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e /l/j L:!,inceton Review 
®
Mathematics Subject Test
4th Edition
Proven techniques for a higher score. •
Fulllength practice GRE
•
Easytounderstand subject review
Mathematics Subject Test
of precalculus, abstract algebra, and calculus
•
Review questions in every chapter with detailed answer explanations
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�
/Jl1e
a3 + a 4
= 2
By inspection, we can see that a = 1, so J 1 (2) = 1.
Example 1.5
=>
a3 + a= 2
Let A be the open interval (1, 1) and define a bijective function/ on A by the equation: X f(x )= 1 X Find a formula for J1(x) that works for every real x. 2
Solution:
Given the equation y =
� , we interchange 1x
x and y and solve for y:
yX=1yz xy2 +yx= 0 vf1 +4 x 2 y= 1 ± 2x
8 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
(*)
There are two problems with the form of this result. The first problem is that it specifies two functions because of the ±. The second is that x = 0 is excluded from the domain of this function (because of the 2x in the denominator); but since 0 is in the range off [because J (O) = 0], it must be in the domain of j1• Let's start by solving the first problem. Notice that �1 + 4x2 > W = 2 l xl, so if we use the form of (*) with the minus sign, then x>O
�
1 �1 + 4x 2 < 2x
�
1 �1 + 4x2 < 1 2x
�
1�1 + 4x2 > 1 2x
and xl 
{

where [x] is the greatest integer::; x. Solution:
We can get the graph of the equation y = 2  x2 by inverting the graph of the basic parabola y = x2• and then translating upward by 2 units to get y = x2 + 2. The graph of the greatest integer function, y = [x], has a break at each integern, since for every x such that n  1 ::; x (y + 1)2 = 16 => y = 3, 5 Thus, the circle crosses the yaxis at the points (0, 3) and (0, 5).
PRECALCULUS
+ 13
(b) Let L denote the tangent line for which we're looking. Then L must be perpendicular to the ra dius BC . Because the slope of BC is m Be
1(1)  1 62 2
the slope of L must be the negative reciprocal, 2. Since mL = 2 and L passes through the point B: (6, 1), the equation of the line is y  1 = 2(x 6) which we can write as y = 2x + 13. When x = 0, the value of y is 13, so L crosses the yaxis at the point (0, 13). 

ELLIPSES
By definition, an ellipse is the set of points in the plane such that the sum of the distances from every point on the ellipse to two given fixed points (the foci) is a constant. (And to avoid a degenerate case, the constant sum must be greater than the distance between the foci.) The standard equation of an ellipse centered at the originwith axes parallel to the coordinate axesis:
x2 y 2 2 + 2 = 1 a b The longer symmetry axis of the ellipse is called the major axis (on which the foci are located), and the shorter one is the minor axis. The endpoints of the major axis are called the vertices. The eccentricity of an ellipse is a number (denoted e) between 0 and 1 that measures its "flatness." The closer e is to zero, the more the ellipse resembles a perfect circle; as e increases to 1, the ellipse flattens out. By comparing the standard equations for the circle and the ellipse, we notice that the only difference is that while the x2 and y2terms always have identical positive coefficients for a circle, these terms have different positive coefficients for an ellipse. Therefore, a circle can be transformed into an ellipse (or vice ver2 2 sa) by changing the scale on one of the axes. For example, we can turn the circle x2 + y 2 = 1 into the ellipse a a x2 2 a 2 + yb 2 = 1 by replacing every point (x, y) on the circle by the point (x, by). a The following diagrams summarize the basic characteristics of the ellipse, where the cases a b and a < b are considered separately.
>
2 y2 x2 + = 1, a>b a b2 foci: (±c, 0), where c = ,Ja 2  b 2 vertices: (±a, 0) major axis length = 2a minor axis length = 2b c eccentricity: e = a
14 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
y
(a, 0)
(0, b) b (c, 0)
•. __
a (c, 0)
(0, b)
(a, 0)
X
y
(0, b)
x2 l = 1, a< b +2 a b2 foci: (0, ±c), wherec = Jb2  a2 vertices: (0, ±b) major axis length = 2b minor a xis length = 2a c .. eccentnoty: e = b
(0, c)
\_
(0, b)
Example 1.10
The sum of the distances from every point on an ellipse to the points (±2, 0) is equal to 8. Find the positive value of x such that the point (x, 3) is on this ellipse.
'� ·
Solution:
Since the foci are on the xaxis and their midpoint (which is the center of the ellipse) is the origin, the equation of the ellipse has the form
x2 y2 + =1 . a2 b2 where b. In this case, the sum of the distances from every point on the ellipse to the foci is equal to 2a. Since we have So the fact that tells us that:
a>
2a = 8,
a = 4.
c=2
b = Ja2  c2 = J42  22 = 2.J3 So the equation of the ellipse is:
x2 y2 = 1 + 16 12 Substituting y
= 3,
we solve for x:
x2 + 32 = 1 1 6 12
==>
==>
==>
x = ±2
Since the question asks for the positive value of x, the answer is x = 2.
HYPERBOLAS By definition, a hyperbola is the set of points in the plane such that the difference between the distances from every point on the hyperbola to two fixed points (the foci) is a constant. (And to avoid a degenerate case, the constant difference must be smaller than the distance between the foci.) Unlike a parabola, circle, or ellipse, a hyperbola is not a single curve; instead, it consists of two separate curves called branches. A hyperbola is also different from these other curves in that it has asymptotes, which are lines that its branches approach but never touch, as the magnitudes of x and y increase.
PRECALCULUS + 1 S
The line that contains the foci is called the focal axis; the midpoint of the segment joining the foci is the center of the hyperbola; and the points at which the branches of the hyperbola intersect the focal axis are called the vertices. The standard equations of the hyperbolas centered at the origin with either the x or the yaxis as their focal axis are:
The diagrams below summarize the basic characteristics of these standard hyperbolas. Notice that the hyperbolas of the first family do not intersect the yaxis, and those of the second family do not intersect the xaxis.
x2 y2 = 1 a2 b2 foci: (±c, 0), where c = .Ja2 + b2 vertices: (±a, 0) b a symptotes: y = ± x a

·
y=
.
y
)�a x
.
···

•.•
• •
y=
(a, b)
(a,
b).····
�::::__ �>·'f
··
!!.x
(�·
_..
• • • • (c, 0) ' ' (c, 0) • •• • ��..� ..��+x .
....
. ...... _'_'(a,::::� · ··�(a,::�:�b). ·· ... b)
. . .... ...�osymptot"�·····... I .
____
.
y
y2 x2 = 1 b2 a2 foci: (0, ±c),
'

vertices:
(0, ±b)
asymptotes:
Example
16 +
1.11
where c = .Ja2 + b2
b y = ±x a
'
'
'
c.
' ' " '
' '
' X '
The foci of a hyperbola are at ( ±3, 0), and the difference between the distances from every point on the hyperbola to the foci is 2. What's the equation of this hyperbola?
CRACKING THE GRE MATHEMATICS SUBJECT TEST
Solution:
Since the foci are on the xaxis and their midpoint (the center of the hyperbola) is the origin, the equation of the hyperbola has the form:
xz l  = 1 az bz
(±a, 0). The difference between the distances from the vertex (a, 0) to the two foci, (c, 0) and (c, 0), is [a  (c) ] (c  a) = 2a. Because we're told that this equals 2, we know that a = 1 . Finally, since c = .Ja2 + b2, we can solve for b: The vertices of this hyperbola are

b =� c2  a2 =� 32  f = 2fi
Therefore, the equation of this hyperbola is:
POLYNOMIAL EQUATIONS
A polynomial in the variable x is an expression o f the form a ,,x'' + a ,_1 x"1 + + a 1 x + a0, where n i s a non negative integer and the coefficients ai are constants. If a, :t 0, the integer n is called the degree of the polynomial and a, is called the leading coefficient. (No degree is defined for 0, the zero polynomial.) A polynomial of degree 2is called a quadratic polynomial, and the roots of the quadratic equation · · ·
ax2 + bx + c = 0
are given by the wellknown quadratic formula:
b ± .Jb 2  4ac 2a The quantity under the square root sign, 11 = b2  4ac, is called the discriminant of the polynomial; it determines the nature of the roots. If 11 0, the equation has two distinct real roots; if 11 < 0, the roots are distinct complex conjugates; and if 11 = 0, the equation has exactly one (double) root. In this section, we'll X=


>
also look at more general polynomial equations.
THE DIVISION ALGORITHM, REMAINDER THEOREM, AND fACTOR THEOREM Let p(x) and d(x) be polynomials. If d(x) is not identically zero, the division algorithm guarantees that dividing p(x) by d(x) gives unique polynomials q(x) and r(x), such that p(x)
= q(x)d(x) + r(x)
where either r(x) is identically zero or the degree of r(x) is less than the degree of d(x). The polynomial q(x) is called the quotient, and r(x) is the remainder. In particular, assume that the degree of p(x) is at least 1, and let d(x) = x  k, a polynomial of degree 1. If p(x) is divided by x  k, we get p(x) = q(x)
•
(x  k)
+
r
where r is a constant. Substituting x = k into this equation gives us p(k) = r, which says that when p(x) is divided by x  k, the remainder is p(k). This is the Remainder theorem. From this, we can conclude that x = k is a root of the polynomial equation p(x) = 0 if and only if x  k is a factor of p(x); this is the Factor
theorem.
PRECALCULUS + 1 7
THE FuNDAMENTAL THEOREM oF ALGEBRA AND RooTs OF PoLYNOMIAL EouATIONS The Fundamental Theorem of Algebra states that every polynomial equation
a X" + a11 1x"1 + II
···
+ a 1X + a0 = 0
(*)
of degree n � 1 has at least one root (real or complex). By using the division algorithm repeatedly, every polynomial of degree n � 1 can be written as the product of unique, linear (degree 1) factors,
a x" + a11_1x"1 + . . . + a1 x + a0 = a (x k1 )(x  k ) ... (x  k) 2 " " where each k; is a constant (real or complex). If the factor x  k; appears on the righthand side exactly m; times, then x  k; is a factor of multiplicity m;, and the polynomial equation ( *) has x = k; as a root of
multiplicity m;. So every degree n polynomial equation has exactly n roots, where a root of multiplicity m.I
is counted m I. times. Let p(x) be a polynomial of degree n � 1, with real coefficients. There are several results that restrict the possibilities when searching for the roots of the polynomial equation p(x) = 0. The Root Location Theorem
If p(x) is a real polynomial, and real numbers a and b can be found such that p(a) < 0 and p(b) 0, then there must be at least one value of x between a and b such that p(x) = 0. (This is a consequence of the Intermediate Value theorem, which you'll see in Chapter
>
2.)
The Rational Roots Theorem
If the coefficients of
P(x) = a11x" + an 1x"1 + . . . + a1x + a0 
are integers, then the
Rational Roots theorem says that if the equation p(x) = 0 has any rational roots,
then, when expressed in lowest terms, they must be of the form x = � , where s is a factor of the constant
t
term (a0) and t is a factor of the leading coefficient (aJ The Conjugate Radical Roots Theorem
If the coefficients of
P(X) = a11X" + an 1x"1 + .. + a1X + a0 ·
are rational, then the Conj ugate Radical Roots theorem states that if the equation p(x) = 0 has a root of the form x = s + t;;;_ (where ;;;_ is irrational), then the equation must also have the conjugate radical, x = s  t;;;_ , as a root. That is, irrational roots of rationalcoefficient polynomial equations must occur in conjugate radical pairs. The Complex Conjugate Roots Theorem
If the coefficients of
P(X) = anX" + an 1x"1 + .. + a1X + a0 ·
are real, then the Complex Conj ugate Roots theorem states that if the equation p(x) = 0 has a complex root of the form x = s + ti, then the equation must also have the complex conjugate, x = s  ti, as a root. That is, complex roots of realcoefficient polynomial equations must occur in complex conjugate pairs.
18 +
CRACKING THE GRE MATHEMATICS SUBJECT TEST
SuM AND PRODUCT OF THE RooTs For the polynomial equation
a X" + a111 x"1 1T
+
·.
. +aX 1
+ a = 0 (a 0
:;t
JJ
0)
the sum and product of the n roots can be written in terms of the coefficients, as follows: sum of the roots = 
a,_l a,

a
product of the roots = ( 1)" _g_
a,
In particular, if the polynomial is monic (which means that the leading coefficient, a,, is equal to 1), then the sum of the roots is a,_1 , and the product of the roots is (1)"a0.
I
I
Example 1.12 Solution:
Find the monic cubic polynomial p(x) that has real coefficients such that the equa tion p(x) = 0 has 1 and  i as roots.
2
2(2 +
Since p(x) is a real polynomial, the fact that i is a root implies that the complex conjugate, i, is also a root. So both x i) and x i) are factors of p(x). And since 1 is a root, p(x) also has x  1 as a factor. Since p(x) is monic and has a degree of 3, it must be true that p(x) is the product of these linear factors:
2+
(2 
p(x) = [x 
=
(2  i)] [x  (2 + i)] [x  1 ] [(x  2) + i][(x  2)  i][x  1] [(x2  4x 4)  (1)][x 1] (x2  4x + S)(x  1)
=
x3  5x2 + 9x  5
=
=
Example 1.13 Solution:
+

bx = 0, the sum of the roots is 3, and the sum of the For the equation squares of the roots is 1. Find the numerical value of
x2 + + c
c.
r1 + + c c, r 2 + r22 = (r + r2)2 2r r2 = (b)2  2c = b2  2c b 3 (3)2  2c 1, c 4. b2  2c 1. b = 3,
First, notice that if and r are the roots of the quadratic equation x2 bx = 0, then the 2 sum of the roots is b and the product of the roots is so the sum of the squares of the roots is: I
I
1
So for this question, we have = and comes = which gives =
=
Since
the second equation be
PRECALCULUS + 1 9
Example 1.14 Solution:
The equation 3x4  7x3 + 5x2  7x + 2 = 0 has exactly two rational roots, both of which are positive . Find the larger of these two roots.
Since this polynomial has integer coefficients, we can apply the Rational Roots theorem to determine the set of possible rational roots. The factors of the constant term, a = 2, are ±1 0 and ±2, and the factors of the leading coefficient, a4 = 3, are ±1 and ±3, which means that the rational roots are among the following:
Since we're told that the equation has exactly two positive rational roots, we only need to consider the four possibilities 1, mial, we find that x = �
that is our answer.
3
� , 2, and � . Substituting each of these into the polyno
3
3
and x = 2 satisfy the equation. The larger of these roots is 2, and
LOGAR ITHMS Logarithms are exponents. Given the equation bY = x, the exponent is y, which means that the logarithm is y. More precisely, we'd say that y is the logarithm base b of x, and write y = logb x. The laws of logarithms follow directly from the corresponding laws of exponents. In the equations below, b is a positive number that's not equal to 1. •
•
logb x = y means bY = x
The function y = 1ogb x is the inverse of the exponential function y = bx. The domain of the function f(x) = logbx is x 0, and the range is the set of all real numbers. If b 1, the function is increasing (see the diagram below); if 0 < b < 1, the function is decreasing.
>
>
y
(0, 1 )
20 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
y=x
•
(logb a)(log. x) = logbx (this is the changeofbase formula, with a :t 1)
The two most important bases for logarithms are b = 10 [because we use a base10 (decimal) number system] and b = e, where e is an irrational constant, approximately equal to 2.718. The selection of this seemingly unusual number is based on considerations in calculus (which we'll review in the next chapter) and is so important that the function j(x) = log, x is called the natural logarithm function. On the GRE Math Subject Test, the "e" is understood, so log x means log,x. It's important to be aware of this, since in many precalculus and calculus textsand on calculatorslog x denotes log10x and the abbreviation for log, x is ln x.
Example 1.15 Solution:
Solve for x: 4x = 2x + 3
Since 4x = (22)" = 22r = (2x)2, the equation is equivalent to (2x)2  2x  3 = 0, which is quadratic in 2x. The quadratic formula gives: 2 x 1 ± .JD = 2 Since 2x cannot be negative, we have to disregard the negative value on the righthand side and conclude that:
Therefore, x = log
Example 1.16 Solution:
2
1 + 2.JD
= log
2
r:;: (1 + vt13) 1.
If x2 + y2 = 14xy, then log[k(x + y)] value of k.
=
� (log x + log y) for some constant k. Find the
+
Adding 2xy to both sides of the first equation gives (x y)2 = 16xy, which is equivalent to 1 [  (x + y)F = xy. Taking the log of both sides of this equation gives: 4 2 (log x + log y ) 2 log[ ..!.. (x + y)] = log x + log y => log l ( x + y) = l 4 4
[
J
1. Therefore, k = 
4
PRECALCULUS + 2 1
Example 1.17 Solution:
Simplify [log:ry(xY) )[ 1 + logXy] .
Since 1 = logx x, the second factor, 1 + logx y, is equal to:
logx X + logx y = logx xy
Applying the changeofbase formula, (log" b)(logb c) = log" c, we find that: [logx xy] [logx (xY)] = logx (xY) = y y
TRIGONOMETRY Although trigonometry began as the study of triangles, the usefulness of the trigonometric functions now extends far beyond this simple geometric form. We'll begin our study by reviewing the definitions of the trig functions, first with acute angles in right triangles and then with arbitrary angles and real numbers.
TRIG fuNCTIONS OF AcuTE ANGLES The classical definitions of the six trig functionssine, cosine, tangent, cosecant, secant, and cotangent use the lengths of the sides of a right triangle. In the figure below, triangle ABC is a right triangle with its right angle at C. The lengths of the legs BC and AC are denoted a and b, respectively, and the length of the hypotenuse AB is denoted c. sin A = cos A = tan A =
opp hyp
adj hyp opp
adj
=�
c
=
� c
=�
b
B
hyp c csc A =  = opp a sec A =
hyp
b b cot A =  = opp a
adj ad·J
c
=�
A ,c__
a
...L_j c
_ _ _ _ _ _ _ _
b
Angles A and B are complementary (that is, the sum of their measures is 90°), and notice that sin A = cos B. That is, the sine of B's complement (namely, A) is cos B, which is where the name "cosine" comes from (complement's sine). The same is true for the other pairs of cofunctionstangent and cotangent, secant and cosecant. You'll also notice certain reciprocal relationships among the functions. Cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. Finally, all of the trig functions can be written in terms of sine and cosine: The reciprocal relationships take care of secant and cosecant, and it's easy to see that tangent is the ratio of sine to cosine, and cotangent is the ratio of cosine to sine. Two special right triangles allow us to determine the numerical values of the trig functions of the angles 30°, 45°, and 60°.
22 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
30
2 1
1
1
sin
30° = cos 60° = _!2
sin
2 esc 60° = sec 30° = 
esc 30° = sec 60° = 2
csc45° = sec45° = J2
60° = cos 30° = J3 2
J3 tan 60° = cot 30° = J3 tan 30° = cot 60° = J3 1
TRIG FUNCTIONS OF ARBITRARY ANGLES The definitions above work only for acute angles (that is, ones whose measures are between 0° and 90°), but they can be extended to any size angle. In the xyplane, consider the ray along the positive xaxis whose initial point is the origin, 0. This ray can be rotated in the plane either clockwise or counterclockwise any number of degrees to form an angle, e, with the positive xaxis as its initial side. By convention, counter clockwise rotation results in a positive angle, and clockwise rotation gives a negative angle. Choose any point P: (x, y)except the originon the terminal side of the angle and let r be the distance from 0 to P. Then the values of the six trig functions of e are as follows: y
r
sin e = }L
csc 8 = 
X r
sec 8 = 
r
cos O = tan 8 = }L
X
y
r X X cot 8 = 
X
X
y
If the terminal side of e coincides with one of the coordinate axes, either x or y will be zero, and exactly two of the trig functions (tan, esc, sec, or cot) will be undefined. For example, if e = 90°, then at every point P on the terminal side, the value of x is zero, so tan e and sec e are undefined. Notice also that the trig functions of arbitrary angles can be positive, negative, or zero. Since r is always positive, the sign of each trig function depends on the sign of x or y (or both). For example, if the terminal side of e lies in the second quadrantas it does in the illustration abovethen x is negative and y is positive. So sin e and esc e will be positive, but the values of the other four trig functions of e will be negative.
PRECALCULUS + 23
TRIG F UNCTIONS OF R EAL N UMBERS alternate but equivalent definition of the trig functions utilizes the unit circle in the xyplane. This circle has a radius of 1, is centered at the origin, and has the equation x2 + y2 = 1 . Let 8 be any real number and measure the arc of length 181 along the circle, starting at the point A: (1, 0). If 8 is positive, travel along the circle in the counterclockwise direction; if is negative, measure a length 18[ clockwise. If P: (x, y) denotes the endpoint of the arc, then the trig functions of 8 are defined as follows. An
8
y
length of arc AP = 8 ,...
1
sin O = y
csc 8 = y
cos O = X
sec 8 = x
tan 8 = � X
1
A:
X
(1, 0)
cot O = y .
umt
.
1
ore e
/
These definitions are the same as those preceding if we interpret
r
8 as the radian measure of the cen
tral angle. Recall that if the central angle in a circle of radius subtends an arc of length s, then the radian measure of the angle is defined as � . the circle, and its arc length is s =
In particular, if the central angle measures
180°, it subtends half
21 (2nr) = nr; thus, the radian measure of this angle is
8
nr r r
= � =
= 1t.
This gives the conversion between radian measure and degree measure:
1t radians = 180°
Using this equation, we have the following correspondences for some special small angles:
Memorizing these four makes it easier to determine the radian measure of larger special angles; for example, the radian measure of a
150° angle is
S n , since 150° = 5(30°). Throughout the remainder of this
6
chapter and in the review questions, we'll use radian measure exclusively.
TRIG I DENTITIES AND FoRMULAS In this section, we'll give you a list of some of the most important trig identities and formulas; these are equations that involve trig functions that hold true for any values of the argument(s) for which both sides are defined. All of these identities can be proved using the definitions given above. Remember that for any number except 1, the notation sink x denotes (sin x)k; the same is true for the other trig functions.
k
24
+
CRACKING THE GRE MATHEMATICS SUBJECT TEST
Fundamental I dentities

cot e = cos O sin8
sin e
tan e = 
cos O
csc 8 =
1  sin 8
sec 8 =
1  cos 8
OppositeAngle Identities
sin(6) =  sin 6 cos(6) = cos 6 (These identities say that sine is an odd function and cosine is an even function.) Pythagorean I dentities
sin26 + cos26 = 1
Addition and Subtraction Formulas
sin(a + �) = sin a cos � + cos a sin �
cos(a + �) = cos a cos �  sin a sin �
sin( a  �) = sin a cos �  cos a sin �
cos(a  �) = cos a cos � + sin a sin �
) tan a + tan B tan( a+ pA = '1  tan a tan B B ) tan a  tan ' tan(a  Ap = 1 + tan a tan B

DoubleAngle Formulas
sin 26 = 2 sin 6 cos 6 tan 28 =
2 tan8
1  tan 2 8
ComplementaryAngle ( Reduction ) Formulas
sin HalfAngle Formu las
)
( �  8) = cos 8
. 8 = ± 1  cos8 sm 2 2
cos
)
( �  8) = sin 8
8 cos � = ± 1 + cos 2 2
e
sin8 
tan  = 2 1 + cos 8
PRECALCULUS + 2S
Example 1.18
Solution:
What is the exact value of tan �12 ? 1t
1t
1t
'
Since 12 = 3. 4 , we can use the formula for tan( � to find that: mn � �mn � tan � =tan ( �  �) = 1+tantan3 4 = .J31+ .J31 = 1+.J3J31 • 11 .J3J3 = 1+213.J3  3 = 2 J3 �12 = .!_2 (2:), Another way to get thi s resul t woul d be to noti c e that 6 so the halfangle formula for tangent gives: a )
12 = 1+ J32 = 2+1.J3 = 2+1J3 . 22 .J3J3 =2  J3 "=" "
j Example 1.19 Solution:
If sin .!_3 what's the value of sec 20? We first find the value of cos 28:. cos 2e 1 2 sin26 12 (} J = � Therefore, since secant is the reciprocal of cosi1 ne, 1 9 sec28= cos28 == 7 7 e=
,
=
=

26 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
9
PERIODICITY OF THE TRIG FUNCTIONS
Ax functionfis said to be periodic if a constant exists, such thatf(x f(x) for all x (such that both x and are in the domain of f). It's clear from the definitions that all of the trig functions are periodic with 2rt, since 2rt corresponds to one complete rotation,alsosoworks the terminal side of 2rt coincides with the for the tangent and cotangent functions. terminal side of However, it turns out that Therefore, these two functions are said to be periodic with their periodthe smallest positive value of such that f(x f(x) always holdsequal to rt; the other four trig functions are periodic with period 2rt. +
k=
k
k
+
e.
+
k) =
e
k = 1t
+
k,
k) =
GRAPHS OF THE TRIG F UNCTIONS
From the definitions, we can determine the domain and range of each of the six trig functions. The functions sin x and cos x are defined for every real x, and the range of each of these func tions is the set of y such that I Y I � that is, the closed interval The function esc x is defined for all values of x for which sin x :t 0, that is, for all x except mul tiples of the range of this function consists of all y such that I Y I � 1 . The function sec x is defined for all values of x for which cos x :t 0, that is, for all x except odd multiples of .:!. rt; the range of this function consists of all y such that I Y I � The function tan x is defined for all values of x for which cos x :t 0, that is, for all x except odd multiples of 1:. the range of this function consists of all real numbers y. 2 The function cot x is defined for all values of x for which sin x :t 0, that is, for all x except multiples of rt; the range of this function also consists of all real y. •
•
•
•
•
1;
[1, 1].
rt;
1.
2
rt; y
y
y = tan x
y
PRECALCULUS + 27
THE I NVERSE TRIG FUNCTIONS
Suppose we wanted to find all real numbers x such that sin x = _!2 . We could immediately choose x = 6 or x = 6 , but there are infinitely many values of x that satisfy this equation (each equal to one of the aforementioned values, plus or minus any multiple of 2n). The function sin xis not onetoone, because for every value of yin the range of sin x, there's more than one x, such that sin x = y. So, if we want to define an inverse function for sine, we must first restrict the domain of sin x to obtain a onetoone function. The standard way to do this is to define a new function, denoted Sin x, that's identical to sin except that its do main is restricted to the closed interval [ � , �] .This function onetoone (and onto), so it has a well defined inverse, which is denoted by sin1x or arcsin x. [Remember that sin1x is not the same as (sin x)1. The former is the notation for a number whose sine is x; the latter is the reciprocal of sin x.] Therefore, arcsin xis the unique number (or angle) between 1t2 and 2 (inclusive), whose sine is x. Similarallmethods recorded the resularets inusedthe folto ldefi owinneg thetablein: verses of the other trig fw1ctions; for simplicity, we have �
Sn
x
is
I
�
Function
Domain
Range
arcsinx jxj�1 [ 1 , 1] arccosx jxj�l ([O, n] ) arctanx all x arccsc x j xj�1 [� , �] except arcsecx j xj�1 (0, n] except � arccotx all x (0, n) Example 1.20 Express the following in simplest form, in terms of x: sin(2 arctan x) Solution: If = arctan x, then, as the following diagram shows: sm = .Jx 2x+ 1 and cos .Jx 21 + 1  2! 2! 2' 2
e
.
e
0
e
== = =
X
1 28 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
=
=== =
Therefore, sin(2 arctan x) = sin2e =X2sin8cose1 =2·
·
2x
Evaluate: arccos "52 + arccos "103 Let a = arccos J52 then cos a = J52 , and: sill a �1  c�' Jl (Js J Jl � }s Next, let � = arccos "10�3 ; then cos �= "10�3 ,and: r;:
Example 1.21 Solution:
=
= �x2 + 1 = �x2 + 1
�
;
a
o
o
o
o
In order to evaluate a+�' we wil first determine sin(a+�): sin(a+�) = sin a cos �+ cos a sin � 1 3 2 1 = J5 .flO + J5 .flO =
1 fi
Since both "25 and "103 are greater than "21� ,both a= arccos "52 and �= arccos "103 are less than 4 This means that their sum is less than 2 so the fact that sin(a+�)= "2� tells us that a+�= arcsin '12� = 4 r;:
r;:
�
2: .
�
2: ,
2: .
PRECALCULUS + 29
CHAPTER 1 R EVIEW QU ESTIONS
Compl ete the folandlowiexplnganati revioewns.questions using the techniques outlined in this chapter. Then, see Chapter for answers 1. Find the sum of the roots of the equation Jx  1 + J2x 1 = x. (E) 6 (C) 4 (D) 5 2 1
8
(A)
2.
(B)
Determine the set of positive values of x that satisfy the following inequality: 1 1 1 x1 x  2 1 (B) (0, 2 ) u (l, 2)
   >  X
(0, 1) u ( J2 , 2) (D) (0, J2 ) u �2 , 2) (A)
(E) (1, J2 ) u (2, 2 J2 )
(
3.
Solveforx: lx+11lxl + 2lx  11 = 2x1 (C)
4.
X= 1,2
+
I
X
+
+
Letfbe a realvalued function whose inverse is given by the equation: jl(x) = x(1 x2) (1 x2) What's the value ofJ(J1(/(2)))? 2 (D) 2 1 (C) 1 +
(A)
30
(D) X =  21 1, 2 (E)
�1
Leting fbecorrectla functi on suchJ(nthatf(n2) in terms 1) = 1  [f(n)F for all nonnegative integers n. Which of the follow y expresses off(n)? (A) 2[f(n)F 2f(n)  2[f(n)F (C) 2f(n) + 2[f(n)F (E) 2[f(n)F [f(n)P (D) 2[f(n)F  [f(n)P (B)
5.
(C) ( 21 , 1) u ( J2 , 2 J2 )
(B)
+ CRACKING THE GRE MATHEMATICS SUBJECT TEST
+
(E)
7
6.
Let f, g, and h be realvalued functions defined for all positive x such that: lf g)(x) = h)(x) Iff(x) x 1 and g(x) = J; , what is h(x)? (B) .Jx1 x2 1 (D) x Fx + 1 J; + 2
(A) 7.
=
o
+
(C)
What' s the equation of all points the xyplane that are equidistant from the points (1, 4) and (5, 2)? 2xy = 3 (B) X  y 1 X+y=3 (D) y = x2 4x + 1 (E) (x 2)2 (y  1)2 = 18 in
=
(A)
8.
(C)
+
Whi c h of the fol l o wi n g best descri b es the graph of the equati o n x2 y2 2x + 4y + 5 = in the xyplane? circle (B) parabola ellipse (D) line (E) point +
(A)
9.
(g o
0
(C)
Let C be the curve in the xyplane described by the equation x2 + 4y2 16. If every point (x, y) on C is replaced by the point ( _!2 x, y), what is the area enclosed by the resulting curve? 16 (D) 81t (B) 4n l61t 8 =
(A)
(C)
(E)
10. poiEverynts?point on the parabola y .J2x 1 is equidistant from the yaxis and which of the following (D) (2, (�I (� (B) (1, (E) ( �2 I 0) 2 2 One of the foci of the hyperbola y2 = ( � J 1 is the point .fi ). Find 1 1 1 (D) 1 (B) J2 (E) " 3 '2 2J2 =
(A)
0)
(C)
0)
11.
+
(A)
12.
1
(C)
0)
0)
(0,
a.
r::;
p(Whix) c=h0?one of the following polynomials p(x) has the property that .J3  .fi is a root of the equation 2x2 + 6x + 3 (B) .x3 2x 6 (A)
+
PRECALCULUS +
31
1, by x2  1. 1. Find the1,remainder when p(x) is divided
13. When the polynomial p(x) is divided by x  it leaves a remainder of and when p(x) is divided by x + 1, it leaves a remainder of
(A) 1
14.
(B) 0
(C)
X
(E) 2x
(D) x
Given that p(x) is a real polynomial of degree � 4 such that one can find five distinct solutions to the equation p(x) = 5, what is the value of p(S)?
�4
� 1
w o
(E) Cannot be determined from the information given
10
(A) 2  b2
=
1 0
+
Bx + = are the squares of the roots of the equation , which of the following expresses B in terms of b?
15. If the roots of the equation x2 x2 + bx +
�5
(B)
1  b2
16. Find the largest value of b such that
(D) b2
1
+ bi satisfies the equation
x3  3x2 + 6x  4 = 0
1
given that every root of this equation has the form + bi (where b is real).
(A) 1
(B) fi
(C) J3
(E) 3
(D) 2
17. If a and x are positive numbers and A = a2, express the following in its simplest form in terms of x:
(A) 2x
(B)
x2
(C)
J;
(E) x ./x
18. What are the roots of the following equation?
(A) 1, e2
(B)
1, );
32 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
(log x)2 = 2 log x (C) 1, ezt
(D) all x
>0
(E) all real x
19. The hyperbolic sine function, denoted sinh, is defined by the equation: e
.
Find a formula for sinh1 x.
1)
x  x e

sinh x =
2
1)
(C) log(x �)
(B) log(1 + �x 2 + (E) · log(.Jx2 + 1  x )
(A) log(1  �x2 + (D) log(x � )
20. The hyperbolic cosine function, denoted cosh, is defined by the equation:
2 If the hyperbolic tangent function, tanh, is defined by tanh x = sinh x find a formula for tanh1 x.
1 x1 (A) log 2
x+1
1 1
x1 (B) log 2 x+1 2
cosh x
1 x1
1 + 1 x (C) log 2 (D) .!_ log x + 2 1x
1
2
(E )
1+X 1 1og2 1x
21. Let x be the real number such that sin(sin x) = .!. and 2 < x < 3. What's the value of cos(sin x)? (A)
Jl  (�J
(B)
J1 (�)'
Jl  (�) 2
(C)
1111 is just slightly greater than 353.64 1t.) (C) 1111 (A) 11 (B) 111
'
(D)
J3
2
(E )
J3
2
22. Which one of the following is in the domain of the function f(x) = log( sin x)? (You may use the fact x
that
(D)
11,111
(E) None of these
23. Simplify tan(2 arcsin .!. ). 3
J2 (A) 2 9
(B)
J2 3
(C) 3
4
J2 (D) 4 7
(E)
6
S
PRECALCULUS + 33
(
�) 1. (B) 1
24. Simplify csc 2 arccot (A)
1
(D)
1t
4
(E) �
2
25. Determine the exact value of the sum arctan 1 + arctan 2 + arctan 3. (A) 2:
2
34 +
(B) 1t
CRACKING THE GRE MATHEMATICS SUBJECT TEST
(C) 327t
1t
(D) 1 4
(E)
1t
1 2
Calculus I I NTRODUCTION A full fifty percent of the questions on the GRE Math Subject Test involve calculus, which includes differential equations. We'll review singlevariable calculus in this chapter, multivariable calculus in Chapter 3, and differential equations in Chapter
4.
LI MITS OF SEQUENCES Let's begin our review of limits by looking at the limit of a sequence. Although a sequence is formally defined as a function that's defined on the set of positive integers, we usually think of a sequence as an infinite, ordered list of terms:
(x11),
For now, we'll restrict the values of the terms to real numbers. We say that a sequence, approaches (or has) a limit, L, if for every £ > no matter how small, there's an integer, N, such that the difference between x and L is less than £ for every n > N. If a value of L exists that satisfies this criterion, then we say the sequence is convergent, and that it converges to L; this is written \ 1 L or lim X11 = L. If there's no real number L with this property, that is, if the limit does not exist, the sequence is said to be divergent. II
0,
3S
( 2n 1 ) . The first few terms of this sequence
For example, let's consider the sequence (x), where xn =
rz
2  (�), n it's clear that as increases, the quantity .!_ decreases to zero, so the terms x should approach a limit of 2. We prove this n conjecture as follows: Let 0 be given. Then, if ! , every term x with n will be within of 2; thus, the limit of this sequence is indeed equal to 2. There are two types of divergent sequences that are often described as though they have a limit: 1. If, for any choice of M 0 (no matter how large), we have X11 M for every n greater than some (which will, in general, depend on M), then we say that the sequence diverges to infinity ( ) are x1
=
1, x2 =
� , x3
2
=
� , . . . . If w:e rewrite this equation for x as xn = 3
II
f.
>
N
2.
II
N>
f. >
rz
II
>N
f.
>
oo
and write X11 � oo, or lim X11 = oo, even though oo doesn't denote a real number. For example, the sequence (x11) with X11 = diverges to oo
2n  1
.
0
n
If, for any choice of M > (no matter how large), we have X11 < M for every greater than some N (which will, in general, depend on M), then we say that the sequence diverges to minus infinity and write xn � oo or lim x11 = oo. As an example, the sequence (xn) with xn = diverges to oo. Of course, there are divergent sequences that don't fall into either of these two latter categories. For example, the sequence (x11) with X11 = diverges; the terms simply oscillate between 1 and 1, so the sequence doesn't approach a unique limiting value.
2n  n2
(1 )"
One more important definition: A sequence (x11) is said to be monotonic if it's increasing (xn � xn 1 for every from some point on) or decreasing (x11 ;::: X11 for every from some point on). The sequences are increasing, and is decreasing, but ((1)11) is not monotonic because it .!. ) and oscillates.
(2  n
n
(2n  n2)
(2n  1)
n
+ 1
+
Some of the most useful rules about convergent sequences are summarized below: Every convergent sequence is bounded; that is, there exists a positive number, M, such that the absolute value of every term of the sequence is no greater than M. [The converse of this state ment is not true; for example, the sequence (x) with X11 = is bounded, but not convergent.]
1.
2.
(1 )"
If a sequence is monotonic and bounded, then it's convergent.
3. If k is a constant, and (an) converges to A, then (ka11) � kA. 4. If (a 11 ) converges to A and (b11) converges to B, then
(an + bn) � A + B,
5.
(5J_)bn
k (b) If lkl > 1, then ( ��� ) � 0.
(a  b ) � A  B (anbn) � AB, and II
�
(a) If is a positive constant, then
36 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
A B
II
1
(assuming that B :t
( �k)
� 0.
0).
( c c
6. Assume that (a ) and ) are sequences that converge to the same limit, L. If(b ) is a sequence such that a � b � for every n > N, then (b ) also converges to L. This is sometimes called the Sandwich (or Squeeze) theorem, since the terms of the sequence (b) are sandwiched between those of(a ) and ). n
II
II
II
(c
n
11
II
n
n
7. If a = f(n), then the sequence (a) converges to L if f(x) converges to L as x � oo (which may be decided using L'Hopital's rule). Rule 7 is included in the list for completeness, but we'll postpone illustrating its use until we've re viewed L'Hopital's rule (later in this chapter). n
Example 2.1
In each case, show that the sequence (x11) is convergent:
4n 3  n2 + 5n
(a) xn = 3 2n + 6n z
(b) xn =
(c) x = 11
I
L
11
�n + k  J;; (k is a constant)
_1·23 _·45 ·6·_·· ..:_·( 2· n(2n) _:_1 (:2) _ _ ·
_ _ ·
_ _ _
)
I
(d) xn = (lr
(e)
x = (cos n11)en
� �� ·
_
Solution:
_j
(a) Dividing both the numerator and denominator of the expression for X11 by n3 and applying rules 4 and 5, we see that this sequence converges to 2:
5 4  1 + n n2 11 2 + 6  n n3
7
40+0 = 2 2+00
(b) Notice that if k is negative, then the sequence will start with the first integer, n, that's greater than or equal to k (otherwise .Jn + k would not be real). A sequence can start at any n, and neither its convergence nor its limit will be affected. We can rewrite the terms x as follows: X11
=
( \In + k  \InI) ,;
•
rn+k + fn = 1 ,;\ln + k + \ln
=
,;
}";; � . Because the sequence � n + k + n \In � ), the sequence l converges to zero (rule 3). Now
Now, for all n 2 k, we have .Jn + k + fn 2 fn , so
( l) converges to zero (rule Sa, with k
k \ln + k + \l1n
"
X1 1
=
�
apply the Sandwich theorem (rule 6), with the sequence (a,) every term of which equals zero, and
the sequence
(c ) with c II
II
=
1
k
\Jn
. Since a � x � c for every n 2 k, we know that the sequence (x ) II
II
II
II
also converges to zero.
CALCULUS I + 37
�, x2 ( �) ( �) %, x3 ( � ) ( �) ( �) �!, . . . . Since x x [ (2n + 1)/(2n + 2)] and [ (2n + 1)/(2n + 2)] 1, we see that x x , so the sequence is decreasing; and because all the terms are positive, the sequence is bounded below by zero. Since the sequence is monotonic and bounded, rule 2 assures us that it converges. The sequence (c ), with c � , converges to zero (rule Sa, with 2), so the sequence a ) with a 1 also converges to zero (rule 3). Since a x ::::; c for every n, the sandwich theorem
(c) The first few terms of tlUs sequence are x1 = n+
(d)
1 =
n
•
n
n
=
n
n2
=
=
1
(d) For every value of x such that 0 � x < 1, we have 1 � x  1 < 0, so [x  1 ] = 1. Therefore:
lim [x  1] = lim (1) = 1 x�l
x�1
LIMITS OF f UNCTIONS AS
X � ±oo
We can also look at the behavior of a function f(x) as x increases (or decreases) without bound. To mimic the £8 definition given above, we say that f(x) 7 M as x 7 = if, for every £ > 0 (no matter how small), we can find a positive number 8 such that x > 8 implies I J(x)  Ml < £. On the other hand, we say that J(x) 7 m as x 7 oo if, for every £ > 0 (no matter how small), we can find a negative number 8, such that x < 8 implies I J(x )  m l < £.
Example 2.3 Find the value of each of these limits (if they exist): 2xz  x + 1 (b) 2x z  x + 1 (c) lim (arctan x) lim (a) lim 2 3 X>� + X4� x>� + X
4
X
Solution: (a) Dividing the numerator and denominator by
4
(d)
lim ! x>0 X
_jI
r gives us:
1 2  x1 + 2 2x x + 1 x2 2  + = 2 = lim lim X>� x 2 + 4 4 1+ 1+
00 0 2 x 2x2  x + 1 would have 2 as a horizontal asymptote. X2 + 4 x>�
. tells us that the graph of y = This
(b) Dividing the numerator and denominator by x3 gives us:
y=
2  12 + 1 2 x +1 = + = 2x x x x3 = lim lim 4 1+ x.� x3 + 4 x.� 1+x3 2 This tells us that the graph of y = 2x  x + 1 would have y = 0 (the xaxis) as a horizontal asymptote. x3 + 4
00 0 0 0
4 0 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
 1t
(c) We know, from the graph of y = arctan x, that lim (arctan x) =  :
2
X> �
y
1t
y
2
= arctan x
�• x 1t
2
(d) As the graph below shows, we can write lim lim x>0
(!)X
= oo
and
lim
(!)X =
oo;
but
(!)X does not exist, not only because the left and righthand limits are not the same, but X>0
also because they're both infinite!
X>0+
y
CONTI N UOUS F U N CTIONS Intuitively, a realvalued function f is continuous if the graph of the equation y = f(x) contains no breaks. To make this more precise, we say that a function f is continuous at a if limx J(x) = f(a). Notice that this definition involves checking three things. First, f must be defined at x = a, so that j(a) actually exists; second, the limit ofj(x) as x approaches a must exist (and be finite); and third, this limit must be equal to the value off at x a. Iff is continuous at every point in its domain, then we simply say that f is continuous. Let's look at some examples. The functions whose graphs are shown below are continuous at all points except at x = but the reason that f fails to be continuous at x = is different in each case. _,
=
1,
y
y= f(x )
x
r+..1
f(x) = xx2 11
y
3
{
1
�= g(x )
g(x) = Xx ++ 12 ififxX �< 11
y
{
•
y= h(x)
h(x) = x3 + 1 ififxX = 11 :;t:
CALCULUS I + 4 1
The function f is discontinuous at x = 1 because f is not defined at this point. The function g is discon tinuous at = 1 because, even though g is defined at x = 1, limx 1 g(x) does not exist (since the lefthand limit at 1 is not equal to the righthand limit at 1, as we saw above). Finally, the function h is discontinu ous at x = 1 because, although h is defined at = 1 and limx .... 1 h(x) exists (it's equal to the value of h at x = 1 is 3, so limx ; 1 h(x) 7: h(1). The list below summarizes some important properties of continuous functions.
x
....
x
2),
1. The following functions are continuous everywhere: Every constant function,
f(x) = k
11 Every polynomial function,f(x) = a11X + a11_/'1 + Every exponential function, J(x) = P (with k >
0)
·
·
+ a ? + a0
2. The following functions are continuous everywhere they're defined: Every function of the form
f(x) = x', where r is a rational number
The logarithm function, f(x) = log x The trig functions
3. If the functions f and g are both continuous at a, then each of the following functions is also continuous at a: J+ g J g
f
Jg
g
(provided that g(a) 7:
0)
4. Iff is continuous at a and g is continuous at f(a), then the composite function g
x
at = a.
Example 2.4
What value must we choose for k so that the function
ifx = 2 is continuous everywhere?
42 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
of is continuous
Solution:
Regardless of the value of k, this function is continuous for every x 7: 2 since, for all such x, f(x) is the quotient of two polynomials (which are everywhere continuous), where the polynomial in the denominator is never equal to zero. So the question is, What should k be for this function to be continuous at x = 2 ? By definition, f is continuous at 2 if limx 2 f(x) = f(2), which gives us: __,
x3  8 =k X>2 X  2
lim
lim (x  2)(x 2 + 2x + 4) x>2
Example 2.5
{
X2 2 + 2x + 4) = k lirn(x X>2 22 + 2 . 2 + 4 = k
12 = k
(0, 1) as follows:
Let f be the function defined on the interval I =
0
f ( x) = 1 n
L
=k
if x
is irra tional
if x =
m
n
(in lowest terms, m and n are integers with n > 0)
Show that f is continuous at every irrational point in I and discontinuous at every rational point in I.
0
Let a be an irrational number in J, and let £ > be given. We want to show that there exists a positive number 8 such that: Jx  aJ < 8 => j(x) < £
····
Solution:
Choose a positive integer q such that .! is less than £ . Then, in any open interval centered
q
+ 2 + + (q  1 ) = q ( q  1 ) positive 2 rational numbers less than 1 and of the form m , where m and n are positive integers and
at a and contained within I, there are fewer than
1
···
n
n is less than q. From this finite list of rational numbers, choose the onecall it rthat's
closest to a, and let 8 = J r aJ . Then, within the open interval (a  r, a + r), the value of f(x) will be less than _! , so f(x) will certainly be less than £. This establishes that f is continuous 
q
at every irrational a in J. To show that f is discontinuous at every rational a in 1
number a
=
J,
choose an arbitrary rational
E. in this interval, where p and q are positive integers and E. is in lowest
q
q
terms. If we can find a sequence (x) that converges to a such that the sequence (f(x)) does not converge to f(a), then we will have established that f is not continuous at a. To do this, simply consider the sequence (x ) where x, ,
=a
� For every integer n greater than
n
.
CALCULUS I + 43
1� , the terms of the sequence are in
av2
(0, 1) and increase monotonically to a. Since every
term of this sequence is irrational, the value of f(x) is 0 for every n, which means that the
0.
0
sequence (j(xn )) converges trivially to Since (j(x11 )) j but f(a) =
f is not continuous at a.
1

q
i:
0, we conclude that
THEOREMS CONCERNING CONTINUOUS FUNCTIONS There are several important theorems of calculus in which continuity plays a central role. In each of
these theorems, continuity of a realvalued function on a closed interval of the real line is an essential hypothesis; that is, if the function is not continuous on a closed interval, then none of the statements be low is necessarily true. Iff is defined on a closed interval [a, b ], then continuity at the lefthand endpoint, a, means that limx ) a+ f(x) = f(a), and continuity of f at the righthand endpoint, b, means that limx ) b f(x) = f(b).
The Extreme Value theorem: Iff is a function that's continuous on a closed interval [a, b], then f attains an absolute minimum value, m, at some point c E [a, b], and an absolute maximum value, M, at some point d E [a, b]. That is, there exist points c and d in [a, b] such that f(c) � f(x) � f(d) holds true for every x E [a, b]. y M

y = f(x)
Bolzano's theorem: Ifj is a function that's continuous on a closed interval [a, b] such thatf(a) and f(b) have opposite signs, then there's a point c between a and b such that f(c) =
0.
y y j(x) =
44 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Balzano's theorem is generalized in the foHowing theorem: The Intermediate Value theorem: Let fbe a function that's continuous on a closed interval [a, b]. Let be the absolute minimum value of f on [a, b], and let M be the absolute maximum value off on [a, b]. Then, for every number such that M, there is at least one value of c E [a, b] such that f(c) m
=
Y
Y.
y
m�Y�
y = f (x)
Prove or give a counterexample to the following statement: Iffthereis exicontistsnauouspoinont the[0,in1]terval [0, 1] and if 0 � f(x) � 1 for every x E [0, 1], then such thatf(c) The statement is 1,true,thenandthe conclfactusioint ihass immedi a name:ate, soBrouwer's fixedpoint theorem. If f(O) or i f /(1) i n what fol lows, we'll assume that <j(O) 1 and f(1) < 1. If we define a new function g by the equation g(x) f(x)  x, then gofisgatconti1 insuous on1, whi [0, 1]. The value of gat 0 is equal to f(O), which is positive, and the value f(1)negatiovne. means By Balzthatano'f(c)s theorem, there must be a poi n t c E [a, b] such that g(c) Thisclhastis equati c or, equi v al e ntl y , f(c) c. 
Example 2.6
cE
Solution:
0
=0
�
=
= c.
in
=
0�
= 0.
= 0
=
THE D E R IVATIVE
For a realvalued functionf of a single real variable, the derivative offis the functionr whose value at is f(x +h) f(x) f'(x) = lim h ilifnthie through s limit exithests.poiGeometri cally, the difference quotient, [f(x + h) f(x)] !h, gives the slope of the secant n ts (x,f(x)) and (x + h,f(x + h)). If the limit above exists, then ash� 0, this secant line becomes the tangent line to the graph off at the point (x, f(x)): X
h>0
CALCULUS I + 45
y
secant line
y = f(x)
and let h � 0 y
tangent line
The value of the derivative, f'(x), gives the slope of the tangent line, which is, by definition, the slope of the curve at x. To illustrate this definition, consider the function f(x) = :2. Its derivative is found in this way:
j(x + h)  j(x) f'(x) = lim h>0 h (x + h)z  xz = lim 11>0 h h(2x + h) = lim h>0 h = lim (2x + h) h>0 = 2x This tells us, for example, that the slope of the tangent line to the curve y = x2, at the point where x = 1, 1. is f'(1) = • 1 = so the equation of this tangent line is y = Not all functions have derivatives. For example, the absolutevalue function, f(x) = lxl , is not differ entiable at x because the limit
2 2, 0,
2x
=
1
does not exist (since the lefthand limit is 1, but the righthand limit is + ) Loosely speaking, a continuous function will fail to have a derivative at a point where its graph has a sharp comer (like the absolutevalue function) or a cusp; that is, functions with derivatives have smooth graphs.
46 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
.
y
y
��. x
"'does derivative not exist at x =
0
Other examples of functions that do not have derivatives can be given by exploiting the following im portant fact about differentiability: Iff is not continuous at a, then f is not differentiable at a. For example, the functionf whose value at every rational point on the real line is and whose value at every irrational point is 1, is continuous nowhere, so it's differentiable nowhere. It's important to notice that the examples illustrated above show us that the converse of the fact given above about differentiability and continu ity is not true: The functions y = l x l and = V are continuous at x = but they're not differentiable there. So differentiability implies continuity, but continuity by itself doesn't guarantee differentiability. We've used a prime to denote the derivative of a function. Another very common notation for deriva tives of functions is called Leibniz notation. The difference quotient used in the definition of the derivative is a change in f divided by a change in x; that is, we can write:
0
0,
y
f '(x) = limo d.t.
f(x + �)  f(x) = lim t¥ = df d.t.o � dx �
Therefore, the derivative of f with respect to x can be written either as f'(x) or, in Leibniz notation, as df/ dx. We think of the symbol d/ dx as the differentiation operator; that is, it says to differentiate whatever follows it. For example (d/ dx)(r) = 2x. The prime notation and the Leibniz notation for derivatives can be used interchangeably. When we want to find the derivative of a function, we usually use a table of standard derivatives and rules that make the computation easy, rather than going back to the definition every time. Following is a list of rules, all of which can be proved directly from the definition of a derivative. These rules must be memorized for the Math Subject Test. 1. Derivative of a sum
GRE
The derivative of a sum is the sum of the derivatives: (j+ g)'(x) = f'(x) + g'(x)
2. Derivative of a constant times a function: (kj)'(x) = kf'(x)
3. Derivative of a product The product rule says that: (jg)'(x) 4. Derivative of a quotient f ' The quotient rule says that:
() g
=
f(x)g'(x) + f'(x)g(x)
(x) =
g(x)f'(x)  f(x)g '(x) [ g(x)] z
CALCULUS I
+ 47
5. Derivative of a composite function
The chain rule says that: if u)'(x) = j'(u(x)) • u'(x) o
6. Derivative of an inversefunction
The inversefunction rule says that if rl is the inverse off, and f has a nonzero derivative at xo, thenr1 has a derivative at y0 = f(x0), and (f1 ) '(y0 ) = 1  .
f'(xo )

The chain rule is easy to remember when it's written using Leibniz notation. Iff is a function of u and u is a function of x, then the derivative of the composite function, f u, with respect to x, is: o
df = df du dx du dx
This equation makes it look like the du's cancel out of the fractions, even though the expressions in this equation aren't fractions. The rule for differentiating an inverse function is also easy to remember in Leibniz notation. If the derivative of y = f(x) is written dyI dx and the derivative of the inverse function, x = g(y), is written dx/dy, then
dx = 1 dy dy dx
which looks like a simple identity if we think of the derivatives as simple fractions. Of all of these rules, the chain rule is arguably the most widely used, and we can write it in an alterna tive (and more compact) form. In Leibniz notation, the chain rule is
df du d [f(u(x))] dx du dx
= •
which we can simplify, using the notation of differentials. To illustrate this, consider the function u(x) = :?, whose derivative is du/dx = 2x. If we move the dx to the righthand side of this equationas if du/ dx were a fraction with numerator du and denominator dxwe get du = 2x dx. This says that du, the differential of u, is equal to 2x times dx, the differential of x. principal use of differentials is in providing us with a simpler way of writing derivative formulas. For example, it's easy to show that the derivative of x! (where k is a constant) is equal to kx!1. So, if u is a differentiable function of x, then the chain rule tells us that the derivative of uk is kuk1 (du/ dx). Rather than writing d(uk)!dx = (kuk1 )(du/ dx), we can use differentials and write d(uk) = kuk1du, which means the same thing. In the list below, k is any constant, a is any positive constant, and u is any differentiable function.
A
48 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
d(k) = 0 d(uk) = kuk1du d(e" ) = e"du d(a" ) = Qoga)a"du 1 dQogu) = du u dQoga u) = 1 du ( a :1: ( u loga ) d(sinu) = cosu du d(cosu) = sinu du d(tanu) = sec 2 u du d(cotu) = csc 2 u du d(sec u) = sec u tan u du d(cscu) = cscu cotu du d(arcsinu) =
l�Example
� 
1)
h 1  u2
du d(arct:an u) = 2 1+u 2.7
Find the derivative of each of the following: log( sin2 x) (a) f(x) = x3e x  x  3 (b) g(x) = cosx 3

  ·
 � �
    



r
(c) h(x) = arctanvx 

J
·'�
Solution: (a) f '(x) = x 3 (3x 2ex' ) + 3x 2ex'  1 = 3x 2ex\ 1  x 3 )  1
. log(sm . 2 x) . 2 1 . 2 x) (cosx) 2 sinxcosx  ( smx) . 2 x og( sm 2cos 2 x + sm sm x = g'(x) = (b) cos2x cos 2 xsinx 1 1 (c) h'(x) = 2Tx = 2 1 + ( $) 2 $(1 + x)
Example
Solution:
2.8
What's the equation of the normal line through the origin to the curve y = (x4  1)3 log(x + 1) ?    
First, we'll need to find the slope of the line tangent to the curve at the indicated point; the normal line's slope will then be the opposite reciprocal of the tangent line's slope. Since
CALCULUS I + 49
1 y ' = (x4  1)3  + 3(x4  1) 2 4x 3 log(x + 1) , the slope of the line tangent to the curve x+1 •
•
at the point where x = 0 is y'(O) = 1, which tells us that the slope of the normal line is 1. The equation of the line with slope 1 through the origin is y = x. LINEAR APPROXIMATIONS USING DIFFERENTIALS
Let fbe a continuous function such that j'(a ) :t 0. The line tangent to the curve y = j(x) at x = a provides a good approximation of the graph off near x = a. � Near x = a, the tangent line is y close to the graph of y = f(x).
I,/
0.
a
So, if we want to figure out the value off at a point b close to a, we can instead find the value of y at b on the tangent line; this result will be a good approximation, and its accuracy will improve as b moves closer to a. The equation of the line tangent to the graph of y = j(x) at the point x = a is y = f(a) + j'(aXx a) , so f(b) will be approximately equal to f(a) + j'(a)(b a) . The equation of the line tangent at x = a can be written as y  j(a) = j'(a)(x a) . If we think of y j(a) as 11y, and x  a as 11x, then we have 11y = j'(a)11x as the equation of the tangent line. In the limit as x approaches a, this equation becomes dy = j'(a)dx which contains the differentials dy and dx, and illustrates the Leibniz notation, dy I dx = f'. What's an approximate value for e0·1 ?
Example 2.9 Solution:
If we let f(x) = r!, then we want the value of j(b), where b = 0.1. Since f(a), where a = 0, is easy to compute, and since b is close to a, we'll use the tangent line to the graph of y = f(x) at x = a to approximate the value of f(b) Since j '(x) = r!, we know that j'(a) = f'(O) = e0 = 1, so we have: dy = f '(O)dx � dy = 1 · dx � dy = dx � 11:y !u .
z
Because l1x = b  a = 0.1  0 = 0.1, this result tells us that 11y "' 0.1. Finally, since y at a = 0 is = 1, we know that y at b is approximately 1 + 11y = 1.1. (By the way, a calculator would tell you that e0. 1 1.105, so our approximation of 1.1 is pretty good.)
e0
"'
50 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
IMPLICIT DIFFERENTIATION
An equation of the form y = f(x) defines y as an explicit function of x, since the dependent variable, y, appears all by itself on one side of the equation, and the formula for computing it involves only the independent variable, x. But sometimes an equation that mixes the variables x and y can define a func tion. For example, the equation x2y + y = 2 defines y as a function of x, since for every value of x, there is exactly one value of y that satisfies the equation. In this case, we could solve for y explicitly and get y = 2l(x2 + 1), and the derivative of this function can be determined as we've done before. But the equation x2y5 + y = 2 is a different story. This fifthdegree equation cannot be solved for y ex plicitly in terms of x. Nevertheless, if we plot every point (x, y) that satisfies the equation, we'd get a curve that's the graph of a function of x. Even though we can't solve for y in terms of x, we can still figure out the derivative of this function by using a technique called implicit differentiation. In implicit differen tiation, we differentiate both sides of the equation with respect to x, remembering that y is a function of x. This means that every time we differentiate an expression involving y, the chain rule requires that we multiply by y', and then solve for y'. For this equation, we'd find that: xzl + y = 2 (x2 • 5y4y' + 2xy5) + y' = y'(Sx2y4 + 1) = 2xy 5 t/ = 2xys 5x2l +
0
·
1
Using this formula for the derivative, we can find the slope of the curve at any point. The Implicit Function theorem tells us precisely when an equation of the form f(x, y) = c (where c is a constant) actually defines y as a function of x, so that the formula we'd derive for y' using implicit dif ferentiation gives a result that makes sense. This theorem involves partial derivatives (which we'll review in the next chapter), but for the sake of completeness, we'll state it here. The implicit function theorem states that, if P0 = (x0, y0) satisfies the equation f(x, y) = c and both partial derivatives ft and /y are continuous in a neighborhood of P0, then if the value of fv is not zero at P0, there exists a unique differentiable function, y = g(x), that satisfies both the original equation, j(x, y) = c, and y0 = g(x0). Furthermore, the derivative of this function is given by the equation y' =  fx . fy HIGHERORDER DERIVATIVES
If we differentiate a function y = j(x), we generally get another function, y = j'(x). If we then differentiate this functionthat is, find the derivative of the derivativewe'll get another function, y = j"(x), called the second derivative off The process can continue: The derivative of the second derivative givesj"', the third derivative off, and so on. These higherorder derivatives can be used to give important information about the graph of a function and to classify critical points (as we'll see in the next section), as well as to determine the Taylor polynomials and Taylor series of a function, topics we'll review in the last section of this chapter. In Leibniz notation, the second derivative of y = y(x) is d2y I dx2, the third derivative is d3y I d�, and so on.
CALCULUS I + S 1
Example 2.10
What's the second derivative of the function f(x) = log(log x)?
Solution: We begin by finding the first derivative,
1
f'(x) = lo;x = xl�gx = (xlogxf1 ,
then we differentiate again to get the second derivative: 1 + logx = = ol
f"(x) (xlogxf2[x x + logx] (xlogx)2
C U RVE SKETC H I NG The first and second derivatives of a function y = f(x) can provide us with valuable information about the shape of the graph, if we keep the following geometric facts in mind. PROPERTIES OF THE fiRST DERIVATIVE •
•
• •
At a point where f'(x) > 0, the slope is positive, so the function is increasing. At a point where f'(x) < 0, the slope is negative, so the function is decreasing. At a point where f'(x) = 0, the slope is zero, so the function has a horizontal tangent line. This point is called a critical (or stationary) point off, and often signifies a turning point. At a point where f '(x) does not exist, the function could have a vertical tangent line, or it might not be differentiable at that point. This point is also called a critical point off and can some times signify a turning point for a function. y
f is decreasing for x a fhas a critical point at x = a f is increasing for a x b fhas a critical point at x = b f is decreasing for x > b
o f'(x)< D
f(c) is a local maximum
0 0
0
Furthermore, iff is defined on a closed interval, then the Extreme Value theorem guarantees that fwill actu ally attain an absolute minimum and an absolute maximum on this interval. There are three possibilities for the location of the absolute extrema on a closed interval: An absolute extremum will occur at a point c such that f'(c) = 0, f'(c) fails to exist, or at an endpoint of the interval.
CALCULUS I + 55
�

y = f(x)
on the interval [a, b] y
absolute maximum /                      f(a) relative  ,.. maximum : f(c2)  
:
' '
' ' '
' '
Example 2.12 Solution:
b
The sum of two nonnegative numbers, x and y, is 12. What's the largest possible product of :? and y? What' s the smallest?
Since x + y = 12, we know that y = 12  x; and because x and y are nonnegative, it must be true that 0 � x � 12. Therefore, we first want to maximize the function f(x) = :i(12  x) = 12r  0 on the closed interval [0, 12]. To do this, we first set f' equal to zero to find the critical points: f(x) = 12x2  x3 => f'(x) = 24x  3x 2 � 0 3x(8  x) = O X = 0, 8 Let's use the secondderivative test to classify these points. Since f"(x) = 24  6x, we have f"(O) = 24 0. This means that f has a minimum at x = 0; and f"(8) = 24 < 0, so f has a maximum at x = 8. Because we're finding the extreme values off on a closed interval, we must check the endpoints; at either x = 0 or x = 12, the value off is 0, so these are the abso lute minima, and f(8) = 82 = 256 is the absolute maximum off on [0, 12].
>
•
Example 2.13
Solution:
4
A rectangle in the fourth quadrant of the xyplane has adjacent sides on the coor
dinate axes. If the vertex opposite the origin is on the curve y = log x, what's the maximum area this rectangle can have?
The value of log x is negative on the interval A(x) = x(log x) = x log x.
56 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
(0,
1), so the area of the rectangle is
y y = log x
To find the maximum value of the function A on this interval, we take its derivative, set it equal to zero, and solve: A(x) = x log x
�
set
A'(x) = (1 + log x) = 0 log x = 1 1 ] x=e =e
1 1.) The secondderivative test verifies that this critical point does e indeed give a maximum: (Notice that 0
< 
0) J sin u du =  cos u + c J casu du = sinu + c J csc2 u du = cotu + c J sec u tan u du = sec u + c J cscucotu du =  csc u + c f r:?' du = arcsm. u + c '1/ 1 u 2 f du = arctan u + c 1 + u2 6 0 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
f 4 x dx2
Evaluate each of the following integrals:
Example 2.16
J tan x dx
(a)
(b)
x + 2x + 2
Solution: (a) First, we write tan x as (sin x) /(cos x). Now, if we let u = cos x, then du = sin x dx, and:
f
f
f tan x dx = dx sinx du =  =  log I u I + c =  log I cos xI + c u cos x
(b) The denominator of the integrand is equal to (x2 + 1 )2 and the given integral becomes:
f
(c)
f
f
+
1. If we let u = x2 + 1, then du = 2x dx,
t du x dx x dx  z a rctan u + c  2 arctan(x + 1) + c x + 2x 2 + 2 (x 2 + 1) 2 + 1 u2 +1 _
4
_
_ 1
_
1
2
The doubleangle formula for the cosine function says cos 2x = 1  2 sin2 x; solving this for sin2 x gives sin2 x =
1
2 (1  cos 2x). Therefore, I sin2 x dx = t I ( 1  cos 2x) dx = t I dx  t I cos2x dx
In this last integral, let u
= 2x, so du = 2 dx. This gives us:
J sin2 x dx = t J dx  t J cos 2x dx = 1x  12 J cos u · 1du 2 2 = 12 x  1 sin u + c = 12 x  1 sin2x + c 4 4
I ntegration by Ports
This integration technique is simply the reverse of the product rule for differentiation. The differential of the product of the functions u and v is d(uv) = u dv + v du
which is equivalent to the equation u dv = d(uv)  v du. Therefore,
J u dv = uv  J v du
This is the formula for integration by parts. To illustrate, let's evaluate this integral:
I x log x dx
CALCULUS I + 6 1
x
Any attempt at integration by substitution will be in vain, but, if we let u = log and dv = then we can write du =
1
X dx and v =
x dx,
1  2, and the integration by parts formula gives us:
2
x
J xlogx dx = (logx)(�x2) J (t x2 { � ) dx = � x 2 log x � J x dx � x2 log x � (t x2) + c = ( � x2 )(log x �) + c =
The secret to integration by parts is in choosing which part of the integrand to call u and which part to call dv so that the resulting integral is easier to figure out than the original integral. For example, if we had chosen u = and dv = log dx for the integral above, finding v would have been just as difficult as evaluating the given integral.
x
Trig Substitutions
x
�
�
,�
Integrals that contain the expressions a 2  u 2 a 2 + u 2 , or u 2  a 2 , where a is a positive constant, can be simplified by making a change of variable that involves a trigonometric function. For example, if we're given an integral that contains the expression a2  u 2 , we can make the change of variable u = a sin e1 to get:
�
The square root sign disappears, and that's the point; each of the trig substitutions uses a Pythagorean identity to eliminate the square root. What substitution should we make in each of the three cases? Make this substitution If the integrand contains
�az  uz .Ja 2 + u 2 .Ju 2  a z Let's try this out by evaluating the integral:
u = a sin 8 (and du = a cos 8 d8) u = a tan e (and du = a sec2e de) u = a sec 8 (and du = a sec e tan e d e)
x
If we were to attempt a basic substitution, like letting u = 2 + 1, we wouldn't have du = the integrand. Instead, we let x = tan 8, dx = sec28 de and transform the integral into:
62 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
2x dx left over in
u = sin 8, du = cos 8 de, which gives us: 1 + c =  csc8 +c cos e = J I_ � + c = du J sin2 d8 = e u2 u sin e
To evaluate this last integral, we let
Now, all that we have to do is rewrite this final result in terms of our original variable, � = this triangle is an illustration of our original substitution, triangle below, tan
8 = 1 x;
x.
In the right tan
x = 8:
X
1
Since the triangle tells us that esc
The Method of Partial Fractions
. e = �x2X+ 1 , we can now wnte: f x2�x21 + 1 dx =  � x +c
P(x) , where P and Q are polynomials x, that is, if it's equal to Q(x) (x) as a sum of simpler deg Q), we can often evaluate the integral by first expressing P Q(x)
If the integrand is a rational function of (with deg P
0? ax  1 lim X+0
Solution:
X
..
We notice that this looks like the definition of a derivative. If we let f(x) rivative off at x = 0 is f '(O) = lim h+0
= ax, then the de
J(O + h )  J(O) lim a h  l = h+0 h h
which (except for the choice of the dummy variable) is exactly the same as the limit for a which we're looking. Since f(x) = ax = f" Iog , we know that f'(O) = e x rog a
•
log a l ;o = log a x
and this is our answer. (Note: the value of a derivative off at a point x = a can be written either as f ' (a) or f'i x;a ·)
Example 2.24
Let A denote the area of the region bounded by the curve y =
X
1
the xaxis, and
the vertical lines x = 1 and x = a (where a > 1). In terms of A, what's the area of the region bounded by the curve y =
Solution:
1

1

X
1
the xaxis1 and the vertical lines x = a2
The area we're trying to find can be expressed as the definite integral
fa'a3 X1 dx
which we can simplify as follows: 3
fa'a X1 dx
=
3
log x ] a, = log a 3  log a 2 = 3 log a  2 log a = log a a
The information we're given about A allows us to write
a
A = f 1 dx 1X
so A = log a. Therefore, the area we're trying to find is also equal to A.
78 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Example 2.25 Solution:
What's the derivative of the function
  The differentiation formula   · 


f(x) = x..fx ?

 
·
!_ (xk ) = kxk1 dx
..1..
  ·
I
l

only works when the expression in the exponent is a constant. For the function we're given, however, the expression in the exponent is not a constant, so this differentiation formula cannot be used. Instead we use the fact that uv = e" to rewrite as
f(x)
log "
j(x) = x..[x = efxlog x
and then apply the differentiation formula d(e'')
=
e"du, to find that:
j'(x) = efx logx !___ dx (.[; logx) x x vIx 1 + log x = e f log x x.rx ( log x ) =  1 + •
·
·
•
Fx
(
•
•
1 2 Fx

)
2
L' HQPITAL' S R U LE The list of rules for dealing with limits (given in the first section of this chapter) mentioned L'Hopital's rule, and we'll now discuss it. L'Hopital's rule provides us with one of the most useful techniques for evaluating limits. Let's start by reconsidering Example 2.2(b) from page 39: lim If we attempt to substitute
x
=
X>1
� Fx  1
1 into this expression, we get the meaningless fraction
0
0 ; this is called an
indeterminate form, since it has no unique value. L'Hopital's rule gives us an easy way to handle this
j(x) and g(x) are functions that are differentiable in an open interval, I, containing a (except possibly at a itself) such that g '(x) 0 for all x :t a in I, and J(a) g(a ) 0, then j(x) = lim f '(x) ( ) lim g(x) x>a g '(x) situation; it says that if
::F
=
=
*
x....a
provided that the limit on the right exists. Applying this rule to the limit problem above, we'd differenti ate the numerator and denominator (separately!) and find lim � = lim 1 = J_ = 2 � '\Ir; X_1
x>1
x>1 2Jx _ 2·11
"x 7 a" is replaced by either the lefthand limit "x 7 a" or the righthand limit "x a+. " It also applies when "x 7 a " is replaced by
just as we did earlier. The statement of L'Hopital's rule remains valid if 7
CALCULUS I + 79
"x
7
both
00
oo" or "x 7 =." Furthermore, the rule can take care of the indeterminate form , which arises when 00
f(x) and g(x) become infinite, as x 7 a. And finally, the statement of (*) is still valid even when the
righthand side of (*) is + oo. L'Hopital's rule is truly powerful!
Example 2.26 Find each of the following limits: 1 s in 4 x (b) lim Fx (a) lim x.._.l $, x.._.O X 1 1 lim
x2 x>o 1  cos x
(c) lim
_
(d)
X_,..
(e) lim xl/
1t
x>1+
x( arctan x)
2
Solution: (a) This limit gives us the indeterminate form
0
0 , so we apply L'Hopital's rule:
4 cos 4x = 4 cos 0 = _± = 4 sin 4x lim X>0 X = lim X>0 1 1 1 (b) This problem also gives the indeterminate form
0
0 , and we find that: � .
.[;  1 = lim xl/2  1 x 1/2 = 2 �2lim = lim X>1 rx  1 x>1 X1/3 1 X>1 t x2/3 } 

(c) Once again, we have the indeterminate form
•
r1/2 � = 3 r2/3 = .l.} 2
0
0 ; applying L'Hopital's rule gives us:
2 lim 1 xcos x = lim� ..... sin x x>o
x
o
But notice that the limit on the righthand side is also of the form rule again:
0
0 , so we simply apply L'Hopital's
llin� = 2 = llin�= x>0 sin X cos X>0 cos X �1 = 2 X
(d) The expression in the denominator gives us oo • 0, which is another indeterminate form, but we 0 can easily turn this into a 0 indeterminate form, as follows
lim x�oo
1t
x(2
1
1 
arctan
x)
= lim
xtoo 7t
X
  arctan
2
x
and then apply L'Hopital's rule: 1 1 + X2 lim 'X' = lim __i__ = lim X>� 1 X>� x 2 X>� 1t   arctan x 2 1 + x2 1
80 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Now at this point, we could apply L'H6pital's rule again, but it's easier to simply notice that:
( )
lim
l + x2 +1 = 0+1 = 1 x.� x 2 = lim x.� _!_ x2
(e) This limit problem gives rise to the indeterminate form 100• When faced with a limit of an ex pression like [j(x)]8( x ) that gives any of the indeterminate forms or 0°, the trick is first to figure out the limit of log [f(x)JS< x l = g(x) log j(x). In this case, we would let y = x 11<x'll, so 0 log y = (log x)/(x2  1). The limit of log y as x 7 1 + gives us the familiar  indeterminate form, 0 and L'H6pital's rule tells us that: 1
100, oo0,
.!, y approaches e112 = Fe .
Therefore, since log y approaches
Example 2.27
2
What's the limit of the sequence an =
Solution:
(a11), where
�ogn)2 .j;; .
7
Remember rule 7, given near the beginning of this chapter on page 37: If
an = f(n), then the sequence (a11) converges to L if and only if f(x) converges to L as
x 7
oo (which may be decided using L'H6pital's rule).
Therefore, we determine the value of:
00
Since this limit is of the indeterminate form , we apply L'H6pital's rule: lim
00
Oog x)2 = lim 20og x) · x 1 = lim 4 log x x.� Fx xt� } x l/2 xt� xl/2 00
But the limit on the right is also of the form , so we apply L'H6pital's rule again: lim
00
4 log x = lim 4x1 lim � x;oo x l/2 x;� 1 x 1/2 x>oo x l/2 2 =
=
0
IMPROPER I NTEGRALS Up to now, the only definite integrals we've looked at have been integrals over a bounded interval, [a, b]. But we will now begin to work with improper integrals, one type of which are integrals over unbounded intervals, of the form ) (oo, b ], or ( ) . The definition of such an integral is easy to give. For example,
[a, oo ,
oo,oo
CALCULUS I + 81
r f(x)dx
=�I: f(x)dx
provided that the limit on the right exists; if it does, then we say that the integral converges. Similarly, we define:
J: f(x)dx = �J: f(x)dx
For an improper integral over the entire real line, (oo, oo) , we agree that, if both
J: f(x)dx and r j(x)dx converge for some number c, then J: f(x)dx converges, and J: f(x)dx = J: f(x)dx + r f(x)dx For example, let's determine the value of:
0
� 2
1 + xz f dx o
By definition, we would write:
f� 2 z dx = lim f b 2 dx= lim [ 2 b>� 1 + xz b>� 1+x o
o

arctan x]ob
= 2lim[arctanb] b;
�
=2[�) = 1t
As an example of a divergent improper integral, notice that:
This type of integral is called improper because one or both of the limits of integration are infinite; these are sometimes referred to as improper integrals of the first kind. Another type of integral that's also called improper has finite limits of integration but the integrand becomes infinite at one or both of the limits of integration or at a point between them. (These are improper integrals of the second kind.) For example,
I ==dx 1
I �1  x2 o
x
is an improper integral (of the second kind) because the integrand, 1/ �1  X 2 , goes to infinity as � 1, the upper limit of integration. This kind of improper integral is also defined as a limit; in this case, we would write:
dx= limb>1 f' h dx= limb>1 [arcsinx]' =Iim[ b>1 an:sinb]=�2 f' h 1 _ xz 1 _ xz 0
0
82 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
o
�f dx
f�
Example 2.28
Evaluate each of the following improper integrals: 3 e£ x (a) 0 ..[x dx (b) e x log x (c) 0 ( 2  1)2/3 x
Solution:
f
dx
(a) This integral is improper for two reasons; not only is one of the limits of integration infinite, but the integrand itself becomes infinite at the lower limit of integration. (We might call this an improper integral of the third kind!) First, let's find an antiderivative of the integrand. By letting = j; so that = dx/ 2 f; , we find that:
u
du
Therefore,
f
f
� e fx 
b  fx e
= = blim dx dx I r; >�
0 VX
a?0+ a V X
u
[
which gives us:
e
1 b>� e Jb atO+
1 e Fa
f
1 1]
0 =2
J

f

x, du = dx/x, we find that f dx = du = log u = log(log x) u x log x
(b) By making the substitution = log
�
] = 2 • lim [] = 2 • [
b 2 lim b>� e fx a a?0+
b
� = lim � = lim [ log�og x) ]b = lim [ log�og b) ] = oo X log X
b>�
e
X log X
b>�
e
b>�
This improper integral diverges to infinity.
x
(c) The integrand goes to infinity at = 1, a point within the domain of integration, [0, 3]. Therefore, we write the given integral as the sum
f
J
J
3 x = J x + 3 x (x2  1)2/3 dx (x2 1)2/3 dx (x2  1)2/3 dx 1
o
o
and attempt to evaluate each of the two integrals on the right, which are improper at a limit of integration. First, let's substitute = :2  1, = to find:
du 2x dx J (x2 _x1)2/3 dx = fudu2f3 = lz u1/3 = l (xz  1)1/3 u
J
2
CALCULUS I + 83
Therefore,
and
Since both of these integrals converge, we can conclude that the original improper integral con verges, and:
I N F I N ITE S E R I ES If we're given a sequence (an) = a1, a2, a3, .
. . , then �
� � a n = a1 + a 2 + a 3 + · · · tt=l is called an infinite series. We know how to add a finite number of terms, but how do we find the sum of infinitely many? The answer is to form a sequence associated with the series, the sequence (sn ) of its
partial sums:
51 = al 52 = a l + a2 53 = al + a2 + a3 5n = a1 + a2 + · · · + a
11
If the sequence of partial sums converges to a finite limit, S, then we say the infinite series converges to S; otherwise, the series is said to diverge. Let's look at an example; consider the series:
n=l The first few terms of its corresponding sequence of partial sums are
which seem to be tending to 1. We can prove this by using the formula
1 + x + x2 + · · · + X11 = 1  x"+1 1x

84 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
which holds for every x not equal to 1 . (If x does equal 1, the sum on the left is just equal to n + 1.) In this case, then, the n1h term of the sequence of the series' partial sums is:
s /1 = 1 + (1)2 + · · · + (t)" = Since
1
( 1 rl �(�)
[
]
1 1 = 2 1  ( t r  1 = 1  �"
_!_ 1 0 as n 1 oo, we see that s11 1 1. Therefore, 211
If r is a constant, then the series I, r'' is called a geometric series, and it can be shown that this series converges if and only if l rl < 1 . If this is the case, the following formula holds true for r :f:. 0: 1 I r" = 1n=O
r
Notice that this series starts at n = 0. A series can start at any value of the index n, with no effect on its convergence. However, the sum of a convergent series depends on which terms are included. The geometric series is one of the few series for which we can compute the value of its sum by a simple formula; this won't be the case for most series. Instead, we'll only be concerned with whether a series converges and, for this purpose, we need tests for convergence. The list below gives some of the most common and useful convergence tests for infinite series. 1. If the terms a of the series n verge.
I, all
do not approach zero as n
1 oo,
then the series cannot con
This does not imply that if the terms a n do approach 0, then the series converges. This statement is not 1 true, as the following important example illustrates. Let a =  . Then, although a n 1 0, the series
:L
1
" n
1 1 =1+++··· 2 3 n=l n 00
which is called the harmonic series, can be shown to diverge:
4.
n
.
I, a" converges, then L, ka" converges to k �>" for every constant k. If I, a " I, ka" diverges for every constant k * 0. If L, a" and L, b" both converge, then the series l:(an + bJ converges, and:
2. If
3.
L, I_ = oo
The series
diverges, then
L __!.._ (called the pseries) converges for every p > 1 and diverges for every p � 1 . nP
CALCULUS I + 8S
5.
The comparison test. Assume that 0 � an � b for all n > N. Then: n
�)" converges � La" converges �
�>n diverges 6. The ratio test. Given a series
�)" diverges
La" of nonnegative terms, form the limit: a lim n+] = L nt� a II
Then:
L
1
(Note: If L = 1, then this test is inconclusive.)
7. The root test. Given a series
Then:
L,a" of nonnegative terms, form the limit:
L1
(Note: If
�
�
Lan converges
�>�� diverges
L = 1, then this test is inconclusive.)
8. The integral test. If j(x) is a positive, monotonically decreasing function for x � 1, such that f(n)
= an for every positive integer n, then: �
La"
n= l
Example 2.29
converges
r j (X) dx converges
Which of the
following series converge, and which diverge? n2ooo 1 .. sin2 n L,L, (a) (b) n=O �n + 3 (c) n=l n ! n3 + n
�
..
(e)
86 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
log n �.. 7
..
Solution: (a) Notice that, for every n 2 1, we have:
n n3 + n
Since the series
=
converges (it's the pseries with p
11

diverges (it's a nonzero multiple of the pseries, with p =
2'\/n
diverges by the comparison test.
! ), this series 2
+ 2000 • �l  lim [( n + 1 )2000 • _1_] = 1 . 0 = 0 lnim>� aan,+)  limn>� [ (n(n+1)! n2000 n>� n n+1
(c) Let's apply the ratio test:
1)

Since the value of this limit is less than 1, the ratio test tells us that the series converges. (d) Here, we apply the root test:
lim \Ja" = lim Qogn33)" = lim _lno_g31"3 = 1 3 • lim n31" lim n 31", lim x31x 0
IH�
,c
In order to figure out log Y
"
1t
0
II>�
1
log
II>�
we first find
II>�
using L'H6pital's rule. Letting
y x31x,
m �: .lim Qog y)= lim. 3 logx = lim. �31 = 0 lim y = e0 =1 lim x31x =1 1 limrl>� \Jan, = log3 • lII>im� n3j'n = 1 3 · 1 = 1 3 < 1 ·
log X, a
XtOQ
x�oo
Therefore,
X

H•
=>
x�oo
,c
log
so the series converges.
=>
X ?oo
=
we have
X too
log
f lo�x dx u x, du ( �) dx, dv dx/x2, v �:
(e) We'll apply the integral test here; this means we must investigate the convergence of the improp er integral: � 1
Let's use integration by parts, with
J
log x d x2
x=
X
= log
=
_ log x _
1 d
x
J
x= x2
=
and
=
_ log x + 1 x
CALCULUS I + 87
Therefore,
f
] [
[
f
b log x dx = lim b log x dx = lim  log x + l = im l  log b + l lb+� b+� 1 x2 b+� x b 1 x2 1 �
But, by L'H6pital's rule,
lim log b
]
1
+ 1 = lim b = 0
b>� 1 b>� b so the improper integral converges (to 1). The integral test now assures us that the series will converge, too.
ALTERNATING SERIES Convergence tests 5 through 8 apply only to series whose terms are all nonnegative. But what if some of the terms are negative? One important type of series, which contains infinitely many negative terms, is the alternating series �
L, ( 1) "+1 a, ll=l
in which each all is nonnegative. The series is called alternating since the terms alternate in sign. Loosely speaking, alternating series have a better chance of converging to a finite sum, since the negative sum mands help offset the positive ones. The convergence test for an alternating series is simple: 9.
The alternating series test. The alternating series
L, ( 1)"+1 a, (with an 11=1
� 0 for all n) will con
verge provided that the terms all decrease monotonically, with a limit of zero. Notice that the condition "an decrease monotonically to 0" is not sufficient to ensure that a general series
L,a" converges [remember rule 1 above] , but it is sufficient to guarantee that an alternating series
I ( 1)"+1 a, converges.
Given a series :La, , some or all of whose terms all may be negative, we can form the series I l a, l , all of whose terms are clearly nonnegative. If this latter series converges, then we say the original series converges absolutely. If the original series converges but the latter does not, then we say the original series converges conditionally. Because convergence tests 5 through 8 apply only to series of nonnega tive terms, the following rule is especially helpful for deciding the convergence of series that may contain infinitely many negative terms, but which aren't necessarily alternating:
!n� if n
10. Every absolutely convergent series is convergent. Notice how helpful rule 10 is in the following example: If we let
1
a, 
is prime
� >" converge or diverge? Although this series contains infinitely many negative �
then does the series
n2 ifn
is not prime
11= 2
88 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
terms (because there are infinitely many primes), it is not alternating, so the alternating series test cannot be applied. However, since the series
L 21 is convergent (it's the pseries with p = 2), our series is 11
absolutely convergent, and so, by rule 10, it's convergent.
POWER S E R I ES in x is an infinite series whose terms are a11x", where each a
A power series
� LJ a , x" = a �
0
n=O
+ a1 x + a2 x 2 + a3 x3 + . . .
n
is a constant:
You can think of a power series as a polynomial of infinite degree, but since it's actually an infinite series, the definitions and convergence tests we've studied will apply here. A power series is useful only if it converges, so we'll first determine the values of x for which the series is convergent. Since some of the terms may be negative, we'll find those values of x for which the series is absolutely convergent. Ac cording to the ratio test, the series will converge absolutely if:
Let L x; if L
=
�� �:1 1 a
= oo,
a x"+1 a lim •�+1 = l x l lim � a X11 n>� a
, where we'll allow L
11
= oo.
_!!_:t:1_
If L
=
0, then the power series converges absolutely for all
then the power series converges only for
series converges absolutely for series converges is called the
11
x
=
0; and if L is positive and finite, then the power
l x l < .!.L and diverges for l xl > .!.L . The set of all values of x for which the
interval of convergence, and we now see that every power series in x falls
into one of three categories: 1 . The power series converges absolutely for all x in the interval (R, R ) and diverges for all x such that 1 > R, for some positive number R called the radius of convergence. The value of R is equal to ,
l xl
L
where L is defined as above. (The power series may also converge at one or both endpoints of the intervalthat is, at x
2.
= ±Rbut this must be checked on a casebycase basis.)
The power series converges absolutely for all x; the interval of convergence is (oo, oo) , and we say that the radius of convergence is oo.
3. The power series converges only for x = 0, so the radius of convergence is 0.
Example 2.30
Solution:
What's the interval of convergence for this power series? � 8 2x + x2 +  x3 + x4 + · · · = x" 9 n=l n2
I, 2"
( )2 z n n •  = xI • lim 2 ·  = 2 I x I < 1
According to the ratio test, the series will converge absolutely if:
lim 2"+1 x"+l n>�
( n + 1) 2
I
2"x " I
•�>�
n+1
=>
I x I < l2
CALCULUS I + 89
We'll now check convergence at the endpoints of the interval series becomes
(t, t ). When x t , the =
which converges (it's the pseries, with p = Therefore, the interval of convergence also t includes the point x = . Now, at x = , the series becomes
t
2).

i 2n: ( t)" = i(l)" � n 11=1
n=l
which also converges (because it's an alternating series whose terms are decreasing and approach zero). So we conclude that the interval of convergence of the power series is the closed interval ].
[ t, t
FuNCTIONS DEFINED BY PowER SERIES A function of x can be defined by a power series; in fact, this is where power series are most used. Let
L,a, x" �
be a power series in x, and define a functionfby: 
f(x) = L,a,x" n=O
Notice that, in order for the function to be well defined, the domain off must be a subset of the interval of convergence of the power series, so that the series actually converges. We'll now list a few very important facts about functions defined by power series. Within the interval of convergence of the power series, 1. The function f is continuous, differentiable, and integrable.
2.
The power series can be differentiated term by term, and the resulting power series gives the derivative off; that is: f(x) =
� >,x" n=O
::::}
 d f'(x) = L (a,x" ) = na,x "1 n=O dx n=l
.L,
3. The power series can be integrated term by term, and the resulting power series gives an inte gral off; that is:
90 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Example 2.31
We know that the geometric series
1 � X " =1 +x+ x2 +···= ..t_. 1x " converges for l xl < 1 . Use this series to find a power series expansion for ;()
(a) (b) (c)
g(x) log{l  x), h(x) = x / ( 1  x)2, and k(x) = arctan =
x.
·� � 
Solution:
(a) Once we realize that log(1  x) is an integral of 1/(1  x), we can write:
1 � ( 1) X = £... 1X n ;(J
1 >r+1 = �£... 1 X � log(1  X) = � £.. X
n
�o n + 1
�I n
11
(b) Now, for h(x), we know that 1 /(1  x)2 is the derivative of 1/(1  x), so:
1 = Ix " 1x 00
n=O
1 = x" n 1 (1  x) 2 I n=l 00
�
Multiplying this series by x gives us the power series for x/(1  xl 1 X 1 2 = I nx" � 2 = I nx " 0� 0� oo
oo
�
�
(c) We want to use the fact that arctan x is an integral of 1 /(1 + x2). So, first, we'll take the power series expansion of 1/(1  x) and replace x with x; this gives us:
1_ = I x" _ 1X n=O
Next, we replace x by x2
�
1 = f(x)" 1  (X) n=O
1_ = f(1)" x" _ 1+X n=O
�
and then integrate term by term:
� = f (1)" x2 " 1+X n=O
�
1_ = f (1)" x" _ 1+X
�
n=O
_1_2 = f (1 )" x2 " 1 + X n=O ( 1)" .. =o 2n + 1
arctan x = L  x2"+1 00
Notice that in part (c), we found the power series for 1 /(1 + x2) from the power series for 1/ (1 + x) by replacing x with x2• In general, we can get the power series for f [g(x)] from the power series forf(x) by replacing x with g(x). Other operations that are permitted on power series include adding, subtracting, multiplying, and dividing two power series or multiplying a power series by a polynomial [as we did in part (b), when we multiplied the power series for 1 /(1  x)2 by x to get the power series for xI (1  x)2].
CALCULUS I + 9 1
TAYLOR SERIES We'll complete our study of power series by learning how to generate the power series for a given func tion. Let's assume that a functionf has a power series expansion of the form:
f(x) = :L a11 x" = a0 + a1x + a2 x 2 + a3 x 3 + . . . 11;0 �
Substituting x = we immediately get a0 = J(O). But what are the values of the other coefficients: a1 , a2, etc.? Well, differentiating the equation above gives us + + · · · . If we now substitute = a + 0, we get a1 = f'(O). Differentiating again gives us J"(x) + · · · , so a2 = J"(O) . Continuing + • like this, we can figure out that the following formula gives u� every coefficient, an :
0,
f'(x) = 1 2a2 x 3a3 x2 = 2a2 2 3a3 x
j a" =
t
x
(n) (Q)
n!

These numbers are called the Taylor coefficients off, the power series
is called the Taylor series off and, if J(x) can be represented by a power series, then this is it.
Example 2.32 Solution:
Find the power series expansions for ff and sin
x.
= ff, then J (n)(x) = ex for every n, so: eo 1 a" = J(n)(O) = n! n! = n! Therefore, the power series for ff is: 3 1 11 X2 X · ex = � .L.  x = 1 + x +  +  + · · · 2! 3! n;Q n! If J(x)

�
As for sin x, we have
j(x) = sinx f'(x) = cosx f"(x) = sinx f'"(x) = cos x
=>
a0 = 0 f'(O) = 1 aI = _!_1! = 1 f "(O) = 0 a2 = 0 f/J/(0) = 1 a3 = _ 2_3!
f(O) = O
=>
=>
=>
=>
=>
=>
=>
n
and the cycle repeats for the higherorder derivatives. Therefore, every an with an even is equal to leaving only the coefficients an with an odd we conclude that:
0,
n;
3 X X � ( 1)" x2n+1 = x  11;0 (2n + 1)! 3! + 5!  · · · �
sin x = .L.
92 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
5
The Taylor series we found in the preceding example belong to a list that you should memorize for the test; these are the Taylor series for some of the most common functions encountered in calculus: 1 = 1 + x + x2 + x3 + · · · = �L.J x valid for 1 < x < 1
1X
1 1+X
 =
1
n
n;O
 x + x2  x 3 + · · · = � LJ (1) 11 x n;O
�
II
(1)"+1 x2 x3 log(1 + x) = x   +   · · · = 1x 2 3 n n;J e
X
3 1 X2 X LJ x =1+x++  + · · · = "' 2! 3! 11;o n ! �
valid for 1 < x < 1 II
II
valid for 1 < x � 1 valid for all x
x3 + x5  · · · = L (1) 1 x211+1 sinx = x  3! 5! 11;0 (2 n + 1 ) !
valid for all x
x2 x4 · "' (1) 11 211 COSX = 1   +   · · = LJ X 2! 4! n;O (2 n )!
valid for all x
�
�
lAYLOR POLYNOMIALS
If we truncate a Taylor series, we obtain a Taylor polynomial, which can be used to give us an approxi mation of the value of the Taylor series' function at some x in its interval of convergence. For example, let's use the Taylor series for I! to obtain an approximation of .,[e . If we choose the first four terms of the series, we get the cubic polynomial: Therefore,
(A calculator would tell you that, to three decimal places, .,[e "' 1.649, so our approximation is pretty good.) The error we incur in approximating the value of f(x) by the n th_degree Taylor polynomial f '( O) x + f" ( O ) x 2 + · · · + f ( " ) ( O) X 11 P (x) = f( O) + 1! 2! n! n
is exactly equal to

(n+1 ) ( c ) f (x)  P" (x) = f(n + 1) ! X 11 +1
CALCULUS I + 93
where c is some number between 0 and x. This form of the remainder (as it's called) can be used to find an upper bound on the error. Also, if the expression above is positive, then we know the approxima tion is too low, and if the error is n;gative, then the approximation is too high. Let's figure out a bound for the error in our estimation of ./e . Since we approximated .fe with the Taylor polynomial of degree n = 3, the exact error is equal to
1 2
for some number c between 0 and  . Since e = 2.71 8.. value of l certainly satisfies 1
t
.
< 4, we know, for example, that for 0 < c < 21 , the
< < 2. Therefore, we can be assured that the error is less than: _3_ (!)4 1 "" 0.0052 24 2 192 =
Using the values given above (our approximation, .fe 1.646, and the exact value to three decimal places, .Je 1.649), we can see that our approximation is too low by about 0.003, which is indeed within the bound established in the calculation above. One final note. The power series we've studied have been series in x; that is, the terms contain powers of x. However, we can also obtain a series expansion for a function in powers of(x  a), that is, a series of the form: "'
"'
�
� >n (x  a)" n;Q
In this case, we'd find the Taylor coefficients from the formula:
( a " = f "l(a) I

n.
When a = 0, we obtain the Taylor series in powers of x, which is sometimes referred to as the Maclaurin series, although this oldfashioned terminology is going out of style. If we use a Taylor polynomial in powers of(x  a) to approximate the value of a function at a point x within the interval of convergence of its Taylor series, then the error is ( n+l) f(x)  (x) = f ( ) (x  a)"+1 P"
where c is some number between a and x.
94 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
c (n + 1)!
CHAPTER 2 R EV I EW QU ESTIONS Complete the following review questions using the techniques outlined in this chapter. Then, see Chapter for answers and explanations.
1.
(x11) whose terms are given by the formula X = (cos mt)(sin2 n)
Consider the sequence
..:.__ ; ::::'
n
for each integer (A) 2.
(x)
Let
4
x1 2
(B)
4.
(E) 16
2
(B) 0

x.

(C)
lim x>o
5.
(D) 10
8
Given that this
[x] denote the greatest integer � If n is a positive integer, then � ( l x l  [x J) � (l x l  [x J) = ?
Evaluate the following limit:
(A)
(E) �
x,, = �5X 1 _1 + 6 for every integer n � 2. (C)
6
x
(A)
4n
n � 1. Given that this sequence converges, what is its limit? (C) log 2 (B) 1 (D) fi
Let be the sequence with = and sequence converges, what is its limit? (A)
3.
0
8
2
(E)
2n
(D) _!_2
(E)
1
arcsin x x x3
0
The curve whose equation is
(D) 2n  1
2x2 + 3x  2xy  y = 6
has two asymptotes. Identify these lines.
x = 1 and y = 2 (D) X =  21 and y = X 1
(A)
+
(B) x = 2 and y = 1 1 (E) X =  and y = 1  X 2
(C)
x =  21
and y =
x
CALCULUS I + 95
6.
{ x'  6x + 8 ifx ;t 2 f(x) = 3  2x2 + 2x  4
If the function
�
ifx = 2
is continuous everywhere, what is the value of k? (A)
7.
(B) !2
1
Evaluate the following limit:
(C) !8
8.
_!_
(B)
21t
 31
(E)
1
lim [_!_x2 r 1 t++ t' t at] o
x.o
(A)
(D)
!
(C)
1t
sin
.!2
(D) 1
(E ) �
2
Determine the domain of the following function:
f(x) = arcsinQogf;) (A)
9.
[0, 2e1 ]
1 1] (B) [ 2, e
(C) [e, i]
(D)
[ 2e1 , i]
(E)
[1, i]
Evaluate the derivative of the following function at x = e:
f(x) = arcsinQogf;)
1 eJ3
e (B) 
(A) 
10. For what values of m
(A) (D)
m =
m =
3, b 2 =
2, b 1 =
3e J2 J2 and b will the following function have a derivative for every x? x2 + x  3 �x � 1 f(x) = mx +b ifx > 1 7te
(C) 2
(D) £e
(E)
{
(B)
(E )
96 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
m=
m =
2, b 3 3, b 4 =
=
(C)
m=
1, b = 4
11.
If j(x) is a function that's differentiable everywhere, what is the value of this limit? lim j(x + 3 h 2 )  f(x  h 2 ) h>0 2h2
(B) 2j'(x)
(A) 4j'(x)
(E) 12.

(B) y = X  1
y=X+1
(C)
(D) y = 3x 
2
(E)
X+y= 1
(B) 1
2
=
0
(C)
(D) 1
(E)
2
If j(x) = 2 lx  11 + (x  1) 2 , what is the value ofj'(O)? (A)
15.
2
y= x 3
What is the slope of the tangent line to the curve xy(x + y) x + y4 at the point (1, 1)? (A)
14.
(D) t j '(x)
What is the equation of the tangent line to the curve y = x3  3x2 + 4x at the curve's inflection point? (A)
13.
The limit does not exist.
(C) j'(x)
(B)
4
2
(C) 0
If j(x) =
(D)
2
(E) 4
ex arccos x cos x
then the slope of the line tangent to the graph off at its yintercept is 1t
16.
(B)
1
(B)
0.04
 
1
1
1t (E)  +
=
the change in y? (A)
17.
1t
1 (C) 2 (D) 2 2 Let y � . If x increases from 2 to 2.09, which of the following most closely approximates x3 + 1 (A)
0.08
(C) 0.02
Ifj(1) = 1 andj'(1) = 1, then the value of (A)
1
(B)
0
d
dX
[ ?]
(C) 1
(D)
0.06
(E)
0.09
(E)
3
f (x at x = 1 is equal to Xf (X )
(D)
2
CALCULUS I + 97
1 at x = _!. ? f(x) = 12x 2
18. If n is a positive integer, what is the value of the n 1h derivative of (A) l(n") 2
(D)
(B) l(n') 2 0
_
(E)
n
n"
n!
=
=
19. Letf(x) be continuous on a bounded interval, [a, b] , where a :t b, such that f(a) 1 and f(b) 3, and exists for every x in (a, b). What does the MeanValue theorem say about f ?
j'(x)
(A) There exists a number c in the interval (a, b) such that j'(c) 0 . 0. (B) There exists a number c in the interval (a, b) such that (C) There exists a number in the interval (a, b) such that j'(c) (D) There exists a number c in the interval (a, b) such that j'(c) a). (E) There exists a number c in the interval (a, b) such that (b  a)j'(c ) 2. =
J(c) = = 2. = 2(b=
c
20.
What is the maximum area of a rectangle inscribed in a semicircle of radius a? (A) .fi a2
2
(B)
J3 a2 
1t
(D)
2
� 2v2
a2
21. The following function is defined for all positive x:
j(x)= fzx sint dt X
t
At what value of x on the interval (0, 3 rt ) does this function attain a local maximum? (A) 22.
23.
6
Let f(x) by f ?
(B) �
(D)
(C) � 2
= xkex, where k is a positive constant. For 3
k
J; (B) �kk
(C)
(B)
(C) 1t
Oog k)k
x>
rt
(
E)
2rt
3
0, what is the maximum value attained (D)
c�k J
(
E)
(H
The radius of a circle is decreasing at a rate of 0. 5 em per second. At what rate, in cm2 I sec, is the circle's area decreasing when the radius is 4 em? (A)
24.
�
2
4rt
The function (A) log 4
J(x) = fe,
e2x
(B)
2rt
t log t d t
(D)
98 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
(C)
log 2
2
1
(E)  rt 4
= 0, and a local maximum at x = (D) 1 (E) log 4
has an absolute minimum at x
log 2
1
 rc
25.
Evaluate the following integral: 7 (A) 20
26.
5
(B)
2
7 2
If
(A) 28.
Integrate
f x 2 dx
�.
\11  x
0
t j(x)dx
(A) �(arcsinx  �l x2 ) + c (B) (D)
29.
(B)
1
15
{
(D)
1 60
(E) 7 20
(D)
17 2
(E) 37 2
(D)
2
(E) 3
7
fa"2 [x]dx =
(C) �2
f(x) =
then the value of k for which
_3__
(C)
If [x] denotes the greatest integer � x, then
(A) 27.
1 (B) 60
2(x + 1) ifx � O k(l  xz ) if x > O
=
1 is
(C)
1
! (arcsinx + x�l  x2 ) + c
(C)
�(xarcsinx  �l  x2 ) + c
!(arcsin x  x.J1  x 2 ) + c (E) !(xarcsin x + .J1  x 2 ) + c
What is the area of the region in the first quadrant bounded by the curve y = x arctan x and the line x = 1? 7t n+2 7t + 4 (B) n 4 2 (C) (A) n 4 4 (E) (D) 4 4 4
30. Simplify the following:
exp
Js 3
dx
xz  3x + 2
[Note: Recall that exp x is a standard, alternate notation for er.] (A) � ! (B) 32 (D) �2 3
8
(C)
(E)
�
CALCULUS I + 99
31.
Calculate the area of the region in the first quadrant bounded by the graphs of y = 8x, y = :2, and y = 8.
(A) 12 32.
(B)
8
(C) 6
(D) 16
3
(E) 4
Which of the following expressions gives the area of the region bounded by the two circles pic tured below? r
= 13 sin e
y
�
"
(A) J02 �[(� sin8)2 (3cos W ]de (B) fo6 �(3cos8)2 d8 + JI �(J3sinWd8 (C) J06 �(� sin8 )2d8 + J; �(3 cosW de "
"
(D)
"
"
6
"
"
6
J0312 (3cosWd8+J2� 1(J3sin8f d8 2 3
(E) t3 �(� sine)2 de + J; �(3cose)2 de "
"
3
33.
34.
Let a and b be positive numbers. The region in the second quadrant bounded by the graphs of y = ai and y = bx is revolved around the xaxis. Which of the following relationships between a and b would imply that the volume of this solid of revolution is a constant, independent of a and b?
The region bounded by the graphs of y = i and y = 6  lxl is revolved around the yaxis. What is the volume of the generated solid? (A) 332 1t (B) 91t (C) 81t (E) 136 1t
1 00 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
35.
::::'fi:��r: :�:::: ��: r
(A) 36.
2
(B)
8
of ilie hypocycloid x213 + y'1' = 1 in the first quadrant from ilie
)
3J24
(C ) 1
(D)
sJ28
(E )
J3
(D)
e
(E)
e2
(D)
1
(E) fi
2
What positive value of a satisfies the following equation?
(A)
!
r
(C)
(B) ;{e
e
dx
f ax  = 1 X y
dy
a
Fe
37.
Evaluate the following limit:
38.
1 (B) (C) Fe Fe Let n be a number for which the improper integral (A)
lim (cos x)cot' x X>0
1
2
2

f�. x�od:x)"
converges. Determine the value of the integral. 1 1 (A) __ (B) ! (C) __ n n+1 n1 39.
(D) logn n+1
(E )
(D)
(E ) �
log n n1
Find the positive value of a that satisfies the equation:
r .J dx r .J x dx o
(A)
2J2 1t
(B)
1
a2  x2

o
(C) �
2J2
a2  x 2
J2
2
CALCULUS I + 1 01
40.
�f_ (x2d+ 1?
Which of the following improper integrals converge? X
I.
r xex dx
II.
2f (2d x)z
III.
X
o
(B)
(A) I only (D) I and III only 41.
(E)
II and III only
(C)
II only
(C)
II only
Which of the following infinite series converge?
�
1
Ln=2 n log n
II.
' L 'S(n + 2Xn + 3Xn + 4) �
III.
11=0
(n + 1)3
(B)
(A) I only
(D) 42.
I and II only
I and III only
(E)
I and II only
II and III only
Find the smallest value of b that makes the following statement true: If 0 � a
(A) 1
(B)
< b, then the series
2 log 2
(C) 2
43. Evaluate the following limit: lim
n>�
(A)
23
(B) 21
I 02 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
�
L (n2')z a" converges. 11=1
t k']
t n2  n3
·
( n)!
(D) fi
(E ) 4
(D) 61
(E)
 
.!.
(C) 3
1 12
44. Which of the following statements are true?
I. If an � 0 for every n, then:
then:
II. Ifan � 0 for every n, III. If an � 0 and all+
I
� all
�
=>
I, nan converges 11=1
f., F, converges. �
::::::>
I, a, converges.
for every n, then: 2>� converges �
11=1
11=1
=>
(B) I and III only (E) III only
(A) I and II only (D) II and III only 45. If 1 < x < 1, then
f., an converges
'Lni" �
II= I
�) 1)" all converges. �
11=1
(C)
X
xz (1  x2? x2 (E) (1 + x2)2
(C) (1 + x2)2
·
·
46. The sma11est pos1hve mteger x for wh'1ch the power senes
(A) 4
II only
=
(B)
· ·

(B) 6
(C) 7
� n !(2n)! "
f;t (3n ) ! x does not converge 1s (D)
8
·
(E) 9
47. In the Taylor series expansion (in powers of x) of the function J (x) = e x'x, what is the
coefficient of �?
(A) 7
(B)  3 2
7 (C)  6
(D) 7_ 6
(E) �
2
48. If ki (i = 0, 1, 2, 3, 4) are constants such that x4 = k0 + k 1 (x + 1) + k (x + 1)2 + k3 (x + 1? + k4 (x + 1) 4
is an identity in x, what is the value of k/
(A) 4
(B) 3
(C) 2
2
(D) 3
(E) 4
CALCULUS I + 1 03
49.
If the function f(x) = r! is expanded in powers of x, what is the minimum number of terms of the Taylor series that must be used to ensure that the resulting polynomial will approximate if; to within 1 06?
(B) 4
(A) 3
(D) 6
(C) 5
(E) 7
50. There is exactly one value of the constant k such that lim x>0
(e x2  x 2 lXcos x  1)
xk
is finite and nonzero. What is this value of k, and what is the limit, L?
1
(A) k = 4, L = 2_ 
(D)
1
k = 6, L = 4 
(B)
1
k = 6, L = 4
1
(E) k = 4, L = 2
1 04 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
(C)
k = 4, L = 1
Calculus I I I NTRODUCTION In the previous chapter, we reviewed the calculus of functions of one independent variable. In this chap ter, we'll study the calculus of functions of two or more independent variables. We'll begin by reviewing the analytic geometry of euclidean 3space (R3), then move on to topics that involve the differentiation of functions of several variables, and finish up by introducing topics that involve the integration of func tions of several variables. ANALYTI C GEOMETRY OF R3 If we construct a third copy of the real line and place it perpendicular to the xy plane so that all the ori gins coincide, we have the geometric representation of threedimensional space, R3 ("Rthree"). This third line is called the zaxis, and any point P in this threedimensional space is uniquely specified by its three coordinates, as an ordered triple.
1 05
z

� P = (xo , Yo , z o ) .
;!� y
Yo
.
 ..  .. ..       ..   5'
'
X
Just as the x and yaxes partition the plane into 22 = 4 quadrants, the x, y, and zaxes partition space (sometimes called xyzspace) into 23 = 8 octants. The octant in which all three coordinates are positive is called the first octant; there is no universal agreement on how the other seven octants are numbered. As in the figure above, the first octant is almost always the one that's presented faceon when we view xyzspace. The unit vectorAs i and j which point in the +x and +y directions, respectively, are now joined by a third unit vector, k , which points in the +z direction. So in 3space, we have i = (1, 0, 0), j = (0, 1, 0), = (0, 0, 1 ) and any vector in 3space can be written in terms of these three unit vectors: A A A v = (x,y,z) v = xi + y j + zk
k
When a vector is written like thisthat is, in terms of its componentsthe magnitude (or norm) of the vector, which is its length, is given by:
z
A
A
A
v,= x i + y j + zk 
k
....
zk
A
j �+'+. y
A
X
I 06 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
yj
''
The operations of vector addition, vector subtraction, and scalar multiplication are also easy to per form. If v1 = (x1, yl' z1 ) and v2 = (x2, y2, z2), then
vl v2 = vl + (vz ) = (xl , yl , zl) + (xz , Yz ,  Zz ) = (xi  xz , Y1  yz , 21  zz) where a is a scalar (that is, a number). The distance between two points, P1 = (x1, y1, z1) and P2 = (x2, y2, zz), is the length of the vector, v = Pl2, that connects them: z
v r� .
+.. , 
    ; "'              � ' 
' '
    _ ...
' •
,'
,
y
X Plz = l vl = li(xz  XI , Yz Y1 ' 2z  z1 ) l = �(xz  xj + (yz  yY + (zz  zj THE DoT PRODUCT
One way to multiply two vectors is to form their dot product. The dot product of two vectors, A and B, is defined as A B = ABcos8 •
where 8 is the angle between them. Particularly, notice that the dot product of two vectors is a scalar; for this reason, the dot product is also called the scalar product. The dot product has a variety of uses. For example, consider the vector projection of B onto A, which is denoted proj AB:
B
A
B A
A
CALCULUS I I + 1 07
If 0 � e � 90°, then the scalar projection of B onto A, projAB, is the magnitude of the vector projection; if 90° < e � 1 80°, then proj AB is equal to the negative of the magnitude of the vector projection. In all cases (assuming only that A :;:. 0), we have projAB = B cos 8, so: projA B = (proj A B) A = (proj A B)
A A
This equation can be rewritten neatly using the dot product. Since it's clear from the definition that the dot product of a vector with itself gives the square of its magnitude, A • A = N, and since proj AB = B cos e = (A B) / A, we can write: A·B A proj A B = A•A All we need now is a convenient way to actually calculate the value of the dot product. Well, we no tice that, by definition, the dot product is commutative (that is, A • B is always equal to B • A) and that: •

"
,..
,..
,..
,..
and
,...
i o i=j · j=k · k=l
,..
,..
,..
"
,..
"
i · j= j • k= k o i=O
The dot product is also associative with respect to vector addition; that is, A • (B + C) = (A • B ) + (A • C). Therefore, the dot product of a pair of 3vectors, A = (a1, a2, a3) and B = (b1, b2, b3), is equal to: A • B = (a1 i + a2 j + a3 k) • (b1 i + b2 j + b3 k) = al bl + az bz + a3b3 A
A
A
A
A
A
This simple formula allows us to evaluate the dot product of two vectors even if we don't know the angle between them. Because the dot product of a pair of vectors is equal to the product of their lengths and the cosine of the angle between them, if the angle between the vectors is 90°, then their dot product will be equal to zero. If we agree that the zero vector is perpendicular to every vector, then the converse is also true. This gives us one of the most frequently used properties of the dot product: A l_ B � A · B = O
1 08 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
THE (ROSS PRODUCT
The other principal way to multiply two vectors is to form their cross product. The cross product of two vectors, A and B, is defined as the vector that's perpendicular to the plane containing A and B (in accor dance with the righthand rule, which we'll explain shortly), and whose magnitude is = AB sin e where 8 is the angle between them. The magnitude of the cross product is equal to the area of the paral lelogram determined by vectors A and B: IIA x Bll �
AxB
area of p:ml elog'"� �         , :B
sin 8
A Since the cross product of two vectors is a vector, the cross product is also called the vector product. To figure out the direction of the cross product, we use the righthand rule. To understand this rule, first remember that any two vectors in space (neither of which is a scalar multiple of the other) determine a unique plane; the question is, Which of the two directions perpendicular to this plane is the direction of the cross product? Let the vectors A and B share the same initial point, and imagine placing the wrist of your right hand at this initial point, with your fingers pointing in the direction of vector A and your palm fac ing B. If you close your hand, curling your fingers inward (toward B), then your thumb will point in the direction of A x B. This shows that, unlike the dot product, the cross product is anticommutative; that is, B x A is always equal to (A x B). AxB
One of the most common uses of the cross product is in figuring out a vector normal to a plane. Since, by definition the cross product of two vectors is perpendicular (normal) to the plane that defines them, �ll !"'e p.e�d i,? a �onvenient way to calculate the cross product of two vectors. To do this, we notice that i x i = j x j = k x k = 0 and that: k i x j = k =  (j x i ) j X k = i = (k X j )
"
"'
,....
"
"'

k x i = j =  ( i x k) A
A
A
(�
10 1
CALCULUS II + 1 09
[We use the diagram on the previous page as follows: The cross product of any vector into the next vector (following the arrow) gives the third vector, and the cross product of any vector into the previous one (that is, going against the arrow) gives the opposite of the third vector.] Next, the cross product is associative with respect to vector addition; that is, A x (B + C) = (A x B) + (A x C). So, the cross product of a pair of 3vectors, A = (a 1, a2, a3) and B = (b1, b2, b3), is equal to A x B = (a1 i + a2 j + a3k) x (b1 i + b2 j +b3k) .....
"
"'
"
"
.....
= (a2b3  a3bz )i + (a3b1  a1 b3 )j + (a1 b2  a2 b1)k a a3 aJ a3 j + a J az k = z b2 b3 i  bl b3 bl b2 1 j k aJ az a3 bl b2 b3
=
where the last two equations use the determinant of a matrix, which makes the formula easier to remember. THE TRIPLE ScALAR PRoDucT
One of the higherorder products of vectors is the triple scalar product, (A x B) • C. The absolute value of this product is the volume of the parallelepiped formed by the vectors A, B, and C: I( A x B)
• C I = volume of parallelepipe
A
It can be shown that:
aJ az a3 (A x B) · C = b1 b2 b3
1 1 0 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
r
Example 3.1
Let A = t + k and B = t  2 J + 2 k . (a) . What's the magnitud: of the vector SA  B? (b) Find a vector whose j component is 6, and is orthogonal to both A and B. (c) Find a vector whose magnitude is 6, and is orthogonal to both A and B.
Solution:
(a) Since we find that:
SA  B = 5(1, 0, 1)  (1,  2, 2) = (5, 0, 5)  (1,  2, 2) = (6, 2 , 3)
(b) Let C = (c1, 6, c3) be the vector we're looking for. If C is to be orthogonal (that is, perpendicular) to both A and B, then its dot product with both A and B must be equal to zero. This gives us two equations: A · C=O B • C=0
:=}
:=}
(1, 0, 1) • (c1 , 6, c3 ) = 0 (1, 2, 2) • (c1 , 6, c3) = 0
:=}
:=}
c , + c3 = 0 c1  12 + 2c3 = 0
Solving these simultaneous equations gives us c1 = c3 = 4, so the vector C is (4, 6, 4) = 4i + 6) + 4k. (c) Trying the approach we used in part (b) would not be as simple here. Instead, we'll use the fact that A x B is automatically orthogonal to both A and B, and start by figuring out the cross prod uct: J
k
A x B = 1 0 1 = [0  (2)]i  (2  1)J +(2  0)k = 2i + 3J + 2k 1 2 2 Since any scalar multiple of this vector will also be orthogonal to both A and B, all we need is to find a scalar multiple of A x B that has magnitude 6. Let D = A x B = (2, 3, Since D is a unit D vector (magnitude = 1), the vector
2).
has magnitude 6 and thus satisfies the conditions of the problem.
CALCULUS II + 1 1 1
LINES IN 3SPACE
A line in the plane is determined once we know its slope and a point through which the line passes. Similarly, a line in 3space is determined once we know a point, P0 = (xO' y0, z0), on the line and a vector, v = (v1 , v2, v3), that's parallel to the line; v tells us the line's direction. As the figure below shows, a point P = (x, y, z) will lie on line L if and only if Pl = (x  x0, y  y0, z  z0), the vector from P0 to P, is a scalar multiple of v: z
X
That is, P = (x, y, z) is on L if and only if (x  x0, y  yO' z  z0)
= tv = t(v1, v2, v3)
which is equivalent to the three parametric equations :
We can also eliminate the parameter, t, from this set of equations and write the equation of L in the form: x  xo
__
=
y  yo
__
=
z  zo
__
If any one of the components v1, v2, or v3 is equal to zero, then the fraction containing that component is simply omitted from the equation. These equations are the symmetric equations of the line.
1 1 2 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Example 3.2 Solution:
Where does the line that passes through the points P = (3, 1, 2) and Q = (5, 8, 4) pierce the xyplane?
The line is parallel to the vector PQ (5, 8, 4)  (3, 1, 2) = (8, 9, 2) so the parametric equations for the line are: X = 3 + 8t, y = 1 + 9 t, Z = 2 + 2t The line will intersect the xy plane when z = 0. The value of t that makes z equal to 0 is t = 1. For this t, the values of x and y are x = 3 + 8(1) = 11 and y = 1 + 9(1) 10, so the line crosses the xy plane at the point (11, 10, 0). =
=
PLANES IN 3SPACE
A plane in 3space can be determined once we know a point, P0 = (x0, y0, z0), on the plane and a vector, n = (n1, n2, n3), that is perpendicular (normal) to the plane; n tells us the plane's orientation in space. As the figure below shows, a point P = (x, y, z) will lie on the plane if and only if Pl, the vector from P0 to P, is normal to n : z
X That is, P = (x, y, z) is on the plane if and only if the dot product of Pl and n is zero: (x  x0, y  y0, z  z0) • ( n1, n2, n3) = 0 which can be rewritten as or, more simply, as n1x + n 2y + n3z = d where d nlo + n2y0 + n3z0• It's important to notice that the coefficients of x, y, and z in the equation of the plane are the components of a vector normal to the plane. Planes that are parallel to one of !he coordinate planes have particularly simple equations. Any plane parallel to the xy plane must have = (0, 0, 1 ) as a normal vector, so the equation of such a plane must have the form z = z0. Similarly, any plane parallel to the xz plane has y = Yo as its equation, and any plane parallel to the yz plane has x = x0 as its equation. The three planes x = x0, y = yO' and z = z0 intersect in the point P = (x0, y0, z0), as shown on the following page. =
k
CALCULUS II + 1 1 3
z
X
Find the equation of the plane that contains the points and C (3, 2, 6).
Example 3.3 Solution:
=
A
=
(4,
1, 2), B
=
(1, 5, 4),
The vectors
AB = (1, 5, 4)  (4, 1, 2) = (3, 4, 2) and
AC = (3, 2, 6)  (4, 1, 2) = (7, 1, 4) lie in the plane, so their cross product,
ABxAC =
i j k 3 4 2 7 1 4
= (16 2)i  (12 + 14)j + (3 + 28)k A
"
A
/'\
A
A
= 14i  2j + 25k
gives a vector normal to the plane. Since n = (14, 2, 25) is normal to the plane, the equation of the plane must have the form 14x  2y + 25z = d, for some constant d. Substituting the coordinates of any of the three points into this equation will give us the value of d. Using (4, 1, 2), we find that d = 14(4)  2(1) + 25(2) 104. Therefore, the equation of the plane that contains A, B, and C is 14x  2y + 25z 104. (The coordinates of B and C also satisfy this equation, which verifies it.)
A=
1 1 4 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
=
=
CYLINDERS
In the xy plane, the graph of the equation x2 + y2 = 1 is a circle. In 3space, this equation would describe a right circular cylinder, which has circular cross sections parallel to the xy plane. Because the variable z is missing from the equation, it can assume any value, as the figure below illustrates.
xz + yz
' '
y

 
L ' '
=1
'
t '
'
'
' '

'
v�
 
v
 
y
' '
X
/
'
'
'
'


'
' L ' '
' ' ' '
In general, given a curve, C, a cylinder is a surface in 3space generated by moving a line along C so that it is always parallel to a given line, L (called the generator). The lines that make up the surface of the cylinder are called elements. The figure below shows a cylinder generated by a line perpendicular to the xy plane along a curve C in the xy plane whose equation is f(x, y) = 0: z
f(x, y)
=0
cylinder with elements p arallel to the zaxis
i Y
X
CALCULUS II +
IIS
As the example of the right circular cylinder illustrates, an equation that is missing one of the vari ables x, y, or z represents a cylinder whose generator is parallel to the axis of the missing variable. For example, the equation y2 + 9z2 = 1 describes an elliptic cylinder whose elements are parallel to the xaxis, and, pictured below, z = x2 is a parabolic cylinder with elements parallel to the yaxis. z
y
X
Example 3.4
Solution:
Let C be the curve x = y2 in the xyplane, and let v be the vector What's the equation of the cylinder generated by a line moving along C in such a way that it's always parallel to v?
(1, 2, 1).
Let P0 = (xO' yO' 0) be any point on the curve C. The parametric equations of the line through P0 and parallel to v are: x = x0  y = y 0 + z=
t
2t,
t,
t
Eliminating gives the symmetric equations: X  Xo = y  y o = z 2 1
__

This tells us that every point (x, y, z) on the element through P0 must satisfy the equations: X + Z = X0 y
2z = Yo
Now, since P0 = (x0, y0) is an arbitrary point on the curve C, we know that x0 = y02; substitut ing the pair of equations displayed above into this last equation gives x + z = (y 2z)2 
as the equation of the cylinder.
1 1 6 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
SURFACES OF R EVOLUTION
In the previous chapter, we learned how to figure out the volume of a solid of revolution, obtained by
rotating a curve in the xy plane around a horizontal or vertical line. In this section, we will show how to find equations for surfaces of revolution in 3space. Let's consider the curve z = y2 in the yzplane; what would be the equation of the surface we'd get if this curve were revolved around the zaxis? Let P0 = (0, yO' z0), with Yo > 0 be a point on the curve. Qo =
(0, 0, Zo )   

z • ..._
:_         
_ , . _ _
p
Po = ( O, y o, Zo )
X
Referring to the figure above, we notice that as the point P0 revolves around the zaxis (that is, as we
revolve the curve around the zaxis), it generates a circle of radius Yo centered at Q0 in the plane z = z0•
A point P, at height z0 above the xyplane, will lie on the surface of revolution if Ql is equal to Ql0; that is, if the distance from P to the zaxis is also equal to y0• Since the distance from P to the zaxis is �x2 + y 2 ,
the point P = (x, y, z0) will lie on the surface if �x 2 + y2 is equal to ± Yo· (We include the ± here because
Yo could
be negative.) Now, since z0 = y � , we can conclude that P (x, y, z0) will lie on the surface if: =
Zo
=
(±�Xz + y2 r
This will be true for any height z0 � 0. So if we replace y by ± �x 2 + y 2 in the equation of the curve, we'll get the equation of the surface of revolution: becomes which we can rewrite as z = x2 + y2• In general, if we take the curve f(y, z) = 0 in the yz plane and revolve
it around the zaxis, the equation of the surface we generate is j(± �x 2 + y 2 , z) = 0 , where we use ± to ac count for the possibility that y may be negative (this can be eliminated by squaring the resulting equation).
CALCULUS II + 1 1 7
What if we revolved this same curve around the yaxis? Let Q0 = (0, y0, 0) and, as J;>efore, let P0 = (0, y0, 20) be a point on the curve. As P0 revolves around the yaxis, it generates a circle of radius
Ql0 = 20 in the plane y = y0, centered at Q0 . So a point P = (x, y0, 2) will lie on the surface of revolution if Ql is equal to Ql0; that is, if the distance from P to the yaxis is also equal to 20• Since the distance from P to the yaxis is � 2 + 2 , the point P = (x, y , 2) will lie on the surface if �X2 + 22 is equal to 2 . Since this x
2
0 0 will be true for any y0, the result is that if we replace 2 by �x2 + 22 in the equation of the curve, we'll get
the equation of the surface of revolution: 2 = y2
becomes
In general, if we take the curve f(y, 2) = 0 in the y2 plane and revolve it around the yaxis, the equation of the surface we generate is f(y, ± �x2 + 22 ) = 0 , where we include the ± to account for the possibility that 2 may be negative (although that's not the case in this particular example). The table below lists how the equations for the various surfaces of revolution are found. If the curve
is revolved
around the
then replace
by
to get surface of revolution
f(x , y) = 0
xaxis
y
± jyz + 2 2
f(x, ±jyz + 2 2 ) = 0
yaxis
X
± Jxz + 2 2
f(± Jxz + 2 2 ' y) = 0
xaxis
2
± jyz + 2 2
2axis
X
± jx z + y z
f(x, ± Jyz + 2 2 ) = 0
f(±jx2 + y 2 , z ) = 0
yaxis
z
± Jx 2 + 2 2
f(y , ± Jx 2 + 2 z ) = O
zaxis
y
± jx z + y z
f(±jxz + y z ' 2) = 0
f(x, 2 ) = 0
j(y,2) = 0
Example 3.5
Solution:
1,
If the ellipse x2 + x219 = in the xzplane, is revolved around the zaxis, what's the equation of the resulting ellipsoid surface? Using the table above, we replace x by ±�x2 + y2 in the equation of the curve to obtain the equation of the surface of revolution. This gives us:
1 1 8 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
LEVEL CuRvEs AND LEvEL SuRFACES
The graph of a function z = J(x, y) is the set of all points (x, y, z) in threedimensional space that satisfy the equation. It is often difficult to sketch these graphs, since they may have peaks, valleys, and other features that aren't easy to illustrate on a flat sheet of paper. A simple way to provide some kind of representation of the graph of the function z = j(x, y) is to draw some of its level curves; these lie flat in the xy plane. If c is a constant, thenf(x, y) = c is called the level curve of height c of the function z = f(x, y). Therefore, a level curve is the set of all points (x, y) in the xy plane where the function assumes a given, constant value. Geometrically, we imagine slicing through the surface with the horizontal plane z = c; in general, the intersection of this plane with the surface will be a curve, called a contour curve. Projecting this curve perpendicularly onto the xy plane gives the level curve of height c. Let's look at an example. The graph of the function z = f(x, y) = x2 + y2 was shown on page 117; it's the surface of revolution we get by revolving the curve z y2 around the zaxis. This surface is called a circular paraboloid because its cross sections perpendicular to the zaxis are circles, and its cross sections perpendicular to the xyplane are parabolas. The level curves of this function are described by the equation x2 + y2 = c, where c is a positive constant. (Of course, if c = 0, then the level curve is just a point; the origin.) These curves are circles centered at the origin, which are easy to sketch: =
level curves x2 + y 2
=
y
c
How do you use level curves to reconstruct the actual graph of the function? By imagining that the level curve of height c is actually at height c above the xy plane. (If c is negative, imagine the curve to be J cJ units below the xy plane.) So, for the level curves pictured above, we would imagine the circle x2 + y2 = 1 lifted a distance of 1 unit; the circle x2 + y2 = 4 lifted up by 4 units; and so on, so that these elevated level curves form the surface. Now it isn't difficult to picture the paraboloid in 3space, as on the following page.
CALCULUS II + 1 1 9
z
X
If w = f(x, y, z) is a function of three independent variables, we would need a space of four dimensions for its graph, which is clearly hopeless. In this case, we have no choice but to sketch its level surfaces. If c is a constant, thenf(x, y, z) = c is a level surface on which the value of w is always equal to c. For example, the level surfaces of the function w = x2 + y2 + z2 are concentric spheres centered at the origin, and the level surfaces of the function w = x + y + z are parallel planes with n = (1, 1, 1) as a normal.
Solution:
1
The equation x2 + t y2 z2 = describes a surface called a hyperboloid of one sheet. Sketch some of its level curves.
Example 3.6

To find the level curves, simply set z equal to a constant, c. In this case, then, the equation of the level curve of height c is x2 +t y 2 = c2 + 1 . These are ellipses centered at the origin that grow larger as the magnitude of c increases. z z hyperboloid of one sheet
c
=
±2
X
X
Also notice here that the level curve of height c is the same as the level curve of height c. This means that the surface extends both above and below the xyplane, and, in fact, the xyplane cuts the surface into symmetrical halves. Cross sections parallel to the xyplane are ellipses, while cross sections perpendicular to the xyplane are hyperbolas.
1 20 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
CYLINDRICAL CooRDINATES
In 3space, cylindrical coordinates are nothing more than the polar coordinates r and 8 in the xy plane, along with the usual (cartesian) zcoordinate. To find the cylindrical coordinates of a point P = (x, y, z) in 3space, we drop a perpendicular from P to the xyplane, and let Q be the foot of this segment. (Of course, if P is already in the xyplane, then Q = P.) Now, by definition, if (r, 9) are polar coordinates of Q, then (r, 8, z) are cylindrical coordinates of P. z cylindrical coordinates r".
� P = ( r, 8 , z) :z
x
a....._ .,.. __,;...___ Y

e
r  
:
�
       y        � Q X
The name cylindrical comes from the fact that the surface described by the equation r = c (where c is a nonzero constant) is a right circular cylinder in space. The surface 9 = c (where c is a nonzero constant) is a plane through the origin and perpendicular to the xyplane and, of course, the surface z = c (where c is a constant) is simply a horizontal plane (that is, parallel to the xyplane). SPHERICAL CooRDINATES
To find the spherical coordinates of a point P in 3space, we once again drop a perpendicular from P to the xy plane, and let Q be the foot of this segment. Then we draw line segments from the origin, 0, to both P and Q. The spherical coordinates of the point P are (p, �' 9), where p is the distance OP, � is the angle that the segment OP makes with the positive zaxis, and 9 is the angle that the segment OQ makes with the positive xaxis (the same as the cylindrical coordinate 9). z
spherical coordinates r".
� P = ( p , , 8 )
x
,./ �
:z
/p
o��� y
�e

:
�
        y       � Q X
The equations that relate cartesian coordinates to spherical coordinates can be derived from the figure above. Since LOPQ = �, we have z = p cos � and OQ r p sin �; this equation tells us that: X = r COS 9 = p sin � COS 9
=
and
=
y = r sin 9 = p sin � sin 9
CALCULUS II + 1 2 1
The name
spherical
comes from the fact that the coordinate surface described by the equation
p = c (where c is a positive constant) is a sphere of radius c, centered at the origin. The coordinate surface
� = c (where c is a positive constant) is generally a cone centered on the zaxis, with vertex angle 2c. (The
surface � = 0 is just the positive zaxis, � = � is the xyplane, and � = 1t is the negative zaxis.) And, just
as with cylindrical coordinates, the surface
2 8 = c (where c is a nonzero constant) is a plane through the
origin and perpendicular to the xy plane.
Example 3.7 Solution:
Let S be the sphere that's tangent to the xzplane and whose center is C = (0, 1, 0). Write the equation of S in spherical coordinates. Since S is centered at (0, 1, 0) and tangent to the xzplane, the radius of the sphere must be 1. In cartesian coordinates, the equation of the sphere with center (x0, y0, z0) and radius a is
(x  xo)z + (y Yo)z + (z zo)2 az 
=

so the cartesian equation for S is
x2 + (y  1 ) 2 + z2 = 1
which we can rewrite as x2 + y2 + z2  2y = 0. Because the equation for S in spherical coordinates is:
x2
+
y2 + z2 = p2 and y = p sin � sin e,
p2  2 p sin � sin e = 0
Factoring gives p(p  2 sin � sin e) = 0, so either p = 0 or p = 2 sin � sin e. Since the latter equation includes the former one, we can conclude that the equation for S in spherical coordinates is p = 2 sin � sin e.
PARTIAL D E R IVATIVES Let f(x, y) be a function of two variables. Keep y constant, and consider the limit: lim f(x +
11>0
h , y)  f(x, y) h
If this limit exists, its value is the derivative off with respect to x at (x, y); it's called the partial off with respect to x, and is denoted by:
1 22 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
derivative
Similarly, we can keep x constant and consider: lim
k
If this limit exists, it's called the
>0
f(x, y + k)  f(x, y)
k
partial derivative off with respect to y, and denoted by
z
(x, y) or
f¥ (x, y)
To determine the partial derivative of j(x, y) with respect to, say, x, we treat y as if it were a constant and apply the usual differentiation rules. For example, consider the function: f (x, y) = 2x3 y + xsin y Its partial derivatives with respect to x and y are: and Although we've introduced the idea of a partial derivative for a function of two variables, we can also find partial derivatives of a function of three or more variables, and the rule is the same: Treat every variableexcept the one you're differentiating with respect toas a constant and apply the usual dif ferentiation rules. So, for example, if
then:
GEOMETRIC I NTERPRETATION OF IX AND I y For a function f(x, y) of two variables, what do the partial derivatives !,X and f.y represent geometrically? Remember that for a function of one variable, the derivative is the slope of the line tangent to the curve. When we want to take a partial derivative of f(x, y), we treat one of the variables as a constant. In effect, we consider f as a function of one variable, so the value of the derivative should again be the slope of the line tangent to a curve. Let's look at this a little more closely. To form fx, we treat y as a constant (let's call it y0), so the graph of the function z = f(x, y0) is a curve: It's the intersection of the surface z = f(x, y) with the plane y = y0. The partial derivative off with respect to x gives the slope of the line tangent to this curve. Similarly, the partial derivative with respect to y gives the slope of the tangent line to the curve z = f(x0, y), which is the intersection of the surface z = f(x, y) with the plane x = x0.
CALCULUS II + 1 23
z
tangent line to curve z = f(x, y0) at
surface z = f(x, y )
P = (xo, Yo)
curve z = f(x, Yo)
X
Example 3.8
'
( af )
slope = OX
P = ( x o , yo )
( aJ J
slope = ()y
X
P = ( x o , yo)
What's the equation of the line tangent to the intersection of the surface
z=
arctan xy with the plane x = 2, at the point (2,
'
Solution:
surface z = f(x, y)
tangent line to curve z = f(x0, y) at P = (xo, Yo)

.!_ ,
� )? 4
2

··

Since x is held constant, the slope of the tangent line we're looking for is equal to the value of the partial derivative (d I dy)(arctan xy) at (x0, y0)
� (arctan xy)
�
1+
� ( )'
=>
=
(2,
slo pe =
.!_ ):
: [ ( y )' L 2
1+
l1
=
1
Since the tangent line is in the plane x = 2, this calculation shows that the line is parallel to the vector v = (0, 1, 1), so we can write the equation of the tangent line in parametric form, as follows: 1t 1 x = 2I y =  + tI z =  + t 4 2
HIGHERORDER PARTIAL DERIVATIVES A partial derivative of a function of two or more variables is itself a function, so it may be differentiated with respect to any of the variables. For example, after findingfx and f for a function f(x, y), we may dif � ferentiate the derivatives with respect to either X or y. The derivative of fx with respect to X is denoted:
a2 J dX 2
or
and the derivative of J with respect to y is denoted:
Y
or
1 24 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
The secondorder mixed partialswhich are the derivatives offx with respect to y, and ofJ with respect Y to xare denoted, respectively: and For almost all commonly encountered functions, these mixed partials are continuous, which ensures that they're identical; that is, fxy = Jryx so the order in which the derivatives are taken doesn't matter. To illustrate, let's consider the function: ,
f(x, y) = y log(4x + y 2 ) Its firstorder partial derivatives are
and
y2
2 f = 4 x + y 2 + log( 4 x + y 2 )
Y
and the secondorder mixed partial derivatives off are identical, since
and:
We can also form partial derivatives of orders higher than two; fxxy or Jryxy, for example. Just remember that the order in which the derivatives are taken is specified differently in our two notations for a partial derivative. When using the Leibniz "fraction" notation, we look at the denominator and read the variables from right to left; but when using the subscript notation, we read the subscripts from left to right. For example, if we differentiate f(x, y) first with respect to x, then we differentiate the result with respect to x, and then finish by differentiating the result with respect to y, the resulting thirdorder partial derivative would be written as either:
_fl_
()yoxax
or
fxxy
Notice that by invoking the equality of mixed partials, fn is equal to fxyx or to fvxx y
CALCULUS II + 1 25
THE TANGENT PLANE TO A SURFACE Just as the derivative of a function of one variable can be used to find the equation of a tangent line to a curve, the partial derivatives of a function of two variables can be used to find the equation of a tangent plane to a surface. Let z = f(x, y) be the equation of a surface in 3space, and let fx and J denote, as usual, Y the first partial derivatives of f Then the equation of the tangent plane to the surface at P = (x0, y0, z0) is Z  Zo = fx l r
o
(x  x o ) + JY i r
o
(y  y o )
assuming that fx and J aren't both equal to zero at P. The equation of a surface can also be given in the Y
form f(x, y, z) = c, for some constant c. The difference is that z is not written explicitly in terms of x and
y, as is the case if the surface is described by an equation of the form z = f(x, y). To take care of this situa az az tion, simply use the equation J(x, y, z) = c to find and by implicit differentiation; the equation of ax the tangent plane to the surface at the point P = (x0, y0, z0) is
ay
z  z0 = assuming once again that
Example 3.9
l
P
o
( x  x0 ) +
az
0y
l
o P
( y  y0 )
dz and dz don't both equal zero at P. ax ay
Find the equation of the tangent plane to the surface at the point P =
Solution:
az ax
x2z
(1, 2, 1).
 y3z5
+
xy
=
9
Since the equation of this surface is given in the form J(x, y, z) = c, we use implicit differ entiation to find the partial derivatives. First, we hold y fixed and differentiate implicitly az . w1th respect to x to find  : ax x2 z  y 3 z5 + xy = 9 az az x2  + 2xz  5y 3 z4  + y = 0 ax ax az 2xz  y ax  x2  5y 3z 4
(
)
Next, we hold x fixed and differentiate implicitly with respect to y to find x2z  y 3l + xy = 9
x2
: ( : ]
1 26 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
3 Sy z4
+ 3y2l + x = 0 az _ 3y2 z5  X ay  xz  5y3z 4
az
ay
:
Evaluating each of these at the point P
az ax
l
l
r
=
=
(1, 2, 1) gives
2xz y 2(1)(1) 2 _ 0 _ 3 x2  5y z4 (1,2,1) 1 2  5(23)(1)"1
az 3y 2 z5  x = 4 ;:),, vy X2  5Y3 p
Z
I
and
3(22)(1)5  1
( 1 ,2,l)
1
= 1 2  5(2 3 )(1) 4 = 3
so the equation of the tangent plane at P is z + 1
y  3z = 5
1
= 3 (y  2), or, equivalently:
LINEAR APPROXIMATIONS The equation of the tangent plane to a surface z = f(x, y) at a point P0 (x0, y0) provides a firstorder (linear) approximation of the value of z at points near P0. To be more specific, let's say we wanted to evalu ate fat a point P1 = (x1, y1) that's close to P0• Since P1 is close to P0, the value of z at P1 on the tangent plane is close to the value of z at P1 on the surface. So, if z 1 = j(x1 , y1), we can use the equation of the tangent plane at P0 to say that: =
Example 3.10
What's an approximate value for this expression? (1.1) log [4(0.05) ·· 
Solution:
+ (1.1)2]
 
·
··
____)
·_____
If we let f(x, y) = y log (4x + y2), then we're being asked to approximate the value of f(0.05, 1.1). Notice that the point P1 = (0.05, 1.1) is very close to the point P0 = (0, 1), at which the exact value off is easy to calculate: z0 = f(x0, y0) = 1 log (0 + 12) = 0. So, if we figure out the equation of the plane tangent to the surface z = f(x, y) at the point P0, then the value of f at P1 will be approximately equal to the value of z at P1 on the tangent plane. Earlier in this section, we found the firstorder partial derivatives of f(x, y); they are: fx =
4y 4x + y 2
and
2 .!,Y = 2y 2 + log( 4x + l) 4x + y
Evaluating each of these at the point P0 gives
fr l r = 4 x;y 2 o
=4 (0,1l
a nd
[
2
2y JY I �� = 4x + y 2 + log(4x + y 2 )
so the equation of the tangent plane to the surface at more simply:
]
=2 (O.Il
P0 is z  0 = 4(x  0) + 2(y  1) or,
z = 4x + 2y  2
CALCULUS II + 1 27
Since P1
=
(0.05,
1.1) is close to P0 = (0, 1), the value of z at P1 on the tangent plane, z =
4(0.05)
+ 2(1.1)  2 = 0.4
should give a close approximation to the value of z at P1 on the surface. [Note: A calculator would tell you that (1.1) log [4(0.05) + (1.1)2 ] "" 0.378, so our approximation of 0.4 is pretty good.]
THE CHAIN R uLE FOR PARTIAL DERIVATIVES In the previous chapter, we used the chain rule to differentiate composite functions: Iff is a function of u, and u is a function of x, then we find the derivative off with respect to x from the equation:
df = df du dx du dx

 
For functions of a single variable, this is the only form of the chain rule we need. But in the case of functions of more than one variable, there are many different forms of the chain rule; which one to apply depends on how many variables there are and how they're interrelated. Here we'll describe a method that you can use to differentiate a composite function in any situation. As our first illustration of this method, let's say that we're given a function z = F(u, v) with u = f(x, y) and v = g(x, y). Notice that z depends on the variables u and v, and these variables in tum depend on x and y; so ultimately, z depends on x and y. We call z the dependent variable, u and v the intermediate variables, and x and y the independent variables. Therefore, you could be asked to find the derivative of z with respect to x or to y. The first step is to sketch a diagram of the situation by drawing an arrow from the dependent variable to every intermediate variable, and one from every intermediate variable to every independent variable. Our diagram would look like this:
Think
that v =
z
of each arrow as meaning depends
depends on usince
z =
on. So, for example, the arrow pointing from z to u means
F(u, v), and the arrow from v to y means that v depends on ysince
g(x, y). Now, let's say we are asked to find the derivative of z with respect to x. To answer this ques tion, we use the diagram to find every path from z to x. Each path will give us a product of derivatives,
one derivative for each arrow, and then we add up the products for every path. In this case, there are two
paths from z to x; one going through u and one going through v. The path z 1 u 1 x contains two arrows; the arrow z 1 u gives us the derivative
az , and the arrow u 1 x gives us the derivative au . (Notice that ax au
we have to use partial derivatives here because z depends on more than one variable, as does u.) So, the doublearrow path z 1 u
1 x gives us two derivatives, which we multiply to give: az au au ax
1 28 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
In the same way, the other path from z to x, the doublearrow path z � v � x, gives us: az (Jv (Jv ax
Therefore, the derivative of z with respect to x is the sum of all possible paths from z to x: dZ
dX

=
dZ dU
dU dX
 
+
dZ dv dv dX
 
You can imagine how things can get messy. For example, let's say we were given z = F(u, v, y), where f(v, x) and v = g(x, y). The diagram we'd draw to depict the complicated dependencies in this situa tion would look like this:
u
=
z
/1
/ �
y
Now, let's find the derivative of z with respect to y. First, we need to find all the paths that connect z to
y. There are three: the onearrow path z � y; the twoarrow path z � v � y; and the threearrow path z � u � v � y. Since there are three paths from z to y, we know that the expression for of three terms, one for each path:
02
ay
will consist
OZ = dZ + dZ dv + dZ dU dv
ay ay av ay au av ay


 
  
One of the drawbacks of the partialderivative Leibniz notation is apparent here. Notice that the equation above has
az ay
on the lefthand side and on the righthand side, and it seems like we could cancel
them out (which would leave us with no answer to our question, "What's
02
ay
:
?"). But the problem is that
on the lefthand side means the derivative of the composite function with respect to y (that is, the deriva
tive of z with respect to y taking any dependency of u and we're looking for), while the
az ay
v
on y into account, which is the derivative
on the righthand side means the derivative of z with respect to y, holding
u and v constant. There's a way to remove the ambiguity: We write the variables being held constant as subscripts on each partial derivative. While at first this might seem superfluousafter all, that's how we figure out partial derivatives, by keeping all the variables constant except the one we're differentiating with respect tothis example shows that it's not. So, using this more careful notation, the equation for
az ay
would be written like this:
It looks bad, but it resolves all possible ambiguities.
CALCULUS II + 1 29
Example 3.11
z
Let = F(u, w, x), where u = f(x, y) and w = g(x, y). (a) Find a general expression for
az .
dX
(b) Verify your answer to (a) if F(u, w, x)
g(x, y) = y + xy2.
==
w3  2ux, f(x,
y) 4xy  x + ==
1, and
Solution: (a) First draw a diagram:
z
� w
In order to find
az
x y
dX , we find all paths from z to x. Since there are three such paths, z
z 1 u 1 x, and z 1 w 1 x, we get the following three terms:
1
dZ dZ dZ dU dZ dw ax ax au ax dw ax
= ++

To resolve the ambiguities of these partial derivatives, we can write
or, by labeling the arrows with the names of the functions,
we can also write:
( OXaz ) ( aFOX ) ( OUaF ) y
==
li,W
+
W,X
( )
df + aF ax dw
li,X
ag ax
(b) First, we use the formulas given and write z explicitly in terms of the independent variables x and y:
z = F( u, w, x) w3  2ux (y + xy 2)3  2(4xy  x + l)x = (y + xy 2)3  8x2y + 2x 2  2x ==
==
1 30 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
x,
az ax oz  = 3(y + xy 2 ) 2 (y 2 )  16xy + 4x  2 OX
. ., . From th1s, 1t s easy to determme  :
Now, we can check that our formula from part (a) gives the same result:
( oF ) + ( oF )
of ( oF ) = og (2u)+(2x)(4y 1)+(3w 2 )(y 2 ) + ax au ax aw ax = 2(4xy x 1)2x(4y 1) + 3y 2 (y + xy 2 ) 2 = 16xy + 4x2 +3y2(y + xy 2)2 Example 3.12 Let w = j(x, y, z) yz2 x, where x, y, and z are the following functions of t: x(t) = t2, y(t) = 2t  3, and z(t) = 1  t. Find dw dt at t 1. ll,W
li,X
W,X
+
=
Solution:
=
Our schematic of the variables in this case is
so we can write:
dw of dx of dy + of dz dt ox dt ay dt oz dt

=
 
+

This gives us:
dw (1)(2t) (z2 )(2) (2yz)(1) dt (2t) + 2(1t)2  2(2t 3)(1t)
=
Therefore:
=
+
+
dw l = [(2t) +2(1  t)2  2(2t 3)(1 t)] = 2 1 dt 1=1
�
CALCULUS II + 1 3 1
D I R ECTIONAL DERIVATIVES AND TH E GRADI E NT Consider the surface z = J(x, y), and let P = (x0, y0) be a point in the domam off. What if we were asked to
find the value of the derivative off at P? The derivative should tell us how z changes as we move away from P, but a moment's reflection tells us that the rate at which changes depends on the direction in which we want to move.
z
z
Xo,'
,''
/
... .
'   L
Yo        
,;.    ':'!' = ·'tt: .. ,
·
\:
  �
y
Jf'p',......
X
For example, if we move away from P in the positive x direction, then we know the rate of change of will be the value of at P, and if we move from P in the positive y direction, then the rate of change of will be the value of J at P. But what if we wanted to know how changes if we move from P in some Y other direction? We need a way to figure out a general directional derivative. The easiest way to determine a directional derivative is to first form the following vector, whose com ponents are the firstorder partial derivatives off
z z
fx
z
This vector is called the gradient of J, and is denoted either by grad f or more commonly by Vf (the sym bol V is pronounced del). The rate of change off at the point P in the direction of u (where u is a vector whose initial point is P) is denoted D J� and is called the directional derivative off at P in the direction of u. It's found by evaluating the dot product of the gradient off at P and the unit vector, u:
fx
i
This equation verifies that is just the derivative off in the +x dir�ction (that is, in the direction ), and JY is the derivative off in the +y direction (that is, in the direction j ), since
DJI P = vf lp . i = Ux ,fy )l p • (1 , 0) = fx l p
and
D/l p = Vfi p . J = Ux ' fy )l p • (0,1 ) = !Yip
but the real power of the equation is that i t tells us how f will change a s we move in any direction.
1 32 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
Example 3.13
Let f(x, y) = �81  x2  y 2 Find the rate of change off as we move from the point P = (1, 4) in the direction toward the point Q = (4, 8). •
·
Solution:
·
=



We want to move in the direction of the vector u = PQ = (4 1 8  4) (3, 4). Because this vector has magnitude 5, the unit vector in this direction is u = (t , t) . Since the gradient of f is 
'\1J = {jx ' Jy ) =
( �81 xx
2
�81  x2  y 2 y
y
2
'
,
our equation for the directional derivative tells us that:
J •
(l5 .:!.) 5 '
= ( � ,  t ) (t , t) .
=  12. 40
=
Since DJ l r =  ¥a , the line that's tangent to the surface at the point P {1, 4) and parallel to the vector PQ will descend by !� of a unit as we move from P one unit in the direction u.
Now, let's take a closer look at the equation for calculating a directional derivative:
Let e be the angle between the vectors Vf lr and u; by definition of the dot product, we can write
=
since IIUJ I 1. The maximum value of cos e is 1, which occurs when e = 0, so the maximum directional derivative is f l which occurs when u points in the same direction as Vf �· This property of the gra r dient is important enough to repeat:
IV I
,
The vector Vf points in the direction in which f increases most rapidly, and the magnitude of Vf gives the maximum rate of increase. It's also easy to see that the vector Vf points in the direction in which f decreases most rapidly, and the negative of the magnitude of f gives the maximum rate of decrease. So far, we have discussed directional derivatives and the gradient of functions of two variables, but these ideas carry over to functions of three or more variables . For example, given a function f(x, y, z), its gradient is the vector
V'f
=
of i
ax
+
oayf
:
J
+
of k. az
=
0 means thatf has a local minimum at x = x0• (If j "(x0) = 0, then the test is inconclusive: f could have a local maximum, a local minimum, or nei ther, at x = x0.) We now want to extend these ideas to maximize or minimize a function of two variables. Let's assume that we're given a function y) and asked to find its extreme values. Iff has a maxi
f(x,
mum (or minimum) at P0 = (x0, y0), then the curve z = j(x, y0) has a maximum (or minimum) at x = x0, 'df so = 0 at x = x0; it will also be true that the curve z = j(x0, y) has a maximum (or minimum) at 'df = 0 at y = Yo· Therefore, we say that P0 is a critical point of f(x, y) if the first partial derivay = y0, so
dX
dy
tives off are both equal to zero there: P0 is a critical point ofj(x, y) if:
( dX )Po 'df
=
0 and
( dy ) 'df
Po
=
0
In the onevariable case, a critical point doesn't have to be the location of a maximum or a minimum; perhaps the simplest example of this is the function j(x) = x3 . Although its derivative equals zero at x = 0, the point (0, 0) is neither a local maximum nor a local minimum; it's an inflection point. The same kind of behavior can be exhibited by a function of two variables.
xo
'
              •' '
'
'
Po
X
For example, if, at P0 = (x0, y0), the curve z = j(x, y0) has a maximum at x = x0, but the curve z = j(x0, y) has a minimum at y = y0 (or vice versa), then we say thatj(x, y) has a saddle point at P0• Although P0 is a critical point, a saddle point is neither a local maximum nor a local minimum for the function. Here's the secondderivative test for classifying a critical point, P0 = (x0, y0), of a function f(x, y). Let !l be the determinant of the following 2by2 matrix, which contains the second partial derivatives of f, evaluated at P0:
CALCULUS II + 1 3S
J(x, y),
D.,
(This matrix is called the Hessian matrix of and its determinant, is called the Hessian off We're assuming that the mixed second partials of f are continuous at P0, so fry = fy) Then the following state ments hold: If If If If
D. > 0 fx (P0) < 0, f > 0, f D. > 0 D. < 0, D. = 0,
then attains a local maximum at
and fxx(P0)
then attains a local minimum at P0•
then f has a saddle point at P0• then no conclusion can be drawn (any of these behaviors is possible).
Example 3.15
Locate and classify the critical points of the function:
j(x, y) = x2 Solution:
P0•
and
The first step is to find
+ 1j 
6xy
fx and JY and set both equal to zero:
= 2x6y= 0 JY = 3y2 6x = 0 x = 3y, 3y26(3y) 0 y2 6y=O y(y 6)=0 y=O, 6 = (0, 0) 2 = (18, 6). fxx fry ] det [_� �6] 12y 36 det [ Y fyx Jyy (0, 0) 12(= 36;0) 36 =D.>36;0 D. =< 20,> 0, P2 = (18,D. 6) =12(6)36 fxx fr
From the first equation, we see that gives us:
and substituting this into the second equation =
Therefore, the function has two critical points, P1 we figure out the Hessian off: L1 =
The value of of D. at Example 3.16
at P1 is
and P
=
=
is
since
To classify them,
since and
P1 is a saddle point. The value the point P is a minimum. 2
Let S be the closed square region: S
{(x, y): x 2 and 0 y  f x,y =_x2 + 4_xy _y2_5x =
0
s
s
s
s
2}
Determine the absolute minimum and maximum values on S o f the function: �

( __)_ ___ 
1 36 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
__
__
__ __ ________
____j . I
Solution:
In the last chapter, we learned that every continuous function on a closed interval attains both an absolute maximum and an absolute minimum value; these values can be assumed either at critical points or at the endpoints of the interval. This example illustrates the cor responding property for functions of two variables. Since S is closed (because it contains its boundaries) and bounded (because it's a subset of a circular region of finite radius), any continuous function defined on S is guaranteed to attain an absolute maximum and an ab solute minimum value, and these values may occur either at critical points or at points on the boundary of S. So our first step is to find the critical points. Setting both fx and JY equal to zero gives us: fx = 2x + 4 y  5 =
JY = 4x  2 y = 0
0
Solving this pair of simultaneous equations yields the critical point, P = ( t , 1), which is in the interior of the square S. The value off at P isf( t , 1) =  � . We don't need to classify this critical point because now we'll examine the behavior of f on the boundaries of S. Along the left edge of the square, x is always equal to so j(x, y) = f(O, y) = y2 • On the closed interval � y � 2, this function has an absolute maximum at y = (where its value is and an absolute minimum at y = 2 (where its value is 4). Next, along the top edge, y is always equal to 2, so f(x, y) = f(x, 2) = x2 + 3x  4. This function has a critical point at x = :; , which is on the top edge, but not withinin S, so we just evaluate the function at the endpoints of the top edge: f(O, 2) = 4, and f(2, 2) = 6. Along the right edge, x is always equal to 2, so f(x, y) =j(2, y) = 6 + 8y  y2 . This function has a critical point at y = 4, which is not in S, so we just evaluate the function at the endpoints of the right edge: f(2, 0) = 6, andf(2, 2) = 6. Finally, along the bottom edge, y is always equal to so f(x, y) = f(x, 0) = x2  5x. This function has a critical point at x = t which is not in S, so we just evaluate the function at the endpoints of the bottom edge: f(O, = and /(2, = 6. We can now say that the absolute maximum val ue off on S is attained at the point (2, 2), and the absolute minimum value off on S is 6, at tained at the point (2, [By the way, the secondderivative test tells us that the critical point P = ( t , 1) is a saddle point.]
0,
0
0
0),
0,
6,
0) 0,
0)
0).
MAx/MIN PROBLEMS WITH A CoNSTRAINT Often, a question will ask us to maximize or minimize a function subject to a constraint. For example, "What is the maximum value of the functionf(x, y) = xy subject to the constraint x2 + y2 = 1 ?" The constraint equation limits the possible choices of x and y, so the question really asks, "For all pairs (x, y) such that x2 + y2 = 1, which pair(s) maximize the value of f(x, y) = xy?" One way to solve this problem is to use the constraint equation to rewrite the function to be maximized or minimized in terms of just one variable, and then proceed as we did in the onevariable case. For example, solving this constraint equation for y gives us: y = ± 1  x2
.J
Let's ignore the minus sign for the moment, and substitute y =
.J
.J1  x 2
into f(x, y) = xy:
.J
g(x) = f(x, 1  x 2 ) = x 1  x2 , for 1 � x � l
CALCULUS II + 1 37
To maximize g, we first set its derivative equal to zero and solve for x:
x2 + v1r::? g '(x) = p  x2 = 0 1  x2 x2 + (1  x2) = 0 =± 1 X fi
_,h x2 , we would have found the same critical values of x.) Since g(x) = 0 at both endpoints of the closed interval 1 � x � 1, we can say that g(x) attains a maximum value of .!. at x = and a minimum value of _ .!_ at x =  � Since y = ±,h x2 , we see that there are four cr�tical point�:
�
(Had we used y =
P, (]z, Jz} P, + Jz, Jz}P, (Jz,Jz}P, (]z,Jz) f P1 P4 P2 P3• 2
v2
=
=
=
We can now conclude that, subject to the given constraint, the function attains a maximum value of .! at and and a minimum value of _ ! at and
2
2
THE LAGRANGE MULTIPLIER METHOD Another method of maximizing or minimizing a function subject to a constraint, and one that's espe cially useful if the constraint equation cannot be easily solved for one of the variables, is known as the Lagrange multiplier method. Let's assume that we're asked to find the extreme values of j(x, y) subject to the constraint g(x, y) = c (where c is a constant). If M is an extreme value of !attained at the point = (xO' y0)then the level curve f(x, y) = M and the curve g(x, y) = c share the same tangent line at
P
P.
level curves Y
ofj(x, y) ,
tangent line at P
P.
Now, because these curves share the tangent line at P, they also share the normal line at Since the vector Vf is normal to the curve f(x, y) = M and the vector Vg is normal to the curve g(x, y) = c, one of these vectors must be a scalar multiple of the other; that is,
Vf = 'AVg
P.
for some scalar assuming that Vg :t The scalar A is called the Lagrange multiplier. Its value is unimport ant; it only serves to help us figure out the critical point
0.
1 38 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
values the func J(x, y) = subject to the constraint x2 y2 Sibecomes nce the constraint is x2 y2 = we set g(x, y) equal to x2 y2. The equation Vf = AVg (y, x) = A(2x, 2y) which is equivalent to the pair of equations Use the Lagrange multiplier method to determine the extreme tion xy, + = 1.
Example 3.17 Solution:
+
+
��··· ·
1,
of
·'
y = A • 2x X = A . 2y
Since the firxst equation says that A= 2x and the second equation says that A= 2 y we know that )2x/__ = ,so x2 = y2. (We're assuming here that neither x nor y is 0; if x or y were equal to 2y zero, then the value of the functionf would be zero. We'l see that 0 is neither a maximum nor a minimum value after we the critical points.) Substituting this into the constraint, x2 y2 = 1, gives us x2 x2 = 1, sox=±�. Therefore y ± �, and we have four critical points: Pl = (�, �) , P = ( �' �), P3= ( �,�), P4 = ( �,�) We can now conclude that, subject to the given constraint, the functionfattains a maximum value of !2 at P1 and P4, and a minimum value of _ !2 at P2 and Py just as we found on the previous page. )/__
2_,
find
+
=
+
z

L I N E I NTEGRALS
Whentakiweng plinategrate a thefunctipathonJ(onx) theof onexaxivaris from able from,to b,say,andx the tvalo xu=e ofwethecanintegral interpretas githeviningtegrati on asbounded ce over the area by the curve y = f(x) over this path, y =a
b,
a [a, b]:
height = j(x)
+ area Jra.b]f dx =
CALCULUS II + 1 39
Butt iswecalcanled ala scurve, o integrate functions over other paths, not just straightline paths along the xaxis. The resul contour, or path integral, or, most commonly, a line integral (even though the path does not need to be a straight line). ToC, iseen thehowxythese i n tegral s can be defi n ed, l e t' s begi n by consi d eri n g a functi o n f(x, y) and a smooth curve, plane. curve given parametrically by the vector equation r = r(t) = ((x(t), y(t)) is said to be smooth if the derivative r'(t) is continuous and nonzero. To say that C is piecewise smooth means that iint'tos composed of a finite number of smooth curves joined at consecutive endpoints.] Imagine breaking C n tiny segments, of arc length �s;. LINE I NTEGRALS WITH R ESPECT TO ARC lENGTH [A
y
;• x
Onat P;each choosetheanyfolpoilowintnP;g sum:(xv y), then multiplyf(x;, y)the value of the function byofthetheselengthtiny&vpieandces, form =
II
L f(x; , Y;)&; i=l
Ifintegral the valoffue ofalongthis Csumwithapproaches a fi n i t e, l i m i t i n g val u e as n 7 oo, then the result is called the line respect to arc length, and is written as: What does the value of thi s i n tegral mean geometri c al l y It' s the area of the regi o n whose base i s the ? curve heigcalht curtai aboven each poin(tporti(x, y)onisofgithevenvertiby thecal cylvaliuneder)of thewhosefunctibaseon,if(x,s C and y). That is, iheit'sgthehtCatareaandeachofwhose the verti or fence whose point (x, y) isf(x, y): z
area of curtain X
=
height of curtain f(x, y)
Jcf ds
1 40 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
curve
)
C
=
'r·
y
To actually evaluate this line integral, first parameterize the curve
{
C:
x(t) fora � t � b (ora 7t b) C = yx == y(t)
to bedidirected; thatwriis,tweing thia� nk ofbthein curve as beinegritraced inwe'a defi nitengdithat rec titheon,Weparameter whiconsich wedertcaltherunsl curve thefropositive r ecti o n. By the paramet z ati o n, r e sayi b, so (x(a), y(a)) is the initial point and (x(b), y(b)) is the final point of the curve, which givesm a toa posi tive direction. Now, since = (dx)2 we can write C
A=
(ds)2
C
B= (dyf,
+
where the tivsiegdinriectif theonparameter t increases in the positive direction on C and the sign if t de creaseswein theuse posi on We then have: +
C.
f f ds = f b f(x(t), y(t))dsdt dt a
C
value the first function over the First we parameterize the curve. Since is directed counterclockwise, we can write:
Example 3.18 Solution:
Determine the of the line integral of the f(x, y) = x + y2 uarter circle x 2 + y2 = 4 in uadrant, from (2, 0) to (0, 2). q q C
y
/ t = !l2
(0, 2)
{
C : yx == 2cost 2sint
r
t
fo 0 7 � 7t
arrowtivspeci / posi e directifieson t=O
/ �� x ��(2 , 0)
. . to f' d ds Next, we use the parametenzatwn dt m

:
ds = +�[x'(t)]2 + [y'(t)f dt = �(2sint)2 +(2cost) 2 =2
CALCULUS II + 1 4 1
where we choose the + sign, since in our parameterization, t increases (from 0 to � ) in the 2 positive direction on C. This gives us:
fc J ds = J: J(x(t), y(t)) • �� dt fo111\2cost + 4sin2 t) . 2 dt = 4J 1112 (cos t + 2sin2 t) dt 0 =
4 [ sint + t  � sin2t J:12 = 2(2 + 1t) =
Let's see what the value of this line integral would have been if we had traversed the curve c in the opposite direction; that is, if we had chosen the positive direction for c to be clockwise, from (0, 2) to (2, 0). In this case, we could parameterize the curve as follows:
y (0, 2)
C:
{
t x = 2 cos t for t 1t 7 0 y = 2sin t
7t
/t = 2
�

Notice that the parameter t decreases (from
�
/t = O
+�

(2, 0)
x
to 0) in the positive direction on C, so we
2 would choose the minus sign when finding ds . So we would have had dt ds =  [x'(t)f + [y'(t)f dt = 2
�
and our integral would have been
f f ds C
Jb j(x(t), y(t)) • dsdt dt = J 0 2 (2 cost + 4sin2 t) • (2) dt n/ J 0 2 (2 cost + 4sin2 t) • 2 dt n/ 4 J 1112 (cos t + 2 sin2 t) dt 0 =
a
=
=
= 2(2 + 7t)
just as we found before. This illustrates an important property of line integrals with respect to arc length: The value of f ds does not depend on the orientation or parameterization of C.
fc
1 42 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
{C : yx = y(t)x(t)
Line integrals with respect to arc length can also be easily defined over curves in 3space. is a function defined on a smooth curve If
f(x, y, z)
=
=
for
a�b
z z(t) ds f f ds f b f(x(t), y(t), z(t)) dt dt
then
=
C
where
a
In this case, one interpretation of the value of the line integral involves thinking of C as a curved wire and f(x, y, as its linear density (mass per unit length); then gives the total mass of the wire.
fc f ds
z)
THE LINE INTEGRAL OF A VECTOR FIELD
Now, we'll look at another kind of line integral: the line integral of a vector field. We'll define this line integral in two dimensions; the generalization to three dimensions follows as easily as it did in the case of a line integral with respect to arc length. A vector field is a vectorvalued function. Let D be a region of the plane on which a pair of con tinuous functions, M(x, y) and N(x, y), are both defined. Then the function, F, that assigns to each point (x, in D the vector
xy
y)
F(x, y) = M(x, y)i + N(x, y)j = ( M(x, y), N(x, y)) is a continuous vector field on D . Now, if C is an oriented, piecewise smooth curve in D, we want to define the line integral of F along C. Let
r(t)=x(t)i + y(t)) = (x(t), y(t)),
t
for a 1 b
be a parameterization of the curve C. Then the line integral of the vector field F along C is defined as:
fc F
•
dr J: F(r(t)) r'(t)dt =
•
What's the interpretation of the value of this integral? Consider F a force field, and imagine a particle moving along the curve C. Along each infinitesimal piecespecified by the vector drof the curve C, the field F is approximately constant, so the work, dW, done by F as the particle undergoes the displacement dr is equal to the dot product F dr. •
CALCULUS II + 1 43
y
Adding up (that is, integrating) all these contributions gives the total work done by F as the particle moves along the entire curve:
t
t
W = dW = F · dr To actually evaluate this integral, we proceed as before. First we parameterize the curve,
r(t) = x(t)i + y(t)J = (x(t), y(t) ) , for a �t b then we substitute, to write everything in terms of t :
fc F
•
dr = J: F( r( t )) • r '(t) dt = J: F ( x(t), y ( t )) ( x '( t), y '(t) ) dt = J: ( M(x(t), y ( t )), N (x(t), y (t)) ) ( x '(t), y '(t) ) dt = J: [M(x(t ), y( t ) ) x '(t) + N( x( t), y( t ) )y '(t )] dt •
•
Since dx = x'( t)dt and dy = y'( t)dt, the last equation is also commonly written in the abbreviated form:
fc F
1 44 + CRACKING THE GRE MATHEMATICS SUBJECT TEST
•
dr = fc M dx + N dy
Example 3.19
Evaluate the line integral
fc (7y3 x) dx + (x2 + 2y 1) dy
(a) if C is the portion of the parabola y = x2 from (0, 0) to (1, 1); (b) if C is the straightline path from (0, 0) to (1, 1).
 
 ·
Solution: (a) We can parameterize the curve as follows: C:
=t Jx t h=
2
_j
··
for 0 � 1
With this parameterization, we have dx = dt and dy = 2t dt, so:
fc
This final equation says that the product of A and then, (51  A) must be the inverse of A:
A1
To check this formula, we compute
=>
A) gives the identity matrix; by definition,
= 51  A
and notice that this does indeed match the matrix
given in part (c).
GAUSSIAN E LIMI NATION
Now that we have the machinery of matrix algebra at our disposal, we return to the solution o f linear systems. To solve the linear system
5 x+2y = 3x+5y =
14
we take the coefficients of the unknowns and form the coefficient matrix
1 8 8 + CRAC K I N G T H E G R E MAT H EMAT I CS S U BJ E CT TEST
( 1 52 145) 3
and attach the constants that appear on the righthand sides of the equations as an additional column, producing the augmented matrix:
Now, to eliminate the variable x, for example, we multiply the first row by 3 and add the result to the second row; this operation doesn't change the first row, but it does change the second row:
(10 21 15)
1
The bottom row of this matrix is equivalent to the equation Ox + (1)y = 1 from which we immediately get y = 1 . Plugging y = into the equation that corresponds to the top row, 1x + 2y = 5, gives x = 3. There fore, the solution of the given linear system is (x, y) = (3, 1). In general, Gaussian elimination proceeds as follows: 
,
Step 1: Write down the augmented matrix corresponding to the given linear system.
Step 2: Perform a series of elementary row operations to reduce (transform) the augmented matrix to echelon form. An elementary row operation is one of the following: 1.
Multiplying a row by a nonzero constant. 2. Interchanging two rows. 3. Adding a multiple of one row to another row. A matrix is said to be in echelon form when it is an upper triangular matrix, any zero rows appear at the bottom of the matrix, and the first nonzero entry in any row appears to the right of the first nonzero entry in any higher row. Thus, the goal of this step is to use elementary row operations to produce zeros below the first entry in the first column, then below the second entry in the second column, and so forth. Step 3: Working from the bottom of the echelon matrix upward, evaluate the unknowns using back substitution.
Example 5.3
z = 3 2 3 y 2x3x 5yy z == 29
Solve the following linear system using Gaussian elimination: X
Solution:
+
+ +
!] 5 0 2
The first step is to write down the corresponding augmented matrix:
� �
L I N EAR A L G E B RA + 1 8 9
To get a matrix that has zeros below the first entry in the first column, we perform the fol lowing elementary row operations: add times the first row to the second row (abbrevi ated as + and add times the first row to the third row +
2r1 r2)
2
3
(3r1 r3):
(01 2 3 3]3 0 1
Next, let's interchange rows and (r2 H
2 3 r3):
5 7 9 7
(01 21 3 0 0 38
Finally, add 5 times the second row to the third row (5r2 + r3) :
9
38z = 38,
The matrix is now in echelon form. The bottom row translates to the equation so we see that Working our way up the echelon matrix, the second row corresponds to the equation + 7, so substituting we get Finally, the top row corresponds to the equation + substituting both and into this equation yields Therefore, the solution to the system is
zy= 1. 9z = x = 4. x 2y 3z = 3;
Example
5.4
Solve this system using Gaussian elimination:
2x3x 

X
Solution:
z = 1, z =y 1= 2. y = 2 (x, y, z) = (4, 2, 1).
';.
+
3z
=
4y + Sz
=
2y y
+

The corresponding augmented matrix is:
21 13
z
=
3
9
8
21 13
u 4 j] 2 3 (� 10 14 ;] 5
Performing the series of elementary row operations indicated below,
u
4
5
2r1+r2
_:]
3r1 +r3
5 7
2r2 + r3
25 73 (� 0 0 ;]
we run into a problem. The bottom row of this final matrix corresponds to the equation + which clearly has no solutions! Gaussian elimination stops. This system has no solutions; it is inconsistent.
Ox + Oy Oz = 7,
1 9 0 + CRAC K I N G TH E G R E MATH EMAT I CS S U BJ ECT TEST
Example 5.5
Solve the following linear system: X

+
2x 3 x
Solution:

2y + 3 z z y 4 y + Sz
=
3
=
9 15
=
The corresponding augmented matrix is:
1 J u 4 5 : 3 5 J [� 10 14 j 2
3
1

l
Since the coefficient matrix is the same as it was in the previous example, we'd apply the same elementary row operations:
u 4 5 :J 2
3 1 1

2r1 + r2 3r1 + r3
l
[� 0 0 �J
2
7
2
3
5 7
2r2 + r3
The bottom row of the final matrix corresponds to the equation Ox + Oy + Oz = 0, which places no restriction on the variable z. lt therefore become a free variable; let's call it t. Substituting z = t into the (equation corresponding to the) second row of the echelon matrix, Sy  7z = 3, gives y = i (7t + 3). Finally, backsubstituting both z = t and y = i (7t + 3) into the first row yields x = i (t + 21). Therefore, this system has infinitely many solutions, all of which can be expressed as follows:
(x, y, z) = (i(t + 21), i (7t + 3), t) Example 5.6
Solve this linear system: X

2y + 3 z
2x + y 3x  4 y +

Sol u ti on:
z 5z
=
0
=
0
=
for any t E
R
0
3 1 1 u 4 5 �J
The corresponding augmented matrix is:
2
li N EAR A L G E B RA + 1 9 1
(
(1
(1
Because the coefficient matrix is the same as it was in the previous two examples, we will apply the same elementary row operations:
' � � � 3
4
�J
5 0
2r1+ r2 3 + 3 rt r
)
J 14
3 O 5 7 0 10 0
2
0 0

2r2+r3
)
0 0
3 O 5 7 0 0 0 0
2
J
Once again, we let z = t be the free variable. Substituting z = t into the second row of the echelon matrix, Sy  7z = 0, gives y = i (7t). Finally, backsubstituting both z = t and y = i (7t) into the first row yields x = i (t). Therefore, this system also has infinitely many solutions, all of which can be expressed as follows: (x, y, z) =
( i t , t t, t )
for any t E
R
The preceding two examples illustrate an important general fact about linear systems. The system in Example 5.6 has only O's on the righthand side of every equation; such a system is called homogeneous. Homogeneous systems are always consistent, because there is always at least one solution, namely, all the unknowns equalling zero (this is called the trivial solution). The system in Example 5.5 had the same coefficient matrix, but it was nonhomogeneous. Comparing the general solution of the nonhomogeneous system with the solution of the corresponding homogeneous system reveals their similarity: Solution to nonhomogeneous system:
(x, y, z) = (i t , t t , t) + ( .?t , f , 0 ) for any t E R
Solution to homogeneous system: (x, y, z) =
(i t, t t, t )
for any t E
R
In general, every solution of a consistent, nonhomogeneous system can be expressed in the form x = xh + x, where x = xh is the general solution of the corresponding homogeneous system, and xP is a particular solution of the nonhomogeneous system. Notice the similarity between this theorem and the corresponding theorem for linear differential equations. In the case illustrated above, xh = (  t t,
Example 5.7
and xp =
(.?t,
t , o) .
Find all the solutions of the following linear system: w
2w +
w
Solution:
t t, t)
+ 3z
2x
==
y + z 8x + 2y + 7z X
==

==
5 2 19
The corresponding augmented matrix is:
1 9 2 + C R A C K I N G T H E G R E M AT H EMAT I CS S U B J ECT TEST
1 1 1 J 1 ! u 2
0 3
8
2
7
S
We perform the following elementary row operations to reduce this to echelon form:
u 8 [�
2 0 3 S 1 1 1 2 7 1
!
J
2 0 3 S 5 1 5 1 2 10 2 10 24
[� [�
2r1+r2 rt+r3
J
2 r2+r3
2 0 3 S 5 1 5 12 10 2 10 24 2 0 3 S 5 1 5 12 0 0 0 0
J
J
As in the preceding example, we have a zero row, but in this case there are a total of four unknowns. Since there are two nonzero rows in the echelon matrix, there are really only two constraints on the variables, so the other 4  2 = 2 are free. Let's designate y and z as the free variables, denoting y by t1 and z by t . Substituting these into the second row yields 2 Sx  t1  St2 12 => x i (t1 + St2 + 12)
=
=
and backsubstitution into the first row gives: w
2[i(t1 + 5t2 + 12)] + 3t2 5
=
=>
w
=
i (2t1  5t2  1)
Therefore, all the solutions of the given linear system are given by the equation:
(w, x, y, z) = (i (2t1  St2  1), i (t1 + 5t2 + 12), t1 , t2 ) for any t1, t2 E R
In general, the number of free variables (also called parameters) appearing in the solution of a linear system (that possesses infinitely many solutions) is equal to the number of un knowns minus the number of nonzero rows in the echelon form of the augmented matrix.
SOLVING MATRIX EQUATIONS USING A1 Any linear system can be written in the form Ax = b, where A is the coefficient matrix, x is the column matrix containing the unknowns, and b is the column matrix whose entries are the constants on the right hand sides of the equations in the system. For example, the system of Example 5.3, X 2y + 3z 3 2x + y z 9 3x + Sy 2 can be expressed as the matrix equation Ax = b, where:
A= [ � �l x = [:j , 2 1 3 5
== =
and
=
b UJ LI N EA R A L G E B RA + 1 93
Since the coefficient matrix A is square, it may be invertible. If A is invertible and we had a way to compute AI, then we could determine x by a simple matrix multip'lication, x = A1b, since:
A(A1 b) = (AA1 )b = Ib = b
Therefore, if A is an invertible square matrix, then the system Ax = b is consistent for every choice of the col11mn matrix b, and the unique solution is given by the product x = A1 b. Since figuring out A1 is usually more timeconsuming than simply using Gaussian elimination to solve the original system, this theorem does not offer an improved method of calculation. But if you do happen to know AI, then this theorem immediately provides a quick way to find the solution. Notice, for example, that we can use this theorem to conclude that if A is square and invertible, then the homogeneous system Ax = 0 has only the trivial solution (since x = A1 0 = 0). We'll close this section by giving the formula for the inverse of a 2 x 2 matrix A. The 2 x 2 matrix
(
1 ( a] J 
is invertible if and only if ad  be :1 0; in this case, the inverse of the matrix is: A
1
=
b�
d adbe c adbe
ad be
adbe
=
ad
d
be c
_ _ _

b
You should memorize this result for the GRE Math Subject Test.
Example 5.8
Find the matrix X that satisfies the equation AX = B, where: and
Solution:
B=( 4282
35 53
42 )
64
First, we notice that because A is 2 x 2 and B is 2 x 3, the matrix X must also be x 3. Now, if A is invertible, then by multiplying both sides of the equation AX = B by A1 on the left, we'd get:
2
Since A is 2 x 2, we can use the result above to find A1 • Since ad A is invertible and: A
12 (
1 =  6 4 5
Therefore,
1 9 4 + CRA C K I N G T H E G R E MATH EMAT I CS S U BJ ECT TEST
3
) ( 3 2) =
� i
be = (3)(6)  (4)(5) = 2
:1
0,
VECTOR S PACES The general definition of a vector space reads as follows: A vector space is a set that is closed with respect to two operations, addition and scalar multiplication. The scalars we'll consider are real numbers, so the resulting structure is called a real vector space. The elements of the set are called vectors, even when those objects are not of the traditional, directedlinesegment variety. The xy plane, referred to as R2 ("Rtwo"), is an example of a vector space. Any two vectors in R2 can be added together (componentwise) and the result is again a vector in R2• Thus, the set is closed under addition. Furthermore, multiplying a vector in R2 by a (real) scalar always results in a vector in R2; this means the set is also closed under scalar multiplication. R2 is a rather "big" vector space, however. Much more interesting are proper subsets of R2 that are themselves vector spaces. For example, consider the line y = 2x in R2 . This subset consists of all vectors, or points, of the form (x, 2x). Adding any two such vectors gives a vector of the same type, since:
(x1, 2x) + (x2, 2x2) = (x1 + x2, 2(x1 + x2)) Multiplying a vector in this set by a scalar also gives a vector of the same type:
k(x, 2x) = (kx, 2(kx))
Since this subset is closed under addition and scalar multiplication, it is itself a vector space; it's a sub
space of R2•
Now consider the line y = 2x + 1; it consists of all vectors or points of the form (x, 2x + 1). This set is not a subspace of R2; in fact, it fails to satisfy both requirements. It's not closed under addition, because, for example, although both (1, 3) and (2, 5) are in this set, their sum, (3, 8), is not. It's also not closed under scalar multiplication since (1, 3) is in the set, but the scalar multiple 2(1, 3) (2, 6) is not.
1
=
Because a vector space must be closed under scalar multiplication, every vector space must contain the zero vector. The line y = 2x + does not define a subspace of R2 because, for one thing, it doesn't contain the origin. However, containing the origin doesn't guarantee that a subset is a subspace. Con sider, for example, the closed first quadrant of the xy plane; this is the set of all points (x, y) such that x :::: 0 and y :::: 0. This subset of R2 contains the origin, but it's not a subspace, because although it's closed under addition, it's not closed under scalar multiplication. For example, multiplying the vector (1, 2) in this subset by the negative scalar k = 3 gives the vector (3, 6), which does not lie in the closed first quadrant.
Example 5.9
Let A be a 3 x 3 matrix such that the matrix equation
Ax = (1, 2, 4)T
has infinitely many solutions. Does the set of solutions form a subspace of R3?
Solution:
Let x1 and x2 be two of the solutions of the given equation. Then Ax1 = (1, 2, 4)T and Ax = (1, 2, 4)T, so 2 Ax1 + Ax2 = (2, 4, 8)T � A (x1 + x2) (2, 4, 8)T i: (1, 2, 4)T which shows that x1 + x2 is not a solution of Ax (1, 2, 4)T. Since the set of solutions of Ax = (1, 2, 4)T is not closed under addition, it is not a vector space. Another way to see that the answer to the question is no is simply to notice that the zero vector is not in the solu tion set, because x = 0 clearly doesn't satisfy the given nonhomogeneous equation. Since every subspace of R3 must contain the zero vector, the set of solutions to this equation is 3 not a subspace of R . =
=
LI N EAR A L G E B RA + 1 9 5
THE NuusPACE Let A be an m x n matrix. The set of all solutions of the homogeneous equation Ax = 0 forms a subspace of called the nullspace of A, denoted JY(A). That N(A) actually is a subspace follows from the fact that the sum of any two solutions of A x = 0 is itself a solutionso N(A) is closed under additionand any scalar multiple of a solution is also a solutionso N(A) is also closed under scalar multiplication. In Example 5.6, we found that the solution set of the linear system
R"
x
2y
2x
y 4y +
+
3x

is:
0
+ 3z =
0
= =
z Sz
0
{(i t, ft, t) : t E R}
[ � � �J
This is the nullspace of the system's coefficient matrix: A=
3
Example 5.10
4
5
What is the nullspace of an invertible matrix?
If A is invertible, then the only solution of Ax = invertible n x n matrix is the trivial subspace of
Solution:
0 is x = 0. Therefore, the nullspace of an
R".
LINEAR COMBINATIONS Consider a collection of nvectors, v1 , v2,
•
•
.
,
V111 •
Any expression of the form
kJvJ + k2v2 + . . . + kmvm
where the coefficients k.I are scalars, is called a linear combination of the vectors vI.. The set of all linear combinations of the vectors vi is called their span. For example, consider the vectors i = (1, 0, 0) and j = (0, 1, 0) in The span of and j is equal to the xyplane, since every vector in the xyplane can be expressed as a linear combination of and j and vice versa. The vector v = (1, 1, 1), for example, is not in span I , j I because no linear combination of i and j could equal v. Notice that the span of any collection of nvectors is always a subspace of because it's automatically closed under addition and scalar multiplication; after all, the span contains all linear combinations of the vectors in the spanning set. If all the scalars k.I in the linear combination �
R3.
i
i
i
R"
k1v1
+ k2v2 +
+ kmv"'
are zero, then the result is the zero vector, 0. This is called the trivial combination. If this is the on ly way the vectors vi can be combined to give the zero vector, then the vectors vi are said to be linearly indepen dent. On the other hand, if
1 96 + C R A C K I N G TH E G R E MAT H EMAT I CS S U BJ E CT TEST
is true for some nontrivial choice of the scalars kv then the vectors v; are said to be linearly depen dent. The name arises because in this case at least one of the vectors depends on the others, that is, it can be written as a linear combination of the other vectors in the collection. For example, the vectors v 1 = (1, 2, 3), v2 = (1, 4, 9), and v3 = (1, 1) are dependent since we can find a nontrivial linear com bination that gives 0; one such combination is 2v1  v2 + 3v3, as you may verify. An equivalent way to show that these three vectors are dependent is to notice that we can write, say, v2 as a linear combination of v1 and v3:
0,
Because of this, the span of v1 , v2, and v3 is the same as the _ span of just v1 and v3• The inclu sion of v2 doesn't enlarge the span; any linear combination of v1, v2, and v3 is already a linear com bination of just v1 and v3. (In fact, since v1 can be written in terms of v2 and v3' we could just as well leave out v1 a d just keep v nd v3• Or we co ld ke just v1 and v2 and ignore v3.) T erefore, � span v 2 , v . So any one of these two element span v u v 2 , v  span v 1 , v z 1 span v 1 , v sets is a minimal spanning set for the space span v 1 , v 2 , v 3 . A minimal spanning set for a vector space is called a basis for the space. In other words, a basis is a collection of vectors from the space that not only span the space but are also linearly independent. A basis is just the right size; it contains enough vectors to span the entire space, but not so many as to include unnecessary (dependent) ones. If B is a finite set and a basis for a space V, then the number of vectors in B is independent of the choice of B; this number is called the dimension of V, denoted dim V. If no finite collection of vectors spans V, then V is said to be infinite dimensional and we write dim V = "" ·
{
J
Example 5.11
�
{
�

{
{
J
r
�

J
�
Consider the vectors:
v1 = (1, 2, 3), v2 = (1, 4, 9),
{
(a) Describe the set span v1 , (b) Find a basis for P. (c) What is dim P?
v , v 2
and
J
v3 =
(1, 0, 1)
Solution: (a) The span of v1 , v2, and v3 is the set of all linear combinations of these vectors. Since they are 3 3 3vectors, the span is some subspace of R . A subspace of R must be one of the following: the trivial subspace (!0)), a line through the origin, a plane through the origin, or all of W. Because the vec tors are not all scalar multiples of a single vector, the span is not a line through the origin (and the span is clearly not the trivial subspace, since we're given nonzero vectors). Therefore, if the three 3 given vectors are linearly independent, then the span will be all of R ; otherwise, the span will be a plane through the origin. The vectors are linearly independent if the only scalars k1 , k2, and k3 that satisfy k1v1 k2v2 + k3v3 = 0 are k1 = k2 = k3 = The equation k1v1 + k2v2 + k3v3 = 0 is equivalent to the homogeneous system Ak = 0, where k = (k1, k2, k3)r and:
+
0.
A +� �: �]
L I N EAR A L G E B RA + 1 9 7
(
Applying Gaussian elimination to the system Ak = 0, we get:
1 1 1 2 4 0 0 3 9 1 0
OJ
The bottom row of zeros implies that we can take k3 as a free variable, so there are infinitely many solutions. Since k1 = k2 = k3 = 0 is not the only solution, the vectors v1, v2, and v3 are not linearly independent. We can eliminate, say, v3 from the original set without altering the span. So, the span of v1, v2, and v3 is identical to the span of v1 and v2 alone, which is a plane through the origin. Because v1 and v2 lie in this plane, a vector normal to the plane is any nonzero scalar multiple of the cross product:
1
A
v 1 x v2 =
i
1
j 2 4
k
3 9
=
6i  12j  6k A
A
A
=
6(i + 2j + A
A
k) A
We can therefore take N = (1, 2, 1 ) as a normal vector to the plane, so the equation of the plane is x + + z = d, for some constant d. Since the plane must contain the origin (since the span is a subspace, and any subspace must contain the zero vector), the value of d must be 0. The span of v1, v2, v3 is the plane P in R3 whose equation is x + + z = 0.
2y
2y
(b) A basis for P consists of any two of the three vectors vi' v2, and v3• (c) Since a basis for P contains two vectors, dim P = 2.
THE RANK, CoLUMN SPACE, AND Row SPACE oF A MATRIX
We can view the columns of an m x n matrix as vectors in Rm, and the rows of the matrix as vectors in Rn. The maximum number of linearly independent columns is called the column rank of the matrix, and the maximum number of linearly independent rows is called the row rank. Although the columns are mvectors and the rows are nvectors, it can be shown that, for any matrix, the column rank always equals the row rank; this common value is therefore simply known as the rank of the matrix. The subspace of Rm spanned by the columns of an m x n matrix A is called the column space of A, which we denote CS(A). Since the rank is equal to the maximum number of linearly independent columns, the dimension of CS(A) is equal to the rank of A. What characterizes the column space? That is, how do we decide whether or not a given mvector is an element of CS(A)? Let v1, v2, , v" denote the columns of A. If an mvector b is in CS(A), then there must be some linear combination of the columns of A that equals b; that is, there must be scalars k1 , k2, , k" such that: •
•
•
•
1 9 8 + CRA C K I N G TH E G R E MATH EMAT I CS S U B J ECT T EST
•
•
But this equation can be written in matrix form as
which is Ak = b. So we see that an mvector b is in CS(A) if and only if the equation Ax = b has a solution. The subspace of R" spanned by the rows of an m x n matrix A is called the row space of A, which we denote RS(A). Since the rank is equal to the maximum number of linearly independent rows, the dimen sion of RS(A) is equal to the rank of A. How do we decide whether or not a given nvector is an element of RS(A)? Perhaps the simplest way is to notice that:
RS(A) = CS(N) which says that the row space of A is the column space of AT. So, combining this observation with the criterion given above for membership in the column space, we conclude that an nvector b is in RS(A) if and only if the equation Nx = b has a solution. By definition, the rows of A form a spanning set for RS(A), but if the rows are not linearly independent, they do not form a basis for RS(A). If we reduce the matrix A to echelon form by a sequence of elementary row operations, the nonzero rows that remain give a basis for RS(A), and the number of these rows gives the rank of A. To find a basis for CS(A), we notice that:
CS(A) = RS(N) So, if we reduce AT to echelon form and write the nonzero rows that remain as column vectors, we'll have a basis for CS(A).
Example 5.12
Solution:
For the following matrix, compute its rank, find a basis for its row space, and find a basis for its column space:
A+� � �J
Reducing A to echelon form, we get:
[� � �J 2
0
2
.,., ) [ � � �J
There are only two nonzero rows remaining, so the rank of the matrix is 2. As a basis for the row space of A, we can take the two nonzero rows in the echelon form: basis for RS(A) = {(1, 1, 1), (0, 2, 0)}
L I N EA R A L G E B RA +
1 99
(If it's desired to have basis vectors that were rows in the original matrix, we could take any two of the rows of A.) Notice that since the rank is 2, the number of vectors in a basis for the row space is two; thus, the row space is a 2dimensional subspace of R3 . Now, a simple way to find a basis for the column space of A is to find a basis for the row space of AT: AT =
[ ] 1 1 2 1 1 0 1 1 2
By transposing the vectors in a basis for the row space of AT, get a basis for the column space of A:
{(1,  1,
2), (0,
2, 2) } , we
Notice that dim CS(A) is also equal to 2. The dimension of the column space is always equal to the dimension of the row space; both are equal to the rank of the matrix.
OTHER VECTOR SPACES Although we've been identifying and describing subspacesnamely, the nullspace, column space, and row spacethat arise from matrices and matrix equations, several other vector spaces are commonly encountered. Let n be a positive integer and consider the collection of all polynomials of degree S n. This set is closed under addition and scalar multiplication, so it's a vector space, denoted by P,.. The "vectors" in this space are polynomials. A basis for the polynomial space P" is { 1, x, x 2 , , x" } , which contains n + elements, so dim P,. = n + 1. Other function spaces include the space of all bounded functions, all continuous func tions, or all differentiable functions on a given interval; these spaces are infinitedimensional. The set of all m x n matrices with real entries is denoted Mm x ,.(R); it forms a vector space, since this set is closed under addition and scalar multiplication. The "vectors" here are m x n matrices. The dimension of M"' x ,/R) is equal to mn. To illustrate this, notice that a basis for M x 2(R) consists of these 3 (3)(2) = 6 vectors: • . •
1
DETERMI NANTS Let A be a square matrix. Then the determinant of A, denoted by detA or IAI, is a number associated with A that describes certain properties of A. First, we'll show how to evaluate the determinant of a matrix, then discuss its applications. The determinant is only defined for square matrices, so throughout this section all matrices are square. There are essentially two ways of computing the determinant. The first one defines the
200
+ C R A C K I N G T H E G R E MAT H EMAT I C S S U B J ECT T EST
determinant by giving a specific formula for it, but using this formula is impractical for large matrices. The second way uses properties of the determinant function to provide a stepbystep method for its computation; this method is the one that's actually used in practice. Consider the set n of integers from 1 to n. A permutation of n is a function from n to itself that is both onetoone and onto. [Recall that a function f is onetoone (or injective) if distinct elements in the domain are always mapped to distinct elements in the range, and f is onto (or surjective) if every element in the range has a preimage in the domain.] For example, for the set 4 = 11, 2, 3, 4l, the function cr: 4 7 4 defined by the equations cr(1) = 2, cr(2) = 4, cr(3) = 1, and cr(4) = 3 is a permutation of 4. It maps the elements 1, 2, 3, 4 onto the elements 2, 4, 1, 3; essentially, then, a per mutation of n is just some arrangement of the elements 1 through n. Every permutation of n can be achieved by a sequence of transpositions; the transposition of two ele ments i, j is simply the same two elements in the reverse order: j, i. For the permutation given above, we can achieve the image of cr from the original order 1, 2, 3, 4 by the following sequence of transpositions: 1 , 2, 3, 4 7 2, 1, 3, 4 7
2, 1, 4, 3
7
2, 4, 1, 3
Since it took three transpositions and three is an odd number, the permutation cr is said to be odd. On the other hand, if the number of transpositions in the sequence were even, the permutation would be called even. The set of all permutations of n is denoted 5 11 . 51l contains n! permutations, half of which are even and half of which are odd (if n 2:: 2). The sign of a permutation cr, a number denoted sgn cr, is defined to be + 1 if the permutation is even and 1 if it's odd. Permutations of n are used to define the determinant of an n x n matrix. Recall that the entry in row i and column j of A is denoted a;r The determinant of A is defined to be the following sum det A =
L (sgn cr) alcr(J) a2cr(2)
OE511
· · ·
a 1 a( 11)
where the sum is taken over all n! permutations of n. It's easy to see why this formula is impractical for large matrices. For n = 4, there would be 4! = 24 terms in this sum, for a 5 x 5 matrix, the sum would contain 5! = 120 terms, and so on. For a 2 x 2 matrix, however, the sum is easy to compute. There are only two permutations of the set 2 = /1, 2l, namely the even permutation 1, 2 and the odd permutation 2, 1 . Therefore, the sum that defines the determinant of the 2 x 2 matrix
is: det A =
L (sgn cr) alcr(l)a2cr(2)
cres,
L I N EAR A L G E B R A + 2 0 1
This formula says that, to find the determinant of a 2 x 2 matrix, simply multiply "down" the main diagonal (to form the product a11a22) , multiply "up" the offdiagonal (to form the product a12a21 ) , and subtract:
or, more simply,
det(: !l=adbc.
Memorize this formula!
The determinant of a 3 x 3 matrix can be computed using this formula, and the result isn't too intimi dating; it's det A =
L (sgn cr) a lcr(l) a 2cr(2 ) a 3cr( 3)
cre S3
which can be remembered as follows: Copy the first two columns of A and place them to the right of A. Take the products formed by multiplying "down" and, from their sum, subtract the products formed by multiplying "up": subtract these products
This gives:
A practical method for computing the determinant of large matrices (and which also can be used for 3 x 3 matrices) relies on the properties that actually define the determinant function: 1. The determinant is linear in each row, that is: rt r2 det
= det
+ det
r;2 r,
2 0 2 + CRA C K I N G T H E G R E MATH EMAT I C S S U BJ ECT T EST
and
rJ
det
rJ
r z
krI
r z
= k det
r.I
r111
rni
2. If two rows of a matrix are interchanged, the determinant changes sign. 3. The determinant of the identity matrix is equal to
1.
These properties can be used to describe the effect that the elementary row operations we studied earlier have on the value of the determinant: 4. If a row is multiplied by a constant, then the determinant is multiplied by that constant. 5. Adding a multiple of one row to another row leaves the determinant unchanged.
Remember that we used elementary row operations to reduce a matrix to echelon form. We now simply note the following additional facts about the determinant: 6. The determinant of a triangular matrix is equal to the product of the diagonal entries.
7. If a matrix has a zero row, the determinant is zero. 8. The determinant of the product of two square matrices of the same size is equal to the product of their determinants: det(AB) = (detA)(detB). Using these properties, we can evaluate the determinant of a large matrix with much less computation than would be required by the formula given earlier involving permutations.
Example 5.13
A=
Solution:
� _: [1 2 �]1 (12 2 10] 1
What's the determinant of this matrix?
Let's reduce A to uppertriangular form by a sequence of elementary row operations. First, switch rows 1 and 3, which changes the sign of the determinant: det A =  det
3 4 4
Next, we add twice the first row to the second row and four times the first row to the third row; neither of these operations changes the determinant:
L I N EA R A L G E B RA + 203
Adding
12
det A =  det
[1 21 21] 12 [1 21 21] 0 0
5
times the second row to the third row also doesn't change the determinant: det A =  det
(1)(1)(19) 19.
0
0
0
19
Since the determinant of a triangular matrix is equal to the product o f the diagonal entries, we find that detA = =
Example 5.14 Solution:
3
If A is a ? x matrix whose determinant equals 5, what is the determinant of the matrix 2A?
2,
23. 2.
2.
If only one row of A were multiplied by the determinant would be multiplied by However, the matrix 2A is formed by multiplying every row by Since there are rows, Property 4 states that we multiply the determinant by Therefore: det (2A) =
1)1) 1)
23
detA
=
8(5)
=
3
40
LAPLACE EXPANSIONS Laplace expansion provides a method for computing an n x n determinant in terms of simpler, (n x (n determinants. Let A be an n x n matrix and choose any row, say row i. Let A;i denote the (n x (n  1) matrix obtained by deleting row i and column j from A. The determinant of Aii is called the aii minor, and multiplying the aii minor by gives the aii cofactor, denoted cof(a;) : cof(a;i)
The determinant of A is then equal to:
(1()i+1i)i+i =
det A =
(a;i minor)
n
�>i cof(aii ) j;l
This is called the Laplace expansion by row i. Let's illustrate this with the 3 x 3 matrix:
( � :: ) + a, (1:: :: 1 ) + •, (1:: :: l
The Laplace expansion by the first row is: det A � a,.
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•
C RA C K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
The determinant of A can also be computed by finding the cofactors of the entries in any column from A; the Laplace expansion by column j is: II
det A = L,. ai
cof(a;) i=l i For example, the cross product of a pair of 3vectors, u = (u 1 , u2, u) and v = (v1, v2, v3), is usually calculated by a Laplace expansion by the first row of the following symbolic 3 x 3 determinant: i j k u u u u u u U X V = u ] Uz u 3 = i z 3  j l 3 + k l z v] Vz v3 v2 v3 vl v3 vl v2
Laplace expansions are especially useful when the matrix contains a row (or column) with zeros, since those terms drop out of the sum. Example 5.15
Evaluate the determinant:
1 2 1 0
Solution:
0
3
1
2
0
0
0
1 1
0
1
3
The third column has only one nonzero entry, so we'll perform a Laplace expansion by the third column. The given determinant is equal to: 2 1 3 a13 • (  1)1+ cof (a13 ) = (2) • (1) + 1 3
0
3 0 1 1 1 3
Now this 3 x 3 determinant can itself be computed by using the formula from p. 202 (add the downwardpointing products and subtract the upwardpointing products): [6 + 0 + 0]  [0 + 2 + (9)] = 6  (7) = 13 Therefore, the given 4 x 4 determinant is equal to: (2) (1 ) 1+3 1 3 = 26 You can check that this determinant is correct using Gaussian elimination. •
•
THE ADJUGATE MATRIX
Let A be an n x n matrix. Consider the n x n matrix whose (i, j) entry is the cofactor of the entry aii in A; the transpose of this cofactor matrix is called the adjugate (or classical adjoint) matrix of A, denoted Adj A: AdJ. A = [cof(a. ))T 'I
L I N EA R A L G E B RA + 2 0 S
Why form the adjugate matrix? Because if A is invertible, the following gives a formula for the inverse of A:
This result establishes the following important theorem: A square matrix, A, is invertible if and only if
det A � a.
Example 5.16 Solution:
If A is an invertible n x n matrix with detA in terms of k.
=
k, express the determinant of Adj A
From the equation for k1 in terms of Adj A, we can write: Adj A = (detA) k1 = kk1
Therefore, det (Adj A) = det (kA1 ). Since AA1 = I, we have (detA)(detk1 ) = deti = 1, so detk1 = (detA)1 . Multiplying k1 by k means multiplying every one of the n rows of 1 k1 by k, so the determinant of kA is equal to k" times the determinant of k1 . We conclude that: det(Adj A) = det(kA1 ) = k" det(k1 ) = k"(detA)1 = k"(k1 ) = k"1
CRAMER's RuLE Determinants can be used to give a formulaknown as Cramer's rulefor the solutions of a square linear system. Let A be a square matrix and consider the system Ax = b. Let A. denote the matrix formed I by replacing column j of A by the column vector b. If detA :t 0, then the value of the unknown x.I is given . by the equation:
Using Cramer 's rule, we can establish the following result: If A is a square matrix, the linear system Ax = b has a unique solution for every b if and only if detA :t 0.
Example 5.17
Solution:
Find the value of y
if
2x  3y + 4z = l 4x + 9y =0 7x  2y + 5z = 0
Let's answer this using Cramer's rule. Since we want the value of the second unknown, y, we replace the second column of the coefficient matrix A with the column vector b = (1, 0, O)T to form the matrix whose determinant is the numerator in the Cramer's rule expression:
2 0 6 + C R A C K I N G TH E G R E MATH EMAT I CS S U B J ECT TEST
y=
det A2 = det A
2 1 4 4 0 0 7 0 5 2 3 4 4 9 0 7 2 5
=
20 = 10

134

67
L I N EAR TRANSFORMATIONS Let V and W be vector spaces. A linear transformation (or linear map) i s a function T: V 7 W such that T(x1 + x2) = T(x1 ) + T(x2)
and
T(kx) = kT(x)
for any vectors x1, x2, x in V and any scalar k. If W = V, then a linear transformation T is also called a linear operator. Notice that a linear transformation always maps the zero vector in the domain space to the zero vector in the range space, because T(O + 0) = T(O) + T(O) implies T(O) = T(O) + T(O), so T(O) = 0. Therefore, if a function T doesn't map the zero vector to the zero vector, it can't be a linear transformation. There's a simple and elegant connection between linear transformations and matrices. Let A be an m x n matrix and define a function T: R" 7 R"' by the equation T(x) = Ax for x E R". Then T is a linear transformation, as you can verify. Conversely, if T: R" 7 R"' is a linear transformation, then there's an m x n matrix A such that T(x) = Ax for all x in R". Let T: R" 7 R"' be a linear transformation and let B = b1 , b2 , . . . , b " } be a basis for R". Once the images of the basis vectors are known, the image of any vector in R" is determined. To see why, notice that every vector x in R" can be written as a unique linear combination of the basis vectors:
{
The column vector k = (k1, k2, , kJ is called the component vector of x relative to B, denoted [x] B ' Using the linearity of T, we find that the image of x is: •
•
•
T(x) = T(k1b1 + k2 b 2 + . . · + k"b" ) = k1T(b1 ) + k 2 T(b 2 ) + · · · + k J(b " ) So if we know T(b) for every i, we know T(x) for any x.
Example 5.18 Solution:
If T: R2 7 R2 is the linear transformation that maps (1, 1) to (2, 2), what's the image of (3, 1)?
(3, 4) and
(1, 1) to 2
Since the vectors b, = (1, 1) and b2 = (1, 1) are linearly independent and span R , they form a basis for R2• Using the information given about the images of these basis vectors, we can find the image of x = (3, First, we need to find the components of x relative to this basis; that is, the scalars k1 and k2 such that x = k1b1 + k2b2. This equation is equivalent to the system
1).
kl  k2 = 3 k] + k2 = 1
LINEAR ALGEBRA
+ 207
which gives
k1 = 2 and k2 =
1.
b1  b2, linearity of T gives: T(x) k1T(b1 ) + k2T(b2 )
Since x = 2 =
)6
T(3, 1) = 2T(1, 1 )  T(1, 1) = 2(3, 4)  (2, 2)
= ( 4,
STANDARD MATRIX REPRESENTATIVE If R" and Rm are considered with their standard bases, that is, the bases composed of the standard basis vectors ej = (0, 0, . . . 0, 1, 0, . . . , 0) , which have a 1 in the ith spot and O's elsewhere, then finding a matrix representative, A, for T: R" 7 Rm is easy: Column i of A is the image of er The m x n matrix A is called the standard matrix for T, and T(x) = Ax for every x in R" . . Example
5.19
If T: R3 7 R3 is the linear operator that maps (1, 0, 0) to (1, 21 3), (0, 1, 0) to (1, 1, 1), and (0, 0, 1) to (1, 2, 0), find the standard matrix for T and use it to determine T(3, 4, 5 ) 
Solution:
'
.
The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1) are the standard basis vectors for R3, so the standard matrix for T is formed by making the images of these basis vectors the columns of A:
1 [� Jr531 r61
Therefore, writing x = (3, 4, 5) as a column vector, its image under T is computed by multiplying by A:
T(x) = Ax =
1
1 2 1
0
4
=
20
13
T H E RANK PLUs N uLLITY THEOREM Let T: R" 7 Rm by a linear transformation. The set of all vectors x in R" that get mapped to 0 is called the kernel of T, denoted ker T:
kerT = {x : T(x) = O}
The kernel of T is a subspace of the domain space R"; its dimension is called the nullity of T. The set of all images of T is called the range of T, denoted R(T):
R(T) = {T(x) : x E R"}
208 + CRAC K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
The range of T is a subspace of R111; its dimension is called the rank of T. Notice that if T is given by T(x) = Ax, then the kernel of T is the same as the nullspace of A, and the range of T is the same as the column space of A. The Rank Plus Nullity S states that: nullity(T)' + rank(T) = n = dim (domain T)
If T: R" 7 R" is a linear operator, then the nullity of T is zero precisely when the rank of T is n. It can be shown that T is onetoone when the nullity is zero, and that T is onto when the rank of T is n. Therefore, the linear operator T is onetoone if and only if it's onto, and this happens when T has full rank. So, if T(x) = Ax, then T is onetoone and onto when the rank of A is n, that is, when detA :t 0.
Example 5.20
A= (l 2 3)
Determine the kernel, nullity, range, and rank of the linear transformation T: R3 7 R2 given by the equation T(x) = Ax where: 4 5 6
Solution:
The kernel of T is the nullspace of A, the set of the solutions to Ax gives:
A=
(1
2
3)
4 5 6
(1
2
0 3
3)
=
(1
0
6
0. Rowreducing A
2 3 1 2
)
If x = (x, y, z)Y, then we can take z as a free variable; let z t. The bottom row of the final matrix implies that y + 2t = 0, so y = 2t. Substituting y 2t and z t into the equation corresponding to the first row, x + 2(2t) + 3t = 0, gives x = t. Therefore, =
=
=
kerT = {(1, 2, 1)t: t E R}
Since ker T contains one parameter, it is onedimensional, so the nullity of T is 1. By the Rank Plus Nullity theorem, the rank of T must be 3  1 = 2. Since the range of T is a two dimensional subspace of R2, it must actually be all of R2. A Nou oN INVERSES AND CoMPOSITIONS Let T: R" 7 R" be a linear operator. If T is both onetoone and onto, then T has an inverse, r1, that's
defined as follows: r1(y) x if T(x) = y. If A is the matrix representative for T, then A1 is the matrix representative for T1• Let S: R"' 7 R" and T: R" 7 RP be linear transformations. If [S] is the matrix representative for S, and [T] is the matrix representative for T, then the matrix product [T][S] is the matrix representative for the composition T o S = TS. In fact, the rather involved definition of matrix multiplication was motivated by this result. =
E I GENVALUES AND EIGENVECTORS If we multiply a square matrix A by a compatible nonzero column vector x, the product Ax is generally not a scalar multiple of x. However, if Ax does equal a scalar multiple of x, that is, if Ax = f..x for some scalar
L I N EA R A L G E B RA + 2 0 9
A, then A is called an eigenvalue (or characteristic value) of A, and the nonzero vector x is a correspond ing eigenvector (or characteristic vector). If we interpret the product Ax as the image of x by the linear transformation represented by A, then an eigenvector of A is a nonzero vector x that's transformed by A into a scalar multiple of itself (with the scalar giving the corresponding eigenvalue). Eigenvalues and ei genvectors are defined only for square matrices (or, equivalently, only for linear operators), so throughout this section, A will always represent a square matrix.
To find the eigenvalues and eigenvectors of a matrix A, we must find a nonzero vector x and a scalar A that satisfy the equation Ax = Ax. This equation can be rewritten in the form (A  AI)x = 0, which is homogeneous. As we've already seen, a homogeneous square system has a nonzero solution if and only if the coefficient matrix is noninvertible, which is true if and only if the determinant of the coefficient matrix is zero. In this case, the coefficient matrix is A AI, so the eigenvalues of A are the values of A that satisfy the characteristic equation

det(A  AI) = 0
and the eigenvectors corresponding to each eigenvalue are the nonzero vectors x that satisfy Ax = A.x.
Example 5.21
Find the eigenvalues and eigenvectors of the matrix:
( � �)
A= 
Solution:
First, we form the matrix A  AI:
The determinant of this matrix is the characteristic
polynomial of A:
2 ) ( 1  A. _ _ 4 1 A. 4 = (1  A)(  A.)  (2)
det (A  IJ) = det =
A2  SA + 6
The characteristic equation, /.}  SA + 6 = 0, is equivalent to (A  2)(A  3) = 0, so we see that the eigenvalues of A are A = 2 and A = 3. Now, to find the eigenvectors corresponding to A = 2, we seek nonzero vectors x that satisfy Ax = 2x, or, equivalently, that satisfy (A  2l)x = 0: (A  2/) x = O
=::}
'
( 1 1
2)
2
x=O
=::}
x=k
( 2) 1
Since eigenvectors are nonzero by definition, the eigenvectors corresponding to the eigen value A = 2 are all the nonzero multiples of (2, 1)T.
2 1 0 + C R A C K I N G TH E G R E MATH EMATI CS S U B J ECT TEST
The eigenvectors corresponding to
(A  3I)x = o
�
A = 3 satisfy (A  3I)x 0:
(=� �) x = o
Thus an eigenvector corresponding to the eigenvalue (1, l)T.
Example 5.22 Solution:
=
�
x=k
A=3
G)
is any nonzero multiple of
What can you say about a square matrix A that has zero as an eigenvalue?
If 0 is an eigenvalue of A, then the equation Ax = Ox must have a nonzero solution x (an eigenvector). Since A x = 0 has nonzero solutions if and only if A is noninvertible, we can conclude that if 0 is an eigenvalue of A, then A is not invertible (or, detA = 0).
Example 5.23
22
The trace of a square matrix is the sum of the diagonal entries; it's denoted tr(A). Show that the characteristic polynomial for any x matrix A is:
A2  [tr(A)] A + (detA) Solution:
Let
be an arbitrary
22 x
matrix. Its characteristic polynomial is:
p(A)
= (aAc d 'Ab ) = (aA)(d 'A)bc = 'A2 (a+ d)'A+ (ad be) = 'A2 [tr(A)]'A + A) det
(det
Recall that the sum o f the roots of the monic polynomial equation X" + a n1 x"1 +
· · ·
+ a1 X + a0
=0
22
is equal to a,_r Applying this to the characteristic equation of a x matrix A shows that the sum of the eigenvalues of A is equal to the trace of A. This result can be used to prove that the sum of the eigenvalues of A is equal to the trace of A for any size square matrix. The product of the roots of the monic polynomial equation above is equal to a0• Ap plying this to the characteristic equation of a x matrix A shows that the product of the eigenvalues of A is equal to the determinant of A. This result can be used to prove that the product of the eigenvalues of A is equal to the determinant of A for any size square matrix.
22
L I N EA R A L G E B RA + 2 1 1
E IGENSPACES Let A be an eigenvalue of an n x n matrix A. If we adjoin the zero vector to the collection of all the eigen vectors corresponding to A, the result is a subspace of R" called the eigenspace of A corresponding to A:
Notice that E�.(A) is the nullspace of the matrix A  AI.
Example 5.24
Without computing its characteristic polynomial, explain why the matrix A below has 0 as an eigenvalue, then find a basis for the corresponding eigenspace.
A
=
[ � � �J 


1 1
Solution:
1
·
'
.
Since the second row of the matrix is the opposite of the first row and the third row is equal to the first row, these rows are clearly not linearly independent, so the determinant of the matrix must be zero. Because the product of the eigenvalues always equals the determinant, the fact that detA = 0 implies that A must have 0 as an eigenvalue. The eigenspace corresponding to A = 0 is the set of solutions to Ax = 0; reducing A to echelon form is easy:
2
Since there's only one nonzero row remaining, there's only one constraint on the unknowns; therefore, 3  1 = of them are free variables . Let x = (x, y, z)T, and take y = t1 and z = t2 as the free variables. Substituting these into the equation corresponding to the first row gives so the solutions of Ax =
so
{ (1, 1, O)T , (1,
0,
X  t1 + t2 0 => X = t1  t2 =
0 can be expressed in the form
1 )T }
2 1 2 + CRAC K I N G T H E G R E M ATH EMATI CS S U B J E CT TEST
is a basis for the twodimensional eigenspace E�. = o (A) .
THE CAYLEYHAM I LTON TH EOR EM The remarkable CayleyHamilton theorem states that every square matrix satisfies its own characteristic equation. That is, if p(A.) = det(A  A.I) is the characteristic polynomial of the matrix A, then p(A) = 0. This theorem can be used, for example, to express an integer power of A as a linear polynomial in A.
Example 5.25
If
A = (4 2) 5 3
show that A4
Solution:
=
5A + 6I.
The characteristic polynomial of A is
det(A  >r) =
l
4  >5
1
 2 = >2  >  2 = 0 3  >.
so by the CayleyHamilton theorem, N  A  2I = 0. Therefore, N = A + 21. Multiplying both sides of this equation by A gives N = N + 2A, and using N = A + 2I again, we get N = 3A + 21. Multiplying by A again gives A4 = 3N + 2A = 3(A + 2I) + 2A = SA + 61, as desired.
L I N EA R A L G E B RA + 2 1 3
CHAPTE R S R EVI EW QU ESTIONS
Complete the following review questions using the techniques outlined in this chapter. Then, see Chapter 8 for answers and explanations.
1.
Two distinct solutions, x1 and x2, can be found to the linear system Ax = b. Which of the following is necessarily true?
(A) b = 0. (B) A is invertible. (C) A has more columns than rows. (D) x 1 = x2•
(E) There exists a solution x such that x :f. x1 and x :f. x2.
2.
The solution of the system
is
(x, y, z) =
(A)  t 3.
(a,
a
b, a). If is NOT an integer, what is the numerical value of a + b?
(B)
1
(C)
0
(D)
�
(E)
1
Let A, B, and C be real 2 x 2 matrices, and let 0 denote the 2 x 2 zero matrix. Which of the following statements is/ are true? I. A2 = 0 :=} A = O II. AB = AC :=} B = C 1 III. A is invertible and A = A :=} A = I or A = I
(A)
(C) II and III only
(B) I and III only
I only
(E) None of the above
(D) III only
4.
ax+ayz=1 xayaz=1 axy + az=1
If
(01 then n
(A) 7
1 J  ( 1 O J =( 0 1 1 1 6
= (B) 5
2 1 4 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ ECT T EST
(C) 5
�) (D) 6
(E)
7
5.
If the matrices
[�3 � �J
and
1 2
2
are inverses of each other, what is the value of c?
(A) 3 6.
The vectors v1 = (1, 1, 1), EXCEPT k =
(A) 7.
(B) 2
2
[� � �J
(C) 0
c
(D)
1
2
(E)
3
v2 = (1, 1, 1), and v3 = (1, 1, k) form a basis for R3 for all real values of k (C ) 0
(B) 1
For the matrix
(D)
1
(E) 2
23
[ ] 1
A= 4 5 6 7 8 9
let r denote its rank and d denote its determinant. What is the value of r  d?
(A) 8.
2
(C) 0
(B) 1
If
da then
[
;[ 7J = d
(D)
1
(E) 2
b l q
J
k 2(a  k) p + k det l 2(b  l) q + l = m 2(c  m) r + m
(A) 8d
(B) 6d
(C )
2d
(D) 2d
(E) 8d
L I N EA R A L G E B RA + 2 1 S
9.
For what value of x will the following matrix be noninvertible?
7 6 0 1 5 4 X 0 8 7 0 1 0 0 1 1 (A) 10.
(B) 0
26
(D) 3
(E)
9
b = (12, 11, d)T in the column space of this matrix?
A=[� : !]
(B) 10
(C) 13
(D) 10
(E) 26
What is the dimension of the following subspace of R5?
span
(A) 12.
(C) 1
For what value of d is the vector
(A) 11.
1
1
1 0 0 1 0 0 1 0 0 0 0 1 1 ' 1 ' 0 ' 1 ' 1 ' 1 0 1 0 0 0 1 1 0 0 1 1 0
(B) 2
(C) 3
(D)
4
(E)
5
A square matrix A is said to be symmetric if it equals its own transpose: A = AT. What is the dimen sion of the subspace Sn x n(R) of real symmetric n x n matrices in the space of all real n x n matrices, MII X II ( R) ? (A)
tn
(B) t n(n  1)
2 1 6 + CRAC K I N G THE GRE MAT H EMAT I C S S U BJ E CT TEST
(D) t n(n + 1)
(E) t n!
13.
The set of all points (x, y) in R2 that satisfy the equation X
y
0
y1
1
Yz
1 1 1
=
0
forms
(A)
(B)
an ellipse centered at the origin . . Yz a lme w1th slope 
Yr
(C) a circle with center
(0, y1 ) and radius 1
(D) a line with slope y  y1 2 (E) a parabola with vertex (1, y ) 2
14.
The linear transformation T: R2 � R2 that maps (1, 2) to (1, 1) and (1, 1) to
(A) (1, 2)
15.
(B)
(1,
0)
(C) (2, 1)
(D)
(2, 1)
(0, 1)
to (2, 1) will map
(E) (1, 1)
Let T: R5 � R3 be a linear transformation whose kernel is a threedimensional subspace of R5. The set ! T(x): x E R5) is
(A)
(B )
the trivial subspace
a line through the origin
(C) a plane through the origin (D) all of R3
(E) Cannot be determined from the information given
16.
Define linear operators 5 and T on the xyplane (R2) as follows: 5 rotates each vector 90° coun terclockwise, and T reflects each vector through the yaxis. If 5T and T5 denote the compositions 5 o T and T o 5, respectively, and I is the identity map, which of the following is true?
(A)
5T = I
(B)
5T = I
(C) T5 = I
(D) 5T = T5
(E) 5T = T5
L I N EAR ALG E B RA + 2 1 7
17.
Choose a nonzero vector v = (a, b, c )T in R3 and define a linear operator T: R3 7 R3 by the equation T(x) = v x x, the cross product of v and x. Then T(x) = Ax for every x in R3 if A =
(B) [ �
[ � � 0:] ( D) [� � :]
(A)
18.
b
a
b
a
b
� a
:]1
(c)
[�b � 0:] a
1
If A is an invertible matrix with an eigenvalue of 3 corresponding to the eigenvector x, which of the following statements must be true?
(A) The matrix A 1 has an eigenvalue of
}
corresponding to the eigenvector x.
( B) The matrix A 1 has an eigenvalue of } corresponding to the eigenvector whose entries are the reciprocals of the entries of x.
(C) The matrix N has an eigenvalue of 3 corresponding to the eigenvector x.
(D) The matrix N has an eigenvalue of 6 corresponding to the eigenvector 2x.
(E) The matrix N has an eigenvalue of 3 corresponding to the eigenvector 3x .
19.
The eigenvalues of the matrix
are 4 and b 
1. Find b .
(A) 2
20.
( B)
3
A= G �) (C)
4
( D) 5
(E) 6
(D)
(E) 1 + i
A= (2i2 2; i)
The complex matrix
has which one of the following as an eigenvalue?
(A)
1
(B ) 3
2 1 8 + CRA C K I N G T H E G R E MAT H EMAT I C S S U BJ ECT TEST
(C)
7
Number Theory and Abstract Algebra
I NTRODUCTION In the first part of this chapter, we'll review some fundamental ideas in number theory; you'll need these
not only for the few purely numbertheoretic questions you might see on the GRE, but we'll also use these ideas in many of the examples in Part B, Abstract Algebra. Since the abstract algebra on the GRE Math Subject Test concentrates on group theory, the bulk of this chapter will provide a review of groups and their structure. You'll also be required to know the definitions and fundamental properties of rings and fields, so we'll go over these as well . The area of abstract algebra is usually math majors' first contact with a subject in which the careful proof of theorems is an important part of the course . However, we won't bother with proofs of the theorems; knowing how to use them is much more important than is their derivation. Part B lists the many definitions and theorems that you'll need to know (and there are more than a few!); they're illustrated by an unusually large number of examples.
219
PART A: N UMBER THEORY DIVI S I B I LITY An important subdivision of number theory is the study of the divisibility properties of integers. Let a and
!!_
c such that ac = bthen we say that a divides bor, equivalently, that b is divisible by aand write alb. For example 4 j 8 and  3 1 1 2, but 4 does not divide 9 and 3 does not divide 13. If a divides b, then a is said to be a divisor or factor of b, and b is a multiple of a. There are a few wellknown definitions that are based on divisibility: An integer n is said to be even if 2 divides n; otherwise, n is odd. Furthermore, an integer greater than 1 whose only positive divisors are 1 and itself is called a prime number; and an integer greater than 1 that's not prime is said to be composite. Prime numbers hold a prominent place in number theory, and b be integers (where a :t 0); if
a
is an integerthat is, if there exists an integer
we'll have more to say about them a little later. There are a few ways of telling whether a given integer which are undoubtedly very familiar.
n is divisible by another integer m, some of
An integer n is divisible: •
by 2, if and only if the units digit of n is 0, 2, 4, 6, or 8.
•
by 3, if and only if the sum of the digits of n is itself divisible by 3.
•
by 4, if and only if the number formed by the last two digits of n is divisible by 4.
•
by 5, if and only if the units digit of n is 0 or 5.
•
by 6, if and only if n is divisible by both 2 and 3.
•
by 8, if and only if the number formed by the last three digits of n is divisible by 8.
•
by 9, if and only if the sum of the digits of n is itself divisible by
•
by 10, if and only if the units digit of n is 0.
For example, 1358 is divisible by 2, but 2467 is not. For example, 28554 is divisible by 3 since 2 + 8 + 5 + 5 + 4 = 24 is divisible by 3. For example, 139756 is divisible by 4 because 56 is divisible by 4. For example, 4085 is divisible by 5, but 5804 is not.
For example, 6498 is divisible by 6 since it's even and 6 + 4 + 9 + 8 = 27 is divisible by 3.
9.
For example, 4895064 is divisible by 8 since 064 = 64 is divisible by 8.
For example, 4895064 is divisible by 9 since 4 + 8 + 9 + 5 + 0 + 6 + 4 = 36 is divisible by 9.
For example, 567890 is divisible by 0, but 98765 is not.
There is no simple test for divisibility by 7.
The rule for divisibility by 6 illustrates a basic result about divisibility in general: If n is divisible by a and
b, where a and b are distinct primes, then n is divisible by ab.
2 2 0 + CRAC K I N G T H E G R E MAT H EMATICS S U B J ECT T EST
Thus, if n is divisible by both 2 and 3, then n is divisible by 2 3 = 6. This result is true only if a and b have no common divisors (other than 1). For example, 20 is divisible by both 2 and 4, but 20 is not divisible by 2 4 = 8. The result quoted above can be generalized to any number of prime divisors (not just two), as well as to divisors that are relatively prime. Two nonzero integers are said to be relatively prime if the only positive divisor they have in common is 1 . So, for example, 14 and 15 are relatively prime, but 14 and 16 are not. The generalized version of the result above can be expressed like this: o
o
If n is divisible by a, b, c, . . . , and these divisors are (pairwise) relatively prime, then n is divisible by the product of these divisors.
Example 6.1 Solution:
Is 9876543210 divisible by 30? Since 30 = 3 10, and 3 and 10 are relatively prime, n is divisible by 30 if and only if it's divisible by 3 and by 10. The integer 9876543210 is divisible by 10 since it ends in a zero. Furthermore, since 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 45, which is divisible by 3, we can conclude that this integer is divisible by 3, and thus also by 30. o
THE DIVISION ALGORITHM
If a and b are positive integers (and a :t 0), then the division algorithm tells us that we can find unique integers q and r such that b = qa + r, with 0 � r < a. We call q the quotient and r the remainder, when b is divided by a. This simple fact can be used to determine the greatest common divisor of two integers, in a process called the Euclidean algorithm. We'll return to this after the next section.
PRIMES As we've said, a prime number has no positive divisors except 1 and itself, and the list of prime numbers begins: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, . . . . The sequence of primes is infinite and rather mysterious. For ex ample, there are arbitrarily large gaps in the sequence (for every integer k, we can find k consecutive integers all of which are composite), so we might suspect that primes are rather scarce. Nevertheless, for every integer k > 1, there's always at least one prime between k and 2k, and the series
where pk is the kth prime, diverges. There's no formula for generating all the primes, but it is possible to estimate how many integers, less than or equal to a number x, are prime. Let 1t(x) denote the number of primes that are less than or equal to x; then the prime number theorem says that 1t(x) "' x/log x, and the approximation gets better as x gets larger.
Every integer n greater than 1 can be expressed as a unique product of primes. If n is already a prime, then the "product" simply consists of one factorthe number itself. In addition, we don't consider a rear rangement of the prime factors as a different product; this is a statement of the fundamental theorem of arithmetic. For example, the prime factorization of 3960 is 23 32 5 11. o
o
o
THE GREATEST CoMMON DIVISOR AND THE LEAST CoMMON MuLTIPLE
To find the greatest common divisor (gcd) and the least common multiple (lcm) of two integers, a and b (both greater than 1), first express them in terms of their prime factorizations: and
N U M B E R TH E O R Y A N D A B STRACT ALG E B RA +
221
Notice that we've used the same list of primesp1 through pkin both factorizations; this can always be done for any two integers a and b if we allow for zero exponents. Now, for each i from 1 to k, let:
and Then the greatest common divisor and least common multiple of a and b are given by:
and It's also helpful to know that the equation gcd(a, b) · lcm(a, b) = ab is true for �my two positive integers, a and b.
Example 6.2 Solution:
What are the gcd and lcm of 3360 and 3960? Since 2 3360 = 25 • 3 • 5 • 7 • 11° and 3960 = 23 • 3 • 5 • 7D • 11 we can conclude that gcd(3360, 3960) = 23 • 3 • 5 • 70 • 110 = 120, and lcm(3360, 3960) = 25 • 32 • 5 • 7 • 11 = 110880.
THE EUCLIDEAN ALGORITHM The Euclidean algorithm is a stepwise method to figure out the greatest common divisor of two integers. It's particularly convenient if the integers are large or if finding their prime factorizations is difficult. This algorithm makes repeated use of the division algorithm, and the following schematicin which we find the greatest common divisor of 2002 and 315illustrates the process:
1 20021 = 6 · 13151+ 11 2
/1 1 2/ + 91 // 1 12 = 1 9 1 + 21 // 9 1 = 4 . 21 + 7 / 2 1 = 3 7 + IQJ � gcd
3 15 = 2
•
•
•

this signals that the process STOPS
Notice that each equation after the first one uses the remainder from the preceding equation as the new divisor. The last divisor that gives a remainder of 0 is the greatest common divisor of the two original numbers. Therefore, gcd(2002, 315) = 7.
Example 6.3
What are the greatest common divisor and the least common multiple of 1 020 and 3234?
2 2 2 + C R AC K I N G T H E G R E MATH EMAT I CS S U BJ E CT T EST
'
Solution:
First, we'll use the Euclidean algorithm to find gcd(1020, 3234): 3234 = 3 . 1020 + 174 1020 = 5 . 174 + 150 174 = 1 • 150
+ 24 150 = 6 . 24 + 6 24 = 4 6 + [QJ •

r
this signals that the process
STOPS
gcd
This tells us that gcd(1020, 3234) = 6. Now, using the fact that the product of the greatest common divisor and the least common multiple of two integers is always equal to the product of the number themselves, we find that: lcm( 1 020 ' 323 4)
=
1 020 • 3234 1 020 • 3234 = gcd(1020, 3234) 6
= 549 780
This example shows that it's possible to determine the greatest common divisor and the least common multiple of two integers even if we can't figure out the integers' prime fac torizations.
THE DIOPHANTINE EouAnoN ax + by = c An equation that involves one or more variables is called a Diophantine equation if the solutions sought are restricted to a particular set of numbers, usually integers. To look at a very simple example, the equation x + 2y = 3 has infinitely many real solutions; but if we restrict x and y to be positive integers, the equation has only one solution, (x, y) = (1, 1). If x and y simply need to be integers (not necessarily positive), then x + 2y = 3 has infinitely many solutions; and they're all of the form x = 1 + 2t, y = 1 t, for any integer t. The most famous Diophantine equation of all time is x" + y" = z"; Fermat's last theorem conjectured (without proof) that, if n is any integer greater than 2, then this equation has no solutions in nonzero integers x, y, and z. For over 350 years, proving (or disproving) this conjecture was one of the most outstanding unsolved problems in all of mathematics. The recent (1994) solutiona proof in the affirmativeby Andrew Wiles was frontpage news around the world. But here our goal will be far more modest; we will look for a way to solve the linear Diophantine equation ax + by = c in integers, where a, b, and c are themselves integers (with ab :t 0). First of all, the equation ax + by = c will have no integral solutions if gcd(a, b) does not divide c. How ever, if gcd(a, b) does divide c, then the equation will have infinitely many integer solutions; if we can find even onelet's call it (x1 , y1 )then all the solutions are given by the formulas 
X = XI + (b I d)t
and
y=y 1
(a/d)t
where d = gcd(a, b), and t is any integer. The method for finding a solution follows directly from the Eu clidean algorithm. By working backward with the list of equations generated by the algorithm, we can always write d, the greatest common divisor of a and b, as a linear combination of a and b; that is, we can always find integers, x0 and y0, such that:
N U M B E R TH E O R Y A N D A B STRACT A l G E B R A + 2 2 3
If c
= 0, then all the integer solutions of ax + by = c are given by x = x0 + bt and y = Yo  at. c
can multiply the equation above by
d
(which will be an integer because
equation to have any solutions at all), and get
If c :t
0, we
d must divide c in order for the
a (� x0 ) + b ( � y 0 ) = c � x,
�
y,
and the formulas given above will give us all the integer solutions. Let's illustrate the method by finding all integer solutions to the equation
15x + 49y = 8
1,
Since gcd(15, 49) = this equation will indeed have solutions. (In general, if a and b are relatively prime, then the equation ax + by = c will have integer solutions.) The Euclidean algorithm would produce the following list of equations in search of the greatest common divisor of 15 and 49:
49 = 3 · 15 + 4 15=3 . 4+3 4 = 1 · 3+ 1 3=3 · 1+0 So, working from the secondlinefromthebottom upwards, we can write:
1=4  3  1 = 4 (15  3 4) 1 = 4 . 4  15 = 4 • ( 49  3  15)  15 = 15 . (13)+ 49 . (4) •
•
We work our way back up by rewriting each step of the Euclidean algorithm above and substituting it into our expression. Therefore, the greatest common divisor of and 49 can be written as a linear combination (with integer coefficients) of and 49:
15 15 . (13) + 49 . (4) = 1
15
Then, multiplying both sides by 8 gives us
15 . (104) + 49 . (32) = 8 so one solution of our original equation is x1 = 104 and y1 = 32. Therefore, all the solutions of the equa
tion are given by the formulas
x = 104 + 49t for any integer
t.
2 2 4 + C R AC K I N G TH E G R E MATH EMATICS S U BJ ECT T EST
and
y = 32  15t
CONG R U ENCES
a
Let a, b, and n be integers. If n :1 0, we say that a is congruent to b modulo n if  b is divisible by n, which is written a = b (mod n). For example, 9 = 1 (mod 4) and 15 = 3 (mod 6), but 21 =I= 3 (mod 5). Congruences (mod n) depend on the remainder that's obtained in the division algorithm when the divi sor is n. For example, since 987 = 197 • 5 + 2, we know that 987 = 2 (mod 5). Of course, it's also true that 987 = 7 (mod 5) and that 987 = 3 (mod 5), etc., but the idea of a congruence (mod n) is that we throw out any multiple of n and keep only what's left over. The list below gives several important rules that we can use when working with congruences. The first three of these are similar to the corresponding rules for equalities . Unless otherwise stated, all variables stand for integers, with n restricted to be a positive integer.
1.
If a = b (mod n ) and
2.
If a = b (mod
b e (mod n), then a = e (mod n). =
n), then for any e,
a ± e = b ± e (mod
n)
ae = be (mod n) 3.
If a 1 = b1 (mod a1 ±
a2 = b1
n) and a2 = b2 (mod n), then
±
b2 (mod n)
a 1a2 = b1 b (mod
2
4. 5.
n)
For any positive integer e, the statement a = b (mod b + n, b + 2n, . . . , b + (e  1)n (mod en). If ab = ae (mod
n) is equivalent to the congruences a = b,
n), then
b = e (mod n) if gcd (a, n) = 1 n b = e (mod d ) if d = gcd(a, n) > 1
6. Fermat's Little theorem. If p is a prime and a is an integer, then 1 ap = 1 (mod
p), if p does not divide a
aP = a (mod p), for any integer
a
The rules above can be used to reduce a congruence and solve congruence equations. For example, let's 1 figure out the remainder that results when 31 0 is divided by 7. We could actually compute 3 0, divide by 7, and see what the remainder is, but that's quite a bit of work: Congruences will make this task a lot easier. We're asked to find the value of b such that 31 0 = b (mod 7), with 0 :::; b 7. First, we notice that 32 is congruent to 2 (mod 7). Since 31 0 = (32 )5, we can apply the third 1 equation in rule 3 above and conclude that 3 0 = (32)5 = 25 (mod 7). Now it's easy to figure out that 25 = 32 = 4 1 (mod 7). Because 3 10 = 4 (mod 7), the remainder when 3 0 is divided by 7 is 4.
=>
U3 that consists of the cube roots of unity;
(z  1)(z 2 + z + 1 ) = 0 z = 1 or z = 1 ± N =
2 3 6 + C R AC K I N G T H E G R E MATH EMAT I C S S U B J ECT T EST
2
t±i�
{
Therefore, U3 = 1,  t + i Although this group contains imaginary numbers, and the t i group operation is multiplication, it's isomorphic to the group, Z3, of integers, which has the group op eration of addition (modulo 3): U3 = Z3. Both of these groups have order 3, and we showed earlier that every group of order 3 is isomorphic to Z3. Since the group structure of U3 is identical to that of Z3' we or 0)2 =  t  i (because z3 is cyclic know that u3 is cyclic, and is generated by either 0) 1 =  t + i and generated by either one of its two nonidentity elements). Furthermore, without even working out the multiplication, we know that w� = w2 and w; = W1 because, in the group Z3' the corresponding equations are 1 EB 1 = 2 and 2 EB 2 = 1.
1} .
1,
1
1
THE CLASSI F I CATION OF F I N ITE ABE LIAN GROUPS In this section, we'll learn how to describe every finite Abelian group. First, let groups; on the set
G1 x G2 = {(a, b): a E G1
and
(G1, \) and (G2, *z) be
b E G2}
define a binary operation, *, by the equation
(a1, a2) * (b1, bz) = (a1 \ b1, a2 *z b2)
Then (G 1 x G2, *) is a group, called the direct product of the groups G1 and G2. If G1 and G2 are finite groups, and G1 has order m and G2 order n, then G1 x G2 has order mn. If G1 and G2 are both Abelian, then the notation G1 EB G2 is sometimes used in place of G1 x G2, and the resulting (Abelian) group is called the direct sum of the groups G1 and G2• Both the definition of a direct product (or direct sum) and the result about the order of the direct product can be generalized to any number of groups (not just two). For example, let's consider the direct sum Z2 EB Z4• It consists of the following 2 x 4 = 8 elements:
(0, 0), (0, 1), (0, 2), (0, 3), and (1, 0), (1, 1), (1, 2), (1, 3)
Here's a question: Both Z2 and Z4 are cyclic; is their direct sum, Z2 EB Z4, also cyclic? Since Z2 EB Z4 has order 8, if this group were cyclic, there would have to be an element (a, b) with this property: The smallest integer n such that n(a , b) = (0, 0) is n = 8. However, we can see that, here, this isn't the case. No matter which of the eight elements (a, b) we choose, we'll always have 4(a, b) = (0, 0); that is, the order of every element of this group is no greater than 4. But what about the direct sum Z2 EB Z3? It consists of the following 2 x 3 = 6 elements:
(0, 0), (0, 1) , (0, 2), and I
(1 , 0), (1 , 1),
(1, 2 )
Is Z2 EB Z3 cyclic? Yes; in fact, it's generated by the element (1, 1), as we can check: 1 ( 1 , 1)
=
(1, 1)
2 ( 1 , 1) = (0, 2 ) 3(1, 1) = (1, 0)
4(1 , 1) = (0, 1) 5(1, 1) = (1, 2)
6(1 , 1) = (0, 0)
N U M B E R TH E O R Y A N D A B STRACT A L G E B RA + 2 3 7
Why was Z El1 Z3 cyclic, but not Z2 El1 Z/ It is because 2 and 3 are relatively prime, but 2 and 4 are not. 2 These examples illustrate the following theorem: The direct sum z, El1 zn is cyclic if and only if gcd(m, n) = 1. If this is the case, then, since Zm El1 ZrE has order mn, Zm El1 Zn is isomorphic to Zmn .
We can generalize this result to any number of factors:
The direct sum Z mt EB Z m2 EB EB Z is cyclic if and only if gcd(m ., m.) = 1 for every mk t 1 distinct pair m.I and m1.. If this is the case, then Z ml EB Z m2 EB . · · EB Zmk is isomorphic to Z mtm mk · · ·
2···
For example, z8 ffi z3 El1 z25 is cyclic and isomorphic to z8 . 3 . 25 = Z6\IJ, but the group z8 ffi z3 El1 zlO is not cyclic, since gcd(8, 10) :;:. 1. We're now ready to state the results o f this section: Every finite Abelian group G is isomorphic to a direct sum of the form
pi
ki
where the are (not necessarily distinct) primes and the are (not necessarily distinct) for a given representation of G, positive integers. The collection of prime powers, are known as the elementary divisors of G .
(p/i,
We can also write every finite Abelian group in another way: Every finite Abelian group G is isomorphic to a direct sum of the form
�
where m 1 2, m 1 divides m2, m 2 divides m , , and m 1 _ 1 divides m r The integers m 1 3 , m 1 is unique, and these integers through m1 are not necessarily distinct, but the list m 1, are called the invariant factors of G. •
•
•
•
Example 6.7 Solution:
•
•
How many structurally distinct (that is, mutually nonisomorphic) Abelian groups are there of order 600? Our first step is to find the prime factorization of 600:
600 = 23 • 3 • 52 Therefore, if G is an Abelian group of order 600, there are six possible collections of elemen tary divisors:
1) 2, 2, 2, 3, 5, 5 2) 2, 2, 2, 3, sz 3) 2, 22, 3, 5, 5 4) 2, 22, 3, 52 5) 23, 3, 5, 5 6) 23, 3, 52
2 3 8 + CRAC K I N G T H E G R E MATH EMATICS S U BJ ECT T EST
Each of these lists of elementary divisors gives rise to an Abelian group, as follows:
1)
2)
3)
4) 5)
6)
2, 2, 2, 3, 5, 5 7 Z2 (B Z2 (B Z2 (B Z3 (B Z5 (B Z5 2, 2, 2, 3, 52 7 Z2 (B Z2 (B Z2 (B Z3 (B Z25 2, 22, 3, 5, 5 7 Z2 (B Z4 (B Z3 (B Z5 (B Z5 2, 22, 3, 52 7 Z2 (B Z4 (B Z3 (B Z25 23, 3, 5, 5 7 Z8 (B Z3 (B Z5 (B Z5 23, 3, 52 7 Z8 (B Z3 (B Z25
Therefore, there are six different Abelian groups (up to isomorphism) of order 600. We can also figure out the invariant factors of each of these six groups. In each case, we must express the group in the form Zm1 (B Z m2 (B .. (B Z m1 , where m 1 � 2 and m.1 divides m 1+ 1 • Notice the method we use to construct the invariant factors from the elementary divisors: .
2, 2, 3, 5, 5 7 2
2
2
2, 2, 2, 3, 52 7 2
2
3
3 5 1
1
5 1
22
2, 22, 3, 52 7 2
1
1
z1o (B z6o 23
23, 3, 5, 5 7 5
1
1
22 3
5
1
52
z2 (B z2 (B z 1so
3 5
1
1
z2 (B z 1o (B z3o 2, 22, 3, 5, 5 7 2
2
52 1
z2 (B z3oo
23, 3, 52 7
3 5 1
Zs (B z 12o
23 3 2 5 1
z600
Therefore, the six mutually nonisomorphic Abelian groups of order 600 can be expressed as follows, where in each case, we've written the representation of G in terms of the elementary divisors on the left, and the corresponding representation in terms of the invariant factors on the right:
1) Z2 (B Z2 (B Z2 (B Z3 (B Z5 (B Z5 = Z2 (B Z10 ffi Z 30
2) Z2 (B Z2 (B Z2 (B Z3 (B Z25 = Z2 ffi Z2 (B Z 150 3) Z 2 (B Z4 (B Z 3 (B Z5 (B Z5 = Z 10 (B Z60 4) z2 (B z4 (B z3 (B z25 = z2 (B z300 5) z s (B z3 (B zs (B zs = zs (B z 120 6) zs (B z3 (B z2s = z6oo
N U M B E R TH E O RY A N D A B STRACT A L G E B RA + 2 3 9
Example 6.8 Solution:
Find the order of the element (1, 6, 25) in the group Z2 EEl Z 0 EEl Z:Mr
1
By definition, the order of an element x is the smallest positive integer m such that x"' = e. In this case, since we're using additive notation for our Abelian groups, we want the smallest positive integer m such that m(1, 6, 25) = (0, 0, 0). Clearly, the order of 1 in Z2 is m 1 = 2. Next, the order of 6 in Z 10 is 5, because 5 is the smallest positive integer, m2, such that 6 m2 is a multiple of 10. Finally, the order of 25 in Z30 is 6, because 6 is the smallest positive integer, m 3, such that 25 • m3 is a multiple of 30. (It's worth noting that, in gen eral, if 1 � k < n, then the order of k in Z" is n / [gcd(k, n)], or, equivalently, [lcm(k, n)] /k.) Therefore, in order for m(l, 6, 25) to equal (0, 0, 0), m must be a multiple of m 1 = 2, of m2 = 5, and of m3 = 6. Since we want the smallest positive integer m that has this property, we want the least common multiple (lcm) of 2, 5, and 6; this is m = 30. •
GROUP HOMOMORPH ISMS Let (G, • ) and (G', * ) be groups. A function �: G 7 G' with the property that •
�(a) * �(b) for all elements a and b in G, is called a group homomorphism. A homomorphism provides a direct relationship between the groups' structures. Why? Because we can operate on a and b in G (that is, form a b) and carry the result over to G' (in other words, apply � to a b), or we could carry a and b over to G' first and then operate on them there; the result will be the same. This is pr�cisely what the equation that defines a homomorphism says. A homomorphism that's onetoone (injective) is called a monomorphism; if it's onto (surjective), it's called an epimorphism; and if it's onetoone and onto (bijective), it's called an isomorphism. As we discussed earlier, two groups are said to be isomorphic if they're structurally �(a
b)
=
•
•
identical. This is true if and only if there exists a bijective homomorphism, an isomorphism, between the groups. Finally, a homomorphism from a group to itself is called an endomorphism, and an isomorphism from a group to itself is called an automorphism. The following is a list of some of the most important properties of homomorphisms. Each of the facts below applies to a homomorphism of groups, �: G 7 G':
1.
If e is the identity in G, then �(e) is the identity in G'.
2. If g E G has finite order m, then �(g) E G' also has order m.
3. If a1 is the inverse of a in G, then �(a1 ) is the inverse of �(a) in G'.
4. If H is a subgroup of G, then �(H) is a subgroup of G', where:
(H) = { ( h): h E H} 5. If G is finite, then the order of �(G) divides the order of G; if G' is finite, then the order of �(G) also divides the order of G'.
6.
If H' is a subgroup of G', then
�1 (H') is a subgroup of G, where �1 (H') = {h E G: �(h) E H'}
2 4 0 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ E CT TEST
If �: G 7 G' is a homomorphism of groups, then (e'l, where e' is the identity in G', is a sub groupthe trivial subgroupof G'. By property 6, the inverse image of (e'J is a subgroup of G. This subgroup is given a name: It's called the kernel of �, denoted by ker � : ker <j> = {g E
G: <j>(g) = e'}
A homomorphism is a monomorphism if and only if its kernel is trivial. Now, let's look at some examples on the following page. Let U" be the multiplicative group of the nth roots of unity; this group is cyclic of 2 2 order n and is generated by ro = cos "1t + i sin ; . If we define �: (Z, +) 7 U" by the equation <j>(a) = ro•, show that � is a homomorphism. Is � a monomorphism? an epimorphism? an isomorphism?
Example 6.9
·
Solution:
...J
· ··
The function � is a homomorphism, since: �(a + b) = wa + l• = w " wb = �(a)�(b ) The kernel of this homomorphism is the subgroup: ker � = {a E Z: w" = 1 } = { rnn: m E Z }
The function � is an epimorphism, but not a monomorphism (and thus not an isomor phism), since it's not onetoone (its kernel is not trivial). We also know that � is not an isomorphism, because an infinite group cannot be isomorphic to a finite group.
Example 6.10
Solution:
Let GL(n, R) be the multiplicative group of invertible n by n matrices with real entries. Show that the function �: GL(n, R) 7 (R*, x ) ; given by �(A) = detA, is a homomorphism, and find ker�.
The function � is a homomorphism because:
<j>(AB) = det(AB) = (det A)(det B) = <j>(A)<j>(B) The kernel of this homomorphism is the subgroup: ker � = { A E GL(n, R ): det A = 1 } = SL(n, R )
Example 6.11 Solution:
If p is a prime and �: ZP 7 G' is a nontrivial homomorphism (that is, �(n) :t e' for every n :t 1 in ZP), show that � must be a monomorphism.
The kernel of a homomorphism is always a subgroup, so ker� must be a subgroup of ZP . Therefore, by Lagrange's theorem, the order of ker� must be a divisor of p. Because p is a prime, the order of ker� must be 1 or p. We've already ruled out the possibility that ker� has order p, by stating that � is nontrivial. Therefore, the order of ker� is 1, so ker� must be the trivial subgroup of Z1,; this shows that � is onetoone.
N U M B E R TH E O R Y A N D A B STRACT A L G E B RA + 24 1
Example 6.12 Solution:
Is there a nontrivial homomorphism $:
(Z8, +) t (Z3, +)?
If the answer is yes, then the homomorphic image of Z8 must be a subgroup of Z3 • Since
is nontrivial, the order of $(Z8) is not 1 . By Lagrange's theorem, then, the order of $(Z8) must be 3. However, according to property 5 listed above, the order of $(Z8) must divide the order of Z8. This is a contradiction, since 3 does not divide 8. Therefore, the answer to the question is no.
Example 6.13 Solution:
Let $: (Z 4, +) t (Z8, +) be a homomorphism such that $(1) toone?
=
6. Find (3). Is $ one
First, it's worth noting that we know the image of every n in Z4 since we're told what $ does to 1 . Once we know what a homomorphism of a cyclic group does to a gen erator of the group, we know what it does to every element in the group. In this case, $(3) = $(3 • 1) = $(1 ) = 3 • 6 = 2. Now, by definition,
{
ker <j> = n E Z4 : <j>(n) = 0}
Since $(n) = 6n, ker$ consists of those elements n in Z4 such that 6n is a multiple of 8. Only n = 0 satisfies this, so ker$ is trivial, which implies that is onetoone.
Example 6.14 Solution:
Let <j>: G t G be a function such that ${g) = g1 for every g in G. Show that G is Abelian if and only if $ is an endomorphism.
First, let's assume that G is Abelian. Since the inverse of xy is y1r1 in any group (Abelian or not), we know that $(xy) = (xy)1 = y1r1 • Now, because G is Abelian, y1 r1 = rJy1 , which is equal to $(x)$(y). Since $(xy) = $(x)$(y), $ is a homomorphism of G to itself. To complete the solution, let's now assume that is a homomorphism. Then $(xy) = <j>(x)<j>(y), so (xy)1 = rJy1, which implies y1r1 = r1y1. Taking the inverse of both sides of this last equation, we get (y1r1 )1 = (r1y1 )1 , which is equivalent to xy = yx. This shows that G is Abelian.
Example 6.15
Is the group (R, +) isomorphic to the group (R+, .
Solution:
x )? .
Define a function <j>: (R, +) t (R+, x ) by the equation $(x) = ex, where now e denotes the base of the natural logarithm. We'll show that is an isomorphism. First, $(x + y)
= ex + Y = exeY = $(x)<j>(y)
so $ is a homomorphism. Next, <j>(X) = 1
=>
ex = 1
=> X = 0
=>
ker <j>
= {0}
which shows that is onetoone. Finally, for every x in R +, we have <j>(log x)
= e'og x = x
so is onto. Since is an isomorphism, we have (R, +) = (R+, x ) .
242 +
CRAC K I N G T H E G R E MAT H EMAT I C S S U B J ECT T EST
A subgroup N of G is said to be a normal subgroup if xnx1 E N for every n in N and every x in G. If $: G � G' is a homomorphism of groups, show that ker$ is a normal subgroup of G . ��. Solution: We already know that ken� is a subgroup of G; we're asked to show that it's actually a normal subgroup. Let g be any element in ker�. Since �(g) = e' and �(.r1) = [�(x)]1, the fol lowing is true for every x in G:
Example 6.16
IL_
_ _
. · ····
·

 ·
�.
� ( xgx 1 ) = �( x) • �( g ) • �( x1 ) = � ( x) • e ' • � ( x1 ) = �( x) • [ � ( x )r1 = e '
Therefore, xg.r1 E ker�. Since we've shown that xg.r1 E ker� for every g in ker� and every x in G, we conclude that ker� is, by definition, a normal subgroup of G.
Example 6.17 Solution:
Can an infinite group be isomorphic to one of its proper subgroups?
Consider the infinite, cyclic group ( Z, + ) . For every integer n � 2, the group
nZ
=
{ nk: k E z}
is an infinite, proper subgroup of Z. We can define a function �,: Z � �, (k) = n k, and show that ��� is actually an isomorphism. First, because
nZ by the equation
��� ( k + k2 ) = n (k1 + k2 ) = nk +r zk 2 = ��� (k1 ) + ��� (k2 ) , 1 1 we know that ��� is a homomorphism. Next,
{
ker �, = { k E Z : nk = O} = O } shows that ��� is onetoone. And finally, since every element of nZ is of the form nk for some integer k, and �, (k) = nk, it's clear that ��� is onto as well. Therefore, (Z, +) = ( nZ, +).
Example 6.18
Let G be a group. For a fixed element a in G, define a function �.: G � G by the equation �.(g) = aga1 • (a) Show that �. is an automorphism. (It's called the inner automorphism
induced by a.)
Let Aut(G) denote the collection of all automorphisms of G. According to the operation of function composition, (Aut( G), o) is a group. Show that the set of all inner automorphisrns of G, Inn(G) = {�.: a E G}, is a subgroup of Aut( G). (c) Describe Inn(G) if G is Abelian.
(b)
Solution: (a)
First, the following equation establishes that �. is a homomorphism:
Next, we'll prove that �. is onetoone. We can do this in two ways; we can either notice that �. ( g 1 ) = �. (g 2 )
� ag1 a1 = ag2a1
�
a\ag1 a1 )a = a1 (ag2a1 )a � g1 = g2
N U M B E R T H E O R Y A N D A B STRACT A l G E B RA + 243
or we can show that ker�a = le): .
g E ker �.
¢=>
<J>. (g ) = e
¢=>
aga1 = e
ag = ea
¢=>
¢=>
g=e
Finally, we need to show that �. is onto G. Since G is a group, we know that a1ga E G for every a and g in G. Now, for every g in G, notice that the image of a1ga is g:
. ( a1 ga) = a (a1 ga)a1 = ( aa1 ) g ( aa1 ) = ege = g
Therefore, . is onto, so we're able to conclude that it's an isomorphism of G to itself; that is, it's an automorphism of G. (b) To show that Inn(G) is a subgroup of Aut(G), we need to show that Inn(G) is closed under the group operation, it contains the identity, and it contains the inverse of every element in Inn( G). Let's first establish closure. Let . and <j>b be elements of Inn(G) then, since (. o �Xg) = . (� (g)) = . (bgb1 ) = a(bgb1 )a1 =
(ab)g(b1 a1 ) = (ab)g(abf1 = ab (g)
is true for every g in G, we know that �. o � = �a so Inn(G) is closed under function composition. b b Next, the identity of Aut( G) is clearly the identity map, id: G � G, where id(g) = g for all g in G. But if e is the identity of G, then the inner automorphism induced by e, , is the identity map, because
e ( g ) = ege1 = ege = g
for every g in G. Therefore, Inn(G) contains the identity. Finally, we'll show that if a1 is the inverse of a, then the map �.' is the inverse of �.. The calculation
aga1 = g '
¢::>
ga1 = a 1 g ' ¢::> g = a 1 g 'a ¢::> g = a 1 g '( a1 r1
means that �.(g) = g' g = .,(g'), so .' is indeed the inverse of �.. Thus, Inn( G) � Aut( G). (c) If G is Abelian, then for any a in G, �.(g) = aga1 = aa1g = g for every g in G; so, for every a in G, the inner automorphism �. is just the identity map. Therefore, if G is Abelian, Inn(G) is the trivial subgroup of Aut(G). R I NGS A group is a special binary structure with one binary operation; a ring is a special binary structure with two binary operations. To be more precise, a set R, together with two binary operations (we'll use addition, +, and multiplication, • ), is called a ring if the following conditions are satisfied:
(R, +) is an Abelian group;
(R, • ) is a semigroup (that is, multiplication is associative);
the distributive laws hold; that is, for every a, b, and c in R, we have: •
•
a (b + c) = a b + a
•
c
and
(a + b) c = a •
•
c
+b
•
c
This last requirement says that the two binary operations in the ring are compatible. A point of notation: We don't usually write the product of a and b as a • b; we just write ab. 2 4 4 + C R AC K I N G T H E G R E MATH EMAT I C S S U B J ECT T EST
If the multiplicative semigroup (R, • ) is a monoidthat is, if there's a multiplicative identitythen R is called a ring with unity. The identity of the additive abelian group (R, +) is usually denoted by 0 and, in a ring with unity, the identity of the multiplicative monoid (R, • ) is usually denoted by 1 . We typically require that these identities be distinct; that is, 0 7: 1 . (Although this looks like an obvious statement con cerning the numbers 0 and 1, remember that in a general ring, the additive identity is not necessarily the number 0, and the multiplicative identity is not necessarily the number 1.) If the operation of multiplication is commutative, then we call R a commutative ring. (The term abelian is generally not used to describe a ring in which multiplication is commutative.) If S is a subset of R, and S satisfies the ring requirements with the same operations as R, then we say that (5, +, • ) is a subring of (R, +, • ) . As is the case with groups, it's common to refer to a ring (R, +, • ) simply by R, if the two binary operations are understood. Let R be a ring. The smallest positive integer n such that na = 0 for every a in R (if such an n exists) is called the characteristic of the ring, and we write char R = n. (Remember that if n is a positive inte ger, n a means a + a + . . + a, where the sum contains exactly n summands.) If no such n exists (as, for example, in the infinite rings Z, Q, R, and C), then we say that R has characteristic zero. In the case that char R > 0, we don't need to check that na = 0 for every a in R; it is sufficient to find the smallest positive integer n such that n • 1 = 0. Here are some examples: .
1 . (Z, +, • ) is the simplest example of a ring; it's a commutative ring with unity, called the ring of integers. We mentioned earlier that (Z, • ) is not a group because most elements have no in verse; however, the definition of a ring doesn't require that (R, • ) be a group, only a semigroup. If n is a positive integer and we let nZ = Ink: k E Z), then ( nZ, +, • ) is a subring of Z. Notice that, if n > 1, then nZ is also a commutative ring, but not a ring with unity.
2. The ring of integers is a subring of the rings (Q, +, • ) , (R, +, • ), and (C, +, • ) .
3. Since (Z", + ) is an Abelian group and (Z11, • ) is a semigroup, (Z",
modulo n.
+,
•
) is the ring of integers
4. The set of n by n matrices with entries in R, with the operations of matrix addition and multi plication, (M11(R), +, x), is a ring with unity, a noncommutative ring if n > 1. M"(Q) and M"(Z) are subrings.
{
5. The set R = a + b fi: a, b E Z , together with the usual operations of addition and multiplica tion, is a commutative ring with unity. It's straightforward to check that R is closed under both addition and multiplication, since:
}
(a1 + b1 fi ) + (a2 + b2 fi ) = (a1 + a2 ) + (b1 + b2 ) J2 and
(a1 + b1 J2 )(a2 + b 2 fi ) = (a1 a 2 + 2 b1 b2) + (a1 b2 + a2 b1 ) J2 Notice, however, that the set under multiplication:
C = {a + b fi.: a, b E Z} does not form a ring, since it's not closed
(a1 + b 1 .fi. X a2 + b 2 .fi ) = (a1 a 2) + (a1 b2 + a2 b1 ) .fi + b 1 b2 !4 rt C
N U M B E R TH E O R Y A N D A B STRACT A l G E B RA + 2 4 5
6. With the operations of addition and multiplication in C, the set
Z[i] = { a + bi: a, b E Z and i = /=i}
is a subring of C, called the ring of Gaussian integers.
7. Let R be a ring, and consider the collection of all polynomialsincluding the zero polynomial in the variable (or indeterminate) x with coefficients in R:
R[x] = h + r1 x + r2x2 + . . · + r;, x " : 1j E R}
Then R[x] is also a ring, called the ring of polynomials in x over R. 8. The collection of all functions f: R � R is denoted RR. With the operations of additionwhere (j + g)(x) = f(x) + g(x)and pointwise multiplicationwhere (jg)(x) = f(x)g(x)RR becomes a com mutative ring with unity, called the ring of realvalued functions on R. The additive identity is the constant function 0, and the multiplicative identity is the constant function 1 .
Let R be a ring with unity, and let 5 be a subring with unity of R. Does the unity of S need to be the same as the unity of R?
Example 6.19 Solution:
6,
Let R be the ring Z10 (whose unity is 1) and let 5 be the subset { 0, 2, 4, 8} . It's straight forward to check that 5 satisfies the conditions to be a ring, but notice that the unity of 5 is the element 6, because: 0
•
6
=
0,
2
•
6
=
2,
4
•
6 = 4,
6
•
6
=
6,
and
8
•
6
=
8
Therefore, 5 is a ring with unity 6 that's a subring of the ring R with unity 1 . An element a
of a rin'g R is said to be nilpotent if there's some positive integer, m, such that am = 0. If a and b are nilpotents in a commutative ring R, show that their sum, a + b, must be nilpotent, too. Is this result still true if R is not commutative? Solution:
a and b are nilpotent, there are positive integers, m and n, such that a"' 0 and b" 0. (Notice that this automatically implies that aM = 0 for every integer M � m and that bN 0 for every integer N � n.) Let k 2 max(m, n), and consider the binomial
Since
=
=
=
=
•
expansion (a + b)k. Every term in the resulting polynomial consists of a numerical coef ficient, c;r multiplied by aibi, where i and j are nonnegative integers whose sum, i + j, is always equal to k. (We can write every term in the compact form c;i a; rather than leaving the mixed terms in the form precisely because R is commu tative.) Because of our choice of k, every term of this expansion will equal 0. To see why this is so, let's first answer this question: What if i < m? Then j cannot be less than n, since if it were, we'd have i + j < m + n, which contradicts the fact that i + j = 2 • max(m, n). Since j must be at least equal to n, the term bi will be 0, which makes the entire term a;bi equal to 0. Similarly, if j < n, then it must be true that i � m, so the term aiand thus the term a;biwill be 0. Therefore, for every term in the binomial expansion, we'll have i � m or j � n, so every term will be 0. We've shown that (a + W is 0, so a + b is nilpotent. This result is not necessarily true if R isn't commutative. Consider the ring M2(Z), and take:
bi,
a'' bha''b12 ... a''b",
A=
2 4 6 + CRAC K I N G T H E G R E MATH EMAT I CS S U B J ECT T EST
(�
�)
and
B=(
� �)
Then N = 0 and B2 = 0 (where 0 denotes the zero matrix here), so A and B are both nilpo tent. However, their sum,
A+B = (� �)
is not nilpotent; (A + B) " is never equal to 0, since:
(�
( 0 01 ) �r = �) (� 1
if n is even
if n is odd
Alternatively, we can simply note that A + B is invertible (its determinant isn't zero). If an element is invertible, it can't be nilpotent. [Why not? Well, let's assume that e is invertible in a ring R; then e1 exists and ee 1 = 1. This equation implies (ee1 )" = 1 for every positive integer n, so c"(e1)" = 1. Therefore, c" cannot be 0 for any n; in other words, c can't be nilpotent.]
(i;ample 6.21 Solution:
What's char Z.?
The additive group (Z,, +) is cyclic, with generator 1 . Since Z, contains exactly n ele ments0, 1, 2, . . . , n  1we know that the smallest positive integer m, such that m • 1 = is m = n. Therefore, char ZII = n .
0
RING HoMOMORPHISMs Let (R, +, x) and (R', EB, ®) be rings. A function $: R � R' is called a
following conditions hold for every
a and b in R : <j>(a + b) = <j>(a) EB <j>(b) <j>(a b) = <j>(a) ® <j>(b)
ring homomorphism if both of the
x
Notice that this is similar to the definition of a group homomorphism, except now that we're dealing with rings, we have two binary operations, so $ must preserve both operations. In the list below, we'll give several important facts for any ring homomorphism $: R � R':
1.
{a R : <j>(a) = 0'},
The kernel of a ring homomorphism is the set ker <j> = E where 0' is the additive identity in R'. Just as the kernel of a group homomorphism $: G � G' is always a subgroup of G, the kernel of a ring homomorphism $: R � R' is always a subring of R.
2. The image of R, $(R) 3. The image of this that $(r)
=
{ <j>(r): r
E
R } , is a subring of R'.
0, the additive identity in R, must be 0', the additive identity in R'. It follows from =
$(r) for every r in R, where r is the additive inverse of r.
Now, we'll look at some examples that illustrate the concept of a homomorphism of rings:
1 . The function $: Z � Z , given b y $(k) = the unique integer in Z " that's congruent to k (mod n ) is a ring homomorphism (actually, it's a ring epimorphism, since $ is onto). To show this, let $(k1) = m1 and $(k2) = m 2 • Then it must be true that k1 = a 1 n + m 1 (where is an integer and 0 :::;; m1 < n) and, similarly, k2 = a2n + m 2 (where a2 is an integer and 0 :::;; m2 < n). Adding these two
a1
N U M B E R TH E O R Y A N D A B STRACT A L G E B RA + 247
equations gives k1 + k2 = (a1 + a2)n + (m1 + m2), so �(k1 + k2) is equal to either m 1 + m2 (if m1, + m2 < n ) or the unique integer in z,. that's congruent to ( m 1 + mz) (mod n). But �(k1 ) + �(k2) also equals m1 + m2 (if m1 + m2 < n) or the unique integer in z,. that's congruent to (m 1 + m2) (mod n). This shows that � preserves the operation of addition. Now, by multiplying the equa 2 tions k1 = a1 n + m 1 and k2 = a2n + m2, we get k1 k2 = (a1a2)n + (a1m2 + a2m 1 ) n + (m1m2) , so � (kl2) is equal to m 1 m2 (if m 1 m2 < n) or to the unique integer in z,. that's congruent to m 1 m2 (mod n). But �(k1 )�(k2) also equals m1m2 (if m 1 m2 < n ) or the unique integer in z,. that's congruent to m 1 m2 (mod n). This shows that � also preserves the operation of multiplication. The kernel of this homomorphism is the subring nZ = { nm: m E Z} of Z.
2. Let R be a ring and consider R[x], the ring of polynomials in x over R. Every element in R[x] is a
polynomial, p(x), with coefficients in R. For a fixed element a in R, define a function �.: R[x] 1 R as follows:
. (p(x)) = p(a)
This says that in order to find the image of a polynomial, p(x), under the map �a substitute a for ' x in p(x) and evaluate. The map �. is called the evaluation (or substitution) homomorphism at a. That �. actually is a ring homomorphism follows at once from the equations:
<J>. (p(x) + q (x)) = p(a) + q (a) = . (p(x )) + <J>. (q(x )) and �. (p(x)q(x)) = p(a)q(a) = �.(p(x))�a (q(x))
3. If R is a commutative ring with unity whose characteristic is a prime, p, then the function
� : R 1 R given by �(a) = aP is a ring homomorphism (called the Frobenius endomorphism). P To see why this is so, we need to show that � preserves both addition and multiplication. The P second part is easier, so we'll take care of that first. Since R is commutative, we have:
<j>P (ab) = (ab)P = (ab)(ab) · · · (ab) = (aa · · · a) (bb · · · b) = a P b P = <j>P (a)<j>P (b) p factors p factors p factors Now, let's verify that �p (a + b) = �p(a) + �p (b), which is equivalent to the equation (a + b)P = aP + bP. The binomial theorem says:
where
(�l
(a + W
(l (�) p
� k � p  1, the coefficient

k
p! k ! (p  k ) ! (*)
is divisible by p [since p is prime, there will always be a
factor of p left in the numerator of (*) for 1 248 + CRAC K I N G T H E G R E MATH EMATI CS S U BJ E CT TEST
(]
f=O Pk apkbk
denotes the binomial coefficient:
k
But for 1
=
� k � p  1 ] , so
where mk is an integer. Every term of the form (mk p)apk b k is a multiple of p, so it's equal to 0 in R, because char R = p. This reduces the equation above to (a + b)P = aP + bP. So, although beginning students sometimes make the mistake of writing, for example, (a + W = a3 + b3, this is actually correct in a ring of characteristic 3 (like Z3)! 4. Let (R, +, x ) and (R', EB, ®) be rings. It's possible for �: (R, +) 7 (R', EB) to be a group homomor phism but not a ring homomorphism. Let R = R' = Z, the ring of integers. Then the function �: (Z, +) 7 (Z, +) given by �(m) = 2m is a group homomorphism, but � does not preserve the operation of multiplication, since �(mn) = 2mn, but �(m)�(n) = (2m)(2n) = 4mn.
5.
Let (R, +, x ) and (R', EB, ®) be rings. It's possible for �: (R, x ) 7 (R', ®) to be a group homomor phism but not a ring homomorphism. Let R = GL(2, R), and let R' = R; these are groups under matrix multiplication and ordinary multiplication, respectively. Then the function �: (R, x) 7 (R, • ) given by �(A) = detA is a group homomorphism, but not a ring homomorphism since � does not preserve the operation of addition: In general, det(A + B) :f:. detA + det B.
Example 6.22 Solution:
Describe all the ring endomorphisms of Z.
For any rings R and R', the zero map, z: R 7 R', given by z(r) = 0' for every r in R, is always a ring homomorphism (although certainly not a very interesting one). So now let's assume that f: Z 7 Z is a nonzero ring endomorphism. Since 1 generates the cyclic group Z, �(1) cannot be equal to 0, because if it were, then �(m) would equal 0 for every m in Z [because �(m) = �(1 • m) = �(1)�(m)], contradicting our assumption that f is a nonzero map. Then [�(1 )]2 = �(1)�(1) = �(1 • 1 ) = �(1), but �(1) is not equal to 0. The only nonzero integer n that satisfies n2 = n is n = 1, so we must have �(1) = 1. Now, if m is a positive integer, then �(m) = �(1 + 1 + . . . + 1), with m summands, which gives �(m) = �(1) + �(1) + . . . + �(1) = m�(1) = m • 1 = m; and �(m) = �(1 + 1 + . . . + 1), with m summands, which gives �(m) = �(1) + �(1) + . . . + �(1) = m($(1)) = m. And finally, if m = 0, then �(0) = 0, since any ring homomorphism maps the additive identity to the additive identity. Therefore, for every integer m, we have �(m) = m, so � is the identity map. We conclude that there are only two ring homomorphisms from Z to itself: the zero homomorphism and the identity homomorphism.
Example 6.23
Let I be a subring of a ring R. If rx E I and xr E I both hold for every r in R and every x in I, then I is called an ideal in R. (This substructure of a ring is analogous to a normal subgroup, N, of a group G; see Example 6.16.) (a) If n is a positive integer, verify that n Z is an ideal in Z. (b) Show that
S
=
{(� :}
}
a, b, c e Z is a subring of M,(Z) but not an ideal.
(c) If �: R 7 R' is a ring homomorphism, show that ker� is an ideal in R'.
L
N U M B E R T H E O RY A N D A B STRACT A L G E B RA + 249
Solution: (a) The subring nZ consists of all multiples of n, so every x in nZ is of the form mn for some integer m. Therefore, if r is any integer (that is, any element in Z), we have:
r(mn) = (nn)n :::::} r(mn) E nZ
==>
and ( mn ) r = ( mr) n
By definition, then, nZ is an ideal in Z.
( mn ) r E nZ
0)
(b) 5 contains the additive identity (the zero matrix, by taking a = b = c = and the multiplicative identity (the identity matrix, by taking a = c = 1 and b = It's closed under addition,
0).
Since M2(Z) is a ring with unity, it follows that 5 is a subring. But it's not an ideal. If X E 5, it's not true that XA E Mz(Z) for every A in M2(Z); for example:
� )( �
XA = ( �
�) = G �)
'r' '�
XES
A E M2 (Z)
e
S
This shows that 5 is not an ideal. (c) The kernel of a ring homomorphism is the following subring of R:
ker = { x E R: (x) = 0'} We want to show that rx and xr are both in ker for every x in ker$ and every r in R. The equations (rx)
= <j>(r) (x) = <j>(r) 0 = 0 and <j>(xr) = <j>(x) <j>(r) = 0 <j>(r) = 0 •
•
•
•
verify that this is true. I NTEGRAL DoMAINS Let's look again at Z", the ring of integers modulo n, and take n = 6. Notice that it's possible to choose elements a and b, both of which are nonzero, for which ab = For example, we could take a = 3 and b = 4; then ab = 3 • 4 = 12, which is equal to 0 in Z6. The same thing can happen in a ring of matrices. For instance, in the commutative subring D of M2(Z) that consists of the diagonal matrices, if
0.
0
0
A=(
� �)
and
B=(
� �)
then AB = (where is the additive identity of D, the zero matrix), even though neither A nor B is zero. A commutative ring with unity in which this type of behavior cannot happen is called an integral domain. Normally, we accept the statement "ab = if and only if a = or b = as selfevident, but that's only because we're used to working in the rings C, R, Q, or Z, all of which are integral domains. If a and b are
0
2 5 0 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ ECT TEST
0
0,"
both nonzero elements of a ring but the product ab is nevertheless equal to 0, then a is called a left zero divisor and b is called a right zero divisor. (If the ring is commutative, we don't need to specify left or right.) So, an integral domain is a commutative ring with unity (1 :t 0) that has no zero divisors. Why is it desirable to work in an integral domain? Because the familiar cancellation law,
a :;t 0 and a b = ac
=>
b=c
is only guaranteed when there are no zero divisors. To see why, notice that ab = ac is equivalent to a(b  c) = 0, but this equation could hold if a and b  c are zero divisors. In this case, b  c would not be equal to zero, so b would not be equal to c. However, this situation wouldn't be possible if there were no zero divisors, that is, if a, b, and c were elements of an integral domain. For example, in Z6 (which is not an integral domain), it's true that 2 • = 2 • 4, but 1 :t 4. Also, notice that in the subring D of M2(Z) we considered above, if we choose
1
and
C=
(� �)
then AB = AC, but B :t C. Making sure that the cancellation law does hold is the motivation behind the definition of an integral domain. The reason that 3 and 4 were zero divisors in the ring Z6 is that neither of them is relatively prime to 6; the element 2 in Z6 is also a zero divisor (since 2 • 3 = 0). But the elements 1 and 5 are not zero divisors in Z6• In general, we can say that: A nonzero element m E Z" is a zero divisor if and only if m and prime.
n are not relatively
And this fact leads directly to the following important result: If n is prime, then zll is an integral domain.
Example 6.24 Solution:
Is the ring RR an integral domain?
Consider the functions f: R 7 R and g: R 7 R given by
f (x) = x  lxl
and
g(x) = x + lxl
0
The pointwise product function, Jg, is the zero functionthat is, (jg)(x) = for every x in Reven though neither f nor g is the zero function. Since RR contains zero divisors, it's not an integral domain.
F I ELDS Let a be a nonzero element of a ring R with unity. Since the multiplicative binary structure (R, ) isn't re quired to be a group, a may not have an inverse in R. However, if a does have a multiplicative inversethat is, if it's invertiblethen a is called a unit. (Don't confuse unit with unity: Unity is the unique multiplica tive identity in R (a unit is a nonzero element with a multiplicative inverse.) For example, in the ring Z6, •
the element 5 is a unit since it has a multiplicative inverse (namely, itself, since 5 • 5 = 1 in Z6), but 2 is not a unit; there is no element b in Z6 such that 2 • b = 1. If every nonzero element in R is a unitthat is, if (R*, • ) is a groupthen R is called a division ring. This name arises since in such a ring, every equation N U M B E R T H E O RY A N D A B STRACT A L G E B R A + 2 S 1
of the form a • x = c (with a :f. 0) can be solved for x by multiplying both sides by a1; that is, by "dividing by a." A commutative division ring is called a field. Let's look at some examples. 1.
Q, R, and C are fields, but Z is not. Z is not a field since the only units in Z are 1 and 1 
2. The commutative ring with unity that every nonzero element in 1
�
1
�
Q(.J2) = { a + b.fi: a, b E Q} is a field. All we need to show is
Q(fi) has an inverse in Q(.fi) ; if a and b are not both zero, then:
• a bfi� = a2 bfi2 = 2 a
a + b'J L. a+ b'J 2 a b 'J2 a =
.

a
2b

b � '/ 2L. E Q('J�2) + 2 2 2 L. 2b a 2b 
3. It's not difficult to show that a finite integral domain must be a field. Using this result, it follows
that if p is a prime, then Z is a field (our first example of a finite field). In particular, this means P that every nonzero element in Z is a unit; this would not be true if p weren't prime. P 4. Consider the subset K of M2(R) that consists of the following 2 by 2 matrices:
This set contains the additive identity (the zero matrix)by taking a = b = 0and the multipli cative identity (the 2 by 2 identity matrix)by taking 1 and b = 0. Since K is closed under addition,
( ba ab) + (dc dc) = (
and multiplication,
a=
)
a+c b +d E K (b + d) a + c
(ba a)(dc dc) = ((adac+bc)bd adac +bdbc) E K b
it follows that K is a subring with unity of M2(R). But although matrix multiplication is gener ally not commutative, multiplication in K is commutative, since:
( ba ba) (dc dc) = ((adac+bc)bd adac +bd) = (dc dc) (ba ab) be
Now, if a and b are not both zero, we have
det
( ba ba) = a2 + b2
*
0
=>
(ba ba)
is invertible
which shows that every nonzero element of K is a unit. Therefore, this set of matrices is a com mutative division ringit's a field! In fact, it can be shown that K is structurally identicaliso morphicto C, the field of complex numbers.
2 5 2 + C R AC K I N G T H E G R E MATH EMAT I C S S U BJ E CT T EST
5. Consider the subset H of M2(C) that consists of the following 2 by 2 matrices
where z denotes the complex conjugate of z. Like the set K in the preceding example, we can show that H contains the additive identity and the multiplicative identity, and that H is closed under addition and multiplication. It follows that H is a subring (with unity) of M2(C). Further more, if a and b are not both zero, we have: det
(b �Ja lal2 + lbl2 �
=
:t
0
==>
(b a�J �
is invertible
Therefore, H is a division ring. However, matrix multiplication is not commutative in H, so H is not a field. It's a noncommutative division ring (also called a strictlyskew field); H is known as the division ring of real quatemions.
Example 6.25
An element a of a ring R
is said to be idempotent if a2 = a. (a) Assume that R is a commutative ring with unity 1 :t 0. If T is the set of idempotent elements in R, show that T is closed under multiplication. (b) If R is a division ring, how many idempotent elements does R have? (c) Show that if every element of R is idempotent (whereupon we'd call R a Boolean ring), then R is commutative.
����
Solution:
(a )
·����
1.
First, we notice that there are at least two elements in T, namely 0 and (The fact that F = 1 fol lows from the fact that 1 is the multiplicative identity, so • 1 = 1 . To show that 02 = 0, we notice that 02 = 0 • 0 = 0 • (0 + 0) = 0 • 0 + 0 • 0; adding (0 • 0) to both sides gives 0 = 0 • 0.] Choose any two elements, a and b, in T; our goal is to show that ab is also in T. Consider the product (ab)2 = abab. Because R is commutative, ba = ab, so abab = aabb = aW. These equations imply that (ab)2 = aW. Now, because a2 = a and b2 = b (both a and b are idempotent), we get (abf = ab, which shows that ab is idempotent and thus in T.
1
(b) The equation a2 = a is certainly satisfied by a = 0, so we'll now assume that a :t 0. Since R is a division ring, every nonzero element has an inverse. Multiplying both sides of a2 = a by a1 gives a = 1 . Therefore, the only nonzero element of R that's idempotent is 1 . This tells us that a division ring has precisely two idempotents: 0 and 1 . (c)
Although we're asked to prove something about the multiplicative structure of R, we can nev ertheless exploit its additive structure and the fact that (R, + ) is a group. If x is any element in R, then x + x is in R, so x + x is idempotent, by hypothesis; thus, (x + x)2 = x + x. This equation simplifies to x2 + 2x2 + x2 = x + x, and since x2 = x, canceling and substituting gives us 2x = 0. Now, if a and b are any elements in R, then a + b is in R, so a + b is idempotent: (a + b)2 = a + b, an equation that simplifies to a2 + ab + ba + b2 = a + b; since a2 = a and b2 = b, we get ab + ba = 0. Because 2x = 0 for any x in R, we can take x = ab and write 2ab = 0. Since ab + ba = 0 and 2ab = 0, we conclude that ab + ba = 2ab. Adding ab to both sides gives ba = ab, which shows that R is commutative.
N U M B E R TH E O R Y A N D A B STRACT A L G E B RA + 2 S 3
CHAPTER 6 R EVI EW QUESTIONS Complete the following review questions using the techniques outlined in this chapter. Then, see Chapter 8 for answers and explanations.
1.
There is only one integer, x, between 100 and 200 such that integer pair (x, y) satisfies the equation 42x + 55y = 1 . What's the value of x in this integer pair? (A) 127
2.
(D) 22
(E) 33
(B) 27
(C) 31
(D) 38
(E) 47
(B) 6x  lOy  z
(C) x  y  2z
(D) 7x + l2y + 3z
(E) Sx + 3y  4z
(B) 24
(C) 30
(D) 32
(E) 50
(B) 6
(C) 8
(D) 10
(E) 12
Which one of the following groups is cyclic? (A)
8.
(C) 17
How many generators does the group (Z24' +) have? (A) 2
7.
(B) 11
When expressed in its usual decimal notation, the number 100! (that is, 100 factorial) ends in how many consecutive zeros? (A) 20
6.
(E) 183
If x, y, and z are positive integers such that 4x  Sy + 2z is divisible by 13, then which one of the following must also be divisible by 13? (A) x + l3y  z
5.
(D) 167
Let x 1 and x2 be the two smallest positive integers for which the following statement is true: "85x  12 is a multiple of 19." Then X1 + X2 = (A) 19
4.
(C) 158
Let L be the least common multiple of 1001 and 10101. What's the sum of the digits of L? (All numbers are written in their usual decimal representation.) (A) 6
3.
(B) 148
Z2 x Z4
(B)
Z2 x Z6
(C)
Z3 x Z4
(D)
Z3 x Z6
(E)
Z4 x Z6
If G is a group of order 12, then G must have a subgroup of all of the following orders EXCEPT (A) 2
(B) 3
2 5 4 + CRA C K I N G TH E G R E MAT H E MATI CS S U B J ECT TEST
(C) 4
(D) 6
(E) 12
9.
How many subgroups does the group Z3 E9 Z16 have? (A) 6
10.
11.
R+ :
If S =
{a E
(A)
1 log a
(B) 10
(D)
(C) 12
a :;t: 1 } , with the binary operation
20
(E) 24
defined by the equation a log b = loge b), then (5, • ) is a group. What is the inverse of a E S? e
(B)
e _ _ log a
(C)
•
l a e og
(D)
e log(l/a)
•
(E )
b = a log b (where e lfioga
Which of the following are subgroups of GL(2, R), the group of invertible 2 by 2 matrices (with real entries) under matrix multiplication? T = { A E GL(2, R): det A = 2t
I.
U = { A E GL(2, R): A is upper triangular}
II.
V = { A E GL(2, R): tr(A) = 0 }
III.
Note: tr(A) denotes the (A) I and II only
trace of A, which is the sum of the entries on the main diagonal.
(B) II only
(D)
(C) II and III only
III only
(E) I and III only
q be distinct primes. How many (mutually nonisomorphic) Abelian groups are there of 12. Let p and 2 4
order p q ? (A) 6
(B) 8
(D)
(C) 10
12
(E) 16
13. Let G 3be6the group generated by the elements x and y and subject to the following relations:
2 x2 = y , y = 1, and x1yx = y1• Express in simplest form the inverse of the element z = x yx3y3•
(B) xy2
(D)
(C) xy
yx
14. Let H be the set of all group homomorphisms �: Z3 7 Z6. How many functions does H contain? (A) 1
(B) 2
(D)
(C) 3
4
(E) 6
15. Let G be a group of order 9, and let e denote the identity of G. Which one of the following state ments about G CANNOT be true?
(B)
:t e and x1 = x. There exists an element x in G such that x :t e and x2 x5•
(D)
G is cyclic.
(A) There exists an element x in G such that x
=
(C) There exists an element x in G such that ( x ) has order 3.
(E) G is Abelian.
N U M B E R T H E O RY A N D A B STRACT A L G E B RA + 2 S S
16. Let R be a ring; an element x in R is said to be idempotent if x2 = x. How many idempotent elements does the ring Z 0 contain?
2
(B) 4
(A) 2
(C) 5
(E)
(D) 8
10
17. Which of the following rings are integral domains?
Z EtJ Z Zp' where p is a prime ZP2' where p is a prime
I. II. III.
(A) I and II only
(C) II and III only
(B) II only
(D) III only
(E) I and III only
18. Which one of the following rings does NOT have the same number of units as the other four? (A)
Z EtJ Z
19. How many elements x in the field Z 11 satisfy the equation x12  x10 = 2? (A) 1
(C) 3
(B) 2
(E)
(D) 4
20. Which of the following are subfields of C? I.
K1 =
II.
K2 =
Ill.
K3 =
{a+ b.JI: a, bE Q} {a+ b../2: a, bE Q and ab < .J2} {a+ a, b E Z and i f=1} bi:
=
(A) I only
(B) I and II only
(D) I and III only
(E) None of the K; are subfields of C.
2 S 6 + CRAC K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
(C) III only
5
Additional Topics
I NTRODUCTION In this final chapter, we'll briefly review several other areas that may be covered on the GRE Math Subject Test. We'll look at some fundamental ideas of logic, set theory, combinatorics, graph theory, algorithms, probability and statistics, pointset topology, real analysis, complex variables, and numerical analysis.
LOGI C In mathematics, logic refers to sentential calculus (also known as propositional calculus), which is a logical and welldefined system based on sentences. A sentence is a wellformed expression within a language and a syntax that can be evaluated as true or false using algebraic tools. The sentence "The ball is red," can be evaluated as either true or false, but the sentence "Red is the best color," cannot be. The language of logic includes a set of variables and a set of connectives from which any sentence can be built according to the rules of syntax. On the GRE, capital letters, such as X or Y, are used to represent sentences. A list of connectives follows:
2S7
iff or
H
"if and only if"
,X XAY X vY x�Y
the negation of
XHY
X iff Y
X and Y
X or Y If X, then Y
For sentences
Commutativity:
3.
Associativity: Distribution: de Morgan's Laws:
Solution:
,,X H X X A Y H Y A X and X v Y H Y v X X A(YA Z ) H (X A Y ) A Z and X v(Y v Z) H (X v Y) v Z X v(Y A Z) H (X v Y) A(X v Z) and X A(Yv Z) H (X A Y)v (X A Z) ,(X A Y ) H ,X v ,Y and ,(X v H ,X X � H ,Y � ,X
y)
t\ ,
y
Contradiction:
Example 7.1
X is true iff Y is true.
X, Y, and Z, the following theorems hold:
2.
6.
X is true and Y
are logically equivalent.
X
Double Negation:
5.
This conditional sentence is false only if is false. This biconditional sentence says The two sentences, and Y,
1.
4.
X
Prove the statement X �
y
Y H ,Y � ,X using truth tables.
To prove this theorem, show that the truth table for
,Y � ,X
X�Y
is equivalent to that of
y
x�Y
,X
,Y
,Y � ,X
T
T
F
F
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
X T
Since all values of the truth table are the same for
2 5 8 + C R A C K I N G TH E G R E MAT H E MATICS S U B J ECT TEST
X � Y and Y � ,X, they are logically equivalent. ,
SET TH EORY The idea of a set is perhaps the most fundamental notion in mathematics and cannot be defined in terms of simpler concepts. A set is simply described as a collection of objects (called its elements or members) that can always be definitively defined either as belonging to the collection or not belonging. Let A be a set; if x is an element of A, then we write x E A; if x is not an element of A, we write x � A. For example, the set A of positive even integers less than 10 is written:
A = { 2, 4, 6, 8 } The braces enclose the elements of a set, which are separated by commas. Notice that for this set A, we have, for example, 6 E A, but 7 � A. The set A can also be written as
A = { x: x is a positive even integer less than 1 0 }
which is read, "A is the set of all positive even integers x that are less than 10." The statement after the colon (which stands for such that) describes the condition that an object must satisfy in order to be con sidered an element of the set. Two sets are said to be equal if they contain exactly the same elements. The elements of a set can be anythingnumbers, functions, matrices, etc. In fact, the elements of a set can themselves be sets; in this case, it's common to refer to a collection or family of sets, simply to avoid the phrase "a set of sets." Finally, it's very convenient to include the concep t of the empty set. This is the set that contains no elements whatsoever; it's denoted 0. SuBsETs AND CoMPLEMENTS Let A and B be sets; if every element in A is also in B, then A is a subset of B (and B is said to be a super set of A), and we write A � B. Every set is a subset of itself, and it's agreed that the empty set is also a subset of every set. If A is a subset of B, and A :t B, then A is said to be a proper subset of B, and we write c B , (although some authors use this notation whether A is a proper subset of B or not). The standard way to prove that two sets are equal is to show that each is a subset of the other; that is:
A
A � B and B � A
¢=>
A=B
In many discussions involving sets, it's clear that all the sets are subsets of some universal set, U. For example, in calculus and real analysis, the domain and range of a function are implicitly assumed to be subsets of R, the set of real numbers. When it's clear what the universal set is, we can define the comple ment of a set Adenoted by N or A'as the set of all x in U that aren't in A:
Ac = { x: X E U and x � A } We can also define the difference of B
from A, also called the complement of B relative to A, as:
A  B = { x: x E A and x � B }
A D D IT I O N A L T O P I CS + 2 5 9
UNION AND I NTERSECTION Several other operations can be defined for sets. Given two sets, A and B, we can form their intersection and their union. The intersection of A and denoted is the set that contains all elements that are in both A and B:
B, An B , An B = { E A and E B} An A A disjoint. union A Au B = { E A B} symmetric difference A A� B =(AB)u(B  A) x: x
x
If B = 0, then we say and B are The of contains all elements that are either in or in B (or in both):
of two sets,
A u B,
is the set that
or x E
x: x
The
and B, denoted
and B, is defined by the equation:
A pair of laws relates the concepts of complement, intersection, and union; they're called de Morgan's laws. For two sets, A and B, they are:
(AnB)" = Ac uw (AuBY =N nBc Basically, de Morgan's laws say that the complement of an intersection is the union of the complements, and the complement of a union is the intersection of the complements. Both of these laws can be general ized to any number of sets, and we also have laws that concern set differences:
(An B)C =(AC)n(B C) (Au B)C =(AC)u(BC) Another pair of laws can be used to simplify an expression involving both the intersection and the union of sets; they're known as the distributive laws:
An(BuC)= (AnB)u(AnC) Au(BnC)= (AuB)n(AuC) { A; }i EI . AvI A. { E A. i E I} i E I} i E I A. = { E A
The definitions of intersection and union can be generalized. Let be some indexing set, and consider This notation means that there's a set associated with every i E I. The in the family of sets tersection of the sets in this family is written: n
iEf
1
=
I
x: X
for every
Similarly, the union of the sets in this family is written: U
I
x: x
I
for a t least one
.
Cartesian Products
A B
If and are nonempty sets, we can form a set called their cartesian product, which is the set of ordered pairs given by:
Ax B = {(a, b): a A and b E B} E
For example, R x R is the familiar xyplane. 2 6 0 + C R AC K I N G T H E G R E MATH EMATICS S U BJ E CT T EST
I ntervals of the Real line
The geometric interpretation of the set of real numbers, R, is a line (called the number line or the real line). The xyplane is determined by two copies of the real line that are perpendicular to each other and intersect at the origin (the point corresponding to the number 0). Certain subsets of R, which are of fundamental importance, are called intervals. Let a and b be real numbers, with a < b; then the set of all x such that a < x < b is called the open interval between a and b, and is denoted by (a, b):
(a, b) = { x: a < x < b} Notice that (a, b) is also the notation for an ordered pair, which symbolizes a point in the xyplane (the point at which x = a and y = b). Contextually, it will be clear what (a, b) stands for, so no confusion should result. Other intervals of the real line are denoted as on the following page.
[a, b) = { x: a ::; x < b} (a, b ] = { x: a < x ::; b} [a, b] = {x: a :s; x :s; b} The first two are called halfopen (or halfclosed) intervals, and the third is called a closed interval. If the interval doesn't include one (or both) of its endpoints, we use a parenthesis, "(" or ")", for that endpoint, and if the interval does include one (or both) of its endpoints, we use a bracket, "[" or "]", for that endpoint. The definition of an interval can be generalized to an infinite interval. Formally, we consider the set of extended real numbers, which is the set R with two elements, oo (minus infinity) and oo (or +oo, plus infinity) adj oined : { oo} R { oo} , where by definition, oo < x < oo for every x in R. These symbols are introduced as a notational convenience for infinite intervals; if a and b are real numbers, then we have: 
uu
(oo , b) = {x: oo < x < b} = {x: x < b} (oo, bJ = {x: oo < x � b} = {x: x � b} (a, oo ) = {x: a < x < oo} = { x: x > a} [a, oo ) = {x: a � x < oo} = {x: x 2 a} ( oo oo ) = { X : oo < X < oo} = R 1
Example 7.2
If A = (1, 2] and B = (1, 3) are intervals of R, what are each of the following: AnB AuB AB A�B N
B
N0 . Therefore, R is infinite but not countably infinite; it's said to be uncountable. The cardinality of R is denoted by c, which is called the cardinal number of the continuum. This provides the next "level" of infinity, which is greater than the infinity that describes the cardinality of the set of positive integers. Other sets that are uncountable include: •

The set of all irrational numbers, Qc = R Q
•
The set of all transcendental numbers
•
Any subset of R that contains a (nonempty) interval
(A real number is said to be transcendental if it's not algebraic. The two most familiar examples of transcendental numbers are 1t and e. )
By analogy with the expression for the cardinality of the power set of a finite set, the cardinality of the continuum, c, is also written as 2�" , because R is equivalent to P(Z+). Since the cardinality of the power set of A is always strictly greater than the cardinality of A, we can consider the cardinality of the power set of R, which represents the next "level" up in cardinality, and clearly the process can continue. We can now see that there are infinitely many cardinal numbers, which we can order as follows:
0 < 1 < 2 < . . · < No < C < t < 2 2' < . . . �
finite cardinal numbers
Example 7.4 Solution:
infinite cardinal numbers
Show that any open interval of the form (a, b), where a and b are real numbers and a < b, is uncountable. Since R is uncountable, we'll show that
(a, b) is uncountable by finding a bijection from
(a, b) to R. First, the tangent function, f(x)
( 2 ' 1t)2 7t
=
tan x, is a bijective function from the interval
onto R. Next, the linear function
1t
1t
g(x) = b  a (x a)  2.
is a bijective function from (a, b) to bijection from b) to R .
(a,
( 2 1t)2  7t ,
. Therefore, the composite function f o g is a
GRAPH THEO RY A graph consists of a finite collection of vertices and e d ge s where each edge connects a distinct pair
of vertices. We will only discuss simple graphs whose edges have no orientation; these graphs are called Let G be a graph; then we say V(G) is the set of vertices and E(G) is the set of edges. The order of a graph is l V I , the number of vertices; the size of the graph is l E I , the number of edges. If V(G) and are connected by the edge vertices. The degree of a vertex is b ), we say and b are the number of vertices adjacent to (or the number of edges at vertex A vertex with an odd degree is an vertex, and a vertex with an even degree is an vertex. If every distinct pair of vertices is connected by an edge, the graph is
undirected.
(a,
odd
v complete.
a
adjacent even
a, b E
v) .
v
ADDITIONAL TOPICS + 265
f:
Graphs F and G are isomorphic if there exists a function V(F) 7 V(G) that is both onetoone and onto, and all adjacencies are preserved. A path indicates movement from a start vertex to an end vertex. A path is a sequence of adjacent if every pair of vertices can be connected by at vertices and the connecting edges. A graph is least one path. A path that begins and ends at the same vertex is a cycle. A graph with no cycles is a and if such a graph is connected, it is a tree. (Trees are the most widely used application of graph theory in applied mathematics.)
connected
forest,
Let G be a graph with vertex and edge sets V(G) and E(G). H is a subgraph of G if
1.
2. 3.
V(H) � V(G)
E( H) � E(G)
every edge of H connects two vertices of H
This says that the vertices of H are a subset of the vertices of G, and every pair of adjacent vertices in H also appears in G. If V(H) = V(G), then H is a spanning subgraph of G. In addition, if H is a tree, then H is a spanning tree of G. Example 7.5
X
Graphs and Y are shown below. a. Are the two graphs isomorphic? b. What is the degree of vertex A? c. What is the fewest number of edges you must add to each graph for it to be complete? F
A
B
E
(X)
(Y)
Sol u ti on:
a.
First, note that each graph contains exactly five vertices. The function that maps A to F, B to G, C to H, D to I, and E to ] is both onetoone and onto.
Next, list the edges for each graph. The first graph has edges (A, B), (B, C), (C, D), ID, E ) , and IE, A}. The second graph has edges (F, G), IG, H), {H, I}, /I, J}, and IJ, F}. Since the function pre serves every pair of adjacent vertices, the graphs are isomorphic. b.
Since vertex A has exactly two adjacent vertices (B and E), A is an even vertex.
c.
= 10 edges. Thus, each graph needs 3!2! five more edges to be complete; for example, graph X needs (A, C), /A, D), (B, D), (B, E), and
Using the combination formula, 5 vertices can have /C, E).
266 + CRA C K I N G THE G R E MATH EMATICS S U BJ E CT TEST
5!
ALGOR ITHMS An algorithm is a finite sequence of welldefined and logical steps used to complete a task, usually to
solve a problem. An algorithm has the following properties: 1 . N o instructions can be dependent on the user. All ambiguity i n the process must be removed. 2. The process must end with a solution after a finite number of steps.
3. The process is deterministic; that is, each step unambiguously determines what the next step will be.
Simply stated, an algorithm must receive an input and through a finite process produce a solution. Us ing the same input value will always produce the same solution (unless a step incorporates randomizing). An algorithm can have any sort of notation. An algorithm on the GRE could use any of the symbols commonly used in the section on Logic. Typically, ETS uses algorithms expressed in a programming language (or a flow chart). This language is designed to express the steps, or computations, that tell the computer how to behave. The syntax of a programming language used on the GRE should be simple enough to be understood by any test taker who has taken an undergraduate course in computer science. The following algorithm is used to prove the famous "3x + 1" problem.
Example 7.6
input ( a ) whi l e a >
1
begin
end
if
( a mod
2
a := a else a : = 3 output ( a )
=
I
•
O)
a
2
+
1
When the input value of the algorithm is 3, what is the sequence of outputs produced?
Solution:
Interpret the steps of the algorithm. The input value is a, and a loop is created while a > 1 . For each step, if a = 0 (mod 2), then divide a b y 2 . Otherwise multiply a by 3 and add 1 . Finally, the algorithm will output the new value o f a and reevaluate whether a > 1 before continuing. The outputs of the algorithm will be the following: 10, 5, 16, 8, 4, 2, 1 Because a is not greater than 1, the process ends, and the sequence converges to a solution.
COMB I NATOR ICS Combinatorics is the branch of mathematics concerned with counting. (Although that may sound fairly
easy, the methods of more advanced combinatorics are far from trivial.) Here's an example. Let's say we're trying to get from City A to City C via City If there are two roads between City A and City and three roads from City to City C, how many different journeys are there from A to to C? Since there are two choices for the first route, then three possible choices for the second, the total number of journeys is their product: 2 x 3 = 6. This illustrates a basic principle: • If there are a possible choices for the first decision and b possible choices for the second deci sion, then the total number of possible ways these two independent decisions can be made is the product ab.
B
B.
B
B,
ADDITIONAL TOPICS + 267
•
If there's a third independent decision to be made (with, say, c choices), then the total number of ways to make all three decisions is the product abc, and so on.
Example 7.7
Let's say that the personal code for your ATM card consists of six characters, each of which can be any letter or numerical digit. (a) How many possible personal codes are there? (b) Suppose you want your personal ATM code to have no repeated characters; now how many possible codes are there?
Solution:
(a) There are twentysix letters (A through Z) and ten digits (0 through 9), so there are 26 + 1 0 = 36 possible choices for each character. Since there are six decisions, any of which can use any of the 36 possible characters, the total number of ways the six decisions can be made is 366• (By the way, that's over two billion possible codes.) (b) There are 36 ways to make the first decision, but only 35 for the second (because, to avoid a re peat, you can't use the same choice you made for the first character). Then there are only 34 pos sible choices for the third character, 33 for the fourth, 32 for the fifth, and 31 for the sixth. Thus, the total number of codes that don't have any repeated characters is 36 x 35 x 34 x 33 x 32 x 31 . (This cuts down the total number of codes by roughly onethird.) PERMUTATIONS AND CoMBINATIONS Let's say we're given a set of n objects; in how many different ways can we choose k of them? The answer depends on whether the order in which we select the objects makes a difference. For example, in a horse race that involves eight horses, how many win, place, and show possibilities are there? Well, we want to see how many ways we can choose three horses out of eight, but here order makes a difference; if horses A, B, and C are the ones that come in first (win), second (place), and third (show), it matters whether the finishing order is A, B, C or, say, B, C, A. On the other hand, if you're playing fivecard draw poker, you receive five cards out of a standard deck of 52; the order in which they're dealt to you makes no differ ence. The number of possible poker hands is found by counting the number of ways we can choose five cards out of 52, without regard to order. If the order in which the objects are chosen does make a difference, each complete selection of k ob jects from the original n is called a permutation (the variable r is often used in place of k). If the order is irrelevant, then each complete selection is called a combination. In general, there are always more per mutations than there are combinations. To see why, let's go back to the example of the horse race. Each of the selections
A, B, C
A, C, B
B, A, C
B, C, A
C, A, B
and
C, B, A
is a different permutation, but they're all equivalent to a single combination. Let's figure out the number of ways we can choose k things from n, with regard to order. For our first selection, we can choose any of the given n objects; for the second selection, we can choose any of the remaining n  1 objects; this process continues until, for the kth selection, we can choose any of the remaining n  (k  1) objects. Since these decisions are independent, the total number of permutations of n things taken k at a time is the product:
P(n, k) = n(n  1)(n  2) . . . [n  (k  1)]
2 6 8 + CRAC K I N G TH E G R E MATH EMAT I CS S U BJ E CT TEST
Now, let's figure out the number of ways we can choose k things from n, without regard to order. Let's say we've chosen k objects from a group of n, with regard to order. That is, assume we have a permuta tion of n things taken k at a time. In how many ways can we permute these k objects? The answer is k!. For example, consider the example of the three horsesA, and Cgiven above. Let's say we selected them in the order A then then C. Now that we have these three, we can arrange them in 3! = 6 differ ent orders; these orders are displayed above. Each of these six orders is a different permutation, but they collectively represent just one combination. In general, the number of combinations of n things taken k at a time is equal to the number of permutations of n things taken k at a time divided by k!:
B,
B
2 · (k C( n k) = P(n, k) = n(n  l)(n  ) · · [n   1)] I
''''='_.:..:.
k!
k!
This expression for C(n, k) can be written in another way:
C(n k) = '
n! k !(n  k) !
()
In this form, you should recognize it as the binomial coefficient:
n
n!
k  k !(n  k) !
It's called the binomial coefficient because it's precisely the coefficient that appears in the
theorem:
Example 7.8 Solution:
In a horse race consisting of 8 horses, how many different winplaceshow (the
top three) finishing orders are there?
The order in which the horses finish matters here, so we're looking for the number of per mutations of 8 things taken 3 at a time; this number is: P(8, 3)
Example 7.9 Solution:
binomial
=
8•
7
•
6 = 336
How many 3element subsets does a set containing 9 elements have? The order in which the elements are arranged in a subset is irrelevant. The number of combinations of 9 things taken 3 at a time is:
C(9 3) = I
() 9
=�= 3 3! 6! e
9•8•7 9•8•7 = = 3 4 7 = 84 • • 2 3 3! e
A D D I T I O N A L T O P I CS + 2 6 9
With Repetitions Allowed
Imagine that you have a hat that contains 10 billiard balls, numbered 1 through 10. You're asked to choose three of the balls, but after each selection, you record the number and toss the ball back into the hat. So, for example, you could choose the 5 ball, then the 8 ball, then the 5 ball again. It's clear that the number of permutations and combinations of n things taken k at a time in a situation like this would be different than if repetitions weren't allowed. The formula for the number of permutations in this case is easy:
P(n, k) with repetitions allowed = nk
(n +�  ) 4
Determining a formula for the number of combinations is not as easy:
C(n, k) with repetitions allowed = Example 7.10 Solution:
1
In how many ways can we write the number as the sum of 5 nonnegative integers?
Consider the following setup. You're given 4 balls and seated in front of 5 empty boxes, labeled 1, 2, 3, 4, and 5:
D D D D D 1
2
4
3
5
Your job is to place the 4 balls any way you want into the 5 boxes. You can place more than 1 ball into the same box, but you must drop in all 4 balls. After you've finished, there'll be k1 balls in Box 1, k2 balls in Box 2, etc., where each k; is a nonnegative integer and k1 + k2 + k3 + k4 + k5 The 5tuple (k1, k2, k3' k4, k5) gives a way of writing 4 as the sum of 5 nonnegative integers. For example, let's say you placed the balls in the boxes as follows:
= 4.
CJ CJ D c:J D 2
1
4
3
1
5
This would correspond to the 5tuple (1, 1, 0, 2, 0), and + 1 + 0 + 2 + 0 is a way of writ ing the number 4 as the sum of 5 nonnegative integers. By placing 4 balls in these 5 boxes, you're choosing 4 objects from with repetitions allowed (placing a ball in a box is your way of marking your choice of a box). Therefore, the total number of ways of writing 4 as the sum of 5 nonnegative integers, in which the order of the summands does matter, is the same as the number of ways of choosing 4 objects from 5, in which the order doesn't matter, with repetitions allowed. According to the formula given above, this number is:
5,
5·6· 7 ·8 = 70 ( 5 +4 1) = (48 ) = � = 4 !· 4 ! 4
2 7 0 + C RAC K I N G T H E G R E MAT H E MATI CS S U B J E CT T EST
4!
THE PIGEONHOLE PRINCIPLE This simple principle can be used in a variety of combinatorics and number theory problems. It states:
If n objects are placed into k pigeonholes, and n 2 k + tain more than one object.
1, then at least one pigeonhole must con
Imagine a mail carrier standing in front of five mailboxes. If he has six or more envelopes to place in these boxes, then at least one mailbox is going to get more than one envelope. Here's a more generalized statement of the pigeonhole principle: Let's say you're given n objects, each of which is painted exactly one of c possible colors. Then for any integer k such that k � [(n  1)/ c] + 1 , there must be k objects, all of which are painted the same color.
If this weren't true, then there'd be, at most, k  1 different objects of each of the c colors, which would mean that c(k  1) 2 n; or, equivalently, n � c(k  1). But this would contradict the assumption that k � [(n  1 )/c] + 1, which is equivalent to the statement n 2 c(k  1) + 1 .
Example 7.11
Let A b e a set containing n + 1 integers. Show that it's always possible to choose two numbers, a and b, from A such that a  b is divisible by n.
···...._____________________ _.
Solution:
·· ·····
·· ······ �
�·············· ··
We can partition Z into n subsets A0, A 1 , A 2,
, 
____ __
•
•

•
· ..•..
__....______ . ��� 
, A11_1 such that:
k E A; if and only if k = i (mod
J

n)
Therefore, every integer is in exactly one of these subsets A;. If we now place the n + 1 integers from A into these subsets, the pigeonhole principle tells us that at least one of the sets Alet's call it Awill contain at least 2 integerscall them a and bfrom A Since } a = j (mod n) and b = j (mod n), we know that a  b = 0 (mod n), so a  b is divisible by n. I
Example 7.12
Imagine a set of 10 objects, each of which is painted one of 3 possible colors. Show that there exist 2 disjoint subsets of this set such that each subset consists of 3 objects all painted the same color.
Solution: Let n 1 = 10 and c = 3. Because 3 �
[(n1  1)/c] + 1, our general statement of the pigeonhole
principle assures us that there are at least 3 objects, all of which a.re painted the same color. Choosing exactly 3 of these objects, we get one 3element subset that consists of objects all painted the same color. Now, removing these 3 objects from the original 10 leaves 7 objects, each of which is painted one of 3 possible colors. Now, let n 2 = 7 and c = 3; since 3 � [(n2  1 ) / c] + 1, again we know that there exist at least 3 objects, all of which are painted the same color. This gives us our second 3element subsetwhich is clearly disjoint from the firstall of whose elements are painted the same color.
PROBABI LITY AN D STATISTICS Consider rolling a fair die. It has six sides, all of which are equally likely to turn up. Since there are six possible outcomes, all equally likely, the probability of any one of them happening is
�. 6
To find the probability of an event, simply divide the number of desired outcomes by the total number of possible outcomes.
AD DITIONAL TOPICS + 2 7 1
Example 7.13
Solution:
Each of the first ten positive integers is written on a slip of paper, and the ten slips are tossed into a hat. What's the probability that someone will pull out a prime number?
Assuming that all ten numbers are equally likely to be selected, we just divide the number of desired outcomes by the total number of possible outcomes. Among the first ten positive integers, there are only four primes: 2, 3, 5, and 7. Thus, the probability of selecting a prime from the hat is
Example 7.14
_±_ 10
=
3. . 5
An ordinary deck of playing cards is shuffled, and a card is randomly chosen. What's the probability that the card selected is (a) a heart? (b) red? (c) the jack of hearts? (d) not a face card?1
We'll use the fact that an ordinary deck contains 52 cards, with 4 suits of 13 cards each. 13 (a) Since there are 13 hearts, the probability that the card selected is a heart is _!_ .
Solution:
=
(b) Because two suits are red (hearts and diamonds), there are 13 + 13 . 26 . . ab Ility that a red card IS se1ected IS = . 
52
1.
=
52 4 26 red cards, so the prob
2
__!__ .
(c) Since there's only one jack of hearts, the chances that this particular card is chosen is
52
(d) There are 3 face cards per suit: the jack, queen, and king. Since there are 4 suits (hearts, diamonds, spades, and clubs), there are 3 x 4 = 12 face cards. Therefore, 52  12 = 40 of the cards are
not face cards. We conclude that the probability of not selecting a face card is
Examgle 7.15
Solution:
40
52
=
10 . 13
You deal S cards to yourself from a standard deck The result is 4 diamonds and a club. You discard the club, and draw one more card from the deck What's the probability of getting the flush (that is, of drawing a fifth diamond)?
After you deal yourself the original hand, the deck contains a total of 47 cards, 13  4 = 9 of which are diamonds. Therefore, the probability of drawing one of the remaining 9 dia. mondS IS
Example 7.16 Solution:
9
47 .

Two points, x and y1 are selected at random in the interval [0, 1]. What's the prob
ability that the product xy will be less than
_!_ ? 2
Selecting two points at random in the interval [0, 1] is equivalent to selecting a point (x, y) at random in the square [0, 1] the curve xy
=
x [0, 1]. The following shows this square and the graph of
_!_ ; the shaded region denotes those points (x, y) where xy < _!_ . 2
2 7 2 + C R A C K I N G T H E G R E MATH EMAT I CS S U BJ E CT TEST
2
y + points where xy >
points where xy < � ;=,.. 1
2
�
1
The area of the unshaded region is:
1 1  2 ) ax = [ x  � log x J1 = (1 ;( J x 2
0 )  ( �  � log� ) = H1  log 2 )
Therefore, the area of the shaded region is 1  [ � (1  log 2)] = � (1 + log 2) . Since the square has an area of 1, the ratio of this area to the area of the entire square is � (1 + log 2), so this is the probability that xy < � .
PROBABILITY SPACES We'll need to introduce some more formal terminology and techniques to continue our study of prob ability for the GRE Math Subject Test. Let 5 be a nonempty set, and consider P (5), the power set of 5, which is the collection of all the subsets of 5. A Boolean algebra (or simply an algebra) of sets on S is a nonempty subfamily of the power set of S, E � P(S), that satisfies the following two conditions:
1.
I f A and B are sets in E , then so are A u B and A n B .

2. If A is a set in E , then so is N = 5 A .
Because E is nonempty, i t contains some set, A . B y the second condition, E must also contain S  A. Then, by condition 1, E must also contain S = A u (S  A) and 0 = 5 ) = e213 e4/3 "" 0.25 3 3 3 3 In many cases, the random variable is defined so that the probability that X will equal any one par ticular number is zero; we'd only get nonzero probabilities by finding the probability that the value of X is in some interval. We call such a random variable (and its associated distribution function) continuous. The derivative of the distribution function, fx(t) = F� (t ), is called the probability density function of X (or of Fx). The density function may be continuous, but we only really require it to be nonnegative and integrable, so that and that: Since Fx(t) � 0 as t � =, we get the following important formula: P(X :::; t) = Fx(t) = f�fx( t)dt
In most problems, we're given the density function,fx (t), so it all comes down to this:
To figure out the probability that the random variable, X, will be in some range of values, we integrate the density function over this range, since:
Notice that if X is continuous, the inequality signs on the lefthand side of this equation may be < or �, because the probability that X equals any one number is zero.
Example 7.21
A company hires a
marketing consultant who determines that the length of time
(in minutes) that a consumer spends on the company's Web site is a random vari
fort< 0 fort 2:: 0
able, X, whose probability density function is:
What's the probability that a consumer will spend more than 10 minutes on the company's Web site? Solution:
The value of P(X > 10) is equal to 1  P(X � 10), which we can figure out as follows: P(X > 10) = 1  P(X :::; 1 0) = 1  fx (10) = 1  f 10 fx (t)dt = 1  f 1 0 le116 dt �
0 6= 1  [ e t/6 r = e lD/6 "" 0.19
2 7 8 + CRAC K I N G T H E GRE MATH EMAT I CS S U BJ E CT TEST
Expectation, Variance, and Standard Deviation
If X is a continuous random variable with probability density function fx, we can define several quanti ties that are important in the statistical analysis of X. The expectation (or mean) of X, denoted by E(X) or !l(X), is E(X) = Jl(X) =
f: tfx (t)dt
which gives us, intuitively, the average value of X. For example, if X is equal to the number rolled on a fair diea discrete, rather than continuous random variable, but the idea's the samethe expectation of X is 3.5, which is just the average of the possible values1, 2, 3, 4, 5, or 6that X can have. The variance of X, denoted by either Var(X) or cr 2(X), is: Var(X) = 0"2 (X) =
f:[ t  Jl(XW fx (t)dt
The variance tells us about the spread of values of X from the mean (that is, from the expectation): The smaller the variance, the smaller the probability that the value of a random variable will stray far from its mean. Finally, another way to analyze the deviation of X from its mean is to compute the standard deviation of X. This is simply the (nonnegative) square root of the variance:
�
standard deviation of X: cr(X) = cr 2 (X) This is the reason for the alternate notation, cr2(X), for the variance of X.
Example 7.22
fo
Let X be a random variable whose probability density function is:
f(x) =
if x :::; o i x3 if 0 < x < 2 0 · if x � 2
What's the standard deviation of X?
Solution:
First, we'll need to figure out the mean of X: Jl(X) =
f� tf(t)dt = 502 t _
•
5
[ ]
l t3 dt = .1 2 t 4 dt = l l t 5 2 = �5 0 4 4 0 4 5
Now, we can find the variance:
f:[ t  Jl (X)]2f(t)dt = f \ t  2/ 5 . l t3 dt 0 .l4 f \ t 5 _ 1Q t 3 ) dt 5 t 4 + fli 0 25 = .l4 [ l6 t 6 _ 1Q5 t 5 + 1£.5 t 4 ] 2 0 2 2 l [ t 4 ( l t 2 _ li5 t + l£.5 ) ] 2 6 4 2 2 0 1 6 [ .1 . 2 2 _ jQ5 . 2 + 1£.] = 4 2 25
cr 2 (X) =
4
=
=
.l .
_ JL  75
6
A D D IT I O N A L T O P I CS + 2 7 9
The standard deviation of X is the square root of the variance:
cr(X) = �cr2(X) = /s = 2J2 = 2J2 • J3 = 2/6
�75 sJ3 sJ3 J3 
1s
The Normal Distribution
Statisticians analyze random variables that describe a variety of measurementsthe length of long distance phone calls, the lifetime of machine parts (and of people), the distribution of student grades in a large lecture, etc. Many such measurements obey a probability distribution known as the normal distribution; because of the wide range of applicability of normal distribution, it's the most important distribution function for a random variable. A random variable, X, is said to be normally distributed if its probability density function has the form:
The choice of parameters was no accident; cr is the standard deviation of X, and 11 is its mean. The graph of is the familiar "bellcurve."
fx
/ ¥
f.l  3 cr
f.l  2cr
f.l
y  (t)  cr.�127t_ _
J.l + cr
£ 1x
_
,;:c
_
f.l + 2cr
e ( 1  �)2 / 2cr'
f.l + 3 cr
As is the case with any other density function, the total area under the graph is equal to
1. This graph
is symmetric with respect to the vertical line t = jl, so the line cuts the region below the graph exactly in
half; the area to the left of t = ll is equal to _! , and the area to the right of t = f.l is also _! . To figure out the
2
2
probability that the value of a normally distributed random variable, X, will be in some range of values, we'd integrate the normal density function over the range:
where t1 and t2 are any extended real numbers such that t1 < t2.
2 8 0 + CRAC K I N G T H E G R E MAT H EMATICS S U B J E CT TEST
Since this integral cannot be expressed in closed form in terms of elementary functions, it must be integrated numerically. To put the general density function into a standard formfor which tables of values existwe make the change of variable u = ( t  !l )/ a , and set cr = 1 to obtain the standard normal probability density:
f (u ) 2
=
_1_ eu'/2 r;:v27t
for the standardized normal random variable, Z. This random variable is normally distributed with mean
0 and standard deviation 1, and it's related to the original random variable, X, by the equation: P(tl
< X � tz ) = p
c]:
ll
t 1 . 71)
I
,· .· _.3,,2 _.1+,1,_�f�·::.:�i�:"_:;�.>�·.�· .�· ·1 .71
3� �
2
ADDITIONAL TOPICS + 281
The Normal Approximation to the Binomial Distribution
Example 7.19, we answered a question about a sequence of n Bernoulli trials. A random variable, X, used to describe such an experiment is said to have a binomial distribution; it's a discrete (rather than continuous) random variable and, in this case, if the probability of a success is p and the probability of a failure is q = 1  p, then: In
If n is large, it can be quite difficult to evaluate the righthand side of this equation. Instead, we can use the normal distribution to approximate it. The mean of X, 11, is np, and the standard deviation, cr, is J;;pq . It can be shown that:
P(ai :::;; X :::;; a2) ""
( 2 2 )  ( a  JJ. + l cr
2)
a  JJ.  l I cr
Because it's helpful to memorize «(z) for various values of z, here's a brief table:
z «l>(z)
0 0.5
0.5 0.69
1 0.84
2.5 0.99
2 0.97
1.5 0.93
3 or more "' 1
To handle negative values of z, notice that, by the symmetry of the normal density function, it's always true that (z) = 1  (z).
Example 7.24 Solution:
A fair coin is flipped 100 times. What's the prob.ability of getting between 60 and 70 "heads"?
We're asked to figure out P(60 :::;; X :::;; 70), where X has a binomial distribution with n = 100 and p = q = 2. ; this probability is equal to the value of the expression:
2
We'll approximate this using the normal distribution. The mean of X is 11 = (100)( t ) = 50, and its standard deviation is cr = J1 00(t)(t) = 5 . Therefore:
P(60 :::;; X :::;; 70) ""
(
�
70  0 + t
) ( 
�
60  0  t
= (4.1)  (1.9)
)
The value of (4.1) is virtually equal to 1, so P(60 :::;; X :::;; 70) 1  (1.9). From our brief table above, we can write «1>(1.9) "' «1>(2) "' 0.97 and conclude that: "'
P(60 � X � 70) "' 1  0.97 = 0.03 = 3%
2 8 2 + C R AC K I N G T H E G R E MAT H EMAT I CS S U BJ E CT TEST
POI NTSET TOPOLOGY I NTRODUCTION
The subject of pointset topology provides a rigorous foundation for many of the advanced topics in mathematics. While proving theorems is an important part of most college courses in topology, the GRE doesn't expect you to provide proofs; instead, it will test you on your knowledge of many of the defini tions in basic general topology and the statements of a few theorems. For this reason, we'll concentrate on stating the definitions and theorems and illustrating them with examples. Let X be a nonempty set. A topology on X, denoted by T, is a family of subsets of X, for which the following three properties always hold true: 1 . 0 and X are in
T.
2. If 01 and 02 are in 3.
T, so is their intersection, 01 n 02• If {0 }. is any collection of sets from T, then their union, U 0 , is also in T . l
1E
IEI
.
1
I
Conditions 2 and 3 say that a topology is always closed under finite intersections and arbitrary unions. The sets in T are known as open sets, and this is why we used the letter 0 to denote them. If x E X and N is a subset of X such that there exists an open set 0 with x E 0 � N, then we say that N is a neighbor hood of x. If 0 is an open set, then its complement, oc X  0, is called a closed set. (Note: In this sec tion, we'll never use the alternate notation 0' to denote the complement of 0; this is because the use of a prime is a standard notation for something else in topology.) A set X together with a topology T on X, (X, T), is called a topological space. When the topology on X is understood, it's common to simply refer to "the topological space X," rather than "the topological space (X, T)." There's a rather weak condition that's usually imposed on topological spaces. We say that (X, T ) is a Hausdorff space if for every pair of distinct points, x and y, in X, there exist disjoint open sets, Ox and OY, such that x E Ox and y E 0� . If X is any nonempty set, then the collection { 0, X } is always a topology on X. It's called the indis crete or trivial topology. At the other extreme, the power set of X, P(X), which consists of all subsets of X, is always a topology as well; it's called the discrete topology. If T1 and T2 are topologies on X, then we say that T1 is finer than T2 (or, equivalently, that T2 is coarser than T1) if T1 :::2 T2• Therefore, the discrete topology is the finest topology a set X can have, and the trivial topology is the coarsest. Here's an example. Let X be the set Ia, b, c, df, and let T 1 be the following collection of subsets of X: =
T1 Then
= {0, {a, b}, {c, d}, x }
T 1 is a topology on X. The collection A = {0, { a}, { a, b} , {c, d} , x}
is not a topology on X because, although {a} and { c , d} are in A, their union, if we were to include this subset in A, the resulting collection,
{a, c, d} , is not. However,
T2 = {0, {a}, {a, b}, { c, d}, { a, c, d}, x}
is a topology on X, and one that's finer than
T1 .
A D D I T I O N A l T O P I CS + 283
Let's now turn to a more interesting and useful example. In the set X = R, the set of real numbers, let's declare a subset 0 of R to be open if, for every x in 0, there exists some positive number, £, such that every x' that satisfies the inequality l x  x' l < £ is also in 0. Then the resulting collection of open sets is a topology on R. The empty set is open, since it satisfies the condition vacuously (that is, there's no x in 0 that doesn't satisfy the condition), and the entire set R certainly satisfies the condition (since it contains every x in R). Next, we'll check that this topology satisfies property 2. Let 01 and 02 be open sets, and consider the intersection, 01 n 02• If their intersection is empty, we're done; otherwise, let x be a point in 01 n 02• Then x is in 01, which means that there exists some positive number, £1, such that l x  x 'l < £1 for every x' in 01; similarly, since x is also in 02, there exists some positive number, E2, such that l x  x ' l < £ 2 for every x' in 02. Now, let £ be the smaller of £ and E2; then all those points x' that sat 1 isfy l x  x' l < £ will be in both 01 and 02 and thus in 01 n 02. This tells us that 01 n 02 is open. Finally, let's verify that property 3 is satisfied: If I O;li e 1 is any collection of open sets, let x be a point in Uie pi' By definition, x must belong to at least one of the setslet's call it 0Xin (0.). 1. Therefore, there exists I IE some positive number, E, such that every x' that satisfies the inequality l x  x ' l < £ is also in Ox, and,
consequently, is also in U; e pi' With this topology, R is called the real line. Earlier in this chapter, in the section on Set Theory, we mentioned that the set
(a , b) = { x E
R:
a < x < b}
is called an open interval. We'll now see that it is indeed open in this topology on R. Choose any point x in the interval (a, b ) , and let £ be the smaller of the numbers x  a and b  x; then (a, b) contains every point x' that satisfies the inequality l x  x ' l < £ , so (a, b) is open. In particular, the sets (oo, a) and (b, oo) are open sets, so their union, (oo, a) u (b, oo), is also open. The set [a, b] is the complement of this union, so it's closed in this topology on R, which agrees with our earlier statement that [a, b] is called a closed interval.
THE SUBSPACE TOPOLOGY If (X, T ) is a topological space, we can use T to define a topology on any subset, S, of X as follows: A subset U of S is said to be open in S if U is equal to 0 n S, where 0 is open in X. This topology on S, de noted by T5, is called the subspace (or relative) topology, and S is called a subspace of X. It's important to notice that a subset of S may be open in S (that is, a member of the subspace topology, T 5) but not open in X (that is, not a member of the topology T ). For example, let S = [0, 2) in R. Then the set U = [0, 1 ) is open in S, because, for example,
[0,
1) = (1, 1)
n
[0, 2)
That is, U = 0 n S, where 0 = (1, 1 ) is an open set in R. However, U is not open in R (with its standard topology), because [0, 1 ) contains the point 0, and no matter how small we make the positive number £, the set of all x' such that l x '  01 < £ is not contained in U. But we can say that if S is itself open in X, then every set that's open in S is also open in X.
THE I NTERIOR, ExTERIOR, BouNDARY, LIMIT PoiNTS, AND CLOSURE oF A Sn Let (X, T ) be a topological space, and let A be a (not necessarily open) subset of X. The interior of A, denoted by int(A), is the union of all open sets contained within A, or, equivalently, it's the largest open set contained in A. The exterior of A, denoted by ext(A), is the union of all open sets that do not intersect A. (We say that a set U intersects A if U n A # 0.) The exterior of A can also be described as the interior of the complement of A. The boundary of A, denoted by bd(A), is the set of all x in X such that every
284 + C R A C K I N G T H E G R E MATH EMATICS S U BJ E CT T EST
open set containing x intersects both A and the complement of A. A point x in X is called a limit (or accumulation or cluster) point of A if every open set that contains x also contains at least one point of A, other than x. The set of all the limit points of A is called the derived set of A and is denoted A' (not to be confused with the complement of A). Finally, the closure of A, denoted by cl(A), can be defined in two equivalent ways: It's equal to the union int(A) u bd(A), and it's also equal to the union A u A'. The set A is closed if and only if it contains all of its boundary points and all of its limit points. Now, let's look at some examples to illustrate all these definitions. 1. Let A be the subset (1, 2] in R: 2
1
Then we have: int(A) = (1, 2) ext(A) = (<X>, 1) u (2, oo) bd(A) = (1, 2) A' = [1, 2] cl(A) = [1, 2] 2. Let A be the subset (0, 1) u (1, 2) in R: 0
2
1
In this case: int(A) = (0, 1) u (1, 2) = A
ext(A) = (<X>, 0) u (2, oo) bd(A) = (0, 1, 2 ) A' = [0, 2] cl(A) = [0, 2]
3. Let Z be the set of integers in R: •
2
•
1
•
0
•
1
•
2
A D D I T I O N A L T O P I CS + 2 8 S
Then we have: int(Z) = 0 ext(Z) = U11 e z(n  1, n) bd(Z) = Z Z' = 0 cl(Z) = Z
4. Let A be the subset (0, 1) u 121 u [3, 4] in R:
o��or�•��•e4•�3 4 o 1 2
Then we have: int(A) = (0, 1)
u
(3, 4)
ext(A) = (oo, 0) U (1, 2) bd(A) = { 0, 1,
U
(2, 3) U (4, oo)
2, 3, 4}
A ' = [0, 1]
u
[3, 4]
cl(A) = [0, 1]
u
/ 2) u [3, 4]
Notice here that 2 is a boundary point, but not a limit point, of A. In order for 2 to be a limit point of A, every open set containing 2 would need to contain a point of A other than 2 itself. But the open set
(%, %)
, for example, contains no point of A except 2.
BASIS FOR A TOPOLOGY Let X be a nonempty set, and let B be a collection of subsets of X that satisfies the following properties: 1 . For every x in X, there is at least one set B in B such that x E
2. If B 1 and B are sets in B and x E 2 xE
B3 � B 1 n B2•
B.
B1 n B2, then there exists a set B3 in B such that
The collection B is called a basis, and the sets in B are known as basis elements. A basis is used to generate a topology, T , on X, as follows. We say that a subset 0 of X is openthat is, 0 belongs to the topology generated by Bif, for every x in 0, there exists a basis element, B, such that x E B � 0. An equivalent way to describe the topology generated by the basis B is to say that the topology consists of all possible unions of basis elements.
2 8 6 + C R A C K I N G T H E G R E MATH EMATICS S U BJ E CT T EST
For example, the collection B = !(a, b) � R: a < bl of all open intervals of R is a basis for the standard topology on R (that is, the topology we described above). We can define another topology on R by using a different basis. Let:
B' = {[a, b): a, h E R a nd a < b} The topology generated by this basis is called the lowerlimit topology on R. The interval [0, 1) is open in the lowerlimit topology, but not in the standard topology, on R. Also, the lowerlimit topology is finer than the standard topology; in fact, if we let T denote the standard topology on R, and TL the lower limit topology, then T L ;;:;:> T , but T L :t T , which shows that T L is strictly finer than T . Every subset of R that's open in T is open in T L' but not vice versa.
THE PRooucr ToPOLOGY If (X, T J and (Y, T y) are topological spaces, there's a standard way to define a topology on the cartesian product, X x Y. If we let
then the topology generated by this basis B is called the product topology on X x Y.
Example 7.25 L
Which of the following subsets of R2 = R x R are open in the product topology? (a) The interior of the unit circle (boundary not included) (b) The line y = x x (c) The set
(1, 2] (1, 2)
Solution:
'
� 
Notice that sets of the form (xa, xb) x (y", yb)called product topology on R2.
open rectanglesare
open in the
(a) The interior of the unit circle is open in W; the following diagram shows a typical point con tained within an open set: boundary of circle not included
� .. ... ,
' '
.... ..
..

. . . ..
.. ...
;:� :
.. .
\
.
,, .. .. ..
If the boundary of the circle
 
' ' ' '
_ _ _ ..
were included, the region would be a closed set.
(b) For no point p on the line can we find an open rectangle containing p that is a subset of the line. Therefore, the line is not an open set in R2. (c)
Because the given set contains part of its boundary, the set cannot be open. (It's not closed either, because it doesn't contain all of its boundary points.)
A D D I T I O N A L T O P I CS + 2 8 7
CoNNECTEDNEss Let (X, T ) be a topological space. If there exist disjoint, nonempty open sets, 01 and 0 , such that 01 u 02 = X, then X is said to be disconnected. On the other hand, if T contains no pair of2 subsets of X that are disjoint and whose union is X, then X is said to be a connected space. If 5 is a subspace of X, then 5 is disconnected if it's possible to find two open subsets of X, A and B, such that 5 � A u B, and both A n 5 and B n 5 are disjoint and nonempty; otherwise, it's connected. For example, the real line, R, is a connected space, and every interval in R is connected; in fact, the intervals are the only connected subspaces of R. Our intuition tells us that the subspace 5
(; )
,
= (0, 1) u [2, 4] is
disconnected, because it consi st of two separate pieces. We can check that this is true by using the definition 9 above. Let A = (0, and B = 2 , 2 which are both open; then 5 � A u B, and both A n S and B n S are
1)
disjoint and nonempty. Here's a trickier example: Is 5 =
[2, 4) a connected subspace of R in the lowerlimit
topology? Well, the set 5 seems to be one piece and is connected as a subspace of R in its standard topol ogy, but it's not connected as a subspace of R in the lowerlimit topology. If we let A =
[2, 3) and B = [3, 4),
we notice that A and B are open and disjoint, and that their union is 5. The following can be used to identify connected spaces:
1. If A and B are connected and they intersect, then their union is also connected. (This result also holds true for any number of connected sets, as long as their intersection is nonempty.)
2.
Let A be a connected set, and let B be any set such that
3. The cartesian product of connected spaces is connected.
A � B � cl(A ). Then B is connected.
x2,
Let X be a topological space with the property that any two points, x1 and in X can be joined by a continuous path. This means that there exists a continuous function p: [0, 1] X such that p(O) = x1 and p(1) = x2• Then X is said to be pathconnected, and any pathconnected space is connect ed. Interestingly, the converse is not true. There are connected spaces that are not pathconnected. Statement 3 can be used to show that R2 and R3 are connected, and statement 4 can be used to show that any convex set (such as the interior of a circle in R2 or a sphere in R3) is connected.
4.
COMPACTNESS Let (X, T) be a topological space. A covering of X is a collection of subsets of X whose union is X; an open covering is a covering that consists entirely of open sets. If every open covering of X contains a finite subcollection that also covers X, then X is said to be a compact space. Let's look at some examples to illuminate this rather abstract definition.
1. 2. 3.
The real line, R, is not compact, because the open covering A = {A": n E Z} , where n + contains no finite subcollection that can cover all of R. A" = (n
 1,
1) ,
Any topological space X endowed with the trivial topology is compact, because the only open sets are 0 and X, and the finite collection lXI covers X.
The O,P en interval I = A = tA" :
(0, 1) in R is not compact. Consider the collection of open subsets = n 3, 4, . . . } , where A" = ( !; , 1  !;) . Every point of I is in some An, so A is an open
covering of I. However, no finite subcollection of A can also cover I. To see why this is, assume that there was a finite subcollection of A that covered I; of the intervals A" in this finite subcollection, choose the one with the largest value of n; let's call it AN. Then AN contains all the other intervals A " in this subcollection, but does not contain any of the points in the subinterval (0, }:; ] or in the subinterval [1  }:; , 1). This shows that the entire interval I cannot be covered by any finite subcollection of A .
2 8 8 + C R A C K I N G T H E G R E MATH EMATICS S U BJ E CT TEST
4. In contrast to the preceding result, the closed interval I = [0, 1] in R is compact. To see why, let { a } be an open covering of I, and define b as the largest number in I such that the interval
0
[0, b] can be covered by a finite subcollection of
{ 0a } . (To be more precise, b should be defined
as the least upper bound of the set of x's such that [0, x] can be covered by a finite subcollec
{0a } . ) If we can show that b is not less than 1, then we'll have shown that the entire interval [0, 1] can be covered by a finite subcollection of {oa } , which would prove that [0, 1] is compact. So, suppose that b < 1; now we'll derive a contradiction. Since { 0 a} covers all of I, tion of
there must be at least one set in the covering that contains b; let's choose one and call it O . Since O
B
B
is open, it must contain an open interval of the form (b  £, b + £) for some positive number £.
Now, because we defined b as the largest x in [0, 1] for which [0, x] can be covered by a finite � subcollection of { Oa } , the number b  � is less than b, so the interval [0, b  ] can certainly be covered by a finite subcollection of
{0a } . But then simply adjoining the set Ob to this finite 2
subcollection gives us a finite subcollection that covers [0, b + tion of b.
2
�
2
], which contradicts our defini
The following can be used to identify compact spaces: 1. Let X be a compact topological space, and let 5 be a subset of X. If 5 is closed, then it's compact. The converse (compact subsets are closed) is true if X is Hausdorff.
2. The cartesian product of compact spaces is compact. We'll close this section with what is perhaps the most commonly used result in identifying a com pact set. First, a couple of definitions. We define the norm of a point x = (x1 , x2, , x,) in R" as the real number: .
.
•
This is just the magnitude of the vector joining the origin, 0, to the point x. A subset A of R" is said to be
bounded if there exists some positive number M such that the norm of every point in A is less than M. (This means that A is entirely contained within the sphere of radius M, centered at the origin.)
3. HeineBorel theorem: A subset of R" (in the standard topology) is compact if and only if it's both closed and bounded. Notice that with this result, we could immediately say that the interval (0, 1) is not compact (because it's not closed), but that the interval [0, 1] is compact.
METRIC SPACES Let X be a nonempty set, and let d: X x X 7 R be a realvalued function defined on ordered pairs of points in X. This function d is said to be a metric on X if the following properties hold true for all x, y, and z in X:
d(x, y) � 0, and d(x, y) 0 i f and only i f x = y 2. d(x, y) = d(y, x) 3. d(x, z) � d(x, y) + d (y, z) 1.
=
A D D I T I O N AL T O P I CS + 289
We call d(x, y) the distance between x and y. A set X, together with a metric on X, is called a space. If £ is a positive number, then the set:
metric
Bd (x, E ) = { x ' E X: d(x, x ') < c} is called an £ball, the (open) ball of radius £ centered on x. Every metric space can be given the structure of a topological space. The collection of all £balls,
is a basis for a topology on X, called the metric topology (induced by d). In this topology, a subset 0 of X is open if and only if every point in 0 is the center of some £ball entirely contained within 0; that is, 0 is open if for every x in 0, there's some positive number, Ex, such that Bd(x, £) � 0. 2 The euclidean spaces R, R , R3, etc. are all metric spaces. If we define the euclidean metric on R" as
it can be shown that the metric topology induced by the euclidean metric, as well as by the simpler metric, cr( x , x') = max (called the square ogy on R".
{ l x 1  x; l , l x 2  x � l ,
l
. . . , x ,  x;,
l}
metric, because the £balls Bcr(x, E) in R2 are actually squares), gives the standard topol
CONTINUOUS fUNCTIONS We've already discussed the idea of what it means for a function f: R 1 R to be continuous; in this section, we'll generalize the definition to functions between arbitrary topological spaces. Let's first review the definition of continuity for a function f: R 1 R. We say that f is continuous at the point x0 if, for every £ > 0 (no matter how small), there exists a 8 > 0 such that:
Now, let's generalize this to metric spaces. Let (X1, d1) and (X2, d2) be metric spaces. A function
f: X1 1 X2 is said to be continuous at the point x0 if, for every £ > 0 (no matter how small), there exists a 8 > 0 such that:
Iff is continuous at every x0 in X1 , then we simply say that f is continuous. Notice that in R, the euclidean metric is d(x, x ' ) = l x  x 'l , so this definition is the same as the previous one, in the case of a function f: R 1 R. Based on our discussion of the metric topology, we can formulate the definition of a continuous func tion between metric spaces in another way. We can say that f: X1 1 X2 is continuous at the point x0 if for 1 every open set, 0, containingf(x0), the inverse image, F (0), is an open set containing x0• Now, let 0 be an open set in X2 that contains f(x0), and choose an £ball, Bd2 (f(x0 ), £), centered on f(x0) and contained in 0. F1 Bd2 (f(xo ), is open and contains Xa, so it must contain some open halt Bd (xo, D), centered on l x0• Therefore, for any x such that d1 (x, x0) < 8, we have d2 (f(x), f(x0 )) < £ , which matches the condition for continuity given above.
(
£))
2 9 0 + C R AC K I N G T H E G R E MAT H EMAT I CS S U BJ ECT T EST
T 1) F1 (0)
We're now ready for the final generalization. Let (X1, and (X2, T 2) be topological spaces; we say that the function (or map) f: X1 7 X2 is continuous if, for every open set, 0, in X2 (that is, for every E T2 ), the inverse image, is open in X1 (that is, E T1 ). In particular, f is continuous at the point x0 if for every open set 0 in X2 containing f(x0), there's an open set 01 containing x0 such that j(01) � 0. If the topology of the range space, T2, is generated by a basis, B , then, to check whether f: (X1, T1) 7 (X2, T2) is continuous, it's sufficient to check that the inverse image of every basis element B E B is open in X1.
F1 (0) ,
0
To illustrate this abstract definition, our first few examples will be of realvalued functions defined on R, with which you're comfortable. 1. Let f: R
7
R be the function: f(x) =
{2+2
if x
, which is not periodic), the functions cos z and sin z are evidently periodic with period 2n; that is, cos (z + 2n) = cos z and sin (z + 2n) = sin z. However, not all the properties of cos x and sin x (for real x) are the same as those of cos z and sin z (for complex z). One notable example is the fact that both cos x and sin x are bounded ( J cos xJ � 1 and J sin xJ � 1 for every real x), but the functions cos z and sin z are unbounded (that is, J cos zJ and J sin zJ can assume any nonnegative real value). The definitions of the other trig functions follow from those for cos z and sin z: 1 1 1 sin z cos z tan z =  ' cot z =  =  ' sec z =  ' and csc z =  cos z tan z sin z sin z cos z
Notice that these equations are just like their real counterparts. Example 7.29 Solution:
Does there exist a complex number z such that sin z = 2? \
Of course, there exists no real number x such that sin x = 2, but there does exist a complex number z such that sin z = 2in fact, there are infinitely many. Using the definition of sin z, the equation sin z = 2 becomes: A D D I T I O N A L T O P I CS + 303
iz
iz
e e 2i = 2
e iz  eiz = 4i e 2iz  1 = 4ieiz
Since this equation is quadratic in the quantity eiz, the quadratic formula gives us:
(4i) ± �(4if + 4 = 4i ± i . 2J3 = (2 ± 'Jr;;3 )z. 2 2 Therefore, iz = log [ (2 ± .J3 )i J, which implies z = i log [ (2 ± .J3 )i J, so: e
iz
=
ilog [ (2 ± J3)i J = i [ log(2 ± .J3) + i ( f + 2/crt ) J, for any k (f + 2/crt )  ilog(2 ± J3), for any k Z
z=
=
E
E
Z
THE HYPERBOLIC FUNCTIONS
The hyperbolic cosine (cosh, rhymes with gosh) and the hyperbolic sine (sinh, pronounced sinch) are defined for real x by the equations: e x  ex e x + ex cosh x = and sinh x = 
2
2
These functions have names similar to the trigonometric functions because they have many similar properties. For example, cosh 0 = 1 and sinh 0 = 0, just as cos 0 = 1 and sin 0 = 0; cosh(x) = cosh x and sinh(x) = sinh x, just as cos(x) = cos x and sin(x) = sin x; and the identity cosh2 x  sinh2 x = 1 is simi lar to the Pythagorean trig identity cos2 x + sin2 x = 1. One important difference between the hyperbolic functions and their "corresponding" trig functions is that, while 1 ::; cos x ::; 1 for all real x, the hyperbolic cosine satisfies cosh x � 1 for all real x; and although 1 ::; sin x ::; 1 for all real x, the value of sinh x can be any real number. Thus, both cos x and sin x are bounded, but cosh x and sinh x are unbounded. The definitions of the hyperbolic sine and cosine are exactly the same if x is allowed to be a complex variable, z. It's apparent that these definitions look remarkably like the definitions of cos z and sin z for complex z. The relationships between them are summarized in the following pairs of formulas: cos{iz) = cosh z cosh(iz) = cos z
and
i sinh(iz) = i sin z
sin{iz) = sinh z
With these formulas, we can derive equations that can be used to figure out the values of cos z and sin z, where z = x + yi, in terms of x and y.
304 + CRAC K I N G TH E G R E MATH EMAT I CS S U BJ E CT T EST
For example: +
=
cos z = cos(x yi) cos x cos(yi)  sin x sin(yi) cosxcoshy  i(sinxsinhy) =
and
sin z = sin(x + yi) = sin x cos(yi) cos x sin(yi) = sin x cosh y i(cos x sinh y) +
+
These equations can be used to derive the following (where we use the alternate notation, *, to denote complex conjugation): =
i coszl 2 (cosz)(cosz)* = [cos x cosh y  i(sin x sinh y )][cos x cosh y i(sin x sinh y )] cos2 xcosh2 y + sin2 xsinh2 y = cos2 x(1 + sinh2 y) + (1  cos2 x)(sinh2 y) cos2 x + sinh2 y +
=
=
and
lsinzl 2 = (sinz)(sinz)* = [sinxcoshy + i(cosx sinhy)][sin xcoshy  i(cosxsinhy)] = sin2 x cosh 2 y + cos 2 x sinh2 y = sin2 x(1 + sinh2 y) (1  sin2 x)(sinh2 y) = sin2 x + sinh2 y +
Since sinh2y is unbounded, these equations verify our assertion in the last section that are unbounded.
l sinzl
Ieos zl and
THE DERIVATIVE OF A fUNCTION OF A (OMPLEX VARIABLE A function j(z), that accepts a complex number, z, as input is called a function of a complex variable. The function f can be expressed in terms of realvalued functions, as follows: Letting z = + we can write = = For example, the function j(z) = can be written as:
z2 j(z) j(x + yi) u(x, y) + iv(x, y) . j(z) = (x + yi)2 = (x2  y2) i (2xy) A function j(z) is
u(x,y)
+
said to be differentiable at the point z0 if lim h;0
j(zo
x yi,
....__,..__..
v(x,y )
+ h)  j(zo ) h
exists; if this is the case, the derivative off at z0 is denoted by j'(z0). It's important to understand that h is also a complex variable, so the statement "h 1 0" means that the complex number h approaches the origin. Because a nonzero complex number h can approach the origin from infinitely many directions, the existence of the limit above demands that the difference quotient approach a finite value regardless of the A D D I T I O N AL T O P I CS + 305
way that h t 0. This is a very stringent requirement, much stronger than the corresponding statement that h t 0 where h is a real variable (because in this latter case, h has only two avenues of approach to zero: from the right or from the left). Consequently, differentiability of a function of a complex variable has powerful repercussions (some of which we'll mention below), results that do not apply to functions of a real variable. Let's look at a couple of examples. The function f(z) = z is not differentiable at the point z = 0, since
j(O + h)_:_"''j(O) = h h h and lim ( h /h) does not exist. To see why, notice that if h approaches the origin along the real axis, then h = h, so lim ( h /h) = lim (h/h) = lim 1 = 1; but, if h approaches the origin along the imaginary axis, then h h, so lim ( h /h) = lim (h/h) lim (1) 1 However, we can show that the functionf(z) = z2 is differentiable everywhere; the difference quotient is: j(z + h)  j(z) = (z + h) 2  z2 = 2zh + h 2 = h(2z+ h) = 2z + h h h h h .::._ ...;_ _
h+0
=
h+0
h+0
h+0
h+0
h+0
=
h+0
=

.
Therefore, no matter how h approaches 0, we have
lim f(z + h)h  j(z) = lim (2z + h) = 2z IHO
h+0
so j'(z) = 2z . Notice that the derivative ofj(z) = z2 is j'(z) = 2z, just as the derivative of the realvalued function f(x) = x2 is j'(x) = 2x . In fact, all the differentiation formulas listed in Chapter 2 hold true in the case of a complex variable. Example 7.30 Solution:
Iff(z) = z  arctan z, solve the equation f '(z) = i .
Since
1  ) , we have: f'(z) = 1  ( 1 + z2 4 1 1 =  1 = 1 1 + z2 3 1 + z2 3 =>
3
=>
z2 = 4
=>
z = ±2i
THE CAUCHYRIEMANN EQUATIONS
Let f(z) = f(x + yi) = u(x, y) + iv(x, y). If f is differentiable at z = x + yi, then the limit
u(x + Re h,y + lmh)  u(x,y) +lim i[v(x + Re h,y + lm h) v(x, y)] li....mo f(z + h)h  j(z) = lim ....o Re h + i Im h Re h + i lm h IHO
�z
�z
exists, regardless of how h approaches the origin. If we let h approach zero along the real axis, then Im h = 0, and the righthand side of the equation above becomes:
h,y) u(x,y) + lim i[v(x +Reh,y)  v(x,y)] = au + i av lim u(x+Re Re dX dX h Re h
ReiHO
3 0 6 + CRAC K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
ReiHO
Alternatively, if we let h approach zero along the imaginary axis, then Re h = 0, and the righthand side of the equation that defines j '(z) can be written as:
i[v(x,y +Imh)v(x,y)] = ! au + av av i au l i lim u(x,y+ limh)u(x,y) m + lm h i ay ay ay ay i lm h =
lm ii>0
lm il>0
The two expressions we've obtained must be identicat since they're both equal to f' (z) ; therefore,
au .av av .au +z=zax ax ay ay auax  avay axav  auay
which implies that:
a nd
These are called the CauchyRiemann equations, and both must be satisfied if j(z) is to be differentiable. Conversely, if the CauchyRiemann equations are satisfied and the four partial derivatives are continuous throughout an open set containing z0, then the function f is differentiable at the point z0 . For example, we showed that the function j(z) = :Z is not differentiable at the origin . Using the CauchyRiemannequations, we could establish this result (and something more) withlittleeffort. The equation j( + gives us = and = Since
x yi) xyi
u(x, y) x v(x, y) y. au = 1 ay =1 ax the first of the CauchyRiemann equations, ux vY, is never satisfied, so j(z) is differentiable If f( z) f(x + yi) = u(x, y) + iv(x, y) is differentiable throughout some open set then the Cauchy Riemann equations hold true Because ()ujax = avjay and auj(}y avjax, we have =
but
dV
= :Z
=
=
=
.
so:
�> :x (� ) = :x (� ) = ::�
am
0,
nowhere.
�� = � (� ) = � (: ) =:;x = ::�
Any function of(x, y) that satisfies this equation (which is called Laplace's equation) throughout some open set 0 is said to be harmonic in 0. Using the same method, we could use the CauchyRiemann equa tions to show that the function also satisfies Laplace's equation; that is:
v(x, y)
a2v a2v +=0 ax2 ay2
Therefore, ifj(z) is differentiable throughout some open set, 0, in the plane, then the real and imaginary parts of Jthe functions and respectivelyare harmonic in 0.
u(x, y) v(x, y),
A D D IT I O N A L T O P I CS + 3 0 7
ANALYTIC fUNCTIONS
If f(z) is differentiable at z0 and at every point throughout some open set in the complex plane contain ing z0, then we say that f(z) is analytic at the point z0. In general, if f(z) is differentiable throughout some open set 0 in the complex plane, then f(z) is said to be analytic in 0. If f(z) is analytic everywhere in the complex plane, then j(z) is said to be an entire function. Examples of entire functions include all polynomial functions, as well as the functions ez, cos z, and sin z. A rational function of zthat is, a function of the form f(z) p(z)/ q(z), where p and q are polynomialsis analytic at all points for which q(z) :t 0. Iff(z) is analytic, its component functions, u(x, y) and v(x, y), must have continuous partial derivatives of all orders. Furthermore, ifj(z) is analytic at some point z0, then its derivative, f'(z) , is also analytic at z0; this result immediately implies that the derivatives of all orders of j(z) are analytic at z0! There's no such result for functions of a real variable. For example, consider the realvalued function: =
for x < 0 for x :2: 0 Then f(x) is differentiable everywhere, and
:2:0
for x < 0 for x
so f'(x) = lxl . However, the second derivative off(x) does not exist at x = 0, since we know that j'(x) = l xl is not differentiable at x = 0. Example 7.31
Solution:
Let f(z) be the function f(z) = 2xy + i(x2 + y2). Show that there exist infinitely many points in the complex plane at which f is differentiable (and describe those points) but that f is analytic nowhere.
The component functions of f(z) are u(x, y) = 2xy and v(x, y) = x2 + y2 • The partial deriva tives of these functions are u = 2y, uy = 2x, v = 2x, and vy = 2y . We notice that the partial derivatives are continuous everywhere, so f will be differentiable wherever the CauchyRiemann equations are satisfied. The first CauchyRiemann equation, ux = vY, becomes 2y = 2y, which is satisfied everywhere. The second CauchyRiemann equation, uY = vx, becomes 2x = 2x, which is satisfied if and only if x = 0. The line x = 0 is the imaginary axis, so f is differentiable at every point on the imaginary axis, and for these points we have: X
X
f'(z) = ux + ivx = 2y + i(2x) 2y =
Despite the fact that f is differentiable at infinitely many points, it is analytic nowhere. By definition,/ is analytic at a point z0 if it's differentiable at every point throughout an open set containing z0• However, every open set in the complex plane (for example, every open rectangle) that contains a point on the imaginary axis, contains points not on this axis. Be cause f is not differentiable throughout any open set in C, it's analytic nowhere.
3 0 8 + CRAC K I N G T H E G R E MAT H E MATI CS S U BJ ECT TEST
CoMPLEX LINE I NTEGRALS Let C be a smooth path in the complex plane, which is given parametrically by the equation
from t = a to t = b (where t is real). If f(z) is a continuous function along C, then we have:
z(t) = x(t) + iy(t),
fc J( z)dz = J: f(z(t)) z'(t)dt •
For example, let's integrate the function f(z) = z2  iz along the path z(t) = x(t) + iy(t) that connects the origin to the point 1 + 2i, given by:
{yx == t2f
for 0�1
Since f(x + yi) = (x + yi)2  i(x + yi) = (x2  y2 + y) + (2xy  x)i, we get:
fc J(z )dz = J: f( z(t)) z '(t)dt •
= f� f(t + 2fi) . f,(t + 2t2i)dt = J� { [t 2 (2t2 f + 2t2 ] + [2t(2t2 )  t]i } . (1 + 4ti)dt = f)(3t2  4t4 ) + ( 4e  t)i](1 + 4ti)dt = J� [(7t2 20t 4 ) + i(16t 3  t  16t5)]dt 1 = [ (Z·t3 3 4t5) +i(4t4  12 e  .§.t6 3 )] =  .2. + .2.i 0
6
3
Alternatively, we could have invoked the fundamental theorem of calculus and written
fc J( z)dz = J: f( z(t)) z '(t)dt = J: F'(z(t))dt = F(z(t)) J: = F(z(b))  F(z(a)) where F(z) is an antiderivative off(z). Forf(z) z2  iz, it's clear that we can take F(z) t z3  t iz2, so 1 I f( z)dz = 3 z3  l2 iz2 ] +zi •
c
1
=
=
0
= t (1 + 2i)3  t i(1 + 2i)2 = i (1 + 2i)2 [ 2(1 + 2i) 3i J = i (3 + 4i)(2 + i) = t(10+ 5i) = .2.+.2.i 3 6
which is the same result as before.
A D D IT I O N A L T O P I CS + 309
THEOREMS CONCERNING ANALYTIC fUNCTIONS
In this section, we'll list some of the most important results in complex analysis, which you should know for the GRE Math Subject Test. 1 . Cauchy's Theorem: If J(z) is analytic throughout a simply connected, open set, D, then for every closed path, C, in D, we have:
In particular, ifj(z) is an entire function, then � f (z)dz is equal to 0 for any closed path in the c plane. 2. Morera's Theorem: This result provides a sort of converse to Cauchy's theorem. If f(z) is contin uous throughout an open, connected set 0 in the complex plane, and �c f(z)dz = 0 for every closed curve C in 0, then f(z) is analytic in 0. 3. Cauchy's Integral Formulas: If f(z) is analytic at all points within and on a simple, closed path, C, that surrounds the point z0, then:
Furthermore, the nth derivative, J R, "=0 where the radius of convergence, R, is given by: R=
�
�
lim ' ja" l fl'too
,
lim .zfaJ is nonzero and finite
if _ , ,�
�
= 0 , then the series If � = oo , then the power series converges onl� for z 0, and if � converges for every z. More generally, the power series L a, (z  z0 )'' converges absolutely for all z that satisfy l z  z0 1 < R and diverges for all z such that l z  z0 1 > R, where R is given as above. If � then the power series converges only for z zO' and if � 0 , then the series converges for every z. Every functionf(z) that's analytic in an open, connected subset, 0, of the complex plane can be expand ed in a power seriesits Taylor serieswhose disk of convergence lies within 0. That is, if z0 is any point within 0, and R is a positive number such that the disk of radius R centered at z0 is contained in 0, then =
n;::: Q
�
=
=
\1faJ
= oo,
is valid in the open disk lz  z0 j < R. Not only can every analytic function be expanded in a Taylor series; the converse is also true. That is, for every z in its disk of convergence, a Taylor series gives an analytic function. SINGULARITIES, PoLEs, AND LAURENT SERIES
Let z0 be a point in the complex plane, and R a positive number. The set of all z such that 0 < jz  z0 I < R is called the punctured open disk (of radius R) centered at z0. If a function f(z) is not analytic at a point z0, but is analytic at some point in every punctured disk centered at z0, then z0 is said to be a singularity of f(z). If z0 is a singularity of j(z), but f(z) is analytic at every point in some punctured open disk centered at z0, then we call z0 an isolated singularity of f(z). Let's now assume that f(z) has an isolated singularity at the point z0• If there's a positive integer n such that
j (z) = (zg(zz ) )" 0 with g(z) analytic in a nonpunctured open disk centered at z0 and g(z0) :t 0, then the singularity z0 is called a pole of order n. A pole of order 1 is called a simple pole, a pole of order 2 is called a double pole, and so forth. However, if we can't write f(z) in this form, for any positive integer n, then z0 is said to be an essential singularity.
A D D I T I O N A L T O P I CS + 3 1 1
If a function f(z) has a singularity, it may not be possible to expand it in a Taylor series. For example, the function f(z) = 1/(1  z) can be expanded in a Taylor series within the disk jzj < 1, but it can't be expanded in a Taylor series in the disk lzl < 2, because this larger disk contains the point z = 1, which is a singularity of f(z), and the sum of a Taylor series is always an analytic function throughout its disk of convergence. To provide a valid series representation for functions that have singularities, we need to generalize the notion of a Taylor series. An annulus is the region between two concentric circles; in the complex plane, an annulus centered at z0 is described by a pair of inequalities of the form R1 < lz  z0l < R2. (We'll allow R1 to be zero, and we'll also allow R2 to be oo, so not only can an annulus be a finite ring, it could be a punctured disk of finite radius, the entire complex plane minus a closed disk, or the entire complex plane minus a single point, z0.) If f(z) is analytic in some annulus centered at z0, then j(z) can be expanded in a Laurent series, which is a series of the form: �
+
�
(z z " (z z �>n=l n  or �>"  oY siJ�ar (or pnncipal)
analytic part
part
The coefficients of the nonnegative powers of(z  z0)those in the analytic partare the usual Taylor coefficients, which can be calculated from the formula:
The coefficients of the negative powers of(z  z0)those in the singular (or principal) partare called Laurent coefficients. If C is a simple, closed, positively oriented curve in the annulus, then these coef ficients can be computed using the equation: a n
1 =2 · 1tl
f (z f(z) c
Zo
)11+1
dz (n � 0)
In practice, we don't usually figure out the Laurent coefficients from this formula; instead, we derive them by algebraic manipulations of the Taylor series. The Laurent series can be used to identify poles and essential singularities. If the singular part of the Laurent series contains at least one term, but only a finite number, then z0 is a pole. In particular, if k is the largest integer such that ak :;:. 0 but a_" = 0 for all n > k, then z0 is a pole of order k. On the other hand, if the singular part of the Laurent series contains infinitely many terms, then z0 is an essential singularity. Let's look at an example. The functionf(z) = 1/(1  z) has a singularity at z = 1, so it cannot be expanded in a Taylor series in, say, the disk jzj < 2. Within the disk jzj < 1, the Taylor series is valid, and we have 2 1  = 1 + z + z + z 3 + · · · , for I z I < 1 1z
which is the wellknown formula for the geometric series. Now, to avoid the singularity at z = 1, we expand f(z) = 1/(1  z) in a Laurent series in the region 1 < j zj < oo. To do this, we first write: 1 = 1 =1 1 · 1  z z (�  1) ; 1  �
3 1 2 + CRAC K I N G T H E G R E MATH EMAT I C S S U B J ECT TEST
lzl > 1, then � �� < 1, so the following expansion is valid: 2 2 1 � z =  � · 1� l =  � [ 1+ (� ) + (� ) + ··· ] =  �  ( �)  (�t  · · · , for lzl > 1 Since there are infinitely many terms in the Laurent series, z = 1 is an essential singularity.
Now, we notice that, if
z
Example 7.32
Find the Laurent series expansions of the function
f(z) = ( X4z z 1 z 3)2
that are valid in the annuli (a) 1 < lzl < 3 (b) o < 1 31 < 2 Solution:
z
The partial fraction decomposition of f(z) is:
1 + 6 1 + 4z = (z  1)(z 3)2 z 1 z 3 (z  3i �
�

(a) For
lzl > 1, we can write:
Next, for
l z l 3, the following expansion is valid: 1=3  3 ( 11 1. )  1 L� ( )"  L� y.;T z
0), and e is never equal to zero. We can check that the critical point x = k gives a maximum since: :t
j'(x) = xkl ex (x + k) :=} f"(x) = xkl e x (1) + ( x + k) • :x (xkl ex ) :=} f"(k) = e� ek < o
23.
Therefore, the maximum value attained by f is f(k) = (; r The area of a ci r cl e of radi u s r is given by the formula A(r) = rtr2. Differentiating this with respect to time, we find that: =
A
k'l!"
dA = 2nr dr dt dt So, if drIdt = 0.5 em/ sec,dAthe value of dA/dt when r = 4 em is: dt = 2n(4 cm)(0.5 em/sec) = 4n cm2/sec 24. B
Appl calculyuins gandthethefinalchairesuln rult quoted e tell usinthat:the solution of Example 2.18, the fundamental theorem of e"
f(x) = L tlogtdt :=} j'(x) = [e2x log(e2x )] • 2e2x  [ex log(ex )] • ex = 4xe4x  xe2x = xe2x (4e2x  1)
Setting this equal to 0, we find the critical points:
:=} X = 0 or 4e2x = 1 :=} x = 0 or x = t log t = log t = lo g 2 so we iconcl The tellsmusum.thatWef hascouland veriabsolfyuthite smibynimchecki um atngxthat= 0, f"(x) givesquesti a localonmaxi s negatiudevethatat xx=l= loogg2:2 j'(x) = xe2x (4e2x  1) :=} f "(x) = (xe2x )(8e2x ) + (4e2x  1) • :x (xe2x ) :=} j "(log 2) = (log2)(e21og2 )2(8) + 0 = tlo g2 j'(x) � 0 :=} xe2x (4e2x  1)
=
0
tan e
We now express the area of the shaded region, r =
33. C
3 cos e
= J3
=>
e = .::.3
y
in polar coordinates, as the sum of these two definite integrals: area= J n13 �(J3 sine?de + Jnnn12 �(3 cos0)2 d0 The parabola y = ax2 and the line y = bx intersect at x = 0 and at x = ab . From the figure below, 0
y
weevalseeuatithat the vol u me of the sol i d of revol u ti o n can be obtai n ed by the washer method by ng the definite integral as on the following page. S O LU T I O N S TO T H E C H APTER R E V I EW Q U ESTI O N S + 3 S 1
volume= J 0 n [(bx)2  (ax2l]dx = nf0 (b2x2  a2x4 )dx 5 Jo 3 = n [lbzx _ lazx 5 3  b/a
 b/a
 b/a
= .l..15 n • if this resul t is to lbey posia constant, independent of a both and b,a andthenbtheare ratiposiotb5ive).a3 That mustisi,tselwef ' beNow, a constant ( n ecessari t i v e, si n ce we r e tol d that must have = ka3, for some positive constant k. = x2 and the graph of y =  l xl intersect at (±2, 4). From the figure below, which The incluparabol des theaequati ons of the boundaries in the first quadrant solved for x explicitly in terms of Jt.. aJ
I
b5
34. A
6
y
y,
(2, 4)
(2, 4)
x ={Y
weatinseeg thethatfolthelowivolngupaimerofofthedefisolniitde ofintegral revolus:tion can be obtained by the disk method by evalu volume= fo4 n [(.jY)2] dy + J: n [(6  yl] dy = 1tJ04 y dy + 1tJ: + y2)dy = 1t [t l J: + 1t [  6l + t l J: = 81t + �1t3 = 332 1t Note: One way to avoid evaluating the second integral above (the more complicated of the two), and thus providing a shortcut to the answer, is to notice that the portion of the solid of revolution from = 4 to = 6 is a right circular cone with base radius 2 and height h 2, so its volume is = t nr2 = t 1t • 22 • = t 1t , just as we found above. (36  l2y
36y
y
Vcore
y
h
3 S 2 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ E CT TEST
2
r=
=
35.
A First, we use the equation of the hypocycloid to find y': x zf3 + y zf3 = 1
y = (1  x 2f3 )3/2 ' y = � (1  x 2f3 ?fz (f x 1/3 ) = x1/3 (1  x 2f3 )1/2
=>
=>
•
Then, using the formula for arc length, we get: s=
sx=x=ab �1 + (y')2 dx = J1/1 8 \j11 + [ x1/3 (1  x2f3 )1/2 ] 2 dx = J1/1 8 �1 + x 213 (1  x2f3 ) dx = J11/8 �x213 dx =
JI1/8 x1/3 dx
= 2 x z/3 ]1/1 8 = t(1  (t?/3 ] = .2.8 36. D
[.2
Simplifying the lefthand side, we find that: ae dx fe xfax dy = 1 r x �::y]� = l fae dx = 1 xlogx log�og x)J:' = 1 log�og ae)e = 1 loga = e 
a
y
e
a=e
S O L U T I O N S TO TH E C H APTER R E V I EW Q U ESTI O N S + 3 5 3
37. B
z The given limit is of the indeterminate form 1�. So we set y = (cos x)cot x , giVmg 0 log y = (cotzx)log(cos x) = [log(cos x)] /(tan2x), which is of the indeterminate form 0 . Applying L'H6pital' s rule to this expression, we get: lim log(cos x)
x>0
tan X
2
Since log y 7
 sin x lim =
�
X>0 2 tan X sec 2 X
 sin x  cos 2 x cos x = lim x>0 1 sin X 2 X>0 2 cos x cos 2 x
lim =
=
1
2
1
 2 , we can conclude that y 7 e1 12; that is: z lim (cos x)cot x = e1/2 = _!_ x>0 .j";
38. C
J
We can simplify the integral dx/ xQ.og xY by making the substitution u = log x, du = dx/x:
J
dx xQ.og x) "
=
f
du
7
=
+ { ;+1 U  n 1 if n :;t 1
log u
if n = 1
;+1 Q.og xf" = {logQ.og x)
+1 if n :;t 1
if n = 1
Since log x 7 oo as x 7 oo, we see that the improper integral converges if and only if n + 1 < 0, that is, if and only if n > 1. In this case, we find that:
f� e
39. E
]
b 1 dx m [ 1 fl,.og X ) n +1 ] _ 1 lim [ _ 1 _ 1 li>� b "1  1   11+1 ( 0  1 )  n1 Q._ e  11+1 b>� 1 n + b xQ._og x) " og )
Xa
a.
Both of the integrals in this question are improper integrals of the second kind, since each integrand goes to infinity at the upper limit of integration, x = However, as we'll see, both integrals converge. The integral on the lefthand side of the given equation is simplified by making a trig substitution: = sin 6, dx = a COS 6 dS:
a2
The integral on the righthand side of the given equation is simplified by making a simpler substitution: u =  x2, du = 2x dx: u �_ d_ x=d=x= 1 u 1 1z � u = _u 112 = ,./a 2 = = x2 r=z 2 ,./a  x2 .[;;_
J
Therefore, the equation
J
simplifies to:
3 5 4 + CRAC K I N G T H E G R E MAT H EMAT I C S S U BJ ECT TEST
J(
_
_
_
arcsin (; x)J: = [ .Ja2  x2 I arcsin 1 = l al a = + "'2 Since we're asked for the positive value of a that satisfies the equation, we take a = To evaluate the integral in I, make the substitution X = tan e, dx = sec2 e de dx =J sec 2 8 d8 = sec2 8 d8 = _:!!!_ = J cos e 8 f (x2 + 1)2 tan2 8 + 1)2 J (sec2 8)2 J sec2 8 [
40. B
2
(
�.
d
= I t(l + cos28) de = te + tsin28 = l8 2 + .lsin8cos8 2 = 2 arctan x + 2(x2X+ 1) l
where we've used the triangle
=
1
to calculate the product sin 8 cos 8 = (x/.Jx2 + 1 ) (1/.Jx2 + 1 ) = x/(x2 + 1) . Now, since x =  arctan a  a = 1 dx = arctanx+ a� 2 2 2 fa_(x +1) { 2(x2 + 1) lo �{ 2 2(a2 + 1) l 2 (  2)  0 = 4 and dx:x = arctan b + b = 1  + 0 = + x (x:2,. .+ 1?  = arctan 2(b2 + 1) l 2 ( 2 ) 4 2(x2 + 1) l { 2 f� {2 we conclude that the given integral converges, and, in fact, we have .
1
n
n
.
b
0
lim 1
0
b >
1t
1
lim
1

b >
7t
1t
7t
notalthough its actualfor thivalsuquesti e. on we're concerned only with the convergence of the improper integral, Tov = eval uate the integral in II, we use integration by parts, with u = x, du = dx, and ex, dv = ex dx: x
X+1 I xexdx = x(e ) I (e ) dx = xe  ex = e (x + 1) =  7 x

x
x
S O LU T I O N S TO T H E C H APTER R EV I EW Q U ESTI O N S + 3 5 5
Therefore, the given improper integral converges, since whereFiwenalfound lim"__uate. J(bthe 1)integral eb] = infromIII, weonemake applithecatisubsti on of tL'utiHoopin utal='s2rulx,e. du = dx: ly, to eval f 2dxx)2 _ udu2 ( u2 ) du u1 _ _!_  21 J J ( We now see that the given2 improper integral divberges, because: ] lim [1 _ _!_] dx lim [1 J (2 x)2  b72 2 x  b>2 2 2 Thus, Alonlternati y the ivneltegral y, yous icann I andsimplIIyconverge. use estimates: 1 = 12 1 ( x 2 + 1 r x4 + 2x + 1 Since the series L� n1 converges (it's the pseries, with p = 4 > 1): cos4 (arctann) 1 revery n � cos4 (arcta n) converges nvn n nvn As for the series in II, the integral test shows that it diverges, since: f = dx log�og x) f� = dx lim . [ log�og x) Jb = 2 b log >� l o g 2 The series in also diverges, because the terms do not approach as +
I
0



_
0
A
n=l
�
��
41
X
E
_ _
x
_
0
b
oo
5
fo
=>
X
III
42.
u
_
_
41.

_
=>
L...
X
X
11=1
n
41
oo
0
n 7 oo:
Therefore, the series in I is the only one among the three that converges. Let's apply the ratio test to the given series: xn+l = lim {[(n + 1)!f an+l • (2n)! } 11>� xn n>� [2(n+1)] ! (n ! f a" + 1/ =a • lim, _,�(2n +(n1)(2n + 2) 2n + 1 =a • limn>� 4n2n2 ++6n+2 =a•41 lim
3 5 6 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ E CT TEST
order for the series to converge, this limit must be less than 1; therefore, the series converges if 0 1, that is, if 0 4. Thus, we take b = 4. The expressi o n we' r e asked to si m pl i f y i s the l i m i t of a Ri e mann sum, whi c h we can compute by evaluating its corresponding definite integral. Since In
43. D
:s;
fa
�
II>�
>H�
2
6�
6�
.
= 7.
47. C
ex2 x ,
2
 1 ex x ex2 x
2ex2 x
2 = ex x
ex2 x
2
2

7 ,
7
7 6
If,
and
so
= = (1 + X2 + ...)(1 x +! X2 7; X3 + ... ) =1(1x+!x2 7; x3 + .. ·) +x2 (1x+!x2 7; x3 + . (1 x + t x2 7; x33 + + (x2 x3 + . . .) + ... = 1x+.J.x2 2 _zx + from which we can see directly that the coefficient of x3 is  i . ex2 .:
ex2 e x
=
· · ·
6
..
. ·)+ . . .
)
·
S O L U T I O N S TO TH E C H A PT E R R EV I EW Q U ESTI O N S
+ 3S9
48. A
The righthand side of the given equation is the Taylor expansion of the function f(x) = .x4 in powers of(x + 1). The coefficient of(x + 1)3 is therefore:
k3
If f(x) = .x4, we have
f'"(1) 3!
.::._'_ '
f"(x) = 12x2,
f'(x) = 4x3 ,

=
f"'(x) = 24x, . . .
so f"'(1) = 24(1) = 24 , which gives us:
49. C
k3
=
/"'(1) 24 24 =  =  = 4 3! 3! 6
We're looking for the smallest integer, n, such that the nthdegree Taylor polynomial, P,(x), for f(x) = !?' approximates � = e0·2 = /(0.2) to within 1�. The difference betweenf(0.2) and P,(0.2) is
/(0.2)  p" (0.2) =
f( n+1 ) (c ) (n + l) !
(0.2)"+1
where c is some number between 0 and 0.2. Therefore, we want to find the smallest integer n such that:
f (n+1 ) (c ) .::.___;_ .__ _.:_ ( 0. 2 )"+1 < 1 o� ( n + 1)! Because j5 X
X+
� 5>0 X
x + 6  5x
>o X (x  2..:... X x.:  '3) '__ _ _ >0 X
The solution set of the inequality is equivalent to the set of xvalues for which (x  2)(x 3) X
is positive. The critical values are 0, 2, and 3. Eliminate (B) and (E). There are four important intervals which we must investigate: ( 0), (0, 2), (2, 3), and (3,oo) . The solution set is the (x  2Xx  3) > 0: (0, 2) and (2, 3). two intervals for which oo,
X
18. B
19. E
20.
C
The given formula is an implication, and the contrapositive of an implication says that the ne gation of the conclusion implies the negation of the hypothesis. So, , (.C vD) 7 .(A vB) . Using DeMorgan's Laws and double negation: C .D 7 ,A .B 1\
1\
A set in IR 2 lR x lR consists of a rectangle in the coordinate plane. If the set is closed, it con tains every boundary point of the rectangle. The only answer choice that contains the boundary points along the lines x = 2, x = 5, y = 1, and y = 3 is choice (E).
=
Since Log z = log l z l
+
i Arg z, we plug in (eJ) for z.
= log (e3) + i(1t) =3+
1ti
3 9 6 + C R A CK I N G T H E G R E MATH EMAT I CS S U B J E CT TEST
21. C
A tautology is a formula in logic that is true for every possible truth value of its sentences. Choice (A) states that the double negation of a sentence A is logically equivalent to the sentence itself, which is always true. Choice (B) is a tautology called the law of the excluded middle because the sentence A and its negation .A must have opposite truthvalues. Choice (D) is one of DeMorgan's Laws, which proves it a tautology. Finally, choice (E) is the Distributive Property for sentential logic, so the statement is always true regardless of the values of sentences A, B, and C.
22. C
Two graphs are isomorphic if there exists a function f : V(G) � V(H) that is bijective, and all adjacencies are preserved. Graph I has four vertices with the following edges: AB, BC, and CD. Although there exists a bijective function f such that f(A) = E, f(B) = F, J(C) = G, and f(D) = H, adjacencies are not preserved; for example, there is no edge EF. There also exists a bijective function g from Graph I to Graph III such that g(A) = g(n, g(B) = g(K), g(C) = g(L), and g(D) = g(M). Since the adjacent vertices are preserved for this function, Graphs I and III are isomorphic.
23. A
First note that card A = 2, card B = 4, and card C = 3. Since card P(C) = 2card then the cardinal ity of set D is 23 (4) = 32. Thus card P(D) = 232• The only answer choice with cardinality 232 is P[P(A) P(C)]. (Notice that the cardinality of the quantity in the brackets is 22(23) = 32.) c,
x
24. E
You can express the right side of the equation as a power of e and equate the exponents: e2z
=
ei(n/2+2mt)
2z = i(n / 2 + 2nrt) z=
25. E
26. B
( � )( � + 2nn ) , for n
E
Z
Let the arbitrary complex number z = x + iy. Then z = x  iy, and
(x  iyf = (x + iyf x2  2ixy  y2 = x2 + 2ixy  y2 4ixy = 0 So the solution is either x = 0 or y = 0, two lines which represent the real and imaginary axes. To find the harmonic conjugate of v = x  3x2y + y3, differentiate v with respect to y. v = 3x2 + 3y2 = ux. Now we can integrate ux to find the harmonic conjugate: u = x3 + 3xy2 + g(y). We still need to find g(y), which we can do by differentiating our current function u with respect to y. So uy = 6xy + g'(y) vx, but we can also use our original function v to find an equivalent expression vx = 1 + 6xy. Setting 6xy + g'(y) equal to 1 + 6xy says thatg'(y) = 1, and so g(y) = y. Therefore, the harmonic conjugate u(x, y) = x3 + 3xy2  y. =
27.
D
Recall that z2 + 1 = (z  i)(z + i), so the integrand f(z) has simple poles at z = 3, z = i, and z = i. The curve C encloses only the poles at z = i and z = i, so the value of the integral (ac cording to the residue theorem), is equal to 2 ni * [Res(i, f) + Res(i f)]. Because both poles are ,
simple, we can easily find the residues using the formula:
S O L U T I O N S TO T H E C H A PT E R R E V I EW Q U ESTI O N S + 3 9 7
z
Res( .,j)
=
z + 1 l = lim [ z + 1 . l iH [( z  z ) • (z+3)(z 2 + 1) Hi (z+3Xz+z)
lim
'
i+ 1 2 +6i i+ 1 2 6i 2+6i 2  6i 1  2i  10 z+1 Res ( i, J ) = lim (z  i) z + 12 l= l.in1.1 (z+3)(z +1� z . . l(z+3)(z2 i) =
•
H•
[
]
•
=
=
Therefore,
1 i  2  6i 1i  2  6i 1 + 2i 10
 2 + 6i
· 
2
+ 6i
z+ � dz = 2ni* [ Res(i,j)+Res(i, f) ] JC (z + 3)(z + 1) 1  2i 1+2i ]  2 1tl. * [ +10 10 2 .
=
28. E
 1tl
5
The function 8 is essentially a distance function between two ordered pairs on the realvalued coordinate plane. Choice (A) says that distance cannot be negative, which is true. Choice (B) says the distance from x to z is either equal to or less than the sum of the distances from x to y and from y to which is also true. Choice (C) says that distance does not depend on the direc tion traveled, so the distance from x to y is equal to the distance from y to x. Choice (D) says that the distance between x and y is zero if x and y are the same point. Finally, choice (E) says that the distance from x to is equal to the sum of the distances from x to y and from y to A quick counterexample will show that this is not true:
z,
z
z.
X
y 398 + C R A C K I N G T H E G R E MATH EMATICS S U B J E CT TEST
z
29. B
The order that the players are chosen matters, so the number of permutations of 8 out of the 15 · · p 1ayers position
1 5 ! s· . mce we then choose any 1 of the 5 pitchers, mu1tip1y the IS· ( 1 51 5 !8) ! = 7!
result by 5 to get the answer. 30. A
Write the complex number (
r=
�( J3 f + 1 2 2 (1) =
1 e = tan 
So
J3 + 1 ) in the form z = re;e .
J3
J3 + 1 = 2e; 561t , and now we can find the sixth power of z.
(i  J3t
=
( 2e; 561t J = 26e5ni
z
Now express z in the polar form: = r(cos e + i sin 8)
= 26(cos 5n + i sin 5n)
=
31.
C
32. E
26
To identify which set is compact, first consider the HeineBorel theorem says that a subset of JR" (in the standard topology) is compact if and only if it's both closed and bounded. Eliminate (B) and (D) because they are not bounded. Since answer choices (A) and (E) only use strictly less than or greater than symbols, neither of these sets is closed. By definition, we know that iff is a probability density function for X, then j (x)dx 1 ·
J
=
Integratingj(x), we have: 1
=
8 J f + c dx = x; 0
4
+ ex
]8
0
=
64 + 6c 4
Therefore, c =  3 . 33. B
An unoriented incidence matrix is an organizational tool used to represent the rela tionships between the vertices and edges of an undirected graph. The dimensions of the matrix are lVI x l EI , so there are four rows (for four vertices) and five columns (for five edges). An entry bii = 1 if vertex v; and edge e; are incident and equals 0 otherwise. Since e1 is incident with v1 and v4, the first column should have 1's in the first and fourth rows only. Eliminate (A) and (C). e2 is incident with v1 and v2, so the second column should have 1's in the first two rows only. Eliminate (D). e5 is incident with v1 and v3' so the fourth column should have 1 's in the first and third rows only. Eliminate (E).
S O LU T I O N S TO TH E C H A PT E R R E V I EW Q U ESTI O N S + 3 9 9
34. C
The exterior of A, denoted by ext(A), is the union of all open sets that do not intersect A. Elimi nate (B) and (D) because they are not union� of open sets. ext(A) is the union of three regions that constitute the complement of A: (oo, 0), ( 1, 4), and (6,oo).
35.
The order that employees are placed into a committee is not important even though it
A
( �) . ( �) . (: ) . m = 369,600
matters what committee the employee is placed into. The number of combinations is
36. D
The Cauchy integral forrnula says f(z0 ) = 2�; fc ;�:� dz , for which z0 is in the interior of (. We can apply this formula becausef(z) h(3)
37. D
comntitlees.
= fc ��i dz
=
=
= 3z2 + z  4 is analytic and z0 = 3 is inside the circle l z l = 5. Then:
2ni f(3) 52ni
Represent the situation with a Venn diagram:
A
B
c
( A n B) u
C , statement I is true. Eliminate (B), (C), Since the set shown above is a subset of and (E). Since statement III is not in any remaining answer choices, it must have been false. (The set in statement III is a subset of the one above, so it is not guaranteed that x will be an element of this smaller set.) Finally, notice that the set above is also a subset of so statement II is true.
(A u B) ,
4 0 0 + C R A C K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
38. E
The problem asks for the probability that at least one of the variables is greater than 2. That's the same as 1 minus the probability that both variables are less than or equal to 2. Since X and Y are independent, we can multiply the complements of their individual probabilities.
( 3�� ) ( 424� )
So: P(at least one > 2) = 1  P(X
=1
:S
2) P(Y
1
.
:S
2) 1
5 = 1 
6
=
1
6
39.
A
A circuit is a sequence of adjacent vertices that starts and ends at the same vertex and does not repeat any edges. Eliminate choice (D) because it does not begin and end with the same vertex. You can eliminate choice (B) because CA is not an edge of the graph. Choice (C) repeats the edges CD and AD. Finally, there is no edge connecting vertex E to vertex B, so eliminate (E). (The answer is only one of many cycles that exist in this graph.)
40.
D
· The order of a permutation is the minimum value of such that for every element a E S, n O' ( a ) = a (that is, that applying the permutation times returns the original element). First notice that cr(1) = 5, and cr(S) = 1, which says that the elements 1 and 5 both have order 2. Now apply the permutation to another element and iterate: cr(2) = 6, cr (6) = 4, cr ( 4) = 3, and cr( 3) = 2. The order of these elements is 4. Thus, since cr 4 (a) for all elements in this set, and it is the least such value that will do so, the order of the permutation is 4.
41. E
First, use the contrapositives of the last two statements so that you have four implications: A 7 C , B 7 D , D 7 C , and C 7 A . Now draw a diagram that represents these implications:
n
A
n
B
() ) C
D
�
Statement I is true if we can follow a path from sentence A to sentence D. We can see from the diagram that D 7 A , but not the other way around. Eliminate (A) and (D). Next, to prove statement II, we must show that A H C . Since both A 7 C and C 7 A , statement II is true. Before evaluating statement III, write the contrapositive of the statement: B 7 A . Although there is no direct implication from sentence B to sentence A in the diagram above, we can see a transitive implication: B 7 D 7 C 7 A . Thus statement III is true.
42. D
If M = (a, b), [a, b), (a, b], or [a, b], then the Lebesque measure (denoted !l(M) ) is simply the length of the interval, b  a. In this case M is a finite union of disjoint, finite intervals, so then !l(M) is the sum of the interval lengths. !l(M) = [3  (1)] + (8  7) = 5.
S O LUTI O N S TO T H E C H APTER R EV I EW Q U ESTI O N S + 40 1
43.
D
44. D
Suppose the four tables are W, X, Y, and Z. Since the question asks which must be true, we can quickly break Statement I with a counterexample. Let tables W, X, and Y have 15 flowers each; then table Z has 16 flowers, and no table has 13 or fewer flowers. Eliminate (B), (C), and (E). Since both answer choices include Statement II, we can assume it must be true. Looking at Statement III, suppose the florist only placed 30 flowers at tables W and X combined. Then 31 flowers are at tables Y and Z, so Statement III cannot be broken and must be true. The question asks us to find P( 40 :::; X :::; 50) , where X has a binomial distribution with
k wok
n = 100 and p = q = .!. ; this probability is equal to the expression: 2
kf ( 100 ) ( � ) ( � ) =40 k
2
2
This expression is far too difficult to calculate on the GRE, so we will approximate using the binomial distribution. The mean of X is cr
=
11
= np = 100
j:;;pq = �lOO(f)(f) = 5. Therefore:
( � ) ( � )
(� ) = 50. The standard deviation is
t P(4 o ::; x ::; 5 o) z SO O + t  40  0 =
(O.l)  (0.1)
The table in the chapter shows the closest approximation of ( 0.1) is ( 0) z 0.5 . We also know that (
z ) = 1  ( z ) , so ( 2.1) = 1  ( 2.1), which is very close to 0.03.
Therefore:
P(40 :::; X :::; 50) z 0.5  0.03 = 0.47 = 47% 45.
c
4!
At first this problem may seem like a simple permutation: _ ! = 24 ways, but it's not. Since (4 4) the arrangement into which you're placing the four people has no start or end (it's circular), then orientation should not matter. If we call our four customers A, B, C, and D, then these two arrangements should not be counted separately: A B C D and D A B C.
o �fuesameasQ
A
D
B
A
B
C
C
D
As you can see from the diagram above, in both arrangements A is sitting with 0 to the left, B to the right, and C across. Therefore, we must divide our original computation by the number of possible locations for the first object placed (i.e., the number of seats!). There are 24 4 6 ways. +
4 0 2 + C R A C K I N G T H E G R E MAT H EMAT I C S S U B J ECT TEST
=
46. C
We can rewrite e iz + e iz
2

=3
cos
z to solve this equation for z:
Now make a substitution u = eiz and apply the quadratic formula: u 2  6u
+
1 =0
U = 6 ± �362  4

u
=
3 ± 2J2
=
eiz
Take the complex logarithm of each side to solve for z: iz Log ( 3 ± 2 J2 ) =
=  i Log ( 3 ± 2J2 ) z =  i [log ( 3 ± 2J2 ) i( 2krt )], for any k z = 2krt  i log ( 3 ± 2 J2 ), for any k z
+
47. E
E
Z
E
z
Begin by drawing a diagram of the relationship between sets A and B:
From this diagram, we can clearly see that the union of sets A and B is simply set B, so Statement is true and eliminate (D). Statement II says that the set of all elements in A that are not in B is empty, and we can see that there cannot exist an element in A unless it is also an element of B. Eliminate (A) and (C). Finally, notice that the set C  B is contained within the larger region created by C  A. Therefore, C  B � C  A, and the answer is (E). I
48. D We are given thatj(z) is analytic, so u(x, y) and v(x, y) are harmonic conjugates. So ux = 5 = vY . Integrating with respect to y, we get v(x, y) 5y + h(x), where v contains an unknown function in x. Now we turn to the other CauchyRiemann equation by evaluating vx h '(x) uY. But we already know that uy 3, so h(x) 3x + c. At this point we know that v(x, y) 5y + 3x + c, and we have an initial point to find c: v(4, 1) 7 5(1) + 3(4) + c
=
=
=
= = = Therefore, v(x, y) = 5y + 3x  10. So v(3, 2) = 9. c =
=
=
1 0
S O LUTI O N S TO T H E C H APTER R E V I EW Q U ESTI O N S + 403
49. B
While the value of the variable a is greater than 1, the first step of each iteration is to divide a by 2 if a 0 (mod 2). This says that the value of a will be halved if a is even. Eliminate (A) and (E). Otherwise, if a is odd, multiply a by 3 and add 1. This eliminates choice (C). Finally, notice that a" is determined by evaluating whether a,_1 is even or odd, which eliminates (D).
50. D
First, notice that the algorithm does not output until after the first iteration. Eliminate choices (A) and (C). Since 17 is odd, the first output will be 3(17) + 1 52. The second output will be 52 2 = 26, so eliminate (B). The algorithm will converge to 1 eventually according to the sequence in choice (D).
=
+
404 + C R A CK I N G T H E G R E MAT H EMAT I CS S U BJ E CT TEST
=
·
Practice Test
MATHEMATICS TEST (RESCALED) Time170 Minutes 65 Questions
Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. In each case, select the one that is the best of the choices offered and then mark the corresponding space on the answer sheet.
Computation and scratchwork may be done in this examination book. [***On the real exam, each lefthand page of the booklet will have test questions and each righthand p age will have a blank scratchwork area. For this practice test only, you may use sep arate scratch paper.***] Note: In this examination:
( 1 ) All logarithms with an unspecified base are natural logarithms, that is, with base e .
(2)
The set of all x such that
(3 ) The symbols Z, Q,
1.
2.
I f x2
C
x
�b
[a, b ] .
is denoted by
R, and denote the sets of integers, rational numbers, real numbers, and comp lex numbers, respectively.
+ x = 3, then
:xGID�x ) =
(A)
a �
x  (log x)2
x4 + x =
(B) 4(3  x)
x2  2 log x
(C)
3(x  1 )
(C)
x1
( D)
6 (2  x)
2x  1
(x log x)2
(D )
2x2 log x
(C)
1
( D)
2
Let A = 0, a , where 0 is the empty set. If of the set A u B ?
B
(x log
x)2
(B)
2x2 log x
(E) 3x
(E)
2x  1 (xlog x)2
3. 1
(A)
4.
(B) _!
2
2
{
(A)
4
}
(C ) 6
(B) 5
is the set of all subsets of
(D )
GO ON TO THE NEXT PAGE.
4 0 6 + CRA C K I N G T H E G R E MATH EMAT I CS S U BJ E CT T EST
7
(E) oo A, what's the cardinality (E) 8
5.
For how many values of (A) 0
6.
(B) 1
(E) 4
(E) 5
r
Let f(xl y) = xy(x  y) where x(tl u ) = t(t  u ) and y( tl u) = u(u  t) . What is the value of at t the point (t1 u) (11 1)?
If
=
I
(B)
4
J: tan x dx = 2 1 then b could equal (B) arccos(2e)
a
(C) 2
(D) 4
(E) 8
(C) arcsec 2
(D) arcsec2 e
(E)
Let be the smallest positive value of x at which the function j(x) point. What is the value of j( a ) ? (A)
10.
(D) 3
(D) 4
(C) 3
(B) 2
(A) arccos 2 9.
8x + 1 ) dx equal to zero?
=
(A) 8
8.
(C) 2
+
At how many points does the graph of the curve y x5 + x3 + x  2000 cross the xaxis? (A) 1
7.
a is the integral J�' (12x2
J2 4
(B)
�2
(C)
J2 2
(D) 1
=
arcsec(e2)
(cos x2)(sin x2) has a critical
(E) J2
Find the equation of the integral curve that passes through the point (21 3) of the differential equation 2x y = zy 1 1
(E)
=
2(x2  1) ty3  y
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PRACT I C E TEST + 4 0 7
11. Let i = (1, 0, 0), } = (0, 1, 0), and k (0, 0, 1). If b and care vectors such that f b f = 3, b c = 5, l cl = 7, and b c = 8} + 4k, which of the following vectors satisfies the equation a · [(a + c) b] = 1 2 ? (B) a = i  } + 2k (A) a = 2 i + }  2k (C) a = i  2) + k (D) a = 2 i  } + k a = i 2}  2k 12. Ifatedonesolarchid? of the curve y = sin xis revolved around the xaxis, what's the volume of the gener 2 (B) (C) 21t (A) 4 2 13. If is a positive constant, what is the maximum value of the following function? logx f(x) = =
x
x
(E)
·
+
2
�
�
a
x"
(A) _.!._ ae
(B)
1 aa 
(C)
a'
(D) e"
(E)
e
"2
a
14. Find the family of all nonzero solutions of the equation d2y ( dy )2 = dx2 dx 1 (A) y =  + c (B) y = c + log(c1  x) 2 x+c 2 (E) y = c log( c1 + x) (D) y = c + log(c1 + x) 2 2 15. For each integer n > 1, let a" = 1/(log n). Which of the following statements is/are true? I. The sequence ) converges. II. The series Ln=2 converges. III. The series I,n=2 (1)" a� converges. (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III I
(a
�
a
n
�
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408 + C R A C K I N G T H E G R E MATH EMAT I C S S U B J E CT TEST
16. Let X = r cos e and y = r sin e. Which of the following expressions is equal to derivative of e with respect to r, keeping x constant? (A) r  1
17.
tan e
(A)  n
(B) 2n
lim (x + e3x )1/x = x�O (A) 0
(B) 1
2
19.
(D)
r  1
cot
(D) 1
�
(C) 3.
e
(B) _fe
1
aJ I  of
0
Xp

;}., vy
(D)
e
p
(C)
e
(D) Sn
(C) 37t
Let j(x, y) = xY, for x > 0. Then (A)
21.
tan e
If y' = x  y and y(O) = 1, then y(1) = (A)
20.
r  1
1 + cot2 (arcsin 2: ) = 4 3
18.
(B)
.
(�e ) r
(E)
(E)
x
I
the partial
r  1
esc e
16 n
z
+ e3
( �)  1
(E)
e
(E)
e'
at the pomt P = (e, y0), where y0 = e
(D)
e2
For any real number x, let [x] denote the unique integer, n, such that n � x < n + 1. Evaluate the integral f2bb (X  [X]) dx for b = 3
(A) 
4
21 .
(B)
7 8

5 (C) 4
(D) 98
(E) does not exist
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PRACT I C E TEST + 4 0 9
22.
In SL(2, Z), the multiplicative group of all 2 by 2 matrices with integer entries whose determinant is 1, what is the order of the cyclic subgroup generated by the following element?
(A) 2
23.
(B) 1tn(1 + log 1t) (E) 0
(D) 1
A sixsided die (whose faces are numbered 1 through 6, as usual) is known to be counterfeit: The probability of rolling any even number is twice the probability of rolling any odd number. What is the probability that if this die is thrown twice, the first roll will be a 5 and the second roll will be a 6? (A)
1
2 81
J
l/2
1/2
18
2 27

(E) Cannot be determined from the information given
dx = 1x
(A) log 2
26.
(C)
(B) 
(D) 1 9 25.
""
Determine the slope of the curve y = xn  1tx at the point (1t, 0). (A) 1tn(1  log 1t)
24.
(E)
(D) 6
(B) 3
2
(B) log 3
(C) 3 log 2
(D) 2 log 3
(E) 3 log 3
Let C be the circle in the yzplane whose equation is (y  3)2 + z2 = 1. If C is revolved around the zaxis, the surface generated is a torus. What is the equation of this torus? (A) x 2 + y 2 + i + 8 = 6y
(B) (x2 + l + z2 )2 = 8 + 36(x2 + z2 )
(C) (x2 + y 2 + z2 + 8)2 = 36(x2 + i ) (D) (x2 + y2 + z 2 + 8)2 = 36(x2 + l ) (E) (x2 + y2 + z2 + 8)2 = 36(y 2 + i )
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4 1 0 + CRA C K I N G T H E G R E MAT H EMAT I C S S U B J E CT TEST
27. Find the general solution of the equation d 4y + 2d 3y + 5d2y = 0 dx 4 dx 3 dx2 (A) y = c1 + c2x + e2x (c3 cos x + c 4 sinx) (B) y = c1 + c2x + e2x (c3 cos x + c4 sinx) (C) y = c1 + c2x + ex(c3 cos2x + c4 sin2x) (D) y = c1 + c2 x + e x (c3 cos 2x + c4 sin2x) (E) y = c1 + e x (c2 cos2x + c3 sin2x)
28.
A = [ � � �J
Let f : R3 � R3 be the function defined by the equation f(x) = Ax, where
7 8 9
Which of the following best describes the image off ?
(A) A point (B) A line through the origin (C) A plane through the origin (D) The union of a plane through the origin and a line through the origin, with the line perpen dicular to the plane (E) All of R3
29.
J: (x log x) dx = (A) 1
(B)
1
2
(C)
1 4
(D) 1
(E) does not exist
30. Define a binary operation on Z, the set of integers, by the equation m
•
n = m + n + mn. Which of
the following statements is/ are true about the binary structure (Z, ) ? I. This structure is not a group since the operation is not associative. II. This structure is not a group since there is no identity element. III. This structure is not a group since not all elements have an inverse. IV. This structure is a group. •
(A) I and II only (D) III only
(B) I and III only (E) IV only
(C) II and III only
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P R A CT I C E TEST + 4 1 1
31.
Given that
(A)
32.
fo� ex'dx
i .j;,
=
t .J;, , what's the value of the following integral?
(B)
Ia� �dx t .J;,
(C)
(E)
{ n
x2 xy2z z2 = 1. The plane tangent to S at the point 1 X= (B) X= y, z = 1 y x 1 = 2(y 1) = z
Let S be the surface whose equation is (1, 1, 1) contains which of the following lines?
x = y = z (D) y = X= 1 (A)
(E)
Z,
33.
.J;,
Define a function
+

(C)
1
Z,
=
f(x) for all positive x by the equation f(x) = Jxx'' Oog t)dt
Find the critical point(s) of this function.
3 (C) X= 2 X= 94 ' 1 (B) x 1 (D) f(x) Ln�=O (n(n !)1Y/2 2 x n , what is true of the value of j
C
7

1
u
= x + c1
=>
U=
0.) Integrating once more gives y:
y y = J dy = d dx = J u dx = dx 15.


Let u = dy/dx. Then the given equation becomes du/dx = u 2, which is separable:
n 7
J( J 1  
x + c1
1
X + C1
dx =  log(x + c1 ) + c2
_
7
Since log n oo as oo, we know that a" = 1 /(log n ) 0, so the sequence (a,) converges (to 0). Eliminate (D). Applying the comparison test, the series does not converge, since for all n ?: 2, =>
log n < n
1
log n
�>��
1
>
n
and the harmonic series, L � , diverges. Eliminate (B) and (E). Finally, the series converges, because it's an alternating series and the terms
a "2
L (1)" a�
1 Qog n ) 2
= :
decrease monotonically with a limit of 0. 16.
C
If x is held constant, then r cos e = c. Differentiating both sides of this equation with respect to r, we find that: r •
17.
E
[
r cos e = c
 sin e •
The Pythagorean identity 1
+
��] +
cos e = 0
ae cas e = = r1 cot e ar r sin e
cot2 e = csc2 e implies that
1 + coe (arcsin 1 ) = csc 2 (arcsin 1) =
18.
E
()
.
1 . 2 s m (arcsm � )
.
_1_  1 6 ( � ) 2 1t2
_! log (x + e3x); we'll first find the limit of(log y) as x approaches Let y = (x + eJx) llx, so log y = x 0 using L'H6pital's rule:
3x lim 1og( x + e ) X>0
Since (log y)
X
1 _ 3 • (1 + 3e 3x )
= lim x + e x X>0
7 4, we conclude that y
1
7
3x 1 3 4 = = lim 1 + 3e = � X + e 3x 0 + 1 x>0
e4. PRACT I C E TEST ANSWERS A N D EXPLANATI O N S + 4 2 S
19. C
The equation y' = x  y can be written in standard linear form as y' + y = x. The general solution of the associated homogeneous equation, y' + y = 0, is y = cex. We can also see that y = x  1 is a particular solution of the given nonhomogeneous equation. Therefore, the general solution of the nonhomogeneous equation is y = cex + x  1. To satisfy the initial condition y(O) =
substitute x = 0 and y = 1, which gives 1 = c  1, so c 2 y = 2ex + x  1. This tells us that y(1) = 2e1 =  .
=
1, we
2. Therefore, the solution of the IVP is
e
20. C
Ifj(x, y) = xY, then dj/dx = yxY1• Rewriting xY as ey log x, we find that: a+ _ 'J =
a ylogx (e ) = e ylogx • log x = x Y log x
dy dy
Therefore:
Y 1 Yoe o 1 Yo e
= eYo =1
Yo = e
21. B
The figure below shows the graph of the function y = x  [x], and the area under this graph from x = 1 to x = 1 is shaded. y
1
2
1
2
The area from x =  } to x 0 is equal to the area from x = 1 to x = 0, ( } )(1)(1), minus the area from x = 1 to x =  1 , ( 1 )( 1 )( 1 ); this gives }  t = � . Adding this to the area from x = 0 to x = 1, which is equal to ( 1 )(1)(1) = 1 , we get: =
5
1
1/2
4 2 6 + CRACK I N G T H E G R E MAT H EMAT I C S S U BJ E CT TEST
( x  [ x]) dx = l + l2 = Z. 8
8
22.
B
If a is an element of a group G, then the order of the cyclic subgroup generated by a is equal to the smallest positive integer n such that a" = e, where e is the identity of the group (if such an n exists). The identity of Z) is the 2 by identity matrix I. Since
SL(2, 2 A2 = (� �)( _� �) = ( � �) A3 =A2 · A = (� �)(_� �)=(� �) =I
we see that 23.
A
(A) =
3.
Since 1tx = e' 1og n, we find that:
y ' =x nxn1 (log n). y'(n) = 1t • nn1  n"(logn) = n"(1log n)
Therefore, if y = 1tX, then (7t, 0), we simply evaluate y ' at
xn
24.
A
1t:
1, 2, 4,
According to the information given in the question, the probability of rolling a 3, 5, or 6 is p, 2p, p, 2p, p, or 2p, respectively. Since one of these numbers must be rolled, it must be true that:
Therefore, 9p =
25.
=
To find the slope of the curve y = xn  1tx at
 1tx
1,
so p =
p + 2p + p + 2p + p + 2p = 1 5 .!. (.!.) ( ) _3_81 . 1 =  + _2_ 1x2 1 +x 1x
.!. . The probability of rolling a 9
is p =
a 6 is 2p = � , so the probability of rolling a 5 then a 6 is 9
B
9
9
, and the probability of rolling � = 9
x
First, find the partialfraction decomposition of the integrand: l 2
Integrating gives us:
Therefore:
l
1 dx = J _i_ dx + J _i_ dx J1x2 1 +x 1x = � log(1 + x)� log(1 x) = ll2 og 1+x 1x 2 � og 1 + x l 11 = � [ log f log =[ dx J11 /22 � f] 1x 112 1x = ±(log 3 log t) = ±(2log3) =log3 l
2
2
P RA CT I C E TEST ANSWERS A N D EX PLANAT I O N S + 4 2 7
26.
D
If a curve j(y, z)
= 0 in the yz plane is revolved around the zaxis, the equation for the resulting
surface of revolution is obtained by replacing y with
±�x2 + y2
•
In this case, (y
 3)2 + z2 1
becomes:
=
(±�x2 + y2  3 r + 22 = 1
[(x2 +y2 ) ± 6�x2 +y2 +9] + 22 = 1 xz + yz + 22 + 8 = ±6�x2 + y2 (xz + yz + 22 + 8)2 = 36(x2 + i ) 27.
D
Replacing d"y I dx" by m " transforms the differential equation into the algebraic equation m4 + 2m 3 + m 0, which we solve as follows:
5 2=
m 4 +2m 3 +5m 2 = 0 m2(m2 + 2m + 5) = 0 m = 0, 2 ± �22 2 4 1 5 = 0 (double root), 1 ± 2i •
•
'' 
The part of the solution of the differential equation corresponding to the double root of 0 is c1 + c2x, and the part corresponding to the conjugate complex roots ± 2i is ex(c cos 2 + c4 sin 2x . 3 Therefore, the general solution of the given differential equation is:
1
y = c1
28. C
(A = 1 52 6)
x
)
+ c2x + ex(c3 cos 2x + c4 sin 2x)
(01
)
The dimension of the image off is equal to the rank of A. The following elementary row opera tions show that rank A = 2:
3
4
7 8
9
4r1 added to r2 7r1 added to r3
)
0
2
3
6
3 6 12
Since the image of f is a twodimensional subspace of origin.
4 2 8 + CRAC K I N G T H E G R E MAT H EMATICS S U B J E CT TEST
R3, it must be a
plane through the
29.
C
1 Integration by parts, with u = log x, du =  dx, v = � x2, and dv = x dx, yields: X
J (x log x) dx
=
t x2 log x  J (t x2 )
_l
I
= t x 2 log x  t x dx
= l x 2 log x
2
Therefore:
I: x log x) dx (
=
4x
(�)
2 +c
[t x2 log x ± x2 J:
(
[
)
dx

(
)
= O  l4  Jim l x2 log x  o x.0 2
]
To evaluate the limit, we use L'Hopital's rule:
So we can now write:
J: (x log x) dx = ( 0 i) [ � ( t x 2 log x ) 0 J 

=
30.
D
( o ±) [ o o ] 


The binary operation is associative, as we can check: ?
(m • n) • q � m • (n • q) ?
(m + n + mn) • q ;, m • ( n + q + nq)
(m + n + mn) + q + (m + n + mn)q = m + (n + q + nq) + m(n + q + nq) ./
This binary structure does have an identity, namely 0, since m•O=m+O+m o O=m
m.
O•m=O+m+O o m =m
and
are true for all However, not all integers have an inverse with respect to this binary opera tion. For example, 1 has no inverse, for if we try to solve 1 • = 0, we get 1•n
=
0
�
1+n +1 o n
n.
=
0
�
2n
n
=
1
which is not satisfied by any integer Since not every element has an inverse, this binary structure is not a group. (It is a monoid, however. As an additional instructive exercise, show that the only elements in this binary structure that do have an inverse are 0 and 2.)
P RACT I C E TEST ANSWERS A N D EXPLANATI O N S + 4 2 9
31. C
32. A
Let x
= u2, so dx = 2u du. Applying this change of variable, we find that:
The surface S is a level surface of the function f(x, y, z) = x2  xy2z + z2• The gradient of this function, evaluated at the point P, will be normal to S, and, therefore, tangent to the plane at P. Since
VJ = Vx ' Jy ' fz )
we have
VJj
= ( 2x  y 2z, 2xyz, xy2 + 2z )
P=(l,l,l)
= (1, 2, 1). Since this is normal to the plane tangent to S at P, the equation
of the tangent plane must have the form x  2y + z = d, for some constant, d. Since this plane passes through the point P = (1, 1, 1), the value of d must be 0, so the equation of the tangent plane is x  2y + z = 0. Every point of the form (t, t, t) t(1, 1, 1) lies in this plane; but these
=
points describe the line x = y = ;, so this line lies in the plane.
33.
A
The critical points off are found by determining the points at which the derivative is equal to 0. In this case, the fundamental theorem of calculus and the chain rule tell us that: d �og t)dt f '(x) = dx = �og(x3 )] • fx (x3 )  �og(x2 )] • :x (x2 )
f x3 x2
= 9 x2 log x  4x log x = (x log xX9x  4)
=
Setting this equal to 0, we get x 0, x = 1, or x = i . Since f is defined only for positive x, we 9 ignore x 0, and conclude that the critical points occur at x i and x 1. 9
=
34. D
=
From the formula for the Taylor coefficients, we know that: J
rankA = 2
This question asks us to maximize the value of j(r, 8) = r sin 8 subject to the constraint This can be done by the Lagrange multiplier method, as follows: g (r, 8) = r2 
4cos28 = 0 .
Vfa = A'Vg (f, , fa ) = A(g, , ga ) (sine, rcose) = A(2r, 8sin2e) Eliminating A from this last equation gives us
rcose = 2r 8sin 2e
sin e

cos 8 = 8 sin 8 sin 28. cos cos 8 = sin 8 sin 28.
since both of these expressions are equal to A. Cross multiplying, we get 2 r 2 We can now use the c�nstraint equation, y2 = cos 2 e to reduce this to 28 and now solve this equation as follows:
4
I
cos 2 Ek:os 8= sin fuin 2 e 1  = tan 28 tan 8 2 tan e 1 tan e 1 tan 2 8 2 tan2 8= 1 tan2 8 tan S= ± )J
From this last equation, we take e = 7t
7t
 .
6
Substituting this into the constraint equation gives
r2 = 4 cos 2 6 = 2, so r = .J2 . We now conclude that the maximum value of y = r sin e is
Ymax =
.J2 sin !!:_ =
6
.J2
2
.
4 3 4 + CRAC K I N G T H E G R E MATH EMAT I CS S U BJ E CT T EST
44.
C
A number A. is an eigenvalue of a matrix A if and only if det(A  A.I) = 0. Therefore, if 1 is an eigenvalue of A, it must be true that:
det(A  I) = det This equation implies that:
(
)
t1 1t 0 2t  1 1
=0 + =0 t = 0, 1
(t  1)(2t  1)  (1  t) (t  1)[(2t  1) 1]
( � �)
Since t can only equal 0 or 1, the given matrix must be:
A=
or
A=
G �)
In the first case, the eigenvalues are 1 and 1, while in the second case, the eigenvalues are 1
and 2. (An easy way to calculate the eigenvalues of a 2 by 2 matrix is to use the fact that the determinant of the matrix always gives the product of the eigenvalues, and the trace of the ma trixthat is, the sum of the entries on the diagonalalways gives the sum of the eigenvalues. So, for the first possibility for the matrix A, the product of the eigenvalues must be 1 and their sum must be 0; therefore, the eigenvalues must be 1 and 1. We can figure out the eigenvalues of the second possibility for A in the same manner.) Therefore, another eigenvalue of A could be 1 or 2, but only 2 is listed among the choices. 45.
B
x cos y, then the integrand, M dx + N dy = (et sin y) dx + (et cos y) dy is an exact differential, since MY = cos y = Nx ' Therefore, F = (Mx, NY) is equal to the gradient If we let M(x, y) = e" sin y and N( , y)
=
e'
e"
of some function, f(x, y), and the line integral is independent of the path. In particular, a func tion f whose gradient is (e" sin y, e" cos y) is f(x, y) = ex sin y. Using the parametric equations (and the interval of the parameter, t) given in the ques
)
tion, we see that the curve C begins at the point A = (0, 0) and ends at the point B = (1, 2:. . 2 Therefore:
fcMdx+Ndy = fc F dr = fc 'Vf dr = J� 'Vf dr = f(B)  f(A) = [ex sm. y JB;(l,Jt/2) =e •
•
•
A;(Q,O)
P RACTI C E TEST A N S W E R S A N D EX PLANAT I O N S + 43S
46.
E
n elements. Then the number of 2element subsets is the bino mial coefficient ; , and the number of 3element subsets is ; . We first want to find the value of n such that ; = + 14. Perhaps the easiest way to do is to write down Pas cal's triangle, which is a triangular array of numbers such that the kth entry in the n th row (where we start counting from 0 in both cases) is equal to : . The rows always begin and end with
()
()
Assume that X contains exactly
( ) (�)
this
()
1, and every other entry is equal to the sum of the entries to its left and right in the row above. 1 1
1
1 1 1 1 1 1
6
21
= 21
4
( ;)
1 1 1
5
15
6 21
35
35
n = 7, since:
G)
10 20
15
We're looking for the row in which the
Since X contains
3
10
5
this happens in the row for
1
3 4
6
7
2
7
1
entry is 14 more than the
and
G)
1
(;)
= 35
n = 7 elements, the number of 4element subsets of X is:
4 3 6 + C R A C K I N G T H E G R E MATH EMAT I CS S U BJ ECT T EST
(:)
= 35
entry. We see that
47.
C
First, we'll find where the curves intersect: (sin x + cos x)4 = (sin x cos x)4 (sin x + cos x ) 2 = (sin x cos x ) 2 2 sin X COS X = 2 sin X COS X
4 sin x cos x = 0 2 sin 2x = 0
Since we want the smallest positive value of x at which the curves intersect, we take 2x = 1t, so
x = 2: ; this is a. Therefore, the area between the curves from x = 0 to x = a = 2: is: 2 . 2 "12 4 [(sin x + cos x)  (sin x  cos x)4] dx A= 0 "1 = 2 [(sin x + cos x? + (sin x  cos x) 2 ][(sin x + cos x ) 2  (sin x  cos x? ] dx 0 "1 = 2 [2 sin2 x + 2 cos 2 x ][2 sin x cos x  ( 2 sin x cos x)] dx
J J fo n/2 = 4 J (2 sin x cos x ) dx 0 = 4 [ sin2 X r2 =4
m
C
The ring Z111 is a field if and only if is a prime. Choice (A) leads to the set containing Z , Zy 2 Z , Z5, and Z6, three of which are fields (Z , Z3, and Z5). Choice (B) gives the rings Z4, Z5, Z6, Z7, 4 Z , two of which are fields. Choice (C)2 gives the rings Z6, Z7, Z , Z9, and Z 0, exactly one of and 8 8 1 which (Z7) is a field, so this is our answer.
49. D
Assume that the sequence converges and let x denote its limit. Then both X11 7 x and X11+1 7 x as n 7 oo, so the recursion relation becomes, in the limit:
48.
x=
l ( x + �)

x2 + 2 2x 2 2 2x = x + 2 x2 = 2 X=
x = ± fi If x1 is negative, the recursion relation implies that every term X11 is negative; on the other hand,
 fi
if x1 is positive, then every term X11 is positive. Therefore, we conclude that the sequence (xJ converges to
if x1
0.
P RACTI C E T EST A N S W E R S A N D EX PLANAT I O N S + 4 3 7
50. B
The prime factorization of 72 is 2 x 2 x 2 x 3 x divisors for an Abelian group of order 72 reads:
3. Therefore, the list of possible elementary
2, 2, 2, 3, 3 2) 2, 2, 2, 32 3) 2, 22, 3, 3 4) 2, 22, 32 5) 23, 3, 3 6) 23, 32 1)
This list implies that there are 6 nonisomorphic Abelian groups of order 72. 51.
C
By definition, if the probability density function of a random variable, X, isfit), then the expecta
tion (or mean) of X is �t(X)
=
f: t
We'll evaluate the integral J t
v =
•
•
fx (t)dt . In this case, then, we have !J.(X) =
e{t1)/2 dt
rt
using integration by parts, with
2 eUll/2, and dv = e c ,/zx + 1 dx +(2xy + Jy' y ) dy = fJ, [! (2xy + Jy'  y )  � (,/zx + l J] dy dx = JL 2y dydx
4 4 2 + C R A C K I N G T H E G R E MAT H E MAJI CS S U BJ E CT TEST
Using the figure below: y 3 2 (0, 1)
y = 2x + 1
(
( 1 , 3) c (1, 1)
y=1
X
1
we can evaluate the double integral using iterated integrals:
JL 2y ay ax = s: fx+l 2y ay ax = J: [l J:X+1 dx = J: [(2x + 1)2 1 J dx = J: (4x2 + 4x) dx = [fx3 + 2x z J: _ lQ 3
62. E
We can show that choices (A), (B), (C), and (D) are all false with a single counterexample: Con sider the characteristic function on Q, the set of rationals; this is the function defined by:
f(x) = { 10 ifx if x
E
�
Q
Q
This function is equal to zero almost everywhere; that is, the set of points at which the function is not zero (the set Q) has measure zero, because Q is a countable set. Therefore, this function is Lebesgue integrable (and the value of the integral is zero). This function is not Riemann integra ble (so choice (A) is false). This function is discontinuous and nondifferentiable everywhere, so choices (C) and (D) are false; and since f is nondifferentiable on the entire real line, it's certainly nondifferentiable on every countably infinite of R, so choice (B) is false. The answer must be (E).
PRACT I C E TEST A N S W E R S A N D E X P L A N AT I O N S + 443
63. B
The ratio test says that a series of positive terms, "" .L,.; an , will converge if lim, 7 co(an+1 I an) Applying the ratio test to the given series, we find that:
! ( 3 n) [[ 3(n(n +1)1?x2 (n x2l n } (1 3n + 3)(3n + 2)(3n + 1) 27 2
nl lim an+l = lim [(n + 1)!f x2( + ) . n>� an n>� + ]!
{
�
= X From this, we find that:
lim an+l n>� an
64.
B