Calculus: A modern, rigorous approach

Calculus: A modern, rigorous approach Horst R. Beyer Louisiana State University (LSU) Center for Computation and Technology (CCT) 330 Johnston Hall Baton Rouge, LA 70803, USA & Louisiana State University (LSU) Department of Mathematics 148 Lockett Hall Baton Rouge, LA 70803, USA July 3, 2007

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Dedi ated to the Holy Spirit

Contents Contents

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1 Calculus I 1.1 Basics . . . . . . . . . . . . . . . . . . . . . 1.1.1 Elementary Mathematical Logic . . . 1.1.2 Sets . . . . . . . . . . . . . . . . . . 1.1.3 Maps . . . . . . . . . . . . . . . . . 1.2 Limits and Continuous Functions . . . . . . . 1.2.1 Limits of Sequences of Real Numbers 1.2.2 Continuous Functions . . . . . . . . 1.3 Differentiation . . . . . . . . . . . . . . . . . 1.4 Applications of Differentiation . . . . . . . . 1.5 Riemann Integration . . . . . . . . . . . . .

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5 . 5 . 5 . 14 . 23 . 36 . 36 . 55 . 79 . 94 . 140

2 Calculus II 2.1 Techniques of Integration . . . . . . . . . . . . . . . 2.1.1 Change of Variables . . . . . . . . . . . . . 2.1.2 Integration by Parts . . . . . . . . . . . . . . 2.1.3 Partial Fractions . . . . . . . . . . . . . . . 2.1.4 Approximate Calculation of Integrals . . . . 2.2 Improper Integrals . . . . . . . . . . . . . . . . . . 2.3 Series of Real Numbers . . . . . . . . . . . . . . . . 2.4 Series of Functions . . . . . . . . . . . . . . . . . . 2.5 Analytical Geometry and Elementary Vector Calculus 2.5.1 Metric Spaces . . . . . . . . . . . . . . . . . 2.5.2 Vector Spaces . . . . . . . . . . . . . . . . . 2.5.3 Conic Sections . . . . . . . . . . . . . . . . 2.5.4 Polar Coordinates . . . . . . . . . . . . . . . 2.5.5 Quadric Surfaces . . . . . . . . . . . . . . . 2.5.6 Cylindrical and Spherical Coordinates . . . . 2.5.7 Limits in Rn . . . . . . . . . . . . . . . . . 2.5.8 Paths in Rn . . . . . . . . . . . . . . . . . .

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163 163 163 173 186 196 202 223 248 292 292 298 323 330 338 346 351 356

3 Calculus III 3.1 Vector-valued Functions of Several Variables . . . . . . . 3.2 Derivatives of Vector-valued Functions of Several Variables 3.3 Applications of Differentiation . . . . . . . . . . . . . . . 3.4 The Riemann Integral in n-dimensions . . . . . . . . . . . 3.4.1 Applications of Multiple Integrals . . . . . . . . . 3.5 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . 3.6 Generalizations of the Fundamental Theorem of Calculus . 3.6.1 Green’s Theorem . . . . . . . . . . . . . . . . . . 3.6.2 Stokes’ Theorem . . . . . . . . . . . . . . . . . . 3.6.3 Gauss’ Theorem . . . . . . . . . . . . . . . . . .

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375 375 394 423 446 460 471 483 486 502 512

4 Appendix 4.1 Construction of the Real Number System . . 4.2 Lebesgue’s Criterion of Riemann-integrability 4.3 Properties of the Determinant . . . . . . . . . 4.4 The Inverse Mapping Theorem . . . . . . . .

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529 529 542 546 563

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References

566

Index of Notation

569

Index of Terminology

571

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1 Calculus I 1.1 Basics 1.1.1 Elementary Mathematical Logic Definition 1.1.1. (Statements) A statement (or proposition) is an assertion that can determined as true or false. Often abstract letters like A, B, C, . . . are used for their representation. Example 1.1.2. The following are statements: (i) The president George Washington was the first president of the United States , (ii) 2 + 2 = 27 , (iii) There are no positive integers a, b, c and n with n ¡ 2 such an cn . (Fermat’s conjecture)

bn



The following sentences are no statements: (iv) Which way to the Union Station? , (v) Go jump into the lake! Definition 1.1.3. (Truth values) The truth value of a statement is denoted by ‘T’ if it is true and by ‘F’ if it is false. Example 1.1.4. For example, the statement 9

16  25

(1.1.1)

is true and therefore has truth value ‘T’, whereas the statement 9

16  26

is false and therefore has truth value ‘F’. Also, Example 1.1.2(i) is true, Example 1.1.2(ii) is false, and Example 1.1.2(iii) is very likely true. (Wiles 1995) 5

Definition 1.1.5. (Connectives) Connectives like ‘and’, ‘or’, ‘not’, . . . stand for operations on statements. Connective ‘not’ ‘and’ ‘or’ ‘if . . . then’ ‘. . . if and only if . . . ’

Symbol

Name Negation Conjunction Disjunction Conditional Bi-conditional

^ _ ñ 

Example 1.1.6. For example, the statement ‘It is not the case that 9

16  25’

is the negation (or ‘contrapositive’) of (1.1.1). It can be stated more simply as 9 16  25 . Other examples are compounds like in the following Example 1.1.7. (i) Tigers are cats and alligators are reptiles , (ii) Tigers are cats or (tigers are) reptiles , (iii) If some tigers are cats, and some cats are black, then some tigers are black , (iv) 9

16  25 if and only if 8

15  23 .

Definition 1.1.8. (Truth tables) A truth table is a pictorial representation of all possible outcomes of the truth value of a compound sentence. The connectives are defined by the following truth tables for all statements A and B. A T T F F

B T F T F

A F F T T

A^B T F F F

A_B T T T F

6

AñB T F T T

AB T . F F T

Note that the compound A _ B is true if at least one of the statements A and B is true. This is different from the normal usage of ‘or’ in English. It can be described as ‘and/or’. Therefore, the statement 1.1.7 (ii) is true. Also, the statements 1.1.7 (i) and 1.1.7 (iv) are true. Also, note that from a true statement A there cannot follow a false statement B, i.e., in that case the truth value of A ñ B is false. This can be used to identify invalid arguments and also provides the logical basis for so called indirect proofs. Note that valid rules of inference do not only come from logic, but also from the field (Arithmetic, Number Theory, Set Theory, ...) the statement is associated to. For instance, the equivalence 1.1.7 (iv) is concluded by arithmetic rules, not by logic. Those rules could turn out to be inconsistent with logic in that they allow to conclude a false statement from a true statement. Such rules would have to be abandoned. An example for this is given by the statement 1.1.7 (iii). Although the first two statements are true, the whole statement is false because there are no black tigers. In the following, the occurrence of such a contradiction is indicated by the symbol . Note that the rule of inference in 1.1.7 (iii) is false even if there were black tigers. Example 1.1.9. (Inconsistent rules) Assume that the real numbers are part of a larger collection of ‘ideal numbers’ for which there is a multiplication which reduces to the usual multiplication if the factors are ? real. Further, assume that for every ideal number z there is a square root z , i.e., such that ? 2 z z , which is identical to the positive square root if z is real and positive. Finally, assume that for all ideal numbers z1 , z2 , it holds that

?z z  ?z  ?z . 1 2 1 2

Note that the last rule is correct if z1 and z2 are real. Then we arrive at the following contradiction:


1 

?

a ? ? ? 
1 2 
1 
1  p
1qp
1q  1  1 .

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Hence an extension of the real numbers with all these properties does not exist. A simple example for an indirect proof is the following Example 1.1.10. (Indirect proof) Prove that there are no integers m and n such that 2m 4n  45 . (1.1.2) Proof. The proof is indirect. Assume the opposite, i.e., that there are integers m and n such that (1.1.2) is true. Then the left hand side of the equation is divisible without rest by 2, whereas the right hand side is not. Hence the opposite of the assumption is true. This is what we wanted to prove. Example 1.1.11. Calculate the truth table of the statements 

pA ñ B q ^ pB ñ C q ñ pA ñ C q (Transitivity) ,   pA _ B q ^ pA ñ C q ^ pB ñ C q ñ C (Proof by cases) , p B ñ Aq  pA ñ B q (Contraposition) . (1.1.3) Solution: A T T T T F F F F

B T T F F T T F F

C T F T F T F T F

AñB T T F F T T T T

B

ñ C pA ñ B q ^ pB ñ C q T F T T T F T T

T F F F T F T T

8

AñC T F T F T T T T

pA ñ B q ^ pB ñ C q ñ pA ñ C q T T T T T T T T A T T T T F F F F

B T T F F T T F F

C T F T F T F T F

A_B T T T T T T F F

AñC T F T F T T T T

pA _ B q ^ pA ñ C q ^ pB ñ C q



B

ñ C pA ñ C q ^ pB ñ C q T F T T T F T T

T F T F T F T T 

pA _ B q ^ pA ñ C q ^ pB ñ C q ñ C

T F T F T F F F

A T T F F

A F F T T

B T F T F

T T T T T T T T

B F T F T

B

ñ

AñB T F T T

A

T F T T

9

p

B

ñ

Aq  pA ñ B q T T T T

The members of (1.1.3) are so called tautologies , i.e., statements that are true independent of the truth values of their variables. At the same time they are frequently used rules of inference in mathematics, i.e., for all statements A, B and C it can be concluded from the truth of the left hand side (in large brackets) of the relations on the truth of the corresponding right hand side. Example 1.1.12. (Transitivity) Consider the statements (i) If Mike is a tiger, then he is a cat, (ii) If Mike is a cat, then he is a mammal, (iii) If Mike is a tiger, then he is a mammal. Statements (i), (ii) are both true. Hence it follows by the transitivity of ñ the truth of (iii) (and since ‘Mike’, the tiger of the LSU, is indeed a tiger, he is also a mammal). Example 1.1.13. (Proof by cases) Prove that n

|n
1 | ¥ 1

(1.1.4)

for all integers n. Proof. For this, let n be some integer. We consider the cases n ¤ 1 and n ¥ 1. If n is an integer such that n ¤ 1, then n
1 ¤ 0 and therefore

|n
1 |  n 1
n  1 ¥ 1 . If n is an integer such that n ¥ 1, then n ¥ 1 and therefore n |n
1|  n n
1  2n
1 ¥ 2
1  1 . n

Hence in both cases (1.1.4) is true. The statement follows since any integer is ¤ 1 and/or ¥ 1. Example 1.1.14. (Contraposition) Prove that if the square of an integer is even, then the integer itself is even. 10

Proof. We define statements A, B as ‘The square of the integer (in question) is even’ and ‘The integer (in question) is even’ , B corresponds to the statement

respectively. Hence

‘The integer (in question) is odd’ , and

A corresponds to the statement ‘The square of the integer (in question) is odd’ .

Hence the statement follows by contraposition if we can prove that the square of any odd integer is odd. For this, let n be some odd integer. Then there is an integer m such that n  2m 1. Hence n2

 p2m

1 q2

 4m2

4m

1  2 p2m2

2mq

1

is an odd integer and the statement follows. Based on the result in the previous example, we can prove now that there is no rational number whose square is 2. Example 1.1.15. (Indirect proof) Prove that there is no rational number whose square is 2. Proof. The proof is indirect. Assume on the contrary that there is such a number r. Without restriction, we can assume that r  p{q where p, q are integers without common divisor different from 1 and that q  0. By definition,  2 p p2 2  2. r  q q2 Hence it follows that p2

 2q2 11

and therefore by the previous example that 2 is a divisor of p. Hence there is an integer p¯ such that p  2¯ p. Substitution of this identity into the previous equation gives 2¯ p2  q 2 . Hence it follows again by the previous example that 2 is also divisor of q. As a consequence, p, q have 2 as a common divisor which is in contradiction to the assumption.

Problems 1) Decide which of the following are statements. a) b) c) d) e) f) g) f) g) h) i)

Did you solve the problem? Solve the problem! The solution is correct. Maria has green eyes. Soccer is the national sport in many countries. Soccer is the national sport in Germany. During the last year, soccer had the most spectators among all sports in Germany. Explain your solution! Can you explain your solution? Indeed, the solution is correct, but can you explain it? The solution is correct; please, demonstrate it on the blackboard.

2) Translate the following composite sentences into symbolic notation using letters for basic statements which contain no connectives. a) Either John is taller than Henry, or I am subject to an optical illusion. b) If John’s car breaks down, then he either has to come by bus or by taxi. c) Fred will stay in Europe, and he or George will visit Rome. d) Fred will stay in Europe and visit Rome, or George will visit Rome. e) I will travel by train or by plane.

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f) Neither Newton nor Einstein created quantum theory. g) If and only if the sun is shining, I will go swimming today; in case I go swimming, I will have an ice cream. h) If students are tired or distracted, then they don’t study well. i) If students focus on learning, their knowledge will increase; and if they don’t focus on learning, their knowledge will remain unchanged. 3) Denote by M , T , W the statements ”Today is Monday”, ”Today is Tuesday” and ”Today is Wednesday”, respectively. Further, denote by S the statement ”Yesterday was Sunday”. Translate the following statements into proper English. a) b) c) d) e) f)

M Ñ pT _ W q , SØM , S ^ pM _ T q , pS Ñ T q _ M , M Ø pT ^ p W qq _ S , pM Ø T q ^ pp W q _ S q .

4) By use of truth tables, prove that a) b) c) d) e) f) g) h) i) k) l)

p Aq  A , pA ^ B q  pB ^ Aq , pA _ B q  pB _ Aq , pA  B q  pB  Aq , pA ^ B q  p Aq _ p B q , pA _ B q  p Aq ^ p B q , pA Ñ B q  p Aq _ B , A ^ pB ^ C q  pA ^ B q ^ C , A _ pB _ C q  pA _ B q _ C , A _ pB ^ C q  pA _ B q ^ pA _ C q , A ^ pB _ C q  pA ^ B q _ pA ^ C q .

for arbitrary statements A, B and C. 5) Assume that

a  pb

cq  a  b

c

for all real a, b and c is a valid arithmetic rule of inference. Derive from this a contradiction to the valid arithmetic statement that 0  1. Therefore, conclude that the enlargement of the field of arithmetic by addition of the above rule would lead to an inconsistent field.

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6) Prove indirectly that 3n

2 is odd if n is an odd integer.

7) Prove indirectly that there are no integers m ¡ 0 and n ¡ 0 such that m2
n 2

1.

8) If a, b and c are odd integers, then there is no rational number x such that ax2 bx c  0. [Hint: Assume that there is such a rational number x  r{s, where r, s  0 are integers without common divisor. Show that this implies the equation rpar bsq 
cs2 which is contradictory.] 9) Prove that there is an infinite number of prime numbers, i.e., of natural numbers ¥ 2 that are divisible without remainder only by 1 and by that number itself. [Hint: Assume the opposite and construct a number which is larger than the largest prime number, but not divisible without remainder by any of the prime numbers.] 10) Prove by cases that

|x
1 |
| x

2| ¤ 3

|x
1 | | x

2| ¥ 3

for all real x. 11) Prove by cases that

for all real x. 12) Prove by cases that

|a|  b |b| for all real numbers a, b such that b  0.   a  

13) Prove by cases that if n is an integer, then n3 is of the form 9k where k is some integer and r is equal to
1, 0 or 1.

r

14) Prove that if n is an integer, then n5
n is divisible by 5. [Hint: Factor the polynomial n5
n as far as possible. Then consider the cases that n is of the form n  5q r where q is an integer and r is equal to 0, 1, 2, 3 or 4.]

1.1.2 Sets In the following we adopt the naive definition of sets given by the founder of set theory, the German mathematician Georg Cantor (1845-1918): 14

‘Eine Menge ist eine Zusammenfassung bestimmter, wohlunterschiedlicher Dinge unserer Anschauung oder unseres Denkens, welche Elemente der Menge genannt werden, zu einem Ganzen.’ or translated Definition 1.1.16. (Sets) A set is an aggregation of definite, different objects of our intuition or of our thinking, to be conceived as a whole. Those objects are called the elements of the set. This implies that for a given set A and any given object a it follows that either a is an element of A or it is not. The first is denoted by a P A , and the second is denoted by a R A . The set without any elements, the so called ‘empty set’, is denoted by φ. Example 1.1.17. Examples of sets are The set of all cats , The set of the lowercase letters of the Latin alphabet , The set of odd integers . Definition 1.1.18. (Elements) The following statements have the same meaning a is in the set A , a is an element of the set A , a is a member of the set A , aPA. Given some not necessarily different objects x1 , x2 , . . . , the set containing these objects is denoted by

tx1 , x2, . . . u . In particular, we define the set of natural numbers N , the set of natural numbers N without 0 , the set of integers Z and the set of integers Z without 0 by 15

Definition 1.1.19. (Natural numbers, integers) N : t0, 1, 2, 3, . . . u , N : t1, 2, 3, . . . u , Z : t0, 1,
1, 2,
2, 3,
3 . . . u , Z : t1,
1, 2,
2, 3,
3, . . . u . Another way of defining a set is by a property characterizing its elements, i.e., which is shared by all its elements, but not by any other object:

tx : x has the property P pxqu . It is read as: ‘The set of all x such that P pxq’. In this, the symbol ‘:’ is read

as ‘such that’. In particular, we define the set of rational numbers Q, the set of rational numbers Q without 0 , the set of real numbers R and the set of real numbers R without 0 by Definition 1.1.20. (Rational and real numbers) Q : tp{q : p P Z ^ q P N ^ q  0u , Q : tp{q : p P Z ^ q P N ^ q  0u , R : tx : x is a real numberu , R : tx : x is a non-zero real numberu . Definition 1.1.21. (Subsets, equality of sets) For all sets A and B, we define A € B : Every element of A is also an element of B and say that ‘A is a subset B’, ‘A is contained in B’, ‘A is included in B’ or ‘A is part of B’. Finally, we define A  B : A € B ^ B € A  A and B contain the same elements . Here and in the following, wherever meaningful, the symbol ‘:’ in front of other symbols is read as ‘per definition’. 16

Example 1.1.22. For instance,

t1, 1, 2, 3, 5u € t1, 1, 2,?3, 5, 8, 13u , ? ? t1, 1, 2, 3, 5u € trp1 5qn
p1
5qns{p2n 5q : n P Nu , t1, 2, 3, 3, 5, 1u  t1, 2, 3, 5? u, ? ? t1, 1, 2, 3, 5, . . . u  trp1 5qn
p1
5qns{p2n 5q : n P Nu . In particular, we define subsets of R, so called intervals , by Definition 1.1.23.

ra, bs : tx P R : a ¤ x ¤ bu , pa, bq : tx P R : a   x   bu ra, bq : tx P R : a ¤ x   bu , pa, bs : tx P R : a   x ¤ bu rc, 8q : tx P R : x ¥ cu , pc, 8q : tx P R : x ¡ cu p
8, dq : tx P R : x   du , p
8, ds : tx P R : x ¤ du for all a, b P R such that a ¤ b and c, d P R. We define the following operations on sets. Definition 1.1.24. (Operations on sets) For all sets A and B, we define (i) their union A Y B, read: ‘A union B’, by A Y B : tx : x P A _ x P B u (ii) and their intersection A X B, read: ‘A intersection B’, by A X B : tx : x P A ^ x P B u . If A X B

(iii)

 φ, we say that A and B are disjoint. the relative complement of B in A, AzB, read: ‘A without B’ or ‘A

minus B’, by

AzB : tx : x P A ^ x R B u . 17

y

y

x

x

A

B

Fig. 1: Two subsets A and B of the plane.

A

AÜB

AÝB

B

Fig. 2: Union and intersection of A and B. The last is given by the blue domain.

18

A\B

Fig. 3: The relative complement of B in A.

(iii) their cross (or Cartesian / direct) product A  B, read: ‘A cross B’, by A  B : tpx, y q : x P A ^ y P B u , where ordered pairs px1 , y1q, px2 , y2q are defined equal,

px1 , y1q  px2 , y2q , if and only if x1  x2 and y1  y2 . In addition, if C is some set, we define the Cartesian product of A, B and C by A  B  C : tpx, y, z q : x P A ^ py

P B ^ z P C qu

where ordered pairs px1 , y1, z1 q, px2 , y2 , z2 q are defined equal,

px1 , y1, z1q  px2 , y2, z2q , if and only if x1  x2 , y1  y2 and z1  z2 . In particular, we define R2 : R  R , R3 : R  R  R . 19

y 3

2

A

1

x

B

2 1

1

2

3

3

2 z

A´B 1

0

1

1 2 3

2

3

y

x

Fig. 4: Subsets A of the real line and B of the plane and their cross product.

20

x

Example 1.1.25.

t1, 2, 3, 5, 8, 13u Y t1, 3, 4, 7, 11, 18u  t1, 2, 3, 4, 5, 7, 8, 11, 13, 18u t1, 2, 3, 5, 8, 13u X t1, 3, 4, 7, 11, 18u  t1, 3u t1, 2, 3, 5, 8, 13uzt1, 2, 3, 5u  t8, 13u , t1, 2, 3, 5, 1uzt1u  t2, 3, 5u , t1, 2u  t1, 3, 4u  tp1, 1q, p1, 3q, p1, 4q, p2, 1q, p2, 3q, p2, 4qu . The naive Definition 1.1.16 of sets leads to paradoxa like the one of ZermeloRussel (1903): Assume that there is a set of all sets that don’t contain itself as an element: S : tx : x is a set ^ x R xu .

Since S is assumed to be a set, either S P S or S P S. From the assumption that S P S, it follows by the definition of S that S R S . Hence it follows that S R S. From S R S, it follows by the definition of S that S P S . Hence there is no such set. Bernard Russell also used a statement about a barber to illustrate this principle. If a barber cuts the hair of exactly those who do not cut their own hair, does the barber cut his own hair? So a more restrictive definition of sets is needed to avoid such contradictions. For this, we refer to books on set theory. In the following such paradoxa will not be important role because we don’t use the full generality of Definition 1.1.16 and deal with a far reduced class of sets. Problems 1) For each pair of sets, decide whether not the following sets are equal: A : t
2, 3u, B : t3,
2u Y φ, C : t
2, 3u Y tφu, D :

tx P R : x2
x
6  0u, E : tφ,
2, 3u, F : t
2, 3,
2u, G : t
2, φ, φ, 3u . 21

2) Simplify

t
2, 3u Y tt
2u, t3uu Y t
2, t3uu Y tt
2u, 3u . 3) Decide whether

t1, 3u P t1, 3, t1, 7u, t1, 3, 7uu . Justify your answer.

4) Let A : tφ, t1u, t1, 3u, t3, 4uu. Determine for each of the following statements whether it is true or false. a) 1 P A , b) t1u € A , c) t1u P A, d) t1, 3u € A , e) tt1, 3uu P A , f) φ P A , g) φ € A , h) tφu € A .

5) Give an example of sets A, B, C such that A A R C.

P B and B P C, but

6) Sketch the following sets A : tpx, y q P R2 : x

B : tpx, y q P R : 2x 2

1  0u ,

y 3y

5  0u ,

C : tpx, y q P R : x y  1u, D : tp0, 1qu , E : tp
1,
1qu, F : tp0,
1qu, G : tp
1, 0qu, H : tp2,
3qu , 2

2

2

I : tp
4, 1qu, J : tx P R :
4 ¤ x ¤ 2u, K : t0u, L : t1u

into a xy-diagram and calculate A X B, A X C, pA X B q X C, A X pB X C q, B XpJ  Lq, C XpJ  K q, AzB, B zA, B Y E, pC Y F qY G. 7) Let A, B and C be sets. Show that a) b) c) d) e)

If A € B and B € C, then A € C , AYB BYA , AXB BXA , A Y pB Y C q  pA Y B q Y C , A X pB X C q  pA X B q X C ,

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f) g) h) i)

A Y pB X C q  pA Y B q X pA Y C q , A X pB Y C q  pA X B q Y pA X C q , C zpA Y B q  pC zAq X pC zB q , C zpA X B q  pC zAq Y pC zB q .

1.1.3 Maps Definition 1.1.26. (Maps) Let A and B be non-empty sets. (i) A map (or mapping) f from A into B, denoted by f : A Ñ B, is an association which associates to every element of A a corresponding element of B. If B is a subset of the real numbers, we call f a function. We call A the domain of f . If f is given, we also use the notation Dpf q for the domain of f . (ii) For every x under f .

P A, we call f pxq the value of f at x or the image of x

(iii) For any subset A 1 of A, we call the set f pA 1 q containing all the images of its elements under f , f pA 1 q : tf pxq : x P A 1 u ,

(1.1.5)

the image of A 1 under f . In particular, we call f pAq the range or image of f . (iv) For any subset B 1 of B, we call the subset f
1 pB 1 q of A containing all those elements which are mapped into B 1 , f
1 pB 1 q : tx P A : f pxq P B 1 u , the inverse image of B 1 under f . In particular if f is a function, we call f
1 pt0uq  tx P A : f pxq  0u , the zero set of f .

23

(v) For any subset A 1 of A, we define the restriction of f to A 1 as the map f |A 1 : A 1 Ñ B defined by f |A 1 pxq : f pxq for all x P A 1 .

Example 1.1.27. Define f : Z Ñ Z by f pnq : n2 for all n P Z. Moreover, let g be the restriction of f to N. Calculate f pZq, f pt
2,
1, 0, 1, 2uq, f
1pt
1, 0, 1uq, f
1pt6uq, g
1 pt
1, 0, 1uq . Solution: f pZq  tn2 : n P Nu , f pt
2,
1, 0, 1, 2uq  t0, 1, 4u , f
1 pt
1, 0, 1uq  t
1, 0, 1u , f
1 pt6uq  φ , g
1 pt
1, 0, 1uq  t0, 1u . Example 1.1.28. Define f : Df (a) f pxq 

?

x

Ñ R and g : Dg Ñ R such that

2 for all x P Df

(b) g pxq  1{px2
xq for all x P Dg and such Df and Dg are maximal. Find the domains Df and Dg . Give explanations. Solution: In case (a) the inequality x

2¥0

p x ¥
2q

has to be satisfied in order that the square root is defined. Hence Df : tx P R : x ¥
2u

?

and f : Df Ñ R is defined by f pxq : x 2 for all x P Df . In case (b) the denominator has to be different from zero in order that the quotient is defined. Because of x2
x  xpx
1q  0  x P t0, 1u , 24

we conclude that Dg : tx P R : x  0 ^ x  1u

and that g : Dg

Ñ R is defined by gpxq : 1{px2
xq for all x P Dg .

Definition 1.1.29. (Graph of a map) Let A and B be some sets and f : A Ñ B be some map. Then we define the graph of f by: Gpf q : tpx, f pxqq P A  B : x P Au .

Example 1.1.30. Sketch the graphs of the functions f and g from Example 1.1.28. Solution: See Fig. 5 and Fig. 6. Example 1.1.31. Find the ranges of the functions in Example 1.1.28. Solution: Since the square root assumes only positive numbers, we conclude that f pDf q € ty : y ¥ 0u . Further for every y

P r0, 8q, it follows that a y2
2 2  y

and hence that

ty : y ¥ 0u € f pDf q and, finally, that f pDf q  ty : y ¥ 0u. Further, for x   0 or x ¡ 1, it follows that xpx
1q ¡ 0 and hence that g pxq ¡ 0. For 0   x   1, it follows that

2




¤ x
12
14  xpx
1q   0 and hence that g pxq ¥
4. Hence it follows that 1 4

ty : y ¡ 0u Y ty : y ¤
4u € gpDg q . Finally, for any real y such that py ¡ 0q _ py ¤
4q, it follows that 

g

1 2

and hence that g pDg q € ty : y

1 y

1 4



y

¡ 0u Y ty : y ¤
4u. 25

y 2

1.5

1

0.5

-2

-1

1

2

x

Fig. 5: Gpf q from Example 1.1.28.

y

4 2 -1

-0.5

0.5

1.5

-2 -4 -6 -8 -10

Fig. 6: Gpg q from Example 1.1.28.

26

2

x

Definition 1.1.32. (Injectivity, surjectivity, bijectivity) Let A and B be some sets and f : A Ñ B be some map. We define (i) f is injective (or one-to-one) if different elements of A are mapped into different elements of B, or equivalently if f pxq  f py q ñ x  y for all x, y P A. In this case, we define the inverse map f
1 as the map from f pAq into A which associates to every y P f pAq the element x P A such that f pxq  y. (ii) f is surjective (or onto) if every element of B is the image of some element(s) of A: f pAq  B . (iii) f is bijective (or one-to-one and onto) if it is both injective and surjective. In this case, the domain of the inverse map is the whole of B. Example 1.1.33. Let f and g be as in Example 1.1.27. In addition, define h : Z Ñ Z by hpnq : n 1 for all n P Z. Decide whether f, g and h are injective, surjective or bijective. If existent, give the corresponding inverse function(s). Solution: f is not injective (and hence also not bijective), nor surjective, for instance, because of f p
1q  f p1q  1 , 2 R f pAq . g is injective because if m and n are some natural numbers such that g pmq  g pnq, then it follows that 0  m2
n2 and hence that

mn

 pm
nqpm _ 27

m 
n

nq

and therefore, since g has as its domain the natural?numbers, that m  n. The inverse g
1 : g pAq Ñ A is given by g
1 plq  l for all l P g pAq. g is not surjective (and hence also not bijective), for instance, since 2 R g pAq. h is injective because if m and n are some natural numbers such that hpmq  hpnq, then it follows that 0m

1
pn

1q  m
n

and hence that m  n. h is surjective (and hence as a whole bijective) because for any natural n we have hpn
1q  n. The inverse function h
1 : Z Ñ Z is given by h
1 pnq  n
1 for all n P Z.

Theorem 1.1.34. Let A and B be sets and f : A Ñ B be a map. Further, define for every y P B the corresponding intersection Gf y by Gf y : Gpf q X tpx, y q : x P Au . Then

(i) f is injective if and only if Gf y contains at most one point for all y P B. (ii) f is surjective if and only if Gf y is non-empty for all y

P B.

(iii) f is bijective if and only if Gf y contains exactly one point for all y P B.

Proof. (i) The proof is indirect. Assume that there is y P B such that Gf y contains two points px1 , y q and px2 , y q. Then, since Gf y is part of Gpf q, it follows that y  f px1 q  f px2 q and hence, since by assumption x1  x2 , that f is not injective. Further, assume that f is not injective. Then there are different x1 , x2 P A such that f px1 q  f px2 q. Hence Gf f px1 q contains two different points px1 , f px1 qq and px2 , f px1 qq. (ii) If f is surjective, then for any y P B there is some x P A such that y  f pxq and hence px, y q P Gf y . On the other hand, if Gf y is non-empty for all y P B, then for every y P B there is some x P A such that px, y q P Gf y and hence, since Gf y is part of Gpf q, that y  f pxq. Hence f is surjective. (iii) is an obvious consequence of (i) and (ii). 28

y 2

1.5

0.5

-2

-1

1

2

x

-1

Fig. 7: Gpf q from Example 1.1.28 and parallels to the x-axis.

y

2 -1

-0.5

0.5

1.5

2

x

-4

-8 -10

Fig. 8: Gpg q from Example 1.1.28 and parallels to the x-axis.

29

Example 1.1.35. Apply Theorem 1.1.34 to investigate the injectivity of f and g from Example 1.1.28. Solution: Fig. 7, Fig. 8 suggest that f is injective, but not surjective and that g is neither injective nor surjective. Example 1.1.36. Show that f and the restriction of g to tx : x ¥ 1{2 ^ x  1u, where f and g are from Example 1.1.28, are injective and calculate their inverse. Solution: If x1 , x2 are any real numbers ¥
2 and such that f px1 q  f px2 q, then ? ? x1 2  x2 2 and hence x1

2  x2

2

and x1  x2 . Hence f is injective. Further, for every y in the range of f there is x ¥
2, such that ? y x 2 and hence

x  y2
2 .

Therefore

f
1 py q  y 2
2

for all y from the range of f . Further, if x1 and x2 are some real numbers ¥ 1{2 different from 1 and such that 1

x21 then


x1

 x2
1 x 2

px1
x2 qpx1

,

2

x2
1q  0

and hence x1  x2 . Hence the restriction of g to tx : x ¥ 1{2 ^ x  1u is injective. Finally, if y is some real number in the range of this restriction, then y is in particular different from zero and y

 x2 1
x , 30

hence



x


and

Therefore

x

1 2

1 2

f
1 py q 

2

 y1

1 2

1 y

1 4 1 . 4 1 y

1 4

for all y from the range of that restriction of g. Definition 1.1.37. (Composition) Let A, B, C and D be sets. Further, let f : A Ñ B and g : C Ñ D be maps. We define the composition g  f : f
1 pB X C q Ñ D (read: ‘g after f ’) by

pg  f qpxq : gpf pxqq x P f
1pB X C q . Note that g  f is trivial, i.e., with an empty domain, for instance, if B X C φ. Also note that f
1 pB X C q  A if B € C.



Example 1.1.38. Calculate f  f , h  h, f  h and h  f , where f , h are defined as in Example 1.1.27, Example 1.1.33, respectively. Solution: Obviously, all these maps map from Z into itself. Moreover for every n P Z:

pf  f qpnq  f pf pnqq  f pn2 q  pn2q2  n4 , ph  hqpnq  hphpnqq  hpn 1q  pn 1q 1  n 2 , ph  f qpnq  hpf pnqq  hpn2 q  n2 1 , pf  hqpnq  f phpnqq  f pn 1q  pn 1q2  n2 2n 1 . Note in particular that h  f  f  h. 31

Example 1.1.39. Let A and B be sets. Moreover, let f : A Ñ B be some injective map. Calculate f
1  f . Assume that f is also surjective (and hence as a whole bijective) and calculate also f  f
1 for this case. Solution: To every y P f pAq, the map f
1 associates the corresponding x P A which satisfies f pxq  y. In particular, it associates to f pxq the element x for all x P A. Hence f
1  f

 idA , f  f
1  idf pAq

where for every set C the corresponding map idC : C

Ñ C is defined by

idC pxq : C for all x P C. Further, if f is bijective, f pAq  B and hence f

 f
1  idB .

Theorem 1.1.40. (Graphs of inverses of maps) Let A and B be sets and f : A Ñ B be an injective map. Moreover, define R : X  Y Ñ Y  X by Rpx, y q : py, xq for all x P A and y

P B. Then the graph of the inverse map is given by Gpf
1 q  RpGpf qq . Proof. ‘€’: Let py, f
1py qq be an element of Gpf
1 q. Then y P f pAq and f
1 py q P A is such that f pf
1 py qq  y. Therefore pf
1 py q, y q P Gpf q and py, f
1pyqq  Rpf
1pyq, yq P RpGpf qq . ‘’: Let pf pxq, xq be some element of RpGpf qq. Then f
1 pf pxqq  x and hence

pf pxq, xq  pf pxq, f
1pf pxqq P Gpf
1q . 32

y 2

1

-2

-1

1

2

x

-1

-2

Fig. 9: Gpf q, Gpf
1 q from Example 1.1.28 and the reflection axis.

Example 1.1.41. Apply Theorem 1.1.40 to the graph of the function f from Example 1.1.28 to draw the graphs of its inverse. (See Example 1.1.36.) Solution: See Fig. 9.

Problems 1) Find f pr0, π {2sq , f
1 pt1uq , f
1 pt3uq , f
1 pr0, 2sq . In addition, find the maximal domain D € R that contains the point π {8 and is such that f |D is injective. Finally, calculate the inverse of the map h : D Ñ f pDq defined by hpxq : f pxq for all x P D. a) f pxq : 2 sinp3xq , x P R , b) f pxq : 3 cosp2xq , x P R , c) f pxq : tanpx{2q{3 , x P tx P pp2k
1qπ, p2k 1qπ q : k

P Zu . 1 for x P R

2) Define f : R Ñ R and g : Rzt
1u Ñ R by f pxq : x and g pxq : px 1q2 {px 1q for x P Rzt
1u. Is f  g?

33

y

y

1

y 1

2

1

-1

1

2

x

1

x

-1

-2

-1

1

2

-2

x

-1

Fig. 10: Subsets of R2 . Which is the graph of a function? 3) Let f : Df Ñ R be defined such that the given equation below is satisfied for all x P Df and such that Df € R is a maximal. In each of the cases, find the corresponding Df , the range of f , and draw the graph of f : a) b) c) d) e) f) g) h)

f pxq  x2
3 , ? f pxq  1{ x , f pxq  1{p1
xq , f pxq  x2 |x| , f pxq  x{|x| , f pxq  |x|1{3 , f pxq  |x2
1| , a f pxq  sinpxq .

4) Which of the subsets of R2 in Fig. 10 is the graph of a function? Give reasons. 5) Find the function whose graph is given by a) b) c)

px, yq P R2 : x2 y x px, yq P R2 : x  y{py px, yq P R2 : y2 6xy

(

10 , (

1q , 9x2

0

(

.

6) In each of the following cases, find a bijective function that has domain D and range R and calculate its inverse.

 tx P R : 1 ¤ x ¤ 2u, R  tx P R : 3 ¤ x ¤ 7u ,  tx P R :
1 ¤ x ¤ 1u, R  tx P R : x ¥ 3u . Define f : Df Ñ R and g : Dg Ñ R such that a x
1 x2
9 , g pxq : 2 f pxq : x
3

a) D b) D 7)

a)

34

for all x P Df , x P Dg , respectively, and such Df and Dg are maximal. Find the domains and ranges of the functions f and g. Give explanations. b) If possible, calculate pf  g qp5q and pg  f qp5q. Give explanations. c) f is injective (= ‘one to one’). Calculate its inverse. 8) Is there a function which is identical to its inverse? Is there more then one such function?

9) Define f : R Ñ R, g : R Ñ R and h : R Ñ R by f pxq : 1

x , g pxq : 1

x

x2 , hpxq : 1
x

for every x P R. Calculate

10)

pf  f qpxq , pf  gqpxq , pg  f qpxq , pg  gqpxq , pf  hqpxq , ph  f qpxq , pg  hqpxq , ph  gqpxq , ph  hqpxq , rf  pg  hqspxq , rpf  gq  hspxq for every x P R. Define f : R Ñ R, g : R Ñ R and h : tx P R : x ¡ 0u Ñ R by f pxq : x a , g pxq : ax , hpxq : xa for every x in the corresponding domain, where a P R. For each of these functions and every n P N , determine the n-fold composition with itself.

11) Define f : R Ñ R by f pxq : r 1

p2
xq1{3 s1{7 , gpxq : cosp2xq

for every x P R. Express f and g as a composition of four functions, none of which is the identity function. In addition, in the case of g, the sine function should be among those functions.

12) Let A and B be sets, f : A Ñ B and B1 , B2 be subsets of B. Show that f
1 pB1 Y B2 q  f
1 pB1 q Y f
1 pB2 q ,

f
1 pB1 X B2 q  f
1 pB1 q X f
1 pB2 q . 13) Express the area of an equilateral triangle as a function of the length of a side.

35

14) Express the surface area of a sphere of radius r its volume.

¡ 0 as a function of

15) Consider a circle Sr1 of radius r ¡ 0 around the origin of an xydiagram. Express the length of its intersections with parallels to the y-axis as a function of their distance from the y-axis. Determine the domain and range of that function.

16) From each corner of a rectangular cardboard of side lengths a ¡ 0 and b ¡ 0, a square of side length x ¥ 0 is removed, and the edges are turned up to form an open box. Express the volume of the box as a function of x and determine the domain of that function.

17) Consider a body in the earth’s gravitational field which is at rest at time t  0 and at height s0 ¡ 0 above the surface. Its height s and speed v as a function of time t are given by 1 sptq  s0
gt2 , v ptq 
gt , 2

where g is approximately 9.81m{s2 . Determine the domain and range of the functions s and v. In addition, express s as a function of the speed and determine domain and range.

1.2 Limits and Continuous Functions 1.2.1 Limits of Sequences of Real Numbers For the study of local properties of real-valued functions, like continuity and differentiability, we need to develop the notion of limits of sequences of real numbers. Definition 1.2.1. Let x1 , x2 , . . . be a sequence of elements of R and x P R. Then we define lim xn  x n

Ñ8

if for every ε ¡ 0, there is a corresponding n0 such that for all n ¥ n0 :

|xn
x|   ε ,

i.e., from the n0 -th member on, all remaining members of the sequence are within a distance from x which is less than ε. 1 In this case, we say that the 1

As a consequence, only finitely many members have distance ¥ ε from x.

36

2 1.75 1.5 1.25 1 0.75 0.5 0.25 10

Fig. 11: pn, pn

20

30

40

50

n

1q{nq for n  1 to n  50 and asymptotes.

sequence x1 , x2 , . . . is convergent to x. Note that in this case for ε ¡ 0

|xn|  |xn
x

x| ¤ |xn
x|

|x| ¤ ε |x|

for all n P N , apart from finitely many members of the sequence, and hence that x1 , x2 , . . . is bounded, i.e., that there is M ¥ 0 such that |xn | ¤ M for all n P N . In case that the sequence is not convergent to any real number, we say that it is divergent. Example 1.2.2. Let a be some real number and xn : a for all n Then lim xn  a . n

P

N.

Ñ8

Example 1.2.3. Investigate whether the following limits exist. (i) lim

n

Ñ8

37

1

n n

(1.2.1)

2 1.5 1 0.5 10

20

30

40

50

n

-0.5 -1 -1.5 -2

Fig. 12: pn, p
1qn pn

1q{nq for n  1 to n  50 and asymptotes.

50

40

30

20

10

10

Fig. 13: pn, pn2

20

30

40

50

n

1q{nq for n  1 to n  50 and an asymptote.

38

(ii)

lim p
1qn 

n

1

n

Ñ8

(iii) lim

n

n

n2

Ñ8

1

,

(1.2.2)

.

n

(1.2.3)

Solution: Fig. 11, Fig. 12 and Fig. 13 suggest that the limit 1.2.1 is 1, whereas the limits 1.2.2, 1.2.3 don’t exist. Indeed n

1

n

lim

Ñ8

n

1.

(1.2.4)

For the proof, let ε be some real number ¡ 0. Further, let n0 be some natural number ¡ 1{ε. Then it follows for every n P N such that n ¥ n0 :  n 1   n

  1


 n1 ¤ n1   ε . 0

and hence the statement (1.2.4). The proof that (1.2.2) does not exist proceeds indirectly. Assume on the contrary that there is some x P R such lim p
1qn 

n

Then there is some n0

Ñ8

n

P N such that    

p
1q  n

1

n

1

n

n

x.

  x


  14

for all n P N such n ¥ n0 . Without restriction of generality, we can assume that n0 ¥ 4. Then it follows for any even n P N such that n ¥ n0 :

|x
1| 

 n 1   n

¤ 14

 12

1 4


x






1   n 1 ¤ n  n

39




  x

1 n

¤ 14

1 n0

and for any odd n P N such that n ¥ n0 :

|x

1| 

¤ 14

1 4

  n 1   n





x





1   n 1 ¤
n n 




  x

1 n

¤ 14

1 n0

 12 ,

and hence we arrive at the contradiction that 2  |x
1
px

1q| ¤ |x
1|

|x

1| ¤

1 2

1 2

1.

Hence our assumption that (1.2.2) exists is false. The proof that (1.2.3) does not exist proceeds indirectly, too. Assume on the contrary that there is some x P R such n2 1 lim x. nÑ8 n Further, let ε be some real number ¡ 0. Finally, let n0 be some natural number ¥ |x| ε. Then it follows for n ¥ n0 that  2 n 1   n

  x

  n 



x



1  n
x n

1 n

¡ n
x ¥ |x|

ε
xε .

Hence there is an infinite number of members of the sequence that have a distance from x which is greater than ε. This is a contradiction to the existence of a limit of (1.2.3). Hence such a limit does not exist. The alert reader might have noticed that Def 1.2.1 might turn out to be inconsistent with logic, and then would have to be abandoned, if it turned out that some sequence has more than one limit point. Part piq of the following Theorem 1.2.4 says that this is impossible. In particular, this theorem says that a sequence in R can have at most one limit point (in part (i)), that the sequence consisting of the sums of the members of convergent sequences in R is convergent against the sum of their limits (in part (ii)), that the sequence consisting of the products of the members of convergent sequences in R is convergent against the product of 40

their limits (in part (iii)), and that the sequence consisting of the inverse of the members of convergent sequences in R is convergent against the inverse of that limit if it is different from zero (in part (iv)). Theorem 1.2.4. (Limit Laws) Let x1 , x2 , . . . ; y1 , y2, . . . be sequences of elements of R and x, x¯, y P R. (i) If then x¯  x.

lim xn

Ñ8

 x and

n

lim xn

 x and

n

n

(ii) If n

then

Ñ8

Ñ8

(iii) If lim xn

n

Ñ8

then

 x and

lim xn  yn

n

Ñ8

(iv) If lim xn

n

Ñ8

then

Ñ8

 x¯ ,

lim yn

y ,

Ñ8

yn q  x

lim pxn

n

lim xn

y.

lim yn

n

Ñ8

y ,

xy .

 x and x  0 ,

1 nÑ8 xn lim

 x1 .

Proof. ‘(i)’: The proof is indirect. Assume that the assumption in (i) is true and that x  x¯. Then there is n0 P N such that for n P N satisfying n ¥ n0 :

|xn
x|   12 |x¯
x|

and |xn
x¯|  

1 |x¯
x| . 2

Hence it follows the contradiction that

|x¯
x|  |x¯
xn

xn
x| ¤ |x¯
xn | 41

|xn
x|   |x¯
x| .

Hence it follows that x¯  x. ‘(ii)’: Assume that the assumption in (ii) is true. Further, let ε ¡ 0. Then there is n0 P N such that for n P N with n ¥ n0 : |xn
x|   2ε and |yn
y|   2ε and hence

|xn

yn
px

y q| ¤ |xn
x|

|yn
y|   ε .

‘(iii)’: Assume that the assumption in (iii) is true. Further, let ε ¡ 0 and δ ¡ 0 such that δ pδ |x| |y |q   ε. (Obviously, such a δ exists.) Then there is n0 P N such that for n P N with n ¥ n0 :

|xn
x|   2δ

and |yn
y |  

δ . 2

Then

|xn  yn
x  y|  |xn  yn
xn  y xn  y
x  y| ¤ |xn|  |yn
y| |xn
x|  |y| ¤ |xn
x|  |yn
y| |x|  |yn
y| |xn
x|  |y|   ε . ‘(iv)’: Assume that the assumption in (iv) is true. Further, let ε ¡ 0 and δ ¡ 0 such that 1{p|x|p|x|
δ qq   mint|x|, εu. (Obviously, such a δ exists.) Then there is n0 P N such that for n P N satisfying n ¥ n0 : ||xn|
|x| | ¤ |xn
x|   δ , and hence also and

|xn| ¡ |x|
δ ¡ 0   1  x

n






1  |xn
x|  x  |xn |  |x|

  p|x||x
n
δqx||x|   ε .

42

Theorem 1.2.5. Let x1 , x2 , . . . and y1 , y2, . . . be sequences of real numbers such that |xn| ¤ yn

for all n P N. Further, let

lim yn

Ñ8

0.

lim xn

0.

n

Then n

Proof. Let ε that

Ñ8

¡ 0. Since y1, y2, . . . is convergent to 0, there is n0 P N such |xn| ¤ yn  |yn|   ǫ

for all n ¥ n0 . Hence it follows that x1 , x2 , . . . is convergent to 0. Theorem 1.2.6. (Limits preserve inequalities) Let x1 , x2 , . . . and y1 , y2, . . . be sequences of elements of R converging to x, y P R, respectively. Further let xn ¤ yn for all n P N. Then also x ¤ y. Proof. The proof is indirect. Assume on contrary that x follows the existence of an n P N such that, both, x
xn

¤ |xn
x|   12 px
yq

and

yn
y

¡

y. Then it

¤ |yn
y|   12 px
yq .

and hence the contradiction x
y

¤x
y

yn
xn

 x
y

.

Hence x ¤ y. Definition 1.2.7. (Cauchy sequences) We call a sequence x1 , x2 , . . . of real numbers a Cauchy sequence if for every ε ¡ 0 there is a corresponding n0 P N such that |xm
xn |   ε

for all m, n P N satisfying m ¥ n0 and n ¥ n0 . 43

x 1

0.8

0.6

0.4

0.2

10

20

30

40

50

n

Fig. 14: (n, xn ) from Example 1.2.8 for n  1 to n  50.

Example 1.2.8. Define x1 : 0, x2 : 1 and xn

2

:

1 pxn 2

xn

1

q

for all n P N . Show that x1 , x2 , . . . is a Cauchy sequence. Solution: First, it follows for every n P N that xn 2 is the midpoint of the interval In between xn and xn 1 given by In  [xn , xn 1 ] if xn ¤ xn 1 and In  [xn 1 , xn ] if xn ¡ xn 1 . Further, xn

2


xn 1  12 pxn

xn

1

q
xn 1 
12 pxn 1
xnq .

Hence it follows by the method of induction that I1 xn

1

p
1qn
1 


xn

2n
1

As a consequence, if ε ¡ 0 and n0 follows for m, n P N satisfying m

 I2  I3 . . . and that

.

P N is such that 21
n   ε, then it ¥ n0 and n ¥ n0 that xm P In and

44

0

0

therefore that

|xm
xn | ¤ 2n1
1   ε . 0

Hence x1 , x2 , . . . is a Cauchy sequence. See Fig. 14. Theorem 1.2.9. Every convergent sequence of real numbers is a Cauchy sequence. Proof. For this, let x1 , x2 , . . . be a sequence of real numbers converging to some x P R and ε ¡ 0. Then there is n0 P N such that

|xn
x|   ε{2 for all n P N satisfying n ¥ n0 . The last implies that

|xm
xn |  |xm
x
pxn
xq| ¤ |xm
x| |xn
x|   ε for all n, m P N satisfying n ¥ n0 and m ¥ n0 . Hence x1 , x2 , . . .

is a

Cauchy sequence.

That every Cauchy sequence of real numbers is convergent is a deep property of the real number system. This is proved in the Appendix, see the proof of Theorem 4.1.11 in the framework of Cantor’s (1872) construction of the real number system by completion of the rational numbers using Cauchy sequences. Theorem 1.2.10. (Completeness of the real numbers) Every Cauchy sequence of real numbers is convergent. Proof. See the proof of Theorem 4.1.11 in the Appendix. In the following, we derive far reaching consequences of the completeness of the real numbers. Theorem 1.2.11. (Bolzano-Weierstrass) For every bounded sequence x1 , x2 , . . . of real numbers there is a subsequence, i.e., a sequence xn1 , xn2 , . . . that corresponds to a strictly increasing sequence n1 , n2 , . . . of non-zero natural numbers, which is convergent. 45

Proof. For this let x1 , x2 , . . . be a bounded sequence of real numbers. Then we define S : tx1 , x2 , . . . u . In case that S is finite, there is a subsequence x1 , x2 , . . . which is constant and hence convergent. In case that S is infinite, we choose some element xn1 of the sequence. Since S is bounded, there is a ¡ 0 such that S € I1 : r
a{4, a{4s. At least one of the intervals r
a{4, 0s, r0, a{4s contains infinitely many elements of S. We choose such interval I2 and xn2 P I2 such that n2 ¡ n1 . In particular I2 € I1 . Bisecting I2 into two intervals, we can choose a subinterval I3 € I2 containing infinitely many elements of S and xn3 P I3 such that n3 ¡ n2 . Continuing this process, we arrive at a sequence of intervals I1 , I2 , . . . such that I1  I2  . . . and such that the length of Ik is a{2k for every k P N . Also, we arrive at a subsequence xn1 , xn2 , . . . of x1 , x2 , . . . such that xk P Ik for every k P N . For given ε ¡ 0, there is k0 P N such that a{2k0   ε. Further, let k, l P N be such that k ¥ k0 and l ¥ k0 . Then it follows that xk P Ik0 , xl P Ik0 and therefore that |xk
xl | ¤ a{2k0   ε . Hence xn1 , xn2 , . . . is a Cauchy sequence and therefore convergent according to Theorem 1.2.10. For the following, the Bolzano-Weierstrass theorem will be fundamental. In particular, it will be applied in the proofs of the important theorems, Theorem 1.2.26, Theorem 1.2.35 and Theorem 2.5.57. Theorem 1.2.12. Let x1 , x2 , . . . be an increasing sequence of real numbers, i.e., such that xn ¤ xn 1 for all n P N, which is also bounded from above, i.e., for which there is M ¥ 0 such that xn ¤ M for all n P N. Then x1 , x2 , . . . is convergent. Proof. Since x1 , x2 , . . . is increasing and bounded from above, it follows that this sequence is also bounded. Hence according to the previous theorem, there is a subsequence, i.e., a sequence xn1 , xn2 , . . . that corresponds

46

to a strictly increasing sequence n1 , n2 , . . . of non-zero natural numbers, which is convergent. We denote the limit of such sequence by x. Then,

¤x for all n P N . Otherwise, there is m P N such that xm ¡ x. If nk P N is such that nk ¥ m, then xn ¥ xm ¡ x for all k P N such that k ¥ k0 . This implies that lim xn ¥ xm ¡ x . k Ñ8 xn

0

0

k

k

Further, for ε ¡ 0, there is k0 such that

|xn
x|   ε k

for all k P N such that k n ¥ nk0 that

¥ k0. Hence it follows for all n P N satisfying

|xn
x|  x
xn ¤ x
xn  |xn
x|   ε . k0

k0

Therefore, x1 , x2 , . . . is convergent to x. Corollary 1.2.13. Let x1 , x2 , . . . be an decreasing sequence of real numbers, i.e., such that xn 1 ¤ xn for all n P N, which is also bounded from below, i.e., for which there is a real M ¥ 0 such that xn ¥ M for all n P N. Then x1 , x2 , . . . is convergent. Proof. The sequence
x1 ,
x2 , . . . is increasing, bounded from above and therefore convergent to a real number x by the previous theorem. Hence x1 , x2 , . . . is convergent to
x.

47

x 0.5

0.4

0.3

0.2

0.1

10

20

30

40

50

n

Fig. 15: (n, xn ) from Example 1.2.14 for n  1 to n  50.

Example 1.2.14. Show that the sequence x1 , x2 , . . . defined by x1 : 1{2 and 1  3 . . . p2n
1q xn : 2  4 . . . p2nq

for all n P N zt1u is convergent. Solution: The sequence x1 , x2 , . . . is bounded from below by 0. In addition, xn

1

 22n pn

1 xn 1q

¤ xn

for all n P N and hence x1 , x2 , . . . is decreasing. Hence x1 , x2 , . . . is convergent according to Corollary 1.2.13. See Fig 15. Definition 1.2.15. Let S be a non-empty subset of R. We say that S is bounded from above (bounded from below) if there is M P R such that x ¤ M (x ¥ M) for all x P S.

48

Theorem 1.2.16. Let S be a non-empty subset of R which is bounded from above (bounded from below). Then there is a least upper bound (largest lower bound) of S which will be called the supremum of S (infimum of S) and denoted by sup S (inf S). Proof. First, we consider the case that S is bounded from above. For this, we define the subsets A, B of R as all real numbers that are no upper bounds of S and containing all upper bounds of S, respectively, A : ta P R : There is x P S such that x ¡ au , B : tb P R : x ¤ b for all x P S u . Since S is non-empty and bounded from above, these sets are non-empty. In addition, for every a P A and every b P B, it follows that a   b. Let a1 P A and b1 P B. Recursively, we construct an increasing sequence a1 , a2 , . . . in A and a decreasing sequence b1 , b2 , . . . in B by an bn

1

1

: :

#

#

pan

bn q{2

an bn pan

bn q{2

if pan if pan

if pan if pan

bn q{2 P A bn q{2 P B ,

bn q{2 P A bn q{2 P B

for every n P N . According to Theorem 1.2.12, both sequences are convergent to real numbers a and b, respectively. Since,

 pb1
a1 q{2n
1 for all n P N , it follows that a  b. In the following, we show that b  sup S. For every x P S, it follows that x   bn for all n P N and hence that x ¤ b. Hence b is an upper bound of S. Let ¯b be an upper bound of S such that ¯b   b. Then there is n P N such that ¯b   an . Since an is no bn
an

upper bound for S, the same is also true for ¯b. Therefore, b is the smallest upper bound of S, i.e., b  sup S. Finally, we consider the case that S is bounded from below. Then
S : t
x : x P S u is bounded from above. Obviously, a real number a is a lower bound of S if and only if
a is an upper bound of
S. Hence
supp
S q is the largest lower bound of S, i.e., inf S exists and equals
supp
S q. 49

Example 1.2.17. Prove that there is a real number x such that x2 Solution: For this, we define S : ty



2.

P R : 0 ¤ y 2 ¤ 2u .

Since 0 P S, S is a non-empty. Further, S does not contain real numbers y ¥ 2 since the last inequality implies that y 2
2  py
2qpy 2q 2 ¥ 2. Hence S is bounded from above. We define x : sup S. In the following, we prove that x2  2 by excluding that x2   2 and that x2 ¡ 2. First, we assume that x2   2. Then it follows for n P N that 

x

1 n

2

 x2
2 Hence if n ¥ (2x


2  x2
2 2x

1 n

2x n

1 n2

¤ x2
2

2x n

1 n

.

1){(2
x2 ) it follows that 

x

1 n

2

¤2

and therefore that x (1{n) P S. As a consequence, x is no upper bound for S. Second, we assume that x2 ¡ 2. Then it follows for ε ¡ 0 that

px
εq2
2  x2
2
2εx ε2 ¥ x2
2
2εx . Hence if ε   (x2
2){(2x), it follows that px
εq2 ¡ 2 . As a consequence, x is not the smallest upper bound for S. Finally, it follows that x2  2. Note that according to Example 1.1.15, x is no rational number. In the following, we define the exponential function using sequences using an elementary approach by O. Dunkel, 1917 [14] which does not use Bernoulli’s equation. 50

2.74

2.73

2.72

2.71

2

4

6

8

10

12

14

n

Fig. 16: pn, xn q, pn, yn q from Lemma 1.2.18 and pn, eq for n  1 to n  15.

Lemma 1.2.18. Let x P R. Define


xn : 1


x p2n q x
p2n q , y :  1
n 2n 2n

for all n P Z. Then for all n P N such 2pn
1q

¡ |x|: 0   xn
1 ¤ xn ¤ yn ¤ yn
1   xn  y

and

n

  1

2

x
¤ 4m 2

.

Proof. For this let n P N be such that m : 2pn
1q


1


1


x 2 2m x 2 2m

1 1


x m x m 51

(1.2.5)

x2 4m2 x2 4m2

¥1 ¥1


(1.2.6)

¡ |x|. Then x m x m

¡0, ¡0

and hence

0   xn
1

¤ xn and 0   yn ¤ yn
1 .

Finally, it follows that yn
xn

 






1


x
2m

1 2m #

x 2m 2m +







x 2 x
2m 1
1
2m 2m




x
2m x2  1
1
2m 4m2

1
1






and hence xn



1


x2 4m2

0





1


x2 4m2

2m



1





1


x2 4m2

2m
1 

¤ yn and (1.2.6).

Note that the sequence y1 , y2, . . . in Lemma 1.2.18 is a decreasing and bounded from below by 0 and hence convergent according to Theorem 1.2.13. Hence we can define the following: Definition 1.2.19. We define the exponential function exp : R Ñ R by


exppxq : ex : lim 1
n

Ñ8

x
p2n q 2n

for all x P R. Then we conclude Theorem 1.2.20. (i) x

e and ex

¡ 0 for all x P R.

 nlim Ñ8

52




1

x p2n q 2n

(ii)


1

x p2n q 2n

x¤ 1

¤e ¤ x




1


for all x P R such |x|   1 and all n P N. (iii) for all x, y

ex

P R.

y

x
p2n q 2n

¤ 1
1 x

(1.2.7)

 exey

Proof. From (1.2.6), it follows for every x P R: xn lim 1 nÑ8 yn and hence by the limit laws Theorem (1.2.4) that xn lim yn  lim  nlim x nÑ8 nÑ8 yn Ñ8 n

and by (1.2.5) and Theorem 1.2.6 that ex ¡ 0 for all x P R. Further, if |x|   1, it follows from (1.2.5) and by Theorem 1.2.6 the estimates (1.2.7). Finally, if y P R and n P N is such that m : 2n ¡ maxt4|x|, 4|y |, 2|x||y |u, then  

m  y m x m 1 m 1 m hm m  1 m 1 xmy where

hm :

xy m x is such that |hm |   1. Hence by (1.2.7) 1

hm

¤



1

hm m

m

y

¤ 1
1h

,

m

and it follows by Theorem 1.2.4 and Theorem 1.2.6 that ex ey ex y

 nlim Ñ8



1

53

hm m

m

1.

Problems 1) Below are given the first 8 terms of a sequence x1 , x2 , . . . . For each find a representation xn  f pnq, n  1, . . . , 8, where f is an appropriate function. a) b) c) d) e) f) g) h) i) j) k) l)

2, 4, 6, 8, 10, 12, 14, 16, 2, 4, 8, 16, 32, 64, 128, 256,
1, 1,
1, 1,
1, 1,
1, 1, 1, 3, 6, 10, 15, 21, 28, 36,
1, 3{4,
5{7, 7{10,
9{13, 11{16,
13{19, 15{22, 2, 0, 2, 0, 2, 0, 2, 0, 5{7, 0, 7{9, 0, 9{11, 0, 11{13, 0, 1, 1, 4{6, 8{24, 16{120, 32{720, 64{5040, 128{40320, 0, 1, 0,
1, 0, 1, 0,
1, 0, 1, 0, 1, 0,
1, 0,
1, 0, 1, 0, 0, 0,
1, 0, 1, [0, 0, 0,
1,] 0, 1, 0, 0, 0, 1, 0, 1, [0, 0, 0, 1].

2) Prove the convergence of the sequence and calculate its limit. For this use only the limit laws, the fact that a constant sequence converges to that respective constant and the fact that lim p1{nq  0 .

Ñ8

n

Give details. a) b) c) d) e) f) g)

xn xn xn xn xn xn xn

: 1 p1{nq, n P N , : 5 p
2q p1{nq 3 p1{nq2 , n P N , : r1 p
4q p1{nqs{r2 3 p1{nq2 s, n P N , : 3{n2 , n P N , : p2n
1q{pn 3q, n P N , : p3n2
6n
10q{p7n2 3n
5q, n P N , : p3n2
6n
10q{p7n3 3n
5q, n P N .

3) Determine in each case whether the given sequence is convergent or divergent. Give reasons. If it is convergent, calculate the limit.

54

a)

xn :

c)

xn : p
1qn 1


e)

p
1qn

b)

xn :

d)

xn :

xn : sinpnπ q

f)

xn : sin

g)

xn :

h)

xn :

n2 n2 1

i)

xn :

j)

xn :

n2 n n3 1

n 1 n 1 n



n n2 1 n3 n2 1

for every n P N .

n

1

p
1qn n

nπ 2



cospnπ q




4) The table displays pairs pn, sn q, n  1, . . . , 10, where sn is the measured height in meters of a free falling body over the ground after n{10 seconds and at rest at initial height 4m.

p1, 3.951q p2, 3.804q p3, 3.559q p4, 3.216q p5, 2.775q p6, 2.236q p7, 1.599q p8, 0.864q

.

Draw these points into an xy-diagram, where the values of n appear on the x-axis and the values of sn on the y-axis. Find a representation sn  f pn{10q, n  1, . . . , 10, where f is an appropriate function, and predict the time when the body hits the ground.

5) The table displays pairs p2n{10, Lnq, n  1, ..., 8, where 2n{10 is the pressure in atmospheres (atm) of an ideal gas (, at constant temperature of 20 degrees Celsius,) confined to a volume which is proportional to the length Ln . The last is measured in millimeters (mm).

p0.2, 672q p0.4, 336q p0.6, 224q p0.8, 168q p1.0, 134.4q p1.2, 112q p1.4, 96q p1.6, 84q Draw these points into an xy-diagram, where the values of n appear on the x-axis and the values of Ln on the y-axis. Find a representation Ln  f p2n{10q, n  1, . . . , 8, where f is an appropriate function, and predict L10 .

1.2.2 Continuous Functions The next defines the continuity of function at a point by its property to commute with limits taken at that point. 55

Definition 1.2.21. (Sequential continuity) Let f : D Ñ R be a function and x P D. Then we say f is continuous in x if for every sequence x1 , x2 , . . . of elements in D from lim xν

ν

Ñ8

it follows that lim f pxν q  f

ν

x




Ñ8



lim xν

ν

Ñ8

r f pxqs .

If f is not continuous in x, we say f is discontinuous in x. Also we say f is continuous if f is continuous in all points of its domain D. Example 1.2.22. (Basic examples for continuous functions.) Let a, b be real numbers and f : R Ñ R be defined by f pxq : ax

b

for all x P R. Then f is continuous. Proof. Let x be some real number and x1 , x2 , . . . be a sequence of real of numbers converging to x. Then for any given ε ¡ 0, there is n0 P N such that for n P N with n ¥ n0 :

|a|  |xn
x|   ε and hence also that

|f pxnq
f pxq|  |axn and

b
pax

bq|  |axn
ax|  |a|  |xn
x|   ε

lim f pxn q  f pxq .

n

Ñ8

An example for a function which is discontinuous in one point.

56

Example 1.2.23. Consider the function f : R Ñ R defined by f pxq : for x  0 and f p1q : 1. Then 1 lim nÑ8 n 1 n

lim f

n

Ñ8



1  0 and nlim
Ñ8 n



but

x |x|

 1 and



lim f

n

Ñ8




n1

0,


1 .

Hence f is discontinuous at the point 1. See Fig. 17. Such discontinuity is called a ‘jump discontinuity’. An example for a function which is nowhere continuous is the following. discontinuous Example 1.2.24. (A function which is nowhere continuous) Define f : R Ñ R by # 1 if x is rational f pxq : 0 if x is irrational for every x P R. For the proof that f is everywhere discontinuous, let xP? R. Then x is either rational or irrational. If x is?rational, then xn : 2{n for every n P N is irrational. (Otherwise, 2  npxn
xq is a x rational number. ) Hence lim f pxn q  0  1  f pxq ,

n

Ñ8

and f is discontinuous in x. If x is irrational, by construction of the real number system, see Theorem 4.1.11 (i) in the Appendix, there is a sequence of rational numbers x1 , x2 , . . . that is convergent to x. Hence lim f pxn q  1  0  f pxq ,

n

Ñ8

and f is discontinuous in x also in this case. 57

Remark 1.2.25. The statement lim f pxq  b ,

x

Ña

for some real-valued function f , x varying in the real domain of f and some real numbers a and b, is per definition equivalent to the statement: For every sequence x1 , x2 , x3 , . . . in the domain of f converging to a it follows that lim f pxn q  b . n

Ñ8

In addition, the statement lim f pxq  b

x

Ñ8

is per definition equivalent to the statement: For every sequence x1 , x2 , x3 , . . . in the domain of f which is such that for every n P N there are only finitely many members ¤ n, it follows that lim f pxn q  b .

n

Finally, the statement x

Ñ8

lim f pxq  b ,

Ñ
8

is per definition equivalent to the statement: For every sequence x1 , x2 , x3 , . . . in the domain of f which is such that for all n P N there are only finitely many members ¥
n, it follows that lim f pxn q  b .

n

Ñ8

One main application of continuous functions is given by the following: Theorem 1.2.26. (Existence of maxima and minima of continuous functions on compact intervals) Let f : ra, bs Ñ R be a continuous function 58

y

0.5

-1

-0.5

1

0.5

x

-0.5

Fig. 17: Graph of f from Example 1.2.23.

y

0.4

0.3

0.2

0.1

0.2

0.4

0.6

0.8

1

Fig. 18: Graph of f from Example 1.2.27.

59

x

where a and b are real numbers such that a   b. Then there is x0 such that f px0 q ¥ f pxq p f px0 q ¤ f pxq q

P ra, bs

for all x P ra, bs.

Proof. For this, in a first step, we show that f is bounded and hence that sup f pra, bsq exists. In the final step, we show that there is c P ra, bs such that f pcq  sup f pra, bsq. For this, we use the Bolzano-Weierstrass theorem. The proof that f is bounded is indirect. Assume on the contrary that f is unbounded. Then there is a sequence x1 , x2 , . . . such that f pxn q ¡ n

(1.2.8)

for all n P N. Hence according to Theorem 1.2.11, there is a subsequence xk1 , xk2 , . . . of x1 , x2 , . . . converging to some element c P ra, bs. Note that the corresponding sequence is f pxk1 q, f pxk2 q, . . . is not converging as a consequence of (1.2.8). But, since f is continuous, it follows that f pcq  lim f pxnk q k

Ñ8

Hence f is bounded. Therefore let M : sup f pra, bsq. Then for every n P N there is a corresponding cn P ra, bs such that

|f pcnq
M |   n1 .

(1.2.9)

Again, according to Theorem 1.2.11, there is a subsequence ck1 , ck2 , . . . of c1 , c2 , . . . converging to some element c P ra, bs. Also, as consequence of (1.2.9), the corresponding sequence f pck1 q, f pck2 q, . . . is converging to M and by continuity of f to f pcq. Hence f pcq  M and by the definition of M: f pcq  M ¥ f pxq for all x P ra, bs. By applying the previous reasoning to the continuous function
f , it follows the existence of a c 1 such that


f pc 1q ¥
f pxq 60

and hence also

f pc 1 q ¤ f pxq

for all x P ra, bs.

A simple example of a function which is discontinuous in one point and does not assume a maximal value is: Example 1.2.27. Define f : r0, 1s Ñ R by f pxq :

"

if 0 ¤ x   1{2 if 1{2 ¤ x ¤ 1 .

1 x2 px
1q2

See Fig. 18. Another important application is given by the so called ‘intermediate value theorem’: Theorem 1.2.28. (Intermediate value theorem) Let f : ra, bs Ñ R be a continuous function, where a and b are real numbers such that a   b. Further, let f paq   f pbq and γ P pf paq, f pbqq. Then there is x P pa, bq such that f pxq  γ . Proof. Define

S : tx P ra, bs : f pxq ¤ γ u .

Then S is non-empty, since a P S, and bounded from above by b. Hence c : sup S exists and is contained in ra, bs. Further, there is a sequence x1 , x2 , . . . in S such that |xn
c| ¤ n1 (1.2.10) for all n P N. Hence x1 , x2 , . . . is converging to c, and it follows by the continuity of f that lim f pxn q  f pcq . n

Ñ8

Moreover, since f pxn q ¤ γ for all n P N, it follows that f pcq ¤ γ. As a consequence, c  b. Now for every x P pc, bs, it follows that f pxq ¡ 61

y 3

2

1

-1

-0.5

0.5

1

x

-1

Fig. 19: Graph of f from Example 1.2.30.

γ because otherwise c is not an upper bound of S. Hence there exists a sequence y1 , y2 , . . . in pc, bs which is converging to c. Further, because of the continuity of f lim f pyn q  f pcq n

Ñ8

and hence f pcq ¥ γ. Finally, it follows that f pcq that c  a and c  b.

 γ and therefore also

Corollary 1.2.29. Let f : ra, bs Ñ R be a continuous function where a and b are real numbers such that a   b. Moreover, let f paq   0 and f pbq ¡ 0. Then there is x P pa, bq such that f pxq  0. Example 1.2.30. Define f : R Ñ R by f pxq : x3

x

1

for all x P R. Then by Theorems 1.2.37, 1.2.39 below, f is continuous. Also, it follows that f p
1q 
1   0 and f p0q  1 ¡ 0 62

and hence by Corollary 1.2.29 that f has a zero in p
1, 0q. See Fig. 19. Remark 1.2.31. Note in the previous example that the value (0.375) of f in the mid point
0.5 of r
1, 0s is ¡ 0. Hence it follows by Corollary 1.2.29 that there is a zero in the interval r
1,
0.5s. The iteration of this process is called the ‘bisection method’. It is used to approximate zeros of continuous functions. Theorem 1.2.32. Let n be a natural number and a0 , a1 , . . . , a2n be real numbers. Define the polynomial p : R Ñ R by ppxq : a0

   a2nx2n x2n 1 for all x P R. Then there is some x P R such that f pxq  0. a1 x

Proof. Below in Example 1.2.40, it is proved that p is continuous. Further, define x0 : 1 maxt|a0 |, |a1|, . . . , |a2n |u . Then 

¤ |a0| |a1|  |x0|    |a2n|  |x0|2n
a0 a1x0    a2nx2n 0 2n 1 ¤ px0
1q  p1 x0    x2n
1   x02n 1 0 q  x0 and hence ppx0 q ¡ 0. Also a0 a1 p
x0 q    a2n p
x0 q2n ¤ |a0 | |a1 |  |x0 |    |a2n |  |x0 |2n 2n 1 ¤ px0
1q  p1 x0    x2n
1  
p
x0q2n 1 0 q  x0 and hence pp
x0 q   0. Hence according to Theorem 1.2.28, there is x P r
x0, x0 s such that f pxq  0. The ‘converse’ of Theorem 1.2.28 is not true, i.e., a function that assumes all values between those at its interval ends is not necessarily continuous on that interval. This can be seen, for instance, from the following Example.

63

y 1

0.5

0.2

0.4

0.6

x

-0.5

-1

Fig. 20: Graph of f from Example 1.2.33.

Example 1.2.33. Define f : r0, 2{π s Ñ R by f pxq : sinp1{xq for 0   x ¤ 2π and f p0q : 0. Then f is not continuous (in 0), but assumes all values in the in the interval rf p0q, f p2{π qs  r0, 1s. Note also that f has an infinite number of zeros, located at 1{pnπ q for n P N . Theorem 1.2.34. Let f : ra, bs Ñ R be a continuous function, where a and b are real numbers such that a   b. Then the range of f is given by f pra, bsq  rα, β s for some α, β

(1.2.11)

P R such that α ¤ β.

Proof. Denote by α, β the minimum value and the maximum value of f , respectively, which exist according to Theorem 1.2.26. Then for every x P rα, β s α ¤ f pxq ¤ β . 64

Further, let xm , xM P ra, bs be such that f pxm q  α and f pxM q  β, respectively. Finally denote by I the interval rxm , xM s if xm ¤ xM and rxM , xm s if xM   xm . Then the restriction f |I of f to I is continuous and, according to Theorem 1.2.28 (applied to the function
f |I if xM   xm ), every value of rα, β s is in its range. Theorem 1.2.35. Let f : ra, bs Ñ R, where a, b P R are such that a   b, be continuous and strictly increasing, i.e., for all x1 , x2 P ra, bs such that x1   x2 it follows that f px1 q   f px2 q. Then the inverse function f
1 is continuous, too. Proof. From the property that f is strictly increasing, it follows that f is also injective. Further, from Theorem 1.2.34 it follows the existence of α, β P R such that the range of f is given by rα, β s and hence that f
1 : rα, β s Ñ ra, bs . Now let y be some element of rα, β s and y1 , y2, . . . be some sequence of elements of rα, β s that is converging to y, but such that f
1 py1 q, f
1 py2q, . . . is not converging to f
1 py q. Then there is an ǫ ¡ 0 along with a subsequence yn1 , yn2 , . . . of y1 , y2 , . . . such that 
1 f y nk



p q
f
1pyq ¥ ǫ

(1.2.12)

for all k P N . According to the Bolzano-Weierstrass’ Theorem 1.2.11, there is a subsequence ynk1 , ynk2 , . . . of yn1 , yn2 , . . . such lim f
1 pynkl q  x

l

Ñ8

(1.2.13)

for some x P ra, bs. Hence it follows by the continuity of f that lim ynkl

l

Ñ8

 f pxq

and y  f pxq, since ynk1 , ynk2 , . . . is also convergent to y, but from (1.2.12) it follows by (1.2.13) that x  f
1 py q 65

which, since f is injective, leads to the contradiction that y

 f pxq

.

Hence such y and sequence y1 , y2 , . . . don’t exist and f
1 is continuous. Definition 1.2.36. Let f1 : D1 Ñ R, f2 : D2 Ñ R be functions such that D1 XD2  φ. Moreover, let a P R. Then we define pf1 f2 q : D1 XD2 Ñ R (read: ‘f plus g’) and a  f1 : D1 Ñ R (read: ‘a times f ’) by

pf1

f2 qpxq : f1 pxq

f2 pxq

for all x P D1 X D2 and

pa  f1 qpxq : a  f1pxq for all x P D1 .

Theorem 1.2.37. Let f1 : D1 Ñ R, f2 : D2 Ñ R be functions such that D1 X D2  φ. Moreover let a P R. Then it follow by Theorem 1.2.4 that (i) if f1 and f2 are both continuous in x continuous in x, too,

P

D1

X D2 , then f1

f2 is

(ii) if f1 is continuous in x P D1 , then a  f1 is continuous in x, too. Definition 1.2.38. Let f1 : D1 Ñ R, f2 : D2 Ñ R be functions such that D1 X D2  φ. Then we define f1  f2 : D1 X D2 Ñ R (read: ‘f1 times f2 ’) by pf1  f2 qpxq : f1 pxq  f2pxq

for all x P D1 X D2 . If moreover Ranpf1 q D1 Ñ R (read: ‘1 over f1 ’) by

€ R, then we define 1{f1

p1{f1 qpxq : 1{f1 pxq for all x P D1 . 66

:

Theorem 1.2.39. Let f1 : D1 D1 X D2  φ.

Ñ R, f2 : D2 Ñ R be functions such that

(i) If f1 and f2 are both continuous in x continuous in x, too.

P

D1

X D2, then f1  f2 is

(ii) If f1 is such that Ranpf1 q € R as well as continuous in x P D1 , then 1{f1 is continuous in x, too. Proof. For the proof of (i), let x1 , x2 , . . . be some sequence in D1 which converges to x. Then for any ν P N

X D2

|pf1  f2qpxν q
pf1  f2qpxq|  |f1pxν qf2pxν q
f1pxqf2 pxq|  |f1pxν qf2 pxν q
f1 pxqf2 pxν q f1 pxqf2pxν q
f1 pxqf2pxq| ¤ |f1pxν q
f1 pxq|  |f2pxν q| |f1pxq|  |f2pxν q
f2 pxq| ¤ |f1pxν q
f1 pxq|  |f2pxν q
f2pxq| |f1pxν q
f1pxq|  |f2pxq| |f1pxq|  |f2pxν q
f2 pxq| and hence, obviously, lim pf1  f2 qpxν q  pf1  f2 qpxq .

ν

Ñ8

For the proof of (ii), let x1 , x2 , . . . be some sequence in D1 which converges to x. Then for any ν P N

|p1{f1qpxν q
p1{f1 qpxq|  |1{f1pxν q
1{f1pxq|  |f1pxν q
f1pxq|{r |f1pxν q|  |f1pxq| s and hence, obviously, lim p1{f1 qpxν q  p1{f1 qpxq .

ν

Ñ8

67

Example 1.2.40. Let n P N and a0 , a1 , . . . , an be real numbers. Then the corresponding polynomial of n-th order p : R Ñ R defined by ppxq : a0

a1 x



an xn

for all x P R, is continuous. Proof. The proof is a simple consequence of Example 1.2.22, Theorem 1.2.37 and Theorem 1.2.39. Example 1.2.41. Explain why the function f pxq :

x3

2x2 x 1 x2
3x 2

(1.2.14)

is continuous at every number in its domain. State that domain. Solution: The domain D is given by those real numbers for which the denominator of the expression (1.2.14) is different from 0. Hence it is given by D

 Rzt1, 2u .

Further, as a consequence of Example 1.2.40, the polynomials p1 : R Ñ R, p2 : D Ñ R defined by p1 pxq : x3 2x2 x p2 pxq : x2
3x 2

1,

for all x P R and x P D, respectively, are continuous. Since p2 pRq € R , it follows by Theorem 1.2.39 that the function 1{p2 is continuous. Finally from this, it follows by Theorem 1.2.39 that p1 {p2 is continuous.

Theorem 1.2.42. Let f : Df Ñ R, g : Dg Ñ R be functions and Dg be a subset of R. Moreover let x P Df , f pxq P Dg , f be continuous in x and g be continuous in f pxq. Then g  f is continuous in x.

Proof. For this, let x1 , x2 , . . . be a sequence in Dpg  f q converging to x. Then f px1 q, f px2 q, . . . is a sequence in Dg . Moreover since f is continuous in x, it follows that lim f pxν q  f pxq . ν

Ñ8

68

y 3

2

1

-2

2

3

x

-1

-2

-3

Fig. 21: Graph of sin, arcsin and asymptotes.

Finally, since g is continuous in f pxq it follows that lim pg  f qpxν q  lim g pf pxν qq  g pf pxqq  pg  f qpxq .

ν

Ñ8

ν

Ñ8

Example 1.2.43. Show that f : R Ñ R defined by f pxq : |x| for all x P R, is continuous. Solution: Define the polynomial p2 : R Ñ R by p2 pxq : x2 for every x P R. According to Example 1.2.40, p2 is continuous. Then f  s2 p2 , where s2 denotes the square-root function on r0, 8q, which, by Theorem 1.2.35, is continuous as inverse of the strictly increasing restriction of p2 to r0, 8q. Hence f is continuous by Theorem 1.2.42. Example 1.2.44. The functions sin : R Ñ R and exp : R Ñ R are continuous. Show that arcsin : r
1, 1s Ñ r
π {2, π {2s, cos : R Ñ R, arccos : 69

y 3

2

-3

-2

2

3

x

-1

-2

-3

Fig. 22: Graph of cos, arccos and asymptotes.

y 3

2

1

-3

-2

-1

1

2

x

-1

-2

-3

Fig. 23: Graph of tan, arctan and asymptotes.

70

y 3

2

1

-3

-2

-1

1

2

3

x

-1

-2

-3

Fig. 24: Graph of exp, ln.

r
1, 1s Ñ r0, πs, tan : p
π{2, π{2q Ñ R, arctan : R Ñ p
π{2, π{2q and the natural logarithm function ln : p0, 8q Ñ R are continuous. Solution: Since the restriction of sin to r
π {2, π {2s and exp are in particular increasing, their inverses arcsin and ln are continuous according to Theorem 1.2.35. Further since


cospxq  sin x

π 2

for all x P R, the cosine function is continuous as composition of continuous functions according to Theorem 1.2.42. Further, the restriction of cos to r0, π s is in particular increasing and hence its inverse arccos continuous according to Theorem 1.2.35. Also, tan : Rz tk  π pπ {2q : k P Zu Ñ R defined by sinpxq tanpxq  cospxq for every x P Rz tk  π pπ {2q : k P Zu is continuous according to Theorem 1.2.39 as quotient of continuous functions. Finally, the restriction of 71

F x 1 tanHxL

sinHxL

x A

cosHxL

B

C

D

Fig. 25: Sketch for Example 1.2.45.

tan to p
π {2, π {2q is in particular increasing and hence its inverse arctan continuous according to Theorem 1.2.35. Example 1.2.45. (Removable singularities) Define f : R Ñ R by f pxq 

sinpxq x

for every x P R and f p0q  1. Then f is continuous. Proof: By Theorem 1.2.39, the continuity of sin and the linear function p : R Ñ R, defined by ppxq : x, x P R, see Example 1.2.22, it follows the continuity of f in all points of R . The proof that f is also continuous in x  0, follows from the following inequality (compare Fig 26):   sin x   x



p q
1 ¤ 1
1 ,  cospxq

(1.2.15)

for all x P p
π {2, π {2qzt0u. For its derivation and in a first step, we assume that 0   x   π {2 and consider the triangle ADF in Fig 25, in particular the areas ApABF q, ApACF q and ApADF q of the triangles ABF , ACF and ADF , respectively. Then we have the following relation: ApABF q ¤ ApACF q ¤ ApADF q 72

y 2.5

2

1.5

1

0.5

-1.5

-1

-0.5

0.5

1

1.5

x

Fig. 26: Graphs of f (red) and h (blue) from Example 1.2.45.

and hence

and

x 1 sinpxq cospxq ¤ 2 2

¤ tan2pxq

sinpxq 1 ¤ . x cospxq From this follows, by the symmetries of sin, cos under sign change of the argument, the same equality for
π {2   x   0. Hence for x P p
π{2, π{2q zt0u: 1 sinpxq
1
1¤ x cospxq and sinpxq 1 1

1 ¤ 1
cospxq ¤ x cospxq cospxq ¤

and hence finally (1.2.15). Now since h : p
π {2, π {2q Ñ R defined by hpxq :

1 cospxq 73


1 ,

for all x P p
π {2, π {2q is continuous, it follows by (1.2.15) and Theorem 1.2.5 the continuity of f also in x  0. Definition 1.2.46. (Limits at infinity) Let f : ra, 8q Ñ R be some continuous function, where a ¡ 0. Further let L be some real number. We define lim f pxq  L x

Ñ8

if the transformed function f¯ : r0, 1{as Ñ R defined by f¯pxq : f p1{xq for all x P p0, 1{as and f¯p0q : L is continuous in 0. In this case, we call the parallel y  L to the x-axis a ‘horizontal asymptote of Gpf q for large positive x’. Analogously, we define for every continuous function f : p
8,
as Ñ R lim f pxq  L x

Ñ
8

if the transformed function f¯ : r
1{a, 0s Ñ R defined by f¯pxq : f p1{xq for all x P r
1{a, 0q and f¯p0q : L is continuous in 0. In this case, we call the parallel y  L to the x-axis a ‘horizontal asymptote of Gpf q for large negative x’. Example 1.2.47. Consider the function f : r1, 8q Ñ R defined by f pxq 

x2
1 x2 1

for all x P r1, 8q. See Fig. 27. Then the transformed function f¯ : r0, 1s Ñ R, defined by 1
x2 f¯pxq : 1 x2 74

y

0.5

-10

-5

5

10

x

-1 Fig. 27: Gpf q and asymptote for Example 1.2.47.

for all x P r0, 1s, is continuous and hence since f¯pxq  f p1{xq for all x P p0, 1s, it follows that lim f pxq  1 .

x

Hence y Fig. 27.

Ñ8

 1 is a horizontal asymptote of Gpf q for large positive x.

Example 1.2.48. Find the limits

?

?

2x2 1 2x2 1 , lim . lim xÑ
8 3x
5 xÑ8 3x
5

Solution: Define f : tx P R : x  5{3u Ñ R by f pxq :

?

2x2 1 3x
5

75

See

y 2 1

2

4

6

8

x

-1 -2 Fig. 28: Gpf q and asymptotes for Example 1.2.48.

for all x P R ^ x  5{3. Then the transformed functions f¯ corresponding to the restrictions of f to r1, 8q and p
8,
1s are given by the continuous functions ? 2 x 2 x ¯ f pxq :  |x| 3
5x2 for all x P r0, 1s and

f¯pxq :

x |x|

?

 3 2
5xx2

for all x P r
1, 0s, respectively, and hence

?

2x2 1 lim xÑ8 3x
5



?

2

?

2x2 1 2 , lim xÑ
8 3x
5 3

See Fig. 28.

Problems

76




?

2 . 3

1) Show the continuity of the function f . For this, use only Theorems 1.2.37, 1.2.39, 1.2.42 on sums, products/quotients, compositions of continuous functions, and the continuity of constant functions/the identity function idR on R. a) b) c) d) e)

f pxq : x 7 , x P R , f pxq : x2 , x P R , f pxq : 3{x , x P R , f pxq : px 3q{px
8q , x P Rzt8u , f pxq : px2 3x 2q{px2 2x 2q , x P R .

2) Assume that f and g are continuous functions in x f p0q  2 and lim r2f pxq
3g pxqs  1 . Calculate g p1q.

x

 0 such that

Ñ0

In the following, it can be assumed that rational functions, i.e., quotients of polynomial functions, are continuous on their domain of definition. In addition, it can be assumed that the exponential function, the natural logarithm function, the general power function, the sine and cosine function and the tangent function are continuous. 3) For arbitrary c, d P R, define fc,d : R Ñ R by fc,d pxq :

$ 2 ' &1 x

{p 1q if x P p
8, 1q cx d if x P r
1, 1s ' %? 4x 5 if x P p1, 8q

for all x P R. Determine c, d such that the corresponding fc,d is everywhere continuous. Give reasons.

4) For arbitrary c P R, define fc : r0, 8q Ñ R by fc pxq :

#

x sinp1{xq if x P p0, 8q c if x  0

for all x P r0, 8q. Determine c such that the corresponding fc is everywhere continuous. Explain your answer.

P R, define fk : r
1{3, 8qzt1u Ñ R by #? ? 3x 1
2x 2 if x P r
1{3, 8qzt1u x
1 fk pxq : k if x  1

5) For every k

For what value of k is fk continuous? Give explanations.

77

.

6) Define the function f : R Ñ R by f pxq : x4 10x
15 for all real x. Use your calculator to find an interval of length 1{100 which contains a zero of f (i.e, some real x such that f pxq  0). Give explanations. 7) Determine in each case whether the given sequence has a limit. If there is one, calculate that limit. Otherwise, give arguments why there is no limit. xn :

a)

?

1


n

?n

,

xn :

b)

? ?n n 1

for all n P N. 8) Find the limits. 

a) lim e { nÑ8

,

c) lim

,

1 n

cospnq n lnpnq d) lim nÑ8 n e) lim n1{n

Ñ8

n

Ñ8


n

f) lim

a

Ñ8


n

b) lim cos

Ñ8

n

,

Hint: Use d)

,

1{3 1{3 Ñ8 n
pn 1q

Hint: Use that a
b  h

Ñ0

π

?

n

p1

n



Hint: Use that lnpnq ¤ 2 n for n ¥ 1 .



g) lim

h) lim

1

,

6n
n

n2

n

hq2{3
1 h

,

a3
b 3 for all a  b a2 ab b2 ,

,

Hint: Use the hint in g) .

9) Calculate the limits. 



17n 4 , b) lim tan a) lim sin nÑ8 xÑ2 n 5 2 x
8x 15 . c) lim xÑ5 x
5 In each case, give explanations. 10) Find the limits

a) limxÑ8 rx{px

1qs ,

78



3x 5x

2 7



,

b) c) d) e) f) g) h) i)

limxÑ
8 rx{px 1qs , limxÑ8 rpsin xq{xs , limxÑ8 rp3x3 2x2 5x 4q { p2x3 x2 ? limxÑ8 p x{ 1 x2 q , ? limxÑ
8 p x{ 1 x2 q , ? ? limxÑ8 p 3x2 2x { 2x2 5 q , ?2 ?2 limxÑ8 p x ? 2 3
x ?1 q ,2 4x 5
x 2q. limxÑ
8 p x

11) Define f : R Ñ R by

f pxq :

#

x

5qs ,

x if x is rational 0 if x is irrational

for every x P R. Find the points of discontinuity of f .

12) Define f : R Ñ R by f pxq : 0 if x  0 or if x is irrational and f pm{nq : 1{n if m P Z and n P N have no common divisor greater than 1. Find the points of discontinuity of f . 13) Let f and g be functions from R to R whose restrictions to Q coincide. Show that f  g.

14) Let D € R, f : D Ñ R be continuous in some x P D and f pxq ¡ 0. By an indirect proof, show that there is ε ¡ 0 such that f pxq ¡ 0 for all x P D X px
ε, x εq.

15) Let a, b P R such that a   b and f : ra, bs Ñ ra, bs. By use of the intermediate value theorem, show that f has a fixed point, i.e., that there is x P ra, bs such that f pxq  x. 16) Use the intermediate value theorem to prove that for every a ¥ 0 there is a uniquely determined x ¥ 0 such that x2  a. That x is ? denoted by a.

1.3 Differentiation In 1589, using inclined planes, Galileo discovered experimentally that in vacuum all bodies, regardless of their weight, shape, or composition, are uniformly accelerated in exactly the same way, and that the fallen distance s is proportional to the square of the elapsed time t: 1 sptq  gt2 2 79

(1.3.1)

for all t P R, where g  9.81m{sec2 is the gravitational acceleration. This result was in contradiction to the generally accepted traditional theory of Aristotle that assumed that heavier objects fall faster than lighter ones. Consider the average speed of such a body, i.e., the traveled distance divided by the elapsed time, during the time interval rt, t hs, if h ¥ 0, rt h, ts, if h   0, respectively, for some t, h P R: spt

hq
sptq t h
t



g rpt 2h

hq

2


t s 2

g p2ht 2h

2

h

qg



t

h 2



.

Hence it follows by Example 1.2.22 that lim

h

Ñ0

spt

hq
sptq t h
t

 gt ,

which suggests itself as (and indeed is the) definition of the instantaneous speed v ptq of the body at time t: v ptq : lim h

Ñ0

spt

hq
sptq t h
t

 gt .

A geometrical interpretation of v ptq is the slope of the tangent to Gpsq at the point pt, sptqq, see Fig. 29. Definition 1.3.1. Let f : pa, bq Ñ R be a function where a, b P R such that a   b. Further, let x P pa, bq and c P R. We say f is differentiable in x with derivative c if for all sequences x0 , x1 , . . . in pa, bqztxu which are convergent to x: f pxn q
f pxq lim c. nÑ8 xn
x In this case, we define the derivative f 1 pxq of f in x by f 1 pxq : c . Further, we say f is differentiable if f is differentiable in all points of its domain pa, bq. In that case, we call the function f 1 : pa, bq Ñ R associating 80

sHtL @mD 4

3

2

sH0.8L-sH0.4L

1 0.8-0.4

0.2

0.4

0.6

0.8

1

1.2

t @secD

Fig. 29: Gpsq, secant line and tangent at p0.4, sp0.4qq.

to every x P pa, bq the corresponding f 1 pxq the derivative of f . Higher order derivatives of f are defined recursively. If f pkq is differentiable for k P N , we define the derivative f pk 1q of order k 1 of f by f pk 1q : pf pkq q 1 , where we set f p1q : f 1 . In that case, f will be referred to as pk 1qtimes differentiable. Frequently, we also use the notation f 2 : f p2q and f 3 : f p3q .

Theorem 1.3.2. Let f : pa, bq Ñ R be a function where a, b P R such that a   b. Further, let f be differentiable in x P pa, bq. Then f is also continuous in x.

Proof. Let x0 , x1 , . . . be a sequence in pa, bq which is convergent to x. Obviously, it is sufficient to assume that x0 , x1 , . . . is a sequence in pa, bqztxu. Then it follows by the limit laws Theorem 1.2.4 that f pxn q
f pxq nÑ8 xn
x

lim pf pxn q
f pxqq  lim

n

Ñ8

and hence that

lim f pxn q  f pxq .

n

Ñ8

81

 nlim px
xq  0 Ñ8 n

Example 1.3.3. Let c P R, n P N and f, g : R Ñ R be defined by f pxq : c , g pxq : xn for all x P R. Then f, g are differentiable and f 1 pxq  0 , g 1 pxq  nxn
1 for all x P R.

Proof. Let x P R and x0 , x1 ,    be a sequence of numbers in Rztxu which is convergent to x P R. Then: lim

ν

Ñ8

f pxν q
f pxq xν
x

 νlim 00. Ñ8

P N: p g pxν q
g pxq xν qn
xn  x
x xν
x ν n
1 n
2  pxν q pxν q x   

Further, for any ν

xν xn
2

xn
1

xxn
2

xn
1

and hence by Example 1.2.40: g pxν q
g pxq ν Ñ8 xν
x lim

 xn
1

xn
2 x



 nxn
1 .

Example 1.3.4. The exponential function is differentiable with exp 1 pxq  exppxq for all x P R.

82

Proof. First, we prove that exp is differentiable in 0 with derivative e0  1. For this, let h1 , h2 , . . . be some sequence in Rzt0u which is convergent to 0. Moreover, let n0 P N be such that |hn |   1 for n ¥ 0. Then for any such n: ehn
e0 ehn
p1 hn q .
e0  hn
0 hn We consider the cases hn ¡ 0 and hn   0. In the first case, it follows by (1.2.7) and some calculation that 0¤

ehn


p1

hn q

hn

3
hn

¤ h4n 

1


 hn 2 2

¤ 124 hn  3hn .

Analogously, it follows in the second case that h ¤ 1
hnh ¤ e
hp1 n

hn

n

n

Hence it follows in both cases that  h e n  

hn q




p1

hn q  

hn

¤ h4n

.

¤ 3|hn|

and therefore by Theorem 1.2.5 that ehn
p1 nÑ8 hn lim

hn q

0.

Now let x P R and x1 , x2 , . . . be some sequence in Rztxu which is convergent to x. Then exn
ex xn
x


e e  x

x

exn
x
r1 pxn
xqs , xn
x

and hence it follows by Theorem 1.2.4 and the previous result that exn
ex nÑ8 xn
x lim

 ex

and therefore the statement of this Theorem. 83

y 1

-1

-0.5

0.5

1

x

Fig. 30: Graph of the modulus function. See Example 1.3.6.

Example 1.3.5. The sine function is differentiable with sin 1 pxq  cospxq for all x P R.

Proof. Let x P R and x1 , x2 , . . . be some sequence in Rztxu, which is convergent to x. Further define hn : xn
x, n P N. Then it follows by the addition theorems for the trigonometric functions sinpxn q
sinpxq sinpx hn q
sinpxq  xn
x hn  sinpxq  cosphhn q
1 cospxq  sinhphn q n n  2 
sinpxq  h2n sinhph{n2{2q cospxq  sinhphnq n n and hence by Example 1.2.45 and Theorem 1.2.4 that sinpxn q
sinpxq nÑ8 xn
x lim

84

 cospxq .

y 1 0.5

-1

-0.5

0.5

1

x

-1 Fig. 31: Graph of f from Example 1.3.7.

Example 1.3.6. The function f : R Ñ R defined by f pxq : |x|

for all x P R, is not differentiable in 0, because lim

n

Ñ8

 



 


n1 
0 
1  lim  n1 
0  1 . nÑ8 1
0
n1
0 n

See Fig. 30. Example 1.3.7. The function f : R Ñ R defined by f pxq : x1{3

for all x P R, is not differentiable in 0, because the sequence  1 1 3 n 1 n

{


01{3  n2{3
0

has no limit for n Ñ 8. See Fig. 31.

85

Theorem 1.3.8. (Sum rule, product rules and quotient rule) Let f, g be two differentiable functions from some open interval I into R and a P R. (i) Then f

g, a  f and f  g are differentiable with

pf gq 1pxq  f 1 pxq g 1 pxq , pa  f q 1pxq  a  f 1pxq pf  gq 1pxq  f pxq  g 1pxq gpxq  f 1pxq for all x P I. (ii) If f is non-vanishing for all x P I, then 1{f is differentiable and  1 1 f 1 pxq p xq 
f rf pxqs2 for all x P I. Proof. For this let x P I and x1 , x2 , . . . be some sequence in I ztxu which

is convergent to x. Then:

|pf ¤

g qpxq
pf 1 pxq |xν
x| 1 |f pxν q
f pxq
f pxqpxν
xq| |xν
x| g qpxν q
pf

g 1 pxqqpxν


xq|

|gpxν q
gpxq
g 1pxqpxν
xq| |xν
x|

and

|pa  f qpxν q
pa  f qpxq
ra  pf 1 qpxqspxν
xq| |xν
x| 1  |a|  |f pxν q
f px|xq

fx|pxqpxν
xq| ν

and hence lim

ν

Ñ8

|pf

g qpxν q
pf

g qpxq
pf 1 pxq |xν
x|

86

g 1 pxqqpxν


xq|  0

and

|pa  f qpxν q
pa  f qpxq
ra  pf 1 qpxqspxν
xq|  0 . ν Ñ8 |xν
x| lim

Further, it follows that

|pf  gqpxν q
pf  gqpxq
pf pxq  g 1 pxq gpxq  f 1pxqqpxν
xq| |xν
x| 1 ¤ |f pxν q
f px|xq

fx|pxqpxν
xq|  |gpxq| ν 1 |f pxq|  |gpxν q
gpx|xq

gxp|xqpxν
xq| ν |f pxν q
f pxq|  |gpx q
gpxq| ν |x
x| ν

and hence that |pf  gqpxν q
pf  gqpxq
pf pxq  g 1pxq lim ν Ñ8 |xν
x| 0.

g pxq  f 1 pxqqpxν


xq|

If f is does in any point of its domain I, it follows that   1  f pxν q



 1 r p qs  f pxqpxν
xq ¤ |xν
x| |f pxν q
f pxq
f 1pxqpxν
xq| 1  |f pxq|2 |xν
x| |f pxν q
f pxq|2 |f pxν q|  |f pxq|2  |xν
x|

and hence that lim

ν

Ñ8


f p1xq

  1  f pxν q


f p1xq

Finally, since x1 , x2 , . . . follows.

1 f x

2



 1 r p qs  f pxqpxν
xq 0. |xν
x| and x P I were otherwise arbitrary, the theorem 1 f x

2

87

Example 1.3.9. Let n P N and a0 , a1 , . . . , an be real numbers. Then the corresponding polynomial of n-th order p : R Ñ R, defined by ppxq : a0

for all x P R, is differentiable and for all x P R.

p 1 pxq : a1

a1 x





an xn

nan xpn
1q

Proof. The proof is a simple consequence of Example 1.3.3 and Theorem 1.3.8. Theorem 1.3.10. (Chain rule) Let f : I Ñ R, g : J Ñ R be differentiable functions defined on some open intervals I, J of R and such that the domain of the composition g  f is not empty. Then g  f is differentiable with for all x P Dpg  f q.

pg  f q 1  g 1 pf pxqq  f 1 pxq

Proof. For this let x P Dpg  f q and x1 , x2 , . . . be some sequence in Dpg  f qztxu which is convergent to x. Then:

|pg  f qpxν q
pg  f qpxq
pg 1 pf pxqq  f 1 pxqqpxν
xq| ¤ |xν
x| |gpf pxν qq
gpf pxqq
g 1 pf pxqqpf pxν q
f pxqq| |xν
x| 1 |g pf pxqqpf pxν q
f pxq
f 1pxqpxν
xqq| |xν
x|

and hence, obviously,

|pg  f qpxν q
pg  f qpxq
pg 1 pf pxqq  f 1 pxqqpxν
xq| ν Ñ8 |xν
x| 0. Finally, since x1 , x2 , . . . and x P Dpg  f q were otherwise arbitrary, the lim

theorem follows.

88

Example 1.3.11. The cosine and the tangent function are differentiable with cos 1 pxq 
sinpxq for all x P R and

tan 1 pxq  for all x P Rz

π 2

Proof. Since

kπ : k

1 cos2 pxq

1

tan2 pxq

(

PZ

.


π 2 for all x P R, it follows by Examples 1.3.5, 1.3.3 and Theorem 1.3.8 (i.e., the ‘sum rule’) and Theorem 1.3.10 (i.e., the ‘chain rule’) that cos is differentiable with derivative cospxq  sin x




cos 1 pxq  cos x for all x P R. Further, because of tanpxq  (

π 
sinpxq 2

sinpxq cospxq

for all x P Rz π2 kπ : k P Z , it follows by Examples 1.3.5 and Theorems 1.3.8 (i.e., the ‘Quotient Rule’) that tan is differentiable with derivative cospxq  cospxq
sinpxq  p
sinpxqq cos2 pxq tan2 pxq

tan 1 pxq 

1 for all x P Rz

π 2

kπ : k

(

PZ

.

89

 cos12pxq

Definition 1.3.12. Let f : U Ñ R be a function in several variables, where U is a subset of Rn , n P Nzt0, 1u. In particular, let i P t1, . . . , nu, x P U be such that the corresponding function f px1 , . . . , xi
1 , , xi 1 , . . . , xn q is differentiable at xi . In this case, we say that f is partially differentiable at x in the i-th coordinate direction, and we define:

Bf pxq : rf px , . . . , x , , x 1 i
1 i Bxi

1 , . . . , xn

qs 1pxiq .

If f is partially differentiable at x in the i-th coordinate direction at every point of its domain, we call f partially differentiable in the i-th coordinate direction and denote by B f {B xi the map which associates to every x P U the corresponding pB f {B xi qpxq. Partial derivatives of f of higher order are defined recursively. If B f {B xi is partially differentiable in the j-th coordinate direction, where j P t1, . . . , nu, we denote the partial derivative of Bf {Bxi in the j-th coordinate direction by

B2f Bxj Bxj

.

Such is called a partial derivative of f of second order. In the case j we set B2f : B2f . Bx2i BxiBxi Partial derivatives of f of higher order than 2 are defined accordingly.

Ñ R by f px, y q : x3 x2 y 3
2y 2

Example 1.3.13. Define f : R2

for all x, y

P R. Find

Bf p2, 1q Bx

and 90

Bf p2, 1q . By

 i,

Solution: We have f px, 1q  x3 for all x, y

x P R,

x2
2 and f p2, y q  8

P R. Hence it follows that Bf px, 1q  3x2 Bx

4y 3
2y 2

2x ,

Bf p2, yq  12y2
4y , By

P R, and, finally, that Bf p2, 1q  16 and Bf p2, 1q  8 . Bx By Example 1.3.14. Define f : R3 Ñ R by f px, y, z q : x2 y 3z 3x 4y 6z 5 for all x, y, z P R. Find Bf px, y, zq , Bf px, y, zq and Bf px, y, zq Bx By Bz for all x, y, z P R. Solution: Since in partial differentiating with respect to y

one variable all other variables are held constant, we conclude that

Bf px, y, zq  2xy3z Bx Bf px, y, zq  x2 y3 Bz for all x, y, z

3,

Bf px, y, zq  3x2 y2z By

6,

P R.

Problems

91

4,

1) By the basic definition of derivatives, calculate the derivative of the function f . a) f pxq : 1{x , x P p0, 8q , b) f pxq : px
1q{px 1q , x P Rzt
1u , ? c) f pxq : x , x P p0, 8q .

2) Calculate the slope of the tangent to G(f) at the point p1, f p1qq and its intersection with the x-axis. a) f pxq : x2
3x 1 , x P R , b) f pxq : p3x
2q{p4x 5q , x P Rzt
5{4u , c) f pxq : e3x , x P R .

3) Calculate the derivatives of the functions f1 , . . . , f8 with maximal domains in R defined by a) f1 pxq : 5x8
2x5 b) f3 ptq : p1

f4 pxq : 3e

3t

2

6 , f2 pθq : 3 sinpθq 5t

4

qpt

2

4 cospθq ,

8q ,

rsinpxq 6 cospxqs , 3t
2t 5 5 cospϕq , f6 pϕq : c) f5 ptq : t3 8 tanpϕq 2 4 sinp7tq d) f7 pxq : sinp3{x q , f8 ptq : e . x

4

,

4) A differentiable function f satisfies the given equation for all x from its domain. Calculate the slope of the tangent to Gpf q in the specified point P without solving the equations for f pxq.

? ? pf pxqq2  1 , P  p
1{ 2 , 1{ 2 q , ? ? b) px
1q2 r x2 pf pxqq2 s
4x2  0 , P  p1 2,1 2q, a f pxqr2
f pxqs  arccosp1
f pxqq , c) x ? ? P  ppπ {4q
p1{ 2 q, 1
p1{ 2 qq . a) x2

Remark: The curve in b) is a cycloid which is the trajectory of a point of a circle rolling along a straight line. The curve in c) is named after Nicomedes (3rd century B.C.), who used it to solve the problem of trisecting an angle.

5) Give a function f : R Ñ R such that

a) f 1 ptq  1 for all t P R and such that f p0q  2 , b) f 1 ptq 
2f ptq for all t P R and such that f p0q  1 ,

92

c) f 1 ptq 
2f ptq

3 for all t P R and such that f p0q  1 .

6) Let I be a non-empty open interval in R and p, q f : I Ñ R. Show that f 2 pxq

pf 1 pxq

for all x P I if and only if f¯2 pxq





p2 4

P R.

Further, let

qf pxq  0


q f¯pxq

for all x P I, where f¯ : I

Ñ R is defined by f¯pxq : epx{2 f pxq

for all x P I. 7) Newton’s equation of motion for a point particle of mass m moving on a straight line is given by mf 2 ptq  F pf ptqq

¥

0

(1.3.2)

for all t from some time interval I € R, where f ptq is the position of the particle at time t, and F pxq is the external force at the point x. For the specified force, give a solution function f : R Ñ R of (1.3.2) that contains 2 free real parameters. a) F pxq  F0 , x P R ,where F0 is some real parameter , b) F pxq 
kx , x P R ,where k is some real parameter .

8) Newton’s equation of motion for a point particle of mass m ¥ 0 moving on a straight line under the influence of a viscous friction is given by mf 2 ptq 
λf 1 ptq (1.3.3) for all t P R, where f ptq is the position of the particle at time t, and λ P r0, 8q is a parameter describing the strength of the friction. Give a solution function f of (1.3.3) that contains 2 free real parameters.

9) For all px, y q from the domain, calculate the partial derivatives pBf {Bxqpx, yq, pBf {Byqpx, yq of the given function f . a) f px, y q : x4
2x2 y 2 3x
4y 1 , px, y q P R2 , b) f px, y q : 3x2
2x 1 , px, y q P R2 , c) f px, y q : sinpxy q , px, y q P R2 .

93

10) Let f : R Ñ R and g : R Ñ R be twice differentiable functions. Define upt, xq : f px
tq g px tq for all pt, xq P R2 . Calculate

Bu pt, xq , Bu pt, xq , B2 u pt, xq , B2 u pt, xq Bt Bx B t2 Bx2 for all pt, xq P R2 . Conclude that u satisfies B2 u
B2 u  0 Bt2 Bx2 which is called the wave equation in one space dimension (for a function u which is to be determined).

1.4 Applications of Differentiation Theorem 1.4.1. (Necessary condition for the existence of a local minimum/maximum) Let f be a differentiable real-valued function on some open interval I of R. Further, let f have a local minimum/maximum at some x0 P I, i.e, let f px0 q ¤ f pxq for all x such that x0
ε   x   x0

{

f px0 q ¥ f pxq ε, for some ε ¡ 0. Then

f 1 px0 q  0 , i.e, x0 is a so called ‘critical point’ for f . Proof. If f has a local minimum/maximum at x0 sufficiently small h P R that 1 rf px0 h

P I, then it follows for

hq
f px0 qs

is ¥ p¤q 0 and ¤ p¥q 0, for h ¡ 0 and h   0, respectively. Therefore by Theorem 1.2.6, f 1 px0 q is at the same time ¥ 0 and ¤ 0 and hence, finally, equal to 0.

94

y 1.15 1.1 1.05

0.95 -1

-0.6 -0.4 -0.2

0.2 0.4

x

Fig. 32: Gpf q from Example 1.4.2.

Example 1.4.2. Find the critical points of f : R Ñ R defined by

for all x equation

P R.

f pxq : x4

x3

1

Solution: The critical points of f are the solutions of the 0  f 1 pxq  4x3

3x2

 x2p4x

3q

and hence given by x  0 and x 
3{4. See Fig. 32. Note that f has a local extremum at x 
3{4, but not at x  0. Hence the condition in Theorem 1.4.1 is necessary, but not sufficient for the existence of a local extremum. Example 1.4.3. Find the maximum and minimum values of f : r
π, π s Ñ R defined by f pxq : x
2 cospxq for all x P r
π, π s. Solution: Since f is continuous, such values exist according to Theorem 1.2.26. Those points, where these values are assumed, can be either on the boundary of the domain, i.e., in the points
π 95

y 5 4 3 2 1

-3

-2

-1

2

3

x

-1 -2

Fig. 33: Gpf q from Example 1.4.3.

or π, there f assumes the values 2
π and 2 π, respectively, or inside the interval, i.e., in the open interval p
π, π q. In the last case, according to Theorem 1.4.1 those are critical points of the restriction of f to this interval. The last are given by π 5π x
,
6 6 since f 1 pxq  1 2 sinpxq for all x P p
π, π q. Now f


π


6 



π

6

?

3





, f


5π6



 ?

?

3


5π 6

and hence the minimum value of f is
pπ {6q
3 (assumed inside the interval) and its maximum value is π 2 (assumed at the right boundary of the interval). See Fig. 33.

96

Theorem 1.4.4. (Rolle’s theorem) Let f : ra, bs Ñ R be continuous where a, b P R are such that a   b. Further, let f be differentiable on pa, bq and f paq  f pbq. Then there is c P pa, bq such that f 1 pcq  0. Proof. Since f is continuous, according to Theorem 1.2.26 f assumes its minimum and maximum value in some points x0 P ra, bs and x1 P ra, bs, respectively. Now if one of these points is contained in the open interval pa, bq, the derivative of f in that point vanishes by Theorem 1.4.1. Otherwise, if both of those points are at the interval ends a, b it follows f paq ¤ f pxq ¤ f pbq  f paq

for all x P ra, bs and hence that f is a constant function and by Example 1.3.3 that f 1 pcq  0, for every c P pa, bq. Hence in both cases the statement of the theorem follows. Example 1.4.5. Show that f : R Ñ R defined by f pxq : x3

x

1

for all x P R, has exactly one (real) zero. (Compare Example 1.2.30.) Proof. f is continuous and because of f p
1q 
1   0 and f p0q  1 ¡ 0

and Corollary 1.2.29 has a zero x0 in p
1, 0q. See Fig. 19. Further, f is differentiable with f 1 pxq  3x2 1 ¡ 0

for all x P R. Now assume that there is a another zero x1 . Then it follows by Theorem 1.4.4 the existence of a zero of f 1 in the interval with endpoints x0 and x1 . Hence f has exactly one zero. Theorem 1.4.6. (Mean value theorem) Let f : ra, bs Ñ R be a continuous function where a, b P R are such that a   b. Further, let f be differentiable on pa, bq. Then there is c P pa, bq such that f pbq
f paq b
a

97

 f 1pcq .

y

fHbL

fHaL

a

c

x

b

Fig. 34: Sketch for the Mean Value Theorem 1.4.6.

Proof. Define the auxiliary function h : ra, bs Ñ R by hpxq : f pxq
for all x with

P ra, bs.

f pbq
f paq b
a

 px
aq
f paq

Then h is continuous as well as differentiable on pa, bq f pbq
f paq b
a 0. Hence by Theorem 1.4.4 there is

h 1 pxq  f 1 pxq


for all x P pa, bq and hpaq c P pa, bq such that

 hpbq 

h 1 pcq  f 1 pcq


f pbq
f paq b
a

0.

Theorem 1.4.7. Let f : pa, bq Ñ R be differentiable, where a, b P R are such that a   b. Further, let f 1 pxq  0 for all x P pa, bq. Then f is a constant function. Proof. The proof is indirect. Assume that f is not a constant function. Then there are x1 , x2 P pa, bq satisfying x1  x2 and f px1 q  f px2 q. Hence it 98

follows by Theorem 1.4.6 the existence of c P px1 , x2 q such that f px2 q
f px1 q x2
x1

 f 1pcq  0

and hence that f px1 q  f px2 q. Hence f is a constant function. Example 1.4.8. (Energy conservation) Newton’s equation of motion for a point particle of mass m ¥ 0 moving on a straight line is given by mf 2 ptq  F pf ptqq

(1.4.1)

for all t from some non-empty open time interval I € R, where f ptq is the position of the particle at time t, and F pxq is the external force at the point x. Assume that F 
V 1 , where V is a differentiable function from an open interval J  f pI q. Show that E : I Ñ R defined by E ptq :

m 1 pf ptqq2 2

V pf ptqq

(1.4.2)

for all t P I is a constant function. Solution: It follows Theorem 1.3.8, Theorem 1.3.10 and (1.4.1) that E is differentiable with derivative E 1 ptq  mf 1 ptqf 2 ptq

V 1 pf ptqq  f 1 ptq  f 1 ptq rmf 2 ptq
F pf ptqqs  0

for all t P I. Hence according to Theorem 1.4.7, E is a constant function. In physics, its value is called the total energy of the particle. As a consequence, the finding of the solutions of the solution of (1.4.1), which is second order in the derivatives, is reduced to the solution of (1.4.2), which is only first order in the derivatives, for an assumed value of the total energy. Theorem 1.4.9. Let f : ra, bs Ñ R be continuous where a, b P R are such that a   b. Further, let f be differentiable on pa, bq and such that f 1 pxq ¡ 0 ( f 1 pxq ¥ 0 ) for every x P pa, bq. Then f is strictly increasing ( increasing ) on ra, bs. 99

Proof. Let x and y be some elements of ra, bs such that x   y. Then the restriction of f to the interval rx, y s satisfies the assumptions of Theorem 1.4.6, and hence there is c P px, y q such that f py q  f pxq

f 1 pcqpy
xq ¡ f pxq

p ¥ f pxq q

.

Example 1.4.10. Show that the exponential function exp : R Ñ R is strictly increasing. Solution: By Example 1.3.4 and Theorem 1.2.20 it follows that exp 1 pxq  exppxq ¡ 0 for all x P R. Hence it follows by Theorem 1.4.9 that exp is strictly increasing. Hence there is an inverse function to exp which is called the natural logarithm and is denoted by ln. See Fig. 24. Example 1.4.11. Show that (i) ex for all x P p0, 8q. (ii) ex for all x P p0, 8q.

¡1

¡x

(1.4.3)

1

(1.4.4)

(Compare Theorem 1.2.20.) Proof. Define the continuous function f : r0, 8q Ñ R by f pxq : ex
1 for all x P r0, 8q. Then f is differentiable on p0, 8q with f 1 pxq  ex ¡ 0 for all x P p0, 8q. Hence f is strictly increasing according to Theorem 1.4.9, and (1.4.3) follows since f p0q  e0
1  0. Further, define the continuous function g pxq : ex
1
x for all x P r0, 8q. Then g is differentiable on p0, 8q with g 1 pxq  ex
1 ¡ 0 for all x P p0, 8q where (1.4.3) has been applied. Hence (1.4.4) follows by Theorem 1.4.9 since g p0q  e0
0
1  0. 100

From Example 1.4.10 and (1.4.4), it follows by Theorem 1.2.28 that expp r0, 8q q  r1, 8q and hence by part (iii) of Theorem 1.2.20 that the range of exp is given by p0, 8q which therefore is also the domain of its inverse function ln. As a consequence, exp is a strictly increasing bijective map from R onto p0, 8q. See Fig. 24. Example 1.4.12. Show that lnpa  bq  lnpaq

lnpbq

for all a, b ¡ 0. Solution: For a, b ¡ 0, it follows by Theorem 1.2.20 that lnpa  bq  ln elnpaq  elnpbq



elnpaq

 ln

p q   lnpaq

ln b

lnpbq .

Lemma 1.4.13. (An ‘energy’ inequality for solutions of a differential equation) Let p, q P R. Further, let I be some open interval of R, x0 P I and f : I Ñ R satisfy the differential equation f 2 pxq

p f 1 pxq

q f pxq  0

for all x P I. Finally, define E pxq : pf pxqq2

pf 1pxqq2

for all x P I. Then for all x P I 0 ¤ E pxq ¤ E px0 q ek|x
x0 | , where

k : 1

2|p|

|q| .

Proof. Since f is twice differentiable, E is differentiable such that E 1 pxq  2f pxqf 1 pxq

2f 1 pxqf 2 pxq

101

 2f pxqf 1pxq
2 r p f 1 pxq q f pxq s f 1pxq  2 p1
qqf pxqf 1pxq
2 p pf 1 pxqq2 for all x P I. Hence E 1 is continuous and satisfies

|E 1pxq| ¤ 2 p1 |q|q |f 1 pxq| |f pxq| 2 |p| pf 1pxqq2 ¤ p1 |q|q pf pxqq2 pf 1pxqq2 2 |p| pf 1 pxqq2 ¤ kE pxq for all x P I, where it has been used that 2 |f 1 pxq| |f pxq| ¤ pf pxqq2 pf 1 pxqq2 . As a consequence,


kE pxq ¤ E 1 pxq ¤ kE pxq for all x P I. We continue analyzing the consequences of these inequalities. For this, we define auxiliary functions Er , El by Er pxq : e
kx E pxq , El pxq : ekx E pxq for all x P I. Then Er1 pxq  e
kx pE 1 pxq
kE pxqq ¤ 0 , El1 pxq  ekx pE 1 pxq

kE pxqq ¥ 0

for all x P I. Hence Er is decreasing, which is equivalent to the increasing of
Er , and Er is increasing. Hence it follows by Theorem 1.4.9 that E pxq ¤ E px0 q ekpx
x0 q

 E px0 q ek|x
x | 0

for x ¥ x0 and that E pxq ¤ E px0 q ekpx0
xq for x ¤ x0 .

102

 E px0q ek|x
x | . 0

Theorem 1.4.14. Let p, q P R. Further, let I be some open interval of R, x0 P I and y0 , y01 P R. Then there is at most one function f : I Ñ R such that f 2 pxq p f 1 pxq q f pxq  0 (1.4.5) for all x P I and at the same time such that

f px0 q  y0 , f 1 px0 q  y01 .

Ñ R be such that f 2 pxq p f 1 pxq q f pxq  f¯2 pxq p f¯1 pxq q f¯pxq  0 for all x P I and f px0 q  f¯px0 q  y0 , f 1 px0 q  f¯1 px0 q  y01 . Then u : f
f¯ satisfies u 2 pxq p u 1 pxq q upxq  0 for all x P I and upx0 q  u 1 px0 q  0 . Hence it follows by Lemma 1.4.13 that upxq  0 for all x P I and hence that f  f¯. Proof. For this, let f, f¯ : I

Definition 1.4.15. We define the hyperbolic sine function sinh, the hyperbolic cosine function cosh and the hyperbolic tangent function tanh by  1 x 1 x e
e
x , coshpxq : e 2 2 sinhpxq , tanhpxq : coshpxq

sinhpxq :

for all x metric

P R.

e
x



,

Obviously, sinh, tanh are antisymmetric and cosh is sym-

sinhp
xq 
sinhpxq , cosp
xq  coshpxq , tanhp
xq 
tanhpxq 103

y

3 2

-2

-1

1

2

x

-1 -2 -3

Fig. 35: Graphs of the hyperbolic sine and cosine function. y

0.5

-2

-1

1

2

x

-0.5

Fig. 36: Graphs of the hyperbolic tangent function and asymptotes given by the graphs of the constant functions on R of values 1 and
1.

104

for all x P R. Also these functions are differentiable and, in particular, sinh 1

 cosh ,

cosh 1

 sinh

similarly to the sine and cosine functions. Another resemblance to these functions is the relation cosh pxq
sinh pxq  2

 14



2

e
x
ex

ex

1 x e 4 e
x



2 2 e
x
ex
e
x

ex

e
x

ex
e
x

for all x P R. In particular, this implies that tanh 1 pxq : for all x P R.

cosh2 pxq
sinh2 pxq cosh2 pxq

Theorem 1.4.16. Let p, q the unique solution to





 14 2 e
x 2 ex  1

 1
tanh2pxq  cosh12pxq

P R, D : pp2{4q
q and x0 , y0, y01 P R. Then

f 2 pxq

pf 1 pxq

qf pxq  0

satisfying f px0 q  y0 and f 1 px0 q  y01 is given by 

f pxq  y0 e
ppx
x0 q{2 coshpD 1{2 px
x0 qq
py  0 D
1{2 y01 sinhpD 1{2 px
x0 qq 2

for x P R if D

¡ 0, 
py  0 y01 px
x0 q f pxq  e
ppx
x q{2 y0 2 for x P R if D  0 and  f pxq  y0 e
ppx
x q{2 cosp|D |1{2px
x0 qq
 |D|
1{2 py20 y01 sinp|D|1{2px
x0 qq for x P R if D   0. 0

0

105

Proof. For this, we first notice that for a function h : R Ñ R satisfies h 2 pxq

ph 1 pxq

for all x P R if and only if



¯h 2 pxq

q


qhpxq  0

p2 4



(1.4.6)

¯ pxq  0 h

(1.4.7)

¯ : R Ñ R is defined by for all x P R, where h ¯ pxq : epx{2hpxq h

(1.4.8)

¯ is twice differenfor all x P R. Indeed, it follows by Theorem 1.3.8 that h tiable if and only if h is twice differentiable and in this case that


¯ 1 pxq  epx{2 h 1 pxq h

p hpxq , 2

¯ 2 pxq  epx{2 h 2 pxq h

p h 1 pxq



for all x P R. The last implies that 



p2 hpxq 4







p2 ¯ p2 hpxq hpxq  epx{2 h 2 pxq p h 1 pxq q
4 4 

p2 px{2 e q
hpxq  epx{2 ph 2 pxq ph 1 pxq qhpxqq  0 4

¯ 2 pxq h

for all x P R if and only if (1.4.6) is satisfied for all x hpx0 q  y0 and h 1 px0 q  y01 if and only if
¯ 1 px0 q  py0 ¯ px0 q  y0 epx0 {2 , h h 2

P R.

In addition,



y01 epx0 {2 .

(1.4.9)

For the solution of (1.4.7) and (1.4.9), we consider three cases. If D : pp2{4q
q ¡ 0, then a solution to (1.4.7) and (1.4.9) is given by ¯ pxq  y0 epx0 {2 coshpD 1{2 px
x0 qq h 106

D
1{2


py



y01 epx0 {2 sinhpD 1{2 px
x0 qq

0

2 for x P R. If D  0, then a solution to (1.4.7) and (1.4.9) is given by
py 0 ¯ pxq  y0 epx0 {2 y01 epx0 {2 px
x0 q h 2 for x P R. If D   0, then a solution to (1.4.7) and (1.4.9) is given by

¯ pxq  y0 epx0 {2 cosp|D |1{2px
x0 qq h
|D|
1{2 py2 0 y01 epx0{2 sinp|D|1{2px
x0 qq for x P R. Hence, finally, it follows by (1.4.8) and by Theorem 1.4.14 the statement of this theorem. Theorem 1.4.17. (Derivatives of inverse functions) Let f : ra, bs Ñ R be continuous where a, b P R are such that a   b. Further, let f be differentiable on pa, bq and such that f 1 pxq ¡ 0 for every x P pa, bq. Then the inverse function f
1 is defined on rf paq, f pbqs as well as differentiable on pf paq, f pbqq with 1 1 f
1 py q  1
1 (1.4.10) f pf py qq for all y

P pf paq, f pbqq.

Proof. By Theorem 1.4.9, it follows that f is strictly increasing and hence that there is an inverse function f
1 for f . Further, by Theorem 1.2.35 f
1 is continuous, and by Theorem 1.2.34 it follows that f pra, bsq  rf paq, f pbqs and hence that f
1 is defined on rf paq, f pbqs. Now let y P pf paq, f pbqq and y1 , y2 , . . . be a sequence in pf paq, f pbqqzty u which is convergent to y. Then f
1 py1 q, f
1py2 q, . . . is a sequence in pa, bqztf
1 py qu which, by the continuity of f
1 , converges to f
1 py q. Hence it follows for n P N that f
1 pyn q
f
1 py q yn
y





f pf
1 pyn qq
f pf
1 py qq f
1 pyn q
f
1 py q


1

and hence by the differentiability of f in f
1 py q, that f 1 pf
1 py qq by Theorem 1.2.4 the statement (1.4.10). 107

¡ 0 and

Example 1.4.18. Calculate the derivative of ln, arcsin, arccos and arctan. Solution: By Theorem 1.4.17, it follows that ln 1 pxq 

1

exp 1 plnpxqq

 exppln1 pxqq  x1

for every x P p0, 8q, arcsin 1 pxq 

a

1

sin 1 parcsinpxqq

1  cosparcsin pxqq

?

1

1

, 1
x2 1
sin parcsinpxqq 1 1 arccos 1 pxq  
1 cos parccospxqq sinparccospxqq 1 
a 
? 1 2 2 1
x 1
cos parccospxqq 2

for all x P p
1, 1q and arctan 1 pxq  for every x P R.

1

tan 1 parctanpxqq

 p1

1 tan qparctanpxqq 2

1

1 x2

Example 1.4.19. In terms of Kruskal coordinates, the radial coordinate projection r : Ω Ñ p0, 8q of the Schwarzschild solution of Einstein’s field equation is given by r pu, v q  h
1 pu2
v 2 q for all pv, uq P Ω, where h : p0, 8q Ñ p
1, 8q is defined by hpxq :


x

2M


1



ex{p2M q

for all x P p0, 8q. Here Ω : tpv, uq P R2 : u2
v 2 108

¡
1 u ,

y 2

1

1

2

3

x

-1 Fig. 37: Graph of the auxiliary function h from Example 1.4.19.

and M ¡ 0 is the mass of the black hole. In addition, geometrical units are used where the speed of light and the gravitational constant have the value 1. Finally, h is bijective and h
1 is differentiable. Calculate

Br , Br . Bv Bu

for all pv, uq P Ω [7]. Solution: For this, let pv, uq conclude by Theorem 1.4.17 that

P Ω. In a first step, we

Br pu, vq 
2v  ph
1q 1pu2
v2 q 
2v  r h 1ph
1pu2
v2 qq s
1 Bv 
2v  r h 1prpu, vqq s
1 , Br pu, vq  2u  ph
1q 1pu2
v2q  2u  r h 1ph
1pu2
v2qq s
1 Bu  2u  r h 1prpu, vqq s
1 .

Since h 1 pxq 

1 x{p2M q e 2M

1
x 2M 2M 109


1



ex{p2M q

 4Mx 2  ex{p2M q

for every x ¡ 0, this implies that

Br pu, vq 
8M 2 v  e
r{p2M q pu, vq , Bv r 
r{p2M q

Br pu, vq  8M 2 u e pu, vq . Bu r Definition 1.4.20. (General powers) For every a responding power function by

P R, we define the cor-

xa : ealn x for all x ¡ 0. By Theorem 1.3.10, this function is differentiable with derivative a aln x e x

 xa  eln x

pa
1qln x  a  eln x  epa
1qln x  a  xpa
1q x

for all x ¡ 0. Example 1.4.21. Show that for every a ¡ 0 lnpxq  

1 a px
1q a

(1.4.11)

for all x ¡ 1. (See Exercise 1.2.2 for an application of the case a Solution: Define the continuous function f : r1, 8q Ñ R by f pxq :

1 a px
1q
lnpxq a

for all x ¥ 1. Then f is differentiable on p1, 8q with f 1 pxq 

1 a aln x 1  e
x a x

 x1

ealn x
1



¡0

for x ¡ 1 and f p1q  0. Hence (1.4.11) follows by Theorem 1.4.9. 110

 1{2.)

A verbalization of the estimate (1.4.11) is that ‘ lnpxq is growing slower than any positive power of x for large x ’. An important consequence of Theorem 1.4.6 is a generalization given by Taylor’s theorem. Definition 1.4.22. If m, n we define n ¸



P N such that m ¤ n and and am , . . . , an P R,

ak : am

am

. . . an .

1

k m

Note that, as a consequence of the associative law for addition, it is not necessary to indicate the order in which the summation is to be performed. Further, obviously, n ¸



pak

bk q 





n ¸

ak



k m





k m

and λ

n ¸



ak



bk

k m n ¸



k m

n ¸



λ ak

k m

for every λ P R and bm , . . . , bn P R. In addition, we define for every n P N the corresponding factorial n! recursively by 0! : 1 , pk

1q! : pk

for every k P N . Hence in particular, 1! 5!  120 and so forth.

1qk!

 1, 2!  2, 3!  6, 4!  24,

Theorem 1.4.23. (Taylor’s theorem) Let n P N , I be a non-trivial open interval and f : I Ñ R be n
times differentiable. Finally, let a and b be two different elements from I. Then there is c in the open interval between a and b such that f pbq 


f pkq paq

n¸1



k 0

k!

pb
aqk 111

f pnq pcq pb
aqn . n!

(1.4.12)

Ñ R by n¸
1 f pkq pxq g pxq : f pbq
pb
xqk

Proof. Define the auxiliary function g : I

k!



k 0

for all x P I. Then it follows that g pbq  0 and moreover that g is differentiable with n¸
1 f pk 1q pxq 1 pb
xqk g pxq 
k! k 0 pnq 
f pxq pb
xqn
1


f pkq pxq

n¸1

pk
1q! pb
xq k 1

pn
1q!

for all x P I. Define a further auxiliary function h : I hpxq : g pxq




for all x P I. Then it follows that hpaq tiable with h 1 pxq 


b
x b
a

n




k 1

Ñ R by

g paq

 hpbq  0 and that h is differen-

f pnq pxq p b
xqn
1 pn
1q!

n

pb
xqn
1 gpaq pb
aqn

for all x P I. Hence according to Theorem 1.4.4, there is c in the open interval between a and b such that 0  h 1 pcq 


f pnq pcq pn
1q pn
1q! pb
cq

n

pb
cqn
1 gpaq pb
aqn

which implies (1.4.12). Taylor’s Theorem 1.4.23 is usually applied in the following form: Corollary 1.4.24. (Taylor’s formula) Let n P N , I be a non-trivial open interval of length L and f : I Ñ R be n
times differentiable. Finally, let x0 P I and C ¥ 0 be such that

|f pnqpxq| ¤ C 112

for all x P I. Then

   f x 

p q



f pkq px q 0

n¸1

k!



k 0

   x0 k  

n

px
q ¤ CL n!

.

for all x P I. Remark 1.4.25. The polynomial pn
1 pxq :


f pkq px q 0

n¸1



k 0

k!

px
x0 qk

for all x P R in Corollary 1.4.24 is called ‘ the pn
1q-degree polynomial of f centered at x0 ’. In particular, it follows (for the case n  2) that: p1 pxq  f px0 q

f 1 px0 q px
x0 q

for all x P R which is also called the ‘linearization or linear approximation of f at x0 ’ and 2 |f pxq
p1 pxq| ¤ CL 2 if C ¥ 0 is such that |f 2pxq| ¤ C for all x P I. In applications, one often meets the notation f pxq  p1 pxq saying that f and p1 are approximately the same near x0 . If the error can be seen to be ‘negligible’ for the application, this often leads to a replacement of f by its linearization. Example 1.4.26. Calculate the linearization p1 of f : r
1, 8q Ñ R defined by ? f pxq : 1 x 113

y 1.25 1.2 1.15 1.1 1.05 0.1

0.2

0.3

0.4

0.5

x

Fig. 38: Graphs of f and p1 from Corollary 1.4.26.

for all x P r
1, 8q at x  0, and estimate its error on the interval r0, 1{2s. Solution: f is twice differentiable on p
1, 8q with f 1 pxq 

1 p1 2

Hence p1 is given by for all x P R Because of

xq
1{2 , f 2 pxq 
p1 pxq  1  1  1 4

p

1 p1 4

xq
3{2 .

1 x 2 

 1 xq
3{2  ¤ 

4

for all x P r0, 1{2s, it follows from (1.4.25) that the absolute value of the relative error satisfies |p1pxq
f pxq| ¤ 1 |f pxq| 32

for all x P r0, 1{2s.

114

Theorem 1.4.27. Let f : pa, bq Ñ R be twice differentiable on pa, bq, where a, b P R are such that a   b, and such that f 2 pxq ¡ 0 (f 2 pxq   0) for all x P pa, bq. Then f pxq ¡ f px0 q

f 1 px0 qpx
x0 q

for all x0 , x P pa, bq such that x0

p f pxq   f px0 q

f 1 px0 qpx
x0 q

q

 x, i.e., ‘f is convex’ (‘f is concave’). Proof. First, we consider the case that f 2 pxq ¡ 0 for all x P pa, bq. For this, let x0 P pa, bq and x P pa, bq be such that x ¡ x0 . According to Theorem 1.4.6, there is c P px0 , xq such that f pxq
f px0 q  f 1pcq . x
x0 By Theorem 1.4.9, it follows that f 1 is strictly increasing on rx0 , xs and hence that f pxq
f px0 q  f 1 pcq ¡ f 1 px0 q x
x0 and that

f pxq ¡ f px0 q

f 1 px0 qpx
x0 q .

f px0 q
f pxq x0
x

 f 1pcq   f 1px0 q

(1.4.13)

Analogously for x P pa, bq such that x   x0 , it follows that there c P px, x0 q such that f px0 q
f pxq  f 1pcq x0
x and such that f 1 strictly increasing on rx, x0 s and hence that

which implies (1.4.13). In the remaining case that f 2 pxq pa, bq, application of the previous to
f gives


f pxq ¡
f px0 q
f 1 px0qpx
x0 q

and hence for all x0

f pxq   f px0 q

P pa, bq and x P pa, bqztx0 u.

f 1 px0 qpx
x0 q

115

  0 for all x P

y

30 25 20 15 10 5

-1

1

2

3

x

Fig. 39: Graphs of exp along with linearizations around x  1, 2 and 3.

Example 1.4.28. The exponential function exp is convex because of exp 2 pxq  exppxq ¡ 0 for all x P R. See Fig. 39. Example 1.4.29. Find the intervals of convexity and concavity of f : R Ñ R defined by f pxq : x4 x3
2x2 1 for all x P R. Solution: f is twice continuously differentiable with f 1 pxq  4x3

 12



x



3x
4x , f 2 pxq  12x2

6x
4  12 x

2

1 4

 

19 48



x

1 4




2



19 48

for all x P R. Hence f is convex on the intervals 


8,
14






19 48



, 116


14

19 ,8 48



1 1 x
2 3



y 4

2

-2

-1

x

1 -2

Fig. 40: Graph of f from Example 1.4.29 and parallels to the y
axis through its inflection points.

and concave on the interval 


14




19 1 ,
48 4



19 48

.

Definition 1.4.30. Let f : pa, bq Ñ R be differentiable on pa, bq, where a, b P R are such that a   b. We call f convex (concave) if f pxq ¡ f px0 q

f 1 px0 qpx
x0 q

for all x0 , x P pa, bq such that x0

p f pxq   f px0 q

f 1 px0 qpx
x0 q

q

 x.

The following theorem gives another useful characterization of a function defined on interval I € R to be convex. Such function is convex if and only if for every x, y P I such that x   y the graph of f |px,yq lies below the straight line between px, f pxqq and py, f py qq. 117

y

x Fig. 41: Graph of a convex function (black) and secant (blue). Compare Theorem 1.4.31.

Theorem 1.4.31. Let f : pa, bq Ñ R be differentiable on pa, bq, where a, b P R are such that a   b. Then f is convex if and only if f pz q   f pxq for all x, y, z

pz
xq f pyyq

fxpxq



 f pyq
py
zq f pyyq

fxpxq



P pa, bq such that x   z   y.

Proof. If f is convex, we conclude as follows. For the first step, let x, y P pa, bq be such that x   y. As a consequence of the convexity of f , it follows that f py q ¡ f pxq and hence that

f 1 pxqpy
xq , f pxq ¡ f py q f 1 pxq  

f py q
f pxq y
x

f 1 py qpx
y q

  f 1pyq . This is true for all x, y P pa, bq be such that x   y. Not that this implies that f 1 is strictly increasing. For the second step, let x, y, z P pa, bq be such 118

that x   z   y. By the mean value theorem Theorem 1.4.6, it follows the existence of ξ P px, y q such that f py q
f pxq y
x

 f 1 pξ q .

¤ z, it follows by help of the first step that f py q
f pxq f py q
f pz q  f 1 pξ q ¤ f 1 pz q   y
x y
z

In the case that ξ

and hence that f pz q   f py q
py
z q

f py q
f pxq . y
x

¤ ξ, it follows by help of the first step that f py q
f pxq f pz q
f pxq  f 1 pξ q ¥ f 1 pz q ¡ y
x z
x

In the case that z

and hence that f pz q   f pxq

pz
xq f pyyq

fxpxq .

On the other hand, if f pz q   f pxq

pz
xq f pyyq

fxpxq

for all x, y, z P pa, bq such that x   z this, note that the previous implies that f pz q
f pxq z
x



 f pyq
py
zq f pyyq

fxpxq

  y, we conclude as follows.

For

  f pyyq

fxpxq   f pyyq

fz pzq .

In the following, let x, y, z, ξ P pa, bq be such that x follows from the assumption that f pz q
f pxq z
x



 z ξ 

  f pξξq

fxpxq   f pyyq

fxpxq . 119

y. It

Ñ x that f pξ q
f pxq f py q
f pxq f 1 pxq ¤ .   ξ
x y
x

From this, it follows by taking the limit z

This implies that f 1 pxq  

f py q
f pxq y
x

and therefore that f py q ¡ f pxq

f 1 pxqpy
xq .

(1.4.14)

Also, it follows from the assumption that f py q
f pxq y
x

  f pyyq

fz pzq   f pyyq

fξ pξ q .

Ñ y that f py q
f pz q f py q
f pxq   ¤ f 1pyq . y
x y
z

From this, it follows by taking the limit ξ

This implies that

f py q
f pxq y
x

  f 1pyq

and therefore that f pxq ¡ f py q

f 1 py qpx
y q .

Since (1.4.14) is true for all x, y P pa, bq such that x following for x, y P pa, bq such that y   x f pxq ¡ f py q

f 1 py qpx
y q .

Finally, from this and (1.4.15), it follows that f pxq ¡ f py q for all x, y

f 1 py qpx
y q .

P pa, bq such that x  y. 120

(1.4.15)

  y, we conclude the

A typical example for the application of Theorems 1.4.27, 1.4.31 is the following. Example 1.4.32. Show that sinpxq ¥ 2x{π

(1.4.16)

for all x P r0, π {2s. Solution: By application of Theorem 1.4.27, it follows that the restriction of
sin to p0, π q is convex. According to Theorem 1.4.31, this implies that

p1{nq
1
sinpxq ¤
sinp1{nq rx
p1{nqs  psin π {2q
p1{nq for all x P r1{n, π {2s, where n P N . By taking the limit n Ñ 8, this leads to
sinpxq ¤
2x{π

for all x P p0, π {2s. From the last and the fact that (1.4.16) is trivially satisfied for x  0, it follows the validity of (1.4.16) for all x P r0, π {2s. Corollary 1.4.33. (Sufficient condition for the existence of a local minimum/maximum) Let f be a twice continuously differentiable real-valued function on some open interval I of R. Further, let x0 P I be a critical point of f such that f 2 pxq ¡ 0 (f 2 pxq   0). Then f has a local minimum (maximum) at x0 . Proof. Since f 2 is continuous with f 2 px0 q ¡ 0 (f 2 px0 q   0), there is an open interval J around x0 such that f 2 pxq ¡ 0 (f 2 pxq   0) for all x P J. ((Otherwise there is for every n P N some yn P I such that |yn
x0 |   1{n and f 2 pyn q ¤ 0 (f 2 pyn q ¥ 0). In particular, this implies that limnÑ8 yn  x0 and by the continuity of f 2 also that limnÑ8 f 1 pyn q  f 2 px0 q. Hence it follows by Theorem 1.2.6 that f 2 px0 q ¤ 0 (f 2 px0 q ¥ 0). )) Hence it follows by Theorem 1.4.27 that f pxq ¡ f px0 q (f pxq   f px0 q) for all x P J ztx0 u.

121

y

0.8

0.2 -6

-4

-2

2

4

6

x

Fig. 42: Graph of f from Example 1.4.34.

Example 1.4.34. Find the values of the local maxima and minima of f pxq : ln



5 4

sin pxq



2

for all x P R. Solution: f is twice continuously differentiable with f 1 pxq 

5 4

sinp2xq 2 cosp2xq , f 2 pxq  5 2 sin pxq sin2 pxq 4




sin2 p2xq 5 4

sin2 pxq

for all x P R. Hence the critical points of f are at xk : kπ {2, k for each k P Z: 2p
1qk f 2 pxk q  5 . sin2 pxk q 4

2

P Z and

Hence it follows by Theorem 1.4.33 that f has a local minimum/maximum of value lnp5{4q at x2k and of value lnp9{4q at x2k 1 , respectively, and each k P Z. 122

Another important consequence of Theorem 1.4.6 is given by Cauchy’s extended mean value theorem which is the basis of L’Hospital’s rule Theorem 1.4.36 for computing indeterminate forms. Theorem 1.4.35. (Cauchy’s extended mean value theorem) Let f, g : ra, bs Ñ R be continuous functions where a, b P R are such that a   b. Further, let f, g be continuously differentiable on pa, bq and such that g 1 pxq  0 for all x P pa, bq. Then there is c P pa, bq such that f pbq
f paq g pbq
g paq

 fg 1ppccqq . 1

(1.4.17)

Proof. Since g 1 is continuous with g 1 pxq  0 for all x P pa, bq, it follows by Theorem 1.2.28 that either g 1 pxq ¡ 0 or g 1 pxq   0 for all x P pa, bq and hence by Theorem 1.2.35 that g is either strictly increasing or strictly decreasing on pa, bq. Since g is continuous, from this also follows that g pbq  g paq. Define the auxiliary function h : ra, bs Ñ R by hpxq : f pxq
f paq


f pbq
f paq g pbq
g paq

 pgpxq
gpaqq

for all x P ra, bs. Then h is continuous as well as differentiable on pa, bq such that f pbq
f paq 1 h 1 pxq  f 1 pxq
 g pxq g pbq
g paq for all x P ra, bs and hpaq  hpbq  0. Hence according to Theorem 1.4.4, there is c P pa, bq such that h 1 pcq  f 1 pcq


f pbq
f paq 1  g pcq  0 g pbq
g paq

which implies (1.4.17). Theorem 1.4.36. (Indeterminate forms/L’Hospital’s rule) Let f : pa, bq Ñ R and g : pa, bq Ñ R be continuously differentiable, where a, b P R are such that a   b, and such that g 1 pxq  0 for all x P pa, bq. Further, let lim f pxq  lim g pxq  0

x

Ña

x

Ña

123

(1.4.18)

or let |f pxq| ¡ 0 and |g pxq| ¡ 0 for all x P pa, bq as well as 1

Ña |f pxq|

lim

x

Finally, let

1  xlim 0. Ña |g pxq|

(1.4.19)

f 1 pxq xÑa g 1 pxq lim

exist. Then

f pxq xÑa g pxq lim

f 1 pxq  xlim . Ña g 1 pxq

(1.4.20)

Proof. Since g 1 is continuous with g 1 pxq  0 for all x P pa, bq, it follows by the Theorem 1.2.28 that either g 1 pxq ¡ 0 or g 1 pxq   0 for all x P pa, bq and hence by Theorem 1.2.35 that g is either strictly increasing or strictly decreasing on pa, bq. First, we consider the case (1.4.18). Then f and g can be extended to continuous functions on ra, bq assuming the value 0 in a. Now, let x0 , x1 , . . . be a sequence of elements of pa, bq converging to a. Then by Theorem 1.4.35 for every n P N there is a corresponding cn P pa, xn q such that f 1 pcn q f pxn q  . g pxn q g 1 pcn q Obviously, the sequence c0 , c1 , . . . is converging to a, and hence it follows that f pxn q f 1 pcn q f 1 pxq lim 1 lim 1 (1.4.21)   lim nÑ8 g pcn q xÑa g pxq nÑ8 g pxn q and hence, finally, that (1.4.20). Finally, we consider the second case. So let |f pxq| ¡ 0 and |g pxq| ¡ 0 for all x P pa, bq, and in addition let (1.4.19) be satisfied. Further, let x0 , x1 , . . . be some sequence of elements of pa, bq converging to a and let b 1 P pa, bq. Because of (1.4.19), there is n0 P N such that    1   f pb 1 q      1 and  g pb q    1 .   f px q   g px q  n n

124

for all n P N such that n ¥ n0 . Then according to Theorem 1.4.35 for any such n, there is a corresponding cn P pxn , b 1 q such that f pxn q
f pb 1 q g pxn q
g pb 1 q

1  fg 1 ppccnqq . n

Hence it follows that f pxn q g pxn q

1 1
ggppxbnqq 1 1
ffppxbnqq



1  fg 1 ppccnqq n

and since c0 , c1 , . . . is converging to a by (1.4.19) and Theorem 1.2.4, it follows the relation (1.4.21) and hence, finally, (1.4.20). Example 1.4.37. Find

lim x lnpxq .

x

Ñ0

Solution: Define f pxq : lnpxq and g pxq : 1{x for all x P p0, 1q. Then f and g are continuously differentiable and such that g 1 pxq 
1{x2  0 for all x P p0, 1q. Further, |f pxq|  | lnpxq| ¡ 0, |g pxq|  |1{x|  1{|x| ¡ 0 for all x P p0, 1q. Finally, (1.4.19) is satisfied and f 1 pxq xÑ0 g 1 pxq lim

 xlim p
xq  0 . Ñ0

Hence according to Theorem 1.4.36: lim x lnpxq  0 .

x

Ñ0

Example 1.4.38. Determine lim xe
x .

x

Ñ8

Solution: Define f py q : 1{y and g py q : expp1{y q for all y Then f and g are continuously differentiable and such that g 1 py q

125

P p0, 1q. 
y
2

expp1{y q  0 for all y P p0, 1q. Further, |f py q|  1{|y | ¡ 0, |g py q| expp1{y q ¡ 0 for all y P p0, 1q. Finally, (1.4.19) is satisfied and f 1 py q y Ñ0 g 1 py q lim

 ylim Ñ0

1

e1{y



0.

Hence according to Theorem 1.4.36: lim xe
x

x

Ñ8

1{y  ylim 0. Ñ0 e1{y

Example 1.4.39. Calculate lim x2 e
x .

x

Ñ8

Solution: Define f py q : 1{y 2 and g py q : expp1{y q for all y P p0, 1q. Then f and g are continuously differentiable as well as such that g 1 py q 
expp1{yq{y2  0 for all y P p0, 1q. Further, |f pyq|  1{y2 ¡ 0, |gpyq|  expp1{yq ¡ 0 for all y P p0, 1q. Finally, (1.4.19) is satisfied and by Example 1.4.38 f 1 py q y Ñ0 g 1 py q lim

2{y  ylim 0. Ñ0 e1{y

Hence according to Theorem 1.4.36: lim x2 e
x

x

Ñ8

1{y 2  ylim 0. Ñ0 e1{y

Remark 1.4.40. Recursively in this way, it can be shown that lim xn e
x

x

Ñ8

0.

for all n P N. That the condition that g 1 pxq  0 for all x P pa, bq in Theorem 1.4.36 is not redundant can be seen from the following example. 126

Example 1.4.41. For this define f pxq :



2 x

sin

2 x

, g pxq :



 

2 x

2 x

sin

e sinp1{xq

for all x P p0, 2{5q. Then f and g are continuously differentiable and satisfy 1 xÑ0 |f pxq| lim

Since

1  xlim 0. Ñ0 |g pxq|

f pxq g pxq

 e
sinp1{xq

for all x P p0, 2{5q, pf {g qpxq does not have a limit value for x Ñ 0. Further, it follows that 

 

2 f 1 pxq 
2 1



2 4 1 
cos 2 cos 2 x x x x 

 

1 1 2 2 e sinp1{xq sin g 1 pxq 
2 cos x x x x

,  

4 cos

1 x

and hence that f 1 pxq g 1 pxq for all

2

x P p0, 2{5q

4x cosp1{xq e
sinp1{xq x sinp2{xq 4x cosp1{xq "

z p2k

2

1qπ

:k

PN

*

.

We notice that lim

x

Ñ0 2

4x cosp1{xq e
sinp1{xq x sinp2{xq 4x cosp1{xq

0.

This does not contradict Theorem 1.4.36 since g 1 has zeros of the form 2{pp2k 1qπ q, k P N. Hence there is no b ¡ 0 such that the restrictions of f and g would satisfy the assumptions in Theorem 1.4.36. 127

The following example shows that in general the existence of f pxq xÑa g pxq lim

does not imply the existence of f 1 pxq lim 1 . xÑa g pxq Example 1.4.42. For this, define f pxq : x sinp1{x2 q expp
1{xq , g pxq : expp
1{xq for all x ¡ 0. Then

f pxq xÑ0 g pxq

lim f pxq  0 , lim g pxq , lim

x

Ñ0

x

Ñ0

0.

Further, f 1 pxq  g 1 pxq 

 1  2 2 x p x 1 q sin p 1 { x q
2 cos p 1 { x q expp
1{xq , x2 f 1 pxq 1 p
1 { x q , exp  xpx 1q sinp1{x2 q
2 cosp1{x2 q x2 g 1 pxq

for all x ¡ 0. Hence f 1 {g 1 does not have a limit value for x Ñ 0. Lemma 1.4.43. (Contraction Mapping Lemma on the real line) Let T : ra, bs Ñ R be such that T pra, bsq € ra, bs where a, b P R are such that a   b. In addition, let T be a contraction, i.e., let there exist α P r0, 1q such that |T pxq
T pyq| ¤ α  |x
y| (1.4.22) for all x, y P ra, bs. Then T has a unique fixed point, i.e., a unique x ra, bs such that T px q  x . 128

P

Further,

and

|x
x | ¤ |x 1

T αpxq|

(1.4.23)

lim T n pxq  x

(1.4.24)

n

Ñ8

for every x P ra, bs where T n for n P N is inductively defined by T 0 : idra,bs and T k 1 : T  T k , for k P N. Proof. Note that (1.4.22) implies that T is continuous. Further, define the hence continuous function f : ra, bs Ñ R by f pxq : |x
T pxq| for all x P ra, bs. Note that x P ra, bs is a fixed point of T if and only if it is a zero of f . By Theorem 1.2.26 f assumes its minimum in some point x P ra, bs. Hence 0 ¤ f px q ¤ f pT px qq  |T px q
T pT px qq| ¤ α |x
T px q|  α f px q and therefore f px q  0 since the assumption f px q  0 leads to the contradiction that 1 ¤ α. If x¯ P ra, bs is a fixed point of T , then

|x
x¯|  |T pxq
T px¯q| ¤ α  |x
x¯| and hence x¯  x since the assumption x¯  x leads to the contradiction that 1 ¤ α. Finally, let x P ra, bs. Then |x
x |  |x
T pxq|  |x
T pxq T pxq
T pxq| ¤ |x
T pxq| |T pxq
T pxq| ¤ f pxq α  |x
x | and hence (1.4.23). Further from (1.4.23) and n

|T npxq
x |  |T npxq
T npxq| ¤ αn|x
x | ¤ 1 α
α f pxq , it follows (1.4.24) since limnÑ8 αn

 0.

129

Example 1.4.44. (Babylonian method of approximating roots by rational numbers) Let a ¡ 0 and N P N be such that N 2 ¡ a. Finally, define ? T : r a, N s Ñ R by 1
a T pxq : x 2 x ? for all x P r a, N s. Then lim T n pN q 

n

Ñ8

?

a.

(1.4.25)

Proof. First, we note that

?

T p aq 

?

?

a , T pN q 

1
N 2

a  N N

and ? hence that a is a fixed point of T . Further, T is twice differentiable on p a, N q with derivatives T 1 pxq 

?

 1
a 1 a 1
2  2 x2
a ¡ 0 , T 2 pxq  3 2 x 2x x

¡0

N q. Hence T, T 1 are strictly increasing according to Theofor all x P p a,? ? rem 1.2.35, T pr a, N sq € r a, N s and 0 ¤ T 1 pxq ¤

1
a 1 1
2   . 2 N 2

In particular, it follows by Theorem 1.4.6 that

?

|T pxq
T pyq| ¤ 12  |x
y|

for all x, y P r a, N s. By Lemma 1.4.43, it follows that T has a unique ? fixed point, which hence is given by a, and in particular (1.4.25).

?

For instance for 2, we get in this way (with N : 2) the first five approximating fractions 3 17 577 665857 886731088897 , , , , 2 12 408 470832 627013566048 130

y 14

-2

12

x2

x1

x0

x

Fig. 43: Graph of f from Example 1.4.45 (a  2) and Newton steps starting from x0

 4.

with corresponding errors (according to (1.4.23)) equal or smaller than 1 1 1 1 1 , , , , . 6 204 235416 313506783024 555992422174934068969056 The Newton-Raphson method is used to find zeros of the equation f pxq  0, where f : I Ñ R is a differentiable function on a non-trivial open interval I of R. The method is iterative. Starting from an approximation xn P I to such a root, the correction xn 1 is given by the zero of the linearization around xn : f pxn q f 1 pxn qpx
xn q , x P R and hence by

xn assuming f 1 pxn q

1

 xn
ff 1ppxxnqq n

 0, thereby essentially replacing the function f by its

linearization around xn .

131

Example 1.4.45. Let a ¡ 0. Define f : R Ñ R by f pxq : x2
a for all x P R. Then xn for xn

1

 xn


f pxn q f 1 pxn q



1 2



xn

a xn



 0 which is the iteration used in Example 1.4.44.

Theorem 1.4.46. (Newton-Raphson) Let f be a twice differentiable realvalued function on a non-trivial open interval I of R. Further, let I contain a zero x0 of f and be such that f 1 pxq  0 for all x P I and in particular such that    f pxqf 2 pxq     f 12 pxq  ¤ α for all x P I and some α P R satisfying 0 ¤ α   1. Then lim T n pxq  x0

n

for all x P I where Finally, for all x P I.

Ñ8

T pxq : x


(1.4.26)

f pxq . f 1 pxq

|x
x0 | ¤ |x 1

T αpxq|

(1.4.27)

Proof. First, it follows that T is differentiable with derivative T 1 pxq  for all x that

f pxqf 2 pxq f 12 pxq

P I and that x0 is a fixed point of T . By Theorem 1.4.6 it follows      T pxq
T px0 q   T pxq
x0        x
x  1 x
x0 0 132

y

0.5

-1

-0.5

0.5

1

x

-0.5 -1 Fig. 44: Zero of f from Example 1.4.47 given by the x
coordinate of the intersection of two graphs.

for all x P I different from x0 and hence that

|T pxq
x0 | ¤ |x
x0 | (1.4.28) for all x P I. Now let ra, bs, where a, b P R such that a   b, be some closed subinterval of I containing x0 . Then it follows by (1.4.28) that T pra, bsq € ra, bs and by Theorem 1.4.6 that    T pxq
T py q   ¤α   x
y for all x, y P ra, bs satisfying x  y and hence that |T pxq
T pyq| ¤ α|x
y| for all x, y P ra, bs. Hence by Lemma 1.4.43, the relations (1.4.26) and (1.4.27) follow for all x P ra, bs. 133

Example 1.4.47. Find an approximation x1 to the solution of x0

 cospx0 q

such that |x0
x1 |   10
6 . Solution: Define f : R Ñ R by f pxq : x
cospxq for all x P R. Then f is infinitely often differentiable with f 1 pxq  1 sinpxq , f 2 pxq  cospxq f pxqf 2 pxq cospxq  px
cospxqq  1 2 f pxq p1 sinpxqq2 where only in the last identity it has to be assumed that x is different from π {2 2kπ for all k P Z. Further f


π

6


π 6

?

3 2

 0,

f


π

4

 π4
?1 ¡ 0 , 2

and hence according Theorem 1.2.28, f has a zero in the open interval I : pπ {6, π {4q. Also f 1 pxq  1 for all x P I. Further, 

f f2 f 12

sinpxq ¡ 1

1

sinpπ {6q  3{2 ¡ 0

pxq  3 cosppx1q

and 3 cospxq x sinpxq
2x ¥ 3 cos


π

x sinpxq
2x sinpxqq2


π π π
sin 2 6 6 4

4

 ?3
5π ¡0 12 2

and hence f f 2 {f 12 is strictly increasing on rπ {6, π {4s as a consequence of Theorem 1.4.9. Therefore, 1

9 27

?



3π



cos

π 6



p1




cos  sin π6 q2

134

π 6

π 6



2   f pfxq12fpxpqxq

  and

cos

π 4



p1




cos  sin π4 q2 π 4

π 4





?

π
2 2 ? 8 6 2

p q p q   1
9
?3 π   α : 1   1 pq 27 3

  f x f 2 x     f 12 x 

for all x P I. Starting the iteration from 0.7 gives to six decimal places 0.739436 , 0.739085 with the corresponding errors 0.000527006 , 4.08749  10
8 . Hence the zero x0 of f in the interval I agrees with x1

 0.739085

to six decimal places. That there is no further zero of f can be concluded as follows. Since the derivative of f does not vanish in the interval p
π{2, π{2q, it follows by Theorem 1.4.4 that there are no other zeros in this interval. Further, for |x| ¥ π {2 p¡ 1q there is no zero of f because | cospxq| ¤ 1 for all x P R. The quantity U2

pU1
U2 q x20

is the ground state energy of a particle in a finite square well potential with U3  U1 , γ  0, KL  2. See [31].

Problems 1) Give the maximum and minimum values of f and the points where they are assumed. a) f pxq : x2 5x b) f ptq : t3 6t2

7 , x P r
5, 0s , 9t 14 , t P r
5, 0s ,

135

c) d) e) f) g)

f psq : s4 p8{3qs3
6s2 1 , s P r
5, 5s , f ptq : 4pt
3q2  pt2 1q , t P r
1, 4s , f pxq : p9x 12q{p3x2
4q , x P r
1, 0s , f pxq : px2 x 1q expp
xq , x P r
0.3, 1.5s , ? f pxq : expp
x{ 3 q cospxq , x P r0, 8q .

2) Consider a projectile that is shot into the atmosphere. If v ¥ 0 is the component of its speed at initial time 0 in the vertical direction, its height z ptq above ground at time t ¥ 0 is given by z ptq  vt
gt2 {2, where g  9.81m{s2 is the acceleration due to gravity and it is assumed that z p0q  0. Calculate the maximal height the projectile reaches and also the time of its flight, i.e., the time when it returns to the ground. 3) Reconsider the situation from previous problem, but now with inclusion of a viscous frictional force opposing the motion of the projectile. Then z ptq  α rpv αg q  p1
expp
t{αqq
gts, where it is again assumed that z p0q  0. Here α  m{λ, where m ¡ 0 is the mass of the projectile, and λ ¡ 0 is a parameter describing the strength of the friction. Calculate the maximal height the projectile reaches and also the time of its flight, i.e., the time when it returns to the ground.

4) Let a ¡ 0 and b ¡ 0. Find an equation for the straight line through the point pa, bq that cuts from the first quadrant a triangle of minimum area. State that area. 5) Let a ¡ 0 and b ¡ 0. Find an equation for the straight line through the point pa, bq whose intersection with the first quadrant is shortest. State the length of that intersection. 6) Find the maximal volume of a cylinder of given surface area A ¡ 0.

7) From each corner of a rectangular cardboard of side lengths a ¡ 0 and b ¡ 0, a square of side length x ¥ 0 is removed, and the edges are turned up to form an open box. Find the value of x for which the volume of that box is maximal. 8) A rectangular movie screen on a wall is h1 -meters above the floor and h2 -meters high. Imagine yourself sitting in front of the screen and looking into the direction of its center. Measured in this direction, what distance x from the wall will give you the largest viewing angle θ of the movie screen? [This is the angle between the straight lines that connect your eyes to the lowest and the highest points on the

136

screen.] Assume that the height of your eyes above the floor is hs meters, where hs   h1 .

9) Imagine that the upper half-plane H : R  p0, 8q and the lower half-plane H
: R p
8, 0q of R2 are filled with different ‘physical media’ with the x
axis being the interface I. Further, let px1 , y1 q P H , px2 , y2 q P H
. Light rays in both media proceed along straight lines and at constant speeds v1 and v2 , respectively. According to Fermat’s principle, a ray connecting px1 , y1 q and px2 , y2 q chooses the path that takes the least time. Show that that path satisfies Snell’s law, i.e., sinpθ1 q{ sinpθ2 q  v1 {v2 , where θ1 (θ2 ) is the angle of the part of the ray in H Y I (H
Y I) with the normal to the x
axis originating from its intersection with I. 10) For the following functions find the intervals of increase and decrease, the local maximum and minimum values and their locations and the intervals of convexity and concavity and the inflection points. Use the gathered information to sketch the graph of the function. If available, check your result with a graphing device. a) f psq : 7s4
3s2 1 , s P R , b) f ptq : t4 p8{3qt3
6t2 3 , t P R , c) f pxq : 4px
3q2  px2 1q , x P R .

11) For the following functions find vertical and horizontal asymptotes, the intervals of increase and decrease, the local maximum and minimum values and their locations and the intervals of convexity and concavity and the inflection points. Use the gathered information to sketch the graph of the function. If available, check your result with a graphing device. a) f pxq : x{p1 x2 q , x P R , ? b) f pxq : x2 1
x , x P R , ? ? c) f pxq : p9x 12q{p3x2
4q , x P Rzt
2{ 3, 2{ 3u .

12) Calculate the linearization of f around the given point. a) b) c) d) e) f)

f pxq : p1 xqn , x ¡
1 , around x  0 , where n P R , f pxq : lnpxq , x ¡ 0 , around x  1 , f pϕq : sinpϕq , ϕ P R , around ϕ  0 , f pϕq : tanpϕq , ϕ P p
π {2, π {2q , around ϕ  0 , f pxq : sinhpxq : pex
e
x q{2 , x P R , around x  0 , f pϕq : lnrp5{4q cosp3ϕqs , ϕ P R , around ϕ0  3π {4 ,

137

g) f pxq : p3x2
x 5q{p5x2 6x
3q , x 5x2 6x
3  0u , around x  1 .

P Rztx P R :

13) Show that a) b) c) d) e) f) g)

p1 p1

xqn ¡ 1 nx for all x ¡ 0 and n ¥ 1 , xqn   1 nx for all x ¡ 0 and 0   n   1 , ln x ¤ x
1 for all x ¡ 0 , sinpϕq   ϕ for all ϕ ¡ 0 , tanpϕq ¡ ϕ for all ϕ P p0, π {2q , sinhpxq : pex
e
x q{2 ¡ x for all x ¡ 0 . ln x ¥ px
1q{x for all x ¡ 0 .

14) Calculate
a x a
x , b) lim 1
, xÑ8 xÑ8 x x tanpxq x tanpxq , d) lim lim , xÑ0 1
cospxq xÑ0 x

 sinpxq 1 1 , f) lim
, lim ? xÑ0 x xÑ0 x sinpxq lnpxq r lnpxq sn 2 , lim , h) lim xÑ8 xÑ8 x x lnpxq , j) lim xx , k) lim xa{ lnpxq , lim xÑ0 xÑ0 xÑ1 tanpπxq


a) lim c) e) g) i) l)

1

lim r sinpxq s tanpxq , m) lim x1{x ,

xÑ8 p q n) lim x , o) lim r cosp1{xq sx , xÑ8 xÑ0 x

Ñ0

sin x

p)

lim

x

Ñ0

cosp3xq
cosp2xq , q) x2

where n P N, a P R.

1

cospπxq

Ñ1 x2
2x 1 ,

lim

x

15) Explain why Newton’s method fails to find the zero(s) of f in the following cases. a) f pxq : x2
x
6 , x P R , with initial approximation x  1{2 , b) f pxq : x1{3 , x P R .

16) A circular arch of length L ¡ 0 and height h ¡ 0 is to be constructed, where L{h ¡ π.

138

a) Show that x : L{p2rq, where r ¡ 0 is the radius of the corresponding circle, satisfies the transcendental equation cospxq  1


2h x. L

b) Assume that L{h  7. By Newton’s method, find an approximation x0 to x such that |x0
x|   10
6 . 17) The characteristic frequencies of the transverse oscillations of a string of length L ¡ 0 with fixed left end and right end subject to the boundary condition v 1 pLq hv pLq  0, where v : r0, Ls Ñ R is the amplitude of deflection of the string and h P R, is given by ω  x{L, where x (1.4.29) tan x 
hL [9]. Assume hL  1{3, and find by Newton’s method an approximation x0 to the smallest solution x ¡ 0 of (1.4.29) such that |x0
x|   10
6. 18) The characteristic frequencies of the transverse vibrations of a homogeneous beam of length L ¡ 0 with fixed ends are given by ω  rEJ {pρS qs1{2 px{Lq2 , where coshpxq cospxq  1 ,

(1.4.30)

E is Young’s modulus, J is the moment of inertia of a transverse section, S is the area of the section, ρ is the density of the material of the beam, and coshpy q : pey e
y q{2 for all y P R [34]. By Newton’s method, find an approximation x0 to the smallest solution x ¡ 0 of (1.4.30) such that |x0
x|   10
6 .

19) (Binomial theorem) Let n P N . Define f : p
1, 8q Ñ R by f pxq :

n ¸



k 0

for all x P defined by

for every k

n k x k

p
1, 8q, where the so called ‘binomial coefficients’ are



n 0



: 1 ,



n k

:

1 n  pn
1q    pn
pk
1qq k!

P N . 139

a) Show that

p1

xqf 1 pxq  nf pxq

for all x P p
1, 8q. b) Conclude from part a) that

f pxq  p1

xqn

for all x P p
1, 8q. c) Show the binomial theorem, i.e., that

px

y qn



n ¸



k 0

for all x, y

P R.



n k n
k x y k

1.5 Riemann Integration For motivation we go back to the start of Section 1.3 to the discussion of Galileo’s results on bodies in free fall near the surface of the earth. There starting from the fallen distance sptq (1.3.1) at time t, we determined the instantaneous speed v ptq of the body at time t as the derivative v ptq  s 1 ptq  gt ,

where g  9.81m{sec2 is the acceleration of the earth’s gravitational field. We now investigate the reverse question, how to calculate sptq from the instantaneous speeds between times 0 and t. There are two main approaches to this problem. The first uses that v ptq  s 1 ptq and concludes that s is the anti-derivative of v such that sp0q  0 and hence (by application of Theorem 1.4.7) is given by (1.3.1). A second approach leading on integration might proceed as follows. For every t ¡ 0 and n P N , it follows that sptq
sp0q 

n¸1 








k 0

s

1

k n



t


s



140



k t n

vHtL @msecD

10 8 6 4 2

0.2

0.4

0.6

0.8

1

1.2

t @secD

Fig. 45: S6 p1.2q is given by the shaded area under Gpv q.




s

n¸1



k 0






1  k t
s nk t   k 1 t
k t  S ptq : n¸ v t n n n n n
nk t k 0

k 1 t n k 1 t n

which leads to sptq  lim n

Ñ8










n¸1



v

k 0





k t n

gt2 n
1 n  nlim Ñ8 n2 2

t n





 nlim Ñ8



n
1 lim nÑ8 2n

n
1 gt2 ¸ k n2 k0



gt2



 12 gt2 .

Note that the sum Sn ptq has the geometrical interpretation of an area under the Gpv q, see Fig. 45. Below the limit 

lim

n

will be denoted by

Ñ8






n¸1



v

k 0

»t 0

v pτ q dτ 141



k t n



t n

such that

sptq 

»t 0

v pτ q dτ

which is the relation between instantaneous speed and distance traveled between times 0 and t (assuming that distance is zero at time t  0) satisfied for the motion in one dimension in general. Definition 1.5.1. (i) Let a, b P R be such that a ¤ b. We define the lengths of the corresponding intervals pa, bq, pa, bs, ra, bq, ra, bs by lppa, bqq  lppa, bsq  lpra, bqq  lpra, bsq : b
a . A partition P of ra, bs is an ordered sequence pa0 , . . . , aν q of elements of ra, bs such that a  a0

¤ a1 ¤    ¤ aν  b

where ν is an element of N . A partition P 1 of ra, bs is called a refinement of P if P is a subsequence of P 1 . (ii) A partition P  pa0 , . . . , aν q of a closed interval I of R induces a division of I into (in general non-disjoint!) subintervals I






ν¤1



Ij , Ij : raj , aj

1

s , j  0, . . . , ν .

j 0

The size of P is defined as the maximum of the lengths of these subintervals. In addition, we define for any bounded real-valued function f on I the lower sum Lpf, P q and upper sum U pf, P q corresponding to P by: Lpf, P q :




ν¸1



inf tf pxq : x P Ij u lpIj q ,

j 0

142

U pf, P q :




ν¸1



suptf pxq : x P Ij u lpIj q .

j 0

Example 1.5.2. Consider the interval I : r0, 1s and the continuous function f : I Ñ R defined by f pxq : x for all x P I. P0 : p0, 1q , P1 : p0, 1{2, 1q are partitions of I. The size of P0 is 1, whereas the size of P1 is 1{2. Also, P1 is a refinement of P0 . Finally, Lpf, P0 q  0  1  0 , U pf, P0 q  1  1  1 , 1 1 1 1 , Lpf, P1 q  0    2 2 2 4 1 1 1 3 1  U pf, P1 q   2 2 2 4 and hence Lpf, P0 q ¤ Lpf, P1 q ¤ U pf, P1 q ¤ U pf, P0 q . Lemma 1.5.3. Let f be a bounded real-valued function on a closed interval I of R. Further, let P, P 1 be partitions of I, and in particular let P 1 be a refinement of P . Then Lpf, P q ¤ Lpf, P 1 q ¤ U pf, P 1 q ¤ U pf, P q .

(1.5.1)

Proof. The middle inequality is obvious from the definition of lower and upper sums given in Def 1.5.1(ii). Further, let P  pa0 , . . . , aν q be a partition of [a,b] where ν P N and a0 , . . . , aν P ra, bs. Obviously, for the proof of the remaining inequalities it is sufficient (by the method of induction) to assume that P 1  pa0 , a11 , a1 , . . . , aν q, where a11 P I is such that a0

¤ a11 ¤ a1 , 143

and where we simplified to keep the notation simple. Then Lpf, P 1 q
Lpf, P q  inf tf pxq : x P ra0 , a11 su  lpra0 , a11 sq inf tf pxq : x P ra11 , a1 su  lpra11 , a1 sq
inf tf pxq : x P ra0, a1su  lpra0 , a1sq ¥ inf tf pxq : x P ra0, a1su  tlpra0, a11 sq lpra11 , a1sq
lpra0, a1squ  0 . Analogously, it follows that U pf, P 1 q
U pf, P q ¤ 0 and hence, finally, (1.5.1). Theorem 1.5.4. Let f be a bounded real-valued function on the interval ra, bs of R where a and b are some elements of R such that a ¤ b. Then supptLpf, P q : P

P P uq ¤ inf ptU pf, P q : P P P uq . (1.5.2) Proof. By Theorem 1.5.3, it follows for all P1 , P2 P P that Lpf, P1 q ¤ Lpf, P q ¤ U pf, P q ¤ U pf, P2 q , where P P P is some corresponding common refinement, and hence that supptLpf, P1 q : P1 P P uq ¤ U pf, P2 q and (1.5.2). As a consequence of Lemma 1.5.3 and since every partition P of some interval of R is a refinement of the trivial partition containing only its initial and endpoints, we can make the following definition. Definition 1.5.5. (The Riemann Integral) Let f be a bounded real-valued function on the interval ra, bs of R, where a and b are some elements of R such that a ¤ b. Denote by P the set consisting of all partitions of ra, bs. We say that f is Riemann-integrable on ra, bs if supptLpf, P q : P

P P uq  inf ptU pf, P q : P P P uq . 144

In that case, we define the integral of f on ra, bs by »b a

f pxq dx : supptLpf, P q : P

In particular if f pxq graph of f by

P P uq  inf ptU pf, P q : P P P uq .

¥ 0 for all x P ra, bs, we define the area A under the A :

»b a

f pxq dx .

Example 1.5.6. Let f be a constant function of value c P R on some interval ra, bs of R where a and b are some elements of R such that a ¤ b. Then by induction, all lower and upper sums are equal to c  pb
aq. Hence f is Riemann-integrable and »b a

f pxq dx  c  pb
aq .

Example 1.5.7. Consider the function f : ra, bs Ñ R defined by f pxq : x , for all x P ra, bs where a and b are some elements of R such that a ¤ b. For any n P N , define the partition Pn of ra, bs by Pn :



b
a ,...,a n

a, a

n  pb
aq n



b

.

Calculate Lpf, Pn q and U pf, Pn q for all n P N . Show that f is Riemannintegrable over ra, bs and calculate the value of »b a

f pxq dx .

Solution: We have: I



n¤1 






j 0

a

j pb
aq ,a n 145

pj

1qpb
aq n



and L pf, Pn q 

n¸1 




a



j 0

j pb
aq n





b
a n

n
1 p b
aq2 ¸ j  a  pb
aq n2

 a  pb
aq pb
n2aq  n2 pn
1q  a  pb
aq pb
2 aq 2

U pf, Pn q 

n¸1 






a

1qpb
aq n

pj

j 0

n
1 b
aq2 ¸ p  a  pb
aq pj n2



Hence

1 n



Ñ8

n



j 0

1


1 n



,

pb
aq2  n pn n2

2

1q

,

lim L pf, Pn q  lim U pf, Pn q 

n



 b
n a

1q  a  pb
aq

j 0

 p b
aq2 1  a  pb
aq 2



2

Ñ8

1 2

 pb2
a2 q .

As a consequence,

and

supptLpf, P q : P

P P uq ¥ 12  pb2
a2 q

inf ptU pf, P q : P

P P uq ¤ 12  pb2
a2q

and hence by Theorem 1.5.4 supptLpf, P q : P

P P uq  inf ptU pf, P q : P P P uq  12  pb2
a2 q .

Hence f is Riemann-integrable and »b a

x dx 

1 2

 pb2
a2q .

146

Theorem 1.5.8. Let f, g be bounded and Riemann-integrable on the interval ra, bs of R where a and b are elements of R such that a ¤ b and c P R. Then f g and cf are Riemann-integrable on ra, bs and »b a »b a

pf pxq

g pxqq dx 

cf pxq dx  c

»b a

»b a

f pxq dx

»b a

g pxq dx ,

f pxq dx .

If f is in addition positive, then »b a

f pxq dx ¥ 0 .

Proof. First, it is easy to see that for every subinterval J of I : ra, bs inf tf pxq : x P J u inf tg pxq : x P J u ¤ inf tpf g qpxq : x P J u ¤ suptpf gqpxq : x P J u ¤ suptf pxq : x P J u suptg pxq : x P J u . and hence that Lpf, P q Lpg, P q ¤ Lpf ¤ U pf, P q U pg, P q .

g, P q ¤ U pf

g, P q

and »b a

¤

f pxq dx »b a

f pxq dx

»b a

g pxq dx ¤ Lpf »b a

g, P q ¤ U pf

g pxq dx .

for every partition P of I. Hence supptLpf g, P q : P inf ptU pf g, P q : P P P uq exist and satisfy inf ptU pf

g, P q : P

g, P q

P P uq  supptLpf 147

P P uq,

g, P q : P

P P uq 

»b a

f pxq dx

»b a

g pxq dx .

Further for every subinterval J of I, it follows that inf taf pxq : x P J u  a inf tf pxq : x P J u , suptaf pxq : x P J u  a suptf pxq : x P J u and hence that Lpaf, P q  aLpf, P q , U paf, P q  aU pf, P q for every partition P of I as well as inf tf pxq : x P J u ¥ 0 and hence

Lpf, P q ¥ 0

for every partition P of I if f is in addition positive. Hence, finally, the Theorem follows. Corollary 1.5.9. Let f, g be Riemann-integrable on the interval ra, bs of R where a and b are elements of R such that a ¤ b, and in addition let f pxq ¤ g pxq for all x P ra, bs. Then »b a

f pxq dx ¤

»b a

g pxq dx .

Definition 1.5.10. (Sets of measure zero) A subset S of R is said to have measure zero if for every ε ¡ 0 there is a corresponding sequence I0 , I1 , . . . of open subintervals of R such that the union of those intervals contains S and at the same time such that lim

n

n ¸

Ñ8 k0

lpIk q   ε .

148

Remark 1.5.11. Note that any finite subset of R and also any subset of a set of measure zero has measure zero. Example 1.5.12. Every countable subset S of R is a set of measure zero. Proof. Since S is countable, there is a bijection ϕ : N Ñ S. Let ε 0 and define for each n P N the corresponding interval In : pϕpnq ε{2n 3 , ϕpnq ε{2n 3q. Then for each N P N: N ¸



lpIn q  ε 

n 0

N  n ¸ 1



n 0

2

2



and hence

N ¸

lim

N

N

ε 1
12  1
4

Ñ8 k0

lpIk q 

ε 2

1 2

1

 



ε 2

1


 N

1 2

1

¡




 ε.

Theorem 1.5.13. (Lebesgue’s criterion for Riemann-integrability) Let f : ra, bs Ñ R be bounded where a and b are some elements of R such that a   b. Further, let D be the set of discontinuities of f . Then f is Riemann-integrable if and only if D is a set of measure zero. Proof. See the proof of Theorem 4.2.6 in the Appendix. Remark 1.5.14. A property is said to hold almost everywhere on a subset S of R if it holds everywhere on S except for a set of measure zero. Thus, Theorem 1.5.13 states that a bounded function on a non-trivial bounded and closed interval of R is Riemann-integrable if and only if the function is almost everywhere continuous. Example 1.5.15. Many functions important for applications have integral representations which are crucial for the derivation of special properties. For instance, for any n P Z the corresponding Bessel function of the first kind Jn satisfies Jn pxq 

1 π

»π 0

cospx sin θ
nθq dθ 149

y 1

-20

20

x

-0.5

Fig. 46: Graph of J0 .

for all x P R and is the solution of the differential equation x2 f 2 pxq

xf 1 pxq

px2
n2 qf pxq  0 ,

for all x P R. By Corollary 1.5.9, it follows the simple estimate

|Jnpxq| ¤

1 π

»π 0

| cospx sin θ
nθq| dθ ¤

1 π

»π 0

dθ  1

for all x P R. Theorem 1.5.16. (Additivity of upper and lower Integrals) Let f : ra, bs Ñ R be bounded where a and b are some elements of R such that a ¤ b and c P ra, bs. Then supptLpf, P q : P

P P uq  supptLpf |ra,cs, P q : P P Pra,cs uq supptLpf |rc,bs , P q : P P Prc,bs uq , inf ptU pf, P q : P P P uq  inf ptU pf |ra,cs , P q : P P Pra,cs uq inf ptU pf |rc,bs , P q : P P Prc,bs uq , where Pra,cs , Prc,bs denote the set consisting of all partitions of ra, cs and rc, bs, respectively. 150

Proof. For this, let P1  pa0 , . . . , aν q P Pra,cs and P2 Prc,bs , where ν, µ are some elements of N , and P : pa0 , . . . , aν , aν 1 , . . . , aν

µ

 paν

1 , . . . , aν µ

qP

q

the corresponding element of P. Then Lpf, P q  Lpf |ra,cs , P1 q U pf, P q  U pf |ra,cs , P1 q

Lpf |rc,bs , P2 q , U pf |rc,bs , P2 q .

Now let ε ¡ 0. Obviously because of Lemma 1.5.3, we can assume without restriction that P is such that ε supptLpf, P q : P P P uq
Lpf, P q ¤ , 3 ε supptLpf |ra,cs , P q : P P Pra,cs uq
Lpf, P1 q ¤ , 3 ε supptLpf |rc,bs , P q : P P Prc,bs uq
Lpf, P2 q ¤ . 3 Then also   sup L f, P : P

pt p q P P uq
supptL pf |ra,cs, P q : P P Pra,cs uq
supptLpf |rc,bs, P q : P P Prc,bs uq ¤ ε .

Analogously because of Lemma 1.5.3, we can also assume without restriction that P is such that ε U pf, P q
inf ptU pf, P q : P P P uq ¤ , 3 ε U pf, P1 q
inf ptU pf |ra,cs , P q : P P Pra,cs uq ¤ , 3 ε U pf, P2 q
inf ptU pf |rc,bs , P q : P P Prc,bs uq ¤ . 3 Then also   inf U f, P : P

pt p q P P uq
inf ptLpf |ra,cs, P q : P P Pra,cs uq
inf ptLpf |rc,bs, P q : P P Prc,bsuq ¤ ε . 151

Corollary 1.5.17. (Additivity of the Riemann Integral) Let f : ra, bs Ñ R be Riemann-integrable where a and b are some elements of R such that a ¤ b, and c P ra, bs. Then »b a

f pxq dx 

»c a

f pxq dx

»b c

f pxq dx .

Proof. The statement is a simple consequence of Theorem 1.5.13 and Lemma 1.5.16. Theorem 1.5.18. Let f : ra, bs Ñ R be Riemann-integrable where a and b are some elements of R such that a   b. Then F : ra, bs Ñ R defined by F pxq :

»x a

f ptq dt

for every x P ra, bs is continuous. Furthermore, if f is continuous in some point x P pa, bq, then F is differentiable in x and F 1 pxq  f pxq . Proof. For x, y

P ra, bs, it follows by the Corollaries 1.5.17, 1.5.9 that

|F pyq
F pxq|  if y

¥ x as well as that |F pyq
F pxq| 

» y     f t dt 

¤ M  |y
x|

 » x     f t dt  

¤ M  |y
x|

x

y

pq

pq

if y   x, where M ¥ 0 is such that |f ptq| ¤ M for all t P ra, bs, and hence the continuity of F . Further, let f be continuous in some point x P pa, bq. Hence given ε ¡ 0, there is δ ¡ 0 such that

|f ptq
f pxq|   ε for all t P ra, bs such that |t
x|   δ. (Otherwise, there is some ε ¡ 0 along with a sequence t0 , t1 , . . . in ra, bs such that |f ptnq
f pxq| ¥ ε and 152

|tn
x|   1{n for all n P N. Then t0, t1 , . . . is converging to x, but f pt0 q, f pt1 q, . . . is not convergent to f pxq. ) Now let h P R be such that |h|   δ and small enough such that x h P pa, bq. We consider the cases h ¡ 0 and h   0. In the first case, it follows by Theorem 1.5.13 and Corollary 1.5.17, 1.5.9 that    » x h  »x  1 F x h F x  f x   f t dt f t dt  h h a a   » x  »x h »x  1   t dt t dt t dt f x f f f  h a x a  »x h  »x h 1  1  f t f x dt f t f x dt ε . h h

p

q
p q
p q  pq

 

x

pq


pq


r p q
p qs

¤

x

pq

pq

  f x 


pq


pq

| p q
p q| ¤

Analogously, in the second case it follows that    » x h   »x F x h  1  F x     f x t dt t dt f x f f  h   h a a  » x h   »x h »x 1   f t dt f t dt f t dt f x  h a a x h   »x »  1 x  1  f t f x dt f t f x dt ε .  h h

p

q
p q
p q 






pq


x h

pq


pq

pq


pq

pq


r p q
p qs

¤| |

x h


pq

| p q
p q| ¤

Hence it follows that h

lim

Ñ0,h0

F px

hq
F pxq h

 f pxq

and that F is differentiable in x with derivative f pxq. Remark 1.5.19. Note that because of Theorem 1.5.13, the function F in Theorem 1.5.18 is differentiable with derivative f pxq for almost all x P pa, bq. Theorem 1.5.20. (Fundamental Theorem of Calculus) Let f : ra, bs Ñ R be bounded and Riemann-integrable where a and b are some elements of 153

R such that a   b. Further, let F be a continuous function on ra, bs as well as differentiable on pa, bq such that F 1 pxq  f pxq, for all x P pa, bq. Then »b a

f pxq dx  F pbq
F paq .

Proof. Let ε ¡ 0 and P  pa0 , . . . , aν q be a partition of ra, bs where ν is an element of N . By Theorem 1.4.6 for every j P t0, 1, . . . , ν
1u, there is a corresponding cj P raj , aj 1s such that

q
F paj q  F 1 pcj qpaj 1
aj q , where we define F 1 paq : f paq and F 1 pbq : f pbq. Hence F paj

F pbq
F paq 

1




ν¸1



rF paj 1q
F paj qs 

j 0

and




ν¸1



f pcj qpaj

1


aj q .

j 0

Lpf, P q ¤ F pbq
F paq ¤ U pf, P q .

Hence supptLpf, P q : P P P uq ¤ F pbq
F paq ¤ inf ptU pf, P q : P P P uq  supptLpf, P q : P

Example 1.5.21. Calculate

»π

7 sin 0

Solution: By


x

3

f pxq : 7 sin

P P uq .

dx .
x

3 for all x P r0, π s, there is defined a continuous and hence Riemann-integrable function on r0, π s. Further by F pxq :
21 cos 154


x

3

y 4 2

-4

-2

2

4

x

-2 -4 Fig. 47: Graph of the greatest integer function and anti-derivative.

for all x P r0, π s, there is defined a continuous function on r0, π s which is differentiable on p0, π q such that g 1 pxq  f pxq for all x P p0, π q. Hence by Theorem 1.5.20 »π

sin 0


x

3

dx 
21 cos


π

3

21 cosp0q  21


21 2

 212 .

Example 1.5.22. A simple number theoretic function is the greatest integer or floor function defined by rxs : n for all x P rn, n 1q and n P N. Calculate » » x

0

rys dy

0

, x

rys dy

for all x ¥ 0 and x   0, respectively. Solution: Note that the greatest integer functions is almost everywhere continuous and hence according to Theorem 1.5.13 also Riemann-integrable on any closed interval of R. For every n P N and every x P rn, n 1q, it follows by Corollary 1.5.17 and

155

Theorem 1.5.20 that »x 0



»x

0

n

rys dy  rys dy


n¸1 » k 1



k 0



»n

rys dy 

npx
nq 

k dy

x

n pn
1q 2

rys dy 

 npn  npn

k

npx
nq

n

1

npx
nq  n x


»n

1

x

rys dy

1
xq 1
xq


»0 n 1


1 » k ¸ 

k dy

k n 1 k

n pn 2

2

1q  n



 rxs

n dy n



x


1

rxs

2

.

P Z such that n ¤
1 and every x P

rys dy 

1



»x

rys dy

k

k 0

Analogously, it follows for every n rn, n 1q that »0



k 0




n¸1

 

k




n¸1 » k 1

»n

n dy x

 npn 1

n 2

1


1 » k ¸ 

k n 1 k

1
xq


1 ¸



1

rys dy

k

k n 1


x 
rxs

x


1

rxs

2

.

See Fig. 47. A basic method in the evaluation of integrals with trigonometric integrands consists in the application of the addition theorems for sine and cosine. Example 1.5.23. Calculate »π 0

sinpmθq sinpnθq dθ ,

where m, n P N . Solution: By help of the addition theorem for the cosine function, it follows that cosppm

nqθq  cospmθq cospnθq
sinpmθq sinpnθq , 156

cosppm
nqθq  cospmθq cospnθq

sinpmθq sinpnθq ,

and hence that sinpmθq sinpnθq  for all θ »π 0

1 r cosppm
nqθq
cosppm
nqθq s 2

P R. This leads to

sinpmθq sinpnθq dθ



1 2

»π 0

r cosppm
nqθq
cosppm

 2pm1
nq r sinppm
nqθq sπ0
2pm1 nq r sinppm if m  n and »π 0

sinpmθq sinpnθq dθ



1 2

 π2
2pm1 nq r sinppm

»π 0

r 1
cosppm

nqθq sπ0

nqθq s dθ

nqθq sπ0

0

nqθq s dθ

 π2

if m  n.

Example 1.5.24. Find the solutions of the following (‘differential’) equation for f : R Ñ R: f 1 pxq  e2x sinp3xq (1.5.3) for all x P R. Solution: If f is such function, it follows that f is continuously differentiable. Hence it follows by Theorem 1.5.20 that f pxq
f px0 q 





»x x0

f 1 py q dy 

1 2y 1 e
cosp3y q 2 3

where x0

x

x0

»x x0

pe2y

sinp3y qq dy

 12 e2x
13 cosp3xq
12 e2x

P R and x ¡ x0 . Hence f pxq 

1 2x 1 e
cosp3xq 2 3 157

c,

0

1 cosp3x0 q , 3

for all x P R, where c  f p0q


1 2

1 3

 f p0q
16 .

On the other hand if c P R and fc : R Ñ R is defined by fc pxq :

1 2x 1 e
cosp3xq 2 3

c

for all x P R, then it follows by direct calculation that fc satisfies (1.5.3) for all x P R. Hence the solutions of the differential equation are given by the family of functions fc , c P R. Note that c  f p0q
p1{6q. Hence for every c P R, there is precisely one solution of the differential equation with ‘initial value’ f p0q  c. The same is true for initial values given in any other point of R. Example 1.5.25. Find the solutions of the following differential equation for f : R Ñ R. f 1 pxq af pxq  3 (1.5.4) for all x P R, where a P R. Solution: If f is such function, it follows that f is continuously differentiable. Further, by using the auxiliary function h : R Ñ R defined by hpxq : eax

for every x P R, it follows that

phf q 1pxq  hpxqf 1pxq h 1pxqf pxq  eaxf 1pxq  eaxrf 1pxq af pxqs  3 eax for all x P R. Hence it follows by Theorem 1.5.20 that phf qpxq
phf qpx0 q 

»x

3 eay dy x0

aeax f pxq

 a3 eax
a3 eax

and therefore that

phf qpxq  phf qpx0 q

3 p1
eax0 q a 158

3 ax pe
1q a

0

for x ¡ x0 , where x0

P R. From this, we conclude that phf qpxq  a3 peax
1q

c

and f pxq  for all x P R, where c is defined by

3 p 1
e
ax q a

c e
ax

 f p0q. On the other hand if c P R and fc : R Ñ R

3 p1
e
axq c e
ax a for all x P R, then it follows by direct calculation that fc satisfies (1.5.4) for all x P R. Hence the solutions of the differential equation are given by the family of functions fc , c P R. Note that c  f p0q. Hence for every c P R, there is precisely one solution of the differential equation with ‘initial value’ f p0q  c. The same is true for initial values given in any other point of R. fc pxq :

Problems 1) Calculate »3

a) 2 »2

c)

px2

pe2x
3xq dx


1

»2

e) 1 »1

g)

0 π 2

{ 1

»

0 »π

{

π 2

3

b) »2

,

d)


3x

,

dx ,

h)

1 2

x


3x x3

2

sinpx{2q cospx{2q dx dx

4 sinpxq cos2 pxq dx 5

?

3x

f) »5

1

sinpπxq dx π

»4

dx

9 t2{3 dt ,

0


3



x 2x2

i) j)

?4x

»2

7q dx ,

5x

159

5

dx

, ,

, »3

,

2 dx

,

k) 1

|3x
4| dx

,

»3

l)




1



»2

m)


2

1 |x
1| 3


2

5 |x
1||x 4

»5

n)

| 5x
3 |2

x

» 3π

| sinpx{2q| dx

o) 0 » 2π

s) 0

dx

4|x

, 1|

1| |x ,



dx

p)

,

2 | dx , » π{6


π{6

| cosp3xq| dx

» 2π

sinpmθq sinpnθq dθ

,

cospmθq cospnθq dθ

,

q) 0 » 2π



r) 0

,

sinpmθq cospnθq dθ

,

where m, n P N .

2) Define f : R Ñ R by

? ? p xq f pxq : 3{2 ' %? 3 p1
xq $ ' & 3 1

if x ¤
1{2 if
1{2   x   1{2 if 1{2 ¤ x

for every x P R. Calculate the area in R2 that is enclosed by the graph of f and the x-axis. Verify your result using facts from elementary geometry. Use the result to calculate the area enclosed by a hexagon of side length 1. 3) Calculate the area in R2 that is enclosed by the graphs of the polynomials p1 pxq :
1

p7{2q x
x2 , p2 pxq : 4
p7{2q x

where x P R. 4) Calculate the area in R2 that is enclosed by the curve C : tpx, y q P R2 : y 2
4x2 5) Show that cospxq ¤ 1
for all x P r0, π {2s.

160

x2 π

4x4

 0u .

x2 ,

6) Find the solutions to the differential equation for f : R Ñ R. a) f 1 pxq
3f pxq  x{2 , x P R , b) f 1 pxq 3f pxq  ex{4 , x P R , c) 2f 1 pxq
f pxq  3 e
x , x P R .

7) Consider the following differential equation for f : R Ñ R. f 2 pxq  3x

4

for all x P R. a) Find the solutions of this equation. b) Find that solution which satisfies f p0q  1 and f 1 p0q  2. c) Find that solution which satisfies f p0q  2 and f p1q  3. 8) Calculate a0 : bk : for all k

» 2π

1 2π

0

» 2π

1 π

0

f pxq dx , ak :

1 π

» 2π 0

cospkxqf pxq dx ,

sinpkxqf pxq dx

P N .

a) f pxq :

#


1

if x P r0, π s , 1 if x P pπ, 2π s

b) f pxq : x for all x P r0, 2π s , c) f pxq : 

#

x 2π
x

if x P r0, π s . if x P pπ, 2π s

Remark: These are the coefficients of the Fourier expansion of f . The representation f pxq  lim

#

Ñ8 a0

n

lim

n ¸



rak cospkxq

bk sinpkxqs

k 1

is valid for every point x P r0, 2π s of continuity of f .

161



+

9) Calculate the area in R2 that is enclosed by the ellipse C :

"

2 px, yq P R : xa2 2

where a, b ¡ 0.

y2 b2

1

*

,

10) Calculate the area in R2 that is enclosed by the branches of hyperbolas *

"

y2 x2 C1 : px, y q P R : y ¥ 0 ^ 2
2  1 , a b " * 2 p y
c q x2
 1 , C2 : px, y q P R2 : y ¤ c ^ a2 b2 2

where a, b ¡ 0 and c ¡ a.

11) Let a, b P R be such that a   b. Further, let f : ra, bs Ñ R be positive, i.e., such that f pxq ¥ 0 for all x P ra, bs, and assume a value ¡ 0 in some point of ra, bs. Show that »b a

f pxq dx ¡ 0 .

12) Let a, b P R be such that a   b. Further, let f : ra, bs Ñ R and g : ra, bs Ñ R be bounded and Riemann-integrable. Show the following Cauchy-Schwartz inequality for integrals: » b a

2

f pxqg pxq dx

¤

» b

 »

f pxq dx

b

2

a



g pxq dx 2

a

.

In addition, show that equality holds if and only if there are α,β P R satisfying that α2 β 2  0 and such that αf βg  0. Hint: Consider » b

as a function of λ P R.

a

r f pxq

λ g pxq s2 dx



13) Newton’s equation of motion for a point particle of mass m moving on a straight line is given by mf 2 ptq  F pf ptqq

¥

0

(1.5.5)

for all t P R, where f ptq is the position of the particle at time t, and F pxq is the external force at the point x. For the specified force, calculate the solution function f of (1.5.5) with initial position f p0q  x0 and initial speed f 1 p0q  v0 , where x0 , v0 P R.

162

a) F pxq  0 , x P R , b) F pxq  F0 , x P R ,where F0 is some real parameter .

14) Newton’s equation of motion for a point particle of mass m ¥ 0 moving on a straight line under the influence of a viscous friction is given by mf 2 ptq 
λf 1 ptq (1.5.6) for all t P R, where f ptq is the position of the particle at time t, and λ ¡ 0 is a parameter describing the strength of the friction. Calculate the solution function f of (1.5.6) with initial position f p0q  x0 and initial speed f 1 p0q  v0 , where x0 , v0 P R. Investigate, whether f has a limit value for t Ñ 8.

15) Newton’s equation of motion for a point particle of mass m ¥ 0 moving on a straight line under the influence of low viscous friction, for instance friction exerted by air, is given by mf 2 ptq 
λ pf 1 ptqq2

(1.5.7)

for all t P R, where f ptq is the position of the particle at time t, and λ ¡ 0 is a parameter describing the strength of the friction. Find solutions f of (1.5.7) with initial position f p0q  x0 and initial speed f 1 p0q  v0 , where x0 , v0 P R. 16) Consider a projectile that is shot into the atmosphere. According to Newton’s equation of motion, the height f ptq above ground at time t P R satisfies the equation mf 2 ptq 
g
λ pf 1 ptqq2

(1.5.8)

for all t P R, , where g  9.81m{s is the acceleration due to gravity and λ ¡ 0 is a parameter describing the strength of the friction. Find solutions f of (1.5.8) with initial height f p0q  z0 and initial speed component f 1 p0q  v0 , where z0 , v0 P R. 2

2 Calculus II 2.1 Techniques of Integration 2.1.1 Change of Variables Theorem 2.1.1. (Change of variables) Let c, d P R are such that c   d. Further, let g : rc, ds Ñ R be continuous, non-constant, such that g pcq ¤ 163

g pdq and continuously differentiable on pc, dq with a derivative which can be extended to a continuous function on rc, ds. Finally, let I be an open interval interval of R containing g prc, dsq and f : I Ñ R be continuous. Then » »

pq

g d

pq

g c

f pxq dx 

d

c

f pg puqq  g 1 puq du .

(2.1.1)

Proof. In the following, we denote by g 1 the extension of the derivative of g |pc,dq to a continuous function on rc, ds. We define G : rc, ds Ñ R by Gpuq :

»u c

u f pg pu¯qq  g 1 pu¯q d¯

for all u P rc, ds. By Theorem 1.5.18 it follows that G is continuous as well as differentiable on pc, dq with G 1 puq  f pg puqq  g 1 puq for all u P pc, dq. Further, we define F : rx0 , x1 s Ñ R by F pxq :

»x x0

f px 1 q dx 1


» gpcq x0

f px 1 q dx 1

for all x P rx0 , x1 s, where x0 , x1 P I are such that x0 is smaller than the minimum value of g and x1 is larger than the maximum value of g, respectively. By Theorem 1.5.18 it follows that F is continuous as well as differentiable on px0 , x1 q with F 1 pxq  f pxq for all x P px0 , x1 q. Hence it follows by Theorems 1.2.42, 1.3.10 that F is continuous as well as differentiable on pc, dq with

g

pF  gq 1puq  f pgpuqq  g 1puq  G 1puq for all u P pc, dq. From Theorem 1.4.7 and F pg pcqq  Gpcq  0, it follows that F  g  G and hence by Corollary 1.5.17 also (2.1.1). 164

Example 2.1.2. Calculate

»x a

dy

y2

0

a2

for every x ¡ 0 where a ¡ 0. Solution: Define g : p
π {2, π {2q Ñ R by g pθq : a  tanpθq for all θ P p
π {2, π {2q. Then g is a bijective as well as continuously differentiable such that g 1 pθq  a  p1

tan2 pθqq

P p
π{2, π{2q. The inverse g
1 is given by
x g
1 pxq : arctan a for all x P R. By Theorem 2.1.1 for all θ

»x

a

y2

0



dy

» g
1 pxq 0

 2

a

dθ cospθq

» gpg
1 pxqq

pq

g 0

 ln



a

dy

y2

1

 2

» g
1 pxq

a

sinpg
1 pxqq cospg
1 pxqq

0



 ln

g 1 pθq dθ pgpθqq2 a2

a 

x a

1

x2 a2



.

Example 2.1.3. Calculate »2 0

?

9
x2 dx .

Solution: Define g : p
π {2, π {2q Ñ p
3, 3q by g pθq : 3 sinpθq for all θ P p
π {2, π {2q. Then g is a bijective as well as continuously differentiable such that g 1 pθq  3 cospθq for all θ

P p
π{2, π{2q. The inverse g
1 is given by
x g
1 pxq  arcsin 3 165

for all x P p
3, 3q. By Theorem 2.1.1 »2 0



?



dx 

» gparcsinp2{3qq

» arcsinp2{3q a » arcsinp2{3q 0

?

pq

g 0

9
pg pθqq

0

 92 

9


x2

p1



9 2 arcsin 2 3  

9 2 arcsin 2 3

2

9
x2 dx

 g 1pθq dθ  9

cosp2θqq dθ

» arcsinp2{3q 0

cos2 pθq dθ



1 sinp2 arcsinp2{3qq 2  ? 2 cosparcsinp2{3qq  5 3

9 arcsin 2



2 3

.

Example 2.1.4. Calculate »5 4

x
4 

?

x2
9 dx .

Solution: Define g : p0, π {2q Ñ p3, 8q by g pθq : 3 cospθq for all θ P p0, π{2q. Then g is a bijective as well as continuously differentiable such that sinpθq g 1 pθq  3  cos2 pθq for all θ

P p0, π{2q. The inverse g
1 is given by



3 g
1pxq  arccos

x

for all x P p3, 8q. By Theorem 2.1.1 »5 4



? x
4  x2
9 dx 

» gparccosp3{5qq

» arccosp3{5q

a

p{q

arccos 3 4

p

p { qq

g arccos 3 4

x
4 

?

x2
9 dx

pgpθqq
4  pgpθqq2
9  g 1pθq dθ 166

»

arccosp3{5q  19 cospθq sin2 pθq dθ arccosp3{4q ?    271 sin3parccosp3{5qq
sin3 parccosp3{4qq  271 rp4{5q3
p 7{4q3 s . Example 2.1.5. Calculate

» π{2 0

dθ . 4 cospθq

5

Solution: Define g : R Ñ p
π, π q by

g pxq : 2 arctanpxq

for all x P R. This is a standard substitution to transform a rational integrand in sin and cos into a rational integrand. Then g is bijective as well as continuously differentiable such that g 1 pxq 

2 1

for all x P R. The inverse g
1 is given by

x2

g
1 pθq : tan pθ{2q

for all θ » π{2 0

 

P p
π, πq. By Theorem 2.1.1 5

»1 0

0

  g p0q 5 »1 g 1 pxq dx 
  0 4 2 cos2 gp2xq
1 5

5

»1

» gp1q

dθ 4 cospθq

2 1

Note that for all x P R.

x2

5

dx 4

2 1 x2

cospg pxqq 

dθ 4 cospθq


1



2

»1 0

»1 0

g 1 pxq dx 5 4 cospg pxqq



4 dx

x2

g 1 pxq dx

p p qq
1

1 tan2

 9

g x 2

2 arctan 3

1
x2 2x , sinpg pxqq  . 2 1 x 1 x2

167



2



1 3

.

y

2

-1

1

2

3

x

-2

Fig. 48: Graphs of solutions of the differential equation (2.1.2) in the case that a  1 with initial values
π,
π {2, π {2 and π at x  0. Compare Example 2.1.6.

Example 2.1.6. Find solutions of the following differential equation for f : R Ñ R with the specified initial values. f 1 pxq  a sinpf pxqq

(2.1.2)

for all x P R, where a ¡ 0, f p0q P p0, π q. Solution: If f is such function, it follows that f is continuously differentiable. Since f p0q P p0, π q, it follows by the continuity of f the existence of an open interval c, d P R such that c   0   d and such that f p[c, d]q € p0, π q. Since a ¡ 0 and the sine function is ¥ 0 on the interval [0, π], it follows from (2.1.2) that f px1 q  f px0 q

 f px0 q

» x1 x0

f px1 q
f px0 q  f px0 q

» x1 x0

f 1 pxq dx

a sinpf pxqq dx ¥ f px0 q

for all x0 , x1 P [c, d] such that x0 ¤ x1 . In addition, the restriction of f to [c, d] is non-constant since the sine function has no zeros on p0, π q. Hence 168

we conclude from (2.1.2) by Theorem 2.1.1 for x P [c, d] that apx
cq 

»x

a du 

»x

f 1 puq du  sinpf puqq

» f pxq

dθ . c c f pcq sinpθ q Further, it follows by use of the transformation g from the previous Example 2.1.5 and Theorem 2.1.1 that » gptanpf pxq{2qq

» f pxq

»

tanpf pxq{2q dθ g 1 pxq   dx g ptanpf pcq{2qq sinpθ q tanpf pcq{2q sinpg pxqq 

» tanpf pxq{2q dx tanpf pxq{2q   ln tanpf pcq{2q . tanpf pcq{2q x

dθ f pcq sinpθ q

Hence it follows that apx
cq  ln which leads to





tanpf pxq{2q tanpf pcq{2q 

f pcq f pxq  2 arctan tan 2

From (2.1.3), we conclude that 

tan

f pcq 2







(2.1.3) 

ep
q . ax c



f p0q e
ac  tan



2

Substituting this identity into (2.1.3) gives 

f pxq  2 arctan tan





(2.1.4)

f p0q ax e 2

. 

.

On the other hand, for every c P p
π, π q, it follows by elementary calculation that f : R Ñ R defined by 
c  ax e f pxq : 2 arctan tan 2 for all x P R satisfies (2.1.2) and f p0q  c. As a side remark, note that for every k P Z the constant function of value kπ is a solution of (2.1.2). In addition, if f is a solution of (2.1.2), then for every k P Z also fk : R Ñ R defined by fk pxq : f pxq 2πk q for every x P R is a solution of (2.1.2). 169

Theorem 2.1.7. Let f be a bounded Riemann-integrable function on ra, bs where a and b are some elements of R such that a   b. Then »b a

f pxq dx 

»
a


b

f p
xq dx .

Proof. Define f
: r
b,
as Ñ R by f
pxq : f p
xq for all x P r
b,
as. Then f
is bounded, and for any partition P  pa0 , . . . , aν q of ra, bs, where ν P N , a0 , . . . , aν P ra, bs, P
: p
aν , . . . ,
a0 q itis a partition of r
b,
as, and in particular Lpf, P q  Lpf
, P
q, U pf, P q  U pf
, P
q. Analogously, for any partition P  pa0 , . . . , aν q of r
b,
as, where ν P N , a0 , . . . , aν P r
b,
as, P
: p
aν , . . . ,
a0 q is a partition of ra, bs, and in particular Lpf
, P q  Lpf, P
q, U pf
, P q  U pf, P
q. Hence the set consisting of the lower sums of f is equal to the set of lower sums of f
and the set consisting of the upper sums of f is equal to the corresponding set of upper sums of f
. Example 2.1.8. Calculate »1


1

3 sinp2xq dx .

Solution: By Theorem 2.1.7, it follows that »1


1

3 sinp2xq dx 

and hence that

»1


1

»1


1

3 sinp
2xq dx 


»1


1

3 sinp2xq dx

3 sinp2xq dx  0 .

A variation of the foregoing reasoning is given in the following example Example 2.1.9. Calculate »π 0

x sin2 pxq dx . 170

Solution: First by Theorem 2.1.7, it follows that »π

x sin pxq dx 

»0

p
xq sin p
xq dx 


2

0

»0

2


π


π

x sin2 pxq dx .

Further, it follows by Theorem 2.1.1 and Example 1.5.23 that




»0

x sin pxq dx 


»π

2


»π




π

»π

y sin py q dy 2

0

and, finally, that

0

py
πq sin2 py
πq dy

π 0

»π 0

sin py q dy 


»π

2

x sin2 pxq dx 

0

y sin2 py q dy

π2 2

π2 . 4

Example 2.1.10. Show that »π

»π

sinpxq dx  2
π x

0

sinpxq dx . x

Solution: By Corollary 1.5.17 and Theorem 2.1.7, it follows that »π

»

»

0 π sinpxq sinpxq sinpxq dx  dx dx x
π» x
π » x 0 »π π π sinp
xq sinpxq sinpxq dx dx  2 dx . 
x x x 0 0 0

Problems 1) Calculate the value of the integral. For this, if the antiderivative of the integrand is not obvious, use a suitable substitution. »1

a) 0

»1

c)

»1

p2x 1q { dx , b) 1 2


1

x p3x2

1q1{2 dx ,

0

»3

d)

171

1q1{2 du

u p2u

2

?

s 2s2

3

ds

, ,

»5

e) 3

px
2q
2  sin

»3

g) »

1 x

i) 0

?

tanpθq dθ 3

3

»6 »

o) 3 »2

q) s) u)

b

3

2

?dx2 2

x

x

»2

dx »1

,

u e
u

»2

,

{

x1 3

1

?x dx

,

,

»2

p

1 »π 2

?dx 2 , r) 2 5
x 1 x 0 »πa 1 sinp2θq dθ , t) 0 » π {2 cos4 pθq dθ 4
π{2 sin pθq cos4 pθq

x2

1

x2

{

dx
2x 4q3{2

,

dx

4

?dx2 x

» π{2

x

2

,

,

9

dθ sinp3θq

?x

,

2

n)

p)

2 du , u
12

dx

j)

x1{2

l)

u2

{2 du , »4

x P r0, π {2q ,

4u


2

0

»3


4

f) 2

h)

?x dx

»4b

m)

,



x

x
2

sinp xq ?x dx ,

k)

3 7



2

,

dθ sinpθq 2 cospθq

0

,

.

2) Let a P R, f : r
a, as Ñ R be Riemann-integrable and g : R Ñ R be Riemann-integrable over every interval rb, cs, where b, c P R are such that b ¤ c. Show that a)

»a


a

f pxq dx  0

if f is antisymmetric, i.e., if f p
xq r
a, as. b)

»a


a

f pxq dx  2

»a 0


f pxq for all x P

f pxq dx

if f is symmetric, i.e., if f p
xq  f pxq for all x P r
a, as. c)

»c b

f pxq dx 

»c

τ

b τ

f pxq dx

if b, c P R are such that b ¤ c and f is periodic with period τ ¥ 0, i.e., if f px τ q  f pxq for all x P R.

172

3) Calculate the area in (
8, 0 ]2 that is enclosed by the strophoid C :

px, yq P R2 : pa
xq y2
pa

xq x2

0

(

,

where a ¡ 0. 4) Find solutions of the following differential equation for f : R with the specified initial values. f 1 pxq  2 cospf pxqq

ÑR

3

for all x P R, f p0q P [
π, π q.

Remark 2.1.11. The solution of problem n) from 1) illustrates the wellknown fact that one should never blindfoldly rely on computer programs. In Mathematica 5.1, the command Integraterpx^ 2
2x

4q^ t3{2u, tx, 1, 2us

gives the output 1 p68 16

27 Logr3sq

which is incorrect. 2.1.2 Integration by Parts Theorem 2.1.12. (Integration by parts) Let f , g be bounded Riemannintegrable functions on ra, bs, where a and b are elements of R such that a   b. Further, let F, G be continuous functions on ra, bs which are differentiable on pa, bq and such that F 1 pxq  f pxq and G 1 pxq  g pxq for all x P pa, bq. Then »b a

F pxqg pxq dx  F pbqGpbq
F paqGpaq


»b a

f pxqGpxq dx .

Proof. First as a consequence of Theorem 1.5.13, f G and F g are both Riemann-integrable as products of Riemann-integrable functions. Moreover, F G is continuous and differentiable such that pF Gq 1 pxq  f pxqGpxq 173

F pxqg pxq for all x P pa, bq, and f G F g is Riemann-integrable by Theorem 1.5.8 as a sum of Riemann-integrable functions. Hence by Theorem 1.5.20 »b a

f pxqGpxqdx

»b a

F pxqg pxqdx 

 F pbqGpbq
F paqGpaq .

»b a

f pxqGpxq



F pxqg pxq dx

Example 2.1.13. Calculate »π 0

x cosp3xq dx .

Solution: Define F, G, f, g : r0, π s Ñ R by F pxq : x , g pxq : cosp3xq , f pxq : 1 , Gpxq :

1 sinp3xq 3

for all x P r0, π s. Hence by Theorems 2.1.12, 1.5.20: »π 0

1 x cosp3xq dx 
3

»π 0

sinp3xq dx 

1 1 2 cosp3π q
cosp0q 
. 9 9 9

Example 2.1.14. Calculate »π 0

ex sinp2xq dx .

Solution: Define F, G, f, g : r0, π s Ñ R by 1 F pxq : ex , g pxq : sinp2xq , f pxq : ex , Gpxq :
cosp2xq 2 for all x P r0, π s. Then by Theorem 2.1.12, »π

e sinp2xq dx  x

0

1 p 1
eπ q 2 174

1 2

»π 0

ex cosp2xq dx

(2.1.5)

To determine the last integral, define F, G, f, g : r0, π s Ñ R by F pxq : ex , g pxq : cosp2xq , f pxq : ex , Gpxq :

1 sinp2xq 2

for all x P r0, π s. Then by Theorem 2.1.12, 1 2

»π

1 e cosp2xq dx 
4

»π

x

0

0

ex sinp2xq dx .

(2.1.6)

and hence by (2.1.5), (2.1.6) finally: »π 0

ex sinp2xq dx 


2 π pe
1q . 5

Example 2.1.15. Calculate »e 1

lnp4xq dx .

Solution: Define F, G, f, g : r1, es Ñ R by F pxq : lnp4xq , g pxq : 1 , f pxq : for all x P r0, es. Then by Theorem 2.1.12, »e 1

lnp4xq dx  p1

ln 4q  e
ln 4


»e 1

1 , Gpxq : x x

dx  pe
1q  ln 4

Example 2.1.16. Calculate In :

»π 0

sinn pxq dx

for n P N . Solution: For n  1, 2, we conclude that »π 0

sinpxq dx  r
cospxqs

π 0

2,

175

»π 0

sin2 pxq dx

1.



»π



1 1 1 r 1
cosp2xqs dx  x
sinp2xq 2 2 2

0

π 0

π .

For n ¥ 3, we conclude by partial integration that In





»π

sin pxq dx 

»π

n

0

0

sinn
1 pxq sinpxq dx

(π sin
pxqr
cospxqs 0


»π

n 1

 pn
1q  pn
1q

»π »0π 0

0

pn
1q sinn
2pxq cospxqr
cospxqs dx

sinn
2 pxq cos2 pxq dx sinn
2 pxqr1
sin2 pxqs dx  pn
1qpIn
2
In q

and hence that In

 n
n 1 In
2 .

Hence we conclude by induction that I2k

1

 2  23    2k2k 1 ,

I2k

 π  12    2k2k
1

for all k P N zt0, 1u. this result, leads on Wallis’ product representation of π which will later be used in the calculation of the Gaussian integral. Theorem 2.1.17. (Wallis’ product representation of π) lim 4pk

k

1q

Ñ8



2 3



2k 2k 1

2

π .

Proof. In this, we are using the notation from the previous example. Since 0 ¤ sinpxq ¤ 1 for all x P r0, π s, it follows that sinn

1

pxq  sinpxq sinnpxq ¤ sinnpxq

for all x P r0, π s and hence that In

1



»π

n 1

sin 0

pxq dx ¤ 176

»π 0

sinn pxq dx  In

3.3

3.2

Π 3.1

10

20

30

40

50

n

Fig. 49: Sequences a1 , a2 , . . . and b1 , b2 , . . . from the proof of Wallis product representation for π, Theorem 2.1.17, that converge to π from below and above, respectively.

for all n P N . As a consequence,

   2k2k 1  I2k 1 ¤ I2k  π  12    2k2k
1 pk
1q ¤ I2k
1  2  23    22k
1

2

2 3

and 2

2 3

pk
1q  2k  2k    2k2k 1  21  43    22k
3 2k
1 2k 

2

 ak : p4k 2q    2k2k 1 ¤ π pk
1q  2  4    2pk
1q  2k ¤ 2  23    22k
1 1 3  2k
3 2k
1  2 2 2pk
1q  bk : 4k 3    2k
1 2 3

177

1 1

y 1.05 1.04 1.03 1.02 1.01 10

20

Fig. 50: Graph of pp0, 8q Ñ R, x ÞÑ Γpx for every n P N. See Theorem 2.1.18.

P N zt0, 1, 2u. Further, 

2 4k 6 2pk 1q 1  

30

40

?

50

x

1q px{eq
x { 2πx q. Note that Γpn

1q  n!

for k ak ak

bk 1 bk bk ak



4k



2

4pk 1q 4k

4k 4k 2



2k



3

2

 8k 8k 2

16k 16k

8 6

¡1,

2

4k  4k4k  1, 2 4k 1

2 1  p2k 1q  1 1

2k 2k 1

2k 2k

2

8pk 1q2 p4k 2qp2k 3q

2k

2k

for all k P N zt0, 1, 2u. Hence the sequences a3 , a4 , . . . and are convergent, as increasing sequence that is bounded from above by π and decreasing sequence that is bounded from below by π, respectively, and converge to the same limit π. Essentially as an application of Wallis’ product formula, we prove Stirling’s asymptotic formula for factorials which is often used in applications. 178

Theorem 2.1.18. (Stirling’s formula) n! ? nÑ8 n lim


n
n

e



?

2π .

(2.1.7)

Proof. First, we notice that ln is concave since ln2 pxq 
1{x2 x ¡ 0. Hence it follows by Theorem 1.4.31 that »x

1

lnpy q dy ¥

x


x ln



1

x



1

»x x

lnpxq



1 2

x

x

py
xq ln





ln

1

x



x





1

x

 lnpxq

dy

x

 12 r lnpxq

  0 for all

lnpx

1qs

for all x ¡ 0. In addition, it follows from the Definition 1.4.30 of the concavity of a differentiable function that y
x , lnpy q ¤ lnpx x

lnpy q ¤ lnpxq where x ¡ 0 and y »x

1

x

»x x

1

y
px 1 q , x 1

1q

¡ 0, and hence that

lnpy q dy ¤ lnpxq
1 lnpy q dy ¤ lnpx

1

1q
1

 lnpxq

1 2x 1


2px

1

1 , 2x

 lnpx 1q

1q


2px

1

and »x

1

x

lnpy q dy ¤

 12 r lnpxq

1 2

lnpx



lnpxq

lnpx

1qs

1 4



1 x

1q

1 2x


x11




2px

1



1q

.

Hence it follows that 0¤

»x x

1

lnpy q dy


1 r lnpxq 2

lnpx

179

1qs ¤

1 4



1 x


x



1 1

1q

for all x ¡ 0 and hence that 0¤

»n

 14

1



lnpy q dy


1


1 n






1¸ r lnpkq 2 k1 n 1

lnpk

1qs ¤

n 1




1¸ 4 k1

1 k


k



1 1

¤ 14 .

Therefore, we conclude that the sequence S1 , S2 , . . . , where Sn : »n

»n 1

lnpy q dy





1¸ r lnpkq 2 k1 n 1

lnpk

1qs

 r y lnpyq
y sn1
lnpn!q lnp2nq 1 ?
 n lnpnq  n lnpnq
pn
1q
lnpn!q 2  1 ln n!n ne



lnpy q dy
lnpn!q

lnpnq 2

for every n P N , is increasing as well as bounded from above and therefore convergent to an element of the closed interval form 0 to 1{4. Hence it follows also the existence of n! ? nÑ8 n lim


n
n

e

which will be denoted by a in the following. For the determination of its value, we use Wallis’ product. According to Theorem 2.1.17 

2

2pk 1q p2k k!q4  klim 2pk 1q     lim k Ñ8 2k Ñ8 1 p2k 1q rp2k q!s2 p2k k!q4  klim Ñ8 p2k 1q rp2kq!s2   
k 4  ?  2k 2k 2 2
p4k 1q k p2k k!q4 1 k ? e 2k  klim Ñ8 p2k 1q rp2kq!s2 e k     
k 4 ?  2k 2k 2 a2 k pk!q4 1 k ? e 2k  klim 4 . Ñ8 2p2k 1q rp2kq!s2 e k

π 2

2 3

2k 2k 1

180

Hence it follows that a 

?

2π and, finally, (2.1.7).

Definition 2.1.19. If I is some non-empty index set and ai i P I, the symbol ¹ ai

P I for every

P

i I

denotes the product of all ai , where i runs through the elements of I. Note that, as a consequence of the associative law for multiplication, the order in which the products are performed is inessential. Example 2.1.20. Show that for every x P R sin cos


πx



2
πx

2



¹ πx x2 lim 1
2 2 nÑ8 k1 4k n

 nlim Ñ8

n ¹



1




k 0

p2k

x2



,

1 q2

.

(2.1.8)

Solution: For this, we define for every n P N a corresponding In : R Ñ R by In pxq :

» π{2

cospxtq cosn ptq dt

0

for every x P R. In particular, this implies that I0 pxq 

#

π 2

1 sinpπx{2q{pπx{2q

if x  0 , if x  0

for x R t
1, 1u I1 pxq 





» π{2 0

cospxtq cosptq dt 

1 sinppx 1qtq 2 x 1

» π{2 0

sinppx
1qtq x
1

1 rcosppx 2

π{2 0

181

1qtq

 cos1 p
πxx{22q ,

cosppx
1qtqs dt

for x P t
1, 1u I1 pxq 





» π{2 0

1 sinp2tq 2

1 t 2

and hence I1 pxq 

#

In the following, let x integration that x In pxq  x 2

n

» π{2 0

n

» π{2

0

π{2

» π{2 0



1 r1 2

0

π {4 cospπx{2q{p1
x2 q

P R.

cosp2tqs dt

 π4

T For n

if x P t
1, 1u . if x R t
1, 1u

P N z t0, 1u, we conclude by partial

cosn ptq x cospxtq dt  r x cosn ptq sinpxtqs0π{2

sinptq cosn
1 ptq x sinpxtq dt 

0

» π{2

» π{2

cos ptq dt  2






n sinptq cosn
1ptq cospxtq 0π{2 

cosn ptq
pn
1q sin2 ptq cosn
2 ptq cospxtq dt



n n cosn ptq
pn
1q cosn
2 ptq 0  n2 Inpxq
npn
1qIn
2pxq . Therefore, it follows that In
2 pxq 

x2 In
2 pxq  1
2 In
2 p0q n From this, it follows by induction that I0 pxq I0 p0q



cospxtq dt

n2
x2 In pxq npn
1q



and hence that







In pxq . In p0q

I2n pxq ¹ x2 1
I2n p0q k1 p2kq2 n

182



,

I1 pxq I1 p0q



I2n I2n

n  pxq ¹ x2 1
p2k 1q2 1 p0q k 1

1

for every n P N . In the following, we show that In pxq nÑ8 In p0q lim

1.

(2.1.9)

For this, we note that

| cospxtq
1|  | cosp|x|tq
1| 

»  |x|t    0

   sin s ds 

r
p qs ¤ |x|t

for t ¥ 0. Hence it follows for every n P N that

|Inpxq
Inp0q|  ¤ |x|

» π{2 0

»  π{2  cos xt   0

   n 1 cos t dt  » π{2

r p q
s

t cosptq cosn
1 ptq dt ¤ |x|

pq

sinptq cosn
1 ptq dt 

0

|x| , n

where it has been used that t cosptq ¤ sinptq for 0 ¤ π {2. Hence it follows (2.1.9) and, finally, (2.1.8). For this, note that the second relation in (2.1.8) is trivially satisfied for x P t
1, 1u. Example 2.1.21. Let m be some natural number Define F, G, f, g : R Ñ R by F py q : py 2 Gpy q : y for all y »x a

¥ 1, a P R and c ¡ 0.

c2 q
m , g py q : 1 , f py q :
2my  py 2

c2 q
pm 1q ,

P R. Then by Theorem 2.1.12 for every x ¡ a

dy py2 c2qm

 px2

x


c2 qm pa2 183

a

»x

c2 qm

2m a

y 2 dy py2 c2qm

1

 px2

x

q»
pa2

c2 m


2mc2

x

a

»x

a

q

c2 m

dy c2 qm

p

y2

2m a

p

y2

dy c2 qm

1

and hence it follows the recursion (or ‘reduction’) formula »x a

py2

dy c2 qm



1

1 x a  2mc
2 px2 2 2 m cq pa c2qm »x 2m
1 dy , 2 2 2mc c2 qm a py



which is used in the method of integration by decomposition into partial fractions below. Example 2.1.22. Show that

|Jnpxq| ¤ π2  n2
x x2

for all n P N and x P R such that 0 ¤ x   n. Solution: Define F pθq : f pθq : for all θ

1 , g pθq : px cos θ
nq  cospx sin θ
nθq , x cos θ
n x sinpθq px cos θ
nq2 , Gpθq : sinpx sin θ
nθq

P r0, πs. Then by Theorem 2.1.12, Jn pxq 


π1

1 π

»π 0

and hence

|Jnpxq| ¤

1 π

»π 0

cospx sin θ
nθq dθ

x sinpθq px cos θ
nq2  sinpx sin θ
nθq dθ , »π 0

2 x x sinpθq . dθ   2 2 px cos θ
nq π n
x2 184

Problems 1) Calculate the value of the integral. In this, where applicable, n P N . »3

» π {2

4t e
5t dt , b) 2

a) 0 »π

c) 0

»1

e) 0

»π

g) 0

ϕ  r sinp2ϕq

0

e
ϕ cosp2ϕq dϕ

, ,

f) »3

,

lnp2xq dx , x2

» 1{?2 1

x2 arctanp3xq dx x sinpnxq dx

»2

d)

h) 1

3 cosp7ϕq s dϕ ,

0

lnp2x2

xn lnpxq dx

1q dx

,

.

2) Derive a reduction formula where the integral is expressed in terms of the same integral with a smaller n. In this n P N , a P R, x ¥ a and, where applicable, m P N, b, c P R . »x

»x

sin py q dy n

a) »ax

y n eby dy

c) a »x

e) a »x

g) a

,

b) »x

,

d) a

y n cospby q dy

,

a

cosn py q dy

y n sinpby q dy »x

f) a

e
cy sinpby q dy

,

, ,

y m rlnpy qsn dy

»x

h) a

,

e
cy cospby q dy

3) Let I be some non-empty open interval of R, h : I a, b P I be such that a   b.

.

Ñ R a map and

a) If h is twice differentiable on I and such that hpaq  hpbq  0, show that »b a

hpxq dx 

1 2

»b a

px
bqpx
aqh 2 pxq dx .

b) If h is four times differentiable on I and such that hpaq hpbq  h 1 paq  h 1 pbq  0, show that »b a

hpxq dx 

1 24

»b a

px
bq2 px
aq2 h pivq pxq dx .

185



[Remark: Note that if h  f
p, where f : I Ñ R is twice, four times differentiable, respectively, and p : I Ñ R is a polynomial function of the order 1, 3, respectively, then h 2  f 2 , h pivq  f pivq , respectively. In connection with the above formulas, this fact is used in the estimation of the errors for the Trapezoid Rule / Simpson Rule for the numerical approximation of integrals. See Section 2.1.4.]

4) Let a, b P R be such that a   b and f, g : ra, bs Ñ R be restrictions to ra, bs of twice continuously differentiable functions defined on open intervals of R containing ra, bs. In addition, let f paq  f pbq  0 and g paq  g pbq  0. a) Show that »b a

g pxqf 2 pxq dx 

»b a

g 2 pxqf pxq dx .

b) In addition, assume that f and g solve the differential equations


f 2 pxq U pxq f pxq  λ f pxq ,
g 2 pxq U pxq gpxq  µ gpxq , where U : ra, bs Ñ R is continuous and λ, µ P R are such that λ  µ. Show that »b a

f pxqg pxq dx  0 .

2.1.3 Partial Fractions Lemma 2.1.23. Let P, Q : R Ñ R be polynomials of degree m, n P N , respectively, where m   n. Finally, let a1 , . . . ar be the (possibly empty) sequence of pairwise different real roots of Q, where r P N, and let m1 , . . . , mr be the sequence in N consisting of the corresponding multiplicities. (i) There are s P N along with (possibly empty and apart from reordering unique) sequences pbr 1 , cr 1 q, . . . , pbr s , cr s q of pairwise different elements of R  p0, 8q and mr 1 , . . . , mr s in N such that Qpxq  qn  px
a1 qm1 . . . px
ar qmr 186



 px
br 1 q2

cr

mr

1

1



. . . px
br

s

q2

cr

mr

s

s

for all x P R where qn is the coefficient of the n
th order of Q. (ii) There are unique sequences of real numbers A11 , . . . , A1m1 , . . . , Ar1 , . . . , Armr and pairs of real numbers pBr 1,1 , Cr 1,1 q, . . . , pBr 1,mr 1 , Cr 1,mr 1 q, . . . , pBr s,1, Cr s,1q, . . . , pBr s,mr s , Cr s,mr s q, respectively, such that P pxq Qpxq

 xA
11a

   px
A1ma qm . . . 1 1 Ar1 Arm    x
ar px
ar qm Br 1,m x Br 1,1 x Cr 1,1    px
b q2 c rpx
b q2 1

1

r

r

Cr cr Cr cr

r 1

r 1

r 1

Br s,1 x Cr s,1 px
br sq2 cr s

r 1

   rpBxr
s,mb qx2 r s r s

1,mr mr 1

1

s,mr mr s

s s

s

s

...

1

for all x P Rzta1 , . . . , ak u. Proof. See Function Theory. Corollary 2.1.24. Let P, Q, m, n; a1 , . . . ak , m1 , . . . , mr , pb1 , c1 q . . . , pbn
k , cn
k q, mr 1, . . . , mr s, A11 , . . . , A1m1 , . . . , Ar1, . . . , Armr and pBr 1,1, Cr 1,1q, . . . , pBr 1,mr 1 , Cr 1,mr 1 q, . . . , pBr s,1, Cr s,1q, . . . , pBr s,mr s , Cr s,mr s q as in Lemma 2.1.23. Then by F pxq : A11 lnp|x
a1 |q
  


A1m 
1 px
a qm
1

1 m1

Ar1 lnp|x
ar |q
  


1

Arm 
1 px
a qm
1

1 mr

Br 1,1 lnrpx
br 1 q2 cr 1 s 2  br 1 Br 1,1 Cr 1,1 x
br arctan cr 1 cr 1 187

1

1

r

r

r

1

...

... ...

Br 1,1 2p1
mr

q

1

rpx
br 1q cr 1sm
1 pbr 1 Br 1,1 Cr 1,1q  Fr 1 pxq . . . Br s,1 lnrpx
br s q2 cr s s 2 

br s Br s,1 Cr s,1 x
br s arctan cr

Br s,1 2p1
mr

pbr

1

2

r 1

cr

s

s

s Br s,1

q

1

rpx
br sq cr ssm Cr s,1q  Fr s pxq 2

...

s

r s


1

for all x P Rzta1 , . . . , ak u, there is defined an anti-derivative F of P {Q. Here Fr 1 , . . . , Fr s : R Ñ R denote anti-derivatives satisfying 1 Fr1 l pxq  2 rpx
br l q cr lsmr l

for all x P R and l  2, . . . s. Note that such functions can be calculated by the recursion formula from Example 2.1.21. Example 2.1.25. Calculate

»2 0

4 x2


9 dx .

Solution:

»2 »2 »2  1 1 4 4 2 dx  dx 
dx 2 3q x
3 x 3 0 x
9 0 px
3qpx 0 3  23 plnp|2
3|q
lnp|2 3|qq
23 plnp|
3|q
lnp|3|qq 
23 lnp5q , where it has been used that for any function f , a, b P R such that a  b and any x from the domain of f such that f pxq R t
a,
bu, the following holds

pf pxq

1

aqpf pxq

bq

 b
a 1



1

f pxq

a




1 f pxq



b

.

(2.1.10)

This identity is of use in the application of the method of integration by partial fractions to more complicated situations. 188

Example 2.1.26. Calculate »3 0

Solution: »3 0

 

3x x2

4 2x

2

dx .

»3

»

3 3 2x 2 1 dx dx dx  2 2 x 2x 2 2x 2 1 q2 1 0 2 x 0 px 3 3 lnp32 2  3 2q arctanp3 1q
lnp2q
arctanp1q 2 

2 3 17 π arctanp4q
. ln 2 2 4

3x

4

Example 2.1.27. Calculate »2 1

x2

p

1

1 q2

x2

dx .

Solution: Considering the symmetry of the integrand, by Lemma (2.1.23) there are A, B, C P R such that 1

x2 px2

A  1 q2 x2

B x2

for all x  0. Hence for all x P R 1  Apx2

1 q2

Bx2 px2

1q

1

px2

C

1 q2

(2.1.11)

 pA B qx4 p2A B C qx2 A and hence A  1, B 
1 and C 
1. Hence it follows by the recursion Cx2

formula from Example 2.1.21 that »2 1

 

1

x2 px2 1 π 2 4 1 π 2 4

»2

»

»

2 2 1 1 1 dx
dx
dx  dx 2 2 2 2 1q 1 1 q2 1 x 1 x 1 px »2
arctanp2q
px2 1 1q2 dx 1 

» 1 2 1 1 2 1
arctanp2q
2 5
3
2 x2 1 dx 1

189

y 1

-4

-2

2

x

4

-1

Fig. 51: Graph of the antiderivative F of f pxq : 1{p1 Compare Example 2.1.28.

3
π 2 4

 157


arctanp2q

x4 q, x P R, satisfying F p0q  0.



.

Another way of arriving at the decomposition (2.1.11) is by help of the identity (2.1.10) which leads on 1

x2 px2

1 q2 1

 x2px2

 x2 1q




1



1



1



1 x2




1 x2 px2 1q x2 1 x2 1 1 1 1 

2 2 2 2 2 px 1q x x 1 px 1q2



1 1

for all x P R . Example 2.1.28. Calculate »x a

dy , 1 y4

where a P R and x ¥ a. Solution: Since x4 1 ¡ 0 for all x according to Lemma 2.1.23 there are b, c, d, e P R such that 1

y4

 py2

by

cq  py 2 190

dy

eq

P

R,

(2.1.12)

 y4 dy3 ey2 by3 bdy2 bey cy2 cdy ce  y4 pb dqy3 pc e bdqy2 pbe cdqy ce for all y P R. This equation is satisfied if and only if b d  0 , c e bd  0 , be cd  0 , ce  1 . From the first equation, we conclude that d 
b which leads to the equivalent reduced system

d 
b , e

c  b2 , bpe
cq  0 , ce  1 .

The assumption that b  0 leads to e 
c and
1 
ce  c2 . Hence it follows that b  0. Therefore, the second equation of the last system leads to the equivalent reduced system d 
b , b2

 2c , e  c , c2  1 ? ? which has the solution c  e  1 and?b  2,?d 
2. (The other remaining solution c  e  1 and b 
2, d  2 results in a reordering of the factors in (2.1.12). Hence it follows that ? ? 2 y 1q  py 2
2 y 1q 1 y 4  py 2 for all y P R. Note that, the last could have also been more simply derived as follows ? 1 y 4  1 2y 2 y 4
2y 2  py 2 1q2
p 2y q2 ? ?  py2 2 y 1q  py2
2 y 1q valid for all y P R. Further, according to Corollary 2.1.24 there are uniquely determined A, B, C, D P R such that 1 1 for all y

y4



?

Ay y2

B 2y 1




P R. In particular, this implies that 1 1
Ay? B   4 4 1 y 1 p
y q y2
2 y 191

?

Cy

y2

D 2y 1

1


Cy?

y2

(2.1.13)

D 2y

1

 2
Cy? y

D 2y

y2

1


Ay? B
2y 1

for all y P R. Since A, B, C and D are uniquely determined by the equations (2.1.13) for every y P R, it follows that C 
A and D  B. Hence we conclude that there are uniquely determined A, B P R such that 1 1 for all y

y4



?

Ay y2

B 2y 1

P R. In particular, 1

and hence B

1 1

04


Ay? B y2
2 y 1

 2B

 1{2. Also

 1 1 14  A p?1{2q
A ?p1{2q ?  2 2 ? 2
 2

? 2
2 1 2 2 1 2 2

A 2  2 A  2 A 2 2 ? and hence A  2{4. We conclude that ?  ?  1 2y 2 2y 2
1 ? 4 2 ? 1 y4 y 2 y 1 y2
2 y 1 ?  ?  2y 1 2y
1 1 ? 4 2 ?
y 2 y 1 y2
2 y 1  ?  ? ? 2 2 2 ? ? 2 2 4 2y
1 2y 1 1 1 for all y P R. Hence it follows that ?   2 ?

 2 ? »x 2 2x 1 dy x x
2x ? ?  ln ln
4 y 8 2a 1 a2 a2
2 a a 1 1 2

192

2 2

1 1



?

 ? ? 2 arctanp 2 x 1q
arctanp 2 a 1q ?4   ? ? 2 arctanp 2 x
1q
arctanp 2 a
1q . 4

Remark 2.1.29. The previous example gives an illustration of the wellknown fact that one should never blindfoldly rely on computer programs. In Mathematica 5.1, the command Integrater1{p1 gives the output

? ?1 p
2ArcTanr1
2xs 4 2 ? 2x x2 sq Logr1

2ArcTanr1

x^ 4q, xs

?

2xs
Logr
1

?

2x
x2 s

which is incorrect. A first inspection of the last formula reveals that the argument of the first natural logarithm function is becoming negative for large x such that the logarithm is not defined. This gives a first indication that the expression is incorrect. Comparison with the result from Example 2.1.28 shows that the sign of that argument has to be reversed. Example 2.1.30. Find solutions of the following differential equation for f : R Ñ R with the specified initial values. f 1 pxq  af pxqp1
f pxqq

(2.1.14)

for all x P R, where a ¡ 0, f p0q P p0, 1q. Solution: If f is such function, it follows that f is continuously differentiable. Since f p0q P p0, 1q, it follows by the continuity of f the existence of an open interval c, d P R such that c   0   d and such that f p[c, d]q € p0, 1q. Since a ¡ 0 and the function af p1
f q is ¡ 0 on the interval [c, d], it follows from (2.1.2) that f px1 q  f px0 q

f px1 q
f px0 q  f px0 q 193

» x1 x0

f 1 pxq dx

y 1.2 0.8 0.5

-4

-3

-2

-1

1

2

3

4

x

Fig. 52: Graphs of the solutions f0 , f1{4 , f1{2 , f3{4 and f1 of (2.1.14) in the case that a  1. Compare Example 2.1.30.

y

1 -0.5 -1

0.5

1

x

-9

Fig. 53: Graphs of the solutions f2 and f4 of (2.1.14) in the case that a Example 2.1.30.

194

 1.

Compare

» x1

 f px0 q

a f pxqp1
f pxqqdx ¥ f px0 q

x0

for all x0 , x1 P [c, d] such that x0 ¤ x1 . In addition, the restriction of f to [c, d] is non-constant since the function pR Ñ R, x ÞÑ axp1
xqq has no zeros on p0, 1q. Hence we conclude from (2.1.2) by Theorem 2.1.1 for x P [c, d] that apx
cq 



» f pxq 

pq

f c

 ln



»x c

a du 

1 u

1

»x



1
u

c

f 1 py q dy f py qp1
f py qq 

du  ln

1
f pcq f pxq  f pcq 1
f pxq

and hence that f pcq
ac ax e e 1
f pcq This implies that

and hence that





u



» f pxq

f pxq

1
u

du f pcq up1
uq

pq

f c

 1
f pfxpqxq  f p1xq

f 1pxq 1  1
1f pxq
1 . f pcq e
ac 1
f pcq f p0q eax 1
f p0q

 1
f pf0pq0q

 1
1f pxq
1 .

Finally, this leads on f pxq  1


1

 ax

eax

. pq e 1
f p0q ax e
pq f p0q  On the other hand, for every c P R , it follows by elementary calculation 1

f 0 1 f 0

that the function fc defined by fc pxq :

$ &

eax c eax 1
c ax % axe 1
c e c

for x P R if 0   c ¤ 1

for x P Rzta
1 lnppc
1q{cqu if c ¡ 1 or c   0 195

satisfies (2.1.14). Note also that f0 , defined as the constant function of value zero on R, is a further solution of (2.1.14) such that f0 p0q  0.

Problems 1) Calculate the integral. »2

a)

3u


2

»4

c) 3 »3

e) 2 »1

g)

u2

»4

i) 0

3

x3

x2 6x2


2

n) 0 »3

o) 0 »2

p) 1

,

3x 4x2

x3

8

4 x

0

3x 1 dx x3
7x
6 »3

dx

,

f) »

dx

4

2x 1 dx x2
6x 9

b)

d)

3 12x

x3 4x x4
2x2

x4

,

»1

3x 5 dx x4 4x2 3


1{2

»1

du

0

x2

»0

m)

du

u2

» 1{2

k)

u
12

u3


1 x3

»2

2

,

h)


3

»4

,

j) 2

5 dx 1

,

»1

l)

2x2 1
9x2 11x
4 dx

x2

6

1

x4

2x3

3x2

4x

2

x4

2x3 x2 3x3 5x2

4 9x

6

,

x2 3x
1 dx x3
2x2
7x
4

x2
1 dx x4 4x2 4 3x

x4 0

x3 x 1 x4
3x3
3x2 7x

1 3

,

4x3

7 6x2

4x

,

,

1

dx ,

x3 3x2 1 dx x4
15x2 10x 24

,

,

dx , dx , dx .

2.1.4 Approximate Calculation of Integrals Theorem 2.1.31. (Midpoint Rule) Let a, b P R be such that a   b, f : ra, bs Ñ R be bounded and twice differentiable on pa, bq such that |f 2pxq| ¤ K for all x P pa, bq and some K ¥ 0. Then

196

y 40 30 20 10

1.2

1.4

1.6

1.8

x

2

Fig. 54: Midpoint approximation.

(i)

» b   f x dx  a



p q
f

a

b



2

  a 

K pb
q ¤ 24 pb
aq3 .

(ii) In addition, let n P N , h : pb
aq{n and ai : a i P t0, . . . , nu. Then »  b   f x dx  a

pq





n¸1



f


a

ai 2

i

i 0

   1 h 

i  h for all

K pb
aq ¤ 24 n2

Proof. (i) By Taylor’s Theorem 1.4.23

|f pxq
p1 pxq| ¤ for all x P pa, bq where p1 pxq : f



a

b 2



K 2



x




a f1 197

b 2

a

b

2

2



x


a

b 2



3

.

for all x P R is the first-degree Taylor-polynomial of f centered around pa bq{2. Further, »b a

p1 pxq dx  f

f f

 

a

b 2

a

b 2





a

b



2

pb
aq

1 f1 2

pb
aq



a f1



a

b

b



2





2

»b

x


a

a

b 2

x


a

b 2

2 b   



dx

a

pb
aq .

In addition, » b   f x dx  a »b

pq


¤ 

»b a

  p1 x dx

2

pq

K a b x
2 a 2 K p b
aq3 . 24

¤

dx 

»b a

K 6

|f pxq
p1pxq| dx 

x


a

b 2

3 b   

a

(ii) is a simple consequence of (i). Theorem 2.1.32. (Trapezoid Rule) Let I be some non-empty open interval of R, f : I Ñ R be twice continuously differentiable and a, b P I be such that a   b. In particular, let |f 2 pxq| ¤ K for all x P pa, bq and some K ¥ 0. Then (i)

» b   f x dx  a

pq


f paq

f pbq 2

  a 

K  pb
q ¤ 12 pb
aq3 .

(ii) In addition let n P N , h : pb
aq{n and ai : a i P t0, . . . , nu. Then »  b   f x dx  a

pq



f pa q i

n¸1



i 0

f pai 2

198

i  h for all



1

q  h ¤ K pb
aq3 

12

n2

.

y 40 30 20 10

1.2

1.4

1.6

1.8

2

x

Fig. 55: Trapezoid approximation.

Proof. Define f pbq
f paq b
a

ppxq : f paq for all x P R and h : f and »

 px
aq


p. In particular, it follows that hpaq  hpbq  0 b f paq f pbq ppxq dx   pb
aq . 2

a

By partial integration, it follows that »b a

hpxq dx 

1 2

»b a

px
bqpx
aqh 2pxq dx  1 2

»b a

px
bqpx
aqf 2pxq dx

and hence that » b     h x dx   a

pq

¤

1 2

»b a

pb
xq px
aq |f 2pxq| dx 199

¤

K 2

»b a

K pb
xq px
aq dx  12 pb
aq3 .

(ii) is a simple consequence of piq. Theorem 2.1.33. (Simpson’s Rule) Let h ¡ 0, I be some open interval of R containing r
h, hs, f : I Ñ R be four times continuously differentiable and |f pivq pxq| ¤ K for all x P p
h, hq and some K ¥ 0. Then » h   f x dx 


h

1 rf p
hq 3

pq


4f p0q

Proof. Define "

ppxq :

1 rf phq 2

f p
hqs
f p0q

for all x P R and g : f »h


h

ppxq dx 

 13 rf p
hq

"

*

2

 hx2

f phqs 

  h

K 5 ¤ 90 h .

1 x r f phq
f p
hqs  2 h

f p0q


p. Then gp
hq  gp0q  gphq  0 and 1 rf phq 2

4f p0q

f p
hqs
f p0q

*

 2h 3

f p0q  2h

f phqs  h .

By partial integration, it follows that »0


h

px

»h

hq p3x
hq g pivq pxq dx 3

»h

 px
hq p3x 3

0

and hence that

hq rg pivq pxq

» h      g x dx  
h 

pq



0

¤

K 36

»h 0

px
hq3 p3x

hq g pivq pxq dx

g pivq p
xqs dx  72

ph
xq3 p3x 

h K 3 p h
xq5
h ph
xq4  36 5 0

200

»h

hq dx

K 5 h .  90


h

g pxq dx

y 50 40 30 20 10 1.2

1.4

1.6

1.8

2

x

Fig. 56: Simpson’s approximation.

Corollary 2.1.34. Let I be some non-empty open interval of R, f : I Ñ R be four times continuously differentiable and a, b P I be such that a   b. In particular, let |f pivq pxq| ¤ K for all x P pa, bq and some K ¥ 0. Finally, let n P N , h : pb
aq{n and ai : a i  h for all i P t0, . . . , nu. Then »  b   f x dx  a

pq


n
1  1¸ f pai q 6 i1

K pb
aq ¤ 2880 n4

4f ppai

ai

1

q{2q

f pai



1

q 

   h 

5

.

Proof. The corollary is a simple consequence of Theorem 2.1.33.

Problems 1) Calculate the integral. In addition, evaluate the integral approximately, using the midpoint rule, the trapezoidal rule and Simpson’s

201

rule. In this, subdivide the interval of integration into 4 intervals of equal length. Compare the approximation to the exact result. »1

a) 0 »1

c) 0

»1

du p1 uq2

,

b) 0

p1

2x dx x2 q2

,

3u2 p1 u3q2 du .

2) By using Simpson’s rule, approximate the area in R2 that is enclosed by the Cartesian leaf C : tpx, y q P R2 : 3

?

2 py 2
x2 q

2 x px2

3y 2 q  0u ,

where a ¡ 0. In this, subdivide the interval of integration into 4 intervals of equal length. Compare the approximation to the exact result which is given by 1.5. 3) The time for one complete swing (‘period’) T of a pendulum with length L ¡ 0 is given by T

a

 L {g π
k 2 I pk q , 2 1
k

? 2

where I pk q 

»1

?


1 1
k2 u2

? 2 ? 1
2u ? 2 2  du , 1
k 1
k u

θ0 P p
π, π q is the initial angle of elongation from the position of rest of the pendulum, k : | sinpθ0 {2q|, and where g is the acceleration of the Earth’s gravitational field. By using Simpson’s rule, approximate T for θ0  π {4. In this, subdivide the interval of integration into 4 intervals of equal length.

2.2 Improper Integrals Definition 2.2.1. (Improper Riemann integrals) (i) Let a P R, b P R Y t8u be such that a   b and f : ra, bq Ñ R be almost everywhere continuous. Then F : ra, bq Ñ R, defined by F pxq : 202

»x a

f py q dy

for every y P ra, bq, is a continuous function by Theorem 1.5.18. We say that f is improper Riemann-integrable if there is L P R such that »x

lim F pxq  lim

x

Ñb

x

Ñb

f py q dy  L .

a

In this case, we define the improper Riemann integral of f by »b a

»x

f py q dy  lim x

Ñb

a

f py q dy .

(ii) Let a P R Y t
8u, b P R be such that a   b and f : pa, bs Ñ R be almost everywhere continuous. Then F : pa, bs Ñ R defined by »b

F pxq :

x

f py q dy

for every y P ra, bq is a continuous function by Theorem 1.5.18. We say that f is improper Riemann-integrable if there is some L P R such that » lim F pxq  lim

x

Ña

x

Ña

b

x

f py q dy  L .

In this case, we define the improper Riemann integral of f by »b a

f py q dy  lim x

Ña

»b x

f py q dy .

(iii) Let a P R Y t
8u, b P R Y t8u such that a   b and f : pa, bq Ñ R be almost everywhere continuous. Then we say that f is improper Riemann-integrable if, both, f |pa,cs and f |rc,bq are improper Riemannintegrable for some c P pa, bq. In this case, we define »b a

f pxq dx :

»c a

f pxq dx

»b c

f pxq dx .

Note that as a consequence of the additivity of the Riemann integral, 1.5.17, this gives a well-defined definition of the symbol Theorem ³b f pxq dx. a 203

Remark 2.2.2. Note that, in case that the integrand has an extension to a continuous function on the respective closed interval, the improper Riemann integral is equal to the Riemann integral of that extension. Example 2.2.3. Define fα : Show that

pp0, 1s Ñ R, x ÞÑ 1{xα q for every real α.

(i) fα is improper Riemann-integrable for every α   1 and »1

dx xα

0

 1
1 α .

(ii) fα is not improper Riemann-integrable for every α ¥ 1.

Solution: For α P Rzt1u and ε P p0, 1q, it follows that »1 ε

and for α  1 that

dx xα

»1 ε

 1 1

ε α




1 α


lnpεq

dx x

and hence the statements. Example 2.2.4. Define fα : Show that

p r1, 8q Ñ R, x ÞÑ 1{xα q for every real α.

(i) fα is improper Riemann-integrable for every α ¡ 1 and »8 1

dx xα

 α
1 1

(ii) fα is not improper Riemann-integrable for every α ¤ 1.

Solution: For α P Rzt1u and x ¡ 1, it follows that »x 1

dy yα

1
α  x 1

α 1

204

and for α  1

»x 1

dy y

 lnpxq

and hence the statements. Theorem 2.2.5. Let a P R, b P R Yt8u be such that a   b and f : ra, bq Ñ R be almost everywhere continuous. Finally, let G : ra, bq Ñ R be defined by » Gpxq :

x

|f pyq| dy for all x P ra, bq and be bounded. Then, f and |f | are improper Riemanna

integrable.

Proof. Let pbn qnPN be a sequence in ra, bq which is convergent to b. Since G is bounded and increasing, suptRan Gu exists. As a consequence for given ε ¡ 0, there is some c P ra, bq such that suptRan Gu
Gpcq   ε. Hence it also follows that suptRan Gu
Gpxq   ε for all x P rc, bq. Then there is n0 P N such that bn ¥ c for all n P N satisfying n ¥ n0 . Therefore it also follows that | suptRan Gu
Gpbn q|   ε for n P N satisfying n ¥ n0 and hence finally lim Gpbn q  suptRan Gu . n

Ñ8

Hence |f | is improper Riemann-integrable and »b a

|f pxq| dx  suptRan Gu .

Further, for every x P ra, bq, it follows that »x a

»x

p|f pyq|
f pyqq dy ¤ 2 |f pyq| dy ¤ 2 suptRan Gu . a

Hence |f |
f and therefore also f is improper Riemann-integrable.

205

y

y 1

24

0.5 6 2 1

2

3

4

5

x

1

2

3

4

5

x

Fig. 57: Graphs of the gamma function Γ (left) and 1{Γ.

Example 2.2.6. Show that fy : pp0, 8q Ñ R, x ÞÑ e
x xy
1 q is improper Riemann-integrable for every y ¡ 0. Hence we can define the gamma function Γ : p0, 8q Ñ R by Γpy q : for all y

»8

e
x xy
1 dx

0

P p0, 8q. Solution: Let y ¡ 0. For every ε ¡ 0, it follows that »1 ε

e
x xy
1 dx ¤

»1 ε

xy
1 dx ¤

1 y

and hence by Theorem 2.2.5 that fy |p0,1s is improper Riemann-integrable and that »1 1 e
x xy
1 dx ¤ . y 0 Further, hy : pr1, 8q Ñ R, x Ñ e
x{2 xy
1 q has a maximum at x0 : maxt1, 2py
1qu. Hence it follows for every R ¥ 1 that »R 1

e
x xy
1 dx ¤ hy px0 q

»R 1

206

e
x{2 dx ¤ 2hy px0 q e
1{2

and by Theorem 2.2.5 that fy |r1,8q is improper Riemann-integrable. Note for later use that Γp1q 

»1

»R

e
x dx

lim

Ñ8

R

0

1

 Rlim 1
e
R q  1 . p Ñ8

e
x dx  lim RÑ8

»R

e
x dx

0

Example 2.2.7. Show that Γpy

for all y

¡ 0 and hence that

1q  y  Γpy q

(2.2.1)

1q  n!

(2.2.2)

Γpn

for all n P N. Solution: By partial integration it follows for every y ε P p0, 1q and R P p1, 8q that »R ε

e
x xy dx  e
ε ε y
e
R R y

and hence (2.2.1). Since Γp1q induction.

1

»R

y

¡ 0,

e
x xy
1 dx

ε

0! , from this follows (2.2.2) by

Example 2.2.8. ( Gaussian integrals, I ) Show that fm,n : p r0, 8q Ñ 2 R, x ÞÑ xm e
nx {2 q is improper Riemann-integrable for all m P N, n P N . In particular, show that I : N  N Ñ R defined by I pm, nq :

»8

2 xm e
nx {2 dx

0

for all m P N, n P N satisfies I pm

2, nq 

for all m P N, n P N and, in particular, I p2k

1, nq 

1

m

2k k! , nk 1 207

n

I pm, nq

(2.2.3)

I p2pk for all k

1  3    p2k nk 1

1q, nq 

1q

 ?1n I p0, 1q

(2.2.4)

P N. Solution: First, it follows for n P N and x ¥ 1 that

»x 0

1 2 y e
ny {2 dy 
1


2 1
e
nx {2

n

»x 0

¤

0

2 e
ny {2 dy

n

0



p
nyq e
ny {2 dy 
n1

2 e
ny {2

2

x 0

,

»1

2 e
ny {2 dy 

»1



»x

»x

2 e
ny {2 dy

»x

0

2 e
ny {2 dy

1 2 ye
ny {2 dy

0

and hence by Theorem 2.2.5 that f0,m and f1,m are improper Riemannintegrable as well as that I p1, nq 

»8

2 y e
ny {2 dy

0

Further, according to Example 1.4.11, ex Hence it follows that x

e

¥e
1

»x y

x

e dy 0

 n1 .

(2.2.5)

¥ 1 and ex ¥ x for all x ¥ 0.

¥

»x

2

y dy 0

 x2

for all x ¥ 0 and in this way inductively that m

ex for all x ¥ 1 and m x ¥ 0 that »x

y 0





m 2

P N.

¥ xm!

In addition, it follows for m

1 2 e
ny {2 dy 



n1 ym

1

e


ny 2

{2

x 0

n

»x

ym 0

1 n 208

»x 0

1

P N, n P N and

p
nyq e
ny {2 dy

pm

2

2 1q y m e
ny {2 dy


n1 xm Since,

1

2 e
nx {2

  1 m   nx

n

»x

2 y m e
ny {2 dy .

(2.2.6)

0






we notice that

1

m

1

m! x
nx2 {2 2  e
nx {2  ¤ , e n

1 m x
xÑ8 n lim

1

2 e
nx {2

0.

Hence it follows from (2.2.6) inductively the improper Riemann-integrability of fn,m for all m P N and n P N as well as the validity of (2.2.3) for all m P N and n P N . Further, it follows from (2.2.5) and (2.2.6) by induction that I p2k for all k

1, nq 

2k k! , I p2pk nk 1

1q, nq 

1  3    p2k nk 1

1q

I p0, nq

P N and, finally, as a consequence of »x

1 2 e
ny {2 dy  ?

n

0

» ?n x

2 e
u {2 du

0

that I p2pk

1q, nq 

1  3    p2k nk 1

1q

 ?1n I p0, 1q .

Example 2.2.9. ( Gaussian integrals, II ) Together with Wallis’ product representation of π from Theorem 2.1.17, the application of the results of Example 2.2.8 allow the calculation of I p0, 1q as follows. Employing the notation of Example 2.2.8, in a first step, we conclude for m P N, n P N that 0 

»x

2 y py
tq e
ny {2 dy 

m

0


2t

»x

ym

2

»x

ym

1

2 e
ny {2 dy

0

2 e
ny {2 dy

»0 x

ym

t2 0

209

2

2

2 e
ny {2 dy

and hence that » x

y

m 2

2 e
ny {2 dy

» x

y

0




» x

y

m 1

2 e
ny {2 dy

2

0

for all x ¥ 0 and, finally, I pm

m 2

2 e
ny {2 dy

0

¡0

a

I pm, nq I pm

1, nq  



2, nq .

In particular, since according to (2.2.3) I pn

1, nq  I pn
1, nq , I pn

2, nq 

1

n n

I pn, nq ,

we conclude that I pn

1, nq  

a

I pn, nq I pn

  I pn 2, nq , I pn, nq   I pn
1, nq I pn a

2, nq 

n n

1, nq  I pn

1

I pn

2, nq

1, nq

and hence that I pn, nq   I pn In particular, the case n  2k 2k k! p2k 1qk 1  I p2k  1  p32k   p12kqk 1 1q k 1  2p2k pk1qk1q2!

1, nq   I pn

1, where k 1, 2k

 ?2k1

P N, leads to

1q   I p2k 1

210

2, nq .

2, 2k

I p0, 1q   I p2k

1q 3, 2k

1q

and hence to d

2k 4pk d

 

1 ? 2 k 1q

2  4    2k 3    p2k 1q

1

3 ? 2 k 1

1 2k  2q 2k

2k 4pk

2

  I p0, 1q 2  4    2pk 1q 3    p2k 3q

Finally, taking the limit k Ñ 8 in the last expression and applying Wallis’ product representation of π (2.2.4) leads to I p0, 1q 

»8 0

2 e
y {2 dy 

Example 2.2.10. Show that Γp1{2q 

π . 2

(2.2.7)

?

π.

Solution: For this, let ε, R ¡ 0. By change of variables, it follows that »R ε

1 2 e
y {2 dy  ?

» R2 {2

2

ε2

{2

x
1{2 e
x dx

and hence by taking the limits that

π 2

 ?1 Γp1{2q . 2

Example 2.2.11. ( Beta function, I ) Show that fx,y : pp0, 1q Ñ R, x ÞÑ tx
1 p1
tqy
1 q is improper Riemann-integrable for all x ¡ 0, y ¡ 0. Hence we can define the Beta function B : p0, 8q2 Ñ R by B px, y q :

»1 0

tx
1 p1
tqy
1 dt

211

40

2

z 20

1.5

0 0

1

y

0.5 0.5

1 x

1.5 2 0

Fig. 58: Graph of the Beta function.

for all x ¡ 0, y ¡ 0. Solution: For this, let x ε, δ P p0, 1{2q. Then » 1{2 ε

¤ x1

t
p1
tqy
1 dt ¤ 2

» 1
δ

{

1 2

 y2

x 1

» 1{2

1 2

 x   x 1{2 2t 2 1
x t dt ¤ 
ε x 1

x

ε

 x
1

¡ 0, y ¡ 0. In addition, let

ε

x

2

,

t
p1
tqy
1 dt ¤ 2 x 1

 y

1 2




δ y ¤ y1

» 1
δ

{

1 2

p1
tq


y 1



 y
1

1 2



y
2p1
y tq

1
δ

{

1 2

.

Hence it follows by Theorem 2.2.5 that fx,y |p0,1{2s and fx,y |p1{2,1q are improper Riemann-integrable and that » 1{2 0

1 t
p1
tqy
1 dt ¤ x 1

x

 x
1

1 2

»1

,

{

1 2

212

1 t
p1
tqy
1 dt ¤ x 1

y

 y
1

1 2

.

As a consequence, fx,y is improper Riemann integrable and satisfies »1

1 t
p1
tqy
1 dt ¤ x 1

0

 x
1

1 2

x

1 y

 y
1

1 2

.

Example 2.2.12. ( Beta function, II ) Show that lim y x B px, y q  Γpxq

y

for all x Then

¡ 0. » 1
δ ε

For this, let x

¡ 0, y ¡ 2.

In addition, let ε, δ

P p0, 1{2q.

tx
1 p1
tqy
1 dt

 y
1 1



(2.2.8)

Ñ8

» py
1qp1
δq 

s

x
1 

1


s

y
1

y
1 y
1 py
1qε 

y
1 » py
1qp1
δq 1 s x
1 ds . py
1qx py
1qε s 1
y
1

ds

Further,

»  

y
1 » py
1qp1
δq  py
1qp1
δq  s   x
1 x
1
s ds s 1 s e ds    py
1qε  y 1 py
1qε  

y
1   » py
1qp1
δq  s  x
1
s  s e 1 1 es  ds .   y 1 py
1qε





¤









We consider the auxiliary function h : r0, 8q Ñ R by 

hpsq : 1
1
for all s P r0, 8q. Then hp0q p0, 8q with derivative h 1 psq 

s

s

y
1

y
1

es

 0, h is continuous and differentiable on

y
1



1


s

y
1

213

y
2

es

¡0

for 0   s   y
1. Hence it follows for s P r0, y
1s that

|hpsq|  hpsq   ¤ ¤

»s

»s 0



u

y
1

1


y
2

u

y
1 

eu du



exp u py
2q ln 1
du y
1 y
1   

»s u u u u du  du exp u
py
2q exp y
1 y
1 0 y
1 0 y
1 e s2 , 2py
1q u

u

0 »s

where the case a that

 1 of (1.4.11) has been used.

» py
1qp1
δq

  x
1
s  s e 1 

py
1qε » py
1qp1
δq

¤

py
1qε

sx
1 e
s






1


s

Hence it follows further

y
1

y
1

  s e  ds 

e s2 e ds ¤ Γpx 2py
1q 2py
1q

2q .

From the previous, we conclude that

|py
1qxB px, yq
Γpxq| ¤ 2py e
1q Γpx and hence that

2q

lim y x B px, y q  Γpxq .

y

Ñ8

Example 2.2.13. ( Beta function, III ) Show that B px, y q 

Γpxq Γpy q Γpx y q

for all x, y ¡ 0. Solution: For this, let x, y by use of partial integration that B px, y

1q 

y B px x

214

(2.2.9)

¡ 0. In the first step, we show 1, y q .

(2.2.10)

For this, let ε, δ » 1
δ

P p0, 1{2q. Then

1
δ  1 x
y y t p1
tq t p1
tq dt 

x

ε

y x

x 1

ε

 x1 rp1
δqxδy
εxp1
εqy s

y x

» 1
δ ε

» 1
δ ε

tx p1
tqy
1 dt

tx p1
tqy
1 dt

which implies (2.2.10). Further, it follows that » 1
δ ε



» 1
δ

t
p1
tqy dt x 1

» 1
δ ε

ε

tx p1
tqy
1 dt

tq t
p1
tqy
1 dt 

p1
t

x 1

» 1
δ

tx
1 p1
tqy
1 dt

ε

and hence that B px, y

1q

B px

1, y q  B px, y q .

As a consequence, we obtain from (2.2.10) the equation B px, y q  B px, y which results in

1q

nq 

px

1q 

1, y q 

x

y y

B px, y

1q

B px, y q . x y By induction, we conclude from (2.2.11) that B px, y

B px, y

B px y

(2.2.11)

y  py 1q    py n
1q B px, y q y q  px y 1q    px y n
1q

for every n P N . In particular,

1  2    pn
1q , y  py 1q    py n
1q 1  2    pn
1q y, nq  px yq  px y 1q    px y

B py, nq  B px

215

n
1q

,

(2.2.12)

where it has been used that B pz, 1q  for every z

»1 0

tz
1 dt 

1 z

¡ 0. Hence it follows that B py, nqB px, y nq B px, y q  B px y, nq

x y  x  y n n n B py,nnxq pyyB pxnq y,B pnx,q y

nq

.

From this follows (2.2.9) by taking the limit n Ñ 8 and applying (2.2.8). Note that, as a consequence of (2.2.8) and the first identity of (2.2.12), we arrive at Gauss’ representation of the gamma function Γpxq  lim pn n

Ñ8

 nlim Ñ8 x  px

1qx B px, n

nx n! 1q    px

1q  lim nx B px, n n

Ñ8

1q

nq

for every x ¡ 0. Theorem 2.2.14. (Gauss’ representation of the gamma function) For every x ¡ 0 nx n! Γpxq  lim . (2.2.13) nÑ8 x  px 1q    px nq As an application of Gauss’ representation of the gamma function, we prove the reflection formula for Γ. Theorem 2.2.15. (Euler’s reflection formula for the gamma function) The equation π Γpxq Γp1
xq  (2.2.14) sinpπxq

holds for all 0   x   1.

216

y

y

10

4 3 2

-1.5 -0.5

1

2

3

4

x 1 -4

-1.5

1

2

3

4

x

-1

-10

Fig. 59: Graphs of the extensions of the gamma function Γ (left) and 1{Γ to negative values of the argument.

Proof. For this, let 0   x   1. Then it follows by (2.1.8) that Γpxq Γp1
xq 





Ñ8 x  px n1
x n!

lim

n

nx n! 1q    px



nq



Ñ8 p1
xq  p2
xq    rpn

1q
xs pn!q n lim  nÑ8 p1
x2 q    pn2
x2 q pn 1q
x 1 1 πx π   lim   . 2 2 x nÑ8 1
x x sin p πx q sin p πx q    1
2 2 1 n

lim

n



 x1  x1

2

Remark 2.2.16. Note that the reflection formula (2.2.14) can and is used to extend the gamma function to negative values of its argument. See Fig. 59. Example 2.2.17. Show Legendre’s duplication formula for the gamma function 1 Γp2xq  ? 22x
1 Γpxq Γpx p1{2qq (2.2.15) π 217

for all x ¡ 0. Solution: For this, let x ¡ 0 and ε, δ

P p0, 1{2q. Then

»1

Γpxq Γpxq Γp2xq

 B px, xq  rtp1
tqsx
1 dt 0

Further, it follows by change of variables that » 1
δ ε



rtp1
tqs

» p1{2q
δ 

1 4


dt 

x 1


u

» 1
δ "

ε

x 1




2

t


du  2


2 2x


p1{2q » 1
2δ 1
2x 2 p1
v2qx
1 dv 2ε
1 » 1
2δ  21
2x p1
v2qx
1 dv ε

2


0 » 1 2δ




1 2x 0

p1
v q

1 2

1 2

» p1{2q
δ ε

»0






1 2




p1{2q



t


1
p2uq2

1 2

*x
1

x
1

dt du



p1
v q


dv

p1
v q


dv .

2 x 1





2ε 1 » 1 2ε


dv

2 x 1

0



2 x 1

(2.2.16)

Further, it follows by change of variables that »b

p1
v q


dv  1

» b2

2 x 1

a

2

a2

y
1{2 p1
y qx
1 dy ,

where 0   a   b, and hence by taking the limit a Ñ 0 that »b

p1
v q


dv  1

» b2

2 x 1

0

2

0

y
1{2 p1
y qx
1 dy .

Hence it follows from (2.2.16) that » 1
δ ε

rtp1
tqsx
1 dt

 2
2x

»

p1
2δq2

0

y
1{2p1
y qx
1 dy

218

» p1
2εq2 0

y
1{2p1
y qx
1 dy



and by taking the limits that Γpxq Γpxq Γp2xq

 21
2xB p1{2, xq  21
2x ΓΓppx1{2qp1Γ{p2xqqq .

Example 2.2.18. Show that » π{2 0



Γ µ2 1 Γ 2 Γ µ2 ν

sin pθq cos pθq dθ  ν

µ

for all µ, ν ¡
1{2. Solution: For this, let µ, ν Then it follows by change of variables that » 1
δ ε





ν 1 2

1

(2.2.17)

¡
1{2 and ε, δ P p0, 1{2q.

tpµ
1q{2 p1
tqpν
1q{2 dt

» arcsinp?1
δ q

2 sinpθq cospθq rsin pθqs ? arcsinp ε q » arcsinp?1
δ q 2 sinµ pθq cosµ pθq dθ ? arcsinp ε q 2

pµ
1q{2 r1
sin2 pθqspν
1q{2 dθ

and hence by taking the limits that Γ



µ 1 Γ ν21 2  Γ µ2ν 1



B



µ 2

1 ν ,

1



2

2

» π{2 0

sinµ pθq cosν pθq dθ .

Problems 1) Show the existence in the sense of an improper Riemann integral and calculate the value. In this, if applicable, s, a ¡ 0. »1

a) 0

c)

»1

lnpxq dx

,

e
sx dx

,

»8

b) 0

d)

»8

0

»8

e) 0

x lnpxq dx ,

0

e
sx sinpaxq dx ,

e
sx cospaxq dx , f)

219

»8 0

?x

e


dx ,

g)

»8 0

»8

e


?x

?x

dx

,

»8

h)


8

x expp
x2 q dx ,

8 dx dx i) , j) , 4 x3 3 0 1 0 x »8 »8 dx dx , , l) k) x 2
x e e x a2
8
8» »8 8 dx dx m) , n) 2 3 2 x 5x 6 x 2x 3x 0 0 »

6

.

2) The radial part of the ‘wave function’ of an electron in a bound state around a proton is given by Rnl : p0, 8q Ñ R, where n P N , l P t0, . . . , n
1u are the principal quantum number and the azimuthal quantum number, respectively [32]. Calculate the expectation value xry of the radial position of the electron in the corresponding state given by ³8 3 2 xry  ³08 r2 Rnl2 prq dr  a , r Rnl prq dr 0 where a  0.529  10
8 cm is the Bohr radius. a) R10 prq  2 e
r

?

,

b)

R20 prq 

?

2
r
r{2 1
e 2 2

6
r{2 , re 12 ? 

2 3 2 2 2 d) R30 prq  1
r e
r{3 r 9 3 27 

8 1 e) R31 prq  ? r
r2 e
r{3 , 6 27 6 4 f) R32 prq  ? r2 e
r{3 81 30 c) R21 prq 

,

,

for all r ¡ 0. 3) The ‘wave function’ of a ‘harmonic oscillator’, i.e., the ‘wave function’ of a point particle of mass m ¡ 0 under the influence of a linear restoring force, is given by ψn : R Ñ R, where n P N is the principal quantum number [2]. Calculate the expectation value xxy of the position of the mass point in the corresponding state given by

xxy 

³8


8 x ψn pxq dx . 2
8 ψn pxq dx

³8

220

2

y

y 0.2

0.5 0.4 0.3

0.1

0.2 0.1 1

2

3

4

r

y

2

4

6

8

10

12

4

8

12

16

20

24

4

8

12

16

20

24

r

y

0.2 0.1

0.1

0.05

2

4

6

8

10

12

r

y

r

y

0.1

0.1

0.05

0.05

4

Fig. 60: Graphs of Problem 2.

8

12

16

20

24

r

r

pp0, 8q Ñ R, r ÞÑ r2 Rnl2 prqq corresponding to a) to f).

221

Compare

y

y

0.6

0.4

0.2 0.1 -4

-2

2

4

x

-4

-2

y

-2

4

2

4

x

y

0.4

-4

2

0.4

2

4

x

-4

-2

x

Fig. 61: Squares of the wave functions of a harmonic oscillator. Compare Problem 3

222

[In this, a  pmω {~q1{2 , ω  pk {mq1{2 , k ¡ 0 is the spring’s constant and ~ is the reduced Planck’s constant.]

a)

ψ0 pxq 

b) ψ1 pxq  c)

ψ2 pxq 

d) ψ3 pxq 

   

?aπ

1{2

? 2 π ?

2

1{2

a

a

e
a

1{2

8 π a ? 48 π

{

x2 2

,

2axe
a

2

{

x2 2

,

p4a2 x2
2q e
a x {2 2

1{2

2

,

p8a3 x3
12axq e
a x {2 2

2

for all x P R. 4) The time for one complete swing (‘period’) T of a pendulum with length L ¡ 0 is given by T

2

d

L g

»1


1

du

a

p1
u2 qp1
k2 u2q

,

where θ0 P p
π, π q is the initial angle of elongation from the position of rest of the pendulum, k : | sinpθ0 {2q|, and where g is the acceleration of the Earth’s gravitational field. Show that the corresponding integral exists in the improper Riemannian sense. Split the integrand into a Riemann integrable and an improper Riemann integrable part, where the last leads on an integral that can easily be calculated. In this way, give another representation of T that involves only a proper Riemann integral.

2.3 Series of Real Numbers Definition 2.3.1. Let x1 , x2 , . . . be a sequence of elements of R. We say that x1 , x2 , . . . is summable or convergent if the corresponding sequence of partial sums 1 ¸



k 1

xk ,

2 ¸



xk ,

k 1

3 ¸



k 1

223

xk , . . .

(2.3.1)

is convergent to some real number. In this case, the sum of x1 , x2 , . . . is denoted by

8 ¸ 

xk .

(2.3.2)

k 1

Otherwise, we say that x1 , x2 , . . . is not summable or divergent. The sequence in (2.3.1) is also called a series and in case of its convergence a convergent series with its sum denoted by (2.3.2). In case of its divergence, that series is called divergent. Example 2.3.2. Let x P R. Show that the so called geometric series 1 ¸



2 ¸

xk ,



k 1

3 ¸

xk ,

k 1



xk , . . .

k 1

is convergent if and only if |x|   1. In that case,

8 ¸

xk



k 0

 1
1 x .

Solution: For this, we define Sn :

n ¸



xk

k 0

for every n P N. Note that in the case x  1, it follows that Sn  n and hence the divergence of the corresponding the geometric series. For x  1, it follows that x  Sn



n ¸



xk

1

k 0



n¸1



xk

 Sn
1

k 1

and hence that

 1 1

x x

n 1

Sn

224

.

xn

1

5

4

3

2

1

10

20

30

40

50

n

Fig. 62: Partial sums of the harmonic series and graphs of ln and 2
1

[2 lnp2q]
1 ln.

As a consequence, the series of partial sums is convergent if and only if |x|   1, and in this case

8 ¸ 

xk

k 0

1 .  nlim Sn  Ñ8 1
x

Example 2.3.3. Show that the harmonic series 1 ¸ 1

 k

,

k 1

2 ¸ 1

 k

3 ¸ 1

,

k 1

is divergent. For every n P Nzt0, 1u 2 ¸ 1 n



k 1

¥

¥ k

2 ¸ 1

2

 k

 k

k 1

2 ¸ 1 0

0  2

k 1

2 ¸ 1

0

 k

, ...

k 1

2 ¸ n



1 k k 2n
1

k 21

2 ¸ 1 2

2  2



k 21

225

2 ¸ n

1 2n k 2n
1

¥ 1 p22
21q  212    p2n
2n
1q  21n  1 1  1  1 n
1  n 1 2

2

2

2

and hence the divergence of the harmonic series. Remark 2.3.4. Note that because series are sequences of partial sums, we can apply the Limit Laws of Theorem 1.2.4 to series. Theorem 2.3.5. (Integral test) Let f : r1, 8q Ñ R be positive decreasing and almost everywhere continuous. Then f p1q, f p2q, . . . is summable if and only if f is improper Riemann-integrable. In this case, »8 1

as well as

f pxq dx ¤

8 ¸ 

f pk q ¤ f p1q

»8 1

k 1

»8 m 1

8 ¸

f pxq dx ¤

f pk q ¤



k m 1

for every m P N .

»8 m

f pxq dx

f pxq dx

(2.3.3)

(2.3.4)

Proof. For this, we define the auxiliary function g : r1, 8q Ñ R by g p1q : f p2q as well as g pxq : f pk 1q for all x P pk, k 1s and k P N . Then »n m

1

g pxq dx 

n »k ¸



k m

1

k

n¸1

g pxq dx 



f pk q

k m 1

for every m, n P N such that m ¤ n. If f is improper Riemann-integrable, it follows because of |g | ¤ f and by Theorem 2.2.5 that g is improper Riemann-integrable and hence that f p1q, f p2q, . . . is summable and

8 ¸ 

f pk q ¤

»8

k m 1

226

m

f pxq dx

y 10 8 6 4 2 2

1.5

3

2.5

s

Fig. 63: Graphs of ζ (black), 1{p1
sq (blue) and s{p1
sq (red).

for every m P N . If on the other hand f p1q, f p2q, . . . is summable, we define the auxiliary function h : r1, 8q Ñ R by hpxq : f pk q for all x P rk, k 1q and k P N . Then »x m

f py q dy ¤

»x m

hpy q dy ¤

8 ¸ 

f pk q

k m

for every m P N and x P r1, 8q. Hence it follows by Theorem 2.2.5 that f is improper Riemann-integrable and that

8 ¸ 

k m

f pk q ¥

»8 m

f py q dy .

for every m P N . Remark 2.3.6. Note that (2.3.4) can be used to estimate remainder terms of the sequence.

227

2.5

2

1.5

1

0.5

10

20

30

40

50

n

Fig. 64: Partial sums of the series from Example 2.3.8 for the case p  1.

Example 2.3.7. (Riemann’s Zeta function ) Show that by ζ psq :

8 1 ¸ s  n

n 1

for every s P p1, 8q there is defined a function ζ : p1, 8q Ñ R. This function is called Riemann’s zeta function. Solution: For every s P p1, 8q the corresponding function fs : r1, 8q Ñ R defined by fs pxq : 1{xs for every x ¥ 1 is positive decreasing and continuous and by Example 2.2.4 improper Riemann-integrable. Hence the statement follows from Theorem 2.3.5. In addition, it follows by (2.3.3) that

 s
1 1

»8 1

fs pxq dx ¤ ζ psq ¤ 1

Example 2.3.8. Let p defined by

¥

»8 1

fs pxq dx 

s

s
1

.

1. Determine whether the sequence a2 , a3 , . . . an :

n ¸

1 k plnpk qqp k 2 228

for every n P Nzt0, 1u is convergent or divergent. Solution: For this, we define the auxiliary function h : r2, 8q Ñ R by hpxq : x  plnpxqqp for every x ¥ 2. Then h is strictly positive, strictly increasing and continuous and hence f : r1, 8q Ñ R defined by f pxq : 1{rpx 1qplnpx 1qqp s for every x ¥ 1 is positive, strictly decreasing and continuous. Further for p  1: »n dx  lnplnpn 1qq
lnplnp2qq 1q lnpx 1q 1 px for every n P N . Using that lnplnp2m qq  lnpm lnp2qq for every m P N , it follows that f is not improper Riemann-integrable. Hence it follows by Theorem 2.3.5 the divergence of the corresponding sequence a2 , a3 , . . . . For p ¡ 1 it follows that »x

py

1

dy 1qplnpy

1   rplnpx 1qq 1
p p

1qq1
p
plnp2qq1
p s

for every x ¥ 1 and hence the improper Riemann-integrability of f and by Theorem 2.3.5 the convergence of the corresponding sequence a2 , a3 , . . . . Theorem 2.3.9. (Comparison test) Let x1 , x2 , . . . and y1 , y2 , . . . be sequences of positive real numbers. Further, let xn ¤ c yn for all n P tN, N 1, . . . u where c ¥ 0 and N is some element of N. If y1, y2, . . . is summable, then x1 , x2 , . . . is summable, too. Proof. If y1 , y2 , . . . is summable, the sequence of partial sums of x1 , x2 , . . . is increasing and bounded from above and hence convergent. Example 2.3.10. Show that the sequence a1 , a2 , . . . defined by an :



n ¸ 1



 k


lnpnq

k 1

for all n P N is convergent. See Fig. 62. Solution: For this, we define an auxiliary sequence b1 , b2 , . . . by bn :



n ¸ 1

k k 1




lnpn

1q  an

lnpnq
lnpn 229

1q  an



ln



n n

1

for all n P N . Then





n 2
ln bn 1
bn  n 1 n 1

»1 »1 1 n 1 1 1
dx  n 1 x n 1 n 1 0 x 0 1

and hence

for all n since

0 ¤ bn

1


bn ¤ pn

1

x n

1

dx

1 q2

P N. Therefore b1 , b2, . . . is increasing and bounded from above bn




n¸1

 b1



8 1 ¸

pbk 1
bk q ¤ b1

2  k

k 1

k 1

for all n P N z t0, 1u. Hence b1 , b2 , . . . and a1 , a2 , . . . are convergent. The constant    n ¸ 1
lnpnq γ : lim nÑ8 k k 1 is known as Euler constant. Presently, it is not yet known whether it is rational or irrational. Since lnpn

¤1

1q 

»n




1

n¸1 » k 1



k 1

it follows that

k

1

dx x

dx x



n »k ¸



k 1 k

»n

1

0 ¤ ln

1

1



dx x 1

n

dx x

1



n

¤



k 1

1 k




n¸1

 k

k 1

1 1

lnpnq ,

¤ an ¤ 1

for every n P N z t0, 1u and hence that 0 ¤ γ given by 0.5772156.

230

n ¸ 1

¤ 1. To 7 decimal places it is

Example 2.3.11. Show Weierstrass’ representation of the gamma function 1 Γpxq

 xe

γx

n
¹

lim

n

Ñ8 k1

1

x
x{k e k

(2.3.5)

for every x ¡ 0, where γ is Euler’s constant. Solution: For this, let x ¡ 0. According to (2.2.13), Γpxq is given by Γpxq  lim n

Ñ8 x  px

nx n! 1 q    px

 x
1 nlim nx Ñ8 nq

n ¹

1

 1

x k





k 1

Further, nx

 exppx lnpnqq  exp

 exp

 

x lnpnq


Hence it follows that

 

n ¸ 1

x lnpnq




k k 1

n ¹



n ¸ 1



k k 1

exp

n ¸ x

 k

k 1

ex{k .

k 1

 

¸ 1 Γpxq  x
1 lim exp x lnpnq
nÑ8 k k 1 n ¹ ex{k  x
1 e
γx nlim Ñ8 k1 1 xk n



n ¹ ex{k

 1

k 1

x k

which implies (2.3.5). Theorem 2.3.12. (Limit comparison test) Let x1 , x2 , . . . and y1 , y2, . . . be sequences of positive real numbers. Further, let xn nÑ8 yn lim

1.

(Note that this implies that yn ¡ 0 for all n P tN, N 1, . . . u, where N is some element of N.) Then x1 , x2 , . . . is summable if and only ify1 , y2, . . . is summable. 231

Proof. Since limnÑ8 pxn {yn q  1, there is N 1 2 for all N P N such that n rem 2.3.9. Example 2.3.13. Let p defined by for all n P n P N that

P N such that

¤ xyn ¤ 32 n

¥ N.

Hence the statement follows by Theo-

  1. Determine, whether the sequence a1 , a2, . . . an :

1 np N , is summable. Solution: Since p

  1, it follows for every

1 1 ¥ p n n  for all n P N and hence by Theorem 2.3.9 and Example 2.3.3 the divergence of a1 , a2 , . . . .

Example 2.3.14. Let p defined by

  1.

Determine whether the sequence a2 , a3 , . . .

an :

n ¸

1 k  plnpk qqp k 2

for every n P Nzt0, 1u is convergent or divergent. Solution: Since p it follows for every k ¥ 3 that

  1,

plnpkqq1
p ¥ 11
p  1 and hence that

1 k  plnpk qqp

¥ k  ln1 pkq .

Hence it follows by Theorem 2.3.9 and Example 2.3.8 that the sequence a2 , a3 , . . . is divergent.

232

Example 2.3.15. Determine whether the sequence a1 , a2 , . . . defined by an :

3n2 pn5

n 1 2q1{2

for all n P N is summable. Solution: Define bn :

3 n1{2

for all n P N . Then b1 , b2 , . . . is not summable according to Example 2.3.13. In addition, it follows that

1

an nÑ8 bn lim

and hence by Theorem 2.3.12 also that a1 , a2 , . . . is not summable. Theorem 2.3.16. (Summation by parts) Let x1 , x2 , . . . and y1 , y2, . . . be sequences of real numbers and S1 , S2 , . . . be the sequence of partial sums of x1 , x2 , . . . . Then n ¸

xk yk



 pSn

cqyn

1


pSm
1

n ¸

cqym


k m



pSk

cqpyk

1


yk q

k m

for all m, n P N such that n ¥ m and all c P R, where we define S0 : 0. Proof. It follows for all m, n P N , that n ¸



xk yk

k m



n ¸



k m



n ¸



pSk
Sk
1qyk 

k m

Sk y k


n ¸

n ¸





Sk y k


k m

Sk y k

1

Sn y n

1

 Snyn 1
Sm
1ym




Sk pyk

1

k m

233




k m 1


Sm
1ym

k m n ¸




n¸1


yk q

Sk y k

1

 Snyn 1
Sm
1ym
 Snyn 1
Sm
1ym


n ¸



pSk

cqpyk

k m n ¸



n ¸

1
yk q



c pyk

1


yk q

k m

pSk

cqpyk

1


yk q

cyn

1


cym

k m

 pSn

cqyn

1
pSm
1

cqym


n ¸



pSk

cqpyk

1


yk q .

k m

Theorem 2.3.17. (Dirichlet’s test) Let x1 , x2 , . . . be a sequence of real numbers such that its partial sums form a bounded sequence and y1 , y2, . . . be a decreasing sequence of real numbers such that limkÑ8 yk  0. Then the sequence x1 y1 , x2 y2 , . . . is summable,

8 ¸ 

xk yk

 M1 y1

k 1

8 ¸ 

pSk
M1 qpyk
yk 1q

(2.3.6)

k 1

and for every n P N

  8  ¸    xk yk    



¤ pM2
M1 q  yn

1

,

(2.3.7)

k n 1

where M1 , M2 P R are a lower bound and upper bound, respectively, of the partial sums of x1 , x2 , . . . . Proof. For this let S1 , S2 , . . . be the sequence of partial sums of x1 , x2 , . . . and M1 , M2 P R be lower and upper bounds, respectively. Then by Theorem 2.3.16 n ¸



xk yk

 pSn
M1 qyn

1

n ¸

M1 y1



k 1

pSk
M1 qpyk
yk 1q

k 1

as well as 0¤

n ¸



pSk
M1 qpyk
yk 1q ¤

k 1

n ¸



pM2
M1qpyk
yk 1q

k 1

234

 pM2
M1 qpy1
yn 1q ¤ pM2
M1 q y1 for all n P N . Therefore the sequence 1 ¸



2 ¸

pSk
M1 qpyk
yk 1q, pSk
M1 qpyk
yk 1q, . . . 

k 1

k 1

is increasing as well as bounded from above and hence convergent. Therefore, since limkÑ8 yk  0, it follows the summability of x1 y1 , x2 y2 , . . . and p2.3.6q. Finally, it follows for every n P N that

8 ¸ 

xk yk


pSn
M1 qyn

8 ¸ 1

k n 1

pSk
M1 qpyk
yk 1q



k n 1

and hence

8 ¸  8 ¸

xk yk

¥
pSn
M1 qyn 1 ¥
pM2
M1 qyn

xk yk

¤
pSn
M1 qyn

k n 1



k n 1

8 ¸ 1



1

pM2
M1 qpyk
yk 1q

k n 1

 pM2
Snqyn 1 ¤ pM2
M1 qyn

1

and (2.3.7). Example 2.3.18. Let s defined by

¡ 0.

Determine whether the sequence a1 , a2 , . . . an

1qn
1 p
: ns

for all n P N is summable. Solution: Define xn : p
1qn
1 , yn :

235

1 ns

y 10 5

1.5

2

2.5

3

s

-5 -10 Fig. 65: Graph of an extended Riemann’s zeta function ζ.

for all n P N . Then the partial sums S1 , S2 , . . . of x1 , x2 , . . . oscillate between 0 and 1 and y1 , y2 , . . . is decreasing as well as convergent to 0. Hence by Theorem 2.3.17 a1 , a2 , . . . is summable and

8 ¸ 

ak

k 1



8 ¸ 

k 0

 

8 ¸



p2k



k 0

p2k

1

1 qs

1

1 qs

p2k


p2k 1

2 qs

1



2 qs


21
s  ζ psq  p1
21
sq  ζ psq

if, in addition, s ¡ 1. Note that the last formula can and is used to define ζ on p0, 1q. See Fig. 65. Theorem 2.3.19. (Abel’s test) Let x1 , x2 , . . . be a summable sequence of real numbers and y1 , y2 , . . . a decreasing convergent sequence of real num-

236

bers. Then the sequence x1 y1 , x2 y2 , . . . is summable and

8 ¸ 

xk yk

 M1 y1

k 1



8 ¸ 

xk
M1

k 1

where M1



8 ¸

y  nlim Ñ8 k



pSk
M1 qpyk
yk 1q

k 1

P R is a lower bound of the partial sums of x1 , x2, . . . .

(2.3.8)

Proof. For this, let S1 , S2 , . . . be the sequence of partial sums of x1 , x2 , . . . and M1 , M2 P R be lower and upper bounds, respectively. Further, let M3 , M4 P R be lower and upper bounds, respectively, of y1 , y2, . . . . Then by Theorem 2.3.16 n ¸



xk yk

 pSn
M1 qyn

1

n ¸

M1 y1



k 1

pSk
M1 qpyk
yk 1q

k 1

as well as 0¤

n ¸



pSk
M1 qpyk
yk 1q ¤

k 1

n ¸



pM2
M1qpyk
yk 1q

k 1

 pM2
M1 qpy1
yn 1q ¤ pM2
M1 q pM4
M3 q for all n P N . Therefore, the sequence 1 ¸



2 ¸

pSk
M1 qpyk
yk 1q, pSk
M1 qpyk
yk 1q, . . .

k 1



k 1

is increasing as well as bounded from above and hence convergent. Finally, it follows the summability of x1 y1 , x2 y2 , . . . and p2.3.8q by the limit laws for sequences. Definition 2.3.20. (Absolute summability) A sequence x1 , x2 , . . . of real numbers is said to be absolutely summable if the corresponding sequence |x1|, |x2|, . . . is summable. It is called conditionally summable if it is summable, but |x1 |, |x2|, . . . is not. 237

Theorem 2.3.21. Any absolutely summable sequence of real numbers is summable. Proof. For this, let x1 , x2 , . . . be some absolutely summable sequence of real numbers. Then x1 |x1 |, x2 |x2 |, . . . is a sequence of positive real numbers and n ¸



pxk |xk |q ¤ 2

k 1

n ¸



|xk | ¤ 2

k 1

8 ¸ 

|xk |

k 1

for all n P R. Hence x1 |x1 |, x2 |x2 |, . . . and therefore by the limit laws for sequences also x1 , x2 , . . . is summable. Example 2.3.22. The examples of the harmonic series Example 2.3.3 and the alternating harmonic series, i.e, the case s  1 in Example 2.3.18, show that not every summable sequence is absolutely summable. Example 2.3.23. Determine whether the sequence sinp1q sinp2q sinp3q , , ,... 12 22 32 is absolutely summable. Solution: For every k   sin k   k2



p q  ¤ 1  k2

(2.3.9)

P N

.

Hence it follows by Example 2.3.7 and Theorem 2.3.9 that the sequence (2.3.9) is absolutely summable. Theorem 2.3.24. Let x1 , x2 , . . . be a summable sequence of real numbers. Then for every ε ¡ 0, there is some N P N such that   n ¸    xk    



¤ε

k m

for all m, n P N satisfying n ¥ m ¥ N. 238

Proof. For this, let ε such that

¡ 0. Since x1 , x2, . . . is summable, there is N P N  m ¸  xk  




8 ¸

   xk  

¤ 2ε

k 1 k 1  for all m P N satisfying m ¥ N. Hence it follows for all m, n P N such

that n ¥ m ¥ N   n ¸    xk    



k m



 n ¸  xk  



k 1

1 that







m ¸1



k 1

   xk  

¤

 n ¸  xk  






k 1

8 ¸ 

k 1

   xk  

 8 ¸  xk  



k 1







m ¸1



k 1

   xk  

¤ ε.

Corollary 2.3.25. Let x1 , x2 , . . . be a summable sequence of real numbers. Then lim xn  0 . n

Ñ8

Theorem 2.3.26. (Ratio test) Let x1 , x2 , . . . be a sequence of real numbers. (i) If there are q

P p0, 1q and N P N such that   xn   x

n



1 



¤q

for all n P N such that n ¥ N, then x1 , x2 , . . . is absolutely summable. (ii) If there is N

P N such that

  xn   x

n



1 



¥1

for all n P N such that n ¥ N. Then x1 , x2 , . . . is not summable. Proof. ‘(i)’: For this, let q P p0, 1q and N P N be such that |xn 1| ¤ q |xn | for all n P N satisfying n ¥ N. Then it follows by induction that

|xn| ¤ |xN |  qn
N 239

for all n P N such that n ¥ N. Hence it follows by Example 2.3.2 and Theorem 2.3.9 the absolute summability of x1 , x2 , . . . . ‘(ii)’: For this let N P N be such that |xn 1 |{|xn | ¥ 1 for all n P N satisfying n ¥ N. Then it follows by induction that

|xn| ¥ |xN | for all n P N satisfying n ¥ N and hence since xN  0 that x1 , x2 , . . . is not converging to 0. Hence it follows by Corollary 2.3.25 that x1 , x2 , . . . is not summable. Example 2.3.27. Find all values x P R for which the sequence x0 x1 x2 , , ,... 0! 1! 2!

is summable. Solution: For x  0, the corresponding sequence is obviously absolutely summable. For x P R and n P N, it follows that  n 1   x  n!  lim  nÑ8 n 1 ! xn 

p

q

|x|  0  nlim Ñ8 n 1

and hence by Theorem 2.3.26 the absolute summability of the corresponding sequence. Theorem 2.3.28. (Root test) Let x1 , x2 , . . . be a sequence of real numbers.

P r0, 1q and N P N such that |xn|1{n ¤ q P N satisfying n ¥ N, then x1 , x2 , . . .

(i) If there are q

for all n summable.

(ii) If there is N

is absolutely

P N such that

|xn|1{n ¥ 1 for all n P N satisfying n ¥ N, then x1 , x2 , . . . is not summable. 240

Proof. ‘(i)’: For this, let q P r0, 1q and N P N be such that |xn |1{n all n P N satisfying n ¥ N. Then it follows that

¤ q for

|xn| ¤ qn

for all n P N satisfying n ¥ N and hence by Example 2.3.2 and Theorem 2.3.9 the absolute summability of x1 , x2 , . . . . ‘(ii)’: For this let N P N be such that |xn |1{n ¥ 1 for all n P N satisfying n ¥ N. Then it follows that |xn| ¥ 1

for all n P N such that n ¥ N and hence that x1 , x2 , . . . is not converging to 0. Hence it follows by Corollary 2.3.25 that x1 , x2 , . . . is not summable. Example 2.3.29. Determine whether the sequence

p
1q2  rlnp12qs2 , p
1q3  rlnp13qs3 , . . .

is summable. Solution: For n P Nzt0, 1u, it follows that   lim  nÑ8

p
1q  n

1{n

 1  n rlnpnqs 

1 0  nlim Ñ8 lnpnq

and hence by Theorem 2.3.28 the absolute summability of the corresponding sequence. Example 2.3.30. Note that in the case of the sequence a1 , a2 , . . . defined by 1 an : s n  for all n P N , where s ¡ 0, that neither the ratio nor the root test can be applied, since   an lim  nÑ8 a

n s n

lim n {

n

Ñ8



1 



s

 1s  1  nlim es lnpnq{n  e0  1 . Ñ8 

 nlim Ñ8

n

n

241

1

1.1

1

0.9

10

20

30

40

50

n3

0.7

0.6

0.5

Fig. 66: Partial sums of the alternating harmonic series and its reordering from Example 2.3.31

Example 2.3.31. (Rearrangement of a conditionally convergent series) Define a1 , a2 , . . . by 1 1  p
1q4k
4  a3k
2 : 4k
3 2p2k
1q
1 1 1 a3k
1 :  p
1q4k
2  4k
1 2p2k q
1 1 1 a3k :
 p
1q2k
1  2k 2k  for every k P N . Its ninth partial sum is given by 1 1 1 1 1 1 1 1 . 1


3 2 5 7 4 9 11 6 Obviously, a1 , a2 , . . . is a rearrangement of the alternating harmonic sequence which is only conditionally summable. The ninth partial sum of the alternating harmonic series is given by 1 1 1 1 1 1 1 1 1
.


2 3 4 5 6 7 8 9 242

Obviously, it follows that

8 ¸ 

p
1qk   1
1 2

k

k 1

1 3

 56 .

Further, because of a3k
2 for every k

a3k
1

a3k

P N, it follows that


3  2kp4k 8k
3qp4k
1q ¡ 0

3n ¸



ak

k 1

¡ 56

for every n P N . So either a1 , a2 , . . . is not summable (!) or

8 ¸ 

ak

¡ ¡ 5 6

k 1

8 ¸ 

k 1

p
1qk . p!q k

Theorem 2.3.32. (Rearrangements of absolutely convergent series) Let x1 , x2 , . . . be an absolutely summable sequence of real numbers. Further, let f : N Ñ N be bijective. Then the sequence xf p1q , xf p2q , . . . is also absolutely summable and

8 ¸ 

xk

k 1



8 ¸ 

xf pkq .

(2.3.10)

k 1

Proof. First, it follows that the sequence of partial sums of |xf p1q |, |xf p2q |, . . . ° is increasing with upper bound k80 |xk | and hence convergent. Hence |xf p1q |, |xf p2q|, . . . is absolutely summable. Further, let ε ¡ 0. By Theorem 2.3.24, there is N P N such that for all n, m P N satisfying n ¥ m ¥ N, it follows that n ¸



|xk | ¤ ε .

k m

243

y 1 0.8 0.6 0.4

-3

-2

-1

1

2

3

x

Fig. 67: Graphs of the first five functions of the series from Example 2.4.1(i).

P N such t1, . . . , N u € tf p1q, . . . , f pNf qu . Hence it follows for every n P N satisfying n ¥ maxtN, Nf u: Since f is bijective, there is Nf

 n ¸  xf pkq  



k 1




n ¸



k 1

   xk  

¤ε.

Hence it follows also (2.3.10).

Problems 1) Express the periodic decimal expansion as a fraction. a)

0.9 ,

b) 0.3 ,

244

c) 0.377 .

y 1

-2

x

1

-1 Fig. 68: Graphs of the first 4 functions of the series from Example 2.4.1(ii).

y

-3

-1

1

3

x

-1 Fig. 69: Graphs of the derivatives of the first 4 functions of the series from Example 2.4.1(ii).

245

y 1.4 1.2 1 0.8 0.6 0.4 0.2 0.2

0.4

0.6

0.8

1

x

Fig. 70: Graphs of the first 10 functions of the series from Example 2.4.1(iii). 2) Calculate a)

8 31 ¸

p

p
1qn q

c)

,

2n

 8 ¸

n 1

 npn

1

3q

n 1

,

8 1 ¸

b)

8 ¸

 4

n 1

 n pn

d)

n 1



1 n


n

1 1qpn



1

3q

,

4 .

3) Determine whether the sequence a1 , a2 , . . . is absolutely summable, conditionally summable or not summable. a) ak a) ak c) ak


1 k

 k

4 5

,

 plnpk 1 1qqk  arctanpkq , {

k4 3

1

e) ak

 p
3q pk!1

g) ak

 p
1qk  ke2

,

b)

k

31{k

,




2k 1

d) ak

k2 q

k

 51{k , p
1qk a 

b) ak


1{k 2

 p2k2
1q! , ? ? k 3
k ak  k

,

f)

,

h) ak

246

 p
1qk  k2 k

3

, ,

2 3  p
1qk  k2k
k 3 2 , j) ak  e3kk{2 p2kq! , l) a  kk , ak  k p3kq! k! 

k rlnpkqs3 , 1 , l) ak  ak  1 k k2

ak

i) k) k)

where k

P N .

4) Determine the values q a3 , a4 , . . .

,

¥ 1 for which the corresponding sequence

ak :

1 , k lnpk q rlnplnpk qqsq

P t3, 4, . . . u, is summable. Give reasons for your answer. Define a4k : a4k 1 : 1, a4k 2 : a4k 3 :
1 for every k P N. k

5)

Determine whether the sequence

?ak , 3 k 7 k P N, is absolutely summable, conditionally summable or not summable. Give reasons for your answer. 6) Estimate the error if the sum of the first N terms is used as an approximation of the series. a) a)

8 1 ¸ n2 n1

8 ¸ 

n 1

, N

p
1qn

3 1

n

, N

,

8 ¸

1 n r ln pnqs2 , N n2

b)

7

,

b)

8 ¸

p
1qn



n2

n 1

9

1

, N

,

 14

.

7) Calculate the sum correct to 3 decimal places a) a)

8 1 ¸ n4 n1

8 ¸ 

,

b)

p
1qn p2n1 q!

8 ¸

1 5 lnpnq n n2 ,

n 1

b)

8 ¸ 

,

p
1qn

n 1

1

n! n2n

.

8) A rubber ball falls from initial height 3m. Whenever it hits the ground, it bounces up 3{4-th of the previous height. What total distance is covered by the ball before it comes to rest?

247

9) If a1 , a2 , . . . is sequence of real numbers such that lim an

Ñ8

n

0,

does this imply the summability of the sequence? Give reasons for your answer. 10) Give an example for a convergent sequence of real numbers a1 , a2 , . . . and a divergent sequence of real numbers b1 , b2 , . . . satisfying lim

an

Ñ8 an

1

n

1,

lim

bn

Ñ8 bn

1

n

1.

11) Give an example for a convergent sequence of real numbers a1 , a2 , . . . and a divergent sequence of real numbers b1 , b2 , . . . lim pan q1{n

Ñ8

n

1,

lim pbn q1{n

Ñ8

n

1.

12) Assume that a1 , a2 , . . . is a summable sequence of real numbers. Show that the sequence a21 , a22 , . . . is also summable.

2.4 Series of Functions Example 2.4.1. (Examples of limits of sequences of functions) (i) Define the sequence of infinitely often differentiable functions f1 , f2 , . . . by x2n fn pxq : 1 x2n for all n P N and x P R. Then




lim fn

n

Ñ8

pxq : nlim f pxq  Ñ8 n

$ ' &0 1 '2

%

1

if |x|   1 if x  t
1, 1u if |x| ¡ 1

and hence limnÑ8 fn is not a continuous function. (ii) Define the sequence of differentiable functions g1 , g2 , . . . by gn pxq : 248

sinpnxq n

for all n P N and x P R. Then




lim gn

n

Ñ8

pxq : nlim g pxq  0 Ñ8 n

for all x P R and hence limnÑ8 gn is a differentiable function. On the other hand because of gn1 pxq  cospnxq for all n P N and x P R, the limit of g11 pxq, g21 pxq, . . . does not exist for any x P R and 1  lim gn1 p0q 




lim gn

1

p0q  0 .

Ñ8 nÑ8 Hence the sequence g11 , g21 , . . . of derivatives of g1 , g2 , . . . does not converge pointwise to the derivative of limnÑ8 gn . n

(iii) Define the sequence of continuous functions h1 , h2 , . . . by hn pxq : nx p1
x2 qn for all n P N and x P r0, 1s. Then




lim hn

n

Ñ8

pxq : nlim h pxq  0 Ñ8 n

for all x P r0, 1s defines a continuous function. On the other hand, »1 0

hn pxq dx 

for every n P N and hence 1 2

 nlim Ñ8

»1 0

hn pxq dx 

1 n  2 n 1

»1




lim hn

0

n

Ñ8

pxq dx  0 ,

i.e., the limit of the integrals of h1 , h2 , . . . over r0, 1s is different from the integral of its limit function over r0, 1s. 249

y 1 0.5 -2

-1

1

x

2

Fig. 71: A uniform neighbourhood of size 1{4 around sinp2xq.

Definition 2.4.2. (Uniform convergence) A sequence f1 , f2 , . . . of functions on some non-empty subset T of R is said to be uniformly convergent to some function f : T Ñ R, if for every ε ¡ 0 there is some N P N such that |fnpxq
f pxq|   ε for all x P T and all n P N such that n ¥ N.

Theorem 2.4.3. (Uniform limits of continuous functions are continuous) Let f1 , f2 , . . . be a sequence of functions on some non-empty subset T of R which is uniformly convergent to some function f : T Ñ R. Further, let all f1 , f2 , . . . be continuous at some point x0 P T . Then f is continuous at x0 , too, i.e., lim lim fn pxq  lim

n

Ñ8 xÑx0

x

Ñx 0






lim fn

n

Ñ8

pxq .

Proof. For this, let a1 , a2 , . . . be some sequence in T which is convergent to x0 and ε ¡ 0. Then for every m, n P N

|f panq
f px0q|

250

 |f panq
fm panq fmpanq
fm px0 q fm px0q
f px0q| ¤ |f panq
fm panq| |fmpanq
fmpx0 q| |fmpx0 q
f px0q| . (2.4.1) Since f1 , f2 , . . . is uniformly convergent to f , there is m0 P N such that |fm pxq
f pxq| ¤ 3ε 0

for all x that

P T . Further, since fm

0

is continuous in x0 , there is N

P N such

|fm panq
fm px0 q| ¤ 3ε for all n P N satisfying n ¥ N. Hence it follows by (2.4.1) that |f panq
f px0 q| ¤ ε for all n P N such that n ¥ N. 0

0

Theorem 2.4.4. (A simple limit theorem for Riemann integrals) Let f1 , f2 , . . . be a sequence of almost everywhere continuous functions on ra, bs, where a, b P R are such that a ¤ b, which is uniformly convergent to some almost everywhere continuous function f : ra, bs Ñ R. Then »b

lim

n

Ñ8

a

fn pxq dx 

»b




lim fn

a

n

Ñ8

pxq dx 

»b a

f pxq dx .

Proof. For this, let ε ¡ 0. Since f1 , f2 , . . . is uniformly convergent to f , there is n P N such that

|fnpxq
f pxq| ¤ ε for all x P ra, bs. Hence » b   fn x dx 

pq
¤ pb
aq ε . a

»b a

  f x dx

pq

¤

»b a

251

|fnpxq
f pxq| dx ¤

»b

ε dx a

Theorem 2.4.5. Let f1 , f2 , . . . be a sequence of continuous functions on ra, bs, where a, b P R are such that a ¤ b, such that f1 px0 q, f2px0q, . . . is convergent for some x0 P ra, bs. Further, let the restriction of each fn , n P N , to pa, bq be differentiable with a derivative that can be extended to a continuous function fn1 on ra, bs. Finally, let the sequence f11 , f21 , . . . be uniformly convergent to some continuous function g : ra, bs Ñ R. Then f1 , f2 , . . . is uniformly convergent to a continuous function f : ra, bs Ñ R whose restriction to pa, bq is differentiable with derivative given by g |pa,bq . Hence in particular,

for all x P pa, bq.

lim fn1 pxq  nÑ8




lim fn

n

1

Ñ8

pxq

(2.4.2)

Proof. By Theorem 1.5.20, it follows that fn pxq 

»x a

fn1 py q dy

fn paq

for all n P N and x P ra, bs. Further, from this follows by the convergence of f1 px0 q, f2 px0 q, . . . , the uniform convergence f11 , f21 , . . . to g and Theorem 2.4.4 the pointwise convergence of f1 pxq, f2 pxq, . . . to some f : ra, bs Ñ R and f pxq 

»x a

g py q dy

lim fn paq

n

Ñ8

for all x P ra, bs. Further, from this follows by Theorem 1.5.18 and the continuity of g that f is continuous with its restriction to pa, bq being differentiable with derivative given by g |pa,bq . Finally, from

|fnpxq
f pxq| ¤

»x a

|fn1 pyq
gpyq| dy |fnpaq
nlim f paq| Ñ8 n

valid for every n P N and x P ra, bs, the uniform convergence of f11 , f21 , . . . to g and the convergence of f pa1 q, f pa2 q, . . . to limnÑ8 fn paq, it follows the uniform convergence of f1 , f2 , . . . to f . 252

Remark 2.4.6. Note that Example 2.4.1(ii) shows that only the assumption of a uniform convergence of the sequence f1 , f2 , . . . alone in Theorem 2.4.5 does not guarantee the validity of (2.4.2) in general. Theorem 2.4.7. (Weierstrass test) Let f1 , f2 , . . . be a sequence of functions on some non-empty subset T of R for which there is a summable sequence M1 , M2 , . . . of positive real numbers such that

|fnpxq| ¤ Mn for all x P T and n P N . Then the series S1 , S2 , . . . defined by Sn :

n ¸



fk

k 1

for all n P N is uniformly convergent to a function S : T Ñ R on T . Also is the series S1 pxq, S2 pxq, . . . absolutely convergent to S pxq for all x P T . Proof. For this let x P T . Then by Theorem 2.3.9, it follows the absolute summability ° of f1 pxq, f2 pxq, . . . . Hence we can define S : T Ñ R by S pxq : 8 k 1 fk pxq for all x P T . Further for given ε ¡ 0, there is N P N such that   n ¸  Mk  






k 1

8 ¸ 

k 1

  Mk  

¤ε

for all n P N satisfying n ¥ N. For such n, it follows that  n ¸  fk x  



k 1

   S x 

p q
p q ¤

8 ¸ 

|fk pxq| ¤

k n 1

8 ¸ 

Mk

¤ε

k n 1

for all x P T and hence the uniform convergence of S1 , S2 , . . . to S. Example 2.4.8. Calculate

8 xn ¸  n

n 1

253

for all x P p
1, 1q. Solution: For this, define xn 1 n 1

fn pxq : xn , gn pxq : for all n P N and x P R. Then

|fnpxq| ¤ pmaxt|a|, |b|uqn , |gnpxq| ¤ pmaxt|a|, |b|uqn 1 for all n P N and x P ra, bs where a, b P R are such that
1   a ¤ b   1.

Hence by Theorem 2.4.7, it follows the uniform convergence of Sf n :

n ¸



fk |ra,bs , Sgn :

k 0

n ¸



gk |ra,bs

k 0

for n Ñ 8 as well as the absolute summability of f0 pxq, f1 pxq, . . . and g0 pxq, g1 pxq, . . . for all x P p
1, 1q. Further, it follows by Theorem 2.4.4

8 bk ¸ 

k 0




ak 1  lim » b S pxq dx  fn nÑ8 k 1 1

a

»b




Ñ8

pxq dx 

8 xn ¸

8 xn ¸

1

 n

1

lim Sf n

a

»b

n

a

dx

1
x

 ln



and hence, finally,

 n

n 1



n 0


lnp1
xq

for all x P R. Example 2.4.9. Calculate

for all x P p
1, 1q.

8 ¸ 

nxn

n 1

254

1
a 1
b



Solution: For this define fn pxq : xn , gn pxq : nxn
1 for all n P N and x P R. Then

|fnpxq| ¤ pmaxt|a|, |b|uqn , |gnpxq| ¤ n  pmaxt|a|, |b|uqn
1 for all n P N and x P ra, bs where a, b P R are such that
1   a ¤ b   1.

Hence by Theorem 2.4.7, it follows the uniform convergence of Sf n :

n ¸



n ¸

fk |ra,bs , Sgn :

k 0



gk |ra,bs

k 0

for n Ñ 8 as well as the absolute summability of f0 pxq, f1 pxq, . . . and g0 pxq, g1 pxq, . . . for all x P p
1, 1q. Hence it follows by Theorem 2.4.5 that

8 ¸ 

k 0

kx


k 1

 nlim S pxq  lim Sf1 n pxq  Ñ8 gn nÑ8




lim Sf n

n

Ñ8

1

pxq  p1
1 xq2

for all x P pa, bq. Hence, finally, it follows that

8 ¸ 

n 1

nxn

 p1
x xq2

for all x P p
1, 1q. Example 2.4.10. Show that ζ psq 

1 Γpsq

»8 0

xs
1 dx ex
1

for all s ¡ 1. Solution: For this, let s ¡ 1. According to (1.4.4) ex
1 ¡ x 255

for all x P p0, 8q and hence xs
1 ex
1

  xs
2

for all x P p0, 1s. Further, an easy calculation shows that ex
1 ¡ ex{2 for all x ¥ 1 and hence that xs
1 ex
1

  xs
1 ex{2

for all x ¥ 1. Hence by Examples 2.2.3, 2.2.6 and Theorem 2.2.5, it follows the improper Riemann integrability of f : pp0, 8q Ñ R, x ÞÑ xs
1 {pex
1qq. Further, define for k P N fk pxq : xs
1 e
pk 1qx for every x P R. Then it follows for ε, R P R such that 0   ε   R and any x P rε, Rs |fk pxq| ¤ Rs
1e
pk 1qε . Hence by Theorem 2.4.7, it follows the uniform convergence of Sf n :

n ¸



fk |rε,Rs

k 0

for n Ñ 8 to f |rε,Rs as well as the absolute summability of f0 pxq, f1 pxq, . . . for all x P p0, 8q. Hence it follows by Theorem 2.4.4 that »R ε

8 »R

¸ xs
1  dx ex
1 k 0

ε

xs
1 e
pk 1qx dx

8 1 » kR ¸  ks  ys
1e
y dy ¤ Γpsq ζ psq kε k 1 256

and hence that

»8 0

xs
1 dx ¤ Γpsq ζ psq . ex
1

On the other hand, n ¸ 1



k 1

 ks

» kR

y
e
y dy ¤ s 1

kε

for every n P N and hence Γpsq

n ¸ 1



k 1

¤ ks

and, finally,

»8

Γpsq ζ psq ¤

0

0

xs
1 dx ex
1

xs
1 dx ex
1

0

»8

»8

xs
1 dx . ex
1

Example 2.4.11. There is a fairly elementary way to show that ζ p2q 

π2 . 6

(2.4.3)

For this, we define for every n P N a corresponding Sn : R Ñ R by Sn pxq :

1 2

n ¸



cospkxq

k 1

for every x P R. In a first step, it follows that Sn pxq 

sinrp2n 1q x{2s 2 sinpx{2q

for all x P Rzt2πk : k P Zu and every n P N . The proof proceeds by induction of n P N . First, it follows by the addition theorem for the sine function that 2 sin


x

2

S1 pxq  2 sin


x 1

2

2

cospxq 257



 sin


x

2

2 sin


x

2

cospxq

 sin


x

2

sin

sin


x

2


x

2

cospxq

cospxq
cos cos


x

2


x

2

sinpxq

sinpxq  sin



3x 2



for every x P R and hence the validity of the statement in the case n  1. Further, if the statement is true for some n P N , then it follows by the addition theorem for the sine function that
x
x
x Sn 1 pxq  2 sin Sn pxq 2 sin cos rpn 1qxs 2 sin 2 2 2  

 sin p2n 2 1q x sin x2 cos rpn 1qxs
cos x2 sin rpn 1qxs  
x
x p 2n 3q x sin cos rpn 1qxs cos sin rpn 1qxs  sin 2 2 2 for every x P R and hence the validity of the statement where n is replaced by n 1. Hence the statement holds for all n P N . In the following let n P N . Then »π 0

 π2 4

»π

xSn pxq dx  n  ¸ x

2

π 4



k 1 n ¸



k 1



k

0

n »π ¸

x dx 2

sinpkxq

1 cospkxq k2

π

π 0

0



k 1 0 n ¸


k1 k 1



π2 4

x cospkxq dx »π 0

n ¸





k 1

Further, by integration by parts, it follows that »π 0

xSn pxq dx 

»π

»π

x 0

sinrp2n 1q x{2s dx 2 sinpx{2q

 sinxp{x2{2q sinrp2n 1q x{2s dx π 0 2 x{2 
2n 1 cosrp2n 1q x{2s sinpx{2q 0 258

sinpkxq dx

p
1qk
k2

1 k2



.

»π

1 2n

1

0

sinpx{2q
px{2q cospx{2q cosrp2n sin2 px{2q

1q x{2s dx

Note that in this derivation it has been used that x

Ñ0 sinpxq

lim

x

1,

sinpx{2q
px{2q cospx{2q sin2 px{2q

lim

x

Ñ0

0

which follows from an application of L’Hospital’s theorem, Theorem 1.4.36. As a consequence, the function pp0, 2π q Ñ R, x ÞÑ px{2q{ sinpx{2q and its derivative have uniquely determined extensions to continuous functions on r0, 2πq. Further, by using the last fact, it follows that » π  x 2 cos x 2 sin x 2  cos 2n 2  sin x 2 0  »π  sin x 2 x 2 cos x 2    dx  sin2 x 2

¤

0

p { q
p { q p { q rp p{q p { q
p { q p { q p{q »π

and hence that lim

n

Ñ8

0

  1 x 2 dx

q {s

xSn pxq dx  0 .

Hence it follows that π2 4

8  ¸ 

p
1qk
k2

k 1

1 k2



0.

The last implies that ζ p2q 

8 ¸

p
1qk



k2

k 1


ζ p22q

π2 4


ζ p2q

π2 4

and hence p2.4.3q.

259

2

8 ¸

1 p2kq2 k 1

π2 4

y 1

-3

-2

-1

1

2

3

x

Fig. 72: Graph of the auxiliary function h in Example 2.4.12.

Example 2.4.12. (A continuous nowhere differentiable function) In the first step, we define an auxiliary function h : R Ñ R by hpx

2k q : |x|

for all
1 ¤ x   1 and k for all x, y P R. mintx, y u. Then

for all v

P Z. This implies that |hpxq
hpyq| ¤ |x
y| For the proof, let x, y P R and n P »v

hpv q 

n

g puq du ,

P R, where g : R Ñ R is defined by g py

2k q :

#

if
1 ¤ y   0 1 if 0 ¤ y   1


1

260

Z such that n

 

y

y

0.7

-1

1

-0.5

0.5

1

x

-1

-0.5

y

-0.5

Fig. 73: Graphs of pR Example 2.4.12.

1

0.5

1

x

y

1.4

-1

0.5 2

0.5

1

x

-1

-0.5

x

Ñ R, x ÞÑ °nk1 p3{4qk hp4k xqq for n  1, 2, 3 and 10. Compare

261

for all
1 ¤ y

  1. Hence if x ¤ y

|hpxq
hpyq|  and if y

¤x

» y     g u du   x

pq

» x     g u du  

|hpxq
hpyq| 

y

pq

¤

¤

»y x

»x y

|gpuq| du  y
x  |x
y|

|gpuq| du  x
y  |x
y| .

In the next step, we define f : R Ñ R by f pxq :

8 ¸

 n



n 0

for all x P R. Since

3 4

hp4n xq

   

 n

 n  3  h 4n x  4

p q¤

3 4

for all x P R, the summability of the sequence p3{4q, p3{4q2 , . . . and the continuity of pR Ñ R, x ÞÑ p3{4qn hp4n xqq for every n P N , it follows by Theorems 2.4.3, 2.4.7 that f is continuous. In the following, we show that f is nowhere differentiable. For this, let x P R and m P N . Define δm :

#

4
m {2
4
m{2

and

if p4m x, 4m x p1{2qq contains no integer if p4m x
p1{2q, 4mxq contains no integer

γmn :

hp4n px

δm qq
hp4n xq δm

for every n P N . When n ¡ m, 4n δm  4n
m {2 is an even integer which implies that γmn  0. When n ¤ m, it follows that   h 4n x  

|γmn| : p p



δm qq
hp4n xq  | 4n px ¤ δm

262

δm q
4n x| δm

 4n .

We conclude that

n    °8  f x δm   n0 34 h 4n f x       δm     8  3 n m  n ¸  ¸  3     γmn   γmn   n0 4  n0 4   m  n m  n ¸ ¸ 3 3  n 4n γmn 4  n0 4 4 n0

p





p px

q
p q 

δm qq
δm

°8



n 0

 3 n 4



hp4n xq   






Hence the sequence

f px

p


    

q¥

δ1 q
f pxq f px , δ1

m ¸



3n

n 0

 12

3m

1


1



.

δ2 q
f pxq , ... δ2

is unbounded and hence not convergent, but lim δm

Ñ8

m

0.

Therefore, f is not differentiable in x. Example 2.4.13. (A space-filling curve) Let h : r0, 2s Ñ R be some continuous function such that hptq  0 for all t P r0, 1{3s Y r5{3, 2s and hptq  1 for all t P r2{3, 4{3s. We consider the 2-periodic continuous extension of this function to the whole of R which will also be denoted by h. Then we define f1 , f2 : R Ñ R by f1 ptq :

8 h 32k
2 t ¸

p



k 1

2k

8 hp32k
1 tq q , f ptq : ¸ 2 2k 

k 1

for all t P R. By Theorem 2.4.7, it follows that both series converge pointwise absolutely as well as uniformly on R and hence by Theorem 2.4.3 that f1 and f2 are continuous. In the following, we show that f pr0, 1sq  r0, 1s2 where f : R Ñ R2 is the continuous curve defined by f ptq : pf1 ptq, f2 ptq for all t P R. For this, let x, y P r0, 1s and x

8 x ¸ k k  2

, y

k 1

263



8 y ¸ k

k  2

k 1

y

0.8

0.6

0.4

0.2

0.2

0.4

0.6

x

0.8

Fig. 74: Curve from Example 2.4.13 resulting from truncating the sums after k

 4.

their binary representation where x1 , x2 , . . . and y1 , y2 , . . . in t0, 1u. Then we define t P r0, 1s by 8 t ¸ k , t2 3k k 1 where t2k
1 : xk and t2k : yk for all k n P N that 

hp3n tq  h 2



k 1

In case that tn

1



8 t ¸ k




3k n

 0, 2

2



8 ¸



k n 1

8 t ¸ n 

k 1

and hence hp3n tq  0  tn 2 3

h



P N . Then it follows for every

¤2

1

k

3k

tk k 3 n

8 1 ¸

¤2



k 2

3k






k 1

k

3k

¤2

264

8 1 ¸ 3k k 1

2

8 t ¸ n 

k 1

 13

and in case that tn

8 t ¸ n

h



1

 1,

1

3k

 k

.

and hence hp3n tq  1  tn 1 . As a consequence, f1 ptq 

8 h 32k
2 t ¸

8 t 8 ¸ q¸ 2k
1  xk  x ,

p

k

2  8 hp32k
1 tq ¸

k 1

f2 ptq 



k 1

2k

k  2 8 t ¸

k  2

k 1





k 1

2k 2k

k 1



8 y ¸ k

k  2

y .

k 1

Note in particular that f p0q  p0, 0q and f p1q  p1, 1q. Lemma 2.4.14. Let a0 , a1 , . . . be a sequence of real numbers such that the set t|a1|, |a2|1{2, |a3|1{3, . . . u is unbounded. Then the sequence a0 , a1 x, a2 x2 , . . . is not summable for every non-zero real x. Proof. For this, let x be some non-zero real number. Then also

t|a1x|, |a2x2 |1{2, |a3x3 |1{3, . . . u is unbounded and hence there exists for every N P N some n P N such that |anxn | ¥ N n . Hence the sequence an xn , n P N, does not converge to zero and therefore is also not summable by Corollary 2.3.25.

Theorem 2.4.15. (Power series) Let x0 P R, a0 , a1 , . . . be a sequence of real numbers which contains infinitely many non-zero members and is such that the set t|a1|, |a2|1{2, |a3|1{3, . . . u is bounded from above. Finally, let N 

rN : sup |aN |

P N and

{ , |a N

1 N

265

1

|

{p

q, . . . (

1 N 1


1

.

(i) Then the sequence a0 , a1  px
x0 q, a2  px
x0 q2 , . . . is absolutely summable for every x P px0
rN , x0 rN q, and the series of polynomials defined by Sn pxq :

n ¸



ak  px
x0 qk

k 0

for every x P R is uniformly convergent on rx0 every r 1 satisfying 0 ¤ r 1   rN .

r 1 s for


r 1, x0

(ii) The number r :



lim sup |aN |

N

{ , |a N

1 N

Ñ8

1

|

{p


1

q, . . . (

1 N 1

,

where we set 1{0 : 8, is called the radius of convergence of the ‘power series’ S0 , S1 , . . . . For any x P R such that |x
x0 | ¡ r, S0 pxq, S1 pxq, . . . is divergent. (iii) The power series S0 , S1 , . . . and S01 , S11 , . . . have the same radius of convergence. Proof. ‘(i)’: First, it follows by the definition of rN that

|an|1{n ¤ r1

N

for all n P N satisfying n ¥ N. Further let x P px0
rN , x0

rN q, then

|anpx
x0 qn|1{n  |an|1{n|x
x0 | ¤ |x
r x0 |   1 N

for n P N satisfying n ¥ N, and hence it follows by Theorem 2.3.28 that the sequence a0 , a1  px
x0 q, a2  px
x0 q2 , . . . is absolutely summable. Further, let r 1 be such that 0 ¤ r 1   rN . Then it follows for every n P N, N 1, . . . and every x P rx0
r 1 , x0 r 1 s that

|ak px
x0 q

k

|  |ak |  r 1k 266

¤

 1 k r

rN

and hence, obviously, by Theorem 2.4.7 the uniform convergence of S0 , S1 , . . . on rx0
r 1 , x0 r 1 s. ‘(ii)’: First, it follows that r is well-defined since 1{r1 , 1{r2 , . . . is decreasing and bounded from below by 0 and hence convergent to some positive real number. In the case that this number is different from zero, it follows that lim sup |aN |1{N , |aN

N

Ñ8

1

Hence, if x P R is such that |x
x0 | n P N such that |an|1{n ¡

|1{pN

q, . . . (  1 .

1

r

¡ r there is an infinite number of 1

|x
x0 | and hence the sequence an px
x0 qn , n P N, does not converge to zero and

therefore is also not summable by Corollary 2.3.25. ‘(iii)’: First, we note that the series S01 , S11 , . . . is convergent for some x P R if only if pidR
x0 q S01 , pidR
x0 q  S11 , . . . is convergent in x and hence that both power series have the same radius of convergence. Further for k P t3, 4, . . . u, it follows by (1.4.11) that

|ak |1{k ¤ elnpkq{k  |ak |1{k  pk  |ak |q1{k ¤ e  |ak |1{k and that expplnp3q{3q, expplnp4q{4q, . . . is decreasing and convergent to 1. Hence, obviously, it follows that the convergence radii of pidR
x0 q  S01 , pidR
x0 q  S11 , . . . and S0 , S1 , . . . are the same. Corollary 2.4.16. Let x0 , a0 , a1 , . . . and r be as in Theorem 2.4.15. Then f : px0
r, x0 r q Ñ R defined by f pxq :

8 ¸ 

ak px
x0 qk

k 0

for all x P px0
r, x0

r q is infinitely often differentiable with derivative

f pnq pxq 

8 ¸

k!

 pk
nq!

k n

267

ak px
x0 qk
n

and in particular

pnq  f n!px0q for every n P N and x P px0
r, x0 r q. Further for every a, b P R such that a ¤ b and ra, bs € px0
r, x0 r q, an

»b a



f pxq dx 

8 ¸ 

»b

ak a

k 0

8 ¸

px
x0 qk dx

ak  pb
x0 qk k 1 k 0

1


pa
x0 qk

1



.

Proof. The statement is a simple consequence of Theorems 2.4.15, 2.4.5 and 2.4.4. Theorem 2.4.17. (Abel’s theorem) Let x0 P R, a0 , a1 , . . . , r be as in Theorem 2.4.15 and f : px0
r, x0 r q Ñ R be defined by f pxq :

8 ¸ 

ak px
x0 qk

k 0

for all x P px0
r, x0

r q. Further, let a0 r 0 , a1 r 1 , . . . be summable. Then

x

lim f pxq 

Ñx0

r

8 ¸ 

ak r k .

k 0

Proof. For this, let S
1 , S0 , S1 , . . . be the sequence of partial sums of a0 r 0 , a1 r 1 , . . . , S
1 : 0 and S :

8 ¸ 

ak r k .

k 0

Then it follows that n ¸



k 0

ak px
x0 qk



n ¸



ak r k


x

k 0

268


x0 k r



n ¸



pSk
Sk
1q

k 0

 Sn


x


x

r


x0 n




r

for every x P px0
r, x0


x0 k

1





x
x0 ¸
x
x0 k Sk r r k 0 n 1

r q, n P N and hence by Theorem 2.3.9




f pxq  1


8 x
x0 ¸
x
x0 k Sk . r r k 0

Further if M ¡ 0 is some upper bound for S1 , S2 , . . . , it follows for given ε ¡ 0, n0 P N such that |Sn
S | ¤ ε{2 for all n P tn0 , n0 1, . . . u and x P px0
r, x0 r q satisfying "

r

r x0
x   min r, 2n0 pM

that

|f pxq
S | 

 
  1 




|S |q  ε

*



8
x
x k  x
x0 ¸  0 p Sk
S q   r r k 0



k 8 | x
x0 ¸ x
x0 | |Sk
S | ¤ 1
r r k 0
¤ n0 pM |S |q 1
x
r x0 2ε ¤ ε


for all n P tn0 , n0

1, . . . u.

Example 2.4.18. By Examples 2.3.18,2.4.8, it follows that the sum of the alternating harmonic series is given by

8 ¸ 

n 0

p
1qn 1  lnp2q . n 269

y 1.1 1 0.9 0.8 0.7 0.6 0.5

10

20

30

40

50

n

Fig. 75: Partial sums of the alternating harmonic series and Graph of the constant function of value lnp2q.

y 1 0.8 0.6 0.4 0.2 2

6

-0.2 -0.4 Fig. 76: Graphs of J0 , J1 and J2 .

270

10

x

P r0, 8q.

Example 2.4.19. Let ν differential equation

Find a solution fν : R

xfν2 pxq

p2ν 1q fν1pxq for all x P R and such that fν p0q  1{Γpν

Ñ

R of the

xfν pxq  0

(2.4.4)

1q. Solution: We assume that fν has a representation as a power series around 0 fν pxq 

8 ¸

ak xk



(2.4.5)

k 0

for all x P p
r, r q, where is some sequence of real numbers a0 , a1 , . . . with corresponding convergence radius r ¡ 0, which are to be determined. Then it follows by Corollary 2.4.16 that 0  xfν2 pxq



8 ¸

 8 ¸

1q fν1 pxq

p2ν

k pk
1q ak x


k 1

k 0





k pk

2ν q ak xk
1

k 1

 p2ν

1q a1

8 ¸ 

xfν pxq

p2ν

8 ¸

1q



8 ¸

k ak x


k 1



k 0

8 ¸

ak xk



ak xk

k 0

1

k 0

rpk

2qpk

2ν

2q ak

2

ak s xk

k 0

which is satisfied for all x P p
r, r q if a0 for every k

1,

a1

0,

ak

2


pk

P N or explicitly if p
1{4qk , a2k  k!  Γpν k 1q 271

ak 2qpk 2ν

a2k

1

0

2q

1

1



for all k

P N. Since

  a2pk 1q x2pk lim  k Ñ8 a x2k

q 

1

2k

 

x2 1qpk

 klim Ñ8 4pk

ν

1q

0

foe every x P R, it follows by Theorem 2.3.26 that the convergence radius of the corresponding power series is infinite and hence that fν : R Ñ R defined by (2.4.5) has the required properties. In terms of fν , the so called Bessel function Jν of the first kind and of order ν is given by Jν pxq :


x ν

2

 fν pxq 



8
x ν ¸

p
1qk k! Γpν k k 0

2

1q

x2 4

k

(2.4.6)

for all x P p0, 8q. By (2.4.6), (2.4.4), it follows that Jν satisfies the differential relation x2 Jν2 pxq

xJν1 pxq

px2
ν 2 qJν pxq  0 ,

for all x P p0, 8q. Theorem 2.4.20. (Cauchy product of series) Let a0 , a1 , . . . , b1 , b2 , . . . be absolutely summable and summable, respectively, sequences of real numbers. We define n ¸

cn :



ak bn
k

k 0

for all n P N. Then c1 , c2 , . . . is summable and

8 ¸ 

k 0

ck





8 ¸ 

 

ak

k 0

8 ¸







bk

.

k 0

Proof. For this, let A0 , A1 , . . . , B0 , B1 , . . . and C0 , C1 , . . . be the sequence of partial sums of a0 , a1 , . . . , b1 , b2 , . . . and c1 , c2 , . . . , respectively. Further, let βn : Bn
B for every n P N where B :

8 ¸ 

k 0

272

bk .

In a first step, it follows by induction that



Cn

n ¸



ak Bn
k

k 0

and hence that Cn



n ¸



ak pB

n ¸

βn
k q  An B



k 0

for every n P N. Since, limnÑ8 βn such that |βn | ¤ ε for all n P tN, N   n ¸    an
k βk    



¤

k 0

¤M

n ¸




an
k βk

k 0

N ¸

 0, for given ε ¡ 0 there is N P N 1, . . . u. Hence for such n n ¸

|an
k | |βk |



k 0

|ak |

ε

k n N

8 ¸ 



|an
k | |βk |

k N 1

|ak |

k 0

where M ¡ 0 is such that |βk | ¤ M for all k by Theorem 2.3.24, there is N 1 P N such that n ¸




P tN, N

|ak | ¤ ε

k n N

for all n P tN

N 1, N

N1

  n ¸    β a  n
k k   



1, . . . u. Hence for such n

¤



M

k 0

8 ¸ 

|ak |

k 0

Example 2.4.21. Define an : bn : 273

n ?p
1q

n

1



ε.

1, . . . u. Further,

and cn :

n ¸



ak bn
k

k 0

for all n Further,

P N. Then a0 , a1, . . . and b1 , b2 , . . . are conditionally summable.

n ¸ p
1 qk 1qn
k p
n ? cn :  ?n
k 1  p
1q  a k 1 pk k 0 k 0 n ¸

 p
1qn 

n ¸



k 0

1

b n 2

and hence

1

2




n 2


k

1 1qpn
k

1q

2

|cn| ¥ 2pnn

1q 2 for all n P N. As a consequence, limnÑ8 |cn | is not summable.

 0 and therefore c1, c2, . . .

Example 2.4.22. (Arithmetic series) Show that for all m, n P N n ¸



r k pk

1qpk

m
1q s 

k 1

1 1

m

n pn

1qpn

mq . (2.4.7)

Solution: For this, let n P N. In a first step, it follows that for all |x|   1 n! p1
xqn

1



8 k ¸

p



k 0

nq! k!

xk ,

including the absolute summability of the series. The proof proceeds by induction over n P N. For the case n  0, this follows from Example 2.3.2. If the statement is true for some n P N, we conclude by Theorem 2.4.15 and Corollary 2.4.16 that

8 pk nq! 8 ¸ pn 1q!  ¸ k
1 k x  pk p1
xqn 2 k1 k! k 0 274

1q

pk

pk

n

1q! k x 1q!

8 k ¸

p



n 1q! k x k!



k 0

for all |x|   1, including the absolute summability of the last series, and hence the validness of the statement where n is replaced by n 1. Further, it follows by Theorem 2.4.20 that 1

n! 1
x p1
xqn



8 ¸ k ¸ pl 

k 0

for |x|   1. Since,



l 0

1

n! 1
x p1
xqn

1



nq!



8 ¸ 



xk

p

k 0



l  pl

l 1

8 pk 1 pn 1q!  ¸  1 n 1 p1
xqn 2 pn



1q    pl

p

nq!

k ¸ l



l 0

l!



xk

xk

l!

for |x|   1, it follows by Corollary 2.4.16 that k¸1

nq! k!



k 0



8 k ¸

n
1q 

 n 1 1 pk

k ¸



pl



k 0

1q  pl

n 1 q! k x 1q k!

2 q    pl

nq

l 0

n 1q! k!

 n 1 1 pk 1q  pk 2q    pk n 1q for all k P N and hence (2.4.7) for all m, n P N .

From (2.4.7), we can

iteratively determine the sum of the arithmetic series Sm pnq :

n ¸

km

k 1  N . We carry the procedure through for m

for all m, n P m  1, according to (2.4.7)

S1 pnq 

n ¸



k 1

k

 12 npn

275

1q .

 1 to 4.

For

for all n P N . For m  2, according to (2.4.7) S2 pnq

S1 pnq 

n ¸



r kpk

1q s 

k 1

1 npn 3

1qpn

2q

and hence S2 pnq 

1 npn 3

1qpn

2q


1 npn 2

1q 

1 npn 6

1qp2n

1q

for all n P N . For m  3, according to (2.4.7) S3 pnq

3S2 pnq

2S1 pnq 

n ¸



r kpk

1qpk

2q s 

2qpn

3q


1 npn 2

5n

6
4n
2
4q

k 1

1 npn 4

1qpn

1qp2n

1q

and hence S3 pnq 

1 npn 4


npn

1q 

 14 n2pn

1qpn 1 npn 4

1qpn2

1 q2

for all n P N . For m  4, according to (2.4.7) S4 pnq

6S3 pnq

11S2 pnq

6S1 pnq 

n ¸



r k  pk

1 q    pk

k 1



1 n  pn 5

1q    pn

4q

and hence S4 pnq 

1 n  pn 5

1q    pn

4q


3 2 n pn 1q2 2
116 npn 1qp2n 1q
3npn 1q  301 npn 1q  r 6pn 2qpn 3qpn 4q
45npn 1q
55p2n 1q
90 s 276

3q s

2q

 301 npn 1q  r 3p2n3 18n2 52n  301 npn 1q  r 3p2n3  301 npn

1q  r 3pn2

 301 npn

1qp2n

48
15n2
15n
30q
55p2n 3n2

37n

18qp2n

n

1qp3n2

18q
55p2n 1q
55p2n

1q s

1q s 1q s

3n
1q

for all n P N . As a consequence, we arrive at the following results. S1 pnq  S3 pnq 

1 1 npn 1q , S2 pnq  npn 1qp2n 1q , 2 6 1 2 1 n pn 1q2 , S4 pnq  npn 1qp2n 1qp3n2 4 30

3n
1q

for all n P N . Theorem 2.4.23. (Taylor expansions) Let I be a non-trivial open interval, a P I and f : I Ñ R be infinitely often differentiable and such there are M ¥ 0 and N P N such that

|f pnqpxq| ¤ M n for all n P tN, N

1, . . . u. Then f pxq 

8 f pkq a ¸ 

k 0

for all x P I.

p q px
aqk k!

(2.4.8)

Proof. By Theorem 1.4.23 for every x P I and n P tN, N 1, . . . u, there is some cx,n in the closed interval between a and x such that    f x 

p q



f pkq paq

n¸1



k 0

k!

   a k 

px
q 

 pnq f cx,n   n!



p q  |x
a|n ¤ pM  |x
a|qn  n!

277

and hence

   lim f x nÑ8 

p q


Corollary 2.4.24. Let I, a 2.4.23. Then



k!

P I, f

: I

k 0

f 1 pxq  for all x P I and F : I


f pkq paq

n¸1



px
q  0 .

Ñ R and M ¥ 0 as in Theorem 1q paq px
aqk

8 f pk ¸ k 0

   a k 

k!

Ñ R, defined by

F pxq :

8 f pk
1q a ¸ 

k 1

p q px
aqk k!

for all x P I, is an anti-derivative of f . Proof. Obviously without restriction, we can assume that M, N ther, let g : f 1 . Then g is infinitely often differentiable and n

¥ 1. Fur-



|gpnqpxq|  |f pn 1qpxq| ¤ M n 1  M 2 ¤ M 2 n for all n P tN, N 1, . . . u and x P I. Hence it follows by Theorem 2.4.23 that 8 f pk 1q paq ¸ g pxq  px
aqk (2.4.9) for all x P I. Further, let c, d F : pc, dq Ñ R by F pxq :



k 0

1

2

k!

P I be such that c   a   d. Then we define

»x c

f py q dy


»a c

f py q dy

for every x P pc, dq. Then F is infinitely often differentiable with its first derivative given by the restriction of f to the interval pc, dq. Further, F paq  0 and |F pnqpxq|  |f pn
1qpxq| ¤ M n
1 ¤ M n 278

y 6 5 4 3 2 1 0.25 0.5 0.75 1 1.25 1.5 1.75

x

Fig. 77: Graphs of the exponential function and corresponding Taylor polynomials of orders 0, 1 and 2 around 0.

for all n P tN, N rem 2.4.23 that

1, . . . u and x

P pc, dq. Hence it follows by Theo8 f pk
1q paq ¸ F pxq  (2.4.10) px
aqk k!



k 1

for all x P pc, dq.

Example 2.4.25. By Theorem 2.4.23 it follows that x

e



8 xk ¸

,

 k! 8 ¸

k 0

cospxq 





p
1qk  p2kx

k 0

sinpxq  for all x P R.

2k

p
1qk  px2kq! ,

8 ¸

k 0

2k 1

279

1q!

y 1 0.5

0.5

1

2

x

2.5

-0.5 -1 Fig. 78: Graphs of the cosine function and corresponding Taylor polynomials of orders 0, 2, 4 around 0.

y 2 1.5 1 0.5 0.5

1

1.5

2

3

x

-0.5 Fig. 79: Graphs of the sine function and corresponding Taylor polynomials of orders 1, 3, 5 around 0.

280

y 1 0.8 0.6 0.4 0.2 0.25 0.5 0.75 1 1.25 1.5 1.75

x

Fig. 80: Graphs of the error function, an associated asymptote and corresponding Taylor polynomials of orders 1, 3, 5 around 0.

Example 2.4.26. Find the power series expansion around zero of the error function defined by » 2 x
y2 e dy erfpxq : ? π 0 for all x ¥ 0. By Example 2.4.25, it follows that e
y

2



8 ¸ 

2k

p
1qk  yk!

k 0

for all y P R including the absolute summability of this sum and also the uniform convergence of the sequence of functions S0 , S1 , . . . defined by Sn py q :

n ¸



p
1q 

k 0

for every n

P

N and y

P

k

y 2k k!

R on every closed subinterval of R. Hence it

281

follows by Theorem 2.4.4 erfpxq 

 for all x ¥ 0.

?2π

»x 0

»x¸ 8

2 2 e
y dy  ?

π

8

k ¸ ?2π p
k!1q k 0



»x

y 2k dy 0

2k

p
1qk  yk!

 8 2 ¸

dy

0 k 0

 ?π

p
1qk x2k 1 p2k 1q k! k 0

Example 2.4.27. Find the first three terms in the Taylor expansion of f pxq : eax cospxq

for all x P R where a P R. Solution: By Examples 2.4.20,2.4.25, it follows that the first three terms in the Taylor expansion of f are given by c0

c2 x2 ,

c1 x

where

 a0b0 , c1  a0b1 a1b0 , c2  a0b2 a1b1 a2 b0 , a0  1, a1  a, a2  a2 {2, b0  1, b1  0, b2 
1{2, and hence by a2
1 2 x 1 ax c0

2

for all x P R.

Example 2.4.28. (Binomial series) Let ν

p1

xq

ν



8 ¸ 

n 0

P R. Show that



ν n x n

for all x P p
1, 1q if ν R N and all x P p
1, 8q if ν P N. The coefficients in the series are called ‘binomial coefficients’. They are defined by 

ν 0

: 1 ,



ν n

:

1 ν  pν
1q    pν
pn
1qq n! 282

for every n P N . Note that 

ν n

Γpν 1q 1q  Γpν
n

 Γpn

1q

for all n P N satisfying n   ν 1. The series is called a binomial series. Note that in case that ν P N, the series terminates since 



ν ν

k

0

for all k P N . In this case, the series coincides with a finite sum. Solution: First, we notice that there is n0 P N such that 

ν n

0

for n P N satisfying n ¥ n0 if and only if ν P N. In this case, the power series coincides with a finite sum and its convergence radius is therefore infinite. In the case that ν R N, it follows that     

ν



xn

n 1  ν xn n

   

1

 pn

n!

  ν ν 1  1 ! ν ν 1

q



 p
q    pν
nq  |x|  p
q    pν
pn
1qq 

 |νn
n1| |x| ¤ n n 1 |x| for all x P p
1, 1q and n P N satisfying n ¥ ν.

Hence it follows by the ratio test, Theorem 2.3.26, that the series is absolutely summable for every x P p
1, 1q and not summable for every |x| ¡ 1. As a consequence, in this case, the convergence radius of the power series is equal to 1. In the following, we define I : p
1, 8q if ν P N, I : p
1, 1q if ν R N and f : I Ñ R by 8 ν

¸ f pxq : xn n n0 for all x P I. Then,

p1

xqf 1 pxq  p1

xq

8 ¸ 

n 1

283



n

ν xn
1 n



8 ¸



8

¸ ν n  n nν xn
1 xn n n1 n1

 8 8 ν

¸ ¸ ν n n  pn 1q n 1 x xn n n0 n0





8 ¸ 

pn

1q

n 0

for every x P I. In particular,

p0

1q





ν 0





ν

ν 0 0

ν n

n

1

n



1

 





ν 1

xn

νν



ν 0

and

pn

1q





ν

n

1



n

ν n

1q

 pn

pn

1

1q!

ν  pν
1q    pν
nq

1 ν  pν
1q    pν
pn
1qq n! 

1  rpν
nq ns n! ν  pν
1q    pν
pn
1qq  ν nν . n

Hence it follows that

xqf 1 pxq  νf pxq

p1

for all x P I. The last is equivalent to 

id I q
ν f

p1

1

(2.4.11)

pxq  0

for all x P I. Then it follows by Theorem 1.4.7 that p1 id I q
ν f is a constant function and, since f p0q  1, that f pxq  p1 xqν for all x P I. Example 2.4.29. Show that Jν pxq 

?π Γ

 x ν 2

ν

1 2



»π 0

284

cospx cos θq sin2ν θ dθ

for all x ¡ 0 and ν ¥ 0. Solution: For this, let ν ¥ 0. Then it follows by use of the power series expansion of cos and Theorem 2.4.4 that »π 0

cospx cos θq sin2ν θ dθ 

8 ¸







k 0

» π ¸ 8

p
1qk x2k cos2k θ sin2ν θ p2kq! k 0

0

p
1qk x2k » π sin2ν θ cos2k θ dθ p2kq! 0



dθ

.

Further, it follows by change of variables and (2.2.17) that »π 2ν

2k

sin θ cos θ dθ 0

» π{2 0





» π{2

sin2ν θ cos2k θ dθ

0 1 2



1 2



{

π 2

0

0




sin2ν θ¯

» π{2

sin2ν θ cos2k θ dθ

2k

sin θ cos θ dθ

» π{2

sin2ν θ cos2k θ dθ

» π{2

»π 2ν

π 2




cos2k θ¯

π ¯ dθ 2

cos2ν θ¯ sin2k θ¯ dθ¯

0

 Γ νΓpk νΓ k 1q For k  0, it follows by Legendre’s duplication formula (2.2.15) for Γ that    ? 1
2k Γ ν 12 Γp2kq Γ ν 12 Γ k 12  π 2 Γpk ν 1q Γpkq Γpk ν 1q   ? 1
2k Γ ν 12 2k Γp2kq ?
2k Γ ν 12 p2kq!  π 2 Γpk ν 1q k! π2 Γpk ν 1q 2k Γpk q and hence that »π 0

sin2ν θ cos2k θ dθ 

?

π 2
2k

Γ ν Γpk ν

1 2



p2kq!

1q k!

P N, where, as usual, 0! : 1. This leads to 8 ?  1 ¸ p
1qk 2ν cospx cos θq sin θ dθ  π Γ ν 2 k! Γpk ν

for all k »π 0



k 0

285


x 2k

1q 2

and hence to

 x ν 2

?π Γ 

ν

 1

»π 0

2

8
x ν ¸ 2

cospx cos θq sin2ν θ dθ

p
1qk k! Γpk ν k 0


x 2k

1q 2

 Jν pxq ,

where the last equality is a consequence of (2.4.6).

Problems 1) Find the interval of convergence of the given series a)

8 ¸

xn

,

b)

8 ¸

xn 1 n2 n0

 1 n 8 ¸ xn , d) c) 3n pn 1q n0 8 p3x
2qn ¸ n 0

e)

5n

 8 ¸

n 0

g)

xn lnpn 2q n0

,

8 x ¸

p?
1qn

 n 1 8 n! ¸  xn , , f) n 10 n0 8 ¸ xn

,

h)

,

n 0

 plnpn

n 0

2qqn

for real x. 2) Find the Taylor series of f around x0 and the corresponding convergence radius and interval of convergence. a) f pxq : 4x{p1 2x
3x2 q , x P Rzt
1{3, 1u ; x0 b) f pxq : sinpxq , x P R ; x0  π {4 , c) f pxq : sin2 pxq , x P R ; x0  π {4 , d) f pxq : lnp1
xq , x   1 ; x0  0 , e) f pxq : sinhpxq , x P R ; x0 f) f pxq : coshpxq , x P R ; x0 g) f pxq : x3

3x

7 , x P R ; x0

h) f pxq : 3 , x P R ; x0 x

0 0

0

286

,

, ,


3

,

0

,

i) j)

f pxq : 1{p1

f pxq : 1{p1

x 3

x

x2 q , x P R ; x0

q, xPR;

x0

0

0

,

,

2 k) f pxq : e
x {2 , x P R ; x0  0 ,

l)

f pxq : lnp1
x2 q , x P R ; x0

0

.

3) Find the MacLaurin series of f pxq :

1 1

x2

,

x P R, and use the result to determine the MacLaurin series of arctan. Finally, show that π 4



8 ¸

p
1qn

 2n 1

.

n 0

4) The MacLaurin series for f : p
1, 1q Ñ R defined by f pxq : lnp1
xq for all x P too slowly.

p
8, 1q is not useful for computation since converging

a) Find the MacLaurin series for g : p
1, 1q Ñ R defined by g pxq : ln



1 x 1
x



for all x P p
1, 1q. Also, find the convergence radius and interval of convergence of the series. b) Show that the error of truncating the series after n P N terms is equal or smaller than 2 2n

2n 1

x 1 1
x2

for x P p
1, 1q. c) Compute lnp2q to four decimal places by using the series obtained in a). Show the accuracy of your result by using the estimate from bq.

287

5) By use of the Cauchy product of series, determine the MacLaurin series of f . a) f pxq : p1
xq
2 , x   1 , b) f pxq : lnp1 xq{p1 xq , x ¡
1 c)

f pxq : rlnp1

xqs , x ¡
1 .

,

2

6) By use of the Cauchy product of series show that exppxq  exppy q  exppx for all x, y

yq

P R.

7) Calculate the first k nonzero terms in the Taylor expansion of f around x0 . a)

f pxq : ecospxq , x P R ; x0

0, k3

,

b) f pxq : cospxq{p1
xq , |x|   1 ; x0  0 , k  6 , ? c) f pxq : expp x q , x P r0, 8q ; x0  1 , k  3 , 8) Use the Taylor series of the sine function around π {4 to approximate its value at 470 degrees correctly to five decimal places. 9) Evaluate

» 1{2 0

1

dx x6

to four-decimal-place accuracy by using a suitable power series expansion. Give reasons for the validity of your calculation. 10) Calculate the leading first four digits of »1 0

cospe
u q du

by using a suitable power series expansion. Give reasons for the validity of your calculation.

11) Define f : R Ñ R by

f pxq :

#

0 if x ¤ 0 . expp
1{xq if x ¡ 0

a) Show that f is infinitely often differentiable.

288

y

0.5

-1

1

3

5

x

Fig. 81: Graphs of f from Problem 11 and the constant function of value 1 which is an asymptote for large positive values of the argument. b) Calculate the MacLaurin series of f and show that it does not converge to f pxq for any x ¡ 0.

12) By a power series expansion around x  0, find a solution of the differential equation satisfying the given boundary conditions. Determine the convergence radius of the series. a) for all x P R; b) for all x   1; c) for all x P R; d)

f 2 pxq

2f 1 pxq

f pxq  0

f p0q  0 , f 1 p0q  1 .

p1
xq2 f 2 pxq
2f pxq  0 f p0q  f 1 p0q  1 . xf 2 pxq

2f 1 pxq

xf pxq  0

f p0q  1 .

xf 2 pxq
f 1 pxq

289

p3
xqf pxq  0

for all x P R; 13) Let a, b ¡ 0 and c ¡ a

f 2 p0q  2 . b.

a) Find the convergence radius r of the Gauss hypergeometric series 8 Γpcq ¸ Γpa nqΓpb nq xn ΓpaqΓpbq n0 Γpc nq n!

for x P R. b) Show that the corresponding hypergeometric function f , which is generally denoted by the symbol ‘F pa, b; c; q’ in the literature, satisfies the hypergeometric differential equation x p1
xqf 2 pxq

rc
pa

b

1qxsf 1 pxq
abf pxq  0 ,

x P p
r, rq. c) Show that arctanpxq , x 1 lnp1 xq , F p1{2, 1; 3{2, x2q  2x 1
x lnp1 xq F p1, 1; 2,
xq  , x

F p1{2, 1; 3{2,
x2q 

for 0   |x|   r1{2 , 0   |x|   r, respectively.

14) Let a, b ¡ 0.

a) Find the radius r of convergence of the confluent hypergeometric series 8 Γpbq ¸ Γpa nq xn Γpaq n0 Γpb nq n!

for x P R. b) Show that the corresponding confluent hypergeometric function f , which is generally denoted by the symbol ‘M pa, b, q’ in the literature, satisfies the confluent hypergeometric differential equation x f 2 pxq

pb
xqf 1 pxq
af pxq  0 ,

x P p
r, rq.

290

y 15

10

5

-2

-1

1

2

x

-5

-10

Fig. 82: Graphs of Hermite polynomials H0 , H1 , H2 , H3 . c) Show that M pa, a, xq  ex , M p1, 2, 2xq  ex for all x P p
r, rq, 0   |x|   r, respectively.

sinhpxq x

15) Let n P N. By a power series expansion around x  0, find a solution Hn : R Ñ R of Hermite’s differential equation f 2 pxq
2xf 1 pxq

2nf pxq  0 ,

x P R, satisfying Hn p0q  p
1qn{2

n! 1 pn{2q! , Hn p0q  0

if n is even and Hn p0q  0 , Hn1 p0q  2 p
1qpn
1q{2

n! rpn
1q{2s!

if n is odd. Determine the convergence radius of the associated power series around x  0.

16) Let ν P R. By a power series expansion around x  1, find a solution of Legendre’s differential equation

p1
x2 qf 2 pxq
2xf 1 pxq 291

ν pν

1qf pxq  0 ,

y 4

2

-2

-1

1

2

x

-2

Fig. 83: Graphs of Legendre polynomials P0 , P1 , P2 , P3 . x P p
1, 3q, satisfying

f p1 q  1 .

Determine the convergence radius of the associated power series around x  1. What happens if ν P N?

2.5 Analytical Geometry and Elementary Vector Calculus 2.5.1 Metric Spaces Definition 2.5.1. A metric space is a pair pM, dq consisting of a non-empty set M, whose elements we shall call points, and a (‘distance’- or ‘metric’-) function d : M  M Ñ R such that for all p, q, r P M (i) dpp, q q ness’)

¥ 0 and dpp, qq  0 if and only if p  q, (‘Positive definite-

(ii) dpp, q q  dpq, pq for all p, q (iii) dpp, q q ¤ dpp, r q

P M,

dpr, q q for all r

292

(Symmetry)

PM

(Triangle inequality).

Lemma 2.5.2. Cauchy-Schwarz inequality Let n pb1 , . . . , bn q P Rn. Then   n ¸    aj bj   j 1 

¤



n ¸



1{2 



a2j

n ¸



j 1

P N and pa1 , . . . , anq, 1{2

b2j

.

(2.5.1)

j 1

In addition, if bj  0 for some j P t1, . . . , nu, then equality holds in (2.5.1) if and only if aj  pC {B q  bj for all j  1, . . . , n where B :

n ¸



, C :

b2j

j 1

n ¸



aj bj .

j 1

Proof. In addition, define A :

n ¸



a2j .

j 1

Then it follows that 0¤

n ¸



pBaj
Cbj q 

j 1 2

2

 AB
2BC

n ¸



B 2 a2j
2BCaj bj

C 2 b2j



j 1 2

2

C B

 B pAB
C 2q

and hence (2.5.1) in case B  0. In the remaining case B  0, it follows that b1      bn  0 and hence also (2.5.1). Further if bj  0 for some j P t1, . . . , nu, then equality holds in (2.5.1) if and only if aj  pC {B q  bj for all j  1, . . . , n. Example 2.5.3. Let n P N . Show that pRn , dq where en : Rn r0, 8q is the usual Euclidean distance function defined by en px, y q :

g f¸ f n e xj



p
yj q2

j 1

293

 Rn Ñ

y

q

p

r

x

Fig. 84: The square of the distance between two points p, q in the plane is given by the sum of squares of the distances between p, r and between r, q.

q

s

p

r

Fig. 85: The square of the distance between two points p, q in space is given by the sum of squares of the distances between p, r, r, s and s, q.

294

r

q

p

Fig. 86: According to the triangle inequality, the distance between the points p, q is smaller than the sum of the distances between p, r and between r, q.

for all x  px1 , . . . , xn q, y  py1 , . . . , yn q P Rn , is a metric space. Solution: The positive definiteness and symmetry of en are obvious. Further, it follows by Lemma 2.5.2 that n ¸

n ¸

penpx, yqq2  pxj
yj q2  pxj
zj 



j 1 n ¸

 pxj
zj q 

j 1

¤ pen px, zqq2 ¤ pen px, zqq2 and hence that

2

zj


yj q2

j 1

2

n ¸



pxj
zj qpzj
yj q

j 1

 n ¸  2 xj j 1

   yj  

n ¸



pzj
yj q2

j 1

p
zj qpzj
q penpz, yqq2

2en px, z qen pz, y q

penpz, yqq2  pen px, yq

en px, y q ¤ en px, z q

for all x  px1 , . . . , xn q, y

en pz, y q

en pz, y qq2

 py1, . . . , ynq P Rn and z  pz1 , . . . , znq P Rn. 295

y 3

2

1

-3

-2

-1

1

x

-1

Fig. 87: Circle of radius 2 and center p
1, 1q.

4

y 2 0 2

0 z -2 -4 -2 0 x

2 4

Fig. 88: Sphere of radius 3 centered at the point p1, 2,
1q.

296

Example 2.5.4. Let n P N and a  pa1 , . . . , an q P Rn . Find an equation whose solution set is given by the sphere Srn paq of radius r ¡ 0 with center a. Solution: A sphere of radius r ¡ 0 and center a contains precisely those points x  px1 , . . . , xn q P Rn which have Euclidean distance r from a. Hence Srn paq is given by S n paq 

#

n ¸

px1, . . . , xnq P Rn : pxj
aj q2  r2

+

.



j 1

In particular, Sr1 paq is called a circle of radius r around a.

Problems 1) Which of the following sets are circles? Find center and radius where this is the case. a) b) c) d) e) f) g)

tpx, yq P R2 : x2 tpx, yq P R2 : x2 tpx, yq P R2 : x2 tpx, yq P R2 : x2 tpx, yq P R2 : x2 tpx, yq P R2 : x2 tpx, yq P R2 : 4x2

y2 y y

2

x
2y

2y
3x

2

3x

2

6x

y

 0u

2y

,

1  0u ,

4  0u ,

1  0u ,


5y 4  0u , y
12x
8y 43  0u 4y 2 2x
8y 1  0u y

2

2

, .

2) Find a function whose zero set is a circle with center p
3, 2q passing through p2, 1q.

3) Find a function whose zero set is a circle with center pa,
aq passing through the origin, where a P R. 4) Find a function whose zero set is a circle passing through all three points p
1, 2q, p1, 0q and p3, 2q.

5) Decide whether the points p0, 3q, p2, 0q and p4, 1q lie on a circle. If yes, find its center and radius.

6) Find functions whose zero sets are circles with center p1, 2q that touch a) the x-axis, b) the y-axis.

297

7) Find the intersection S1 X S2 where

 tpx, yq P R2 :
3
3x 2x2 S2  tpx, y q P R2 : 3
2x
3x2

S1

4y

 0u ,  0u .

2y 2

4y
3y

2

8) Which of the following sets are spheres? Where this is the case, find center and radius. a) b) c) d) e) f) g)

tpx, y, z q P R3 : x2 y2 z 2
2x 3y z  0u , tpx, y, z q P R3 : x2 y2 z 2
4x 5  0u , tpx, y, z q P R3 : x2 y2 z 2
6x
y
z
1  0u , tpx, y, z q P R3 : x2 y2 z 2 x y
z 1  0u , tpx, y, z q P R3 : x2 y2 z 2 3x y 2z  21u , tpx, y, z q P R3 : x2 y2 z 2  3x
y
2z u , tpx, y, z q P R3 : 3px2 y2 z 2 q
2x
z  4u .

9) Find a function whose zero set is sphere with center p1,
1, 2q and radius 3. What is the intersection of the sphere and the xz-plane? 10) Find a function whose zero set is a sphere that passes through the point p2, 3, 1q and is centered in p3, 1, 1q.

11) Find functions whose zero sets are spheres with center p1, 3, 2q that touch a) the xy-plane, b) the yz-plane, c) the xz-plane.

12) Show that the spheres

 tpx, y, z q P R3 : 49px2 S2  tpx, y, z q P R3 : 49px2

S1

y2 y

2

z 2 q
32x z

2

q
20x

8y

26z

26y
10z

8  0u , 2  0u

have only one point in common, and find its coordinates.

2.5.2 Vector Spaces In the following, the notion of a vector will be introduced. Physically, a quantity which has magnitude, direction and a point of attack is described by a vector. Examples are by forces, speed, acceleration, momentum, angular momentum, torque and so forth. For the moment, we discard the last property, namely the point of attack in the definition of a vector. 298

Fig. 89: Equivalent oriented line segments. See Definition 2.5.5

. Definition 2.5.5. Let n P Nzt0, 1u. Then we define

# from p to q : p, q S : tAll oriented line segments pq

P Rn u .

Further, we define on S the relation  by

#  rs # pq for p, q, r, s P Rn if

 Tappq and s  Tapqq for some a  pa1 , . . . , an q P Rn where the translation Ta : Rn Ñ Rn is r

defined by

Ta pxq : px1

a1 , . . . , xn

an q

for every x  px1 , . . . , xn q P Rn . Note that such a is unique and, in particular, that s  Tpq1
p1 ,...,qn
pn q pr q . 299

Theorem 2.5.6. Let n P Nzt0, 1u. Then  is an equivalence relation, i.e., reflexive, symmetric and transitive. Proof.  is reflexive: For this, let p, q P Rn . Then p  Tp0,...,0q ppq and q  #  pq. #  is symmetric: For this, let p, q, r, s P Rn Tp0,...,0q pq q and hence pq # # and pq  rs. Then there is a  pa1 , . . . , an q P Rn such that r Hence

 Tappq

and s  Ta pq q .

p  Tp
a1 ,...,
an q pr q and q

 Tp
a ,...,
a q psq #  pq. #  rs #  is transitive: For this, let p, q, r, s, t, u P Rn and pq # and rs # #  tu. Then there are a  pa1 , . . . , an q, b  pb1 , . . . , bn q P Rn such and rs

that

r

 Tappq

n

1

and s  Ta pq q

as well as such that t  Tb pr q and u  Tb psq . Hence t  Tpa1

b1 ,...,an bn

q ppq and u  Tpa1

b1 ,...,an bn

q pq q

#

#  tu. and pq Definition 2.5.7. Let n P Nzt0, 1u. We define:

# ] corresponding to (i) For arbitrary p, q P Rn , the equivalence class [ pq # pq by # ] : trs # : rs #  pq # , r, s P Rnu . [ pq Every such equivalence class is called a vector. (ii) The set of all vectors by

# ] : p, q S/ : t[ pq 300

P Rn u .

s v

r u q t p

Fig. 90: Vector addition. See Definition 2.5.7 (iii)

.

uHΛL

u qHΛL t q

p

Fig. 91: Scalar multiplication. See Definition 2.5.7 (iv)

. 301

# ] [ rs # ] of [ pq # ] and [ rs # ] as (iii) For arbitrary p, q, r, s P Rn , the sum [ pq # # ]. Then there is a unique v P Rn such follows. For this, let tu P [ pq #  P [ rs # ], and we define that uv #

# ] [ pq

# ] : [ tv ] . [ rs

# ] can be represented as Note that every element of [ pq #



Ta ptqTa puq for some a P Rn . Then

#

(2.5.2)



# ] Ta puqTa pv q P [ rs and

#



#

Ta ptqTa pv q P [ tv ] .

# ] This shows that [ pq

# ] is well-defined. [ rs

# ] as follows. (iv) For every p, q P Rn and λ P R, the scalar multiple λ.[ pq # # For this, let tu P [ pq ]. Then #

# ] : [ t pt λ.[ pq 1

λpu1
t1 q, . . . , tn



λpun
tn qq ] .

# ] can be represented in the form of (2.5.2) Since every element of [ pq for some a P Rn , it follows that #

ta pta1

λpua1
ta1 q, . . . , tan



λpuan
tan qq

 T# aptq Tappt1 λpu1
t1 q, . . . , tn λpun
tnqqq P [ t# pt1 λpu1
t1 q, . . . , tn λpun
tnqq ] # ] is wellwhere ta : Ta ptq, ua : Ta puq. This shows that λ.[ pq

defined.

# ]| of [ pq # ] as follows. For this, (v) For arbitrary p, q P Rn , the length |[ pq # # let tu P [ pq ]. Then a

# ]| : pu
t q2    pu
t q2 , |[ pq 1 1 n n 302

i.e., as the Euclidean distance of the points t and u. Since every # ] can be represented as (2.5.2) for some a P Rn , it element of [ pq follows that a

pa ua1
ta1 q2    puan
tan q2  pu1
t1 q2    pun
tn q2 # ]| is well: Ta ptq, ua : Ta puq. This shows that |[ pq

where ta defined. Vectors of length one are called unit vectors.

# ]  (vi) For arbitrary p, q, r, s P Rn the scalar product (or dot product) [ pq # ] of [ pq # ] and [ rs # ] by [ rs # ]  [ rs # ] : 1 [ pq



2

# ] |[ pq

# ]|2
|[ pq # ]|2
|[ rs # ]|2 [ rs



.

Note that according to the Law of Cosines 

# ], [ rs # ]q , ?p[ pq # ]|  0, |[ rs # ]|  0 and where the angle ?p[ pq # ], [ rs # ]q P r0, π s if |[ pq # ]  [ rs # ]  |[ pq # ]|  |[ rs # ]|  cos [ pq

# ] and [ rs # ] is defined by the angle between represenbetween [ pq tatives of both equivalence classes originating from the same point. Vectors with a vanishing scalar product are called orthogonal to each other. (vii) The bijective map ι : Rn

Ñ S/ by # ιpxq : [ Ox ] .

for all x P Rn where O denotes the origin defined by O : p0, . . . , 0q P Rn .

P Rn and arbitrary λ P R, x y : ι
1 pιpxq ιpy qq , λ.x : ι
1 pλ.ιpxqq , |x| : |ιpxq| , x  y : ιpxq  ιpy q .

(viii) For arbitrary x, y

303

Theorem 2.5.8. Let n P Nzt0, 1u. Then (i) x

y

 px1 , . . . , xnq py1, . . . , ynq  px1

and

y1 , . . . , xn

yn q

a.x  a . px1 , . . . , xn q  pa  x1 , . . . , a  xn q

for all x, y

P Rn and a P R.

(ii) pRn , , .q is a real vector space with 0 : p0, . . . , 0q as neutral element and for each x P Rn with
x : p
x1 , . . . ,
xn q as corresponding inverse element, i.e, the following holds:

y x px yq z  x py x 0x x p
xq  0 x

y

(Addition is commutative),

zq

(Addition is associative), (0 is a neutral element),

(
x is inverse to x)

as well as 1.x  x, pabq.x  a.pb.xq, pa

a.px

y q  a.x

for all x, y

bq.x  a.x

b.x,

a.y

P Rn and a, b P R.

(iii) The sequence of n vectors e1 , . . . , en , defined by e1 : p1, . . . , 0q , . . . , en : p0, . . . , 1q , is a basis of Rn , i.e., for every x P Rn , we have x  x1 .e1



xn .en

and the coefficients x1 , . . . , xn in this representation are uniquely determined. I.e., if x  x¯1 .e1    x¯n .en 304

for some x¯1 , . . . , x ¯n

P R, then x¯1  x1 ,

. . . , x¯n

 xn .

The sequence e1 , . . . , en is called the canonical basis of Rn . Proof. ‘(i)’: For this, let x a P R. Then x

y : ι
1 pιpxq


 px1 , . . . , xnq, y  py1, . . . , ynq P
#

ιpy qq  ι
1 [ Ox ]

#  # ] [ Tx pO qTx py q ]  ι
1  ι
1 [ Ox  ι
1 ιppx1 y1, . . . , xn ynqq  px1 and




#

Rn and

#

[ Oy ]


#



[ OTx py q ]

y1 , . . . , xn




yn q

 ι
1 [ O# pλx1, . . . , λxnq ]  ι
1 ιppλx1 , . . . , λxnqq  pλx1 , . . . , λxnq .

λ.x : ι
1 pλ.ιpxqq  ι
1 λ.[ Ox ]

(ii) and (iii) are trivial consequences of the definitions and the algebraic properties of the real numbers. Theorem 2.5.9. Let n P Nzt0, 1u. Then (i) for every x P Rn

|x| 

b

x21



x2n .

(ii) The absolute value satisfies the defining properties of a norm on Rn , i.e.,

|x| ¥ 0 and |x|  0 if and only if x  0 (Positive definiteness), |a.x|  |a|  |x| (Homogeneity), |x y| ¤ |x| |y| (Triangle inequality) for all x, y P Rn and a P R. 305

Proof. ‘(i)’: For this, let x  px1 , . . . , xn q P Rn . Then

#

|x| : |ιpxq|  |[ Ox ]| 

b

x21



x2n .

‘(ii)”: The positive definiteness and homogeneity of the absolute value are straightforward consequences from the definitions. The triangle inequality follows from the corresponding property of the metric en . For this, let x  px1 , . . . , xn q, y  py1 , . . . , yn q P Rn . Then

|x

y |  en pO, x

y q ¤ en pO, xq

en px, x

y q  |x|

|y| .

Theorem 2.5.10. Let n P Nzt0, 1u. Then (i) for all x, y

P Rn :

xy

n ¸





xk yk .

k 1

(ii) This product satisfies the defining properties of a scalar product on a real vector space, i.e, xy

yx (Symmetry), px yq  z  x  z y  z (Additivity in the first variable), pa.xq  y  apx  yq (Homogeneity in the first variable), x  x ¥ 0 and x  x  0 if and only if x  0 (Positive definiteness) for all x, y P Rn and a P R. As a consequence, it satisfies the impor-

tant Cauchy-Schwarz inequality

|x  y| ¤ |x|  |y| .

for all x, y P Rn . In particular, in the case that y in (2.5.3) if and only if x

xy |y|2 . y .

306

(2.5.3)

 0, equality holds (2.5.4)

 px1 , . . . , xnq and y  py1, . . . , ynq P

Proof. ‘(i)’: For this, let x Then x  y : ιpxq  ιpy q 

 12 |x

1 2



#

#



n ¸

#

#

|rOxs rOys|
|rOxs|
|rOys|

y |2
|x|2
|y |2

 12



2

rpxk

2

2

yk q2
x2k
yk2s 

k 1

Rn .

n ¸



xk yk .

k 1

‘(ii)’: The symmetry, additivity, homogeneity in the first variable, and positive definiteness of the dot product are obvious. Let a : |y |2 and b :
x  y. For the case y  0, inequality (2.5.3) is trivially satisfied. If y  0, then |y | ¡ 0 and 0 ¤ pa.x b.y q  pa.x b.y q  a2 |x|2 2ab px  y q b2 |y |2  |y|4|x|2
2|y|2px  yq2 px  yq2|y|2  |y|2  r|x|2|y|2
px  yq2s , and hence it follows (2.5.3). In particular, equality holds in (2.5.3) if and only if (2.5.4) is true. Note that the basis e1 , . . . , en of Rn is in particular orthonormal with respect to the Euclidean scalar product, i.e., it satisfies

and for all i, j

ei  ej

0,

if i  j pOrthogonalityq

ei  ej

1,

if i  j pNormalizationq

P t1, . . . , nu.

Theorem 2.5.11. Let n P Nzt0, 1u. Then (i)

|x y|2  |x|2 |y|2 for all orthogonal x, y P Rn .

307

(2.5.5)

y

y’

z x

z’

Fig. 92: Orthogonal projections y 1 , z 1 of y and z, respectively, onto the direction of x.

(ii) For every x P Rn zt0u and every y Px py q P Rn such that

P

Rn , there is a unique vector

|y
Pxpyq|  mint|y
λ.x| : λ P Ru .

Px py q is called the orthogonal projection of y onto the direction of x. In particular, it is given by yx Px py q : |x|2 .x , y
Px py q is orthogonal to x and

a

|y
Pxpyq|  |x1|  |x|2  |y|2
px  yq2 . Proof. ‘(i)’: Let x  px1 , . . . , xn q, y Then x  y  0 and hence

|x

y |2 

n ¸



pxk

yk q2



k 1

n ¸



k 1

 py1, . . . , ynq P Rn be orthogonal. n ¸

x2k



k 1

‘(ii)’: First, it follows for every λ P R that 

y




yx |x|2 . x

yk2

2

n ¸



xk yk

 |x|2 |y|2 .

k 1

 x  y  x
py  x|xqp|x2  xq  0 308

and hence by piq that   y 

  y x   x2

|y
λ.x| ¥


for all λ P R and

  y 

|y
λ.x| 


yx |x|2


λ






λ . x2  ||

  y 

Hence



yx |x|2 . x

|y
λ.x| 
 2  y  x    y
|x|2 . x 2

2  . x

.



y  x  |x|2 . x



y  x  1 a 2 2 2 . x  |x|2  |x|  |x|  |y|
px  yq

if and only if

λ

yx |x|2 .

Example 2.5.12. Let n P Nzt0, 1u and Opqr be a parallelogram where p, q, r are points in Rn . Show that its area A is given by A

a

|a|2  |b|2
pa  bq2  |a|  |b|  sinpαq # # where a  [ 0p ] and b  [ 0r] and α : ?pa, bq P r0, π s. Solution: See

Fig. 93. If a are multiples of one another, A vanishes. This is consistent with the above formula since in that case |a  b|  |a|  |b| according to Theorem 2.5.10. In the remaining cases let P pbq denote the orthogonal projection of b onto the direction of a. The areas of the triangles Orr 1 and pqq 1 are identical. Hence the area of Opqr is given by A  |a|  |b
P pbq|  |a|  a

 |a|2  |b|2
pa  bq2 .

  b 






a  b  1 a 2 2 2 .a a |   | |a|2  |a|  |a|  |b|
pa  bq

309

r’

q’

r

q

b-PHbL

b

Α O

a

p

PHbL

Fig. 93: Calculation of the area of the parallelogram Opgr. See Example 2.5.12.

Note that |a|2 |b|2
pa  bq2 ¡ 0 according to the Cauchy-Schwarz inequality Theorem 2.5.10. Further, since a  b  |a|  |b|  cospαq , it follows that A

a

|a|2  |b|2  p1
cos2 pαqq  |a|  |b|  sinpαq .

In the following, we motivate the definition of a vector product a  b for a  pa1 , a2 , a3 q, b  pb1 , b2 , b3 q P R3 which are assumed not to be multiples of each other. Natural candidates for the definition of a  b are vectors that are at the same time orthogonal to a and b. Orthogonal vectors to a are given by α . p
a2 , a1 , 0q β . p0,
a3 , a2 q

where α, β P R. In this, we assume in addition that a2  0 which excludes that p
a2 , a1 , 0q and p0,
a3 , a2 q are multiples of each other. The condition b  r α . p
a2 , a1 , 0q

β . p0,
a3, a2 q s  0 310

leads to

pa1b2
a2 b1 q α pa2 b3
a3b2 q β  0

which is satisfied if α


γ a2

 pa2b3
a3 b2 q , β  aγ  pa1b2
a2 b1 q 2

P R. Then α . p
a2 , a1 , 0q β . p0,
a3 , a2 q  γ . pa2 b3
a3 b2 , a3 b1
a1 b3 , a1 b2
a2 b1 q . for some γ

To restrict the final parameter γ, we calculate the square of the norm of this vector for γ  1. In this, we drop all the additional restricting assumptions on a, b P R3 made above. This gives

pa2b3
a3 b2 q2 pa3 b1
a1 b3q2 pa1 b2
a2b1 q2  a22b23 a23 b22
2a2b2 a3 b3 a23b21 a21 b23
2a1 b1a3 b3 a21 b22 a22b21
2a1b1 a2b2  |a|2|b|2
pa1b1 a2 b2 a3 b3 q2  |a|2|b|2
pa  bq2 which according to Example 2.5.12 is the square of the area of the parallelogram determined by a, b if a, b P R3 zt0u. This suggests the following definition. Definition 2.5.13. For all a, b a  b P R3 by

P R3, we define the corresponding product

a  b : pa2 b3
a3 b2 , a3 b1
a1 b3 , a1 b2
a2 b1 q . A simple calculation shows that a  pa  bq  b  pa  bq  0 and by the foregoing a

|a  b|  |a|2|b|2
pa  bq2  |a|  |b|  sinpαq , where α : ?pa, bq P r0, π s, which according to Example 2.5.12 is the area of the parallelogram determined by a, b. 311

a x b b a

Fig. 94: Vector product of two vectors a and b.

Remark 2.5.14. Let a, b P R3 zt0u. Then it follows by Example 2.5.12 that a and b are parallel if and only if ab0 . Theorem 2.5.15. Let a, b, c, d P R3 and λ P R. Then (i) e1  e2 (ii) (iii) (iv) (v) (vi) (vii)

 e3 , e3  e1  e2 , e2  e3  e1 , a b 
b a, pλ . aq  b  λ . pa  bq , pa bq  c  a  c b  c , a  pb  cq  c  pa  bq , a  pb  cq  pa  cq . b
pa  bq . c , pa  bq  pc  dq  pa  cq  pb  dq
pa  dq  pb  cq . 312

Proof. The relations (i) to (iv) are obvious. ‘(v)’: a  pb  cq  a  pb2 c3
b3 c2 , b3 c1
b1 c3 , b1 c2
b2 c1 q  a1b2 c3
a1b3 c2 a2 b3c1
a2 b1 c3 a3 b1 c2
a3 b2 c1  c1  pa2 b3
a3 b2 q c2  pa3 b1
a1 b3 q c3  pa1 b2
a2 b1 q  c  pa  bq . ‘(vi)’: a  pb  cq  pa2 pb  cq3
a3 pb  cq2 , a3 pb  cq1
a1 pb  cq3 , a1 pb  cq2
a2 pb  cq1 q  pa2 pb1 c2
b2c1 q
a3 pb3 c1
b1 c3q, a3 pb2 c3
b3 c2 q
a1 pb1 c2
b2 c1 q, a1 pb3 c1
b1 c3 q
a2 pb2 c3
b3 c2 qq  pa2 c2b1 a3 c3b1
a2 b2 c1
a3 b3 c1, a3 c3b2 a1 c1b2
a3 b3 c2
a1 b1 c2, a1 c1 b3 a2 c2 b3
a1 b1 c3
a2 b2 c3 q  pa1 c1b1 a2 c2b1 a3 c3b1
a1 b1 c1
a2 b2 c1
a3 b3 c1, a3 c3 b2 a1 c1 b2 a2 c2 b2
a2 b2 c2
a3 b3 c2
a1 b1 c2 , a1 c1 b3 a2 c2 b3 a3 c3 b3
a3 b3 c3
a1 b1 c3
a2 b2 c3 q  pa  cq . b
pa  bq . c . ‘(vii)’:

pa  bq  pc  dq  pa2b3
a3 b2 , a3b1
a1 b3 , a1b2
a2 b1 q  pc2d3
c3 d2, c3d1
c1 d3, c1d2
c2 d1q  a2b3 c2 d3 a3 b2 c3d2
a2 b3 c3d2
a3 b2 c2d3 a3 b1 c3 d1 a1 b3 c1 d3
a3 b1 c3 d1
a1 b3 c3 d1 a1 b2 c1 d2 a2 b1 c2 d1
a1 b2 c2 d1
a2 b1 c1 d2  a2c2 b3 d3 a3 c3b2 d2 a3 c3b1 d1 a1 c1 b3d3 a1c1 b2 d2 a2 c2b1 d1
pa2 d2b3 c3 a3 d3b2 c2 a3d1b1 c3 a1 d1b3 c3 a1 d1b2 c2 a2 d2b1 c1 q  pa  cq  pb  dq
a1 c1b1 d1
a2 c2b2 d2
a3c3 b3 d3 313


pa2 d2b3 c3 a3 d3b2 c2 a3d1b1 c3 a1 d1b3 c3 a1 d1b2 c2  pa  cq  pb  dq
a1 d1b1 c1
a2 d2b2 c2
a3d3b3 c3
pa2 d2b3 c3 a3 d3b2 c2 a3d1b1 c3 a1 d1b3 c3 a1 d1b2 c2  pa  cq  pb  dq
pa  dq  pb  cq .

a2 d2 b1 c1 q a2 d2 b1 c1 q

a x b

PHcL c b

a

Fig. 95: The volume of a parallelepiped with sides a, b and c is given by |a  pb  cq|.

Example 2.5.16. Show that the volume V of the parallelepiped with sides a, b, c P R3 is given by the absolute value of the scalar triple product a  pb  cq V

 | a  pb  cq| p | c  pa  bq|  | b  pc  aq| q .

Solution: The volume is equal to the length of the orthogonal projection P pcq of c onto the direction of a  b times the area of the parallelogram 314

with sides a, b. Compare Fig 95. Hence V

 | c |apab|bq|  |a  b|  | c  pa  bq|

where it is assumed that a and b are not scalar multiples of each other. In that case the V vanishes which is consistent with the above formula since in that case also a  b  0. Definition 2.5.17. (The determinant function) Let n P N . (i) For every pk1 , . . . , kn q P Zn , we define n ¹

spk1 , . . . , kn q :

  

sgnpkj
ki q

i,j 1,i j

where the signum function sgn : R Ñ R is defined by sgnpxq :

$ ' & 1

0
1

' %

if x ¡ 0 if x  0 if x   0 .

In this, we use the convention that the empty product is equal to 1. Note that this definition implies that spk1 , . . . , kn q  0 if the ordered sequence pk1 , . . . , kn q contains two equal integers. Also, note for the case of pairwise different integers k1 , . . . , kn that spk1 , . . . , kn q  1 if the number of pairs pi, j q P t1, . . . , nu2 such that i   j and kj   ki is even, whereas spk1 , . . . , kn q 
1 if that number is odd. (ii) For every ordered n-tuple pa1 , . . . , an q of elements of Rn , we define a corresponding determinant a11    a1n      detpa1 , . . . , an q :           an1    ann 315

:

n ¸



spk1 , . . . , kn q a1k1    ankn .

k1 ,...,kn 1

Note that according to the remark in (ii), the sum in this definition has only to be taken over n-tuples k1 , . . . , kn which are permutations of 1, . . . , n. Further, note that this definition leads in the case n  1 to detpa1 q  a1 and in the case n  2 to a1 a2  a1 b2
a2 b1 detpa, bq  b1 b2 for every pa, bq such that a, b P R2 . Note that in the last case

r detpa, bq s2  |a|2|b|2
pa  bq2 . Hence the area A of the parallelogram with sides a and b is given by A  | detpa, bq | . Finally, in the case n  3, this definition leads to detpa, b, cq a1 a2 a3 b2 b3 b1 b3  b1 b2 b3  a1  c2 c3
a2  c1 c3 c1 c2 c3

b1 b2 a3  c1 c2

(2.5.6)

 a1b2 c3
a1b3 c2 a2 b3 c1
a2 b1 c3 a3 b1 c2
a3 b2 c1  a  pb  cq  a1b2 c3 a2b3 c1 a3 b1 c2
pa3 b2 c1 a1 b3 c2 a2 b1 c3q for every pa, b, cq such that a, b, c P R3 . Note that detpa, b, cq  a  pb  cq . Hence the volume V of the parallelepiped with sides a, b and c is given by V

 | detpa, b, cq | . 316

v x0 O

Fig. 96: Line corresponding to x0 and v, see Definition 2.5.19.

Remark 2.5.18. Formally, for ease of remembrance, we can write e1 e2 e3 a  b  a1 a2 a3 b1 b2 b3 thereby utilizing the representation of the determinant in terms of minors given in (2.5.6). Definition 2.5.19. (Lines and Planes) Let n P Nzt0, 1u and x0 x03 q, v P Rn .

 px01 , x02 ,

(i) We define a corresponding line by

tx0

t.v

P Rn : t P Ru .

(ii) In addition, let n  3 and w P Rn be such that v, w are not multiples of one another. Then we define a plane corresponding to x0 , v, w by

tx0

t.v

s.w 317

P Rn : t, s P Ru .

v w

x0

O

Fig. 97: Plane corresponding to x0 ,v and w, see Definition 2.5.19.

Note that this set is equal to

tx P Rn : n  px
x0 q  0u  tx P Rn : n1x1 n2 x2 n3 x3
pn1x01 n2 x02 n3 x03 q  0u where n  pn1 , n2 , n3 q is some normal vector to v and w, i.e., some non-trivial multiple of v  w. Example 2.5.20. Calculate the distance d between the two lines L1 : tp4, 3, 1q and

L2 : tp1, 0,
3q

t.p1, 1, 2q P R3 : t P Ru t.p1, 1, 2q P R3 : t P Ru .

Solution: L1 and L2 are parallel. Therefore d is given by the distance of some point on L1 , like p4, 3, 1q, from L2 , which by Theorem 2.5.11 is given by the length of (note that p4, 3, 1q
p1, 0,
3q  p3, 3, 4q)

p3, 3, 4q
16  rp3, 3, 4q  p1, 1, 2qs.p1, 1, 2q  23 .p1, 1,
1q . ? Hence d  2 3{3. 318

Example 2.5.21. Let x0 , v, w P R3 such that v, w are not multiples of one another. Finally, let E be the corresponding plane and u P R3 . Show that the distance dpu, E q of u from E is given by dpu, E q 

|pu
x0q  pv  wq| . |v  w|

Solution: For this, let n :

|v  w| .pv  wq , u0 : u
x0 . Then it follows for every t, s P R that |u0
t.v
s.w|  |pu0  nq.n u0
pu0  nq.n
t.v
s.w| ¥ |pu0  nq.n|  |u0  n| . Also because u0
pu0  nq.n is normal to n, it follows that there are t, s P R such that t.v s.w  u0
pu0  nq.n. Hence dpu, E q : mint|u
x| : x P E u  |u0  n| . 1

Example 2.5.22. Find the distance d between the planes x y
3z 
1 and 2x 2y
6z  0.5. Solution: First, since the normals n1  p1, 1,
3q and n2  p2, 2,
6q are multiples of each other these planes are parallel. Hence d is given by the distance of some point on L1 , like p0, 0, 1{3q, from the second plane. Therefore d





?1 p1, 1,
3q  rp0, 0, 1{3q
p0, 0,
1{12qs  445 11

?

11 .

Example 2.5.23. Let x01 , x02 , v, w P R3 and in particular v, w be no multiples of one another. Calculate the distance d of the (‘skew’) lines L1 : tx01

t.v

P R3 : t P Ru .

319

and

L2 : tx02

s.w

Solution: For all t, s P R,

|x01

t.v
px02

P R3 : s P Ru .

s.w q|  |x01
x02

t.v

p
sq.w| .

Hence d is equal to the distance of the plane

tx01
x02

t.v

s.w

P R3 : t, s P Ru

from the origin, i.e, by d

|px01
x02 q  pv  wq| . |v  w| Problems

# ]q, of the vector that cor1) Determine the representative, i.e., ι
1 p[ pq #  responds to the oriented line segment pq between the points p and q given by a list of their coordinates with respect to a Cartesian coordinate system. In addition, calculate the distance between p and q. a) b) c) d) e) f) g) h)

p  p1, 2q , q

 p4, 7q , p  p1, 2q , q  p3,
4q , p  p1, 3q , q  p
1,
2q , p  p2, 4q , q  p
3, 4q , p  p1, 3, 2q , q  p3, 4, 7q , p  p1, 2, 2q , q  p3, 1,
4q , p  p3, 1, 2q , q  p
1, 5,
2q , p  p7, 1, 4q , q  p
3, 9, 4q . 3b, 3a
4b, a  b, the angle θ between a and b,

2) Calculate |a|, 2a a vector of length one in the direction of a, the orthogonal projection of a onto the direction of b and, if at all possible, a  b of the vectors a and b. a)

a  p3, 5q , b  p1, 3q ,

320

b) a  p7, 1q , q  p1,
3q , c) p  p6, 9q , q  p
2,
4q ,

d) p  p1, 5q , q  p
2, 6q , e) p  p2, 3, 1q , q  p3, 7, 7q ,

f) p  p1, 3, 1q , q  p4, 2,
3q , g) p  p5, 2, 2q , q  p
2, 6,
3q , h) p  p9, 2, 3q , q

 p
5, 8, 2q

.

3) Calculate detpa, bq, detpa, b, cq, respectively. a) a  p4, 0q , b  p7, 0q ,

b) a  p8, 3q , b  p
2, 6q , c) a  p
8, 9q , b  p5,
2q ,

d) a  p4, 0, 1q , b  p2, 0,
1q , c  p3, 0, 9q , e) a  p3, 1, 4q , b  p1, 9,
1q , c  p2,
3, 1q , f) a  p4,
2, 3q , b  p3, 5, 2q , c  p4, 4, 1q .

4) Find suitable vectors x0 and v such that L  tx0

tv : t P Ru ,

where a)

L  tpx, y q P R2 : 3x

b) L  tpx, y q P R : 4y 2

c)

L  tpx, y q P R : 7x

4y

 3u

2

3y

d) L  tpx, y, z q P R : x 3

e) f)

L  tpx, y, z q P R : 3z 3

L  tpx, y, z q P R : 5x 3

 5u ,

 0u

3y

4z

 7 ^ 9x 6y

, .

 5 ^ 3x 2y

z

3  0 ^ 8y

9y

 0u 6z

12z ,

 0u

5) Find suitable vectors x0 , v and w such that P

 tx0

P

 tpx, y, z q P R3 : x
3y

tv

sw P R3 : t, s P Ru ,

where a)

321

2z

 1u

 1u

,

.

,

 tpx, y, z q P R3 : 3x  5u P  tpx, y, z q P R3 : 2x
9y

b) P

,

c)

z

 0u

.

6) Calculate the distance between the lines L1 and L2 .

 tpx, y, z q P R3 : 12x
3y 2z  4 ^ x y z  0u , L2  tpx, y, z q P R3 : 4x
y
z  5 ^ 9x 3x 4z  7u , L1  tpx, y, z q P R3 : x
3y
z 
1 ^ y
z  0u , L2  tpx, y, z q P R3 : 2x y 18z  4 ^ 5x 3y  1u , L1  tpx, y, z q P R3 : 9x 7y
2z  8 ^ x
y
z  0u , L2  tpx, y, z q P R3 : 22x 4y
14z  19 ^ 28x 7z  12u

L1

a) b) c)

7) Calculate the distance of the point p from the plane P . a) p  p5, 12, 3q , P b) c)

 tpx, yq P R3 : x 2y
3z  9u , p  p0, 0, 0q , P  tpx, y q P R3 : x y z  1u , p  p6, 7, 2q , P  tpx, y q P R3 : 18x
y 9z 
3u

.

8) Let A, B and C be the vertices of a triangle. Find the representative # # # of [ AB ] [ BC ] [ CA ]. 9) Show that the line joining the midpoints of two sides of a triangle is parallel to and one-half the length of the third side. 10) Show that the medians of a triangle intersect in one point which is called the centroid. 11) Show that the perpendicular bisectors of a plane triangle intersect in one point which is called the circumcenter. 12) Show that the altitudes of a triangle, i.e., the the straight lines through the vertexes which are perpendicular to the opposite sides, intersect in one point which is called orthocenter. 13)

a) Show that vectors a, b P R2 are linearly dependent, i.e., such that there are α, β P R such that α a β b  0 and α2 β 2  0, if and only if detpa, bq  0. b) Show that vectors a, b, c P R3 are linearly dependent, i.e., such that there are α, β, γ P R such that α a β b γ c  0 and α2 β 2 γ 2  0, if and only if detpa, b, cq  0.

322

.

Fig. 98: Parabola and corresponding directrix and focus. Compare Example 2.5.24.

14) (Cramer’s rule in two and three dimensions)

a) Let a, b, c P R2 . Show that the of equation

pa  x, b  xq  c has a unique solution x P R2 if and only if detpa, bq  0. For that case express the solution x only in terms of detpc, bq, detpa, cq and detpa, bq. The result is called ‘Cramer’s rule’ in

two dimensions. b) Let a, b, c, d P R3 . Show that the of equation

pa  x, b  x, c  xq  d has a unique solution x P R3 if and only if detpa, b, cq  0. For that case express the solution x only in terms of detpd, b, cq, detpa, d, cq, detpa, b, dq and detpa, b, cq. The result is called

‘Cramer’s rule’ in three dimensions.

2.5.3 Conic Sections A parabola is a subset of the plane consisting of those points that are equidistant from a given line, called its directrix, and a given point, called its focus. The point bisecting the distance of the focus and the directrix is called its vertex; it is on the parabola. The infinite line through the vertex which is perpendicular to the directrix is called the axis of the parabola. 323

Example 2.5.24. (Parabolas) Find a function whose zero set is a parabola P with vertex in the origin, focus in the upper half-plane at p0, pq and axis given by the y
axis of a Cartesian coordinate system. Solution: As a consequence of the assumptions the directrix is given by y 
p. Hence px, yq P P if and only if a

x2  x2  x2

py
pq2  |y p| py
pq2  py pq2  4py

and therefore

x0

if p  0, i.e., in this case the parabola is given by y
axis and 2

y if p  0. Hence

P

if p  0 and

x  4p

 tpx, yq P R2 : x  0u "

P

 px, yq P R

:y


2

if p  0.

x2 4p

0

*

An ellipse is a subset of the plane consisting of those points for which the sum of the distances from two points, called foci, is constant. Because of the triangle inequality for the Euclidean distance, that constant is greater or equal than the distance of the foci. If the constant is non-zero, the ratio between the distance of the foci and the constant is called the eccentricity of the ellipse. The line connecting the foci of an ellipse is called eccentric line and its midpoint the center of the ellipse. Example 2.5.25. (Ellipses) Find a function whose zero set is an ellipse E with foci at p
c, 0q and pc, 0q and constant 2a where a ¥ c ¥ 0. Solution: px, yq P E if and only if d1

d2

 2a

324

(2.5.7)

Fig. 99: Ellipse and corresponding foci. Compare Example 2.5.25.

a

a

px cq2 y2 and d2 : px
cq2 y2 . In case that where d1 : a  c  0, this is equivalent to x  y  0, i.e., the ellipse is given by the origin. In the following, let a ¡ 0. Then 2a pd1
d2 q  d21
d22

and

 px

c q2

y 2
px
cq2
y 2

d1
d2

 2ca x .

 4cx

(2.5.8)

Hence (2.5.7) is equivalent to (2.5.7), (2.5.8) and therefore to c c x , d2  a
x . (2.5.9) d1  a a a We consider two cases. In case that a  c, equations (2.5.9) are equivalent to |x| ¤ c and y  0, i.e., in this case, the ellipse is given by the line r
c, cs  t0u. In case that a ¡ c, equations (2.5.9) are equivalent to y2

x2 a2 and the condition

1

a2
c2

(2.5.10)

2

|x| ¤ ac

.

(2.5.11)

Now the assumption |x| ¡ a2 {c and (2.5.10) lead to the contradiction that 0¡1


a2 c2

2

¡ a2 y
c2

325

.

Fig. 100: Hyperbolas, corresponding foci and asymptotes (dashed). Compare Example 2.5.26.

Hence (2.5.10) implies (2.5.11), and (2.5.9) is equivalent to (2.5.10). Hence E if a  c  0,

E

if a ¡ 0, c  a and E



 tp0, 0qu

 tpx, yq P R2 :
c ¤ x ¤ cu

"

2 px, yq P R : xa2 2

y2

a2
c2

1

*

if a ¡ 0, a ¡ c. Note that E is a circle of radius a if c  0. A hyperbola is a subset of the plane consisting of those points for which the difference of the distances from two points, called foci, is constant. Because of the triangle inequality for the Euclidean distance, the absolute value of that constant is smaller or equal than the distance of the foci. If the constant is non-zero, the ratio between the distance of the foci and its absolute value is called the eccentricity of the hyperbola. Example 2.5.26. (Hyperbolas) Find a function whose zero set is a hyperbola H with foci at p
c, 0q and pc, 0q, where c ¥ 0, and constant 2a such that |a| ¤ c. Solution: px, y q P H if and only if d1
d2

 2a

326

(2.5.12)

a

a

where d1 : px cq2 y 2 and d2 : px
cq2 y 2 . We consider two cases. In case that c  a  0, equation (2.5.12) is satisfied by all px, yq P R2, i.e., the hyperbola is given by the whole plane. In case that c ¡ 0, a  0, (2.5.12) is equivalent to x  0 and y P R, i.e., the hyperbola is given by the y
axis. In case that c ¡ 0, a  0, it follows that 2a pd1

d2 q  d21
d22

 px

c q2

d1

 2ca x .

and that d2

y 2
px
cq2
y 2

 4cx (2.5.13)

Hence (2.5.12) is equivalent to (2.5.12), (2.5.13) and therefore to d1

 ac x

a , d2

 ac x
a .

(2.5.14)

 c2
a2

(2.5.15)

Equations (2.5.14) are equivalent to c2
a2 2  x
y2 a2 and the condition

x |a| . (2.5.16) ¥ a c In case that a P t
c, cu, (2.5.15) and (2.5.16) are equivalent to x ¤
c and y  0, x ¥ c and y  0, respectively, i.e., the hyperbola is given by the respective half-lines. In case that |a|  c, equations (2.5.15) and (2.5.16) are equivalent to y2 x2
1 (2.5.17) a2 c2
a2 and a2 x¥ c if a ¡ 0 and a2 x¤
c

327

if a   0, respectively. The assumption 0 ¤ x lead together with (2.5.17) to the contradiction

  a2 {c or
a2 {c   x ¤ 0

x2 a2
c2 
1   ¤0. c2
a2 a2 c2 y2

Hence (2.5.15) and (2.5.16) are equivalent to (2.5.17) and x and (2.5.17) and x   0 if a   0. We conclude that H if c  a  0, if c ¡ 0, a  0, if c ¡ 0, a 
c, if c ¡ 0, a  c,

 R2 ,

H

 tp0, yq P R2 : y P Ru ,

H

 tpx, 0q P R2 : x ¤
cu

H

 tpx, 0q P R2 : x ¥ cu

#

H if c ¡ a ¡ 0 and H

¡ 0 if a ¡ 0

 px, yq P R2 : x  a #

 px, yq P R2 : x  a

1

1

y2

+

c2
a2

y2

+

c2
a2

if c ¡
a ¡ 0. Remark 2.5.27. Note that from an analytical algebraic point of view conics are zero sets of second order polynomials in the coordinates of Cartesian coordinate systems in the plane. Later on in Section 2.5.5 Quadrics will be defined as corresponding sets in three-dimensional space.

328

Problems 1) Find the vertex, focus and the directrix of the parabola. a) c) d) f)

tpx, yq P R2 : y2
3x  0u , b) tpx, yq P R2 : y2 
x{2u tpx, yq P R2 : y 5  2px 3q2 u , tpx, yq P R2 : y  x2
xu , e) tpx, yq P R2 : y  2
x2 u tpx, yq P R2 : y  x2 3x 1u .

2) Find a function whose zero set coincides with P .

a) P is the parabola with focus p1, 3q and directrix tpx, y q P R2 : x 2y 1  0u, b) P is the parabola with focus p4, 3q and directrix tpx, y q P R2 : x
2y  1u, c) P is the parabola with focus p
1,
2q and directrix tpx, y q P R2 : 2x y  3u.

3) Find the location of the foci and the eccentricity of the ellipse. a) b) c) d) e) f)

tpx, yq P R2 : x2 2y2  6u , tpx, yq P R2 : 5x2 11y2  10u , tpx, yq P R2 : 2x2 4y2  5u , tpx, yq P R2 : 3x2 2y2  1u , tpx, yq P R2 : 6x2 7y2  4u , tpx, yq P R2 : y  px2 {3q py2 {4q  p1{2qu

.

4) The lines connecting the foci of the following ellipses are parallel to the x-axis. Find the location of their foci and their eccentricities. a) b) c)

tpx, yq P R2 : 3x2 tpx, yq P R2 : x2 tpx, yq P R2 : 4x2

2y 2
3x
p4y {3q  p1{36qu ,

3y 2
4x 2y

2

12y 
12u ,

8x
12y

21  0u .

5) Find a function whose zero set is an ellipse of eccentricity 2 and foci at p
1,
1q, p2, 2q. 6) Find the location of the foci and the eccentricity of the hyperbola. a)

tpx, yq P R2 : 2x2
y2  5u 329

,

,

b) c) d) e) f)

tpx, yq P R2 : 7x2
9y2  9u , tpx, yq P R2 : 3x2
5y2  4u , tpx, yq P R2 : 4x2
y2  2u , tpx, yq P R2 : 7x2
4y2  1u , tpx, yq P R2 : y  px2 {4q
py2 {2q  p1{4qu

.

7) The lines connecting the foci of the following hyperbolas are parallel to the x-axis. Find the location of their foci and their eccentricities. a) b) c)

tpx, yq P R2 : 2x2
4y2
p2x{3q 4y  p35{18qu tpx, yq P R2 : 3x2
y2
12x
4y 
7u , tpx, yq P R2 : 2x2
3y2 4x 18y  30u .

,

8) Find a function whose zero set is a hyperbola of eccentricity 4 and foci at p
1, 1q, p2,
2q. 9) Show that the given set is an ellipse "

2t 1
t2 ,b a 1 t2 1 t2



PR :tPR ( pa cospθq, b sinpθqq : θ P R , b) where a ¡ 0 and b ¡ 0. a)

*

2

,

10) Show that the given set is a hyperbola. "

a) b)



*

2t 1 t2 ,b P R2 :
1   t   1 , a 2 1
t 1
t2

* " a , b tanpθq : θ P p
π {2, π {2q , cospθq

pa coshptq, b sinhptqq : t P R where a ¡ 0 and b ¡ 0. c)

(

.

2.5.4 Polar Coordinates Example 2.5.28. (Polar coordinates) Define g : p0, 8q  p
π, π s Ñ R2 z tp0, 0qu 330

y p

r sinHjL

r

j

r cosHjL

x

O

Fig. 101: Polar coordinates r, ϕ of a point p in the plane. r is the Euclidean distance of O and p. Compare Example 2.5.28.

by

g pr, ϕq : pr cos ϕ, r sin ϕq

for all r

P p0, 8q, ϕ P p
π, πs. Then g is bijective with the inverse g
1 : R2 z tp0, 0qu Ñ p0, 8q  p
π, π s

given by " a 2 ax

p p x2 for all px, y q P R2 z tp0, 0qu. g
1 px, y q 

a

y 2 , arccospx{ a x2 y 2 qq if y ¥ 0 2 2 2 y ,
arccospx{ x y qq if y   0

Example 2.5.29. Find a parametrization of the ellipse E :

"

x2 px, yq P R : a2 2

y2 b2

1

*

where a, b ¡ 0. Solution: Define the scale transformation f : R2 f px, y q : pax, by q for all x, y

P R. Then

f pS 1 p0qq  E . 331

Ñ R2 by

Employing polar coordinates for parametrization, S 1 p0q is given by S 1 : Hence E

pcos ϕ, sin ϕq P R2 :
π ¤ ϕ ¤ πs

(

 pa cos ϕ, b sin ϕq P R2 :
π ¤ ϕ ¤ πs

(

. .

Example 2.5.30. (Polar representation of parabola with focus in the origin) Let p ¡ 0. Show that Pp

 pr cos ϕ, r sin ϕq P R2 : r ¡ 0 ^
π   ϕ ¤ π ^ r p1 cos ϕq  2pu

(2.5.18)

is a parabola with focus at the origin and directrix given by the parallel through the y-axis through the point p2p, 0q. Solution: For this, denote by Pp parabola with focus at the origin and directrix given by the parallel through the y-axis through the point p2p, 0q. Then px, y q P Pp if and only if a

x2

y2

The equation

a

 px
2pq2  |x
2p| .

a

x2

implies that

x
2p 

(2.5.19)

 x
2p

y2 a

x2

y2

¥x

and hence that
2p ¥ 0 which is in contradiction to the assumptions. Hence this equation has no solution in R2 . Therefore (2.5.19) is equivalent to a x2 y 2 x  2p and Pp

 tpx, yq P R2 :

a

x2

y2

x  2pu .

Finally, since g from Example 2.5.28 is bijective and p0, 0q R Pp , we conclude (2.5.18). Note that as a consequence of the foregoing, (2.5.19) is equivalent to x2

y2

 p2p
xq2  x2
4px 332

4p2

and hence to

 2

y

a

p pp
xq .

Example 2.5.31. (Polar representation of ellipses with focus in the origin) Define for a ¡ 0 and 0 ¤ ε   1 the corresponding ellipse Ea,ε with center p
aε, 0q, foci at p
2aε, 0q, p0, 0q and excentricity ε by Ea,ε :

"

px, yq P R

2

px :

aεq2

y2 a2 p1
ε2 q

a2

1

*

.

Show that

 pr cos ϕ, r sin ϕq P R2 : r ¡ 0 ^(
π   ϕ ¤ π ^ r p1 ε cos ϕq  ap1
ε2 q . (2.5.20) Solution: In Example 2.5.25, we showed for a ¡ 0 and 0 ¤ c   a that the following equations are equivalent for px, y q P R2 Ea,ε

x2 a2 and

a

px

c q2

y2

a2
c2

y2

1

a

px
cq2

y2

y2 a2 p1
ε2 q

1

 2a .

Hence if a ¡ 0 and 0 ¤ ε   1, then also the equations

px

aεq2

a2 and

a

px

2aεq2

y2

 2a


a

x2

y2 .

are equivalent for px, y q P R2 . The last equation is equivalent to

px

2aεq2

y2






2a


since the equation a

px

2aεq2

y2



333

a

a

x2

2

x2

y2

y 2
2a

(2.5.21)

leads by use of the triangle inequality to

|px, yq
p
2aε, 0q|  |px, yq|
2a ¤ |px, yq
p
2aε, 0q|
2ap1
εq and therefore has no solution in R2 since a ¡ 0 and ε   1. Further, (2.5.21) is equivalent to x2

y2

4a2 ε2

4aεx

 x2

a

y 2
4a x2

4a2

y2

which is equivalent to a

x2

εx  ap1
ε2 q .

y2

As a consequence, we arrive at the representation !

Ea,ε

 px, yq P R

2

:

a

x2

y2

εx  ap1
ε

2

)

q

Finally, since g from Example 2.5.28 is bijective and p0, 0q conclude (2.5.20).

.

R

Ea,ε , we

Example 2.5.32. (Polar representation of hyperbola with focus in the origin) Define for a   0 and ε ¡ 1 the corresponding hyperbolas Ha,ε with center p0,
aεq, foci at p0, 0q, p0,
2aεq and excentricity ε by Ha,ε :

"

px, yq P R

2

:x¤

px aεq2 a 2 p ε
1q ^ ε a2




y2 a2 pε2
1q

1

*

.

Show that

 pr cos ϕ, r sin ϕq P R2 : r ¡ 0 ^(
π   ϕ ¤ π ^ r p1 ε cos ϕq  ap1
ε2 q . (2.5.22) Solution: In Example 2.5.26, we showed for a   0 and c ¡
a that the following equations are equivalent for px, y q P R2 Ha,ε

x2 a2

2


c2 y
a2  1 334

y 3

2

-3

-1

x

1

-2

-3

Fig. 102: Parabola, ellipse, hyperbola and asymptotes corresponding to the parameters p  1, ε  1{2 and ap1
ε2 q  1, ε  3{2 and ap1
ε2 q  1, respectively. Compare Examples 2.5.30, 2.5.31 and 2.5.32.

together with the condition that x¤
and

a

px

c q2

y2


a2 c

a

px
cq2

y2

 2a .

Therefore also the equations

px
cq2
a2

y2

c2
a2

1

together with the condition that x¤ and

a

x2

y2


1 2 pc
a2 q c

a

px
2cq2 335

y2

 2a .

are equivalent. Hence if a   0 and ε ¡ 1, the equations

px

aεq2 a2

2


a2pεy2
1q  1

together with the condition that x¤ and

a

a 2 pε
1q ε

y 2
2a 

x2

a

px

2aεq2

y2 .

are equivalent. The last equation is equivalent to

px

2aεq2

2


a

y2



y2

 px

x2

y 2
2a

(2.5.23)

since the equation 2a


a

x2

a

2aεq2

y2

has no solution in R2 since a   0. Further, (2.5.23) is equivalent to x2

y2

4a2 ε2

4aεx

a

 x2

y 2
4a x2

4a2

y2

which is equivalent to a

x2

εx  ap1
ε2 q .

y2

As a consequence, we arrive at the representation !

Ha,ε

 px, yq P R

2

:

a

x2

y2

εx  ap1
ε

2

)

q

Finally, since g from Example 2.5.28 is bijective and p0, 0q conclude (2.5.22).

Problems

336

.

R

Ha,ε , we

1) Sketch the image of the set under g from Example 2.5.28 on polar coordinates.

tpr, ϕq P Dpgq : r   3u , b) tpr, ϕq P Dpg q : r ¡ 2u , c) tpr, ϕq P Dpg q :
π {2 ¤ ϕ ¤ π {2u , d) tpr, ϕq P Dpg q : 3π {4 ¤ ϕ ¤ 5π {6u , e) tpr, ϕq P Dpg q :
3π {4 ¤ ϕ ¤
π {4u , f) tpr, ϕq P Dpg q : 1   r   2 ^ π {6 ¤ ϕ ¤ π {3u , g) tpr, ϕq P Dpg q : 0   r   1 ^
π {3 ¤ ϕ ¤
π {6u . Find a function whose zero set coincides with g pC q, where g is the a)

2)

transformation from Example 2.5.28 on polar coordinates. In addition, sketch g pC q. a) C  tpr, ϕq P Dpg q : r  4u , b) C  tpr, ϕq P Dpg q : 1 r r 2 cospϕq

sinpϕq s  0u ,

c) C  tpr, ϕq P Dpg q : r  2 cospϕqu , d) C  tpr, ϕq P Dpg q : r  1{ r 2
cospϕq su , e) f) g)

 tpr, ϕq P Dpgq : r  sin2 pϕqu , C  tpr, ϕq P Dpg q : r2 sinp2ϕq  1u , C  tpr, ϕq P Dpg q : r2  2 sinpϕqu . C

3) Find a function whose zero set coincides with g
1 pC q, where g is the transformation from Example 2.5.28 on polar coordinates. a) b) c) d) e) f) g)

 tpx, yq P R2 : x  3u , C  tpx, y q P R2 : 3x
2y  7u , C  tpx, y q P R2 : 3x2 y 2  9u , C  tpx, y q P R2 : y
4x2
1  0u , C  tpx, y q P R2 : 8x2
4y 2
1  0u C  tpx, y q P R2 : 3xy  4u , C  tpx, y q P R2 : 2x2
3x y 2  1u C

, .

4) Show that g from Example 2.5.28 on polar coordinates is bijective by verifying that g
1 pg pr, ϕqq  pr, ϕq

337

Fig. 103: Example of a parabolic cylinder. Compare Example 2.5.33. for all pr, ϕq P Dpg q and

g pg
1 px, y qq  px, y q

for all px, y q P R2 z tp0, 0qu.

2.5.5 Quadric Surfaces Quadric surfaces are zero sets of second order polynomials in three space variables x, y, z, like conics are zero sets of such polynomials in two variables. All quadrics are unique only up to rigid transformations in space. In the following, we briefly mention the most important normal forms of quadrics. Example 2.5.33. For every plane curve C, the set C  R is called a cylinder. Examples are: (i) The parabolic cylinder ZP :

px, y, zq P R3 : x2
4py  0 338

(

Fig. 104: Example of an elliptic cylinder. Compare Example 2.5.33.

Fig. 105: Example of a hyperbolic cylinder. Compare Example 2.5.33.

339

Fig. 106: Example of an ellipsoid. Compare Example 2.5.34.

where p ¡ 0. The intersection of ZP with every parallel plane to the xy-plane is a parabola. (ii) The elliptic cylinder ZE :

"

2 px, y, zq P R : xa2

y2 b2

3

1

*

where a, b ¡ 0. The intersection of ZE with every parallel plane to the xy-plane is an ellipse. (iii) The hyperbolic cylinder ZH :

"

2 px, y, zq P R : xa2 3




y2 b2

1

*

where a, b ¡ 0. The intersection of ZH with every parallel plane to the xy-plane is a hyperbola.

340

Fig. 107: Example of an elliptic paraboloid. Compare Example 2.5.35.

Example 2.5.34. The surface E :

"

2 px, y, zq P R : xa2 3

y2 b2

z2 c2

1

*

,

where a, b, c ¡ 0, is an ellipsoid with half-axes a,b and c. The intersection of E with a plane parallel to a coordinate plane is an ellipse, a point, or the empty set. E may be viewed as a ‘deformed’ sphere, because it is the image of S 2 p0q under the scale transformation f : R3 Ñ R3 defined by f px, y, z q : pax, by, cz q for all px, y, z q P R3 . Example 2.5.35. The surface EP :

"

px, y, zq P R : zc 3

341



x2 a2

y2 b2

*

,

Fig. 108: Example of a hyperbolic paraboloid. Compare Example 2.5.36.

where a, b, c ¡ 0, is called an elliptic paraboloid. The intersection of EP with a parallel to the xy-plane is an ellipse, a point or the empty set. The intersection of EP with a plane containing the z-axis is a parabola. The surface looks similar to a ‘saddle’ and is therefore often called a ‘saddle surface’. Example 2.5.36. The surface HP :

"

px, y, zq P R : zc 3



x2 a2




y2 b2

*

,

where a, b, c ¡ 0, is called an hyperbolic paraboloid. The intersection of HP with a parallel to the xy-plane is a hyperbola. The intersection of HP with a plane containing the z-axis is a parabola. Example 2.5.37. The surface EC :

"

2 px, y, zq P R : zc2 3

342



x2 a2

y2 b2

*

,

Fig. 109: Example of an elliptic cone. Compare Example 2.5.37.

where a, b, c ¡ 0, is called an elliptic cone. The intersection of EC with a parallel to the xy-plane is an ellipse, with midpoint given by its intersection with the z-axis, or a point called its vertex. The intersection of EC with a plane containing the z-axis are two straight lines crossing in the vertex. Example 2.5.38. The surface H1 :

"

2 px, y, zq P R : zc2 3



x2 a2

y2 b2


1

*

,

where a, b, c ¡ 0, is called a hyperboloid of one sheet. The intersection of H1 with a parallel to the xy-plane is an ellipse with midpoint given by its intersection with the z-axis. The intersection with a plane containing the z-axis consists of two hyperbolas. Example 2.5.39. The surface H2 :

"

2 px, y, zq P R : zc2 3

343



x2 a2

y2 b2

*

1

,

Fig. 110: Example of a hyperboloid of one sheet. Compare Example 2.5.38.

Fig. 111: Example of a hyperboloid of two sheets. Compare Example 2.5.39.

344

where a, b, c ¡ 0, is called a hyperboloid of two sheets. The intersection of H2 with a parallel to the xy-plane is an ellipse with midpoint given by its intersection with the z-axis, a point or the empty set. The intersection with a plane containing the z-axis consists of two hyperbolas. Problems 1) Describe and sketch the surface. a) b) c) d) e)

tpx, y, z q P R3 : 2y2 3z 2  9u , tpx, y, z q P R3 : z 6x2  1u , tpx, y, z q P R3 : x
2y2 
3u , tpx, y, z q P R3 : xz  12u , tpx, y, z q P R3 : 3x2
5y2  7u .

2) Find the intersections of the surface with the coordinate planes. In this way, identify the surface and sketch it. a) b) c) d) e) f) g)

tpx, y, z q P R3 : 2x2
3y2 z 2  4u , tpx, y, z q P R3 : 9x2 3y2 5z 2  12u , tpx, y, z q P R3 : x2
2y2
4z 2  3u , tpx, y, z q P R3 : y2  6x2 4z 2u , tpx, y, z q P R3 : 4x2 3z 2  2yu , tpx, y, z q P R3 : 4z 2
3x2
2y  0u , tpx, y, z q P R3 : x2 4y2 3z 2 1  0u .

3) Identify the surfaces. a) b) c) d)

tpx, y, z q P R3 : z 2  4u , tpx, y, z q P R3 : 2y2
3z 2  0u , tpx, y, z q P R3 : x2
4xy 4y2  2u , tpx, y, z q P R3 : px
2yq2  2px z q2 u

.

4) Find a function whose zero set consists of all points that are equidistant from p0, 0, 1q and the coordinate plane

tpx, y, z q P R3 : z 
1u .

Identify the surface.

345

5) Find a function whose zero set consists of all points whose distance from the z-axis is 3-times the distance from the xy-plane. Identify the surface. 6) Show that through every point of the the surfaces go two straight lines that are contained in that surface. a) The elliptic cone EC , b) the hyperbolic paraboloid HP , c) the hyperboloid of one sheet H1 . 7) (Conic sections) Let α P r0, π {2s and C be the circular cone defined by ( C : px, y, z q P R3 : z 2  x2 y 2 . a) Find a function whose zero set coincides with the cone Cα resulting from a C by clockwise rotation in the yz-plane around p0, 0, 1q and about the angle α. The symmetry axis of that cone is given by

tp0, t sinpαq, 1

t cospαqq : t P Ru

and its vertex by

p0,
sinpαq, 1
cospαqq . b) Find the intersection of Cα with the coordinate plane parallel to the xy-plane through p0, 0, 1q. Classify those curves.

2.5.6 Cylindrical and Spherical Coordinates Example 2.5.40. (Cylindrical coordinates) Define g : p0, 8q  p
π, π s  R Ñ R3 z pt0u  t0u  Rq by

g pr, ϕ, z q : pr cos ϕ, r sin ϕ, z q

for all pr, ϕ, z q P p0, 8q  p
π, π s  R. Then g is bijective with inverse

g
1 : R3 z pt0u  t0u  Rq Ñ p0, 8q  p
π, π s  R 346

z

p

y

O

j

r q r×sinHjL r×cosHjL x

Fig. 112: Cylindrical coordinates r, ϕ, z of a point p in space. q is the orthogonal projection of p onto the xy-plane, r is the Euclidean distance of O and q, ϕ the angle of the line from O to q with the x-axis. Compare Example 2.5.40.

given by a

" a 2 ax

y 2 , arccospx{ a x2 y 2 q , z q p p x2 y2 ,
arccospx{ x2 y2 q , zq for all px, y q P R3 z pt0u  t0u  Rq. g
1 px, y, z q 

if y ¥ 0 if y   0

Example 2.5.41. Find a parametrization of the cylinder Z1

 px, yq P R2 : x2

y2

1

(

.

Solution: Employing cylindrical coordinates, Z1 is given by Z1

 tpcospϕq, sinpϕq, zq : ϕ P p
π, πs, z P Ru

Example 2.5.42. (Spherical coordinates) Define g : p0, 8q  r0, π s  p
π, π s Ñ R3 ztp0, 0, 0qu 347

.

z

p y r r×cosHΘL Θ O

j

r×sinHΘL q

x

Fig. 113: Spherical coordinates r, θ, ϕ of a point p in space. r is the Euclidean distance of O and p, θ the angle between the line from O to p and the z-axis, q the orthogonal projection of p onto the xy-plane, ϕ the angle of the line from O to q with the x-axis. Compare Example 2.5.42.

by

g pr, θ, ϕq : pr sin θ cos ϕ, r sin θ sin ϕ, r cos θq

for all pr, θ, ϕq P p0, 8q  r0, π s  p
π, π s. Then g is bijective with inverse

g
1 : R3 ztp0, 0, 0qu Ñ p0, 8q  r0, π s  p
π, π s given by a

"

p|r| , arccospz{|r|q , arccospx{ a x2 y 2qq if y ¥ 0 2 2 p|r| , arccospz{|r|q ,
arccospx{ x y qq if y   0 for all px, y, z q P R3 ztp0, 0, 0qu. In analogy with the situation on the globe, for px, y, z q P R3 ztp0, 0, 0qu the second and third component of g
1ppx, y, z qq can be called the longitude, co-latitude, respectively of px, y, z q. Note that for a point on the northern hemisphere π {2 minus its co-latitude g
1prq 

348

gives its latitude, whereas for a point on the on the southern hemisphere the latitude is given by difference of its co-latitude and π {2. Example 2.5.43. Find a parametrization of the ellipsoid E :

"

2 px, y, zq P R : xa2 3

y2 b2

z2 c2

1

*

where a, b, c ¡ 0. Solution: E is the image of S 2 p0q under the scale transformation f : R3 Ñ R3 defined by f px, y, z q : pax, by, cz q for all px, y, z q P R3 . Employing spherical coordinates, a parametrization of S 2 p0q is given by S 2 p0q 

psinpθq cospϕq, sinpθq sinpϕq, cospθqq P R3 : ϕ P p
π, πs, θ P r0, π su .

Hence E

 pa sinpθq cospϕq, b sinpθq sinpϕq, c cospθqq P R3 : ϕ P p
π, πs, θ P r0, π su . Problems 1) Describe the image of the set under g from Example 2.5.40 on cylindrical coordinates. a) b) c) d) e) f) g)

tpr, ϕ, z q P Dpgq : r   3 ^
1   z   1u , tpr, ϕ, z q P Dpgq : r ¡ 2 ^ 0   z   3u , tpr, ϕ, z q P Dpgq : 1   r   2u , tpr, ϕ, z q P Dpgq :
π{2 ¤ ϕ ¤ π{2 ^ z  1u , tpr, ϕ, z q P Dpgq : 3π{4 ¤ ϕ ¤ 5π{6 ^ z ¤ 0u , tpr, ϕ, z q P Dpgq :
3π{4 ¤ ϕ ¤
π{4 ^ z ¥ 1u , tpr, ϕ, z q P Dpgq : 0   r   1 ^
π{3 ¤ ϕ ¤
π{6u 349

.

2) Find a function whose zero set f : U Ñ R coincides with g pS q, where g is the transformation from Example 2.5.40 on cylindrical coordinates. In addition, sketch g pS q. S

a)

 tpr, ϕ, z q P Dpgq : r  3u

,

b) S  tpr, ϕ, z q P Dpg q : z  2ru , c) S  tpr, ϕq P Dpg q : z  3r sinpϕq

d) S

3)

12u ,

 tpr, ϕq P Dpgq : z 4r  1u , e) S  tpr, ϕq P Dpg q : 6z
2r2
3  0u , f) S  tpr, ϕq P Dpg q : z 2 3  5r2 u , g) S  tpr, ϕq P Dpg q : r  2 cospϕqu . Find a function whose zero set coincides with g
1 pS q, where g is the 2

2

transformation from Example 2.5.40 on cylindrical coordinates.

 tpx, y, z q P R2 : 2x
6y z  1u , S  tpx, y, z q P R2 : x2 y 2  4u , S  tpx, y, z q P R2 : x2 y 2 3z 2  2u , S  tpx, y, z q P R2 : 2x2 2y 2  9z u , S  tpx, y, z q P R2 : x2 y 2
2z 2  5u , C  tpx, y q P R2 : 4x2 4y 2
z 2 1  0u C  tpx, y q P R2 : 8x2 y 2
3z 2  0u .

a) S b) c) d) e) f) g)

,

4) Describe the image of the set under g from Example 2.5.42 on spherical coordinates.

tpr, ϕ, θq P Dpgq : r   3u , b) tpr, ϕ, θq P Dpg q : r ¡ 2u , c) tpr, ϕ, θq P Dpg q : 1   r   8u , d) tpr, ϕ, θq P Dpg q : 0 ¤ θ ¤ π {4u , e) tpr, ϕ, θq P Dpg q : π {6 ¤ θ ¤ π {4u , f) tpr, ϕ, θq P Dpg q : r P r1, 2s ^ θ P rπ {6, π {3s ^ ϕ P rπ{6, π{3su . Find a function whose zero set coincides with g pS q with g from Example 2.5.42 on spherical coordinates. In addition, sketch g pS q. a) S  tpr, ϕ, z q P Dpg q : r
5  0u , a)

5)

350

b) S  tpr, ϕ, z q P Dpg q : ϕ  π {6u , c) S  tpr, ϕ, z q P Dpg q : θ  π {4u ,

d) S  tpr, ϕ, z q P Dpg q : r  6 cospθqu , e) S  tpr, ϕ, z q P Dpg q : r sinpθq  4u , f) g)

 tpr, ϕ, z q P Dpgq : r cospθq  2u , S  tpr, ϕ, z q P Dpg q : r2 cosp2ϕq sin2 pθq  1u

S

.

6) Find a function whose zero set coincides with g
1 pS q, where g is the transformation from Example 2.5.42 on spherical coordinates. a) b) c)

 tpx, y, z q P R2 : x2 S  tpx, y, z q P R2 : x2 S  tpx, y, z q P R2 : px2 S

y2

z 2
2y

 0u

,

 3u , y q  4z 2 px2
y 2 qu

y

2

2 2

.

7) Show that g from Example 2.5.40 on cylindrical coordinates is bijective by verifying that g
1 pg pr, ϕ, z qq  pr, ϕ, z q for all pr, ϕ, z q P Dpg q and g pg
1 px, y, z qq  px, y, z q for all px, y, z q P R3 z pt0u  t0u  Rq. 8) Show that g from Example 2.5.42 on spherical coordinates is bijective by verifying that g
1 pg pr, ϕ, θqq  pr, ϕ, θq for all pr, ϕ, θq P Dpg q and

g pg
1 px, y, z qq  px, y, z q

for all px, y, z q P R3 ztp0, 0, 0qu.

2.5.7 Limits in Rn Definition 2.5.44. Let x1 , x2 , . . . be a sequence of elements of Rn and x P Rn . We define lim xm  x

Ñ8

m

351

x

y

z

Fig. 114: A sequence in space is convergent if and only if all its coordinate projections converge. Compare Theorem 2.5.45.

if for every ε ¡ 0 there is a corresponding m0 such that for all m ¥ m0 :

|xm
x|   ε . The following theorem says that a sequence of elements in Rn is converging to some x P Rn if and only if for all i P t1, . . . , nu the corresponding sequence of its i-th components converges in R to the i-th component of x. In this way, the question convergence or non-convergence of a sequence in Rn , n ¥ 2, is reduced to the question of convergence or non-convergence of sequences of real numbers. Theorem 2.5.45. Let x1 , x2 , . . . be a sequence of elements of Rn and x Rn . Then lim xm  x

Ñ8

m

if and only if lim xmj

for all j

 1, . . . , n.

Ñ8

m

352

 xj

P

Proof. First, we note that max |yj | ¤ |y| ¤ |y1|



j 1,...,n

. . . |yn|

for all y P Rn .



Hence if limmÑ8 xm m ¥ m0

¡

x, ε

0 is given and m0 is such that for all

|xm
x|   ε , then also for every j P t1, . . . , nu and every m ¥ m0 |xmj
xj |   ε

and hence also

 xj . On the other hand, if for every j P t1, . . . , nu lim xmj  xj , mÑ8 lim xmj

Ñ8

m

ε ¡ 0 is given and for every j P t1, . . . , nu the corresponding m0j is such that for every m ¥ m0j |xmj
xj |   nε , then it follows for every m ¥ m01 m02 . . . that

|xm
x|   ε , and hence that lim xm

Ñ8

m

x.

Example 2.5.46. Calculate 

lim

n

Ñ8

sinpnq n2 2 , 2 , n n 1 n 353



.

Solution: 

sinpnq n2 2 lim , 2 , nÑ8 n n 1 n  p0, 1, 0q .







sinpnq n2 2 , lim 2 , lim lim nÑ8 n nÑ8 n 1 nÑ8 n



The following corollary says that a sequence in Rn can have at most one limit point (in part (i)), that the sequence consisting of the sums of the members of convergent sequences in Rn is convergent against the sum of their limits (in part (ii)) and that the sequence consisting of scalar multiples of the members of a convergent sequence in Rn converges against that scalar multiple of its limit. Corollary 2.5.47. Let x1 , x2 , . . . ; y1 , y2 , . . . be sequences of elements of Rn ; x, x ¯, y P Rn and a P R. (i) If then x ¯  x.

lim xm

 x and

m

lim xm

 x and

m

Ñ8

m

(ii) If

Ñ8

m

then

lim pxm

Ñ8

m

lim xm

 x¯ ,

lim ym

y,

Ñ8

Ñ8

ym q  x

(iii) If lim xm

Ñ8

m

then lim a.xm

Ñ8

m

Problems

354

x,  a.x .

y.

1) If existent, calculate the limit of the sequence pxn , yn , f pxn , yn qq, n P N . Otherwise, show non-existence of the limit. Where applicable, a P R. a) b) c) d) e) f) g) h)

1 a 2xy 2 , px, y q P R2 z t0u , yn : , f px, y q : 2 n n x y2 1 a 2xy 2 , px, y q P R2 z t0u xn : , yn : , f px, y q : 2 n n x y4 1 1 2xy 2 , px, y q P R2 z t0u xn : 2 , yn : , f px, y q : 2 n n x y4 1 a xy , px, y q P R2 z t0u xn : , yn : , f px, y q : 2 n n x y2 1 a x y xn : , yn : , f px, y q : 2 , px, y q P R2 z t0u n n x y2 1 a x2 xn : , yn : , f px, y q : 2 , px, y q P R2 z t0u n n x y2 1 a x xn : , yn : , f px, y q : 2 , px, y q P R2 z t0u n n x y2 1 a x2 y 2 xn : , yn : , f px, y q : 3 , n n x y3 px, yq P R2 z tpx,
xq : x P Ru xn :

1 e1{n x2 y 2 , , yn :
, f px, y q : 3 n n x y3 px, yq P R2 z tpx,
xq : x P Ru

i)

xn :

j)

xn :

1 a x3 , yn : , f px, y q : 2 n n x 2 2 px, yq P R z tpx,
x q : x P Ru

y3 , y

1 e1{n x3 , yn :
2 , f px, y q : 2 n n x 2 2 px, yq P R z tpx,
x q : x P Ru

k) xn :

l)

1 a x3 , yn : , f px, y q : n n x 2 px, yq P R z tpx,
xq : x P Ru

xn :

y3 , y

y3 , y

1 a x2 y 2 , , yn : , f px, y q : 4 n n x y4 px, yq P R2 z t0u

m) xn :

355

1 a x2 y 2 , , yn : , f px, y q : 2 n n x y2 px, yq P R2 z t0u .

n) xn :

2) Prove Corollary 2.5.47.

2.5.8 Paths in Rn Definition 2.5.48. Let n P N . (i) A path is a map u : I Ñ Rn from some non-empty subinterval I of R into Rn . The range of a path is frequently called a curve. (ii) A path u : I Ñ Rn is called continuous if all corresponding coordinate projections ui : I Ñ R, defined by ui ptq : puptqqi for all t P I and i P t1, . . . , nu, are continuous. (iii) A path u : I Ñ Rn is said to be differentiable in some inner point t0 P I, i.e. some point t0 P I for which there is some ε ¡ 0 such that pt0
ε, t0 εq € I, if all corresponding coordinate projections ui : I Ñ R, i P t1, . . . , nu, are differentiable in t0 . In this case, we define its derivative in t0 by u 1 pt0 q : pu11 pt0 q, . . . , un1 pt0 qq and its tangent vector in t0 by

pupt0q, u 1pt0 qq P R2n . Example 2.5.49. Calculate the derivative and the tangent vector field v of the path u : R Ñ R3 defined by uptq : pcosptq, sinptq, tq 356

v

O

Fig. 115: Tangent vector v at a point of a helix. Compare Example 2.5.49.

for all t P R. Solution: u is differentiable since all its component functions are differentiable in the sense of Calculus I. Hence u 1 ptq : p
sinptq, cosptq, 1q for all t P R and the corresponding tangent vector field v is given by v ptq : ppcosptq, sinptq, tq, p
sinptq, cosptq, 1qq for all t P R. See Fig. 2.5.49. Theorem 2.5.50. Let l, m, n i.e., such that

P N, λ : Rl  Rm Ñ Rn be a bilinear map,

λpα.x β.y, z q  α.λpx, z q β.λpy, z q , λpx, α.z β.w q  α.λpx, z q β.λpx, w q 357

for all x, y P Rl , z, w P Rm and α, β P R. Further, let I be a non-void open interval of R and u : I Ñ Rl , v : I Ñ Rm be differentiable paths. Then the path λpu, v q : I Ñ Rn defined by

rλpu, vqsptq : λpuptq, vptqq for all t P I is differentiable, and

rλpu, vqs 1ptq  λpu 1ptq, vptqq

λpuptq, v 1 ptqq

for all t P I. m l Proof. For this, let el1 , . . . , ell , em 1 , . . . , em , be the canonical basis of R and Rm , respectively. It follows by the bi-linearity of λ that

λpx, z q 

l ¸ m ¸

 

pxj zk q . λpelj , emkq

j 1k 1

for all x P Rl , z

P Rm . Hence rλpu, vqsi 

l ¸ m ¸

 

rλpelj , emkqsi uj vk

j 1k 1

is differentiable by Theorem 1.3.8 with derivative

rλpu, vqsi1ptq 

l ¸ m ¸

rλpelj , emk qsi puj1 ptq vk ptq

j 1 k 1 1  rλpu ptq, vptqq λpuptq, v 1ptqqs

uj ptq vk1 ptqq

i

for all t P I and i P t1, . . . , nu. Theorem 2.5.51. Let n P N , I,J be non-void open intervals of R, u, v : I Ñ Rn differentiable paths, f : I Ñ R and g : J Ñ R be differentiable. Then

358

Ñ Rn, defined by pu vqptq : uptq vptq for every t P I, is differentiable and pu vq 1ptq  u 1ptq v 1 ptq for all t P I. f.u : I Ñ Rn , defined by pf.uqptq : f ptq.uptq for every t P I, is differentiable and pf.uq 1ptq  f 1 ptq.uptq f ptq.u 1ptq for all t P I. u  v : I Ñ R, defined by pu  vqptq : uptq  vptq for every t P I, is differentiable and pu  vq 1ptq  u 1ptq  vptq uptq  v 1 ptq for all t P I. if n  3, then u  v : I Ñ R3 , defined by pu  vqptq : uptq  vptq for every t P I, is differentiable and pu  vq 1ptq  u 1ptq  vptq uptq  v 1ptq for all t P I.

(i) u

(ii)

(iii)

(iv)

v:I

359

€ I, then u  g : J Ñ R is differentiable and pu  gq 1ptq  g 1 ptq.u 1pgptqq for all t P J.

(v) if Ran g

Proof. ‘(i)-(iv)’ are consequences of Theorem 2.5.50. ‘(v)’: It follows by Theorem 1.3.10 that pu  gqi  ui  g is differentiable with derivative

rpu  gqis 1ptq  ui1pgptqq  g 1 ptq  rg 1ptq.u 1pgptqqsi for all t P J, i P t1, . . . , nu. Example 2.5.52. Let r be a twice differentiable path (the trajectory of a point particle parametrized by time) from some non-void open interval I of R into R3 and satisfying m.r 2 ptq  0 for all t P I (Newton’s equation of motion without external forces), where m ¡ 0 (the mass of the particle). Then
m

2

v2

1

ptq  m r 1ptq  r 2ptq  0

for all t P I where v : r 1 (the velocity field of the particle) and v 2 : v  v. Hence it follows by Theorem (1.4.7) that the function m 2 v 2 (the kinetic energy of the particle) is constant (‘is a constant of motion’).

360

Example 2.5.53. (Kepler problem, I) Let r be a twice differentiable path (the trajectory of a point particle parametrized by time) from some nonvoid open interval I of R into R3 zt0u satisfying m . r 2 ptq 


γmM |rptq|3 . rptq

(2.5.24)

for all t P I (Newton’s equation of motion for a point particle under the influence of the gravitational field of a point mass located at the origin.), where m, M, γ ¡ 0 (the mass of particle, the mass of the gravitational source, the gravitational constant). Show that the total energy E : I Ñ R of the system, the angular momentum L : I Ñ R3 and the Lenz vector A : I Ñ R3 defined by m 1 γmM r ptq  r 1 ptq
2 |rptq| , Lptq : rptq  rm . r 1 ptqs  m . rptq  r 1 ptq ,

 γmM 1 Aptq : m . r ptq  Lptq
|rptq| . rptq

E ptq :

for every t P I are constant. Solution: It follows by Theorem 2.5.51, (2.5.24) and Theorem 2.5.15 (vi) that E 1 ptq  m r 1 ptq  r 2 ptq L 1 ptq  m . r 1 ptq  r 1 ptq for all t P I and

γmM rptq  r 1 ptq 0, |rptq|3 m . rptq  r 2 ptq  0

γmM rptq  r 1 ptq γmM 1 . rptq
3 |rptq| |rptq| . r ptq rptq  rm . r 1 ptqs
m . rr 1 ptq  r 2 ptqs . rptq

a 1 ptq  r 2 ptq  Lptq

 r 2ptq  1
γmM |rptq| . r ptq  m . rr 1ptq  r 2ptqs . rptq
m . rrptq  r 2ptqs . r 1ptq 361


m . rr 1ptq  r 2ptqs . rptq
γmM . r 1 ptq  0 |rptq| for all t P I, where a : I

Ñ R3 is defined by aptq : m
1 . Aptq

for all t P I. Hence it follows by Theorem (1.4.7) that E, L and A are constant. In the following, we derive necessary consequences of these conservation laws. In this, we denote by E, L and A the corresponding constants and L : |L|, A : |A|. In particular, we assume that A  0, L  0 and denote for t P I by θptq P p
π, π s is the ‘polar’ angle between A and rptq. Then it follows by Theorem 2.5.15 (v) that A|rptq| cospθptqq  A  rptq  m rptq 



γmM r 1 ptq  L
. rptq



|rptq|

 m rptq  r r 1ptq  L s
γm2M |rptq|  m L  r rptq  r 1ptq s
γm2M |rptq|  L2
γm2 M |rptq| and hence that

|rptq|  r 1

ε cospθptqq s  p

for every t P I, where ε :

A L2 , p : .  γm2 M γm2 M

In addition, it follows by Theorem 2.5.15 (v), (vii) for t P I that A2 m2





γmM r 1 ptq  L
. rptq



γmM r 1 ptq  L
. rptq

 |rptq| |rptq| 2γmM  pr 1ptq  Lq  pr 1ptq  Lq
|rptq| . rptq  pr 1ptq  Lq

1  L2 |r 1ptq|2
r L  r 1ptq s2
2γmM |rptq| . L  prptq  r ptqq 362



γ 2 m2 M 2 γ 2 m2 M 2

L



2

2E m

2γM |rptq|



2


2γML |rptq|

γ 2 m2 M 2

 γ 2m2 M 2

2EL2 m

d

and hence that ε As a consequence,

2EL2 . γ 2 m3 M 2

1

$ ' &

 1 ε 1 ' % ¡1

if E if E if E

 0 0 ¡0.

By its definition, L is orthogonal to r 1 ptq for every t for for every t0 , t1 P I satisfying t0   t1 that L  rrpt1 q
rpt0 qs 

» t1 t0

P I. Hence it follows

L  r 1 ptq dt  0 .

Hence the motion of the particle proceeds in a plane S with normal vector n3 : L
1 .L . In the following, we make the natural assumption that θptq assumes all values in p
π, π s. Then S contains the origin. This can be seen as follows. By assumption, there are t0 , t1 P I such that θpt1 q 


π π , θpt2 q  2 2

and hence L  rpt0 q  L  rpt1 q  0 , A  rpt0 q  A  rpt1 q  0 , |rpt0 q|  |rpt1 q| . Therefore, we conclude, by noting that L  A  0, that rpt0 q 
rpt1 q and hence that

rpt0 q

1 prpt1q
rpt0 qq  0 P S . 2 363

Therefore, it follows from Examples 2.5.30, 2.5.31, 2.5.32 that Ranprq are conics in S with one focus in the origin. In particular, the conic is an ellipse, parabola or hyperbola if E   0, E  0 or E ¡ 0, respectively. Note that the previous was derived from the assumption of the the existence of a solution of (2.5.24) with the prescribed properties. Indeed, that existence can be proved, and this is done, for instance, in courses in theoretical mechanics. Example 2.5.54. (Kepler problem, II, Levi-Civita’s transformation) We continue the discussion from the previous discussion and present LeviCivita’s ingenious method to transform (2.5.24) into a form whose solutions are obvious. In the first step, we introduce a new time variable. For this, let t0 P I and I0 : pt0 , 8q X I. We define a time function τ : I0 Ñ R by »t dt1 τ ptq : 1 t0 |rpt q|

for all t P I0 . Then τ is strictly increasing, and hence according to Theorem 1.4.17, the restriction in its image on its range, given by an open interval J0 , has a differentiable inverse which will be denoted by the symbol τ
1 in the following. In particular, we define ξ : r  τ
1 .

Then ξ1



ξ2

 |ξ |2

and hence

τ
1

1

r 1  τ
1

r 2  τ
1





 τ 1 1τ
1 |ξ | 1

r 2  τ
1

r 1  τ
1

r 1  τ
1





 |ξ |2

 |ξ |

r 1  τ
1

r 2  τ
1





,

|ξ | 1 ξ 1 |ξ |

1  |ξ1|2 ξ 2
||ξξ||3 ξ 1 .

Hence it follows from (2.5.24) that

|ξ | ξ 2
|ξ | 1 ξ 1 364

γM ξ

0

(2.5.25)

and E







m 1 2 γmM |r |
|r| 2

 τ
1  m2 ||ξξ ||2
γmM |ξ | 12

.

(2.5.26)

For the next step, we assume that the r and hence also ξ assume values in the x, y-plane, only. Note that the discussion in the previous example indicates that it is sensible to search for such solutions. For the solutions of (2.5.25), we make the ansatz

 u2
v2 , ξ2  2uv , where u : J Ñ R, v : J Ñ R are twice differentiable functions that are to ξ1

be found. Then

|ξ |  u2 v2 , |ξ | 1  2uu 1 2vv 1 , ξ11  2uu 1
2vv 1 , ξ21  2u 1 v 2uv 1 , |ξ 1|2  p2uu 1
2vv 1q2 p2u 1v 2uv 1q2  4pu2 v2qpu 1 2 v 1 2 q ξ12  2uu 2
2vv 2 2u 1 2
2v 1 2 , ξ22  2u 2v 4u 1 v 1 2uv 2 . Substitution of the ansatz into (2.5.25) leads to v 2 qp2uu 2
2vv 2 2u 1 2
2v 1 2 q
p2uu 1 2vv 1 qp2uu 1
2vv 1 q γM pu2
v 2 q  2pu2 v 2 qpuu 2
vv 2 q 2pu2 v 2 qpu 1 2
v 1 2 q
4pu2u 1 2
v2 v 1 2 q γM pu2
v2q

pu2





 2pu2 v2qpuu 2
vv 2q γM
2pu 1 2 v 1 2 q pu2
v2 q  0 2pu2 v 2 qpu 2 v 2u 1 v 1 uv 2 q
4puu 1 vv 1 qpu 1 v uv 1 q 2 γM uv    2pu2 v2qpu 2v uv 2 q 4 pu2 v2 qu 1v 1
puu 1 vv 1qpu 1v uv 1q 2 γM uv

 2pu2

v 2 qpu 2 v

uv 2 q



γM


2pu 1 2



v 1 2 q 2uv

0

and hence to 2pu2

v 2 qupuu 2
vv 2 q



γM


2pu 1 2

365



v 1 2 q upu2
v 2 q

v 2 qv pu 2v

uv 2 q






2pu 1 2 v 1 2 q 2uv 2  0  (   pu2 v2q 2pu2 v2qu 2  γM
2pu 1 2 v 12q u  0 γM
2pu 1 2 v 1 2 q 2u2 v 2pu2 v 2 qupu 2 v uv 2 q  
2pu2 v2 qvpuu 2
vv 2q
γM
2pu 1 2 v 1 2q vp(u2
v2 q  pu2 v2q 2pu2 v2qv 2 γM
2pu 1 2 v 1 2 q v  0 2pu2

γM

and hence to 2pu2

2pu2



v 2 qu 2


2pu 1 2  γM
2pu 1 2

v 2 qv 2

γM



v 1 2q u , 

v 1 2q v .

Substitution of the ansatz into (2.5.26) leads to E 2m

12

 uu2

v12 v2


2puγM 2 v2q

which leads to the system of equations u2


E E u  0 , v2
v 2m 2m

0.

The solution of the last equations are given by Theorem 1.4.16. In the following we define the arc length of a curve as a limit of the lengths of inscribed polygons. The length of any such polygon should be smaller than the length of the path, since intuitively we expect straight lines to be the shortest connection between two points. This suggests the following definition. Definition 2.5.55. Let n P N and u : I Ñ Rn be a path where I is some non-empty closed subinterval of R. We say u that is rectifiable, nonrectifiable if the set #




ν¸1



|uptj q
uptj 1q| : P  pt0 , . . . , tν q P P

µ 0

366

+

y 0.2

0.2

0.4

0.6

1

x

-0.2 -0.4

Fig. 116: Graph of the non-rectifiable continuous path u from Example 2.5.56.

is bounded or unbounded, respectively. In case u is rectifiable, we define its length Lpuq by Lpuq  sup

#




ν¸1



|uptj q
uptj 1q| : P  pt0, . . . , tν q P P

+

.

µ 0

Example 2.5.56. (A non-rectifiable continuous path) Define u : r0, 1s Ñ R2 by 

 π uptq : t, p1
tq cos 2p1
tq

for all t P p0, 1s and uptq : p0, 0q. Then u is a continuous path. For every n P N , we define a partition pt0 , . . . , t2n 1 q of r0, 1s by t0 : 0 , t2k
1 : 1
for k

 1, . . . , n and

t2n

1 1 , t2k : 1
2p2k
1q 4k 1

: 1 .

367

Then 2n ¸



µ 0

¥ Hence

#


p
q





ν¸1



|upt2k
1q
upt2k q|

k 1

n  ¸  1   2 2k 1 k 1



n ¸

|uptj q
uptj 1q| ¥



n 1  1 ¸ 1 . ¥  4k  2 k1 k

|uptj q
uptj 1q| : P  pt0 , . . . , tν q P P

+

µ 0

is unbounded, and u is non-rectifiable. Theorem 2.5.57. (Uniform continuity) Let f : ra, bs Ñ R, where a, b P R are such that a   b, be continuous. Then f is uniformly continuous, i.e., for every ε ¡ 0 there is some δ ¡ 0 such that for all x, y P ra, bs it follows from |x
y | ¤ δ that |f pxq
f py q| ¤ ε.

Proof. The proof is indirect. Assuming the opposite, there is some ε ¡ 0 for which the statement is not true. Hence for every n P N , there are xn , yn P ra, bs such that |xn
yn | ¤ 1{n and at the same time such that |f pxn q
f pyn q| ¡ ε. According to the Bolzano-Weierstrass’ Theorem 1.2.11, there are subsequences xn1 , xn2 , . . . of x1 , x2 , . . . converging to some element x P ra, bs and ynk1 , ynk2 , . . . of yn1 , yn2 , . . . converging to some element y P ra, bs. Hence it follows by the continuity of the modulus function (see Example 1.2.43), the continuity of f , Theorem 1.2.4 and Theorem 1.2.6 that x  y and |f pxq
f py q| ¥ ε. Theorem 2.5.58. Let n P N , a, b P R be such that a ¤ b, u : ra, bs Ñ Rn be continuous and differentiable on pa, bq such that its derivative on pa, bq can be extended to a continuous path u 1 on ra, bs, such a path will be called a C1 -path in the following, then u is rectifiable and Lpuq 

»b a

|u 1ptq| dt .

368

Proof. For this, let ν by Theorem 1.5.20,

P N, pt0, . . . , tν q P P and µ P t1, . . . , ν
1u. Then n ¸

|uptµq
uptµ 1q|  2



|ukptµq
uk ptµ 1 q|  2

k 1

 n » tµ ¸    tµ



1

2 

 uk1 ptq dt . 

k 1

By Theorem 2.5.2,  2 » »  n n » tµ 1  tµ 1 ¸ tµ 1 ¸   uk1 t dt uk1 s ds uk1 t dt   tµ  tµ tµ k 1 k 1  » 2 1{2 » n  tµ 1  tµ 1 ¸   1  uk t dt u 1 t dt .   tµ  tµ

pq

¤



pq

pq



pq

| p q|



k 1

Hence

 n » tµ ¸    tµ

1



k 1

and

as well as

2   uk1 t dt 

pq

¤

» tµ

|uptµq
uptµ 1 q| ¤


ν¸1



tµ

» tµ

1

tµ

|uptµq
uptµ 1q| ¤

µ 1

2

1

»b a

|u 1ptq| dt

|u 1ptq| dt |u 1ptq| dt .

Hence u is rectifiable and Lpuq ¤

»b a

|u 1ptq| dt .

For the proof of the opposite inequality, let ε ¡ 0. Since u 1 is continuous, it follows by application of Theorem 2.5.57 to its component functions the existence of δ ¡ 0 such that for all s, t P ra, bs

|u 1psq
u 1ptq| ¤ ε 369

if |s
t| ¤ δ .

P N, pt0, . . . , tν q P P of size ¤ δ, µ P t1, . . . , ν
1u. Then |u 1ptq|  |u 1ptq
u 1ptµ q u 1ptµq| ¤ |u 1ptµq| ε for all t P rtµ , tµ 1 s. Hence Let ν

» tµ tµ



1

|u 1ptq| dt ¤ p|u 1ptµq|

»  tµ    tµ »  tµ    tµ

1

ru 1ptq    1 u t dt 

εq  l p rtµ , tµ

u 1 ptµ q
u 1 ptqs »  tµ    tµ

   dt 

1

sq

ε  l p rtµ , tµ

1

sq

   dt 

ru 1ptµq
u 1ptqs ε  l p rtµ , tµ 1 s q ? ¤ |uptµq
uptµ 1 q| p1 n q  l p rtµ, tµ 1 s q  ε ¤

1

pq

1

where integration of vector-valued functions is defined component-wise. Hence »b a

|u 1ptq| dt ¤

and, finally,




ν¸1



|uptµq
uptµ 1q| p1

?

n q  ε ¤ Lpuq

p1

?

nqε

µ 1

»b a

|u 1ptq| dt ¤ Lpuq .

Theorem 2.5.59. (Invariance of the length under reparametrizations) Let n P N , a, b P R be such that a ¤ b, u : ra, bs Ñ Rn be a C1 -path, c, d P R such that c ¤ d, g : rc, ds Ñ ra, bs be continuous, increasing (not necessarily strictly) such that g pcq  a, g pdq  b, differentiable on pc, dq with its derivative on pc, dq being extendible to a continuous function on rc, ds. Then u  g is a C1-path and Lpu  g q  Lpuq . 370

For this reason, we define the length LpRan uq of the curve Ran u by LpRan uq : Lpuq if u is in addition injective. Proof. First, u  g is continuous, differentiable on pc, dq with its derivative on pc, dq having the continuous extension g 1 .pu 1  g q. Hence u  g is a C1 -path, and it follows by Theorem 2.1.1 that Lpuq 



»d c

» gpdq

pq

g c

|u 1ptq| dt 

»d c

|pu 1  gqpsq| g 1psq ds

|g 1psq.pu 1  gqpsq| ds 

»d c

|pu  gq 1psq| ds  Lpu  gq .

Example 2.5.60. Calculate the length of the circle Sr1 p0q of radius r ¡ 0 around the origin. Solution: An injective parametrization of the part of Sr1 p0q in the upper half-plane is given by the C1 -path u : r0, π s Ñ R2 defined by upϕq : pr cospϕq, r sinpϕqq for every ϕ P r0, π s. Since Lpuq 

r it follows that

»π 0

»π 0

|u 1pϕq| dϕ 

»π

|p
r sinpϕq, r cospϕq| dϕ

0

dϕ  πr , L Sr1 p0q



 2πr .

Problems

371

Ñ Rn . upxq : px, 4x 7 q , x P I : r0, 1s , uptq : p2t3 , 3t2 q , t P I : r2, 5s , upxq : px, 2x4 p16x2 q
1 q , x P I : r1, 2s , upxq : px, x2{3 q , x P I : r2, 3s , upxq : px, 128px5 {15q p8x3 q
1 q , x P I : r1, 3s , upθq : pθ, ln cos θq , θ P I : rπ {8, π {4s , upsq : ps, cosh sq , s P I : r1, 8s , upθq : p2θ, cosp3θq, sinp3θqq , θ P I : r0, π s , ? uptq : pt2 {2, 2 t, lnptqq , t P I : r2, 7s , uptq : pt, cosh t, sinh tq , t P I : r0, 4s , upv q : p2v 3 , cos v v sin v, v cos v
sin v q , v P I : r0, π {2s , uptq : p2e
t , e
t sin t, e
t cos tq , t P I : r3, 4s .

1) Calculate the length of the path u : I a) b) c) d) e) f) g) h) i) j) k) l)

2) Calculate the length of the curve C. a)

C : tpx, y q P R2 : x2{3

y 2{3

 9 ^ 1 ¤ x ¤ 3u ,
3 ^ 0 ¤ y ¤ 2u , 2 3 2 C : tpx, y q P R : y
4x  0 ^ 3 ¤ x ¤ 4u , C : tpx, y q P R2 : 1 px4 {3q
xy  0 ^ 2 ¤ x ¤ 3u , C : tpx, y q P R2 : 8y 2  9px
1q2 ^ x ¥ 1 , 0 ¤ y ¤ 1u , ? C : tpx, y q P R2 : y
x p3
2xq  0 ^ 0 ¤ x ¤ 4u , C : tpx, y q P R2 : xy  px4 {2q 24 ^ 1 ¤ x ¤ 5u .

b) C : tpx, y q P R : x  2y { 2

c) d) e) f) g)

3 2

3) A cycloid is the trajectory of a point of a circle rolling along a straight line. Calculate the length of the part of the cycloid

tpapt
sin tq, ap1
cos tq : t P Ru between the points p0, 0q and p2πa, 0q, where a ¡ 0.

4) An astroid is the trajectory of a point on a circle of radius R{4 rolling on the inside of a circle of radius R ¡ 0. Calculate the length of the part of the astroid

tpR cos3 t, R sin3 tq : t P Ru between the points pR, 0q and p0, Rq. 372

y 1 0.5 Π

2Π

x

Fig. 117: A cycloid.

y 1

0.5

-1

-0.5

0.5

-0.5

-1

Fig. 118: An astroid.

373

1

x

y

1

1 €€€€€€ 2

1 €€€€€€ 2

1

x

3 €€€€€€ 2

1 - €€€€€€ 2 -1

Fig. 119: A cardioid. 5) A cardioid is the trajectory of a point on a circle rolling on the inside of a circle of the same radius. Calculate the length of the cardioid

tpa cos ϕp1 where a ¡ 0.

cos ϕq, a sin ϕp1

cos ϕqq : ϕ P r0, 2π qu ,

6) Consider all real-valued functions on [0,1] such that f p0q  1, f p1q  1 and that are continuously differentiable on the interval p0, 1q with a derivative that has a continuous extension to [0,1]. Find that function whose Graph is shortest. Give reasons for your answer. 7) Let b ¡ a ¡ 0. Consider all C 1 -paths u : [0,1] up0q  pa, 0q, up1q  pb, 0q and such that

Ñ R2 such that



uptq  rptq cos ϕptq , rptq sin ϕptq , for every t P [0,1], where r : [0,1] Ñ R and ϕ : [0,1] Ñ R are continuous, continuously differentiable on the interval p0, 1q with derivatives that have continuous extensions to [0,1]. Characterize the shortest paths. What is the common range of all these paths?

8) Let a, b, c, d P R such that a2 b2  1 and c2 d2  1. Consider all C 1 -paths u : [0,1] Ñ R2 on the sphere of radius 1 around the origin

374

such that up0q  pa, 0, bq, up1q  pc, 0, dq and such that



uptq  sinpθptqq cospϕptqq , sinpθptqq sinpϕptqq , cospθptqq ,

for every t P [0,1], where r : [0,1] Ñ R, θ : [0,1] Ñ R and ϕ : [0,1] Ñ R are continuous, continuously differentiable on the interval p0, 1q with derivatives that have continuous extensions to [0,1]. Characterize the shortest paths. What is the common range of all these paths?

3 Calculus III 3.1 Vector-valued Functions of Several Variables Definition 3.1.1. (Vector-valued functions of several variables) A vectorvalued function is a map from a non-trivial subset of Rn into Rm where n P N and m P N zt1u. A function of several variables is a map f from a non-trivial subset D of Rn into Rm for some n P N zt1u and m P N . A vector-valued function of several variables is a vector-valued function and/or a function of several variables. In accordance with Definitions 1.1.26, 1.1.29, for such a function, we define Definition 3.1.2.

(i) The domain of f by: Dpf q : D .

(ii) The range of f by: Ranpf q : tf pxq : x P D u (iii) The Graph of f by: Gpf q : tpx, f pxqq : x P D u . (iv) The level set (or contour) of f corresponding to some c P Rm by: f
1 pcq : tx P D : f pxq  cu . 375

y

12

-12

12

-12

Fig. 120: Range of γ1 .

Examples are: Example 3.1.3.

(i) γ1 : R Ñ R2 defined by γ1 ptq : pcosptq, sinptqq

for every t P R, (ii) γ2 : R Ñ R3 defined by γ2 ptq : pcosptq, sinptq, tq for every t P R, (iii) f3 : R2 zt0u Ñ R defined by f3 pxq : 1{|x| , for every x P R2 zt0u, 376

x

x 0

-1

1

10

z 5

0 -1

1

0

y

Fig. 121: Range of γ2 .

3 2

z 2 1

1

0 -2

0 -1 -1

0 x

1 2 -2

Fig. 122: Truncated graph of f3 .

377

y

2

y

1

0

-1

-2 -2

-1

0 x

1

2

Fig. 123: Contour map of f3 . Darker colors correspond to lower values of f3 .

(iv) f4 : R3 zt0u Ñ R defined by

f4 pxq : 1{|x|

for every x P R3 zt0u.

Example 3.1.4. Consider the expression a

for px, y q P R2 .

36
9x2
4y 2

(i) Evaluate the in the point px, y q ? ? value of this expression Solution: 36
9
16  11.

 p1, 2q.

(ii) Find and sketch the domain of the function g that is induced by the expression and has maximal domain. Solution: That domain is the subset of R2 consisting of all those px, y q P R2 for which the square root in the definition is defined. Hence it is given by those px, y q P R2 satisfying px{2q2 py{3q2 ¤ 1 . 378

y

2

1

-1

x

1 -1

-2

Fig. 124: Dpg q

Geometrically, this set consists of the inner part plus boundary of an ellipse centered around the origin with half axes 2 and 3. (iii) Find the range of g. Solution: Ranpg q px, yq P Dpgq, we have:

 r0, 6s. (Proof: For every

0 ¤ 36
9x2
4y 2 and hence also

0¤

a

¤ 36

36
9x2
4y 2

¤6.

Therefore, Ranpg q € r0, 6s. In addition for every z 

g

1? 36
z 2 , 0 3



P r0, 6s,

z .

Hence it follows also that Ranpg q  r0, 6s and, finally, that Ranpg q  r0, 6s.) Analogous to the corresponding definition in Calculus I, the next defines continuity of a vector-valued function of several variables at a point by its property to commute with limits taken at that point. 379

y

0.5

-1

-0.5

0.5

1

x

-0.5

Fig. 125: Graph of f from Example 3.1.6.

Definition 3.1.5. Let f : D Ñ Rm be a vector-valued function of several variables and x P D. We say f is continuous in x if for every sequence x1 , x2 , . . . of elements in D from lim xν

ν

Ñ8

it follows that lim f pxν q  f

ν

Ñ8

x






lim xν

ν

Ñ8

r f pxqs .

Otherwise, we say f is discontinuous in x. Moreover, we say f is continuous if f is continuous in all points of its domain D. Otherwise, we say f is discontinuous. Example 3.1.6. Consider the function f : R Ñ R defined by f pxq : 380

x |x|

for every x P Rzt0u and and f p0q : 1. Then 1 lim nÑ8 n



1  0 and nlim
Ñ8 n



but lim f

n

Ñ8

1 n

 1 and



lim f

n

Ñ8



0,


n1




1 .

Hence f is discontinuous at the point 1. See Fig. 125. Example 3.1.7. Consider the function of several variables f5 : R2 zt0u R defined by x2
y 2 f5 pxq : 2 x y2

Ñ

for all x P R2 zt0u. Then

f5 px, 0q  1 , f5 p0, y q 
1 for all x, y P Rzt0u, and hence there is no extension of f5 to a continuous function defined on R2 . Note that for every real a f5 px, axq 

1
a2 1 a2

for all x P Rzt0u. Hence for every b P r
1, 1s, there is a real number a such that lim f5 px, axq  b . x

Ñ0,x0

Example 3.1.8. (Basic examples of continuous functions.) Let n P N . (i) Constant vector-valued functions on Rn are continuous as a consequence of Theorem 2.5.45.

381

2

1

2

y

1 0.5 z 0 -0.5 -1 -2

0

1 0

-1

y

-1 -1

0 x

-2

1

-2

2 -2

-1

0 x

1

2

Fig. 126: Graph and contour map of f5 . In the last, darker colors correspond to lower values.

(ii) For i P t1, . . . , nu, define the projection pi : Rn i-th component by pi pxq : xi

Ñ R of Rn onto the

for all x  px1 , . . . , xn q P Rn . Then pi is continuous as a consequence of Theorem 2.5.45.

In the case of functions of one real variable, one main application of continuity was the fact that continuous functions defined on bounded and closed intervals assume a maximum value and minimum value. Similar is true for continuous functions of several variables. Such functions assume a maximum value and minimum value on so called ‘compact’ subsets, defined below, of their domain. Again, as in the case of functions of one real variable, this property is a simple consequence of the Bolzano-Weierstrass theorem for sequences in Rn . The last is a simple consequence of its counterpart Theorem 1.2.11 for sequences of real numbers. Theorem 3.1.9. (Bolzano-Weierstrass) Let n P N and x1 , x2 , . . . be a bounded sequence in Rn , i.e., for which there is M ¡ 0 such that |xk | ¤ M for all k P N . Then there is a subsequence, i.e., a sequence xn1 , xn2 , . . . 382

that corresponds to a strictly increasing sequence n1 , n2 , . . . of non-zero natural numbers, which is convergent in Rn . Proof. Since x1 , x2 , . . . is bounded, the corresponding sequences of components x1k , x2k , . . . , k  1, . . . , n are also bounded. Therefore, as a consequence of an n-fold application of Theorem 1.2.11 and an application of Theorem 2.5.45, it follows the existence of a subsequence xn1 , xn2 , . . . , where n1 , n2 , . . . is a strictly increasing sequence of non-zero natural numbers, which is convergent in Rn . Definition 3.1.10. (Open, closed and compact subsets of Rn ) Let n P N . (i) A subset U of Rn is called open if for every x ball of some radius ε ¡ 0 around x’

P U there is an ‘open

Uε pxq : ty P Rn : |y
x|   εu

which is contained in U. In particular, φ , Rn , and every open ball of radius ε ¡ 0 around x P Rn is open. Obviously, arbitrary unions of open subsets of Rn and intersections of finitely many subsets of Rn are open. (ii) A subset A of Rn is called closed if its complement Rn zA is open. In particular, the so called ‘closed ball of radius ε ¡ 0 around x P Rn ’ Bε pxq : ty P Rn : |y
x| ¤ εu

and the sphere of radius ε centered at x Sε pxq : ty P Rn : |y
x|  εu are closed. As a consequence of the last remark in (i), arbitrary intersections of closed subsets of Rn and unions of finitely many closed subsets of Rn are closed. In particular, we define for every subset S of Rn its corresponding closure S¯ as the intersection of all closed subsets of Rn that contain S. Hence S¯ is the smallest closed subset of Rn that contains S. 383

(iii) A subset K of Rn is called compact if it is closed and bounded, i.e., it is closed and contained in some open ball UR p0q of some radius R ¡ 0 around the origin. Example 3.1.11. Let a, b P R be such that a ¤ b. Then the interval pa, bq is bounded and open. The interval [a, b] is compact. Theorem 3.1.12. Let n P Rn and S € Rn . Then the closure S¯ of S consists of all x P Rn for which there is a sequence x1 , x2 , . . . of elements of S that is convergent to x. ¯ there are two cases. In case that x P S, the constant Proof. If x P S, sequence x, x, . . . is a sequence in S that is converging to x. If x R S and U is some open subset of Rn that contains x, it follows that U also contains a point of S. Otherwise, it follows that S¯ z U

 S¯ X p Rn z U q

is a closed subset of Rn that contains S and hence that S¯  S¯ z U ¯ In particular by applying the previous to U1{ν pxq which implies that x R S.  for every ν P N , we obtain a sequence x1 , x2 , . . . of elements of S that is convergent to x by construction. On the other hand, if x P Rn is such that there is a sequence x1 , x2 , . . . of elements of S that is convergent to x and A is a closed subset Rn that contains S, it follows that x P A. Otherwise, x is contained in the open set Rn z A and there is ε ¡ 0 such that Uε pxq € Rn z A. Hence if ν P N is such that |xν
x|   ε, it follows that xν P Rn z A and hence that xν R S. As a consequence, x is also contained in S¯ which is the intersection of all closed subsets A of Rn that contain S. Example 3.1.13. Let a, b P R be such that a intervals pa, bq and [a, b] is given by [a, b].

384

¤ b. Then the closure of the

Theorem 3.1.14. (Existence of maxima and minima of continuous functions on compact subsets of Rn ) Let n P N , K € Rn a non-empty compact subset and f : K Ñ R be continuous. Then there are (not necessarily uniquely determined) xmin P K and xmax P K such that f pxmax q ¥ f pxq , f pxmin q ¤ f pxq for all x P K. Proof. For this, in a first step, we show that f is bounded and hence that sup f pK q exists. In the final step, we show that there is c P K such that f pcq  sup f pK q. For both, we use the Bolzano-Weierstrass theorem. The proof that f is bounded is indirect. Assume on the contrary that f is unbounded. Then there is a sequence x1 , x2 , . . . in K such that f pxn q ¡ n

(3.1.1)

for all n P N. Hence according to Theorem 3.1.9, there is a subsequence xk1 , xk2 , . . . of x1 , x2 , . . . converging to some element c P K. Note that the corresponding sequence f pxk1 q, f pxk2 q, . . . is not converging as a consequence of (3.1.1). But, since f is continuous, it follows that f pcq  lim f pxnk q . k

Ñ8

Hence f is bounded. Therefore, let M : sup f pK q. Then for every n P N there is a corresponding cn P K such that

|f pcnq
M |   n1 .

(3.1.2)

Again, according to Theorem 3.1.9, there is a subsequence ck1 , ck2 , . . . of c1 , c2 , . . . converging to some element c P K. Also, as consequence of (3.1.2), the corresponding sequence f pck1 q, f pck2 q, . . . is converging to M. Hence it follows by the continuity of f that f pcq  M and by the definition of M: f pcq  M ¥ f pxq 385

for all x P K. By applying the previous reasoning to the continuous function
f , it follows the existence of a c 1 P K such that


f pc 1q ¥
f pxq . Hence it follows that for all x P K.

f pc 1 q ¤ f pxq

In the case of functions of one real variable, it was shown that a continuous function f : [a, b] Ñ R, where a, b P R are such that a   b, which is differentiable on the open interval pa, bq assumes its extrema either in a critical point in pa, bq or in the points of a or b. Similar is true for functions of several variables. For this reason, we define the notion of inner points and of boundary points of subsets of Rn , n P N . Definition 3.1.15. (Inner points and boundary points of subsets of Rn ) Let n P N and S € Rn . (i) We call x P S an inner point of S if there is ε ¡ 0 such Uε pxq € S. In particular, we call the set of inner points of S the interior of S and denote this set by S  . Obviously, S  is the largest open set that is contained in S. (ii) We call x P Rn a boundary point of S if for every ε ¡ 0 the corresponding Uε pxq contains a point from S and a point from Rn z S. Hence a boundary point of S cannot be an inner point of S. We call the set of boundary points of S the boundary of S and denote this set by B S. Example 3.1.16. Let a, b P R be such that a ¤ b. Then the interior of the intervals pa, bq and [a, b] is given by pa, bq. The boundary of pa, bq and [a, b] is given by ta, bu. We note that the closure of pa, bq, i.e., [a, b], is the union of the interior of pa, bq, i.e., pa, bq, and the boundary of pa, bq, i.e., ta, bu. The last is true for every subset of Rn , n P N .

386

Theorem 3.1.17. (Decomposition of the closure of subsets of Rn ) Let n P N and S € Rn . Then S¯  S  Y B S . ¯ If x P B S, then for every ν Proof. First, we note that S  € S € S. there is xν P S such that |xν
x|   1{ν. Hence lim xν

ν

Ñ8

P N

x

¯ Hence we showed that S¯  S  Y B S. If x P S, ¯ then either and x P S.  x P S or x P B S. Otherwise, there is ε ¡ 0 such that Uε pxq is contained in Rn z S. In this case, there is no sequence x1 , x2 , . . . of elements of S that ¯ Hence if follows that S¯ € S  Y B S is convergent to x and hence x R S. and finally that S¯  S  Y B S. Definition 3.1.18. Let f1 : D1 Ñ Rm , f2 : D2 Ñ Rm be vector-valued functions of several variables such that D1 X D2  φ. Moreover, let a P R. We define pf1 f2 q : D1 X D2 Ñ Rm and a.f1 : D1 Ñ Rm by

pf1

f2 qpxq : f1 pxq

f2 pxq

for all x P D1 X D2 and

pa.f1 qpxq : a.f1 pxq for all x P D1 . Theorem 3.1.19. Let f1 : D1 Ñ Rm , f2 : D2 Ñ Rm be vector-valued functions of several variables and such that D1 X D2  φ. Moreover, let a P R. Then by Corollary 2.5.47: (i) If f1 and f2 are both continuous in x continuous in x, too.

P

D1

X D2, then f1

(ii) If f1 is continuous in x P D1 , then a.f1 is continuous in x, too. 387

f2 is

Definition 3.1.20. Let f1 : D1 Ñ R, f2 : D2 Ñ R be functions of several variables such that D1 X D2  φ. We define f1  f2 : D1 X D2 Ñ R by

pf1  f2qpxq : f1pxq  f2 pxq for all x P D1 X D2 . If moreover Ranpf1 q € R , we define 1{f1 : D1 Ñ R

by

p1{f1 qpxq : 1{f1 pxq

for all x P D1 .

Theorem 3.1.21. Let f1 : D1 Ñ R, f2 : D2 variables such that D1 X D2  φ. (i) If f1 and f2 are both continuous in x continuous in x, too.

Ñ R be functions of several P

D1

X D2, then f1  f2 is

(ii) If f1 is such that Ranpf1 q € R as well as continuous in x P D1 , then 1{f1 is continuous in x, too. Proof. For the proof of (i), let x1 , x2 , . . . be some sequence in D1 which converges to x. Then it follows for every ν P N that

X D2

|pf1  f2qpxν q
pf1  f2 qpxq|  |f1pxν qf2pxν q
f1pxqf2 pxq|  |f1pxν qf2pxν q
f1pxqf2 pxν q f1 pxqf2pxν q
f1 pxqf2pxq| ¤ |f1pxν q
f1pxq|  |f2pxν q| |f1pxq|  |f2pxν q
f2pxq| ¤ |f1pxν q
f1pxq|  |f2pxν q
f2pxq| |f1pxν q
f1pxq|  |f2pxq| |f1pxq|  |f2pxν q
f2pxq| and hence, obviously, that lim pf1  f2 qpxν q  pf1  f2 qpxq .

ν

Ñ8

For the proof of (ii), let x1 , x2 , . . . be some sequence in D1 which converges to x. Then it follows for every ν P N that

|p1{f1qpxν q
p1{f1qpxq|  |1{f1pxν q
1{f1 pxq| 388

 |f1pxν q
f1pxq|{r |f1pxν q|  |f1pxq| s and hence, obviously, that lim p1{f1 qpxν q  p1{f1 qpxq .

ν

Ñ8

Definition 3.1.22. Let f : Df Ñ Rm and g : Dg Ñ Rp be vector-valued functions of several variables and Dg be a subset of Rm . We define g  f : Dpg  f q Ñ Rp by Dpg  f q : tx P Df : f pxq P Dpg qu and for all x P Dpg  f q.

pg  f qpxq : gpf pxqq

Theorem 3.1.23. Let f : Df Ñ Rm , g : Dg Ñ Rp be vector-valued functions of several variables and Dg be a subset of Rm . Moreover, let x P Df , f pxq P Dg , f be continuous in x and g be continuous in f pxq. Then g  f is continuous in x. Proof. For this, let x1 , x2 , . . . be a sequence in Dpg  f q converging to x. Then f px1 q, f px2 q, . . . is a sequence in Dg . Moreover, since f is continuous in x, it follows that lim f pxν q  f pxq .

ν

Ñ8

Finally, since g is continuous in f pxq, it follows that lim pg  f qpxν q  lim g pf pxν qq  g pf pxqq  pg  f qpxq .

ν

Ñ8

ν

Ñ8

389

Example 3.1.24. In the following, we conclude that f5 is continuous. For this, we define the projections pi : R2 zt0u i-th component by pi pxq : xi

Ñ R of R2 zt0u onto the

for every x  px1 , x2 q P R2 zt0u. By Theorem 2.5.45, pi is continuous, i.e., continuous in every point of its domain R2 zt0u. We have the following representation of f5 : f5

 rp1  p1 pp
1q.p2q  p2 s  p1{rp1  p1

p2  p2 sq .

Hence the continuity of f5 follows by application of Theorems 3.1.19, 3.1.21. Note that another way of concluding the continuity of the second factor 1{rp1  p1

p2  p2 s

is by means of Theorems 3.1.19(i),3.1.21(i) and 3.1.23, using the continuity of the function pRzt0u Ñ R, x ÞÑ 1{xq known from calculus in one real variable.

Problems 1) Find the maximal domain Dpf q of f such that a) b) c) d) e) f) g) h) i)

f px, y q 

a

f px, y q  5

1
2x2
y 2 , a


p3x
2yq2 , f px, y q  p1{px 1qq p1{y 2 q , f px, y q  lnpx 3y q , f px, y q  arccosp2xq lnpxy q , a f px, y q  px2 y 2
3qp1
x2
y 2 q , a f px, y q  p x
y 2 q
1{2 , a ? ? y 2
z 1 , f px, y, z q  x
3 f px, y, z q  arccospxq arccospy q
arccosp1
z q , 390

j)

f px, y, z q  lnpxyz q , a

k) f px, y, z q  1
x2
2y 2
4z 2 , l) f px, y, z q  arcsinpx 3y
6z q for all px, y q P Dpf q or px, y, z q P Dpf q.

2) Find the maximal domain Dpg q, level sets and range of g such that a) b) c) d) e) f) g) h) i) j) k) l)

g px, y q  x
2y ,

g px, y q  2x3
3y ,

g px, y q  y {x2 ,

g px, y q  3x2
5y 2

g px, y q  4x

2

2y

2

g px, y q  px{3q
2y g px, y q  3{rpx

g px, y q 

a

2

7,


1 , 2

6,

1qpy
1qs ,

x
3y ,

g px, y q  px 3q{py
1q , g px, y, z q  x
5y 2z
3 , g px, y, z q  6x2

2y 2

g px, y, z q  x2
y 2

z2
3 ,

4z 2

9

for all px, y q P Dpg q or px, y, z q P Dpg q. In addition, for the cases a) - i), draw a contour map showing several curves and sketch G pg q.

3) Where is the function h : D Ñ R continuous and why? In particular, decide whether h is continuous or discontinuous at the origin p0, 0q. Give reasons. a)

x2 xy 2 for px, y q P D : R2 z t0u , x2 y 2 hp0q : 1 ,

hpx, y q :

3xy 2 for px, y q P D : R2 z t0u , x2 y 2 hp0q : 0 ,

b) hpx, y q :

c)

xy 2 for px, y q P D : R2 z t0u , x2 y 4 hp0q : 0 ,

hpx, y q :

391

d) hpx, y q :

e)

xy

x2 hp0q : 0 , x hpx, y q : 2 x hp0q : 1 ,

y2

y for px, y q P D : R2 z t0u , y2

hpx, y q :

y2

g) hpx, y q :

y

f)

x2 hp0q : 1 ,

for px, y q P D : R2 z t0u ,

x2 hp0q : 0 ,

y2

for px, y q P D : R2 z t0u ,

y2

for px, y q P D : R2 z t0u ,

x2 y 2 for px, y q P D : R2 z tpx,
xq : x P Ru , x3 y 3 hp0q : 0 ,

h) hpx, y q :

i)

hpx, y q :

x3 x2 hp0q : 0 ,

y3 for px, y q P D : R2 z tpx,
x2 q : x P Ru , y

j)

hpx, y q :

x3 x hp0q : 0 ,

y3 for px, y q P D : R2 z tpx,
xq : x P Ru , y

k) hpx, y q :

x2 y 2 for px, y q P D : R2 z t0u , y4

x4

hp0q : 0 , l)

hpx, y q :

hp0q : 0 , m)

x2 y 2 for px, y q P D : R2 z t0u , y2

x2

xy xz yz for px, y, z q P D : R3 z t0u , hpx, y, z q : a 2 2 2 x y z hp0q : 0 ,

n) hpx, y, z q : hp0q : 0 ,

o) hpx, y, z q :

x2

xyz for px, y, z q P D : R3 z t0u , y2 z 2 xy

x2

yz y2

z2

for px, y, z q P D : R3 z t0u ,

392

hp0q : 1{2 . 4) For every n P N , show that p | | : Rn

Ñ R, x ÞÑ |x| q is continuous. z t0u Ñ R by

P R. Further, define f : R |x|p |y|q for px, yq P D : R2 z t0u , f px, y q : 2 x
xy y 2 f p0q : 0 . 2

5) Let p, q

Find necessary and sufficient conditions on p and q such that f is continuous. 6) Sketch the subsets of Rn and determine whether they are bounded, unbounded, open, closed and compact. In addition, determine there interior, closure and boundary. a) The intervals I1 : p3, 4q , I2 : r1, 2q , I3 : p
1, 3s , I4 : r1, 3s ,

I5 : p1, 8q , I6 : r0, 8q , I7 : p
8,
3s , I8 : p
8, 1s , b) the 2-dimensional intervals I9 : tpx, y q :
1   x   4 ^ 0   y   1u , I10 : tpx, y q : 0 ¤ x ¤ 4 ^
3 ¤ y ¤
1u , I11 : tpx, y q : 0   x ¤ 4 ^
3 ¤ y I12 : tpx, y q : x ¡ 0 ^ y  
3u ,

¤
1u ,

I13 : tpx, y q : x ¥ 1 ^ y ¥ 4u , I14 : tpx, y q : x  
1 ^ y ¥
2u , c) the sets S1 : tpx, y q P R2 : xy

S2 : tpx, y q P R : 9x

¡ 1u ,

  36u , S3 : tpx, y q P R : x
y ¤ 1u , S4 : tpx, y q P R2 : 3x2 y 2 ¡ 3u , S5 : tpx, y q P R2 : x2 y 2 ¤ 5u , S6 : tpx, y q P R2 : 2px
1q2 y 2 ¤ 3u , S7 : tpx, y q P R2 : x y 2 ¤ 2u , 2 2

2

2

393

4y 2 2

S8 : tpx, y, z q P R3 : x2

2y 2

S9 : tpx, y, z q P R :
x 3

2

S10 : tpx, y, z q P R : 4x 3

2

S11 : tpx, y, z q P R3 : 9x2 S12 : tpx, y, z q P R3 : x2

3y

¤ 4u , 2z 2 ¤ 1u , z 2 ¡ 2u , 4z 2 ¥ 4u , z 2   9u .

z2 2

y2 3y 2 4y 2

7) Let n P N . a) Show that the union of any number of open subsets of Rn and the intersection of a finite number of open subsets of Rn are open. b) Show that the intersection of any number of closed subsets of Rn and the union of a finite number of closed subsets of Rn are closed. c) Give an example of an intersection of non-empty open subsets of R which is non-empty and closed. d) Give an example of a union of non-empty closed subsets of R which is open.

€ Rn , where n P N . Show that B S is closed.

8) Let S, T

a) b) Show that S is closed if and only if B S c) Show that B S  BpRn zS q. d) Show that S¯  T . e) Show that S Y T  S¯ Y T¯. f) Show that S X T € S¯ X T¯.

€ S.

g) Give an example that shows that in general S X T

 S¯ X T¯.

9) Let n, m P N and f : Rn Ñ Rm . Show that f is continuous if and only if f
1 pU q is open for every open subset U of Rn .

3.2 Derivatives of Vector-valued Functions of Several Variables In the following, we define derivatives of such functions as linear maps. The motivation for that definition can be taken from Taylor’s formula for

394

functions in one variable. For every twice continuously differentiable function f from some open interval I into R, we have f py q  f pxq

f 1 pxqpy
xq

1 2 f pz qpy
xq2 , 2

for all x, y P I where z is some element in the closed interval between x and y. Hence it follows for every sequence y1 , y2, . . . in Rztxu which is convergent to x that

|f pyν q
f pxq
f 1 pxqpy
xq|  0 . ν Ñ8 |yν
x| lim

In this note that the map from R to R which associates to every z value f 1 pxq  z is linear. Definition 3.2.1. (Linear maps) Let n, m say that λ is linear if λpx

yq  λpxq

P R the

P N and λ : Rn Ñ Rm . We

λpyq , λpαxq  αλpxq

for all x, y P Rn and α P R. Since in that case λpxq  λ



n ¸





xj enj



j 1

n ¸



xj λp

enj

q

j 1

n m ¸ ¸

 

Λij xj em i

i 1j 1

m n m where en1 , . . . , enn and em 1 , . . . , em denote the canonical basis of R and R , respectively, and for every i  1, . . . , m, j  1, . . . , n, Λij denotes the component of λpenj q in the direction of em i , such λ is determined by its n values on the canonical basis of R . On the other hand, obviously, if

pΛij qpi,jqPt1,...mut1,...nu is a given family of real numbers, then by λpxq :

m ¸ n ¸

 

i 1j 1

395

Λij xj em i

for all x P Rn , there is defined a linear map λ : Rn Ñ Rm . Interpreting the elements of Rn and Rm as column vectors and defining the m  n matrix Λ by  λ11     λ1n

        Λ :     λm1        

   

  

λmn

    

 λ1n x1 Æ    Æ   ÆÆ Æ   ÆÆ     Æ

     λmn xn

Æ Æ Æ , Æ

the last is equivalent to 

       λpxq : Λ  x       λm1 

    

λ11





where the multiplication sign denotes a particular case of matrix multiplication defined by

pΛ  xqi :

n ¸



Λij xj

j 1

for every every x P Rn and i  1, . . . , m. In this case, we call Λ the reprem sentation matrix of λ with respect to the bases en1 , . . . , enn and em 1 , . . . , em . Definition 3.2.2. A vector-valued function of several variables f from some open subset U of Rn into Rm is said to be differentiable in x P U if there is a linear map λ : Rn Ñ Rm such that for all sequences x1 , x2 , . . . in U ztxu which are convergent to x:

|f pxν q
f pxq
λpxν
xq|  0 . ν Ñ8 |xν
x| lim

Remark 3.2.3. Note that from the differentiability of f in some x also follows the continuity of f in x.

396

P U, it

Ñ R by: f6 px, y q : 2x2

Example 3.2.4. Define f6 : R2

y2

for all x, y P R. Then f6 is differentiable, in particular, at the point p1, 1q. This can be seen as follows: For x, y P Rzt1u, we calculate: f6 px, y q  2x2 y 2  f6 p1, 1q 2x2 y 2
3  f6p1, 1q 2rpx
1q2 2px
1qs py
1q2  f6p1, 1q 4px
1q 2py
1q 2px
1q2

2py
1q py
1q2 .

Hence f6 px, y q
f6 p1, 1q
4px
1q
2py
1q  2px
1q2

py
1q2

and

|f6px, yq
f6p1, 1q
4px
1q
2py
1q|  |px, yq
p1, 1q| 2 2px
1q py
1q2 ¤ 2|x
1| |y
1| . a px
1q2 py
1q2 Hence for every sequence x1 , x2 , . . . in R2 ztp1, 1qu which is convergent to p1, 1q: |f6pxν q
f6 p1, 1q
4pxν
1q
2pyν
1q|  0 . lim ν Ñ8 |xν
p1, 1q| As a consequence, a linear map λ : R2 Ñ R satisfying the conditions of Definition 3.2.2 is given by

λpxq : 4x

2y

for all x  px, y q P R2 . The plane (see Fig. 127) z x, y

 f6 p1, 1q

λpx
1, y
1q  4x

2y
3 ,

P R, is called the tangent plane of the Graph of f6 at the point p1, 1q. 397

y -2

-1

0

1 2

10 5 z 0 -5 -10 -15 -2

-1 0 1 2

x

Fig. 127: Graph of f6 together with its tangent plane at (1,1).

Definition 3.2.5. A vector-valued function of several variables f from some open subset U of Rn into Rm is said to be partially differentiable in the i-th coordinate, where i P t1, . . . , nu, at some x P U if for all j P t1, . . . , mu the corresponding real-valued function in one real variable fj px1 , . . . , xi
1 , , xi 1 , . . . , xn q is differentiable at xi in the sense of the Calculus I. In that case, we define:

Bf pxq : prf px , . . . , x , , x , . . . , x qs 1px q, . . . , 1 1 i
1 i 1 n i Bxi rfmpx1 , . . . , xi
1, , xi 1, . . . , xm qs 1pxiqq .

If f is partially differentiable at x in the i-th coordinate direction at every x P U, we call f partially differentiable in the i-th coordinate direction and denote by B f {B xi the map which associates to every x P U the corresponding pB f {B xiqpxq. Partial derivatives of f of higher order are defined recursively. If B f {B xi is partially differentiable in the j-th coordinate direction, where j P t1, . . . , nu, we denote the partial derivative of B f {B xi in 398

the j-th coordinate direction by

B2f Bxj Bxj

.

Such is called a partial derivative of f of second order. In the case j we set B2f : B2f . Bx2i BxiBxi Partial derivatives of f of higher order than 2 are defined accordingly.

Ñ R by f7 px, y q : x3 x2 y 3
2y 2

Example 3.2.6. Define f7 : R2

for all x, y

P R. Find

Bf7 p2, 1q Bx

and

Bf7 p2, 1q . By

Solution: We have f7 px, 1q  x3 for all x, y

x P R,

y

x2
2 and f7 p2, y q  8

P R. Hence it follows that Bf7 px, 1q  3x2 Bx

4y 3
2y 2

2x ,

Bf7 p2, yq  12y2
4y , By

P R, and, finally, that Bf7 p2, 1q  16 Bx

and

399

Bf7 p2, 1q  8 . By

 i,

Example 3.2.7. Define f : R3

Ñ R by

f px, y, z q : x2 y 3z

3x

4y

6z

5

P R. Find Bf px, y, zq , Bf px, y, zq and Bf px, y, zq Bx By Bz for all x, y, z P R. Solution: Since in partial differentiating with respect to for all x, y, z

one variable all other variables are held constant, we conclude that

Bf px, y, zq  2xy3z Bx Bf px, y, zq  x2 y3 Bz for all x, y, z P R.

3,

Bf px, y, zq  3x2 y2z By

4,

6,

Note that differently to a function that is differentiable in a point of its domain, a function that is partially differentiable in a point of its domain is not necessarily continuous in that point.

Ñ R by xy {px2 y 2 q f px, y q : 0

Example 3.2.8. Define f : R2 #

Then

if px, y q P R2 z t0u if px, y q  0 .

a{n2 a  nÑ8 p1 nÑ8 a2 q{n2 1 a2 for every a P R and hence f is discontinuous in the origin. On the other hand, lim f p1{n, a{nq  lim

h

lim

Ñ0,h0

and hence

f ph, 0q
f p0, 0q h

0,

h

lim

Ñ0,h0

f p0, hq
f p0, 0q h

Bf p0, 0q  Bf p0, 0q  0 . Bx By 400

0

1

0.5 0.5 1

0

y

z

0

-0.5 -1

0

-0.5

y

0 -1

x

-1

1 -1

-0.5

0 x

0.5

1

Fig. 128: Graph and contour map of f from Example 3.2.8. In the last, darker colors correspond to lower values.

There are two loose ends here. The notion of the approximating linear map in the definition of differentiability whose uniqueness we still don’t know, and the notion of partial differentiability whose usefulness has not been demonstrated yet. Both will be clarified by the next Theorem 3.2.9. Let f be a vector-valued function of several variables from some open subset U of Rn into Rm . Furthermore let f be differentiable in x and λ : Rn Ñ Rm be some linear map such that for all sequences x1 , x2 , . . . in U ztxu which are convergent to x:

|f pxν q
f pxq
λpxν
xq|  0 . ν Ñ8 |xν
x| lim

Then f is partially differentiable at x in the i-th coordinate with λpei q 

Bf pxq Bxi

for all i P t1, . . . , nu. In particular, it follows that λpyq  y1  for all y P Rn .

Bf pxq    Bx1 401

yn 

Bf pxq , Bxn

Proof. Let i P t1, . . . , nu and t1 , t2 , . . . be some null sequence in R . Then the sequence x1 , x2 , . . . , defined by xν : x

tν .ei

for all ν P N , converges to x. Also its members are contained in U for large enough ν. For such ν, it follows that

|f pxν q
f pxq
λpxν
xq| ν Ñ8 |xν
x| |f px1, . . . , xi
1 , xi tν , xi 1, . . . , xnq
f pxq
tν .λpei q|  νlim Ñ8 |tν |    f px1 , . . . , xi
1 , xi tν , xi 1 , . . . , xn q
f pxq   lim 
λpe q  0 lim

ν

Ñ8 

i

tν



Hence, since this is true for all null sequences t1 , t2 , . . . in R , the statements of the theorem follows. Definition 3.2.10. Let f be a vector-valued function of several variables f from some open subset U of Rn into Rm . In addition, let f be differentiable in x P U, and let λ be as in Definition 3.2.2. According to Theorem 3.2.9, λ is uniquely defined by the property stated in Definition 3.2.2. (i) We define the derivative f 1 pxq of f at x by f 1 pxq : λ . According to Theorem 3.2.9 λ is given by: f 1 pxqpyq  y1 

Bf pxq    Bx1 402

yn 

Bf pxq , Bxn

for all y P Rn . Note that if the elements of Rn , Rm are interpreted as column vectors, the representation matrix of f 1 pxq with respect to the canonical bases of Rn and Rm is given by  B f1

Bx1 pxq

       f 1 pxq       Bf pxq  Bx m 1

    

    

    

B f1 Bxn pxq

 ÆÆ  ÆÆ  Bf pxq Bx m n

where f1 , . . . , fm are the component functions of f .

Ñ Rm defined by p1 pyq : f pxq f 1 pxqpy
xq  f pxq py1
x1 q  BBxf pxq    pyn
xn q  BBxf pxq

(ii) We call the function p1 : Rn

1

n

for all y P Rn , the Taylor polynomial of f of total degree ¤ 1 at x. (iii) If f is in addition real-valued, we call the graph of p1 the tangent plane of Gpf q at the point x. Important special cases are: Example 3.2.11. (i) Let f be a vector-valued function of several variables from some open interval I in R into Rm which is differentiable at some t P I. Then

rf 1ptqsp1q  ppf1q 1ptq, . . . , pfmq 1ptqq where the derivatives on the right hand side are in the sense of Calculus I. (ii) Let f be a function of several variables from some open subset U of Rn into R which is differentiable at some point x P U. Then

rf 1 pxqspyq  rpy  ∇qf spxq 403

for all y P Rn , where

f B f B p∇f qpxq : Bx pxq, . . . , Bx pxq 1 n

and for every y P Rn .



rpy  ∇qf spxq : y  p∇f qpxq

Example 3.2.12. (Basic examples of differentiable functions) Let n, m P N . (i) Constant vector-valued functions on Rn are differentiable with zero derivative. (ii) Any linear map from Rn into Rm is differentiable and its derivative is given by the same linear map at any x P Rn . An important sufficient criterion for differentiability. Theorem 3.2.13. Let f be a function of several variables from some open subset U of Rn into R. Moreover let f be partially differentiable in all coordinates, and let those partial derivatives define continuous functions on U. Then f is differentiable. Proof. For x P U and y P U ztxu, it follows by the mean value theorem for functions of one real variable Theorem 1.4.6 that f pyq
f pxq  f py1, y2, . . . , ynq
f px1 , y2, . . . , ynq f px1 , y2 , . . . , yn q
f px1 , x2 , . . . , yn q

  

f px1 , x2 , . . . , xn
1 , yn q
f px1 , x2 , . . . , xn q 404

 BBxf pc1, y2, . . . , ynq  py1
x1q 1 Bf px , c , . . . , y q  py
x q n 2 2 Bx2 1 2    Bf px , x , . . . , x , c q  py
x q n
1 n n n Bx 1 2 n

where for each i P t1, . . . , nu the corresponding ci is some element of the closed interval between xi and yi . Hence

Bf pxq
  
py
x q  Bf pxq n n Bx1  Bxn   BBxf pc1, y2, . . . , ynq
BBxf pxq py1
x1 q 1 1  Bf px , c , . . . , y q
Bf pxq py
x q n 2 2 Bx2 1 2 Bx1    Bf px , x , . . . , x , c q
Bf pxq py
x q n
1 n n n Bx 1 2 Bx

f pyq
f pxq
py1
x1 q 

n

and

  f y

p

n



q
f pxq
py1
x1 q  BBxf1 pxq
  
pyn
xn q  BBxfn pxq |y
 x|  q
BB p q . . .   B f q
Bx pxq n

  f f  x  x x1 , c2 , . . . , yn x1 2   f   x x1 , x2 , . . . , xn
1 , cn

B p B B p Bn

405

,

1

0.5

1

y

1 z

0

0 -1 -1

0

-0.5

y

0 -1

x

-1

1 -1

-0.5

0 x

0.5

1

Fig. 129: Graph and contour map of f from Example 3.2.14. In the last, darker colors correspond to lower values.

Hence, obviously, by the continuity of the partial derivatives of f , it follows the differentiability of f in x and f 1 pxqpyq  y1 

Bf pxq    Bx1

yn 

Bf pxq , Bxn

for all y P Rn . Finally, since x was otherwise arbitrary, it follows the differentiability of f on U. Example 3.2.14. (A continuous and partially differentiable function that is not differentiable) Define f : R2 Ñ R by f px, y q :

#

p3x2y
y3q{px2 0

y2q

if px, y q P R2 z t0u if px, y q  0 .

As a consequence of Theorem 3.2.13, the restriction of f to R2 z t0u is differentiable. In addition, since

|3x2y
y3|  |y|  |3x2
y2| ¤ 3|y|px2 406

y 2q

for every px, y q Further,

P

R2 z t0u, it follows that f is continuous at the origin.

f ph, 0q
f p0, 0q h f p0, hq
f p0, 0q lim hÑ0,h0 h h

and hence

lim

Ñ0,h0

0,


h 
1  hÑlim 0,h0 h

Bf p0, 0q  0 , Bf p0, 0q 
1 . Bx By

We lead the assumption that f is differentiable in the origin to a contradiction. If f is differentiable in the origin, it follows by Theorem 3.2.9 that f 1 p0qphq  h1

Bf p0q Bx

h2

Bf p0q 
h 2 By

for every h  ph1 , h2 q P R2 . Hence it follows for every sequence h1 ph11 , h12q, h2  ph21 , h22q, . . . in R2 z t0u that is convergent to 0 that 

| f phν q hν2 | 1  3h21ν h2ν
h32ν lim  νlim  ν Ñ8 Ñ8 |hν |  |hν | |hν |2 4h21ν |h2ν | 0.  νlim Ñ8 |h |3



  h2ν 

ν

But in the case that h1ν : h2ν : 1{ν for all ν 4h21ν |h2ν | ν Ñ8 |hν |3 lim



?

P N, it follows that

2.

Hence f is not differentiable in the origin. Compare Fig. 129 which indicates that there is no tangential plane to Gpf q in the origin. Note that

Bf px, yq  8xy3 Bx px2 y2q2 for px, y q P R2 z t0u. Hence B f {B x is discontinuous in the origin, and

Theorem 3.2.13 is not applicable to f .

407

Definition 3.2.15. We say that a real-valued function defined on some open subset U of Rn is of class C p for some p P N , if it is partially differentiable to all orders up to p, inclusively, and if all those partial derivatives define continuous functions on U. This includes partial derivatives of the order zero, i.e., that function itself is continuous. A real-valued function defined on some open subset U of Rn is said to be C 8 if it is of class C p for all p P N . Remark 3.2.16. As a consequence of the previous definition, the Theorem 3.2.13 can be restated as saying that every real-valued function which is defined on some open subset of Rn and which is of class C 1 is also differentiable. Definition 3.2.17. (Gradient operator) Let n P N . We define for every real-valued function f which is defined as well as partially differentiable in all coordinate directions on some open subset U of Rn 

B f B f p∇f qpxq : Bx pxq, . . . , Bx pxq 1 n

(3.2.1)

for all x from its domain. We call the map ∇ which associates to every such f the corresponding ∇f , the gradient operator.

Ñ R by f8 px, y q : x3 x2 y 3
2y 2

Example 3.2.18. Define f8 : R2

for x, y

P R. Find the second partial derivatives of f8 .

Solution: We have for all x, y

P R:

Bf8 px, yq  3x2 2xy3 , Bf8 px, yq  3x2 y2
4y , Bx By 2 2 B f8 px, yq  6x 2y3 , B f8 px, yq  6x2 y
4 , Bx2 By2 B2f8 px, yq  6xy2 , B2f8 px, yq  6xy2 . BxBy ByBx 408

We notice that the second mixed partial derivatives in the last example were identical. This is true for a large class of functions. Theorem 3.2.19. (Schwarz, 1843 - 1921) Let f be some real-valued function on some open subset of Rn which is of class C 2 . Then

B 2f  B 2 f BxiBxj Bxj Bxi

(3.2.2)

P t1, . . . , nu. Proof. If i  j the statement is trivially satisfied. For i  j, x P U and sufficiently small hi , hj  0, it follows by the mean value theorem

for i, j

for functions of one variable Theorem 1.4.6 that there are si , ti in the open interval between xi and xi hi and sj , tj in the open interval between xj and xj hj such that 

f px

hj .ej hi .ei q
f px hi .ei q
f px hj .ej q
f pxq   f f B B  hi Bx px si.ei hj .ej q
Bx px si .eiq i i 2 f B  hihj  Bx Bx px si.ei sj .ej q j i   f px hi.ei hj .ej q
f px hj .ej q
f px  hi.ei q
f pxq  hj BBxf px tj .ej hi .eiq
BBxf px tj .ej q j j 2 B f  hihj  Bx Bx px ti.ei tj .ej q . i j

Since hi , hj are otherwise arbitrary, from this and the continuity of

B2f , B2f BxiBxj Bxj Bxi follows (3.2.2) and hence, finally, the theorem.

409

1 2

z

0 1 -1 -2

0

y

-1 -1

0 x

1 2 -2

Fig. 130: Graph of f from Example 3.2.20.

Example 3.2.20. Define f : R2 f px, y q :

#

Ñ R by

xy px2
y 2 q{px2 0

y 2 q if px, y q P R2 z t0u if px, y q  0 .

Then

Bf p0, yq  lim f ph, yq
f p0, yq  lim y h2
y2 
y , hÑ0,h0 hÑ0,h0 h2 Bx h y2 Bf px, 0q  lim f px, hq
f px, 0q  lim x x2
h2  x hÑ0,h0 hÑ0,h0 x2 By h h2 for all x, y

P R. Further, B2f p0, 0q  lim 1  Bf ph, 0q
Bf p0, 0q  1 , hÑ0,h0 h B y BxBy By  2 B f p0, 0q  lim 1 Bf p0, hq
Bf p0, 0q 
1 hÑ0,h0 h B x ByBx Bx 410

2

1

1

0

0

y

y

2

-1

-1

-2

-2 -2

-1

0 x

1

-2

2

-1

0 x

1

2

Fig. 131: Contour maps of B f {B x and B f {B y from Example 3.2.20. Darker colors correspond to lower values.

and hence

B2f p0, 0q  B2f p0, 0q . BxBy ByBx

Note that

B2f px, yq 
4xy3px2
3y2q Bx2 px2 y2q3 for all px, y q P R2 z t0u. The function that associates the right hand side of the last equation to every px, y q P R2 z t0u cannot be extended to a

continuous function on R2 . Hence f is not of class C 2 and Theorem 3.2.19 is not applicable to f . Definition 3.2.21. (Laplace operator, Laplace equation) Let n P N . We define for every real-valued function f which is defined on some open subset U of Rn and twice partially differentiable in every coordinate direction △f :

B2f Bx2i i1 n ¸

.

(3.2.3)

We call the map △ which associates to every such f the corresponding △f the Laplace operator. In particular, if such f is mapped into the zero 411

y 1

0

1 -1 5

0 z

y

0.5

0

-0.5 -5 1 -1

0

-1 -1 1

x

-0.5

0 x

0.5

1

Fig. 132: Graph and contour map of f from Example 3.2.22. In the last, darker colors correspond to lower values.

function defined on the domain of f , f is called a solution of the Laplace equation △f  0 . Note that the zero on the right hand denotes the zero function on the domain of f . Example 3.2.22. (A solution of Laplace’s equation) Define f : R2 zt0u Ñ R by x f px, y q : 2 x y2 for all px, y q P R2 zt0u. Then

Bf px, yq  x2 y2
2x2  y2
x2 , Bf px, yq 
2xy Bx px2 y2q2 px2 y2q2 By px2 y2q2 B2f px, yq 
2xpx2 y2q2
4xpy2
x2 qpx2 y2q Bx2 px2 y2q4 2 2 2 2 2
x2 , 
2xpx pyx2q
y2x2qp32y
2x q 
2x p3y x2 y 2 q3 412

,

B2f px, yq 
2xpx2 y2q2
8xy2px2 y2q By2 px2 y2q2 2 2 2 2 2 B2f px, yq 
2xpxpx2 y yq 2
q38xy 
2x pxx2
3y 
y 2 q3 Bx2 for all px, y q P R2 zt0u. Hence f is a solution of Laplace’s equation. The differentiation of vector-valued function of several variables follows rules analogous to the case known from Calculus I. So there is a sum rule, a rule for scalar multiples, a product rule, a quotient rule and a chain rule. Theorem 3.2.23. (Rules of differentiation) Let f, g be two differentiable vector-valued function of several variables from some open subset U of Rn into Rm and a P R. (i) Then f

g and a.f are differentiable and

pf gq 1pxq  f 1pxq for all x P U.

g 1 pxq , pa.f q 1 pxq  a.f 1 pxq

(ii) If f, g are both real-valued, then f  g is differentiable and

pf  gq 1pxq  f pxq.g 1 pxq

g pxq.f 1 pxq

for all x P U.

(iii) If f is real-valued and non-vanishing, then 1{f is differentiable and  1

1 f

for all x P U.

pxq 
rf p1xqs2 .f 1 pxq

Proof. For this, let x P U and x1 , x2 , . . . be some sequence in U ztxu which is convergent to x. Then:

|pf

g qpxν q
pf

g qpxq
pf 1 pxq |xν
x| 413

g 1 pxqqpxν


xq|

1 1 ¤ |f pxν q
f px|xq

fxp| xqpxν
xq| |gpxν q
gpx|xq

gxp|xqpxν
xq| ν

and

ν

|pa.f qpxν q
pa.f qpxq
ra.pf 1qpxqspxν
xq| |xν
x| 1  |a|  |f pxν q
f px|xq

fxp| xqpxν
xq| ν

and hence lim

ν

and

Ñ8

|pf

g qpxν q
pf

g qpxq
pf 1 pxq |xν
x|

g 1 pxqqpxν


xq|  0

|pa.f qpxν q
pa.f qpxq
ra.pf 1qpxqspxν
xq|  0 . ν Ñ8 |xν
x| lim

If f and g are real-valued, it follows that

|pf  gqpxν q
pf  gqpxq
pf pxq.g 1pxq gpxq.f 1 pxqqpxν
xq| |xν
x| 1 ¤ |f pxν q
f px|xq

fxp| xqpxν
xq|  |gpxq| ν 1 |f pxq|  |gpxν q
gpx|xq

gxp|xqpxν
xq| ν |f pxν q
f pxq|  |gpx q
gpxq| ν |x
x| ν

and hence that |pf  gqpxν q
pf  gqpxq
pf pxq.g 1 pxq lim ν Ñ8 |xν
x| 0

g pxq.f 1 pxqqpxν

If f is real-valued and non-vanishing, it follows that   1  f pxν q


f p1xq

1 r p qs2 .f pxqpxν |xν
x| 1 f x

414




xq

¤


xq|

 |f pxν q
f px|xq

fxp| xqpxν
xq| ν 2 |f pxν q
f pxq| |f pxν q|  |f pxq|2  |xν
x| 1

1 |f pxq|2

and hence that lim

ν

Ñ8

  1  f pxν q




1 f x

pq

1 f x



r p qs

 1 2 .f pxqpxν
xq

|xν
x|

0.

Hence, since f 1 pxq

g 1 pxq , a.f 1 pxq , f pxq.g 1 pxq

g pxq.f 1 pxq ,


rf p1xqs2 .f 1pxq are all linear maps, x1 , x2 , . . . and x the theorem follows.

P U were otherwise arbitrary, finally,

Theorem 3.2.24. (Chain rule) Let f : U Ñ Rm , g : V Ñ Rl be differentiable vector-valued functions of several variables defined on some open subsets U of Rn and V of Rm , respectively, and such that the domain of the composition g  f is non trivial. Then g  f is differentiable with

pg  f q 1  g 1pf pxqq  f 1 pxq for all x P Dpg  f q. Proof. For this, let x P Dpg  f q and x1 , x2 , . . . be some sequence in Dpg  f qztxu which is convergent to x. Then:

|pg  f qpxν q
pg  f qpxq
pg 1pf pxqq  f 1 pxqqpxν
xq| ¤ |xν
x| |gpf pxν qq
gpf pxqq
g 1pf pxqqpf pxν q
f pxqq| |xν
x| 415

|g 1pf pxqqpf pxν q
f pxq
f 1pxqpxν
xqq| |xν
x| and hence, obviously,

|pg  f qpxν q
pg  f qpxq
pg 1pf pxqq  f 1 pxqqpxν
xq| ν Ñ8 |xν
x| 0. lim

Hence, since

g 1 pf pxqq  f 1 pxq

is a linear map, x1 , x2 , . . . and x finally, the theorem follows.

P Dpg  f q were otherwise arbitrary,

Definition 3.2.25. Let n P N , f : Df Ñ R, g : Dg Ñ Rn be functions of several variables such that Df X Dg  φ. We define the product function f.g : Df X Dg Ñ R by

pf.gqpxq : pf pxq g1pxq, . . . , f pxq gnpxqq P Df X Dg , where g1, . . . , gn : Dg Ñ R are the component

for all x functions of g.

By application of Theorem 3.2.9, we get as a corollary the chain rule for partial derivatives. Corollary 3.2.26. Let f : U Ñ Rm , g : V Ñ Rl be differentiable vectorvalued functions of several variables defined on some open subsets U of Rn and V of Rm and such that the domain of the composition g  f is non trivial. Then for each x P Dpg  f q, i P t1, . . . , nu:

Bpg  f q pxq  Bf1 pxq. Bg pf pxqq    Bfm pxq. Bg pf pxqq . Bxi Bxi Bx1 Bxi Bxm

416

Proof. By Theorem 3.2.24 and Theorem 3.2.9, it follows that

Bpg  f q pxq  rpg  f q 1pxqspe q  rg 1pf pxqq  f 1pxqspe q i i Bxi 

f1 B fm B 1 1 1  g pf pxqqpf pxqpeiqq  g pf pxqq Bx pxq, . . . , Bx pxq i i f1 B g fm B g B B  Bx pxq. Bx pf pxqq    Bx pxq. Bx pf pxqq i 1 i m and hence the corollary. Example 3.2.27. (Polar coordinates) Let f : R2 Ñ R be differentiable. Calculate all partial derivatives of first order of f¯ : R2 Ñ R defined by f¯pr, ϕq : f pr cos ϕ, r sin ϕq for all pr, ϕq P R2 . Solution: We define g : R2

Ñ R2 by g pr, ϕq : pr cos ϕ, r sin ϕq for all pr, ϕq P R2 . Then g is differentiable and f¯  f  g. Hence we get by

Corollary 3.2.26 that

Bf¯pr, ϕq  cos ϕ Bf pr cos ϕ, r sin ϕq sin ϕ Bf pr cos ϕ, r sin ϕq Br Bx By ¯ Bf pr, ϕq 
r sin ϕ Bf pr cos ϕ, r sin ϕq r cos ϕ Bf pr cos ϕ, r sin ϕq Bϕ Bx By

for all pr, ϕq P R2 . Solving the previous system for the partial derivatives of f leads to the more useful formula

Bf pr cos ϕ, r sin ϕq  cos ϕ Bf¯pr, ϕq
sin ϕ Bf¯pr, ϕq Bx Br r Bϕ ¯ Bf pr cos ϕ, r sin ϕq  sin ϕ Bf pr, ϕq cos ϕ Bf¯pr, ϕq By Br r Bϕ

for all r

P R z t0u and ϕ P R.

417

Problems 1) Calculate the partial derivatives of f : D Ñ R, and in this way conclude the differentiability of f . In addition, calculate f 1 p1, 2q and the Taylor-polynomial of total degree ¤ 1 (‘Linearization’) at p1, 2q. a)

f px, y q : 4x3

2y 3
3xy for px, y q P D : R2 ,

c)

f px, y, z q : xy

e)

f px, y q : arccospx{

b) f px, y q : 8x3
5x2 y 2 yz

7y 3 for px, y q P D : R2 ,

xz for px, y, z q P D : R3 ,

d) f px, y q : xy for px, y q P D : tpx, y q P R2 : x ¡ 0u , f)

a

y 2 q for px, y q P D : R2 z t0u ,

x2

f px, y q : arctanpy {xq for px, y q P D : tpx, y q P R2 : x  0u ,


g) f px, y q : ln x

a



x2

y2

for px, y q P D : R2 z t0u ,

h) f px, y, z q : exyz for px, y, z q P D : R3 , i)

f px, y, z q : xyz for px, y, z q P D : tpx, y, z q P R2 : x ¡ 0u .

2) Find a function whose zero set coincides with the tangent plane to the surface at the point p. a) b) c) d) e) f) g) h) i)

S1 : tpx, y, z q P R3 : x

 3u , p  p1, 1, 1q , S2 : tpx, y, z q P R : xyz  2u , p  p1, 2, 1q , S3 : tpx, y, z q P R3 : x2 y 2  2u , p  p
1,
1, 1q , S4 : tpx, y, z q P R3 : x2 y 2
z  0u , p  p0, 0, 0q , S5 : tpx, y, z q P R3 : x2 y 2
z 2  1u , p  p1, 1,
1q , S6 : tpx, y, z q P R3 : x2 y 2 z 2  3u , p  p1, 1, 1q , S7 : tpx, y, z q P R3 : x2 y 2
2z 2 
1u , p  p1, 1, 1q , S8 : tpx, y, z q P R3 : x2 3xy 2 4z
1  0u , p  p1, 2,
3q , S9 : tpx, y, z q P R3 : sinpxyz q  1{2u , p  p1, π, 1{6q , y

z

3

3) Use the chain rule to calculate

Bg Bu p2, 1q

and

418

Bg Bv p2, 1q ,

where

g pu, v q : f



?uv , 1 ln
u

, 2 v for all u, v ¡ 0 and f is a differentiable function such that

for all x, y

P R.

Bf px, yq  y , Bf px, yq  x , Bx By

4) Let f be a differentiable function with partial derivatives

Bf px, yq  x , Bf px, yq 
y Bx By

P R. Define g pϕ, θq : f pcos ϕ  p2 cos θq, sin ϕ  p2 for all ϕ, θ P R. By using the chain rule, calculate Bg pϕ, θq Bϕ for all ϕ, θ P R.

for all x, y

5) Use the chain rule to calculate

Bg p1, π{6q Br

where

and

cos θqq

Bg p1, π{6q , Bϕ

g pr, ϕq : f pr cos ϕ, r sin ϕq

for all r, ϕ P R, and f is a differentiable function such that

6)

Bf px, yq  3x2
y , Bf px, yq  3y2
x Bx By for all x, y P R. Let f : R3 Ñ R be a differentiable function satisfying Bf px, y, z q  x , Bf px, y, z q  y , Bf px, y, z q 
z Bx By Bz for all x, y, z P R. Define the function g by g pr, θ, ϕq : f pr sin θ cos ϕ, r sin θ sin ϕ, r cos θq for all r, θ, ϕ P R. By using the chain rule, calculate Bg p1, π{4, 0q . Bθ 419

7) Let U be a non-empty subset of Rn , where n P N . Further, let f, g : U Ñ R be partially differentiable, I a non-empty open interval of R such that I  Ranpf q, h : I Ñ R differentiable and a P R. Show that a) b) c) d) e)

∇pf g q  ∇f ∇g , ∇pa .f q  a .∇f , ∇pf g q  f .∇g g .∇f , ∇f k  k f k
1 .∇f , k P N , ∇pf {g q  p1{g q .∇f
pf {g 2 q .∇g , if f
1 pt0uq  φ , ∇ph  f q  ph 1  f q .∇f .

8) Let f : R Ñ R and g : R Ñ R be twice differentiable functions. Define upt, xq : f px
tq g px tq for all pt, xq P R2 . Calculate all partial derivatives of u up to second order and conclude that u satisfies B2 u
B2u  0 . Bt2 Bx2 The last is called the wave equation in one space dimension (for a function u which is to be determined). 9) Determine whether f is a solution of Laplace’s equation. If applicable, a, b P R. a)

f px, y q : a ex cospy q

b ex sinpy q for px, y q P D : R2 ,

c)

f px, y q : p1{2q lnpx2

y 2 q for px, y q P D : R2 z t0u ,

b) f px, y q : x3
3xy 2 for px, y q P D : R2 ,

d) f px, y q : x sinpx e)

f px, y q : arctan



yq

y cospx

x y 1
xy



for px, y q P D : tpx, y q P R2 : xy

 1u , f) f px, y, z q : e for px, y, z q P D : R3 , g) f px, y, z q : a e
5x sinp3y q cosp4z q b e
5x cosp3y q cosp4z q for px, y, z q P D : R3 , h) f px, y, z q : x3
2xy 2
xz 2 for px, y, z q P D : R3 , a i) f px, y, z q : 1{ x2 y 2 z 2 for px, y, z q P D : R3 z t0u . Let f, g : R Ñ R be differentiable, but otherwise arbitrary. Define u : R  R3 Ñ R by 1 upt, xq : |x| r f p t
|x| q gp t |x| q s xyz

10)

y q for px, y q P D : R2 ,

420

for all t P R  pR3 z t0uq. Calculate all partial derivatives of u up to second order and verify that u satisfies

B2 u
△u  0 , B t2

(3.2.4)

where

p△uqpt, xq : r△upt, qspxq for all pt, xq P R  R3 . The equation (3.2.4) is called the wave

equation in three space dimensions (for a function u which is to be determined).

11) (Transformation of the Laplace operator into polar coordinates) Define g : R2 Ñ R2 by g pr, ϕq : pr cos ϕ, r sin ϕq for all pr, ϕq P R2 . Further, let u : R2

Ñ R be of class C 2 . Then 2 2 p△uqpgpr, ϕqq  BBru¯2 pr, ϕq 1r BBur¯ pr, ϕq r12 BBϕu¯2 pr, ϕq for all r P R , ϕ P R, where u ¯ : u  g. 12) (Transformation of the Laplace operator into cylindrical coordinates) Define g : R3 Ñ R3 by g pr, ϕ, z q : pr cos ϕ, r sin ϕ, z q for all pr, ϕ, z q P R3 . Further, let u : R3

Ñ R be of class C 2 . Then 2 p△uqpgpr, ϕ, z qq  BBru¯2 pr, ϕ, z q 1r BBur¯ pr, ϕ, z q 1 B 2 u¯ B2u¯ pr, ϕ, z q p r, ϕ, z q r2 B ϕ2 Bz 2

for all r P R , pϕ, z q P R2 , where u ¯ : u  g.

13) (Transformation of the Laplace operator into spherical coordinates) Define g : R3 Ñ R3 by g pr, θ, ϕq : pr sin θ cos ϕ, r sin θ sin ϕ, r cos θq for all pr, ϕ, z q P R3 . Further, let u : R3

Ñ R be of class C 2 . Then 2 p△uqpgpr, ϕ, z qq  BBru¯2 pr, ϕ, z q 2r BBur¯ pr, ϕ, z q 421

2

1 0.5 2

0

y

z

0 1

-0.5 -2

0

-1

y

-1 -1

0 x

-2

1

-2

2 -2

-1

0 x

1

2

Fig. 133: Graph and contour map of f from Problem 13. 1 2 r sin2 θ

for all r P R , θ

14)



B2 u¯ pr, ϕ, z q Bϕ2

B2 u¯ pr, ϕ, z q B θ2  Bu¯ pr, ϕ, z q sin θ cos θ Bθ sin2 θ

P R ztkπ : k P Zu and ϕ P R, where u¯ : u  g. Define f : R Ñ R by # x2 y {px4 y 2 q if px, y q P R2 z t0u f px, y q : 0 if px, y q  0 . 2

a) Show that f is discontinuous at the origin. b) Show that f is partially differentiable at the origin into every direction, i.e., f ph.uq
f p0q lim hÑ0,h0 h exists for every pu1 , u2 q P R2 such that u21

15) As in Example 3.2.14, define f : R f px, y q :

#

2

p3x2 y
y3 q{px2

0

u22

 1.

Ñ R by y 2 q if px, y q P R2 z t0u if px, y q  0 .

Show that f partially differentiable at the origin into every direction, i.e., f ph.uq
f p0q lim hÑ0,h0 h exists for every pu1 , u2 q P R2 such that u21

422

u22

 1.

1

0.5

z

1

y

1 0

0 -1 -1

0

-0.5

y

0 -1

x

-1

1 -1

-0.5

0 x

0.5

1

Fig. 134: Graph and contour map of f from Problem 16. 16) Define f : R2

Ñ R by

f px, y q :

#

x3 {px2 0

y 2 q if px, y q P R2 z t0u if px, y q  0 .

a) Show that f is not differentiable in the origin. b) Show that f  g is differentiable for any differentiable path in R2 that passes through the origin.

3.3 Applications of Differentiation Definition 3.3.1. (Directional derivatives) A function of several variables f defined on some open subset U of Rn is said to be differentiable in the direction of some unit vector u P Rn at some x P U if the auxiliary function h : I Ñ R, defined by hptq : f px t.uq

for every t P I and some open interval I around 0, is differentiable at 0 in the sense of Calculus I. In this case, we define:

Bf pxq : h 1 p0q . Bu

Theorem 3.3.2. Let n P N , f be some differentiable function defined on some open subset U of Rn and u P Rn be some unit vector. Then f is 423

differentiable in the direction of u at all points of U and

Bf pxq  p∇f qpxq  u  cospαq |p∇f qpxq| Bu for all x P U where α denotes the angle between p∇f qpxq and u. Proof. For this, let x P U and define the auxiliary function g : I Ñ Rn by g ptq : x t.u for every t P I and some open interval I of R around 0 such that Ranpg q € U. According to Example 3.2.12, g is differentiable, and moreover according to Example 3.2.11 its derivative is given by

rg 1ptqsp1q  u for all t P I. Hence it follows by Theorem 3.2.24 that f  g is differentiable and pf  gq 1ptq  rf 1 pgptqqs u  p∇f qpxq  u where Example 3.2.11(ii) has been used. Hence the theorem follows. Remark 3.3.3. (Interpretation of the gradient) If the gradient vector p∇f qpx0q of f in x0 P U is non vanishing, then (i)

|p∇f qpx0q|
1.p∇f qpx0 q

and


|p∇f qpx0q|
1.p∇f qpx0q

are the directions of steepest ascent and steepest descent of f at x0 , respectively. The rate of the ascent and descent is given by

|p∇f qpx0q|

and

respectively.

424


|p∇f qpx0q| ,

(ii) Moreover, p∇f qpx0 q is perpendicular to the level set (or contour) of f at x0 . Hence the equation of the tangent plane to this set is given by p∇f qpx0 q  px
x0 q  0 and its normal line through x0 by x0

λ.p∇f qpx0 q

where λ P R. The next important step will be the derivation of Taylor’s formula in several variables. For this, we need the following lemma. Lemma 3.3.4. Let f be a real-valued function defined on some open subset U of Rn and of class C r for some r P N . In addition, let x P U, h P Rn and I be some open interval of R around 0 such that x t.h P U for all t P I. Finally, define g : I Ñ U by g ptq : x t.h for all t P I. Then

pf  gqprq  rph  ∇q r f s  g . The proof proceeds by induction. For r 

Proof. 1, it follows by Remark 3.2.16 that f is differentiable. Moreover, obviously, g is differentiable. Hence by Example 3.2.11 and the chain rule,Theorem 3.2.24, it follows that

pf  gq 1ptq  f 1pgptqq prg 1ptqsp1qq  f 1 pgptqqphq  rph  ∇qf spgptqq for all t P I and hence the statement for r  1. Now assume that the statement is valid for some s P N such that 1 ¤ s ¤ r
1. Then it follows by Remark 3.2.16 that ph  ∇q s f is differentiable and by the analogous arguments applied in the first step that

pf  gqps 1q  rph  ∇q s

425

1

fs  g .

Theorem 3.3.5. (Taylor’s formula) Let f be a real-valued function defined on some open subset U of Rn and of class C r for some r P N . Moreover, let x0 P U and h P Rn such that the x0 t.h P U for all t P r0, 1s. Then there is τ P p0, 1q such that f px0

hq  f px0 q

1 rph  ∇qf spx0q 1!

...

(3.3.1)

1 rp h  ∇q pr
1q f spx0 q rph  ∇q r f spx0 τ.hq . pr
1q! r! Proof. First, since x0 t.h P U for all t P r0, 1s and U is an open subset of Rn , there is some open interval I from R containing r0, 1s and such that x0 t.h P U for all t P I. Hence we can define g : I Ñ U by g ptq : x0 t.h for all t P I. Since f is of class C r , it follows by Lemma 3.3.4 that the real-valued function f  g of one variable is r times continuously 1

differentiable. Hence it follows by Taylor’s formula for functions of one variable, Theorem 1.4.23, that there is some τ P p0, 1q such that

pf  gqp1q  pf  gqp0q 1 pf r!

1 pf 1!

 gq 1p0q    pr
1 1q! pf  gqpr
1qp0q

 gqprqpτ q .

Finally, from this follows (3.3.1) by application of Lemma 3.3.4. Theorem 3.3.6. (Estimate of the remainder in Taylor’s formula) Let f, U, r, x0 , h be as in Theorem 3.3.5. Moreover, let K be a bound for all partial derivatives of f on U of order r. Finally, define the remainder term by 1 Rr px0 hq : rph  ∇q r f spx0 τ.hq . r! Then there is a number C P N depending only on r and n such that

|Rr px0

hq| ¤

426

CK r |h| . r!

(3.3.2)

Proof. Obviously, ph  ∇q r f is of the form ¸

ph  ∇q r f  i1



r    hin B i x B. . f. B i 1

ci1 ...in hi11



n

n

1

in r

xn

where the numbers ci1 ...in come from a multinomial expansion. Hence

|ph  ∇q r f | ¤ CK |h|r where

¸

C : i1





ci1 ...in

in r

depends only on n and r. Hence it follows (3.3.2). Definition 3.3.7. Let f, U, r, x0 , h and τ be as in Theorem 3.3.2. Then we call the function pr
1 : Rn Ñ R defined by pr
1 pxq :f px0 q

1 rppx
x0 q  ∇qf spx0q 1!

pr
1q! rppx
x0 q  ∇q 1

...

pr
1q f spx

0

q

for all x P Rn , the Taylor polynomial of f of total degree ¤ r
1 at x0 and Rr pxq :

1 rppx
x0 q  ∇q r f spx0 r!

its remainder term at x  x0

τ.px
x0 qq

h.

Example 3.3.8. Define f9 : R2 zt0u Ñ R by f9 px, y q :

xy x2

y2

for all px, y q P R2 zt0u. Calculate the Taylor polynomial of f9 of total degree ¤ 2 at p1, 1q, and give an estimate of its remainder term at the point p1.1, 1.2q. 427

Solution: Obviously, f9 is of class C 8 on R2 zt0u. As a consequence of Schwarz’s Theorem 3.2.19 and the symmetry of f9 under exchange of coordinates, there is only 1 ‘independent’ first order partial derivative as well as 2 independent second order and third order partial derivatives of f9 , respectively. In particular,

Bf9 px, yq  y py2
x2 q , B2f9 px, yq 
x4 6x2y2
y4 , Bx px2 y2q2 BxBy px2 y2q3 B2f9 px, yq 
2xy p3y2
x2 q , B3f9 px, yq  2x5
28x3 y2 18xy4 Bx2 px2 y2q3 Bx2By px2 y2q4 B3f9 px, yq 
6y x4
6x2 y2 y4 , (3.3.3) Bx3 px2 y2q4 for all px, y q P R2 zt0u. Hence we have for x0 : p1, 1q, small enough h P R2 and some τ P r0, 1s: 1 1 f9 px0 hq  f9 px0 q rp h  ∇qf9 spx0 q rph  ∇q 2f9 spx0q 1! 2! 1 rph  ∇q 3 f9spx0 τ.hq 3!  f9px0 q BBfx9 px0q hx BBfy9 px0q hy  2 f9 1 B 2f9 B 2 p x0 q hx 2 2 2 Bx BxBy px0 q hxhy  12
14 phx
hy q2 R3 px0 hq , where  1 B 3f9 R3 px0 hq  p x0 τ.hq h3x 3 6 Bx 3 B f9 px τ.hq h h2 B3f9 px 3 x y BxBy2 0 By3 0

B2f9 px q h2  By2 0 y 3 f9 B 3 2 px0 Bx By  τ.hq h3y .

R3 px0

hq

τ.hq h2x hy

Hence the Taylor polynomial p2 of f9 of total degree ¤ 2 at p1, 1q is given by 1 1 1 1 p2 px, y q 
ppx
1q
py
1qq2 
px
y q2 2 4 2 4 428

for all px, y q P R2 . Further for px, y q on the line segment between x0 and x1 : p1.1, 1.2q, it follows that  3  f9   x3 x, y







B p q  
6y x4
6x2y2 y4  ¤ 6|y| x4 6x2 y2 y4   B px2 y2q4  px2 y2q4 4 2 2 4 ¤ 6  1.2 p1.1q 6p1.1q16p1.2q p1.2q  6.29645    3  4 2 2 4 4    B f9 x
6x y y 6x2 y 2 y 4   
6x   ¤ 6|x| x p x, y q    By3 px2 y2q4  px2 y2q4 4 2 2 4 ¤ 6  1.1 p1.1q 6p1.1q16p1.2q p1.2q  5.77174  3   5  5  B f9   2x
28x3 y 2 18xy 4  28|x|3y 2 18|x|y 4    ¤ 2|x| p x, y q  B x2 B y    px2 y2q4 px2 y2q4 5 3 1.2q2 18  1.1p1.2q4 ¤ 2p1.1q 28p1.1q p16  6.12151  3    5 5  B f9   2y
28y 3x2 18yx4  28|y |3x2 18|y |x4   ¤ 2|y |  p x, y q  B xB y 2    px2 y2q4 px2 y2q4 5 3 1.1q2 18  1.2p1.1q4 ¤ 2p1.2q 28p1.2q p16  5.94662 and hence that 

|R3 px0 hq| ¤ 16 p0.1q3  6.29645 3  p0.1q2  0.2  6.12151  3  0.1  p0.2q2  5.94662 p0.2q3  5.77174 ¤ 0.03 . Definition 3.3.9. (Local minima and maxima) Let n P N and f be some real-valued function defined on some open subset U of Rn . Then we say that f has a local minimum, maximum at x0 P U if there is a open ball Uǫ px0 q around x0 such that f pxq ¥ f px0 q 429

for all x P Uǫ px0 q and

f pxq ¤ f px0 q

for all x P Uǫ px0 q, respectively.

Theorem 3.3.10. (Necessary condition for the existence of a local minimum/maximum) Let n P N , and let f be a differentiable real-valued function on some open subset U of Rn which has a local minimum/maximum at x0 P U. Then Bf px q  0 , i P t1, . . . , nu , Bxi 0 or, equivalently, x0 is a critical point for f , i.e., f 1 px0 q  0 . Proof. If f has a local minimum (maximum) at x0 i P t1, . . . , nu and sufficiently small h P R that

P

U, it follows for

1 f px01 , . . . , x0pi
1q , x0i h, x0pi 1q , . . . , x0n q h 
f px01 , . . . , x0pi
1q , x0i, x0pi 1q , . . . , x0nq is ¥ p¤q 0 and ¤ p¥q 0, for h ¡ 0 and h   0, respectively. Therefore

Bf px q Bxi 0

is at the same time ¥ 0 and ¤ 0 and hence, finally, equal to 0. Example 3.3.11. Define the differentiable function f10 : R2 f10 px, y q : x2
y 2 for all px, y q P R2 . Then

Bf10 px, yq  2x , Bf10 px, yq 
2y Bx By 430

Ñ R by

2 1

z 1 0

0.5

-1 -1

0

y

-0.5 -0.5

0 x

0.5 1 -1

Fig. 135: Graph of f10 .

for all px, y q P R2 and hence p0, 0q is a critical point of f10 , but, obviously, not a local minimum or maximum. It is a so called ‘saddle point’. Note that graph of f is a quadric, namely a hyperbolic paraboloid. Since hyperbolic paraboloids look similar to saddles, these are also often called ‘saddle surfaces’. Lemma 3.3.12. (Sylvester’s criterion) Let n P N , A  pAij qi,j Pt1,...,nu be a real symmetric n  n matrix,i.e., such that Aij  Aji for all i, j P t1, . . . , nu. Then A is positive definite, i.e., ¸



Aij hi hj

¡0

i,j 1,...,n

for all h P Rn zt0u, if and only if all leading principal minors detpAk q, k  1, . . . , n, of A are ¡ 0. Here Ak : pAij qi,j Pt1,...,ku , k

P t1, . . . , nu .

Proof. See the proof of Theorem 4.3.8 in the appendix. 431

Example 3.3.13. For the real symmetric matrix 



1 2 0  2 5 3 A : 0 3 11 it follows that detpA1 q  detp1q  1 ¡ 0 , detpA2 q  detpA3 q 

         



1 2  15
221¡0 2 5  1 2 0 2 5 3 0 3 11

and hence that for all h P R ¸



     

 1  5  11
3  3  1
11  2  2  2 ¡ 0 ¸



Aij hi hj

i,j 1,2,3

zt0u. Note that this can also be seen directly from Aij hi hj  h1 ph1 2h2 q h2 p2h1 5h2 3h3 q h3 p3h2 3

i,j 1,2,3

 h21 ¡0

¡0

4h1 h2

5h22

11h23

6h2 h3

 ph1

2h2 q2

ph2

3h3 q2

11h3 q 2h23

for all h P R3 zt0u. Theorem 3.3.14. (Sufficient condition for the existence of a local minimum/maximum) Let n P N and f be a real-valued function on some open subset U of Rn which is of class C 2 . Finally, let x0 be a critical point for f . Then f has a local minimum/maximum in x0 if all leading principal minors of its Hessian matrix at x0 

2 B f H px0 q : BxiBxj px0 q i,jPt1,...,nu are ¡ 0/all leading principal minors of
H px0 q are ¡ 0.

432

Proof. First, since x0 is a critical point of f , we conclude by Taylor’s formula, Theorem 3.3.5, (together with Theorem 3.2.19) that for every h from some a sufficiently small ball Uε p0q, ε ¡ 0, around the origin f px0

hq  f px0 q

1 rp h  ∇q 2 f spx0 2

1 ¸ pH px0 2 i,j 1,...,n

 f px0q

τ.hq

τ.hqqij hi hj

(3.3.4)

where τ P r0, 1s. Now, if all leading principal minors of H px0 q are ¡ 0/all leading principal minors of
H px0q are ¡ 0, and since all leading principal minors of the Hessian of f define continuous functions on U, ε can be chosen such that also all leading principal minors of the Hessian of f are ¡ 0/all leading principal minors of
H px0q are ¡ 0 at all points from Uε px0 q. Hence it follows from (3.3.4) and by Lemma 3.3.12 that f px0

hq ¥ f px0 q pf px0

hq ¤ f px0 qq

for all h P Uε p0q and hence, finally, the theorem.

Example 3.3.15. The function f9 : R2 Ñ R, see Example 3.3.8, is of class C 2 and has a critical point in p1, 1q. By (3.3.3), the negative of the Hessian matrix of f9 in p1, 1q is given by 



1{2
1{2


1{2 .
H p1, 1q  1{2 Hence the principal sub-determinants of
H p1, 1q are given by 1{2 and 0,

and hence Theorem 3.3.14 is not applicable. Nevertheless, f9 has even a global maximum at p1, 1q because f9 p1, 1q  1{2 and xy x2

y2

¤

1 x2  2 x2

y2 y2

 12

for every px, y q P R2 zt0u. This example demonstrates that the assumptions of Theorem3.3.14 for the existence of local minimum/maximum are not necessary. 433

0.5 0.45 z 0.4 0.35 0.3

1.4 1.2 1

0.6 0.8

0.8

1 x

y

1.2

0.6 1.4

Fig. 136: Graph of f9 .

The next example shows how to combine Theorem 3.1.14 and Theorem 3.3.10 to find global minima/maxima of functions of several variables. Example 3.3.16. Find the length, width and height of a parallelepiped of given area A ¡ 0 and maximal volume. Solution: The volume V and area A of a rectangular box of length x width y ¡ 0 and height x ¡ 0 are given by: V

 xyz , A  2pxy

xz

¡ 0,

yz q ,

respectively. Hence if applicable, we have to find the global maximum of the function V : r0, 8q  r0, 8q Ñ R defined by V px, y q :



xy x

y

A 2


xy

for every px, y q P pr0, 8q  r0, 8qqzt0u, and V p0, 0q : 0 . 434



4

2

z

00

0 2

2

y

x 4

4

Fig. 137: Graph of the constraint surface for A  6.

Note for later application of Theorem 3.1.14 that, obviously, V is of class C 8 on p0, 8q  p0, 8q as well as continuous on DpV qzt0u. In addition because of 1 px y q2 1 xy ¤  px yq x y 2 x y 2 for every px, y q P DpV qzt0u, it follows the continuity of V in p0, 0q and hence, finally, the continuity of V . In addition, note that V px, y q ¤

1 A2 a 4 x2 y 2

(3.3.5)

for every px, y q P Dpf qzt0u. This is obvious for xy ¥ A{2 because in this case V ¤ 0, whereas for xy ¤ A{2, px, y q  p0, 0q it follows that A
 V px, y q  2

A 2


xy x



A2 4

y 435


x2y2 ¤ A2  4

1 x

y

1 0.5 z 0 -0.5 -1 0

2 1.5 1

y

0.5 0.5

1 x

1.5 2 0

Fig. 138: Graph of V for the case A  6. 2

¤ A4

a

1

x2

y2

.

In the next step, we determine the critical point of V on p0, 8q  p0, 8q. The partial derivatives of V are given by

BV px, yq  y2 A2
x2
2xy , BV px, yq  x2 A2
y2
2xy Bx px yq2 By px yq2 for x, y P p0, 8q. Hence the critical points of V on p0, 8q  p0, 8q are given by the solutions of the system x2

2xy


A 2

0,

y2

2xy


which has the unique solution x0 : y0 : 436

A . 6

A 2

0

In px0 , y0q the volume V assumes the value V px0 , y0q 



A 6

3{2

Now we define the subset C of R2 by C : Dpf q X BR p0q , R :

.

(3.3.6)

27 ? A. 4

Then C is in particular closed and bounded and hence compact. According to Theorem 3.1.14, V assumes a maximum value on C. Since V vanishes on both axes and because of (3.3.6) and (3.3.5), it follows that V does not assume its maximum on the boundary of C. Hence V assumes its maximum in the inner part of C and hence it follows by Theorem 3.3.10 and the previous analysis that this happens in the point px0 , y0 q. Finally again because of (3.3.5), it follows that V assumes its (global) maximum in px0 , y0q and that its maximum value is given by (3.3.6). Note that this implies that the box is a cube. In the following, we indicate a derivation of a necessary condition for the existence of constrained minima/maxima like the previous one. For this, let f be a real-valued function on some open subset U of Rn and of class C 1 , and let S : tx P U : f pxq  0u . (the constraint surface) be such that

p∇f qpxq  0 for every x P S. Finally, let g : U Ñ R be differentiable, and assume that the restriction g |S of g to S has a minimum/maximum in p P S. Then it follows for any differentiable path γ : I Ñ S through p, where I is some open interval around 0 and γ p0q  p, that g  γ has a minimum/maximum

in 0 and hence by Calculus I, the Chain Rule Theorem 3.2.24 and Example 3.2.11 that

pg  γ q 1p0q  p∇gqppq  γ 1 p0q  0 . 437

HÑfLHpL

p

S

Fig. 139: Sketch of the constraint surface S.

Hence alike p∇f qppq, p∇g qppq is orthogonal to the pn
1q
dimensional tangent space of f at p, and hence there is a so called ‘Lagrange multiplier’ λ P R such that p∇gqppq  λ.p∇f qppq . Example 3.3.17. For this, we consider again the situation from Example 3.3.16. The constraint surface S is given by the zero set of f : U Ñ R, where U : p0, 8q  p0, 8q  p0, 8q, defined by f px, y, z q : xy

xz

yz


A 2

for every px, y, z q P U. Note that

p∇f qpxq  py z, x z, x yq  0 for every px, y, z q P U. The function V : U Ñ R, defined by V px, y, z q : xyz 438

for every px, y, z q P U, is to be maximized on S. According to the previous analysis, there is a real λ such that

p∇V qpxq  pyz, xz, xy q  λ.p∇f qpxq  λ.py Hence it follows that λ  0 and 1 1 1 1 1 1    λ y z x z x and therefore that

xy

z, x

z, x

yq .

1 y

z

and, finally by using the constraint equation, that xy

z

A 6

which is identical to the result of Example 3.3.16. More generally, the following is true Theorem 3.3.18. (Lagrange multipliers) Let n, m P N and g, f1, . . . , fm be functions of class C 1 defined on some open subset U. Finally, assume that the restriction g |S of g to the constraint surface S, defined by S : tx P U : f1 pxq      fm pxq  0u , assumes a minimum/maximum value in p P S. Then there are ‘Lagrange multipliers’ λ0 , . . . , λm P R that are not all 0 and such that λ0 .p∇g qppq

λ1 .p∇f1 qppq

. . . λm .p∇fm qppq  0 .

Proof. First, we consider the case that g |S assumes a minimum value in p. For this, let ε0 ¡ 0 such that the closed ball Bε0 ppq is contained in U. In addition for every M ¡ 0, we define an auxiliary function hM : Uε0 ppq Ñ R of class C 1 by hM pxq : g pxq
g ppq

|x
p|2

M

m ¸



k 1

439

fk2 pxq

for every x P Uε0 ppq. In a first step, we conclude that for every 0   ε ¤ ε0 , there is M pεq ¡ 0 such that hM pεq pxq ¡ 0 for all x P Sε ppq. Otherwise, there is 0   ε M ¡ 0 such that hM pxq ¡ 0

for all x P Sε ppq. Hence for such ε and any N such that hN pxN q ¤ 0

¤ ε0 for which there is no P N, there is xN P Sεppq (3.3.7)

or, equivalently, such that m ¸



fk2 pxN q ¤


k 1

1  g pxN q
g ppq N



ε2 .

(3.3.8)

Therefore, as a consequence of the boundedness and closedness of Sε ppq and by application of Bolzano-Weierstrass’ Theorem 3.1.9, it follows the existence of a strictly increasing sequence N1 , N2 , . . . of non-zero natural numbers such that the corresponding sequence xN1 , xN2 , . . . is convergent to some x P Sε ppq. By performing the limit in (3.3.8), it follows that x belongs to the constraint surface S and hence that g px q ¥ g ppq. But, the last implies that hM px q ¥ ε2 for every M ¡ 0 and hence that (3.3.7) cannot be valid for every N P N . Hence for the second step, let 0   ε ¤ ε0 and M pεq ¡ 0 be such that hM pεq pxq ¡ 0 for all x P Sε ppq. Then there is xε λm pεqq P Rm 1 such that λ0 pεq. rp∇g qpxε q

P Uεppq and a unit vector pλ0pεq, . . . ,

2.pxε
pqs

m ¸



k 1

440

λk pεq.p∇fk qpxε q  0 .

This can be proved as follows. By Theorem 3.1.14, the restriction of hM pεq to Bε ppq assumes a minimum value in some point xε P Bε ppq. Since hM pεq ppqq  0, it follows that xε P Uε ppq and that p∇hM pεq qpxε q  0

p∇gqpxεq

2.pxε
pq

2M pεq.

m ¸



fk pxε q.p∇fk qpxε q  0

k 1

which implies the above statement. In the last step, we choose a sequence ε1 , ε2 , . . . in the open interval between 0 and ε0 s which is convergent to 0. In particular, we choose it such that the corresponding sequence

pλ0pε1q, . . . , λmpε1qq, pλ0pε2q, . . . , λmpε2qq, . . . is convergent to a unit vector pλ0 , . . . , λm q in Rm 1 . This is possible as a

consequence of Bolzano-Weierstrass’ Theorem 3.1.9. Since lim xεk

k

Ñ8

 xε ,

we conclude that λ0 .p∇g qppq

m ¸



λk .p∇fk qppq  0 .

k 1

Finally, if g |S assumes a maximum value in p, then
g |S assumes a minimum value in p and the statement of the theorem follows by application of the just proved result to
g |S . Example 3.3.19. Let n P N ,pakl qk,lPt1,...,nu be a family of real numbers and g : Rn Ñ R be defined by g pxq :

n ¸



akl xk xl

k,l 1

for all x P Rn . Since S1n p0q  f
1 p0q, where f : Rn f pxq : |x|

2


1 
1

n ¸



i 1

441

Ñ R is defined by

x2i

for all x P Rn , is compact, the restriction of f to S1n p0q assumes a minimum and a maximum. Let x a point where f assumes an extremum. Then it follows by Theorem 3.3.18 the existence of real λ0 , λ1 such that λ20 λ21  0 and λ0 p∇g qpxq λ1 p∇f qpxq  0 . Since

n ¸



akl xk xl



k,l 1

f pxq  for all x P Rn , it follows that

p∇gqpxq 

n ¸





pa1k



n ¸



k,l 1



alk xk xl

k,l 1



1 pakl 2

ak1 qxk , . . . ,

k 1

n ¸



alk xl xk

k,l 1

and hence



n ¸

n ¸



alk q xk xl

pank

akn qxk



, p∇f qpxq  2x ,

k 1

it follows that x satisfies the following system of equations n ¸

λ0



paik

aki qxk

2λ1 xi

0,

k 1

for i  1, . . . , n. Further, since x  0, it follows that λ0 that the last system is equivalent to n  ¸ 1

2



k 1

for i  1, . . . , n, where

paik



aki q xk

 0 and hence

 λ xi ,

λ :


λ1 . λ0 By introducing matrix notation, the last system is equivalent to 

            a ¯n1  a¯11

    

    









 a¯1n x1 x1 Æ   Æ   ÆÆ    Æ  Æ Æ  λ.  Æ   ÆÆ     Æ  Æ       a¯nn xn xn 442

(3.3.9)

where a ¯kl : pakl alk q{2 for k  1, . . . , n, l  1, . . . , n and the multiplication sign on the left hand side of the last equation denotes matrix multiplication. By Theorem 4.3.6 from the Appendix, it follows that λ satisfies a a¯12    a¯1n ¯11
λ a¯21 a ¯22
λ    a¯2n        0       a ¯n1 a ¯n2    a¯nn
λ which is leads on a polynomial equation for λ. After solving that equation and substitution of the calculated values for λ into (3.3.9), the solutions of the remaining system can be easily found.

Problems 1) Find the rate of change of f : D Ñ R at the point p in the direction of v. In addition, find the direction of steepest ascent / steepest descent of f in p and the associated rates. a) f px, y q : x2
2xy 3y 2 for all px, y q P D : R2 , p  p1, 2q, v  p2,
1q , b) f px, y q : y cospxy q for all px, y q P D : R2 , p  p0, 2q, v  pcospπ {3q, sinpπ {3qq , c) f px, y q : x expp
2px2 y 2 qq for all px, y q P D : R2 , p  p1, 0q, v  p1, 3q , d) f px, y q : lnpx2 y 2 q for all px, y q P D : R2 zt0u, p  p1, 1q, v  p
3, 3q , e) f px, y, z q : xy yz xz for all px, y, z q P D : R3 , p  p1, 2,
1q, v  p1, 1, 1q , f) f px, y, z q : 5x2
3xy xyz for all px, y, z q P D : R3 , p  p3, 4, 5q, v  p1, 1,
1q , g) f px, y, z q : xyz px{y q py {z q pz {xq for all x ¡ 0, y ¡ 0, z ¡ 0, p  p2, 1, 4q, v  p
1,
1,
1q .

2) Decide whether the matrix is symmetric and in case whether it is positive definite. A1 :



1 3 3 7



, A2 :



4 2

443



2 6

, A3 :



5
3




3 2

,

A4 :



2 4








4
1

3 , A5 :  1
1

4 2 A6 :  2 3
5 4




1
1 ,

1 3
1



5 3 4 , A7 :  3 6
1

5

3 3
1




1
1 . 9

3) Decide which values of k P R make the matrix positive definite. A1 : A4 :

 

6 1 9k k



k A6 : 
5 8



, A2 :

1 2k

k
k




5





3 k k 2 

, A3 :

10 2 , A5 :  2 9 5k 3







5k 3 , 7

k 4k



4k 1

,



8 6 k 2 12 4 , A7 : k 5 7k . 4 2k 2 7k 14

4) Calculate the Taylor polynomial of f : D Ñ R of total degree ¤ 2 at p, and estimate the corresponding remainder term on B.

a) f px, y q : sinpx y q for all px, y q P R2 , p  p0, 0q, B  tpx, yq P R2 : |x| ¤ 1 ^ |y| ¤ 1u , b) f px, y q : ex y for all px, y q P R2 , p  p0, 0q, B  tpx, y q P R2 : |x| ¤ 1 ^ |y | ¤ 1u , c) f px, y q : p1 x y q1{2 for all px, y q P R2 such that y ¥
p1 xq, p  p0, 0q, B  tpx, y q P R2 : |x| ¤ 1{2 ^ |y | ¤ 1{2u , d) f px, y q : xy for all x ¡ 0, y P R, p  p1, 1q, B  tpx, y q P R2 : |x
1| ¤ 1{10 ^ |y
1| ¤ 1{10u .

5) Find the maximum and minimum values, so far existent, of f : D R and the points where they are assumed. If applicable, a, b P R. 4y q
5x2
y 2 for px, y q P D : R2 , x b) f px, y q : xy 1

y for px, y q P D : R2 , 2 c) f px, y q : 2
2x 5x2 2y  p4 x 5y q

a)

f px, y q : xp2




for px, y q P D : R2 ,

d) e)

f px, y q : x3

f px, y q : x4

xy

y 3 for px, y q P D : R2 ,

y 4
2x2

444

4xy
2y 2

Ñ

for px, y q P D : R2 ,

pa3 {xq pa3 {yq for px, y q P D : tpx, y q P R2 : x ¡ 0 ^ y ¡ 0u , g) f px, y q : x3 y 3
9xy 27 for px, y q P D : tpx, y q P R2 : 0 ¤ x ¤ 4 ^ 0 ¤ y ¤ 4u , h) f px, y q : x4 y 4
2x2 4xy
2y 2 for px, y q P D : tpx, y q P R2 : 0 ¤ x ¤ 2 ^ 0 ¤ y ¤ 2u , i) f px, y q : e
px y q pax2 by 2 q for px, y q P D : R2 , where a, b ¡ 0 , j) f px, y, z q : xyz p4a
x
y
z q for px, y q P D : R3 , k) f px, y, z q : px3 y 3 z 3 q{pxyz q for px, y, z q P D : tpx, y, z q P R3 : x ¡ 0 ^ y ¡ 0 ^ z ¡ 0u , l) f px, y, z q : rx{py z qs ry {px z qs rz {px y qs for px, y, z q P D : tpx, y, z q P R3 : x ¡ 0 ^ y ¡ 0 ^ z ¡ 0u . Find the maximum and minimum values of g : D Ñ R on the set(s) f)

f px, y q : x2

xy

2

6)

y2

2

S and the points where they are assumed. Give reasons for the existence of such values. a) b) c) d) e) f) g)

f px, y q : x2

2xy

y 2 for px, y q P D : R2 ,

on S : tpx, y q P R2 : x2
2x

 0u , f px, y q : x 2y for px, y q P D : R2 , on S : tpx, y q P R2 : x4 y 4  1u , f px, y q : x2 y 2 for px, y q P D : R2 , on S : tpx, y q P R2 : 3  px2 y 2 q 2xy  1u , f px, y q : x2 xy y 2 for px, y q P D : R2 , on S : tpx, y q P R2 : x2 y 2  1u , f px, y q : xy for px, y q P D : R2 , on S : tpx, y q P R2 : x2 y 2  1u , f px, y, z q : xyz for px, y, z q P D : R3 , on S : tpx, y, z q P R3 : x2 y 2 z 2  3u , f px, y, z q : x2 2y 2 3z 2 for px, y, z q P D : R3 on S1 : tpx, y, z q P R3 : x2 y 2 z 2  1u , 2

y2

2

445

,

S2 : tpx, y, z q P R3 : x

 0u , y z for px, y, z q P D : R3 , h) f px, y, z q : x on S1 : tpx, y, z q P R3 : x y z  0u , S2 : tpx, y, z q P R3 : px2 y 2 z 2 q2  x2 2y 2 4z 2 u , i) f px, y, z q : sinpx{2q  sinpy {2q  sinpz {2q for px, y, z q P D : tpx, y, z q P R3 : x ¡ 0 ^ y ¡ 0 ^ z ¡ 0u , on S : tpx, y, z q P R3 : x y z  π u . Let p ¡ 0. Determine the triangle with largest circumscribed area 2

7)

2

2y

3z

2

and perimeter 2p.

8) Determine the point inside a quadrilateral V with minimal sum of squares of distances from the corners. 9) Determine the point inside a quadrilateral V with minimal sum of distances from the corners. 10) Determine the triangle with maximal sum of squares of side lenghts and corners on a circle. 11) Let p P R3 z t0u. Determine the plane of largest distance from the origin among all planes through p.

12) Let a ¡ b ¡ c ¡ 0 and E :

"

2 px, y, z q P R : xa2 3

y2 b2

z2 c2

1

*

.

Find the point of E that has largest distance from the origin.

3.4 The Riemann Integral in n-dimensions Definition 3.4.1. (i) Let a, b P R be such that a ¤ b and ra, bs be the corresponding closed interval in R. A partition P of ra, bs is an ordered sequence pa0, . . . , aν q of elements of ra, bs, where ν is an element of N, such that a  a0 ¤ a1 ¤    ¤ aν  b .

A partition P 1 of ra, bs is called a refinement of P if P is a subsequence of P 1 . 446

(ii) Let n P N . A closed interval I of Rn is the product of n closed intervals I1 , . . . , In of R:

 I1      In . We define the volume v pI q of I as the product of the lengths lpIi q of the intervals Ii , i P t1, . . . , nu v pI q : lpI1 q . . . lpIn q . A partition P of I is a sequence pP1 , . . . , Pn q consisting of partitions Pi of Ii , i P t1, . . . , nu. I

P induces a division of I into (in general non-disjoint!) closed subintervals I






ν¤ 1 1 j1

0




...

ν¤ n 1 jn

0

Ij1 ...jn ,

Ij1 ...jn : ra1j1 , a1pj1 1q s      ranjn , anpjn 1q s , j1  0, . . . , ν1 ; . . . ; jn  0, . . . , νn where Pi  pai1 , . . . , aiνi q, νi P N , i P t1, . . . , nu. The size of P is defined as the maximum of all the lengths of these subintervals. Hence, we define for any bounded real-valued function f on I the lower sum Lpf, P q and upper sum U pf, P q corresponding to P by Lpf, P q : U pf, P q :

ν1 ¸



j1 0 ν1 ¸



j1 0

... ...

νn ¸



jn 0 νn ¸



inf tf pxq : x P Ij1 ...jn u v pIj1 ...jn q , suptf pxq : x P Ij1 ...jn u v pIj1 ...jn q .

jn 0

447

Example 3.4.2. Consider the closed interval I : r0, 1s  r0, 1s in R2 and the continuous function f : I Ñ R defined by f px, y q : x for all px, yq P I. P0 : pp0, 1q, p0, 1qq , P1 : pp0, 1{2, 1q, p0, 1{2, 1qq are partitions of I. Also is P1 a refinement of P0 . Finally, Lpf, P q  0  1  0 , U pf, P q  1  1  1 ,  2

1 Lpf, P 1 q  0  2

 2

1 1 U pf, P 1 q   2

2

1 2



1

 2



1 2

1 2

1 2

1

 2

 2

1 2

0

1 2

1 2

 2



 2

1 2

 14 ,

1 2

 34

 2

and hence Lpf, P q ¤ Lpf, P 1 q ¤ U pf, P 1 q ¤ U pf, P q . Lemma 3.4.3. Let n P N and I  I1    In be a closed interval of Rn , Further, let P  pP1 , . . . , Pn q, P 1  pP11 , . . . , Pn1 q be partitions of I and in particular let P 1 be a refinement of P . Then Lpf, P q ¤ Lpf, P 1 q ¤ U pf, P 1 q ¤ U pf, P q .

(3.4.1)

Proof. The middle inequality is obvious from the definition of lower and upper sums given in Def 3.4.1(ii). Obviously for the proof of the remaining inequalities, it is sufficient (by the method of induction) to assume that there is i0 P t1, . . . , nu such that Pi1  Pi for i  i0 , for simplicity of notation 1 , a11 , . . . , a1ν q, where a 1 P I1 is we assume i0  1, and P11  pa10 , a11 1 11 such that 1 ¤a . a10 ¤ a11 11 Here we again simplified for notational reasons. Then Lpf, P 1 q
Lpf, P q  448

y 1

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1

x

Fig. 140: Sketch of P0 .

y 1

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

Fig. 141: Sketch of P1 .

449

1

x

ν2 ¸



j2 0

...

νn ¸



1 s      ra , a inf tf pxq : x P ra10 , a11 njn npjn 1q su

jn 0

 vpra10, a111 s      ranj , anpj 1q sq 1 , a s      ra inf tf pxq : x P ra11 11 nj 1  vpra11, a11 s      ranj , anpj 1q sq
inf tf pxq : x P ra10 , a11 s      ranj(  vpra10, a11 s      ranj , anpj 1q sq ¥ inf tf pxq : x P ra10 , a11s      ranj , anpj 1qsu  vpra10, a111 s      ranj , anpj 1q sq 1 , a s      ra , a v pra11 11 npj 1q sq nj (
vpra10, a11 s      ranj , anpj 1q sq  0 . n

n

n

n

n

n

n

n

n

n

n

, anpjn 1q su

n

, anpjn 1q su

n

n

n

n

Analogously, it follows that U pf, P 1 q
U pf, P q ¤ 0 and hence, finally, (3.4.1). Theorem 3.4.4. Let f be a bounded real-valued function on some closed interval I of Rn , n P N . Then supptLpf, P q : P

P P uq ¤ inf ptU pf, P q : P P P uq . (3.4.2) Proof. By Theorem 3.4.3, it follows for all P1 , P2 P P that Lpf, P1 q ¤ Lpf, P q ¤ U pf, P q ¤ U pf, P2 q , where P P P is some corresponding common refinement, and hence that supptLpf, P1 q : P1 P P uq ¤ U pf, P2 q and hence (3.4.2). As a consequence of Lemma 3.4.3 and since every partition P of some closed interval I of Rn is a refinement of the trivial partition containing only the coordinates of the initial and endpoints, we can make the following definition. 450

Definition 3.4.5. (The Riemann integral, I) Let n P N , f be a bounded real-valued function on some closed interval I of Rn , and denote by P the set consisting of all partitions of I. We say that f is Riemann-integrable on I if supptLpf, P q : P P P uq  inf ptU pf, P q : P P P uq . In that case, we define the integral of f on I by » I

f dv : supptLpf, P q : P

P P uq  inf ptU pf, P q : P P P uq .

Example 3.4.6. Let f be a constant function of value a P R on some closed interval I of Rn . Then it follows induction that all lower and upper sums are equal to a  v pI q. Hence f is Riemann-integrable and »

f dv I

 a  vpI q .

Example 3.4.7. Consider the closed interval I : r0, 1s2 of R2 and the function f : I Ñ R defined by f px, y q : x  y for all x, y

P R. For every n P N , define the partition Pn of I by Pn :



n 1 0, , . . . , n n



n 1 , 0, , . . . , n n



.

Calculate Lpf, Pn q and U pf, Pn q for all n P N . What is the value of »

f dv ? I

Solution: We have: I







j j 1 1





n¤1 n¤1

j1 0 j2 0

n

,

1 n

451







j2 j2 1 , n n



and





jj 1 2



2  n

n¸1 n¸1

L pf, Pn q 

j1 0 j2 0



2 1
n p n
1 q n4 2

U pf, Pn q 







pj1

n¸1 n¸1





 n4

pn 2

1q

Hence

2

 14



1


1 n

1q

1 n

1





1qpj2 n2

j1 0 j2 0

1
n



1 4



1 n2





1 n4






n¸1



j1

2

Ñ8

,

1 n2





1 n4

2

n

j2

j2 0

n ¸



j1 1



j1

n ¸



j2 1

.

lim L pf, Pn q  lim U pf, Pn q 

n



j1 0






n¸1

Ñ8

1 . 4

As a consequence

and

supptLpf, P q : P

P P uq ¥ 14

inf ptU pf, P q : P

P P uq ¤ 14

and hence by Theorem 3.4.4 supptLpf, P q : P

P P uq  inf ptU pf, P q : P P P uq  14 .

Hence f is Riemann-integrable and »

f dv I

Note that the product

» 1

 14 .

» 1

x dx



y dy

0

0

gives the same value. 452



j2

Theorem 3.4.8. Let n P N and f, g be Riemann-integrable on some closed interval I of Rn and a P R. Then f g and af are Riemann-integrable on I and »

I

pf

g q dv 

»

»

»

f dv

g dv ,

I

I

af dv I

a

»

f dv . I

If f is in addition positive, then »

f dv I

¥0.

Proof. First, it is easy to see that for every subinterval J of I inf tf pxq : x P J u inf tg pxq : x P J u ¤ inf tpf g qpxq : x P J u ¤ suptpf gqpxq : x P J u ¤ suptf pxq : x P J u suptg pxq : x P J u and hence that Lpf, P q Lpg, P q ¤ Lpf ¤ U pf, P q U pg, P q and

»

»

f dv I

g dv I

¤ Lpf

g, P q ¤ U pf

g, P q ¤ U pf

for every partition P of I. Hence supptLpf g, P q : P P P uq exist and satisfy inf ptU pf »

g, P q : P

»

f dv I

g, P q ¤

g, P q

»

g, P q : P

P P uq  supptLpf

»

f dv I

P P uq, inf ptU pf

g, P q : P

g dv . I

Further for every subinterval J of I, it follows that inf taf pxq : x P J u  a inf tf pxq : x P J u , 453

g dv I

P P uq 

suptaf pxq : x P J u  a suptf pxq : x P J u and hence that Lpaf, P q  aLpf, P q , U paf, P q  aU pf, P q for every partition P of I as well as that inf tf pxq : x P J u ¥ 0 and hence, if f is in addition positive, that Lpf, P q ¥ 0 for every partition P of I. Hence, finally, the Theorem follows. Corollary 3.4.9. Let n P N , f, g be Riemann-integrable on some closed interval I of Rn , and in addition let f pxq ¤ g pxq for all x P I. Then »

f dv I

¤

»

g dv . I

Definition 3.4.10. (Negligible sets) A subset K of Rn is said to be negligible if for every ε ¡ 0 there exists a finite number of closed intervals I1 , . . . , Iν of Rn whose union contains K and which is such that v pI1 q



v pIν q   ε .

Example 3.4.11. Any interval of Rn with at least one side of vanishing length is negligible. Remark 3.4.12. Obviously, negligible subsets are bounded, the closure of negligible subsets is negligible and a finite unions of negligible subsets are also negligible. Theorem 3.4.13. Let m, n P N , B be a bounded subset of Rm and U an open subset of Rm containing B. Finally, let n ¡ m and f : U Ñ Rn be of class C 1 . Then f pB q is negligible. 454

Proof. See [33], XX, §2, Proposition 2.2. Example 3.4.14. Sr1 p0q  tpx, y q P Rn : x2

y2

 ru ,

where r ¥ 0, is a negligible subset of R2 . Proof: For this, define f : p
2π, 2πq Ñ R2 by f ptq : pr cosptq, r sinptqq for every t P p
2π, 2π q. Then f is of class C 1 and Ranpf q  Sr1 p0q. Hence according to Theorem 3.4.13, Sr1 p0q is a negligible subset of R2 . Theorem 3.4.15. (Existence of Riemann integrals) Let n P N . (i) Let f be a bounded real-valued function on some closed interval I of Rn . Moreover, let f be continuous in all points of I except from points in negligible subset of I. Then f is Riemann-integrable on I. (ii) If g is some function on I such that f pxq  g pxq for all x P I except from points in negligible subset of I, then g is Riemann-integrable on I and » » f dv  g dv . I

I

Proof. See [33], XX, §1, Theorem 1.3. Example 3.4.16. Let f : tpx, y q P R2 : x2 y 2 ¤ r u Ñ R be some continuous function, where r ¥ 0. Define fˆ : r
r, r s2 Ñ R by fˆpx, y q :

"

f px, y q for px, y q P Dpf q 0 for px, y q P r
r, r s2 z Dpf q.

Then fˆ is everywhere continuous except possibly on Sr1 p0q which is according to Example 3.4.14 a negligible subset of R2 . Hence according to Theorem 3.4.15, fˆ is Riemann-integrable on r
r, r s2. 455

Theorem 3.4.17. (Fubini’s Theorem for Riemann integrals) Let m, n P N and I, J be closed intervals of Rm and Rn , respectively. Further, let f : I  J Ñ R be Riemann-integrable on I  J. Finally, let f px, q be Riemann-integrable on J for all x P I, except on a negligible subset of I. Then the function on I which associates to every x P I the value »

J

f px, yq dy

is Riemann-integrable on I and »

f px, yq dx dy 

» »



I J

I



J

f px, yq dy dx .

Proof. See [33], XX, §3, Theorem 3.1. Example 3.4.18. Let r

¡ 0. Define f : r
r, rs2 Ñ R by f px, y q : 1

if x2 y 2 ¤ r and 0 otherwise. According to Example 3.4.16, f is Riemannintegrable on r
r, r s2, and we conclude by Theorem 3.4.15 that »

f px, y q dx dy 

r
r,rs2 

 

» ?r2
x2

»r 


r

?

?




r2




x r 2
x2

x2



» r » r


r


r

dy dx  2



f px, y q dy dx

?

»r


r

r 2 arcsinpx{r q

r 2
x2 dx

r


r

 πr2

which is the area of a circular disk of radius r. Note that this result can be achieved only by knowledge of the values of f on Br2 p0q. Hence it appears natural to define »

pq

Br2 0

dx dy :

»

r
r,rs2 456

f px, y q dx dy

since f is the unique extension of the constant function of value 1 on Br2 p0q to a function on r
r, r s2 which is constant of value zero on r
r, r s2 zB12 p0q. Also it is obvious that if g is an analogous extension of the constant function of value 1 on Br2 p0q to some interval I  Br2 p0q, it follows that »

I

g px, y q dx dy



»

r
r,rs2

f px, y q dx dy

as it should be since the symbol »

pq

dx dy

B12 0

does not contain any reference to an interval or an extension of the integrand. Definition 3.4.19. ( The Riemann integral, II ) Let n P N , Ω be a bounded subset of Rn whose boundary is negligible and f be a bounded function on Ω. In addition, let I  Ω be an interval, and let fˆ : I Ñ R defined by fˆpxq :

#

f pxq if x P Ω 0 if x P I z Ω .

be Riemann integrable. Then we define » Ω

f dv :

»

fˆ dv . I

For the proof that this definition is independent of the interval I, we refer to the final part of [33], XX, § 1 on admissible sets and functions. In addition, as a particular case when f is constant of value 1, we define the n-dimensional volume V of Ω by V :

»

dv . Ω

457

Theorem 3.4.20. (Change of variable formula) Let n P N and I be a closed interval of Rn contained in some open subset U. Moreover, let g : U Ñ Rn be continuously differentiable with a continuously differentiable inverse. Finally, let f be a Riemann-integrable function over g pI q. Then »

»

pq

f dv

g I

where det g 1 : U

 pf  gq | det g 1| dv , I

Ñ R is defined by

pdet g 1 qpxq : det



Bgi pxq

Bxj i,j1,...,n



for all x P U. Proof. See [33], XX, §4, Corollary 4.6. Example 3.4.21. (Polar coordinates) Define g : p0, 8q  p
π, π q by g pr, ϕq : pr cos ϕ, r sin ϕq

Ñ R2

for all pr, ϕq P p0, 8qp
π, π q. Then g is continuously differentiable (with Ranpg q  R2 zp
8, 0s  t0u) with a continuously differentiable inverse g
1 : R2 zp
8, 0s  t0u Ñ R2 given by g
1 px, y q 

" a 2 ax

p p

x2

a

y 2 , arccospx{ a x2 y 2 qq if y ¥ 0 2 2 2 y ,
arccospx{ x y qq if y   0

for all px, y q P Ranpg q  R2 zp
8, 0s  t0u. In particular, 

 pdet g 1 qpr, ϕq :  cos ϕ

sin ϕ

for all pr, ϕq P p0, 8q  p
π, π q.

458




r sin ϕ   r r cos ϕ 

Example 3.4.22. (Cylindrical coordinates) Define g : p0, 8qp
π, π q R Ñ R3 by g pr, ϕ, z q : pr cos ϕ, r sin ϕ, z q for all pr, ϕ, z q P p0, 8q  p
π, π q  R.

Then g is continuously differentiable (with Ranpg q  R3 zp
8, 0s  t0u  R) with a continuously differentiable inverse g
1 : R3 zp
8, 0st0u R Ñ R3 given by a

" a 2 ax

y 2 , arccospx{ a x2 y 2 q , z q p 2 2 p x y ,
arccospx{ x2 y2 q , zq for all px, y, z q P R3 zp
8, 0s  t0u  R. In particular,    cos ϕ
r sin ϕ 0    pdet g 1qpr, ϕ, zq :  sin ϕ r cos ϕ 0   r g
1 px, y, z q 



0

0

if y ¥ 0 if y   0

1 

for all pr, ϕ, z q P p0, 8q  p
π, π q  R. Example 3.4.23. (Spherical coordinates) Define g : p0, 8q  p0, π q p
π, πq Ñ R3 by



g pr, θ, ϕq : pr sin θ cos ϕ, r sin θ sin ϕ, r cos θq

for all pr, θ, ϕq P p0, 8q  p0, π q  p
π, π q.

Then g is continuously differentiable (with Ranpg q  R3 zp
8, 0s  t0u  R) with a continuously differentiable inverse g
1 : R3 zp
8, 0st0u R Ñ R3 given by a

"

x2 y 2qq if y ¥ 0 p|r| , arccospz{|r|q , arccospx{ a 2 2 p|r| , arccospz{|r|q ,
arccospx{ x y qq if y   0 for all px, y, z q P R3 zp
8, 0s  t0u  R. In particular,    sin θ cos ϕ r cos θ cos ϕ
r sin θ sin ϕ    pdet g 1 qpr, θ, ϕq :  sin θ sin ϕ r cos θ sin ϕ r sin θ cos ϕ   r2 sin θ   cos θ
r sin θ 0 g
1prq 

459

1

z

1 0 y

-1

0 -1

1

0 x

Fig. 142: For a  0, b  1 and f pz q : z for all z from Example 3.4.24 is a solid cone of height 1.

P [0, 1] , the volume of revolution S

for all pr, ϕ, z q P p0, 8q  p
π, π q  R. 3.4.1 Applications of Multiple Integrals Example 3.4.24. (Volume of a solid of revolution) Let a, b P R such that a   b, f : [a, b] Ñ [0, 8q be a continuous function whose restriction to pa, bq is continuously differentiable and S :

y 2 q1{2

px, y, zq P R3 : 0 ¤ px2

¤ f pzq ^ z P ra, bs

(

.

Note that S is rotational symmetric around the z-axis and can be thought of as obtained from a region in x, z-plane that is rotated around the z-axis. The volume V of S is given by V

π

»b a

f 2 pz q dz .

This can be proved as follows. For this, we define f : S Ñ R by f pxq : 1 for all x P S. As a constant map, f is continuous. We notice that B S is 460

given by the union of A1 : tpx, y, z q P R3 : a   z   b ^ x2 y 2
f 2 pz q  0u , A2 : tpx, y, aq P R3 : x2 y 2 ¤ f 2 paqu , A3 : tpx, y, bq P R3 : x2 y 2 ¤ f 2 pbqu . Further, A1 is the image of the map f1 : C 1 defined by

p
2π, 2πq  pa, bq Ñ R3 of class

f1 pϕ, z q : pf pz q cos ϕ, f pz q sin ϕ, z q for every pϕ, z q P p
2π, 2π qpa, bq, A2 is a subset of the image of the map f2 : p
1, 1 f paqq  p
2π, 2π q Ñ R3 of class C 1 defined by f2 pr, ϕq : pr cos ϕ, r sin ϕ, aq for every pr, ϕq P p
1, 1 f paqqp
2π, 2π q and A3 is a subset of the image of the map f3 : p
1, 1 f paqq  p
2π, 2π q Ñ R3 of class C 1 defined by f3 pϕq : pr cos ϕ, r sin ϕ, bq for every pr, ϕq P p
1, 1 f paqq  p
2π, 2π q. Hence it follows that B S is negligible. Further, if I is some closed interval such that I  S, it follows that # 1 if x P S fˆpxq : 0 if x P I z S . is continuous except in points from a negligible subset of R3 . Therefore fˆ is Riemann-integrable. Hence we can apply the Theorem of Fubini to conclude that V



»

dxdydz S



» b » a



p q p0q

Bf2 z

461

dxdy dz

π

»b a

f 2 pz q dz .

Example 3.4.25. (Total mass, center of mass and inertia tensor of a mass distribution) If ρ : V Ñ r0, 8q is the mass distribution (mass density) of a solid body occupying the region V in R3 , its total mass M, center of mass ~rC and inertia tensor pIij qi,j Pt1,2,3u are defined by: M : ~rC I11 : I33 : and

 » »V V

»

»

ρ dxdydz , V

»

xρ dxdydz,

yρ dxdydz,

V

py

2

px2



»

zρ dxdydz

V

,

V

z qρ dxdydz , I22 :

»

2

V

px2

z 2 qρ dxdydz ,

y 2qρ dxdydz Iij :


»

xi xj ρ dxdydz V

if i  j, if existent. In the integrands, x, y, z, x1 : x, x2 : y, x3 : z denote the coordinate projections of R3 . Example 3.4.26. (Probability theory) A function ρ : Ω subset Ω of Rn such that » ρ dv  1

Ñ r0, 8q from a

Ω

can be interpreted as a joint probability distribution for the random variables x1 , . . . , xn on the sample space Ω. The elements of Ω are called sample points and represent the possible outcomes of experiments. The probability P tpx1 , . . . , xn q P D u for the event that in an experiment px1 , . . . , xn q lies in a subset D € Ω is given by P tpx1 , . . . , xn q P D u  462

»

ρ dv , D

if existent. The mean or expected value E pf q for the measurement of a random variable f : Ω Ñ R in an experiment is defined by »

E pf q :

f ρ dv , Ω

if existent.

Problems 1) Evaluate the following iterated integrals. » 2 » 1

a) 0 »3

b)



px

2y q dx dy ,

2

0

» 5


3

» 4 »




y2 4 2

c) 3

1



px

2y q dx dy , 

dy px yq2 dx ,

» π{2 » 3 cos ϕ

d)


π{2

e) f)

r sin ϕ dr dϕ ,

0

0

?1

» 1 "» 1
x » 1
x
y » 1 #» ?1
x2 0



2

0

» 1 "» 1 » 1 0

2

0

2) Calculate

0

z



*

dy dx , *

xyz dz dy dx ,

» ? 1
x
y

0

2

g) 0



dz x y

0

2



+

dz a dy dx . 1
|px, y, z q|2

»

x dxdy , D

where D € R2 is the compact set that is contained in the first and fourth quadrant as well as is bounded by the y-axis and

tpx, yq P R2 : x Sketch D.

463

y 2
1  0u .

3) By using polar coordinates, calculate » T

where T :

!

px

y q dxdy ,

pr cos ϕ, r sin ϕq P R2 : 0   r ¤ 1 ^ π6 ¤ ϕ ¤ π3

)

.

Sketch T and g
1 pT q, where g is the polar coordinate transformation. 4) Calculate

»

x dxdydz , E

where E : 5) Calculate

!

px, y, z q P [0, 8q3 : x

y

z 2

1

)

.

»

y dxdy , D

where D € R2 is the area of the triangle with corners p0, 0q, p1{2, 0q and p1{2, 1q. Sketch D. 6) By using polar coordinates, calculate »

xy dxdy , T

where T is the compact subset of R2 that is bounded by the coordinate axes and !

p x, p1
x2 q1{2 q P R2 : 0 ¤ x ¤ 1

)

.

Sketch T and g
1 pT q, where g is the polar coordinate transformation. 7) Calculate

»

z dxdydz , E

where E is the compact subset of R3 in the first octant that is bounded by the coordinate surfaces and

tpx, y, z q P R3 : x Sketch E.

464

2y

3z

 1u .

8) Calculate

»

x2 dxdy , D

where D is the compact subset of R2 that is contained in the first and fourth quadrant as well as is bounded by the y-axis and

tpx, yq P R2 : x

y
2  0u ,

tpx, yq P R2 : x
y
2  0u .

Sketch D. 9) By using polar coordinates, calculate »

xy dxdy , T

where T is the compact subset of R2 that is contained in the first quadrant as well as is bounded by both coordinate axes and

tpx, yq P R2 : x2 y2  4u . Sketch T and g
1 pT q, where g is the polar coordinate transformation. 10) Calculate

»

z dxdydz , E

where E € R3 is the compact set contained in the first octant which is bounded by the coordinate surfaces and

tp1, y, z q P R3 : y P R ^ z P Ru , tpx, y, z q P R3 : z

2y

 2u .

Sketch E. 11) Calculate

»

x2 y dxdy , D

where D is the compact subset of R2 that is contained in the upper half-plane as well as is bounded by the x-axis and

tpx, yq P R2 : x2 Sketch D.

465

y

 4u .

12) By using polar coordinates, calculate »

x dxdy , T

where T is the compact subset of R2 that is contained in the first quadrant as well as is bounded by both coordinate axes and

tpx, yq P R2 : x2 y2  1u , tpx, yq P R2 : x2 y2  4u . Sketch T and g
1 pT q, where g is the polar coordinate transformation. 13) Calculate

»

z 2 dxdydz , E

where E is the compact subset of R3 that is contained in

tpx, y, z q P R3 : z ¥ 0u and is bounded by

tpx, y, z q P R3 : x2

y2

z
9  0u .

Sketch E.

14) Calculate the volume of solid ellipsoid with half-axes a, b, c ¡ 0. 15) Calculate the center of mass and the inertia tensor of a solid hemisphere

tpx, y, z q P R3 : x2

pz
rq2 ¤ r2 ^ 0 ¤ z ¤ ru for a mass distribution which is constant of value ρ0 ¥ 0. y2

16) Calculate the center of mass and the inertia tensor for of a solid cone of height h ¥ 0

tpx, y, z q P R3 : a2 px2 y2 q ¤ z 2 ^ 0 ¤ z ¤ hu , where a ¡ 0, for a mass distribution which is constant of value ρ0 ¥

0.

17) (Buffon’s needle problem) A needle of length L ¡ 0 is thrown in a random fashion onto a smooth table ruled with parallel lines separated by a distance d ¡ L. For simplicity, associate to all lines a common orientation. Denote by x P r0, d{2s the minimal distance of the center of the needle to the lines and by θ P r0, π s the angle

466

0.8

0.8

0.6

0.6 y

1

y

1

0.4

0.4

0.2

0.2

0

0 0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

x

0.6

0.8

1

x

Fig. 143: Density maps of ρp1,1,1q p, , 0.5q, ρp1,2,1q p, , 0.5q. See Problem 18. between the direction of the needle and the direction of the lines. Under the assumption that x and θ are uniformly distributed, the joint probability distribution ρ : r0, d{2sr0, π s Ñ r0, 8q of x, θ is given by 2 ρpx, θq  πd for all px, θq P r0, d{2s  r0, π s. a) Determine the set S € r0, d{2s  r0, π s corresponding to all events that cause the needle to intersect a ruled line. b) Calculate the probability pS of the last event given by pS



» S

ρpx, θq dxdθ .

18) A point particle of mass m ¡ 0 is confined to a cube r0, 1s3, but without influence from other forces. In a quantum mechanical description, the probability distributions for the position of the particle in basic stationary states of energy 2 2

E

 π2m~ |k|2 ,

where ~ is the reduced Planck constant, are given by ρk px, y, z q : 8 sin2 pk1 πxq sin2 pk2 πy q sin2 pk3 πz q for all x, y, z P r0, 1s, where k  pk1 , k2 , k3 q of so called ‘quantum numbers’.

467

P N3 is a collection

0.8

0.8

0.6

0.6 y

1

y

1

0.4

0.4

0.2

0.2

0

0 0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

x

0.6

0.8

1

x

Fig. 144: Density maps of ρp2,1,1q p, , 0.5q, ρp2,2,1q p, , 0.5q. See Problem 18. a) For every k  pk1 , k2 , k3 q P N3 , calculate the expectation values xxy, xy y, xz y for the components of the position of the particle

xxy  xyy  xz y 

»

» »

r0,1s3

r0,1s3 r0,1s3

x ρk px, y, z q dxdydz , y ρk px, y, z q dxdydz , z ρk px, y, z q dxdydz .

b) For every k P N3 , calculate the standard deviation σ1 , σ2 , σ3 for the expectation values from part a) σ12



σ22



σ32



»

r0,1s3 » r0,1s3 » r0,1s3

px
xxyq2 ρk px, y, z q dxdydz , py
xyyq2 ρk px, y, z q dxdydz , pz
xz yq2 ρk px, y, z q dxdydz .

c) Calculate and compare the probability of finding the particle in the volume r0, as3 that includes a corner and in rp1
aq{2, p1 aq{2s3 around the center of the cube, where 0   a   1. 19) Calculate the volume of the compact subset of R3 that is bounded by the given surfaces. a)

S1

 t px, y, z q P R3 : x2 468

y2

zu

,

1

-1

1

-1

Fig. 145: Graphical depiction of S from Problem 15 for the case R  1.

 t px, y, z q P R3 : x4 y4  2 px2 y2 q u , S3  t px, y, z q P R3 : z  0 u , S1  t px, y, z q P R3 : x2 y 2 z 2  4 u , S2  t px, y, z q P R3 : px2 y 2 q2  4 px2
y 2 q u S1  t px, y, z q P R3 : x2 y 2 z 2  9 u , S2  t px, y, z q P R3 : x2 y 2  3 |x| u . S2

b) c)

,

20) (Generalized polar coordinates)

a) Let a, b, α ¡ 0. Define g : p0, 8q  p0, π {2q Ñ R2 by g pr, ϕq : par cosα ϕ, br sinα ϕq

for all pr, ϕq P p0, 8q  p0, π {2q. Find the range of g. Show that the restriction of g in range to Ranpg q is a continuously differentiable bijection with a continuously differentiable inverse. In particular, calculate that inverse and detpg 1 q. b) Calculate the area of S : tpx, y q P R2 : |x|1{2

469

|y|1{2 ¤ R1{2 u

1

z 0

1

1 0 y

-1 1

0 x

Fig. 146: Graphical depiction of S from Problem 16 for the case R  1. for R ¡ 0 by use of suitable generalized polar coordinates from part a). 21) (Generalized spherical coordinates)

a) Let a, b, c, α ¡ 0. Define g : p0, 8qp0, π {2qp0, π {2q Ñ R3 by g pr, θ, ϕq : par sinα θ cosα ϕ, br sinα θ sinα ϕ, cr cosα θq

for all pr, θ, ϕq P p0, 8qp0, π {2qp0, π {2q. Find the range of g. Show that the restriction of g in range to Ranpg q is a continuously differentiable bijection with a continuously differentiable inverse. In particular, calculate that inverse and detpg 1 q. b) Calculate the volume of S : tpx, y, z q P R3 : |x|1{2

|y|1{2 |z |1{2 ¤ R1{2 u

for R ¡ 0 by use of suitable generalized spherical coordinates from part a).

470

2

y

1

0

-1

-2 -2

-1

0 x

1

2

Fig. 147: Direction field corresponding to Couette flow between counter-rotating concentric cylinders of radius 1 and 2. See Example 3.5.1.

3.5 Vector Calculus Any vector-valued function of several variables f from a non-trivial subset D of some Rn into some Rm can be interpreted as a vector field assigning vectors to points in space. Example 3.5.1. Define v : tpx, y, z q P R3 : 1 ¤ x2 v px, y, z q :



y2

¤ 4u Ñ R3 by

x2 y 2
2 x2 y 2
2
y x2 y2 , x x2 y2 , 0



for all px, y, z q P R3 satisfying 1 ¤ x2 y 2 ¤ 4. v is the velocity field of a viscous incompressible flow, a so called ‘Couette flow’, between concentric cylinders of radius 1 and 2 rotating at the same rate, but in counterclockwise and clockwise direction, respectively. Example 3.5.2. Define E : R3 zt0u Ñ R3 by E px, y, z q : a x2 471

1 y2

z2

px, y, zq

1

z 0

1 0 y -1

-1 -11

1

0 x

Fig. 148: Electrical field created by a negative unit charge. See Example 3.5.2

for all px, y, z q P R3 zt0u. E describes the electrical field created by a negative unit charge. Example 3.5.3. (Motivation for the definition of curve integrals.) For this, let F be a continuous map from some open subset U in R3 into R3 (the ‘force field’) and r be a twice continuously differentiable map from some open interval I of R into U (the trajectory of a point particle parametrized by time) which satisfies m r 2 ptq  F prptqq

for every t P I (Newton’s equation of motion) where m the particle). Then
m

¡ 0 (the mass of

1

ptq  m r 1ptq  r 2ptq  r 1ptq  F prptqq 2 for every t P I, where v : r 1 (the velocity field of the particle), and hence
m

2

v

2

v2



pt1 q



m

2

v

2



pt0 q  472

» t1 t0

r 1 ptq  F prptqq dt .

The right hand side of the previous equation describes the difference of the ‘kinetic energies’ of the particle at t1 and t0 . Further, if F is in addition ‘conservative’ , i.e., if there is some V : U Ñ R (a ‘potential function’) of class C 1 such that F 
∇V , the we conclude by the chain rule » t1 t0

r 1 ptq  F prptqq dt 





» t1

» t1 t0

r 1 ptq  ∇V prptqq dt

t0

pV  rq 1 ptq dt  V prpt0qq
V prpt1qq

and hence that the function (‘the total energy of the particle’) m 2 v 2

V

r

is constant (‘Energy conservation’). Definition 3.5.4. Let n P N , F be a continuous map from some open subset U of Rn into Rn , a, b P R such that a ¤ b and r : I Ñ Rn be a regular C 1 -path in U, i.e., the restriction of a continuously differentiable map from some open interval Iˆ  I into U. Then we define the curve integral of F along r by » r

F  dr :

»b a

r 1 ptq  F prptqq dt .

Remark 3.5.5. Note that we don’t demand that r is necessarily injective. A simple example for a regular C 1 -path which is not injective is given by r : [
1, 1] Ñ R2 defined by rptq : pt2 , 1q for every t P [
1, 1]. This path begins at the point p1, 1q, moves on to p0, 1q from where it returns to p1, 1q. 473

2

1

y

0

-1

-2 -2

-1

0 x

1

2

Fig. 149: Direction field associated to F from Example 3.5.6 and Ranprq for a  1.

Example 3.5.6. Define F : R2 zt0u Ñ R2 by F px, y q :




x2

y

, y 2 x2



x y2

for all px, y q P R2 zt0u and the parametrization of the circle of radius a ¡ 0 around the origin by r : r0, 2π s Ñ R2 by rptq : a.pcos t, sin tq

for all t P R. Then »

r

 

F  dr : » 2π 0 » 2π 0

» 2π 0

r 1 ptq  F prptqq dt

p
a sin t, a cos tq 



t a cos t ,
a sin 2 a a2

dt  2π .

Note that this result does not depend on the radius a. 474



dt

Theorem 3.5.7. (Invariance under reparametrization) Let n P N , F be a continuous map from some open subset U of Rn into Rn and r : ra, bs Ñ Rn be a regular C 1 -path in U. Further, let g : ra, bs Ñ rc, ds be continuously differentiable with a continuously differentiable inverse (i.e, there is an extension gˆ : I1 Ñ I2 of g, where I1 , I2 are open intervals of R such that I1  ra, bs, I2  rc, ds and such that gˆ is continuously differentiable with a continuously differentiable inverse) and such that g 1 pxq ¡ 0 for all x P ra, bs. Then » » r

F  dr 



r g

F  dr .

Proof. By the change of variable formula (for example, see Theorem 3.4.20) and the chain rule, it follows that » r



F  dr : »d c

»b a

r 1 ptq  F prptqq dt 

»d

pr  gq 1psq  F ppr  gqpsqq ds 

c

r 1 pg psqq  F prpg psqqq g 1 psq ds

»

F  dr .



r g

Example 3.5.8. (Change of orientation/inverse path) Let n, F and r like in Definition 3.5.4. Moreover, define the inverse path r
to r by r
ptq : rpa b
tq for all t P ra, bs. Then it follows by the change of variable formula (for example, see Theorem 3.4.20) and the chain rule that » r


F  dr 





»b

»b a

a

r 1 pa

»b a

r
1 ptq  F pr
ptqq dt

b
tq  F prpa

r 1 ptq  F prptqq dt 


Definition 3.5.9. 475

» r

b
tqq dt  F  dr .

(i) A piecewise regular C 1 -path r is a sequence pr1 , . . . , rν q of regular C 1 -paths r1 , . . . , rν , where ν P N , with coinciding endpoints of ri and starting points of ri 1 for each i P t1, . . . , ν
1u. Further, we say that r is closed if the endpoint of rν coincides with the initial point of r1 . (ii) Further we define the curve integral along such r of a continuous vector field F : U Ñ Rn , where U is some open subset of Rn containing the ranges of all ri , i P t1, . . . , ν u, by »

r

F  dr :

»

r1

F  dr



»

rν

F  dr .

Theorem 3.5.10. Let n P N , F be a continuous map from some open subset U of Rn into Rn and r be a piecewise regular C 1 -path in U from x0 to x1 . Finally, let V : U Ñ R be of class C 1 and such that F  ∇V . Then »

r

F  dr  V px1 q
V px0 q .

Proof. Since r is a piecewise regular C 1 -path in U from x0 to x1 , there are ν P N along with regular C 1 -paths r1 : ra1 , b1 s Ñ U, . . . , rν : raν , bν s Ñ U such that r1 pa1 q  x0 and rν pbν q  x1 . Then it follows by the chain rule that »

r



F  dr  » b1 a1

» b1

»

r1

F  dr



r11 ptq  p∇V qpr1 ptqq dt

»

rν

F  dr

 » bν

» bν aν

rν1 ptq  p∇V qprν ptqq dt

 pV  r1 q 1 ptq dt    pV  rν q 1 ptq dt a a  V pr1pb1 qq
V pr1pa1qq    V prν pbν qq
V prν paν qq   V prν pbν qq
V pr1pa1 qq  V px1 q
V px0q . 1

ν

476

Theorem 3.5.11. (Necessary conditions for the existence of a potential) Let n P N , F  pF1 , . . . , Fn q be a map of class C 1 (i.e., all F1 , . . . , Fn are of class C 1 ) from some open subset U of Rn into Rn , and let V : U Ñ R be of class C 2 and such that F  ∇V . Then

BFi
BFj  0 , i, j P t1, . . . , nu , i  j . Bxj Bxi For every i, j P t1, . . . , nu such that i  j, it follows by Schwarz’s

Proof. Theorem 3.2.19 that

BFi  B2V  B2V  BFj Bxj Bxj Bxi BxiBxj Bxi

.

Remark 3.5.12. For the cases n  2, 3, the condition (3.5.11) is equivalent to the vanishing of the so called rotational field curlF of F : curl F :

$ B F2 & Bx


BBFy

if n  2

1

p By
BBFz , BBFz
BBFx , BBFx
BBFy q

% B F3

2

1

3

2

1

if n  3 .

Example 3.5.13. Let F be as in Example 3.5.6. Then F is of class C 1 and

BFx px, yq  BFy px, yq  y2
x2 By Bx x2 y 2

for all px, y q P R . But, by Theorem 3.5.10 and the result of that Example, it follows that there is no V : UR p0qzt0u Ñ R of class C 1 such that F px, y q  p∇V qpx, y q for all px, y q P UR p0qzt0u, where R ¡ 0. Hence the condition stated in Theorem 3.5.11 is not sufficient for the existence of a potential. Theorem 3.5.14. (Sufficient conditions for the existence of a potential, Poincare Lemma) Let n P N , U be an open subset of Rn which is starshaped with respect to some x0 P U, i.e., such that for all x P U also the 477

line segment tx0 t.px
x0 q : t P r0, 1su is contained in U. Further, let F  pF1 , . . . , Fn q : U Ñ Rn be of class C 1 and such that

BFi
BFj  0 Bxj Bxi for every i, j P t1, . . . , nu such that i  j. Then there is a potential V U Ñ R of class C 2 such that F  ∇V . Proof. Define V : U Ñ R by V pxq :

» rx

Hi pxq :

»1 0

n ¸

F  dr 



:

pxi
x0i qHipxq ,

i 1

Fi px0

t.px
x0 qq dt ,

where rx ptq : x0 t.px
x0 q for all t P r0, 1s, for all x P U. Now let x P U. Since U is open there is d ¡ 0 such Ud pxq € U. Then by Taylor’s formula Theorem 3.3.5, it follows for all h P Ud p0q that »1

BFi px t.px
x q τ h.e q dt 0 j Bxj 0 0 for some τ P r0, 1s. Now rj : r0, 1s  r0, ds Ñ U defined by rj pt, sq : x0 t.px
x0 q s.ej , pt, sq P r0, 1s  r0, ds 1 rHipx h

h.ej q
Hi pxqs 

t

is obviously continuous and hence its image compact, since its domain is compact, too. Since BFi : U Ñ R Bxj is continuous, it is in particular uniformly continuous on Ran rj . Hence for any ε ¡ 0 there is δ ¡ 0 such that   Fi   x x2



B p q
BFi px q   ε Bj Bxj 1  478

P Ran rj and |x2
x1|   δ . In particular for h P R such that |h|   δ, it follows that » 1 »1  BFi px  t B Fi px0 t.px
x0 q τ h.ej q dt
t  Bxj Bxj 0 0 0   2ε whenever x1 , x2

t.px
x0 qq

  dt

and hence also that  1  Hi x h

r p

h.ej q
Hi pxqs


»1 0

BFi px t Bxj 0

t.px
x0 qq

  dt

 ε. Since ε ¡ 0 is arbitrary otherwise, it follows that Hi is partially differentiable in the j-th coordinate direction with partial derivative given by

BHi pxq  » 1 t BFi px Bxj Bxj 0 0

t.px
x0 qq dt , x P U .

Moreover, analogous reasoning shows that

B Hi Bxj is continuous. Hence V is of class C 1 . In particular, »1 n BV pxq  ¸ BFi px p xi
x0i q t Bxj Bxj 0 0 i1 »1 0 n ¸



i 1

Fj px0

pxi
x0i q

t.px
x0 qq dt

t.px
x0 qq dt »1

t 0

BFj px Bxi 0

479

t.px
x0 qq dt

»1 0

Fj px0

 tFj px0

t.px
x0 qq dt

t.px
x0 qq

1   

 Fj pxq , j P t1, . . . , nu

0

and hence, finally, it follows also that V is of class C 2 . Remark 3.5.15. For the case n  2, the statement of Theorem 3.5.14 is also true for the more general case of an open simply-connected U. For the proof see [33], XVI, §5, Theorem 5.4. Example 3.5.16. (i) Any open convex subset of Rn , i.e, any open subset S of Rn such that for all x, y P S also tx t.py
xq : t P r0, 1su € S, like Rn itself and any open ball in Rn , is star-shaped with respect to any of its elements.

Ñ R3 by F px, y, z q : py 2 z 3 , 2xyz 3 , 3xy 2z 2 q ,

(ii) Define F : R3

x, y, z

PR.

F is in particular of class C 1 and curl F

0.

Since R3 is star-shaped with respect to the origin, there is a potential V : R3 Ñ R of class C 2 such that F  ∇V which is unique up to some constant field. Integration of the corresponding equations gives V px, y, z q  xy 2 z 3 , x, y, z

Problems

480

P R.

1) Calculate

» r

F  dr .

Note that the paths in d)-g) all start and end at the same points. a) Fpx, y q : p
y, 2xq , x, y P R , rptq : pt, t2 q , t P r0, 1s , b) Fpx, y q : p3y,
4xq , x, y P R , rptq : pt2 , t3 q , t P r0, 2s ,

c) Fpx, y q : px2
3xy, xy

y 2 q , x, y

rptq : pcos t, sin tq , t P r
π, π s ,

d) Fpx, y q : p2xy,
x2 q , x, y

PR,

e) Fpx, y q : p2xy,
x2 q , x, y

PR,

rptq : p2t, tq , t P r0, 1s ,

PR,

rptq : p2t, t { q , t P r0, 1s , 1 2

f) Fpx, y q : p2xy,
x2 q , x, y

g)

h) i)

PR, r : pr1 , r2 q , r1 ptq : p0, tq , t P r0, 1s , r2 psq : ps, 1q , s P r0, 2s , Fpx, y q : p2xy,
x2 q , x, y P R , r : pr1 , r2 q , r1 ptq : pt, 0q , t P r0, 2s , r2 psq : p2, sq , s P r0, 1s , Fpx, y, z q : py
z, z
x, x
y q , x, y, z P R , rptq : pcos t, sin t, tq , t P r0, 2π s , Fpx, y, z q : py, z, xq , x, y, z P R , rptq : pR cos α cos t, R cos α sin t, R sin αq , t P r0, 2π s , αPR .

2) If possible, find a potential function V : DpF q Ñ R of class C 1 for F , where DpF q denotes the domain of F . Otherwise, give reasons why there is no such function. a) Fpx, y q : p1 y, 1 xq , x, y b) Fpx, y q : px, 2y q , x, y P R , c)

Fpx, y q : p2xy

y2

PR

2xy 2 , x2

,

2xy

d) Fpx, y q : pe cos y,
e sin y q , x, y x

x

481

PR

2x2 y q , x, y ,

PR

,

2

y

1

0

-1

-2 -2

-1

0 x

1

2

Fig. 150: Direction field of associated to F from Problem 4 and Ranprq for a  1. e) f) g)

Fpx, y, z q : py 2 z 2 , 2xyz 2 , 2xy 2 z q , x, y, z

PR , Fpx, y, z q : px, x z y, 3x y z q , x, y, z P R , Fpx, y, z q : |px, y, z q|
3px, y, z q , px, y, z q P R3 z t0u 2

2

.

3) Let n P N zt0, 1u, f : Rn zt0u Ñ R be continuous. Define F : Rn zt0u Ñ Rn by F pxq : f pxq.x for all x P Rn zt0u. Calculate

» r

F  dr ,

where r is a regular C 1 -path whose range is part of Srn p0q for some r ¡ 0.

4) Define F : R2 zt0u Ñ R2 by F px, y q :



2 2
px2 2xyy2 q2 , pxx2
yy2 q2

for all px, y q P R2 zt0u.

482



a) Calculate

» r

F  dr ,

where a ¡ 0 and r : r0, 2π s Ñ R2 is given by rptq : a.pcos t, sin tq for all t P R. Note the difference of the result to that of Example 3.5.6. b) If possible, find a potential function V : R2 zt0u Ñ R of class C 1 for F . Otherwise, give reasons why such function does not exist. c) Calculate » r

F  dr ,

where r is any regular C 1 -path that assumes values in R2 zt0u and has initial point p and end point q.

5) As in Example 3.5.6, define F : R2 zt0u Ñ R2 by F px, y q :




x2 y y2 , x2 x y2



for all px, y q P R2 zt0u. Further, let a ¡ 0, r : r0, 1s Ñ R2 zt0u a regular C 1 -path such that rp0q  rp1q  p
a, 0q and such that the y-component of r is   0 on p0, εq and ¡ 0 on p1
ε, 1q for some ε ¡ 0. a) Find a potential V : R2 zp
8, 0s Ñ R of class C 1 for the restriction of F to R2 zp
8, 0s Ñ R. b) Calculate » r

F  dr .

3.6 Generalizations of the Fundamental Theorem of Calculus In the following, we consider generalizations of the fundamental theorem of calculus theorem 1.5.20 to functions of several variables. For this, we need to define the orientation of n-tuples of vectors in Rn . 483

b

Α O

Fig. 151: Since 0 Example 3.6.2.

a

  α   π, the pair of vectors pa, bq in R2 is positively oriented.

See

Definition 3.6.1. (The orientation of n-tuples of vectors in Rn ) Let n P N , pa1 , . . . , an q be an n-tuple of vectors in Rn . Then we say that pa1 , . . . , an q is positively oriented, negatively oriented if detpa1 , . . . , an q ¡ 0 and

detpa1 , . . . , an q   0 ,

respectively. Note that exchanging the order of two elements in a positively oriented n-tuple leads to a negatively oriented n-tuple and vice versa. In particular, since detpe1 , . . . , en q  1 ¡ 0 the n-tuple pe1 , . . . , en q consisting of the canonical basis e1 , . . . , en of Rn is positively oriented.

Example 3.6.2. If a  pa1 , a2 q P R2 is non-vanishing and b P R2 is in the same direction as the rotation of a in counterclockwise (= mathematically positive) sense around the origin by the angle α P p0, π q, then b  λ.pa1 cospαq
a2 sinpαq, a1 sinpαq 484

a2 cospαqq

a x b

b

a

Fig. 152: The triple of vectors pa, b, a  bq in R3 is positively oriented. See Example 3.6.3.

for some λ ¡ 0 and a1 a2 detpa, bq  λ a1 cospαq
a2 sinpαq a1 sinpαq a2 cospαq

 λ |a|2 sinpαq ¡ 0 and hence the pair pa, bq in R2 is positively oriented. This fact is often used to decide whether a given pair of vectors in R2 is positively oriented.

Example 3.6.3. If a, b P R3 are vectors that are not multiples of one another, it follows by Remark 2.5.14 and Definition 2.5.17 that detpa, b, a  bq  detpa  b, a, bq  |a  b|2

¡0

and hence that the triple pa, b, a  bq in R3 is positively oriented. In applications, this is often used for the construction of positively oriented triples in R3 .

485

3.6.1 Green’s Theorem We start with Green’s theorem for images of rectangles under certain differentiable maps. The basis for its proof is given by the following Lemma. Lemma 3.6.4. Let V be a non-empty open subset of R2 and F  pF1 , F2 q : V Ñ R2 be differentiable. Further, let g  pg1 , g2 q : Dpg q Ñ R2 be defined and of class C 2 on a non-empty open subset Dpg q of R2 and such that g pDpg qq € V . Then

B pF  gq Bg1 pF  gq Bg2 
B pF  gq Bg1 pF  gq Bg2  2 Bx  1 By 2 By By 1 Bx Bx  BBFx2
BBFy1  g  detp g 1 q .

Proof. The proof proceeds by a simple calculation using the chain rule in the form of Corollary 3.2.26 and Schwarz’s theorem 3.2.19.

B pF  gq Bg1 pF  gq Bg2 
B pF  gq Bg1 pF  gq Bg2  2 2 Bx 1 By By By 1 Bx Bx 2 2  BpFB1x gq BBgy1 pF1  gq BBxBg1y BpFB2x gq BBgy2 pF2  gq BBxBg2y 2 2
BpFB1y gq BBgx1
pF1  gq BByBg1x
BpFB2y gq BBgx2
pF2  gq BByBg2x  BpFB1x gq BBgy1 BpFB2x gq BBgy2
BpFB1y gq BBgx1
BpFB2y gq BBgx2



   F1 g1 F1 g2 B g1 B B B B  Bx  g Bx  g Bx By B y  BF2  g Bg1  BF2  g Bg2  Bg2 Bx Bx  By Bx  By 
BBFx1  g BBgy1 BBFy1  g BBgy2 BBgx1  



 B B F2 g1 F2 g2 B g2 B B
Bx  g By By  g By Bx 486

y

gHIL

x

Fig. 153: Illustration for the proof of Green’s theorem, Theorem 3.6.5.





  F1 g2 B g1 F2 g1 B g2 F1 B B B g2 B g1 B B B  By  g Bx By  g
 g Bx Bx By By By Bx 


BBFx2  g BBgy1 BBgx2  BBFx2  g
BBFy1  g  detp g 1 q . 

Theorem 3.6.5. (Green’s theorem for images of rectangles) Let a, b, c, d P R such that a   b and c   d and I : ra, bs  rc, ds, I0 : pa, bq  pc, dq. Further, let U  I be an open subset of R2 , g : U Ñ R2 be twice continuously differentiable such that the induced map from U to g pU q is bijective with a continuously differentiable inverse and such that detpg 1 q ¡ 0. Finally, let V  g pI q be an open subset of R2 and F  pF1 , F2 q : V Ñ R2 be continuously differentiable. Then 

»

p q

g I0

BF2
BF1 dxdy  » F  dr Bx By r 487

(3.6.1)

for any piecewise C 1 -parametrization r of the boundary of g pI0q which is
of the same orientation as the piecewise C 2 -path prc , rb , r
d , ra q where rc pxq : g px, cq , rb py q : g pb, y q , rd pxq : g px, dq , ra py q : g pa, y q

for all x P ra, bs and y

P rc, ds.

Proof. In a first step, we consider the set g pI0q. Since g is twice continuously differentiable with a continuously differentiable inverse, g pI0 q is a bounded open subset in R2 . Further, the restriction of

BF2
BF1 Bx By

to g pI0 q is bounded. In addition, it follows by Theorem 3.4.13 and Theorem 3.4.15 that the extension of this function to a function, defined on a closed subinterval J of R2 containing g pI0 q and assuming the value zero in the points of J z g pI0q, is Riemann-integrable. Hence by Theorem 3.4.20, it follows in a second step that 

»

p q

g I0

BF2
BF1 dxdy  » Bx By I



0

BF2
BF1  g detpg 1q dxdy Bx By

and hence by the previous Lemma 3.6.4 that

BF2
BF1 dxdy Bx By g pI q  » B B B g1 g2  Bx pF1  gq By pF2  gq By dxdy I   » B g1 g2 B B
By pF1  gq Bx pF2  gq Bx dxdy . I 

»

0

0

(3.6.2)

0

Further, by Fubini’s Theorem 3.4.17 and the fundamental theorem of calculus Theorem 1.5.20, it follows that »

I0

B pF  gq Bg1 pF  gq Bg2  dxdy 2 Bx 1 By By 488

(3.6.3)

y

W

a

b

x

Fig. 154: Illustration for the proof of Green’s theorem, Theorem 3.6.7.

 B B g1 g2  pF1  gqpb, yq By pb, yq pF2  gqpb, yq By pb, yq dy c  »d g1 g2 B B
pF1  gqpa, yq By pa, yq pF2  gqpa, yq By pa, yq dy »d

c

and

B pF  gq Bg1 pF  gq Bg2  dxdy (3.6.4) 1 2 Bx Bx I By  »b g1 g2 B B  pF1  gqpx, dq Bx px, dq pF2  gqpx, dq Bx px, dq dx a  »b B B g1 g2
pF1  gqpx, cq Bx px, cq pF2  gqpx, cq Bx px, cq dx . a

»

0

Finally, (3.6.1) follows from (3.6.2), (3.6.3) and (3.6.4).


Remark 3.6.6. For the orientation of the piecewise C 2 -path prc , rb , r
d , ra q note that the region g pI0 q is bounded and hence that there is outward point489

ing unit normal for every point on its boundary, apart from the corner points g pa, cq, g pb, cq, g pb, dq and g pa, dq. Outward pointing vectors are given by 

B B g1 g2 vc px, cq 
px, cq,
By px, cq , B y 

g1 B g2 B vb pb, y q  pb, yq, Bx pb, yq , B x 

B g1 B g2 px, dq, By px, dq , vd px, dq  B y 

g1 g2 B B va pa, y q 
Bx pa, yq,
Bx pa, yq

for every x P pa, bq and y P pc, dq. In particular, as a consequence of the assumption that detpg 1 q ¡ 0 in the previous Theorem 3.6.5, it follows that detpvc px, cq, rc1 pxqq  detpvb pb, y q, rb1 py qq  detpvd px, dq, rd1 pxqq  detpvapa, yq, ra1 pyqq ¡ 0 for all x P pa, bq and y P pc, dq. Hence the orientation for the piecewise C 1 parametrization r of the boundary of g pI0q in Theorem 3.6.5 has to be such that the outward unit normal and the tangent vector in a point of the boundary of g pI0 q are positively oriented in every point of the boundary, apart from a finite number of points. This orientation is indicated in Fig. 153. Theorem 3.6.7. (Green’s theorem for regions bounded by graphs) Let a, b P R be such that a   b, f1 : ra, bs Ñ R and f2 : ra, bs Ñ R restrictions of twice continuously differentiable functions defined on open intervals containing ra, bs. In addition, let f1 , f2 be such that f1 pxq   f2 pxq for all x P pa, bq.1 Further, let Ω : tpx, y q P R2 : a   x   b ^ f1 pxq   y 1

  f2pxqu .

Note that we do not demand that f1 paq   f2 paq or that f1 pbq   f2 pbq. As a consequence, in Fig. 154, each of the line segments of the boundary of Ω that are parallel to the y-axis can consist of one point, only.

490

In particular, let Ω be such that there is a 0   δ   pb
aq{2 such that the corresponding sets Ω Xppa, a δ q Rq and Ω Xppb
δ, bq Rq are convex. Finally, let F  pF1 , F2 q : V Ñ R2 be continuously differentiable where V is an open subset of R2 containing Ω and its boundary. Then: »  Ω

BF2
BF1 dxdy  » F  dr Bx By r

(3.6.5)

for any piecewise C 1 -parametrization r of the boundary of Ω which is of the same (‘mathematically positive’, ‘counterclockwise’) orientation as the
piecewise C 2 -path pr1 , rb , r
2 , ra q r1 pxq : px, f1 pxqq , rb pλq : pb, f1 pbq λ [f2 pbq
f1 pbq]q , r2 pxq : px, f2 pxqq , ra pλq : pa, f1 paq λ [f2 paq
f1 paq]q

for all x P ra, bs and λ P r0, 1s.

Proof. For this, define the open subset U of R2 by U : pa, bq  R and g : U Ñ U by g px, λq : px , f1 pxq

λ [f2 pxq
f1 pxq]q

for all px, λq P U. In particular, g is bijective, of class C 2 with an inverse of class C 2 given by 

g
1 px, y q  x , for all px, y q P U.

y
f1 pxq f2 pxq
f1 pxq



detpg 1 px, λqq  f2 pxq
f1 pxq ¡ 0

for all px, λq P U. Further,

g ppa, bq  p0, 1qq  Ω ,

(3.6.6)

and hence Ω is an open subset of R2 . The validity of (3.6.6) can be seen as follows. First, for px, λq P pa, bq  p0, 1q, it follows that f1 pxq   f1 pxq

λ [f2 pxq
f1 pxq]   f1 pxq 491

f2 pxq
f1 pxq  f2 pxq

and hence that g px, λq P Ω. Second, for px, y q P Ω, it follows that 0 

y
f1 pxq f2 pxq
f1 pxq

f1 pxq   ff2ppxxqq

f pxq  1 2

1

and hence that g
1 px, y q P pa, bq  p0, 1q. In the following, let 0   δ   pb
aq{2 be such that the corresponding sets Ω X ppa, a δq  Rq and Ω X ppb
δ, bq  Rq are convex. Further, let 0   ε   δ, Iε : ra ε, b
εs  r0, 1s and I0,ε : pa ε, b
εq  p0, 1q. Then U  Iε and V  g pIε q. Hence it follows by Theorem 3.6.5 that 

»

p

g I0,ε

q

BF2
BF1 dxdy  » Bx By r

F  drε

(3.6.7)

ε


where rε is the piecewise C 2 -path pr1,ε , rb
ε , r
2,ε , ra

ε

q given by

r1,ε pxq : px, f1 pxqq , rb
ε pλq : pb
ε, f1 pb
εq λ [f2 pb
εq
f1 pb
εq]q , r2,ε pxq : px, f2 pxqq , ra ε pλq : pa ε, f1 pa εq λ [f2 pa εq
f1 pa εq]q for all x P ra ε, b
εs and λ P r0, 1s. In the following final step of the proof, we show that (3.6.5) follows from (3.6.7) by performing the limit ε Ñ 0. For this, let fˆ1 , fˆ2 : pa δ 1 , b
δ 1 q Ñ R twice continuously differentiable extensions of f1 and f2 , respectively, for some 0   δ 1 . Then by
gˆpx, λq : x , fˆ1 pxq λ [fˆ2 pxq
fˆ1 pxq] for all px, λq P pa
δ 1 , b δ 1 q  R, there is defined a twice continuously differentiable extension of g, and hence it follows by Theorem 3.4.13 and Theorem 3.4.15 that the extension of 



BF2
BF1  , Bx By  Ω 492

to a function that is defined on a closed subinterval J of R2 containing Ω and assuming the value zero in the points of J z Ω, is Riemann-integrable. Further, it follows that »   F2    gpI0,ε q x

B
BF1 B By

where M1



dxdy


»  Ω

   dxdy  



BF2
BF1 Bx By

¤ 2Mε

¡ 0 denotes the maximum of BF2
BF1 Bx By

on some closed subset that is contained in V and at the same time contains Ω. Hence, 

»

lim

ε

Ñ0 gpI0,ε q

BF2
BF1 dxdy  »  BF2
BF1 dxdy . Bx By Bx By Ω

Further, »  »    F drε F dr   rε r  » aε   1   F x, f x 1, f x dx 1 1   a » 1   [f2 aε f1 aε ] F2 aε , f1  0 »1



¤




p



p qq  p

p qq

p q
p q p




0

» aε   F x, f2 x 

paεq

[f2 paq
f1 paq] F2 pa, f1 paq

» b    1  F x, f1 x  1, f x dx 1   b » 1ε   [f2 bε f1 bε ] F2 bε , f1 bε  0 »1

p

p qq  p

p q
p q




0

p qq p

p

a

p q

[f2 pbq
f1 pbq] F2 pb, f1 pbq 493



 qq  p1, f 1 pxqq dx

p

2



λ [f2 paε q
f1 paε q] q dλ λ [f2 paq


  f1 a ] dλ

» b   F x, f2 x  bε

p

p

pqq

2

λ [f2 pbε q
f1 pbε q] q dλ λ [f2 pbq




 qq  p1, f 1 pxqq dx

  f1 b ] dλ

pqq




where r : pr1 , rb , r
ε, bε : b
ε. In the following, 2 , ra q and aε : a we estimate the individual terms of the last sum. First, »   F x, f1 x 

p

I



 qq  p1, f 1 pxqq dx ¤

p

»



1

I

{ ,

|F px, f1pxqq|  |p1, f11pxqq| dx

¤ εM2p1 q  »  1 p p qq  p1, f2 pxqq dx ¤ |F px, f2pxqq|  |p1, f21pxqq| dx I I 2 1{2 ¤ εM2p1 M4 q for every interval I € ra, bs of length ε. Here M2 ¥ 0 denotes the maximum of |F | on some closed subset that is contained in V and at the same time contains Ω; M3 ¥ 0 denotes the maximum of the restriction of |fˆ11 | to ra, bs; M4 ¥ 0 denotes the maximum of the restriction of |fˆ21 | to ra, bs. M32 1 2

»   F x, f2 x 

Second, it follows by use of Taylor’s Theorem 3.3.5 that » 1   [f2 aε  0 »1

p q
f1paεq] F2 paε, f1 paεq




0

[f2 paq
f1 paq] F2 pa, f1 paq

¤ |f2paεq
f1 paεq| 

λ [f2 paε q
f1 paε q] q dλ λ [f2 paq


»1

  f1 a ] dλ

pqq

|F2paε, f1paεq λ [f2paεq
f1 paεq]q
F2pa, f1 paq λ [f2paq
f1 paq]q|dλ |f2paεq
f2 paq
f1paεq f1 paq| »1  |F2pa, f1paq λ [f2 paq
f1paq]q|dλ

¤

!

0

0

M2 pM3

M4 q



M7 pM5

M6 q 1

pM3

M4 q2

1{2 )

and that

» 1   [f2 bε  0 »1

p q
f1pbεq] F2 pbε, f1pbεq




0

[f2 pbq
f1 pbq] F2 pb, f1 pbq 494

λ [f2 pbε q
f1 pbε q] q dλ λ [f2 pbq


  f1 b ] dλ

pqq

ε

¤ |f2pbεq
f1pbεq| 

»1

|F2pbε, f1 pbεq λ [f2pbεq
f1 pbεq]q
F2pb, f1 pbq λ [f2 pbq
f1 pbq]q|dλ |f2pbεq
f2 pbq
f1pbε q f1pbq| »1  |F2pb, f1 pbq λ [f2 pbq
f1 pbq]q|dλ

¤

!

0

0

M2 pM3

M4 q

M7 pM5



M6 q 1

{)

 2 1 2

pM3 M4 q ε. Here Taylor M5 ¥ 0 denotes the maximum of |f1 | on ra, bs; M6 ¥ 0 denotes the maximum of |f2 | on ra, bs; M7 ¥ 0 denotes the maximum |∇F2| on some closed subset that is contained in V and at the same time

contains Ω. As a consequence, it follows that »

lim

ε

Ñ0

rε

F  drε



»

r

F  dr

and hence, finally, (3.6.5). Remark 3.6.8. Note that in the previous Theorem 3.6.7, the assumption of convexity Ω X ppa, a δ q  Rq for some 0   δ   pb
aq{2 is redundant if f1 paq  f2 paq, and the assumption of convexity Ω X ppb
δ, bq  Rq for some 0   δ   pb
aq{2 is redundant in the case f1 pbq  f2 pbq. Remark 3.6.9. Also, note that the generalizations of Green’s theorem to a larger class of regions of R2 can be achieved by considering such which can be dissected into regions that satisfy the demands of Theorem 3.6.5 or Theorem 3.6.7. Green’s theorem is then applied to the individual pieces. In this, cuts are traversed twice, but in opposite directions such that their contribution cancels in the sum. For such a case, see Example 3.6.11. Example 3.6.10. Let a, b be strictly positive real numbers such that a   b and * " x2 y 2  1 U : px, y q : 2 a b2 be the interior of the ellipse with half-axes a and b around the origin. Then r : r
π, π s Ñ R2 , defined by rpϕq : pa cospϕq, b sinpϕqq for all ϕ P 495

t Τ

T A Ξ-Τ

Ξ-HΤ-TL

Ξ

Ξ+HΤ-TL

Ξ+Τ

x

Fig. 155: Domain of integration in Example 3.6.9.

r
π, πs is a parametrization of this ellipse with positive orientation. Finally, define the vector field F : R2 Ñ R2 by F px, y q : 1{2.p
y, xq for all px, yq P R2. Then it follows by Theorem 3.6.7 »

dxdy U

 πab .



»

r

F  dr 

»

1 π p
b sin t, a cos tq  p
a sin t, b cos tq dt 2
π

In this way the area of the inner of the ellipse can be calculated by evaluation of a curve integral. Example 3.6.11. (An energy inequality for a wave equation in one space dimension) We consider a function u : U Ñ R of class C 2 that satisfies the wave equation B2u
B2u V u  0 , (3.6.8) Bt2 Bx2 where V : U Ñ R is continuous, assumes only positive values, i.e., RanpV q € r0, 8q, and is such that

BV  0 . Bt 496

In this, U is a non-empty open subset of R2 . Then the functions ǫ, j defined by ǫ :

1 2



B u 2 Bt





B u 2 Bx

V u2

, j :

Bu Bu Bx Bt

satisfy

B ǫ  B u B 2 u B u B 2 u V u B u  B u  B 2 u
V u

Bt Bt Bt2 Bx BtBx Bt Bt Bx2 B u B 2 u V u B u  B u B 2 u B u B 2u  B j . Bx BxBt Bt Bt Bx2 Bx BxBt Bx Hence we conclude the conservation law

Note for later use that

Bj
Bǫ  0 . Bx Bt

(3.6.9)

ǫpx, tq ¥ |j px, tq| .

(3.6.10)

for all px, tq P U. In physical applications, ǫ is called the energy density (corresponding to u) and j is called the energy flux density (corresponding to u). Integration of ǫp, tq over an interval of R gives the energy of u that is contained in that interval at time t P R. The function j describes the flow of that energy. In the following, we derive an important consequence of (3.6.9). For this, let pξ, τ q P R  p0, 8q, and let the area enclosed by the triangle with corners pξ
τ, 0q, pξ τ, 0q and pξ, τ q be contained in U. We integrate (3.6.9) over the subarea enclosed by the trapezoid with corners pξ
τ, 0q, pξ τ, 0q, pξ pτ
T q, T q and pξ
pτ
T q, T q, where 0 ¤ T   t. We will show that the energy content at time T in the interval rξ
pτ
T q, ξ pτ
T qs is equal or smaller than the energy content at time 0 in the interval rξ
τ, ξ τ s » ξ pτ
T q ξ


pτ
T q

ǫpx, T q dx ¤ 497

»ξ




τ

ξ τ

ǫpx, 0q dx .

(3.6.11)

Indeed, it follows that 0

»  A

Bj
Bǫ dxdt  ¸3 »  Bj
Bǫ dxdt  » pǫ, j q  dr , Bx Bt Bx Bt r i1 A i


where r is the piecewise C 2 -path pr1 , r2 , r
3 , r4 q r1 py1 q : py1 , 0q , r2 pλq : pξ τ
λT, λT q , r3 py3 q : py3 , T q , r4 pλq : pξ
τ λT, λT q for all y1 P rξ
τ, ξ τ s, y3 P rξ
pτ
T q, ξ pτ
T qs and λ P r0, 1s. Note that in this, A is dissected into the area A1 enclosed by the triangle with corners pξ
τ, 0q, pξ
pτ
T q, 0q, pξ
pτ
T q, T q, the area A2 enclosed by the rectangle with corners pξ
pτ
T q, 0q, pξ pτ
T q, 0q, pξ pτ
T q, T q, pξ
pτ
T q, T q, the area A3 enclosed by the triangle with corners pξ pτ
T q, 0q,pξ τ, 0q,pξ pτ
T q, T q, and apply Green’s Theorem 3.6.7 to these surfaces. The cuts are traversed twice, but in opposite directions such that their contribution cancels in the sum as indicated in Fig. 155. Further, we conclude that 0

»ξ

τ




ξ τ »1

ǫpx, 0q dx


» ξ pτ
T q ξ


pτ
T q

ǫpx, T q dx

»1 0

pǫ, j qpr2pλqq  p
T, T q dλ


pǫ, j qpr4pλqq  pT, T q dλ 0

and hence that »ξ

τ




ξ τ

ǫpx, 0q dx


»1

» ξ pτ
T q ξ


pτ
T q

ǫpx, T q dx

 pǫ, j qpr2 pλqq  pT,
T q dλ 0 »1

¥ pǫ, |j |qpr2pλqq  pT,
T q dλ 0

»1 0

pǫ, j qpr4 pλqq  pT, T q dλ

»1 0

498

pǫ, |j |qpr4pλqq  pT,
T q dλ ¥ 0 ,

where in the last step (3.6.10) has been used. Hence it follows (3.6.11). As an application of the energy inequality (3.6.11), we assume that v : U Ñ R is another solution of (3.6.8) such that upx, 0q  v px, 0q , for all x P rξ
τ, ξ

Bu px, 0q  Bv px, 0q Bt Bt

τ s. Then u
v is a solution of (3.6.8) such that

pu
vqpx, 0q  0 , BpuB
t vq px, 0q  0 for all x P rξ
τ, ξ τ s and hence the corresponding energy density vanishes at time 0 on rξ
τ, ξ τ s. As a consequence of (3.6.11) and the positivity of the energy density, it follows that the same is true at time T on rξ
pτ
T q, ξ pτ
T qs. Since this is true for every t P r0, τ q and since u
v is continuous, it follows that u and v coincide in every point from the closed area that is enclosed by the triangle with corners pξ
τ, 0q, pξ τ, 0q and pξ, τ q. Note that this triangle is isosceles with a right angle and π{4 radian angles at the corners pξ
τ, 0q, pξ τ, 0q. In addition, note that

pξ, τ q  p r pξ
τ q pξ

τ q s {2,

r pξ

τ q
pξ
τ q s {2 q .

As a consequence, we have the following result. Theorem 3.6.12. (Uniqueness of the solutions of a wave equation in one space dimension) Let U be an non-empty open subset of R2 and u : U Ñ R, v : U Ñ R be of class C 2 and such that

B 2 u
B 2u V u  B 2 v
B 2 v V v  0 , Bt2 Bx2 Bt2 Bx2 where V : U Ñ R is continuous, assumes only positive values, RanpV q € r0, 8q, and satisfies BV  0 . Bt 499

i.e.,

Further, let

Bu px, t q  Bv px, t q Bt 0 Bt 0 for some t0 P R and all x from some closed interval ra, bs of R, where a, b P R are such that a   b. Then upx, tq  v px, tq for all px, tq from the upx, t0 q  v px, t0 q ,

closed area that is enclosed by the isosceles right triangle with corners

pa, t0 q , pb, t0 q , ppa bq{2, t0 pb
aq{2q . Proof. For the case t0  0, the result was proved in the previous example. If t0  0, then U0 : tpx, t
t0 q : px, tq P U u, V0 : pU0 Ñ R, px, tq ÞÑ V px, t t0 qq, u0 : pU0 Ñ R, px, tq ÞÑ upx, t t0 qq, v0 : pU0 Ñ R, px, tq ÞÑ v px, t t0 qq satisfy the assumptions of the theorem for the case t0  0. Hence it follows that upx, t t0 q  u0 px, tq  v0 px, tq  v px, t t0 q for all px, tq from the closed area that is enclosed by the triangle with corners

pa, 0q , pb, 0q , ppa bq{2, pb
aq{2q . Hence it follows that upx, tq  v px, tq for all px, tq from the closed area that is enclosed by the triangle with corners

pa, t0 q , pb, t0 q , ppa

bq{2, t0

pb
aq{2q .

Definition 3.6.13. (Parametric surfaces) Let p P N . A C p -parametric surface (in R3 ) is a pair pS, rq consisting of a subset S of R3 and an injective map (‘parametrization’) r of class C p from some open subset of R2 into R3 with range S. To pS, rq there is an associated normal field given by

Br  Br . Bx By Hence if px0 , y0 q is such that npx0 , y0q  0, the tangent plane to S rpx0 , y0 q P S is given by npx0 , y0 q  ppx, y, z q
rpx0 , y0 qq  0 . n :

500

in

Example 3.6.14. (i) Let p P N and f be a function of class C p defined on some nonempty open subset U of R2 . Then pGpf q, rf q is a C p -parametric surface where rf px, y q : px, y, f px, y qq for all px, y q P U. The corresponding normal field n is given by

B B f f npx, y q 
px, y q,
px, y q, 1 , Bx By px, yq P U, and the tangent plane at Gpf q in a point px0 , y0, f px0 , y0qq 

is given by

 f px0 , y0q BBfx px0 , y0q  px
x0 q BBfy px0 , y0q  py
y0q for all px, y q P R2 . The last is identical to the definition given in z

Definition 3.2.10.

(ii) Denote by S 2 the sphere of radius 1 centered at the origin. Then pS 2 zp
8, 0s  t0u  R, rq is a C p-parametric surface for every p P N . Here rpθ, ϕq : psin θ cos ϕ, sin θ sin ϕ, cos θq for all θ P p0, π q, ϕ given by θ

P p
π, πq. The corresponding normal field n is npθ, ϕq  sin θ . rpθ, ϕq ,

P p0, πq, ϕ P p
π, πq.

(iii) Denote by Z 2 the cylinder of radius 1 with axis given by the z-axis. Then pZ 2 zpp
8, 0s  t0u  Rq, rq is a C p -parametric surface for every p P N where rpϕ, z q : pcos ϕ, sin ϕ, z q 501

for all ϕ P p
π, π q, z P R. The corresponding normal field n is given by npϕ, z q  pcos ϕ, sin ϕ, 0q , ϕ P p
π, π q, z

P R.

(iv) Denote by T 2 the torus obtained by rotating around the z-axis the circle of radius r ¡ 0 in the y, z plane centered at the point p0, R, 0q, where R ¡ r. Then 

T2 z



p
8, 0s  t0u  Rq Y S 1pR
rq  t0u , r is a C p -parametric surface for every p P N where rpϕ, θq : pcos ϕ pR r cos θq, sin ϕ pR r cos θqr sin θq for all ϕ, θ P p
π, π q. The corresponding normal field n is given by npϕ, θq  pR r cos θq.pr cos ϕ cos θ, r sin ϕ cos θ, r sin θ
Rq for all ϕ, θ P p
π, π q.

3.6.2 Stokes’ Theorem Example 3.6.15. (Motivation for the definition of the flux of a vector field across a surface) Consider a constant flow v  pvx , vy , vz q (length / time) of a fluid with constant mass density ρ (mass / volume) across a rectangle R of lengths △y, △z in the y, z-plane. Imagine R to be part of a closed surface, such that the outer normal to R is given by n : ex . Then the change of mass inside the volume after time △t due to flow across R is given by ρvx △t△y△z .

Note that it is negative if vx   0 because of our use of the outer normal. ‘Inflow’ pvx   0q is counted negatively, whereas ‘outflow’ pvx ¡ 0q is counted positively. The rate of change of mass in the volume due to flow across R is given by 

B r Br ρvx △y△z  ρv  n dydz  ρv  By  Bz dydz , R R »

»

502

(3.6.12)

1 z

0

2

-1 0

y

-2 0

-2

x

2

Fig. 156: Torus corresponding to r

 1 and R  2.5.

where the parametrization rpy, z q : p0, y, z q for all py, z q from the projection of R into the py, z q-plane has been used. In addition, note that in the special case ρ  1,   r   y




1  B r B r

B B r  vn B  Bz  . By  Bz

it follows from (3.6.12) that

»   r   y R

,



B  Br  dydz , B Bz 

coincides with the area of R. Definition 3.6.16. (Flux of a vector field across a C 1 -parametric surface) Let pS, rq be a C 1 -parametric surface and F : S Ñ R3 be a continuous vector field on S. Finally, let 

B r Br pF  rq  Bx  By

503

1 Dz 0 0

2 1 v Dt

1 Dy

2 3

Fig. 157: Fluid volume flown through R after time △t △y  2m, △z  1m.

0

 7sec for v  p0.5, 0, 0q m/sec,

be Riemann-integrable. Then we define the flux of F across S by

B B r r F  dS : F prpx, y qq  Bx px, yq  By px, yq dxdy . S Dprq

»



»

In particular, we define the area A of S as the flux (if existent) corresponding to the special case that F coincides with the unit normal field induced by r. Hence A is defined by A : if

»

pq

D r

  r   x x, y



B p q  Br px, yq dxdy ,  B By   r   x



B  Br  B By 

is Riemann integrable. Theorem 3.6.17. (Invariance under reparametrization) Let pS, rq be a C 1 -parametric surface and F : S Ñ R3 be a continuous vector field on S 504



B r Br pF  rq  Bx  By

such that

is Riemann-integrable. Moreover, let V be an open subset of R2 , g : V Ñ Dpf q be continuously differentiable with a continuously differentiable inverse and such that detpg 1 q ¡ 0. Then pS, r  g q is a C 1 -parametric surface and 

Bp Bp r  gq r  gq F ppr  g qps, tqq  ps, tq  Bt ps, tq dsdt B s V 

» r r B B  F prpx, y qq  (3.6.13) Bx px, yq  By px, yq dxdy .

»

pq

D r

Proof. By the chain rule for partial derivatives Corollary 3.2.26, it follows that

Bpr  gq ps, tq  Bg1 ps, tq. Br pgps, tqq Bg2 ps, tq. Br pgps, tqq Bs Bs Bx Bs By Bpr  gq ps, tq  Bg1 ps, tq. Br pgps, tqq Bg2 ps, tq. Br pgps, tqq Bt Bt Bx Bt By and hence that

Bpr  gq ps, tq  Bpr  gq ps, tq Bs  Bt

B r Br 1  detpg ps, tqq Bx  By pgps, tqq for all ps, tq P V where g1 , g2 are the component maps of g. Hence it follows (3.6.13) and finally the theorem follows by the change of variable formula Theorem 3.4.20. Example 3.6.18. Calculate the flux of the vector field F px, y, z q : pz, x, 1q, px, y, zq P R3 , across the surface S : tpx, y, z q P R3 : z

¥ 0, x2

505

y2

z

 1u .

1 z

0 0 y 0 x

Fig. 158: Sketch of S from Example 3.6.18.

Solution: Define

rpx, y q : px, y, 1
x2
y 2q

for all px, y q P R2 such that x2 y 2 ¤ 1. Then pS, rq is a C 2 -parametric surface and the corresponding flux across S is given by

r r B B F  dS  F prpx, y qq  px, yq  By px, yq dxdy B x S Dprq »  p1
x2
y2, x, 1q  p2x, 2y, 1q dxdy

»



»

pq

D r



»

pq

D r

π

2xp1
x2
y 2q

»1»π 0


π

2xy

2r 2 p1
r 2 q cospϕq



1 dxdy 

r 3 sinp2ϕq drdϕ  π .

Example 3.6.19. Let f be a function of class C 1 defined on an non-empty bounded open subset U of R2 . Then pGpf q, rf q is a C 1 -parametric surface, 506

where

rf px, y q : px, y, f px, y qq

for all px, y q P U. The corresponding normal field n is given by

f f B B npx, y q 
px, y q,
px, y q, 1 , Bx By If |n| : U Ñ R is Riemann integrable, the surface area A of 

px, yq P U. Gpf q is given by

A

» a

1

U

|p∇f qpx, yq|2 dxdy .

Example 3.6.20. (Area of a surface of revolution) Let a, b P R such that a   b, f : ra, bs Ñ r0, 8q be a continuous function with a finite set Nf of zeros which is the restriction of a continuously differentiable function defined on a open interval of R containing ra, bs and S :

px, y, zq P R3 : px2

y 2 q1{2

 f pzq ^ z P ra, bs

(

.

Note that S is rotational symmetric around the z-axis and can be thought of as obtained from a curve in x, z-plane that is rotated around the z-axis. An injective parametrization of class C 1 of S zN, where (

N :

px, 0, zq P R3 : x 
f pzq ^ z P ra, bs Y tpx, y, aq P R3 : px2 y2q1{2  f paqu Y tpx, y, bq P R3 : px2 y2q1{2  f pbqu Y tp0, 0, zq P R3 : z P Nf u , is given by r : p
π, π q  ppa, bqzNf q Ñ R3 defined by rpϕ, z q : pf pz q cos ϕ, f pz q sin ϕ, z q for all pϕ, z q P p
π, π q  ppa, bqzNf q. In particular, Br pϕ, zq  p
f pzq sin ϕ, f pzq cos ϕ, 0q , Bϕ 507

Br pϕ, zq  pf 1pzq cos ϕ, f 1pzq sin ϕ, 1q , Bz Br pϕ, zq  Br pϕ, zq  pf pzq cos ϕ, f pzq sin ϕ,
f pzqf 1 pzqq Bϕ Bz and hence



  r   ϕ ϕ, z

B p q  Br pϕ, zq  f pzq 1 pf 1pzqq2 1{2  B Bz for all pϕ, z q P p
π, π q  ppa, bqzNf q. Since    Br  B r     Bϕ Bz  is Riemann integrable over p
π, π q  ppa, bqzNf q, we define the area of S by

»

  r   ϕ ϕ, z



B p q  Br pϕ, zq dϕdz A :  By Dprq B »b    2π f pzq 1 pf 1pzqq2 1{2 dz . a

Theorem 3.6.21. (Stokes’ theorem) Let pS, rq be a C 2 -parametric surface and F be a vector field of class C 1 defined on an open set containing S.1 In particular, let r be such that (i) Its domain Ω is a non-empty bounded open subset of R2 for which Green’s theorem is valid, i.e., there is a piecewise C 1 -path α : I Ñ R2 defined on some non-empty closed interval of I € R and traversing the boundary of Ω such that Green’s identity » 

Bf2
Bf1 dxdy  » f  dα (3.6.14) Bx By Ω α is valid for every continuously differentiable f  pf1 , f2 q : V Ñ R2 defined on some open subset V of R2 containing Ω and its boundary.

1

Note that we do not make any further assumptions on the normal field.

508

y

S

1 W

z x

0 0 y 0

¶W

¶S

x

Fig. 159: Illustration for the proof of Stokes’ theorem, Theorem 3.6.21.

(ii) r is the restriction to Ω of a map ˆ r of class C 2 defined on an open subset containing Ω and its boundary. »

Then

S

curl F  dS 

» γ

F  dγ ,

where γ is any piecewise C 1 -parametrization of B S : Ranpˆ r  αq which is of the same orientation as ˆ r  α. Proof. In the following, for simplicity of notation, we denote ˆ r by the sym1 bol ρ. First, since ρ  α is a piecewise C -path, it follows that »



ρ α

F  dpρ  αq 



i 1

α

rpFi  ρq ∇ρi s  d α

   * ρi ρi B B B B  Bx pFi  ρq By
By pFi  ρq Bx dx dy . i1 Ω Further, it follows for every i P t1, 2, 3u that B pF  ρq Bρi 
B pF  ρq Bρi   BpFi  ρq Bρi
BpFi  ρq Bρi Bx i By By i Bx Bx By By Bx 3 » ¸

"

3 » ¸

509





   B B B B B B Fi ρ1 Fi ρ2 Fi ρ3 B ρi  Bx  ρ Bx  ρ Bx  ρ Bx By B x2 B x3 1



    Fi ρ1 Fi ρ2 Fi ρ3 B ρi B B B B B B
Bx  ρ By Bx2  ρ By Bx3  ρ By Bx 1 

and by using

Bρ  Bρ Bx  By

ρ2 B ρ3 B ρ3 B ρ2 B ρ3 B ρ1 B ρ1 B ρ3 B ρ1 B ρ2 B ρ2 B ρ1 B  Bx By
Bx By , Bx By
Bx By , Bx By
Bx By that

B pF  ρq Bρ1 
B pF  ρq Bρ1  Bx  1 B y By 1 Bx   BBFx1  ρ BBρx2 BBFx1  ρ BBρx3 BBρy1 2 3



   F1 ρ2 F1 ρ3 B ρ1 B B B B
Bx  ρ By  ρ By Bx B x3 2  





B B F1 ρ Bρ F1 ρ Bρ B B 
Bx  ρ Bx  By Bx  ρ Bx  By 2

3

3

B pF  ρq Bρ2 
B pF  ρq Bρ2  Bx  2 B y By 2 Bx   BBFx2  ρ BBρx1 BBFx2  ρ BBρx3 BBρy2 1 3



   F2 ρ1 F2 ρ3 B ρ2 B B B B
Bx  ρ By  ρ By Bx B x3 1





  B B B B F2 ρ Bρ F2 ρ Bρ  Bx  ρ Bx  By
Bx  ρ Bx  By 1

3

3

B pF  ρq Bρ3 
B pF  ρq Bρ3  Bx 3 By By 3 Bx 510

, 2

, 1



  B B B B F3 ρ1 F3 ρ2 B ρ3  Bx  ρ Bx  ρ Bx By B x2 1



   F3 ρ1 F3 ρ2 B ρ3 B B B B
Bx  ρ By  ρ By Bx B x2 1





  B B B B F3 ρ Bρ F3 ρ Bρ 
Bx  ρ Bx  By Bx2  ρ Bx  By 1 1 2 

Hence by using that 

B F3 B F2 B F1 B F3 B F2 B F1 curl F  Bx2
Bx3 , Bx3
Bx1 , Bx1
Bx2

,

we arrive at

B pF  ρq Bρi 
B pF  ρq Bρi *  Bx i By By i Bx i1  





B B F1 ρ Bρ F1 ρ Bρ B B
Bx  ρ Bx  By  ρ Bx  By B x3 2 3





2   BF2  ρ Bρ  Bρ
BF2  ρ Bρ  Bρ Bx1  Bx By 3  Bx3  Bx By 1  BF3  ρ Bρ  Bρ
BBFx3  ρ BBxρ  BBρy Bx2 Bx By 1 1 2

  rpcurl F q  ρs  BBxρ  BByρ . 3 ¸

"

Hence, finally, it follows that

ρ Bρ B F  dpρ  αq  rpcurl F q  ρs   dx dy B x By ρα Ω »  curl F  dS .

»



»

S

511

.

Example 3.6.22. Let S and F be as in Example 3.6.18. A simple calculation gives that F px, y, z q  pcurl Aqpx, y, z q for all x, y, z

P R, where

Apx, y, z q : for all x, y, z



2 2
y,
z2 , x2



P R. Hence it follows by Theorem 3.6.21 that »

S

F  dS 

»

BS

A  dr .

The boundary B S is given by the circle of radius 1 in the x, y-plane centered at the origin. Note that rpB S q  ιpB S q

where ι is the inclusion of R2 into R3 given by ιpx, y q : px, y, 0q for every px, y q P R2 . Therefore, a C 1 -parametrization of B S satisfying the assumptions of Theorem 3.6.21 is given by rptq : pcos t, sin t, 0q for all t P r
π, π s. Hence »

S

F  dS  »π

»π


π




sinptq, 0, cos2 ptq{2  p
sinptq, cosptq, 0q dt

 sin ptq dt 
π π



»π

2

π 

 1 1 sinp2tq p 1
cosp2tqq dt  π
4
π 2
π

which is identical to the result of Example 3.6.18. 3.6.3 Gauss’ Theorem Gauss’ theorem can be considered as a generalization of Green’s theorem to 3 space dimensions. We start with Gauss’ theorem for images of cuboids under certain differentiable maps. The basis for its proof is given by the following Lemma. 512

Lemma 3.6.23. Let V be a non-empty open subset of R3 and F  pF1 , F2 , F3 q : V Ñ R3 be differentiable. Further, let g  pg1 , g2 , g3 q : Dpg q Ñ R3 be defined and of class C 2 on a non-empty open subset Dpg q of R3 and such that g pDpg qq € V . Then

B pF  gq   Bg  Bg  B pF  gq   Bg  Bg  Bx  By Bz  By Bz Bx B pF  gq  Bg  Bg  rpdiv F q  gs  detpg 1q , Bz Bx By

where

div F :

BF1 BF2 BF3 Bx By Bz

.

Proof. The proof proceeds by a simple calculation using the chain rule in the form of Corollary 3.2.26 and Schwarz’s theorem 3.2.19. In the following, we indicate partial derivatives in the coordinate directions x, y, z by the index , x , , y and , z , respectively. In first step, we prove that

pg,y  g,z q,x pg,z  g,xq,y pg,x  g,y q,z  0 . For this, we note that g,y  g,z  pg2,y g3,z
g3,y g2,z , g3,y g1,z
g1,y g3,z , g1,y g2,z
g2,y g1,z q g,z  g,x  pg2,z g3,x
g3,z g2,x , g3,z g1,x
g1,z g3,x , g1,z g2,x
g2,z g1,x q g,x  g,y  pg2,x g3,y
g3,x g2,y , g3,x g1,y
g1,x g3,y , g1,x g2,y
g2,x g1,y q . Hence

pg,y  g,z q1,x pg,z  g,xq1,y pg,x  g,y q1,z  pg2,y g3,z
g3,y g2,z q,x pg2,z g3,x
g3,z g2,xq,y pg2,x g3,y
g3,x g2,y q,z  g2,yx g3,z
g3,yx g2,z g2,y g3,zx
g3,y g2,zx g2,zy g3,x
g3,zy g2,x g2,z g3,xy
g3,z g2,xy g2,xz g3,y
g3,xz g2,y g2,x g3,yz
g3,x g2,yz 0, pg,y  g,z q2,x pg,z  g,xq2,y pg,x  g,y q2,z 513

 pg3,y g1,z
g1,y g3,z q,x pg3,z g1,x
g1,z g3,xq,y pg3,x g1,y
g1,x g3,y q,z  g3,yx g1,z
g1,yx g3,z g3,y g1,zx
g1,y g3,zx g3,zy g1,x
g1,zy g3,x g3,z g1,xy
g1,z g3,xy g3,xz g1,y
g1,xz g3,y g3,x g1,yz
g1,x g3,yz 0, pg,y  g,z q3,x pg,z  g,xq3,y pg,x  g,y q3,z  pg1,y g2,z
g2,y g1,z q,x pg1,z g2,x
g2,z g1,xq,y pg1,x g2,y
g2,x g1,y q,z  g1,yx g2,z
g2,yx g1,z g1,y g2,zx
g2,y g1,zx g1,zy g2,x
g2,zy g1,x g1,z g2,xy
g2,z g1,xy g1,xz g2,y
g2,xz g1,y g1,x g2,yz
g2,x g1,yz 0. Since according to the chain rule Corollary 3.2.26

pF  gq,x  g1,x.pF,x  gq pF  gq,y  g1,y .pF,x  gq pF  gq,z  g1,z .pF,x  gq

g2,x .pF,y  g q g2,y .pF,y  g q g2,z .pF,y  g q

g3,x .pF,z  g q , g3,y .pF,z  g q , g3,z .pF,z  g q ,

it follows in a second step that

rpF  gq  pg,y  g,z qs,x rpF  gq  pg,z  g,xqs,y rpF  gq  pg,x  g,y qs,z  pF  gq,x  pg,y  g,z q pF  gq,y  pg,z  g,xq pF  gq,z  pg,x  g,y q  rg1,x.pF,x  gq g2,x.pF,y  gq g3,x.pF,z  gqs  pg,y  g,z q rg1,y .pF,x  gq g2,y .pF,y  gq g3,y .pF,z  gqs  pg,z  g,xq rg1,z .pF,x  gq g2,z .pF,y  gq g3,z .pF,z  gqs  pg,x  g,y q  pF,x  gq  rg1,x.pg,y  g,z q g1,y .pg,z  g,xq g1,z .pg,x  g,y qs pF,y  gq  rg2,x.pg,y  g,z q g2,y .pg,z  g,xq g2,z .pg,x  g,y qs pF,z  gq  rg3,x.pg,y  g,z q g3,y .pg,z  g,xq g3,z .pg,x  g,y qs . Further, it follows that g1,x .pg,y  g,z q g1,y .pg,z  g,x q g1,z .pg,x  g,y q  g1,x.pg2,y g3,z
g3,y g2,z , g3,y g1,z
g1,y g3,z , g1,y g2,z
g2,y g1,z q 514

g1,y .pg2,z g3,x
g3,z g2,x , g3,z g1,x
g1,z g3,x , g1,z g2,x
g2,z g1,x q g1,z .pg2,x g3,y
g3,x g2,y , g3,x g1,y
g1,x g3,y , g1,x g2,y
g2,x g1,y q  pdetpg 1 q, 0, 0q , g2,x .pg,y  g,z q g2,y .pg,z  g,x q g2,z .pg,x  g,y q  g2,x.pg2,y g3,z
g3,y g2,z , g3,y g1,z
g1,y g3,z , g1,y g2,z
g2,y g1,z q g2,y .pg2,z g3,x
g3,z g2,x , g3,z g1,x
g1,z g3,x , g1,z g2,x
g2,z g1,x q g2,z .pg2,x g3,y
g3,x g2,y , g3,x g1,y
g1,x g3,y , g1,x g2,y
g2,x g1,y q  p0, detpg 1q, 0q , g3,x .pg,y  g,z q g3,y .pg,z  g,x q g3,z .pg,x  g,y q  g3,x.pg2,y g3,z
g3,y g2,z , g3,y g1,z
g1,y g3,z , g1,y g2,z
g2,y g1,z q g3,y .pg2,z g3,x
g3,z g2,x , g3,z g1,x
g1,z g3,x , g1,z g2,x
g2,z g1,x q g3,z .pg2,x g3,y
g3,x g2,y , g3,x g1,y
g1,x g3,y , g1,x g2,y
g2,x g1,y q  p0, 0, detpg 1 qq and hence, finally, that

rpF  gq  pg,y  g,z qs,x rpF  gq  pg,z  g,xqs,y rpF  gq  pg,x  g,y qs,z  rdivpF q  gs detpg 1 q . Theorem 3.6.24. (Gauss’ theorem for images of cuboids) Let a1 , b1 , a2 , b2 , a3 , b3 P R be such that ai   bi for i  1, 2, 3 and I : ra1 , b1 sra2 , b2 s ra3, b3 s, I0 : pa1, b1 q  pa2, b2 q  pa3 , b3q. Further, let U  I be an open subset of R3 , g : U Ñ R3 be twice continuously differentiable such that the induced map from U to g pU q is bijective with a continuously differentiable inverse and such that detpg 1 q ¡ 0. Finally, let V  g pI q be an open subset of R3 and F  pF1 , F2 , F3 q : V Ñ R3 be continuously differentiable. Then »

p q

divF dxdydz

g I0

»

Sb2

F  dS



» Sa 3

»

Sa 1

F  dS

F  dS

» Sb3

515

»

Sb1

F  dS

F  dS ,

»

Sa 2

F  dS (3.6.15)

z

gHIL

y x

Fig. 160: g pI q and some outer normal vectors. Illustration for the proof of Gauss’ theorem, Theorem 3.6.24.

516

where pSa1 , ra1 q, pSb1 , rb1 q, pSa2 , ra2 q, pSb2 , rb2 q, pSa3 , ra3 q, pSb3 , rb3 q are C 2 -parametric surfaces defined by ra1 py, z q : g pa1, z, y q , rb1 py, z q : g pb1 , y, z q , ra2 px, z q : g px, a2 , z q , rb2 px, z q : g pz, b2 , xq , ra3 px, y q : g py, x, a3q , rb3 px, y q : g px, y, b3q for all x P pa1 , b1 q, y

P pa2, b2 q and z P pa3, b3 q and Sa : Ranpra q , Sb : Ranpra q , Sa : Ranpra q , Sb : Ranprb q , Sa : Ranpra q , Sb : Ranprb q . Proof. In a first step, we consider the set g pI0q. Since g is twice continuously differentiable with a continuously differentiable inverse, g pI0 q is a bounded open subset in R3 . Further, the restriction of divF to g pI0q is 1

1

1

1

3

3

3

3

2

2

2

2

bounded. In addition, it follows by Theorem 3.4.13 and Theorem 3.4.15 that the extension of divF to a function, defined on a closed subinterval J of R3 containing g pI0q and assuming the value zero in the points of J z g pI0q, is Riemann-integrable. Hence by Theorem 3.4.20, it follows in a second step that »

p q

divF dxdydz

g I0



» I0

rpdivF q  gs detpg 1q dxdydz

and hence by the previous Lemma 3.6.23 that »

p q

divF dxdydz

g I0



  B g Bg B  Bx pF  gq  By  Bz dxdydz I » B pF  gq   Bg  Bg  dxdydz Bz Bx  I By   » B pF  gq  Bg  Bg dxdydz . Bx By I Bz »

0

0

0

Finally, from this follows (3.6.15) by Fubini’s Theorem 3.4.17 and the fundamental theorem of calculus Theorem 1.5.20. 517

Remark 3.6.25. Since the region g pI0 q is bounded, there is an outward pointing unit normal for every point on its boundary, apart from the corner points g pa1 , a2 , a3 q, g pa1 , b2 , a3 q, g pa1 , a2 , b3 q, g pa1 , b2 , b3 q, g pb1 , a2 , a3 q, g pb1 , b2 , a3 q, g pb1 , a2 , b3 q and g pb1 , b2 , b3 q. Outward pointing vectors are given by 

g1 B g2 B g3 B va pa1 , y, z q 
pa1, y, zq, Bx pa1 , y, zq, Bx pa1 , y, zq , B x 

B g1 B g2 B g3 vb pb1 , y, z q  Bx pb1 , y, zq, Bx pb1 , y, zq, Bx pb1 , y, zq ,

Bg1 px, a , zq, Bg2 px, a , zq, Bg3 px, a , zq , va px, a2 , z q 
By 2 By 2 By 2

 Bg1 px, b , zq, Bg2 px, b , zq, Bg3 px, b , zq , vb px, b2 , z q  By 2 By 2 By 2

g1 B g2 B g3 B va px, y, a3 q 
px, y, a3q, Bz px, y, a3q, Bz px, y, a3q , B z 

B g1 B g2 B g3 vb px, y, b3 q  Bz px, y, b3q, Bz px, y, b3q, Bz px, y, b3q , 1

1

2

2

3

3

for every x P pa1 , b1 q, y P pa2 , b2 q and z P pa3 , b3 q. Normal vectors on the boundary corresponding to the parametrizations ra1 , rb1 , ra2 , rb2 , ra3 , rb3 are given by

B g Bg na pa1 , y, z q : pa1 , y, zq ,  B z By

 g Bg B nb pb1 , y, z q : pb1 , y, zq ,  B y Bz

 B g Bg px, a2, zq ,  na px, a2 , z q : B x Bz

 g Bg B nb px, b2 , z q : px, b2 , zq ,  B z Bx 

B g Bg na px, y, a3q : By  Bx px, y, a3q , 

1

1

2

2

3

518



B g Bg nb px, y, b3 q : Bx  By px, y, b3q P pa1 , b1q, y P pa2, b2 q and z P pa3, b3 q. 

3

for every x In particular, as a consequence of (2.5.6) and the assumption in the previous Theorem 3.6.24 that detpg 1 q ¡ 0 , it follows that the orthogonal projections of the outgoing vectors onto the corresponding normal vectors are everywhere ¡ 0. Hence the parametrizations of the boundary surfaces Sa1 , Sb1 , Sa2 , Sb2 , Sa3 , Sb3 in Theorem 3.6.24 has to be such that the corresponding normal vectors point out of g pI0 q in every point of the boundary, apart from finitely many points, as is indicated in Fig. 160. Remark 3.6.26. Generalizations of Gauss’ theorem to a larger class of regions of R3 can be achieved by considering such which can be dissected into regions that satisfy the demands of Theorem 3.6.24. Gauss’ theorem is then applied to the individual pieces. In this, integration over cuts are performed twice, but with normal fields in opposite directions such that their contribution cancels in the sum. We will not give such generalizations in the following. The regions considered in the following examples and problems satisfy the requirements of such more general theorems. Example 3.6.27. Calculate » S

F  dS ,

using Gauss’s Theorem, where


F px, y, z q : xy, y 2

exz , sinpxy q 2



, x, y, z

PR

and S is the surface of the region E in the first octant bounded by the parabolic cylinder z  1
x2 and the planes y  0, z  0 and y z  2. Solution: By Gauss’ and Fubini’s Theorem, it follows » S

F  dS 

»

3y dxdydz E



» 1 » 1
x2 » 2
z


1 519

0

0



3y dy dxdz

1 0.5

z

-1

0

-0.5

0.5 y

1

0

x

0.5 1

Fig. 161: Sketch of V .

» 1 » 1
x2



3 2
1



1 1 3 1 7x
x3
x5
x7  2 5 7
1



0

p2
zq dz



2

dx 



»

1 1 p 7
3x2
3x4
x6 qdx 2
1

.  184 35

Example 3.6.28. (An energy inequality for a wave equation in two space dimensions) We consider a function u : U Ñ R of class C 2 that satisfies the wave equation B2u
△u V u  0 , (3.6.16) Bt2 where V : U Ñ R is continuous, assumes only positive values, i.e., RanpV q € r0, 8q, and is such that

BV  0 . Bt

In this, U is a non-empty open subset of R3 . In addition, we define for every partially differentiable, twice partially differentiable f : U Ñ R and 520

Ñ R2 p∇f qpt, x, yq : r∇f pt, qspx, yq , div F : rdiv Fpt, qspx, yq , p△f qpt, x, yq : r△f pt, qspx, yq respectively, for all pt, x, y q P U. Then the function ǫ and the vector field j partially differentiable F : U

defined by

ǫ :

1 2



Bu 2 |∇u|2 Bt



V u2

, j :

Bu ∇u Bt

satisfy 



Bǫ
div  Bu ∇u  B 1  Bu 2 |∇u|2 V u2
div  Bu ∇u

Bt Bt Bt 2 Bt Bt

 2  BBut BBtu2 p∇uq  ∇ BBt u V u BBut
∇ BBt u  p∇uq
BBut △u 

u B2u B  Bt Bt2
△u V u  0 . Hence we conclude the conservation law div j


Bǫ  0 . Bt

(3.6.17)

Note for later use that ǫpx, y, tq ¥ |jpx, y, tq| .

(3.6.18)

for all px, y, tq P U. In physical applications, ǫ is called the energy density (corresponding to u) and j is called the energy flux density (corresponding to u). Integration of ǫp, tq over subsets of R2 (if the corresponding integral exists) gives the energy of u that is contained in that subset at time t P R. The vector field j describes the flow of that energy. In the following, we derive an important consequence of (3.6.17). For this, let px, y, tq P R3 521

T t

Τ t-T y

-t

Η -Ht-TL

-Ht-TL x Ξ

t-T t

-t

Fig. 162: Sketch of the domain of integration in Example 3.6.28.

be such that t ¡ 0. Further, let T P r0, tq. We define the solid backward
with apex px, y, tq by characteristic cone SCx,y,t


: SCx,y,t

pξ, η, τ q P R3 : τ ¤ t
|px
ξ, y
ηq|

(

.

Further, we assume that U Then

»

p q

Bt2
T x,y


X r0, T s  R2  SCx,y,t

ǫpuqpT, q dξdη ¤

»

p q

Bt2 x,y



.

ǫpuqp0, q dξdη .

(3.6.19)

This can be shown as follows. It follows from (3.6.17) by Gauss’ Theorem that 0



»


SCx,y,t

Xp r0,T sR2 q

Bǫ
∇  Bu ∇u  pξ, η, τ q dξdηdτ Bt Bt 522

 

B u ǫ,
∇u  v dS Bt Xp r0,T sR qq » ǫpT, q dξdη
ǫp0, q dξdη 

»


BpSCx,y,t

»

p q

Bt2
T x,y

»

2


Cx,y,t

p q

Bt2 x,y



Xp r0,T sR2 q

B u ǫpuq,
∇u  v dS Bt

where v denotes the outer unit normal field on the boundary surface

B


X SCx,y,t

r0, ts  R2




X of SCx,y,t


is given by f : R2 A parametrization of Cx,y,t

r0, ts  R2



.

Ñ R3 defined f pξ, η q : pξ, η, t
|px
ξ, y
η q|q

for every pξ, η q P R2 . In particular, vpf py qq 



?1 1, |px
xξ,
yξ
ηq| , |px
yξ,
yη
ηq| 2



for every pξ, η q P R2 . In addition, it follows for such pξ, η q that

?



 u B 2 2 ǫ,
∇u  v pζ q Bt

2  |p∇uqpζ q|2 V pζ q pupζ q2  BBut pζ q
|px
ξ,2y
ηq| BBut pζ q px
ξ, y
ηq  p∇uqpζ q   

2  Bu  u B 2 2  ¥ Bt pζ q |p∇uqpζ q| V pζ q pupζ q
2  Bt pζ q |p∇uq|pζ q ¥ 0 ,

where ζ : f pξ, η q. Hence it follows (3.6.19). As an application of the energy inequality (3.6.19), we assume that v : U Ñ R is another solution of (3.6.8) such that upx, y, 0q  v px, y, 0q ,

Bu px, y, 0q  Bv px, y, 0q Bt Bt

523

for all px, y q P Bt px, y q. Then u
v is a solution of (3.6.16) such that

pu
vqpx, y, 0q  0 , BpuB
t vq px, y, 0q  0 for all x P Bt px, y q and hence the corresponding energy density vanishes at time 0 on Bt px, y q. As a consequence of (3.6.19) and the positivity of the energy density, it follows that the same is true at time T on Bt
T px, y q. Since this is true for every T P r0, tq and since u
v is continuous, it follows that u and v coincide in every point of


X SCx,y,t

r0, ts  R2



.

As a consequence, we have the following result. Theorem 3.6.29. (Uniqueness of the solutions of a wave equation in two space dimensions) Let px, y, tq P R3 be such that t ¡ 0. We define the solid
with apex px, y, tq by backward characteristic cone SCx,y,t


: SCx,y,t

pξ, η, τ q P R3 : τ ¤ t
|px
ξ, y
ηq|

Further, let U be an open subset of R3 such that U

 SCx
X

R2  r0, ts

(

.



P C 2 pU, Rq be such that B2u
△u V u  B2v
△v V v , Bt2 Bt2 Bu px, y, 0q  Bv px, y, 0q upx, y, 0q  v px, y, 0q , Bt Bt for all px, y q P Bt px, y q. In this, p△uqpt, x, yq : r△upt, qspx, yq respectively, for all pt, x, y q P U, and V : U Ñ R is continuous, assumes only positive values, i.e., RanpV q € r0, 8q, and satisfies BV  0 . Bt
Then u and v coincide on SCx,y,t X p R2  r0, tsq. and u, v

524

Problems 1) Decide whether the tuple of vectors is positively or negatively oriented

pp1, 2q, p3, 4qq , b) pp1,
1q, p0, 1qq , pp
3, 7q, p2, 1qq , d) pp9,
4q, p10,
3qq , e) pp0, 4, 2q, p8, 1, 3q, p9, 12, 5qq , f) pp
2,
2, 14q, p3, 1, 1q, p2, 9, 9qq , g) pp1,
1, 2q, p4, 3, 1q, p8, 2,
5qq , h) pp2, 7, 9q, p
1,
2, 4q, p
8, 8, 1qq . Let U be a non-empty open subset of R3 ; f : U Ñ R, v : U Ñ R3 and w : U Ñ R3 be partially differentiable; f1 : U Ñ R, v1  pv1x , v1y , v1z q : U Ñ R3 and also v2 : U Ñ R3 be of class C 2 . a) c)

2)

Show that

a) rot p∇f1 q  0 , b) div prot v1 q  0 , c) div p∇f1 q  △f1 , d) rot pf.vq  f.prot vq

3)

p∇f q  v , e) div pf.vq  f  pdiv vq p∇f q  v , f) rot prot v1 q  ∇pdiv v1 q
△v1 , g) div pv  wq  w  prot vq
v  prot wq , where △v1 : p△v1x , △v1y , △v1z q . Let a, b, c, d P R be such that a   c   d   b. Further, let f1 : pa, bq Ñ R and f2 : pa, bq Ñ R be twice differentiable and such that f1 pxq   f2 pxq for all x P pa, bq. Find a twice continuously differentiable bijective g : R  pa, bq Ñ R  pa, bq whose inverse is twice continuously differentiable and which is such that

g pr
1, 1s  rc, dsq  tpx, y q P R  rc, ds : f1 py q ¤ x ¤ f2 py qu . 4) Calculate the area of Gpf q of f : U

Ñ R defined by U : tpx, y q P R2 : x2 y 2   1u and f px, y q : 3 3x 7y for all px, y q P U .

5) Let D be the open subset of R2 that is bounded by the triangle with corners p0, 0q, p1, 0q, p1, 1q. Calculate the area of

tpx, y, z q P R3 : 3x2

7y
z

 0uXtpx, y, z q P R3 : px, yq P Du .

525

y 1

0.5

-1

-0.5

0.5

1

x

-0.5

-1

Fig. 163: An astroid. 6) Calculate the area of

tpx, y, z q P R3 : x2

y2 z 2

 1uXtpx, y, z q P R3 : x2

y2

 xu .

7) Calculate the surface areas of the tori from Example 3.6.14. 8) By calculation of a curve integral, find the compact area that is bounded by the astroid where a ¡ 0.

t pa cos3 t, a sin3 tq P R2 : t P r0, 2πq u ,

9) By calculation of a curve integral, find the compact area that is bounded by the cardioid

t pa cos t p1 where a ¡ 0.

cos tq, a sin t p1

cos tqq P R2 : t P r0, 2π q u ,

10) By calculation of a curve integral, find the compact area that is bounded by the folium of Descartes where a ¡ 0.

t px, yq P R2 : x3 526

y 3
3axy

 0u ,

y

1

1 €€€€€€ 2

1 €€€€€€ 2

1

x

3 €€€€€€ 2

1 - €€€€€€ 2 -1

Fig. 164: A cardioid.

y 2

1

-2

-1

1

-1

-2

Fig. 165: A folium of Descartes.

527

2

x

11) Use Stokes’ theorem to calculate the surface integral » S

where

curlA  dS ,

Apx, y, z q : pz, x, y q

P R and S : t px, y, z q P R3 : x2

for all x, y, z

y2

z
4 0^z

¥ 0u .

For this, assume a normal field with positive z-component. Sketch S. 12) Use Stokes’ theorem to calculate the surface integral » S

where

curlA  dS ,

Apx, y, z q : p2yz, 0, xy q ,

P R and S : t px, y, z q P R3 : x2

for all x, y, z

y2
z 2

 9 ^ 0 ¤ z ¤ 4z u .

For this, assume a normal field pointing away from the z-axis. Sketch S. 13) By using Stokes’ theorem, calculate » S

where for all x, y, z

curlA  dS ,

Apx, y, z q : p2y, 3x,
z 2 q ,

P R and S is the closed upper half surface of the sphere t px, y, z q P R3 : x2 y2 z 2  9 u .

For this, assume a normal field with positive z-component. 14) Use Gauss’ theorem to calculate the surface integral » S

A  dS ,

528

where

Apx, y, z q : px2 ,
xy, y 2 q

for all x, y, z P R and S is the compact region in the first octant bounded by the coordinate planes and

t px, y, z q P R3 : x

2y

z

 1u .

Sketch S. 15) Use Gauss’ theorem to calculate the surface integral » S

where

A  dS ,

Apx, y, z q : p3x2 ,
6xy, z 2 q

for all x, y, z P R and S is the compact region in the first octant bounded by the coordinate planes and

t px, y, z q P R3 : x  2 u , t px, y, z q P R3 : z

y2

 1u .

Sketch S. 16) By using Gauss’ theorem, calculate » S

where

A  dS ,

Apx, y, z q : p2xy

z, y 2 ,
x
3y q

for all x, y, z P R and S is the compact region in the first octant bounded by the coordinate planes and

t px, y, z q P R3 : 2x

2y

z

 6u .

4 Appendix 4.1 Construction of the Real Number System Already the ancient Greeks discovered that there was a need to go beyond rational numbers. For instance, they found that there is no rational number 529

to measure the length of the diagonal d of a square with sides of length 1. By the Pythagorean theorem that length satisfies the equation d2  2. In Example 1.1.15, we proved that this equation has no rational solution which was also known to the Greek’s of that time. Still, they did not develop the concept of real numbers. In its final form, that concept was developed only in the 19th century. In the following, we construct the real number system following an approach by Georg Cantor (1872) as completion of the rational number system. For this, in a first step, we identify Q with a space containing equivalence classes of Cauchy sequences of rational numbers. Definition 4.1.1. (Cauchy sequences in Q) Let x  x1 , x2 , x3 , . . . be a sequence of rational numbers. We say that x is a Cauchy sequence if for every rational ε ¡ 0 there is a corresponding n0 P N such that

|xm
xn |   ε for all m, n P N such that m ¥ n0 and n ¥ n0 . Such a sequence is nec-

essarily bounded by a some rational number since this leads in the special case ε  1 to

|xk | ¤ maxt|xl | : l  1, . . . , n0u |xk
xn xn | ¤ maxt|xl | : l  1, . . . , n0 u |xk
xn | |xn | ¤ 1 |xn | maxt|xl | : l  1, . . . , n0 u for k P N such that k ¥ n0 and |xk | ¤ maxt|xl | : l  1, . . . , n0 u ¤ 1 |xn | maxt|xl | : l  1, . . . , n0 u for k P N such that k ¤ n0 . We denote the set of all such sequences by the symbol C. For x, y P C, we define sequence x y an x  y of rational 0

0

0

0

numbers by

x y : x1 y1 , x2 y2 , x3 y3 , . . . x  y : x1 y1 , x2 y2 , x3 y3 , . . . . 530

0

0

Since

|px yqm
px yqn|  |xm
xn ym
yn| ¤ |xm
xn| |ym
yn| |px  yqm
px  yqn|  |xmym
xnyn|  |xmpym
ynq pxm
xnqyn| ¤ |xm|  |ym
yn| |yn|  |xm
xn| ¤ Cx|xm
xn | Cy |ym
yn| where Cx , Cy are rational bounds for x and y, respectively, it follows that x y P C and x  y P C. Finally, we define for every x P C a corresponding sequence
x P C by
x :
x1,
x2 , . . . .

Definition 4.1.2. We define an equivalence relation ‘’ on C as follows. We say that x, y P C are equivalent and denote this by x  y if for every rational ε ¡ 0 there is n0 P N such that

|xn
yn|   ε for all n P N such that n ¥ n0 . Indeed, ‘’ is reflexive since for every x P C and every rational ε ¡ 0 it follows that |xn
xn|  0   ε for all n P N such n ¥ 1 and hence that x  x. Also, ‘’ is symmetric, since for x, y P C such that x  y and rational ε ¡ 0 there is n0 P N such that

|xn
yn|   ε

for all n P N such that n ¥ n0 . This implies that

|yn
xn|   ε

for all n P N such that n ¥ n0 and hence that y  x. Finally, if x, y, z P C are such that x  y and y  z and ε is some rational number ¡ 0, it follows the existence of n0 P N such that

|xn
yn|   ε{2 , |yn
zn|   ε{2 for all n P N such that n ¥ n0 . Hence |xn
zn|  |xn
yn yn
zn| ¤ |xn
yn| |yn
zn | ¤ ε for all n P N such that n ¥ n0 . Therefore it follows that x  z. 531

Lemma 4.1.3. Let x x¯  x.

P C and x¯ be a subsequence of x.

Then x¯

P C and

Proof. Since x¯ is a subsequence of x, there is a strictly increasing sequence n1 , n2 , . . . of elements of N such that x¯  xn1 , xn2 , . . . and such that nk ¥ k for all k P N . Since x P C, for rational ε ¡ 0, there is n0 P N such that |xm
xn |   ε for all m, n that

P N such that m ¥ n0 and n ¥ n0 . In particular, this implies

|xk
xk |   ε for all m, n P N such that m ¥ n0 and n ¥ n0 since the last implies that km ¥ m ¥ n0 and kn ¥ n ¥ n0 . Hence it follows that x¯ P C. Also, it m

n

follows that

|x¯n
xn |  |xk
xn|   ε for all n P N such that n ¥ n0 since the last implies that kn ¥ n ¥ n0 . Definition 4.1.4. (Cantor real numbers) For every x P C, we define the n

associated Cantor real number as the equivalence class [x] defined by [x] : ty : y

P C ^ y  xu .

Also, we define the set C of Cantor real numbers by C : t [x] : x P C u . For [x], [y] P C , where x, y corresponding product by [x]

P C, we define a corresponding sum and a

[y]  [x

y] , [x]  [y]  [x  y] .

Indeed, this is possible, since it follows for x¯, y¯ and [¯ y ]  [y] that [¯ x

y¯]  [x

P C satisfying [¯x]  [x]

y] , [¯ x  y¯]  [x  y] . 532

This can be seen as follows. First, since [¯ x]  [x] and [¯ y ]  [y], it follows that x¯  x and that y¯  y. Hence for every rational ε ¡ 0 there is n0 P N such that |x¯n
xn |   ε{2 , |y¯n
yn|   ε{2 for all n P N such that n ¥ n0 . For such n, it follows that

|px¯ y¯qn
px yqn|  |x¯n
xn ¤ |x¯n
xn | |y¯n
yn|   ε and therefore that

px¯

y¯q  px

y¯n
yn |

yq .

Also, if Cx , Cy¯ are rational bounds for x and y¯, respectively, and ε is some rational number ¡ 0, then there is n0 P N such that Cy¯|x¯n
xn |   ε{2 , Cx |y¯n
yn |   ε{2 for all n P N such that n ¥ n0 . For such n, it follows that

|px¯  y¯qn
px  yqn|  |x¯ny¯n
xn yn|  |px¯n
xn qy¯n ¤ |y¯n|  |x¯n
xn | |xn|  |y¯n
yn| ¤ Cy¯|x¯n
xn | and hence that

xn py¯n
yn q| Cx |y¯n
yn |   ε

px¯  y¯q  px  yq .

Finally, we define the embedding ι of Q into C by ιpq q  [q, q, . . . ]

P Q. It is an obvious consequence of the definitions that ιpq q¯q  ιpq q ιpq¯q , ιpq q¯q  ιpq q  ιpq¯q , for all q, q¯ P Q. Theorem 4.1.5. pC , , q is a field, i.e., the following holds for all x, y, z P for all q

C:

533

(i)

p [x]

(ii) [x] (iii) [x] (iv) [x] (v) (vi) (vii) (viii)

[y] q

[z]  [x]

[y]  [y]

ιp0q  [x] ,

p [y]

[z] q , (Associativity of addition)

[x] ,

[
x]  ιp0q ,

(Commutativity of addition) (Existence of a neutral element for addition) (Existence of inverse elements for addition)

p [x]  [y] q  [z]  [x]  p [y]  [z] q , (Associativity of multiplication) (Commutativity of multiplication) [x]  [y]  [y]  [x] , [x]  ιp1q  [x] , (Existence of a neutral element for multiplication) If [x]  ιp0q, then there is w P C such that [x]  [w]  ιp1q , (Existence of inverse elements for multiplication)

(ix) [x]  p [y]

[z] q  [x]  [y]

[x]  [z] .

(Distributive law)

Proof. The validity of the statements (i)-(vii) and (ix) is an obvious consequence of the analogous laws for rational numbers and the definition of the addition and multiplication on C . For the proof of (viii), let x P C such that [x]  ιp0q. As a consequence, it is not true that for every rational ε ¡ 0 there is n0 P N such that |xn|   ε (4.1.1)

for all n P N such that n ¥ n0 . Therefore, there is a rational δ ¡ 0 and for which there is no n0 P N such that (4.1.1) is valid for all n P N such that n ¥ n0 . This implies the existence of a strictly increasing sequence n1 , n2 , . . . of natural numbers such that

|xn | ¥ δ k

for all k P N . According to Lemma 4.1.3, x¯ : xn1 , xn2 ,    P C and x¯  x. The last implies that [¯ x]  [x]. Therefore, we can assume without restriction that |xn| ¥ δ 534

for all n P N . We define w : 1{x1 , 1{x2 , . . . . Then

|wm
wn| 

  1  x

m






1  |xm
xn |  xn  |xm |  |xn |

Hence if ε is rational such that ε ¡ 0 and n0

¤ |xm δ
2 xn |

P N is such that

|xm
xn |   δ2ε

for all n P N satisfying n ¥ n0 , then also

|wm
wn| ¤ ε

for all n ιp1q.

P N such that n ¥ n0 . As a consequence, w P C and [x]  [w] 

In the next step, after preparation by a Lemma, we define an order relation ‘ ’ on C . Lemma 4.1.6. Let x, y n0 P N such that

P

C be such that there is a rational ε xn

¤ yn
ε

x¯n

¤ y¯n
ε¯

¡

0 and

for all n P N such that n ¥ n0 . Further let x¯, y¯ P C be such that x¯  x and y¯  y. Then, there are a rational ε¯ ¡ 0 and a n ¯ 0 P N such that for all n P N such that n ¥ n ¯0.

Proof. Since x¯  x and y¯  y, it follows the existence of N

P N such that x¯n
xn ¤ |x¯n
xn |   ε{4 , yn
y¯n ¤ |y¯n
yn |   ε{4 for all n P N such that n ¥ N. Hence it follows for n P N satisfying n ¥ maxtn0 , N u that ε ε
ε x¯n
  xn ¤ yn
ε   y¯n 4 4 and hence that

where ε¯ : ε{2.

x¯n

¤ y¯n
ε¯ 535

As a consequence of the previous lemma, it is meaningful to define the following. Definition 4.1.7. For [x], [y] P C , we say that [x] is smaller than [y] and denote this by [x]   [y]

if there are a rational ε ¡ 0 and n0

P N such that xn ¤ yn
ε

for all n P N such that n ¥ n0 . Further, we say that [x] smaller than [y] P C and denote this by

P C is equal or

[x] ¤ [y] if [x]   [y] or if [x] absolute value |[x]| by



[y]. Finally, we define for every [x] #

|[x]| :
[x] [x]

¤ q2 that

ιpq1 q ¤ ιpq2 q

and that

|ιpq1q|  |q1| . Theorem 4.1.8. Let [x] , [y] P C . Then (i)

C its

if ιp0q ¤ [x] if [x]   ιp0q .

It is an obvious consequence of the definitions that for q1 , q2 in the case q1   q2 that ιpq1 q   ιpq2 q , in the case q1

P

ιp0q   ιp0q ;

(ii) if [x]  ιp0q , then either ιp0q   [x] or [x]   ιp0q ; 536

P Q it follows

(iii) if ιp0q   [x] and ιp0q   [y] , then ιp0q   [x]

[y] , ιp0q   [x]  [y] .

Proof. ‘(i)’: The proof is indirect. Assume that ιp0q   ιp0q. Then there is a rational ε ¡ 0 such that 0  
ε. ‘(ii)’: For this, let [x]  ιp0q . Further, assume that both [x]   ιp0q and ιp0q   [x]. Then it is not true that there is a rational ε ¡ 0 and n0 P N such that

¤
ε

xn

for all n P N such that n ¥ n0 . Hence for every rational ε n0 P N , there is n P N such that n ¥ n0 and xn

¡ 0 and every

¡
ε .

Therefore, there is a subsequence x¯ of x such that

¡
n1

x¯n

for every n P N . Since x¯ P [x] according to Lemma 4.1.3, we can assume without restriction that 1 xn ¡
n  for every n P N . Further, since ιp0q   [x], it is not true that there is a rational ε ¡ 0 and n0 P N such that 0 ¤ xn
ε for all n P N such that n ¥ n0 . Hence for every rational ε n0 P N , there is n P N such that n ¥ n0 and xn

 ε.

Therefore, there is a subsequence x¯ of x such that x¯n

  n1

537

¡ 0 and every

for every n P N . Since x¯ P [x] according to Lemma 4.1.3, we can assume without restriction that
n1   xn   n1 for every n P N . Obviously, this implies that x  0, 0, . . . and hence that [x]  ιp0q. Hence it follows that either ιp0q   [x] or [x]   ιp0q are true or that both of these inequalities are true. The last implies that there are a rational ε ¡ 0 and n0 P N such that 0 ¤ xn
ε for all n P N satisfying n ¥ n0 and also that there are a rational δ m0 P N such that xn ¤
ε

¡ 0 and

for all n P N satisfying n ¥ m0 . (iii) For this, let ιp0q   [x] and ιp0q [y] . Then, there are a rational ε ¡ 0 and n0 P N such that

 

ε ¤ xn , ε ¤ yn for all n P N satisfying n ¥ n0 . Hence it follows for all n P N satisfying n ¥ n0 that 2ε ¤ xn yn , ε2 ¤ xn yn and finally that ιp0q   [x]

[y] , ιp0q   [x]  [y] .

Theorem 4.1.9. Let [x], [y] be elements of C such that [x] there is q P Q such that [x]   ιpq q   [y] . Proof. Since [x]   [y], there are a rational ε ¡ 0 and n0 xn

¤ yn
ε 538

  [y]. Then (4.1.2)

P N such that

for all n P N satisfying n such that m0 ¥ n0 and

¥ n0 . Further, since x, y P C, there is m0 P N

|xm
xn |   ε{4 , |ym
yn|   ε{4 for all n P N satisfying n ¥ m0 . We define q : pxm follows for all n P N satisfying n ¥ m0 that 0

0

0

q
xn

 12 pxm

yn
q

 yn
12 pxm

0

ym0 q
xn 0

 xm
xn 0

ym0 q  yn
ym0

ym0 q{2. Then it

1 ε p ym0
xm0 q ¡ 2 4 1 ε pym0
xm0 q ¡ 4 . 2

Hence it follows (4.1.2). Definition 4.1.10. Let [x1 ], [x2 ], . . . be a sequence of elements of C and [x] P C . (i) We call [x1 ], [x2 ], . . . a Cauchy sequence if for every [ε] that ιp0q   [ε] there is a corresponding n0 P N such that

P C such

|[xm ]
[xn ]|   [ε] for all m, n P N such that m ¥ n0 and n ¥ n0 . (ii) We define

lim [xn ]  [x]

n

Ñ8

if for every [ε] P C such that ιp0q n0 P N such that for all n ¥ n0 :

  [ε] there is a corresponding

|[xn]
[x]|   [ε] .

(4.1.3)

In this case, we say that the sequence [x1 ], [x2 ], . . . is convergent to [x]. Theorem 4.1.11. 539

(i) (The rational numbers are dense in the real numbers) Let [x] be some element of C . Then lim ιpxn q  [x] .

n

Ñ8

(ii) (Completeness of the real number system) Every Cauchy sequence in C is convergent. Proof. ‘(i)’: For this, let [ε] rational δ ¡ 0 such that

P C such that ιp0q   [ε].

Then there is a

ιp0q   ιpδ q   [ε]{2 as a consequence of Theorem 4.1.2. Further, it follows for m P N that ιpxm q
[x]  [xm
x1 , xm
x2 , . . . ] and, since x P C, the existence of n0

P N such that |xm
xn|   δ for all m, n P N satisfying m ¥ n0 and n ¥ n0 . Hence it follows for such

m, n that


2δ

δ


δ   xm
xn   δ  2δ
δ

and therefore that


ε  
2 ιpδq ¤ ιpxm q
[x] ¤ 2 ιpδq   ε . This implies that

|ιpxm q
[x]|   ε for all m P N satisfying m ¥ n0 . ‘(ii)’: For this, let [x1 ], [x2 ], . . . be a Cauchy sequence in C . In addition, for every n P N , let qn P Q be such that

[xn ]   ιpqn q   [xn ]

540

ιp1{nq .

Such qn exists according to Theorem 4.1.2. In the following, we will show that lim [xn ]  [q] n

where

Ñ8

q : q1 , q2 , . . . .

First, we show that q P C. For this, let δ such that n0 ¡ 4{δ and such that

¡ 0 be rational and n0 P N be

|[xm ]
[xn ]|   ιpδq{2 for all m, n P N satisfying m ¥ n0 and n ¥ n0 .

For such m and n, it

follows that

|ιpqmq
ιpqn q|  |ιpqm q
[xm ] [xm ]
[xn] [xn]
ιpqn q| ¤ |ιpqm q
[xm ]| |[xm ]
[xn]| |[xn]
ιpqnq|   ιp1{mq ιp1{nq |[xm ]
[xn]|   ιpδq and hence also that

|qm
qn|   δ . Further, let [ε] P C be such that ιp0q   [ε]. Since according to (i) lim ιpqn q  [q] , nÑ8

it follows the existence of n0 P N such that ιp1{n0 q   [ε]{2 and such that for all n ¥ n0 : |ιpqnq
[q]|   [ε]{2 . This also implies that

|[xn ]
[q]|  |[xn]
ιpqn q   ιp1{nq p[ε]{2q   [ε] .

ιpqn q
[q]| ¤ |[xn ]
ιpqn q|

541

|ιpqnq
[q]|

4.2 Lebesgue’s Criterion of Riemann-integrability Apart from notational changes and additions, we follow the lucent Sect. 7.26 of [4] in the proof of Lebesgue’s criterion of Riemann-integrability, Theorem 1.5.13. Theorem 4.2.1. Let S0 , S1 , . . . be a sequence of subsets of measure zero of R. Then the union S of these subsets has measure zero, too. Proof. Given ε ¡ 0, for each k P N there is a sequence Ik0 , Ik1 , . . . of open subintervals of R such that union of these intervals contains Sk and at the same time such that lim

n

n ¸

Ñ8 m0

lpIkm q  

ε

The sequence of all intervals Ikl , where k, l all these intervals contains S and lim

n

n ¸

Ñ8 k0

lpIϕpkq q   lim l

where ϕ : N Ñ N2 is some bijection.

.

2k 1

P N, is countable; the union of

l ¸

ε

Ñ8 k0 2k

1

ε

Definition 4.2.2. (Oscillation of a function) Let I be some non-trivial interval of R and f : I Ñ R be some bounded function. Then we define for every non-trivial subset S of I the oscillation Ωf pS q of f on S by Ωf pS q : suptf pxq
f py q : x P S ^ y

P Su .

(Note that set in the previous identity is bounded from above by suptf py q : y P I u
inf tf py q : y P I u. In addition, note that Ωf pS q is positive.) Further, we define for each x P I the oscillation ωf pxq of f at x by the limit ωf pxq : lim Ωf ppx
δ, x δ

Ñ0

δq X I q

of the decreasing function that associates to every δ Ωf ppx
δ, x δ q X I q. 542

¡

0 the value of

Theorem 4.2.3. Let I be some non-trivial interval of R and f : I Ñ R be some bounded function and x P I. Then f is continuous in x if and only if ωf pxq  0. Proof. First, we consider the case that f is continuous in x. Then for every n P N there are xn , yn P px
1{n, x 1{nq X I such that

|f pxnq
f pynq
Ωf ppx
1{n, x

1{nq X I q| ¤

1 . n

Hence f px1 q
f py1q
Ωf ppx
1, x 1qXI q, f px2 q
f py2q
Ωf ppx
1{2, x 1{2q X I q, . . . is a null sequence. Since both sequences x1 , x2 , . . . and y1 , y2 , . . . are converging to x and since f is continuous in x it follows by Theorem 1.2.4 that ωf pxq  0. Finally, we consider the case that ωf pxq  0. Assume that f is not continuous in x. Hence there is some ε ¡ 0 along with a sequence x1 , x2 , . . . in I ztxu which is convergent to x, but such that

|f pxnq
f pxq| ¥ ε . Hence

Ωf ppx
δn , x

δn q X I q ¥ ε

for all n P N , where δn : 2|xn
x| for all n P N . Since δ1 , δ2 , . . . is converging to 0 it follows that ωf pxq ¥ ε. Hence f is continuous in x. Theorem 4.2.4. Let f : ra, bs Ñ R be bounded, where a and b are some elements of R such that a   b. Further, let ωf pxq   ε for every x P ra, bs and some ε ¡ 0. Then there is δ ¡ 0 such that for every closed subinterval I of ra, bs of length smaller than δ it follows Ωf pI q   ε . Proof. First, it follows from the assumptions that for each x is some δx ¡ 0 such that Ωf ppx
δx , x

δx q X ra, bsq   ε . 543

P ra, bs there

The family of sets px
δx {2, x δx {2q, where x P ra, bs, is an open covering of ra, bs and hence by the compactness of ra, bs there are x1 , x2 , . . . , xn P ra, bs, where n is some element of N, such that ra, bs is contained in the union of px1
δx1 {2, x1 δx1 {2q, px2
δx2 {2, x2 δx2 {2q, . . . , pxn
δxn {2, xn δxn {2q. Now define δ : mintδx1 {2, δx2 {2, . . . , δxn {2u and let I be some closed subinterval of ra, bs of length smaller than δ. Further, let k be some element of t1, 2, . . . , nu such that pxk
δxk {2, xk δxk {2qX I  φ. Then, I € pxk
δxk , xk δxk q , since lpI q   δ, and hence Ωf pI q   ε.

Theorem 4.2.5. Let f : ra, bs Ñ R be bounded, where a and b are some elements of R such that a   b. Further, let ε ¡ 0. Then Jε : tx P ra, bs : ωf pxq ¥ εu is a closed subset of ra, bs.

Proof. Let x be some element of the complement ra, bszJε . Then ωf pxq   ε and hence there is some δ ¡ 0 such that Ωf ppx
δ, x δ q X ra, bsq   ε. In particular it follows for every element y P px
δ, x δ q X ra, bs that ωf py q   ε and as a consequence that px
δ, x δ q X ra, bs is contained in ra, bszJε. Hence is ra, bszJε open in ra, bs and therefore Jε a closed subset of ra, bs. We prove now Theorem 1.5.13: Theorem 4.2.6. (Lebesgue’s criterion for Riemann-integrability) Let f : ra, bs Ñ R be bounded, where a and b are some elements of R such that a   b. Further, let D be the set of discontinuities of f . Then f is Riemann-integrable if and only if D is a set of measure zero. Proof. First, assume that D is not of measure zero. Then D is non-empty and by Theorem 4.2.3 it follows that ωf pxq ¡ 0 for every x P D. Hence D



8 ¤ 

J1{n .

n 1

544

(4.2.1)

Since the union in (4.2.1) is countable, by Theorem 4.2.1 it follows the existence of some n P N such that J1{n is not a set of measure zero. Hence there is some ε ¡ 0 such that the sum of the lengths of the intervals corresponding to any covering of J1{n by open intervals is ¥ ε. Now let P be some partition of ra, bs with corresponding closed intervals I0 , I1 , . . . , Ik , where k P N. Further, denote by S the subset of t0, 1, . . . , k u containing only those indexes j P t0, 1, . . . , k u for which the intersection of the inner of Ij and J1{n is non-empty. Then the open intervals corresponding to Ij , j P S cover J1{n , except possibly for a finite set, which is a set of measure zero. Hence the sum of their lengths is ¥ ε. U pf, P q
Lpf, P q

 ¥

k ¸



j 0

¸

P

rsuptf pxq : x P Ij u
inf tf pxq : x P Ij us  lpIj q rsuptf pxq : x P Ij u
inf tf pxq : x P Ij us  lpIj q

j S

¥ n1

¸

P

j S

lpIj q ¥

ε n

and hence f is not Riemann-integrable. Finally, assume that D is a set of measure zero and consider again (4.2.1). Further, let n P N such that 1{n   pb
aq{2. Then J1{n is by Theorem 4.2.5 compact and has measure zero as a subset of a set of measure zero. Hence there is a covering of J1{n by a finite number of open intervals for which the corresponding sum of lengths is smaller than 1{n. Without restriction we can assume that those intervals are pairwise disjoint. Denote by An the union of those intervals. Then the complement Bn : ra, bszAn is the union of a finite number of closed subintervals of ra, bs. Let I be such subinterval. Then ωf pxq   1{n for each x P I and hence by Theorem 4.2.4 there is a partition of I such that Ωf pI 1 q   1{n for any induced subinterval I 1 . All those partitions induce a partition Pn of ra, bs. Now consider some refinement P P P of Pn with corresponding closed intervals I0 , I1 , . . . , Ik , where k P N. Further, denote by S the subset of t0, 1, . . . , k u containing only those indexes j P 545

t0, 1, . . . , ku for which Ij X J1{n  φ. Then ¸

R

rsuptf pxq : x P Ij u
inf tf pxq : x P Ij us  lpIj q

j S

¤ n1 ¸

P

¸

R

lpIj q ¤

j S

b
a , n

rsuptf pxq : x P Ij u
inf tf pxq : x P Ij us  lpIj q

j S

¤ pM
mq

¸

P

lpIj q ¤

M


m ,

n

j S

where M : suptf pxq : x P ra, bsu and m : inf tf pxq : x hence b
a M
m U pf, P q
Lpf, P q ¤ . n Therefore

P P uq ¤ U pf, P q ¤ Lpf, P q b
a ¤ supptLpf, P q : P P P uq b
a nM
m .

inf ptU pf, P q : P

P ra, bsu, and

M n


m

Since this is true for any n P N such that 1{n   pb
aq{2, from this it follows by Theorem 3.4.4 that f is Riemann-integrable.

4.3 Properties of the Determinant Lemma 4.3.1. (Leibniz’ formula for the determinant) Let n pa1, . . . , anq be an n-tuple of elements Rn. Then (i) detpa1 , . . . , an q 

¸

P

σ Sn

546

signpσ q a1σp1q    anσpnq

P N and

where Sn denotes the set of permutations of t1, . . . , nu, i.e., the set of all bijections from t1, . . . , nu to t1, . . . , nu and signpσ q : spσ p1q, . . . , σ pnqq 

n ¹

  

sgnpσ pj q
σ piqq

i,j 1,i j

for all σ P Sn . Note that signpσ q  1 if the number of pairs pi, j q P t1, . . . , nu2 such that i   j and σpj q   σpiq is even, whereas signpσq 
1 if that number is odd. (ii) signpσ q  for all σ

P Sn .

(iii) for all τ, σ

P Sn .

signpτ

σ pj q
σ piq j
i i,j 1,i j n ¹

 σq  signpτ q  signpσq

(iv) In addition, let n ¥ 2. Further, let τ be a transposition, i.e., an element of Sn for which there are elements k, l P t1, . . . , nu such that k  l and such that τ piq  i for all i P t1, . . . , nuztk, lu, τ pk q  l and τ plq  k. Then there is σ P Sn such that τ

 σ  τ0  σ
1

where τ0 P Sn is the transposition defined by τ0 piq  i for all i P t1, . . . , nuzt1, 2u, τ0 p1q  2 and τ0 p2q  1. Note that from this it follows by (iii) that signpτ q  signpσ  τ0  σ
1 q  signpσ q  signpτ0 q  signpσ
1 q  signpτ0 q 
1 . (v) In addition, let n ¥ 2. For every σ P Sn , there is k sequence τ1 , . . . , τk of transpositions in Sn such that σ

 τ1      τk . 547

P

N and a

(vi) For every i P t1, . . . , nu, define a¯i :

n ¸



aji ej

j 1

where e1 , . . . , en is the canonical basis of Rn . Then a11    an1      ¯1 , . . . , a ¯n q        detpa1 , . . . , an q detpa      a1n    ann a11    a1n            .      an1    ann Proof. ‘(i)’: The statement of (i) is a direct consequence of Definition 2.5.17 and the definitions given in (i). ‘(ii)’: For this let σ P Sn and denote by m the number of pairs pi, j q P t1, . . . , nu2 such that i   j and σ pj q   σ piq. Then n ¹

  

pσpj q
σpiqq

i,j 1,i j







n ¹

   p q σpj q

pσpj q
σpiqq p
1qm

i,j 1,i j,σ i

 p
1q

m

n ¹

  

n ¹

   p q σpiq

|σpj q
σpiq|

i,j 1,i j,σ j

|σpj q
σpiq|  p
1q

m

i,j 1,i j

n ¹

  

pj
iq

i,j 1,i j

where the last equality uses the bijectivity of σ. Hence it follows that signpσ q  p
1qm



σ pj q
σ piq . j
i i,j 1,i j n ¹

548

‘(iii)’: First, it follows from (ii) that

pτ  σqpj q
pτ  σqpiq j
i i,j 1,i j n n ¹ σ pj q
σ piq pτ  σqpj q
pτ  σqpiq  ¹  σ pj q
σ piq j
i i,j 1,i j i,j 1,i j   n ¹ pτ  σqpj q
pτ  σqpiq  signpσq .  σ pj q
σ piq i,j 1,i j

signpτ

 σq 

n ¹

pτ  σqpj q
pτ  σqpiq σ pj q
σ piq i,j 1,i j n ¹





pτ  σqpj q
pτ  σqpiq  σ pj q
σ piq i,j 1,i j,σpiq σpj q n ¹ pτ  σqpj q
pτ  σqpiq  σ pj q
σ piq i,j 1,i j,σpj q σpiq





n ¹



pτ  σqpj q
pτ  σqpiq  σ pj q
σ piq i,j 1,i j,σpiq σpj q n ¹ pτ  σqpj q
pτ  σqpiq  σ pj q
σ piq i,j 1,i¡j,σpiq σpj q n ¹ pτ  σqpj q
pτ  σqpiq  σ pj q
σ piq i,j 1,σpiq σpj q n ¹ τ pj q
τ piq  signpτ q  j
i i,j 1,i j 

n ¹

where the last two equalities use the bijectivity of σ. Hence, finally, it follows that signpτ  σ q  signpτ q  signpσ q . 549

‘(iv)’ For this, let k, l P t1, . . . , nu be such that k  l and such that τ piq  i for all i P t1, . . . , nuztk, lu, τ pk q  l and τ plq  k. Further, let σ be some element of Sn such that σ p1q  k and σ p2q  l. Then σ  τ0  σ
1 pk q  σ  τ0 p1q  σ p2q  l , σ  τ0  σ
1 plq  σ  τ0 p2q  σ p1q  k

and for i P t1, . . . , nuztk, lu

σ  τ0  σ
1 piq  σ  σ
1 piq  i .

‘(v)’: If σ coincides with the identity transformation on t1, . . . , nu, then σ  τ  τ for any transposition τ P Sn . If σ differs from the identity transformation on t1, . . . , nu, then there is i1 P t1, . . . , nu such that σ piq  i for all i P t1, . . . , i1
1u, where we define t1, . . . , 0u : φ, and σ pi1 q  i1 . The last implies that σ pi1 q ¡ i1 . We define the transposition τ1 P Sn by τ1 pi1 q : σ pi1 q, τ1 pσ pi1 qq : i1 and τ piq : i for all i P t1, . . . , nuzti1 , σ pi1 qu. Then σ1 : τ1  σ satisfies σ1 piq  i for all i P t1, . . . , i1 u. Continuing this process, we arrive after at a sequence of transpositions τ1 , . . . , τk in Sn , where k is some element of N , such that idt1,...,nu Then σ

 τk  . . . τ1  σ .

 τ1
1      τk
1  τ1      τk .

‘(vi)’: It follows by (i), (iii) that detpa1 , . . . , an q 

  

¸

P

σ Sn

¸

P

σ Sn

¸

P

¸

P

signpσ q a1σp1q    anσpnq

σ Sn

signpσ q aσ
1 pσp1qq σp1q    aσ
1 pσpnqq σpnq signpσ
1 q aσ
1 p1q 1    aσ
1 pnq n signpσ q ¯a1σp1q    a¯nσpnq



¸

P

σ Sn

signpσ q aσp1q 1    aσp1q n

 detpa¯1, . . . , a¯nq .

σ Sn

550

Theorem 4.3.2. (Properties of the determinant) Let n P N , e1 , . . . , en the canonical basis of Rn , pa1 , . . . , an q an n-tuple of elements of Rn , i P t1, . . . , nu, ai1 P Rn, α P R and j P t1, . . . , nu such that j ¡ i. Then (i)

detpe1 , . . . , en q  1 ,

(ii) ai1 , . . . , an q  detpa1 , . . . , ai , . . . , an q detpa1 , . . . , ai1 , . . . , an q , detpa1 , . . . , α ai , . . . , an q  α detpa1 , . . . , an q , detpa1 , . . . , ai

(iii) if n ¥ 2, then detpa1 , . . . , ai , . . . , aj , . . . , an q 
detpa1 , . . . , aj , . . . , ai , . . . , an q , (iv) if n ¥ 2 and ai

 aj , then detpa1 , . . . , ai , . . . , aj , . . . , an q  0 .

Proof. ‘(i)’: detpe1 , . . . , en q 

 sp1, . . . , nq 

n ¸



spk1 , . . . , kn q e1k1    enkn

k1 ,...,kn 1 n ¹

  

sgnpj
iq  1 .

i,j 1,i j

‘(ii)’: detpa1 , . . . , ai



n ¸



ai1 , . . . , an q

spk1 , . . . , kn q a1k1 . . . paiki

k1 ,...,kn 1

551

aik1 i q . . . ankn

n ¸



spk1 , . . . , kn q a1k1 . . . aiki . . . ankn



k1 ,...,kn 1 n ¸



spk1 , . . . , kn q a1k1    aik1 i . . . ankn

k1 ,...,kn 1

 detpa1, . . . , ai, . . . , anq

detpa1 , . . . , ai1 , . . . , an q ,

detpa1 , . . . , α ai , . . . , an q n ¸





spk1 , . . . , kn q a1k1 . . . pα aqiki . . . ankn

k1 ,...,kn 1 n ¸

α



spk1 , . . . , kn q a1k1 . . . aiki . . . ankn

k1 ,...,kn 1

 α detpa1, . . . , ai, . . . , anq . ‘(iii)’: For this, we define the n-tuple pb1 , . . . , bn q of elements of Rn by bk : ak if k P t1, . . . , nuzti, j u, bi : aj and bj : ai . Then it follows by Lemma 4.3.1 (iv) that

detpa1 , . . . , ai , . . . , aj , . . . , an q n ¸

  



k1 ,...,kn 1 n ¸



k1 ,...,kn 1 n ¸



spk1 , . . . , ki , . . . , kj , . . . , kn q a1k1 . . . aiki . . . ajkj . . . ankn spk1 , . . . , ki , . . . , kj , . . . , kn q b1k1 . . . bjki . . . bikj . . . bnkn spk1 , . . . , kj , . . . , ki , . . . , kn q b1k1 . . . bjkj . . . biki . . . bnkn

k1 ,...,kn 1 n ¸






spk1 , . . . , ki, . . . , kj , . . . , kn q b1k1 . . . biki . . . bjkj . . . bnkn

k1 ,...,kn 1


detpa1 , . . . , aj , . . . , ai, . . . , anq . ‘(iv)’: The statement of (iv) is simple consequence of (iii). 552

Theorem 4.3.3. (Uniqueness of the determinant) Let n P N and w be a map which associates to every n-tuple of elements of Rn a real number. In particular, let w be such that (i) for the canonical basis e1 , . . . , en of Rn w pe1 , . . . , en q  1 , (ii) for every n-tuple pa1 , . . . , an q of elements of Rn , i P t1, . . . , nu, ai1 Rn and α P R

P

(iii) if n ¥ 2, for every n-tuple pa1 , . . . , an q of elements of Rn , i t1, . . . , nu, ai1 P Rn and j P t1, . . . , nu such that j ¡ i

P

ai1 , . . . , an q  w pa1 , . . . , ai , . . . , an q w pa1 , . . . , ai1 , . . . , an q , w pa1 , . . . , α ai , . . . , an q  α w pa1 , . . . , an q , w pa1 , . . . , ai

w pa1 , . . . , ai , . . . , aj , . . . , an q 
w pa1 , . . . , aj , . . . , ai , . . . , an q .

Then w

 det.

Proof. For this, let pa1 , . . . , an q be an n-tuple pa1 , . . . , an q of elements of Rn . Then it follows by (ii),(iii) that w pa1 , . . . , an q 



¸

P

n ¸



a1k1 . . . ankn w pek1 , . . . , ekn q

k1 ,...,kn 0

a1σp1q . . . anσpnq w peσp1q , . . . , eσpnq q .

σ Sn

Further, it follows by Theorem 4.3.1 (v), (iii) and (i) that w peσp1q , . . . , eσpnq q  detpeσp1q , . . . , eσpnq q and therefore, finally, that w pa1 , . . . , an q  detpa1 , . . . , an q .

553

Theorem 4.3.4. (Bases of Rn ) Let n P N .

(i) Let r P N and v1 , . . . , vr be basis of Rn , i.e., a sequence of vectors in Rn which is such that for every w P Rn , there is a unique r-tuple pα1, . . . , αr q of real numbers such that w



r ¸



αk vk .

k 1

Then r

 n.

(ii) In addition, let r P N and v1 , . . . , vr P Rn be no basis Rn , but be linearly independent, i.e., such that the equation r ¸



αk vk

0

k 1

for some real α1 , . . . , αr implies that α1

     αr  0 .

Then there are m P N and vectors w1 , . . . , wm in Rn such that v1 , . . . , vr , w1 , . . . , wm is a basis of Rn . (iii) Let v1 , . . . , vn a basis of Rn .

P Rn be linearly independent. Then v1, . . . , vn P Rn is

Proof. ‘(i)’: Since v1 , . . . , vr and the canonical basis e1 , . . . , en of Rn are both bases, there are real numbers αik , i P t1, . . . , r u, k P t1, . . . , nu and βjl , j  t1, . . . , nu, l P t1, . . . , r u such that vi



n ¸



αik ek , ej



k 1

for every i P t1, . . . , r u and j vi



n ¸



k 1

αik ek



r n ¸ ¸

 

r ¸



βjl vl

l 1

P t1, . . . , nu. For such i, j, it follows that

αik βkl vl , ej

k 1l 1



r ¸



l 1

554

βjl vl



r n ¸ ¸

 

k 1l 1

βjl αlk ek

and hence that n ¸



αik βki

1,

k 1

r ¸



βjl αlj

1.

l 1

This implies that r



n r ¸ ¸

 

αik βki

i 1k 1



r n ¸ ¸

 

βjl αlj

n.

j 1l 1

‘(ii)’: Since the canonical basis e1 , . . . , en of Rn of Rn is a bases, it follows that every element of Rn can represented as a linear combination of the vectors v1 , . . . , vr , e1 , . . . , en . In a first step, we consider the sequence of vectors v1 , . . . , vr , e1 . If e1 is the linear combination of v1 , . . . , vr , then we drop e1 from the sequence v1 , . . . , vr , e1 , . . . , en and still every element of Rn can represented as a linear combination of the vectors from the remaining sequence. Otherwise, we keep e1 in the sequence. Note that in this case v1 , . . . , vr , e1 are linearly independent. Continuing this process, we arrive at a sequence of vectors w1 , . . . , wm in Rn , where m P N , such that v1 , . . . , vr , w1 , . . . , wm is linearly independent and such that every element of Rn can represented as a linear combination of its members. This also implies that v1 , . . . , vr , w1, . . . , wm is a basis of Rn . ‘(iii)’: The proof is indirect. Assume that v1 , . . . , vn is no basis of Rn . Then by (ii) v1 , . . . , vn can be extended to a basis by adding a non-zero number of vectors from Rn . That basis has at least n 1 members which contradicts (i). Definition 4.3.5. (The determinant of a linear map) Let n A : Rn Ñ Rn be linear, i.e. such that Apx

y q  Apxq

P

N and

Apy q , Apαxq  αApxq

P Rn and α P R. Then, obviously, by w pa1 , . . . , an q : detpApa1 q, . . . , Apan qq for every n-tuple pa1 , . . . , an q of elements of Rn , there is given a map w for all x, y

satisfying the conditions (ii) and (iii) in Theorem 4.3.3. Hence according 555

to that theorem, w is the multiple of det. In the following, we call the corresponding factor the determinant of A and denote it by detpAq. By definition, it follows that detpApa1 q, . . . , Apan qq  detpAq  detpa1 , . . . , an q for every n-tuple pa1 , . . . , an q of elements of Rn and hence that detpAq  detpApe1 q, . . . , Apen qq where e1 , . . . , en is the canonical basis of Rn . Theorem 4.3.6. Let n Then

P N and A : Rn Ñ Rn, B : Rn Ñ Rn be linear.

(i) detpA  B q  detpAq  detpB q, (ii) A is bijective if and only if detpAq  0. Proof. For this, let e1 , . . . , en be the canonical basis of Rn . ‘(i)’: From Definition 4.3.5, it follows that 

detpA  B q  det

pA  B qpe1 q, .. . , pA  B qpenq   det ApB pe1qq, . . . , ApB penqq  detpAq det B pe1 q, . . . , B penq  detpAq detpB q .

‘(ii)’: If A is bijective, then detpA
1 q detpAq  detpA
1  Aq  detpidRn q  detpe1 , . . . , en q  1 and hence detpAq  0. On the other hand, if detpAq  detpApe1 q, . . . , Apen qq  0 and α1 , . . . , αn are real numbers such that n ¸



αk Apek q  0 ,

k 1

556

then α1  such that

    αn  0.

Otherwise, there is i

Apei q 
and hence

P t1, . . . , nu and αi  0

n ¸

αk Apek q α i k 1,k i

detpAq  detpApe1 q, . . . , Apen qq  0 .

Hence the vectors Ape1 q, . . . , Apen q are linearly independent and constitute a basis of Rn . Therefore, for every v P Rn , there are real α1 , . . . , αn such that  



v

n ¸

αk Apek q  A



k 1

n ¸



αk ek

k 1

and hence A is surjective. Finally, we show that A is also injective. For this, assume that there are v, w P Rn such that Av  Aw. Then 0  Apv
w q  A



n ¸





pvk
wk qek 

k 1

n ¸



pvk
wk qApek q .

k 1

Since Ape1 q, . . . , Apen q are linearly independent, this implies that vk for every k P t1, . . . , nu and hence that v  w. Definition 4.3.7. (Linear maps) Let n, m P N and A : Rn

 wk

Ñ Rm .

(i) We say that A is linear if Apx for all x, y

y q  Apxq

Apy q , Apαxq  αApxq

P Rn and α P R.

(ii) If A is linear, Apxq  A



n ¸



j 1



xj enj



n ¸



j 1

557

xj Ap

enj

q

m ¸ n ¸

 

i 1j 1

Aij xj em i

m n where en1 , . . . , enn and em 1 , . . . , em denote the canonical basis of R m and R , respectively, and for every i  1, . . . , m, j  1, . . . , n, Aij denotes the component of Apenj q in the direction of em i , such A is determined by its values on the canonical basis of Rn . On the other hand, obviously, if

pAij qpi,jqPt1,...mut1,...nu is a given family of real numbers, then by Apxq :

n m ¸ ¸

 

Aij xj em i

i 1j 1

for all x P Rn , there is defined a linear map A : Rn Ñ Rm . Interpreting the elements of Rn and Rm as column vectors and defining the m  n matrix MA by 

MA :

    

    

A11

  

Am1

    

    

    

Amn

    

    

    

    

A1n

  



Æ Æ Æ , Æ

the last is equivalent to 

Apxq : MA  x 

    

A11

  

Am1

A1n

  

Amn

 Æ Æ Æ Æ

x1



 ÆÆ   ÆÆ      

xn

where the multiplication sign denotes a particular case of matrix multiplication defined below

pMA  xqi :

n ¸



j 1

558

Aij xj

for every x P Rn and every i  1, . . . , m. In this case, we call MA the representation matrix of A with respect to the bases en1 , . . . , enn m and em 1 , . . . , em . (iii) If A is linear with representation matrix MA , l P N and B : Rm Rl is linear with representation matrix MB , it follows that l ¸ m ¸

pB  Aqpxq  B pApxqq  

m l ¸ ¸

 

Bik

i 1k 1

n ¸



Akj xj eli

 

Bik pApxqqk eli

i 1k 1 n l ¸ ¸



j 1

Ñ

 

i 1j 1



m ¸





Bik Akj xj eli

k 1

for every x P Rn and hence that the representation matrix MBA of B  A is given by  °m

MBA



    



k 1 B1k Ak1

  

°m



k 1 Blk Ak1

    

    

    

°m

    

 B1k Akn

k 1

°m

  



Æ Æ Æ . Æ

 Blk Akn

k 1

For this reason, we define the matrix product MB  MA of MB and MA such that MB  MA  MBA . Hence  °m

MB  MA :

    



k 1 B1k Ak1

°m

  

 Blk Ak1

k 1

    

    

    

    

°m

 B1k Akn

k 1

°m



559

Æ Æ Æ . Æ

k 1 Blk Akn

(iv) If A is linear with representation matrix MA and m the determinant detpMA q of MA by detpMA q : detpAq .

  



 n, we define

In this case, it follows by Theorem 4.3.1 (vi) that detpMA q  detpApen1 q, . . . , Apenn qq A11   n n ¸ ¸  n n Ai1 ei , . . . , Ain ei    det  i1 i1 A1n A11    A1n            .      An1    Ann

    

    

 An1        Ann

If in addition B : Rn Ñ Rn is linear with representation matrix MB , it follows by Theorem 4.3.6 (i) that detpMB  MA q  detpMB q  detpMA q . Lemma 4.3.8. (Sylvester’s criterion) Let n P N , A  pAij qi,j Pt1,...,nu be a real symmetric n  n matrix,i.e., such that Aij  Aji for all i, j P t1, . . . , nu. Then A is positive definite, i.e., ¸



Aij hi hj

¡0

i,j 1,...,n

for all h P Rn zt0u, if and only if all leading principal minors detpAk q, k  1, . . . , n, of A are ¡ 0. Here Ak : pAij qi,j Pt1,...,ku , k

P t1, . . . , nu . First, we derive an auxiliary result. For this let n ¥ 2, detpAn
1 q 

Proof. 0 and let α1 , . . . , αn
1 be some real numbers. We define a linear map T : Rn Ñ Rn by T phq : h
hn .pα1 , . . . , αn
1 , 0q 560

for every h P Rn . In particular, T is bijective with inverse ¯q  h ¯ T
1 ph

¯ n .pα1 , . . . , αn
1 , 0q h

¯ q. Then ¯ P Rn . Further, let h ¯ P Rn and h : T
1 ph for every h n ¸



¯2 Ann h n





Aij hi hj

i,j 1

n¸1



i 1



2

Ann ¯n 2h



i 1

2

¯i Ain ph








¯i h




n¸1





Aij αi αj



i,j 1




n¸1

Aij hi hj

¯i Aij ph










i,j 1

i,j 1

n¸1

Ain αi

n¸1

Ain hi hn

¯ n αi qh ¯n h




n¸1 i 1




n¸1

n¸1 i 1




2




Ann h2n

¯ n αi qph ¯j h




n¸1

¯2 h n



¯ n αj q h

¯ ih ¯j Aij h

i,j 1



Ain

αj Aij

j 1

Since detpAn
1 q  0, the column vectors of An
1 are linearly independent, and hence there is a unique n
1-tuple pα1 , . . . , αn
1 q of real numbers such that




n¸1



αj Aij

Ain

0

j 1

for all i P t1, . . . , n
1u. Therefore by choosing these α1 , . . . , αn
1 it follows that n ¸



¯2 Aij hi hj  bnn h n

i,j 1



¯ ih ¯j Aij h

i,j 1

where bnn : Ann




n¸1




2

n¸1






Ain αi

i 1

n¸1



i,j 1

561

Aij αi αj .

P R,

(4.3.1)

As a consequence, A is positive definite if and only if 

A¯ :

A11

  

      A1 n
1

0

     

     

An
1 1

  

0



  

Æ Æ Æ Æ Æ Æ 0

An
1 n
1 0 bnn

is positive definite, Note that by Leibniz’s formula Theorem 4.3.1 (i), it follows that detpA¯q  bnn detpAn
1 q . Further, if M denotes the representation matrix of T , then n ¸



Aij hi hj

i,j 1

 



n ¸



¯ ih ¯j A¯ij h



i,j 1 n ¸



A¯ij Mik Mjl hk hl

i,j,k,l 1 n ¸



n ¸



i,j 1 n ¸





A¯ij pT phqqi pT phqqj Mki A¯kl Mlj hi hj

i,j,k,l 1

¯ qij hi hj , pM tAM

i,j 1

where

M t : pMji qi,j Pt1,...,nu ,

¯ is symmetric, and hence, since M t AM ¯ A  M t AM Therefore, we conclude that detpA¯q  bnn detpAn
1 q  [ detpM q ]
2 detpAq .

(4.3.2)

With the help of the auxiliary result, the proof of the theorem proceeds by induction over n. The statement of the theorem is obviously true in the case n  1. In the following, we assume that it is true for some n P N 562

and consider the case where n is increased by 1. If A is positive definite, it follows, in particular, that ¸



Aij hi hj

¡0

i,j 1,...,n

for all h P Rn zt0u and therefore also that An is positive definite. As a consequence, according to the inductive assumption, the leading principal minors detpAk q, k  1, . . . , n are ¡ 0. Further, it follows by (4.3.1) that bnn ¡ 0 and hence by (4.3.2) that detpAq ¡ 0. On the other hand, if all leading principal minors of A are ¡ 0, it follows by the inductive assumption that An is positive definite and by (4.3.2) that bnn ¡ 0. Therefore, it follows by (4.3.1) that A is positive definite.

4.4 The Inverse Mapping Theorem Theorem 4.4.1. (Banach fixed point theorem for closed subsets of Rn ) Let n P N , B be a non-empty closed subset of Rn , and let f : B Ñ B be a contraction, i.e., let there exist α P r0, 1q such that

|f pxq
f pyq| ¤ α  |x
y| for all x, y that Further,

and

(4.4.1)

P B. Then f has a unique fixed point, i.e., a unique x P B such f px q  x .

|x
x | ¤ |x 1

f pαxq|

(4.4.2)

lim f ν pxq  x

ν

Ñ8

for every x P B where f ν for ν and f k 1 : f  f k , for k P N.

P N is inductively defined by f 0 : idB

563

Proof. Note that (4.4.1) implies that f is continuous. Further, define F : B Ñ R by F pxq : |x
f pxq| for all x P E. Now let x P B. Then it follows that

|f

ν µ1

pxq
f pxq| ¤ ν µ

1 1 µ¸






|f

ν k 1

pxq
f pxq| ¤ ν k

k µ ν

1 1 µ¸






αν

k

F pxq

k µ

¤ 1 α
α F pxq

for all ν, µ, µ1 P N such that µ1 ¥ µ. Hence the components of pf ν pxqqν PN are Cauchy sequences. Hence it follows by Theorem 1.2.10, Theorem 2.5.45 and the closedness of B that this sequence is convergent to some x P B. Further, it follows by the continuity of f that x is a fixed point of f . In addition, if x¯ P B is some fixed point of f , then

|x
x¯|  |f pxq
f px¯q| ¤ α  |x
x¯| and hence x¯  x since the assumption x¯  x leads to the contradiction that 1 ¤ α. Finally, let y be some element of B. Then it follows that |y
x |  |y
f px q|  |y
f pyq f pyq
f pxq| ¤ |y
f pyq| |f pyq
f pxq| ¤ F pyq α  |y
x| and hence (4.4.2). Lemma 4.4.2. Let n P N , U be a non-empty open subset of Rn and f : U Ñ Rn be of class C 1 such that f p0q  0, f 1 p0q  idRn . Finally, let r ¡ 0 be such that Br p0q € U and 0   ε   1 such that n  ¸  fi  max  x x iPt1,...,nu



B p q
Bfi pyq ¤ ?ε Bj Bxj  n j 1 for all x, y P Br p0q. Then for every y P Brp1
εq p0q, there is a unique x P Br p0q such that f pxq  y. 564

Proof. For this, let y sponding gy pxq by

P Brp1
εq p0q. We define for every x P Br p0q a corregy pxq : x
f pxq

y.

By Taylor’s formula Theorem 3.3.5, it follows for every x t1, . . . , nu and some 0   τ   1 that

P

Br p0q, i

P

|fipxq
xi |  |fipxq
fip0q
x  (∇fi)p0q|  |x  (∇fi)pτ xq
x  (∇fi)p0q| ¤ |x|  |(∇fi)pτ xq
(∇fi )p0q|  n  ¸  B fi  B f i ¤ |x|  Bx pτ xq
Bx p0q ¤ ?rεn 

j 1

j

and hence that

j

|f pxq
x| ¤ rε .

The last implies that

|gy pxq| ¤ |x
f pxq| |y| ¤ rε rp1
εq  r for all x P Br p0q and hence that gy : Br p0q Ñ Br p0q. Furthermore, it follows for all x1 , x2 P Br p0q, i P t1, . . . , nu and some 0   τ   1 that |gyipx1 q
gyipx2 q|  |fipx1 q
x1  (∇fi)p0q
pfipx2 q
x2  (∇fi )p0qq|  |fipx1 q
fi px2q
px1
x2 q  (∇fi)p0q|  |px1
x2 q  (∇fi )px1 τ px2
x1 qq
px1
x2 q  (∇fi )p0q| ¤ |x1
x2 |  |(∇fi)px1 τ px2
x1qq
(∇fi)p0q| ¤ ?εn |x1
x2 | and hence that

|gy px1 q
gy px2 q| ¤ ε |x1
x2 | .

Since ε   1, this implies that gy is a contraction and therefore has unique fixed point x P Br p0q according to Theorem 4.4.1. Obviously, x satisfies f pxq  y. 565

Theorem 4.4.3. (Inverse mapping theorem) Let n P N , U be a nonempty open subset of Rn , f : U Ñ Rn be a C 1 map and x0 P U such that f 1 px0 q is bijective. Then there are open subsets of Ux0 € U, Vf px0 q € Rn containing x0 and f px0 q, respectively, and such that f |Ux0 defines a bijection onto Vf px0 q whose inverse is a C 1 map.

References [1] Abramowitz M and Stegun I A (eds.) 1984, Pocketbook of Mathematical Functions, Harri Deutsch, Thun. [2] Alonso M, Finn E J 1967, 1967, 1968, Fundamental university physics, Vols. I - III, Addison-Wesley, Reading, Mass. [3] Anon E 1969, Note on Simpson’s rule, Amer. Math. Month., 76, 929-930. [4] Apostol T M 2002, Mathematical analysis, 2nd ed., Narosa Publishing House, New Delhi. [5] Ayres F, Mendelson E 1999, Calculus, McGraw-Hill, New York. [6] Demidovich B (ed.) 1989, Problems in mathematical analysis, 7th printing, Mir Publishers, Moscow. [7] Beyer H R 2007, Beyond partial differential equations, Springer Lecture Notes in Mathematics 1898, Springer, Berlin. [8] Bronson R 1989, Matrix operations, McGraw-Hill, New York. [9] Budak B M, Samarskii A A, Tikhonov A N 1964, A collection of problems on mathematical physics, MacMillan, New York. [10] Cheney W 2001, Analysis for applied mathematics, Springer, New York. [11] Coleman A J 1951, A simple proof of Stirling’s formula, Amer. Math. Month., 58, 334-336. [12] Coleman A J 1954, The probability integral, Amer. Math. Month., 61, 710-711. [13] Drager L D, Foote R L 1986, The contraction mapping lemma and the inverse function theorem in advanced calculus, Amer. Math. Month., 93, 52-54. [14] Dunkel O 1917, Discussions: Relating to the exponential function, Amer. Math. Month., 24, 244-246. [15] Enderton H B 1977, Elements of set theory, Academic Press, New York.

566

[16] Erdelyi A, Magnus W, Oberhettinger F, Tricomi F G (eds.) 1953, Higher transcendental functions, Vol. I, McGraw-Hill, New York. [17] Fischer G 2003, Lineare Algebra, 14th ed., Vieweg, Wiesbaden. [18] Ford J 1995, Avoiding the exchange lemma, Amer. Math. Month., 102, 350-351. [19] Gelbaum B R, Olmsted J M H 2003, Counterexamples in analysis, Dover Publications, New York. [20] Giesy D P 1972, Still another elementary proof that 45, 148-149.

°

1{k 2

 π2 {6, Math. Mag.,

[21] Goldrei D 1996, Classic set theory, Chapman & Hall, London. [22] Goursat E 1904, 1916, 1917, A course in mathematical analysis, Vols. I, II, III, Ginn, Boston. [23] Guenter N M, Kusmin R O 1971, Aufgabensammlung zur hoeheren Mathematik, Vols. I, II, DVW, Berlin. [24] Haaser N B, Sullivan J A 1971, Real analysis, Van Nostrand, New York. [25] Halmos P R 1958, Finite-dimensional vector spaces, Van Nostrand, New York. [26] Hamilton N, Landin J 1961, Set theory and the struture of arithmetic, Allyn and Bacon, Boston. [27] Hille E 1964, 1966, Analysis, Vols. I, II, Blaisdell, New York. [28] Margaris A 1990, First order mathematical logic, Dover Publications, New York. [29] McShane E J 1973, The Lagrange multplier rule, Amer. Math. Month., 80, 922-925. [30] Mendelson E 1988, 3000 solved problems in calulus, McGraw-Hill, New York. [31] Messiah A 1999, Quantum mechanics, Dover Publications, New York. [32] Landau L D, Lifschitz E M 1979, Lehrbuch der theoretischen Physik, Band III, 9. Aufl., Quantenmechanik, Akademie-Verlag, Berlin. [33] Lang S 1997, Undergraduate analysis, 2nd ed., Springer, New York. [34] Lebedev N N, Skalskaya I P, Uflyand Ya S 1966, Problems in mathematical physics, Pergamon, Oxford. [35] Lipschutz S 1998, Set theory and related topics, 2nd ed., McGraw-Hill, New York. [36] Lipschutz S, Lipson M L 2001, Linear algebra, 3rd ed., McGraw-Hill, New York. [37] Loomis L 1974, Calculus, Addison-Wesley, Reading, Mass.

567

[38] Konnully A O 1968, Relation between the beta and the gamma function, Math. Mag., 41, 37-39. [39] Rudin W 1976, Principles of mathematical analysis, 3rd ed., McGraw-Hill, Singapore. [40] Salas S, Hille E, Etgen G 2003, Calculus: One and several variables, 9th ed., Wiley, New York. [41] Samuels S M 1966, A simplified proof of a sufficient condition for a positive definite quadratic form, Amer. Math. Month., 73, 297-298. [42] Stewart J 1999, Calculus: Early transcendentals, 4th ed., Brooks/Cole Publishing Company, Pacific Grove. [43] Stoll R R 1963, Set theory and logic, Freeman, San Francisco. [44] Toeplitz O 1963, The calculus: A genetic approach, University of Chicago Press, Chicago. [45] Venkatachaliengar K 1962, Elementary proofs of the infinite product for sin z and allied formulae, Amer. Math. Month., 69, 541-545. [46] Whittaker E T, Watson G N 1952, A course of modern analysis, 4th ed. reprint, Cambridge University Press, Cambridge. [47] Wrede R, Spiegel M R 2002, Theory and problems of advanced calculus, McGrawHill, New York.

568

Index of Notation , not, 6

^ , and, 6 _ , or, 6 ñ , if . . . then, 6 , . . . if and only if . . . , 6 P , belongs to, 15 R , does not belong to, 15

φ , empty set, 15 N , t0, 1, 2, . . . u, 15 N , t1, 2, . . . u, 15 Z , t. . . ,
2,
1, 0, 1, 2, . . . u, 15 Z , t. . . ,
2,
1, 1, 2, . . . u, 15 : , such that, 16 Q , rational numbers, 16 Q , non-zero rational numbers, 16 R , real numbers, 16 R , non-zero real numbers, 16 € , subset, 16 : , per definition, 16 ra, bs , closed interval, 17 pa, bq , open interval, 17 pa, bs , half-open interval, 17 ra, bq , half-open interval, 17 rc, 8q , unbounded closed interval, 17 pc, 8q , unbounded open interval, 17 p
8, ds , unbounded closed interval, 17 p
8, dq , unbounded open interval, 17 Y , union, 17 X , intersection, 17 z , relative complement, 17  , Cartesian product, 17 R2 , R  R, 19 R3 , R  R  R, 19 Dpf q , domain of f , 23 f pxq , value of f at x, 23 f pxq , image of x under f , 23 f pA 1 q , image of A 1 under f , 23 f
1 pB 1 q , inverse image of B 1 under f , 23 f |A , restriction of f to A 1 , 24 1

Gpf q , graph of f , 25 f
1 , inverse map, 27  , composition, 31 idC , identity map on C, 32 limnÑ8 , limit, 36 sup , supremum, 48 inf , infimum, 48 exp , exponential function, 52 ex , exponential function, 52 limxÑa , limit, 58 limxÑ8 , limit, 58 limxÑ
8 , limit, 58 f1 f2 , sum of functions, 66 af1 , multiple of a function, 66 f1  f2 , product of functions, 66 1{f1 , quotient of functions, 66 f 1 pxq , derivative of f in x, 80 f 1 , derivative of f , 81 f pkq , k-th derivative of f , 81 f 2 , 2nd order derivative of f , 81 f 3 , 3rd order derivative of f , 81 sinh , hyperbolic sine, 103 cosh , hyperbolic cosine, 103 tanh , hyperbolic tangent, 103 ° n km , Sum from k  m to n, 111 n! , n factorial, 111 n , binomial coefficient, 139 ³bk a f pxq dx , integral of f over ra, bs, 144 ± , product symbol, 181 °8 k 1 xk , sum of a sequence, 224 ν , binomial coefficient, 282 #n , oriented line segment between p and q, pq 297 Srn paq , sphere of radius r around a in Rn , 297 # , oriented line segment between p and q, pq 298 # ] , vector associated to pq, # 300 [ pq # # [ pq ] [ rs ] , sum of vectors, 302

569

# ] , scalar multiple of a vector, 302 λ.[ pq # |[ pq ]| , length of a vector, 302 # ]  [ rs # ] , scalar product of vectors, 303 [ pq a  b , vector product of vectors in R3 , 311 sgn , signum function, 315 Uε pxq , open ball of radius ε centered at x, 383 Bε pxq , closed ball of radius ε centered at x, 383 Sε pxq , sphere of radius ε centered at x, 383 f1 f2 , sum of functions, 387 a.f1 , multiple of a function, 387 f1  f2 , product of functions, 387 1{f1 , quotient of functions, 387 g  f , composition of functions, 389 Bf Bx1i , partial derivative, 398 f pxq , derivative of f in x, 402 ∇ , gradient operator, 402 C p , continuously partially differentiable up to order p, 408 C 8 , C p for all p P N , 408 ∇ , gradient operator, 408 △ , Laplace operator, 411 f.g ³ , product of functions, 416 ³I f dv , integral of f on I, 451 ³r F  dr , curve integral of F along r, 473 F  dS , flux of F across S, 504 S Sn , permutation group, 547 sign , signum function, 547

570

Index of Terminology Rn

polar, 330, 417, 421, 458 spherical, 347, 421, 459 Cramer’s rule, 322 distance of a point from a plane, 319 distance of two lines, 318, 319 distance of two planes, 319 ellipse, 333 hyperbola, 334 lines, 317 metric space, 292 parabola, 332 planes, 317 quadrics, 338 cylinder, 338 ellipsoid, 340, 349 elliptic cone, 342 elliptic cylinder, 340, 347 elliptic paraboloid, 341 hyperbolic cylinder, 340 hyperbolic paraboloid, 342 hyperboloid of one sheet, 343 hyperboloid of two sheets, 343 parabolic cylinder, 338 saddle surface, 341 triangle centroid, 322 circumcenter, 322 orthocenter, 322 triangle inequality, 292 vector spaces, 298 norms, 305 vectors, 298, 300 addition, 300 length, 300 linear independence, 322 orientation , 483 orthogonality, 300 scalar multiplication, 300 scalar product, 300

addition, 300, 304 bases, 554 canonical basis, 304 Cauchy-Schwarz inequality, 292, 306 length, 300, 305 linear independent vectors, 554 orthogonal projection, 307 Pythagorean theorem, 307 scalar multiplication, 300, 304 scalar product, 300, 306 subset boundary, 386 boundary point, 386 bounded, 383 closed, 383 closure, 383, 384 compact, 383 convex, 480 curve, 356 inner point, 386 interior, 386 negligible, 454 open, 383 simply-connected, 480 star-shaped, 477 volume/area, 457 triangle inequality, 305 Analytical geometry area of a parallelogram, 309 conic sections, 323, 346 ellipse, 324, 331 hyperbola, 326 parabola, 323 coordinates, 330 cylindrical, 346, 421, 458 generalized polar, 469 generalized spherical, 470

571

scalar triple product, 314 unit vector, 300 vector product, 311 volume of a parallelepiped, 314 Applications n-dimensional volume, 457 area of a parallelogram, 309, 316 area of a parametric surface, 503 area of a surface of revolution, 507 area under a graph, 144 arithmetic series, 274 astroid, 372, 526 average speed, 80 Babylonian method, 129 Bessel functions, 149, 184, 269, 284 Beta function, 211 cardioid, 372, 526 Cartesian leaf, 202 center of mass, 461, 466 circular arch, 138 conchoid of Nicomedes, 92 confluent hypergeometric functions, 290 conservation law, 497, 521 constant of motion, 360 Couette flow, 471 cycloid, 92, 372 differential equation, 93, 101, 102, 157, 168, 173, 193 electric field of a point charge, 472 energy conservation, 99, 473 energy inequality, 101, 497, 522 error function, 281 Euler constant, 229 Fermat’s principle, 137 floor function, 155 folium of Descartes, 526 force field, 99, 472 conservative, 99, 473 potential function, 99, 473 total energy of a point particle, 99, 473

572

Fourier coefficients, 161 free fall, 55, 136 with low viscous friction, 163 with viscous friction, 136 gamma function, 205, 211, 213, 216, 217, 230 Gaussian integrals, 207, 209 ground state energy, 135 Hermite polynomials, 291 hypergeometric functions, 290 ideal gas law, 55 inertia tensor, 461, 466 instantaneous speed, 80 Kepler problem, 360 angular momentum, 360 energy, 360 Lenz vector, 360 Levi-Civita’s transformation, 364 kinetic energy, 360, 473 Laplace equation, 411, 420 Laplace operator, 411 cylindrical form, 421 polar form, 421 spherical form, 421 largest viewing angle, 136 latitude, 348 Legendre functions, 291 length of a curve, 370 longitude, 348 Mathematica 5.1 error, 173, 193 Newton’s equation of motion, 93, 99, 162, 163, 360, 472 partial differential equation, 93, 420 particle paths, 360 period of a pendulum, 202, 223 probability theory, 462 Buffon’s needle problem, 466 quantum theory confined particle, 467 harmonic oscillator, 220 hydrogen atom, 220

Riemann’s zeta function, 227, 235, 255, proof by cases, 10 257 proposition, 5 Schwarzschild black hole, 108 rule of inference, 10 Snell’s law, 137 statement, 5 Stirling’s formula, 178 tautology, 10 strophoid, 173 transitivity, 10 total mass, 461 truth table, 6 trajectory of a point particle, 472 truth values, 5 transverse vibrations of a beam, 139 Elementary set theory travel distance, 142 sets, 14 velocity field of a point particle, 472 Cartesian product, 17 volume of a solid of revolution, 460 equality, 16 volume/area of a set, 457 intersection, 17 Wallis product, 176 relative complement, 17 wave equation, 93, 420, 496, 499, 520 subset, 16 types of definition, 15, 16 Curves union, 17 astroid, 372, 526 Zermelo-Russel paradoxon, 21 cardioid, 372, 526 Cartesian leaf, 202 Functions circle, 297 B conchoid of Nicomedes, 92 definition, 211 cycloid, 92, 372 F pa, b; c; q, 290 ellipse, 324, 333 Hn , 291 folium of Descartes, 526 Jn , 184 helix, 356 integral representation, 149 hyperbola, 326, 334 Jν length, 370 integral representation, 284 parabola, 323, 332 power series, 269 parallelogram, 309 M pa, b, q, 290 space-filling, 263 Pν , 291 straight lines, 317 Γ strophoid, 173 Γp1{2q, 211 connection to ζ, 255 Elementary logic definition, 205 compound, 6 duplication formula, 217 connectives, 6 Gauss formula, 216 contraposition, 10 limit of beta function, 213 contrapositive, 6 reflection formula, 216 indirect proof, 8, 11, 79, 98 Stirling’s formula, 178 logical law, 10 Weierstrass formula, 230 negation, 6

573

arccos, 69, 108 arcsin, 69, 108 arctan, 69, 108 cos, 69, 89 infinite product, 181 power series, 279 cosh, 103, 139 exp, 69, 100 convexity, 116 definition, 50 derivative, 82 power series, 279 ln, 69, 100, 108 sin, 69, 84 infinite product, 181 power series, 279 sinh, 103, 137, 138 tan, 69, 89 tanh, 103 erf definition, 281 power series, 281 ζ ζ p2q, 257 definition, 227 extension to p0, 1q, 235 integral representation, 255 antisymmetric, 172 bisection method, 63 concave, 115, 117 continuous, 55 contraction, 128 convex, 115, 117 critical point, 94 definition, 23 derivative, 80 of higher order, 80 determinant, 315, 546, 555 Leibniz formula, 546 differentiable, 80 chain rule, 88

concave, 115, 117 convex, 115, 117 inverse functions, 107 linear approximation, 112 linearization, 112 product rule, 86 quotient rule, 86 sum rule, 86 Taylor’s formula, 112 differential equation, 269 discontinuous, 55 everywhere, 57 fixed point, 79, 128 floor, 155 implicit differentiation, 92 increasing, 99 integral representation, 149 limits at infinity, 74 maximum, 58, 94, 121 minimum, 58, 94, 121 not differentiable, 84, 85 nowhere differentiable, 259 periodic, 172 polynomial, 67, 88 powers, 82, 110 rational, 77 removable singularity, 72 Riemann integral, 144 additivity, 152 Cauchy-Schwartz inequality, 162 change of variables, 163 improper, 202 integration by parts, 173 Lebesgue criterion, 149 linearity, 147 Midpoint rule, 196 partial fractions, 186 positivity, 147 simple limit theorem, 251 Simpson’s rule, 186, 200 Trapezoid rule, 186, 198

574

signum, 315, 546 symmetric, 172 Taylor expansion, 277 uniformly continuous, 368 zero set, 23 Functions of several variables, 375 C 8 , 408 composition, 389 continuous, 379, 381 contours, 375 critical point, 430 derivative, 402 differentiable, 396, 404 chain rule, 415 product rule, 413 quotient rule, 413 sum rule, 413 Taylor’s formula, 425 directional derivative, 423 discontinuous, 379, 380 domain, 375 gradient, 424 graph, 375 level set, 375 maximum, 384, 429, 432 minimum, 384, 429, 432 not differentiable, 406 of class C p , 408 partially derivative, 89, 398 of higher order, 89, 398 partially differentiable, 89, 398, 404 product, 387, 416 quotient, 387 range, 375 Riemann integral, 450 change of variables, 457 existence, 455 Fubini’s theorem, 456 negligible sets, 454 scalar multiple, 387 sum, 387

tangent plane, 402 Taylor expansion, 425 Taylor polynomial, 402 zero set, 23 Infinite products Γ, 216, 230 cos, 181 sin, 181 Wallis product, 176 Maps bijective, 27 bilinear, 357 composition, 31 definition, 23 domain, 23 graph, 25 identity map, 32 image, 23 injective, 27 inverse image, 23 Laplace operator, 411 linear, 395, 555, 557 representation matrix, 395, 557 Nabla operator, 408 one-to-one, 27 one-to-one and onto, 27 onto, 27 paths, 356 continuous, 356 differentiable, 356 helix, 356 length, 366, 368 non-rectifiable, 367 rectifiable, 366 tangent vector, 356 permutation, 546 permutation group, 546 transposition, 546 quadratic form, 441 restriction, 23

575

surjective, 27 Matrices definition, 395, 557 multiplication, 395, 557 symmetric, 431, 560 positive definite, 431, 560 Real numbers Babylonian method, 129 completeness, 539 construction, 529 density of rational numbers, 57, 539 intervals, 17 length, 142 partition, 142 subset bounded from above, 48 bounded from below, 48 infimum, 48 measure zero, 148 supremum, 48 Sequences bounded, 36 bounded from above, 46 bounded from below, 47 Cauchy, 43 convergent, 36, 351 decreasing, 47 divergent, 36 increasing, 46 limit laws, 41, 354 limits preserve inequalities, 43 subsequence, 45 Series, 223 ζ-type, 232 Abel’s test, 236 absolutely summable, 237 alternating harmonic, 235, 269 arithmetic, 277 Binomial series, 282 Cauchy product, 272

comparison test, 229 conditionally summable, 237 convergent, 223 Dirichlet test, 234 divergent, 223 geometric, 224 harmonic, 225, 229 harmonic type, 228, 232 integral test, 226 not summable, 223 of functions, 248 power series, 265 uniform convergence, 250 Weierstrass’ test, 253 ratio test, 239 rearrangement, 242 root test, 240 summable, 223 summation by parts, 233 Surfaces area, 503, 506, 507 cylinder, 338 ellipsoid, 340 elliptic cone, 342 elliptic paraboloid, 341 hyperbolic paraboloid, 342 hyperboloid of one sheet, 343 hyperboloid of two sheets, 343 parallelepiped, 314 planes, 317 quadrics, 338 saddle surface, 341 sphere, 297 Theorems Abel, 268 Binomial, 139 Bolzano-Weierstrass, 45, 382 change of variables, 163, 457 contraction mapping lemma, 128 extended mean value, 123

576

Fubini, 456 fundamental theorem of calculus, 153 Gauss, 515 Green, 487, 490 intermediate value, 61 L’Hospital’s rule, 123 Lagrange multiplier rule, 439 mean value, 97 Newton-Raphson, 132 Poincare lemma, 477 Rolle, 96 Schwarz, 409 Stokes, 508 Taylor, 111, 277 Vector calculus area of a parametric surface, 503 closed path, 475 curve integrals, 473 flux across a surface, 502, 503 Gauss theorem, 515 Green’s theorem, 487, 490 inverse path, 475 parametrized surfaces, 500 normal field, 500 tangent space, 500 piecewise regular C 1 -path, 475 Poincare lemma, 477 potential, 476 regular C 1 -path, 473 Stokes’ theorem, 508 vector field, 471 Vector-valued functions, 375 Vector-valued functions of several variables, 375

577