 0° = 2(sec 0° + cos 00); (viii) sec2 = 1 + 2 tan 0°. 14. Eliminate 0 between the following pairs of equations: (i) x = 2 cos 0, y = 3(1 + sin 0); (ii) x = cos 0 — sin 0, y = cos 0 + sin 0; (iii) x = 2 cos 0 — sin 0, y = cos 0+ sin 8; (iv) x = 3 tan 8, y = 4 sin 0; (v) x = cosec 0— 1, y = cos 0+1. 15. A particle oscillates along the x axis in such a manner that its coordinates at time t seconds after the start of the motion are (sin (cot +e), 0). (i) Where is the particle at the start of the motion? (ii) Between what two points does the particle oscillate? (iii) How long does the particle take to move from one extreme point to the other? 16. If a particle is projected with a velocity V cos ai+ V sin aj under gravity then its position vector at time t is given by r = (Vt cos a) i + (Vt sin a — Igt2)j. If r is written in the form prove that

r = xi + yj, 1 (gx2 = x tan a— - — ssec2 a. 2 V2)

Deduce that, if x, y, g, V are known, there are in general two values of a in the interval 0 < a < fir which satisfy this equation. Under what circumstances is there only one such angle? 17. Prove that the equation

x2 + y2 + z2 = a2

represents the surface of a sphere. Show that the point P whose position vector is r = a cos 0 cos Oi + a sin 0 cos Oj + a sin cbk lies on the sphere, for all values of 8 and O.

95


[6

Is the following converse result true? Values of 0 and ¢ can be found such that any point on the surface has a position vector of the form r = a cos 0 cos g5i+ a sin 0 cos cbj+ a sin qlk.

18. Discuss the possibility of solving the following equations for 0: (i) sin 0 = a+ 1/a; (iii) cost 0 +4 = sin 0 +4 cos 0.

(ii) 2ab cos 0 = a2+b2;

19. What are the maximum and minimum values of the expression 4/(2 + sin x)? Without attempting to solve the equations exactly, state how many values of x in the range 0 < x < 27 satisfy (i) the equation cos x(2+ sin x) = 4; (ii) the equation tan x(2 + sin x) = 4.

3. THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS Since revolutions through angles 27r, 47r, ... about 0 brings the unit vector OP into coincidence with its original position, it follows that, for any integer k and any of the six trigonometric functions f: R f(2kird x) = f(x). -

Thus, a trigonometric function is a periodic function, period 27r (see Exercise 5(a), Question 16). Figure 6.7 shows the graphs of the sine function (continuous line) and cosine function (dotted line), for values of the domain —27r x 27r.

Fig. 6.7 Ex. 6. How do the graphs of sin x and cos x illustrate the identities sin (-Pr— x) E cos x, sin an + x) a" cos x, sin (Zi -— x) —cos x, sin an + x) —cos x? Ex. 7. Sketch roughly the graphs of 2 sin x, sin 2x, sin xj. Figure 6.8 shows the graph of the tangent function (continuous line) and cotangent function (dotted line) for values of the domain — 27r < x < 27r. Recall that the tangent function is undefined at values of x which are odd multiples of 17r but, by taking x sufficiently close to such

96

3]

GRAPHS OF TRIGONOMETRIC FUNCTIONS

values, we make 'tan x I arbitrarily large. A similar remark holds for the cotangent function at values of x which are even multiples of 1-Tr. (A line such as x = in is called an asymptote for the curve y = tan x.)

i I I

il

1%. I\

I \ I I \I

I I I I

—

—IT

2 ; I

I I I

\I I II

.Y

%

—

11

I I \,■

■ I.

... V

I I

1I

I

I \ 1 \

3 iN

—27r

I I

11 ilt It

■0

2. \ I \% 1 I I I

/1 "\

*. I I I

II I1 I \ I

X

ir \I IC III I,I

Fig. 6.8 Ex. 8. How does Figure 6.8 illustrate the identities tan air x) —cot x, tan (171. —x) = cot x, tan air +x) —cot x? —

cot x,

tan (PT+ x)

Ex. 9. Sketch roughly the graphs of tan 3x and !tan xi.

Figure 6.9 shows the secant function (continuous line) and cosecant function (dotted line). It II

YI I I t t t

i 1 t t I

Ii

s t

II I %

I

I

— --I— — —27737r •-•71 ir

—2

it 2

\ 1

tt 1

II

I

Fig. 6.9 Ex. 10. How does Figure 6.9 illustrate the identities sec (Pr + x) sec ag— x) = cosec x,

sec (lir — x)

—cosec x, sec (PT +

— cosec x;

cosec x?

Ex. 11. Sketch roughly the graphs of sec 2x and Icosec xi.

97


[6

4. INVERSE TRIGONOMETRIC FUNCTIONS The sine function has no inverse, because if we take any number z in the range -1 < z 1, we can find any number of values x such that sin x = z. However, if we restrict the domain of the sine function to -17r < x 17T (the range remaining -1 y 1) then the inverse function does exist, for the mapping becomes one to one. The inverse function so defined is called the arcsine function. Thus, arcsine is a function from {x E R: -1 x < 1} to {y e R: - < y < -PT} with the property that arcsin x = y if sin y = x, and conversely. (Notice that we retain x as an element of the domain.) Another notation for arcsin x is sin-' x; the slight drawback to this notation is that it suggests that the unrestricted sine function has an inverse, which is, of course, false. The graph of the arcsine function may be obtained directly from the graph of the sine function for the restricted domain -47r < x < 17r by interchanging the x and y axes. If we retain x as an element of the domain, so that y = arcsin x, the graph is as shown in Figure 6.10. y

y

IT

Fig. 6.10

Fig. 6.11

The arccosine function may be similarly defined, with domain {x e R: -1 x < 1) and range {0 y < 7r}. If y = arccos x, then cos y = x. The graph is shown in Figure 6.11. Again, the arctangent may be defined, with domain R and range -1rr < y < (notice the strict inequality). If y = arctan x, then tan y = x. The graph is shown in Figure 6.12. Ex. 12. Suggest appropriate domains and ranges for arccotangent, arcsecant and arccosecant. Ex. 13. Write down the values of arcsin 2iarccos 1/,/2 and arctan (-1). 98

4]

INVERSE TRIGONOMETRIC FUNCTIONS

Fig. 6.12

Example 6. Simplify cos (arcsin x). Let arcsin x = y; then —1-7r y —lir and sin y = x cost y = 1 sine y = 1 —

—

x2.

Thus cos y = + V(1 — x2); but —17r y frr and so cos y > 0; it follows that cos (arcsin x) = ,/(1 x2). —

Exercise 6(c) 1. Show that the function f defined by f(x) = cos x+ sin x

is periodic and sketch its graph. 2. Draw the graph of the function f defined by

f(x) = sin x + sin 2x for values of x in the interval 0 x 27r. Solve the equation sin x+ sin 2x = 0, giving values of x for which 0 5 x

27r.

3. Draw the graph of y = tan x°, from x = — 20 to x = 70, plotting points at intervals of 10°. Using the same axes and intervals, draw also the graph of y = cos (x+ 10)° from x = 20 to x = 50. Read from your graph the value of x when tan x° = cos (x+10)°. (0 & C '0') 4. It is required to find an angle x such that sin x = where x is measured in radians. Draw on the same diagram the graphs of y = sin x and y = Ix for values of x from 0 to -fir, taking 1 in. (or 2 cm) to represent 112-7r on the scale for x, and 0.2 on the scale for y.

99


[6

Estimate from your graphs the required angle, giving your answer in degrees to the nearest degree. (0 & C '0') 5. Solve graphically the equation tan 2x° = 2 cot x°, giving all solutions x such

that 0° < x° < 180°. 6. Draw the graph of y = 1— 2 sin 60x° for values of x from 0 to 6, using 2 cm as unit on both axes. By drawing a suitable straight line on the same diagram, read the solutions, within the given range of values of x, of the equation (0 & C ` O')

2(1 — 2 sin 60x) = x. 7. Solve graphically the equation

sin x = cos 2x +1 for values of x for which 0 < x
16. Find the value of the second expression when n = 24. It is proposed to revise the salary scale so that it begins at £720 a year and rises to a maximum by 10 equal annual increments. If each annual increment is £d find an expression for the total amount to be received in salary by an employee in n years when n > 11. Find the integral value of d which would make the total amount to be received in twenty years on the new scale as nearly as possible equal to the total amount received in twenty-four years on the old scale. (Cambridge) 18. The series of natural numbers is grouped as follows: (1), (2, 3, 4), (5, 6, 7, 8, 9),

...

(i.e. each bracket contains two integers more than the preceding bracket). (i) Find the total number of integers in the first (n — 1) brackets. (ii) Show that the first number in the nth bracket is n2 — 2n + 2. (iii) Show that the sum of the numbers in the nth bracket is n3 + (n — 1)3 . (iv) If the first number in the nth bracket is denoted by a and the first number in the (n + 1)th bracket is denoted by b show that the sum of the numbers in the nth bracket is exactly divisible by (b — a) and that the quotient is an odd number. (Cambridge)

150

Revision exercise A

1. If the expressions x3 — (a+ 2) x + 2b and 2x3 + ax2 — 4x — b have a common factor x+ 3, find the values of a and b, and find a second common factor. (0 & C '0') 2. Find the mid-point of the line joining the points (-4, 2) and (2, — 6), and show that the equation of the perpendicular bisector of this line is 3x— 4y = 5. Prove that the point (3, 1) is one vertex of a square of which the points (-4, 2) and (2, — 6) are opposite vertices, and find the coordinates of the fourth vertex. (0 & C '0') 3. You are given that the equation 3x5 -9x+5 = 0 has a root approximately equal to 1. By substituting x = 1 + h, and neglecting h2and higher powers show that x = 11 is a closer approximation to this root. 4. (i) The fourth term of an arithmetic series is 55, and the tenth term is 45. Calculate the sum of all the positive terms. (ii) Find, correct to one decimal point, the sum of the first fifteen terms of the geometric series 1 +(1.02)+ (1.02)2 + (1.02)2 + (0 & C '0') 5. In how many distinct ways may the letters of the word SYZYGY be arranged? In how many of these do the three Ys appear together ? (What does the word syzygy mean ?) 6. Without using tables, prove that tan 60° = A rectangle ABCD lies in a horizontal plane and DE is a line of length h drawn vertically upwards from D; EA, EB, EC are joined. If the lines AE, CE make angles of 60°, 45° respectively with the horizontal, express AD and CD in terms of h, and hence calculate the angle which BE makes with the horizontal. (0 & C '0') 7. Find the equation of the line of gradient —1 which passes through the point of intersection of the lines x + 3y— 5 = 0 and 2x —y— 7 = 0. 8. If x2+2x + 2 is a factor of x4 + axe + b, find a and b. Express x4 + 16 as the product of two quadratic factors whose coefficients are real numbers. 9. ABC is a triangle: points Q, R are taken on AC, AB respectively so that AQ = 2QC, 2AR = RB.

Express AB, AC in terms of RQ, BC. 10. If in. < x < 71 and sin x = I, find (i) cos x, (ii) tan x, (iii) cosec x. 151

REVISION EXERCISE A

11. Write down the expansion of (1 + x)5in ascending powers of x. If (1 + x+ x2)5 = 1+ax+bx2+..., find a and b. 12. Given that (2x + 1) is a factor of the expression 6x3 + ax2+ 6x + 1 and hence find the other two linear factors. find a, 13. Three integers (not necessarily distinct) are chosen at random from the set of integers {1, 2, 3, ..., 10). What is the probability that their sum is 10? What is the probability that their sum is 10 or less? 14. By considering a suitable rhombus of side 1 unit, show that, if sin 0 = a, then sin 20 = 2a,/(1 – a2). Deduce that 1 sin 15° = 2V2 • 4r3Er_ [r(r + 1)]2 – [r(r – 1)]2.

15. Show that n

Hence evaluate E r3. r=1

16. How many terms are there in the expansion of (a + b + c)6 ? Can you generalize your result for (a + b + c)n ? 17. Show that it is not possible to find a number which is written abc in the scale of 4 and bca in the scale of 5. 18. Two points, P and Q, on the circumference of a circle of radius 10 cm are such that PQ subtends an angle of 1 radian (57° 18') at its centre. State the length of the minor arc PQ of this circle, and calculate the length of the chord PQ. If the above circle is drawn on the surface of a sphere of radius 15 cm, find the angle which the chord PQ subtends at the centre of the sphere, and hence find the length of the minor arc PQ of the great circle of the sphere which passes (0 &C'0') through P and Q. 19. I take my wife out to buy her a new hat. There is a probability of -126 that I shall approve of the first hat that we are shown, but if I approve, there is only a probability of -A that my wife will agree with my choice. If we both like the hat,

I purchase it. If I dislike the first hat that we are shown, there is a probability of that my wife will like it. If she does like it she naturally overrules my choice but, when I hear the price, there is a t probability that I shall veto the purchase. What is the probability that I buy her the first hat we are shown? If I do buy her the hat, what is the probability that we both like it? 20. Prove that, if n is an integer, none of the numbers 7n + 3, 7n+5, 7n+ 6 can be a perfect square. 21. Find the equation of the plane through the point (1, 1, 1) containing the line z x-3 y- 1 = — 5 2 –4• Find also the direction cosines of the line of intersection of this plane with the plane Oxy.

152

REVISION EXERCISE A 22. The set S consists of all points in a plane whose coordinates relative to a given pair of perpendicular axes are both integers (in the units used). If A, B, C are members of S, prove that the area of the triangle ABC is a rational number of square units. 23. Eliminate 0 between the pair of equations a sec 0+ b cosec 0 = c, a' sec 0 — b' cosec 8 = c'. 24. Solve the inequality

x —1 < 2 x2-3x-4

and illustrate your answer by sketching the graph of x- 1 Y = x2 — 3x — 4 n

E r(r +1) (2r +3).

25. Evaluate

r =1

26. A = {xe R:— 3 < x 2}, B = {x e R: —2 x < 3}, C = {x e R: 1 < x}. Express, in the form {:}, the sets

AnB, AnC', A'nB, AnBnC, Au(BnC'). 27. Show, by drawing a rough sketch, that the equation x = 2 cos x has just one positive solution. By drawing an accurate graph, estimate this value of x as accurately as you can. 28. A five digit number (that is, a number lying between 10000 and 99999 inclusive) is selected at random. What is the probability that it contains either the digit 8 or the digit 9 (or both)? 29. Show that the plane containing the points with position vectors i-2j,

i+j+k, 2i —j+2k

has vector equation r = i 2j+A(3j+k)+t(i+j+2k). -

Find the position vector of the point of intersection of this plane with the line r = 3i+ 5j 5k+ v(2i + 3j k). —

—

30. a, b are rational approximations to the irrational numbers VA, VB. An approximation to ,/(AB) is calculated by using the formulae: (ii) V(AB) -1-[(a + b)2— (A + B)]; (i) ,/(AB) ab; (iii) V(AB) 1[(A + B)— (a — b)2]. The errors in the three cases are respectively E1, E2, E3. Find a relation connecting E2, E3. -

153

REVISION EXERCISE A 31. A bag contains three white and four red balls, another bag contains four white and three red balls. One of the bags is selected at random and two balls are drawn from it. If both balls are red, what is the probability that, of the five balls remaining in the selected bag, two are red? 32. A and B are two points with position vectors a, b relative to a given origin 0. Prove that the equation of the internal bisector of the angle AOB is

r = A(A + 6).

a

(Recall that is a unit vector in the direction of a.) What is the equation of the external angle bisector? Find the position vector of the incentre of the triangle ABO in terms of a = lad, b = obi, c = IABI and the unit vectors a, b. (The incentre of a triangle is the point of intersection of the bisectors of the angles of the triangle.) 33. Evaluate the expressions (i) ,/[1 +.,/{1 +,/(1 + ...)}]; (ii) .,/[2 + ./{2 +.,/(2 + ...)}]. If n is an integer, when is an expression of the form .,/[n + ,l{n+ V(n+ ...)}1 a rational number ? 34. Two cards are missing from a pack. If three cards are drawn at random from the remaining fifty (without replacement) and all are found to be aces, what is the probability that, if a fourth card is drawn, it will also be an ace?

154

9. Mathematical induction 1. A NOTE ON MATHEMATICAL PROOFS In a mathematical proof we proceed from one step to the next by processes of deductive logic; that is, each step is logically implied by the previous step or steps of the proof. In other words, a mathematical argument has a direction. The first step must depend upon certain premises or known theorems which may or may not be stated explicitly. For example, in the traditional form of proof for a geometrical problem, the premises (data) are usually stated before the proof commences and the argument leads from this statement to the final deduction, using previously proved results of elementary geometry. If, however, we are asked to prove that the probability of a double with a throw of two unbiased dice is A, we must commence by asserting that the set consisting of the thirty-six pairs of numbers (1, 1), (1, 2), ..., (1, 6), (2, 1), (2, 2), ..., (6, 6) constitutes a possible outcome space and that we assume that each of these elementary events has a probability of 33.(`The dice are unbiased.') We are then able to deduce the required probability. Since a mathematical argument has a direction, we shall frequently meet the situation in which a sequence of statements p, q, r are strung together in the form ' statement p implies the statement q which, in turn, implies the statement r and so on'. The notation

p q (read ' p implies q') has already been used; its use in mathematical proofs can lead to conciseness of expression.

Example 1. Prove that x2+ y2+ z2— yz— zx— xy 0 for real x, y, z. E = x2+ y2+ z2— yz— zx— xy 2E = (y — z)2+ (z — x) + (x — y)2 2E E

0, for real 0, for real

x, y, z, x, y, z.

The reader must always be on his guard, when writing p q, that this is indeed what he means: a genuine deduction from the immediately preceding statement is implied. One of the principle sources of error among 6

PPM

155

MATHEMATICAL INDUCTION

[9

beginners attempting to prove mathematical results is to confuse the implication p q with the implication q p. p q means that, whenever p is true, q is true; it does not tell us anything about the truth of the statement p if q is true. (We shall not discuss the meaning of the statement p q in the case when p is false; the bibliography refers to books in which problems concerning the use of implication are discussed at length.) Ex. 1. Ìf —1 > 0, then (-1)2> 0, which is true, and so —1 > 0.' Comment. Ex. 2. A class is asked to prove that, if a/b =cid, then (a + b)I(a— b) = (c + d)I(c— d) (a, b, c, d unequal and non-zero). Criticize the following proof: à+bc+d b= c—d (a + b) (c — d) = (a — b) (c + d) a— ac+bc— ad— bd = ac+ ad— bc— bd 2bc = 2ad a/b = c/d which is true and so the original proposition is true.' Ex. 3. Criticize the proof of the identity cosec2 A tan2 A-1 E tan2 A given below : Ìf cosec2 A tan2 A-1 E tan2 A, then 1 sine A

sine A cost A which is manifestly true, since

= 1 + tan2A -i- sec2 A

1 cos A

= sec A.'

If the implication p q is reversible, that is, if p hold, then we may write

q and q p both

p -- q (read ' p implies and is implied by q').

Example 2. Solve the equation Al(2x+1)+ ,lx = 5. ,/(2x+ 1) + Vx = 5 2x+ 1+x+ 2V[x(2x+ 1)] = 25

2V[x(2x + 1)] = — (3x — 24) 4(2x2 + x) = 9x2 — 144x + 576

>. -:. 156

x2 — 148x+ 576 = 0 (x-4)(x-144)=0.

1]

A NOTE ON MATHEMATICAL PROOFS

Thus V(2x + 1) dmix = 5 (x — 144) (x —4) = 0, but two of the steps of the argument are not reversible, and so it is not possible to infer that (x — 144) (x — 4) = 0 V(2x + 1) + Vx = 5. To express it another way, if x = a is a root of V(2x + 1) + Vx = 5, then it is certainly a root of (x —144) (x — 4) = 0, but, if x = a is a root of (x —144) (x — 4) = 0 it may not be a root of V(2x + 1) + Nix = 5. To complete the solution of the given equation we must substitute back the two possible solutions x = 4 and x = 144. It is then seen that x = 4 is a root of the original equation, but that x = 144 is not. Ex. 4. In Example 2, how do you know that x = 4 is the only root of the equation V(2x + 1) + Vx = 5? Ex. 5. Solve Question 6, Exercise 1 c, using the implication signs, and explaining carefully which steps are not reversible. If a result is stated in the form p q, the converse result (if it holds) is q u p. For example, if p and q are defined as follows: p is the statement `the triangle ABC has AB = AC', q is the statement `the triangle ABC has LB = LC', then a well-known theorem of elementary geometry asserts that p q. The converse theorem, q u p, is also true and the two theorems may be combined together in the single two-way implication p q. It is by no means always the case that, if a theorem is true, then its converse is also true. Indeed, if a theorem takes the form p and q r a converse is not clearly defined. In the case in which a theorem p q and its converse q p both hold, we may use the two-way implication in its formulation; alternatively, we may use the phrase 'if and only if' —sometimes abbreviated to 'if '. In proving the truth of a two-way implication it is vital to remember that two separate proofs are needed. For example, referring again to the proposition about the triangle ABC mentioned above, we (i) assume AB = AC and deduce that LB = LC (AB = AC LB = LC or alternatively, AB = AC only if LB = LC);t and t From the purely linguistic points of view, the words 'only if' are somewhat ambiguous; mathematically, however, no ambiguity can arise if we define ' p only if q' to mean ' p q'. 6-2

157


[9

(ii) assume LB = L C and deduce that AB = AC (LB = LC AB = AC or, alternatively, AB = AC if LB = LC).'t Ex. 6. The triangle ABC is right-angled at A if and only if BC2 = CA2+ AB2. What would you assume if asked to prove the 'only if' part of this proposition? Ex. 7. Correct the following statement: 'The integer N (expressed in the denary scale) is divisible by 5 if and only if the units digit of N is 5.' Ex. 8. Correct the following statement: 'Two vectors a and b are equal if and only if lad = Ibl.' Ex. 9. Is it true to say that, if x > 0 then ax > x2 only if a > x? It is often necessary to disprove an implication; that is, to show that the truth of statement p does not imply the truth of statement q. In this situation we writep q (read '17 does not implyq'). Sincep q means that, in all cases in which p holds, q holds too, to show that p 4> q we have merely to exhibit one case in which p holds and q does not hold (a counter example). For example, if we have p is the statement ' x is of the form (6n + 1) 7T13, n integral', g is the statement `sin x = sin 2x' it is a fairly straightforward matter to prove that, if p is true, then g is true; that is, p = q. To disprove the converse, q p, we have simply to find a counter example, e.g. x = 0, which certainly satisfies q, but is not of the form (6n ± 1) n/3. Ex. 10. Prove that x = (6n ± 1) 7r/3

sin x = sin 2x.

Ex. 11. If p is the statement n is an odd number' and q is the statement 'an integer k can be found so that n = 4k +1' prove that q p and disprove the converse result p q. To complete this section we mention one final method of proving that the implication p = q holds. If a statement p is modified by the addition of the word `not', a new statement 'not p', written p' (sometimes — is obtained. For example, if p is the statement 'the integer n is divisible by 3' p' is the statement `the integer n is not divisible by 3'. p' is called the negation of p. t Another way of expressing implications is by using the phrases 'necessary condition' and 'sufficient condition'. A necessary condition for p is q means that p q; a sufficient condition for p is q means that q = p. A necessary and sufficient condition for p is q means that p a q.

158

1]

A NOTE ON MATHEMATICAL PROOFS

The implication p q is equivalent to saying that we cannot have p true without q being true or, using negatives, if q' is true, then p' is true; in terms of implication, q' p'. The argument also works in reverse and thus the two implications p = q and q' = p' are equivalent. Expressed more succinctly

(p = q) 1+ nx, provided x >

-

1, n > 1.

15. The terms of the sequence {u,.} are all positive and s„ = E ur. Prove that r=1

(1+ u1) (1 + u2) (1 + u3) ... (1+ un) > 1+ s. for

166

77

2.

3]

METHOD OF MATHEMATICAL INDUCTION

16. Prove that 5n < rz! for all sufficiently large n. Add such precision as you can to the phrase 'sufficiently large n'. 17. Prove by mathematical induction that the sum of the angles of a convex n-sided polygon is (2n — 4) right-angles. 18. Prove that an n-sided convex polygon has -in(zz— 3) diagonals. 19. A straight line separates a plane into 2 regions; two straight lines separate the plane into 4 regions and so on. If n straight lines (no three concurrent, no two parallel) separate the plane into u„ regions, prove that un= 1-(na + n + 2). 20. If n is a positive integer, prove that 52n-1-2_ 24n — 25

is divisible by 576.

(Cambridge) 1 1 1 f(n) = 1+ — +— + ...+ — 22 3 2 n2

21. If prove that

E [(3r2 + 3r + 1) f(r)] = (n +1)3f(n)— -in(n + 1).

(Cambridge)

22. Prove that n r(r+1)(r+2) r=1

(r+k-1) = n(n + 1) ... (n+ k) k+ 1

23. If al, a2, a3, ..., anare all positive, prove that 11 1 (ai+ a2 + + a„) (-+-+ ...+ —) an a1 a2

n2.

24. Prove that, for any positive integer n, 2n 2n —1

+

2n(2n — 2)

2n(2n — 2) (2n — 4)

+ + ... (to n terms) = 2n. (2n —1) (2n— 3) (2n —1) (2n — 3) (2n — 5)

(0 & C) 25. A motorist estimates that, by travelling along a main road at a certain steady speed, the probability that the next set of traffic lights will be green if the last set was green is aand that the probability that the next set of lights will be green if the last set was red is 4. He sets out one day to test his theory. Prove that, if the first lights he meets are green and if pn = Pr (the nth set of lights is green when he reaches them)

then

Pn

3(a)n-1 (n

% 1).

26. A man repeatedly tosses an unbiased coin, scoring 1 for each head and 2 for each tail. If pnis the probability that his score will ever be n, prove that

Pn = + Dn• 27. n bags, numbered 1 to n, each contain one white and one black ball. A ball is taken at random from bag one and placed in bag two; a ball is then taken —

167


[9

from bag two and placed in bag three and so on, until finally a ball is taken from bag n. Prove that the probability that this ball is white is (

)

1 2 1 + 3n-1 given that the first ball drawn was white.

28. The sequence {un} is defined by the recurrence relation un -5u„_1+ 6u„_2 = 0 (n i 3) and the initial values u1 = 7, u2 = 17. Prove that un= 2n+1+ 3 n.

29. The sequence {un } is defined by the recurrence relation lin +2 + Un +1 — 21in =

0 (n i 1)

and the initial values u1 = 4, u2= — 2. Prove that un= 2— ( — 2)n.

30. The sequence {u.} is defined by the recurrence relation lin +3 ± 2u. +2 — 14+1 — 2un =

0 (n i 1)

and the initial values ul= —1, u2 = 7, u3 =— 7. Prove that u„ = 2 + ( — on (1+ 2n). 31. (The Fibonacci Series.) The sequence {un} is defined by the equations u1 = u2 = 1, un±i = an+ un_1 (n i 2);

prove that un =

,*5

(an — /J")

where a =

1 +0 2

and 13 =

1 — V5 2

are the roots of the quadratic equation x2 = x +1. 32. Prove that, for positive integral n, 32 ft — 5n is divisible by 7 if and only if n

is even.

168

10. Expectation

1. RANDOM VARIABLES In Chapter 7 we considered random experiments described by outcome spaces. With each elementary event we associated a number, its probability, which denoted our degree of belief that the experiment would result in that particular events occurring. The elementary events themselves may be described in various ways: in some cases, it is natural to denote them by a number, e.g. for the fall of a die we could denote our elementary events by the numbers 1, 2, 3, ..., 6; but sometimes no such natural numerical description exists. For example, for a single toss of a coin, the two element set {heads, tails) is the natural choice of outcome space, the elements being labelled by the descriptions ' heads' and `tails'. In order to make a mathematical analysis of random experiments it is helpful to describe the possible outcomes numerically even in those cases where no such 'natural' description exists. Thus, in the case of coin spinning, the event ' tails' could be denoted by the number 0, the event ' heads ' by the number 1. Such values are called values of a random variable. Let us pause here to recapitulate. Suppose we have a random experiment whose possible outcomes are the n elementary events sb s2, sn. Then we may take as our outcome space the set S sz, • • •, sn}• We now attach two numerical ' labels ' to each element si of S: (i) the probability, pi, of that event occurring; (ii) the corresponding value xiof the random variable (see Figure 10.1). To help fix ideas, consider the following examples: 1. An unbiased die is thrown and the score noted. The outcome space with the associated probabilities and values of a possible choice of random variable are shown in Figure 10.2. If the experiment is repeated a number of times, then the sum of the values of the random variable obtained gives us our aggregate score. 2. Two unbiased dice are thrown in an attempt to score a double. Since we are interested in just two possible outcomes, we may take our outcome space, S, as shown in Figure 10.3. Associated probabilities together with a possible choice for values of a random variable are as indicated. In this case, if the experiment is repeated, the sum of the values of the random variables obtained gives us the number of doubles thrown. 169

EXPECTATION

[10

I I 1 .P2=6 A=6 P6=6 1 1 n -1 1 -1 1 P1 = 3 I '3-6 I P 6-6 P1 P2

f t

s 4I 4I 1 1

stssb

I

PIT

I

S

11 I I I I

I

S

1

1

s

, "o I

I

I

I

I

I _L

I I

I 1

I

1 1

Saal l 530 540 s50 ScEl 1 1

1 1

■

1 1

1 1

1 1

1

1 1

I I I I 1 = 1 x3 = 3 I x4 = 5 I x2 = 2 x4 = 4 xa = 6 Fig. 10.2

1 Xn

XL X2

I

Fig. 10.1 r, 1 ri= -6

5 P2 = 6

I I

I I —Not a

Double

double

x1= 1 X2 = 0

Fig. 10.3

3. A simple coin-tossing game is played as follows : A pays 5 pence for the privilege of tossing two unbiased coins. If he gets two heads, B pays him 5 pence and his stake money is returned; if he gets a head and a tail (in either order), his stake money is returned; if he gets two tails, B keeps the stake. A possible choice of outcome space, with associated probabilities attached, is shown in Figure 10.4. In this experiment we are primarily conP2 =

1

1 4

FL =

1

1

F3 = 74

f1 = 74

tI 1 3 3 I 34 1 :1=1 r =1 1

It I I

s1 1 $2 I

Th

I

1

xi. =-I-5

x3 = 0 x4 = 5 Fig. 10.4

x2 =0 170

-

1]

RANDOM VARIABLES

cerned with the financial outcome and our choice of random variable reflects this concern. If the experiment is repeated a number of times, the sum of the values of the random variable obtained denotes A's net gain in pence. With these examples behind us we now make the following definition: a variable whose value is a number determined by the outcome of a random experiment is called a random variable. The discerning reader will have observed that a random variable is a function from a given sample space into a set of numbers, the values of the random variable being the images under this function. The words `random variable', however, are often loosely used to denote the image— a vice to which we ourselves shall succumb from time to time for the sake of brevity. In the same way that a function is denoted by f and the image of z under./ by f(z), so a random variable may be denoted by X and its associated value for the ith element of the outcome space by xi. Ex. 1. A man pays 10 pence to throw two dice. For any double he receives his stake momey back, together with a prize: one pound for a double six and 20 pence for any other double. Suggest a suitable outcome space, probability distribution and random variable for this experiment. Ex. 2. A man insures his life for £1000, paying a premium of EX. Suggest a suitable choice of random variable to describe the situation.

2. EXPECTATION A mass of numerical data often has an indigestible appearance and conveys very little, unless subjected to considerable analysis. A device commonly used to give an overall impression of the data is to determine their average value. For example, if the heights of a hundred boys are measured, the complete set of results may usefully be characterized by giving their average height. We shall now investigate what meaning can be attached to the phrase `the average value of the random variable X associated with a given probability distribution'. If an experiment is repeated N times, where N is a large number, our interpretation of the probability p, as a measure of our confidence in securing the outcome s1leads us to expect roughly pi N occurrences of si, with similar results for s2, s3, etc. Thus we should anticipate a score of x1 on IAN occasions, x2 on IAN occasions and so on, giving an approximate average value of our random variable X for all N

171

[10

EXPECTATION

repetitions as

Nxi +p2Nx2+ N

+p„Nx,,

= Pixi+P2x2+ • • •+Pnxn n

= 1=1 E Pi xi. The larger the value of N, the more confidence we should place in the value of E pi xias an estimate of the average. We are thus led to formulate the i= following definition: The expectation e(X) of the random variable X is defined by the equation e(X) = pi xi• Example 1. An unbiased die is thrown; what is the expected score? Here p, = a for each elementary event, and the associated values of the random variable X are the integers 1, 2, 3, 4, 5, 6. Thus S(X) =

‘ r 1-1 u

=

21

= 3.5. It is perhaps surperfluous to remark that no one sufficiently familiar with an unbiased die would expect a score of 3.5 on any one throw. 3.5 simply represents the best estimate available, prior to the actual experiment, that we can make of the final average score if the die is thrown a number of times; the larger the number of throws, the better we expect our estimate to be. (The reader is strongly recommended to test the accuracy of this forecast if ever he finds time lying heavily on his hands by throwing a die, say a hundred times, and computing his average score.) Example 2. Two players, A and B, play the following game with three coins: A pays a stake of 10 pence and tosses the three coins in turn. If he obtains three heads, his stake is returned, together with a prize of 30 pence; for two consecutive heads, his stake money is returned, together with a prize of 10 pence. In all other cases, B wins the stake money. Is the game fair? We must first consider what is meant by the question ìs the game fair ?' Intuitively it seems reasonable to label a game as ' fair ' if, in the long-run neither side anticipates any considerable financial gain. Mathematically, a game between two players is fair if the expectation of gain for either player is zero. 172

2]

EXPECTATION

To continue with the solution of the problem, we choose as our outcome space the four-element set S = {(HHH), (HHT), (THH), (anything else)}. On the assumption that the coins are unbiased and that the results of the tosses are independent, Pr (HHH) = (Z)3= 1; similarly, Pr (HHT) = Pr (THH) = 8 and Pr (anything else) = 1— a = As our random variable we take the net gain, in pence, for A on each particular elementary event: x, = +30, x2 = + 10, x3= + 10, x4= —10. Summarizing in tabular form, we have: S

HHH

'HHT

P

1

1t.

X

30

10

e(x) =

THH 1

10

Anything else i —10

30+ k.10±i.10-1-1(— 10)

= 0.

The game is therefore, according to our definition proposed above, fair. Ex. 3. An experiment with three possible outcomes sl, s2, s3, has probability distribution {4, -4-, 4} and associated random variables 3, 2, 1. Determine the expectation. Ex. 4. A man pays 1 penny to throw three unbiased dice. If at least one six appears he receives back his stake money together with a prize consisting of the number of pennies equal to the number of sixes thrown. Does he expect to win or lose? Ex. 5. An experiment can result in three possible outcomes, whose probabilities are 1-, 4, i-. A random variable is assigned whose values are respectively x2, — x, 1. Show that, if the experiment is repeated a number of times, the player may possibly finish with a negative score but can anticipate an aggregate score which is positive. Can his expectation be zero ?

173

[10

EXPECTATION

Exercise 10(a) 1. The values of a random variable, together with their associated probabilities for four different experiments, are given in the tables below. Calculate AX) in the four cases. (i) xi

0

1

2

3

4

5

p,

A-

A-

1

IAA

xi

—2

—1

0

I

pi

Ai

AAA

x;

1

2

3

4

5

6

(iv) xi

1

2

3

4

5

6

7

8

9

10

pi

0

a

0

1

0

1

0

1

0

1

2

2. In Question 1, a new random variable Y is constructed so that Y = 2X— 1. If the probability distributions remain the same, calculate AY) in the four cases. Can you generalize your result in any way ? 3. A player pays a certain sum of money to spin two coins. For two heads he receives back 10p, for two tails he receives 2p, for a head and a tail he receives nothing. In all four cases he forfeits his stake money. What should the stake money be for the game to be fair? 4. Two dice are thrown; find the expectation of the higher score showing (or the score of one of them, if they fall alike). 5. If the probability that a man aged sixty will survive another year is 0.9, what premium should he be charged for a life insurance policy of £1000? (If he survives the year, he receives no money back.) 6. X1and X2 are two random variables, each with values 0, 1, 2, 3, ..., 9, and each possessing a uniform probability distribution. Evaluate I Ari- x2I). (i) g(X,— X2); (ii) 7. Two bags each contain ten coloured discs as shown.

174

Blue

Red

Green

Bag I

4

3

3

Bag II

5

3

2

2]

EXPECTATION

A player stakes a certain sum of money for the privilege of drawing two discs, one from each bag. For two discs of the same colour his stake is returned and, in addition, he is awarded a prize of 10p for two reds, 20p for two greens and 25p for two blues. For two discs of different colours he loses his stake. Show that, if the stake money is 8p, he can anticipate gaining in the long run, but that with the stake at 9p he should expect to lose. 8. The game of Question 7 is repeated, but the player now tosses a coin to decide which bag he must choose from: if he tosses a head, he chooses bag I, if a tail, bag II; he then draws a disc at random from the chosen bag, notes its colour and replaces the disc. He repeats the process again and is paid prizes as in the previous question. Determine the minimum stake (to the nearest penny) required to ensure that the player will show a loss in the long run. 9. The game of Question 8 is repeated but the discs are not replaced between draws. Determine the minimum stake (to the nearest penny) required to ensure that the player will show a loss in the long run. 10. The game of Question 7 is repeated, but the player is now required to place his stake after the result of the first draw from bag I is known. If he draws a red disc first time he pays 2p, if a green, 9p, if a blue, 12p. Show that, in the long run, the player expects to win. Show further that the above stakes are the fairest available that still give an advantage to the player, in the sense that, if any one of the stakes were increased by 1p, the bank would then expect to win. 11. A man pays a stake to throw two dice. If he scores a total of 3 or 11, he receives 40p. For a total of 5 or 9 he receives 20p, and for a total of 7, 10p (in each case the stake money being returned too). He loses his stake money for any even score. Show that he expects to win if the stake money is 21p, but that, if the prizes for scoring 5 or 9 and 7 are reversed, he would then expect to lose. 12. Two identical bags contain respectively (i) four fivepenny pieces and twelve tenpenny pieces, (ii) nine fivepenny and seven tenpenny pieces. You are allowed to select a bag and draw a coin at random from it. If the coin you draw is a fivepenny piece, what would be a fair price for you to offer for the bag you did not select ? 13. The path in Figure 10.5 represents a simple maze along which a rat is made to run. It starts at S and has to finish at F. If it makes a mistake at A by turning along AA' it will return to A and be forced by the construction of the maze, to turn towards F, and similarly at each of the other junctions. The probability of taking either of the two paths available at each junction is -. Find the expected number of mistakes the rat will make in running from S to F. A' B' C' D'

S

0 Start

F A BCD

Finish

Fig. 10.5

14. What is the expected number of moves that can be made by (i) a bishop, (ii) a knight, placed at random on an empty chess board? 175

EXPECTATION

[10

15. Two dice are thrown in 'one turn', each turn costing 5p. If a prize of 40p is given for a double six and a prize of 20p for any other double (together, in both cases, with the stake money), determine the loss to a person playing the game one hundred times. 16. A man puts three coins into a bag, deciding at random for each coin separately whether it is to be a fivepenny or a tenpenny piece. Calculate the expected total value of the coins he puts in his pocket. 17. A man puts three £5 notes into one envelope and three E1 notes into a similar envelope. Each year at Christimas, he chooses one envelope at random and gives his nephew a note from it. As soon as either envelope is emptied by his taking the last note from it, the process ends. (i) State the different totals which the nephew may have received when the process ends; (ii) for each of these totals calculate the chance of its occurrence; (iii) deduce that the nephew's expectation of gain is £12.375. (Cambridge adapted)

3. STANDARD DEVIATION; EXPECTATION OF FUNCTIONS OF RANDOM VARIABLES The expectation S(X) of the random variable X is often called the mean

of X and is frequently denoted by the letter (or, occasionally, when it is necessary to avoid ambiguity by referring to the random variable under consideration, //O. ,tt gives us a forecast of the average value obtained for the random variable if the experiment is repeated a great number of times; it does not, however, give us any indication of how the individual results will be spread out. By way of illustration, consider two unbiased dice, one of which has its faces printed in the usual way, but the other has three faces showing one dot and three faces showing six dots. It is readily verified that the expected score for the second die is 3.5, just as it is for the first but clearly, if each die is thrown one hundred times, the separate scores in the two cases will present a very different appearance: the spreads of the distributions are quite different. To measure the spreads of these distributions it might seem plausible to consider a new random variable Y whose value for a particular face equals the difference between the score showing and the mean (commonly called the deviation from the mean) and to calculate the expectation of Y. However, a calculation in the above example shows that such a quantity is quite unsuitable. For the first die, Y takes the values { 2.5, 1.5, 0.5, 0.5, 1.5, 2.5) with corresponding probabilities {a, a, *, -, *,}. For the second die, the values of Y are —2.5, 2.5) with corresponding probabilities {4,

176

3]

For die I, For die II,

STANDARD DEVIATION

6'( Y) = i( — 2.5 — 1-5 — 0.5 + 0-5 + 1-5 + 2-5) = 0. g( Y) =

-

2-5 + 2.5) = 0.

A moment's consideration should show that our obtaining the answer zero in both cases is no coincidence. Indeed, for any distribution, the expectation of the deviation from the mean is zero. Why? To avoid the cancelling out of the positive and negative scores a plausible precaution would be to choose as our new random variable not Y, the deviation from the mean, but Z, the squared deviation from the mean and this does, indeed, prove a very satisfactory measure of spread. Let us see how such a new choice of random variable works in our example of the two dice. For die I, Z takes the values {6.25, 2.25, 0.25} each with associated probability while for die II, Z takes the value {6.25), with associated probability 1. -},

For die I,

AZ) = -A-(6-25 + 2.25 + 0-25) = 2-92.

For die II,

g(Z) = 6.25.

The precise meaning of these two numbers in probabilistic terms will be more fully explained in the next section (see Chebyshev's Theorem); for the moment it suffices to point out that the larger answer obtained for the second die corresponds to the fact that the scores obtained from this die are, on the average, farther from the mean than they are for the normal die. Summarizing the results, we make the following definition: given a random variable X, with mean ,u, the variance, o x, of X is the expectation of the squared deviations from the mean. That is, if a new choice of random variable, Z, is made, where zi = —,az)2, then = e(Z) =

Pi(Xi Max)2.

i=1

The positive square root of the variance, cr„ is called the standard deviation of X; it has the advantage of being measured in the same units as X. The standard deviation (or, equivalently, the variance) has been shown, in an intuitive sense, to give a measure of spread; its deeper significance will be appreciated by the reader only as he gains a fuller grasp of the subject, when it will be seen that standard deviation plays a central part in the development of probability theory. There is no difficulty in extending the concept of expectation to other functions of a random variable. Indeed, given a function f that maps the random variable X into another random variable Y, where yi = f(x,), 177

[10

EXPECTATION we may define the expectation of f(X), e[f(X)], by

.fiX)] = E Ptflxi);

6

with this notation

i=1

cr

= eq(x—it.)2].

Ex. 6. Find the mean and variance of the random variable X whose probability distribution is shown in the table below: X 0

1

Pi

III

2

3 4 I

Ex. 7. Find the variance for the number of heads showing if three unbiased coins are tossed. *Ex. 8. A random variable has mean # and variance a-2. If a constant c is added to each value of the random variable, describe (without embarking upon any detailed working) what difference this will make to (i) the mean, (ii) the variance.

4. SOME THEOREMS CONCERNING THE MEAN AND STANDARD DEVIATION OF A RANDOM VARIABLE Throughout this section we shall assume that, for some random experiment, an outcome space S with known probability distribution is given and a random variable, X, assigned, where e(X) = it and e[Gy- /02] = 0.2. and cr in terms of the Our first theorem enables us to determine and expectation of X— a and (X— a)2. Its value lies in the fact that these latter quantities are frequently a great deal less troublesome to calculate than are P and a' directly from the definition. Theorem 10.1. If a is any number, then (ii) (i) [(X — a)] = — a ; Proof. (i)

e[(X-a)2]

e[(x--a)] = E pi(xi — a) i=i n

n

= EPixi — Epia i=1

i=1

n

= au — a E Pi i=1

n

= — a,

178

since E pi = 1. i=1

= 0-2 +(# — a)2.

SOME THEOREMS

(ii)

a)2] = =

i=i i=i

pi(xi —a)2 Pi[(xi -10+(lt — a)]2 71,

=102 + i=1

— a) Y., Pi(xi — it) 1=1

n

+—

P J=1 since 2(1a— a) and (,u—a)2are common factors and thus may be taken outside the summation; thus ‘[(X—a)2] = o 2+(u—a)2, since E pi = 1 J=1 and E pi(xima) = 0 (from Theorem 10.1(i) putting a = Example 3. Find the mean and variance of the numbers 1, 2, ..., 20, assuming that they are uniformly distributed. We use Theorem 10.1, with a = 10,

e[(y-10)] = -216{-(9+8+...+1)+(1+2+...+9)+10] = 0.5, p = 10.5; e[(x.- 10)2] = 26 [2(12+22+ ... +92) + Hp] = -216[1.9.10.19+100] = 33.5, o.2 = 33.5 — (0-5)2 = 3325. The next theorem determines the mean and standard deviation of a linear function of the random variable. Intuitively, the results are easy to understand: multiplication of every value of X by A clearly results in a proportionate increase in the measures of the average (mean) and spread (standard deviation), while a shift of origin shifts the average by the same amount but has no effect upon the spread. Theorem 10.2. If A, a are any numbers: (i) the mean of the random variable Y = AX+ a is Ap,+ a, (ii) the standard deviation of the random variable Y = AX +a is Ao-. Proof (i)

‘[AX + a)] = E pi(Axi+ a) i-----1

E pi xi+a E pi 1:=1 1:=1 = Alt+a.

=A

179

[10

EXPECTATION

(ii) The mean of Y is AA +a, by (i) and the variance of Y is given by g{[ Y— (Ait a)]2} = e{[(A X+ a) — (Au + a)]2} n

= E Pi[Axi — AA]2 i=1 = A2 EPi(xi -/L)2 i=i

= A2o2 Our final theorem of this section is a result due to the great Russian mathematician P. L. Chebyshev (1821-94). Its importance lies in the fact that it gives us an interpretation for standard deviation in terms of probabilities; that is, it gives us, in terms of A, an upper limit for the probability that a value of the random variable lies further away from the mean than A standard deviations, whatever the associated probability distribution. (Of course, if we are given some information about the probability distribution we can usefully refine the inequality, but this does not detract from Chebyshev's result viewed as a general theorem.) Theorem 10.3 (Chebyshev's Theorem). If A is any positive number, Pr ( ixi—/.1 > Ao-) < 1/A2. Proof. Define the subset A of the outcome space S by A = {xi:I xt —

> A20.21.

> Flo'} =

Now S = A U A' and A fl A' = 0. Denoting by E summations over A

those values of i for which xi E A, and similarly E and E, we have, A'

0-2 =

S

EPi(xi /02

= E Pi(xi -f02 + E Pi(xi A

A'

E pi(xi— 102, since

pi(xi — ,u)2

0,

A'

> E piA20.2, by the definition of A, A = A2,72

E pi. A

Since A2o-2> 0 we may divide both sides of the inequality by 1 -A-.i > E Pi = Pr (xi: I xi 180

> Au).

A.2 0-2:

SOME THEOREMS

4]

Ex. 9. A probability distribution has mean 3 and variance 0-2 = 1-2. Give an approximate limit for the probability Pr (lx-31 > 2u). The probability distribution for the random variable X (X = 1, 2, ..., 5) is given in the following table: X 1

2

3 ' 4

P A 1 4

5

1 A

Find the mean and standard deviation. What is Pr (Ix-31 > 2a)? Comment upon the results obtained and any discrepancy you observe. Ex. 10. Find an expression for g[(X—,u)3] in terms of e(A-3), p and u.

5. PROBABILITY GENERATING FUNCTIONS We now introduce a technique of considerable value in dealing with probability distributions of an integral variable. Given a random experiment the outcome space for which has associated integral random variable X, the expectation of the function t' is called the probability generating function (p.g.f), G(t), for the distribution. Thus

G(t) = S[r] =

i---1

For example, suppose a card is drawn at random from a well-shuffled pack and that we score the face value of the card drawn (an ace scoring 1 and a picture card 10). The probabilities of scoring 1, 2, 3, ..., 9 are each I' s, while the probability of scoring 10 is 4/13. Thus t9)+.4t10. G(t) = -het ± 0+ The quantity t has no particular significance: it is simply used as a `carrier' for the values xiof the random variable. Since the coefficient of t; is pi, the probability of each value of the random variable may be read off if we know the form of the p.g.f. Thus, if we are able to deduce the form of the p.g.f. we have a concise summary of the probability distribution. Theorem 10.4 gives a method of building up the p.g.f. for a complicated distribution but, before we embark upon its enunciation and proof, we must make the following definition: Two random variables X and Y defined on the same sample space are said to be independently distributed if Pr (X = xi and Y = y5) = Pr (X = xi) Pr ( Y = y5). Our next theorem shows that, if X and Y are independently distributed, then the new random variable Z = X+ Y has as its generating function

181

[10

EXPECTATION

the product of the generating functions of X and Y. For simplicity, we shall consider the particular case in which X can take the n+ 1 values 0, 1, 2, ..., n with Pr (X = i) = piand that Y can take the (n+ 1) values 0, I, 2, ..., n, with Pr ( Y = j) = qi. Then Z can take the 2n+ 1 values 0, 1, 2, ..., 2n, with Pr (Z = k) = rk, say. Theorem 10.4. If X and Y are independently distributed random variables taking the values 0, 1, 2, ..., n and if the p.g.fs of X and Y are Gx(t), G,(t) respectively, then the random variable Z = X+ Y has p.g.f. G,(t) where G,(t) = Gs(t) Gy(t). Proof rk = Pr (Z = k)

but

= Pr (X = 0 and Y = k) + Pr (X = 1 and Y = k - 1) + . . . +Pr (X = k and Y 0) = Poqk+Plqk-1±P2qk-2+ +Pkg0, Gs(t) G „(t) = (po+ pi t + t2 + + pfltn)(q0 + qit + q2 t2 + + qfltn) = Pogo+(Poqi+Pigo)t +(Pog2+Plql+P2q0) t2 • • • =Gz(t).

Before giving an example of the application of this theorem, we prove one further result to illustrate the value of developing the theory of p.g.f.s: we show that they may be used to calculate the mean and standard deviation of a distribution. Theorem 10.5. If the random variable X, whose p.g.f. is G(t), has mean ,u and standard deviation o, then, (i) G(1) = 1; (ii) G'(1) = it; (iii) G"(1) = a2 + ft2 Proof (i) G(t) = pr r r=1 n

G(1) = Fi Pr r=1

=

(ii) G'(t) =

1;

E rps tr-i

r=1 n.

G'(1) =

E rPr

r=1

= g(X) = ft;

182

5]

PROBABILITY GENERATING FUNCTIONS

G"(t) = G"(1) =

E r(r —1) pr

r =1

E r(r —1) pr

r=1

= E r 2 pr— r=1 E rpr r=i = 1(X2)- 1(X) = (Cr2 /42) -it,

by Theorem 10.1, with a = 0.

Example 4. Two unbiased dice are thrown. Find the expectation of the total score, and its variance. Let the random variable X denote the score from the first die; the random variable Y the score from the second die. Then we have to calculate the expectations and variance of the random variable Z = X+ Y. Now Gz(1) = *(ti+t2+ + te), GAO = 6 (t1+0+ ... +16), t6)2, .*. Gz(t) = -P6(t1+ t2+ assuming that the dice fall independently. Thus Gat ) = 118(t1 +12 ) +2t + 3t2+ 4t3+ 5t4 + 6t5), Gat) = _ii6(t 1+0+

•

) (2 + 6t + 12t2 + 200+ 300)

•6

+-N(1 +2t+ 3t2 + 4t3 + 5t4 + 6t5)2, Gal) = 7,

Gal) = 476. By Theorem 10.5, p, = 7 #2 +

T2_itt = 4775_, p= 7, 0.2 =

Ex. 11. Three coins are tossed and the number of heads noted. Find the p.g.f. for the resulting distribution and determine pand 0-2.

6. EXPERIMENTS WITH INFINITE OUTCOME SPACES Experiments with an infinite number of possible discrete outcomes are easy to visualize. For example, suppose a coin is spun until a head appears. Take as random variable X, the number of tails obtained before the first 183

EXPECTATION

[10

head is tossed; then X has possible values 0, 1, 2, 3, ... and Pr (X = r) = Notice that

E Pr (X = r) = = 1 11 2

1,

r=0

which is as it should be. Indeed, all the results of Chapter 7 hold for such outcome spaces; detailed proofs will be omitted, because they depend upon the manipulation of convergent series. However, assuming the obvious generalizations of the definitions and theorems, the calculations involved are not difficult. For instance, in the example above

1 . 1

6r(X) = But r=0 X

r.2

2r+

r = (1 — X)-1(IXI

< 1)

rxr-1 = (1 —x)-2,

and so

r=0

assuming that the infinite series may be differentiated term by term (a result always true, in fact, for convergent power series). Thus E rxr+1= x2(1 - X)-2

r=o

( x)

and, putting x =

= 1.

Example 5. A and B alternately throw a die, the game terminating in a win for A if he throws a 1 or a 6, or in a win for B if he throws a 2, 3, 4, 5. Find the probability that A wins and, if he wins, find the average number of throws he takes, given that A commences the play. Pr (A throws 1 or 6 on a particular toss) = Pr (B throws 2, 3, 4 or 5 on a particular toss) = Pr (A wins on (2r +1)th. play) = Pr (A and B fail alternately on first 2r plays) x Pr (A throws 1 or 6 on (2r +1)th. play) = (1)r Pr (A wins) =

(Dr ( ) =

r =0

3

1

3

— 7•

We now have to find the expected number of throws taken by A, given that A wins. We take as our random variable X where X = r if 184

A wins on his rth turn.

INFINITE OUTCOME SPACES

6]

Since the game can go on indefinitely, X may take any positive integral 12 1)r-1 1 value 1, 2, 3, ... Pr (X = rIA wins) = 33' 3 9

(1y-1

6'(X) = E

and

r=1

7 = 9• — 9

1 — D2

(on using the result proved above: E rxr-' = (1 — x)-2). r=0

Ex. 12. A man tosses a coin until a head appears and is paid a number of pence equal to the number of tosses he makes. What is his expectation?

Exercise 10(b) 1. X is a random variable with mean it and standard deviation a. Write down the means and standard deviations of the following random variables: (i) — X; (ii) X+1; (iii) 3X-1; (iv) (X—#)/(r. 2. Calculate the mean and variance for each of the following distributions: (i)

(iv)

X

0

1

2

3

4

5

P

i

1

15

25

1

t

X

1

2

3

4

5

6

P

A

'A

10

T35

T243

1

X

—3

—2

—1

0

1

2

3

4

230

215

110

1-1,3

1

P

As

235

1 4

X

50

100

150

200

250

p

T.%

1

2

1

A

3. Calculate the means and variances for X2in each part of Question 2. 4. A cubical die is so weighted that the probability of obtaining any face is proportional to the score showing on that face. Find the mean and variance of the score obtained. 5. If the random variable X can take the values 1, 2, 3, ..., n, all values being

equally likely, calculate the mean and variance for X. 185

EXPECTATION

110

6. A, B and C repeatedly throw a die, A starting, then B, then C, then A again and so on. The winner is the first to throw a six. What are their respective chances of success? 7. A throws a pair of unbiased dice, B a pair of dice of which one is unbiased and the other is such that the probability of a six is p. If they throw in turn and the winner is the first to throw a double six, find p, given that, when A has the first throw, the game is fair. 8. An unbiased coin is tossed n times. If a head appears, + 1 is scored, if a tail, —1 is scored. Prove that the p.g.f. for the random variable X, where xi = total score after i turns, is given by G(t) = (2t)--"(t2 + 1)". 9. A die is thrown n times. An odd face makes the score showing, an even face scores zero. Determine the p.g.f. and hence the mean score for n throws. 10. Two dice are thrown together and the scores added. What is the chance that the total score exceeds 8? Find the mean and standard deviation of the total score. What is the standard deviation of the score for a single die? (Cambridge) 11. A card is drawn at random from a standard pack and scores the face value of the card (with ace one and picture cards 10 each). Find the mean and variance of the score. If the card is replaced, the pack well shuffled and a second card drawn, find the probability that the total score for both draws is 12. 12. Two unbiased dice are given, one of which has the faces numbered in the usual way 1, 2, ..., 6, but the other has two faces numbered 1, two numbered 3 and two numbered 5. Both dice are thrown and the total score, X, is recorded. Find the meanµ and standard deviation, o, of X. Find also (i) Pr ( I X— tcl > a); (ii) Pr (I X— it I > 2o).

Miscellaneous Exercise 10 1. Two bags contain red and white discs as shown in the table below: Red

White

Bag I

5

15

Bag II

10

10

One of the bags is selected at random and a disc drawn from it proves to be red. If the red discs are now valued at £1 each and the white discs are valueless, what would be a fair price to pay for the remaining discs in the selected bag? 2. (The St Petersburg Paradox.) A coin is spun. If a head is obtained first time you are paid 1p ; if you get a tail followed by a head you receive 2p; for two tails followed by a head 4p, the next prize being 8p and so on. Show that, however much you are prepared to pay to play the game, your expected profit will be positive. 186


6J

Criticise any assumptions you have made and indicate what further knowledge you would require before offering a more realistic 'fair price' for the game. If the banker against whom you are playing starts with a capital of 100p, what would be a fair price for you to offer him before playing the game? 3. A and B alternately throw an unbiased die, A having the first throw. The game terminates when a six is thrown. B agrees to pay A £1 if A throws a six on his first throw, £3 for a six on his second throw, £5 for a six on his third throw, and so on. What would be a fair price for A to offer to play the game? 4. Two men play a game with two dice. A has a true die, whereas B has a die which is biased so that each of the even faces is twice as likely to occur as each of the odd faces. The two players throw their own die and A wins from B the sum of the numbers thrown when the sum is even and the numbers are unequal. B wins the sum from A when it is odd; and the game is drawn when the two numbers are equal. Calculate the expectation of the game to A. 5. X and Y are two random variables defined over the same probability distribution. A new random variable Z is constructed, where Z = X+ Y. Prove that the mean of Z is the sum of the means of X and Y. Is it also true that the variance of Z is the sum of the variances of X and Y? 6. If X is a random variable with mean uand standard deviation a prove that the standard deviation of the random variable (X--#)2 is {6[(x -F)4] - a4}1. 7. n points are marked on a line at 1 cm intervals. If two of the points are chosen at random, find the probability that they will be r cm apart. Deduce that the expected distance apart of the two points is +(n +1). 8. A and B play a game in which A's chance of winning is p, while B's is q, where p+q = 1. They have a contest, the winner being the first to score two consecutive successes. Prove that the expected number of games is (2 +pq) (1- pq)-1

.

9. Xis a random variable such that Pr (X < 0) = 0 and the mean and standard deviation of X are respectively g and a. Prove that, for any k 1, Pr (X < kit)

k 1

Deduce Chebyshev's result, that Pr (it - Aa < X < it+Acr) 1 - j . 10. The random variable Yhas a distribution such that Pr (Y = r) is given by the coefficient of Br-1in pt(1 - 0q)--4 for r t and zero otherwise, where 0 < p < 1, q = 1-p. Find the expectation of Y. (M.E.I.) 11. Xis a random variable that can take all positive integral values 0, 1, 2, 3, .... If pr = Pr (X = r) and qr = E p, CO

s=r+1

7

PPM

187

[10

EXPECTATION and if the p.g.f. for Xis G(t) and H(t) = prove that

qr tr,

r=0

H(t) = {1

— G(t)}

1—t Prove, furthermore, that H(1) = 4u where A is the mean of X, and determine the value of H'(1).

12. A die is thrown n times. If S denotes the total sum obtained, prove that g(S2) = 7n(21n + 5)/12. 13. n identical cards are numbered 1, 2, 3, ..., n, and a random sample (with replacement) of size k is drawn. If the number on the highest card is taken as the value of the random variable X, prove that Pr (X = r) = — (r —1)9Ink and deduce that, for large n, the mean value of X is approximately nk k +1. 14. A and B play a game of golf. At each hole A's chance of winning is a and B's is b. Find the chance, h, that the hole is halved, in terms of a and b. Show that the chance of A being r up after n holes is the coefficient of x to the power (n+ r) in the expansion of (ax2+ hx+ If a = b = h, find the chance that A wins 5 and 4, i.e. that he was either 5 up with 5 to play and then halved the fourteenth hole, or was 4 up with 5 to play and won it. Leave your answer in terms of powers of 3 and binomial coefficients. (M.E.I.) 15. A man consistently stakes a fixed proportion of his available capital on a fair bet, that is, a bet with zero expectation of gain. If he wins, his existing capital is increased by a %, but, if he loses, he decreases it by b %. Show that he must expect to lose in the long run, whatever values a and b might take. 16. Two numbers X and Y between 1 and 100 (inclusive) are selected at random, all possible pairs (X, Y) having equal probabilities. Let Z denote the maximum of X and Y. What is the probability that Z 50? By use of the formulae n

r = -1-n(n +1)

r=1

and

E r2 = 611(n +1) (2n +1)

r=i

or otherwise, show that the mean of Z is just over 67. Find a median of Z. (A median of Z is any number such that Pr (Z and Pr (Z (C.S.) 17. Is it possible to devise a coin tossing experiment with a single unbiased coin to give a probability of+ for some event? If you are given a coin which is biased in some unknown way, show how to devise an experiment in which the probability of success, to be defined, is 188

II. Further vectors

1. THE DOT (SCALAR, INNER) PRODUCT OF TWO VECTORS We have defined the combination of two vectors by the triangle rule, an operation which we called addition and for which we used the sign +. The choice of the word àddition' was not entirely arbitrary, for it transpired that the addition of vectors bore resemblances to the addition of numbers, in the sense that certain algebraic properties of addition in the two cases were strikingly similar. We shall now define a new way of combining two vectors, which we shall call dot multiplication (otherwise variously called scalar multiplication, or the formation of the inner product) and denoted by . . The use of a familiar name and notation is again dictated by the fact that dot multiplication and the ordinary multiplication of two numbers display a strong resemblance to one another. However, the reader should again be on his guard and realize that the two operations are quite different as they operate on different types of quantities. Having issued this caveat, we now make our definition of the dot product of two vectors a and b:

a.b = Ia n Ibl cos 0, where 0 is the angle between the vectors a and b. Note that the multiplication on the right-hand side of this equation is the ordinary multiplication of three real numbers, a I,IbI, and cos 0, of which the first two are positive (or zero). Before proceeding we must clarify one apparent ambiguity in the definition given above: the angle 0 between two vectors is chosen as that angle

b

b Fig. 11.1 7-2

189

FURTHER VECTORS

[11

(between 0° and 180°) through which one of the vectors must be rotated to bring it into coincidence in the same sense as the other vector (see Figure 11.1). It follows that the sign of the dot product is positive if the angle thus defined between the two vectors is acute and negative if the angle is obtuse. We have said that the word ' multiplication ' is used in this context because of the similarity shown between dot products and ordinary products. This similarity we now endeavour to display, but to give added force to our warning against pressing the analogy too far, we start by observing two laws of multiplication of ordinary numbers that do NOT hold for dot products. Dot multiplication does NOT give closure; that is, a.b is not a vector. Dot multiplication is NOT associative; indeed, (a .b).c is not even defined, for (a . b) is a number, not a vector. (On the other hand, we can talk about (a .b) c, which represents a vector in the direction of c and whose magnitude is lad lb I cos 0 times that of c.) However, there is a sort of associative rule that holds: if a, b are two vectors and k is a number,

k (a .b) = (k a) .b = a. (kb) . The proof of this follows immediately from the definitions and may be left as an exercise for the diligent reader. Another reassuring note may be struck by observing that the commutative law holds for dot products

a.b = b . a. Again, the proof of this is immediate from the definition and may, with even more confidence, be left as an exercise for the reader. The final rule for dot products, that of distributivity over vector addition, i.e.

a.(b+c) = a.b+a.c

is more subtle and the proof is not immediately obvious. However, the effort expended in demonstrating this result is amply rewarded, for with it we shall have set up all the apparatus necessary for an effective algebra of vectors. We begin by defining the projection of a vector a in a direction specified by a unit vector u. Suppose the displacement OA represents a; draw the plane n through 0 perpendicular to u. Then, if D is the foot of the perpendicular from A to 7r,

a = OD+DA = OD +pu

(see Figure 11.2). Further, the expression of a as the sum of two vec190

1]

THE DOT PRODUCT

tors, one in the plane it and one in the direction u is unique. For suppose that

a = OD +pu

= OD' +piu,

Fig. 11.2

where D, D' both lie in

IT.

Then, by subtraction

0 = OD — OD' +pu —p'u

= D'D + (p —p') u and so

DD' = (p — p') u.

But this is impossible unless D and D' coincide and p = p', for DD' and (p — p') u lie in different planes and so cannot possibly be equal, unless each is the zero vector. The number p that arises in this construction is called the projection of a in the direction u. We now prove the fundamental result for projections: if the projection of a1in the direction u is Piand the projection of a2 in the direction of u is p2, then the projection of a, + a2 in the direction u is Pi +P2. Suppose, with the obvious notation, that a1 = OD,+ piu,

a2= OD2+ p,u; then

a, + a, = (0D1+ OD2)+ (p1+p2) u.

But OD1+ OD, is certainly a vector in the plane 7T, and so, by the uniqueness of the expression for a1 + a, as the sum of two vectors, one in n and the other in the direction of u, the projection of a, + a, in the direction of u is p, +p,. Having defined the projection of a vector in a given direction we may obtain a simple geometrical interpretation for the dot product of two vectors a and b. Represent a, b by the displacements OA, OB; let C be the foot of the perpendicular from A to OB, and D the foot of the perpendicular from B to OA. Then OC is the projection of a in the direction of b, and OD is the projection of b in the direction of a (see Figure 11.3). 191

[11

FURTHER VECTORS

Now

a .b = 0 A(OB cos (9)

A

= OA.OD and

a . b = OB(OA cos 0) = OB.00,

and thus the value of a . b is the product of the magnitude of either of the vectors with the projection of the other in its direction. Our proof of distributivity for dot multiplication over vector addition has been long delayed but our result, which we shall state as a theorem, now follows very readily and the results we have proved concerning projections have an interest and importance in their own right and were doubly worth pursuing. Theorem 11.1. For any vectors a, b, c,

a.(b+c) = a.b+a.c. Proof Let the projection of b and c in the direction a be pi and p2 respectively. Then a.(11)+c) = lal (pi+ p2) = lal pl+ lalp2 = a.b+a.c (by distributive law for ordinary numbers). Ex. 1. A, B have position vectors a, b referred to some origin 0. If OA is perpendicular to OB, show that a .b = 0. Is the converse result true? *Ex. 2. If a is the position vector of the point A, interpret the scalar product a. a. *Ex. 3. Show that (a— b) .(a—b) = a .a + b.b— 2a . b. If a, b are the position vectors of the points A, B, interpret this equation (i) when OA and OB are perpendicular; (ii) when OA, OB are not perpendicular. Example 1. Prove that, if the sum of the squares of two opposite edges of a tetrahedron is equal to the sum of the squares of A another pair of opposite edges, then the remaining pair of opposite edges are perpendicular. We are given that AB2 +CD2 = BC2 + DA2 and have to prove that ACIDB.

C Fig. 11.4

Take any origin 0 and let the position vectors of A, B, C, D be a, b, c, d. Then, in terms of these position vectors, we are given that

(b— a). (b — a) +(d — c).(d —c) = (c—b).(c —b)+(d — a). (d —a). 192

THE DOT PRODUCT

I]

Now

(b — a) . (b — a) + — . (d — = (c — b). (c — b) + (d — a) . (d — a) b.b-2a.b+a.a+d.d-2c.d+c.c=c.c-2b.c+b.b+d.d-2d.a+a.a = a.b+c.d (c — a) . (b — d) = 0. But c — a * 0, b — d * 0, .*. AC _LBD. Exercise 11(a) 1. Show that a.a—b.b = (a — b). (a + b). Interpret this equation geometrically and discuss the particular case la I = lb I. 2. If a, b are non-zero vectors such that la + b I = are perpendicular.

b I, show that a and b

3. A is a fixed point with position vector a and c is a constant vector. If the variable point P has position vector r where r satisfies the equation

(r — a) . (r — a) = c . c what can be said about the position of P? 4. Prove that b+ c = 0 and lal = Ibl

(a — b). (a— c) = 0.

Interpret this result as a geometrical theorem. 5. 0 XYZ is a rhombus of side a; P is any point of 0 Y (or 0 Y produced). Prove that OP .YP = XP2 — a2. 6. ACBD is a straight line and 0 is a point not on the line. LAOB, LCOD are both right-angles. If OA = a, OB = b, AC = x, CB = y, BD = z, find the position vectors (relative to 0) of C and D. Deduce that CB. BD _ OB2 AC .AD 0A2* 7. ABCD is a tetrahedron in which AB1CD and AC1 BD. Prove that AD IBC and that AB2+ CD2 = AC2+ BD' = AD2+ BC2. 8. If each edge of a tetrahedron is equal to the opposite edge, prove that the line joining the mid-points of any two opposite edges is at right-angles to each of these edges. 9. Prove that, if two of the joins of mid-points of opposite edges of a tetrahedron are at right-angles, the remaining edges are equal. 10. ABCD is a skew quadrilateral (that is, the vertices A, B, C, D do not all lie in the same plane). The mid-points of AC, BD are P, Q respectively. Prove that AB2+BC2+ CD2+ DA2 = AC2+ BD2+ 4PQ2. 193

FURTHER VECTORS

[11

11. ABCD is a tetrahedron; X, Y, Z are the mid-points of AB, AC, AD and P, Q, R are the mid-points of CD, BD, BC. Prove that AB2+ DC2+2PX2 = AC2+ BD2+2Q Y2 = AD2+BC2+ aftZ2. 12. ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that ALP+ AC2+ AD2= GB2+GC2+GD2+3GA2.

2. THE DOT PRODUCT (CONTINUED); COMPONENT FORM If i, j, k form a right-handed set of orthogonal unit vectors, the following equalities immediately follow from the definition of the dot product:

i.i = j. j = k.k = 1,

(1)

j.k = k.i = i.j = O.

(2)

Theorem 11.2. If a = aii-Fa2 j+a3k and b = bii-Fb2 j+b3k, then

a.b = aibi+ a2b2+ a,b3.

Proof: a.b = (chi + a2 j + as k) .(bii+b2 j+ b3k) = aibii •i+a2 b2 j•j +a,b,k•k+a,b2i • j+alb,i •k

+a2 bi ji+a2 b3 j.k+a3bik.i+a3b2k.j = aibi±a2b2+a3b3 byvirtue of equations (1) and (2).

From Theorem 11.2 it follows that

1a12 = a.a = cif+ 4+ 4 and that

cos 0 =

a.b lallbl

(ilk+ a2b2+ a3b3 ,1(al+ 4+ ag),I(bT+M+ bR).

Example 2. A, B, C have Cartesian coordinates (2, 3, 4), ( — 2, 1, 0), (4, 0, 2) respectively. Find: (i) cos LBAC; (ii) the unit vector perpendicular to the plane ABC. x = AB = —41-2j-4k, y = AC = 2i-3j-2k giving But

x.y = =8+6+8 = 6. x.y = Ix' ly1 cos LBAC = V36,/17 cos LBAC.

194

THE DOT PRODUCT

2]

Thus

6 = 617 cos /BAC,

. cos LBAC = 1/V17. Now suppose u = + mj +nk is a unit vector perpendicular to the plane ABC. Then tu x = 0, u . y = 0, — 41— 2m — 4n = 0, 1 21— 3m — 2n = 0, .*. —4m-4n = 0 and so /:m:n = 1: 2:— 2, on solving for the ratios /:m:n. Thus

u = Ai+ 2Aj — 2Ak where A must be so chosen that lul = 1, i.e. A2 + 4A2 + 4A2 = 19 A=

i.e. and so

u=

Ex. 4. If a = i— 3j + k, b = 2i + 2j — k, find a. b. Ex. 5. Show that the two skew lines (x-1 2 x +1 1

y+1z+1 = 2 ' 3 y-2 2

z

—4

are perpendicular. Ex. 6. Find the unit vector perpendicular to the plane A (3, 1, 1), B (2, 5, 0), C (-1, —1, 15). Ex. 7. Deduce the two distance formulae of Chapter 3 using dot products. Ex. 8. Prove vectorially that the angle between two diagonals of a cube is arccos

3. FURTHER COORDINATE GEOMETRY OF TWO DIMENSIONS Throughout this section we shall suppose a right-handed set of axes Ox, Oy is given, their directions being defined by unit vectors i, j. P1(x1, yi), P(x, y) is a P2(x2, y2) ... are given points with position vectors r1, r2 ;

general point with position vector r. 195

FURTHER VECTORS

[11

A line is specified given a point P1lying upon it and its direction; let us suppose its direction is defined by the unit vector u = /i +rnj. Here / = cos 0, and m = cos 02, where 01and 02are the angles made by the vector u with the axes Ox and Oy respectively (senses being taken into account) (see Figure 11.5).

The line through P1(x1, y„) in the direction u thus has the equation

x — x, _y — y, _ r m — (see Figure 11.5, where 01+ in). Since cos 02 = sin 01, this equation may be rearranged in the form sin 0 Y — = cos 011 (x x1) y — Yl = tan 01(x— x1)

Or

agreeing with the form given in Chapter 3. (Notice, however, that the letter m was used differently on that occasion.) Given two lines, one through P1in the direction u1= /,i+ m,j, the other in the direction u2 = /21t +m,j, the angle cr5 between the two lines may be found by computing the dot product u1. u2 u2 =

Ull ludcos ft

= cos cb (since u1, u2are unit vectors) and thus

cos cb = 1112+ m„m2.

(3)

For example, the lines 2x— y +1 = 0, 3x+ 2y— 3 = 0 may be put in the form x _ y— 1 x-1 _y 1 2'—2 3 or, equivalently, by making the denominators components of unit vectors (12 +m2 = 1) x _y-1. x-1 _ y 1/05 — 2/05' — 2/03 — 3/013' 196

3]

FURTHER TWO-DIMENSIONAL GEOMETRY

The angle between the lines is thus given by cos =

1 2 2 3 v5.03+ v5.03

4 = V65 Notice in particular that, from equation (3) the two lines are perpendicular 412+

= 0.

Provided m1* 0, m2* 0 this second condition may be written

4 /2 = —1.

m1 m2

But, as was noted above, 4/m1and 12/m2are the gradients of the two lines; we have thus regained, by an alternative method, the result of Chapter 3.4.

Fig. 11.6

It is frequently required to find the perpendicular distance from a point to a line. Suppose we are given a line through P1(x1, y) in the direction u = /i+mj and a point P2(x2, y2). Let Q be the foot of the perpendicular from P2 to the given line. Then the unit vector in the direction of QP2 is + /j (consider the dot product with u). Thus P2 Q

=

= ([x2 —

rni+ + [Y2 — j).(— mi+ +lj)

= — m(x2 — x0+ 1(y2 — y). Although this gives the required perpendicular distance, it may not be a very convenient form to handle: the equation of a line is generally given in the form ax + by + c = 0 and we shall now deduce, as a theorem, the perpendicular distance from the point P2(x2, y2) to this line.

197

FURTHER VECTORS

[11

Theorem 11.3. The perpendicular distance from the point P2(x2, y2) to the line ax+by+c =0 is ax2+by2+c (a2+ b2) Proof. The given line may be rewritten in the form x+ c/a y

b

—a or, making the denominators components of a unit vector, in the equivalent form x+ c/a b (az b2) — aIV(a2+ b2) Thus the unit vector defining the direction of the line is b. u = v(412 4.b2)

a 4J(a2

b2)

and a perpendicular unit vector is —

a

J(a2 + b2) i+V(a2+ b2) j•

If P1(x1, y,) is some (arbitrary) point of the given line, the perpendicular distance, d, from P2 is given by d= a(x,— x2) b(y,— y2) = V(a2+b2) +V(a2+b2). But axi +by, = — c, since P1lies on the line and hence d—

ax2+by2+c V(a2+ b2) •

For example, the perpendicular distance from the point (1, 2) to the line 2x — 5y + 1 = 0 is 2 — 10 +1 _ 7 .,/29 ,,/29 (The significance of the negative sign is explained below; the magnitude of the perpendicular distance is 7/A/29.) Suppose now that the line joining OP2cuts the line ax+by+c = 0 at P1(see Figure 11.7). Then, if OP1:P1P2 = 1: A, the coordinates of Piare ( x2 Y2 kl+A' 1+AJ• 198

3]


Fig. 11.7

But P, lies on the line ax+ by+ c = 0 and thus ax by2 +c=0 2+ 1+ A 1 +A or

ax2+ by2+ c = —Ac.

From this equation it follows at once that, if ax2+ by2+ c and c have the same sign, A is negative and, if they have opposite signs, A is positive. Since c may be written as a .0 + b .0 + c, we may summarize the result as follows. If, in the expression ax+ by+ c, the substitution of the quantities (i) x = 0, y = 0; (ii) x = x2, y = y2gives two values of the same sign, then P2and 0 lie on the same side of the line (for, in this case, P, divides OP2 externally and so A < 0); if on the other hand, it yields values of opposite signs, P2 and 0 lie on opposite sides of the line. Ex. 9. Find the perpendicular distances of the point (2, — 1) from the lines 3x + 4y + 3 = 0 and 4x — 3y — 6 = 0. What can you deduce about the position of (2, —1) relative to the two given lines? Ex. 10. Sketch the two lines 3x +y— 3 = 0, 3x +y — 6 = 0 and shade in the region determined by the four inequalities x > 0, y > 0, 3x + y — 3 > 0, 3x + y — 6 < 0. Ex. 11. Find the equation of the bisector of the acute angle between the lines x = 0 and 5x + 12y — 60 = 0. Find the incentre of the triangle formed by the lines x = 0, y = 0, 5x+12y — 60 = 0. Ex. 12. a, b, c are constants. Show that, as 0 varies, the straight line x cos 0+y sin 0+acos0+b sin 0 + c = 0 touches a certain circle. Find the centre and radius of the circle.

Ex. 13. Find the equations of the lines parallel to 8x+ 6y+3 = 0, and distant 1 unit from it.

Exercise 11(b) 1. Determine the projection upon (a) the x axis, (b) the line 3x — 4y +1 = 0 of the line segments joining the following pairs of points: (i) (2, 3), (4, 4); (ii) (— 1, 2), (1, 5); (iii) (-1, 2), (2, — 3); (iv) (a, b), (a+ b, a— b). 199

FURTHER VECTORS

[11

2. Find the cosines of the acute angles between the following pairs of lines: (i) x—y+ 1 = 0, 3x—y+2 = 0; (ii) 3x+2y+1 = 0, x-2y-3 = 0; (iii) 4x — y + 1 = 0, 3x +4y = 0; (iv) ax+ by = 0, ax—by = 0. 3. Determine the perpendicular distances between the following points and lines. (Leave your answers in surd form.) (i) (0, 0), x + 4y— 1 = 0; (ii) (-1, 2), 2x— 3y— 4 = 0; (iii) (-2, —3), x— 3y+ 5 = 0; (iv) (h, —k), hx—ky+2hk = 0. 4. Find the distances between the following pairs of parallel lines: (i) x —2y +1 = 0, x-2y+3 = 0; (ii) 2x+y— 3 = 0, 4x+2y+ 3 = 0; (iii) 5x—y-1 = 0, 5x— y+ 3 = 0; (iv) ax+ by + a = 0, ax+ by— b = 0. 5. Find the equations of the pairs of lines, parallel to the following lines, and distant three units from them: (ii) 8x + 6y +5 = 0; (i) 3x — 4y + 1 = 0; (iii) 5x+ 12y + 2 = 0; (iv) 3x + 2y + 1 = 0. 6. Draw rough sketches and shade in the areas determined by the following inequalities: (i) x > 0, y > 0, y < 2, 2x + 3y — 8 0, y > 0, x—y— 1 < 0, 3x+y-7 < 0; (iii) 2x— y > 0, x — 2y < 0, 2x— y + 2 > 0, x— 2y + 4 > 0; (iv) 1x+y1 < 1; lx—Y1 < 1; (v) + y — 11 < 1; 12x — 3y — 11 < 1. 7. Find the equations of the bisectors of the acute angles between the following pairs of lines: (i) x = 0, 3x — 4y = 0; (ii) 3x-4y-1 = 0, 4x-3y-2 = 0; (iii) 5x + 12y +3 = 0, 6x— 8y + 3 = 0. 8. Find the equation of the image by reflection of the line x—y = 4 in the line 2x+y = 1. (London) 9. ABC is a triangle; A is the point (0, 3) and B is the point (— 5, —2). The orthocentre (meet of the altitudes) of the triangle is the point (— 1, 1). Find (ii) the area of the triangle; (i) the coordinates of C; (iii) the tangent of the angle ACB. 10. Show that the reflection of the point (a, ft) in the line y = mx is the point

((1 —m2)a +24 2ma— (1 — m2) fi\ 1+m2

l+m2

f•

A is the point (p, q), B is the reflection of A in the line y = x and C is the reflection of B in the line y = — x. Find the coordinates of C. Show that C is the reflection of A in the line px +qy = 0. (London)

200

31


11. A triangle ABC lies wholly within the first quadrant and has an area of 4-1- sq. units. The equation of one side is 2x— 5y + 23 = 0 and the vertices A and B are (1, 5) and (3, 4) respectively. Find the equations of the other two sides, the angle ABC, and the coordinates of the orthocentre of the triangle. (London) 12. If f(x, y) a--ax + by + c, discuss the changes in value and sign of f(X, Y), when the point P with coordinates (X, Y) moves in the x, y plane, in relation to the line whose equation is f(x, y) = 0. Determine the smallest value attained by the expression 22x+ lly-21, when x and y vary subject to the simultaneous restrictions 3x+4y 12, 2x — y 2.

(Cambridge)

4. FURTHER COORDINATE GEOMETRY OF THREE DIMENSIONS As a straight line in two dimensions may be completely defined by a point and a gradient, so a plane may be defined, given a point lying on it and the direction of its normal. Suppose that a plane passes through the point A (position vector a) and is perpendicular to the directions defined by the unit normal n. Then, if P (position vector r) is any point of the plane,

AP = r a —

(Figure 11.8). But

AP.n = 0, (r—a).n = 0, i.e. (4) r.n = a.n. Equation (4) represents the vector equation of 0 the plane through A with unit normal vector n. Fig. 11.8 If Cartesian axes are taken in the directions i, j, k and n = 11i+ 12 j+ 13k, a = a,i+ a2 j+a3k, r = xi+ yj+ zk, then (4) takes the Cartesian form + 12y+ 13z = 11a1+ 12a2+ 13a3. (5) Thus, equation (5) represents the Cartesian equation of the plane through A(ai, a2, a3) with normal whose direction cosines are 11, /2, /3. Conversely, r.n = p (p constant) (6) represents a plane with unit normal vector n. For r.n is the length of the projection of OP in the direction n, and, if this is constant for all P, P clearly lies on a plane perpendicular to n. Thus, (using components), ax+by+cz = d 201

FURTHER VECTORS

[l1

is the equation of a plane. The unit normal vector is

( v(a2 + b2 + 0)) + (v(a2+ b2 + 0)) :I+ (v(a2 + b2 + 0)) k and it follows from equation (6), that the perpendicular distance from the origin to this plane is the numerical value of d

4/(a2 +b2+0). Ex. 14. Write down the unit vectors perpendicular to (i) x-2y+2z = 0; (ii) 3x—y+z = 6; (iii) 3x+2y+z = 2. Ex. 15. Write down the equation of the planes through the given points and perpendicular to the given vectors: (i) (1, 2, 0), — 3j + k; (ii) (2, —1, — 1), j+k; (iii) (3, 4, 2), 2i+ 3j 5k. —

—

—

Theorem 11.4. The perpendicular distance from the point P1(x1, yl, z1) to the plane ax + by+ cz + d = 0 is ax1+ cz1+ d ,l(a2+62+ c2)

Fig. 11.9

Proof Let 1'2(x, y2, z2) be any point of the plane ax + by + cz +d = 0 and let N be the foot of the perpendicular from P1to the plane (Figure 11.9). If n is the unit vector perpendicular to the plane, then P,N = P2 P1. n = [(x,— x2) i+ (y1—y2) j+(z1—z2) k] • LV(a2+62+ c2) I+,l(a2+62+ e2) j+V(a2+ b2 +c2)°J a(x1— x2)+b(Y1—Y2)+ c(z1— z2)

,l(a2+ 62+ c2) 202

4]

FURTHER THREE-DIMENSIONAL GEOMETRY

But ax,+by2+cz2 = —d, since

P2

lies on the plane. Thus

axi+ byi+czi+d P,N — v(a2+b2+c2) • Again, axi+ by1+ czi+ d may be positive or negative, depending upon which side of the plane P1lies. The result is completely analogous to that for the straight line. We conclude this chapter by giving some worked examples of coordinate geometry in two and three dimensions.

Example 3. Find the complete locus of a point which moves so that its perpendicular distance from 8x—y+ 18 = 0 is twice its perpendicular distance (0 & C) from 7x+ 4y 6+= 0. Let P(x, y) be any point of the plane. If d1, d2represent the perpendicular distances from P to the two lines —t

7x + 4y + 6 8x—y+ 18 ; d2 — ± V65 V65

The required locus is the totality of points P which satisfy d1 = 2d2; that is

{(x, y) : ±

8x—y+18 _ V65

14x+ 8y+ 121 V65 f•

This is equivalent to (0c, y)

8x—y+ 18 14x+ 8y+ 121 V65 I V65 U {(x, y':

8x—y+18 J65

14x+ 8y+ 121 V65 f

= {(x, y):2x +3y —2 = 0) U {(x, y) : 22x+ 7y+ 30 = 0}. The locus is thus the pair of straight lines 2x + 3y — 2 = 0, 22x + 7y + 30 = 0.

Example 4. Prove that the lines x- 3 _ y — 2 _ z- 4 1—1 3 and

x _y+1 z-1 0—1 1

are skew, and find the perpendicular distance between them. 203

FURTHER VECTORS

[11

The given lines may be written x-3 = y-2— z -4 — 1 3 1 x 1

y+ 1 z1 = p, -

0

1

and so a general point P of the first line has coordinates (A+3, A+2, 3A+4) and a general point Q of the second line has coordinates (,u, — 1, +1). The two lines are clearly not parallel, since their direction cosines are different. Thus they are either skew or else they meet in a point. If they meet in a point, we must have .

tA+3= A +2 = —1, 3A+4 = ,u+1,

for some A, A. But solution of the first two equations gives A = —3, A = 0 and these values clearly do not satisfy the third equation. The lines are therefore skew. Now OP = (A+3) i+(A+2) j+(3A+4)k, OQ = pi—j+(,u+l)k, QP = (A—ic+3)i+(A+3) j+(3A—,u+3)k. But the directions of the two given lines are given by the vectors i+j+ 3k and

i+k.f

Thus QP is perpendicular to both lines -4=>

and

—#+3)+(a+3)+3(3A—,a+3) = 0 (A ,u+ 3) —

+ (3a-12+3) = 0.

Solving these two equations gives A = = 1 and so the coordinates of the end-points of the common perpendicular to the two lines are (2, 1, 1) and (1, —1, 2) and the length of the common perpendicular = V[(2 — 1)2 + (1 + 1)2 + (1 — 2)2] = V6. t Since we are going to use the condition for perpendicularity, in which the righthand side of the dot product is zero, there is no need to make these vectors unit vectors since this would only introduce a factor that would divide out.

204

4]


Note: A more elegant method for finding the direction of the common perpendicular to two skew lines will be described when the vector product is introduced. Example 5. Find the image by reflection of the point (11, — 13, 8) in the plane 2x-3y+z+1 = 0. The vector 2i— 3j +k is perpendicular to the given plane and so the line through (11, —13, 8) perpendicular to the plane has equations x-11 _ y+13 _ z —8 _ a 2 —3 — 1 • The coordinates of a general point P of this line are (2A + 11, — 3A — 13, A + 8)

P lies in the plane

and

2(2A + 11)— 3( — 3A — 13) + + 8) + 1 = 0

14A+70 = 0 A = —5.

Thus, the reflection of (11, —13, 8) in the plane is given by A = —10, i.e. (-9, 17, —2).

Exercise 11(c) 1. Prove that the vectors:

a = i+2j+4k, b = 2i-3j+k are perpendicular. Find the angle between the vectors c = 3i+ j + 2k, d = i— 2j —3k. 2. Find the equations of the planes through the given points which are normal to the given vectors: (i) (0, 0, 0),

j + k;

(ii) (1, 2, —3), 2i+ j —3k;

(iii) (-1, 2, —4), 3i—k;

(iv) (2,4, —1), 6i-3j — 2k;

(v) (2, 3, 4), 4i+ j —3k;

(vi) (a, b, c), ai+ bj + ck.

3. Find the cosines of the angles between the following pairs of lines. (Leave your answers in surd form.) x 2

y-1 z+1 x 1 y+2 1 —0 —3 — —1' -

z-1. —2 '

205

FURTHER VECTORS

Ell

(ii)

x-2 y-3 z-3 x+1 y z-4 1 = -1 = 4 ' -3 = 2 = 1 ;

(iii)

x+1 y+2 z-1 x-1 = y=z+ 1 = = 2 ' -2 3 4 ' 5 0

x y z- a x- a _ y- a _z (iv) = 1 () -1' 1 -1 0. 4. Find the cosines of the angles between the following pairs of planes. (Leave

your answers in surd form.) 2x-y = 0, x+y+z = 0;

(i) 3x-y+z = 0, x = 0;

(iii) 2x -y- z + 3 = 0, 3x -y+ z- 1 = 0; (iv) x+y-4z+1 = 0, 2x-3y+4z+5 = 0; (v) ax+ by = 0, ay- bz = 0; (vi) ax + by- cz = 0, bx-cy+ax = 0. 5. Find the cosines of the acute angles between the following lines and planes. (Leave your answers in surd form.) x-1y-2 z- 3 x+y+z = 0; = 1 = 2 3, x-3y+2 z+1 2x-y- z = 0; (n) -1 = 0 2 '

(iii)

x-2 2

=

y-1 z = -1' 3x-2y+2z+4 = 0; 2

(iv)

x-3 y-4 =z+3 = 4 -3 4

(v)

x-1 y+1 z-2 = = y -z = 0; 1 0 -1'

(vi)

x- a

1

4x-3y-4z = 0;

= y-b=z-c px+qy+rz = 0. m n'

6. Find the unit vectors normal to the planes through the following sets of points, and deduce the equations of the planes: (i) (1, 1, - 1), (2, 3, 1), (5, -1, -13); (ii) (2, 6, 1), (0, 3, 1), (4, 0, -2); (iii) (1, 4, 1), (2, 7, 2), (-3, 0, -1); (iv) (1, 0, -1), (0, -4, - 1), (3, 2, 2); (v) (1, 1, 2), (2, 2, 3), (-2, 1, 11); (vi) (0, c/b, b/c), (c/a, 0, - a/c), (b/a, alb, 0). 206

4]


7. Find the coordinates of points where the line

x-2

y-3 2

1

z+1 –2

meets the planes (i) x–y–z = 0;

(ii) 2x – y– z– 6 = 0; (iv) 2x+ 3y+ z = 0.

(iii) 3x+ 4y +2z– 9 = 0;

8. Find the reflections of the following points in the given planes: (i) (-1, 7, 5),5x–y–z-10 = 0; (ii) (5, –9, –6), 2x–y + 3z– 8 = 0; (iii) (6, 13, –3), 4x –y – 3z + 19 = 0; (iv) (3, 5, 8), 3x–y–z+ 26 = 0; (v) (-5, 11, 6), 5x-6y-2z-27 = 0; (vi) (0, 0, 0), ax+by+cz+d= 0. 9. Find the feet of the perpendiculars from the origin to the lines

x + 1y 2 (iii) i

z- 4 = 1 = –1 ;

x – 7 = y + 1 = z– 4 –1 0' 3

(ii)

x – 3 y – 1 z– = = —5 • 1 –2 1 '

x-3 (iv) y-8 z-3 = = 3 –4 –2 •

10. Find the foot of the perpendicular from (7, –1, 2) to the line x-9 _ y – 5 = z-5 , 1 3 5 11. Where does the line x-2 y-7 1 3

z- 2

cut the plane x–y+z = 0? Find the image by reflection of the line in the given plane. 12. Find the image by reflection of the line

x-10 9

y-4 –1

z- 2 0

in the plane 3x–y+ 2z– 2 = 0. 13. Find the image by reflection of the line

x 1 in the line

y-3 –1

z+ 6 5

x _ y-3 _ z+6 2

1 – 1 207

[11

FURTHER VECTORS 14. Prove that the following pairs of lines are skew: x y-3 1 x-5

3

1

z = -1'

y-8 7

z—2 —1 •

Find the direction ratios of the common perpendicular and determine the shortest distance between the two lines. 15. ABCD is a rectangle and 0 is a point on the normal at C to the plane of this rectangle. AB = a, AD = b, and CO = h. P is a point on AO and the line through P in the plane AOB which is perpendicular to AO meets AB at M. If AP = x, show that PM = xAl(b2+ h2)1 a ; AM = x .,1(a2+ b2+ h2)1 a . Prove that the cosine of the acute angle between the planes OAB and OAD is abla(a2+ h2) (b2+ h2)].

(London)

16. Define the scalar product of two three-dimensional Euclidean vectors u, v. Deduce an expression for the angle between the vectors u = u1i+u2 j+u3k, v = v1i+v2 j+v3k, in terms of u1, u2, u3, v1, v2, v3. A regular tetrahedron has vertices 0, A, B, C, where 0 is the origin, and A, B, C have position vectors with respect to 0 given by OA = —i+j, OB = ai+bj, OC = pi+ qj + rk. Find the numerical values of a, b, p, q, r given that a > 0 and r > 0.

(SMP)

17. Show that the line L given by

x+1 y -1 z+1 5

3

2

is the intersection of the planes 3x— 5y+ 8 = 0 and 2y-3z-5 = 0. Show that every plane containing the line L can be expressed in the form A(3x — 5y + 8) +,u(2y —3z— 5) = 0. How should A, it be chosen in order to ensure that the plane is perpendicular to the plane 5x — y+2z = 2? -

208

4]


Hence or otherwise obtain the equation of the orthogonal projection of the line L on the plane 5x -y +2z = -2, expressing your answer in vector form, x = ta+ b.

(M.E.I.)

18. Prove that the equation of the plane which cuts off intercepts a, b, c on the axes of x, y, z respectively is x z -+-= 1. a b +-c The foot of the perpendicular from the origin to a plane is P(2, -1, 2). Find the equation of the plane. If the plane meets the axes of x, y, z at A, B, C respectively, prove that AP is perpendicular to BC, and find the angle between AP and CP. (J.M.B.) 19. Two planes, iT1and 77.2, have equations 2x+y+z = 1 and 3x+y -z = 2, respectively. Prove that the plane 773, which is perpendicular to n.„ and contains the line of intersection of 7r1and 7r2has the equation x -2z = 1. Points P and Q lie on the planes ir1and 7T3respectively and the line PQ is perpendicular to Tr2. If the coordinates of P are (-2, 4, 1), find the coordinates of Q. Determine the angle between the line PQ and the perpendicular from P to the line of intersection of the three planes. (J. M.D.) 20. The straight line whose equations are x -2 y z +1 = = -2 1 2 meets the plane x + 2y - 2z = 8 at B, and A is the point (2, 0, -1) on the line. The foot of the perpendicular from A to the plane is C. Find (i) the coordinates of B and C; (ii) the length of AC. Show that the sine of the acute angle between BA and BC is The line AC is produced to D so that AC = 2CD. Find the coordinates of D. (J.M.B.)

Miscellaneous Exercise 11 1. The perpendicular distance from the origin to a straight line 1 is of length p and makes an angle a with the positive x axis. Prove that the equation of the line may be taken in the form x cos a +y sin a = p. What is the equation of the parallel line through the point P1(x1, yi)? Deduce the perpendicular distance from P1to the given line. 2. Find the equations of the straight lines which bisect the angles between the straight lines 3x- 4y- 11 = 0, 12x + Sy - 2 = 0. (0 & C) 209

[11

FURTHER VECTORS 3. Find the equation of the bisector of the acute angle between the lines 4x + 3y — 12 = 0, 12x + 5y — 60 = 0.

(0 & C)

4. Sketch the triangle formed by the lines 3x— 4y — 4 = 0, 12x— 5y + 6 = 0, 7x+ 24y — 56 = 0, and verify by calculation and reference to your sketch that the point (1, 1) is the (0 & C) centre of the inscribed circle. 5. A straight line is drawn through A(h, k) and P(x, y) so that AP makes an angle 0 with the positive direction of the x axis, and AP = r. Prove that x—h _y—k cos e — sin 0 — r.

Three vertices of a square are E(2, 2), F(— 2, 2) and G(— 2, — 2); a straight line of gradient ais drawn from A(— 3, — 1) to meet FG at P and EF at Q. Use the formulae in the first part of the equation to find the length of PQ. Find also the radius of the circle with centre at the origin to which APQ is a (0 & C) tangent. 6. Points A and B lie on the same side of a line 1, C is the optical image of A in 1, and BC meets 1 at P. Prove that 1 is a bisector of LAPB. Given that 1 has the equation x +3y = 5, and that the coordinates of A and B are (1, —2) and (— 11, 2) respectively, verify that these points lie on the same side of 1. Prove that the coordinates of C are (3, 4), and find the coordinates of P. (0 & C) 7. Prove that the equation of the circle, centre (12, 13), radius 7, is x2 + y2 — 24x— 26y + 264 = 0. Two sides of a triangle are 5x-12y+5 = 0, 12x— 5y+ 12 = 0 and the incentre is the point (12, 13). Prove that the third side touches the circle x2 + y2 — 24x— 26y + 264 = 0.

(0 & C)

8. Prove that the perpendicular bisectors of the sides of a triangle are concurrent (at the circumcentre) and that the altitudes of a triangle are concurrent (at the orthocentre). 9. Prove that the circumcentre S, the centroid G and the orthocentre H of a triangle ABC have position vectors (referred to any origin 0) which satisfy the relation h+2s-3g = 0. Deduce that SGH is a straight line and determine the ratio SG: GH. Prove further that the mid-point, N, of the line SH is the centre of the circle through the mid-points of the triangle ABC. What is the radius of this circle? 10. OABC is a tetrahredron in which OA is perpendicular to BC and OB is perpendicular to AC. PQR is a triangle such that A is the mid-point of QR, B is the mid-point of RP and C is the mid-point of PQ.

210

4]


Taking 0 as the origin, express the position vectors of P, Q, R in terms of those of A, B, C. Hence prove that OP = OQ = OR. If D is the foot of the perpendicular from 0 to the plane ABC, prove that D is the circumcentre of the triangle PQR. 11. ABCDA'B'C'D' is a cube with edges AA', BB', CC', DD' and diagonal AC'. Show that B'C is perpendicular to the plane ABC'. Find a line in the figure perpendicular to the plane ACC'. What is the angle between the planes ABC' and ACC'? 12. In a tetrahedron PQRS, the edges PQ, RS are perpendicular to the faces PRS, PQS respectively; L is the mid-point of PS and M is the mid-point of QR. Prove that PQ2+ RS2 = QR2 — PS2, PM = SM = 1QR, 4LM2= PQa + RS2.

(0 &C)

13. Prove that there is one and only one line which joins two given skew lines and is perpendicular to each of them. Two fixed skew lines AL, BM have a common perpendicular AB, and the angle between them is 0. Prove that LM2 = AL2+ BM2-2AL . BM cos 0. Prove that, if 0 = -PT and the points L and M vary on the fixed skew lines so that LM is constant, the locus of the mid-point of LM is a circle. (0 & C) 14. The common perpendicular of two skew lines 1 and l' meets them at A and A' respectively. Points P and P' are taken on land 1' respectively so that AP+ A'P' is constant, where the sense of AP and A'P' is taken into account. Show that the locus of the mid-point M of PP' is a straight line in, and describe its relation to 1 and 1'. If AP — A'P' is constant, show that the locus of M is another straight line m', and describe the relation of m and m'. (0 & C) 15. Find the reflection of the line x -7 y+ 2 z = = —2 1 —1 3 in the plane x—y—z+ 2 = 0. 16. Prove that, given two skew lines and a point 0 not lying on either of them, that just one transversal may be drawn through 0 to cut each of the lines. Prove that the lines x+ 5 _ y— y-7 z . 1 —1 2 x-4 y+3 z+2 and = = 2 0 —1 are skew and find the equation of the common transversal that passes through the origin. 17. Obtain the equation of a plane in the vector form n . r = p (n a unit vector), explaining precisely what you mean by each of the three symbols n, r, p. 211

FURTHER VECTORS

[11

Prove that the length of the perpendicular to the plane from the point S with position vector s is In .s – pi; and find the position vector t of the mirror image T of S in the plane—that is, of the point T such that TS is perpendicular to the plane and bisected by it. (M.E.I.) 18. Referred to a given system of rectangular coordinates in space of threedimensions, the points A, B have coordinates (1, 0, 0), ( – 1, 0, 0) respectively. A variable point P has coordinates (x, y, z). Write down the direction-cosines of PA, PB and prove that, if 0 is the angle between them, then cost 0 =

(x2 +y2+ z2– 1)2 [(x2+ y2+ z2+ 1) – 2x] [(x2+ y2 z 2+ 1) + 2.x]•

Deduce that, if P is restricted to lie in the plane z = 0, then (x2+ y2– 1)2 = [(x2+ y2– 1)2+ 4y2] cost 0 and find the equations of the two circles on which P must lie if the angle 0 is kept constant. (M.E.I.) 19. Two planes

x– 3y + 2z = 2, 2x – y – z = 9,

meet in the line 1. Find the equations of (i) the plane through the origin which contains 1, (ii) the plane through the origin which is perpendicular to 1. Find also the coordinates of the reflection of the origin in 1.

(C.S.)

20. Find the value of k such that the line joining the points (– 2, k, – 9), (2, 1, 7) intersects the line joining (-2, –4, 4), (7, 2, 1). What are the coordinates of the point of intersection? 21. A regular tetrahedron ABCD has the face ABC in the xy plane, the origin is the centre of that face, the vertex A is at the point (1, 0, 0) and the vertex D is on the positive half of the z axis. Find the coordinates of B, C, D and of the centre of the tetrahedron, and the direction ratios of the normals to the four faces. Hence or otherwise show that, if any line makes angles a, y, 8 with the faces of a regular tetrahedron, then sine a + sin2/3 + sin2y + sin2 8 22. Prove that the equation of the straight line through the given point A, position vector a, and in the direction of the unit vector b is r = a+tb, where t is the distance from A of the variable point P of the line whose position vector is r. Prove also that the equation of the plane through the point C (position vector c) whose normal is in the direction given by the unit vector d can be expressed in the form r .d = c.d.

212

4]

MISCELLANEOUS EXERCISE 11 The plane through the point C(1, 2, 4) has normal in the direction

d= Find the length of the shortest distance from the point A (3, 4, 5) to the plane. (M.E.I.) 23. To find the position of an underground rock layer a number of vertical borings are made at points on horizontal ground which form a coordinate grid. Results are as follows, the unit of distance both horizontally and vertically being 300 m: (0, 2) (2, 0) (2, 2) (0, 0) Grid-point 0.225 0.162 0.117 0.270 Depth Show that these results are consistent with this part of the rock layer being a plane. Find the (x, y) equation of the line in which this plane when produced would meet the ground, and show that the plane would be inclined at about 31° to the ground. What would you conclude if a boring at (1, 1) gave a depth of about 0.18? (M.E.I.) 24. Calculate the shortest distance between the line of intersection of the planes x-8y+2z+9 = 0, x-2y—z+6 = 0 and the line of intersection of the planes 2x+y+8z-12 = 0, x—y+z-6 = 0; and show that the line which cuts both these lines at right-angles passes through the point (-4, 12, 8). 25. Find the coordinates of the mirror image of the point (p, q, r) in the plane ax+by+cz+d = O. A ray from the origin is reflected successively in the planes x+y—z+l = 0 x —y + 2z— 1 = 0, and and then passes again through the origin. Find the points at which it meets the two planes. 26. Prove that the lines x+1 y+1 z+2 x-1 y+3 = z 1 —2 2 3 1 3 do not intersect. Find the equation of the plane through the origin which contains the first line, and find also the direction cosines of the line through the origin which meets both lines. (Oxford Mod.) 27. The line x// =ylm = zln is reflected in the plane ax+ by+ cz + d = 0. Show that the equation of the resulting line is b2 c 2) y+2bd (a2 + b2 + c2) x+ 2ad b+c2) 1— 2a(al + bm+ cn) =(a2+ b2 c2) m-2b(a1+ bm + cn) (a2+ b2+ c) z +2cd 02+1,2 +C2• n — 2c(a1+ bm+ cn)•

213

[11

FURTHER VECTORS

Hence, or otherwise, find the equation of the plane such that the angle between it and the plane /x + my+ nz = 0 is bisected by the plane ax+ by+ cz = 0. (Oxford Mod.) 28. Write down the equations of the axes Ox, Oy, Oz. What is the equation of the plane containing Ox and the point (a, b, c)? The roof of a rectangular house consists of four inclined planes, each sloping upwards at an angle of 45° to the horizontal. What is the angle between two adjacent faces of the roof? 29. ABCDA' B'C' D' is a cubical box, with faces ABCD, A' B'C' D' and edges AA',

BB', CC', DD'. E is the point on BB' such that BE = lEB'; F is the mid-point of CC'. Find the angles between (i) the line A' B and the plane AEF; (ii) the plane A' BC and the plane AEF. 30. Two lines through the origin have unit direction vectors 11i + mi j + nik and 121+ m2 j + n2 k. Prove that the locus of points equidistant from the two lines is a

pair of planes and deduce the equations of the straight lines which bisect the angles between the given lines. 31. Two straight paths on a plane hillside are at right-angles, and make angles 0 and 0 respectively with the horizontal. If the hillside itself makes an angle a with the horizontal, prove that sine 0 + sin2 ¢ = sine a. Prove also that the acute angle between the projections of the paths on a horizontal plane is arccos (tan 0 tan cb). 32. A line of slope, in an easterly direction, of a plane hillside is inclined at an angle a to the horizontal. A line of slope of the hillside in a southerly direction is inclined at an angle fi to the horizontal. Prove that the actual inclination, 0, of the hillside to the horizontal is 0 = arctan (tang a + tan2 fi)1. A vertical pole of height h is placed on top of the hill. Show that the angle 0 subtended by it at a point distant a down the line of greatest slope through the foot of the pole is given by h tan 0 = a sec 0 + h tan 0.

Find 0, if h = 16, a = 36, a = 30° and ft = 45°.

(London)

33. Let G be the centroid of the acute-angled triangle ABC of circumradius R. Show that AG2+BG2+ CG2 2R2. 34. If 0 is the point on AB such that AO = 20B, and if /3is any point, prove that

AP2+2BP2 — 30P2 is independent of the position of P. What is the locus of a point which moves so that the sum of the squares of its distances from the vertices of an equilateral triangle is constant? 214

12.

Further trigonometry

1. FORMULAE FOR COMPOUND AND MULTIPLE ANGLES In many applications of trigonometry it is essential to be able to deal with expressions such as sin (A+ B) or sin 2A in terms of the trigonometric functions of the simpler angles A and B. Angles such as A+ B are called compound angles; in particular, those like 2A are called multiple angles. Ex. 1. By taking A = B = 30°, show that sin (A + B) * sin A+ sin B in general. Is it possible to find values of A and B such that sin A+ sin B cannot be expressed as the sine of a single angle?

We shall now establish the fundamental compound angle formulae for sin (0+0), cos (0+ cb). The proof depends upon the fact QB that we may express a unit vector OP in the form

cos Oi+ sin 0j, where i, j are two perpendicular unit vectors and 0 is the angle between i and OP measured anticlockwise from i; see Figure 12.1 which represents a circle of unit radius, Fig. 12.1 with OA = i OB = j, OP = p, where p = cos 01+ sin 0j. ,

Similarly, if OQ makes an angle 900 + 9 with i (that is, if LPOQ=90°, measured anticlockwise), then OQ = q = cos (90° +0)i+ sin (90° +0)j = — sin Oi + cos 0j. These expressions for p, q hold for angles of any size and either sense. Theorem 12.1. For any two angles: (i) cos (0 + q5) 72. cos 0 cos 0 — sin 0 sin 0; (ii) sin (0+ 0) sin 0 cos 0+ cos 0 sin 0; (iii) cos (0 — 0) = cos 0 cos 0 + sin 0 sin 0; (iv) sin (0— c6) = sin 0 cos 0 — cos 0 sin 0. Proof. Take points APRBQ as a unit circle, centre 0. OA = i, OB = j; OP = p, OQ = q; OR = r. LAOP = 0, LPOR = cb (see Figure 12.2). 215

FURTHER TRIGONOMETRY

[12

The proof of (i) and (ii) is effected by expressing the unit vector r (a) in terms of i and j directly and (b) in terms of p and q and hence in terms of i and j. QB Thus (a) r = cos (0 + 0) i+ sin (0+ 0) j and

(b) r = cos Op + sin 0q.

But

p = cos Oi + sin Oj

and

q = — sin &i+ cos Oj

Fig. 12.2

(see remarks preceding this theorem). Substituting for p and q in (b) this gives

(c) r = cos 0(cos Oi+ sin 0j) + sin 0(— sin Oi+ cos OD = (cos 0 cos 0 — sin 0 sin 95) i+ (sin 0 cos 95+ cos 0 sin 95) j. But, in a plane, the expressions for the components of a vector r in the two directions i and j are unique (see Chapter 2) and so, comparing (a) and (c), (i) cos (0 + 0) = cos 0 cos 95— sin 0 sin 0; (ii) sin (0 + 0) = sin 0 cos 95+ cos 0 sin 95. Formulae (iii) and (iv) now follow immediately, on writing — 0 for 0 and recalling that sin (— 0) — sin 0, cos (— 0) = cos 95, (iii) cos (0- 0) = cos 0 cos 0 + sin 0 sin 0; (iv) sin (o— 0) = sin 0 cos 0 — cos 0 sin 0. Ex. 2. Express sin 15°and cos 15°in surd form by writing 15° = 45° — 30°. Ex. 3. Express cos 165° and sin 105° in surd form.

A- and sin y = s find sin (x +y) and cos (x +y) (i) when x and y are both acute angles; (ii) when x is acute and y is obtuse; (iii) when x and y are both obtuse angles.

Ex. 4. If sin x =

Ex. 5. Express cos (A+ B) cos (A B) in terms of sin A and sin B. —

—

The results of Theorem 12.1 are important and should be committed to memory. They give rise immediately to the following results which we shall also state in the form of a theorem:

Theorem 12.2. For any angles 0, 0 for which all of the expressions appearing are defined tan 0+ tan 0 (v) tan (o+0) = 1 — tan 0 tan 0 ' _ tan 0 — tan 0 (vi) tan (0— 0) = 1 +tan 0 tan 0' 216

1]

COMPOUND AND MULTIPLE ANGLES

(vii) (viii) (ix)

sin 20 = 2 sin 0 cos 0; cos 20 = cos' 0— sine 0 = 2 cos20 —1 = 1-2 sin,0; tan 0 tan 20 = 2 1 — tang

Proof (v) The right-hand side is not defined if 0 or 0 = (2k +1) 17T; neither side is defined if 0+¢ = (2k+ 1) in. For any other values of 0 and 0 tan (0 + 0) =

sin 0 cos 0 + cos 0 sin 0 by formulae (i) and (ii); cos 0 cos 0 — sin 0 sin 0

divide top and bottom of the fraction on the right-hand side by cos 0 cos 0 (a non-zero expression, by the restrictions on 0 and 0). (vi) As for (v) by using (iii) and (iv); (vii) set 0 = cb in (ii); (viii) set 0 = cb in (i), and recall that sin' +cost 1; (ix) set 0 = 0 in (v). *Ex. 6. What values of 0 and cb must be excluded in (a) formula (vi); and (b) formula (ix)? Example 1. Find an expression for cos 75° in surd form. Method (i). cos 75° = cos (30° + 45°) = cos 30° cos 45°—sin 30° sin 45° 1 11 = — 2 V2 — 2.V2 V3 —1 2V2 • Method (ii).

cos 150° = 2 cos275° —1, --b/3 = 2 cost 75° — 1,

or

4-20 = cos' 75°. 8

Thus

cos 75° =

—1 2V2 •

(We reject the negative square root since cos 75° > 0.) Example 2. Prove the identity

cos 0 —sin 0 — sec 20—tan 20 (0 + (2n+1) 177). cos 0+ sin 0 217

FURTHER TRIGONOMETRY

[12

The R.H.S. is defined, by the restriction on 0. R.H.S. =

L.H.S. —

1 sin 20 cos 20 cos 20 1— sin 20 cos 2 0 ' (cos 0—sin 0)2 (sine 0 + cos2 0, since 0 + (2n +1) *ir) cos2 0— sine 0' cos20+ sine 0 2 sin 0 cos 0 cos 20 1— sin 20 cos 20 -

The formulae we have proved are also useful in dealing with combinations of inverse trigonometric functions, as is shown in the next example. Example 3. Find the value of arctan 2 + arctan 3. Write x = arctan 2; y = arctan 3, then rr

< x < 17r, err < y < -1-rr and so +7T

det A from the set of all 2 x 2 matrices with real elements to the set of real numbers. Similarly, we have mappings A -> det A from the set of all 2 x 2 matrices with rational/integral elements to the set of all rationals/integers. The rule for determining the value of det A must be carefully memorized: products of the elements are formed diagonally (the leading diagonal first) and then subtracted. Thus 1 2

3 4

2 -1 -3 -2

= -2;

= -7;

cos a sin a = 1. - sin a cos a

Ex. 11. Evaluate: (i)

3 1

(ii)

0 1

1 2

-2

1

(iii)

1 -2 3 1

The construction of the determinant of a 2 x 2 matrix has thus given us a test for singularity. We now define determinants of higher orders; although the definition will no doubt appear complicated, the motivation for their introduction should be clear—we shall use them to extend the result of Theorem 14.1. If A is the 3 x 3 matrix (all an a n a21

a22 a23

a31 a32 a3 then the determinant of A written all an a13 a21

a22 a23

a31

a32 a 33

or, more briefly, det A (or IA l)

is defined by det A =

a22 a23

a32 a33

- a12

a21 a23 + a13 a21 a22 32 a31 a 33 a31 a

= an aa22 a a13 a22 a31 a33 — all a23 a32 - a12 a21 a33 +a12 a23 a31 +a13 a21 a32 It is simply verified by direct calculation that det A may be expressed in the two alternative forms det A = -1121

au a13 - a23 au a12 a12 a13 + a22 a31 a 33 a31 a 32 a32 a33 267

[14

MATRICES 2

or a12

a13

a22

a23

an a12 an a13 + a33 a21 a23 a21 a22 These three forms give the expansion of the determinant by rows. The determinant may also be expanded by columns; for example, expanding by the first column det A = a31

det A = all

— a32

a22 a23

a32 a33

The term

— a21

a12 a13 a12 a13 + a31 23 a32 a 33 a22 a

a23 a32 a 33

a22

is called the minor of a„; similarly, the minor of a12 is a21 a23 a31 a 33 (see expansion by first row) while the minor of a32 is als azl a23 (from the expansion by the third row). To find the minor of any term, write down the determinant and cross-out the row and column containing the term under consideration; the required minor is the 2 x 2 determinant that remains. See Figure 14.1, where the minor of a2, is being sought: it is all

12 au a 32 a31 a

or (a a32 —

a31).

an 012 an — —021— an— -023- -33 a32 /2

Fig. 14.1

Fig. 14.2

To find the sign to be attached to each minor in the expansion of a determinant, draw the chess-board pattern, as shown in Figure 14.2, starting with + along the leading diagonal. Thus, if we were expanding by 268

2]

DETERMINANTS

the second row, or by the third column, we should include the term — a23

all a 12 a31 a32

A minor, together with its correct sign as given in Figure 14.2, is called a cofactor, the cofactor of the term ad being written A. Thus A22 =

all a31

a13 = a33

a33 —

an;

an a13 = a13a21— a11a23. a21 a2.3 In terms of cofactors, the expansion of the determinant may be written A32 = —

det A = all A11 +a12 Al2 a13 A13 (first row), det A = Al2 a22 A22 ± a32 A32(second column), etc.

or

Determinants of orders 4, 5, 6, ... may be defined similarly. Thus, for the 4 x 4 determinant all alt a13 a24

a21 a22 a23

a33 a 34

a31 a32 a41

a42 a43 a44

each cofactor is a 3 x 3 determinant, obtained by rejecting the row and column in which the corresponding element stands and attaching the required sign, obtained from the chess-board pattern, to the remaining determinant. Thus A22 =

an a13 a14 a31 a 33 a43 a 44

all

a12

a21

a22

a31

a32

a14 a34

The expansion of this determinant by the third column is a,A13 ± a, A23 ± a3,A33 + a43 A43. Ex. 12. Evaluate the determinants:

1 0 5 (i)

(iv)

2 1 6 1 1 3

1

1 2

1 1

1 —

1 2

(ii)

0 3 3 1

0 4 2 3

1 2 1

2 3

1

(v)

—1 0 4

(iii)

—1 1 2 1 1 3 —2 4 9

a h g h b f g

f

269

MATRICES 2

[14

3. PROPERTIES OF DETERMINANTS We shall now develop some simple properties of determinants, restricting ourselves to the 3 x 3 case for simplicity, although the results and proofs all extend to n x n determinants. First observe that the interchange of rows and columns does not affect the value of a determinant: for we may expand in the first case by the first row and in the second case by the first column and, in either case, the corresponding cofactors remain the same. Thus, any property possessed by the rows of a determinant is possessed equally well by the columns. For brevity, we shall denote the determinant an a12 a13 a22 a23 a31 a32 a33

Property 1. an +x

+y

a21

a22

a31

a32.

±z a, a33

a13

For L.H.S.

by A.

z x y a21 a22 a23 a31 a32 a33

a13 a23 a21 a22 a31 a32 a33 au.

= (a„+ x) An+ (an+ y) Al2 ± (an + Z

)

A13 =

R.H.S.

Property 2. If two rows (or columns) are identical, det A = 0. For, if two rows are the same, expanding by the third row the corresponding cofactors are all zero. Property 3. If an elementary row operation is performed on A, the determinant of the new matrix is equal to the product of det A and the determinant of the corresponding elementary matrix. Case (i): the interchange of two rows (columns) If, for example, we interchange rows 2 and 3, on expansion by row 1 the cofactors all change sign but remain the same in magnitude; but 1 0 0 0 0 1 = —1. 0 1 0 The result may similarly be verified for any other interchange of two rows (columns). Case (ii): the multiplication of a row (column) by a non-zero constant. ca„ can can a21 a22 a23 a31 a32 a 33 by expanding by the first row. 270

= c det A,

3]

PROPERTIES OF DETERMINANTS

But the corresponding elementary matrix is c 0 0 c 0 0 0 1 0 and 0 1 0= c. 0 0 1 0 0 1 Case (iii): the addition to any one row (column) a constant multiple of another row (column). Suppose, for example, that we add to row 1 c times row 2 all +ca21 + ca22 a13 +ea, a22 a23 a21 a23 = det A + c a21 a22 a23 a21 a22 a31 a32 a33 a33 a31 a32 by properties 1 and 3 (case (ii)) = det A, by property 2. But the corresponding elementary matrix is (1 c 0) 1 c 0 0 1 0 and 0 1 0= 1, 0 0 1 0 0 1 on expanding by the first column. Other possibilities are dealt with in a precisely similar manner. Property 3 may clearly be extended to a sequence of elementary row (column) operations on A. Property 4. det (AB) = det A det B. Write B = (bi;), then det AB aub1, + anb21 + b31 a n + b 22 +anb,2 /513 + a12 b23 + b33 a221721+ a23b31 a21b12 + a22b22 + a23 b32 b13 + a22 b23 a23b33 anbil+ a32 b21+ a33 a31b12 + a32 b22 + a33b32 a 31/513+ a32 b23+ a3,b33 (

But this may be expressed, by an extension of property 1, as the sum of twenty-seven determinants, twenty-one of which are zero. (For example, one of the vanishing determinants would be an bn allbn a13b33 an an a13 a21 b11 a21b12 a23 b33 = bll b12 b33 a21 a21 a23 = 0 by property 2.) a31b11 a 31b12 a 33b33 a31 a31 a33 We are left with six determinants which do not have two identical columns. However, using properties 1 and 3 we get det AB = det A (b11b22 b33 — b11 b23 b32 b12 b23 b31 — b12 b21 b33 b13 b21 b32 — bnb,2 b3i) = det A det B. 271

MATRICES 2

[14

One simple but useful result that follows immediately from property 2 can now be derived. It will be recalled that a„A„+a„A„+a„A„ = det A and similarly for other rows and columns. Now suppose we multiply each element of some row (or column) by the cofactors of the corresponding elements of another row (or column); for example all A21 + a12 A22 + a13A23.

Now

an a12 a13 33 all A21 a12 A22 +a13A23 = a31 a32 a an a12 a13

= 0 (r„ = r3)•

A similar result holds for any other combination of rows (or columns). Such an expansion is called an expansion by alien cofactors and we have demonstrated that expansion by alien cofactors gives the value zero. Example 3. Evaluate the determinant 8 4 12 9 3 3 5 15 10 8 4 12 9 3 3 5 15 10

2 1 3 = 4.3.5 3 1 1 1 3 2

(removing factors)

2 = 60 1 1

r2 = r2 —rl

1 0 3

3 —2 2

2 1 7 =60 1 0 0 1 3 4 1 7 3 4

= — 60

c3' = c3+ 2c, (expanding by second row)

= (— 60) (-17) = 1020. Example 4. Express the determinant 1

1

1

a2 b2 c 2

be ca ab as the product of factors. 272

3]


If we set b = c, the given determinant is zero (two columns identical) and so (b — c) must be a factor. By symmetry, (c — a) and (a — b) must also be factors. But the expansion of the determinant clearly gives an expression of degree 4, in a, b, c and so the remaining factor must be linear and symmetrical, i.e. (a + b + c). 1

1

a2

b2

1 c2

= A(b — c) (c — a) (a — b) (a + b + c).

be ca ab By comparing the term in ab3on both sides, the value of A is clearly 1. Ex. 13. Evaluate the following determinants:

(i)

3 —1 1

(iii)

11 2 16

7 —4 1 —3 ; 3 —1 3 —3 15

5 2 4

(ii)

2 1 4 2 —5 —1

—1 1 ; —2

(iv)

8 13 3 1 14 8 19 55 22

With the properties of determinants that we have just proved we are almost in a position to extend our theorem on the necessary and sufficient condition for a matrix to be non-singular to matrices of order higher than two. However, before we do so, we shall introduce a further concept, the adjugate (or adjoint) of a matrix, since there is an intimate connection between the adjugate and inverse of a non-singular matrix. Given any square matrix A, where 1(an. a12 an 23 , A = a21 a 22 a a31 a 32 a 33 the adjugate (or adjoint) of A, written adj A, is defined by

All

A21

A31

adj A = (An A22 A32)

9

A13

An A33

that is, the adjugate is obtained by substituting for each element its cofactor and transposing (see Miscellaneous Exercise 13, Question 7).

Theorem 14.2. For any square matrix A A (adj A) = (adj A)A = (det A) I. 273

MATRICES 2

[14

Proof (3 x 3 matrices) an a12 ai3 A (adj A) = (21 a22 a23 a3, 32 a3,

(All

)

A21

A

A31 a)

Al2

A22

A32

A13

A23

A3

=

0

0

0 A 0

0 0I, A

since each term in the product is either an expansion of the determinant, or an expansion by alien cofactors; = AI, where A = det A. A precisely similar result holds for (adj A) A. Corollary. If det A + 0, det (adj A) = (det A)2 (3 x 3 matrix). Proof Take determinants of both sides of A (adj A) = (det A) I and divide both sides by det A. It can be shown that, in fact, det A = 0 det (adj A) = 0 and so the restriction that det A must be non-zero may be dropped. If det A + 0 we have now obtained an inverse for A, namely adj A since A. adj A= I det A' det A We note further that multiplication of this inverse with A is commutative. To obtain a complete picture about inverses we need to clear-up two final points: (i) Can a matrix A have an inverse even if det A = 0? (ii) If det A + 0, can we find an inverse of A other than adj A/det A? Theorem 14.3. If A is any 3 x 3 matrix, we have: the matrix A is non-singular'- det A + 0. Proof (i). To prove the implication =. we show that det A = 0 A is singular (see Chapter 9). det A = 0 = for any matrix B, det (AB) = det A det B = 0 AB + I, since det I = 1 A is singular; (ii) det A + 0 adj det A Aexists A is non-singular, by Theorem 14.2. 274

31


Theorem 14.4. _ adj A det A is the unique inverse of the non singular matrix A. Proof. Suppose L is a left inverse other than A-1; that is, suppose that LA = I, L A. Then -

L = LI =- L(AA-1) = (LA) A-1 =

= A-1,

which contradicts the assumption that L A-1. Similarly for right inverses.

Exercise 14(b) 1. Evaluate the following determinants: 1 3 (ii) (i) 2 4 ; (iii)

6 5

(iv)

5 4 ;

2

-7

-1

3

-2

3

-4

-5

2. Evaluate the following determinants: (i)

-4 7 4 3 1 2 5 -9 -4

(iv)

4 3 1 3 4 2 ; 10 8 3

(vii)

1 3 5 2 7 1 ; 5 16 19

(x)

112 129 104 67 78 62 99 114 92

;

(v)

(viii)

1 2 3 12 13 14 ; -7 -3 1

(iii)

3 2 5

3 7 4 2 7 3 ; 5 4 9

(vi)

-4 2 7

6 -2 5

(ix)

7 17 16 13 33 30 10 21 23

5 3 4

7 5 ; 6

4 7 17

8 1 3

-5 -3 2

3. Determine which of the following matrices are non-singular: 10 7 24 9 6 10 1 3 (i) (3 2

5); 1

16 6 31 (iii) (17 9 29); 11 5 20

(ii) (23 14 29); 6 4 1 -1 (iv) ( 3 11

8 1 -7

6 2). 0 275

[14

MATRICES 2

4. Find the adjugate of each of the following matrices and, in each case, evaluate the product of the matrix with its adjugate. Write down the determinant of each matrix. 5 1 —2 1 2 1 5 4 ; (ii) — 2 3 7 5); 2 3 6 (i) (5 13 9 4 11 5 (iv) (1 4 2) . 1 2 1

1 3 1 (iii) (3 10 5); 1 4 5

5. Find the inverses of the following matrices by the adjugate-determinant method: 2 5 1 ( 1 —2 1 (i) 3 —1 5 ; (ii) (1 0 3); —1 4 0 3 1 2 2 4 1 (iv) (1 4 2) . 1 2 1

3 7 2 (iii) (2 6 3); 1 4 2

6. Find the values of a which make the following matrices singular: (i)

1 3 7 4 2 5 ; (5 5 a

(ii)

7 13 1 —1 a 3 ; — 3 7 2

(iii)

( 1 —3 —2 6 4 —11

7. Factorize the following determinants: (i) 8.

1 1 1 a b c a2 b2 c2

1 1 1 a b c a3 b3 c3

;

2 1 2 7 A= (0 1 2), B = (1 1 2 8 0

(iii)

a b c a2 b2 c2 be ca ab

0 1 2

0 0. —1

2 1 —2

1 0, 1

Solve for X the matrix equation A2 +AX = B. 9.

3 4 4 A= (2 1 5) , 1 8 8 1 —2 —1 -- 1

1 C= (2 276

0 B = (1 3

2 1 3 1 2, D= 2 2 1 . 3 0 1 2

5 a). 16


3] Solve for X the matrix equation

AX—B = CX+ D. 2 A= 1 0

10.

1 1 2

—4 3 , B= —1

—4 11 —16

2 —5 7

1 —3 5

Solve for X the matrix equation B-1XB = A.

Miscellaneous Exercise 14 1. Prove that 11 11 )

(-1 0) (0 01

1 0) (a b) 1) (1 1 c d

c d\

b)•

Deduce that the rows of a 2 x 2 matrix can be interchanged by operations which add multiples of one row of the matrix to the other, together with operations changing the sign of a row. Find a 2 x 2 matrix X such that ba =

(a c

2. Do 2 x 2 matrices A, B with integer entries, exist such that (a) AB = 0, BA * 0? (b) AB = BA = 0, A * 0, B * 0, A * B? (c) AB = BA * 0, A, B * I, A * B? In each case, if your answer is 'yes', justify it by giving examples of suitable A, B. (M.E.I.) 3. If

a= (b) a so that the transpose a' is given by

a' = (a b c) write down the 3 x 3 matrix aa' and prove that the determinant of this matrix has value zero. Obtain the analogous result when a 0) a .(1, 1 . (SMP) c 2 4. If A is any 3 x 3 matrix and P is a non-singular 3 x 3 matrix, prove that det (A —

= det (B— AI), where B = 277

MATRICES 2

[14

5. S is a skew-symmetric matrix if S+S' = 0; A is an orthogonal matrix if AA' = I. If S is a given skew-symmetric matrix and A, B given orthogonal matrices, prove the following results: (i) det A = ± 1; (ii) AB is orthogonal; (iii) (I— S) (I+ S)-1is orthogonal. 6. If

A

la = V)

I),

prove by induction that an An =

b aan:Il (a * 0). 0 1

()

Prove also that the nth power of the matrix B=

CO

0[ nib) 1) is (1 b

Do these results hold if n = —1? 7. Prove that b2c2+ a2d 2 bc+ ad 1 c2a2+ b2d2 ca+ bd 1 = (b — c) (c — a) (a — b) (a — d) (b — d) (c — d). a2b2+ c2d 2 ab + cd 1 Also prove that al x x x x a2 x x x x a3 x x x x a4 8. If

1 ). = (al — x) (a2— x) (a3— x) (a4— x) (1 + x 7--1.(1, — x (0 & C)

a h g M= (h b f) g f c

and det M rr 0, prove that, if A is a root of the equation A+AHG H B+ A F = 0, G F C where A is the cofactor of a, etc., then det M/A is a root of the equation (x+ a) (x + b) = h2. 9. Factorize the determinant 1+a2+a4 1+ab+a2b2 1+ac+a2c2 1+ ab+ a2b2 1+62+64 1+ bc+b2c2 1+ac+a2c2 1+bc+b2c2 1+ c2+c4 278

(0 & C)

(0 & C)

3]


10. Prove that 1 1 1 xy yz zx = —16xyz(y— z)(z— x) (x— y). (0 & C) (y + z — x)4(z + x — y)4(x + y — z)4 11. If

(a 0 0 d 0 b 0 e A= 00 c f d e f x

where a, b, ..., fare real and a > b > c, prove that the equation det (A— xI) = 0 has real roots.

(0 & C adapted)

12. If A and B are square matrices such that AB = 0, prove that either A = 0 or B = 0 or det A = det B = 0. 13. Two matrices A and B are said to commute if AB = BA. Let

1o)\ •

A= (2

prove that every matrix B that commutes with A can be expressed in the form B = AA where A and it are scalars and I is the unit matrix. Obtain expressions of this form for A2 and A. (M.T.) 14. Prove that, if a is a 3 x 3 matrix, with adj a = A and det a = a, that aA = Aa = aI. Assuming that a * 0, prove that: (i) det A = a2; (ii) X = A is the only solution of the matrix equation aX = (iii) adj A = aa. Which, if any, of the above results, are still valid if the restriction on a is removed? 15. If A is a 3 x 3 matrix with an* 0, and if the cofactor of every element of A is zero (that is, if adj A = 0), prove that A has the form

(

an a12 a13 kale ka12 ka13 • /an 1 a12 Ian

16. What is meant by the statement that multiplication of real numbers is associative and is distributive over addition? Prove that, if crib bo co(i, j = 1, 2, 3) are real numbers, then 3 ( 3

E

E

3

aiabafl)

fl=1 cc = 1

=

3

aja 1 E bafi c fli).

a=1

fi = 1

279

MATRICES 2

[14

Deduce that multiplication of 3 x 3 real matrices is associative and that, if A is a 3 x 3 real matrix, then the notation A3, A4, A5, ... may be interpreted unambiguously. If the 3 x 3 real matrices B and A2 + A + I are non-singular (I is the identity matrix) why would the notation A/B be ambiguous, but A A2 +A+I not so?

(M.E.I.)

17. The non-singular matrix B has the property BB' = B'B, where B' is the transpose of B. Prove that B'B-1= B-113'. Prove also that, if C = B-113', then CC' is the identity matrix. Find B', BB', B-1and C when 2 2 1 1 2 . B= —2 (M.E.I.) 1 —2 2

280

15. Linear equations

1. LINEAR EQUATIONS IN TWO UNKNOWNS; INTRODUCTION Given perpendicular coordinate axes Ox, Oy in a plane, any linear equation connecting the two variables x and y is the equation of some straight line in the plane; that is, the linear equation defines the set of points (x, y) comprising the line. Two such sets of points will have as their intersection a single point, in general, whose coordinates may be obtained by solving the equations simultaneously, or, in other words, by requiring the defining properties of the sets of points comprising the two lines to hold simultaneously. For example, the pair of straight lines 12x+5y = 1, 1 x —3y = 6 intersect at the point (3, —1), a result derived by solving the simultaneous equations in the usual way. Consider now the pair of straight lines j2x+3y = 1, 12x + 3y = 2. It is immediately obvious algebraically that these equations have no solutions. Geometrically, they represent parallel straight lines, with no finite point of intersection. Again, consider the 'pair' of straight lines x + 2y = 1, 12x + 4y = 2. The second equation is a thinly disguised re-write of the first so any point of the single line represented has coordinates satisfying the equation. In fact, setting y = A and solving the first equation for x, we see that (1 —2A, A) is a solution, for all values of A. Since we may assign values of the single parameter A arbitrarily, our equations are said to have one degree of freedom (or to be a one parameter system). Geometrically, we have a pair of coincident straight lines. The reader may feel this last case somewhat frivolous, and its extensive 281

LINEAR EQUATIONS

[15

discussion pedantic. However, if the coefficients are unknown, this possibility must be remembered. To summarize; given two equations in two unknowns, one of three possibilities occurs: (i) The equations have a unique solution. (The two lines are distinct and intersect.) (ii) The equations have no solution. (The two lines are distinct and parallel.) (iii) The equations have one degree of freedom. (The two lines coincide.)

Example 1. Solve the equations x+ y = 2, ax+2y = b. Eliminating y, we have

(2— a) x = 4— b.

Case (i). If a + 2, we have the unique solution

b —2a , Y = 2— a Case (ii). If a = 2, we have 0.x = 4 —b. Sub-case (iia). If b + 4, no solution for x exists. Sub-case (iib). If b = 4, x may take any value, y then being determined 4— b

x= 2

from the equation x+y = 2. Thus, the complete solution of the equations takes the form

a * 2: x =

—2a

2— a' Y = 2—a ; a = 2, b * 4: no solution;

a = 2, b = 4: one-parameter solutions (A, 2— A). The reader would be wise to consider the meaning of this solution geometrically. The reader is probably so accustomed to solving pairs of simultaneous equations that it may not have occurred to him to ask what happens if he is given three (or more) equations in two unknowns. From geometrical considerations it should be fairly clear that, in general, no values of x and y satisfy all three equations simultaneously. However, consider the system x+2y = 5, 3x— y = 1, x— 5y = —9. Solving the first two equations we get x = 1, y = 2 and these values cer282

11

TWO UNKNOWNS; INTRODUCTION

tainly do satisfy the third equation. Geometrically, these three equations represent three straight lines through the point (1, 2) (see Figure 15.1).

3x —y = Fig. 15.1

It will be recalled that, if

ai x+bly+ = 0, chx + 1,2y + c2 = 0 are two non-parallel straight lines, then any line through their point of intersection has the form

gai x +b,y+ ci)+ A(a2 x+b2y+ c2) = 0. Thus, in the particular case we are considering, any straight line through the point (1, 2) has an equation of the form

,u(x + 2y— 5) + A(3x — y — 1) = 0. Setting Aldu = — z we obtain the third equation. This example illustrates the general result, that, if three equations in two unknowns have a solution, then any one equation is a linear combination of the other two (see Chapter 2.4).

2. LINEAR EQUATIONS IN TWO UNKNOWNS; CONTINUED We consider pairs of equations of the form

Writing

1

aux+ anY = bb} a21x+ a22Y = b2.

(1)

(61) x = (x) b2 b Y the equations (1) may be rewritten in the matrix form 21) A = (a11 a 22 a21 a

Ax = b. 10

PPM

(2) 283

[15

LINEAR EQUATIONS

If we regard A as the matrix of a linear transformation T, equation (2) tells us that the image of the point P, with position vector

(x), is the

point B, with position vector (b1). Thus, the problem of solving a pair 2

of simultaneous equations may be reinterpreted as the problem of finding the point P whose image, under a given linear transformation, is a given point B. If A is non-singular, that is, if det A * 0, we have, for any point B, Ax = b -4=> A-1Ax = A-1b

x = A-13

and we see that the unique point P, whose position vector is A-lb, has image B under the linear transformation T. Ex. 1. Show that A-1is the matrix of a linear transformation S. [You must show that, for every point Q with position vector y, the image S(y) exists and is unique and that S(Ay+,uz) = AS(y)+ ,uS(z).] The linear transformation S so defined is called the inverse of the transformation T; S and T satisfy the relations S[T(x)] = x; T[S(y)] = y for all x, y.

As an example of the solution of equations whose matrix, A, is nonsingular, consider 1 3x — 2y = 4, 12x+ 5y = 1. Write A = (2 3 51' b = (1 4) then det A = 19 = 0 and A-1 =119_ 2 5 2 3)

and we have

(32 - 25) (;) = (1) (x y) = 119 (-2 52 3) CO

ix\

= 1

\ y)

19

22 \

k

The solution may be stated in the form x = y = — 9 or, in terms of the linear transformation T whose matrix is A, the unique point --h) maps into (4, 1). 284

2]

TWO UNKNOWNS; CONTINUED

Ex. 2. Solve the equations

3x — 2y = 4, 2x+3y = 1

by using the inverse matrix method.

Suppose now that det A = 0. Before discussing the existence of a point P whose image is B under the linear transformation T, consider the effect of T upon the unit square OIMJ. For a general transformation M' J

M

Fig. 15.2

(det A 0) 0I'M'J' is a parallelogram whose area is det A (see Chapter 13)— depicted in Figure 15.2, but, if det A = 0, the parallelogram collapses and 0, I', J', M' are collinear as shown in Figure 15.3. J

M

Fig. 15.3

*Ex. 3. If det A = 0, show that every point P of the plane maps into a point of the line OF. Since, in this latter case, T maps every point P into a point of the line 01', a point B not lying on the line cannot be the image of any point under T: the equations Ax = b are inconsistent. However, if B does lie on 0/', we may write the position vectors of J', I', B as r, mr, cr respectively; that is Aj = r, Al = mr, b = cr. Then, P(x, y) is a point which maps into B cr = A(xi +yj) cr = x(Ai) + y(Aj) cr = (rnx+ y) r ra P lies in the straight line y + mx = c, and infinitely many points map into B, by the reversibility of the argument. 10-2

285

LINEAR EQUATIONS

[15

To summarize for the case det A = 0: (i) the equations Ax = b have no solution (are inconsistent; equations with one or more solutions are said to be consistent).

there is no point which maps into B under T B does not lie on the line 01'; (ii) the equations Ax = b have infinitely many solutions -4* there are infinitely many points P which map into B under T B does lies on the line 01'. Ex. 4. Show that the equation ax = b is uniquely soluble only if a + 0, in which case, the unique solution is x = alb. Discuss the case a = 0, showing that failure to obtain a unique solution can occur in two different ways. Ex. 5. What can you say about the equation Ax = b if T(I) = T(J) = 0? (T is the linear transformation with matrix A.) Exercise 15(a) 1. Solve, using the inverse matrix, the following pairs of simultaneous equations: I 70x + 105y = 12, (i) 3x— y = 1, 13x+2y = 1, (iii) (ii) 5x— 7y = 5. 2x+3y = 19; 2x+7y = 29;

A

2. Solve for x the following equations, commenting upon any special cases that arise in the two cases: (ii) ax+ b = bx+ c.

(i) ax + a2 = b2— bx;

3. Solve for x, y the following pairs of equations, commenting upon any special cases that arise for particular values of the coefficients a: 3x+ ay = —2, (ii\ f x+ 2y = 4, f 3x+ ay = 1, ax+ 12y = 2. 3x +2y = 3; 2x+ay = 2a;

{

4. Solve for x, y the following pairs of equations, commenting upon any special cases that arise for particular values of the coefficients a, b: f ax+ by = 2, t12x-4y = a+ b;

(11)

f ax+ 3y = 6; x — by = 2;

(iii) ax+by = 1,

tbx+ ay = — 1.

5. Discuss the solution of the following simultaneous equations for various values of a: x+ y = 1, 2x+ay = 5, [ax+ y = —3. 6. Discuss the solution of the following simultaneous equations for various values of the coefficient a: fax +3y = 4, 3x— ay = 6, x-2y = a.

286

2]

TWO UNKNOWNS; CONTINUED

7. Discuss fully the solution of the following simultaneous equations for the unknowns x, y: f ax+by=c, la2x+b2y = c2. 8. Under what conditions do the homogeneous equations fai x+biy = 0, ia2 x+b2y = 0 have a non-trivial solution (that is, a solution other than x, y = 0) ? Solve the equations fully in the case where non-trivial solutions exist. 9. Solve for x, y the equations: f x cos 0 +y cos cb = cos a, x sin 0+y sin q5 = sin a. 10. Sketch the set, E, of points in the plane whose coordinates satisfy the inequality x+y > 0. The linear transformation, T, has matrix A where A= (1 2) 5 0 Sketch the image set T(E). 11. T is a linear transformation whose matrix is A, where A = (2

1) •

If E is the set of points whose coordinates satisfy the inequality y > 0, sketch the image set T(E).

3. SYSTEMS OF LINEAR EQUATIONS IN THREE UNKNOWNS; INTRODUCTION Given perpendicular axes Ox, Oy, Oz in space, a linear equation connecting the three variables x, y, z is the equation of some plane. Two linear equations represent two planes which may be parallel (same unit normal vector) but otherwise intersect in a straight line. In general, three planes intersect in a unique point and thus, in general, three equations in three unknowns possess a unique solution; see Figure 15.4. For example, the three planes

(x+ y+ z = 0, ix+2y+3z = —3, lx+ y— z = 4,

Fig. 15.4

have in common the single point (1, 1, —2)—a result readily derived by successively eliminating two of the variables. 287

LINEAR EQUATIONS

[15

However, just as in the case of two lines, complications may occur. Consider the equations x+ y+3z = 1, x 2y+ z = 1, 2x— y+4z = 2. If we write L1 x+y+3z- 1, L2 E. x-2y+z-1, L3 2x—y+4z-2, -

then it is readily seen that L3 = 4+4; thus, these three equations represent three planes through a common line (see page 62) and so possess infinitely many solutions (one degree of freedom); see Figure 15.5. Again, the system 2x+y+3z = 6, x+y+4z = 1, Fig. 15.5 x— z= 1, has no solution, for by subtracting the first two equations we obtain x — z = 5, which is inconsistent with the third equation. Geometrically, x — z = 5, which is a plane through the line of intersection of the two planes 2x +y+ 3z = 6 and x+y+ 4z = 1, is parallel to the third plane x—z = 1 and thus the line of intersection of the first two planes is parallel to the third plane. A similar result would hold whatever line of intersection was chosen and so the three planes form a triangular prism of uniform cross-section; see Figure 15.6.

Fig. 15.7

Fig. 15.6

Another possibility is illustrated by the system x+ y+ z = 1, 2x+2y+2z = 3, x+ y+3z = 1, which clearly has no solution, the first pair of equations being inconsistent. Geometrically, the first two planes are parallel to one another but not to the third plane; see Figure 15.7. .(

288

3]

THREE UNKNOWNS; INTRODUCTION

Again, we may have equations in the form:

x+y+z = 1, x+y+z = 2, _( x+y+z = 3. The equations are manifestly inconsistent and thus have no solution. Geometrically, they represent three parallel planes. Finally, we may have two or three equations which reduce to the same equation, in which case we have one or two degrees of freedom—geometrically, two or three coincident planes. (Or, possibly, two coincident planes and the other plane parallel—giving no solution.) To summarize: given three equations in three unknowns one of four possibilities may occur: (i) The equations have a unique solution (no two of the planes are parallel and the planes do not have a common line of intersection). (ii) The equations have no solution (either at least two of the planes are parallel or the three lines of intersection are parallel but not coincident or two of the planes coincide and the third is parallel). (iii) The equations have a one-parameter solution (the planes have a common line of intersection). (iv) The equations have a two-parameter solution (the planes all coincide). The situation is more complicated than in two dimensions, as was to be expected but, if the reader keeps the various geometrical possibilities in his mind, he should avoid confusion.

Example 2. Solve completely the equations 3x+y+ z = a, (1) 2x+y— z = 4, (2) 5x+y+bz = 1. (3) +2z = a -4, (1)—(2) gives x (1)—(3) gives — 2x + (1 — b)z = a -1, (5 — b)z = 3(a -3). .*. 3(a — 3) If b+ 5, z = 5 —b and substitution back gives {

x=

15+5a-12b+2ab —a— 2 ab+4b ' Y= 5—b 5—b —

corresponding to the case of unique solution. If b = 5, a * 3 there is no finite solution for z — we have a triangular prism of planes (clearly no two of the planes are parallel). 289

LINEAR EQUATIONS

[15

If b = 5, a = 3 the equations are consistent: putting x = A in (1) and (2) we have 2y = 7 — 5A, 2z = —1— A and the line of intersection of the first two planes is x _ y — z _ 2 — 5 —1 — By substitution, the general point (2p,, i- 5u, --Imo of this line lies on plane (3) for all it. Since no pair of the planes coincide, we have three planes with a common line of intersection and the general solution may be taken as x =

Zu, Y = z =- 4 —ii

(or any comparable form). If we are given four equations in three unknowns, we have, geometrically four planes and these will not, in general, have a common point. However, it is not impossible for four equations in three unknowns to be consistent; consider, for example, the four planes given by the equations

/

x+ y+ z = 0, 2x— 3y+ z = —3, x+ 3y+ z = 2, 3x+ 14y+ 4z = 9,

all of which contain the point (1, 1, — 2). In practice, the best way of tackling such a system is to solve three of the equations and check whether or not the fourth equation is satisfied.

Ex. 6. Show that the system of equations x+ y+ z = 2, 2x-5y+ z = —1, x+2y+3z = —1, 3x— 3y + 2z = 2 is consistent.

4. SYSTEMS OF LINEAR EQUATIONS IN THREE UNKNOWNS; CONTINUED As in the case of two unknowns, we may write the system

+ ai2Y+ an z = f aux anx+ a22 Y +a23z = b2, l

(3)

1a31x+a32y+a33z = b3,1

in the matrix form 290

Ax = b.

(4)

4]

THREE UNKNOWNS; CONTINUED

Regarding A as the matrix of a linear transformation T, equation (4) tells us that the image of the point P, whose position vector is (x) Y, is the point B, whose position vector is (b) b3 Again, the problem of solving a set of simultaneous equations may be interpreted as the problem of finding the point P whose image, under a given linear transformation, is a given point B. If det A + 0 we have, as in section 2, Ax = b -4.> A-1AX = A-lb

x = A-lb;

the unique point P, whose position vector is A-1b, has image B under the linear transformation T. Ex. 7. Show that A-1is the matrix of a linear transformation S—the inverse of the transformation T.

As an example of the solution of equations whose matrix, A, is nonsingular consider x+ y+z = 4, x— y—z = —2, x+2y—z = 1, {

A

1 1 1 i 1 —1 — ; A-1 = = (1 2 —1

and we have

0(by theusual es inversion pro+ H-

yx) = (4. 0 A- 1 0) (_ 24 (z

—+

1)

/1 = The solution is x = 1, y = 1, z = 2 or, in terms of the linear transformation whose matrix is A, the point (1, 1, 2) maps into the point (4, —2, 1). 291

LINEAR EQUATIONS

[15

The analysis of the case in which det A = 0 follows the same pattern as the two-dimensional case and is pursued in the following series of exercises. Ex. 8. Show that, under the linear transformation T whose matrix is given by a, a2 a 3) 3 1)1 b2 b A = (c1 C2 Cs the images of the points I, J, K, whose position vectors are i, j, k, are I', J', K', whose position vectors are respectively the first, second and third columns of A. Ex. 9. Show that, under T, the unit cube, three of whose edges are 04 OJ, OK, transforms into a parallelepiped (a figure all of whose faces are parallelograms). Ex. 10. Show that the volume of the parallelepiped obtained in Ex. 8. is det A. (This result is necessary for the remaining exercises in this section, but its proof is rather hard and may be omitted at a first reading. The best proof depends upon the triple scalar product, which will be met in Book 2.) Ex. 11. If det A = 0, show that 0, I', J', K' are coplanar. What does this tell us about the columns of A? Ex. 12. If the point D, with position vector d, is the image of a point P under a linear transformation whose matrix is singular, what can be said about (i) the point D; (ii) the vector d? Ex. 13. Show that, if det A = 0, and if D lies in the plane OI'J'K', then all the points of a certain line map into D. Ex. 14. In the extreme case in which 0, I', J', K' are collinear, what can be said about the columns of A? If Ax = d has a solution in this case, what can be said about (i) the point D; (ii) the vector d?

In an actual example, det A should be calculated and the various cases in which det A = 0 should be treated by reverting to the original equation. Matrix methods are admirably suited to studying the structure of systems of equations but rarely constitute an efficient procedure for their solution in individual cases. Example 3. Discuss the solution of the system of linear equations

x+ay+ z = b, ax+3y+ z = 1, {

5x+8y+3z = 1

for various values of a and b. (1 a 1) A = a 3 1 and so det A = — 3a2+ 13a — 14. 5 8 3 292

4]

THREE UNKNOWNS; CONTINUED

Thus det A = 0 if a = 2 or 3. Case (i). a * 2, a * 3. The equations have a unique solution for all values of b. Case (ii). a = 2. The equations reduce to x+2y+ z = b, (1) 2x+3y+ z = 1, (2) 5x + 8y + 3z = 1. (3) Eliminating z between (2) and (3), and between (1) and (2),

x+ y . x+ y

(4) = 2, = 1—b. (5)

Thus, the equations are consistent if and only if b = — 1, in which case we have a one-parameter solution x = A, y = 2 — A, z = A-5 (on using equations (4) and (1)). Case (iii). a = 3. The equations reduce to 3x + 7y + 3z = 3b, (6) 7x + 9y + 3z = 3, (7) 5x + 8y + 3z = 1. (8)

{

(6) + (7)— 2(8) gives 0 = 3b+3-2. Thus, the equations are consistent if and only if b = --I, in which case we have a one-parameter solution x = A, y = 2(1 — A), z = 3(11A — 15) (from equations (7) and (8)). 5. HOMOGENEOUS EQUATIONS We now consider the system of linear equations an x+ any + anz = 0,

a21x+a22y+a23 z = 0, an x+ any+ a33z = 0,

.[

or

Ax

=

O.

Such a system is said to be homogeneous. If A is regarded as the matrix of a linear transformation T we seek these points which are mapped into the origin under T. It is easy to see that the origin maps into the origin under any linear transformation; that is, that x = 0 is always a solution for a set of homogeneous equations—we call this the trivial solution—but it is of more interest to look for non-trivial solutions x + 0. 293

LINEAR EQUATIONS

[15

Ex. 15. Show that the linear transformation T whose matrix is 3 7 1) A = (1 1 3 1 3 1 -

x

maps all points of the line

2

_y_

-

z

—1 1

into the origin. What does this tell us about the system of homogeneous equations

1

3x+7y+ z = 0, x- y - 3z = 0, x+3y+ z = 0?

We conclude this chapter by proving a necessary and sufficient condition for the existence of non-trivial solutions of the homogeneous equations Ax = 0. Theorem 15.1

Ax = 0 has a non-trivial solution a det A = 0. Proof. (i) If det A + 0, A-1exists and x = 0 is the only solution. Thus, if Ax = 0 has a non-trivial solution, then det A = 0, and we have proved that a necessary condition for the existence of a non trivial solution is det A = 0 (the implication (ii) If det A = 0 and all the cofactors are zero then, by the result proved in Miscellaneous Exercise 14, Question 15, the rows of A are in proportion and the three equations reduce to a single equation, which has a two-parameter solution. If det A = 0 and at least one cofactor, say A„, is non-zero, then -

an An + ai2,412 + ai3 A13 = 0 (det A = 0), a21An+a22Al2 4- a23A13 = 0,} expansion by alien cofactors. a31A11+17,2 Al2+ 433 A13 = 0, We have thus constructed a solution

x = An, Y= Al2, z = Al3 which is non trivial, since A11 + 0. The condition det A = 0 is sufficient to ensure the existence of a nontrivial solution (the implication -

Exercise 15 (b) 1. Solve the equations

{3x+ y+ z = -2, 2x+2y+3z = 8, x+3y+2z = 6

(i) by successive elimination; (ii) by using the inverse matrix. 294

5]

HOMOGENEOUS EQUATIONS

2. Calculate the inverse of the matrix (-3 11 7) —1 4 3 2 —7 —5 and hence solve the equations —3x+11y+7z = a, — x+ 4y+3z = b,

2x— 7y-5z = c (i) when a = 4, b = 1, c = —2; (iii) when a = 1, b = —6, c = 0. 3. Solve the equations

(ii) when a = 1, b = 1, c = —1;

x— 2y+ z = 1, 3x+2y-3z = 7, 15x-2y— z = 9

and interpret your result geometrically. 4. Show that the matrix

/3 —1 2) 3 —1 A = (2 1 —15 10

is singular. Discuss the solution of the equations 13x— y+ 2z = 4, (i) -(2x+ 3y— z = 1, x-15y+10z = 8;

(3x— y+ 2z = 4, (ii) 2x+ 3y— z = 1, x-15y+ 10z = 9,

and interpret your results geometrically. 5. Show that the equations 13x—y+ z = 3, j ax+y+ z = 4, 8x—y+3z = b

1

are consistent and have a unique solution, provided a + 2. Discuss the solution of the equations for the case a = 2. 6. Discuss the solution of the equations

x+y+z = 1, ax+ay+2z = 1, 13x+3y+(a+1)z = b for various values of a, b. Interpret your various results geometrically. 7. A = (2 1

1 —4

—1), B = (3 1/ 5

—

—1 —1

2 . —2)

A maps the point P into the point Q and B maps the point Q into the point R, whose coordinates are ( -5, 0, 0). Find the coordinates of P. 295

[15

LINEAR EQUATIONS

8. Discuss the solution of the equations 13x+2y +az = 2a, ax+ y +6z = 15, x+2y+3z = 8

l

for various values of a. Interpret your results geometrically. 9. Find the value of a for which the homogeneous equations .( 6x + 2y— z = 0, 3x + ay —2z = 0, 3x-2y+ z = 0 have a non-trivial solution, and find solutions for this case. 10. Prove that, if det A = 0 and if Ax, = b and Axe = 0, then xi + Ax2 is a solution for the equation Ax = b, for all values of A. Find a solution of the equations x+ y+ z = a, 3x-5y— z = b, 2x-8y-3z = c

{

when a = b = c = 0 and hence write down the general solution of the equations when a = 1, b = 3, c = 2. 11. A linear transformation T has matrix 3 1 A = (1 4 2 —3

2) 1. 1

Show that T maps all points of the line x y 7 1

z —11

into the origin, and find the line of points that maps into the point (6, 6, 0). What is the relation between these two lines ? 12.

A

a, b1 c1 2 • a2 b 2 c 3 C3 = (a3 b

Alis the cofactor of al, etc., and det A = A. If ai x+biy+ciz = d1, a2 x+b2y+c2z = d2, a3 x+b3y+c3z = d3, prove that

Lx = d1 A1+d2 A2+d3 A3.

If A = 0, what does the value of the expression diAl+ d2 A2 + d3 A3 tell you about the solution of the given system of equations? 296

5]

HOMOGENEOUS EQUATIONS

13. Prove that, if

di a, b1 c1 \ A = (a2 b 2 c o), d = d, a3 b3 C3/

(),

x) x= ( z

and if det A * 0, then the unique solution of the equation Ax = d is given by x d1 b1 c1 d2 b2 c2 d3 b3 c3

z a1 b1 d1 1 a2 b2 d2 = det A • a3 b3 d3

Y a1 d1 c1 a2 d2 c2 a3 d3 c3

14. The linear transformation, T, has matrix A where 3 —1 A=( 5

1 2 —1

2 — 1) . 4

E is the set of points in space whose coordinates satisfy the inequalities x > 0, y > 0, z > 0, x+y+ z < 1. Determine the image set T(E).

Miscellaneous Exercise 15 1. Find the inverse of the matrix ( 1 —1 2 Given the equations

0 2 —4

3 —1 . 1

2x1-4x2 + x, = a, +3x3 = b, xi — xi+ 2x3 -- x3 = c,

find the solutions (i) when a = 1, b = 1, c = —1;

(ii) when a = — 1, b = 2, c = 0. (M.E.I.)

2. Let A be a 3 x 3 real matrix and b a three-rowed column vector. It is proposed to solve the equation Ax = b and a particular solution x = xois noted. Prove that any other solution may be written in the form x = xo +u, where u is a solution of the equation Ax = 0. Prove conversely that any vector xo+u is a solution of the equation, where xois a fixed particular solution of the equation and u is any vector such that Au = 0. Interpret geometrically the equation and its solutions if A is singular (i) when xoexists, and (ii) when there is no such particular solution xo. (M.E.I.) 297

LINEAR EQUATIONS

[15

3. If M=

2 3 6 6 2 —3), 3 —6 2

(

form the product MM' and show that M' = 49M-1. Without multiplying out, state the product M'M, giving reasons for your answer. Hence, or otherwise, find the solution of the equations: 2x1+3x2 + 6.; = 1, 6x1 +2x2 -3x3 = 1, 3x1-6x2 +2x3 = 2. (M.E.I.) 4. The simultaneous equations x+ 2y = 4, 2x— y = 0, 3x+ y = 5 may be written in matrix form as /2 1—2 1\ ix\

/4\ or AX = B.

li

\y)= '21 5)

Carry out numerically the procedure of the following three steps: (i) A'AX = A'B; (ii) (AAA)-1A'AX = (A'A)-1A'B; (iii) IX = (x = (A'A)-1A'B. Y Verify that the values of x, y so found do not satisfy all the original three equations. Suggest a reason for this. Under what circumstances will the procedure given above, when applied to a set of three simultaneous equations in two variables, result in values which satisfy the equations? (SMP) 5. By systematic elimination, find values of A, B, C in terms of a, b, c, such that x+ y+ z = 1,

ax+ by+ cz = a+ b+ c, la2x+b2y+c2z = bc+ ca+ab, rx+y+ z = 1, if and only if iAy+ Bz = b+ c Cz = a2+b2. Solve the equations, assuming a, b, c are all different. Describe geometrically the configuration of planes, in three-dimensional Euclidean space, with equations: x+ y+ z = 1, x+2y+2z = 5, x+4y+4z = 8. (SMP) 298

5]


6. Solve the simultaneous equations

x+ y+ z = 3, x+2y +3z = 6, x+3y+kz = 4+k (i) when k rr 5; (ii) when k = 5, giving the general solution. 7. If A is the matrix (a b

c d

(SMP)

and X is the non-zero matrix

that, if AX = AX, where A is a number, real or complex, then

(x) show

Y

A2(a+ d)A + (ad— bc) = 0. Show that A itself satisfies this quadratic equation, in the sense that A2 — (a+ d) A + (ad— bc)I = 0, where I is the matrix

(1 0) and 0 is the matrix (0 0 \O O) 01

(Oxford)

8. By considering det (A — Al), extend the result of Question 7 to a 3 x 3 matrix A. 9. Factorize the determinant

1 1 a b c a2 b2 c 2

1

Show that, if no two of a, b, c are equal, the equations x+ y+ z = a,

ax+ by+ cz = ab, a2x+b2y+ c2z = abc, have unique solutions for x, y, z, and find them. Discuss the special cases (i) a = b * c; (ii) b = c * a;

(iii) a = b = c.

(M.T.)

10. Prove that the only value of A allowing a real non-trivial solution of the simultaneous equations — z = Ax, x

y+ z = Ay, 4x + 2y — z = Az, is A = 1.

(M.T.)

11. If A, B are 3 x 3 matrices and x a three-rowed column vector such that there exist numbers A, it such that Ax = Ax, Bx = px, prove that there exists a number v such that ABx = vx. Prove further that, if x1, x2, x3are three linearly independent column vectors with this property, then A and B are commutative for multiplication. 299

[15

LINEAR EQUATIONS 12. Write the two sets of three equations a.ixi+ai2 x2 +as3x3= c, bilY1+ bi2Y2+ bz3 Y3 =

Xi

(1 =

1, 2, 3)

in matrix form, and prove that they can be solved uniquely for xi, x2, x3, Y1 Y2 y3 if and only if det AB * 0, where A and B are the 3 x 3 matrices of coefficients in the two sets of equations. Show that the equations ,

,

xi+ 2x, = 2, xi+ x2 + x2 = 1, 3x2 —x3 = k, 3Y1+ Y2 + 43/3 = X19

Y1 + 2312 — 3Y3 = x2, Y1 + 5312 —23/3 = Xa

are inconsistent if k * 7, and find the most general solution for xi ... y3 if (M.T.) k = 7. 13. Prove that, if k

0, the system of equations 2x+ y = a, x+ ky— z = b, y +2z = c

has a unique solution (x, y, z) for every choice of (a, b, c). Show also that, when k = 0, the system is consistent if and only if (a, b, c) satisfy a certain linear relation, and find this relation. Verify that the system is consistent when k = 0 and (a, b, c) = (1, 1, — 1) and find an expression for the general solution of the system in this case. (M.T.) 14. Find, for all values of the parameter A, the number of solutions of the equations x+2y+ Az = 0, 2x+ 3y — 2z = (M.T.) Ax+ y+ z = 3.

300

16. Discrete probability distributions

1. INTRODUCTION: THE UNIFORM DISTRIBUTION In Chapter 7 we discussed the concept of an outcome space for a random experiment. To each elementary event we ascribed a probability and the entire set of probabilities was described as a probability distribution for the outcome space. In this chapter we shall enumerate various possible probability distributions that find frequent use as mathematical models for random experiments. We shall generally define our outcome space in terms of an associated random variable X (see Chapter 10), and we shall thus be able to talk about the mean and variance of the distribution; however, it must be remembered that, for example, the mean is defined as e(X) and so, if a new random variable, Y, is chosen, the mean naturally changes, too. Unless there is any possibility of confusion, we shall refer to the mean and variance of X as ,tt and cr2respectively. Suppose we have an outcome space consisting of n elementary events, with associated random variable, X, where X = {1, 2, 3, ..., n}. Perhaps the simplest assumption we can make about the n possible outcomes is that each one is as likely to occur as any other. Thus we take Pr (X = r) =

(1)

Equation (1) defines the uniform distribution for n possible outcomes. The uniform distribution is, in a way, fundamental, for we may often subdivide the elements of an outcome space in such a way that the new outcome space may reasonably be given a uniform distribution. This is not always possible, however; a simple counter-example might be the fall of a biased coin. The mean, ,u, of the random variable X with a uniform distribution is given by = (X) n 1

=Er r=1

.-

n

= +(n+1).

(2) 301

DISCRETE PROBABILITY DISTRIBUTIONS

[16

Again, the variance, o-2, is given by cr2 = 61(X — ,a)2] = [X2] — ,u2 r2.1— n = =

1)2

+1) (2n + 1)— i(n + 1)2 1)•

(3)

The reader will find a large number of examples on the uniform distribution at the end of Chapter 7. Ex. 1. A man spins a coin and throws a die. For a head he scores —1, for a tail +1, and this he adds to the score showing on the uppermost face of the die. The values of the random variable, X, where X = {0, 1, 2, ..., 7} represent his total score. Why do you think a uniform distribution would be an unsuitable mathematical model for this eight-point space? Suggest a way of subdividing the sample space so that a uniform distribution would be suitable.

2. THE BINOMIAL DISTRIBUTION The simplest possible type of random experiment is one that has just two possible outcomes, which we may conveniently designate success and failure. An experiment of this type is called a Bernoulli trial. (J. Bernoulli, 1654-1705, one of a family of distinguished mathematicians, whose famous work on probability, the Ars Conjectandi, was published posthumously in 1713.). Some examples of Bernoulli trials are (i) a coin is tossed : does it show heads or tails ?; (ii) a person is tested for disease: has he got the disease or not ?; (iii) the height of a person is measured: is it less than six feet or not ?; (iv) a marksman fires at a target: does he hit or miss the bull? If the probabilities of the two possible outcomes of a Bernoulli trial are p (success) and q (failure) (where p+q = 1) and n independent repetitions of the trial are made, we may define the random variable X as the number of successes obtained in the n trials and calculate p(r) = Pr (X = r), r= 1, 2, 3, ..., n. Method I. Suppose first we assign some specific order in which we should obtain the r successes: for sake of argument, suppose the first r trials result in success and the final (n — r) trials in failure. Then, by independence, 302

2]

THE BINOMIAL DISTRIBUTION

the probability of obtaining this sequence is prqn-r. But there are

(1 r

different sequences possible. Thus

p(r) = Pr (X = r) = (n) prqn-r.

(4)

Method II. The p.g.f. for a single Bernoulli trial is given by c`°(tr)

= p. tl+q .t° = pt+q.

Thus, the probability generating function for n independent Bernoulli trials is given by G(t) = (pt+q)n. Thus

p(r) = coefficient of t rin the expansion of (pt +q)"

= n) nr qn-r .. (r '

The probability distribution defined by equation (4) is called the binomial distribution. It is easily verified that (4) does indeed define a valid probability distribution. For Pr (X = r) > 0 and also i (1) prq"-r = (p+q)n, by the Binomial Theorem;

r=0

r

= 1, since p+q = 1. It finds a wide field of application as a probability model but the reader must always take care to remember the binomial distribution applies only if we have n independent repetitions of a Bernoulli trial. A binomial distribution is known completely if the two parameters n and p are given; the phrase `the binomial distribution arising from n repetitions of a Bernoulli trial with probability of success p' is often abbreviated to B(n, p).

Example 1. Six unbiased dice are thrown. What is the probability of securing three or more sixes? The dice may be assumed to fall independently and so the random variable X (= number of sixes showing) has a binomial distribution with probability * of success on each of the six Bernoulli trials. Thus Pr (X?.... 3) = 1— p(X < 3)

= 1- NY + CO

V 0 + (2) (:)402)

0.06. 303


[16

We next derive expressions for the mean, ,u, and variance, 0-2, of the number of successes in a binomial distribution. We shall give two methods, the first one using the definition (4) directly, the second employing the probability generating function (and using Theorem 10.5). Method I. By definition ,u = "f (X)

r=()

=

r

pr qn-r

r

n (n-1) nrqn-r,

since

r-1 r

r= 1

= np E

n-1) —1

r=1 (r

= np(p +q)n-1-,

n! n (n-1\ r!(n—r)! r r-11

r

pr-1 qn-r by the Binomial Theorem,

= np, since p +q = 1. Again o2 = e[(X—,u)2] = e[x2]—,a2, by Theorem 10.1, = En r 2 (12)prqn-r_ n2-p2, using the first part, r r=0 =

nr (n-1) r n-r 2 2 p q —n p

r=1

r—1

= r=2n (n-1) p2

(

)pr-2qn-r r—2 np

(r -- 1) pr-ign-r

_ n2p2,

r=1

on writing r = (r-1)+1, = n(n — 1) p2np — n2p2, since p + q = 1, = npq. Thus we arrive at the important result: the mean and variance of the number of successes in a binomial distribution B(n, p) are given by ,u = np, Cr2 = npq. Method II. Since the p.g.f. for the binomial distribution is given by G(t) = (pt +q)n we have G'(t) = np(pt

G"(t) = n(n — 1) p2(pt +q)n---2(n ?„-. 2).

Putting t = 1, A = np, 0.2 ,u2— A = n(n— 1)p2and the results follow.

304

(5)

2]


Example 2. A very large number of balls are in a bag, one-eighth being black and the rest white. Twelve balls are drawn at random. Find (i) the probability of drawing three black balls and nine white balls; (ii) the probability of drawing at least 3 black balls; (iii) the expected number of black balls in the sample; (iv) the most likely number of black balls in the sample. Drawing a ball from the bag and noting its colour may be regarded as a Bernoulli trial (a black ball drawn being a success, p = 1). Since a ball is not replaced before its successor is drawn, the twelve Bernoulli trials constituting our sample are not strictly independent; however, since the number of balls in the bag is very large, we may assume that Pr (black ball drawn) remains sensibly constant at *; that is, each trial is independent of what has previously occurred. Thus the random variable X = number of black balls contained in a sample of 12 has a binomial distribution B(12, *). (i) Pr (X = 3) = (132) (8)3

(ii) Pr (X

3) =

0429; 1 _ (G)12 + (12 1) (87)11 (1 8) ± (12) (87)1° (8 1)2) 0182;

(iii) (iv) Write

e(X) = np = 1.5. p(r) = Pr (X = r) 12

Then

(12) r

nrn12-r .

r+1 /7\11-r

p(r +1) _ kr+ 181 k8) (12 r) (18)T 8)12-r p(r) (7 12! r!(12 —r)! ii\ (r+1)! (11 — r)! 12! \ 1) 12 r 7(r +1)• P( +1) r)1) < 1 if r 1 p(r) 12-0 12 p(1) = p(0) = — p(0) 7 7(0 +1) -

—

Thus and so

is the maximum of the p(r); that is, the most likely number of black balls is 1. 305


[16

[Notice that the ratio deduced in Example 2 (iv) enables us to express

p(r) in terms of po. Thus : Pi = 7 1 Po,

P2 = -11*P] =

HPO,

P3 = 2*P2 = iMPO3 etc. This is a particularly useful device for calculating binomial probabilities if a hand calculating machine is available.]

Ex. 2. Find the values of Pr (X = r) in terms of Pr (X = 0) for the binomial distribution B(4, D. Ex. 3. If fifteen dice are thrown what is (i) the expected number of sixes showing; (ii) the most likely number of sixes showing?

Ex. 4. Calculate the mean and variance for the binomial distribution B(9,1). Ex. 5. In a family of four children, what is the probability of there being two boys and two girls? What is the probability that the eldest two are boys and the youngest two girls? What are the odds against having all four children of the same sex?

Exercise 16(a) 1. A coin is tossed four times. Find the probability that heads appear (i) at the first two tosses, followed by two tails; (ii) just twice in the four throws; (iii) at least twice. 2. 10 % of the very large number of articles produced by a machine are faulty. What is the probability that a random sample of ten articles will (i) be free of faulty articles; (ii) contain more than two faulty articles?

3. From a packet containing a large number of seeds, 40 % of which are advertised to give red flowers and the others white, 10 plants are produced. What is the probability (i) that all the plants have red flowers; (ii) that all the plants have white flowers; (iii) that half the plants have red flowers and half white? 4. 10 % of the very large number of articles produced by a machine are faulty.

If articles are taken at random and tested, how many articles will be tested, on average, before the first faulty article is found? What is the probability that the testing procedure will have to go on longer than this before the first faulty article is found? 5. (i) In a trial, eight coins are tossed together. In one hundred such trials how

many times should one expect to obtain three heads and five tails? 306

2]


(ii) If 8 % of articles in a large consignment are defective, what is the chance that a sample of thirty articles will contain fewer than three defectives? (0 & C) 6. A battery of four guns is firing on to an enemy emplacement. It is reckoned

that each gun should score on the average one direct hit in every five shots, and that three direct hits are needed to destroy the emplacement. If each gun fires one shell, calculate the probability that the emplacement will be destroyed. With new gun crews it is reckoned that two of the guns should score one direct hit in every three shots and that the other two guns should score one direct hit in every four shots. If each gun now fires one shell, calculate the probability that the emplacement will be destroyed. (Cambridge) 7. Nine unbiased dice are thrown. Find p(r), the probability that r sixes appear, and hence determine the value of p(r+1)1p(r). Find (i) the expected number of sixes; (ii) the most likely number of sixes; (iii) the probability of obtaining more than one six. 8. (i) In a binomial distribution where the probabilities of the occurrence of an event are given by the terms of the expansion of (p+q)m, the mean it and the standard deviation o of the distribution are given by the formulae ,u = mp; o = ✓(mpq)• Prove these formulae in the case where m = 3. (ii) Two men A and B play a game in which A should win eight games to every seven won by B. If they play three games, show that the probability that A will win at least two games is approximately 0.55. (Cambridge) 9. Playing a certain 'one-arm bandit', which is advertised to 'increase your

money tenfold ', costs 5p a turn; the player is returned 50p if more than eight balls out of a total of ten drop in a specified slot. The chance of any one ball dropping is p. Determine the chance of winning in a given turn, and for p = 0.65, calculate the mean profit made by the machine on five hundred turns. Evaluate the proportion of losing turns in which the player comes within one or two balls of winning (p = 0.65). (Cambridge) 10. Prove that, in the binomial distribution B(2n, 1), the probability of scoring an even number of successes is 1. 11. An experiment consists of tossing an unbiased coin twelve times and counting the number of heads obtained (X). If the mean and variance of X are ,u

and o2respectively, find the value of

(i) Pr (I X-1u1 > a); (ii) Pr (1X—#1 > 2o).

3. THE GEOMETRIC AND NEGATIVE BINOMIAL DISTRIBUTIONS We must now consider two further probability distributions associated with the independent repetitions of Bernoulli trials. However, instead of repeating the trial a fixed number of times and asking how many successes 307


[16

have been obtained, we reverse the process by asking how long we must go on repeating the trial until a stipulated number of successes is obtained. First, suppose we wish to calculate the probability that the first success occurs at the rth repetition of a Bernoulli trial, the probability of success in any one trial being p. (As in Section 2, we assume that each trial is independent of what has previously occurred.) We take as our random variable X, where X = number of trials up to and including the trial which results in the first success. There is no upper limit to the value X may take: we can theoretically continue obtaining failures for ever, although the probability of doing so steadily decreases. Now Pr (X = r) = Pr (initially (r -1) failures, followed by a success), i.e. Pr (X = =qr 1p. (6) Equation (6) defines the geometric distribution for the random variable X = {1, 2, 3, ...}. It is easily seen that this is a valid probability distribution, for co qr-1 p =__ P E 1—q r=1 = 1. The mean, ,u, and variance, o-2, of the random variable X for a geometric distribution may be derived without undue difficulty: we give two possible methods. CO

Method I.

g(X)= Erpqr-1, r= 1 qS(X) = E rpqr = r=1

Subtract:

E (r —1) pqr-1.

r= 2

co pg(X) = p+ E pqr--1 r=2

Pq = P+1—q = 1, since p+q = 1,

,u = S(X) = llp. Again

e(X2)

= E r2pqr-1, r=1

qe(x 2) =

E r2p—r y = E (r-1)2 pqr-1.

r=1

308

r= 2

BINOMIAL DISTRIBUTIONS

3]

Subtract:

pe"(X2) = p +2 E rpqr-1 — pqr-1 r= 2

r=2

P + 2(14 P)-177 = P + 2{(1/P) — P} — (1— P) = (2/p)— 1, o.2 = e(X2)—#2 2 1 1 = P —P2 = qip2• Thus, for the geometric distribution, It = 1

=p2 q -

(7)

Method II. The p.g.f. for the geometric distribution is given by

G(t) =

or)

= E pqr-i tr r=1

= pt/(1—qt). Thus

G'(t) = 111(1 — qt)2, G°(t) = 2pq/(1—qt)3

and results (7) follow on substituting t = 1 and using Theorem 10.5. The geometric distribution is a suitable probability model in a number of practical examples; for example, in inverse sampling, articles are tested until a faulty article is found.

Example 3. 5 % of the output of a certain machine is faulty. Articles are taken at random from the output and tested, the process stopping when the first faulty article is obtained. Find (i) the expected duration of this sampling process; (ii) the probability that the process will terminate before the expected value is reached. We shall assume that the total output of the machine is large and so the individual Bernoulli trials of testing articles are independent, each with probability of success (that is, of finding a faulty article) equal to -AT. Adopting the geometric distribution as a suitable mathematical model, we have (i) mean length of sampling process = 1/(-21,,-) = 20; 309


(ii) Pr (X < 20) = 1— Pr (X

[16

20)

oo

= 1-r=E20 (2o)r-1(20) = 1— Kit)"(2-top[1-1t] 0.623. The concept underlying the geometric distribution may be generalized by asking how many Bernoulli trials will be required up to and including the kth success. Taking as our random variable X = {1, 2, 3, ...} we observe that, if X = r, the rth outcome must be the kth success, probability p, and the previous (k 1) successes are binomially distributed B(r 1, p). Thus r —1) (p, r k) Pr (X = r) = -

—

—

P K-1 P q

= (k —1) ple qr—k

k).

(8)

Equation (8) defines the negative binomial distribution; to appreciate the reason for this name and also to facilitate the verification that this is indeed a valid probability distribution, we first observe that X= where Xiis the number of trials after the (i— 1)th success up to and including the ith success. Thus, the p.g.f. for a negative binomial distribution is the kth power of the p.g.f. for a geometric distribution, i.e.

yk. G(t) =._ pktko G(t) = ak tk ak & k+iT „+2&k+2 • • • — qt

Writing

we have for r k,

Pr (X = r) = ak+r,

E Pr (X = r)

s=0

r=k

ak+.

= G(1) = 1. Furthermore,

G'(t) = kpktk-11(1—qt)k±1, k(k —1+2qt) Pk ik-2 1 qtyc-1-2

G"kt —/

and it follows that 310

= p,

kg' =

(9)

3]

BINOMIAL DISTRIBUTIONS

Ex. 6. Devise an experiment for which you feel the negative binomial distribution would constitute a suitable mathematical model. Ex. 7. Dice are thrown in succession until two sixes have appeared; find (i) the probability that two dice in all are thrown; (ii) the probability that more than four dice are needed; (iii) the expected number of throws required.

4. THE POISSON DISTRIBUTION Another distribution of common applicability associated with an infinite outcome space and random variable

X = {0, 1, 2, 3, ...} is the Poisson distribution, defined by the equation Pr (X = r) = e-a . a—r (a > 0).

!

Since

a

E Pr (X = r) = e—a E rtr =0. -

r =0

(10)

= e—aea

=1 and Pr (X = r) > 0, for all r, equation (10) does constitute a valid probability distribution. We shall show that the Poisson distribution arises as the limit of a binomial distribution in which n-->- cc and p--->0, in such a way that np = a remains constant. Thus the Poisson distribution will form a suitable probability model, not only to situations where the binomial distribution applies directly and n is very large and p very small but also to situations where the binomial distribution is not directly applicable. For example, consider requests for trunk call connections made at a telephone exchange. Suppose that, for the period 7.00 a.m. to 8.00 a.m. they average out at three and suppose further that the request takes a negligible time to make. Dividing the hour up into a large number of short time intervals (say 720 of 5 seconds each), the probability of there being a call in any one of the intervals may be taken as 7T1T), on the assumption that the calls arrive independently throughout the hour, and the probability of two calls in the interval is negligible. Thus we have 720 independent repetitions of a Bernoulli trial (Ìs there a call or not in a given 5 second interval ?') with small probability of success p = rzt-c), and the Poisson distribution may be regarded as a good mathematical model to employ. We must now verify our original assertion that a binomial distribution

311


B(n, p) tends to a Poisson distribution as n -k oo and p that np = a, a > 0. First, for B(n, p)

[16

0 in such a way

Pr (X = r) = (nr) pr qa-r n(n— 1) (n — 2) ... (n —r + 1) P r! 1 1 (

\ 11

n1 k

- 1

- 2

\ ... (1

n1

— r

1

n

)

19)n (1 P)-r

ar

k

r!

pyr, since np = a.

Since r is a fixed number, the product of the r factors in the numerator of the first fraction tends to 1 as n -> oo, and (1 — p)-r tends to 1 as p 0. Furthermore, (1 — clI n—> e-a t n e-aar Pr (X = r)_> and we have r! as required. The Poisson distribution is completely described if we are given the value of a; thus, a Poisson distribution is a one-parameter distribution and a Poisson distribution with parameter a may be referred to as P(a). The mean, it, and the variance, a.2, of the Poisson distribution P(a) may be found by the direct computation of €(X) and ‘(2(2). Thus = g(X) =fi r.

ar

r=0 ao

= 2.E

ar-1

=1 (r — 1)! . a e-a

Similarly,

= a. 0.2 = 60(A72)__ 1,2 = a l as above.

For the Poisson distribution P(a), p, = a, o-2 = a, i.e. the mean and variance of the Poisson distribution P(a) are both equal to a. t For the proof of this and other results concerning the exponential function, see one of the books on calculus mentioned in the bibliography.

312

41

THE POISSON DISTRIBUTION

Ex. 8. Prove that, if the random variable X = {0, 1, 2, ...} has equal mean and variance, this does not necessarily mean that X has a Poisson distribution. (Hint: consider a suitable uniform distribution.) Example 4. Traffic accidents are reported in a certain town at an average rate of four in a week. Estimate the probabilities: (i) of a given week being accident free; (ii) of there being three or fewer accidents in a given week; (iii) of there being more than four accidents in a given week. To produce reasonable estimates of these probabilities we have first to set up a mathematical model of the situation. The assumption that accidents occur independently of one another is arguable but, without more detailed data, it would be difficult to make a more plausible assumption. With this hypothesis, a Poisson distribution for the number, X, of accidents per week would appear to be a reasonable model, for the reason outlined earlier. Thus we have (i) Pr (X = 0) = e-4 0.018; (ii) Pr (X 5 3) = e-4(1 + 4 +124 -I- cl) 0.433;

(iii) Pr (X > 4) = 1 — e-4(1 + 4 +41 -Eci ,f 44 . 1) 0-371.

Exercise 16(b) 1. The random variable X can take values 0, 1, 2, 3, .... Given that X has a Poisson distribution, mean 2, calculate the probabilities Pr (X = 0), Pr (X = 1), Pr (X = 2), Pr (X = 3), Pr (X = 4), Pr (X = 5), Pr (X > 5). 2. Samples of forty articles at a time are taken periodically from the continuous production of a machine and the number of samples containing 0, 1, 2, ... defective articles are recorded in the following table: No. defective per sample No. of samples

0

1

2

3

4

5

6

Total

30

23

27

14

4

2

0

100

Find the mean number of defectives per sample. 313


[16

Assuming that this is the mean of the population and that the Poisson distribution applies, find the chance of: (a) a sample containing four or more defectives; (b) two successive samples containing between them four or more defectives. (0 & C) 3. The following table shows the results of recording the telephone calls handled at a village telephone exchange between 1.00 p.m. and 2.00 p.m. on each of a hundred weekdays (e.g. on thirty-six days no such calls were made): Calls

0

1

Days

36

35

2 22

3 7

4 or more 0

Assuming that calls arrive independently and at random, estimate (i) the mean m of the corresponding Poisson probability distribution; (ii) the probability that if the operator is absent for ten minutes no call will be missed; (iii) the probability that if the operator is absent for ten minutes two or more calls will be missed. (Cambridge) 4. In an examination 60 % of the candidates pass but only 4 % obtain distinction.

Use the binomial distribution to calculate the chance that a random group of ten candidates should contain at most two failures. Use the Poisson distribution to calculate the chance that a random group of fifty candidates should contain more than one distinction. (Cambridge) 5. The road accidents in a certain area occur at an average rate of one per two days. Calculate the probability of 0, 1, 2, ..., 6 accidents per week in the district. What is the most likely number of accidents per week ? How many days in a week are expected to be free of accidents? (M.E.I.) (Note. This question is best attempted using a hand calculating machine, if one is available.) 6. Explain briefly what is meant by a Poisson distribution and show that for such a distribution the mean is equal to the variance. In a bakery 3600 cherries are added to a mixture which is later divided up to make 1200 small cakes. (i) Find the average number of cherries per cake. (ii) Assuming that the number of cherries in each cake follows a Poisson distribution, estimate the number of cakes which will be without a cherry and the number with five or more cherries. (0 & C) 7. 10 % of the output of screws from a machine are faulty. If screws are taken at random from the output until two faulty screws are found, what will be the average number of screws tested? What is the probability that precisely this number of screws will be needed? 8. 20 % of the butterflies in a district are of type A. If a random sample of size 10 is taken, find the probability (i) that there will be just two butterflies of type A in the sample; (ii) that the tenth butterfly caught will be the second one of type A.

314

4]

THE POISSON DISTRIBUTION

9. Mass-produced articles are taken at random from a batch and tested until a faulty article is found. If the twenty-first article proves to be the first defective, is this at variance with the assumption that 10 % of all the articles are faulty? 10. If X is distributed according to a Poisson distribution with mean A, write down the probability that X = r. How is this probability modified if the values X = 0 are unobservable? Prove that the mean is now A/(1 — e-A), and find the second moment].about X = 0. (Oxford) 11. Evaluate Pr (X = r+ 1)/Pr (X = r) for the Poisson distribution P(a). If the numbers of misprints on pages of an uncorrected proof have a Poisson distribution with mean 2.7, what is the most likely number of misprints to be found on a given page? 12. The average proportion of bad eggs in an egg packing station is one in 2000. The eggs are packed in boxes containing six eggs each. (i) Evaluate to two significant figures the probability that a box contains one bad egg. (ii) A housewife complains if she obtains two or more boxes, with one bad egg each per hundred boxes. What is the probability that she complains.? (M.E.I.) 13. During World War II, 537 flying bombs fell on south London. The distribution of the number of hits in 576 areas, each of 0.25 km2, is given in the table. Compare the actual frequency with the theoretical frequency obtained by assuming that the aim was effectively random and followed a Poisson distribution with the same average number of hits. No. of hits 0 1 2 3 4 5 6 or more Frequency 229 211 93 35 7 1 0 (M.E.I. adapted)

5. SAMPLING INSPECTION SCHEMES Suppose a machine produces a large number of articles and it is required to keep a check upon the number of articles produced which do not attain some prescribed standard; for convenience, we shall describe such articles as 'faulty'. The most obvious method is to check each article individually, but such a process suffers from several drawbacks; for example (i) the process of testing might be very costly; (ii) the process would almost certainly be time consuming; (iii) by the very nature of the test, the article tested might be destroyed (consider, for example, the testing of photographic flash bulbs). A more economic approach is to take a random sample from the total output, to test each article of the sample and to deduce, from a probabilistic argument, the quality of the whole output. The process of selecting a random sample is not as straightforward as it might appear; we shall, t The second moment about X = 0 is II PPM

e (r) . 315


[16

however, here assume that such a sample has been drawn, and concentrate upon the second part of the problem: inferring the quality of the population. A simple approach would be to take a sample of size n, test each article and reject the batch as sub-standard if, say, more than m of the articles prove faulty. We begin by assuming that we have some percentage, 100p say, as an upper limit to the number of faulty articles which may be allowed before the manufacturing process is stopped and corrected for any fault.

Example 5. Samples of size 20 are taken from a large batch of articles produced by a machine; if more than two faulty articles are discovered the batch is withdrawn, otherwise it is accepted. What is the probability that, if the machine is producing 5 % of faulty articles, the batch will be accepted? If the proportion of faulty articles rises to 10 % of the total output, what is the probability the batch will now be rejected? Since the batch is large, we may assume that the removal of twenty articles does not sensibly alter the proportion of defectives and so the binomial distribution B(20, 0.05) may be taken as a suitable mathematical model for the experiment, where our random variable is the number, X, of defectives in a sample of size 20. (i) If Pr (article faulty) = -2-10-, then Pr (not more than 2 faulty articles in 20) /19\20+20 /19\19( 1 0 .1 (19\ 18 12 (20) — k2o)20) k 20) +21.29 k20/ (20) 0.925. (ii) If Pr (article faulty) = -A-, then Pr (more than 2 faulty articles in 20) = 1(190)20_ ( 9 )19 ( 1 ) 20.19 ( 9 )" ( 1 )2 10 20 10 10 1.2 10 0.649. Thus, if only 5 % of the articles produced are faulty, the chance of accepting the batch is about 921 %, whereas, if the number of faulty articles increases to 10 %, there is a 65 % chance of rejecting the batch. Ex. 9. Comment upon the efficiency of the test given in Example 5. Suggest ways in which you feel it could be improved.

The testing procedure exhibited above is an example of a single sampling inspection scheme. Any sampling inspection scheme is open to two types of error: (I) A batch with an acceptable number of faulty articles may be rejected. (II) A batch with an unacceptable number of faulty articles may be accepted. 316

5]

SAMPLING INSPECTION SCHEMES

If a = Pr (error of Type I), A = Pr (error of type II), then, in Example 1 above, a 0.075, A 0.351, if we regard 5 % faulty as acceptable, but 10 % faulty as unacceptable. It can be shown that, if the size of the sample is held fixed, but the standard of rejection is altered, either a increases and Qdecreases or vice versa. Thus, there is an unavoidable margin of error in any inspection scheme; the decision whether to minimize a or A depends upon such external arguments as whether it is economically more desirable to withdraw good batches or allow sub-standard batches on to the market. Of course, if the sample is increased, more accurate information about the population may be obtained and both a and A may be decreased.

Ex. 10. What objections are there to increasing the sample size in order to reduce a and ft? If the probability, p, that an individual article is faulty is now regarded as a variable, and the testing procedure is defined (that is, the sample size and number of faulty articles required for rejection are given), the probability P of accepting the batch is a function of p. The curve obtained by plotting P against p is called the operating characteristic for the testing plan. For example, with the test outlined in Example 5, the operating characteristic has the following shape: P 1.0

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1.0

p

Fig. 16.1 11-2

317


[16

From this graph, the probability of accepting (or rejecting) a sample given the proportion of defectives in the population may be read off. An increase in the sample size causes the operating characteristics to fall off more steeply, thereby improving the test. Other schemes are possible, besides single sampling schemes. For example one can devise double sampling schemes as outlined in the next example. Example 6. A sample of size 8 is taken from a large batch of articles produced by a machine. If the sample contains no faulty article the batch is accepted, if more than one, it is rejected; if, however, the sample contains a single faulty article, a second sample of eight articles is taken. The batch is now accepted if this second sample is free of faulty articles, otherwise it is rejected. Find (i) the probability of accepting a batch containing 1% of faulty articles; (ii) the probability of rejecting a batch containing 10 % of faulty articles; (iii) the average size of sample tested in case (ii).

0.2

0.4

0.6

0.8

1.0

p

Fig. 16.2

Since the batch is large, the binomial distribution constitutes a suitable mathematical model. Thus, if p is the probability of a randomly chosen article being faulty, the probability, P, of accepting the batch is given by 318

HYPOTHESIS TESTING

6]

P = (1—p)8 + [8(1 — P)' p] (1 — p)8 = (1 —p)8 + 8p(1 —p)15 ; (i) ifp = 1/100, P 0.99; (ii) if p = 10/100, 1 —P 0.55; (iii) a first sample of size 8 is always taken; the probability that a second will be required is 8(0.9)7(0.1)cz.,' 0.382 and so the expected sample size is 8+8 x 0.382

11.

As in the case of single sample inspection schemes, the operating characteristics curve can be drawn in this case too; see Figure 16.2. Ex. 11. What advantage does a double sampling scheme possess over a single sampling scheme? Are there any disadvantages?

6. HYPOTHESIS TESTING Probability theory finds a further application in the testing of statistical hypotheses. Consider the following problem. Example 7. A coin is tossed ten times and shows nine heads. Is this sufficient evidence to support the claim that the coin is biased? In order to attack this problem, we adopt the approach made in testing scientific theories: we make a hypothesis and then consider whether or not observational evidence supports this hypothesis. In problems of statistical inference, our hypothesis is generally called the null hypothesis (because it often takes the form of an assumption about the absence of bias in the population): in this example, we shall take as our null hypothesis the statement the coin is unbiased; that is, Pr (heads on any toss) = 2. Having set up our null hypothesis we may ask the question 'Is the probability of obtaining a result as bad as or worse than, the observed result so small that we are reluctant to ascribe it to chance ?' If it is, we agree to reject our null hypothesis in favour of some alternative hypothesis. To complete the framework of our solution we must further decide what numerical value to prescribe to the indefinite phrase so small' ; following common practice, we shall define p = -216as a small probability (the so-called 5 % significance level'); in other cases it might be more reasonable to take p = o o (` 1% significance level '). Returning to the given example, our null hypothesis, H,, is given by 11,:p = 1 we may take as our alternative hypothesis H1:p .

319


[16

Our question now takes the form ' What is the probability of obtaining a result as bad as, or worse than, nine heads with an unbiased coin ?' Since we are comparing the hypothesis 1-10:p = 2 with the alternative hypothesis 111:p -I, we are simply testing for absence of bias and thus we have no reason to distinguish between an apparently abnormal number of heads or tails; it follows that a result às bad as, or worse than, nine heads' is to be interpreted as the result 'nine or ten heads or tails '. We adopt as our probability model for the situation the binomial distribution. Thus we have Pr (9 or 10 heads or tails/Ho) = 2[(2)1°+

warm

47.

Since this is less than -A, we infer that the result is significant at the 5 % level: we have reason to believe that the coin is biased. Sometimes a test of a statistical hypothesis is constructed before the observational evidence is obtained. For instance, in our last example, we might have compared the null hypothesis: Ho :P = with the alternative hypothesis: and so decided what results we should require before accepting or rejecting Ho. In Example 7, our null hypothesis took the form of a statement about the probability of an event; more generally, a null hypothesis is a statement about the probability distribution of a population. We conclude this chapter with a further example, in which our null hypothesis is a statement about the mean of a random variable. Example 8. Batteries are sent out by manufacturers in batches of five hundred. On average, 2.4 batteries per batch are faulty. How many batteries would we need to find faulty in a particular batch before we could adduce evidence that more faulty batteries were appearing than could be ascribed to pure chance? Since we are here concerned only with the question of finding more than the expected number of faulty batteries, we take as our null hypothesis 1/0: mean number of batteries per batch = 2.4, and as our alternative hypothesis I/1: mean number of batteries per batch > 2-4. We adopt as a mathematical model of the situation the Poisson distribution (there is a large number of batteries in every batch and the prob320

HYPOTHESIS TESTING

6]

ability of any particular battery being faulty is small), with random variable X, where X = r when there are r faulty articles in a batch. Thus we have

Pr (X = rIH0) = e-2.4 x

(2.4)r

r!

and we wish to find the least integer n such that Pr (X > nIH0) < 0.05. Equivalently we have to find the least integer n such that Pr (X < nIH0)

0.95.

By direct computation Pr (X < 4114) = e-2.4 (1+2.4+ 2.2412 + 2.3413 + 2.414) 0.0907(1 + 2.4+2.88+ 2.304 +1.382) 0.90, Pr (X < 5IHo) = Pr (X < 4IH0)+ Pr (X = 51H0) 0.96. Thus, n = 5: if we were to find 5 or more faulty batteries in a batch, we should have reason for rejecting our null hypothesis H, in favour of our alternative hypothesis H1.

Exercise 16(c) 1. A machine is believed, on average, to produce 0.1% of faulty articles. Estimate, to 2 D.P., the probability of finding a batch of five hundred articles free of defectives (i) by using a binomial distribution; (ii) by using a suitable Poisson distribution. A batch of five hundred is tested and found to contain two faulty articles. Comment. 2. A large batch of manufactured articles is accepted if either of the following conditions is satisfied: (i) a random sample of 10 articles contains no defective article; (ii) a random sample of 10 articles contains one defective article and a second random sample of ten is then drawn which contains no defective articles. Otherwise the batch is rejected. If, in fact, 5 % of the articles in a batch to be examined are defective, find the chance of the batch being accepted [(0.95)10 = 0.5987]. (0 & 3. From a batch of manufactured articles a sample of ten is taken and each article is examined. If two or more articles are found to be defective the batch is 321


[16

rejected; otherwise it is accepted. Show that, if p is the proportion defective in a batch and P its chance of being accepted,

P = (1 —P)9(1+9P). Find an expression for P if it is now decided to modify the scheme so that when one defective is found in the sample a second sample of ten is taken and the batch rejected if this second sample contains any defectives. In the second case, what will be the average number sampled per batch over a large number of batches when p = 0.05? (M.E.I.) 4. In a certain inspection scheme a sample of fifty items is selected at random from a very large batch and the number of defectives is recorded. If this number is more than three the batch is rejected; if it is less than three the batch is accepted. If the number of defectives is exactly three a further sample, this time of twenty-five items, is taken and the batch is rejected if there is more than one defective in the second sample but accepted otherwise. If the proportion of defective items in the batch is 1 % determine the values of the following probabilities: (i) that the batch is accepted as a result of inspection of the first sample; (ii) that a further sample has to be taken and the batch is accepted as a result of inspection of that sample; (iii) that the batch is rejected. Take (0.99)48as 0.6173. (M.E.I.) 5. State the formula for the probability that a variable following a Poisson distribution of mean m takes the value r. Prove that the variance of r is m. Past experience has shown that the number of defective items produced in a shift by a certain machine is a Poisson variable of mean 4. A new employee in his first shift produced six defectives. Is this clear evidence that he is operating the machine inefficiently? (Cambridge) 6. The probability P(r) that there will be r damaged tomatoes in a crate can be taken as

P(r) =

mr r!

where m is the expectation of r. Over a large number of crates the value of in has been found to be 10. In the first crate from a new supplier the value of r was 4. Test whether this is significant evidence that the value of m for this supplier is less than 10; explain carefully the logic of your argument. (Cambridge)

7. A die is thrown six times. If a score of six is made on three of these occasions, have you any reason for believing that the die is biased? 8. A cubical die with faces marked 1 to 6 is thrown n times. Show that on the hypothesis that the die is unbiased the chance that the face marked 4 will appear uppermost not more than once is p, where

P 322

1,1+5\ 15\n =

k

5

k6f

6]

HYPOTHESIS TESTING

If n = 40 and the face marked 4 comes uppermost exactly once, test whether the hypothesis that the die is unbiased is contradicted (a) at the 1 % significance level; (b) at the 0.1 % significance level. (Cambridge) 9. A bag contains ten balls, each of which is either black or white, but otherwise

the balls are indistinguishable one from another. Three balls are drawn without replacement, and all are found to be white. Test the hypothesis Ho : there is an equal number of black and white balls in the bag. The three balls are now replaced and three more are drawn; these are found to be two white and one black. Test the hypothesis Ho again. 10. A sampling inspection scheme is operated by taking a random sample of size

10 from each large batch of a product. The batch is rejected if more than one defective is found and otherwise the batch is accepted. Plot the operating characteristic of the given plan. Explain the applications of operating characteristics when choosing a suitable plan. (M.E.I.) [You are advised to use a hand calculating machine for the first part, if you have one available.] 11. A double sampling inspection scheme is devised as follows: a sample of size 15 is drawn from a batch of articles; if the sample contains none or one faulty article the batch is accepted, if more than three, it is rejected. If it contains two or three faulty articles a second sample is taken; the batch is now rejected if this second sample contains more than one faulty article. If the batch contains 100p % faulty articles, find (i) the probability that the batch will be accepted; (ii) the value of p which gives the largest expected sample size; (iii) the size of this largest expected sample. 12. An assembled instrument contains two critical components A and B. Sample tests show that we may expect one in ten of A and one in eighteen of B to be

defective. Estimate and compare the costs of the following inspection plans, per hundred fully tested instruments: (i) to test every component before assembly at a cost of 2p for each A tested and 3p for each B tested; or (ii) to test every instrument after assembly, if this test adds nothing to the cost but making good a defective instrument costs on average 24p. (M.E.I.)

Miscellaneous Exercise 16 1. A pack of cards is cut and the suit of the exposed card noted; the pack is then well shuffled and the process repeated. Find (i) the probability that a spade will appear for the first time on the fourth cut; (ii) the average number of cuts required before the first spade appears; (iii) the average number of cuts required to expose cards of all four suits. 2. Show that the probability of r successes in n independent trials is the coefficient of tr in the expansion of (q + pt)n , where p is the chance of success in a single trial and q = 1—p. Prove that the mean number of successes is np.

323


[16

Samples, each of eight articles, are taken at random from a large consignment in which 20 % of the articles are defective. Find the most likely number of defective articles in a single sample and the chance of obtaining precisely this number. If a hundred samples of eight are to be examined, calculate the number of samples in which you expect to find three or more defective articles. (0 & C) 3. Two coins are identical in appearance, but one is unbiased while the other gives, on the average, heads three times as often as it does tails. One of the coins is taken at random and tossed four times. If two heads and two tails appear, what is the probability that it is the unbiased coin? 4. Explain briefly what is meant by a Poisson distribution of rare events and its relation to the binomial distribution. Prove that the mean of the distribution is equal to its variance. A shopkeeper's sales of washing machines are four per month on the average. Assuming that the monthly sales fit a Poisson distribution, find to what number he should make up his stock at the beginning of each month so that his chance of running out of machines during the month will be less than 4 %. (0 & C) 5. A machine depending for its energy upon four complexes of solar cells, each complex functioning independently of the others, will work provided one of the four complexes is working. Each complex has probability p of failing. The machine is redesigned to have six complexes and will function if two of the six complexes are working. Is the new design an improvement on the old?

6. A and B play N games, each of which must result in a definite win to one or other player. A's chance of success in any one game is p. For his rth win, A receives from B £r, for his sth loss, he gives B £s. Find A's expected gain. 7. A coin is tossed repeatedly an even number, 2n, times. Show that, whatever the bias (provided Pr (H) = 0 or 1) the probability of obtaining the same number of heads as tails decreases as n increases. 8. In lawn tennis a set is won by the first player to win six games, except that if the score reaches 5-5 the set is won by the first to lead by two games. Two players have chances respectively p and q of winning in any game (p+q = 1); games may be treated as independent. Find the chance that a set lasts exactly 2n + 2 games (n 5). (Cambridge) 9. The number of eggs laid by an insect has Poisson distribution with mean it. If the probability that an individual egg survives is p, show that the number of eggs surviving has Poisson distribution, and determine its mean value. (You may assume that the survival of an egg is independent of the fate of the other eggs laid.) 10. An experiment has probability p of success. In n independent trials, pois the probability of an even number of successes (pa = 1). Prove that

Pn-(1-2P) P.-1 = P. If f(t) = i po,r, prove that n=0

f(t) = [1- (1- p) t]l{(1- t) [1 - (1-2p) t]}.

Deduce that

324

Pn = i0 + (1 -2P)ni.

61


11. Two machines, A and B, produce large numbers of articles, with 10 % of those from A and 20 % of those from B defective. Machine B produces 50 % more articles than machine A. A batch of ten taken from one of the machines contains two defective articles. What is the probability it came from A? (Give to 1 D.P.) 12. In a simplified probability model of the service in a barber's shop, it is supposed that all haircuts take exactly six minutes and that a fresh batch of customers arrives at six-minute intervals. The number of customers in a batch is described by a Poisson probability function, the mean number being three. Any customer who cannot be served instantly goes away and has his hair cut elsewhere. The shop is open for forty hours a week. Calculate the theoretical frequencies with which batches of 0, 1, 2, 3, 4, 5 and more than 5 customers will arrive. The proprietor reckons that it costs him £25 a week to staff and maintain each chair in his shop, and he charges 25p for each haircut. Calculate his expected weekly profit if he has (i) three; (ii) four; (iii) five chairs. (SMP) 13. A sample of n coins is drawn at random from a large collection in which a fraction p are pennies. What is the probability that just r of the n coins are

pennies ? If the probability that a penny is a Queen Elizabeth one is q, what is the probability that there are exactly s Queen Elizabeth pennies among the r pennies of this sample? Write down the probability that a sample of n coins will contain s+ k 10p pieces, only s of which are Queen Elizabeth ones, and calculate the sum of these probabilities for all possible values of k. (C.S.) 14. A bag contains a large number of red, white and blue dice in equal numbers. If n are drawn at random, show that the probability P(n, r) of drawing exactly r red dice is equal to the term containing (-Dr(4)n-rin the expansion of ef + ip. If r dice are thrown, find the probability Q(r, s) of throwing exactly s sixes. If n dice are drawn from the bag and the red dice drawn are thrown, show that the probability of throwing exactly s sixes is n-s

I P(n, s+ t) Q(s+ t, s)

t=o

and prove that this is equal to a term in a binomial expansion. Explain why a binomial distribution is obtained. (C.S.) 15. A random sample of size 10 is taken from a batch of a thousand components and one defective is found. (i) What is the largest possible percentage of defectives in the batch? (ii) p is the smallest proportion of defectives in the batch such that the probability of obtaining not more than one defective is not less than 95 %. Find the value of p to two decimal places by trial of suitable values of p. (iii) Find the most likely percentage of defectives in the batch (i.e. such that the probability of obtaining one defective in the sample is a maximum). (M.E.I.)

325


[16

16. In sampling inspections of batches of manufactured articles a random sample

of twenty is taken; if none or one defective occurs in the sample the batch is accepted, if three or more defectives occur the batch is rejected. If two defectives occur a second random sample of twenty is taken and if in the combined sample of forty less than four defectives occur the batch is accepted; otherwise it is rejected. Assuming that the proportion p of defectives in a batch is sufficiently small for the Poisson distribution to apply, show that the chance P of the batch being accepted is given by P = e-2" (1+20p) (1 + 200p2 e-2°P). (i) Find the chance of a batch which is 2 % defective being rejected. (ii) Find the chances of batches which are respectively 5 and 10 % defective being accepted. (iii) Sketch the operating characteristic curve for the inspection scheme. Determine the average sample size per batch ifp = 0.05. (0 & C) 17. The number of a certain type of organism in a given volume of water has Poisson distribution with mean 2. A test, applied to indicate absence of the organism has a 90 % chance of success if the organism is in fact absent, but also indicates absence in 10 % of those cases in which they are present. If the test is applied and indicates absence, what is the probability that the water is free of the organism? 18. The probability that a source emits r a-particles in a given time is propor-

tional to ArIr!, where is a constant. Obtain the constant of proportionality and calculate the mean of r. The probability of the same source emitting s fl-particles in the same time is proportional to vs/s!. Assuming that the two types of particle are emitted independently, write down the probability that r a-particles and s fl-particles are given off in this time and show that the probability that a total of n particles of the two types are emitted is ,-(1.44-v) + 07,1n !. (Cambridge) 19. The probability that any randomly chosen rat from the colony used in a

certain laboratory will show a certain undesirable characteristic is p; it is known that the value of p for the colony is either 0.4 or 0.6 and it is desired to set up a sampling scheme to decide which value is correct. The procedure is to take rats one at a time at random and test for the presence of the characteristic; after n rats have been tested, let r be the number with the characteristic. Show how to determine limits L„ and U„ such that for known n Pr (r -C. L„Ip = 0.6)

a and Pr (r

Unip = 0.4)

where a and fi are specified. Use the table of partial binomial sums below to determine Ln and Un for n = 5, 7, 9, 11, 13, 15, 17 where a and fl are less than 0.1 and as close to it as possible. Mark these values on a plot of n — r against r, and explain how such a diagram could be used in the decision procedure. 326

61


Partial binomial

RIM

5

7

9

11

13

15

17

3

317

580

768

881

942

973

988

4

87

290

517

704

831

909

954

5

10

96

267

467

647

783

874

6

19

99

247

426

597

736

7

2

25

99

229

390

552

8

4

29

98

213

359

9

0

6

32

95

199

10

1

8

34

92

11

0

1

9

35

Table of 103 x E (n) (0.4)$ (0'6)n-3.

(Cambridge)

s=r S

20. At a certain seed testing station it is found that a proportion 0.4 of a certain type of seed is fertile. By accident the remaining stock of this seed (whose total amount is very large) is completely mixed with an equal quantity of a second type of seed which is believed to be completely infertile. If this latter assumption is true, what is the probability that a seed taken at random from the mixture will germinate? Each of seven pots is planted with two seeds taken at random from the mixture. Six pots eventually produce one or more plants each. Is this result consistent, at the 5 % level of significance, with the infertility postulate for the second type of seed? (Cambridge) 21. Derive the probability p(r) of obtaining exactly r successes in n independent trials, the probability of a success being p at each trial. Determine the mean and variance of this distribution. Find an expression for [p(r)]/[p(r I)] and hence or otherwise find the conditions that rff, must satisfy if p(r„) is such that no other value of p(r) is greater than p(rm). Show that the mode of a binomial distribution differs from the mean by less (M.E.I.) than unity. —

327

Revision exercise B 1. Find the equation of the line L1through the origin with gradient 1, and also the equation of the line L2 perpendicular to L1and passing through the point (4, -1). 2. In a large university, one-third of the men and one-quarter of the women read science. If four men are selected at random, what is the probability that not more than one reads science? If two men and two women are selected at random what is the probability that not more than one of the four reads science? What is the probability that, in a mixed group, one man reads science and the other man and the two women read something else? 3. A square lamina ABCD of side a is held with the corner A on a horizontal plane. The feet of the perpendiculars from B, C, D on to the plane are B', C', D' and the angles B' AB, D' AD are a, /3. The angle B'AD' is O. Prove that (i) cos 0 = - tan cc. tan )3; sine a sin2 fl)1; (ii) the area of the triangle B'AD' is -1a2(cos2 a cost (iii) the inclination of the lamina to the horizontal can be expressed in the form (0 & C) arccos [cos (cc+ fi) cos (a-/3)]t. 4. If a + -14(4n +1)7r, find x if sin (x+ a) = cos (x- a). What can you say about x if a = 1(4n +1)70 5. Find the sum to infinity, S, of the geometric series 1

1

1

1 + -+ + A/10 10 1000

If SN denotes the sum of the first N terms of this series, find the least value of N such that S and SN are the same correct to 3 D.P. 6. A tennis match usually consists of either three or five sets, and ends when one side has won a majority of the sets. If the probability of a side winning a set is p, and if the result of each set is independent of any previous results, show that the probability of a match going its full legth is 2pq in the case of a three-set match and 6p2q2in the case of a five-set match (q = 1-p). Show that the first probability is always greater than the second, if p 0 or 1. 7. The function f: R -> R+ is defined by f(x) = Ix+1I+ I2x-11+ Sketch the graph of f and determine the least value of f(x). 8. Call a matrix of the form 328

cx

xy)

REVISION EXERCISE B

which is symmetrical about both diagonals 'super-symmetrical'. If A, B, ... are super-symmetrical 2 x 2 matrices and m, n, ... are non-negative integers, prove that AmBn ... is super-symmetrical. Does this result hold for 3 x 3 matrices? 9. Factorize into linear factors the expression 2x2 + 5xy— 3y2— 3x+ 5y — 2 and describe geometrically the set of points {(x, y) : 2x2 + 5xy — 3y2— 3x + 5y — 2 = 0}. Describe geometrically the sets of points (i) {(x, y): x2 — 2y2 = 0); (ii) {(x, y) : x2 + 2y2 = 0). 10. i, j, k are unequal positive integers and x, y, z are real numbers. Prove that x(j— k) + y(k — i) + z(i—j) = 0 x, y, z are respectively the ith, jth, kth terms of an arithmetic sequence. 11. Prove by mathematical induction, or otherwise, that n.13+ (n —1) .23+ (n — 2) .33+ ... +1 . n3= A n(n +1) (n + 2) (3n2+ 6n + 1). (Oxford) 12. Find the inverse of the matrix ( 5 3 7 3 4 6. —1 2 1 Hence solve the simultaneous equations 5x+3y+7z = a, ) 3x+4y+6z = b; 1—x+2y+ z = c; (i) when a = —1, b = 2, c = — 3; (ii) when a = b = 1, c = 2. 13. What are the first three terms in the expansion of (1 — 2x)24in ascending

powers of x? Find the value of 0.9824, correct to 4 D.P. 14. Show that, whatever value is chosen for k, the equation

(3 + 5k) x+ (2— 7k) y + (5 —4k) = 0 represents a straight line through a fixed point A. Find the particular line of the system (i) which passes through the origin; (ii) which is parallel to the y axis. 15. The polynomial P(x) leaves a remainder of x+1 on division by x2 -2 and the polynomial Q(x) leaves a remainder of 2x+3 on division by x2— 2. New polynomials R(x) P(x)+ Q(x) and S(x) E P(x) Q(x) are defined. Find the remainder (i) when R(x) is divided by x2— 2;

(ii) when S(x) is divided by x2— 2. 329

REVISION EXERCISE B 16. If A is the point (a cos a, b sin a) and B is the point (a cos /3, b sin /3), show that the equation of the line AB is fi

x a+ fi y a+13 sin Q cos = cos 2 b 2

2

What is the connection between a and /3 if AB passes through the origin? 17. Prove that the number of spheres that can be drawn to pass through three given points and touch a given plane is 2 or 1 or 0, explaining how the three cases arise. 18. OA, OB, OC are three concurrent straight lines lying in one plane. P is a point outside the plane such that the angles PO A, POB, POC are equal. Prove that PO is perpendicular to the plane. 19. Find the matrix of the linear transformation which reflects all points in the plane x+y+z = 0. 20. ABCDA'B'C'D' is a parallelepiped, with opposite faces ABCD, A'B'C'D' and edges AA', etc., AB = BB' = B'C' , and the mid points of BB', A' D', A'B' are respectively F, G, H. Find the ratio in which the plane AD' H divides the line FG. -

21. Two boxes each contain one hundred cards, numbered 1 to 100. A card is taken from each box, the numbers on the two cards noted, and they are then returned to their respective boxes. If the process is repeated fifty times, find the probability that at no stage a pair of cards bearing the same number will be drawn. How many draws are needed for the probability to be >1? 22. Prove that lg 15 is irrational. (lg 15 means log„ 15.) Is it true that lg x is irrational a x is not a power of 10? 23. Evaluate the following determinants: b— c c— a a— b (x-p)2 (y-p)2 (z-p)2 I 0, (i) c— a a— b b — c ; (11) a—b b— c c— a (X r)2 (y— r)2 (z— r)2 24. Define the projection of a vector in a given direction. ABCD is a quadrilateral in which LA = LC = 90°. The feet of the perpendiculars from B, D to AC are X, Y respectively. Prove that AX = CY. 25. In a large crate of oranges, 100p % are bad. A random sample of ten oranges produces two bad ones. On the assumption that this is the most likely number of bad oranges to find in the sample, what can you say about the possible values of p? 26. ABCD is a square lying in the plane 2x— y — 2z = 5. If A has coordinates (1, 1, —2) and C has coordinates (5, 5, 0), find the coordinates of B and D. 27. The function, f, defined by f(x) = sin x+ cos x has for its domain the set of real numbers {x e R: 0 < x < zr}. Find the range off.

330

REVISION EXERCISE B 28. Factorize the expression ax2+(ac+ b — c) x+ c(b— c) into two linear factors. Hence factorize the expressions (i) a+b—c+bc+ca—c2; (ii) a — b + c+bc— ca— c2; (iii) a3b2+ a2bc + ab2— abc + bc — c2. 29. Describe geometrically the transformation of three-dimensional space represented by premultiplying the position vector of the point (x, y, z) by the matrix 0 1 — 12)).

—0 Which points are left unaltered by this transformation?

(O.S.)

30. When A and B play chess the chance of either winning a game is always + and the chance of the game being drawn is always 4. Find the chance of A winning at least three games out of five. If A and B play a match to be decided as soon as either has won two games, not necessarily consecutive games, find the chance of the match being finished in ten games or less. (J.M.B) 31. Solve the equations (i) 2 sine x+ sin 2x = 0; (ii) sin x = cos (-Pr— x); giving all solutions in the interval 0 32. Solve the inequality

x ‘. 2ir in each case.

x2 -1 1 x2 -4 5

and illustrate your solutions by means of a sketch of the curve x2—1 Y — x2-4 ' 33. The base AB of a triangle ABC is fixed and K is a fixed point on AB. The vertex C of the triangle moves so that the perpendicular distances of K from CA and CB are always equal in length. Prove that, in general, the locus of C is a circle through K. What is the exceptional case? (Cambridge) 34. A pitcher is taken daily to a well and back. Its chance of being broken on an outward trip is pi. If it survives the outward journey it has a further chance p2 of being broken on the return. Show that on any day its change of being broken is P, where P = Pi+P2 —PiP2. Show that the chance that the pitcher will survive for at least n days is Q", where Q = (1—p1) (1 —p2), and find the chance it will survive (n-1) days but be broken on the nth day. If two such pitchers are each taken independently to the well and back daily, prove that the chance that they survive for exactly the same number of days is P1(1+ (C.S.) 331

REVISION EXERCISE B 35. Find the image by reflection of the point (4, 3, 1) in the plane 3x+2y+3z+ 1 = 0. 36. A and B are two towers, B being four miles due east of A. The true bearings of a flagpole C from A and B are a° east of north and a° west of north respectively; the true bearings of a second flagpole D from A and B are (cc + fi)° east of north and (oc— j3)° west of north respectively. Draw a sketch-plane to indicate the positions of A, B, C, D. Assuming that A, B, C and D are on level ground, prove that D is 4 sine f3 cosec 2a (0 & C) miles south of C and 2 sin 213 cosec 2a miles east of C. 37. If ax4+ 3x3+ bx2+ cx+ 2 is exactly divisible by (x+2) (x2 -1), find the values of a, b, c. 38. Sketch the graph of the function defined by f(x) = x+ lxi (x e R). In a separate diagram, shade in the set of points satisfying simultaneously the three inequalities y > x+ Ixl, y...-- Ixl—x, x—y+l > 0. What is the maximum value of the expression E -= x+y+1 if x and y are subject to the three given inequalities? 39. Prove that, if A, B are two non-singular 3 x 3 matrices, then AB is a nonsingular matrix and (AB)-4 = B-1A-1. If 1 0 0 63 5 1 0 0 E, = 0 0 1 , E, = 0 1 0 A= ( 1 1 1 —3 4 — 2 \0 1 0 / find A-1, Ell, EV-. What are the inverses of the following matrices: (i)

6 3 -3 4 (2 5

5 —2); 2

6 3 (ii) (— 3 4 5 12

5 — 2) 5

40. Using the formula for tan (A+ B) in terms of tan A and tan B, show that, if A, B, C are the angles of a triangle, then tan A + tan B+ tan C = tan A tan B tan C. Calculate, in degrees and minutes, the angles of a triangle ABC if tan A: tan B: tan C = 1:2: — 6.

(Cambridge)

41. Find the equation of the plane containing the origin, the point (1, 1, 1) and the point (3, 1, — 1). Find also the direction ratios of the line of intersection of this plane with the plane x+y—z— 4 = 0. 332

REVISION EXERCISE B 42. In a game of tennis one point is scored either by A or by his opponent B. The winner of the game is the player who first scores four points, unless each player has won three points, when ' deuce' is called and play proceeds until one player is two points ahead of the other and so wins. If A's chance of winning any point is and B's chance is calculate the chance of (i) A winning the game without `deuce' being called; (ii) a similar win by B; (iii) ' deuce ' being called. If `deuce' is called, prove that A's subsequent chance of winning the game is I. Deduce that A's chance of winning the game is nearly six times that of B. (Cambridge) 43. Prove that ( ) = 2 n. r=0 r

If A is a square matrix such that A2= A, express (A +I)" in the form ecA +fil. 44. Use the method of mathematical induction to prove that (k+1)! (k+2)!+ (k+n-1)! (k+n)! k!+ + ... + = 1! 2! (n — 1)! (k + 1) (n — 1)! . 45. Show that the equations (k+2)x+2y+3z = 7, x+(k+2)y+z = 0, {

5x+2y+(6—k) z = 13

have a unique solution for all but three values of k. Discuss the solution of the equations in the three exceptional cases. 46. Prove the formula cos 3A = 4 cos3 A-3 cos A. (Formulae for cos 2A and sin 2A may be assumed.) Substitute x = -1-+ cos 8 in the equation 8x3— 12x2 +1 = 0 and, with the aid of the above formula, solve the resulting equation in 8, giving values between 0° and 180°. Hence find the three roots of the cubic equation in x, correct to two decimal places. (Cambridge) 47. AlA2 ... Anis a regular polygon of side 1 unit. Two (distinct) vertices are selected at random. Taking as random variable X, the shorter distance measured along the perimeter between the two points, find g(X) (ii) when n is even. (i) when n is odd; 48. When n is a positive integer the coefficient of an-rbrin the binomial expansion of (a+ Onis denoted by (1I. Write down an expression for (1 and prove r r

that

(n-1 for r = 2, 3, ..., n. r — 1) By comparing the binomial expansions, or otherwise, prove that, when a > 0, b > 0, (a + b)n — a" < nb(a + b)"-1. (0 &C) 1.

86 1. x < 0 or x > 1. 3. x < 0 or

4. x < -2 or

x > 1.

x>

-

1.

5. 1 < x < 3.

6. x < i or x > 1.

7. x < -1 or 1 < x < 2.

8. x > 3 or

9. -1 < x < 1.

10. x < 1 or 2 < x < 3 or x > 4.

x = 2.

11. -2 < x < -1 or 1 < x < 11. 12. -3 < x < -2 or -1 < x < 1. 13. x < -2 or 0 < x < 1 or

x > 2.

14. x < -5 or -3 < x < 0 or 1 < x < 2 or 15. x < - 1(,/13 + 1) or 16.

x > 5.

x > -1-,(V13 - 1).

/3 < x < -V2 or V2 < x < 813.

- 8

17. x < 11 or x ,.. 3.

18. x > 2f.

19. x < -1 or x > 11.

20. x < -8/2 or x > V2.

21. x < 0 or 1 i < x < 2.

22. x < -11 or 31- < x < 5.

351

ANSWERS

CHAPTER 6 PAGE IT IT n 577 377 577 1177 87 Ex. 1. - - 6'3'4' 12' 4'4' 6 •

Ex. 2. 45°, 150°, 120°, 40°, 105°, 36°, 48°.

Exercise 6(a) 117r 577- 197r

777 577 777 377 777

88 1.

-

-

4' 6 'T' 6 •

9'4' 12' 6' 6'

2. 30°, 60°, 135°, 75°, 210°, 810°, 54°, 900°, 50°, 63°. 3. 0.445, F244, 2.552.

4. 27.5°, 37.2°, 124.9°.

6. 6080 feet—the 'nautical mile'.

7. 18.5 cm.

8. (i) 10.5 cm; (ii) 43.3 cm2; (iii) 9.1 cm2.

10.

9. 2r sin 0, nr0190.

1.7

m, 8 m2.

89 11. 9.6 cm, 26.5 cm2.

12. (i) 216°; (ii) 47.1 cm2; (iii) 6.24 cm.

Exercise 6(b) 94 1. -1, -1, 2, -V3/3, -V2/2, 2, 0, V3/3, -V2, -2V3/3.

2. A/2/2, -1, -2, -V2/2, V3/3, -2, V3/2, -2V3/3, -V3/3, -V3/2. 3. -0.5736, -0.6157, -0.8391, 3.0716, 0.3420. 27r 477. 7T 777 117r t..‘37r 777 (iii) T, (iv) 6 , 6; ' 6 ; °I 4 4 77 27r 477 577 137r 177 . 37 77 s , 3 , T ,3; (vi) , j, ,(vii) 7, 4 ,

A ( .1777 `11 6 (v)

(viii)

7T 57r 137r 177r 2577 2977. 577 , . —•' (x) 0, gin, 7r, la, 27r; 18 18' 18 ' 18 ' 18 ' 18 ' (ix) 3

77 777

.. 77 n 2rr 577

(xi) 6 , -w, (xii)

6,

,

-y-, w.

577

5. (i)

,‘ -

- 71. '

-

n,

6 ' -6' "' 11.;

(iv) -n. 0,77; (v) - ' 8'11. (viii) no solution; (ix)

352

7;

,

(vi)

7T -

577

(...\

7T

(Ill)

ìii - 6 ' -6;

2; (xxj\)

n n -

-

2 , 2'

37r .717

..

(vu) 577

rr n 577 .

- 2' 6 ' 6 ' 777 7T 5n 117r - — - , , —12' 12 12 12

7T 77 377 777

- 8 ' -8' 4 ' 8, i•

ANSWERS PAGE

94 6. (i) 3, 1; (ii) 4, 0; (iii) 5, 1. 7. (i) -2A/2/3, - A/2/4; (ii) V5/5, -20/5; (iii) - 2A/2; (iv) - A/15/15, -4A/15/15; (v) -4,J15/15, - A/15/15. 95 11. - ii. H.

12. 3 ± 6A/10/10.

13. (i) 45, 123.7, 225, 303.7; (ii) 70.5, 289.5; (iii) 45, 225, 153.4, 333.4; (iv) 38.2, 141.8; (v) 36.9, 143.1, 216.9, 323.1; (vi) 30, 150; (vii) no solution; (viii) 0, 63.4, 180, 243.4, 360. 14. (i) 9x2 + 4(y- 3)2 = 36; (ii) x2 +y2 = 2; (iii) 2x2 -2xy+ 5y2 = 9; (iv) x2(16 -y2) = 9y2; (v) y(2 -y) (1 + x)2 = 1. 16. a =

15. (i) (sin e, 0); (ii) (± 1, 0); (iii) n/co seconds. 96 19. 4, 11. (i) 0; (ii) 2. 98 Ex. 12. arccot: R arcsec: {x e R: jxj

{y e R:0 < y < 7T}; 1) {y e R: 0 y

TT

Ex. 13. fir, 1n, -

Exercise 6(c) 99 2. 0, in, 7T, Ir 2n.. ,

100 5. 35.3, 144.7.

3. 35-2.

4. 65°.

6. 0.4, 3.32, 5.14.

7. 0.90, 2.25.

15. (i) A/3/2; (ii) V3/2; (iii) A/3/3; (iv) - A/2/2; (v) 1; (vi) 3; (vii) 4; (viii) - A/3/3. 16. (i) x/A/(1 - x2); (ii) x/A/(1+ x2).

17. V2/2.

18. x, Ix'

19. x2+ y2 = 1.

1.

Miscellaneous Exercise 6 1. 0.967. 101 2. 2a2(ka-V3/4).

3. 11, -1.

4. 0.73, 2.41, 3.99, 5.43.

5. 54.4 cm. (i) 12.25 cm; (ii) 75.5'; (iii) 13.2 cm. 6. 0.

7. sin x # ± sin y

8. ar1 5 T, 2 ;7570 or ( g, AlT).

9. (x2 - y2)2 = 16xy.

102 11. 0.88n, 0.78n, 0.69n, 0.60n, 0.50n. 12. 0.46n, 1.63n, 2.31n, 3.95n, 4n. 13. 27777 + n7T + cc,

14. (i) i(3n+ 1) n.; (ii) 3(3n± 1) 77 or 2kn.

16. 0.39n, 2.3 cm. 353

ANSWERS

CHAPTER 7 PAGE

103 Ex. 1. 120, 720, 5040. 104 Ex. 2. 12, 60, 720.

Ex. 3. 120, 6.

105 Ex. 4. 10, 20.

Ex. 5. 1.

Ex. 7. 364, 165.

106 Ex. 8. 210.

Ex. 9. 51°.

Exercise 7(a) 2. 870.

1. 120.

4. 216.

107 3. 665280.

5. 120, 48, 30.

7. 5005, 420420. 8. 34650.

9. 120, 85. 12. in!, 840.

10. (i) 48!/(9!39!); (ii) 48! 16/(9! 39!). 13. 30, 12.

6. 495, 135.

14. 19958400.

15. ln(n -3).

16. 90. 18. ln(n- 1).

108 17. n!/(p! q! r!), n!/{3(n-2p)! (p!)2}.

109 Ex. 10. {HH, HT, TH, TT}, {0H, 1H, 2H}, {coins fall the same, coins fall differently}. 112 Ex. 20.

Ex. 22. A.

Ex. 21. 112-.

Ex. 23. (i) 132(391)2/(27! 52!); (ii) 13 (39!)2/(27! 51!). Ex. 24. (i)

(ii)

(iii) 4.

Exercise 7(b) 1. 64, 20, 6• 113 2 ,614, 64

,

39 1.1

uy,

1 664, 614,

4• W.

3. (i) i; (ii) A.

5• 4-34.

6. Pr (0) = A, Pr (r) = (10-0150, r = 1 to 9. 8.

(i)

7. A-.

(ii)

9. {2, 3, 4,

20},

o,

• • •,

-11(4, rh, • • • , Au} ;

10. 115 Ex. 26. las. Ex. 29.

Ex. 27. 6 1 .

ta.

119 Ex. 31. (i) i; (ii) -2-g; (iii) A. 354

Ex. 28. (i) 4; (ii) 1; (iii) I; (iv) 4.

Ex. 34. fa

ANSWERS

Exercise 7(c) PAGE

119

... 1296,

2. 0.784.

1627916•

3.

4. (0 lf, (i0 212; (iii) 292. 120 8. (0

7. It.

6. 9, 81.

(ii) H.

9. 5!, 20, -M. 10. th.

11. 66.

12. At least 12.

16. a

17. 13 times.

121

20. 0.956.

21. (i)

123

Ex. 35. 8, t.

Ex. 36. 118-, *•

13. H.

14. 0.22.

A•

;

Ex. 38. H, y 3 S s, 5 e

124 Ex. 37. -M.

Exercise 7(d) 125

2. 0.1997.

1. 36047.

3. A•

7• A-.

126 6. 1459 4 ;

4. +,

gr.

5.

8. 41, 0, A•

9. pl(p + q - pq).

Miscellaneous Exercise 7 1. (i) P; (ii) 1

,., 1

n!'

(iii) 0.065; (iv) 0.036. 1

0' 2(n 2)!' 3(n -3)!• -

127 3. No difference in either case.

4. Approx. 1. 5. 1.

6. Replacement-h-; no replacement h• 7. Probability of securing no prize greater for B, but this is compensated by greater probability of two prizes. 9. (i) 1; (ii) 9.

10. (2r- 1)/3T-1.

128 11. (i) m/(m + n); (ii) m(m-1)/{(m+n) (m + n (iv) m/(m + n).

-

1)}; (iii) n 1(m + n 1); -

12. 5- • (1) 162,5, 125,

1

125; OD 115, 11225, 14285, 16245;

14. (i) 0.09; (ii) 0.111; (iii) 0.336; (iv) small; (v) no. 16. (pi+ (P2+ A) (p3+ A) =

A2.

17. A, 0, ;; exactly and oppositely synchronized; 1, 0, *; I, 2, -. 18.TVS.

355

ANSWERS PAGE

128 19. A, (1 -q)(1-q4); 4A,(1 q4. 93;

-q) q;

(1 -9) 94;

-9) R2;

20. 1 (n - 1)! (n2- n)! nn-21(n2 -1)!. -

CHAPTER 8 131 Ex. 1. (i) 0, 1, 4, 17; (ii) 2, 5, 16, 65. Ex. 2. (i) 1,2,4,8; (ii) 1, 2, 3, 4; (iii) 0, 4, 21, 100; (iv) 1, 2, 5, 17; (v) 1.5 F4, F41, F414, (approx.). ,

Ex. 3. (i) 3, 6, 9, 12, 15, 18; (ii) -2, 1, 4, 7, 10, 13; (iii) 1, 2, 4, 8, 16, 32; (iv) 0, 6, 24, 60, 120, 210; (v) 0, 2, 0, 2, 0, 2; (vi) i,1, -A-, 2A, -A 4127 ; (vii) 1, 1, 1, 4, 25, 216; (viii) -1, 5, -1, 9, -1, 13. 132 Ex. 5. 1, 1, 1, 1, 25. Ex. 6. (i) 1+4+9; (ii) 0+2+6+12; (iii) 1+6+21+60; (iv) -1+2-3+4-5; (v) 2+3+10+29; (vi) 3+0-1+0+3+8; (vii) 0+1+4+1+f.

133 Ex. 7. (i) 18; (ii) 14; (iii) 100; (iv) 6; (v) 1A; (vi)

-

15;

(vii) 31; (viii) A; 4

4

4

5

Ex. 8. (i) E r; (ii) E (-1)r+1r; (iii) E 2r; (iv) E 27-1; r =1

r=1

r=1

r=1

6

1r 1); (vii) E (-1)'+1; (viii) E r. r-1; (vi) (r+

(v)

r=1

r=1

Ex. 11. en.

r=1

r=1

Ex. 12. 3r -2, 2(3r-1), 4 2r. -

Ex. 13. (i) 46; (ii) 35; (iii) 86; (iv) 40; (v) 100. 134 Ex. 14. (i) E (3r- 2) (3r + 1) (3r+ 4); (ii) E 6r2(4r+ 1); (iii) E (202r-2; r=1

(iv)

r=1

(2r-1)(4r-1)(3r-1)-1; (v)

1)r-F1r{(2r - 1)(3r - 1) (7r -4)1-2.

r=1

r=1

Exercise 8(a) 135 1. (i) 112; (ii)

(iii) 15.

2. (i) E 3r(3r- 2); r=1

(ii)

(-1)r+1r (r + 1) (r+ 2); (iii) r=1

356

r=1

r=1

(-1)r}1(2r-1)2.

ANSWERS PAGE

135 3. (i) 1, 4, 13, 40; (ii) 1, 1, 1, 1; (iii) 2, 22, 24, 28; (iv) 1, 2, 1a, 114.

4. s 2s + 5,2= d. 5. (i) 9r -6; (ii) 1(2r-1); (iii) 20-8r. 6. (i) 51; (ii) 70; (iii) 30. 8. 6- 5r, - 930.

136 11. 40.

7. (i) 2601; (ii) 17045; (iii) -1155.

9. 8r +1, n(4n + 5). 12. 36.

10. 1(5 r + 3), 4n(5n + 11).

14. 2, 4, 6.

17. 111, 20+.

18. 63.

Ex. 15. (i), (ii), (iv) and (vi).

Ex. 16. (i) 8; (ii) 7; (iii) 32.

Exercise 8(b) 138 1. (i) ;(412-1); (ii) -1(220 -1); (iii) (2+ ,/2) (28 -1); (iv) 1[3 -3-48]. 2. 1i, 2. 3. 18, 2i, 41, ... or -18, 2 5. 21.

,

-

6. 29.

7. 27.

9. 1.126x 108km.

139 11. £316.1.

4}, ....

4. 18. 8. 16.

10. 1, 1, 3.

12. £313.07.

13. (i) 19; (ii) 6; (iii) 3.

14. NI .

15. 23/ years.

16. {pa(pn -1)- an(p -1)} (p -1)-2.

17. ur =

18. 4(1- r) (1- en) (1+ r)-1.

140 Ex. 21. in(n + 1) (n+ 2) (n+ 3) (n + 4) (n+ 5). 141 Ex. 22.

1{(n + 1) (n + 2) (n+ 3) (n+

Exercise 8(c) 142 1. (i) 650; (ii) 6084; (iii) 2865; (iv) -}n(4n2 - 1). 2.

A-n(n + 1) (n+ 2) (3n + 13), -&n(n + 1) (n+ 2) (3n +17).

3. &n(n+ 1)(n+ 2) (3n +1). -

5. 1. 2n2(n2_ 1).

4. in(n3 +2n2 + 3n + 10).

7. i -1-(2n + 3) {(n+ 2) (n+ 3)}-1.

143 6. *n(n + 1) (2n + 7). 8.

- e(3n + 1) {(n + 1) (n+

+

9. 288 - M4n + 13) {(n+1) (n+2) (n+ 3) + 4)}' . n(n + 1) (n + 2) (3n +1). 10. (i) n(3n2+ 3n- 2); (ii) 1 (iii) in(n + 1) (3n + 1) (3n - 2); (iv) A-n(n + 1) (n+ 2) (n+ 3) (4n +1). 357

ANSWERS PAGE

143 11. +, -+; in(n +1)2(n+2). -

12. 1(2n +1) (2n + 3) {(n +1) (n+ 14.

-1{n(n+ 0)-1.

15. in(4n2 -1).

144 Ex. 23. 1, 6, 15, 20, 15, 6, 1; 1, 7, 21, 35, 35, 21, 7, 1; 1, 8, 28, 56, 70, 56, 28, 8, 1. 145 Ex. 26. (i) 81x4 -108x3y+ 54x2y2 - 12xy3 + y4; (ii) 64x6 +192x5y+ 240x4y2 + 160x3y3+ 60x2y4 + 12xy6 + y6; (iii) 128x' - 448x6 + 672x2 - 560x4 + 280x3 - 84x2 + 14x - 1; (iv) 243x2 +810x4y+1080x3y2 +720x2y3 +240xy4 +32y2.

Exercise 8(d) 146 1. (i) x6 - 6x2y+15x4y2 - 20x3y3 +15x2y4 - 6xy6 + y6; (ii) 32x6 +40x4y+20x3y2 + 5x2y3+ ixY4 + (iii) 1 + 14x + 84x2 + 280x3 + 560x4 + 672x6 + 448x6 + 128x'; (iv) 64(x4 +18x6y+135x4y2 +540x3y3 +1215x2y4 +1458xy2 +729y4). 2. (i) 54; (ii) 672; (iii) 264; (iv) 19440. 3. (i) 1 + 16x+ 112x2 ; (ii) 1 + 121x + (iv) - y7 + 21y4x- 189y4x2.

1x2 ; (iii) 256(4 - 20x + 45x2);

8 4. ( ) 2nx2r-8,1120. r 147 5. 27.

6. (i) 1.0615; (ii) 0.98411; (iii) 0.99501; (iv) 235.01.

7. 322.

8. 120, 4200.

9. (i) 1 + 4x + 10x2 ; (ii) 1- 6x+ 21x2; (iii) 1 +10x+ 35x2; (iv) 16(4-12x- 9x2). 10. ±1-1%.

11. 2n. (i) 81; (ii) 1.

Miscellaneous Exercise 8 1. (i) 1-(3n - 1); (ii)

3jn(n-1)

2.

A, 19.99998, 9.

148 3. e. 4. {nxn+i (n+ 1) xn +1)(x 1)-2, {n2xn}2 -(2n2 +2n -1) xn-o- + (n+ 1)2 xn- x- 1) (x 1)-3. -

-

-

5. in + 1\

6. 4, 12, 36; 2.3n-1.

kp + 11 149 12. {(1-bn) nYn(1 - b)} (1- b)-2. —

13. (r -1) digits 1, followed by r digits 0 and finally the digit 1. 358

ANSWERS PAGE

149 15. 627500. 150 17. (i) 15n(n+ 83)/2; (ii) 45(19n-40). 18720, 1Odn + 720n- 55d; 30. 18. (n-1)2.

Revision Exercise A 151 1. 5, 3; x

-

1.

5. 120, 24.

2. (-5, - 5).

4. (i) 1110; (ii) 17.3.

6. 40.9°.

7. 7 x +7y 29 = 0. -

8. 0, 4; (x2 + 2V2x + 4) (x2-2 \/2x + 4). 9. 3RQ-2BC, 3RQ - BC.

10. (i)

-

V5/3; (ii) -2A/5/5; (iii) 11. 12. 11, (3x+1)(x+1).

152 11. 5, 15.

16. 28, 1(n + 1) (n+2).

13. 0.036, 012.

-

18. 10 cm, 9.6 cm, 37.25°, 9.75 cm. 19.

1

21. 2x+ 3y+4z-9 = 0; 3A/13/13, -2A/13/13, 0. 153 23. (bc' + b'c)2(ca' - c'a)2= (ab' + a'b)2[(bc' + b' c)2+(ca' - c'a)2 ].

24. 1(5 - V97) < x < -1 or 1(5 + A/97) < x < 4.

25. in(n +1) (n + 2) (3n+ 7). 26. - 2 < x 21, -3 < x < 11, {2 < x < 3}, {1 < x < 2}, {- 3 < x 2}.

28. 1-2R.

27. 1.03.

29. - i -j - 3k.

30. 2E1 = E2 E3. 154 31. 4.

32. r = /4-1;), ab('; +b)/(a+ b+ c).

33. (i) 1(0+1), (ii) 2; 4n+ 1 is a perfect square.

34. 41 .. -§

CHAPTER 9 158 Ex. 6. ABC is right-angled at A. Ex. 8. Not 'if'.

Ex. 7. Not 'only if'.

Ex. 9. Yes.

Exercise 9(a) 160 4. 13.

5. 0 or 4.

10. No solution if a =

-

6. 13.

1 unless b = 0.

15. No: e.g. right-angle at B and M coincident with B. 359

ANSWERS

CHAPTER 10 PAGE

171 Ex. 2. 1000- X,

- X.

173 Ex. 3. 14.

Ex. 4. Lose.

Ex. 5. Yes, if x = 1.

Exercise 10(a) 174 1. (i) 2+; (ii) - -,÷); (iii) 311; (iv) 51. (iii) 6A; (iv) 10.

3. 3p. 175 8. 9p.

2. (i) 41; (ii)- 1+;

4. 13661.

5. £100.

9. 7p.

12. E1.32.

6. 0, 3.3. 13. 2.

14. (i) 8.75; (ii) 5.25.

3I 8 13 15 16 17

176 15. 27-Ip.

16. 22Ip.

17. X

178 Ex. 6. 2, 11.

Ex. 7. 1.

Ex. 8. (i) #+c; (ii) 0.2.

181 Ex. 9. 4, 0.

Ex. 10. g(X)3-3/2cr2-#3.

16p 2 3 3 2 3 3 •

183 Ex. 11. 1(1+ 03, 11-, 1. 185 Ex. 12. 2p.

Exercise 10 (b) 1. (i)

(ii) ft+ 1, a; (iii)

3#- 1, 30.; (iv) 0, 1.

2. (i) 2.81, 4.41; (ii) 3.7, 2.01; (iii) 0.15, 3.53; (iv) 150, 3000. 3. (i) 12.31, 1211; (ii) 15.7, 102.8; (iii) 3.55, 16.75; (iv) 2.55 x 104, 2.83 x 108. 2-(n2- 1). 5. (n+ 1), -i1

4. 41, 2+. 186 6.

9. (t5 + ts + t + 3)n/6n, in.

7.

10. 15 8 7, 2.415, 1.71.

11. il, 1166890, 11659•

,

12. 6.5, 2.36. (i)

(ii) 0.

Miscellaneous Exercise 10 1. £7.33. 187

2. 4p favours player, 4-Ip favours banker.

3. Just over £3. 4. -1*.

5. Yes, if X, Yindependent.

7. 2(n - r)l{n(n-1)}.

10. 0p-1.

11. 3(cr2 +#2 -#).

(\ 113\ + 18\ 113\ .4_ (7\ 113\ k 5 J k2/ k 6 ) \3/ k 7

188 14. 3-141113\ + lk 4 / kl/

6) (183) + (193)). + (4

16. 1, 71. 360

ANSWERS CHAPTER 11 PAGE

192 Ex. 1. Yes, provided a r 0, b 0. Ex. 2. 0A2. Ex. 3. (i) Pythagoras; (ii) Extension of Pythagoras (Cosine Rule).

Exercise 11(a) 193 3. Sphere, centre A, radius lel. 195 Ex. 4. -5.

Ex. 6. (3i + j + k)/411.

199 Ex. 9. 1, 1; lies on angle bisector. Ex. 11. 3x+ 2y - 10 = 0, (2, 2).

Ex. 12. ( a, -

-

b), lc'.

Ex. 13. 8x+ 6y + 13 = 0, 8x+ 6y - 7 = 0.

Exercise 11(b) 1. (i) 2, 1 ; (ii) 2, 1; 1; (iii) 3, I; (iv) b, (3a- 20/5. 200 2. (i) 245/5; (ii) 465/65; (iii) 807/85; (iv) (a2 62)1(a2+ b2)1. -

3. (i) 417/17; (ii) 12413/13; (iii) 6V10/5; (iv) (h+ k)2(h2+ k2)-i. 4. (i) 245/5; (ii) 945/10; (iii) 2426/13; (iv) (a + b) (a2+ b2)-I. 5. (i) 3x - 4y + 16 = 0, 3x - 4y - 14 = 0; (ii) 8x+ 6y + 35 = 0, 8x+ 6y - 25 = 0; (iii) 5x + 12y + 41 = 0, 5x+12y-37 = 0; (iv) 3x+ 2y+ 1 ± 3,/13 = 0. 7. (i) 2x - y = 0; (ii) 7 x 7 y 3 = 0; (iii) 128x+ 16y + 69 = 0. 9. (i) (9, -9); (ii) 521; (iii) +1. 8. x 7y-18 = 0. -

-

-

10. ( p, -q). -

201 11. x+ 2y- 11 = 0, x y + 1 = 0, arccos (8045/145), -

12. 37. 202 Ex. 14. (i)

-

2j+ 2k); (ii) (31 j + k)N11 ; (iii) (3i+ 2j+ k)/,/14. -

Ex. 15. (i) x- 3y+ z = -5; (ii) x - y+ z = 2; (iii) 2x+ 3y- 5z =

-

16.

Exercise 11(c) 205 1. 110.9°. 2. (i) x y+z = 0; (ii) 2x+y-3z = 13; (iii) 3x- z = 1; (iv) 6x-3y- 2z = 2; (v) 4x+ y - 3z = -1; (vi) ax + by + cz = a2+ b2+ c2. -

3. (i) 2470/35; (ii) 428/84; (iii) 2/29; (iv) 1/2. 361

ANSWERS PAGE

206 4. (i) 34/11/11; (ii) V15/15; (iii) V66/11; (iv) 174/58/174; (a2+62+ c2)-1. (v) ab(a2+ b2)'; (vi) lab bc -

-

5. (i) 4/7/7; (ii) 4/105/15; (iii) 1; (iv) 40/41; (v) V3/2. (vi) v{E(qi_pm)2}/v{(l2 +m2 + n3)(p2 q2 r2)}.

6. (i) 1(21- 2j +k), 2x - 2y +z = -1; (ii) -.1(3i -2j +6k), 3x-2y +6z = 0; 2 (iii) — (i+j -4k), x +y -4z = 1; 6 (iv) 211(41- j - 2k), 4x - y - 2z = 6; V26 (31- 4j +k), 3x- 4y+z =1; 26 (vi) (a 21 b2 j + c2k)/4(a4 + b4 +c4), a2x 62y + c2z = 0.

(v)

-

-

207 7. (i) (2, 3. -1); (ii) (4, 7, -5); (iii) (1, 1, 1); (iv) (0, -1, 3). 8. (i) (9, 5, 3); (ii) (7, -10, -3); (iii) (-6, 16, 6); (iv) (-9, 9, 12); (v) (15, 13, 2); (vi) ( 2ad(a2+10 + c2)-1, -26d(a2+62+ c2)-', -2cd(a2+ b2+ c2)-1). -

-

-

9. (i) (1, 1, 3); (ii) (2, 3, 4); (iii) (6, 2, 4); (iv) (6, 4, 1).

10. (8, 2, 0). 12. i(x

-

11. (1, 4, 3), i(x -1) = y-4 = z -3.

1) = - -1(y - 5) = i(z - 2). 13. x = y

-

3 = - (z+ 6).

208 14. 3: -1:2, 4/14. 16. 1(4/3-1), -1-(V3 + 1), 4/3(1- 4/3)16, V3(1 + V3)/6, f4/3.

17. 5A = 2,u, a = 51+ 29j+ 2k, b = -(i+ 5j +k). 209 18. 2x -y+ 2z = 9, arccos t. 241

19. (1, 5, 0), arccos 4/(151).

(i) (4, - 1, -3), (V, 1, - V); OD 1;

- D.

Miscellaneous Exercise 11 1. 1p x, cos a - yisin al. -

2. 3x+ lly+19 = 0, 11x-3y-17 = 0. 210 3. 14x+8y-57 = 0.

5. 3.75, 1.

6. (-4, 3).

9. 1 :2.

211 11. 60°. 16. -ix =

362

15. x+2 = 3(y+5) = z -5. -

;y =

-

z.

17. s+21n.s-p1(±n).

ANSWERS PAGE

212 18. x2 +y2= 1 ± 2y cot 0. 19. (i) x- 5y + 4z = 0; (ii) x+y+z = 0, (6, -2, -4). 20. -11, (1, -2, 3).

21. B(-1-, V312,0), C(-1, - 012, 0), D(0, 0, V2), (0, 0, V2/4); -4:0:V2, 2:±2V3:V2, 0:0:1. 22. 2. 213 23. 5x+12y- 60 = 0.

24. 2V6.

25. x = [p(62+ c2 - a2)-2a(bq + cr + d)] (a2+ 62+c2)-1, etc.; (-i, -1, D. 26. 15x-9y-3z = 0, 1:2:- 1. 27. E[(a2+ 62+ c2) l 2a(a1+ bm+ cn) x] = 0. -

214 28. y = z = 0, z = x = 0, x = y = 0; cy bz = O. 120°. -

29. arcsin 802, arccos 4/V82. 30. Intersection of planes with Z(m2n3 m3n2) x = 0. -

32. 142°.

33. Sphere, centre the centroid.

CHAPTER 12 215 Ex. 1. Yes, e.g. A = B = 60°. 216 Ex. 2. 4(V6- V2), 4(,/6+V2). EX. 4. (1)

a g;

Ex. 3. - i(V6+ V2), i(V6+ V2). (111)

-14, H.

Ex. 5. -2 sin A sin B.

217 Ex. 6. (a) 0 0 * (2k + 1) ; -

(b) 0 *inn ±

219 Ex. 7. (i) ± V10; (ii) ± V5; (iii) 10, 0; (iv) 1 ± V2; (v) 1(2 ± V2). Ex. 8. (i) 0, 270, 360; (ii) 53.1; (iii) 90, 323.1; (iv) 4.9, 220.5. Ex. 9. 1, 7. 220 Ex. 11.

Ex. 10. 1.

-4, 9V2.

Ex. 12. 14, 4. Ex. 13. in, in.

Ex. 14. V2-1. Ex. 15. 24(1 - x2), 2y2 -1.

Ex. 17. arcsin

221 Ex. 18. (1+ 2t- t2) (1 + t2)-1. 223 Ex. 19. (i) 2 sin 2x cos x; (ii) 2 cos 3x cos x; (iii)

-

2 sin 2x sin x;

(iv) V2 sin (4n+ x); (v) V2 sin x. 224 Ex. 20. 0, fir, in-, 7T, '47T, in-, 277.

Ex. 21. 0, in, 77, 177, 27T. 363

ANSWERS

Exercise 12(a) PAGE

224 1. (i) a(/6+ V2); (ii) - f(V6- V2); (iii) - (2 + V3); (iv) 2- V3.

2. (i) 71(,/5 1); (ii) *A/(10+ 2V5); (iii) 1(0+1); -

(iv) - i(/5+ 1) V(10+ 2,0). 225 5. (i) 28.6°, 208.6'; (ii) 120°, 300'; (iii) 37.5°, 217.5'; (iv) 18.3°,1983°. 6. (i) 225°; (ii) 90°, 330°; (iii) 29.5°, 256.7'; (iv) 22.6°, 90°.

7. (i) 0°, 116.6°, 180°, 296.6°, 360°; (ii) 180°; (iii) 45°, 120°, 135°, 225°, 240°, 315°; (iv) 0°, 45°, 90°, 225°, 360°. 8. 'sin

i; 9.3°, 50.7°, 189.3," 230.7°.

10. 4 cos3 0-3 cos 0, 8 cos4 0-8 cost 0+1. 226 13. (i) ± (a2 b2) (2ab)-1; (ii) a = -

b = 61; tan 0 =

16. R = a = 26.6°. (i) 140.5°, 346.3'; (ii) 0°, 63.4°, 180°, 243.4°, 360°. 17. -97.6°, 0°, 90°, 171.3°. 227 18. (Hi r, . 72;7 ), (

127T,

fin")•

Exercise 12(b) 232 1. 76 m.

3. (i) 84; (ii)

233 4. 1012°, 44.8°. 5. 14.8°.

(iii)

6. 9.3 m.

7. 36°

8. 127.8°.

9. (i) h(h2 + 2a2)-4; (ii) h(h2+a2)-1; (iii) 2ah(112 +a9-1; (iv) Al(112 2a2) N(2172+2a2). 10. (i) 35.3°; (ii) 64°. 234 13. 529 cm2.

16. 1.12r.

11. (i) 77 m; (ii) S 46° W.

12. 121 m.

14. (i) 64.9°; (ii) 69.4°; (iii) 31.6°. 17. 1.96, 56.8°.

235 19. arccos {(0-a4-b4 +2a2b2)1(2b2c2+ 2c2a2+ 2a2b2 a4 -

Miscellaneous Exercise 12 -15-; 236 8. art, r!-1- 10, 14. -1-(1 ± J2). 2.

-

a.

4. 2mr < x < (2n+1)7r, n e Z. 13. 0, fir, 7T. (1±, 1Th), ag, 1410.

15. tan la = alb.

237 17. c > 4. (i) 8 = 2nrr ± arccos 5, 0 = 2m7r+arcsin 1; arcsin = (2m+ 1) (ii) 8 = 2nrr ± arccos (iii) 0 = 2n7r± arccos I, q = 2m7r+ arcsin arcsin 5. (iv) 0 = 2nn.± arccos s,95 = (2m+ 1) -

-

364

ANSWERS

CHAPTER 13 PAGE

245 Ex. 9. S(P) = (3, 5), T(P) = (3, 0), U(P) = (6, 5). Ex. 12. Reflection in Oxy plane.

Exercise 13(a) 1. (-5, —10), (t, — 246 2. (-5, —4), (3, —5); (0, 0).

3. The line x — 2y = 0.

6. Reflect in x = y and then project on to x axis. 7. Reflect in x = y and stretch by factor 2. 8. lad — bc1; all points map into line cx— ay = 0. 247 10.

(1- r rs\ .

ii. (1 0 k i) , (1 0 _ 10).

q s ) —

(9 2 1 —12 4—8 18 —1 4 250 Ex. 13. (i) 2 1 2); (ii) (— 1 —3 9); (iii) (12 9 2); 4 5 —1 — 2 —5 —2 12 —1 2 22 16 —8 —1 51 3\ (iv) ( 2 0 3). Ex. 14. (i) ( 7 4 2); (ii) (__ — 5/ ' 9 4 7 6 —4 0

Exercise 13(b) 1. (0 (2 1 5 1 4\ . (i 253 6/i) ' k k -3 (iii) ( 4 7

1 8\ /3 13 8 8 Of ' (iv)1 ‘4 0 16)

0 4 2 2. (i) (4 1 6; (ii) 3 1 1 —6 (iv) ( 10 0 3. (i)

4\ 5 —2/ '

— 5 —2 —1 3 —3-3 ; (iii) —4 7 —8

—4 4 2 8 —1 6; 1 7 —5

4 2 —2 6) . 10 —8

i —104 . 1 4p (ii)

(12612 10) ;

(iii)

5

(-218) 8 '

4. (i) AB — 2AC; (ii) A2 — AC+ BA—BC; (iii) A2 + AB— BA + AC— CA— (B+ C)2. 365

ANSWERS PAGE

5 253 5. ( — 5 8

0

2

6 3 1)' (1 6) •

_ 15

5 6. (i) — 2 8

/8 (iv) (0 5

0 0 1

0 3 ( —5

254 8. (i)

—10 — — 3/

—1 (ii) (-3 3

20 (v) ( 2 —1

8 2 —3

—3 —1 0

3 5 2 4 1); (iii) (-2 0 0); 2/ 4 0 4

0 —18 —6); (vi) (— 6 19 —8

3 8 0 1 4 5 6 — 3 3 5); (ii) ( 5 4 5); (iii) ( 1 0 3 1 —7 1 2 —1

7. (i)

(iv)

3 0 0 1); 2 —5

5 11 —9

2 6 1 ; (v) —11 2 19

—1 6 4 1 ; (ii) ( —1 4

1 9. (i) ( 3 —3

2 —6 4

—3 4 —2

—19 —38 39

—9 —1 ; (vi) —2

2 —1 ; (iii) 5

—1 2 1

3 3 6

—7 —2 —4

1 6 3

—1 3); 6

10 —27 33

—11 —40 43

30 5). 19

—19 —1 —4

6 — 2. 9

2 .. ( —7 —\ 0); (ii) 3 11; —1

—1 0 5 —14 (iii) ( 21 6); (iv) 6 (-2 —17 — 5

— 2\ 1) •

0 0 a 10. (x y z), (b 0 0 ; T preserves lengths; each ± 1. 0 c 0

11. 2x+y z= 0, (-1, 7, 5). —

12. x y+z = 0, (5, 8, 3). —

255 14. (0, 0), (2, 2), (-1, 0).

Miscellaneous Exercise 13 2. Shear; x, z coordinates fixed. 3. (i) (0, 0) only; (ii) line 2x —y = 0; (iii) all points in the plane. 256 5. (0, 0) only; no. 6. (i) (0, 0) only; A= 10: line 7x 5y = 0, A= —2: line x+y = 0; (ii) all points of line x y = 0; A= 6: line 4x +y = 0. -

—

366

ANSWERS PAGE

256 8. AB = BA. 9. A diagonal, provided non-zero elements of D distinct.

257 11. No.

CHAPTER 14 262 Ex. 9. (i) (iii)

—

3

(—

I 2 \-1 —lf'

2\

51'

—7

15 2); (iv) (_

263 Ex. 10. (i)

0 —1

2-3); (ii) —2 6

0 —1

— 0 (iii) ( 1 5 —4); (iv) ( 0 —4 0 —3 2 —1

—5 17

2; —6

-1 3). —3

Exercise 14 (a) 264 1. (i) I 2

k -3

(iv) ( -111,1

—1\

2) ;

-15r 3 ;

-;

(ii)

(v)

Tr.

; (iii))

h

—h-1 \ h j•

k- h+1 2

—2

2. As in Qu. 1. 4. (i) (3 — 8 —4 —1 —1 1 _3 2 I -2

(iii) (- I

131 I — I); (iv) ( 1 1 —1

(_1

5 ; (ii) 1

11

—12

—8 7

9 —8

?, I- —1 -1 -1); (v) (— i 1 -1 -i

—7

5 ; —4 18,9_

—3 1 _13); I

i

0 -1 1 (v1) ( 1 + -1). 3 -1 0 'I —1 265 7. ( —1 2 9. 6.

5 2 —6

—4 0 3),

—5 8. ( 7). 6

v = (-160 5).

—9 10. (-3

10

—1

29

—1 2

14). —35 367

ANSWERS PAGE

267 Ex. 11. (i) 3; (ii) 5; (iii) 7. Ex. 12. (i) 2; (ii) —3; (iii) 0; (iv) —3; (v) abc+2fgh—aft bg2 —ch2. 273 Ex. 13. (i) 12; (ii) —9; (iii) 0, (iv) 1.

Exercise 14(b) 275 1. (i) —2; (ii) —1; (iii) —1; (iv) 22. 2. (i) —30; (ii) 0; (iii) 0; (iv) 1; (v) 24; (vi) —9; (vii) 3; (viii) —12; (ix) 8; (x) 6. 3. (iii) Singular. —2 —5 276 4. (i) ( — 2 4 4 —3

3 —2 , —21, —2; (ii) 1

—14 —14 14 16 16 —16), 0, 0; —27 —27 27

30 —11 5 0 —1 2 (iii) (-10 4 —2), 21, 2; (iv) ( 1 —1 —3), I, 1. 2 —1 1 —2 3 5 —2 10 5. (i) (— 5 1

1 —2

—2); (ii) 5

A(

7 1

15 1 —5); 13 —5

( 0 —6 9 0 —1 2 4 —4). (iii) —4 —1 4 —5); (iv) ( -4 —1 2 —5 4 0 2 6. (i) 12; (ii) 1; (iii) —10.

7. (i) (b— c) (c— a) (a— b);

(ii) (b — c)(c— a) (a — b) (a + b + c); (iii) (b— c) (c— a) (a — b)(bc+ ca+ ab). 8.

5 —10 5— 4 —3 (

—4 0. 0 —9

9.

—3 33 4 0— 8 0). ( 2 — 2 —1

66 36 100 —173 —95 . 277 10. ( —39 147 249 136

Miscellaneous Exercise 14 1. ( 1 10). 1

2. (a) Yes; (b) yes; (c) yes.

ac () a2 ab ac a2 ab 3. (ba b2 bc ; ba b2 +1 bc+2 . ca cb c2 ca cb + 2 c2+4 278 6. Yes.

368

9. (b — c)2(c — a)2(a — b)2.

ANSWERS PAGE

279 13. A2 = 2A +I, A-1= -A+ 2I. 2 280 17. \2 1 1 8

-2 1 1 - 2 , 91, 2 2

2 1

1 2

-2, 2

-7 8 -7 - 4). 4 1

CHAPTER 15 285 Ex. 2. (ff, -A). 286 Ex. 5. True for any x provided b = 0.

Exercise 15(a) 1. (i) (2, 5); (ii) (- 3, 5); (iii) 2. (i) If a * - b, x = b- a; if a = - b, true all x. (ii) If at b, x = (c- b) (a - b)-1; if a = b * c, no solution; if a = b = c, true all x. 3. (i) x = (4+ 3a) (3a- 6)-1, y = 5(2-a)-1, provided a * 2. If a = 2, equations inconsistent. (ii) If a * 4, x 0, y = 2; if a = 4, x = 2(2 -A), y = A, all A. (iii) If a * ± 6, x = 2(6 +a)-1, y = (6 +a)-1 If a = 6, x = A, y = 6(1- 3A); if a = - 6, equations inconsistent. 4. (i)If a * - 3b,x= (b2+ ab + 8)(4a+ 12b)-1,y = (24 - ab- a2)(4a + 126)-1 . If a = -3h , 1,2 *4, equations inconsistent; if b = 2, x = A, y = 3A+ 1; if b = -2, x = A, y = 3A-1. (ii) If ab * -3, x = 6(1 + b) (ab + 3)-1, y = 2(3-a) (ab+ 3)-1. If a = 3, b = -1,x = A, y = 2 - A ; if ab = - 3, a * 3 equations inconsistent. (iii) If a2 * 62, x = (a - b)-1, y = (b - a)-1. If a = b, equations inconsistent; if a+b = 0, x = A, y = (aA-1) a-1. 5. Inconsistent unless a = 3 or -1; if a = 3, x = - 2, y = 3; if a = -1, x = 2, y = -1. 6. Inconsistent unless a = 1, 2 or -3. If a = 1, x =lsl,y = I; if a = 2, x = 2, y = 0; if a = -3, x = y = 287 7. (i) a * 0, b * 0, a * b, x = c(b- c) a-1(b- a)-1, y = c(c- a) b-1(b - a)-1; (ii) a = 0, b * 0, c * 0, equations inconsistent unless b = c, when x = A, y = 1; (iii) a = 0, b * 0, c = 0, consistent, x = A, y = 0; (iv) a * 0, b = 0, c * 0, equations inconsistent unless 369

ANSWERS PAGE

287 a = c, when solution is x = 1, y = A; (v) a * 0, b = 0, c = 0, consistent, x = 0, y = A; (vi) a = b * 0, c * 0, inconsistent unless a = b = c, when solution is x = A, y = 1—A; (vii) a = b = 0, c * 0, inconsistent; (viii) a = b = c = 0, consistent, x = A, y = ,u.

8. al bs —a2bi = 0, x = Aar', y = 9. If 0 * kff +0 (k integral), x = sin (q5— a) cosec (0— 0),

y = sin (0 — a) cosec (0— 0); if 6 = +0, equations inconsistent unless 0 and q5 both differ from a by an integral multiple of 7T. If 0 = a+ 2mn, q5 = a+ 2nn, x = A, y = 1—A; if 0 = a+ 2nvr, = a + (2n+ 1) n, x = A, y = A-1; if 0 = a+ (2m+1)7r, q5 = cz+2mr, x = A, y = A +1; if 0 = a+(2m+1)7r, q5 = a+ 2(n+ 1)n, x = A, y = —1—A. 292 Ex. 12. (i) Lies in plane 01:1'K'; (ii) can be expressed in terms of any two of i', j', k'. Ex. 14. Columns of A proportional; (i) lies on 01'; (ii) Ai'.

Exercise 15(b) 294 1. (-2, 0, 4). 295 2.

(1 6 5 1 1 2I. (i) (0, 1, —1); (ii) (2, 0, 1); (iii) (-35, —5, —7). —1 1 —1

3. Any point of line 1-(x-1) = -}(y + 1) = +(z+ 2). 4. (i) Line A, 1(6— 7A), 1(13-11A); (ii) inconsistent: planes form prism in direction 5i — 7j -11k.

5. b = 10, line; b * 10, inconsistent. 6. a = 2, inconsistent; a * 2, consistent only if b = a +2, giving line. 7. (2, 1, —3). 296 8. a = 5, inconsistent; a = —1, line; unique point otherwise. 9. 4, (0, A, 2A).

10. (A, A, — 2A), (1+A, A, — 2A).

11. 4-(x+ 6) = y = — A-(z— 12), parallel.

297 14. Tetrahedron 7x-6y-5z > 0, x — 2y — z < 0, 9x-8y-7z < 0, 3x-4y-3z > — 2.

Miscellaneous Exercise 15 1

6 3 1), (-2, —1, 1), (4,1, /). I 0 —2 —1 (

1. 5

2. (i) Line or plane; (ii) inconsistent.

370

ANSWERS PAGE

298 3.

- 4).

5. A = b - a, B= c- a, C (b - c) (c- a); x = (b2 + c2) (a - b)-' (c - a)-1, y = (c2 + a2) (a - 6)-1(b - c)-', z = (a2 + b2)(b- c)--1 (c -

299 6. (1, 1, 1), (A, 3 2A, A). -

9. (b - c) (c - a) (a - b), x = ab(b - c) - a)-' (a-b)-1; y = a2(a - b)-', z = ab(c - a)--' ; a = b a = b = c,

300 12. -4, 3, 2, 13. a

-

2b

-

c,

inconsistent unless a = 0; a

b = c,

line;

plane. -

11(1+ A)/7, 5(1 +A)/7, A.

c = 0, (A, 1

14. A * 4-, - 1,

-

2A, A

-

1).

unique; A = 1, inconsistent; A = -1, line.

CHAPTER 16 306 Ex —X•. 2. n

n n 09 1,, 5,01

6 5-,nO.

Ex. 4. F8,1.44.

Ex. 3. (i) 2.5; (ii) 2. Ex. 5. I, 116, 7 to 1.

Exercise 16(a) 1 .

16 s i, ,

, Il

4. 9, 0.387. 307 6. 6]15', M.•

2. 0.348, 0-071.

(ii) 0.566.

5. (i)

7. (9- r) (Sr +5)

9. p9(10 - 9p), £3.52, 46.8%. 311 Ex. 7. (i) +16;

3. 0.000105, 0.00605, 0.201.

(i) 1'5; (ii) 1; (iii) 0.46. 11. (i) 0.388; (ii) 0.0386.

1;4;(iii) 12. Exercise 16(b)

313 1. 0.135, 0.271, 0.271, 0.180, 0.090, 0.036, 0.017. 2. 1.45; (a) 0.06; (b) 0.33. 314 3. (i) 1; (ii) 0.846; (iii) 0.013.

4. 0.167, 0.594.

5. 0.030, 0.106, 0.185, 0.216, 0.188, 0132, 0.077; 3, 4-25. 6. 3; 60, 222.

315 9. No: Pr (X

7. 20, 0.029.

21)

0121.

11. a(r + 1)-1, 2.

8. 0.302, 0.060. 10. (A2 +.1)/(1 -e-A).

12. 0.0030, 0.0379.

13. 226.8, 211.4, 98.6, 30.6, 71, 1.5. 371

ANSWERS

Exercise 16(c) PAGE

321 1. (i) 0.63; (ii) 0.61. Pr (2 or more)

0.09.

2. 0.79.

3. (1 -p)10 {1 +10p(1 -p)3}, 13.15. 322 4. 0.9860, 0.0119, 0.002.

5. No.

6. Pr (r < 41n2 = 10) = 0.029; yes.

7. Pr (3 or more sixes) = 0.062.

8. Yes; no (p 'A,' 1/163). 323 9. Not significant in either case. 11. (i) (1 -p)14(1+14p) {1 + 105p2(1 -p)13 +455p3 (1 p)12}; (ii) 0.164; (iii) 17.5. -

12. £2.25, £1.50.

Miscellaneous Exercise 16 1. (i) 324 3.

76; 00 4; (iii)

H.

2. 1, 0.335; 20.

5. Yes, if p < 1; no otherwise.

4. 8.

6. IN(N+ 1) (2p-1).

8. s(2pq)n (p2+q2).

-

. FP.

9

325 11. 0.3. 12. 20, 60, 90, 90, 67, 40, 33. (i) £157.50; (ii) £167.50; (iii) £160.75. 13. (1 pr(1 p)n-', -

(s+s lc)(s±n k ) (rs)

p"-0q'(1- p)n-s-k (1 - q)k,

(s) (pq)n (1- pq)n-s. 14. Q(r, s) = ( ) s

ars.

15. (i) 99.1%; (ii) 0.03; (iii) 0.1. 326 16. 0.011, 0.870, 0.516; 23.7.

17. 0.585.

18. e-F, u, e-(ii+v)prV3{115!)-1. 327 20. 02; no.

372

21. p(n- r +1)(r - rp)-1, rm= [p(n+1)].

ANSWERS

Revision Exercise B PAGE 4. 3(4k+1) r.

328 1. x 2y = 0, 4x + 2y — 15 = 0. —

7. 3.

5. 9(10+V10), 7.

8. No.

329 9. (2x —y + 1) (x+ 3y-2), two lines; (i) two lines; (ii) origin. —8

11 — 10 12 — 9). (i) (20, 20, —23); 11 10 —13

(_ 137, _5, i 32).

12. 1( — 9

13. 1 — 48x + 1104x2, 0.6158.

14. (i) 37x-27y = 0;

(ii) 31x+27 = 0. 15. (i) 3x + 4; (ii) 5x+7.

1 —2 — 2 1 —2). —2 —2 1

19.3- 2

330 16. a ft = (2k +1) 7. 21. About I, 69.

20. 3:1.

23. (i) 0; (ii) 2(q r) (r p) (p q) (y z) (z —

25. -A- < p
C is defined by f(z) = z*. For what subset of C is f the identity function? Ex. 12. Find: (ii) (i) 13 +

(iii) 11/(1 — DI, (iv) 1cos + j sin B1, (v) 11—cos 0+j sin O.

Ex. 13. The word 'modulus' has, prior to its use in this chapter, been used in the context of real numbers only, to mean Jae. Show that the two uses coincide if R is regarded as a subset of C. *Ex. 14. If z1, z2are two complex numbers, and z2* 0, prove that: (i) (zi+z2)* (ziz2)* = 44, (iii) zi z;' = 14 2, 1)* 1 z2 12 , (v) z1+4 = 2 Re (z1), (iv) Z2

Z8

i z2i

(vi) zl - zi = 2j Im (z,).

The results proved in Ex. 14 are of great importance and should be committed to memory. The concept of modulus enables us, in a sense, to order complex numbers, but this is not entirely analogous to the ordering of real numbers, since different complex numbers can have the same modulus. It is not possible to order the complex numbers in the same way as we order the reals. Exercise

17(a)

1. If z1 = 1 +2j, z2 = 2—j, z3= 4+ 5j, express as complex numbers in the standard form a+ bj: (i) z,+ z2+ z3; (ii) 3z1— z2+ 2z3 ; (iii) z1i2 ; (iv) +2z3 ; (v) z2z3+z3 z1+z1z2; (vi) z2 zs; (vii) (z1— jz2)/(z2 + 21z3) ; (viii)

+ 1--; (ix) (z2 + Z3)3 ; z

1

(x) (Z1— Z2) (Z2 iZ3) ; (xi)

AZ? +:1Z2) ; (xii)

— Z2 Z3)/(Z2 — z3 z1.

2. If z1= cos 01+j sin 01, z2 = cos 02 +j sin 02, show that z1z2 = cos (01+ 02)+ j sin (01+ 02), and find z1jz2. Evaluate zi2 and (z1z2)2. 3. Solve the following equations: (i) x2— 4x + 5 = 0; (ii) 2x2— 2x + 1 = 0; (iii) x2 — 5x+ 7 = 0; (iv) jx2 — 2x-2j = 0.

383

COMPLEX NUMBERS (1)

[17

4. If (5 —12j) = (a+ bj)2where a > 0, find a, b. Find similarly complex numbers which when squared give (i) j; (ii) 3 +4j. 5. Solve the equation x2 — (4+ j) x + 5 —j = 0, giving each root in the form a+ bj. 6. Given that A(3 +2j)— B(1 — j)— (5+2j) = 0, find A and B: (i) if A, B are both real; (ii) if A, B are conjugate complex numbers. 7. If z = 1— cos 0— j sin 0, write down the value of z*. Express z-1as a complex

number in standard form. 8. If a is a real number and z is a complex number, prove that (1 + az) * = 1+ az*. Deduce that, in standard form, (1 + az)-1= (1 + az*)I(1 +2a Re (z)+a21z12).

9. Given z e C, show that z is real a z = z*. 10. What is the conjugate of z*? By considering the product zi z: for any pair of complex numbers z1and z2, prove that zi z: + z:z, is real. What can you say about the complex number z1zz -zIk z2 ? 11. Show, by constructing an example, that non-real numbers a and b may be found so that the quadratic equation x2 —ax+b = 0 has a real root. Is it possible for the equation to have two real roots if a, b are non-real? 12. Form the quadratic equation whose roots are: (i) 2 + j, 2 — j; (ii) 2 — 3j, 2 + 3j; (iii) 2 — j.,./3, 2 +iN/3; (iv) 1 +2j, 2 + j; (v) 4, 1+j. Can you conjecture any general result about the coefficients of a quadratic equation and its roots? 13. By writing x2 + y2in the form x2 — (jy)2, split x2 +y2into linear factors with coefficients in C. Factorize into linear factors with coefficients in C: (i) x2 — 2x +2; (ii) x2 + 4y2; (iii) x2 + 3y2; (iv) j(x2 + x + 1) — 1 ; (v) j x2 — 4xy — 4jy2; (vi) x3 + 1; (vii) x4 +1. (Hint for (vii): the expression may be written (x2+ 1)2— 2x2.) 14. By putting x equal to j in the identity x4-3x3 —x+2 E_ (x2 +1) Q(x)+ ax+b determine the numerical values of a and b. Find the remainder, on division by x2 +1, of (i) x9 +1; (ii) x9+ 1. 15. Evaluate inin the four cases n = 4m, 4m+ 1, 4m+2, 4m + 3 (m integral). Find the sum of the series n

jr

.

r=0

16. If (1 +j)" = a+ bj, prove that a2 + b2 = 2n:

(i) by taking the complex conjugate of each side of the original expression; (ii) by mathematical induction.

384

2]

MANIPULATION OF COMPLEX NUMBERS

17. Evaluate the determinant 2 1 +j 2 1—j —1 4—j 2 3 4+j where j2 = —1. Explain how without evaluation, it could have been concluded that the value (M.E.I.) of the determinant was real.

3. THE ARGAND DIAGRAM

As forecast in Section 1, we may set up a 1-1 correspondence between the set C of complex numbers and the points in a plane by associating the complex number x+yj with the point P whose coordinates are (x, y) referred to a given pair of rectangular coordinate axes. The x axis corresponds to the set R of real numbers, the y axis to the set of pure imaginary numbers, while a general complex number a+ bj, a + 0, b + 0 lies off the axes in one of the four quadrants. Ex. 15. Plot the points corresponding to the complex numbers 1 +2j, 1-2j, —j, +j. What is the geometrical relationship that exists between the points representing z and z*? Demonstrate geometrically the result z z* z e R.

Such a geometrical representation of complex numbers is generally referred to as the Argand diagram (J. R. Argand, 1768-1822)t or as the complex plane. As we saw in Chapter 3, there is a 1-1 correspondence between the set of points P(x, y) and the set of position vectors OP = xi +yj. Thus, the complex number x +yj may be alternatively represented in the Argand diagram by the point P or by the position vector of P. Both representations have their value, and we shall use both freely. *Ex. 16. If P represents the complex number a + bj (P is the affix of the complex number a+ bj) show that la+ bj1 =

Consider now two complex numbers z1 = ai+ bi j and z2 = a2+b2 j, with affixes P1and P2. The sum z1+z2 = (ai+ a2)+(b,+b,) j; but OP1+OP2 = (di+a2) i+ (bi+ b0 t The first exposition of the geometrical treatment of complex numbers was in fact published in 1797 by a Norwegian surveyor, Casper Wessel (1745-1818). For a translation of his paper see The Treasury of Mathematics: 2 by Henrietta Medonick (Pelican).

385

COMPLEX NUMBERS (1)

[17

and thus the position vector representing the sum of two complex numbers is the sum of the position vectors which separately represent the two numbers (see Figure 17.1).

Fig. 17.1

Fig. 17.2

*Ex. 17. Describe the vector representing the complex number z1 — z2. Ex. 18. Mark in the Argand diagram the affixes of the complex numbers 2—j, 2(2—j) and 1(2—j). Interpret multiplication of a complex number z by the real number (i) a > 0, (ii) b < 0, in terms of an operation upon the vector OP representing z in the Argand diagram.

Ex. 18 shows that multiplication of a complex number z by a positive real number a is represented geometrically by an enlargement of OP (possibly by a factor less than 1). Multiplication by a negative real number —a both enlarges OP and rotates it through an angle n.(see Figure 17.2). *Ex. 19. Mark in the Argand diagram the affixes of the complex numbers 3 + 2j, j(3 +2j), — (3 + 2j), — j(3 + 2j). Interpret multiplication of a complex number z by the pure imaginary number j in terms of an operation upon the vector OP representing z in the Argand diagram. Show that multiplication of z by the pure imaginary number bj is represented in the Argand diagram by an enlargement of magnitude I bl followed by (i) an anticlockwise rotation through -flr if b > 0, or (ii) a clockwise rotation through ig if b < 0. Ex. 20. Interpret geometrically the statement j2 = —1.

To interpret geometrically the multiplication of two general complex numbers, it is convenient to introduce a new concept, the argument of a complex number z. Given a complex number z = x+yj, represented by the vector OP, we have x = r cos 0, y = r sin 0, where r = = V(x2+ y2) (see Figure 17.3). The two real numbers, r and 0, determine the complex number z uniquely; conversely r is determined uniquely by z, but there is an infinite 386

THE ARGAND DIAGRAM

3]

number of values of 0 corresponding to z, differing from one another by integral multiples of 2n. Each such value of 0 is called an argument of z. Just one of these values will lie in the interval — 77 < 0 577 , and this value is called the principal argument of z, and is written arg z. If arg z = 0, we may write z = r(cos 0+j sin q).

Fig. 17.3

A complex number may always be stated in this modulus-argument form. For example 1 1+j = V2 (T2

j)

V2(cos

+ j sin in.),

1 — j = V2(cos [ in] +j sin [--Pr]), —1 +j = V2(cos +j sin 47r), —1—j = V2(cos [— in] + j sin [—fir]). (Notice that we must always be careful to select the angle to lie in the range - 77 < arg z n.) Example 1. Express in modulus-argument form the complex numbers (i) — + j, (ii) 2—j, (iii) 1+j cot a (-7T < a < 77 and a + 0). (i) I — V3 +ji = 2 and we have — 4/3+j = 2 (-1+2-j) = 2(cos 61T +j sin fir). (ii) 12— ji = J5 and we have 2 — j = ,15(75 — 5(cos + j sin a), where a is the angle in the range —in < a < 0 such that tan a = — 387

COMPLEX NUMBERS (1)

[17

(iii) First observe that

11+j cot al = V(1 + cot2 a) = lcosec If 0 < a < Ir cosec al = cosec a and we write ,

1+j cot a = cosec a (sin +j cos a) = cosec a (cos 0+j sin 0), where q5 is chosen so that

—
co r=1

r!

xr = (1+x)'-1

for any real n, provided lx1 < 1. The result is also true for x = 1, provided n > —1 and for x = —1 provided n > 0. It is not true for any Ix! > 1, unless n is a positive integer or zero.

417

[18

POLYNOMIALS AND PARTIAL FRACTIONS

(ii)

1 = (1 — x)--1(1 +x)-1 (1 — x) V(1 + x) (1 + x + x2 + x3) (1 —ix+ ix2 --hx3) 1 + x(1 --1-)+ x2(1 -1+ + x3(1 -1+ 8-16) 1 +-ix+8x2+126x3.

Alternatively, we may write 1 — 1 + ax+bx2+cx3. (1—x) V(1 + x) — Then, using Ex. 26 (v), we have 1— + 8x2 — 6x3 :(1 — (1+ax+bx2+cx3),

a-1 = —2, b— a = 8, c—b =

giving

or

a-2, b =

c= -6

as before.

Ex. 28. Solve Example 7 (i) by the method of Example 7(iD. Example 8. By using the binomial expansion for (1 — 2x)- - , estimate V5 and calculate an upper limit for the error in your approximation. Since (1

-

10)-1= 1/V(1) = 1V5 we have (see Ex. 26 (v)) 2 x1

= 1— ( -110 +

(--116)2 I*;,* 5( 110)3 +

1.3.5.7

1 3. 5 .7.9 ( ( 1104+ . 5 !

1 5

1+0.1+0.015+0.0025+0.00044+0.00008. Thus 21/5 1.11802 and V5 2.23604. To estimate the accuracy of our answer we have to obtain an upper limit for the sum of the remaining terms in the series. Now the coefficient, u„ of (A)' in the expansion above is given by

= and we have Thus 418

ur ur_1

1.3.5

. (2r — 1)

r!

2r-1

1 — 2— - < 2.

< 2u,_„

4]

POLYNOMIAL APPROXIMATIONS

and the sum of the remaining terms in the binomial series above is certainly less than 1.3.5.7.9.11

6!

( A)6 [1+ 126+ Ito+ dciv+...]

= 1.3.5.7.9.11 (m) 6!

1 Lr11 - (sum of infinite geometric series)

1.805 x 10-5. Since 4.4 x 10-4and 8 x 10-5are overestimates of the fifth and sixth terms of the series above we may safely conclude that the error in taking 1.11802 as 10 is less than F805 x 10-5and so our answer of 2.23604 for has an error of less than 3.61 x 10-5. Ex. 29. In fact, V5 = 2.236068 (6 d.p.), and our limit for the error in Example 8 appears to have been rather crude. Explain why this is so. How could a more accurate estimate of the error be obtained? Ex. 30. To 4 d.p., J5 = 2.2361, while in Example 8 above, we obtain V5 = 2.2360 (to 4 d.p.) although we subsequently show our error to be less than 5 x 10-5. Explain. The binomial series gives a polynomial approximation to (1 +x)n valid in the neighbourhood of x = 0. If we require a polynomial approximation valid in some other neighbourhood, it is necessary to shift the origin, as shown in our next example. Example 9. Obtain a quadratic approximation to (1+2x)-1in the neighbourhood of x = 3. (1 +2x)_1= [7+ 2(x — 3)]-1 =1 7

+ _221 -1

where y= x 3 -

7 j '

,z 1 7 2y+ 44y92) (1—

since y is approximately zero

= 7 491x 3) + 3 : 3 (X

3)2*

The quadratic approximation is best left in this form but it may be reduced to 4

-33485X ± 3 : X 3 2.

Notice that this is quite different from the quadratic approximation to (1 +2x)-' in the neighbourhood of x = 0: 1 — 2x + 4x2. 419


[18

Exercise 18(c) 1. Obtain cubic polynomial approximations for the following expressions in the neighbourhood of x = 0: 1 1 1 1 • (iii) (i) 1 (ii) (iv) (1 _ .02; (v) — x)3. (1 — 2x)2; 3 —x' 2. Obtain cubic polynomial approximations for the following expressions in the neighbourhood of x = 0. 1 1 (i) ; (ii) V(1 4x); (iii) V(4 x); (iv) ✓ (v) V(21 V(1-2x) (4 x)' x)• 3. Obtain cubic polynomial approximations for the following expressions, in the neighbourhood of x = 0: (i) (1 + x)*; (ii) (1 +2x)-4; (iii) (1— x)-1; (iv) (2 — x)-5; (v) (8— x)}.

4. Obtain cubic polynomial approximations for the following expressions, in the neighbourhood of x = 0: 1 1 1 • (iii) (i) (1 — x) (1 — 2x) ; (x — 2) (x — 3) (1 + x)2 (2 + x)' x(x+ 3) 1+4x (iv) ; (v) (1 + x2) (1+2x) (1 — 2x)2(1 + 2x2) • 5. Obtain quadratic polynomial approximations for the following expressions in the neighbourhood of x = 0: 1 1 1 (i) (ii) • (iii) • V(1 5x + 6x2)' (1+ x) (1 + x)' (1 + x2) V(1— 2x)' —

1(1— x) ' 1 (v) 1+x (1 + x)2V (1 + 2x) •

. vr

6. Obtain quadratic polynomial approximations for: (i) (1+ x)-1in the neighbourhood of x = 1; (ii) (2— x)--1in the neighbourhood of x = — 1; (iii) V(3 +2x) in the neighbourhood of x = —1; (iv) (2+3x)* in the neighbourhood of x = 2; V(3— x) (v) in the neighbourhood of x = —1. 7. Prove that

1

V (1 + x2) + 1 and deduce that, if x is very small, 1 V(1 ± X2)± 1

Al(1+x2)-1 x2

1X2.

8

8. Find, correct to 5 significant figures, the values of (i) (0.998)1/3; (ii) (1.02)112 ; (iii) (4.01)1/2; (iv) (0799)0. 420

4]

POLYNOMIAL APPROXIMATIONS

9. By substituting x = 1/1000 in the expansion of (1 — x)113, find (999)1/3correct to 4 significant figures. 10. Prove that 19.97 < A/399 < 19.98. 11. Prove that, if

E=

x2 2—x-24(1 — x)'

then E = 2— x+2,/(1— x). Deduce that, if x is small, then E is approximately equal to 4x2. 12. If x is so small that x3and higher powers of x may be neglected, express the

function

A/(4+x) 1— 2x+ (1 + 3x)2/3

in the form, a+ bx+ cx2.

(0 & C)

13. Express the function E given by

E=

x+ 3 (2x + 1) (1 + x2)

in partial fractions. Hence prove that, if x is so large that x-4can be neglected, then 5 + 2x

E=

4x3

(0 &

14. Use the binomial expansion to calculate the value of (16.32)04correct to six places of decimals. (O & C)

Miscellaneous Exercise 18 1. The polynomial x6 + Ax6+Ex4 — Axs + Ax2+ Bx+ C is exactly :divisible by xs +1; find A, B, C. 2. If (1 + x+ x2)n

E

a0+ ai x+ a2 x2+ ...+a2n X2n, write down the values of ao

and a2„ and prove that a,. = 2n

a2n-r. 2n

Show that E a,. = 3" and find E (— 1)rar. r =1

If n is an even number, prove that E (— 1)r±1 a27_1 = 0. r=1

3. Find the value of the constant A for which the expression

2x+ 3 —A(x2 +x+ 1) has (x -2) as a factor and find the remaining factor. Hence, or otherwise, put into partial fractions the expression 2x + 3 (x2 + x +1) (x-2)

(0 .& C)

421


4. Prove that

[18

x2+2bx+1 (x—a)(x+2b+a)+a2+2ab+1. 1 (x-2) (x2 + 2bx + 1)

Put into partial fractions (i) when b = (ii) when b = 1.

(0 & C)

5. Find A, B such that, for all values of x other than 1, — —1,

A(x+1)+B x+1 Ax+B (2x— 1) (2x + 1) (2x + 3) (2x-1) (2x + 1) (2x + 1) (2x+ 3). Find the sum to n terms of the series whose rth term is r +1 (0 & C) (2r —1) (2r +1) (2r+3)' 6. Prove that, if —1 < x

1— x + Deduce that

1.3 —

1.2

x2

1.3.5.7 .3.5 x4 + x3 + 1.2.3 1.2.3.4 1

1.3.5.7 1 1.3 1.3.5 + 1+ + + + 4 4.8 4.8.12 4.8.12.16

— (1 + 2x)-112. V2.

7. Sum to infinity the series:

1 1.3 1.3.5 1.3.5.7 3 + 3.6 + 3.6.9+ 3.6.9.12 +..., 1 1.4 1.4.7 1.4.7.10 (ii) 1+ 4 + 4.8 + 4.8.12+ 4.8.12.16 +—; 3 3.9 3.9.15 3.9.15.21 (iii) 1 8 + 8.16 8.16.24+ 8.16.24.32+—; 1.5.9 1.5.9.13 1 1.5 (iv) 1+ + + + +.-. 6 6.12 6.12.18 6.12.18.24 (i) 1+

8. The cubic polynomial P(x) assumes the values 2, 3, 2, 11 respectively for x= 1,2,3,4. By writing P(x) E A+ B(x-1)+ C(x-1) (x— 2)+ D(x— 1) (x — 2) (x-3) findP(x). Show how to fit a quadratic polynomial to coincide with the values of y = sin x at x = 0, in, n. 9. Find a linear approximation to J(3 — x) in the neighbourhood of x = —1,

and interpret your result graphically. Show that V3.9,-1:1 1.975.

10. Find the sixth term and the rth term of the series whose first five terms are 4, 1, 0. 13, 76 on the assumption that the rth term is a polynomial in r of as low a degree as possible.

422

19.

Complex numbers (2)

1. INTEGRAL POWERS OF COMPLEX NUMBERS In Chapter 17 we saw that the multiplication of complex numbers was best expressed in terms of their moduli and arguments: if z = r (cos 0 +j sin 0) and w = s (cos +j sin g5) then zw = r s [cos (0+ 95) +j sin (0+0]. In particular, if z = cos 0 + j sin 0 then z2= cos 20+j sin 20, from which it follows that z3= cos 30+j sin 30 and so on; the general result,

zn = cos nO +j sin no (n n Z+), may be proved by mathematical induction.

Theorem 19.1. (De Moivre's theorem for positive integral exponent.) For any positive integer n (cos 0+j sin 0)" = cos nO +j sin nO. Proof. (cos 0 + j sin 0)n-1 = cos (n — 1) + j sin (n — 1) 0 (cos +j sin OP = [cos (n — 1) + j sin (n — 1) 0] (cos 0+j sin 0) (cos d+j sin OP = cos nO +j sin no But

(cos 0+j sin e)1 = cos 0+j sin 0

and the result holds for all positive integral n, by induction. Ex. 1. Illustrate de Moivre's theorem using the Argand diagram for the cases

n = 2, 3, 4. What can you say about 0 if (cos 0+j sin OP = 1 (n a positive integer)? Ex. 2. Express J3 +j in modulus-argument form and deduce the value of G/3 +jr. De Moivre's theorem may be extended without difficulty to include negative integral exponents. If we define z° to be 1, this enables us to say that (cos 0+j sin 0)" = cos no +j sin no for all integral n.

Theorem 19.2. (DiMoivre's theorem for negative integral exponents.) For any negative integer n, (cos 0+j sin 0)" = cos nO +j sin no. 423

[19

COMPLEX NUMBERS (2)

Proof Write n = —m; then m is a positive integer and (cos 0+j sin 0)m = cos m0 +j sin m0, by Theorem 19.1. Thus (cos -1-j sin 6)n =

1 cos me +j sin m0

= (cos m0 —j sin m0), since 'cos me +j sin m01 = 1, = cos n0+ j sin no. Ex. 3. Illustrate de Moivre's theorem, using the Argand diagram, for the cases n = —2, —3, —4. What can you say about (i) zn+Z-n, (ii) z6 — z-6 where IzI = 1?

Ex. 4. Evaluate (V3 + j)-9. The results proved in Theorems 19.1 and 19.2 are frequently useful in deriving further results. We shall illustrate, in the next examples, some of the techniques most commonly employed. Example 1. Express cos 60 in terms of cos 0 and sin 0. Since cos 60+j sin 60 = (cos 0+j sin 0)6 we have, using the Binomial Theorem, cos 60+j sin 60 = c6 + 6jc5s-15c4s2 -20jc3s3+ 15c2s4 + 6jcs5 — s6, where c = cos 0, s = sin 0. Comparing the real parts of both sides, cos 60 = cos60 —15 cos40 sine 0 +15 cost o sin4 0 — sing 0. Notice that, in Example 1, (i) by comparing imaginary parts we immediately derive an expression for sin 60 in terms of cos 0 and sin 0; (ii) the expression for cos 60 (but not sin 60) may be written as a polynomial in either cos 0 or sin 0. Ex. 5. Show that cos 60 = 32 cos6 0-48 cos4 0+18 cost 0 1. -

Ex. 6. Express tan 60 as a rational function of tan 0. Example 2. Express cos60 in terms of multiple angles. Writing z = cos 0+j sin 0, we have z-1= cos 0 —j sin 0 and thus 2 cos 0 = z+z-1. Then

64 cos6 0 = (z + z--')6 = (z6 + z-6)+ 6(z4+ z-4)+15(z2+ z-2) + 20.

424

I]

INTEGRAL POWERS

But z6+z-6 = (cos 60+j sin 60) + (cos 60—j sin 60), by de Moivre's theorem

= 2 cos 60, and similar results hold for z4 + Z-4and z2+ Z-2. Thus 32 cos6 0 = cos 60+6 cos 40+15 cos 20+10.

*Ex. 7. Suggest a quick check on the accuracy of the answer to Example 2. *Ex. 8. By writing 2j sin 0 = z z-1, express sins 0 in terms of multiple angles. Ex. 9. Evaluate j.cos6OdO. Example 3. Find the sum of the series x sin 0 + x2sin 20 +x2 sin 30+... +xn--1sin (n-1) 0, where x is a real number. The given series is reminiscent of a geometric series, but we have sines of multiple angles, sin rO, rather than powers of sines, sin' 0. De Moivre's theorem suggests a possible way of changing from the multiple angle form into an exponent form. Consider the two series:

C = 1+x cos 0 + x2cos 20+ ... +xn-1cos (n-1) 0, S=

x sin 0 + x2 sin 20+ ... +xn-1sin (n-1) 0.

Multiplying the second series by j and adding, this gives

C+jS = 1 + x(cos 0+j sin 0)+ x2(cos 20+j sin 20)+ xn-1 (eos (n —1) 0 +j sin (n — 1) 0). Thus, writing z = cos 0+j sin 0, C +jS = 1+ xz + x2z2 + + xn-lzn-1, by de Moivre's theorem 1 —xnzn from the formula for the sum of a geometric series 1 — xz ' 1 —xnzn (1 —xz*) 1 —xz • (1 —xz*) 1 — xz* —xnzn +xn-klzn-1Iz12 1 —x(z+z*)+x2 1z1 2 1 — x(cos 0—j sin 0)— xn (cos n0+ j sin nO)+ xn+1(cos (n-1)0 +j sin (n — 1) 0) 1-2x cos 0+x2 since Izj = 1.

425

[19

COMPLEX NUMBERS (2)

Comparing the imaginary parts of both sides we then have S=

x sin 0 xn sin n0+xn+1sin(n 1) 0 1 2x cos 0 +x2 -

-

-

Ex. 10. What is the value of C in Example 3? CO

Ex. 11. If lx1 < 1, find E xr sin r0. r=1

Exercise 19(a) 1. Simplify (ii) (cos +j sin 170-3; (i) (cos -17 . T+ j sin 1704; . - j sin 1706 ; (iv) (sin 3 6 77+j cos *706; (iii) (cos 17 (v) (cos + j sin *703(cos 47+ j sin 47)4 ; (vii) — j Cot *70-4; (vi) (1 +j tan *704; (viii) (1 + cos 20+j sin 20)-4. 2. Express in standard form

(i) (cos +j sin 17012 ; (ii) (1 + j)6 ; (iii) (.s/3 -D4(1 +jV3)6; (iv) (cos 0+j sin 0)n/(cos ¢-j sin O)m. 3. Express in terms of sin 0: (i) cos 40; (ii) sin 50. 4. Express in terms of cos 0: (i) cos 40; (ii) cos 50;

(iii) sin 60 sin 0 •

5. Express tan 40 in terms of tan 0. 6. Prove that cos 70 = 64 cos7 0-112 cos50 +56 cos3 0-7 cos 0.

Write down the seven roots of the equation 64x7 -112x5 +56x3- 7x = 0, and also the seven roots of the equation 64x7 - 112x5 + 56x3-7x-1 = 0. 7. Express, in terms of cosines of multiple angles: (i) cos5 0; (ii) sin4 0; (iii) cos7 0. 8. Express, in terms of sines of multiple angles:

(i) sin5 0; (ii) sin' 0; (iii) sin30 cos 0. 9. Express sin50 cos4 0 in terms of sines of multiple angles. 10. Evaluate .1 cos4Od0 and

J

cos4 0sing 0d0.

11. Sum to n terms the series:

(i) cos x+ cos 2x+ + cos nx; (ii) sin x+ sin 2x + + sin nx. 12. Sum to n terms the series

cos x + 2 cos 2x + 4 cos 3x+ +2n-1cos nx. 426

1]

INTEGRAL POWERS

13. Sum the series 1+ (1 cos 0+ (n) cos 20 + + cos 1 2 14. Find E (r+ 1) sin r0. r=0 15. Find E sin* 0 sin r0.

r=1 16. By considering cos (01+ 02 +03 +04)+j sin (01+ 02 +03+04), express tan (01+82+03+04) in terms of tan 01i tan 02, tan 03, tan 04. n

17. Find E sin (2r+ 1) 0.sin8r+10. r=1

18. If z = cos 0-I-j sin 0, show that z-1/z = 2j sin 0 and zn —11zn = 2j sin no. Express sin6 0 in the form a sin 50+6 sin 30 + c sin 0 and hence solve completely the equation 16 sin6 0 = sin 50. (0 & C)

2. RATIONAL POWERS OF COMPLEX NUMBERS In this section we shall denote a general rational number by p/q, where p, q are integers and q> 0. Given a real number r > 0, there is just one positive qth root of r9, that is, there is just one a > 0 such that rP = a3. We write a = r2vq or a = *Ex. 12. Prove that, if p, q are integral and r > 0, there is only one positive real number a such that ry = a3. (Show that, for b > 0, ry = b4 = b = a.) *Ex. 13. Prove that, if q is odd there is just one real number a such that, for real r, r5= ag and that, if q is even, there are either two or no such numbers. Now let us consider the problem of finding a complex number w such that zr' = w3, where z is the complex number r(cos 0+j sin 0), r > 0. Suppose w = s(cos c6+ j sin 0); then rv(cos p0 + j sin p0) = sq(cos qcb +j sin qq5). (p, q being integers, we may apply de Moivre's theorem to both sides.) Now two non-zero complex numbers can be equal only if they have the same moduli, and arguments differing by an integral multiple of 2rr. Thus ry

sq,

p0+2krr = q95.

427

COMPLEX NUMBERS (2)

[19

Whether q is even or odd, we may take s = rIvg (see Ex. 13) and

p+q2k7r,

0

090 + 21ar\ ). k q / In this expression, k can take any integral value, positive or negative. Since values of k differing by q or any multiple of q give rise to the same w, there are precisely q different values of w given, for example, by taking giving

w = rr1g (cos

(90+21

+j sin

k = 0, 1, 2, ..., (q— 1); in other words, a complex number has exactly q qth roots. Since the real numbers form a subset of the complex numbers, every real number has q qth roots; by Ex. 13 at least (q- 2) of these will be complex. Ex. 14. What are the four fourth roots of: (i) 1; (ii) 4; (iii) —1?

If z = r(cos 0+j sin 0), where 0 = arg z, we shall define z211q to be

eq +j

rIva (cos

q

It is important to realise that zvia is only one of the qth roots of zr'. We shall sometimes use the notation VZ2) for zwq; in particular, Vz = z. *Ex. 15. Under what circumstances do complex number ?

(zP)liq

and

zvig

represent the same

Ex. 16. Verify that the definition given above for z 2)/q yields the correct value for where r > 0 is a real number.

rvig,

Example 4. Find (-1 + j)1and the other fifth roots of the complex number (-1+j). Writing z = —1+ j = V2(cos and

w5 = z,

we have

w=

+j sin in)

21ar

{cos in + 5

+j sin

in+ 21aT1 5 I•

Distinct values for w are given by k = 0, 1, 2, 3, 4; k Thus (— 1 +j)i = 2116(cos +j sin -21570.

0 gives (-1 +Di.

The remaining four fifth roots of (-1+j) are 2-Ncos 2•116(cos 117T +j sin IP), 2i+cr(cosF j sin fin), and 2*(cos 428

j sin E-g), j sin in).

2]

RATIONAL POWERS

Using tables we have (-1+j)*

0.955+0.487j,

and similar approximations may be found for each of the other fifth roots of (-1+j). Ex. 17. Plot the positions, in the Argand diagram, of the affixes of the complex 1 j number V and of its five fifth roots. -V2 + 2

Ex. 18. Find the three cube roots of j, and verify that their sum is zero. Plot their positions in the Argand diagram.

3. THE nTH ROOTS OF UNITY The equation zn = 1 has, by the results proved in the last section, precisely n roots, for 1 = cos 21ar +j sin 2kg, and thus

z = cos

21ar

+j sin

21ar

(k = 0, 1, 2, ... (n — 1)).

These n complex numbers are the nth roots of unity. Writing 27r . 2ir w = cos —+ sing"

n

and applying de Moivre's theorem (cok = cos 2kg+j sin 2kn n n the nth roots of unity may be written 1, co, 6)2, co3,

con-1.

*Ex. 19. Show that the nth roots of unity are represented by the vertices of a regular n-sided polygon inscribed in the unit circle izi = 1. Since con —1 = 0 and (1) + 1, we have, by summing the geometric series in the usual way, the following important result:

1+w1+0+o)3 +...+6,2-1 = 0; that is, the sum of the n nth roots of unity is zero. *Ex. 20. If p is a prime number, and if 6 is any pth root of unity other than 1, show that the complete set of pth roots of unity may be written as 1, 6, 62, ..., 6P-1. Discuss possible generalizations of this result for the case where p is not prime. (If you have difficulty in proving the general result, consider the particular cases p = 3,p = 4.) 429

[19

COMPLEX NUMBERS (2)

Example 5. Prove that (1+ z)n = zn=- Re (z) = -4. (1+ z)n = zn (1+z) = z, where 6 is an nth root of unity other than 1, (1 +z*) = z*g*

(1 +z) (1 +z*) = zz** 1+(z +z*)+zz* = zz* (since g* = 0 Re (z) =

(See p. 383 Ex. 14 (v).)

Exercise 19(b) 1. Find in standard form the three cube roots of: (i) 8; (ii) - 1; (iii) -j; (iv) (1 +j)3. 2. Express in the form a+ bj, giving a, b to 2 significant figures in (ii)-(iv) V(1 +2j); (iii) AA -3-0; (iv) A/(1-3.0. (i) AA3 -4.0;

3. If co is a complex cube root of unity, show that the cube roots of z3 are z, zo), zo)2. Find the three cube roots of - 2 +2j and deduce surd expressions for cos (77112) and sin (n/12). 4. Find, correct to 2 significant figures, the real and imaginary parts of (2-j)1/6

5. Simplify: (i) V(cos 0-j sin Or V(cos 0+j sin 0); (ii) (sin 0-j cos 0)113; cos 0- j sin B\ (iii) ik(cos 30+j sin 301' (iv) (1 +j cot 0)114 ; (v) {(cos 0+j sin 0) (sin 0-j cos 0)}114. 6. Plot in an Argand diagram the four fourth roots of 16 and, on a separate diagram the six sixth roots of 64.

7. Solve the equation z4-z2+ 1 = 0. 8. Solve the equation 1 + z+z2 + z3 + z4+z3 = 0. 9. Solve the equation z3- (j- z)3 = 0. 10. Solve the equation (1 + jz)3- (1 - jz)6 = 0. 11. Show that, if (j-z)n = (jz- 1)n, then z must be a real number, and find all the real numbers satisfying this equation.

430

3]

nTH ROOTS OF UNITY

12. If 1,

6), &2are the cube roots of unity, prove that: (i) (a+ — w2) (a— w + w2) = a2 +3; (ii) (1 + jco — w2) (1 — co+jco2) = 2; (iii) (a+ b) (a+ bay) (a+ bw2) = as + b3; (iv) (a+ b + c) (a+ bw + cw2) (a + bw2+ cw) = aa + + c3—3abc.

13. Describe geometrically the effect of multiplying a general complex number z by w, a complex cube root of unity. Deduce geometrically that, if I zi = 1, then lz+ wzi = 1.

14. 1, 6, 62,r63, 64 are the five fifth roots of unity and a, b two given complex numbers, and if A1, A2, A3, A4, A5are the affixes of the complex numbers a+ b, a+ bE, a+ bg2, a+ bp, a+ b64show that Ai A2 A3A4 213 is a regular pentagon.

15. If w is a complex cube root of unity, prove that a+ b + c is a factor of the determinant A=

ab c a b ca

Show also that A=

a wb w2c (02c a wb to2c a

Hence express A as a product of linear factors and also as a product of a real linear and a real quadratic factor. Factorize into four linear factors the determinant a b c d d a b d a b da

4. COMPLEX POWERS OF COMPLEX NUMBERS

(This section may be omitted at a first reading) If e is the base of the natural logarithms, and if y is a real number, it seems plausible to assume that, provided ei1has a meaning, d(eiy) = j

On this assumption, z = ejv satisfies the differential equation dz 6 = jz. 431

[19

COMPLEX NUMBERS (2)

But w = cos y+j sin y also satisfies this differential equation (by direct verification). Also, when y = 0, z = 1 = w and thus z = w for all real y: eiv = cos y +j sin y.

Since it follows that

= ex . Or,

ex-1111 = ex(cos y +j sin y).

(1)

*Ex. 21. Show that e-iv = cos y — j sin y and deduce that cos y = +e-iv), 1 . sin y = 2j (ew — e-iv). Deduce from these expressions and the infinite series for

e the series expansions for cos y and sin y. Ex. 22. Show that cos y = cosh jy and that j sin y = sinh jy. z2 = 2knj, k an integer.

*Ex. 23. Prove that ev. = eZ=

*Ex. 24. Discuss de Moivre's theorem in the light of the expression of a complex number in the form re. Now consider the equation

z = ew.

Given any non-zero z we can certainly find a w which satisfies this equation.

For example, if w = In Izi +j arg z,

etc = einizi+Jargz = eln Izle; arg z = I ZI (cos arg z+ j sin arg z)

z.

By Ex. 23, any other solution of the equation may be written in the form w = In jzi +j(arg z+21ar). Such an expression is called a logarithm of z; the particular expression with k = 0 is called the principal logarithm of z and is written In z: In z = lnizi arg z. With this definition of the logarithm of a complex number we are in a position to define a complex power of a complex number: zw = ewinz For example,

ji = &In = e-ig 0.208,

432

3] while

COMPLEX POWERS n+i)

(1 + j)i-i =

e(1-i) an A/ 2+1771}

= can

2-1-170+j(i7-1n A/ 2)

= A./2 eig(COS

(47T

— ln V2) + j sin (ig - In ,/2))

2.8 +1.3j. Ex. 25. Find expressions for: (i) In (-1); (ii) In (-j), (iii) (-1)1 ; (iv) 1(1 +Pill. Ex. 26. Verify that the definition given in Section 2 for zylg (pl q rational) is in accordance with the definition given here for a complex power of a complex number.

Exercise 19(c) 1. If a, b, r, s are real numbers and u = reie, v = seie, find: (i) 1 au+ bv1; (ii) an argument of the complex number au+ by. 2. Express in the form a + bj:

(i) el±bri; (ii) er-FIrD. 3. Express in the form a+ bj;

(i) In(V3 - j);

(■ 13 -

4. If z moves once anticlockwise around the unit circle in the Argand diagram, starting at the point -1, describe the motion of the point representing 5. If z is a complex number and sin z, cos z are defined by 1

sin z = (eiz - e-iz), cos z 2j

2

+

prove that: (i) sin2 z+cos2 z = 1; (ii) sin 2z = 2 sin z cos z; (iii) cos ( w- z) = sin z; (iv) cos (z1+ z2) = cos z1cos z2-sin z1sin z2. 6. Show that: (i) cos (x+yj) = cos x cosh y- j sin x sinh y; (ii) sin (x+yj) = sin x cosh y + j cos x sinh y. 7. Show that: tan 1(u+ jv) =

sin u+j sinh v cos u+ cosh v •

If x+ jy = c tan i(u+ jv), express x2+ y2 +c2 in terms of u, v and c. If v and c are positive constants show that the locus of the point (x, y) referred to Cartesian axes is a circle of radius c cosech v. (tan z = sin z/cos z.) (London) 433

[19

COMPLEX NUMBERS (2)

8. Give a sketch of the representation in the Argand diagram of the two sets: A = {z e C: = 1, — 4ir < Im z < irr}; B= {wEC: z e A, ezw—ez+w+1 = 0}.

Miscellaneous Exercise 19 1. Solve the equation z8— z4 + 1 = 0 and mark the positions of the roots in the Argand diagram. 2. Prove that, if n is a positive integer, (cos 0+j sin 0)" = cos n0+ j sin nO. By putting n equal to 5 in this formula, or otherwise, prove that A15-1 sin = 10 4 3. Prove that cos 70 = cos' 0(1 — 21 tang 0+35 tan40 —7 tang 0). Find the real part of (1 + cos 0+j sin 0)n (1 + cos 0 —j sin O)n•

(0 & C)

(London)

4. Express each of the complex numbers z1= (1 + j),‘12, z2 = 4( — 1 +DA/2 in the form r(cos 0+j sin 0), where r is positive. Prove that z? = z2, and find the (0 & C) other cube roots of z2in the form r(cos 0+j sin 0). 5. Solve the equation z4 + 2z2 + 3 = 0, giving the real and imaginary parts of each root correct to 2 significant figures. 6. If a +16' = ,1{(a+jb)(c +jd)} where a, b, c, d, a and ,8 are real, find the value of a2 in terms of a, b, c and d. (London) 7. Prove that the roots of the equation (z + 1)" — (z — 1)n = 0 (n 3) all lie on the imaginary axis. Illustrate the result geometrically in the case n = 3. 8. Given that 1

,,2

1 1

1

cot 1 1 1 1 co

D=

0)2

where w is a complex cube root of unity, prove that D2= — 27. n-1

(0 & C)

n-1

9. Evaluate E cos (a+ r/3) and E sin (a + rfl). r---0

r=0

10. If z = cos 0+j sin 0, show that z+ 1/z = 2 cos 0 and find the corresponding result for z — 11z. Prove that cos" 0 = 118[cos 88+8 cos 60+28 cos 40+56 cos 20+35]. to

Evaluate 434

fin

(cos" 0+ sing 0) d0.

(London)

3]

MISCELLANEOUS EXERCISE

11. If = 1 and arg z = 0 * 0, express in modulus argument form: /(1+z\ (i) 1+ z, (ii) 1—z; (iii) 1-4 w = 4/(1+z) Prove that 1—z lies on a fixed straight line, whatever the value of O. 12. Demonstrate the following results geometrically, where a is a complex cube root of unity: (i) (1+a) (1+ w2) = 1; (ii)

1+a) 1+

a; (iii) Re (1+2a)) = 0.

13. Find the roots of the equation z6 + 1 = 0 in the form cos cb +j sin cb, where q5 is to be determined. Deduce, or prove otherwise, that the roots of 16x6 -20x3 + 5x +1 = 0 are cos 1[(2r+ 1) n] (r = 0, 1, 2, 3, 4). 14. Solve the equation

(London)

j(1—xj)n = (1+ xj)n,

and verify your solution by setting: (i) n = 1; (ii) n = 2. 15. Find la) + 1 + jj and arg (a) + 1 +j) geometrically, where w is that complex cube root of unity with positive imaginary part. Hence, or otherwise, find surd expressions for sin *Tr and cos *n. 16. Prove that: (x+ y+z) (x+ coy + 0z) (x+ co2y+coz) E x3+ y3+ z3—3xyz, where (1) is a complex cube root of unity. Hence, or otherwise, solve the following problems: (i) Prove that the product (x3 + y3 + z3 — 3xyz) (a3 + b3+ c3 — 3abc) is expressible in the form A3+ B3+ C3—3ABC, where A = ax+ by+ cz, B = ay+ bz+ cx, C = az + bx+ cy. (ii) Solve the equation x3-9x+12 = 0. (0 & C) 17. In the determinant ab d a b cdab c d a the cofactors of a, b, c, d are denoted by A, B, C, D respectively (the expansion of A by its first column being aA+ bB+ cC+ dD). 435

COMPLEX NUMBERS (2)

[19

By considering the product AO, where 1 0 0 0 1 0 0 = cot 0 1 0 (03 0 0 1 prove that, if co is any root of the equation x4 -1 = 0, then a+ bcd + co)2+ dws divides into A, the quotient being A +Bw3 + Cco2 +Dw. Hence show that (0 & C) A+B+ C+D = (a + c — b — d) {(a — c)2(b — d)2} . 18. P, Q are the affixes of the complex numbers p, q and E = cos 2171n +j sin 2n/n. Locate the affix of the complex numbers (p—q) 4. If a, b are two complex numbers, prove that the affixes of the complex numbers Z1, Z2, are the vertices of a regular n-sided polygon if and only if (z,.—a)" = b" (r = 1, 2, 3, ..., n). 19. If Z1, Z2, Z3 are the affixes of the complex numbers z3prove that a necessary and sufficient condition for the triangle Z1Z2Z3to be equilateral is that + +

-Z2 23 Z3 - Z2 = 0.

20. Prove that, if n is a positive integer (n > 1) and = cos 2n/n +j sin 2n/n, = 0. then 1+ In the Argand diagram the points A1, A2, ..., A„ are the vertices of a regular polygon inscribed in a circle of radius a with its centre at the origin. The complex numbers represented by the points A1, A2, . .., An are z1, z2, zn. Prove that 4+4+ + zn2= O. The perpendicular distances of the points A1, A2, ..., Anfrom any given line through the centre are d1, d2, ..., dn. Prove that di + ciT + d „2 = na2. (0 & C) 21. If x+ jy = tanh (u +jv) where u, v, x, y are real, find x and y in terms of u and v. Prove that x2+ y2 — 2x coth 2u+ 1 = 0 and x2 + y2 + 2y cot 2v — 1 = 0. If u and y are regarded as variable parameters, and x and y as Cartesian coordinates, describe the relationship between the two families of circles. 22. Prove that sinh

(0 + kb) = sinh 0 cos cb + j cosh 0 sin q5. For all real or complex values of z the sum of the infinite series ao+aiz+ a2z2+ is f(z). Prove that if co is a root of the equation o2 + + 1 = 0, then ao+ a3 x3+ ao x° + + 1{f(x)+ f(6)x)+ f(w2x)}.

436

3]

MISCELLANEOUS EXERCISE

By considering the series for sinh x, prove that 2 x3 X2 X12 x x 31 cos 2 . + -+ — + = - sinh - {cosh 2 3! 9! 15 ! 3 2

(0 & C)

23. A sequence u1, u2 ... is defined by u1= 1 and u„±1= au,, + n + 1 (n 1), where a is independent of n. Find an expression for un when a = 1. By induction or otherwise show that, when a * 1, unis of the form un = Aan+ Bn+ C, where A, B, C are independent of n, and find A, B and C. If now a is a complex mth root of unity and n is a multiple of m, determine the real part of un. (0 & C) 24. Prove that x2" 1

nkn. ) (x- 1) (x + 1) H (x2 - 2x cos - +1 n k=1

and devise a similar expression for x2411- 1. 25. Prove that 2kg cos nO +1 = H {x2 - 2x cos (0+ — +1) . n k=0 Deduce the following results: nkir (i) sin na = 2,n-1 II sin a+-- ; k=0 ntie+2krr 1 . (ii) cos na- cos nfi = 2n-1H {cos a - cos k n)1 k=0 From (i) and (ii) deduce, by logarithmic differentiation, the further results: 1 n-1 ( kg (iii) cot na = - E cot a+ - , a * r-; n n k=0 X2n— 2xn

n-1

(iv) cosec2 nO = - k E -o cosec2 0+- , 0 * -. n n 26. Prove that x2 + y2+ z2 -yz- zx- xy has a linear factor x + coy + 0)2z, where a is a complex cube root of unity. Deduce that, if x * 3, x2+y2 + z2- yz - zx - xy is a factor of (y- z)n + (z- x)"+ (x- y)n.

3 PPMII

437

20.

Mappings in the Argand diagram

The equation of a curve in the Argand diagram is the condition imposed upon the complex number z whose affix, P, is any point of the curve. For example, the equation of a straight line through 0 and containing the point B (represented by the complex number b) is

z = Ab, A real. A straight line not containing 0 is uniquely defined by the foot, A, of the perpendicular from 0 to the line (see Figure 20.1). Suppose A is the affix of the complex number a and that P(z) is any point on the given line.

Then OP = OA + AP or, in complex number notation,

z = a+jAa, A real.

(1)

(Recall that multiplication by j rotates the vector representing a complex number through Pr.) Taking complex conjugates of both sides of (1)

z* = a* — jAa* Eliminating A this gives

a*(z—a)+ a(z* — a*) = 0, or

a* z + az* = c,

(2)

where c (= 21a 1 2) is a real number, as the equation of the line through A and perpendicular to OA. *Ex. 1. By writing a = h+kj and z = x+yj, prove conversely that any equation of type (2) above represents a straight line.

438

STRAIGHT LINES AND CIRCLES Ex. 2. What is the equation of the perpendicular bisector of the line joining the points which represent the complex numbers 0 and 2-3j? *Ex. 3. Show that a*z+ az* = 2klal2, k real, represents a straight line and describe its relationship to the line a*z+ az* = 21a12. Interpret geometrically the quantity Re (a*z). Ex. 4. Where does the line (1 +2j) z+ (1 —2j) z* = 12 cut (i) the real axis; (ii) the imaginary axis? y

X

0 Fig. 20.2

The equation of a circle in the Argand diagram is also easily found. If the centre of the circle is B, corresponding to the complex number b, and if the radius is r, then, for any point P(z) of the circumference, Iz—bI = r,

(z — b) (z* — b*) = r2, which gives, on writing the real number lb12— r2 as d, zz* — b*z — bz* + d = 0 as the equation of the circle centre B and radius [ b I 2— or

(3)

Ex. 5. Show conversely, that for real a = 0 and real c, the equation azz* +b*z+bz* + c = 0 represents a circle, provided ac < 1612. Ex. 6. What is the equation of the circle, centre 1— 2j and radius 3 in the form (3)? Ex. 7. Show that 2zz* + (3 —j) z+ (3+j) z* +1 = 0 is the equation of a circle, centre — +j) and radius V2. Find the centre and radius of the circle with equation: (i) zz* — z(1 — 3j) — z*(1 + 3j) + 6 = 0; (ii) 4zz* — z(2+ 4j) — z*(2 — 4j)+ 1 = 0. 3-2

439

MAPPINGS IN THE ARGAND DIAGRAM

[20

We now consider the images of straight lines and circles under three simple functions f, g, h from the set C into the set C given by

(I) f(z) = w, where w = z fl; (II) g(z) = w, where w = az; (III) h(z) = w, where w = z--1; a, ft being complex constants. (I) w = This is clearly a translation of the whole plane by an amount equivalent to the position vector corresponding to the complex number ft (see Figure 20.3).

Fig. 20.3

Geometrically it is obvious that this function maps a straight line into a parallel straight line, and a circle into an equal circle with its centre translated an amount ft. Analytically,

a*z + az* = c

a*(w— fl)+a(w* fl*) = c a*w+aw* = c+a*ie+afl*. But a*fi +an* = 2 Re (a*fl); thus the right-hand side is real and the equation represents a straight line. Again, for a circle we have lz —101 = r (r real and positive) I w — (b +16)1 = r, which represents a circle of the same radius but with its centre translated by an amount ft. Ex. 8. Explain why the straight lines a* z + az* = c and

a*w+aw* = c+ a*fi +afi* are parallel. 440

STRAIGHT LINES AND CIRCLES

(II) w = az. As shown in Chapter 17 this maps P -›- Q where OQ = lalOP and LPOQ = arg a; that is, it represents an extension by the factor I al followed by an anticlockwise rotation of magnitude arg a (Figure 20.4).

Fig. 20.4

The effect of this mapping upon straight lines and circles may be deduced analytically as follows:

a*z + az* = c

a* (-1+a (-1 =c a a (aa)* w + (aa) w* = clal 2 and

1Z -b1 = r w -b = r a lw-bal = dal.

Thus straight lines are mapped into straight lines and circles are mapped into circles under the transformation w = az.

*Ex. 9 Show that the centre of the z circle maps into the centre of the w circle under this transformation and that the ratio of the two radii is loc1: 1. Ex. 10. Illustrate geometrically the effect of the transformation w = (1 + j) z upon the circle lz- 11 = 1. (III) w = z-1. We must restrict the domain of this function to the whole of C with the

number 0 deleted. To describe the transformation geometrically it is useful to define the inverse of a point P with respect to a circle. Given a 441


[20

circle, centre 0 and radius r, the inverse of the point P is the point P' on OP such that OP .0P' = r 2. *Ex. 11. Show that the transformation w = z-1maps the point P into the reflection in the real axis of the inverse of P with respect to the unit circle Izi = 1 (see Figure 20.5).

Fig. 20.5

Now consider the effect of the transformation w = z-1upon a general straight line and circle. The straight line a*z+ az* = c maps into

that is, into

a* (1\ 4. a(1_\* =

kw

)

k

)

cww* — a*w* —aw = 0.

This represents a circle through 0, if c 0, and a straight line through 0 if c = 0. Thus, a straight line not containing 0 maps into a circle through 0, while a straight line through 0 maps into another straight line through 0. *Ex. 12 Show that a straight line through 0 maps into its reflection in the real axis under the transformation w = z-1. What lines map into themselves under this transformation? *Ex. 13. Show that, under the transformation w = z-1, the straight line through A and perpendicular to OA maps into a circle with its centre on the line OA*, where A, A* are the affixes of the conjugate complex numbers a, a*. Now consider the effect of the transformation w = z-1upon the circle

zz* — b* z — bz* + d = 0. 442

STRAIGHT LINES AND CIRCLES

The image is the set of points defined by -1-

(

)

w,

w

that is, by

(1)* Fd=o,

dww* -Pm,* -bw +1 = 0.

If d = 0 this represents a straight line, otherwise it represents another circle. Thus, a circle passing through 0 maps into a straight line not through 0, while a circle not through 0 maps into another circle not through 0. *Ex. 14. Show that, under the transformation w = z-1, a circle, centre B, which passes through the origin maps into a straight line perpendicular to OB*, where B, B* are the affixes of the conjugate complex numbers b, b*. *Ex. 15. Show that, under the transformation w = z-I, a circle, centre B, which does not pass through the origin maps into a circle with its centre lying on the line OB*, where B, B* are the affixes of the conjugate complex numbers b, b*. *Ex. 16. Given a circle, centre B, not passing through 0, show that the transformation w = z-1does not in general map B into the centre of the image circle. *Ex. 17. Show that a diameter of a circle is mapped onto a diameter of the image circle under the transformations w = z +fl and w = az. Show also that, under the transformation w = z-1-, a diameter through the origin is mapped onto a diameter of the image circle.

To summarize the effect of the mapping w = z--1upon straight lines and circles we have (reading -4- as 'maps into'): (i) straight line through 0 -> straight line through 0. (ii) straight line not through 0 --> circle through 0; (iii) circle through 0 straight line not through 0; (iv) circle not through 0 circle not through 0.

Example 1. If the point z lies on the circle Izi = 1, find the locus of the point w where w = j/(z+j). Method (i) Consider the sequence of transformations: fi: z ---> u = z+j, 1, f2: u -> v = u f2:

w = jv,

upon the circle Izi = 1. Their effects are shown in the following sequence of diagrams (Figure 20.6). (For A, recall the result of Ex. 14.) The locus is seen to be the line w = + Aj or w + w* = 1. 443


0

[20

0

.f;

fa

4

Fig. 20.6

Method (ii)

w(z+j) = j _ j(1— w) z— w Thus

Iz I = 1 I wl = Li(1 -w)I iwi = ww* = (1 — w) (1 — w*) w+ w* = 1, with the same conclusion as before.

Ex. 18. If z moves anticlockwise around the circle Izi = 1, starting at z = j, how does w move along the line w+ w* = 1? Example 2. Show that the transformation w=

2z 3 +j jz — 2 —

maps the unit circle Izi = 1 into a circle with centre on the real axis and radius V2.

444

TRANSFORMATIONS

Method (i) Write

2z — 3 + j 2(z + 2j)— 3 — 3j 3 — 3j =-2J z + 2j ' j(z+ 2j)

w = jz-2

and consider the effect of the successive transformations fi: z t = z+2j, 1 2: t ->u = llt, f3: u v = 3(1 j) u = 3 V2(cos +j sin 41.) u, 47r —

—

f4: v-->w= v 2j -

upon the circle Izi = 1. (Recall the result of Ex. 15 for deducing the effect

fl

0

f3

Fig. 20.7

445


[20

of f2. Notice that, by Ex. 17, the line joining the successive images of ±j remains a diameter of the corresponding circle.) Thus, the image is a circle, centre -H(1 —j)+ (3 +j)] = 2 and radius 11(3+1) — (1-1)1 = 412 + 211 = V2. Method (ii) 2z— 3 +j w=. jz— 2 ' (jz — 2) w = 2z — 3 + j z=

2w-3+j . jw — 2 •

1z1 = 1

Thus

= 12w-3 (jw — 2) ( — jw* — 2) = (2w — 3 + j) (2w* — 3 — j) ww* —2j(w— w*)+ 4 = 4ww*— 6(w+ w*)— 2j(w — w*)+ 10 ww* —2(w+ w*)+ 2 = 0. This equation certainly represents a circle. Furthermore, since the interchange of w and w* does not affect the equation, it is symmetrical about the real axis and thus its centre lies on the real axis. It cuts the real axis at points given by w = w*, i.e. x2 — 4x +2 = 0, x = 2 + ,4,

i.e.

giving a radius of A/2, as before. Ex. 19. With the notation of Example 2, if z moves anticlockwise round the circle IzI = 1, starting at z = j, how does w move around the circle ww*-2(w+w*)+2 = 0? Example 3. The function f: C C is defined by f(z) = (z — 1)2. Prove that the image of the unit circle Izi = 1 under this mapping is a closed curve, consisting of all points w with the property that r = 2(1 + cos 0), where r = lwl and 0 = arg w. Since Izi = 1, we may write z = cos 0+j sin q; then z 1 = cos g 1 +j sin .75 —

-

= —2 sine Icb + 2j sin 115 cos 10 = 2 sin RS( — sin +¢+j cos 115) = 2 sin 295{cos (1-7T + 195) +j sin (-1-77.+10)). Thus w = (z 1)2 = 4 sin210{cos (Or+ cb)+ j sin (v.+0)1 and it follows that —

446

TRANSFORMATIONS

r = 4 sine 10, 0 = n+c-2k1r, where k = —1, 0, 1, according to which value is required to bring 0 in the range —7r < 0 zr. Eliminating qSgives r = 4 cost 0 = 2(1 + cos 0). In the equation r = 2 (1 + cos 0), any value of 0 in the range 0 0 Tr determines a unique value of r and thus a unique complex number r(cos 0+j sin 0). As 0 varies the corresponding complex number moves around a curve P which is clearly closed since r returns to its original value after 0 has turned through an angle of 2n. r = 2(1 + cos 0) is called the polar equation of ; the shape of the curve, known as a cardioid, is shown in Figure 20.8.

Fig. 20.8

Ex. 20. Prove the first part of Example 3 geometrically by constructing the unit circle and the affixes of numbers z —1, (z — 1)2.

Exercise 20 1. The function f has domain {z E C: jzi = 1}. Find the range off in each of the following cases: (i) f(z) = z+ 3; (ii) f(z) = z+j; (iii) f(z) = 2z; (iv) f(z) = jz; (v) f(z) (1+j) z. 2. In Question 1, if the point z is regarded as moving anticlockwise around the circle Izi = 1, starting at z = 1, describe the motion of the image point w = f(z)

in each of the cases (i)—(v). 3. The function f: C --> C is defined by f(z) = z3+ z2+2z + 1. Find the values of z which remain invariant under this mapping and illustrate your answer by reference to the Argand diagram. 4. The function f: C —> C is defined by

f(z) = az2+ bz+ c, where a, b, c, a C. If f(1) = 0, f(0) = 2j, f(j) = 1 +j, show that just one z remains invariant under this mapping, and find its value. What two complex numbers map into 3 + j ? 5. As z moves once anticlockwise around the unit circle I zl = 1, starting at z = 1, describe the motion of the point w where: (i) w = z2, (ii) w = z3; (iii) w = jz4, (iv) w =

6. Show that the affixes of the complex numbers 1, — 1 + V3j), -1-(— 1 — V3j) form the vertices of an equilateral triangle. Describe the effect of the transforma447

[20


tion w = jz +1+ j upon this triangle, illustrating your result by a sketch of the Argand diagram. 7. Answer the same question as in Question 6 for the transformation w = (1+j) z+ 1 +j. 8. Show that, under the mapping defined by w = j/(z —1), the interior of the circle lz— n = 1 is mapped into the exterior of the circlelw I = 1. Show that two points remain invariant under this transformation and locate their approximate positions on an Argand diagram. 9. Show that, under the mapping defined by w = (j — jz)/(1 + z), the interior of the circle 1z1 = 1 is mapped into the half-plane Im (w) > 0. 10. Show that, under the mapping defined by w = (j + 2jz)/(1 —z), the image of the set {z e C: Izl < 1) is the set {w e C: Im(w) > — 11. Find the equation, in the form a* z + az* = b, of the line joining the points z= 1 and z = j. Find the image, under the mappings defined by w = z/(z + 1), of the set of points represented by the interior of the triangle with vertices z = 0, z = 1, z = j. In nos. 12-17 find the image of the circle Iz1 = 1 under the given transformation. If the image is a circle, find its centre and radius. 12. w = .

2

j— 2z

.

15.(1+j) z— 1 w= (1—j) z+j 17. w =

2 +z . . j —z

13. w 16. w=

14. w =

1+ 1 jz —

jz +1 —j . (1+j) z— 1

2z— j 1+z .

18. Prove that, if (z— 8j)/(z+ 6) is purely imaginary, the locus of z in the Argand diagram is a circle with centre at the point 4j — 3 and radius 5. (0 & C) 19. Show that the set of points {z} satisfying arg

b) = 0, a

where 0 is a real number in the interval —77. < 0 < IT and a, b are complex numbers, is represented by the arc of a circle through the affixes A, B of a, b. What is the condition for a point z to lie on the other arc of this circle? 20. Show that, if k is a real number not equal to 1, the set of points {z} satisfying z— a =k z— b is a circle, with its centre the point C on AB such that CA/CB = k2, and that A and B are inverses with respect to this circle. What can you say about this circle (i) if k = 0; (ii) if k is very large? 448

EXERCISE 20 Draw on the same diagram the system of circles obtained for various values of k by taking a = 1, b = —1. What happens if k = 1? 21. Show that the transformation defined by w = 1/(3 + j— z) maps the circle lz— 2— j1 = 1 into the straight line Re (w) = 4. What transformation maps the line Re (z) = 4 into the circle lw-2—j1 = 1? 22. Show that the transformation defined by w = 1+ z2maps the unit circle I z1 = 1 into the unit circle 1w — 11 = 1. Draw a sketch of the Argand diagram, construct the point 1+z2and deduce the above result geometrically. 23. Show that the transformation defined by w = 1/(1 — z)2maps the unit circle Iz1 = 1 into the curve with polar equation 2r(1 + cos 0) = 1. Draw a sketch of this curve, indicating on it the images of the points 1, j, — 1, —j. 24. Prove that if

1 x+jy = A+114,

then the points on the Argand diagram defined by making A constant lie on a circle, and the points defined by making p constant lie on a circle. Prove also that, whatever be the values of the constants, the centres of the two systems of circles obtained lie on two fixed perpendicular lines. 25. In the transformation defined by w = 1/(z—j), describe the motion of the image point w if z moves anticlockwise around the rectangle with vertices 0, b, b+aj, aj, starting at the origin (where a, b are positive real numbers). 26. The points P and Q represent, in the Argand diagram, the complex numbers z and 1/(z2+ 2). The point P describes a quadrant of a circle, from the origin along the real axis to the point z = a, round the arc of the circle Iz1 = a to the point z = ja, and back along the imaginary axis to the origin. Describe the path traced out by Q (i) when a = 1, (ii) when a = V2. 27. Find the transformation of the form az +b w= cz+d' which maps 0 into j, j into 0 and 1 into —1. Show, that, if this particular transformation maps the complex number g.into the complex number n, then it also maps 7 into g. Can you generalize these results in any way? 28. Show that, if the image of the complex number z under the transformation w = z+z-1is real, then Izl = 1. Deduce a geometrical construction for determining the roots of the quadratic equation z2 +az+1 = 0 where a is a real number in the interval —2 < a < 2. Extend your results to deal with the quadratic equation z2 + az + b = 0, where a, b are real numbers such that a2 < 4b, by considering the transformation w = z+ bz-1.

449

21.

Quadratic equations and quadratic functions

1. THE QUADRATIC EQUATION We have shown in Chapter 19 that it is always possible to extract the mth root of a complex number p; that is, it is always possible to solve the equation win — p = O. In particular, if m = 2, the quadratic equation w2 =0

(1)

has roots +vp and —Vp, and the real and imaginary parts of these two numbers may be calculated to any required degree of accuracy. The more general quadratic equation

az2+ bz + c = 0 (a, b, c e C, a ' 0)

(2)

may be reduced to form (1) by the transformation

w = z+ — 2a' a process usually known as 'completing the square':

az2+bz+c = 0 Z

+ -Z a

—

a

(a + 0)

b)2 = b 2— 4ac (z+— 2a 4a2 z—

—b ± V(b2— 4ac). 2a

If b2 = 4ac, only one root is obtained, otherwise equation (2) has two distinct roots. For uniformity it is useful, in the case b2 = 4ac, to say that (2) has two coincident roots, or a repeated root; with this convention, every

quadratic equation has two roots. Ex. 1. Solve the equation (3 +j) z2— 8jz— 6 + 2j = 0. 450

I]

QUADRATIC EQUATION

If the two roots of equation (2) are denoted by a and ,3 then az2+bz+c has linear factors (z a) and (z—,8) and we may write —

az2+ bz + c = a(z - a) (z - ft), az2+bz+c =- az2-a(a+,8)z+aan.

or

a+,3 =

Thus

—

bla

(3)

of = c/a.

and

(4)

Relations (3) and (4) frequently enable us to avoid the explicit solution of a quadratic equation, especially in those cases where we are concerned with symmetrical relations between the roots. (Particularly important cases of this arise in analytical geometry. See Chapter 22.)

Example 1. If the roots of the equation (1 +j) z2-2jz+1 j = 0 are a, fl find: 1 1 (ii) 00 + /62; ——; (i) a+ )3 (iii) a,/?3+a3/3; (iv) a3+,33; a 6 ±fl5 ; (vi) la-161. (v) -

From (3) and (4),

a+ ft =

2j . = 1+j, 1+.1

1—j aft = 1_F (i)

1 a

. —J.

1 a+11 1+ = .i = 1+j. + = fl aft —J

(ii) Since a, i3 both satisfy the given equation: (1 +j) a2 — 2ja + 1 — j = 0, (1 +j) /32 — 213+ 1 —j = 0, whence, by addition, (1 + j) (a' +132) — 2j(a + fi) + 2 — 2j = 0. But a+/3 = 1+j; thus (1 +j) (a2 +,32) = 2j(1 +j)— 2 +2j = —4+4j giving

a2+ fi2 = 4j. 451

QUADRATIC EQUATIONS AND FUNCTIONS

[21

[Alternatively we could write a2 /32= + — 200, but this approach is less easy to generalize to higher powers—see (iv).] afi3 ± Ace = cciAcc2 fi2) = —j(4j), by (ii), = 4. (1 +j) cz2— 2ja + 1 — j = 0

(iv) Since we have

(1+j) a3-2joc2 + (1 —j) a --- 0;

similarly, (1 +j) A3 — 2132 + (1 — j) = 0 whence, by addition, Thus

(1 + j) (a3+,83) — 2j(a2 +132) + (1 — j) (cx + fi) = 0. 3 (1+j) (a +/33) = 2R4i) — (1 — j) (1+j), by 00, = —10,

giving a3+ R3 = —5+5j, on multiplying both sides by 10 —j). [Alternatively, a3+ /33= (a +,q) (0C2— afi + /3) = +13) {(cc +fl)2 — 3afi}, etc.] (v) As in (iv) (1 +j) (a4 +/34) = 2j(a3 +133) — (1 — j) (a2 +/32) = 2j( — 5 + 5j) — (1 —j) (4j) giving Thus,

cc4

= —14-14j = 14.

(1+j) (0:5+ /65) = 2i(cc4+ /69 — (1 — j) (cx3+ A3) = 2j( — 14) — (1 — j) ( — 5 + 5j) = —38j a5+162 = —19-19j.

giving [Alternatively, (co +,62) 0,3 + A3) = cc5 ± n5 + ct2162(ot o‘ p) etc.] (vi)

(a —/3)2 = +fly — 4ocfi = 6j, la — AP = 6, -fil = V6.

Form Example 2. The roots of the equation 2z2-3z+5 = 0 are a and the equations: (i) with roots ce+,u, fi+,u; (ii) with roots Ala, Aft; (iii) with roots a2, 452

QUADRATIC EQUATION

1]

(i) If w = z+p,, then, when 2z2 — 3z+ 5 = 0, w = a +,u or )3+#. Thus, + tt and 13 + ,u are the roots of the equation 2(w — ,u)2 — 3(w — ,u) + 5 = 0, 2w2 — w(3 + 411)+2112 + 3,u+ 5 = 0.

i.e.

Alternatively, a + it and fi+,u are the roots of the equation

w2+Aw+B = 0, where

A = — (a + fl +2,u), B = (0,+,u) (fl + = ctie + ti(oe + fi) + #2. A = i— 2,u, B =

Thus

+ /12,

leading to the same equation as before. (ii) With reasoning similar to that of (i), we make the transformation z —> (1/A) w: the equation 2w2 -3Aw+5A2 = 0 has roots Act, Aft. Alternatively, the equation is

w2+ Cw+ D = 0, where

C=

—

D = A20,ft = P2 etc.

A(a + fl) = —

,

(iii) If w = z2, then either z = wi or z = —

Thus we have

2z2 — 3z + 5 = 0 -44> 2z2 +5 = 3z either 2w + 5 = 30 or 2w+5 = —30 {2w+ 5 —30} {2w+ 5 +30} = 0 4w2 — llw +25 = 0, on multiplying out and rearranging terms; and this last equation has roots co,fl2. Alternatively, the required equation is where giving

E=

w2 —Ew+F = 0, ± fl2 = (ce, + fly _ 2c96% F E = (2)2

-5 ,

F=

ar,

0,2,2 = @fly, etc. 453

QUADRATIC EQUATIONS AND FUNCTIONS

[21

Ex. 2. If a, ft are the roots of the equation 2z2— 3z + 7 = 0, find: 1 1 (i) a2+ P2 ; (ii) a3± /G.; (iii) la —)81; (iv) w o +7 0i .

Ex. 3. If a, ,8 are the roots of the equation 3z2 — 5z+ 3 = 0 form the equation (i) with roots 2a, 2fi; (ii) with roots a — 1, fl— 1; (iii) with roots co, IA 2. QUADRATIC EQUATIONS WITH REAL COEFFICIENTS If in the equation az2+bz+c = 0, a, b, c are all real (and a is non-zero) all the previous results naturally still hold but the additional restriction placed upon the coefficients enables us to deduce further results. In this section it will be assumed, unless explicitly stated otherwise, that a, b, c are real numbers and a + 0. *Ex. 4. If f(x) E ax2+bx+c and if f(a) > 0, f(fl) < 0 (a, ft real) prove that the equation ax2+ bx+c = 0 has a root lying between a and ft. Why is there no corresponding theorem if a, b, c are complex? Theorem 21.1. If the equation az2+bz+c = 0 (a, b, c e R, a + 0) has a complex root a, Im (a) + 0, then its other root is the complex conjugate a*. Since aa2 +ba+c = 0, we have by taking the complex conjugate of both sides, a*(a*)2+ b*a* + c* = 0*. But a* = a, b* = b, c* = c, 0* = 0 and thus a(a*)2 +ba* + c = 0 and the result follows.

*Ex. 5. Explain why the proof of Theorem 21.1 breaks down if a, b, c are not restricted to the set R. Explain also why the condition Im (a) = 0 was added in the enunciation of the theorem. *Ex. 6. Prove that a quadratic equation with real coefficients either has no nonreal root or two distinct non-real roots. Ex. 7. If a, ft are two non-real numbers with the property that Im (a + fi) = Im (afi) = 0, prove that a, ,8 are complex conjugates. There is an analogous result to Theorem 21.1 for irrational roots of a quadratic equation with rational coefficients.

454

REAL COEFFICIENTS

2]

Theorem 21.2. If the equation az2+bz + c = 0 (a, b, c c Q, a + 0) has a root p+q,Ir where p, q, r are rational (q 0) and Vr is irrational, then the other root of the equation is p—qA1r. (Notice carefully the change in conditions for this theorem: a, b, c are now restricted to be rational numbers.) a(p + qVr)2+b(p + qVr)+ c must be of the form P+ Q.,1r, where P and Q are rational. Thus P+ Q AIr = 0, since p+q,/r is a root of the given equation. But this is possible only if P = Q = 0 since Alr is irrational; thus

P— = 0 and it follows that p—qVr is a root of the given equation. Ex. 8. Explain why the proof of Theorem 21.2 breaks down if a, b, c are not restricted to the set Q. Explain also why the condition q 0 was added in the enunciation of the theorem, and where the fact that Jr is irrational is used.

Exercise 21(a) 1. Write down the sum and product of the roots for each of the following equations: (i) z2 — 3z — 7 = 0; (iii) (1 + j) z2— z + (1 — j) = 0; (ii) 2z2 — 4z + 11 = 0; (iv) (z + j)2= (4 — j) z; (v) (z — 1 + j)2+ za = (2jz — 1)2. 2. Write down the equations with roots:

(i) —2, 3; (ii) — (iii) 3 — V5, 3 + V5; (iv)2— V2,3 + 2V2; (v)3 — 4j, 3 + 4j; (vi) y1 —0), 1(1 +jV3); (vii) 1 +2j, 1 —j; (viii) V2 —j, 1 V2j. —

3. Solve the following equations, giving the real and imaginary parts of the roots correct to 2 significant figures. Check your solutions by calculating the approximate product of the roots (using a slide rule): (i) 2z2 — 3z— 7 = 0; (ii) 2z2 — 3z + 7 = 0; (iii) z2 + (1 + j) z+ (1 2j) = 0. —

4. Find a +/, aft, + fl2,(11co+ (11fl) and a4/3+a/34 :

(i) when a, /3 are the roots of the equation z2 — 3z— 9 = 0; (ii) when a, fl are the roots of the equation 3z2 — z — 5 = 0; (iii) when cc, /3 are the roots of the equation z2 — jz (1 j) = 0. —

—

5. oc, ,8 are the roots of the equation 2z2 — 9z— 4 = 0. Find the values of

(i) (1/oe) + (1/,e);

(cch69+( 67a2); (iii)

; (iv) Ice2— fl21•

6. cc, /3 are the roots of the equation 3z2 — 2z— 7 = 0. Find the values of :

(i) (a +k) (fl+k);

(iii) (1 +a)-1+ (1 +fi)-1. (ii) +ft3; 7. cc, /3 are the roots of the equation (z — a)2= 4zb. Find: (i) 0,2+,62; (ii) (1/a) + OA; (iii) l a —fl.

455

QUADRATIC EQUATIONS AND FUNCTIONS 8. The equations

x2 62

[21

a2y2 = a 2b22

y = mx+ c

are solved simultaneously to give two pairs of solutions x = x1, y = Yi and x = x2, y = y2• Find -1(x1+ x2) and i-(Yi + YD. 9. If a, fi are the roots of the equation 2z2 — z— 7 = 0, form the equations with roots: (iv) a2, fi2. (iii) (1/a), (10); (i) a — 1, 16-1; (ii) 10a, 10fi; 10. If a, 16 are the roots of the equation (1 +j) z2 — 2jz + (2 — 3j) = 0, form the equations with roots: (i) a — 2j, ft— 2j; c'e2, (ii) +j) cc, (1 +D fi; 11. If a, fl are the roots of the equation 2z2 — 5z+4 = 0, form the equations with roots: ce 2 + (i) a —fl, fl— cc+ifi, fi+ja; 12. The roots of the equation 4z2 + az— 37 = 0 differ by 1; find the possible values of a. 13. The roots of the equation z2 — az + 9j = 0 are A and jA. Find the possible values of a. 14. The roots of the equation (2—j) z2 + (3 + j) z— 4 — 5j = 0 are a and A'. Form the equations with roots (i) (1/a), (1M); (ii) a*, ,8*. 15. If the roots of the equation az2+ bz + c = 0 are a, le prove that — aw2+bjw+ c = a(jw— a) Ow—M. 16. If a, b are real and non-zero and z2 + az+ b = 0, where z is non-real, prove that IzI = 1 b = 1 and I al < 2. 17. Discuss the application of the method of completing the square to the vector quadratic equation ar.r+b.r+c = 0 (a * 0). 18. The two equations

z2+az+b = 0,

z2+Az+B = 0 have a common root. Prove that b(a — A)2— a(a — A) (b — B)+ (b — B)2 = 0. 19. If one root of the equation az2+ bz+ c = 0 is the square of the other, find a relation connecting a, b, c. 20. If one root of the equation az2+ bz + c = 0 is j times the other, find a relation connecting a, b, c. 21. Prove that, if one of the roots of the equation z2 + za + 1 = 0 has unit modulus then a is real.

456

3]

QUADRATIC FUNCTION 3. THE QUADRATIC FUNCTION

A function f:

R defined by f(x) = ax2+bx + c (a, b, c c R)

is called a (real) quadratic function. Since in this section we shall be primarily concerned with inequalities, we confine our attention to real quadratic functions and the coefficients a, b, c as well as the variable x, will be assumed to be real. We now derive a necessary and sufficient condition forf(x) to be positive for all real values of x.

Theorem 21.3. If a is not zero, then ax2+ bx + c > 0 for all real x a > 0 and b2– 4ac < 0. Proof Write

y = ax2+bx+c.

y = atx2+b x+— b2 +- – b2 1 a 4a2 a 4a2f \ 2 + 4 a c b21 = x+— b a{( 2af 4a2 f .

Then

( 1)

(i) If a > 0 and b2 < 4ac, then y > 0 from (1), since (x+ b )2 0 for all x. (ii) If ax2+bx+ c > 0 for all x, then certainly a > 0, for we may choose x such that ( b)2 4ac – b2 x — > 0.

+2a

4a2

But, if b2 > 4ac, the equation ax2+bx+ c = 0 has real roots, a, ft say, and

ax2+bx+c = a(x– cc) (x – fl), from which it follows that y < 0 if a < x < ft. But we are given that y is always positive; thus b2 < 4ac. *Ex. 9. Give a geometrical demonstration of thetruth of Theorem 21.3. *Ex. 10. Prove that

ax2+ bx + c < 0 for all real x R defined by

foo = 1

x+a

x 2 + x+1

for different values of a. If x2 + ax +1 = 0, f(x) is undefined; excluding this case, if y is a value assumed by the given expression then y(x2+ ax + 1) = x + a, yx2+ (ay —1) x +(y— a) = 0.

Or

Thu sy is a possible image of the function f if this quadratics in x has real roots; that is, if (ay -1)2 > 4y(y — a) or

E

(a2 — 4) y2 + 2ay + 1 0.

To determine what values of y satisfy this inequality we have, by Theorem 21.3, to consider the discriminant of E and the sign of the coefficient of y2. The discriminant of E is A = 4a2 -4(a2 — 4) = 16 and thus, since A > 0, the equation E = 0 has real roots a, in fact, a = (a +2)--i and /3 = (a —2)-4). Thus we have

ft (where,

E = (a2-4) (y —a) (y —fl), provided I al + 2. The coefficient of y2 in E shows us that the critical values of a are ± 2. (i) lal < 2: For E to be greater than or equal to zero, y must lie between a and fl; the range is thus {y e R: a y < fl}. (ii) I al > By a similar argument, the range is {ye R:y

X

Fig. 22.7

The shape of the curve is shown in Figure 22.7. It is seen to consist of two branches (notice that, if (h, k) lies on the curve, so also does (— h, — k)). The point 0 is called the centre of the rectangular hyperbola (see Ex. 24) and the curve is symmetrical about the lines x—y = 0 and x+y = 0. When x = 0, there is no corresponding value of y and vice-versa; in fact, as x ---> 0 through positive values, y ---> oo and as x .-›- 0 through negative values, y ->--co. Thus, the line x = 0 is an asymptote and similarly so is y = 0.

Ex. 24. Show that, if any line through 0 cuts the rectangular hyperbola at P and Q, then 0 is the mid-point of PQ. (This corresponds to a property of the centre of a circle.) Ex. 25. Trace the position of the point P(ct, c/t) as t varies from —co to +co. The chord joining the points P1(ct1, OD and P2(ct2, OD on the rectangular hyperbola has equation c c(t2 — t1) Y t, = cti t2(t — t2) (x — cti) 1 476

4]

RECTANGULAR HYPERBOLA

which reduces to

x + ti to, = c(ti+ 4).

The tangent at P1may be obtained immediately by letting t2 x+ ti y = 2ct1. *Ex. 26. Obtain the gradient of the tangent at P1by differentiating the equation xy = c2. *Ex. 27. Obtain the equation of the chord P1P2by the method of Ex. 18. Ex. 28. Show that the equation of the normal to the curve at P1is trx– y = c(tt – 1)/t,.

Problems concerning the rectangular hyperbola are solved by methods reminiscent of those employed for the parabola. We conclude this chapter with a worked example. Example 3. A rectangular hyperbola has centre 0. Through a. fixed point A lines are drawn to cut the hyperbola at P and Q. Show that the mid-point, M, of PQ lies on another rectangular hyperbola and find its centre and asymptotes. Set up coordinate axes with 0 as origin and the asymptotes of the given rectangular hyperbola as the axes of coordinates, labelled in such a way that one branch of the hyperbola lies in the first quadrant (positive x, positive y). Let A be the point (a, ft) and let PQ be any chord through A, where P is the point (cti, OD and Q is the point (ct,, 02). Then PQ has equation x + ti 4y = c(t, + (as shown above) and thus, since PQ contains the point A, a+ ti12)3 = c(ti+12).

(1)

Let (h, k) be the mid-point of the chord PQ; then 2h = c(t1 + 4), 2k = giving

2h

4+12

c (1 +1\ = c(t, + 4) ti t, ' t1 12

1112 =

c(t,+ t2) h 2k

=

k•

Hence, using (1) we see that h, k are connected by the equation a+k f = 2h, which may be rewritten as (11—P) (k-413) = ice 16• 477

PARABOLA AND RECTANGULAR HYPERBOLA

[22

Thus (h, k) lies on the curve —

—

=

iocfl

(2)

which represents the required locus (or, more accurately,,contains the locus : see Ex. 29). By translating the coordinate axes to pass through the new origin -Ifl), this is seen to represent a rectangular hyperbola, centre (+a, 113), with axes parallel to the axes of coordinates. Thus, M lies on a rectangular hyperbola, centre the mid-point of OA and with asymptotes parallel to those of the given hyperbola. The reader should compare the choice of parameters in Example 3 with that in the similar problem of Example 2. Ex. 29. The locus sought in Example 3 may be definitely only part of the curve (x— -fa) (y— 3-16) = +aft. Show this by taking A as the origin.

Exercise 22(c) 1. The tangent to a rectangular hyperbola at the point P meets the asymptotes at Q and R. Prove that P is the mid-point of QR. 2. An axis of symmetry cuts a rectangular hyperbola at A, A'. P is any point of the hyperbola and N is the foot of the perpendicular from P to AA'. Prove that PN 2 = AN .NA' 3. With the notation of Question 2, if 0 is the centre of the hyperbola and G is the point of intersection of the normal at P with the line AA', prove that ON = NG. 4. With the notation of Questions 2 and 3, if Z is the foot of perpendicular from O to the tangent at P, prove that OZ .0P = 0A2. 5. With the notation of Questions 2 and 3, if the tangent at P meets AA' at T, prove that ON .0T = 0A2. 6. Show that the equation

(x — a) (y — b) = c2

represents a rectangular hyperbola and find its centre and asymptotes. Find the centre and asymptotes of the rectangular hyperbola xy—x+3y-7 = 0. 7. A variable line passes through the point (1, 1) and meets the x and y axes at P, Q respectively. The rectangle OPRQ is completed. Prove that R lies on a rectangular hyperbola, and find its centre and asymptotes. 8. The tangent to a rectangular hyperbola at the point P meets the asymptotes at Q and R. If 0 is the centre of the hyperbola, prove that QR = 20P and that the area of the triangle OQR does not depend upon the position of P. 478

4]

RECTANGULAR HYPERBOLA

9. P is a variable point on a rectangular hyperbola and Q is the foot of the perpendicular from P on to one of the asymptotes. If R divides PQ in a fixed ratio show that R lies on a rectangular hyperbola with the same asymptotes as the original hyperbola. 10. Find the equations of the normals to the hyperbola xy = c2which are parallel to the line x— 4y = 0. 11. The tangents at P, Q to a rectangular hyperbola meet at R. If 0 is the centre of the hyperbola, show that OR bisects PQ. P19 P2, 133, 134 are four points on a rectangular hyperbola. Prove that, if P1P2 is perpendicular to P3P4, then each of the four points is the orthocentre of the triangle formed by the other three.

12.

13. The normal to the rectangular hyperbola xy = c2at the point P1(ct1, c/t,) meets the curve again at P2(ct2, c/t2). Prove that q. t2 = -1 and deduce that the locus of the mid-point M of P1P2has equation 4x3y3 + c2(x2— y2)2 = 0.

Miscellaneous Exercise 22 1. Prove that the line px + qy = 1 is a tangent to the parabola y2 = 4a(a— x) if a(p2+q2) = p. A line 1 and a point 0 not on / are given; P is a variable point on 1 and Q divides OP in a fixed ratio. Prove that the line through Q perpendicular to OP touches a fixed parabola with focus at 0 and directrix parallel to 1. (0 & C) 2. The coordinates of the mid-point of the line joining the points (au2, 2au), (av2, 2av) are (X, 1'). Express u+ v and uv in terms of X and Y. A variable chord of the parabola y2 = 4ax passes through the fixed point (b, 0). Prove that the locus of the mid-point of the chord is a parabola, and find the coordinates of its vertex and focus. (0 & C) 3. The normal at a point P meets the parabola again at Q. Find the length of PQ in terms of the acute angle 0 between the normal and the axis of the parabola. 4. Two parabolas, y2 = 4a(x+ b) and x2 = 4a(y+ b), where a > 0, are given. Prove that each point of intersection lies either on the line y = x or on the line y+x+4a = 0. Hence, or otherwise, prove that, if the parabolas have four real points of intersection, then b > 3a. (0 & C) 5. Lines are drawn through a variable point P of the parabola y2 = 4ax, making angles a and 7r— a with the axis of the parabola. The lines meet the parabola again at Q and R. Prove that: (i) the point of intersection of the tangents at Q and R lies on the parabola y2= 4a(x + 4a cot2 a); (ii) the line QR touches the parabola y2= 4a(x — 4a cot2 a). (0 & C) 479


[22

6. Prove that the line y = mx + aim touches the parabola y2= 4ax for all values of m. Deduce that, if two tangents to a parabola are perpendicular, then their intersection lies on the directrix. 7. Two directed lines meet at A, and points P and 13' are taken on the two lines so that AP+ AP' is constant, sense being taken into account. It is required to prove that PP' touches a fixed parabola. Prove this by first showing that the equations of the lines may be taken in the form y = (x + c) tan a and y = — (x+ c) tan a, where 2a is the angle between the lines and AP+ AP' = 2c sec a, and that the coordinates of P and 13' may be taken in the form (t cos a, t sin a + c tan a) and (— t cos a, t sin a— c tan a). Deduce that PP' touches the parabola y2 = 4cx tang a.

(0 & C)

8. Prove that the mid-points of parallel chords of a parabola lie on a straight line (called a diameter of the parabola) parallel to the axis. Given a parabola traced on paper, find a construction for its focus. 9. If a point is marked on a rectangular sheet of paper, one of whose sides is AB and the paper is then folded in such a way that AB passes through the point, show that the crease will always touch a fixed parabola. 10. The tangents 1, m from a point P to a parabola meet the directrix at L, M respectively. The other tangents to the parabola from L, M meet m,1 at Q, R respectively. Prove that QR passes through the focus, S, and that the angle PSQ is a right angle. 11. A straight line passes through the point Pi(xi, yi) and has gradient m = tan a. Prove that a general point of the line has coordinates (xi + r cos a, +r sin a), where r is a parameter, and that the values of r giving the points of intersection of the line with the parabola y2 = 4ax are the roots of the quadratic equation r2 sin2 0+2r(yisin 0-2a cos 0)+A— 4ax, = 0. Deduce that, if Piis the mid-point of the chord QR, then the gradient of QR is 2a/y1and its equation is yiy-2ax = y1-2ax1. 12. Two chords P1Q1, /32 Q2of a parabola intersect at K. Prove that the value of the fraction (KA.KQI)/(KP2.KQ2) depends on the directions of the two chords, but not upon the position of the point K. [This result, known as Newton's Theorem, holds for all conics (see Chapter 27). Use the analysis of Questions 11 to prove the result for the parabola.] 13. Follow through the work of Questions 11 and 12 to prove Newton's Theorem for the rectangular hyperbola xy = c2.

480

MISCELLANEOUS EXERCISE 14. Three points P(ap2, 2ap), Q(aq2, 2aq), R(ar 2, 2ar) taken on the parabola y2 = 4ax are such that PQ subtends a right angle at R. Show that

(p+r)(q+r) +4 = 0. Show that every chord of the parabola which subtends a right angle at R intersects the normal at R at the same point F. Show also that, as R varies, F describes (London) another parabola. 15. The chord AB joining points A(cti, OD and B(ct2, 02) on the rectangular

hyperbola xy = c2is of constant length 1. Show that, as the position of the chord varies, the centroid G of the triangle AOB, where 0 is the origin, moves on the curve (9xy _4c2) (x2 ± y2) pxy. Find the area of the triangle AOB when the coordinates of G are (c, 2c). (London) 16. Prove that the equation of the tangent PT at the point P(ct, c/t) on the

rectangular hyperbola xy = c2 is x+ t2y = 2ct. The perpendicular to PT from the origin 0 meets PT at Q and the normal at P meets the hyperbola again at R. Prove that:

(i) as P varies, the locus of Q is (x2 + y2)2 = 4c2xy; and (ii) c2.PR = OP3.

(0 & C)

17. Prove that the equation of the chord joining the points P(ct, c/t) and

Q(cT, c/T) on the rectangular hyperbola xy = c2 is x+tTy = c(t+T). Mis the mid-point of PQ and PQ meets the x axis at N. Prove that OM = MN, where 0 is the origin. The line through N parallel to OM meets the hyperbola at Rand S whose parameters are t, and t2; C is the mid-point of RS. Prove that: (i) CM is parallel to the y axis; (0 & C) (ii) t, t, = — tT. 18. Prove that the equation of the tangent at the point (11, k) on the parabola y2= 4ax and the equation of the tangent at the point (H, K) on the rectangular hyperbola xy = c2are, respectively,

ky = 2a(x+ h) and Kx+ Hy = 2c2. Find the coordinates of the point of intersection, P, of the parabola y2 = 4ax and the rectangular hyperbola xy = 4a2A/2 and prove that the tangent to the parabola at P is the normal to the rectangular hyperbola at P. Prove that, if this normal meets the rectangular hyperbola again at Q, then (0 & C) the abscissa of Q is —4a.

481


[22

19. Prove that the normal to the rectangular hyperbola xy = c2 at the point P(ct, t) meets the curve again at Q(— eIt3, — ct 3). The circle on PQ as diameter meets the hyperbola at R; prove that PR passes through the origin. (0 & 20. Given a point P on a rectangular hyperbola, prove that one and only one real chord PQ can be drawn which is normal to the hyperbola at Q. Deduce that there is only one (real) chord AB of the hyperbola which is normal at both A and B and locate the two points A, B. 21. Points P, Q, R, S are taken on the rectangular hyperbola xy = c2. Prove that PQRS is a rectangle if and only if the parameters of the four points are of the form t, — t, —t-2. Deduce that it is impossible to inscribe a square in a rectangular hyperbola. 22. The tangent to a rectangular hyperbola F at the point P meets the asymptotes at Q and R. 0 is the centre of the hyperbola and 0 QSR is a rectangle. If SQ, SR cut P at Q', R' prove that Q'R' touches a second rectangular hyperbola whose asymptotes coincide with those of P. 23. Prove that all chords of a parabola which subtend a right angle at a point P on the parabola pass through a fixed point Q. Q is called the Fregier point of P. If a given chord subtends right angles at two points, P and 13' of the parabola, prove that the join of the two Fregier points, Q and Q', is parallel to the given chord. 24. Find the equation of the tangent to the parabola SI: y2 = 4ax at the point P(at 2, 2at).

If this tangent cuts the rectangular hyperbola S2: xy = k2at real points Q, R and M is the mid-point of QR show that M lies on part of the parabola E: 2y2 + ax = 0. Sketch in the same diagram the curves Si and S2 and the set of all such points M in the case a > 0. Mark in your diagram the line other than x = 0 which is a tangent to both S1and S2.

482

23. Polynomial equations

1. SOME PRELIMINARY OBSERVATIONS The results we have obtained in Chapter 18 generalize very readily to general polynomial equations. In particular, it is possible to find expressions for symmetric functions of the roots of polynomial equations in terms of the coefficients. We shall pursue the question of symmetric functions in Section 2; in this section we shall discuss a number of other results connected with the solution of general polynomial equations of degree n. If we are given a specific equation, say z4 +(3+2j) z2 —(1—j)z+4 = 0 it is not at all obvious whether or not the equation has a solution. Indeed, two questions arise: (i) Does every equation possess a solution? (That is, is there a complex number which satisfies the equation ?) (ii) If a solution exists, can it be found by processes similar to those employed for solving a quadratic equation? The answer to the first question is ' Yes' but the proof of this is rather difficult (but not beyond the understanding of a really enthusiastic pupil : see, for example, Hardy's Pure Mathematics, Appendix II or Courant and Robbins What is Mathematics?, Chapter 5). The result was first proved, like so many other central theorems of mathematics, by C. F. Gauss and is generally known as ' The Fundamental Theorem of Algebra': explicitly, it states that any polynomial equation with complex numbers as coefficients (and in particular, any equation with real coefficients) has a root which is a complex number (R is regarded as a subset of C in this result). We shall content ourselves with assuming the truth of this theorem. The answer to the second question is ' Yes, if the degree of the given equation is less than five' but 'No, in general, if the degree is five or more'. The first part of this result is comparatively easy to prove and is due to a number of Italian mathematicians of the Sixteenth Century; the second part is very much harder and is due to two mathematicians, E. Galois (1811-32) and N. H. Abel (1802-29). (The truly dedicated reader may care to follow the matter up in, say, Birkhoff and Maclane, 'Survey of Modern Algebra. Brief readable accounts of the life and work of Galois and Abel may be found in Bell, Men of Mathematics, Volume 483

[23

POLYNOMIAL EQUATIONS

Ex. 1. Although it is not possible, in general, to solve a given quintic equation except by approximate numerical methods, this is not to say that no quintic equation is soluble. Solve the equation z5- 32j = 0. If P(z) is a polynomial of degree n, and if P(z) has a factor (z-°C)T (but not (z -COr+1) where r is an integer < n, the equation P(z) = 0 is said to have a root of multiplicity r. For example, the quartic equation (z - 1) (z + 2j)3= 0 has a root z = 1 and a root z = -2j, of multiplicity 3. Ex. 2. Solve the cubic equation z3+3z2-4 = 0, given that it has an integral root of multiplicity 2. *Ex. 3. Assuming that every polynomial equation has a root, prove, by induction, that a polynomial equation of degree n has n roots, where a root of multiplicity r is counted as r roots. *Ex. 4. By writing P(z) (z - a)* Q(z) prove that, if P(z) = 0 has a root of multiplicity r, then P'(z) = 0 has a root of multiplicity (r- 1). Prove that the converse is false. *Ex. 5. If P(z) = 0 has a root of multiplicity r > 2, what can you say about the equation P"(z) = 0? As with quadratic equations, additional restrictions upon the coefficients of a polynomial equation enable us to make further assertions about the roots. In particular, if all the coefficients are real we have the following three important results which enable us to decide whether we have an odd or even number of roots and to locate such roots. Theorem 23.1. If all the coefficients of P(z) = aozn + aizn-1 + +an are real and if the complex number z1is a root of P(z) = 0 so also is zr. Proof. Since aoz14 + + an = 0, we have

ao(zt)n + ai(4)n-1+

+

= 0,

on taking the complex conjugate of each side, and noting that aP = a2, since the a,:are real. The result follows immediately. Ex. 6. Given that -1 +j is a root of the equation z4 +3z3 +5z2 +4z+2 = 0, solve the equation completely.

*Ex. 7. Prove that a polynomial equation of odd degree has at least one real root.

484

1]

PRELIMINARY OBSERVATIONS

Theorem 23.2. If all the coefficients of P(z) ao zn +aizn-l+ ...+ an (a, + 0) are real and if the equation P(z) = 0 has no real root, then P(z) is always positive or always negative for all real z. Proof. (The theorem is intuitively obvious from graphical considerations.) By Theorem 23.1 the roots of P(z) = 0 occurs in conjugate pairs: a1Jai, oc2 ± Jf2, ••••

But

[z—(ocr+16A

and thus

[z - (ar

P(z) = ao[(z - 0:02 +

= - arr + fig-

- 2)2 +

and every square bracket is positive for all real z. The sign of P(z) is thus determined by the sign of a,. *Ex. 8. Prove that, if a(' and anhave opposite signs, the equation P(z) = 0 has a real root.

Theorem 23.3. If all the coefficients of P(z) aozn + aizn-l+ ... + an (a, + 0) are real and if x1, x2are real numbers such that P(x1), P(x2) have opposite signs, then P(z) = 0 has an odd number of real roots between x1and x2 (and, in particular, of course, at least one real root). For the purposes of this theorem a root of multiplicity k is counted as k roots. Proof. (Again, the result is intuitively obvious from graphical considerations.) Let cq, ct2, ar be all the real roots of P(z) = 0. Then P(z) = ao(z -GO (z - cc,) (z - cc,) Q(z), where Q(z) is either positive or negative for all real z (by Theorem 23.2). Thus (x1-cc,) (x1- cc2) (xi-ocr) and (x2 -cc) (x2- cc,) ... (x2 - ar) have opposite signs and

[(x1-061) (x2 -cc1)] [(x1-a2) (x2-cc2)]

[(x1-al.) (x2-ar)] < O.

Thus (x1-oci) (x2-cci) is negative in an odd number of cases and the result follows. Ex. 9. Prove that x3 -x+ 3 = 0 has a root between x = -2 and x = -1. We conclude this Section with two worked examples. The first shows how one equation may be transformed into another in such a way that 485

[23


the roots of the two equations bear a given relationship to one another. The second example indicates how calculus methods may be used to solve problems on polynomial equations.

Example 1. If cc, Ay are the roots of the equation 3z3 5z2 +2z+2 = 0 -

form the equation: (i) with roots a 2, /3 2, y 2; -

-

-

(ii) with roots 2a, 2/3, 2y; (iii) with roots oc2, P'2, y2. (i) As in Example 2, Chapter 21, a-2, fl— 2, y-2 are the roots of the equation 3(w + 2)3 — 5(w +2)2 + 2(w + 2)+ 2 = 0. This may be expanded directly. Alternatively, since we are essentially making the transformation z -- w+2 we may rewrite the polynomial 3z3 — 5z2+2z+ 2

a(z 2)3 b(z 2)2+ c(z 2)+d

in the form

—

—

—

and the required equation will be

aw3+bw2+cw+d = 0. By Homer's Method (Chapter 18, Section 1) we have 3

—5 6

2 2

2(2 8

3

1 6

4 14

10 (2

3

7 6

18

(2

3

13

giving the required equation 3w3+13w2 +18w+10 = 0. (ii) In a similar way, to form the equation with roots 2a, 2/3, 2y we make the transformation z > w/2, giving the required equation: —

3(Zw)3 — 5(1w)2 + 2(1w) + 2 = 0. or 486

3w3 — 10w2 + 8w+ 16 = 0.

I]


(iii) If w = z2, then either z= w or z = — w. Thus 3z3 -5z2 +2z+2 = 0 z(3z2+2) =-- 5z2-2 - either

wi(3w + 2) = 5w-2

or

— wi(3w +2) = 5w-2 {5w-2— wi(3w +2)} {5w —2 + wi(3w+2)} = 0 9w3-13w2 +24w-4 = 0,

on multiplying out and rearranging terms, and this equation has roots a2,

,82,

72.

Example 2. Discuss the nature of the roots of the equation 3x4-4x3-12x2+k = 0 for different real values of k.

(2, -32)

Fig. 23.1

Write

P(x) = 3x4-4x3-12x2;

then

P'(x)

12x3-12x2 — 24x

and thus

P'(x) = 0 when x = —1, 0, or 2.

Since P(x) > 0 for sufficiently large x, the values —1, 0, 2 for x yield respectively a minimum, maximum and minimum of P(x). The graph of y = P(x) thus assumes roughly the shape shown in Figure 23.1. 487

[23


Now the given equation may be written as

P(x) =

—

k

and the number of real roots of the equation may be deduced immediately by considering the intersections of the curve y = P(x) and the straight line y = k: k < 0: the equation has two unequal real roots and two complex conjugate roots. k = 0: the equation has three unequal real roots, one of multiplicity 2. 0 < k < 5: the equation has four unequal real roots. k = 5: the equation has three unequal real roots, one of multiplicity 2. 5 < k < 32: the equation has two unequal real roots and two conjugate complex roots. k = 32: the equation has a real root of multiplicity 2 and two conjugate (complex) roots. k > 32: the equation has two pairs of conjugate complex roots. —

Ex. 10. What are the roots of the equation in Example 2 in the cases (ii) k = 5; (iii) k = 32? (i) k = 0;

Exercise 23(a) 1. Form the equation with roots 1, —2, 3 —0, 3+0. 2. Form the equation with roots 1+2j, 1— 2j, — 2, 3. 3. Form the equation with a root —1 of multiplicity 3 together with a pair of roots +j, —j. 4. Solve the equation

z4-2z3 +3z2 — 4z+ 2 = 0,

given that it has a root of multiplicity 2. 5. Use the Remainder Theorem to solve the equation z3-3z2 — 8z +30 = 0, given that it has an integral root. 6. The equation

24z4 +100z3 +126z2 + 27z— 27 = 0

has a root of multiplicity three. Solve the equation completely. 7. If a, fi, y are the roots of the equation z3 — z— 4 = 0, form the equation with roots:

(i) 488

111

(11)

a+1/1+1y +1

, fi , 7

1]


8. If a, /1, y are the roots of the equation

z3 — 3z2 — 3z— 4 = 0, form the equation with roots: (ii) ,x2,/32,y2. (i) 2a, 2fl, 2y; 9. If a, fi, y are the roots of the equation

z3 +jz+1 +j = 0, form the equation with roots: (ii) (1 +Da, (1+Dig, (1+.D7. (i) lot, ifl, •17; 10. If a, fl, y are the roots of the equation 2z3 —z2 + 4z+ 7 = 0, form the equation with roots (i) a+2,/3+2, y+2; (ii) a — 3, ft— 3, 7— 3. 11. If a, /3, y, 8 are the roots of the equation z4 — z— 5 = 0, form the equation: (i) with roots — a, — /3, —y, — 8; (ii) with roots a + 2, t+2, y+2, 8 + 2; (iii) with roots co, 32, y2, 82. 12. If a, /3, y are the roots of the equation 3z3 — 20z2 + 39z— 18 = 0, form the equation with roots a-3, /3-3, y-3. Hence find a, /3, y. 13. Solve the equation

z4 +4z3 +5z2 +2z-2 = 0.

by first increasing the roots by 1. 14. Solve the equation

z3+3jz2— 5z — 3j = 0 by first increasing the roots by j. 15. Show that the cubic equation z3 + 4z2 — 3z+ 25 = 0 has three real roots, one of which is positive and the other two negative. 16. Show that the cubic equation

z3—3z+ k = 0 has three distinct real roots if and only if k is real and —2 < k < 2. 17. Discuss the nature of the roots of the equation 2z3 — 15z2 + 24z+ k = 0 for different real values of k. 18. Discuss the nature of the roots of the equation z4 — 4z2 — 2z2 + 12z+ k = 0 for different real values of k.

489

[23

POLYNOMIAL EQUATIONS 19. Discuss the nature of the roots of the equation 3z4-4z3-24z2 +48z+k = 0 for different real values of k.

20. Find a necessary and sufficient condition that the three roots of the cubic equation z3+3Hz+G = 0 should be real and unequal.

2. RELATIONS BETWEEN ROOTS AND COEFFICIENTS FOR POLYNOMIAL EQUATIONS Consider first the cubic equation ao z3+ai z2+a2z+a3 =0 with roots a, y. We have aoz3+ aiz2+ a2z a3 aaz — a) (z )8) ao Z3aacz + fl+y) z2+ao(fly+ ya+afl) z—aocch and, equating coefficients of z2, z' and z°, a+ )3 -Fy = fly + aa + aft = + a2/a0,

afly = Similarly, for the quartic equation ao z4+a1 z3+a2z2+a3z+a4 = 0, with roots a,

A y, S we have

a0z4 +al z3+ a2 z2+ a3 z+ a4 aaz — cc) (z — fl) — ao z4 — ao(ec- F fl+y +8) z3-Fao(cc, + ccy + cc8 + fly + fl8 +78) z2 — 41678+ 78a + Safl + ccflY) z+ aoafrY8 from which, by equating coefficients, a+fl-Fy+8 = -- ch/ao, cti3+ay+ca3+fly+,88+Y8 = +a2/(10, 3y8 + y8a + Safi +afly = —a31120, afly8 = +a4/ao. 490

RELATIONS BETWEEN ROOTS

2]

In words, the sum of the roots one at a time is — a,lao, two at a time is +adao, three at a time is — adao, and four at a time is adao. For brevity, we denote a + y + 8 by Ea, afl + ay + a8+ fly ,88±y8 by Zaft, etc.; no ambiguity arises provided we know the degree of the equation under consideration (in this case four). Ex. 11. The roots of the equation 2z3 3z2— 5z -13 = 0 ex, ft, the numerical values of /a, Zafi, afiy. y. Write down are —

Ex. 12. The roots of the equation z4 — 4jz+ 1 = 0 are a, A y, 8. Write down the numerical values of /a, Zafl, Zafly, aflya. *Ex. 13. If the coefficients of a cubic equation are all real, it follows that Ea, /aft and fly are all real. Explain how this can be so, even if the equation has complex roots. Illustrate your answer by considering the equation z3— 1 = 0. Now suppose that the general polynomial equation

+ an = 0 2zn-2 ao zn aizn-1 a has roots al, a2, ..., an. Then it may be shown by induction that the results proved previously in the particular cases n = 2, 3, 4 hold generally. Zai = — di /a° Za ja; = +a,lao Eat c6i ak = — as/a,

cha,a, ... an = ( — on anlao. (Let the equation with roots al, a2,

an_, be

aozn-1+bizn-2±b2zn-3+ ••• +bn-i = 0

and consider the identity (z — an) (a0 zn-1+bi Zn-2 b2Z"+... +bn_1)

ao zn +ai zn-l+a2zn-2+ ... +an.) Ex. 14. By considering the equation zn-1 = 0, prove that the sum of the nth roots of unity is zero (n 2).

Example 3. Find the quartic equation with roots 1, 3, We have Ea = — 2,

—

2,

—

4.

Zai3 = 3-2-4-6-12+8 = —13, Eafly = 24+8-12-6 = 14,

afly8 = 24, 491


[23

and the required equation is z4 +2z3 — 13z2 — 14z + 24 = 0. Example 4. If the roots of the equation z3— 4z — 6 = 0 are a, Ay find (i) Ea2, (ii) Ea2,6, (iii) EC43, (iv) Ea4. We have Ea = 0, Ea f3 = —4, a/37 = 6: (i) Ea2= (ECC)2— 2Eafl = 0+8. = 8. (ii) Notice first that Ea2/6 contains six terms: Ecc2i6 = cc2fl + a27+my +fiza +720, 4_ 72/3. Consider the product (Ea) (Eat): each term such as a2,6 occurs just once, but the terms a,67 appears three times, as a .,67, /6.7a and 7. aft. Thus Ea2,6 = EaEaft — 34'7 = —18. (iii) Since a is a root of the given equation, we have a3 — 4a — 6 = 0 and similarly for i6 and y. Adding these three results Ea3 4Ea — 18 = 0. -

Ea3 = 18.

Thus

(iv) By an argument similar to that adopted in (iii), Ea4 — 4Ea2 — 6Ea = 0. Thus on using the result of part (i).

Ea4 = 32,

Example 5. Solve the equation 8z3— 12z2 — 66z + 35 = 0 given that the roots are consecutive terms of an arithmetic sequence. From the equation we have Zoc = i", Eccfl = —V,

ah = —11..

Since the roots are consecutive terms of an arithmetic sequence, we take a = a— d, )3 = a, y = a+d. 492

2]


Then Ea = 3a, Ea/3 = 3a2— d2, giving 3a = 3a2 —d2 = Thus a = 4-, d = +3 and the roots are —21-, 1, 31. Example 6. Solve the simultaneous equations x+y+z = 2, o2 ± y2 + z2 =

.x3+y3 +z3= 116. Let x, y, z be the roots of the cubic equation t 3 —at2+bt—c = 0. Then

a = Ex = 2, b = Exy = 4-{(Ex)2 — Ex2} -= 1-{4 — 30} —13.

To find c, we make use of the identity x3 +y3+

—3xyz = (x+y+ z) (x2+y2 + z2 yz — zx — xy); c = xyz = 4{Ex 3—Ex[Ex2—Eyz]} = 1{116 — 2[30 + 13]) = 10.

Thus x, y, z are the roots of the cubic equation t3-2t2 — 13t — 10 = 0. By observation, this has a root t = —1: +1) (t2 -3t — 10) = 0, i.e. (t+1) (t+2) (t-5) = 0, and x = —1, y = —2, z = 5 (or any equivalent permutation). Example 7. If the roots of the cubic equation z3+3Hz + G = 0 are a, 13, y, form the cubic equation with roots (/l — y)2, (y— a)2, (cc_ fi)2.

w = (/-y)2 w = Eo2

o2

aw = a(Ea)2-2aZafl—a3-2afly (since (Ea)2 = Ea2 +2Eafl), aw = —6Ha—a3+2G (since Ea = 0, Eocfl = 3H), a = 3G1(w + 3H) (since a3= — 3Ha— G) 27G3+9HG(w+3H)2 +G(w+3H)2 = 0 (since a3+3Ha+G = 0) w3+ 18Hw2 +81H2w+27(G2 +4H3) = 0. 493


[23

Thus, (fl—y)2is a root of the equation w3 +18Hw2 + 811/2w+ 27(G2 + 4H3) = 0 and, by symmetry, so also are (y —a)2, (a _i3)2• Note: the equation derived in Example 7 is of importance in studying the nature of the roots of cubic equations—see Section 4.

Exercise 23(b) 1. If a,

y are the roots of the equation z3-4z2 +z+2 = 0,

find: (i) Ea2; (v) Ea2/32; (viii) Eafi(a2 +,82); 2. If a,

(ii) Eat; (iii) Ea4; (vi) (a + 2) (fl + 2) (y + 2); (ix) El/ale;

(iv) Ea2fl; (vii) El/a; (x) E(cc + 08)2.

y, 8 are the roots of the equation z4-4z3 — z + 9 = 0,

find: (i) Ea2; (v) Ea3fi;

(ii) Ea2fl; (vi) I(a+fi+y)2;

(iii) El/a; (vii) E(a+ 1) (13+ 1).

(iv) Ea3;

3. If a, fi, y are the roots of the equation

z3+pz+q = 0, find: (i) Ea4; 4. Solve the equation

(ii) El/a4;

E(ce +fl —27).

2z3 -3z2 — 23z+ 12 = 0,

given that its roots are successive terms of an arithmetic sequence. 5. Solve the equation

32z3 + 48z2 + 6z— 5 = 0,

given that its roots are successive terms of an arithmetic sequence. 6. Solve the equation

24z3 — 382.2 + 19z— 3 = 0,

given that its roots are successive terms of a geometric sequence. 7. Solve the equation

2z3 — 11z2-5z+ 50 = 0,

given that one of the roots is twice a second root. 8. Solve the equations:

x+y+z = — 4, yz+zx+xy = 1, xyz = 6.

494

2]


9. Solve the equations:

x+y+z = 1, x2+ y2+ z2 = 29, x3 +y3+ z3 =—29.

10. Solve the equations:

x+y+ z = 4, x2+y2+z2 = 14, x3 +y3 +z3 = 34.

11. The roots of the equation:

ax3+3bx2+3cx+d = 0 are in arithmetic progression. Express c in terms of a, b and d.

(0 & C)

12. Prove that, if the cubic equation

ax3+bx2+cx+d = 0 has a pair of reciprocal roots (that is, a and 1/a), then a2 —d2 = ac—bd. Verify that this condition is satisfied for the equation 6x3 +11x2 — 24x— 9 = 0 and hence, or otherwise, solve the equation.

(0 & C)

13. If a, ft, y are the roots of the equation

z3-Fpz+q = 0, form the equations with roots: (0 az,ig2; y2;

(ii) fly, yoc, afl; (iv) /3+y —a, y+ec — fl, cc+ fl --Y. (iii) fl +7,, Y + ce, a + fl; 14. If a, Ay are the roots of the equation z3+pz+q = 0,

form the equations with roots: (i) ct3, /33, y3;

(ii) 0:(6+Y), AY +cc), Y(ce+,8)•

15. If a, Ay are three non-zero complex numbers such that a +fl+y = 0, prove that a2 — fly = /62 — ycc = y2— 4'. 16. Given that a, /3, y are the roots of the equation x3+px2+qx+ r = 0, and that a = fly — a2, b = yoc—fl2, c = of — y2, prove that

a = pot+q, b = pfl+q, c = py+q. Hence, or otherwise, prove that

ax+bfl+cy = (a+ b+ c) (a-l-i6+y). 17. If a,

(0 & C)

ft, y are the roots of the equation x3+px2+qx+r = 0, 495

POLYNOMIAL EQUATIONS find the equations whose roots are: a2 — R7, R2 — vcc, — ccfl; What can be said about a, A y if (i) q3 = p3r;

[23 (ii)

ft+y —2oc, y+a-216,

(ii) 2p3-9pq+27r = 0?

(0 & C)

18. By considering the product (/1+ y) (y+a) (a +/3), find a necessary and sufficient condition that two of the roots of the equation z3+pz2+qz+r = 0 (r * 0) should be equal in magnitude but of opposite signs.

3. THE CUBIC EQUATION For practical purposes an algebraic solution of the general cubic is of limited value since, in practice, only approximate numerical solutions are required and these are best found by the iterative methods to be described in Chapter 26. Nevertheless, a brief survey of the method of attack seems worthwhile on at least three counts: having seen a solution to the general quadratic it is natural to wonder whether a similar method holds for the cubic; such a question did occur to mathematicians of the past and the study of the cubic is of considerable historical interest; finally, the question whether general polynomial equations are soluble leads one to some of the central ideas of modern mathematics. The general cubic equation was first solved in the sixteenth century by Tartaglia and published by Cardan in one of the most famous of all acts of mathematical plagiarism. (See Cardan: the Gambling Scholar by 0.0re for a biography of this unsavoury yet fascinating character.) The general quartic soon followed (Ferrari) but, as has been pointed out already in Section 1, the general quintic evaded all attempts at solution, for the reason discovered independently by Abel and Galois in the nineteenth century. An account of the Italian mathematicians of the sixteenth century whose work we have referred to will be found in History of Mathematics by D. E. Smith, Volume 1, Chapter 8. We shall take the general cubic in the form

aoz3+3a1 z2+3a2z+a3 = 0. (1) If we apply the transformation z -÷ (w—a1)1a, we have aoz3+3a1z2+3a2z+a3 = 0 agz3+3a,c4z2+3a2a,a0 z+a3ag= 0 (w—a1)3 3ai(w—ai)2+3a2a0(w—a,)+a,4 = 0 ws + 3w(a2a, -4)+(2a? -3a,a,a,+a,4) = 0 w3+3Hw+G = 0, where H = a2a0 — G = 2(43 — 3a0a,a2+a3ag. 496

CUBIC EQUATION

Thus any cubic equation may be reduced to (2)

w3+3Hw+G = 0

and we shall, from now on, regard this as the general cubic equation which it is our intention to solve. Ex. 15. Show how to reduce the equations: (i) x3 — 9x2 + 22x— 3 = 0, (ii) 2x3 + 6x2 +10x — 1 = 0, to the standard form x3 + 3Hx + G = 0. If the roots of (1) are z1, z2, z3and the roots of (2) are w1, w2, w3, then the geometrical interpretation of the transformation z --> (w — al)la0 is that it transforms the triangle z1z2z3into the similar triangle w1w2 w3which has its centroid at the origin (w1+w2+w3 = 0). (See Figure 23.2.)

W2

Fig. 23.2

Ex. 16. Why are the triangles z1 z2 z3and w1w2 w3similar? Cardan's solution of (2) consists in expressing the roots as the sums of pairs of complex numbers lying at the vertices of two equilateral triangles with centroid at the origin. (There is no 'completing the cube' process in general and thus it is not possible to find a further transformation that maps the triangle w1w2w3into an equilateral triangle.) Since w3 _3wpq ( pg ± 17 3) = (w q) 04, (j) q) (W wq) (see Exercise 19(b), Question 12) we have

w3-3 wpq — (p3+q3) = 0 w = p-Fq or w = cup- - co2q or w = Now write

—pq = H, —(p3+q3) = G;

coq.

(3) (4) 497

[23


if we can find values p, q satisfying (4) then we shall have a complete solution of the cubic equation. From (4), p3and q3are the roots of the quadratic equation Z2 + GZ— H3= O.

(5)

Let a, /6 be the roots of the quadratic equation (5). Take p„ q, as any cube roots of a and 13; then

14+0 = a+161= —G,

Pi qi = aft = — H3. If H = 0, the original cubic is immediately soluble; if H

0, then

(ART = 1, Hl

Pigs

giving

6

—H

'

where e = 1, a) or (o2, andpi and qile are suitable values to take for p and q. In summary, to solve the equation w3+3Hw+ G = 0 (H

0),

let a, ft be the roots (repeated if necessary) of the quadratic equation

z2+ Gz— H3 = 0, and let p = z1, q = 0e where e = ca01(— H). Then p

wp (.02q, (02p +

are the required roots for the cubic equation. For (w

—

+ (w — (cup + w2q)) (w — (Op + cog)) w 3 -3wpq — (pa ±q3),

and p, q have been so chosen that

pq = —H, p3+q3= —G. Example 8. Solve the equation z3— 6z + 6 = 0. We seek p, q such that p3+q3 = 6, —

pq = 2. Thus, p3and q3are the roots of the quadratic Z2 +6Z+8 = 0, 498

(6)

3]

CUBIC EQUATION

from which it follows that p3 = 2, q3 = 4 = 22. Thus

z3-6z+6 = 0 z3 + (2*)3± (21)3- 3Z(2i) (21) = (Z

+

6,21 +co20) (z+co22*+co21) = 0

z = —2*-21 or — co2*— 02* or —co221—cog z = — (2++ 21) or {(2*+2*)+ j(21 —2*)V3}/2. *Ex. 17. Show that w= Z—(HIZ) transforms equation (2) into

Z6 + GZ3 — H3 = 0 the roots of which are p, (op,

q, cog, co2q.

*Ex. 18. With the notation of Ex. 17, show that, to each w correspond two values of Z and that to the triangle w1wiw3formed by the roots of (2) correspond two equilateral triangles.

Fig. 23.3

Suppose now we have determined p and q as two complex numbers. Then, by (6), we have

w1 = p+q, (See Figure 23.3.)

w2 =

(02q, w3

Ex. 19. Show that w2, w, may be found by constructing equilateral triangles on —p, —q as base.

499


[23

From now on we shall suppose that the coefficients of (2) are real: then either one root is real and the other two are conjugate (complex) numbers, or all three roots are real. From (5), since G, H are real, either p3or q3 are both real or else p3and q3are complex conjugates, in which case we must have IpI = lql. Case (i). p3and q3both real (and unequal)—in which case we may take p and q as both real and the geometrical construction (Figure 23.4) shows that, of the solutions of (2), one is real and two are complex conjugates. Case (ii). p3and q3are complex conjugates (and thus p, q may be taken as complex conjugates). In this case, we must have 11,1 = lqi and the geometrical construction (Figure 23.5) shows that all three roots of (2) are real.

F'g. 23.4

Since this exhausts all the possibilities, we have arrived at the famous paradox that, if only one root of (2) is real, then p and q are both real, but, if all three roots of (2) are real, then they arise from complex values of p and q (the irreducible case). Put another way, if all three roots of a cubic equation are real then the working of Cardan's solution necessarily involves complex numbers p and q at the intermediate stages. Ex. 20. Explain geometrically how (i) a root of multiplicity two, (ii) a root of multiplicity three, arise. The expression 500

A = [(w2 — w3) (ws

w2)12

3]

CUBIC EQUATION

is called the discriminant of the equation

w3+3Hw+G = 0. From Example 7,

A = — 27(G2 + 4H3).

Consider now the various cases that may arise (remember that we are considering cubic equations with real coefficients). (i) One real root a, two conjugate complex roots )6' + jy. A = P./Yr (cx — ViY)2 (3

— ec)2

= — 4y2( — a2 — /32 — y2 +24)2

0. Thus, the nature of the roots determines the sign of A. *Ex. 21. Show, conversely, that the sign of A determines the nature of the roots of the real cubic equation w3+ 3Hw+ G = 0.

The case A > 0 leads to three real solutions via complex values of p and q: the consequent working in Cardan's solution is frequently rather 5

PPMII

501


[23

tedious, but can be avoided in this case by employing the trigonometric solution illustrated in Example 10, which employs the identity cos 30 = 4 cos3 0-3 cos 0. (See p. 244.)

Example 9. Solve the equation z3 — 6z+ 4 = 0. Here H = —2, G = 4 and A > 0: there are three real roots. Put z = 2(2)i cos 0. 16,12 coss 8 -12,/2 cos 0 =

—

4

4 cos3 0-3 cos 0 = cos 30 = — giving 0 = in or -,12-n or firr and z = 2 or 21cos -,5gr or 21cos

+PT.

*Ex. 22. Show that if A > 0, the substitution z = 2(— H)i cos 0 will always yield the solution of the cubic equation z3 + 3Hz+ G = 0.

Exercise 23(c) Solve the cubic equations in Questions 1-5, using a trigonometric substitution if possible and otherwise employing Cardan's solution. 1. z 3— 2z + 4 = 0.

2. 4z3 5z + 6 = 0. —

3. z3 — 3z 1 = 0. —

4. 12z3 — 9z-- 2 = 0. 5. z3 — 9z + 12 = 0. 6. Reduce the equation

z3 — 6z2 + 1 8z — 22 = 0 0, to the form z3 + 3Hz+ G = and hence solve it completely.

7. Show that the equation is reduced to the equation

az3+3bz2+3cz+ d = 0 p(w — q)3= q(w — p)3

by the transformation w = az+ b where p, q are the roots (supposed unequal) of the equation

(ac b2) A24" (a2d 3abc + 2b3) A (ac b2)2 = 0. —

—

Solve the equation

—

z3 15z2+ 57z 5 = 0. —

—

8. Solve the equation z3+5z2+5z +4 = 0.

502

—

(0 & C)

3]

CUBIC EQUATION

9. Solve the equation z3 + 5z + 2j = 0. 10. If p, q are real, find the turning points (if any) of the graph of y = x3+px + q. Deduce a necessary and sufficient condition for all the roots of the equation x3+px+q = 0 to be real. Prove that the roots of the cubic equation det (A— AI) = 0 where a hg A= (h a f) g f a are all real. Can two of the roots be equal? 11. (The Quartic Equation.) Show that the general quartic equation z4+4pz3+6qz2+4rz+s = 0 can be reduced to the form

w4+aw2+bw+c = 0. By identifying this equation with the equation fl(w+ (w2 + a)2

show that a is a root of the cubic equation 8a3 4acc2 -8ca + 4ac — b2 = 0. If one of the roots of this equation is aoand the corresponding values of /3, y are flo, yo, show that w4 + aw2+ bw+ c = (w2--fitw+cco — YoflO) (w2 +fit w+ao+YoflO). Solve the quartic equation x4 + 4x— 1 = 0.

Miscellaneous Exercise 23 1. Prove, by graphical methods or otherwise, that a cubic equation (with real coefficients) of the form ax3+bx2+cx+d = 0 has at least one real root. Prove that, if A is real, the equation x3-3x2 +3x+A = 0 is satisfied for one and only one real value of x. (0 & C) 2. By considering two alternative expansions of the expression In {(1+ax) (1 + fix) (1 + yx)} (a, y all real), prove that +fl4 + 74 = p4 A -2q

+4pr + 2q2, where a, y are the roots of the equation x3+px2+qx+r = 0. 3. In the equation x4+ ax3— 20x2+ bx— 576 = 0, 5-2

503


[23

a and b are to be chosen so that two of the roots are equal and the sum of the other two roots is zero. Find all possible values of a and b, supposing them chosen so that the roots (0 & C) of the given equation are all real. 4. Find the value or values of a for which the roots of the equation 2x3 +6x2 +5x+a = 0 are in arithmetical progression.

(0 & C)

5. Given that —2+ jk is a root of the equation 2x4 +8x3 +11x2 +4x+5 = 0, find k and the other three roots of the equation. 6. Given that a, /3, y are the roots of the equation x3 — px2+qx— r = 0, and that sn= an+fln+yn, prove that si = P, s2 =p2- 2q, s3 = ps2 —qs1+3r. Calculate s2 and s3for the equation x3 — 12x2 + 30x-20 = 0. Show that (s2) 4 and (s3)1are both near to 9 and verify that 9 approximates to a root of the equation. (0 & C) 7. If co is a root other than 1 of the equation og = 1, prove that the other roots are (02, idS, (04, (05, (06. Prove that 1+ CO+ 612+ 613+ (04+ (05+(O6 = 0. If a = co+ wo, 11 = co2+ ws, y = w3+ 01, prove that the equation with roots a, ft, y is z3 +z2 -2z— 1 = 0. Hence, or otherwise, find the values of (i) cos tir+ cos fir, + cos lg. (ii) cos PT cos -17r cos tir.

(0 & C)

8. When f(x) and g(x) are given polynomials having no common factor, prove that the values of the constant k for which the equation f(x)— kg(x) = 0 has a repeated root are given by f(a)/g(a) where a is a root of the equation f(x) g'(x) — f /(x) g(x) = 0. Hence, or otherwise, find the values of k for which the equation x3 — 3x2 + 3kx— 1 = 0 has a repeated root, and find also all the roots of the equation for each case. (0 & C) 9. Discuss the reality of the solution of the equation 2x3 +3x2 -36x+k = 0 for different real values of k. 504

MISCELLANEOUS EXERCISE 23 10. If f(x) is a polynomial in x which has a factor (x- CO2, prove that (x- a) is a factor of f'(x). Prove conversely that, if f(x) and f'(x) have a common factor x- a, then (x- a)2is a factor of f(x). Solve the equation 4x 4 + 4x3-31x2-66x - 36 = 0 given that it has a repeated root.

(0 & C)

11. If a, fl, y are the roots of the equation x3+px+q = 0, find the equation with roots ,B/y+71,6% y/oc+ c6/7, ah8+ 16' loc.

(0 & C)

12. The sum of the kth powers of the roots of the equation x4+px2+qx+r = 0 is denoted by sk. Prove that s2 = -2p, ss = -3q. sn+4 +PSn+2 and deduce that

qSn÷i

rs„ = 0 (II

0),

= 2p2 - 4r, s, = 7q(r - p2).

Given that the roots a, b, c, d of the equation are all real and that 5, = 0, prove that q = 0 and deduce that (a+ b) (a + c) (a + d) = 0. (0 & C) 13. Show that, if z = 2(- H)i cos 8, where z is a root of the cubic equation z3 + 3Hz + G = 0, then cos 30 = ,s/(-G2/4H3). Deduce that, if H < 0 and G3+ 4H3< 0 the cubic equation has three real roots. By using the expression cos 30 = -1-(e 310 + e-30), solve the equation when treating separately the cases

G2+ 4H3 > 0, H < 0 and H > 0.

505

24. Vector products and their applications

1. THE VECTOR PRODUCT We have already seen, in Chapter 11, that it is frequently useful to find a vector which is perpendicular to two given vectors; for example, to find the equation of the plane through three given points A, B, C it is sufficient to know the unit normal vector; that is, a unit vector perpendicular to both AB and AC. We therefore introduce a new method of combining two vectors a and b to form a third vector perpendicular to both of them and called their vector product, written a A b (or a x b). It is defined by

a

Ab

= alibi sin 0e,

where 0 is the angle between a and b (see Chapter 11) and e is a unit vector perpendicular to both a and b and with sense that makes a, b, e (in that order !) a right-handed triple (see Chapter 3). The relative positions of a, b, and a A b are shown in Figure 24.1 where a, a A b lie on the page and b points into the page. Fig. 24.1

*Ex. 1. Show that a A a = 0. *Ex. 2. Show that a A Ab = A(a A b). *Ex. 3. Show that, if a and b are parallel, then a A b = 0.

*Ex. 4. Show that, if a A b = 0, then either a and b are parallel or one or other or both of a, b is zero. *Ex. 5. Show that the vector product is anti commutative; that is, that -

anb =—b A a. *Ex. 6. Verify that i A i = j A j = kA k= 0 and that

jAk=—kAj=i, kAi= —iAk=j, iAj=—jAi=k. Ex. 7. a is a vector of magnitude 2 units pointing due east; b has magnitude 3 units

and points N 60° E. "e' is a vertical unit vector. Find a A b and b A a.

506

1]

VECTOR PRODUCT

So far, apart from the result of Ex. 2, the vector product appears remarkably dissimilar to the product of ordinary numbers. However, there is one rule which vector products obey and which, in a sense, justifies the use of the word 'product': it is distributive over vector addition; that is a A (b + c) = (a A b) + (a A c). We shall deduce this very important result in a manner reminiscent of that used to prove the analogous result for scalar products: a . (b + c) = (a . b) + (a . c). In Chapter 11, we saw that the scalar product of a = OA with a unit vector ñ could be interpreted geometrically as the (scalar) projection of a

Fig. 24.2

in the direction fi and it was this fact that enabled us to deduce the distributive law for scalar products. Consider the plane, II, perpendicular to fi and through 0; let the foot of the perpendicular from A to II be D; then the vector OD is called the (vector) projection of a on the plane H (see Figure 24.2). It follows from the definition of the vector product that OD = la A fil, but

OD a A fi,

since OD and a A ft are perpendicular to one another. As in Chapter 11, Section 1, the vector projection of a on H is unique and, if the projection of a, and a2on H are OD, and OD„ then the projection of OB = a, + a, is OC = OD, + OD, (see Figure 24.3 and consider the mid-point of A,A2). In words, the (vector) projection, in a plane, of the sum of two vectors is the sum of the separate projections of the two vectors. 507

VECTOR PRODUCTS

[24

Now consider 9 A (a1+ 22): from what has just been demonstrated, lit A (al + a01 = 101 A aj-F(fi A a2)!. But 9 A (a1+82) is perpendicular to the plane OBC and thus to OC, 9 A a1 is perpendicular to OD1, 19 A a2 is perpendicular to OD2 and thus, by rotating the parallelogram 0 D1CD2through a right angle in the plane H, it follows that ñ A (al + a2) = fi A al -Ffi A a2. B

Fig. 24.3

*Ex. 8. By writing a = Afi, deduce that a A (b+c) = (a A b)+ (a A c). Show also that

(b+c) A a = (b A a)+(c A a).

With the distributive rules behind us, it is an easy matter to express the vector product a A b in terms of components in three mutually perpendicular directions i, j, k. If a = chi+ a2 j+a3k and b = b1i+b2j+b3k, then a A b = (aii+ a2 j + a3k) A (bl i+b,j+ b,k) = (a2 b3 — a, b) i + (a3bi — b3) j + (a, b2— a2bi) k,

on repeated application of the distributive rule and using the results proved in Ex. 6. The result is best remembered written formally as a determinant: aAb =

a1 a 2 b/ b2 b,

Ex. 9. If a = i+2j+3k and b = i—j—k, find a A b. Since a A = lal Ibl sin 0, we see that the magnitude of the vector product gives us twice the area of AOAB (see Figure 24.4). The vector 508

1]

VECTOR PRODUCT

}- (a A b) is often referred to as the vector area of AOAB. (Notice that the vector area of 6.0AB is minus the vector area of AOBA.) More generally, the vector area of the triangle ABC is +(AB A AC). Ex. 10. OABC is a tetrahedron; prove that the sum of the vector areas of the triangles OBC, OCA, 0 AB is the vector area of the triangle ABC.

Fig. 24.4

Example 1. ABCD is a plane quadrilateral whose sides CD, BA intersect at O; if P, Q are the mid-points of the diagonals AC, BD, prove that the area of AOPQ is one-quarter of the area of the quadrilateral ABCD.

Fig. 24.5

Take origin 0 and a = OA, b = Aa, c = OC, d = ,uc. We have

p = 1-(a+ c), q = -1-(Aa+,uc), and the vector area of OPQ = 1-1) A q

= 8(a+ c) A (Aa+uc) = t,(t— A) (a A C) on using the distributive and anti-commutative rules. But since ABCD is a plane quadrilateral, its area is the magnitude of the sum of the vector

509

VECTOR PRODUCTS

[24

areas of LADB and ABDC (notice that we letter the triangles in the same sense). Thus ABCD = -(d A b+b A a+a A d)+-1-(d n c+c Ab+b Ad)

= 2(b A a+a A d+d AC-1-C Ab) =

A (d—b)—c A (d — b)]

= -1-(a—c) A (d b) —

A (UC— Ala) = = 1(ic A) (a A c), and the result follows. —

Ex. 11. The position vectors, relative to some origin 0, of the vertices of a convex polygon PiP2P3...Pnare denoted by Pi P2 n •••, Pn• Extend the definition of the vector area of a triangle to the vector area of a convex polygon and show that the vector area of P1P2P3...Pn is ,

,

i(Pi. A Pz +P2 A Ps + • • • ±Pn--1 A Pn +Pn A PO. Prove the result of Example 1 by considering the vector areas of ABCD and

OPQ.

2. PRODUCTS OF THREE VECTORS If from two vectors, b and c, we form the vector product b A c, we may multiply a by this new vector in two quite different ways, to give either a scalar or a vector: (i) the scalar triple product, a.(b A c); (ii) the vector triple product, a A (b A c). The component form for a. (b n c) is easily found. Recall that x.y = xiy1 +x2y2 +x3y3; thus, since a = a1i+a2 j+a3k and

i bAC =

k

b1 b2 b3 C1

we have

j C2

C3

a1 a2 a3

a.(b A C) = b1 b2 b3 c1

C2

C3

*Ex. 12. Prove that a. (b A c) = b.(c A a) = c.(a n b) (that is, the scalar triple product is unaffected by cyclic interchange of the letters) and that

a. (b A = (a A b).c (the . and A may be interchanged).

510

2]

PRODUCTS OF THREE VECTORS

The scalar triple product a .(b A c) has an interesting and important geometrical interpretation. Suppose ñ is perpendicular to the plane OBC (see Figure 24.6). Then b A c = Sfi, where S is the area of the parallelogram OBDC. Thus a. (b A c) (a.11) S = volume of the parallelepiped formed by a, b, c. D

Fig. 24.6 Ex. 13. Show that a .(b A = 0 p 0, A, B, C, lie in a plane. Ex. 14. Prove that x.(y A z) = (x A y).z without using components (and thus without assuming the distributive law a A (b+ c) = a A b+ a A c). Ex. 15. The fact that x.(y A z) = (x A y).z may be used to verify the distributive law: define d = a A (b+ c)— a A b— a A c and form the product fi.d, where fi is an arbitrary unit vector. Show that u.d = 0. Why does this imply that d = 0?

*Ex. 16. Show that the volume of the tetrahedron OABC is -1-a . (b A c). *Ex. 17. Prove that, if any one of the vectors a, b, c is a scalar multiple of one of the other two, then a .(b A = 0.

The vector triple product a A (b A c) is a little more difficult to express in component form. First observe that, if a is perpendicular to b and also perpendicular to c, then it is parallel to b A c and thus a A (b A c) = 0; let us assume that none of a, b, c is zero and that a is not perpendicular to b and thus that a.b 0. Again, if b = Ac, a A (b A c) = 0; let us assume that b Ac. Since b A c is perpendicular to both b and c and since a A (b A c) is perpendicular to b A c it follows that a A (b A c) lies in the plane of b and c. Thus, a A (b A c) can be expressed uniquely as Ab -Fix; we have to determine A and 1a. By Ex. 17, a .[a A (b A c)] = 0, .*. A(a .b)+,a(a c) = 0,

A

(a . c)

(a. b)

(since a. b

0), 511

[24

VECTOR PRODUCTS and thus

a A (b A = v[(a c) b — (a . b) c].

To find v, consider the i component of both sides:

a2(b1C2 — b2 — a3(b3c1 —b1c3)

= v[(a,c,+ a,c2+ a3c3) . • . b1(a2 C2 ±

a2b 2+ a3b3) ci],

a3c3) — ci(a2 b2 +a3b3) = v[bi(a2 C2 4- a3c3)—ci(a2b2---a3b3)]• .•. v = 1.

Thus we have finally a A (b A c) = (a.c)b—(a.b)c. *Ex. 18. Prove that the vector triple product is not associative; that is

an(bAc)*(aAb)Ac. A (C A a)+c A (a A

Ex. 19. Prove that a A (b A C) b

= 0.

Exercise 24(a) 1. Prove that (a— b) A (a + b) = 2a terms of areas.

A

b and interpret the result geometrically in

2. What can you deduce from the equation r A a = r A b? 3. Prove that a +b+c = 0 b A c= c A a= a A b and interpret the result geometrically. IsittruethatbAc=cAa=aAba+b+c= 0? 4. Find a A b in the following cases: (i) a = 3i—j+k, b = 2i+j+k; (ii) a = 5i+4j+3k, b = 3i+4j+5k; (iv) a = 21— j — k, b = 2i + j + k; (iii) a = i—k, b = j+k; (v) a = ai+ck, b = bj. 5. If a = 3i +j+ 2k, b = i +2j+ 3k, c = i — 3j-4k, find a.(b A c). What can you deduce about the points A, B, C? 6. If a = i+j, b = 2i+j, c = i+k, form the vector products a A b and (a n b) n c and mark the position of the five vectors a, b, c, a A b, (a n b) n c in a rough sketch. 7. The points A, B, C have position vectors a = 4j + 3k, b = 2j — 3k, c = 3i-6k. Find the unit vector perpendicular to the plane 0 AB and deduce the perpendicular distance from C to this plane. 8. With the notation of Question 7, find the area of AOAB and hence the volume of the tetrahedron OABC. 9. If A, B, C are non-collinear points with position vectors a, b, c, prove that the vector b A c+c A a +a A b is normal to the plane ABC. If d is the position vector of the point D and a = 5i+j+k, b = i+j+3k, c = 31+ 4j — k, d = 8i+ 3j + 2k, find: (i) the unit normal to the plane ABC; (ii) the area of the triangle ABC; (iii) the volume of the tetrahedron ABCD.

512

2]

PRODUCTS OF THREE VECTORS

a2 i+a3k, b = b1i+b2 j+b3k, c = volume of the tetrahedron OABC is

10. If a =

c2 j + csk, prove that the

a3 bl b 2 b 3 cubic units.

a1 a2 6

C1

C2 C3

If d = dii + d2 j+ d3k, write down the volume of the tetrahedron ABCD. What is the condition that A, B, C, D should be coplanar. Is this a necessary and sufficient condition?

11. If a, b are perpendicular vectors and if r A a = b, prove that r = Aa +µ(a

b),

where A may be chosen arbitrarily. Find # in terms of lat.

12. If 11... m3are real numbers such that

g+g-pg = 4+14+4= I prove that

and 4rn1+12m2+13m3 = 0

(4 m3 — 4 m2)2 +(13 m, — m3)2+(11m2— 4 m02 = 1.

13. Prove that (a A b).(c n d) = (b . d) (a. c) — (b. c) (a . d). What trigonometric identity follows if 0, A, B, C, D are coplanar and the angles AOB, COD are both right angles? 14. The vectors a = OA, b = OB, c = OC are non-coplanar; prove that a given vector r may be expressed uniquely in the form r = ab A c-Fi3c A a +ya A b and find a, y in terms of a, b, c and r. If r = OP, locate P in terms of 0, A, B, C: (i) if a = /3 = y = 1; (ii) if a = 0, fi = Y = 1. 15. Points Q, R are taken on the sides CA, AB (not produced) respectively of a triangle ABC in such a manner that the triangles ABQ, ARC are equal in area. Prove that RQ is parallel to BC. 16. G is the centroid of the triangle ABC, and E, F are the mid-points of CA, AB respectively. Prove that the quadrilateral AFGE and the triangle GBC have

the same area. 17. ABCD is a skew quadrilateral; the mid-points of the opposite sides BC, DA are P, Q, while the mid-points of the diagonals AC, BD are R, S. Prove that PRQS is a parallelogram whose area is -II AABC — AABDI. 18. ABC is a triangle and points L, M, N are taken on BC, CA, AB respectively

such that

BL _ a CM _ AN MA = 11' NB = v. LC ,

Prove that ALMN = (1 + Apr) AABC and deduce the theorem of Menelaus, that, if LMN is a straight line, kuv = —1.

513

VECTOR PRODUCTS

[24

19. The sides (and sides produced) of the parallelogram ABCD separate the plane into nine regions (see Figure 24.7). Prove that, if a point 0 is taken within region I, II, VI or IX the area of OAC is the sum of the areas of OAB and OAD but that, if 0 is taken within any other region, the area of OAC is the difference of the areas of OAB and OAD. (The reader who has studied mechanics may be able to interpret this result in terms of moments.)

Fig. 24.7 20. ABCD is a tetrahedron. The lengths of AB and CD are a and b and the angle between AB and CD is 0. A plane is drawn parallel to AB and CD and meets the edges CA, CB, BD, DA at P, Q, R, S. Prove that PQRS is a parallelogram and that its area is ab — x(h — x) sin 0, h2

where x is the distance of the plane from AB and h is the distance between AB and CD. Hence, or otherwise, prove that the volume of the tetrahedron is labh sin O.

3. APPLICATIONS TO COORDINATE GEOMETRY Throughout this section we shall assume that an origin 0 and a righthanded set of coordinate axes Ox, Oy, Oz specified by the unit vectors j, k are given. The position vector of the point A... with respect to 0 will, as usual, be denoted by a .... Consider first the line through A in the direction of the unit vector 1 If R is any point on this line, then AR = Ail and thus

AR A = 0, giving the equation of the line in the form (r — a) A 01 = 0.

514

3]

APPLICATIONS TO COORDINATE GEOMETRY

If r = xi +yj + zk, a = aii+ a j + a, k, fl = li+mj+nk, we can transform this to (or from) the familiar coordinate form as follows:

(r—a) A =0 — ai)i+ (y — a2) j+(z — ajk] A (a rnj + nk) = 0

(y — a2)n— (z — a3) m = (z — a3) 1— (x — ajn = (x— ai)m— (y — a2) 1 = 0 100 522

1]

DENSITY FUNCTIONS

(see Figure 25.3). This distribution is again symmetrical about x = 50 and gives a greater probability to points near the centre. To find Pr (E) we must first calculate k. We have Lioo kx(100 — x) dx = 1

k[50x2 -3x3]000 k=

1

3 500000'

Fig. 25.3

(Again, Figure 25.3 is not drawn to scale!) Defining E as before we have: Pr (E) =

so

3

130 500000

x(100 — x) dx

6 x 10-6[50x2 — 3x928 —

P2A- = 0.432.

Ex. 1. Why would you expect the second answer for Pr (E) to be less than the first? Ex. 2. Write down the value of Pr (E) on the assumption that all points of the interval are equally likely. What form does f(x) take in this case?

Ex. 3. If f(x) = kx, 0

x < 100 and f(x) = 0 otherwise, find k and Pr (E).

A mode of a random variable with given density function f is a value of the random variable for which f has a local maximum. The median, m, is the value of the random variable defined by the equation rf(x) dx 523

CONTINUOUS PROBABILITY DISTRIBUTIONS

[25

Similarly, the lower and upper quartile scores L, Q, are defined by

s

Lcc,

fQ

f(x) dx = 1,

j _. f(x) dx =

Ex. 4. Interpret the median and quartiles in terms of probabilities.

2. DISTRIBUTION FUNCTIONS The density function, f, of a continuous probability distribution bears the same relation to probability as mass density does to mass:mass density is the mass per unit volume, probability density is the probability per unit interval. It must be remembered that f(x) itself is not a probability, but 8p = f(x) Sx is. We now introduce a function, closely related to the density function, whose values do represent probabilities. Suppose the domain of our random variable is R and that we are given a density function f: R —> R. Then we know (i) f(x) > 0, +00 (ii) L oo f(x) dx = 1. Let E be the set of real numbers less than, or equal to x, that is, the set of real numbers to the left of x on the real axis: E = fy R: y Then

Pr (E) =

rx

f(x)dx -depends upon x and we may define a function F: R —> R where F(x) = f x f(x) dx = Pr (E). F is called the distribution function for the probability distribution; F(x) is the probability that, given a probability distribution with density f(x), a point chosen at random will be less than x. From the original definition we have (provided f is continuous) dF dx Since f(x)

0 for all x, F(x) increases as x increases. Also we have lim F(x) = 0, lim F(x) = 1, x-.-

which gives the range of F as the set {x e R: 0 x

524

21

DISTRIBUTION FUNCTIONS

Again, given an event E = {x e R: x1< x < x2} we have Pr (E) = j'x'f(x) dx

x,

= F(x2) — F(xi). Since the sets E and {x e R: x < x1} are mutually exclusive, this illustrates the result that the probability of the union of mutually exclusive events is the sum of the probabilities of the separate events. *Ex. 5. If E1 E2 • • • E,, are mutually exclusive events in a continuous outcome space, prove that Pr (ED + Pr (E2) + + Pr (E.) = Pr (Ei U Ea U . • u Example 1. The density function, f, is defined by if x < 0 f(x) = {0 kx(1 — x) if 0 s x < 1 0 if x > 1. Find (i) the value of k; (ii) the distribution function F; (iii) the density and distribution functions, g and G, of the random variable W where 2 W x Sketch the graphs off and F. (i) Since f + f(x) dx = 1,

we have

f:kx(1

—

x) dx = 1 k = 6.

(ii) If x < 0, F(x) = 0; if x > 1, F(x) = 1. If 0 < x 1, F(x) = f: 6x(1 — x) dx = [3x2 — 2x3]o = x2(3 —2x). Thus F is the function F: R --)- {y e R: 0 ( y ..5 1} defined by 0 {if x < 0 F(x) = x2(3 —2x) if 0 .. 1. 1 525


[25

(iii) If w < 0, G(w) = 0; if w > 1, G(w) = 1. If 0 < w 1,

G(w) = Pr {z e R: 0

z < w}

= w(3 — 2,1w) (since x is positive, x = Vw). Thus 0 if w < 0 G(w) = 3w-2w if 0 < w ,< 1 1 if w > 1. dG g dw'

Also, giving

0 if w < 0 g(w) = 3(1 — A/w) if 0 < w < 1 0 if w > 1. The graphs of f, F are shown in Figures 25.4(i) and (ii).

F(x)

(i) Fig. 25.4

Notice in this question that, since Jx increases as x increases, as we move from left to right along the x axis so we move from left to right along the w axis. More generally, if a transformation y = h(x) of a random variable is made, the method of Example 1 may be used directly if y either increases or decreases steadily with x but, if this is not the case, the domain of f should be split into those parts for which it is an increasing function and those for which it is decreasing. (See Example 4.) The value of transforming from one random variable to another is that a complicated form for the density function may be changed into a simpler (or more familiar) density function. F is sometimes referred to as the cumulative distribution function, to emphasize the property it has of increasing from left to right. The graph 526

2]


of F, which usually has a shape very similar to that shown in Figure 25.4, is called a cumulative probability curve (or, occasionally, an ogive curve).

Ex. 6. Explain how the cumulative probability curve may be used to find the median and upper and lower quartiles. In the case of discrete distributions, if we had a random variable X which could take all integral values r in the range 1 r < n, where Pr (X = r) = p„ we defined the expectation of X by

g(X) =

E

rp„

r=1

For continuous distributions, integration takes the place of summation and we define the expectation of the random variable X, whose density function is f, by S(X) = f +: xf(x) dx.

S(X) is usually referred to as the mean of X, and is denoted by u. Example 2. Find the mean of the random variable X whose density function, f, is defined by if x < 0 f(x) ={0 4x(1- x2) if 0 < x < 1 if x > 1. 0 We have = e(X) = f xf(x) dx -00 =

f

4x2(1 - x2) dx (since f(x) = 0 outside the interval 1), 0x51),

_8_ 15•

More generally, if we have a function g: R -> R, the expectation of the function g of the random variable X is defined by

e[g(x)] = 5±: g(x)f(x) dx. Particularly important is the case where

g(x) = (x - ,u)2. 527


[25

The variance of Xis the expectation of the function g so defined: cr2 = 61(x -/02] =

+co

(x - ,a)2 f(x) dx.t

— 03

Since +0°

(x -11)2 f(x) dx =

+co

(x2-2,ux + #2) f(x) dx

+ CO

OP

= co x2f(x)

= we have

+co

dx-2,u f co xf(x) dx+ It2

x2f(x) dx - 2,a .

a2 +112 =

+op

+ OP

f(x) dx

A2.1,

x2f(x) dx,

a result which may be compared with that on page 178. *Ex. 7. Show that the variance of the random variable in Example 2 is given by (72 +

=f 4x3(1 - x2) dx.

Deduce that 0.2 = *Ex. 8. The mean deviation about the mean, 77, is defined as the expectation of the function g where g(x) = Ix- AI. Find 77 if 0, x < -1 f(x) = -1 x 1 0, x > 1.

Exercise 25(a)

1. f(x) = kx-2 if x Pr (x < 2).

1,1(x) = 0 if x < 1; find the value of k and hence find

2. f(x) = ke-2° if x 0, f(x) = 0 if x < 0; find k and hence find Pr (1 < x < 2). 3. If f(x) = k sin irx if 0 x 1 and is zero otherwise, find k and hence find Pr (x > 4. Find the distribution function for each of the density functions in Questions 1-3. .1-The use of o2to denote variance is sometimes insufficiently explicit. If it is necessary to refer to the random variable, X, under consideration, the associated variance may be written or, more commonly, 17.

528

2]


Find the mean, median, mode and variance for the distribution with the density function defined as in Questions 5-8. 5.

O f(x) = O

if x < 0 x) if 0 < x < 2 if x > 2.

6.

if x < 0 (6x +1)/4 if 0 < x < 1 f(x) = {0 if x > 1. 0

7.

O f(x) = {-fir cos rrx 0

8.

f(x) =

if x < — 4 if — i ---5 x ---S. 4 if x > 4.

(0 if x < 0 Ae-Ax if x 0.

9. If the density function, f, of a distribution is given by 0

if x < 0 if 0 < x < 1

.i(x) = 4- if 1 < x < 2 4(3 — x) if 2 < x < 3 0 if x> 3 find the distribution function, F, and sketch the graphs of f and F. Find the density and distribution functions, g and G, of the new random variable Y, where y = x2, and sketch the graphs of g and G. 10. The density function, f, of a probability distribution is given by

f(x) =

0

if x < 0

x

if 0

x 2.

Find the density function, g, of the new random variable, Y, where y = if x 0, y = 0 if x < 0 and sketch the graphs off and g. Find the mean and variance of X and the mean of Y. 11. A probability density function of a random variable Xis defined as follows: x(x— 1) (x-2) for 0 x < 1 f(x) = {A for 1 <x< 3 otherwise, 0 where A is a suitable cons ant. Calculate the expectation is of x. What is the probability that x is less than or equal to A? (M.E.I.) 529

[25

CONTINUOUS PROBABILITY DISTRIBUTIONS 12. A continuous probability distribution has density function f, where

f(x) = a+ bx+ cx2 for 0 5 x 5 1, and f(x) = 0 outside this range. The mean is 2/3 and the variance is 4/45. Find the values of a, b, and c. 13. The probability density function p(t) of the length of life, t hours, of a certain component is given by p(t) = ke-kt (0 < t < oo), where k is a positive constant. Show that the mean and standard deviation of this distribution are each equal to 1/k. Find the probability that the life of a component will be at least to hours. Given that a particular component is already tlhours old and has not failed, show that the probability that it will last at least a further tohours is e-kto. An apparatus contains three components of this type and the failure of one may be assumed independent of the failure of the others. Find the probability that: (i) none will have failed at to hours; (ii) exactly one will fail in the first tohours, another in the next tohours and a (M.E.I.) third after more than 2t0 hours.

14. A random variable X has the distribution function F(x) = 0

(x < 0),

F(x) = kx3 (0 < x .< 2), F(x) = 1

(x

2),

where k is a constant. Find the mean, median and variance of X.

(M.E.I.)

15. A random variable X has probability density given by f(x) = A/(1 + x2) if 0 5 x 5 1, and

f(x) = 0 otherwise.

Find A the mean and the variance of the distribution. Find also the median and the 90th percentile. (The 90th percentile is the value of X for which F(x) = *).)

(0 & C adapted)

3. SOME IMPORTANT CONTINUOUS

DISTRIBUTIONS (i) The uniform (or rectangular) distribution If a continuous random variable can assume all real values between x = a and x = b and if the density function, f, is independent of position within this interval, then the random variable is said to possess a uniform dis530

3]

CONTINUOUS DISTRIBUTIONS

tribution. f and F are thus given, for the uniform distribution over the interval a < x ' b, by 1 ) if x < a 1

f(x) =

if a.....x:c.b b a 0 if x > b;

if x < a x a F(x) = CO b_ a if a < x < b

1

if x > b.

*Ex. 9. Sketch the graphs of f, F as defined above. *Ex. 10. Prove that the mean and variance of the uniform distribution defined over the interval a s x < b are given by

= (a +b), 0.2 = -,12-(a — b)2. The continuous uniform distribution arises as a direct extension of the discrete uniform distribution. As in the discrete case, the assumption of a uniform distribution is often not made explicit in the formulation of problems. This is a weakness in the statement of the problem and the reader must be on his guard. For example, such a statement as à point is chosen at random on the line AB' is likely to imply, in the eyes of an examiner, that a uniform distribution is the required mathematical model, although there may be little physical justification for this. The dangers of a loose enunciation of a problem are illustrated in the following very well-known example, due to J. Bertrand.

Example 3. A chord of a circle is drawn at random. Find the probability that its length will exceed that of the side of the inscribed equilateral triangle: (i) if one end, A, of the chord is regarded as fixed and the other end, P, obeys a uniform distribution on the circumference of the circle; (ii) if the direction of the chord is regarded as fixed and the perpendicular distance, x, of the chord from the centre obeys a uniform distribution over the interval 0 x a, where a is the radius of the circle. (i) Let ABC be an equilateral triangle inscribed in the circle (Figure 25.5). The chord AP will exceed the length AB if P lies on the arc BC and the probability of this occurring, on the assumption that P is distributed uniformly on the circumference of the circle, is 3. (ii) Let AB, CD be the two chords in the given direction which are distant la from the centre of the circle (and thus are of length aV3, the side of the inscribed equilateral triangle). If x is the distance of a random chord 531


[25

from 0 in the given direction, and if x is uniformly distributed in the interval — a < x < a, the chord will exceed a \)3 in length (e.g. PQ, in Figure 25.6) if — a x +a giving a probability of -I-.

Ex. 11. If, in Example 3, the mid-point of the chord is taken to have a uniform distribution over the interior of the circle, prove that the probability that the chord will exceed the length of the side of the inscribed equilateral triangle is 1 4.

Fig. 25.6

Fig. 25.5

Example 4. A point P is taken on a line AB, of length 2a, the distribution being uniform. A rectangle is then drawn with adjacent sides of lengths AP, PB. Find: (i) the expected value of the area of this rectangle; (ii) the distribution function, G, for the random variable, Y, whose value represents this area. Let AP = x and the area of the corresponding rectangle = y. Then y = x(2a — x). Now, iff is the density function of X, the random variable whose value for given P is the length AP, 0 if x < 0 f(x) = - 1/2a if 0 .. 2a

and, by the symmetry of the curve y = x(2a — x) about the line x = a (see Figure 25.7), we have, for any point (x, y) on the curve, Pr (z e R: z < y) = Pr (z E R: z < x)+Pr (z E R: z > 2a— x)

= F(x)+[1— F(2a— x)] x 2a—x = +1 2a 2a x a

But, since

y = 2ax —x2,

we have

x = a— Al(a2 —y)

giving

G(y) = Pr (z e R: z < y)

(x < a),

a— V(a2 —y) a Thus we have

0

if y < 0

G(y) = (1— 41 (1 —;) if 0 < y . -. . a2 1

if y > a2.

The graphs of F and G are shown in Figure 25.8. 6

PPMU

533


[25

Example 5. X, Y are both uniformly distributed between 0 and 1. Values x, y are chosen independently at random and a new random variable, Z, is formed where z = x+y. Show that Z has a triangular distribution (that is, a distribution with density function whose graph is triangular). Like most problems concerned with independent uniformly distributed random variables, it is best to consider this question graphically. Since X, Y are distributed independently, the choice of two numbers, x and y, may be represented by plotting the point R(x, y), all points of the square OABC (see Figure 25.9) being equally likely. If z = x+y, Y A

B

Q

N P

C

>x

Fig. 25.9

for given z all possible points R lie on the line segment PQ, where P is the point (z, 0). For 0 S z 1, PQ cc z; for 1 < z 2, PQ cc (2 — z) and thus the density function, f, of Z, whose value for given z is proportional to z, is triangular. In fact 0 if z < 0 z if 0 z 1 f(x) (2 — z) if 1 < z S 2 0 if z > 2. 534

3]


Ex. 12. Find the distribution function, F, for the random variable Z of Example 5 and sketch its graph. Ex. 13. If, with the notation of Example 5, the random variable W is defined by w = x—y, describe the distribution of W.

(ii) The normal distribution Suppose that shots are fired at the centre, 0, of a large circular target. Set up coordinate axes as shown in Figure 25.10 and let P(x, y) be the point at which one of the shots hits the target. If we suppose that shots aimed at the centre are liable to errors, equally likely to occur above or below Ox and to the left or right of Oy, we may reasonably assume that the probability of x lying between x and x+ Sx, the 'x error', depends upon the

Fig. 25.10

numerical value of x; that is, the probability is of the form cb(x2) Sx. On the assumption of the symmetry of the errors made above, the probability of the 'y error' is 0(y2) Sy and the probability of lying in a small area

SA = Sx Sy around P is 0(x2) 0(y2) SA, on assuming that the errors occur independently. But on these assumptions the probability may also be written 95(x' 2) 0(Y/2) 8A,

(approx.)

where new axes Ox', Oy' are taken with Ox' passing through P, as shown in Figure 25.10. Thus, since x'2 = x2 +y2 and y' = 0 at the point P,

q5(x2) 0(.Y2) = 0(x' 2) 0(Y' 2) = 0(x2+ Y 2) 0(0) and 0 is a function with values satisfying the equation 0(x2) cb(y2) = 1c95(x2+ y2). 6-2

535


[25

A solution of this equation is 0(x2) = ke--"2where A is taken as a positive constant, since it is reasonable to assume that the probability decreases the further we move from 0. The argument above (due to Thompson and Tait) suggests that the function f(x) = ke-Ax2could form a suitable model for the probability density of errors occurring in observations. This indeed turns out to be the case and we shall therefore make the following definition. The random variable X is said to have a normal distribution if its density function, 0(x), is of the form 1 ec(x)

V(277)

The letter 0 is customarily used in place off to denote the density function of the normal distribution; the corresponding distribution function is

rx

e-ix2 dx. I(x) - v(21T) j _ The factor (277)-i appearing in the forms for 0 and' ensures that the requirement r+.

J ¢(x) dx = 1 -.

is met—that is, that 0(x) is a genuine density function. This follows from the well-known integral + e- x2 dx = V(277),

5

a result whose proof will be found in any sufficiently advanced book on calculus. (The reader should not be too discouraged by the remark, attributed to Sir William Thompson, about this integral: 'No-one can call himself a mathematician to whom this result is not obvious! ') The shape of the curve of the normal density function is shown in Figure 25.11. As is to be expected, the curve is symmetrical about x = 0, which clearly gives the mean of the distribution:

e(X) = 0. Furthermore, the variance of a normally distributed random variable X is 1, for 1 f+ , 'r(X) = .2 = x2 e-ix dx V(2n) 1+cO v(27) , xd(e-iz2) -

1

f

V(27r) j

dx, on integrating by parts,

= 1, on quoting the integral result above. 536


3]

Statistical tables (e.g. The Cambridge Elementary Statistical Tables by Lindley & Miller) give values of D(x) for values of x from 0 to about 4. (Since (I)(4) 0.99997 values of 1(x) for x > 4 are very rarely required.) By the symmetry of the curve 95(x) =

(270

e-ix'

about the line x = 0, values of t(x) for x < 0 may be deduced immediately.

Fig. 25.11

The normal distribution is frequently used as a probability model. Suppose that the random variable X has mean it and standard deviation o-; then the new random variable Y defined by Y

—

x— ft o-

has mean 0 and standard deviation 1. If Y is normally distributed it is customary to say that ' X is normally distributed about the mean it with standard deviation o-', sometimes written as X is distributed N(4a, 0)% and that X has been standardized to the normal random variable Y by the given substitution'. For emphasis, Y is often referred to as 'the standard normal distribution N(0, 1)'. Example 6. If X is normally distributed N(0, 1) find (ii) Pr (0.8 < x < 1.4); (i) Pr (x < 1.4); (iii) Pr (x < 1.4); (iv) Pr (-1.4 < x < 0.8). —

If Y is normally distributed N(2, 0.75), find (vi) Pr (1 < y < 3). (v) Pr (y < 0); 537


From statistical tables, (i) Pr (x < 1.4) = 0(1.4) 0.9192 (Figure 25.12). (ii) Pr (0.8 < x < 1.4) = 0(1.4)-0(0.8) p..,,0-9192-0.7881 = 0.1311 (Figure 25.13).

Fig. 25.14

Fig. 25.15

(iii) Pr (x < —1.4) = Pr (x > 1.4) by symmetry, = 1— 0(1.4) 1-0.9192 = 0.0808 (Figure 25.14). (iv) Pr (-1.4 < x < 0.8) = Pr (x < 0.8)—Pr (x < —1.4) = (I)(0.8)— [1 — (1:0(1.4)] 0.7881 — 0.0808 = 0.7073 (Figure 25.15). 538

[25

3]


(v) The random variable Z, where z=

y-2 0.75

has standardized normal distribution. Thus Pr (y < 0) = Pr (z < — 2.67) = 1 — 0(2.67) 1-0-9962 = 0-0038. (vi) Again, Pr (1 < y < 3) = Pr (-1.33 < z < 1.33) = 20(1.33)-1 F8164 — 1 = 0.8164. Example 7. The heights of a large number of schoolchildren are measured correct to the nearest centimetre and the mean and standard deviation of the resulting frequency distribution are calculated and found to have the values 122 cm and 5.2 cm respectively. As a model of the situation it is assumed that the heights, x, of the children are distributed normally about a mean ,a =122 with standard deviation o = 5.2. Calculate the probabilities for each of the class intervals x 105, 105 < x < 110, 110 < x < 115, ..., 130 < x 135, x > 135. Since the heights are measured to the nearest centimetre, the upper limits for x for the class-intervals above are successively 105.5, 110.5, 115.5, ..., 135.5 (there is no upper limit, of course, for the last class). We now standardize our variable, x, by the transformation

x 122 —

Y = 5-2 so that y has the standard normal distribution (# = 0, Cr = 1) and using statistical tables, we find 1(y) for the upper limit of each class-interval. Finally, the probability associated with each class interval may be calculated P = 00'0 — OW. 539

[25


The working is set out in tabular form below: Class

x

>135 130-135 125-130 120-125 115-120 110-115 105-110 < 105

135.5 130.5 125.5 120.5 115.5 110.5 105.5

y

(y)

2.60 1.64 0.68 -0.29 -F26 -2.22 -318

1.0000 0.9953 0.9495 0.7517 0.3859 0.1038 0.0132 0.0007

p 0.0047 0.0458 0.1978 0.3658 0.2821 0.0906 0.0125 0.0007

1.0000

It should be noticed that a continuous distribution has been used here as a mathematical model for a discrete situation (heights measured to the nearest centimetre). Again, it might be objected that, whereas heights could reasonably be expected to fall, say, within the interval 90-180 cm, the normal distribution is defined for all real values of x; however, as has been pointed out earlier, for the standard normal distribution, Pr (iYi > 4) 0.00006 and the two tails of the distribution, beyond y = + 4, may be neglected. As well as appearing as a distribution in its own right, the normal distribution is often used as an approximation to the binomial distribution B(n, p) for large n, provided neither p nor 1 p is too near zero. In Figure 25.16, the probabilities of obtaining the scores 0, 1, 2, ..., 20 in the -

0

5

10

Fig. 25.16

540

15

20

3]


binomial distribution B(20, 3-) are proportional to the lengths of the corresponding vertical lines. The outline shape is strongly reminiscent of the normal curve. Even in a skew case, where p + 1, the outline is still approximately normal : see Figure 25.17, in which the length of the vertical lines are proportional to the probabilities of obtaining the scores 0, 1, 2, ..., 20 in the binomial distribution B(20, 1).

0

1

5

10

15

Fig. 25.17

In fact, it may be shown that, given a binomial distribution B(n, p) where n is large and neither p nor 1 —p is too near zero, then the probability of obtaining a score r, that is, \ kinr)Pr(1—P)n-r3

is approximately equal to a-1times the ordinate of the standard normal curve at the point x = (r — )11)1 o-, where ,a, o are the mean and standard deviation of the corresponding binomial distribution II = np, o = .N1 (npq). Furthermore (and more importantly for applications) we can relate areas under the normal curve to sums of successive terms of the given

541


[25

binomial distribution; in fact, the probability that the variable takes a value in the interval r1 < r < r2 is approximately fc\ r2+ — i\ k 0- / k a where, as before, = np, o- = Al(npq). (The limits r1, r2 are changed to r1 — 1, r2 + zrespectively in order to avoid complications arising out of approximating to a discrete distribution by a continuous distribution; for example, we wish to associate a non-zero probability with a single value, r, of the variable.) The reader is referred to one of the standard texts on probability for a proof of the results outlined above. As a working rule, the approximations give reasonable results if the lesser of the two numbers, the mean number of successes and the mean number of failures is greater than about five. Example 8. A coin is spun 250 times and turns up tails 139 times. Does this provide any evidence of bias? Take as the null hypothesis H: 'the coin is unbiased'. We have to assess the probability of getting a result as bad as, or worse than, 139 tails on the assumption that H is true; that is, we have to determine Pr(139 or more tails IH) + Pr (139 or more heads I H). The calculation of this from the binomial distribution B(250, -D would be prohibitively laborious; we therefore use the normal approximation to the binomial distribution as our model where ,a = 250 x = 125, U2 = 125 x i = 62.5, giving

a- = 7.906. 2 Pr (x > 139),

We seek

where x is distributed B(250, 1), and this is approximately 2 [I

/138.5-125\1 7.906

k

2[1 — 0(1.71)] 0.087. There is thus insufficient evidence of bias at the 5 % level. 542

3]


Note. Tables of the normal distribution show us that Pr (iyi 3 F96)

0.05

and thus a normal variable greater than 1.96 indicates significance at the 5% level. In Figure 25.18 the two shaded areas each contain 2.5 % of the total area under the curve.

Fig. 25.18

One further application of the normal distribution should be mentioned: if x1, x2, xnare independent observations of any random variable X and a new random variable Z is defined by

z = (xi+ x2+ ... xn)117 then, if the mean and variance of Xare respectively # and cr2, the distribution of Z will be approximately normal, with mean # and variance o-21n. (The Central Limit Theorem.) The proof of this result is beyond the scope of this book. (iii) The exponential distribution

The exponential distribution is defined by the density function f(x)

{0 if x < 0, Ae-Ax if x 0,

where A is a positive constant.

*Ex. 14. Show that, for the exponential distribution, ea) = 1/A,

"r(x) = 1/A2.

The result of Ex. 14 suggests a resemblance between the exponential and Poisson distributions. We shall exhibit the relation between the two distributions in the following problem. (See Exs. 16-18.) 543


[25

Consider the calls received at a telephone exchange, the rth call occurring at time t,., where the time, t, is measured from t = 0. We shall assume that the number of calls constitutes a purely random process; that is, the number of calls received in any interval (t, t') is independent of anything that has occurred previously. Furthermore, we shall assume that the purely random process is a stationary process: that is, the number of calls received in the interval (t, t') depends only upon its length, not on its position. Ex. 15. Discuss the validity of the assumptions made above in the light of what you imagine would be a typical exchange.

Suppose now that Pr (no calls are received in the interval (0, t)) = po(t), and let us assume thatpois a differentiable function. Since, by the definition of a purely random process, the probability of receiving no calls in the interval (t, t + at) is independent of receiving no calls in the interval (0, t), po(t +

Put t = at = 0:

at) = Po(t) po(n).

Po(0) = [Po(0)]2

and thus, since MO) # 0, we must have MO) = 1. Again pat +at) = Po(t)Po(8t) po(t+at)-po(t) = po(t)[po(at)-1] po(t+dt)-po(t) = MO [1480-140)] pi,(t+dt)-po(t) ■ P = Pat ) at Let 8t 0:

- o(0) at •

Po(t) = Po(t)-PO(0) Pa(t) = 4e-At,

(1)

where is a constant ( -pa0)) which must be positive since pat) < 1 and

A = 1 since MO) = 1. Result (1) now enables us to prove that if X is the random variable whose value is the length of the time interval up to the first call, then X has an exponential distribution. If F is the distribution for X then, for x 0, F(x) = Pr (length of the time interval up to the first call x) = Pr (at least one call has been made by the time x has elapsed) = 1 -e-ax and thus

f(x) = ale-Ax for x 0.

Since f(x) = 0 for x < 0 it follows that X has an exponential distribution, 544

3]


Ex. 16. If p„(t) is the probability that there are n telephone calls in the time interval (0, t) and if St is sufficiently small for there to be a negligible probability of more than one call in the interval (t, t +St) show that

P.(t+80= (1 — ASO MO+ AP._1(t)St and deduce that

dp. --d7 =

Pa)•

Ex. 17. Prove by induction that the probability, 1,40, defined in Ex. 16, is given by P.(t)=

e-Al(At)n

n!

•

Ex. 18. Show that the distribution of the number, n, of calls received during a fixed time interval is Poisson and account for this in terms of the Poisson distribution being a limiting form of the binomial distribution.

Exercise 25(b) 1. The random variable, X, has uniform distribution in the interval 0 x 0). 2. The line AB has length 10 cm. A point P is taken at random on AB, all points being equally likely; what is the probability that the area of the circle of radius AP will exceed 10 cm2 ? 3. The line AB has length 10 cm. An interval of length 2 cm is marked at random on the line, the positions of the interval being uniformly distributed. What is the probability that it will contain the mid-point of AB? 4. A circular disc of radius 10 cm is placed on a table. Another disc, of radius 3 cm, is now placed on the table so that it is at least partially in contact with the first disc. On the assumption that the permissible positions of the smaller disc are uniformly distributed, what is the probability that it covers the centre of the larger disc? 5. A point P is taken at random on the side AB (and between A and B) of the square ABCD, the positions of P being uniformly distributed. If PC cuts BD at X, find the probability (i) that BX < -113D; (ii) that BX < 1BD. 6. Figure 25.19 shows a square wooden frame B A ABCD with a square hole A'B'C' D' cut symmetrically in it. AB = 50 cm, A'B' = 30 cm. A' A ball of diameter 5 cm is dropped on to the frame. Assuming that its centre falls within the square ABCD and that it is equally likely to meet the plane of the frame at any point within ABCD, what is the probability that the ball will pass straight through the hole without D touching the frame? 7. A point A is marked on the circumference of a circle of radius r and a chord AP is drawn at Fig. 25.19 random, the positions of the point P on the circumference being uniformly distributed. Find the expected length of the chord. B'

'

C'

545


[25

8. Determine the variance of the length AP of Question 7. 9. A point P is marked on the side AB of the square ABCD, the points within AB being uniformly distributed. Find the mean and variance of the area of the triangle APD. 10. ABC is an isosceles triangle right-angled at B. Through A a line is drawn at random to cut BC at P, the angle BAP being uniformly distributed between 0 and lir . If AB = a, show that the expected area of the triangle ABP is an 2) a2/lr and find its variance. 11. The intelligence quotients of 500 schoolchildren are assumed to be normally distributed with mean 105 and standard deviation 12. How many children may be expected: (i) to have an intelligence quotient greater than 140; (ii) to have an intelligence quotient less than 90; (iii) to have an intelligence quotient between 100 and 110? 12. If an unbiased die is thrown 600 times, what is the probability of throwing a six less than 90 times? 13. X is normally distributed with mean 2 and standard deviation 1. Find numbers p, q such that Pr (x > p) = 0.2 and Pr (p < x < q) = 0.1. 14. A multiple choice test has 100 questions, each question having written beside it five answers, only one of which is correct. If the pass mark is 30 %, what is the probability that a student who makes a completely random guess at each answer will pass? 15. Rods are manufactured with a mean length of 202 cm and standard deviation 0.09 cm, the distribution being normal. If rods of length less than 20.1 cm are rejected, what is the probability that a rod that is retained has a length in excess of 20.3 cm? 16. Experience has shown that when a certain machine is functioning satisfactorily it produces capacitors with capacitances which are distributed normally with standard deviation 0.080 icF. Under these circumstances find the mean capacitance if 99% of the output has a capacitance of at least 2 /IF. Find also what proportion of the output will then have a capacitance between 210 and 2.30 F. Tests on a large batch of capacitors just produced reveal that their mean capacitance is 2.20 uF and that 10 % of them have capacitance below 2 AF. What do you deduce about the variability of the capacitance of the capacitors in this batch? (M.E.I.) 17. The average proportion p of insects killed by administration of x units of insecticide is given by (x -Ayer (27r)-1/2 e-t2/2 dt,

=f

where it and o are constants. When x = 10, p = 0.400 and when x = 15, p = 0.900. What dose will be lethal to 50 % of the insect population, on average? 546

EXERCISE 25 If a dose of 17.5 units is administered to each of 100 insects, how many will be expected to die? What is the probability that just two insects will survive? (M.E.I.) 18. Packets are advertised as containing 500 g of sugar. Tests carried out show that 6.7 % of such packets contain a mass greater than 508 g, while 0.6 % have a mass less than 492 g. Estimate the average mass of the contents of a packet on the assumption that a normal distribution constitutes an acceptable mathematical model. 19. A firm advertises that their runner bean seeds give a 95 % germination rate. Of 200 such seeds, 14 fail to germinate. Have you cause for complaint? 20. A man claims that he can forecast rainy or dry weather 48 hours ahead. Careful records are kept over 80 days and his forecast is found to be correct 47 times. Is his claim justified, or is he merely lucky? 21. A 'chance of failure' distribution is given for time t by the probability density function p(t) = (1/a) e-tea (0 < t < co). Show that a is the mean time of failure and that the variance is a2. Two components in a machine have failure time distributions corresponding to means a and 2a respectively. The machine will stop if either component fails and the failures of the two components are independent. Show that the chance of the machine continuing to operate for a time a from the start is e-312. (0 & C)

Miscellaneous Exercise 25 1. AB is a rod of length 2a. A point P is taken on AB, the points being uniformly distributed, and the stick is broken at P. Find the expected value of AP2+PB2. The two parts of the rod are placed on a table in such a way that A and B are aV3 apart. What is the probability that the triangle APB will be obtuse? 2. The mean deviation about p for a distribution with density function f is defined by the equation +co lx-F1 f(x) dx. = Prove that the mean deviation of a random variable distributed normally about with standard deviation a is approximately fisr. Rods are manufactured with a mean length of 18.4 cm and standard deviation 025 cm. If only those rods with a length greater than 18.4 cm are retained, what is their average length on the assumption that the rods were originally normally distributed about 18.4 cm with standard deviation 0.25 cm? 3. If X, Y are independently and uniformly distributed between 0 and 2, find Pr (xy < 1). 4. The side of a square is uniformly distributed between 0 and 1. Find the density and distribution functions for the area of the square and sketch their graphs. Determine the expected value of the area and its variance. 547


[25

5. The probability density function for the life x of a motor car tyre is given by f(x) = Ae-As for x 0, where A = 0.04 and x is measured in units of 1500 km. What is the probability that a single tyre will last more than 30 units? A car has four tyres in use and they are all of the same age. What is the prob(M.E.I.) ability that all of them will need replacing before 30 000 km? 6. A random variable X has the probability density function exp -gx1(cr,13)} 6,0{1+ exp [ - gx/(c•N/3)]}2 for all real x. Show that the distribution is symmetrical about x = 0. Determine the cumulative distribution F(x) = Pr {X < x}. You are given that the variance of X is 0-2. Compare the value of F(o), F(2cr), F(3c•) with the corresponding values for the normal distribution with zero mean (M.E.I.) and variance Gr2. f(x) -

7. A random variable X is distributed normally with expectation ,u and standard deviation cr. Find the mode and median of the distribution. What is the proportion of the population lying between the points of inflexion of the curve of the probability density function? (M.E.I.) 8. The diameters of some machined components are distributed normally with mean 5.00 cm and standard deviation 0.05 cm. Find the expected proportion of the components which will be outside the range 4.925 to 5.075 cm and the ratio of the expected proportion in the range 5.025 < d < 5.050 to the expected proportion in the range 5.050 < d < 5-075. It is desired to adjust the mean of the process so that there are, on average, twice as many components in the range 5.025-5-050 cm as in the range 5.0505.075 cm. Show that this can be done if the mean is adjusted to a value between 5.00 cm and 4.95 cm, and find the value by trial to the nearest 0.01 cm. (M.E.I.) 9. The probability density function of a distribution is given by e—xxA-1 f(x) = (x 0, A an integer > 0). (A - 1)! Find the expectation and variance of X. If AX) = 4a find 6'{(X-,u)3}. Sketch f(x) when A = 2. (M.E.I.) ,

10. A grocer sells bread and can buy batches of 120 loaves. The number of daily customers for bread is distributed normally with mean 100 and variance 100. The net profit on the sale of a loaf is 2.5p and the net loss on an unsold loaf is 3.5p. What is the average daily net profit to the grocer? (M.E.I.) 11. The random variable X has normal distribution. If a new random variable Y is defined by y = x2 show that the distribution function, F, for Y is given by F(Y) = () f o 548

e-x'/2clx.

MISCELLANEOUS EXERCISE 25 Deduce that Y has a X2distribution with one degree of freedom, defined by the density function f, where (y < 0) 0 f(y)

1

v 112 p-v/2

,1(27r)'

.-

0).

12. Two chords are drawn independently at random in a circle. What is the probability that they intersect? 13. Two men, A and B, are allowed a lunch-break of an hour. A is at liberty to leave the office at any time between 12.00 and 12.45, while B may leave the office between 12.45 and 2.00. What is the probability that they will both be out together, on the assumption that the permissible starting times for both A and B are uniformly distributed? 14. A needle is pivoted at the point (0, a) and is rotated. When it comes to rest, the point P at which its axis (produced if necessary) cuts the x axis is marked. The random variable X is defined to have as its value the x coordinate of P. On the assumption that the angle that the needle makes with the y axis is distributed uniformly between ± -17r, prove that X has a Cauchy distribution, defined by the density function f(x) —

a 71(22 +X2)

( co < x < co).

Discuss the existence of e(X) and 17(X) for this distribution. Sketch the graph of the density function for a = 1. 15. The Laplace distribution has probability density function f(x) = C e-Aix' (—co < x < cc) where C is a constant and Ix' denotes the magnitude of x. Find Pr {I A} and the variance of X. (M.E.I.) 16. A famous early example of the ' Monte Carlo' method for solving problems by random numbers was Buffon's determination of 7r. Small rods of length L1 were dropped at random on to a sheet of paper ruled with thin parallel lines a distance L2 (> L1) apart. A ' success ' is recorded when any part of the rod touches a line. The ratio of 'successes' to the total number of tosses is recorded over several thousand tosses. How may 7r be estimated from this ratio? (C.S.) 17. Let X be a random variable uniformly (rectangularly) distributed over the interval 0 < x < 1. Derive the probability density function of the following random variables: (a) Y = X 2—1, (b) Z = sin 7rX. Find the mean and standard deviation of Y and Z. (C.S.) 18. The chance that the customer at the head of a queue completes service in any interval of length St is ItSt, and the chance that a new customer arrives in any interval is .1St, arrivals and departures are independent and A < it. The chance that at time t there are n customers in the queue (including the one being served) is denoted by pn(t). Show that pn(t) = AStp,,,(t— 80+ (1 — — t8t)p„(t — 80+ ttStp„,i(t— St) (n 1) (over) and obtain the corresponding equation for po(t). 549


[25

By considering the forms that these equations take under the steady-state condition pk(t) = pk(t at), where k is zero or any positive integer, or by any other means, obtain and the mean queue size (in terms if A and p) in the steady state. (C.S.) -

19. X is a continuous random variable with mean p and variance o2. A is a posi-

tive constant. Prove Chebyshev's inequality Pr (I X- pl > Acr) < 1/A2. Show how this inequality may be sharpened, in the case A = 2, if the distribution of X is assumed to be (i) normal, (ii) uniform. 20. Spacecraft land on a spherical planet of centre 0. Each is able to transmit messages to, and receive messages from, any spacecraft on the half of the surface of the planet nearest to it. (i) It is known that spacecraft have landed at points A and B on the surface of the planet. Show that the probability that a space craft, landing at random on the planet, will be able to communicate directly with the spacecraft at A and B is (it -

0)12rt

where 0 is the angle AOB. (ii) What is the probability that three spacecraft, all landing at random on the (C.S.) planet, will be in direct contact with each other? 21. Engine crankshafts are manufactured so that the diameters, in centimetres, form a normal distribution with mean 5 and standard deviation 0.03. Crankshafts with diameters less than 4.94 or greater than 5.06 are rejected. The accepted product is classified into three grades of size 4.940-4.988, 4.988-5.012, 5.012-5.060. Show that:

(i) f xe-°" dx = (ii) f x2e-s'2 dx = - x e-°'/2+ I e-x" dx. Hence find the average diameter in each of the three grades and the ratio of the standard deviation in the middle grade to the standard deviation of the unclassified product before any rejection of under- and oversize shafts. (M.E.I.) 22. The police force in a certain district carries out tests on the brakes of auto-

mobiles chosen at random on the road. Each man is required to test 20 cars. Calculate the distribution of the number of cars with defective brakes in sets of 20 cars if the probability that a single car has defective brakes is 10 %. Show the distribution graphically together with a plot of the normal distribution with the same mean and variance. Comment on the relation between the two distributions. (M.E.I.) 23. A population contains two strata I and II in proportions 3, 3respectively. Both strata are exponentially distributed the first with mean 1 and probability density function f(xi) = Al e-A. °. (x1 0);

the second with mean 3 and probability density function f(x2) A2e-A'°2 (x2 0). 550

MISCELLANEOUS EXERCISE 25 Find the mean and standard deviation of the population and the probability that a randomly chosen member of the population is greater than 2. (M.E.I.) 24. Initially a machine is in good running order but is subsequently liable to breakdown. As soon as a breakdown occurs repairs begin. If the machine is in good order at time t then the probability that a breakdown occurs in a small interval (t, t+ d t) is a dt, and if it is under repair at time t the probability that the repair is completed in time (t, t+ d t) is fldt. Let p(t) be the probability that the machine is under repair at time t. Write down an equation relating p(t + dt) to p(t) and hence show that p(t) is a {1 — exp [ — (oc + /3) t]}.

a + fl

(C.S.)

551

26. Numerical solution of equations

1. ACCURACY When equations arise in practical problems, their solutions are generally required only to some given order of accuracy. General algebraic methods of solution (for example, for the cubic equation), even if they are available, are often less suitable than approximate methods for deriving numerical solutions. Thus, to find the real root of the equation x3 — 2.7x— 5.3 = 0 by Cardan's method would require burdensome calculations and no such method exists to solve, for example, the equation x5 — 2.7x— 5.3 = 0. Because of its great practical importance, the estimation of roots of numerical equations has been extensively studied; in this chapter we shall confine ourselves to some of the most elementary methods available: the reader who wishes to study the subject further should consult one of the standard texts on numerical methods (for example, Henrici, Elements of Numerical Analysis). It may be mentioned in passing that this subject allows more scope than most others in elementary mathematics for students to devise their own methods; although these may often prove less efficient than the standard techniques, much profit will be gained by developing them as far as possible—apart from the obvious enjoyment of producing something original. Before considering the solution of equations it is worth pausing to consider some problems connected with the accuracy of calculations. In numerical calculations, errors occur through rounding-off to a given number of significant figures. If a number is to be rounded-off to N digits and the discarded digits form exactly half a unit, round-off to the nearest even digit. For example, to 3 significant figures, 3.864 0.03765

3.86;

21990

3.76 x 10-2;

220 x 104;

21.55

216 x 10.

*Ex. 1. Suggest any advantage you see in adopting the convention above. Can you suggest any reason why even rather than odd digits have been selected?

552

1]

ACCURACY

Once numerical data have been rounded-off, their subsequent use in calculations introduces further errors which will be cumulative. It would be pleasant to be able to calculate exactly the error in any numerical answer. This is usually impossible, but we may be able to produce a positive àcceptance' for the answer; that is, a quantity which we are sure is more than the absolute error in the answer. For example, suppose we round-off a given angle 0, measured in degrees, to 3 significant figures to give 37.2°. The actual angle lies between 37.2°-0.05° and 37.2° + 0.05°. From fourfigure tables, tan 37.2° = 0.7590 and the difference for ±0.05° is 4(0.7618 - 0.7563) 0.0014. Thus

tan 0 = 0.7590 + 0.0014;

in other words, tan 0 is 0.7590 with an acceptance 0-0014. *Ex. 2. If possible inaccuracies in the tangent tables are taken into account, show that tan 0 = 0.7590 ± 0.0015. Now suppose y is a rational approximation to the number x; thus x = y + e,

where e is the error involved in writing y for x. The absolute error is I el while the relative error is I elly (more correctly 161/x, but the two forms are nearly equal for small relative errors and the form Ici ly has the advantage of possessing a known denominator). Usually we do not know the exact error, only an upper limit to the absolute error—the maximum absolute error. For example, if a number is rounded to 4 decimal places, the maximum absolute error is 5 x 10-5. Similarly, we have a maximum relative error. In a calculation in which we know the maximum absolute errors in the given numbers it is important to be able to estimate the maximum absolute error in the answer. Suppose that x1is written as yiwith maximum absolute error I el l (we use modulus signs to emphasize that maximum absolute errors are positive) and that x2 is written as y2with maximum absolute error 1 621 . Then the maximum absolute error in writing y, +y2for x„ + x2 is 1611 + 16 21,

while the maximum relative error is

(I ei +1621)/0).+ YD. Ex. 3. Show that the maximum absolute and relative errors in writing y1-y2 for x1- x2are respectively I 611 + 1 621 and (I elI + 1 621)01-Y2). *Ex. 4. Neglecting terms such as 161 621, show that the maximum absolute and relative errors in writing y1y2 for x1x2are respectively IY1 621 + IY2 611 and I ei/Y11 + 162/Y2 Show also that, writing yily, for x1/x2they are respectively

I ei/Y2 I +

62/Yil and I el/Yi I + 162/Y21.

553

NUMERICAL SOLUTION OF EQUATIONS

[26

Ex. 5. If you have a calculating machine available, calculate: (i) x, + x2, Yi +Y2 ; Yz; (iii) xix2, Y1312; -X2, (iv) X1/x2, Y1/Y2 (6 significant figures) where x1= 2.914, x2 = 0.3472, )7, = 2'9, y2 = 0.35. Ex. 6. What are the maximum absolute and relative errors in writing (i) for xi; (ii) Vyi for Vxi ; (iii) V(1+.Y2) for 1/(x1+ x2)? Before leaving the subject of errors, it is worth noting one particular case in which significant figures may be lost. Consider 3.144-3.097: both of these numbers are given to 4 significant figures, but their difference, 0.047, is correct to only 2 significant figures. It is sometimes possible to modify a calculation in such a way to minimize this loss of accuracy. For example, if it is required to calculate the difference between the values of a function of x for two given values of x that are close together, it may help to use a Taylor expansion. Thus, using four figure tables, sin 31° — sin 30° = 00150, but, by Taylor's expansion, sin 31° = sin 30° + (7r/180) cos 30° — 1(Tr/180)2 sin 30°... = 0.5000 + 0.01512 — 0.00076 and sin 31°— sin 30° 0.01504; again using four figure tables and obtaining a result correct to 4 significant figures. Again, suppose we have to solve the quadratic equation x2 -16x-1 = 0. The roots are

260 16+2

. 16 — V260 and 2

Direct evaluation of these two quantities gives 16.06 and — 0.06 (using four figure tables). However, in obtaining the negative root, we lost significant figures by subtraction. A better approximation may be had by calculating the negative root using the more accurate positive root 16.06 and the known relation that the product of the roots is —1: CC 7..;

1 16-06

0.0627.

Before leaving the subject of significant figures, it is worth noticing that a calculation can be carried out to a high degree of accuracy and yet not furnish a correct result even when corrected to 1 significant figure. For

554

ACCURACY

1]

example, if the correct answer is 0.649 999 9 and our numerical answer is 0-650 000 1 then, to 1 significant figure our answer is 0.7 but the correct answer is 0.6 although our calculation was only 0.000 000 2 in error.

Example 1. Given that V31=5.568 and V30=5.477, find V31 — V30. Direct computation gives V31 — V30 =0.091; but using (V31— V30) (V31 +V30) =1 we have

V31—/30=1/(5568+5477) =0.090539 (from 5 figure reciprocal tables)

and this is in fact correct to 5 S.F. 2. LOCATING THE ROOTS OF THE EQUATION f(x) = 0 Before embarking upon the accurate estimation of a root of the equation f(x) = 0 it is usually necessary first to locate the root roughly. One method of doing this is to draw the graph of y = f(x) and find where it cuts the x axis. Indeed, it is possible, by accurately plotting a succession of graphs of the relevant regions of f(x) on increasing scales, to obtain quite good approximations to roots of given equations. When using a graphical technique it is sometimes convenient to write the equation in a form other than f(x) = 0, e.g. g(x) = x. See, for example, Ex. 7. *Ex. 7. Verify from a freehand sketch that the equation x = 2 sin x has a root lying between Pr and ta. Draw the graphs of y = x and y = 2 sin x accurately on graph paper, taking values of x from -Pr to im and hence obtain a better estimate of this root. Suggest how you could continue the process. If you have a table of sines of angles in radians, show how these may be used to obtain rapidly a good approximation.

The graphical method just mentioned suggests a simple analytic method for locating the presence of a root: if f is continuous and f(a), f(b) have opposite signs, then the equation f(x) = 0 has at least one root lying between x = a and x = b. Ex. 8. Show that the equation

x5 — x —1 = 0 has just one real root and that this root lies between x = 1 and x = 2. Ex. 9. Show graphically that it is possible for roots of the polynomial equation f(x) = 0 to occur between x = a and x = b without f(a), f(b) being of different signs. 555


[26

When a root has been located between x = xo and x = x1, an approximate value for the root may be obtained by linear interpolation; that is, by approximating to the graph of y = f(x) between x = x, and x = x1 by a straight line (see Figure 26.1, in which OA = x0, OB = x1; we take OC as our approximate root, the exact root being OD). Since the equation of the line PQ, in the notation of Chapter 18 with xl— x, = h, is h(y—f0) = 40(x— x0) hf, x = xo — — Afo as an approximation to the required root. we have, putting y = 0,

Fig. 26.1 *Ex. 10. Show from graphical considerations how the computation of 412 /0 enables one to tell whether the approximation by linear interpolation is likely to be too big or too small.

Example 2. Show that the equation x 3+2x —4 = 0 has only one real root, and find its value, correct to 1 decimal place, by linear interpolation. Write f(x) = x 3 +2x— 4; then f ' (x) 3)0 + 2, and f"(x) = 6x. Thus we see immediately that, since the graph off has no maximum or minimum, it can cut the x axis only once, giving just one real root (Figure 26.2). 556

LOCATING ROOTS

2]

By direct computation, f(1) = —1, f(2) = 8 and the real root lies between x = 1 and x = 2, probably nearer to x = 1. By linear interpolation, a second approximation is given by x =1 Since f'(x) > 0, .f"(x) > 0 between x = 1 and x = 2 (see Figure 26.2) the root x = 19 will be an underestimate.

Fig. 26.2 By direct computation again we have

f(1.1) = —0.469, f(12) = +0.128, showing that the root lies between x = 1.1 and x = 12, and our next approximation to the root is 1.1+ 44 3x-A• As before, this is an underestimate and we can be confident, without further working, that x = 1.2 is the value of the required root correct to 1 decimal place. Another valuable method of locating roots of a polynomial equation is based on a consideration of the number of sign changes among the coefficients in the expression ao xn+ai xn-1+...+an. Ex. 11. Prove that, if all of the coefficients are positive, then the equation ao x" F ai xn-1+ ... + an = 0 can have no positive roots. Prove further that, if the coefficients of all odd powers of x are zero and the remaining coefficients are positive, then the equation can have no real roots.

557


[26

The result of Ex. 11 may be extended to the following result, known as Descartes's Rule of Signs: the number of positive roots of a polynomial equation cannot exceed, and has the same parity as, the number of sign changes among the coefficients, reading from left to right. (Two numbers have the same parity if they are both even or both odd.) For example, the equation 2x2 -7x7 -4x6 +3x5 + 2x3+ X2 -5x-4 = 0 + - - + + + - has three sign changes and thus either one or three positive roots. Again, the equation xs - 6x6 - 4x5 + 3x4 - x2 + 2 = 0 + - - + - + T has four sign changes and thus has zero, two or four positive roots. By writing -x for x one may similarly find an upper limit to the number of negative roots. Ex. 12. Show that the equation x7-3x4-x2-1 = 0 has just one real root, by considering separately the possible positive and negative roots.

The proof of Descartes's Rule is not difficult, but is complicated by the number of special cases that have to be considered. The reader who wishes to follow the proof through should try Exs. 13-17; in all these equations we adopt the notation P(x) = ao xn+ai xn-1+ + an_lx + an, where an> 0 and we suppose that P(x) has k sign changes. Ex. 13. Prove that P'(x) has k sign changes if a,„1> 0 and (k- 1) sign changes < 0.

if

Ex. 14. Prove graphically that, if P'(0) > 0, no roots of the equation P(x) = 0 lie between x = 0 and the least positive root of P'(x) = 0 and that if P'(0) < 0, at most one such root of P(x) = 0 exists. Ex. 15. What happens in Exs. 13, 14 if

= 0?

Ex. 16. Complete the proof of Descartes's Rule of Signs using mathematical induction.

558

2]

LOCATING ROOTS

Exercise 26(a) 1. If x is rounded-off to 3 significant figures to give the numerical value 48.7, give the value and range of acceptance for (i) sin x°; (ii) tan x°, using 4-figure tables. 2. If x, y are rounded-off to 3 significant figures, their values are x = 17.2, y = 5.16; give the range of acceptance for (i) x+y; (ii) xy; (iii) 3. Compare the relative accuracy obtained from your square root tables for x = V3 - V2 with that obtained for x = (V3 + V2)-1. If V2 = 1.41421... and J3 = F73205... find as accurately as you can a value for V3 - V2. 4. Given that V130 = 11.4018 and V132 = 11.4891 find as accurately as you can a value for V132 - V130. 5. The triangle ABC is right angled at B. AC, AB are measured to the nearest millimetre, their lengths being found to be 7.4 and 4.4 cm respectively. Use the theorem of Pythagoras to calculate the length of BC, stating what confidence you have in the reliability of your answer. 6. If x = y+ e, use the Taylor expansion to find the absolute and relative error in taking tan x to be equal to tan y. 7. With the notation of Question 6, find the absolute and relative error in taking (1 +x2)-1/2 to be (1 +y2)-112. 8. If x = 4-6 ± 0.05 and f(x) = x3- 2x +1, and if x is taken to be 4.6, find the relative errors in (i) x; (ii) f(x). Comment upon your result. 9. Find graphically, using the method of enlarging scales, the least positive root of the equation cot 7-1 = 1 + x (2 correct to 2 decimal places. 10. Solve graphically the equation 2x = 1 + In x. 11. Solve graphically x3 = 10. 12. Find 21/6, correct to 2 decimal places, by a graphical method. 13. Locate the three roots of the equation 2x3 - 6x- 3 = 0 and find their approximate values, using linear interpolation. 14. Locate the two real roots of the equation x4 +x2 + 10x- 24 = 0 and find their approximate values, using linear interpolation. 15. Use Descartes's Rule of Signs to prove that the equations (i) x6 - 2x3 - 3 = 0; (ii) x4 -x3 + 5x2+2 = 0; (iii) x6 + x4 - 4x3 + 5x + 2 = 0 each have at least two complex roots.

559


[26

16. Use Descartes's Rule of Signs to show that the equation x5 +x+1 = 0 has only one real root, and that this root is negative. Show that this result can also be obtained from simple graphical considerations. 17. What can you say about the reality of the roots of the equation x"+x2 +2 = 0 where n is an integer greater than 2?

3. ITERATIVE PROCESSES FOR SOLVING EQUATIONS In this section we consider the application of iterative methods for finding

numerical solutions of equations; that is, methods which develop successive approximations to a root of a given equation by a simple repetitive process depending upon a recurrence relation. As a first example, consider the equation x2 — 5x— 5 = 0. (A quadratic equation is chosen for simplicity, but the method to be outlined below is applicable to any polynomial equation.) The given equation has roots lying between —1 and 0 and between 5 and 6. We shall denote successive approximations to one of the roots by xo, Xi, X2, . . .. We first seek the negative root: take x, = —1. Since the equation may be rewritten

x = -}x2-1

we try, as a plausible attempt at developing successive approximations, the recurrence relation x, +, = 14.-1. Starting with x, = —1, this gives successive approximations —1, —0.8, — 0.872, —0.847, — 0.857. Thus x4= — 0.857, and the process is seen to be converging (albeit rather slowly) towards the negative root —0.8541 .... Now suppose we adopt the same recurrence relation to find the root lying between 5 and 6: x, = 6 gives 6.2, 6.69, etc. diverging. x, = 5 gives 4, 2.2, — 0.32, — 0.980, — 0.808, etc. and we are clearly converging towards the wrong root! 560

3]

ITERATIVE PROCESSES

To see what has gone wrong let us write = x7 + 6„ where a is the exact root of the given equation and 6, is the error in the estimate Xr. Xr+1 =

gives

. 14

a — er-Fi = 10: — err 1

which reduces to

= on neglecting 4. and recalling that a = ia2 — 1. Thus I e„±11 < I 6,1 only if lal < 2.5; in other words, the absolute errors diminish in this iterative process only if the root to which we are approximating is less than 2.5 in magnitude. Graphically, the solution of 1 2 1 X = TX — means finding the x coordinate of the intersection of the straight line y=x with the parabola

y = 5x2 -1.

Our iterative process consists in starting with a given value of x,

(x0 = — 1) finding the corresponding point on the parabola, moving from there to the line y = x (x1 = +4— 1), and thence to the parabola, then on to the line y = x, again (x2 =14-1), and so on, spiralling in to the root

x

= —0-854... in the cobweb pattern shown in Figure 26.3.

Fig. 26.3

However, if we start at x, = 5, a quite different pattern emerges. The reader should draw the graph and illustrate the various stages of the iterative process: he will see that the path taken moves rapidly across the graph to the negative root. 561

[26


*Ex. 17. Write the equation

x2

in the form

—

5x-- 5 = 0

x = 5+51x

and obtain the recurrence relation x,.+1= 5+ 51x,.. Show that, with xo = 6, this gives x5= 5.8541 which is correct to four decimal places. Show further that, in this case, 5er ier+il

*Ex. 18. Illustrate graphically the iterative process of Ex. 17 by sketching the graphs y= x and y= 5+51x. Ex. 19. Explain geometrically why convergence is more rapid for finding the positive root from xr.+.1 = 5+ 51x, than for finding the negative root from

x„+1 = ix,2—1. Ex. 20. Explain why, in both the iterative processes considered so far, the successive approximations oscillate from side to side of the exact root.

Now consider a polynomial equation rewritten in the form

x = f(x). (We have already seen that there are various ways of doing this: our object now is to choose the best one for a given root.) Suppose we take Xr+1 = f(xr), where xr, xr±iare successive approximations to the exact root a, with errors er, er+i: = Xr er, CC = Xr+1+ er+•

Since a is a root of the given equation, = Acc).

But giving

xr+1 = f(x,), cc — er+1 = f(c6— er)

f(x)— e„ f '(a), on using Taylor's expansion. Thus

er-F1 erAcc). Successive applications of this result give 67+1

" err(Ix) er-i{ '(42 "'

thus, provided If'(a)! < 1, 562

< I eol

GUM

r+1;

ITERATIVE PROCESSES

3]

and e,. decreases in magnitude as r increases. If, however, if(a)I > 1, the error increases as r increases. The reader should now reconsider the two iterative processes for the quadratic equation considered previously, in the light of this analysis. Ex. 21. Explain graphically what happens to the iterative process if f'(a) = 1. *Ex. 22. Illustrate the significance of the condition If(a)1 < 1 for the convergence of the iterative process

= by drawing four graphs with: (0 no) < 1, f"(a) < 0; (iii) f'(a) > 1, f"(a) < 0;

f(x,)

(ii) f'(a) < 1, f"(a) > 0; (iv) f ' (a) > 1, f"(a) > 0.

As we have already seen, the iterative procedure outlined above converges to the required root rather slowly. A more powerful iterative procedure is provided by the Newton — Raphson process. Consider the equation f(x) = 0 and suppose, as before, that a is an exact root, with x0the first approximation and error eo; that is cc = xo + 60. f(a) = 0,

We then have

© f(x0+ co) = 0, and thus, on using a Taylor expansion and regarding e0as being sufficiently small for us to be able to ignore e2, and higher powers, 0.

f(x0)+ eo f'(x0) This yields an expression for the error term °

f(X0) ovided f'(x0) , (p r f (x0 ``"

0),

and we may take as our next approximation xi = x0 fixo) fuo)

.

Repetition of this process leads us to the recurrence relation fixr)

Xr+1 -

r

Example 3. Show that the cubic equation x3— 5x — 8 = 0 has just one real root, and find its value, correct to 2 decimal places. 563


Writing

f(x) x 3 - 5x-8,

we have

f(x) = 3x2 5,

[26

-

and the curve y = .f(x) is seen to have two stationary points, a maximum at x A/1.7, and a minimum at x +V1.7. But — A/1.7 —1.3 and f( 1.3) < 0 therefore the curve cuts the x axis just once and the equation thus has only one real root (which is clearly positive). (See Figure 26.4.) —

Fg. 26.4

The next step in the solution is to locate the root: to do this we substitute integral values for x until we discover a sign change. In order to write down the differences Of, A2f, ... we tabulate our working, writing the values off in a vertical line: x

f

0

—8

1

—12

of

6,2f

oaf

—4 6 2 2

—10

3

4

6 12

14 The root we seek lies between 2 and 3; using linear interpolation we take as our first approximation xo = 2+14 x 1 ^ 2-7. Since 4/(2) > 0, the curve y = f(x) is increasing between x = 2 and x = 3 and since i2f(2) > 0, it is increasing at an increasing rate: we 564

3]

ITERATIVE PROCESSES

deduce that 2.7 is an underestimate of the exact root (see Figure 26.5, which has not been drawn to scale). Since f'(x) = 3x2– 5, f(2.7) xi = 2 7

f (27)

+1.817 1 16.87 2.808.

= 27

Fig. 26.5

The process is now repeated, taking, x1= 2.808 as our second approximation. (The process is much facilitated by the use of a hand-calculating machine: if one is available, recall the method of nested-multiplication for the evaluation of polynomials—see Chapter 18.)

x, = 2'808– = 2.808

f(2.808)

0.1 18.65

= 2.802. Since we may regard this as a more accurate approximation to the required root than 2.808, we take x = 2.80 as the root to 2 decimal places. Ex. 23. Use a calculating machine to obtain the root of the equation

x3 — 5x — 8 = 0 correct to 4 decimal places.

A feature of the Newton—Raphson process is that, since at each stage the value of f(xr) is calculated, a running check may be kept on the residuals; 7

PPMII

565


[26

that is, the values obtained by substituting our successive approximations into the polynomial. (We want the residuals to be zero eventually.) We now investigate, as before, how rapidly the process converges. Suppose that the exact root of the equation f(x) = 0 that we seek is a and that we obtain a sequence of approximations x0, x1, x2, ..., where = xr + cr. From the recurrence relation f(x,)

fury

Xr-14 Xr

a— er-F1 = a

we thus have giving

f(CC

er)

er f'(ce— err

er+1 = er

fex — er) . — er)

Now write

f (x) g(x) = f, , for f'(x)

then

g(a) = 0, provided.r(a) 0

and also, since

0;

g(x) f'(x) = f(x),

we have

g'(x)f'(x)+ g(x)f"(x) = f'(x)

and

glx)f'(x)+2g'(x)f"(x)+ g(x)fm(x) = f"(x).

Thus

gla) = 1, gloc) = — floc) f ' (cc).

From the equation

er-F1 =

Cr ±

er)

we now have, using Taylor's theorem and ignoring powers of er higher than the second, er-F1 Cr + g(a) — er g'(a) + g"(0) f"(C4) 2

2f '(a) r The analysis above shows us that each error is proportional to the square of the error in the preceding term: we deduce that convergence is more rapid in the Newton—Raphson process than in the first iterative method discussed in this Section. (The Newton—Raphson method is a second-order process.) Geometrically, the Newton—Raphson process is equivalent to drawing a sequence of tangents to the curve y = f(x). Let yo = f(xo); then the tangent at the point (x0, yo) on the curve y = f(x) has equation Y Yo = f'(-x0 (x— x0) 566

3]

ITERATIVE PROCESSES

and this meets the x axis at Yo xi = x0 f ,(xo

, fixo)(see Figure 26.6). - " Pxo) '

Fig. 26.6

Ex. 24. Show by graphical considerations, that, once the process starts to converge to a root, it always does so from one side. Try to produce an analytical argument to support this assertion. Ex. 25. Explain by a graphical argument how inaccuracies may arise in the neighbourhood of two nearly equal roots. The Newton-Raphson process involves division by the awkward number f (x7). A simplification is effected by using the von Mises iteration: xr+i

,

—

.,.r

f(xr)

f'(x0)'

where the variable denominator I (xr) is replaced by the constant f (x0). The convergence is less rapid but nevertheless fairly good. Ex. 26 Find the negative root of the equations x2 - 2x- 2 = 0 by Newton-Raphson and von Mises's iteration, taking x0 = -1.

Ex. 27. Interpret the von Mises process graphically. Explain why a very efficient procedure is to use the Newton method strictly for a suitable number of stages and then stick with a constant value off', e.g. f'(x,) after two stages. We conclude this chapter with an example of how an iterative process of a required order may be developed—in this case determining the reciprocal of a number.

7-2

567


[26

Example 4. If 1/a is calculated from the recurrence relation xr+I. = 2x,— axr. prove that

= aet.•

Find also the connection between Cr, er+1 if 1Ia is calculated from the recurrence relation xr-o.= 3xr — 3a4—a24 and suggest a fourth-order iterative process for finding 1Ia. If

xr+i = 2x,— a4 1 xr+ 6, = a 1 ETr = - xr a— er2 = 1 _ 2xr xr2 a2 a 1 aer2 = - —(2x1 —a4) a

we have

2 Xr-Fi + aer = 1 a er+1 =

Similarly, if xr+i =

(167,.

3a4+a24, 1 xr +er =a ,

1

3

3xr 34 3 — a —xr

a2eT = 1 — (3x,— 3axr2 +a24.) a Cr+1 = a24. Now suppose we seek a recurrence relation such that 1

xr+ er = a et. =

1

1

(4xr— 6a4 + 4a24 — a34)

a3er4 =1- — (4x,— 6a4 + 4a24 — a 568

4

(x Cr.

ITERATIVE PROCESSES

3]

Thus, if we set

x„i= 4x,. — 624+ 4a24.— a34,

it follows that

er„ = a3e4,

and we have a fourth-order iterative process for finding 1/a.

Exercise 26(b) Find the real roots of the equations 1 8, using any suitable iterative process. Use linear interpolation to find the first approximation and give your final answer to 3 significant figures. -

1. x3-3 = 0.

2. x3 — 100 ----- O.

3. x6 -5 = 0.

4. x2— 3x — 11 = 0.

5. x3 6x2 + 10x 9 = 0 (1 root).

6. x3 3x2 — 3x-7 = 0 (1 root).

7. x4-7x-12 = 0 (2 roots).

8. x3 + 3x2

—

—

—

—

9x— 16 = 0(3 roots).

9. Find to 4 significant figures the least positive root of the equation x4 — 13x2 — 18x— 5 = 0. 10. Show that the equation 2x6-10x3 + 10x-1 = 0 has two roots between 1 and 2 and find their numerical values, correct to 4 significant figures. 11. Use the Newton—Raphson method to find an approximate value for the least positive root of the equation 3 tan x = 4x (4 decimal places). 12. Find an approximate value of x such that x + ex = 3. 13. Solve, correct to 2 decimal places, the equation 7TX

sin T = 3x-1. 14. Show that the equation (2k+ 1) x3 — k(x +1) = 0, where k is large and positive, has a root near to 1. Find the equation of the tangent to the curve y = (2k + 1) x3 — k(x +1) at the point x = 1. From the equation of this tangent find a better approximation to the indicated root of the original equation. (0 & C) 15. Find the root of the equation sin x = x2 other than x = 0, to 3 decimal places.

569


[26

16. Establish Newton's formula for obtaining a closer approximation to a real root of the equation f(x) = 0. Use this method to find, correct to 3 significant figures, the positive root of the equation (L.) 4 cos x — 2x — 1 = 0. 17. By using Newton's method of approximation, or any other method, find the value of x correct to 3 decimal places for which the expression

x+1 In x has a stationary value.

(L.)

Miscellaneous Exercise 26 1. Find the greatest root of the equation x3 — 3x + 1 = 0, correct to 3 decimal places. 2. The roots of the quadratic equation ax2+ bx — 1 = 0 are calculated from the recurrence relation xr+1

=

1

ax,.+ b•

Interpret this process geometrically, and prove that er+i = — a X2e,. , where X is the exact value of the root being calculated and X = x,.+ 6,.. With the help of reciprocal tables, use this method to solve the equation 5x2 — 3x— 1 = 0. 3. Give a sketch showing the general shape of the graph of y = sec x for values of x from x = 0 to x = VT. Deduce from the graph that large roots of the equation x cos x = 2 are approximately equal to (n+ i) 7r, where n is a large integer; and prove that closer approximations are given by x = (n+}) 7T± — (2n + 1) IT' where the positive sign is taken when n is odd, and the negative sign when n is (0 & C) even. 4. A root of the equation sin3 ix°+cosx°- 400 = 0 is close to 60. Find the value of the root, correct to 0.1.

(0 & C)

sin x between x = 0 and x = 7r. Draw in 5. Draw an accurate graph of y = the same diagram the lines y = mx for m = s. f, s.f, 1. Determine the values of x where these lines cut the graph of y = ix sin x, giving your answers in the form kg, where k is correct to 2 decimal places.

570

MISCELLANEOUS EXERCISE 26 Use the values obtained to draw a separate graph of y = -Pr sin x/x between (0 & C) x = -1-Th and x = 77. ,

6. Verify that x = inr is an approximate solution of the equation cos x = (0 & C) and show that a better approximation is 1.03. 7. If

+ e,.)2 = a, prove that

er2

va = li+x,.)-2kx,. and deduce the recurrence relation Xr+i =

( XT) 2

for finding Ja. Suggest an intuitive argument leading to this recurrence relation and show that the same relation is obtained by applying the Newton—Raphson process to the equation x2— a = 0.

8. Ja is calculated from the recurrence relation a2 + 6a4. +

Xr+1

Show that

= 424(a+ .4) •

er+1 —

4x7(a+ x,.2)

and find V11 correct to 6 decimal places. 9. Develop a recurrence relation for finding al13in which e,.+1= ke;. (where k depends upon r) and hence find 0 to 5 decimal places. 10. Prove graphically, or otherwise, that the equation cos x = mx (m * 0) has one and only one root in the interval — < x < The angle a is defined as that root of the equation cot a = —a which lies between 17r and ir. Prove that, when m lies in the range — sin a < m < 0, the equation cos x = mx has three and only three roots in the interval —<x< By means of careful graphs of the functions cot x and — x, or otherwise, obtain the value of a, giving your answer in radians to 2 significant figures. 11. If a root of the equation f(x) = 0 is obtained by an iterative process in which ker (k constant) show that the expression x,.

— xi-02 Xr —

may be taken as an approximation to the required root. 571


[26

12. Find approximately the root of the equation x4 + 3x3- 7x2 + 18x - 18 = 0 which lies between 1 and 2. 13. Prove, graphically or otherwise, that the equation 1 f(x) 1+ - -tan x = 0 x2 has a root near x = 4 + kir, where k is any large integer. Denoting (k+4) it by K, prove that a better approximation to the root is given by K+ A, where f(K)+ Af ' (K) = 0, A

and that

1 2K2

Find an approximation to the root correct to the term of order K-4. (0 & C) 14. The equation x2 -ax + b = 0 has roots a, /3. Form the equation with roots fi2 and, if the equation with roots ae, /62' is x2 -a„x+b„ = 0, write down the equation with roots a2"-", 162' 1. If a = A.16 where IA I < 1, show how the formation of a sequence of such equations may be used to give successive approximations to ,6 and hence a. Under what circumstances does this method provide a very satisfactory method for solving quadratic equations? Discuss the magnitude of the errors involved and consider the cases of (i) equal, (ii) complex, roots. Discuss the application of this method of solution to equations of degree higher than the second. 15. Prove that the equation ex = Ax has a real solution if and only if A e, but that x ex = # has a real solution for all real values of it. Solve the equation xex = 1 to 5 decimal places. 16. xr, xr-F2 are successive approximations to a root of the equation f(x) = 0 obtained by the Newton-Raphson process. Writing Cr for the correction -f(x0if '(x,), so that Cr+1-f(xr+ Cr)If'(xr+ Cr) show that 1 f" (x,) C„i 2 f'(x,) by using Taylor's result, f(x+ h) f(x)+ hf'(x)+ +h2f"(x), where h is small. Deduce that x7,2- xr±i.

[f(x012f"(x0 (xr)]3 •

17. It is required to find the nth root of the number a by the Newton-Raphson process. Show that this can be done by solving the equation xn+k- axk = 0. Using the analysis of Question 16, show that most rapid convergence is obtained by choosing k = -1(1- n). Use this method to obtain a value for .;/2, carrying out two iterations. 572

27. The ellipse and hyperbola

1. CONICS: FOCI AND DIRECTRICES In Chapter 22 we considered the geometry of the parabola and rectangular hyperbola; both these curves are examples of a type of plane curve called a conic, which we shall now define. Given a point S and a line 1 not containing S, a conic is the set of all points P in the plane of 1 and S such that the ratio of the distance of P from S to the distance of P from 1 is a fixed number. If we call the foot of the perpendicular from P to 1, M, and the fixed ratio e (not to be confused with the base (e) of the natural logarithms) then a conic is the set of points P in the plane containing S and 1, such that SP = e (Figure 27.1). PM The point S is called a focus of the conic; 1 is a directrix. The number e is called the eccentricity of the conic. The type of conic is determined by

Fig. 27.1

Fig. 27.2

the value of e. The reader will no doubt already have noticed that, with e = 1, the conic defined is a parabola. More generally, we define the following types of conic: (i) if 0 < e < 1, the conic is called an ellipse; (ii) if e = 1, the conic is called a parabola; (iii) if e > 1 the conic is called a hyperbola. The shape of the three types of conic are shown in Figures 27.2 (ellipse), 27.3 (parabola) and 27.4 (hyperbola). (The dotted lines in Figure 27.4 are 573

ELLIPSE AND HYPERBOLA

[27

not part of the hyperbola: they are, in fact, asymptotes and are included as an aid towards drawing the curve.)

Ex. 1. Explain why the set of points {(x, y): x2= 4[(x — 1)2 + (y — 1)2]) represents an ellipse.

Fig. 27.4

Fig. 27.3

2. THE ELLIPSE The ellipse has been defined in Section 1 as a conic with eccentricity e < 1. Before attempting to obtain the Cartesian equation of an ellipse, it is worthwhile to consider a little of the geometry of the curve, in order that we may be able to choose the most suitable coordinate axes. Suppose that S is the given focus and 1 the corresponding directrix and let K be the foot of the perpendicular from S on to 1. If e is the given eccentricity, then there are two points A, A' on the line SK which belong to the ellipse—namely the points dividing SK internally and externally in the ratio e: 1 (see Figure 27.5 and notice that A' lies on KS produced, since e < 1). Let 0 be the mid-point of AA' and set OS = s, OK = k. Then, if AA' = 2a,

SA = a— s, A'S = a+ s, AK = k—a, A'K = k+a, and thus, by the definition of A, A' as points of the ellipse,

a — s = e(k — a) a+s = e(k+a). Solving these two equations we have

s = ae, k = ale. 574

2]

ELLIPSE

Let us now take 0 as the origin, OK as the x axis and the perpendicular to this line through 0 as the y axis. Then, by what we have just shown, S is the point (ae, 0) and 1 is the line x — ale = 0. Let P(x, y) be any point of the ellipse; by the definition of the ellipse sp2

= e2pm25

where M is the foot of the perpendicular from P on to 1 (Figure 27.6).

A

0

'

S A

K

Fig. 27.5

Fig. 27.6

Thus, the ellipse is the set of points {(x, y): (x _ae)2 +y2 =. e2(a/e_ x)2}. The equation may be rewritten x2

—

2aex+ a2e2 + y2 = a2 2aex+e2x2 x2(1— e2) +y2 = a2(1._ e2) y2 x2 = 1. 2+ a a2(1 — e2)

If we write

b2 = a2(1— e2), 575

ELLIPSE AND HYPERBOLA the equation becomes

[27

x2 y2 + = 1 -a-2 '

which is called the standard form of equation for the ellipse. *Ex. 2. Show that the ellipse

b2x2+ a2y2 = a2b2 is symmetrical about the x and the y axes and that (with the notation of Figure 27.6) B is the point (0, b) and B' is the point (0, —b). *Ex. 3. Deduce, by an appeal to symmetry, the existence of a second focus S'

(— ae, 0) and second corresponding directrix x+ ale = 0. *Ex. 4. 0 is called the centre of the ellipse. Show that every chord PP' of the ellipse which passes through 0 is bisected at 0.

AA', BB' are respectively the major and minor axes of the ellipse; a is the length of the major semi-axis, b is the length of the minor semi-axis (b < a, hence the word ' minor '). From the original definition of b, we have the important equation connecting a and b: b2 =a2(1 — e2) . Ex. 5. If a circle is regarded as a special case of an ellipse with equal major and minor axes, show that it corresponds to a conic with zero eccentricity. Ex. 6. A chord drawn through a focus perpendicular to the major axis of an ellipse is called a latus rectum of the ellipse. Show that the length of the latus rectum is 2b2/a. Ex. 7. Show that the eccentricity of the ellipse x2 y 2 fs 6 = 1 is sand deduce the coordinates of its foci. Ex. 8. Show, by translating the coordinate axes, that the equation (x 1)2 (y + 2)2 =1 3 2 represents an ellipse with centre (1, — 2). Find the coordinate of its foci and show the eccentricity to be I.A/3. The ellipse arises in nature as the typical orbit of objects moving under the action of an attractive force varying inversely as the square of their distance from a fixed point. For example, planets describe ellipses (with slight variations and discrepancies due to the attractions exerted on each other) with the sun at one focus—hence the use of the letter S for the focus. Similarly, for the motion of satellites around the Earth, the centre of the Earth being situated at a focus (again with a slight discrepancy, this time

576

ELLIPSE

2]

because the Earth is neither exactly spherical, nor uniform). The aesthetically pleasing shape of the ellipse has long been admired: it can be seen, for example, in the elliptic arches of some bridges which with their reflection in still water yield complete ellipses. Ex. 9. The eccentricity of the Earth's orbit is approximately -617, while that of the planet Pluto is 1. Compare the shapes of their orbits. Ex. 10. An astronomical unit (a.u.) is the mean distance of the Earth from the Sun. (It is approximately 500 light-seconds.) At its nearest approach to the Sun, the planet Mercury is distant about 0.308 a.u. from the Sun, while its maximum distance is about 0.466 a.u. Calculate the approximate eccentricity of the orbit. *Ex. 11. Show that x = a cos 0, y = b sin 0 gives a parametric representation for points of the ellipse b2x2 a2y2 = a 2b2. A rod has three points P, Q, R marked on it, where PQ = a, QR = b. If P, Q are constrained to move along two fixed perpendicular lines, show that R moves along the arc of an ellipse. (This is the engineer's paper trammel method for drawing ellipses.) *Ex. 12. Draw in the same diagram the ellipse b2x2 + a2y2 = a2b2 and the circle

x2 + y2 = a2

(the auxiliary circle of the ellipse). Given a point P of the ellipse, let the perpendicular through P to the major axis of the ellipse cut the auxiliary circle at Q. If OQ makes an angle 0 with the major axis, show that Q is the point (a cos 0, a sin 0) and that P is the point (a cos 0, b sin 0). 0 is called the eccentric angle of PQ.

The relationship between the ellipse and its auxiliary circle

b2x2 +a2y2 = a2b2 x2 +y2 = a2

outlined in Ex. 11 and Ex. 12 is worth a little further study. It will be recalled (see Chapter 13) that linear transformations of the plane into itself map straight lines into straight lines and in particular, parallel straight lines into parallel lines. They also map ratios of lengths on a line into the same ratio of corresponding lengths on the image line; in particular they map the mid-point of a line segment into the mid-point of the image line segment. We shall now consider the particular linear transformation, T, defined by the equation x' = x, y' = by/a. Ex. 13. Show that perpendicular lines do not map into perpendicular lines under T, unless a2 = 62. Identify the two linear transformations for which a = b and a = b, and explain why these do preserve perpendicularity. —

577


[27

*Ex. 14. Show that the image of the circle. {(x, y) : x2+ y2 = a2} under T is the ellipse

{(x y b2.0

a2/2 = a2b2}.

It follows from the preceding remarks and the result of Ex. 14 that properties of parallel chords of circles, mid-points of chords of circles, etc. are preserved by T and thus correspond to identical properties of the ellipse. For example, since the mid-points of parallel chords of a circle lie on a straight line through the centre, the same is true for an ellipse. The

Fig. 27.8

line obtained is called a diameter of the ellipse. Furthermore, the two lines parallel to the given chords passing through the intersections P, Q of the corresponding diameter with the circle are tangents: the same is true of the ellipse (see Figure 27.7). Again, in the circle, the centres of chords parallel to PQ define another diameter, RS, such that all chords parallel to RS are bisected by PQ. The same property is thus true of the ellipse: P'Q' and R'S' are called conjugate diameters; they have the property that each bisects all chords parallel to the other (see Figure 27.8). The geometrical interpretation of T is as follows. Given a point P, draw 578

2]

ELLIPSE

PX perpendicular to the x axis to meet it at X (PX is the ordinate of X); then P' is the point on PX such that P' X = (bla)PX. In particular, using the language of Ex. 12, the image of a point P on the auxiliary circle lies on the ellipse (see Figure 27.9).

Fig. 27.9

*Ex. 15. Show that T is a non-singular linear transformation, that is, that T-1 exists. Example 1. LP is the tangent at P to an ellipse, centre 0, and PR is the chord parallel to LO. Show that, if RQ is a diameter of the ellipse, then LO is a tangent. Suppose that the ellipse is taken in standard form and let us denote image elements under T-1(see Ex. 15) by dashes. The corresponding property of the circle (see Figure 27.10) is immediately obvious, either by an appeal to symmetry, or, equivalently, by proving the triangles L'P'O, L'Q'0 congruent. Thus, since the properties with which we are concerned remain invariant under the linear transformation T (under which the image of P' is P, etc.) the result is also true for the ellipse (see Figure 27.11). Notice that no appeal can be made to symmetry in the figure for the ellipse, but since the final result is couched in terms of tangents, diameters, and parallel lines only, the results for ellipse and circle correspond completely. In solving problems on the ellipse it is sometimes necessary to obtain the equations of particular lines, notably the tangent and normal. The quickest 579


[27

way to arrive at these equations is to use calculus, although alternative methods are available.

Fig. 27.11

Fig. 27.10 *Ex. 16. Show that, if

then

b2x2 ± a2y2 = a2b2,

dy_ _ _ b2x (y * 0). dx a2y

Deduce that the equation of the tangent at the point (x1, yi) on the ellipse is

x— x1 = y — y, = a a2y,

— b2x1

and show that this may be rewritten in the form xxi yy . — - +-- =1. a2 b2i *Ex. 17. Show that the equation of the tangent to the ellipse at the point (a cos 0, b sin 0) is x cos 0 y sin 0.1, + b a

and that the equation of the normal is ax sin 0 by cos 0 = (a2 b2) cos 5 sin 0. —

—

*Ex. 18. Show that the line L, with equation

cuts the ellipse

580

x—x1 _ y—yi _ a r — m— ' b2x2+a2y2 = a2b2

2]

ELLIPSE

at points with parameters A„ A2 which are roots of the quadratic equation _ 0=o. A2 _ + m\ +2A krx,+mi ± (1+ I ke 2 2 )

Deduce that the point nxi, Y1) is the mid-point of the chord L if

lx

i +— = ,

2 and that this gives the equation of the chord with mid-point P1as

x— xi= y—y, = p a2y1 —b2x, • Show how to deduce from this the equation of the tangent at the point of the ellipse.

P2(x2, Y2)

The ellipse, like the parabola, has a rich geometry. We shall now prove two of its most famous properties. (Since both are focal properties, neither is derivable from the circle.)

(i) Focal distance property

For any point P on the ellipse with foci S, S' and major axis 2a, SP+S'P = 2a. Take the ellipse in the standard form b2x2 +a2y2 = a2b2 and let M, M' be the feet of the perpendiculars from P on to the two directrices (see Figure 27.12).

Fig. 27.12

Let P(h, k) be any point on the ellipse; then, by the focus—directrix

definition of the ellipse

SP + S'P = e(PM+ M'P) = e[(9 --k)+(q+k)1 e e = 2a. 581


[27

*Ex. 19. Prove that SQ+S'Q < 2a for all points Q within the ellipse and that SR+ S'R > 2a for all points R outside the ellipse. *Ex. 20. Prove a converse of the above result, namely that if S, S' are fixed points and P is any point such that SP+ S'P is constant and greater than SS', then the locus of P is an ellipse with S, S' as foci. (If the constant distance is taken as 2a, show that a unique ellipse exists with S, S' as foci and major axis of length 2a; then use Ex. 19.) *Ex. 21. If a, b are complex numbers such that la— bl < c, where c is a real positive constant, describe the set of points {z C: lz — al + lz —1)1

c}

in the Argand diagram. Ex. 22. Explain the theory underlying the following well-known mechanical construction for an ellipse. Two drawing pins are stuck in a sheet of paper and a loop of cotton is placed loosely around the two pins. The loop is made taut by the point of a pencil which is then made to trace out a curve on the paper, keeping the loop taut at all stages of the construction. (ii) Reflection property

The tangent and normal at any point P of an ellipse bisect the angle SPS'. Take the ellipse in standard form and let P be the point (a cos 0, b sin 0).

With the notation of Figure 27.13, the equation of the normal, PG, is (see Ex. 17) ax sin 0 — by cos 0 = (a2 b 2) cos 0 sin 0,

Fig. 27.13

and thus G is the point

([(a2 — b 2) cos 011a, 0).

Using b2 = a2(1 — e2), this reduces to (ae2cos 0, 0). 582

2]

ELLIPSE

S'G = ae+ae2cos 0, GS = ae2cos 0

Thus and

GS_ ae — ae2cos 0 _ ale — a cos 0 _ PM _ PS S'G — ae+ae2cos 0 — ale+ a cos 0 — M'P — PS'

and so, by the angle bisector theorem for a triangle, PG bisects LSPS' internally. Since Z. GPT is a right angle, it follows immediately that PT bisects LSPS' externally. Ex. 23. ABC is any triangle and P is any point on the external bisector of the angle BAC. Prove that

BP+ CP ...-- BA+ CA, with equality only if P coincides with A. Example 2. Prove that the chord of contact of tangents from the point C(h, k) to the ellipse 2 Y2 +— 1 a b2 is the line

ky , a2 _i__ — I . b2

Deduce that, if the tangents at the extremities of a variable chord through the fixed point D(a, ,8) meet at R, then R lies on the straight line ax fiy 712 4- --bi = 1. From Ex. 16, we know that the equation of the tangent to the ellipse at the point P1(x1, y1) is

xxi yyl i a2 ± b2 ="

Now suppose that P1(x1, y,), P2(x2, y2) are the points of contact of tangents from C(h, k) to the ellipse. Then the coordinates (h, k) satisfy the equations

xx2 yy2 „ a2+ b2 i hx2 ky, , hxi_L kyi , and a2 + b2 =1. a2-T- b2 = ' xxi yy, ., , 2 = ' and a2 ± b2

and we have

—

Thus Pi(xl, y1) and P2(x2, y2) lie on the line

hx ky —+ = l a2 b2 ' which must represent the required chord of contact. If Ace, fi) lies on this chord of contact,

ha kfi

al+ V2 = 1 583

[27


for all positions of the point (h, k). Thus (h, k) lies on the line a2

+

fiy , = b2

Example 3. The normal at the point P to the ellipse E with equation b2x2 a2y2 = a2b2 meets the x axis at H and the y axis at K, and Q is the fourth vertex of the rectangle OHQK. Prove that the locus of Q is a concentric similar ellipse E', but with the minor axis of E' lying along the major axis of E. [Locus questions are usually best attempted using the parametric form (a cos 0, b sin 0): the coordinates of the point whose locus is sought are then obtained in terms of the parameter 0, which may be eliminated by using the identity cos' 0 + sin' 0 = 1 to yield the equation of the locus.] Let P be the point (a cos 0, b sin 0).

Fig. 27.14

2) cos 0 sin 0 Then the normal at P has equation ax sin 0 — by cos 0 = (a2 b (see Ex. 17). Thus we may write down the coordinates of the points H (y = 0) and K (x = 0) and hence of Q (see Figure 27.14): b2

(a2

H:

cos 0, 0) ,

a

a2 b2 sin 0), b

K: (0 '

/a2 b2

a 584

cos 0 '

a22 b2

sin 0).

2]

ELLIPSE

If we call the coordinates of Q(x, y) we have

ax cos 0 = a2 b2

sin 0 =

a2— by b2

and the equation of the locus of Q is b2y2 a2X2 (a2 b2)2 + (a2 b2)2 — 1,

which, by comparison with the equation y2

X2

A.2- B2 = 1' is seen to be an ellipse, E', with

A=

a2 —b2 a2— b2 ;B= a b

The centre of E' is the origin and A: B = b: a; the ellipses E and E' are thus concentric and similar. Furthermore, if a > b, then A < B and the minor axis of E' lies along the x axis, that is, the major axis of E. (Strictly speaking, the first part of our solution is incomplete until we have verified that every point of E' is a point of the locus. The reader may care to supply the details.)

Exercise 27(a) 1. Find the eccentricities of the following ellipses: x2 —

v2

8 6

= 1 1;

X2 (ii) — +y2 = 4

1

(iii) 3x2 +5y2 = 1.

2. Draw a rough sketch of the ellipse x2 5

y2

+9 = 1.

Find its eccentricity, the coordinates of its foci and the equations of the corresponding directrices. 3. Show that the equation 3(x — 2)2 + 4(y + 1)2 = 36 represents an ellipse and find its eccentricity and the coordinates of its foci. 4. Show that the equation

x2 +2y2 +6x-4y+9 = 0

represents an ellipse and find its eccentricity and the coordinates of its foci. 585

[27

ELLIPSE AND HYPERBOLA 5. Show that the equation 4x2 + 3y2 — 16x+ 12y+ 16 = 0 represents an ellipse and find its eccentricity and the coordinates of its foci.

6. The arch of an elliptic bridge is in the form of half of an ellipse. The span is 12.5 m and the maximum height of the arch is F75 m. Find the eccentricity of the ellipse. 7. Find the equation of the tangent to the ellipse x2

y2

y+ = 1 1 at the point (3, — 2). 8. Find the equation of the tangent to the ellipse x2 y 2 —+ — = 1 3 6 at the point (1, 2). Prove that the line x—y = 3 also touches the ellipse. 9. Prove that the line

x — 3y + 7 = 0

touches the ellipse

2x2 +3y2 = 14,

and find the point of contact. Where does the line

3x—y+7 = 0

meet the ellipse.

3x2 +2y2 = 14?

10. An ellipse has foci S, S', minor axis BB' and major axis of length 2a. Prove that BSB'S' is a rhombus of side a. 11. P is a point on the ellipse b2x2 +a2y2 = a2b2 and N is the foot of the perpendicular from P to AA', where A is the point (a, 0) and A' the point (—a, 0). Prove that PN2 b2 A'N.NA — a2. If the tangent at P meets the directrix corresponding to the focus S at T, prove that LPST is a right angle. 12. The feet of the perpendiculars from the foci S, S' of the ellipse b2 x2 +a2y2 = a2b2 to the tangent at the point P are Y, Y'. Prove that Y, Y' lie on the auxiliary circle of the ellipse and that S Y. ,S' Y' = b2. 13. P is a variable point of an ellipse, focus S. Prove that the locus of the midpoint of PS is an ellipse and locate its centre.

586

ELLIPSE

14. The tangents at the points P, Q of an ellipse, centre 0, meet at the point T. Prove that OT bisects PQ. 15. The parallel chords P1 Q1, P2 Q2of an ellipse are bisected by the diameter UV. Prove that P1Q2 and P2 Q1 meet on UV, as also do P1P2and Q1 Q2. 16. UV is a diameter of an ellipse and P is any point on the ellipse. Prove that the diameters parallel to PU, PV are conjugate. 17. With the notation of Question 16, the tangent at U meets PV at T and the tangent at P meets TU at M. Prove that M is the mid-point of TU. 18. The diameter UV of an ellipse bisects the chord PQ. UP and VQ meet at X, UQ and VP meet at Y; prove that XY and PQ are parallel. 19. What is the locus of the mid-points of chords of an ellipse which pass through a common point? 20. UV is a diameter of an ellipse and the tangents at U, V are u, v respectively; the tangent at any point P meets u, v at X and Y. Prove that X and Y lie on conjugate diameters of the ellipse. 21. Prove that tangents at the extremities of a focal chord of an ellipse meet on the corresponding directrix. 22. The tangent to the ellipse

x2 y 2 — + -= 1 a2 b 2

at the point P meets the axes at Q and R. Find the locus of the mid-point of QR. 23. The perpendicular from the centre of an ellipse with focus S to the tangent at a point P meets SP produced at Q. Prove that the locus of Q is a circle, and find its centre and radius. 24. Define geometrically the eccentric angle 0 of a point P on the ellipse x2 y2 a2 ± b2 = 1

(a > b)

and express the coordinates of P in terms of 0. Prove that the equation of the normal at P is ax cos q

by sin 0

= a2 b2.

0 is the centre of the ellipse and QP is the ordinate of P, the normal at P cuts the x axis at N. Show that ba NQ = — cos cb. a

If the normal at P bisects the angle OPQ prove that the eccentricity e satisfies the equation e2(1+ sin2 0) = 1, that OP = ae, and that h/2 < e < 1. (0 & C) 587


[27

25. S, S' are the foci of the ellipse x2 y 2 — a2+ b

-

=

„

2

and P(a cos 0, b sin 95) is a point on the curve. Calculate the lengths PS, PS' in terms of a, e and 95, where e is the eccentricity of the ellipse and verify that PS +PS' is constant. The tangents to the ellipse at P cuts the x axis at T, and the normal at P cuts the x axis at N. Prove that (i) OT. ON = a2e2; (ii) PT/PN = tan 0/(1 e2) where 0 is the centre of the ellipse. (0 & C) —

26. Show that the coordinates of any point P on the ellipse x2 y 2 —+ = 1 a2 62 can be expressed as (a cos 0, b sin 0). Prove that the equation of the tangent to the ellipse at P is x cos 0 y sin 0 = 1. a P and Q are two points on the ellipse, such that 9b has the value 951at P and 0 has the value 02 at Q. If the tangents to the ellipse at P and Q meet on the line ay = bx, prove that or in. (0 & C) 1-1- = 27. Prove that, if a212+ b2m2 = n2, then the line lx+my+n = 0 touches the ellipse

x2 y 2 =1 a2+ 62

and find the coordinates of the point of contact. Find the equations of the common tangents to the two ellipses x2 2 x2 y2 14+ 4 = 1'

23+

3 = 1'

28. If Z is the foot of the perpendicular from the centre of the ellipse b2x2 +a2y2 = a2b2 to the tangent at a variable point P, prove that the locus of Z is the curve (x2 +y2)2 = b2y2 +a2x2.

3. THE HYPERBOLA A hyperbola is a conic with eccentricity e > 1. The analysis of the hyperbola follows closely that for the ellipse and there are many striking similarities between the geometry of the two curves. To obtain the standard form of the equation of the hyperbola we first derive certain geometrical 588

3]

HYPERBOLA

results, just as we did for the ellipse in Section 2. Suppose that S is the given focus and 1 the corresponding directrix and let K be the foot of the perpendicular from S on to 1. There are two points, A, A', on the line SK which belong to the hyperbola—namely, the points dividing SK internally and externally in the ratio e: 1. Since, for the hyperbola, e > 1, 1 will lie between A and A' (see Figure 27.15, and compare with Figure 27.5). Again we take 0 as the mid-point of AA' and set OS = s, OK = k. Then, if AA' = 2a,

SA = s — a, A'S = s+ a,

KA = a—k, A' K = a+k,

A'

0

K

Fig. 27.15

and thus, by the definition of A, A',

s — a = e(a — k), s+ a = e(a+k). Solving these two equations we have

s = ae, k = ale. (Compare these results with the corresponding results obtained for the ellipse in Section 2.) Following closely the corresponding analysis for the ellipse, we take 0 as the origin, OK as the x axis, and the perpendicular to this line through 0 as the y axis. Then, exactly as for the ellipse, S is the point (ae, 0) and 1 is the line x— ale = 0. If P(x, y) is any point of the hyperbola, by the focusdirectrix definition we have

p2 = 6,2pm2,

where M is the foot of the perpendicular from P on to 1. Thus, the hyperbola is the set of points

{(x, y): (x ae)2 + y2 = e2(x ale)2}. —

—

589


[27

The equation may be written x2 — 2aex+ a2e2 + y2 = e2x2 _ 2aex+ a2 x2(e2_ 1) _y2 = a2(6,2 1) x2

y2

a2 a2(e2-1)

= 1.

Since e2 > 1, a2(e2—1) > 0 and we may write b2 = a2(e2_ 0.

Fig. 27.16

(Notice that this is the first slight point of departure from the corresponding analysis for the ellipse. Notice also that, for a hyperbola, b can be greater than a.) With this notation the equation may be written as x2 y2 a2 b 2

'

which is called the standard form of equation for the hyperbola. *Ex. 24. Show that the hyperbola is symmetrical about both the x and y axes and deduce by an appeal to symmetry, the existence of a second focus S'(— ae, 0) and second corresponding directrix. *Ex. 25. Prove that all chords through the centre, 0, are bisected at 0 and that the length of the latus rectum (see Ex. 6) is 2b2/a. The form of the complete curve is shown in Figure 27.17. Notice that, if P(x, y) lies on the hyperbola, v2 x2 = a2 (1 +, ) .... a2, 2

590

3]

HYPERBOLA

and thus no part of the curve lies within the interval -a < x < a. It follows that the points B (0 , b) and B'(0, - b) do not lie on the curve in contrast to the corresponding points of the ellipse. A (a, 0) and A' ( - a, 0) are called the vertices of the hyperbola; AA' is the transverse axis of the hyperbola, while BB' is the conjugate axis (see Ex. 26).

Fig. 27.17

The equation of the hyperbola may be written in the form

ry) (x +1 = 1, a b a b x -y = 1 a b x y. +ab

that is

x y As x, y both tend to co, ( - --) tends to zero and the equation of the curve ab

approximates to

x y - - = 0. a b

This line is thus an asymptote to the hyperbola; by a similar argument so also is the line x

-+- = O. a b

*Ex. 26. Show that the equations of the asymptotes may be written in the form x2 y 2 - - a2 b2 = 0 and prove that the acute angle between them is 2 arctan (61a).

591

[27

ELLIPSE AND HYPERBOLA *Ex. 27. Sketch in the same diagram the two hyperbolas _ y2 x2 y = 1 and c7 a2 b2 2

= —1

and mark in the points A(a, 0), A' ( a, 0), B(0, b), B'(0, b). Two hyperbolas of this form are said to be conjugate; show that conjugate hyperbolas have the same asymptotes. —

—

A rectangular hyperbola was defined in Chapter 22 as the curve which, with a suitable choice of axes, has equation

xy = c2. To justify the use of the word hyperbola in this definition, that is, to show that such a curve has the required focus—directrix property, consider the linear transformation, T, of the plane into itself defined by the equations

1 x' = — (x+y), A/2 1 y' = —(—x+y). y It is not difficult to see that T preserves distance; for, if the matrix of T is A, then

A=

1 1 V2 V2

vr) —sin (-410\ s((cosin (-1 (— in) cos (—lir)/ which represents a rotation through an angle — (see p. 243, vol. 1). It follows that, under T, the image of a curve C has precisely the same appearance and geometrical properties as the curve C itself, but is rotated through an angle of —17r. But, since

1 x =- (x' — y'), 2 Y=

1„ +Y

we see that any point of the set

{(x, y): xy = c2} maps into a point of the set {(x, y): x' 2 —y'2 = 2c2}. 592

3]

HYPERBOLA

This latter set clearly represents a hyperbola, with a2 = b2 = 2c2 and asymptotes {(x', y'): x' ± y' = 0}. Thus, the equation

xy = c2

represents a hyperbola with perpendicular asymptotes : we may therefore redefine a rectangular hyperbola as a hyperbola with perpendicular asymptotes. Ex. 28. Prove that the eccentricity of a rectangular hyperbola is V2 and that the latus rectum is equal to the distance between the vertices of the curve. (A rectangular hyperbola is sometimes called an equilateral hyperbola. All rectangular hyperbolas are similar to one another.) Ex. 29. Show that the foci of the rectangular hyperbola xy = c2 are the points (± c,/2, ± c‘12) and find the equation of the corresponding directrices. Ex. 30. A parametric form for a point on the curve xy =-1-a2 is (a412, al(412)); by considering the linear transformation T, deduce that the parametric form for a point on the curve x2

y2 = a2

(a(lIt+t)12, all/t— t)/2). Examine how this point moves along the curve as t varies between —co and +oo; in particular, explain what happens as t approaches the value zero from below and from above.

is

Ex. 31. Show that x = a sec 0, y = b tan 0 gives a parametric representation for all points of the hyperbola b2x2 — a2y2 = a2b2 and explain how P(a sec 0, b tan 0) moves on the hyperbola as 0 varies from 0 to 27r. In particular, what happens as 0 approaches the values 17r, lir from below and from above? Ex. 32. The circle.

x2+ y2 = a2

is called the auxiliary circle of the hyperbola b2x2_ a2y2 = a2b2 (see Ex. 12). Sketch in the same diagram a hyperbola and its auxiliary circle. P is a point of the hyperbola with positive coordinates, N is the foot of the perpendicular from P to the transverse axis and NQ is a tangent to the auxiliary circle at the point Q with positive coordinates. The angle NOQ = 0. Show that Q is the point (a cos 0, a sin 0) and that P is the point (a sec (- b tan 0).

593


[27

Ex. 33. Show that the equation of the tangent to the hyperbola x2 y 2

a2 — b2 =

, I

at the point P(a sec 0, b tan 0) has equation x sec 0 y tan 0 _ 1 a b — • Corresponding to the property SP+ S'P = 2a for points of an ellipse we have the following property for the hyperbola. Focal distance property If P is any point on a hyperbola, foci S, S', then ISP— S'Pl = 2a. Take the hyperbola in the standard form b2x2 — a2y2 = a2b2 and let M, M' be the feet of the perpendiculars from P on to the two directrices (see Figure 27.18). Let P be the point (h, k); then, by the definition of the hyperbola SP = ePM, S'P = ePM' and

1SP— S'Pl = eIMM'l = 2a.

Fig. 27.18

It is not difficult to see that SP— S'P = — 2a if P lies on the branch of the hyperbola enclosing S, and that SP—S'P = +2a if P lies on the opposite branch. *Ex. 34. Prove that, if S, S' are fixed points and P moves in such a way that SP S'P is constant, then the locus of P is a branch of a hyperbola. —

594

3]

HYPERBOLA

*Ex. 35. Interpret the conditions ISP— S'Pl < 2a and ISP— S'Pl > 2a geometrically. Ex. 36. Devise a mechanical construction for a branch of a hyperbola with given foci and length of transverse axis, based upon the relation SP — S'P = 2a. *Ex. 37. By adapting the method used for the ellipse, prove that the tangent and normal at the point P of a hyperbola with foci S, S' bisect the angle SPS'.

Hyperbolic orbits arise under the action of forces of repulsion varying inversely as the square of the distance from the centre of force, such as arise in the case of electrically charged particles. They also arise under attractive forces, such as gravity, when the energy content of the orbit is too great for it to be an ellipse, such as, for example, comets which orbit the sun only once before retreating into outer space. (Comets such as Halley's comet, which reappear, obviously follow elliptic orbits: the orbits are usually highly eccentric, that is, e 1.) The property I SP— S'Pl = 2a is used in range-finding. If a gun at P is fired and the times taken for the sound to reach two listening posts, S and S', recorded, I SP— S'Pl = 2a may be determined and P lies on the unique hyperbola with foci S, S' and transverse axis 2a. If the same experiment is conducted from S and a further listening post S", a second hyperbola is obtained and P lies at the intersection of the two hyperbolas. A more sophisticated application of the same idea is employed in navigation. [See the article `Sound Ranging' in No. 195 of the Mathematical Gazette (July 1928) by W. Hope-Jones.] The ellipse and hyperbola have very similar geometrical properties— most of the properties peculiar to the hyperbola are associated with its asymptotes. Both the ellipse and hyperbola possess a centre, that is, a point at which all chords are bisected and, for this reason, they are called central conics; the parabola is not a central conic. If we are concerned with a property common to all central conics, we may take the equation of a typical conic in the form ocx2 fly2 = 1 (where a, ft are not both negative). Example 4. Prove that the line y =- mx+c touches the central conic ax2 +fiy2 = 1 (o:061 0)

if and only if

= (ct

+ ftm2)/(cO

)•

Prove that perpendicular tangents to a central conic meet, in general, on a circle, the director circle of the conic, and point out what exceptional cases may arise. 595


[27

The given line and conic meet in points with x coordinates given by

that is, by

ax2 +fl(mx + c)2 = 1, (a +/3m2) X2 + 2ftmcx +(/3c2— 1) = 0.

The line y = mx+ c touches the conic ax2 +fiy2 = 1 , n2m2c2 = (cc ±/6,712) (fl C-2 1) (the condition for double roots of the quadratic equation above)

a fle2 = ± flm2

.4=>.

C2 = (rx + firn2)/(ccie), since a, ft + 0. Now let y = mx+ c be a tangent through the point (h, k). Then we have

k = mh+ c and thus, using the condition proved above for the line to be a tangent,

afl(k — mh)2 = a+4'1;12, amt, which reduces to m2(a8h2ft) —2afihkm +(aflk2— a) = 0. This is a quadratic equation in m (showing that two tangents can, in general, be drawn through the point (h, k)). Let the roots be m1, m2: then mim2 = (afik2 — a)1(afth2— fl). Thus, (h, k) lies on perpendicular tangents if afth2— ft = — (aflk2— a), 1 1 h2+k2 =- + a fi ' and the locus of (h, k) is thus x2 ± y2 = cc 1+73. that is, if

This is a circle provided (1/a)+(l/fl) > 0. This is certainly always true for an ellipse. For a rectangular hyperbola, (1/a)+(1/fl) = 0 and the director circle reduces to a point (the origin); for any other hyperbola, the circle exists provided (1/a) > —(1/fl), that is, in the usual notation, provided a2 > b2: a hyperbola in which the conjugate axis is greater than the transverse axis has no director circle. Since the angle between the asymptotes is 2 arctan (b I a), a hyperbola has no director circle if the angle between those parts of the asymptotes containing the curve is greater than -1-77.. Ex. 38. An elliptic lamina moves in a plane in such a way that it touches each of two fixed perpendicular lines. What is the locus of its centre?

596

3]

HYPERBOLA

Exercise 27 (b) 1. Find the eccentricities of the following hyperbolas: (i) 4x2 -1y2 = 1; (ii) 2x2—y2 = 1; (iii) (x-1)2 — 2(y-2)2 = 1. 2. Sketch roughly the hyperbola +y2

1

and find its eccentricity, the coordinates of its foci and the equations of the corresponding directrices. 3. Sketch roughly the hyperbola ,16(x — 1)2— i(y + 1)2 = 1, find its eccentricity, the coordinates of its foci and the equations of the corresponding directrices. 4. Show that the equation

x2 -4y2 +6x+8y+1 = 0 represents a hyperbola, and find its eccentricity, the coordinates of its foci and the equations of the corresponding directrices. 5. Find the equations of the asymptotes of the hyperbolas (i) 9x2 _ 4y2 = 12; (ii) 3x2—y2 = 1. 6. Find the equations of the asymptotes of the hyperbolas (ii) 2x2 -4y2 — 8y— 5 = 0. (i) x2 — 4y2+ 2x+ 8y— 5 = 0; 7. Find the equation of the hyperbola with asymptotes 2x —y = 0, 2x+y = 0, which passes through the point (2, 3). 8. Find the equation of the hyperbola with asymptotes 3x-2y-9 = 0, 3x + 2y + 3 = 0, which passes through the point (2, — 4). 9. Find the equation of the tangent to the hyperbola 2x2 — 3y2 = 6 at the point (3, —2) and prove that the line x+y + 1 = 0 also touches the curve. 10. Find the equation of the normal to the hyperbola 3x2—y2 = 2 at the point (1, 1) and find the x coordinate of the point where it meets the curve again. 11. Find the equation of the tangent to the rectangular hyperbola x2— y2 = 3 which passes through the point (3, — 3). 12. Prove that a hyperbola intersects a line parallel to one of its asymptotes at just one point. 13. Find the equation of the rectangular hyperbola which has the points (4, 0), (— 4, 0) as foci. 14. If the eccentricity of a hyperbola is e, prove that the eccentricity of the conjugate hyperbola is a/,,/(e2 -1). 8

PPMII

597


[27

15. Prove the results of Exercise 27(a) 11 for the hyperbola 62x2 — a2y2 = a2b2. 16. Prove the result of Exercise 27(a) 12 for the hyperbola b2x2_ a2y2 = a2b2. 17. If P is a variable point on a hyperbola with vertex A, show that the locus of the mid-point of PA is another hyperbola and find its centre and eccentricity. 18. Find the equation of the hyperbola (x2/a2) — (y2/b2) = 1 when the origin is changed to the point whose coordinates are (11, k), and the new axes are parallel to the original axes. The coordinates of a point are given by x = 2 cos20/(2 cos2 0-1), y = 2 tan 20. Show that the locus of the point is a hyperbola, and find the coordinates of the centre. Find the equation of the tangent to the curve at the point 0 = (0 & C) 19. Obtain the equations of the tangent and normal to the hyperbola 4x2 —y2 = 36 at the point P(5, 8). The tangent at P meets the y axis at Q and the normal at P meets the x axis at R. Prove that the area of the triangle PQR is 145 square units. Obtain the coordinates of the point S such that PQRS is a rectangle. (0 & C) 20. Prove that the normal to the hyperbola x2 y 2 i z2— =1 at the point (a sec 0, b tan 95) is given by the equation ax sin 0+ by = (a2+ b2) tan 0. The normal at a variable point P on the hyperbola meets the axes at X and Y; the mid-point of XY is Z. Prove that, if 0 is the centre of the hyperbola: (i) the length of OZ is not less than (a2+b2)12a; (ii) OZ is inversely proportional to the perpendicular distance from 0 to the tangent at P; (iii) if a = b, then Z always coincides with P. (0 & C) 21. The tangent at the point P of the rectangular hyperbola xy = c2 meets the asymptotes at Q, R. Prove that PQ = RP. By considering a linear transformation T with matrix of the form

show that the result above is true also for the hyperbola b2x2_ a2y2 = a2b2.

598

31

HYPERBOLA

Repeat the above process to prove the following more general result: if a chord PQ of a hyperbola meets the asymptotes at P', Q', then PP' = QQ' .

Suggest a further generalization of this result by considering a second rectangular hyperbola xy = d2in place of the asymptotes.

4. POLAR EQUATIONS OF CONICS Given coordinate axes Ox, Oy, the position of a point is uniquely determined by its Cartesian coordinates (x, y). However, other systems of coordinates are available; in Figure 27.19 the position of the point P is determined if, given the point 0 and the line OX, we know the length r = OP and the angle 0 = XOP. (r, 0) are called the polar coordinates of P relative to the origin or pole 0 and the initial line OX. We adopt the usual convention that positive values of 0 are measured in the anticlockwise sense. The polar coordinates of a point are not uniquely determined. For example, consider the point P with Cartesian coordinates (1, — V3) (see Figure 27.20). Taking 0 as origin and Ox as initial line, the polar coordinates of P can be taken in any of the alternative forms (2, —fa), (2, iir), (— 2, ig), etc.

x

0

P Fig. 27.19

Fig. 27.20

If unique polar coordinates are desired, we define the principal polar coordinates of a point to be the ones for which

r > 0,

- 77

< 0 *4 77 .

Generally, however, we allow r to take both positive and negative values. In the same way that a set of points in a plane could be defined by an equation or inequality involving Cartesian coordinates, sets may be defined using polar coordinates. For example, the set

{(r, 0): r = a} 8-2

599


[27

represents a circle, centre 0 and radius a; the set {(r, 0): r cos 0 = a} represents a straight line (traversed twice) perpendicular to the initial line and at a distance a from 0; the set {(r, 0): a < r < b} represents the annulus defined by two concentric circles of radii a and b. Ex. 39. Describe in words and draw sketches of the following sets of points: (i) {(r, 0): r sin B = a}; (ii) {(r, 0): 0 = e}; (iii) {(r, 0): r = a sin a cosec (0-0. (Hint: rewrite the defining equation and think of the Sine Rule.) (iv) {(r, 0): a < r < b, 0 < 0 < (v) {(r, 0): r sec 0 < Ex. 40. Find the polar equation of: (i) the line passing through the points with polar coordinates (1, 0) and (1, in); (ii) the circle through the points with polar coordinates (2, fir) and touching the initial line at the origin; (iii) the circle of radius 1 with centre at the point with polar coordinates (A/2,

Using the relation

x = r cos 0, y= r sin 0

it is a simple matter to rewrite the defining equation of a set of points in the plane in terms of polar coordinates. For example, the ellipse I' may be written in the alternative forms F = {(x, y): b2x2 + a2y2 = a2b2} = {(r, 0): r2(b2 cost +a2 sine 0) = a2b2}. However, it may well happen that a curve is best expressed in polar form without recourse to Cartesian coordinates. Suppose, for example, that we are given a conic of eccentricity e and semi-latus rectum 1. The focusdirectrix definition suggests that it would be reasonable to take the origin, 0, at a focus and the initial line along the major axis. In Figure 27.21, 0 is the focus of the conic, MN is the corresponding directrix and OL = lis the semi-latus rectum. Then, since OP = ePM, we have, But

r = ePM. PM = LN—r cos 0

= lle—r cos 0, giving us r(1 + e cos 0) = 1. 600

4]

POLAR EQUATIONS OF CONICS

This equation is the standard form for the equation of a conic in polar coordinates with the origin at a focus and the initial line along the major axis.

Ex. 41. Sketch the parabola r(1 +cos 0) = 2. Ex. 42. Show that the equation r(2+ cos 0) = 2 represents an ellipse and sketch the curve.

Fig. 27.21

The polar form for a conic is of great value in dealing with properties of focal chords. We conclude this section with an example illustrating its use in this context.

Example 5. PQ, UV are perpendicular focal chords of a rectangular hyperbola, focus S. Prove that: (ii) PQ = UV. (i) IPS.SQI = 1US.SVI, Taking S as origin and the major axis SX as initial line, the polar equation of the rectangular hyperbola is r(1 + ,/2 cos 0)= l. Let P, Q, U, V have polar coordinates

(r1, 0), (r„ + rr), (r3,0 + Pr), (r4, 0+21T). Then, from the equation of the conic we have r1= 1(1 + V2 cos 0)-1, r, = 1(1— V2 cos 0)-1, r3= 1(1— V2 sin 0)-1, r, = 1(1+ V2 sin 0)-1. 601


[27

(Notice that B cannot be an odd multiple of in if the chords PQ, UV are to exist.) Taking sense along each chord into account, this gives

PS = 1(1 + J2 cos 0)-1, SQ = 1(1 V2 cos 0)-1, —

US = 1(1 V2 sin 0)-1, SV = 1(1 +,12 sin 0)-1. PS. SQ = 12(1_2 coo 0)-1 = _ /2sec 20, —

(i)

US. SV = 12(1— 2 sine 0)-1= 12sec 20. . . PS SQ + US . SV = O.

Fig. 27.22

(ii)

PQ = PS+ SQ 1 1 1+ V2 cos 0+ 1— V2 cos 0 21 1-2 cost 0' 1PQ1 = 12/ sec 201,

and similarly

'UV 1 = 12/ sec 201.

Ex. 43. PQ is a focal chord through the focus S of a conic of semi-latus rectum 1. Prove that 1

1

2

PS+SQ =i •

602

4]

POLAR EQUATIONS OF CONICS

Ex. 44. PP', QQ' are perpendicular focal chords of a conic; prove that 1

1

,+ , PP QQ

is constant. Ex. 45. Show that the polar equation of the directrix corresponding to the focus 0 is er cos 0 = 1. Show also that a polar equation of the form 1/r = a cos 0 + b sin 0 represents a straight line. Deduce (with the help of Exercise 27(a), question 11 (second part)), that the equation 1 - = cos (B— cc)+ e cos 0 r

is the equation of the tangent to the standard conic at the point with vectorial angle a. Prove that the tangents at the ends of a focal chord of a conic meet on the corresponding directrix.

5. SECTIONS OF A CONE This section is included for the attention of readers interested in the historical development of the geometry of the conics. The conics, or conic sections, were first studied, as their name suggests, as sections of a right circular cone. Their history extends back to the time of Ancient Greece: they were first extensively studied by Apollonius of Perga (247-205 B.c.) who wrote a treatise on their properties. The focusdirectrix property, which we have made the basis of our definition, was not discovered until later. The various types of conic arise as sections of a cone made by planes making various angles with the axis of the cone. For the purposes of the following definitions, the cone is taken to extend infinitely in both directions from the vertex. (See Figure 27.23.) First observe that any plane perpendicular to the axis of the cone (and not through the vertex) cuts the cone in a circle, which may therefore be regarded as a particular type of conic. If the plane is oblique, not parallel to one of the generating lines of the cone and cuts only one half of the cone, the resulting section is an ellipse (Figure 27.24). If the plane is parallel to one of the generating lines of the cone, the resulting section is a parabola (Figure 27.25). If the plane is oblique and cuts both halves of the cone, but does not 603


[27

pass through the vertex of the cone, the section is a hyperbola (Figure 27.26). Ex. 46. Show how a pair of straight lines arises as a conic section.

Fig. 27.23

Fig. 27.24

Fig. 27.25

Fig. 27.26

The connection between the conic section and the focus-directrix definitions is exhibited in Exs. 47-52. The notation refers to Figure 27.27 (for convenience of drawing, we take an elliptic section and show only one half of the cone). 604

5]

SECTIONS OF A CONE

V is the vertex of a right circular cone with its axis vertical and we consider the section of the cone by the plane H. A sphere may be drawn to touch the plane H at S and also to touch the cone in a circle lying in a horizontal plane H'. The vertical plane containing V and S cuts H and the cone at A and A' and H' and the cone at C and C'. 1 is the line of intersection of H and H' and A' A meets 1 at K. P is any point on the cone lying in the plane H and VP touches the sphere at P'. M is the point of 1 such that PM and AK are parallel, NPQ is a horizontal section, N lying on AA' and Q lying on VA.

Fig. 27.27

Ex. 47. Show that PNKM is a rectangle. Ex. 48. Show that SP = QC and PM = NK. Ex. 49. Show that QC/NK = AC/AK. Ex. 50. Deduce that SP/PM is fixed for a given plane II, and locate a focus and its corresponding directrix for the resulting conic section. Ex. 51. Show how the parabolic cross-section arises. Ex. 52. Show that, both for elliptic and for hyperbolic sections, a second sphere may be drawn to touch the cone and the plane II, giving rise to a second focus and a second directrix.

Miscellaneous Exercise 27 1. Show that the equation x2 +2y2 -2x+12y+8 = 0 represents an ellipse, and find its eccentricity and the coordinates of its centre. Prove that 3x-2y+2 = 0 is a tangent to the ellipse and find the point of contact.

605


[27

2. The tangents drawn from a point P(xi, y1) to the circle x2 + y2 = r2 touch the circle at L, M. Show that the equation of LM is xx,+ yy, = r2. If P moves on a hyperbola of eccentricity e with its centre at the origin, show that LM touches a hyperbola of eccentricity e', where 11 1 = 1. e2 e'2

Draw a figure for the case when e < Al2 and the circle touches the given hyperbola. (London) 3. Show that the equation of the tangent at P(a cos a, b sin a) to the ellipse b2x2 + a2y2 a2b2 is bx cos a+ ay sin oc = ab. The tangent at P cuts the x and y axes at A and B respectively and the normal at P cuts the x and y axes at C and D respectively. Find the ratio PC: PD. If AD and BC meet at E, prove that BE is perpendicular to AD and hence, or otherwise, find the equation of the circle through A, B, E. (London) 4. Prove that the conics

x2 y2 X2 y2 — = 1 and + =1 a2+b2 a2 + A b2 + A have the same foci, whatever value A takes (provided A — a2or — b2). Find the equation of the rectangular hyperbola whose foci coincide with those of the ellipse b2x2 + a2y2 = a2b2. Show further that two ellipses which have the same foci cannot intersect in real points but that if an ellipse and a hyperbola have the same foci, they intersect orthogonally at four real points. 5. P is any point on a rectangular hyperbola, centre 0 and foci S, S'. Prove that OP2 = SP. SI'. 6. P is the point (a cos 0, b sin 0) of the ellipse

b2x2 + a2y2 = a2b2 and Q(a cos 0, a sin 0) is the corresponding point of the auxiliary circle. Prove that the perpendicular distance from S to the tangent at Q to the circle is equal to SP. 7. Write down the equation of (i) an ellipse which has its minor axis along the y axis and touches the x axis; (ii) an ellipse which has its minor axis along the x axis and touches the y axis. Two such ellipses are given. Assuming that they meet in four (real) points, write down the equation of any conic through these four points, and prove that its centre lies on a certain rectangular hyperbola. (0 & C) 8. A tangent to the ellipse

(a, b > 0) meets the ellipse

b2x2 + a2y2 = bX 2

a 2b2

ay2 = ab(a+ b)

at the points P, Q. Prove that the tangents to the second ellipse at P and Q meet on its director circle. 606

MISCELLANEOUS EXERCISE 27 9. Two hyperbolas, S and S', have the equations x2

y2

= x2 y2 T 12 ba

A,

where A. < 1. Prove that the tangent to S at any point P meets S' in two points Q and R which are equidistant from P. If A = —1, prove that the tangents to S' at Q and R meet on S at P', the reflection of P in the origin. (0 & C) 10. The asymptotes of the hyperbola

x2 ya —— — a2 b2 = 1 are 1 and l', and P is a point on the hyperbola. The perpendicular from P to 1 meets / and l' at X and Y, and the perpendicular from P to l' meets 1' and 1 at X' and Y'. By expressing the coordinates of P in parametric form, or otherwise, prove that, for all positions of P, a2b2 (i) PX.PX' = • a2+b2 ' (ii) PX.PY = (iii) PY.PY' =

a2b2

a2— b2; a2b2(a2+b2)

(a2 — b2)2

(0 &

11. Show that, if the point (x1-1-r cos 0, yi+ r sin 0) lies on the central conic

ax2+ by2 = 1, then r satisfies the quadratic equation r2(a cosy 0 +b sine 0)+ 2r(axicos 0 + byi sin 0)+a4+ by;. — 1 = 0. Deduce that, if (xi, y1) is the mid-point of the chord PQ, then PQ has gradient — ax1/by1.

Prove that the locus of mid-points of parallel chords of a central conic is a diameter of the conic. 12. Use the analysis of Question 11 to prove Newton's Theorem for a central

conic: if PQ, RS are two chords intersecting at X, then the ratio PX. XQ RX. XS depends only upon the directions of the chords PQ and RS, and not on their positions. 13. Given the outline of an ellipse, show how you would construct the centre,

the axes, the foci and the directrices. 14. Prove that, if tangents are drawn to a hyperbola from any point of the

conjugate hyperbola, their chord of contact touches the opposite branch of the conjugate hyperbola and is bisected by it. 607


[27

15. Find the equation of the perpendicular bisector of the line joining the points yi), (x2, Y2). A fixed circle has centre C and radius 2a. A is a fixed point inside the circle and P is a variable point on the circumference. Prove that the perpendicular bisector of AP touches the ellipse whose foci are at C and A, and whose major axis is of length 2a. (C.S.) 16. A point P is taken at random inside an ellipse of eccentricity e. Calculate the probability (in terms of e) that the sum of the focal distances of P should be not greater than the distance from a focus to the opposite end of the major axis. (C.S.) (Note: the area of an ellipse with major and minor semi-axes of lengths a and b respectively is rrab—a result easily deduced by the usual calculus methods, or alternatively by considering the effect on areas of the linear transformation T of Section 2 of this Chapter.) 17. E, is a circle, centre 0, radius b, E2 is a circle, centre A radius a (a < b) which touches /1internally. Describe the locus of the centre, P, of a variable circle, Z, which touches Elinternally and E2 externally. If E, is a straight line, E2 is a circle, centre A and radius a which touches Ei and P the centre of a variable circle E which has Elas a tangent and touches E2 externally, describe the locus of P. 18. Prove that at most four normals may be drawn from a point A to a central conic with centre 0. Prove further that, if the normals at the points -P1, -P2, -P3, -P4 on a central conic intersect at A, then -P1-P2-P3-P4 lie on a rectangular hyperbola which passes through 0 and A and has its asymptotes parallel to the axes of the given conic. (The rectangular hyperbola of this question is known as the hyperbola of Apollonius.) 19. Prove that the vector equation (referred to the vertex as origin) of the tangent to the parabola y2 = 4ax at the point p = at2i+2atj is r = p+Au where u is the unit vector (ti+ j)/(1 + t2)i. Prove that the locus of the meets of tangents to the parabola y2 = 4ax which cut at a fixed angle a is a hyperbola of eccentricity sec a. 20. Prove that, if the chord PQ of any conic with focus S, when produced, cuts the corresponding directrix at R, then SR is a bisector of the angle PSQ. Deduce that, if the tangent at P cuts the directrix at T, then the angle PST is a right angle. 21. If X is any point on the tangent at the point P of any conic with focus S, and if H, K are respectively the feet of the perpendiculars from X to SP and the directrix corresponding to S, prove that SH = eXK, where e is the eccentricity.

608

28. Further matrices

Throughout this chapter it is to be assumed, unless explicitly stated otherwise, that the matrix A of a linear transformation T is referred to the base vectors i, j (or i j, k in three dimensions). ,

1. EIGENVALUES AND EIGENVECTORS FOR 2x 2 MATRICES 2 as A Let us write the matrix of the linear transformation, T: R2 R where b1\ A = (a1 a2 b2I • Whatever form T, and therefore A, has, the origin 0 is mapped into itself. In general, no other point remains fixed under T but, if there is a point P, distinct from 0, such that T(P) = P, then all points of the line OP are mapped into themselves. Ex. 1. Prove the assertion that if T maps a point P (other than the origin) into itself, then it maps every point of the line OP into itself. Ex. 2. Prove that, if T maps two points, P and Q, into themselves, where 0,P, Q are distinct and not collinear, then T is the identity transformation, that is, the transformation with matrix

We may now investigate the answer to the question: Even if T does not map any point other than the origin into itself, is there a line 1 through the

origin such that every point of l is mapped into a point of 1? If such a line exists, we say that 1 is an invariant line and that 1 maps into itself under T. Ex. 3. If S is a point such that T(S) e OS, prove that every point of the line OS is mapped into a point of OS. Suppose that S is a point other than the origin, with the property that T(S) E OS; write OS = s. Then As = As and thus

(A — AI) s = 0. Since s 0, we must have det (A — AI) = 0 (see p. 294). This is a quadratic equation in A, the characteristic equation of the matrix A, with, in general,

609

FURTHER MATRICES

[28

two distinct roots, Al and A2, called the eigenvalues of the matrix A. Any non-zero solution, s1, of the homogeneous equation (A — s =

0

is called an eigenvector corresponding to the eigenvalue A1; similarly, we can find eigenvectors corresponding to the eigenvalue A2. Example 1. Find the eigenvalues of the matrix

A = (4 6 2 5) and determine their corresponding unit eigenvectors.

The characteristic equation (41— A det

k

.4.>

2) = 5—A

o

A,— 9A + 8 = 0

A = 1 or A = 8. Consider first the eigenvalue A = 1: if s1is any corresponding eigenvector then, writing (y) (A — I) s1 = 0,

we have that is

(6

4) (y1 — (0 0) •

On solving these equations we obtain x = 2 a, y = — 3,1t, giving as an eigenvector 2,tt\ = — 3i1) • In particular, a unit eigenvector is given by ,

=

k

2[03 \ 3/03) •

All the points of the line

r = ke„

that is, all points of the line

3x + 2y = 0,

map into points of the same line (in fact, in this case, into themselves since A = 1): 3x + 2y = 0 is an invariant line under the transformation. Now consider the eigenvalue A = 8 : if s2is any corresponding eigenvector we have (A — 81) s, = 0, 610

2 x 2 MATRICES

I]

—6 4

that is, giving

x) _ 32) (y

(0 0) ,

v\

s2

= (2v) •

In particular, a unit eigenvector is given by — (1/V5 2/v5) . r=

All points of the line

2x — y = 0,

that is, of the line

map into points of the same line (but, apart from 0, not this time into themselves): 2x —y = 0 is a second invariant line under the transformation. We shall now analyse the linear transformation represented by the matrix A of Example 1 in greater detail. The matrix 32 2) P

—

(

-

formed by taking as its columns the eigenvectors sl

=

2 2) 3) and s2 = (1

(these eigenvectors being chosen for their simplicity, although any nonzero multiples of s, and s2would do just as well) has an important property. Since As, = s, and As2 = 8s2 it follows that the product AP is a 2 x 2 matrix with s, as its first column and 8s2as its second column. Thus 1 0 AP = P( 0 8) But, since det P = 7, P is non-singular and we may multiply both sides of the above equation on the left by P 1to obtain 1 0 P-1AP = ( 0 8) • The matrix A has been reduced to diagonal-form. Notice that the elements of the diagonal are precisely the eigenvalues of A. 611

FURTHER MATRICES

[28

The linear transformation T, with matrix A relative to i, j as base vectors, maps the point with position vector ocsi +A2 where s1, s2are defined above, into the point with position vector ocs1+ 8,6's2; it follows that, if we take s1, s2as base vectors, the matrix of T is

11

\ '. ‘09 8 0)

for

1 1 0\ 1oc\ 1 a\ V) 8) k,e) — ‘8,e)

as required. Thus, when considering the transformation T, it is simpler to express vectors in terms of the eigenvectors s1, s2rather than in terms of the more usual base vectors i, j. For example, the point Q, where

q = 2s1 + is2, maps into the point Q', where

q' = 251+452. Figure 28.1 shows the effect of the transformation T upon Q:

OQ = 2s1 +-1-s2 = OP+PQ, OQ' = 251+452 = OP'+P'Q', where OP' Q'P is a parallelogram. P'

Fig. 28.1

612

1]

2x2 MATRICES

Ex. 4. Calculate the eigenvalues and corresponding unit eigenvectors of the matrix B= ° 3) . -2 1 Hence describe geometrically the linear transformation T which has matrix B relative to the base vectors i, j. Reverting to Example 1, the reduction to diagonal form enables us to calculate powers of A. For example,

= (P-1AP)3 = (P-1AP) (P-1AP) (P-1AP) = P-1A(PP-1) A(PP-1) AP = P-1A3P As p 1 \ RA k0 83) =1/2 /1 0\ /2 - 1 7 k -3 2)2)k0 512) k3 2) =1 2 512\ /2 7k / - 3 1024) k3 2) = 1/1540 1022\ 7 k3066 2051) /220 146 438 293) • More generally we have A 1 2 1\ /1 0 \ /2 - 1 \ = 7 k -3 2) k0 8n) 3 1 4+3.84 -2+2.84\ = 7 (-6+6.8n 3+4.8n)•

0

83)

( 1

Ex. 5. Verify by direct calculation the form given for A" above and also prove the result by mathematical induction. *Ex. 6. If si and s2are eigenvectors corresponding to the distinct real eigenvalues A1, A2 of a matrix A and if P is the matrix with first column s1and second column s2, and assuming that P is non-singular, show that p iAp (A 0 ) 0 A2 -

Eigenvalues need not be real. For example, if

A= -

1) 613

[28

FURTHER MATRICES

the characteristic equation of A is (A — 1)2 + 1 = 0 giving eigenvalues Al = 1 + j, A2 = 1 — j, with corresponding eigenvectors /1\ 1\ k ' In this case, the invariant lines are imaginary. Of the linear transformations of the plane into itself, those which preserve distance are of particular importance—that is, linear transformations T with the property that, if T(P) = PT, j' then OP = OPT for all points P of the plane. In Theorem 28.1 we shall show that the matrix A of such a transformation is of a special type; to this end we make the following definition (which holds for n x n matrices, although we are considering here only 2 x 2 matrices): A is called an orthogonal matrix if AA' = I. *Ex. 7. Show that, if A is an orthogonal matrix then det A = ± 1, A is nonsingular, and A-1= A'. Ex. 8. In the course of the proof of Theorem 28.1 we shall require the fact that the transpose of a product AB of two matrices is the product of the transposes of

A and B in reverse order; that is (AB)' = B'A'• Prove this result (i) when A is 2 x 2 and B is 2 x 1; (ii) when A is 2 x 2 and B is 2 x 2. Show how your proofs may be generalized to cover all cases of conformable matrices A, B up to order 3 x 3 (or rn x n if you can manage it).

Theorem 28.1. Let T be a linear transformation of the plane into itself, with matrix A. Then T is distance preserving (that is, 1011 = 1OPT I for all positions of the point P) if and only if A is orthogonal. First observe that, if OP = r = xi +yj, then

r'r = (xy) () = (x2 +y2) that is, the 1 x 1 matrix r'r represents OP2. Now suppose that P is any point of the plane, that OP = r, and that OPT = rT. Then r, rT = (Ar)' (Ar) = r'A'Ar (by Ex. 8) and thus rT' r, = r'r

-;*

r'(A'A) r = r'r

•

r'(A'A) r = rIr

r'(A'A —I) r = 0. t We use the notation Pr rather than P' here to avoid possible confusion with the notation for the transposed matrix.

614

2 x 2 MATRICES

I]

(i) Suppose A is orthgonal. Then A'A = I and thus err, = r'r for all points P of the plane and T is distance preserving. (ii) Suppose T is distance preserving Then r'(A'A —I) r = 0 for all position vectors r and, in particular, for r, = r2 = j and r3= i +j. Writing b2) =B A'A—I = (Ll u3 b4 we thus have _ 0; bb24) (01) (00) (1 0) ( b:

(0 1) (1

1)

bb2 (bb: 4) (0) (bb13

b42)

0;

(00)

(1) = (0)

b4 b,+b2+b3+b4 = 0.

Thus b1 = b4 = 0 and b2 = — b3 = b say, and we have

B = (_ ° b o ) and A'A =

1 b —b 1) •

But (A'A)' = A 'A and therefore 1 b b 1) = (b1 —b) and b = 0. This shows that A'A = I and thus A is orthogonal. *Ex. 9. Prove that an orthogonal matrix P preserves separations, in the sense

that

IPx—PyI =

*Ex. 10. Prove that, if P is a 2 x 2 orthogonal matrix, then either P=

cos 0 —sin 0\ ksin 6' cos 0/

corresponding to a rotation through an angle 0, or

sin 0\ ‘sin 0 — cos Of

P (COS 0

corresponding to a rotation through — 0 followed by a reflection in the x axis (that is, a reflection in the axis y = x tan 40). *Ex. 11. Prove that the eigenvalues of a 2 x 2 orthogonal matrix P are (i) conjugate complex numbers of modulus 1 if det P = +1, or (ii) the numbers ± 1 if det P = — 1. 615

[28

FURTHER MATRICES

The reduction of a matrix A to diagonal form P-'AP is of particular importance if A is a symmetric matrix (A' = A). We shall show (Theorem 28.5) that, in this case, the reducing matrix P may be taken to be orthogonal. Before proving this, it is necessary to obtain some simple preliminary results. In the following four theorems, A is a 2 x 2 matrix with eigenvalues Al, A2 and corresponding eigenvectors sl, s2and P is the matrix with first column s1and second column s2. Theorem 28.2. If the eigenvalues of the 2 x 2 matrix A are distinct then the matrix P is non-singular. Proof

det P = 0 s1= ks„ where k is a non-zero number, on multiplying by A, As,— kAs2 = 0, A,s, —Aks, = 0 (Ai — A2) sl = 0,

since ks2 =

Al = A2, Thus, since det P = 0 Al =

since s1+ 0. A2,

we have Al +

A2

det P + 0.

Theorem 28.3. If siand s2are unit eigenvectors of A then s1and s, are perpendicular if and only if P is orthogonal. Proof _ (x) s2 _ (x2) Write Y2 xi x2 then P = ( I. Y2 x+ x1X2 ±Y1Y2) pip = (xi Yi\ i yi2 and • 2 X2 xl x2+ y22 kx2 Y2/ \Yi x2' = Y1Y2 (i) Suppose P is orthogonal then

( 4-FA

x1 x2+yi y2 \ = (1 \

ko 1/ 4+y2 giving 4.+A = 4+3,2 = 1, x1x2+Y1Y2 = 0 and thus s1and s2 are perpendicular unit vectors. (ii) If s1and s2are perpendicular unit vectors then \ Xi X2 ±Y1Y2

pip =

+34 kx1 X2 ± Y1372 = (01 0 1)

and P is orthogonal. 616

x1x2+ yiy2) x22 +y22

1]

2 x 2 MATRICES

Theorem 28.4. If A is a 2 x 2 real symmetric matrix + kI then its eigenvalues are real and unequal. Proof. Write A=( ab bc) • Then

det (A — Al) = 0, (a — A) (c — A) — b2 = 0 A2 (a + c) A + (ac — b2)

= 0.

But the discriminant of this quadratic equation is (a+ c)2 — 4(ac b2) = (a — c)2+ 4b2 > 0, since, if b = 0, a + c. Thus, Al and A2 are real and distinct. Theorem 28.5. The reduction to diagonal form P-'AP for a symmetric matrix A + kI may always be effected, and P may be taken to be orthogonal. Proof. Notice first that, since s1and s2are non-zero, s1perpendicular to s2 si s2 = 0. Now suppose that si and s2are unit eigenvectors of A. As, = A,s, sAs„. = Aiss„ on premultiplying by s;, (sAs„)' = A1(sis2), on taking the transpose of each side, ra

-4-> (Ai—

slAs2 = A,sis2, by Ex. 8 and noting that A' = A, A2 Si S2 = A1s1s2, since As2 = A2 s2, s2 = 0 s;.s2 = 0, since Al + A2 by Theorem 28.4.

Thus, s1and s2are perpendicular and, by Theorem 28.3, P is orthogonal. Furthermore, by Theorem 28.2, P-1exists and the theorem is complete. (Notice that orthogonal matrices are always non-singular.) 1 3 1), find an orthogonal matrix P such that Ex. 12. If A = (3 P'AP =

° 2)

and interpret P, P' as matrices of a rotation transformation.

617

[28

FURTHER MATRICES

Ex. 13. A is a symmetric matrix corresponding to the linear transformation T. By writing A = PDP', where P is orthogonal and D is diagonal, give a geometrical interpretation of T. Illustrate your answer with the particular matrix A of Ex. 12.

Exercise 28(a) 1. Find the eigenvalues of the matrix (5 2 98

A=

and the corresponding unit eigenvectors. What lines in the plane map into themselves under the linear transformation T with matrix A? 2. Find the eigenvalues of the matrix

A

2 3\

= (8 0/

and the corresponding unit eigenvectors. What lines in the plane map into themselves under the linear transformation T with matrix A? 3. Find the eigenvalues of the symmetric matrix A=

(23 36\ 36 2/

and hence find an orthogonal matrix P such that P-1AP is diagonal. 4. Find an orthogonal matrix P such that the matrix P-1AP is diagonal, where A is the symmetric matrix 5 —1\ -

5/ •

Give a geometrical interpretation of the linear transformation Twith matrix A. 5. Give a geometrical description of the linear transformation T with matrix A, where (2 2\ A= 1 3/ 6. Give a geometrical description of the linear transformation T with matrix A, where A = 13 4\ k2 7. The 2 x 2 matrix A has the property that A2 +I = 0. Prove that the eigenvalues of A are ± j. 8. A matrix A which has the property A2 = A is said to be idempotent.

618

1]

2x2 MATRICES

If A is a 2 x 2 idempotent matrix, prove that its eigenvalues are either 0 or 1 and that, if A * I, then A is singular. Show that the matrix (3 —2\ A= 3 —2/ is idempotent and give a geometric interpretation of the linear transformation for which A is the matrix. 9. Find A" in the following cases: 6) co A = (-3 10\ .' (ii) A = k- 35 16 ' k- 3 8/ 1-10 18\ ' (iv) A = (iii) A = k -6 11/ ( -10 1 2 8) ' 10. A is a 2 x 2 matrix and the eigenvalues of the matrix A—I are ± j. Prove that det A = 2. Is the converse of this result true? 11. a, b, c, d are unequal non-negative real numbers such that

a+b = c+d = 1. Prove that the eigenvalues of the matrix

(ac are 1 and A, where 0 < IAI < 1. Find A for the matrix A=

1\

and describe geometrically the linear transformation with matrix A" where n is a large positive integer. 12. Find the eigenvalues of the matrix

A and deduce that

If

23 —33\

= (14 —20/

An = 11.2"}1-21 —33.2"+33 7.2"}1-14 —21.2"+22 B = An+An-1-1-An-2 + +A+I

find an explicit form for B. What are the eigenvalues of B?

2. EIGENVALUES AND EIGENVECTORS FOR 3 x 3 MATRICES The results we have proved for eigenvalues and eigenvectors of 2 x 2 matrices hold in large part for 3 x 3 matrices too. As before, we define the 619

[28

FURTHER MATRICES

eigenvalues of the 3 x 3 matrix A to be the roots Al istic equation det (A — Al) = 0,

,

A2, A3 of

the character-

and, corresponding to each Ai, any non-zero vector sisuch that Asi = Aisi is called an eigenvector of A. Geometrically, each real Aigives rise to a real siwhich defines a line through the origin, the points of which all map into points of the same line under the linear transformation for which A is the matrix. In contrast to the two-dimensional case, since a real cubic must possess at least one real root, there must be at least one such line. Ex. 14. Show that, if T(S) = S where S is not the origin, then T maps each point of the line OS into itself. Ex. 15. U, V, W are distinct non-collinear points in space with T(U) = U, T(V) = V, T(W) = W. Prove that, if the plane UVW does not contain the origin, then T is the identity transformation. As in two dimensions, if the 3 x 3 matrix A has three distinct eigenvalues A1, A2, A3 with corresponding eigenvectors s1, s2, s3and if the matrix P, with first column s3, second column s2, and third column 53is non-singular, then P-1AP is a diagonal matrix. For ASi = A1S1,

AS2 = A2 S2, AS3 = A3 S3

Al 0 0 AP=P 0 A2 0 0 0 A3 Al 0 0 ) P-1AP = 0 A2 0 , since P is non-singular. 0 0 A3 If such a diagonalizing procedure exists, A is said to be reducible to a diagonal matrix. Not all 3 x 3 matrices are reducible to a diagonal matrix; the discussion of necessary conditions upon A for it to be so and the possibility of reducing a matrix A which does not satisfy these conditions to a form (the Jordan canonical form) which approximates to a diagonal matrix is beyond the scope of this book: the interested reader should consult one of the more advanced algebra texts mentioned in the bibliography in Volume 1. (See, however, Exercise 28 (b), Question 25.)

*Ex. 16. Given two 3 x 3 matrices A and B, if a non-singular matrix P exists such that B = P-1AP, B is said to be similar to A. 620

2]

3x3 MATRICES Prove that: (i) A is similar to A; (ii) if B is similar to A, then A is similar to B; (iii) if B is similar to A and C is similar to B, then C is similar to A.

Example 2. Find the eigenvalues and corresponding unit eigenvectors of the matrix 1 0 0 10 A= 10 —7 5 8 (7 —

and interpret the results geometrically. Find a matrix P such that P-1AP is diagonal.

The characteristic equation of A is 0 1—A 0 10 —7—A 10 —5 8 — A 7 that is

= 0;

(1 — A) (A2 — A — 6) = 0,

giving the three eigenvalues A = 1, A = — 2, A = 3. (i) A = 1 The equations

10x — 8y + 10z = 0, 7x— 5y +7z = 0,

have solutions x = 1, y = 0, z = — 1 and the corresponding unit eigenvector is 1/V2 el = ( 0 ) . —1/V2 All points of the line x_ y_ z 1 0 —1 map into points of the same line (in fact, in this case, into themselves, since A = 1). (ii) A = — 2 The equations

= 0,

3x

10x 5y +10z = 0, —

7x -5y + 10z = 0,

621

[28

FURTHER MATRICES

have solution x = 0, y = 2, z = 1 and the corresponding unit eigenvector is 0 e2 = (2/V) . 10 x y

All points of the line

o= = 2

z 1

map into points of the same line (but not, apart from the origin, into themselves). (iii) A = 3 The equations

— 2x

= 0,

10x-10y +10z = 0, 7x

—

5y+ 5z = 0,

have solution x = 0, y = 1, z = 1 and the corresponding unit eigenvector is 0 e, = (1A/2) . 1A/2 All points of the line x y z

6= = I map into points of the same line. The matrix

1 P= ( 0 —1 has the required property, that 1 P-1AP = (0 0

i 0 0 2 1 1 1 0 0) —2 0 . 0 3

Exactly as in the two-dimensional case (see Example 1) the geometrical interpretation of the linear transformation T, which has matrix A, relative to i, j, k as base vectors, is facilitated by expressing the position vector of a point in terms of s1, s2 and s3. For the point Q, with position vector 9 = asi + ies2+ 7s3, maps into the point Q' with position vector 9' = asi.— 23S2 + 3Th• 622

2]

3x 3 MATRICES

Since the coefficient of s1is unchanged, the displacement QQ' is seen to be parallel to the plane determined by s2 and s3. Ex. 17. Verify that, in the notation of Example 2, P-1AP is a diagonal matrix. Pursuing the analogy with 2 x 2 matrices, we define a 3 x 3 orthogonal matrix P as a matrix which has the property that PP' = I. *Ex. 18. If P is a 3 x 3 orthogonal matrix, prove that P is non-singular, det P = + 1 and P-1= P'.

*Ex. 19. Follow through the steps of the proof of Theorem 28.1 to show that a 3 x 3 matrix is distance preserving if and only if it is orthogonal. Theorem 28.6. If P is a 3 x 3 orthogonal matrix with det P = + 1, then 1 is an eigenvalue of P. Proof. Since

we have

(P-I)P i = I-P' = -(P-I)', det (P - I) det P' = det (- (P - I)').

But det P' = det P = 1, and also det (- (P - In = det (- (P - I)) = - det (P - I) (since P- I is a 3 x 3 matrix) and thus

det (P - I) = - det (P - I) det (P -I) = 0 1 is an eigenvalue of P.

Corollary. If P is a 3 x 3 orthogonal matrix with det P = - 1, then -1 is an eigenvalue of P. This follows immediately on writing -P for P, since det (-P) = + 1. *Ex. 20. In the case det P = + 1 (P orthogonal), let s be an eigenvector corresponding to the eigenvalue A = 1. The linear transformation T of which P is the matrix: (i) preserves distances between all pairs of points; (ii) leaves all points of the line OS fixed. Show that T may be interpreted geometrically as a rotation about the axis OS. *Ex. 21. Show that, if det P = -1 (P orthogonal), then the corresponding linear transformation may be interpreted as a rotation about the axis defined by the eigenvalue A = -1 followed by a reflection in the origin. 623

FURTHER MATRICES

[28

Ex. 22. Show that the matrix (cos B - sin 0 0 P = sin 0 cos 0 0 0 0 1 is orthogonal and represents a rotation about the z axis through a positive angle O.

Ex. 23. Show that the columns of a 3 x 3 orthogonal matrix P represent three mutually perpendicular unit vectors and interpret this result geometrically. In summary, if P is a 3 x 3 orthogonal matrix representing the linear transformation T, then the columns of P are the components of three mutually perpendicular unit vectors and T represents either a pure rotation, or a rotation followed by a reflection in the origin. Ex. 24. If

1 P=

-2 - 2

-2

1 -2 , -2 -2 1 prove that P is orthogonal and interpret P as the matrix of a linear transformation. *Ex. 25. If u is a unit vector, show that it is possible to construct an orthogonal matrix with u as its first column. *Ex. 26. If P, Q are two 3 x 3 orthogonal matrices, prove that: (i) P--1is orthogonal; (ii) PQ is orthogonal. Interpret both these results geometrically. Ex. 27. Find a result corresponding to Theorem 28.6 for 4 x 4 orthogonal matrices. (Be careful when you take determinants !) We saw earlier (Theorems 28.4 and 28.5) that the eigenvalues of a 2 x 2 symmetric matrix A + kI were always real and distinct and this enabled us to construct an orthogonal matrix P such that P-1AP was diagonal. 3 x 3 symmetric matrices are not quite so amenable: although it can be shown that their eigenvalues are always real, they are not necessarily distinct. Nevertheless, the diagonalization process is still always possible as we shall now show.

Theorem 28.7. Given any 3 x 3 symmetric matrix A it is always possible to find an orthogonal matrix P such that P-1AP is diagonal. Proof. We reduce the problem to the 2 x 2 case. First, if A., is a real eigenvalue (such an eigenvalue must exist, since the characteristic equation is a cubic) with s, the corresponding unit eigenvector, then As1= A1s1and thus Al. cli. a2) AP=P ( 0 /J A , o 7, y 624

2]

3 x 3 MATRICES

where P is an orthogonal matrix with sias its first column and a, ... 72 are constants (see Ex. 25). Premultiplying by P-1we then have Ai oci. oc2 P-1AP = (0 fli 1 32 . 0 7, 7 But (P-1AP)' = P'A'(P-1)' = P'A(P')' = P-1AP (see Ex. 8), since A' = A and P' = P-1. It follows that P-1AP is a symmetric matrix: A1 0 0 P-1AP = (0 A 13) • 0 fi2 72 Now, by Theorem 28.5, there exists an orthogonal matrix (qi 4721 \r, r2/ which reduces the symmetric matrix 1612 fi2) 72 (8 to diagonal form. Thus the matrix

(1 0 Q = 0 q1 0 r, reduces P-1AP to diagonal form: Ai Q-1P-1APQ = (0

0 q2

r2

0 0

it2 0). 0 0 it3

Finally, we may write PQ = R where, by Ex. 26 (ii), R is orthogonal. Ex. 28. Is the converse result true that, if P is orthogonal and P-1AP is diagonal, then A is a symmetric matrix?

The determination of an inverse matrix is generally a burdensome operation, but, if P is orthogonal, we have P-1= P' and thus the diagonalization process for a symmetric matrix is particularly simple to effect. 625

[28

FURTHER MATRICES

3. THE CAYLEY-HAMILTON THEOREM AND POWERS OF MATRICES A matrix A is said to satisfy the equation a0 An-Fa1An-I -Fa2 A71-2+...+an_I A+an = 0, if

ao An+chA'1+ aoAn-2 + +an_iA + an I = 0.

Theorem 28.8. (The Cayley-Hamilton Theorem.) Every square matrix A satisfies its own characteristic equation. Proof (We prove the result for a 3 x 3 matrix A, but the proof clearly generalizes very readily to the n x n case.) If det (A- AI) = ao+ ai A+ a2 A2 -A3, we have to show that aoI+chA+a2A2 -A3 = 0. Now each element of the matrix adj (A - AI), being a cofactor of (A-AI), is a polynomial of degree at most two in A. Thus adj (A - AI) = Co + ACI +A2C2, where Co, C1, C2 are 3 x 3 matrices whose elements do not contain A. But (A-Al) adj (A-AI) = I det (A-AI) (Theorem 14.2) and thus

(A-Al) (Co +ACI+A2C2) = I(ao +chA+a2A2 -A3).

Comparing coefficients of powers of A in this identity

AC() = a0I,

(1)

ACI -00 = a1I,

(2)

AC2-C1 = a2I,

(3) (4)

-Co = -I. Multiplying (2) by A, (3) by A2, and (4) by A3and adding this gives 0 = a0I+a1A+a2A2-A3 and the proof is complete.

*Ex. 29. Verify the Cayley-Hamilton theorem for 2 x 2 matrices by direct substitution. Ex. 30. If

1 0 -4 A= (-1 2 1), 0 0 -3

find A3by applying the Cayley-Hamilton theorem.

626

3]

CAYLEY-HAMILTON THEOREM

Example 3. If A=

find A8.

11 —1

k2

3)'

The characteristic equation of A is A2-4A+5 = 0 and thus by the Cayley—Hamilton theorem A2 — 4A+ 51 = 0. A8 .=_ (A2 — 4A+ 5)f(A)+ccA +ft,

But

where cc, ft are integers. Since A = 2+j =- A2 -4A + 5 = 0 we have (2 +j)8= cc(2 + j) +fl. Using the Binomial Theorem and equating real and imaginary parts, 28 -28.26 +70.24 -28.22 +1 = 2cc+fl

1

8.27-56.25+56.23-8.2

f16[16-112+70-7]+1 116[64 — 112 + 28 — 1]

=a = 2cc+fi =a

f cc = —336 tiG = 145. Thus

A8----- (A2 — 4A + 5)f(A)— 336A +145

and

A8= —336A +1451, since A2-4A+ 51 = 0, = 1-336 k — 672 191 = k — 672

I

—

336\ + (145 0\ —1008) k o 145/ 336\ —863) •

Alternatively, one can 'build up' A8: A2 = 4A-51, as before; A4 = 16A2 — 40A + 251, using the commutative rule, = 16(4A— 51)— 40A + 251 = 24A-551; A8 = 242A2 — 48 . 55A + 5521 = 242(4A — 51)— 48 . 55A + 5521 = —48 .7A— 5(576 — 605) I = — 336A+ 1451, and proceed as before. 627

[28

FURTHER MATRICES

Notice that, by multiplying the Cayley-Hamilton result by Ar we establish a recurrence relation between successive powers of A: Ar+2 4Ar+1 -5Ar.

Exercise 28 (b) 1. Find the eigenvalues of the matrix 3 4 1 A= (2 5 1 2 3 3 and determine corresponding unit eigenvectors. 2. Find the eigenvalues of the matrix (1 -2 1 A = 3 -4 1 . 3 -7 4 What does the fact that zero is one of the eigenvalues tell us about A? Find a matrix P such that P-1AP = D, where D is a diagonal matrix. 3. Find the eigenvalues of the matrix 3

1 -2 0 -2 4 -1 -1 ( and hence write down the value of det A. Find A-1and determine the eigenvalues of A-1. Suggest and prove a general result about the eigenvalues of an inverse matrix.

A= 4

4. Find the eigenvalues of the matrix 1 2 -2 A= 6 4 -6 6 5 -7 ( and corresponding unit eigenvectors. Find a matrix P such that P-1AP = D where D is a diagonal matrix. Show that, if A is the matrix of the linear transformation T, then T maps all points of the plane 2x +y - 2z = 0 into points of the same plane. Find the equations of the other two planes through the origin which have this property. 5. Find the eigenvalues of the symmetric matrix -1 -6 -4 A = 1 (- 6 -2 -2 -4 -2 3 and hence find an orthogonal matrix P such that P-JAP = D, where D is diagonal.

628

EXERCISE 28 6. Find the equations of the planes through the origin, which map into themselves under the linear transformation T with matrix (1 3 —2 A=31—2. 3 4 —5 7. P is a 2 x 2 orthogonal matrix with det P = 1. Show that the eigenvalues of P are ej0 and e-i°, and interpret O. Show that, although the geometrical property of a 2 x 2 orthogonal matrix P is characterized by its eigenvalues if det P = + 1, this is no longer true if det P = 1. —

8. The trace of the matrix A, written tr A, is defined to be the sum of elements of A in the leading diagonal (top left to bottom right). If A is a 3 x 3 matrix with eigenvalues ill A,, A,, prove that ,

tr A = k+A2 +4. Prove also that tr (A + B) = tr A + tr B and that tr (,itA) = u tr (A). 9. A and B are both 2 x 2 matrices. Prove that A and B have the same eigenvalues if and only if both det A = det B and tr A = tr B. (See Question 8 for definition of trace.) 10. Prove that, if A is a non-singular 3 x 3 matrix, then det (A — AI) = det A — A det A tr (A-1)+M tr A— A3. 11. If A, B have a common eigenvector s, with corresponding eigenvalues A, it, prove that s is an eigenvector of (i) A + B; (ii) AB. What are the corresponding eigenvalues? 12. A is a 3 x 3 skew-symmetric matrix (A = — A'). Prove that det (A— AI) = — det (A + AI) and deduce that if A possesses a non-zero eigenvalue a, then — a is also an eigenvalue. If f(A) = det (A2 —AI), prove that — f(A)/A is the square of a linear polynomial in A. 13. If B = P-1AP and s is an eigenvector of A prove that P-ls is an eigenvector of B. 14. If A is an eigenvalue of the 3 x 3 matrix A prove that Ar" is an eigenvalue of An. Deduce that, if A* = a2A2 + alA + a,I, then

Ar = a2 A2+ai A+a0.

15. A matrix A is said to be nilpotent if there exists a positive integer n such that An = 0. Prove that, if A is nilpotent, then all the eigenvalues are zero. 16. Find A4 + A2 + I if A=( 3 —1 9

PPMII

5\ .

—2)

629

FURTHER MATRICES 17. If

[28 A— (12 5 "52)

find A", where n is a positive integer. 18. If A=

3 — 1), 1'

find A3 + 3A2 + 12A. 19. Show that the characteristic equation of the matrix 5 7 3 A= (1 5 2 3 2 1 is

A3-11A2 +15A-1 = 0.

Deduce that A is non-singular and that A-1= A2 -11A+ 151. 20. If A show that and

1 1 2) (0 2 1 , 1 0 2

A3= (5A-1) (A-1) A-1 = (A-31) (A-21).

Deduce explicit forms for A3and A-1. 21. The matrix A is defined by

1 2 3

A= 3 1 2 . 2 3 1 ( Show that

A3 — 3A2 — 16A— 161 = 2I— A

and express (21—A)-1as a quadratic polynomial in A. 22. Show that 1, 2, 3 are the eigenvalues of the matrix (17 —16 8 A= 10 —8 6. —10 11 — 3 Hence, or otherwise, find an explicit form for A". Write down the inverse matrix of A. 23. The 2 x 2 matrix A has equal eigenvalues A. Prove that, for n

A" =

nil"-1A —(n— 1) An1.

Prove further that, if A * 0, this formula is true for all integral n. 630

2

EXERCISE 28 24. If S is a 3 x 3 skew-symmetric matrix, prove that its eigenvalues are 0 and + ja, a real. If P is a 3 x 3 orthogonal matrix and P +I is non-singular, prove that S = (P — I) (P + is skew-symmetric, and that, if A is an eigenvalue of S, then A = P—I—A(P+I) is singular. The eigenvalues of P are 1, 00, e-Je; what are the corresponding eigenvalues of S? 25. The 3 x 3 matrix A has characteristic equation (A— A,)2(A — A,) = 0, where A1, A2 are real and Al= A2. Show geometrically that, if the equations Ax Aix have a plane of solutions, then it is possible to form a matrix P such that 0 0 Al P-1AP = 0 Al 0 . 0 0 A2

If A is the matrix of the linear transformation T, and if 1 is the invariant line through the origin corresponding to the eigenvalue A2, show that every plane containing 1 is mapped into itself by T. If —4 —10 30 A = —3 —5 18 —2 —4 13 show that Tleaves all the points of a certain plane containing the origin fixed and find the equation of this plane. Show also that every plane containing the line x y =z 5 3 2 is mapped into itself by T. Find a diagonalizing matrix P for this particular matrix A. 26. The linear transformation T has matrix A where A is non-singular with an eigenvector d = OD. P is a given point not on OD and T(P) = Q. Show that T maps PQ into itself if PQ is parallel to OD. Show further that, if this is the case, then the linear transformation S with matrix A—I maps the line PQ into the line OD. 9-2

631

FURTHER MATRICES

[28

27. P is a non-singular 3 x 3 matrix, co is a cube root of unity and r is an integer. Prove that, if ((dr 0 0 52 = P-1 0 (V+1 0 P, 0 0 Nr+2 523 = I.

then

Prove conversely, that if SL is a 3 x 3 matrix, with real determinant, having the property

na = I

then NT

0

El = 13-1(0

Nr+1

0

0

J

0 0 P, (dr+2

for some matrix P and integer r. Prove further that, if 52 t I, then the matrix 1+52+522 is singular. 28. Show that the linear transformation T with matrix 1

3

—2

A= -1 5 —2 —1 (

4

—1

leaves just two lines through the origin invariant and find their equations. Explain geometrically why it is impossible to find a matrix P such that the matrix P-1AP is diagonal.

632

29. Further coordinate geometry

1. PARAMETRIC FORMS FOR PLANE CURVES We have seen in Chapters 22 and 27 that the coordinates of points on certain plane curves may be given in terms of parameters. Indeed, we could define, for example, the parabola as the set of all points P with position vectors r = at2i+ 2atj relative to some origin 0, where t is a scalar parameter. More generally a plane curve is completely specified if we are given a parametric representation r = f(t) i+g(t)j, where f, g are two (continuous) functions. We shall also demand that there is a 1-1 correspondence between points of the curve and values of the parameter — with the possible exception of a limited number of multiple points, for which several values of the parameter may yield the same point (see Ex. 5). If f(t), g(t) happen to be algebraic expressions, the resulting analysis is much simplified and we are often able to employ the theory of polynomial equations in problem solving. Ex. 1. Show that an algebraic parametric form for the ellipse bax2 ±a2y2 = a2b2

is

r—

a(1— t2) 2bt i+ j. 1+t 2 l+t 2

What point of the ellipse is excluded by this parametric representation? Find a similar algebraic parametric form for the hyperbola. b2x2— a2y2 = a2b2.

In the following two examples we illustrate some of the methods available for the solution of geometrical problems employing algebraic parameters. In the first example, we define a curve parametrically and deduce a geometric property; in the second, we obtain a geometrical property of the parabola using the standard parametric form. Example 1. The semi-cubical parabola is defined parametrically by r = at2i+ at3j. 633

FURTHER COORDINATE GEOMETRY

[29

Prove that a straight line not parallel to the y axis cuts the curve in either one or three real points and that, if a straight line not through the origin cuts the curve at P, Q, R and the tangents to the curve at these three points cut the curve again at P', Q', R', then P' Q' R' is a straight line. Any line not parallel to the y axis has an equation of the form y = mx + c. This cuts the curve in points with parameters given by the roots of the equation at3= mat2+ c, that is, by

at3—mat2 — c = 0.

Since this is a cubic equation in t, it has either one or three real roots, due regard being paid to their multiplicity. This proves the first result. Now suppose that the parameters of the points P, Q, R are respectively p, q, r. Then, if the equation of the line PQR is y = mx+ c we see that the roots of the cubic equation at 3— mat 2—c = 0 are p, q, r; thus qr+rp+pq = 0. But the line does not contain the origin; thus p, q, r

0 and we have

1/p+ 1/q+ 1/r = 0. Again, since the tangent at P cuts the curve at P', parameterp', we have, as above, (1/P)+(l/P)+(1 /P') = 0, giving,

= — 2/p, similarly 1/q' = — 2/q, and 1/r' = — 2/r and we have (1/p')+(l/q')+(l/r') = —2[(1/p)+(1/q)+(l/r)] = 0.

It follows that p', q', r' are the roots of a cubic equation of the form ats — m' at2— c = 0 (where m' = p' +q' +r' and c' la = p'q'r') and thus the points P', Q', R' lie on the line y = m'x+ c'. The shape of the semi-cubical parabola is shown in Figure 29.1. 0 is called the pole and Ox is the axis of the curve. The Cartesian equation is x3= aye. A connection between the parabola and the semi-cubical parabola is obtained in Example 2. Example 2. Given a parabola and a point C, prove that, in general, either one or three normals may be drawn to the parabola to pass through C but that, if just two such normals may be drawn, then C lies on a certain semi-cubical parabola. 634

PARAMETRIC FORMS y

x

Fig. 29.1

If the normals at three points, P, Q, R on a parabola are concurrent, show that the circumcircle of the triangle PQR passes through the vertex of the parabola. Take the parabola in the standard form y2 = 4ax and let C be the point (h, k). The equation of the normal at the point P(at 2, tat) is tx+y = at3+2at and this passes through C if

at3+(2a—h) t—k = 0. This is a cubic equation in t, with, in general, either one or three real roots, and the first part of the question is proved. If just two normals pass through C, then the cubic equation above has a repeated root and the condition for this (see Chapter 23) is

4 2— /h\ 3± k2 = 27 k

a2

showing that C in this case lies on the curve 4(x — 2a)3= 27ay2. By Example 1, this is a semi-cubical parabola with pole at (2a, 0) and axis coinciding with the axis of the parabola. Now suppose that the normals at P(ap2, 2ap), Q(aq2, 2aq), R(ar2, 2ar) meet at C; then p, q, r are the roots of the equation

at3+(2a—h) t—k = 0 and thus p+q+r = 0. 635


[29

Suppose that the circumcircle of triangle PQR has equation (x-cc)2 +(y -/{)2 = 72•

This cuts the parabola at points with parameters given by the quartic equation (at 2 -a)2 + (2at fi)2 ==

72.

-

But the coefficient of t3in this equation is zero and thus the sum of the roots is zero. Since three of the roots are, by definition p, q and r, and since p+q 1 r = 0, the fourth root must be zero and we have proved the final result. --

Ex. 2. Prove that the normals to the parabola y2= 4ax touch the semi-cubical parabola of Example 2. Draw in the same sketch the parabola y2 = 4ax and the semi-cubical parabola 4(x- 2a)3= 27ay2. (It is worth drawing the parabola accurately and then constructing a large number of normals—which may be done very rapidly using the geometrical property that, in the notation of Exercise 22(b), NG = 2a.) Ex. 3. Draw the parabola y2 = 4ax and shade in the set of all points from which three real normals may be drawn to the parabola. Ex. 4. The Folium of Descartes has Cartesian equation x3 +y3 = 3axy. By considering the intersection of the line y = tx with the folium, show that the curve may be represented parametrically by x=

3at

1+ t 3

y

3at 2 1+ t 3

Taking t = 0 it is seen that the origin lies on the curve and that the x axis is a tangent there; by writing u = t-1, show that the y axis is also a tangent at the origin (which is thus a double point of the curve, that is, a point through which pass two separate branches of the curve). Show that the line x- y = 0 is an axis of symmetry and that the line x+y = a is an asymptote. Sketch the curve. Ex. 5. Using the method of EX. 4, find a parametric form for the curve

y2 = x2(1 + and deduce that no point of the curve has an x coordinate less than -1. Prove also that the x axis is an axis of symmetry for the curve. Show that the origin lies on the curve and that this point arises from two distinct values of the parameter. Deduce that the origin is a double point of the curve and determine the equations of the two tangents there.

636

PARAMETRIC FORMS

1] Exercise 29(a)

1. Show that, if OP = r = at i + at 3j, then P lies on the curve I': x3 = a2y. Prove that any line in the plane cuts r in either one or three real points. The tangents at three points P, Q, R of the curve meet F again at P', Q', R' respectively. Prove that: (i) if the centroid of the triangle PQR lies on the y axis, so also does the centroid of the triangle P' Q'R' ; (ii) if P, Q, R are collinear, so also are P', Q', R'. 2. The cissoid is defined parametrically by r=

at 2 at 3 . 1+ J. 1+0 1+0

Find the Cartesian equation of the cissoid and give a rough sketch of the curve. Prove that, if the chord PQ subtends a right angle at 0, then the mid-point M of PQ lies on a fixed straight line. 3. A curve F is given parametrically by r =

t3+ 1 . 2 1 1+ t

(t

Prove that, if the points with parameters t1, t2, t3are collinear, then 1+t1+t2 +t3 +t1 t2 t3 = 0, and conversely. Show that, given a point P lying on 1', two lines may be drawn through P to touch the curve at Q and R. If QR cuts r again at S find the parameter of the point S in terms of the parameter of P. Show that the roots of the equation t 2+ t — 1 = 0 give the double point of the curve. 4. A curve P is given parametrically by r = (t2 +14)1+02 -1/0j. Obtain a necessary and sufficient condition for the three points P, Q, R of the curve to be collinear. The tangent at P1meets P again at /32, the tangent at P2 meets I' again at P3 and the tangent at P3 meets r again at P4. PI P, meets the curve at Q while P2P4 meets the curve at R. Prove that QR is a tangent to r at Q. 5. A curve is given parametrically by the equations x = a(1 — t2), y = a(t — t 3). Prove: (i) that an arbitrary line lx+ my + na = 0 meets the curve in three points; (ii) that if three points t = t1, t = t2, t = t3are collinear then t2t3+ t3 ti + r2 — 1;

(over) 637


29

(iii) that if t2 and t3satisfy the above equation then the points t = tl, t = t2 t = t3 are collinear. A chord through the point (a, 0) meets the curve again at the points P and Q. Prove that the locus of the middle point of PQ is a curve with parametric equations x = — laT2, y = — 1-T(T 2+2). (0 & C) ,

6. Prove that, in general, three tangents may be drawn from a point C to the cubic curve x3 = a2y and that, if these tangents cut the curve again at P, Q, R then the tangents at P, Q, R are concurrent. 7. A rectangular hyperbola is given parametrically by the equations x = ct, y = c/t. If the four points of the curve with parameters tb t2, t3, t4lie on a circle, show that ht2t3t4, = 1. Show conversely that, ifif _1 t tt t2 t 3 t4 = 1, then the four points lie on a circle. A variable circle passes through the fixed points A, B of a rectangular hyperbola, and meets the hyperbola again at P, Q. Show that the direction of PQ is fixed. 8. P, Q are variable points on the parabola y2 = 4ax such that PQ is parallel to the fixed line x+ ky = 0. The normals to the parabola at P and Q meet at R. Prove that the locus of R is the normal to the parabola at a fixed point on the parabola, and find the coordinates of this point. (0 & C)

9. Find the equation of the normal to the rectangular hyperbola xy = c2 at the point (ct, c/t). The normals to the rectangular hyperbola at the points P, Q, R, S are concurrent; prove that each of these points is the orthocentre of the triangle formed by the other three. 10. The normals at three points P, Q, R of the parabola y2 = 4ax meet at a point N. Prove that the centroid of the triangle PQR lies on the axis of the parabola. If N coincides with P, prove that QR passes through a fixed point (that is, a point whose position is independent of P, Q, R). (0 & C) 11. Find the coordinates of the point P, other than the origin in which the line y = tx meets the curve x3 + y3 = axy. (t is called the parameter of P.) If a line meets the curve in three points whose parameters are t1, t2, t3, prove that ti t2 t3= —1. If Q is any point on the curve with negative parameter, not equal to —1, prove that there are two points P1, P2 of the curve (other than Q) such that the tangents at P, and P2 pass through Q. Prove that OP'and OP2make equal angles (apart from the sense) with either of the coordinate axes. (0 & C) 12. Prove that the equation of the normal to the curve y = x3at the point (t, t3) is x+3t2y = t+30. 638

1]

PARAMETRIC FORMS

By considering the maximum and minimum values of a certain function of t and drawing a rough graph of the function, or otherwise, prove that three normals can be drawn to curve y = x3from a point (0, b) when b > 4/3V3. How many real normals can be drawn to the curve from a point (0, c) when 0 < c < 4/3A/3 ? (0 & C) 13. A circle has a diameter OA of length a, and the tangent at A is 1. A variable line through 0 meets the circle again at Q, and 1 at R; P is the point on OR such that OP = QR. If 0 is the origin and OA is the x axis, show that P has coordinates (at 21(1+t 2), at 31(1+ t3)), where t is a suitable parameter. Prove that, if a line meets the locus of P in three points with parameters t1, t2, t3then t2 t3+ t3 t1+ = 0. Hence, or otherwise, prove that the tangent to the locus at P meets the locus again at the point with parameter — 4t. (0 & C) 14. A point on the curve aye = x3is given parametrically in the form (at 2, at 3). If the points on the curve with parameters t1, t2, t3are collinear, prove that (OD+ (OD+

= 0.

Hence show that, if the tangent at the point with parameter t1meets the curve again at the point with parameter t4, then t1+ 2t4 = 0. Perpendicular lines through 0 meet the curve at P, Q; PQ meets the curve again at Rand S is the point of the curve such that the tangent at S passes through R. Prove that OP, OQ are the bisectors of the angles between OS and Ox. (0 & C) 15. The rectangular hyperbola xy = k2is met by a circle, passing through its centre 0, in four points A1, A2, B1, B2. The lengths of the perpendiculars from 0 to A1 42 and B1B2 are a and b. Prove that ab = k 2. (C. S.) 16. Show that there are three values of t, not necessarily real, for which the point (t2, t3) lies on a given straight line. P, Q and R are distinct points (p2, p3),(q2, q3) and (r2, r3) on the curve y2 = x3. Show that: (i) if these points are collinear then Epq = 0; (ii) if the tangents at these points are concurrent, then Ep = 0; (iii) there are no real points on the curve for which these two conditions co-exist. The tangent at the point P on this curve meets the curve again at P'. Find the ratio in which PP' is divided by the x axis. (London)

2. SURFACES AND CURVES IN IN THREE DIMENSIONS The simplest surface in three-dimensional space is the plane. Given three points A, B, C, the vector equation of the plane through ABC is r = a+A(b — a)±,u(e — a), where A, it are scalar parameters. Notice that we need two parameters to define the surface. 639


[29

Two surfaces intersect in a curve (which need not necessarily be a plane curve). Thus, two planes intersect in a line; the vector equation of the line through A, B is r = a+A(b a), —

where we have the single scalar parameter A. Another familiar surface is the sphere, which is defined as the set of points in three dimensions lying at a fixed distance from a fixed point. If the fixed distance (radius) is c and the fixed point (centre) is A, then the vector equation of the sphere is (r a). (r a) = c2. —

—

Ex. 6. Prove that

r = a + cos 0 cos ¢i+ sin 0 cos cbj+ sin cbk gives a parametric representation of the sphere. (Notice now that we have two parameters; suggest names for them.) *Ex. 7. Prove that the vector equation (r a). (r b) = 0 represents a sphere, and locate the points A and B as points on the sphere. —

—

The Cartesian equation of a sphere, centre A(ai, a2, a3) is thus (x ai)2+ (y— a2)2 +(z— a3)2 = —

which may be rewritten x2 +y2 +z2 +2ux+2vy+2wz+d = 0. Conversely, an equation of the form

x21-Y - 2 +Z2 +2UX+2VY+2WZ+d = 0 may be rewritten as (x+u)2+(y+v)2+(z+w)2 = u2 ±v2 4.14,2—d, and thus represents a sphere, centre (— u, — v, — w), provided u2 v2+ w2 > d. Ex. 8. Show that the equation x2 +y2 +z2 -2x-4y+2z—l0 = 0 represents a sphere, and find its centre and radius.

*Ex. 9. Show that four points in space in general define a unique sphere. What exceptional cases may arise?

The intersection of the sphere oc _ ay+0, 640

±

= e2

2]

SURFACES IN THREE DIMENSIONS

z=0

with the plane

is the curve in the Oxy plane with equation (x — (21)2+ (y — a2)2 = c2— 4 which is seen to be a circle provided c2 > 4 If c2 = 4, the equation represents a single point and, if c2 < 4, the sphere and plane do not intersect. Thus, a sphere and a plane, if they intersect at all, intersect in a circle, which may be of zero radius if the plane is tangential to the sphere. The centre of the circle lies on the perpendicular from the centre of the sphere on to the plane. If the sphere and the plane have vector equations

(r a) . (r a) = c2, r . n = p (where In! = 1), —

—

r = a + An.

this perpendicular will be

To find the value of A which gives the centre of the circle determined by the sphere and the plane we solve

r = a +An and r.n = p, (a + An) . n = p.

giving

Thus, A = p — a . n, since n . n = 1 and the position vector of the centre of the circle is r = a+(p—a.n)n. Ex. 10. Find the centre and radius of the circle determined by the sphere (x— 1)2 + (y — 3)2 + (z — 2)2 = 20

x—y-3z = 3.

and the plane

Now consider two spheres, with equations

(r a) . (r a) = c2, (r b) . (r b) = d2. —

—

—

—

These equations may be rewritten

r . r 2r . a = c2 a . a and r . r 2r . b = d2 b b —

—

—

—

which on subtraction, give

2r.(b a) = c2— d2 —a.a+b.b. —

But this is the equation of a plane, perpendicular to the vector b a, that is, to the line of centres AB and, since a plane and a sphere determine a circle, we see that two spheres, provided they intersect at all, intersect in a circle. —

641

[29


The line

r = a+An (InI = 1) (r — b). (r — b) = c2

cuts the sphere

in points with parameters given by (An + a b). (An + a b) = c2; —

—

that is, by the quadratic equation A2 + 2An (a b)+ (a b). (a b) c2 = 0. —

—

—

—

Thus, a line cuts a sphere in two points (which may be coincident or imaginary). If n is given and we take A to be the mid-point of any chord in the direction defined by n, the two roots A and A2 of the above quadratic equation have the same magnitude but opposite sign; that is, A1+A2 = 0. It follows that n . (a b) = 0 and A lies on the plane n. (r b) = 0 which passes through the centre, B, of the sphere and is perpendicular to n; thus the locus of the mid-points of parallel chords of a sphere is a plane through the centre of the sphere (a diametral plane). —

—

Example 3. Find the equation of the tangent plane to the sphere x2 ± y2 + Z-2 8x-6z-2 = 0 at the point (1, 3, 0). Prove that the plane x-5y + z+20 = 0 is a tangent to the sphere and find its point of contact. Rewriting the equation of the sphere in the form (x — 4)2+ y2+ (z — 3)2 = 27 we see that the centre of the sphere is the point (4, 0, 3). (i) The vector (41+ 3k) (i + 3j) = 3i 3j + 3k is normal to the required tangent plane whose equation is thus —

1(x — 1) that is

—

—

1(y — 3) + 1(z — 0) = 0,

x— y+ z +2 = O.

(ii) The radius of the sphere is V27; but the perpendicular distance from (4, 0, 3) to the plane x— 5y + z +20 = 0 is

4+3+20 V(12 +52 +12)

— V27.

Thus the plane x — 5y z + 20 = 0 is tangential to the sphere. A normal to the given plane is i — 5j +k and thus any point on the 642

2]


radius to the point of contact has coordinates (4+A, — 5A, 3 +A). This lies on the given plane if (4+A)— 5( — 5A)+ (3 +A)+ 20 = 0, that is, if A = —1. This gives the coordinates of the point of contact as (3, 5, 2). The vector equation of the tangent plane at a point T of the sphere, centre A and radius c, is easily derived by the same method as that used in the last example. For a unit vector in the direction AT is (t—a)/c and thus the equation of the tangent plane is (r — a) . (t — a) = c2. Notice that the equation of the tangent plane is obtained from the equation of the sphere by writing t for r in one of the brackets; this simple rule enables us to write down the equation of the tangent plane at any point of the sphere. For example, the equation of the tangent to the sphere x2 +y2 +z2-2x+4y+12z+4 = 0 at the point (1, —3, 0) is x.l+y.(-3)+z.0—(x+1)+2(y-3)+6(z+0)+4 = 0 y-6z+3 = 0.

or

Example 4. Determine whether or not the circles x2+y2+ z2 — 18x-6y-4z+ 14 = 0, x+3y+5z = 0, S2: x2 +y2 +z2 -6x+14y-12z+22

= 0, x+y+z = 2,

are linked (as in the links of a chain). The planes of the two circles meet in the line, 1, with equations x+3y+5z = 0, x+y+z = 2 which reduce to

x-3 _ y +1 _ z 1 — —2 1

and any point on this line has coordinates (3+A, — 1 — 2A , A). Thus 1 meets the given sphere through S1at points given by (3 + A)2 + (— 1 — 2A)2 + A2 — 18(3 + A) + 6(1 + 2A) — 4A + 14 = 0, or giving A = +2 or —2.

6A2— 24 = 0, 643


Again, 1 meets the given sphere through

S2

[29

at points given by

(3 + A)2 + (-1— 2A)2 + A2 — 6(3 + A)— 14(1 + 2A)— 12A + 22 = 0, Or 6A2— 36A = 0 giving A = 0 or 6. Thus, if S, meets 1 at Al(A = —2) and B1(A = +2) and S2 meets 1 at A2(A = 0) and B2 (A = 6), the order of the points on the line is A1A2 B1132 and the circles must therefore be linked.

Another familiar surface in three dimensions is the circular cylinder. If the axis of the cylinder passes through the point A(ch, a2, a3) and is in the direction of the unit vector u = /i+mj+nk, any point P(x, y, z) lies on the cylinder if its perpendicular distance from the axis is a constant, b. Thus, if OA = a, OP = r, we have (see Figure 29.2) (r — a) A ul = b, or in Cartesian form, E[n(y— a2)—m(z— a3)]2 = b2.

Fig. 29.2

If we take the x axis as the axis of the cylinder (and thus 1 = 1, m = n = 0 and a, = a2 = a, = 0) this equation reduces to the much simpler form y2+ z2 = b2. *Ex. 11. Show that a parametric form for a cylinder, radius b, with its axis along the z axis, is r = b cos Oi+ b sin 61+ Ak.

Notice once again that we have two parameters, 0 and A for this surface.

Ex. 12. What does the surface

x2 y 2 a2

+

b2

=1

represent in three dimensions?

A cylindrical spring is a three-dimensional curve lying on the surface 644

2]


of a circular cylinder. The curve is called a helix and is defined parametrically by r = b cos Oi+b sin 0j+ — 27T k where p (the pitch of the helix) and b (the radius of the cylinder on which the helix lies) are constants. Note that the parametric representation of the helix involves only one parameter. *Ex. 13. Interpret geometrically the constant p, the pitch of the helix. Ex. 14. Show that the helix defined above is one of the two curves of intersection of the cylinder x2 +y2 = b2and the corrugated surface x = a cos

2lTz

. Show also

that the other curve formed is an oppositely twined helix.

The right-circular cone is the surface traced out by a variable line through a fixed point 0 (the vertex of the cone) and making a constant angle with a fixed line though 0 (the axis of the cone). If we take 0 as the origin and the unit vector u = /i+ +nk as defining the direction of the axis we have, for any point P of the cone, r.0 = Irl cos a, where r = OP and a is the angle of the cone (Figure 29.3). In Cartesian form, this gives (lx + my + nz)2= (x2+ y2+ z2) cos' a

Fig. 29.3

and we see that the equation of a cone with its vertex at the origin is homogeneous in x, y, z. We saw in Chapter 26 that the section of a plane with a cone is a conic. Example 5. Obtain the equation of the cone with vertex V (2, 1, 2) which touches the sphere x2 ± y2 ± z2 = 1 at the points of a circle on the surface of the sphere. Since the centre of the sphere is the origin, 0, and the cone touches the sphere, VO is the axis of the cone. Let VT be a line on the surface of the —

—

645

[29


cone (a generator of the cone) (see Figure 29.4). Then, if LOVT = a, since OV = 3 and the radius of the sphere is 1, cos a = 232

.

Now consider any point P(x, y, z) on the cone. A unit vector along OV is u = and thus, since VP.0 = IVPI cos a we obtain [1(x — 2) — A-(y + 1) — 1(z + 2)]2 = Ex— 2)2 + (y +

+ (z. + 2)9,

Fig. 29.4

which may be rewritten as 4(x — 2)2 + 7(y + 1)2 + 4(z + 2)2 — 4(y + 1) (z + 2) + 8(z + 2) (x — 2) +4(x-2) (y+1) = 0.

3. CHANGING THE COORDINATE SYSTEM IN A PLANE: ROTATION OF AXES In Chapter 13, a linear transformation of the plane into itself of the form x' = ax+ by, y' = cx+dy was regarded as a mapping of the plane into itself, in which the point P(h, k) is mapped into the point P'(ah+bk, ch+dk), coordinates being referred to the same axes. In particular, the transformation T whose matrix P, is orthogonal, P = (cos 0 —sin 0\ ksin 0 cos Of has the effect of rotating any line OA through the angle 0, measured in a positive sense, into the position OA', where OA = OA'. 646

CHANGES OF COORDINATE SYSTEM

3]

There is, however, an alternative way of looking at an orthogonal linear transformation. Suppose we leave each point of the plane where it is but re-name the coordinates by taking a new pair of perpendicular axes, with the origin still at 0, but with the new axes Ox', Oy' making an angle — 0 with the old axes Ox, Oy (Figure 29.5).

Fig. 29.5

Let OP = r, and suppose OP makes an angle a with Ox. Then

x = r cos a, y = r sin a x' = r cos (a + 0), y' = r sin (a + 0). Thus,

x' = r cos a cos 0 — r sin a sin 0 = x cos 0 — y sin 0, y' = r cos a sin 0 +r sin a cos 0 = x sin 0+y cos 0.

In matrix notation, this means

ix:\

/cos — ksin 0

0\ tx\

cos el k J?)

and we have precisely the orthogonal linear transformation discussed in the previous paragraph. In summary, the linear transformation T, with orthogonal matrix

P = (COS 0 \ sin 0

— sin 0) cos 0

may be regarded either (i) as mapping each point A into another point A', where OA = OA' and AOA' = +0 or (ii) leaving each point in its original 647

[29


position but rotating the axes of coordinates through an angle — (i.e. through 0 clockwise). We shall now show how this new interpretation of an orthogonal matrix may be employed to find the major and minor axes of a central conic. Consider first the equation

ax2+by2 = 1 (a, b not both negative), which represents a conic with its centre at the origin and its major and minor axes along the axes of coordinates. If we rotate the axes through an angle —0, that is, if we rename each point by the rule x' = x cos 0 —y sin 0, y' = x sin 0+y cos 0, x = x' cos 0+y' sin 0,

then

y = —x' sin 0 +y' cos 0, and the new equation of the conic takes the form a(x' cos 0+y' sin 0)2 + b(— x' sin +y' cos 0)2 = 1, or

(a cost B + b sin2 0)x'2 + 2(a — b) x'y' sin 0 cos 0 + (a sine 0 + b cost 0) y2 = 1. We may rewrite this as

a' x' 2+2h' x'y' +b'y'2 =1, which is the typical form of equation of a central conic with its centre at the origin. Now consider the converse problem: given the equation of a central conic in the form ax2+2hxy+by2 = 1, how can we rotate the axes to obtain the equation in the form

a' x' 2+ b'y' 2= 1? The equation ax2+2hxy+by2 = 1 may be rewritten in matrix form as (a h\ ix\ "

or where

kyl

'

u'Au = 1, u = (x) and A = (ha h h).

The problem thus reduces to that of diagonalizing the matrix A for then the equation would become, in the new coordinate system,

(x, y) (0 A, aoj 648

(yx:)

3]


or

A1x'2 +A2y'2 = 1.

But A is a symmetric matrix and, by Theorems 28.3 and 28.4, we can find an orthogonal matrix P such that PAP' = D where

D

Al

\

= (0 A2P

Al and A2 being the eigenvalues of A. (Recall that, for an orthogonal matrix, P-1= P'. Notice too that, to obtain PAP' in diagonal form (rather than P'AP), the matrix P is obtained by transposing the matrix with columns which are the eigenvectors of the matrix A; that is, the eigenvectors appear as the rows.) The fact that P is orthogonal tells us that the new axes, Ox', Oy' will be perpendicular. The details of the transformation are given below. Consider the linear transformation defined by v = Pu, x where v = , the new position vector of the point u = . Then .Y, (;) vv P'v = P-lv = u and the equation transforms into the equation

u'Au = 1

(P'v)' A(P'v) = 1 or, since (P')' = P, and using the results of Ex. 8, Chapter 28, v'PAP'v = 1. But PAP' = D, a diagonal matrix and the equation has been transformed into A02) (y x,') (x' Y) (O1 A1x,2 + 42, = 1.

or

If we define the trace of the matrix A, tr A, as tr A = a+b (see Exercise 28(b), Questions 8-10), the characteristic equation of A, a—A h = 0, h b— A may be written in the form A2 —AtrA+detA = 0. 649


Thus we have

[29

tr A = Ai + A„ det A = A1A2;

but

tr D = Ai + A„

det D = A1k2, and thus both the trace and the determinant of the matrix defining the conic remain invariant under the given linear transformation. *Ex. 15. Show how the invariance of the trace and determinant enables us to deduce immediately the nature (ellipse, hyperbola) of a central conic ax2+2hxy+ by2= 1. Ex. 16. Prove that, if Q is any non-singular 2 x 2 matrix, then tr (Q-1-AQ) = tr A and det (Q-1AQ) = det A. Can these results be generalized for higher-order matrices (the trace still being defined as the sum of the elements in the leading diagonal)?

Example 6. Determine the nature of the central conic 5x2+8xy+11y2 = 1 and find the lengths of its axes. The matrix A of the conic is given by A=

(5 4\ 4 11/ •

Let the eigenvalues of A be A1, A2; then A,+ A, = 5+11 = 16, AlA, = 55 — 16 = 39 giving

= 3, A2 = 13.

The equation of the conic may thus be reduced to the form 3x'2 + 13y'2 = 1. This is seen to be an ellipse, with major semi-axis 1/V3 and minor semiaxis 1/V13. To draw a sketch of the ellipse, showing it in relation to the original axes, we need to find the inclination of the axes of the conic to the axes Ox, Oy. This may be done by obtaining an explicit form for the reducing matrix P. Corresponding to the eigenvalue = 3 we have the unit eigenvector ( 2/V5) ' while corresponding to A2 = 13 we have (1/V5 — 1/5 2/V5) 650

3]


Thus and

13 '

P

2 1)

= =

/2

k

The transformation

—1\

v = Pu

may be interpreted as a rotation of axes clockwise through an angle 0 = arctan

(Figure 29.6).

Fig. 29.6

In other words, in terms of the original coordinate system, the major and minor axes of the conic lie respectively along

x + 2y = 0 and 2x — y = O. *Ex. 17. Show that the equation ax2+2hxy+by2 = 1 (i) represents an ellipse if h2— ab < 0, and a > 0; (ii) represents a hyperbola if h2 — ab > 0; (iii) represents a rectangular hyperbola of a+ b = 0; (iv) represents a pair of parallel straight lines if h2 = ab (h * 0). Ex. 18. Let us write the conic of Ex. 17 in the form x'Ax = 1, where

A=

(a

x = (x ) h' Y); u is a unit direction vector and P is a point with position vector p. The point with position vector p+ Au lies on the conic if (p+Au)' A(p+Au) = 1; 651


[29

show that this reduces to A2u 'Au + 2Au'Ap + p'Ap = 1. If P is the mid-point of a chord parallel to u, prove that u'Ap = 0; what does this tell us about the vector Au? The axes of a central conic (other than a circle) may be defined as the (unique) pair of perpendicular conjugate diameters: use the results proved above to deduce that u is in the direction of an axis if there exists a number k such that Au = ku. Interpret your results in terms of eigenvectors. Exercise 29 (b) 1. Obtain the equation of the sphere: (i) with centre (0, -1, 0) and radius 1; (ii) with centre (2, - 3, -1) and radius 5; (iii) with centre (a, b, c) and radius ,N1(a2+ b2+ c2). 2. Find the centre and radius of the sphere:

(i) with equation x2 + y2+ z2- 6x-2y +2z- 5 = 0; (ii) with equation x2 + y2+ z2 + 4x-8y -2z-60 = 0; (iii) with equation x2 + y2 + z2 -2ax-2by+2cz+2bc-2ca+2ab = 0. 3. Find the equation of the sphere:

(i) through the points (4, 4, 4), (5, 6, 1), (0, -4, 2), (7, 3, 2); (ii) through the points (2, 4, 4), (5, 1, 4), (3, 4, 3), (- 3, -1, 0); (iii) through the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c). 4. Find the centre and radius of the circle:

(i) with equations x2 + y2 + z2 - 12x+ 2y - 10z+ 44 = 0, x- y+z = 6; (ii) with equations x2 +372 + z2 - 18x - lOy + 12z-55 = 0, 2x + 3y - 4z = -1. 5. The position vectors of points A and B relative to some origin 0 are respectively a and b. Show that the locus of a point P which moves in space in such a way that PA2+PB2 = d2, where d is a constant, is a sphere with its centre at the mid-point of AB. Find the radius of the sphere in terms of a, b and d. What can you say about the sphere if d = la-bl? 6. With the notation of Question 5, show that the locus of a point P which moves

in space in such a way that PA2+PB2+PC2 = d2, where C is a point with position vector c, is a sphere and find its centre. Can you generalize this result in any way? 7. Show that the line r = Au, where u is a unit vector, cuts the sphere (r -3i- 2j - 2k). (r -3i - 2j - 2k) = 6 at two points with parameters given by the quadratic equation A2-2Au . (3i + 2j + 2k) + 11 = 0.

652

EXERCISE 29 Deduce that the line through the origin in the direction of the vector i+j+ 3k touches the given sphere. Prove more generally that the line through the origin in the direction of the vector /i+ mj+nk is tangential to the sphere if and only if (31+2m+2n)2 = 1412+m2+n2), and hence show that the equation of the cone with vertex the origin which circumscribes the sphere is (3x+ 2y + 2z)2 = 11(x2"2+z2). 8. Find the equation of the cone obtained by rotating the line xlp = ylq = zlr about the line

x/a = ylb = z/c.

9. A right-circular cone of semi-angle a has its vertex at the origin and contains lines in the directions i, i+j and i +j+k. Show that sect a = 9— 242-246. 10. Show that the equation

yx+ zx+xy = 0

represents a cone, vertex the origin and semi-angle arctan 42. What is the curve of intersection of this cone: (i) with the plane x+y+ z = 1; (ii) with the plane x— 2y+ z = 0; (iii) with the plane x —2y + z = 1? 11. A circle, S, is defined by the vector equation (r — a) . (r — a) — c2 = 0, r . n = p. Show that any sphere passing through the circle S has vector equation (r a) . (r a) c2 +A [r n —p] = 0 —

—

—

and deduce that, if B is a point not on the plane r .n = p and if OB = b, then the vector equation of the sphere through S and containing the point B is (b n —p) [(r a). (r a) —

—

—

—

[(b a) . (b a) —

—

—

(r . n — p) = 0.

Find the centre and radius of the sphere which passes through the point (1, — 1, 1) and contains the circle. x2 +y2 +z2 = 4, x+2y+ z = 1. 12. A is the fixed point with position vector ak, and L, M are variable points with position vectors Ai, A respectively, where A, # are scalar parameters. If P is a point such that the angles LPM, MPA, APL are all right angles show that, whatever the values of A and ,a, P lies on a sphere, centre A and radius a. 13. Find the equation of the tangent to the sphere x2 +y2 +z2 -10x-2y-12z+35 = 0 at the point (4, 2, 1). Prove that the plane x+y+ z = 3 is also a tangent to the sphere and find the point of contact. 653


[29

14. Find the values of k for which the plane 4x+2y+3z = k touches the sphere x 2 +y2 +z2 — 10x— 6y— 10z+ 30 = 0. 15. The point B lies outside the sphere (r — a) . (r — a) — c2 = 0. If OB = b, prove that the length of the tangent from B to the sphere is (b — a) . (b — a) — c2. Prove that the set of points from which the tangents to two non-intersecting spheres are of equal length is a plane which is perpendicular to the line of centres of the two spheres. Deduce that the set of such points for three non-intersecting spheres is a line and that, given four non-intersecting spheres with non-coplanar centres, there is a point B from which the tangents to the four spheres are of equal lengths. Explain how the last result can be used to show that a sphere may be circumscribed about any tetrahedron. 16. Describe the curve of intersection of the surface with equation x2 +y2 —z2 = 0; (i) with the plane z = 1; (ii) with the plane x = 1. What surface does the equation represent? 17. Show that the parametric equation of the helix of pitch 8 which lies on the surface of the cylinder x2 + z2 = 1 is 40 r = cos Oi + — 7T j + sin Ok. Find the coordinates of the point in which the helix cuts the plane /2x — y + 2,/2z = 0.

1

Show also that the helix cuts the plane x+ z = 0 at an infinite number of points, and explain this geometrically. 18. Prove that the mid-points P, Q, R, S, T, U of the edges BC, CA, AB, AD, BD, CD of a tetrahedron ABCD, in which the opposite edges (BC, AD), (CA, BD), (AB, CD) are perpendicular pairs, lie on a sphere. Is the converse result, that if the points P, Q, R, S, T, U lie on a sphere then the opposite pairs of edges are perpendicular, true or false? If the sphere cuts BC again at P', prove that BC is perpendicular to the plane P AD. 19. Two spheres, centres Aland A2 and radii c1and c2intersect in a circle of radius d. If AlA2 = 1, where 12 = c,2+ cL prove that d = c1c211 and find the ratio in which the centre of the circle divides the line A1./12. 20. Prove that there is one and only one sphere which contains a given circle and passes through a given point not in the plane of the circle. Two circles, not in the same plane, meet at points A and B. A plane meets one of these circles at C and D and the other circle at E and F. Prove that: (i) the points C, D, E, F lie on a circle; (ii) the lines AB, CD, EF are either concurrent or parallel. (0 & C) 654

EXERCISE 29 21. Show that, if the ellipse (x2/a2) + (y2/b2) = 1 lying on the plane z = 0 is rotated through four right angles about the x axis, the equation of the surface obtained is x2 y 2 z 2 1 ai+ b2± b2 = '. The surface thus obtained is called an oblate spheroid if a < b and a prolate spheroid if a > b. Both surfaces are particular cases of the ellipsoid. x2 y 2 z 2 -+ = 1. a2 -b2 +c2 By considering sections of this ellipsoid with planes x = p, y = q, z = r, describe the shape of the general ellipsoid and in particular, show that it is a closed surface. 22. Describe the nature of the hyperboloid of one sheet x2

y2 z2 —— a2+b2 c2 = 1

and of the hyperboloid of two sheets 2 z 2 1 x2 y a2— b2— c2= ' • 23. Describe the nature of the elliptic paraboloid x2 y 2 2z —— = a2+ b2 c and of the hyperbolic paraboloid x2 y 2 2z -- — =— . a2 b 2 c 24. Prove that the hyperboloid of one sheet (see Question 22) x2 y 2 z 2 +— = 1 a2 b 2 c 2 contains the line determined by the two planes x+ z a c

A 1 + -1' and A x— -z = 1 — 1' / b ac b•

Call this line a A generator. Prove /also that the surface contains lc generators of the form x z x z ( y Y --- = A 1+ -) , #(-+-) = 1—b a c b ac .

Prove also that any pair of A generators and any pair of it generators are necessarily skew lines but that each A generator meets every ,u generator at a point with coordinates of the form

ta(A+ A b(1—A,u) c(A—A\ 1+A# '

1+4.4 ' 1 +A.,a / ' 655


[29

(Notice that we have obtained a parametric representation of the hyperboloid of one sheet in terms of the two parameters A, p.) 25. (See Questions 23, 24.) Prove that the hyperbolic paraboloid y2 _ 2z a2 b 2 c contains a system of A generators and a system of it generators with the properties that any pair of A generators are skew, any pair of u generators are skew, but each A generator meets every ,u generator in a point of the form (a(A+ ,u), b(A— ,u), 2cAp). 26. Determine the nature of the central conic 7x2 — 8xy + 13y2 = 150 and find its eccentricity and the equations of its axes. 27. Determine the nature of the central conic 6x2 + 5xy — 6y2 = 169 and find its foci and the equations of its axes. 28. Show that the equation x2 +6xy+y2 + 6 = 0 represents a hyperbola and find its eccentricity and the equations of its asymptotes. 29. Two lines, lland 12, intersect at an angle a. Find the locus of the centre of a sphere of given radius which touches both lines.

656

Revision exercise C

1. The polynomials g(x), h(x) both leave the same remainder on division by (x— a). Prove that the polynomial

xg(x) — ah(x) is divisible by (x — a). The polynomial f(x) (degree 4) leaves a remainder of rx+ s on division by (x— b)3. Prove that f"(x) is divisible by (x— b). 2. If z * 1 and

1 +z

w= 1 —z show that IzI = 1 if and only if Re (w) = 0.

3. If a < b = an < Irn for all a, b e R where n e Z, what further conditions does n satisfy? Are these further conditions still necessary if a, b are restricted to positive real numbers? (S. M.P.) 4. The function f: R R is defined by

f(x) = x2— 21x1. Sketch the graph off and find what values of x are invariant under!. 5. Solve the equation

3 sin 0+ 2 sin (60°— 0) = 2

for values of 0 lying between 0° and 360°. 6. A random number table consists of a succession of digits each chosen at random independently from the set 0, 1, 2, ..., 9. Two successive digits are taken from such a table; show that the probability that their sum is 9 is 1/10. Four successive digits are taken from the table. Calculate the probability, p, that the sum of the first two equals the sum of the third and fourth. (M.E.I.) 7. Prove that

a+ ar+ ar2+ ...+arn-' =

a(1 — r") 1 —r '

where r * 1. Prove that in general (1+x+x2+...+x2n)(1—x+x2—... 4.x21 = 1±x2+x4+...-1-x4". State any values of x for which your proof does not apply and obtain the appropriate results in each of these cases. (0 & C) 8. Prove that, if the points 0, A, C are non-collinear, then the position vector with respect to 0 of any point in the plane 0 AC may be expressed uniquely in terms of a = OA and c = OC. (over) 657

REVISION EXERCISE C OABC is a parallelogram, L is the mid-point of AB and T is the point which divides OB in the ratio 3:2. CT meets OA at X and LT meets OC at Y. XY meets OB at Z. Find the ratio OZ:ZB in which Z divides OB. 9. If A = 15 1\ k3 3P prove by induction that 10. If

M

An = 4(6"— 2°) (A — 21) +

= (2 3 , ki 1/\

_ _ (4 N _ 5

—1 — \ ' Q= (21 24)

form the products MQ and NQ. What law, which is true for products in the algebra of real numbers but which is not true for products of matrices, is ex(S.M.P.) emplified by this? 11. Express P

5+j _ 242 and — 2 + 3j

in the form x + jy, where x and y are real. Find also the modulus and the argument of p and of q. Hence write down the (M.E.I.) argument of p + q. 12. Express in partial fractions: 6 — 3x (n) 2x-3 (x — 1) (x+2) ' x(1 +x2)'

... 3x2 — 7x +3 (1 — x)2(1 — 2x) .

13. Two concentric circles have radii 10 and 15 cm. If a point is taken at random in the annulus formed by them (all points of the annulus being equally likely) find the expected distance of the point from the centre of the circles. Explain why you would expect your answer to be greater than 12.5 cm. 14. P is the point (at 2, 2at) on the parabola y2 = 4ax and 0 Q is the chord passing through the origin 0 and parallel to the tangent at P. Find the coordinates of the point of intersection, R, of the tangents at P and Q. Give a reason to show that the locus of R is another parabola. If 5' is the mid-point of 0 Q, prove that PSR is a right-angled triangle, and that (0 & C) the area of triangle PQR is la2t 3. 15. The sets A, B, C, D are defined as follows: A = {x R: 2 < x 1}, B= {x e R: —1 x < 2}, C = {x E R: x 1}, D = {xe R:x ‘. 2}. Express in the form {:}, the sets: (i) A n B; (ii) A' n C; (iii) A n C; (iv) B' n C' n D'; (v) (A n B) U (B n C'). Express the sets —

—

E={xER:x< -1} and F={xER:x 0 and odd; n > 0 only. 4. —1, 0, 3. 5. 8.2°, 90°. 6.0.067. 7. —1, 1. 8.1:1. 658 10. Cancellation. 11. (1/2, —170, (12, in); --4717.. 12. (i) 1/(x-1)-4/(x+2); (ii) (2 +3x)/(1 + x2)— 3/x; (iii) 1/(1 — x)+1/(1 — x)2 + 1/(1 — 2x). 13. 12.67. 14. (2at 2, 3at). 15. (i) { —1 x 1); (ii) {x > 1}; (iii) {x = 1}; (iv) {— 2 < x < —1}; (v) { —1 < x < 1}; E = B' n C', F = B u C'. 659 16. Line x+y = a+ b. 17. (i) H;

699

ANSWERS PAGE 0 0 0) 659 18. (2 0 0 . 0 1 0 20. 7g/12, 5g/4, 23g/12. 22. x < -5 or -3 < x < 7. 21. -1+1k, -1/k. 23. 1/(1- 3x)- 2/(1- 2x); -1- x + x2. 24. Pal, 0. 660 25. Plane perpendicular to OA. 26.1, -2, - 2, 6. 28. I, 0-215. 27. -1, (-1 ± V105)/2. 31. 2:1. 30. 1(6n3- 3n2 -n), -1(6n3 +3n2 -n). 661 33. (i) 60° < 0 < 109.5°, 250.5° < 0 < 300°; (ii) 0° < 0 < 30°, 150° < 0 < 199.5°, 240.5° < 0 < 360°.

662 663 664 665

666

35. A = -2, (1, f, D. 36. (i) (2t+ 03 (3t+ 1)2/432; (ii) W ; (iii) 3.5. 38. (i) 1/V33; (ii) 3/V33. 44. (2, -3, 7), 2. 42. 3. 43. -2 < x < 0, 2 < x < 7. 50. 5.02, 103.26, no (1 % exceed 106.52). 47. A/2, 2A/2. 51. 2, 6. 53. cot 0 -2"" cot (2'W). 54. 3/128. 60. (i) R; (ii) {x E R: V2 < x < V2}. 61. 2 cos-0, -1-0; (2 cos 10)n sinln0. 62. -1, - ,, ; 1-, -1, -2; -1, -1 ±V3-. 65. gw+ g*w* + cz2 =0. 64. 2; 2, 0.89, 2; 0.374. 14 1 0 0) 1 -4 14) -73 -32 66. (1 1 0 , 0 1 -3 , 16 7 -3 . 0 0 5 -2 1 5 -2 1 1 -

-

-

67.t-, 11, 1, •

68. *; *(4a+3b).

667 69. x = -y/5 = z/2. 72. (i) 5i+(i+j)- 5(i +j + k); (ii) (31- 2k)- (21+4j + 3k). 73. (6)5, 1 -(D6 -ay, (6 !) ay, ay-1/6; 4; 6, 66. 75. 0.966. 74. -2i+j+3k, y = 0. 668 77. 16y2 -24xz+7z2 = 0, 4y2 = 36x-63. 78. 1 + 5x/2 + 39x2/8, 5x 2y+ 2 = 0, curve above tangent. 84. (i) (1, -2, 1); (ii) no solution; (iii) (A, -3A, A + 1); 8; 16. 669 83. 1. 85. (- Oln. 670 87. O. 88. (i) Line; (ii) circle; (iii) ellipse; (iv) parabola. 89. c(ct + 13)12, c(ct+,8)1(2c93); cYt2- 2XYt+cX = 0. 90. 63 (3n -1)77 . -

700

ANSWERS PAGE

670 92. x = Aa + (a A b)/Ia12, y = (1 - A)a - (a A 13)i1a1 2. 671 94. pn 1(p" +qn). 97. 2x + 9y + 5z -12 = 0. 672 98. 2.06. 99. 5x + 3y + 4z = 0, y-z = 0, x - y = 0. 100. Half-turn about (0, J3). a3 - a2. 673 102. a4 - a, a. 104. Line in direction 3i+ 2j+ k; plane x+y- 4z = 0. 674 110. -0.96, 2.88, 5.08. 675 112. (11) mrAn-V 1.6) n ;- 0, 0.0001, 0.0014, 0.0090; 7. Y 113. 13025, 164.77, 215.49, 285.65. 114. (N- 1) (N + 4)/(2N3). 676 118. p p2(e2a - e"). 119. F26. 677 121. N 21. )

122. 8N2(N + 1)/(2N+ 1)3. 123. y,

aa

678 128. Integral polynomial in k divisible by k(k + 1). 679 133. 2z = - Re (a), 2z = j Im (a). Rectangular hyperbola. p(z) = 0. -

701

Index

Abel, 483 absolute error, 553 affix, 385 Apollonius, 603 Argand diagram, 385 argument, 386 auxiliary circle of ellipse, 577; of hyperbola, 593 Buffon, 549

C, 380 Cardan, 496 cardioid, 447 Cauchy distribution, 549 Cayley—Hamilton theorem, 626 central limit theorem, 543 characteristic equation, 609 chi-squared distribution, 549 complex numbers, 381 addition of, 381; argument of, 386; conjugate of, 382; division of, 382; equality of, 381; imaginary part of, 382; logarithm of, 432; modulus of, 382; real part of, 382 complex plane, 385 cone, 645 conic, 573 central, 595; conjugate diameters of, 578; diameter of, 578; directrix of, 573; eccentricity of, 573; focus of, 573 conjugate complex number, 382 c. roots of real equation, 484 continuous random variable, 520 cover-up rule, 411 cubic equation, 496 discriminant of, 501; irreducible case, 500 cumulative distribution function, 526 cylinder, 644 de Moivre's theorem, 423 density function, 521 Descartes's rule of signs, 558 diagonal form, 617, 620 diameter, 578 diametral plane, 642

differences, 403 director circle, 595 distribution function, 524 double point, 636 eigenvalue, 610 eigenvector, 610 ellipse, 573 construction of, 577, 582 ellipsoid, 655 equations of curve in Argand diagram, 438; polynomial, 483; quadratic, 450; sum and product of roots, 451, 490 equilateral hyperbola, 593 expectation, 527 exponential distribution, 543 field, 379 focal distance property of ellipse, 581; of hyperbola, 594 folium of Descartes, 636 Fregier point, 482 fundamental theorem of algebra, 380, 483 Galois, 483 generator of cone, 646 Gregory—Newton formula, 406 helix, 645 Horner's method of synthetic division, 402 hyperbola, 573 asymptotes, 591; conjugate axis, 591; of Apollonius, 608; rectangular, 593; transverse axis, 591 hyperboloid, 655 initial line, 599 invariant line, 609 inversion, 441 iterative methods, 560 linear interpolation, 556 log-normal distribution, 676 major axis, 576

703

INDEX mean, 527 mean deviation, 528 median, 523 minor axis, 576 mode, 523 modulus, 382 modulus—argument form, 387 multiple roots, 484 nested multiplication, 402 Newton conic theorem, 480, 607; —Raphson, 563 nilpotent, 629 normal distribution, 535 oblate spheroid, 655 ogive curve, 527 orthogonal matrix, 614 parabola, 465, 573 axis, 465; directrix, 465; family of, 475; focal chord, 468; focus, 465; latus rectum, 467; vertex, 465 paraboloid, 655 parity, 558 partial fractions, 409 polar coordinates, 599; equations, 447 probability density function, 521 prolate spheroid, 655 pure imaginary, 385 purely random process, 544 quartic, 503 quartiles, 524 real quadratic function, 457

rectangular distribution, 530 rectangular hyperbola, 475, 593 reflection property of ellipse, 582; of parabola, 470 relative error, 553 roots, coincident (repeated, multiple), 450, 484 roots of unity, 429 rounding-off, 552 scalar triple product, 510 second-order process, 566 semi-cubical parabola, 633 similar matrices, 620 skew symmetric matrix, 629 sphere, 640 standard normal distribution, 537 stationary process, 544 symmetric matrix, 616 synthetic division, 401 trace, 629 translation of axes, 462 triangle inequalities, 390 triangular distribution, 534 uniform distribution, 530 variance, 528 vector area, 509; product, 506; projection, 570; triple product, 510 volume of parallelepiped, 511; of tetrahedron, 511 von Mises's iteration, 567

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