REAL ANALYSIS
REAL ANALYSIS FRANK MORGAN
AmmiucArr MATHEMATICAL SOCIETY
Providence, Rhode Island
2000 Mathematics Subject Classzficatzon. Primary 26-XX.
Front cover: The cover illustrates how continuous functions can converge nonuniformly to a discontinuous function. It is based on Figure 17.1, page 76. The balls in the background illustrate an open cover, as in the definition of the important concept of compactness (Chapter 9, page 41). , Back cover: The author, in the Math Library outside his office at Williams College. Photo by Cesar Silva. Cover design by Erin Murphy of the American Mathematical Society, on a suggestion by Ed Burger.
For additional information and updates on this book, visit
www.ams.org/bookpages/real
Library of Congress Cataloging-in-Publication Data Morgan, Frank. Real analysis / Frank Morgan. p. cm. Includes index. ISBN 0-8218-3670-6 (alk. paper) 1. Mathematical analysis. I. Title.
QA300.M714 2005 515-dc22
2005041221
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Contents
Preface
vii
Part I. Real Numbers and Limits Chapter
1.
Numbers and Logic
Chapter 2. Infinity Chapter
3.
Sequences
Chapter 4. Functions and Limits
3 9 13
21
Part H. Topology Chapter 5.
Open and Closed Sets
27
Chapter 6.
Continuous Functions
33
Chapter 7.
Composition of Functions
35
Chapter 8. Subsequences
37
Chapter 9.
41
Compactness
Chapter 10. Existence of Maximum
45
Chapter 11.
Uniform Continuity
47
Chapter 12.
Connected Sets and the Intermediate Value Theorem
49
Chapter 13.
The Cantor Set and Fractals
53
v
Contents
vi
Part III. Calculus Chapter 14.
The Derivative and the Mean Value Theorem
61
Chapter 15.
The Riemann Integral
65
Chapter 16.
The Fundamental Theorem of Calculus
71
Chapter 17.
Sequences of Functions
75
Chapter 18.
The Lebesgue Theory.
81
Chapter 19.
Infinite Series E a7
85
Chapter 20.
Absolute Convergence
89
Chapter 21.
Power Series
93
Chapter 22.
Fourier Series
99
Chapter 23.
Strings and Springs
105
Chapter 24.
Convergence of Fourier Series
109
Chapter 25.
The Exponential Function
111
Chapter 26.
Volumes of n-Balls and the Gamma Function
115
Part IV. Metric Spaces Chapter 27.
Metric Spaces
121
Chapter 28.
Analysis on Metric Spaces
125
Chapter 29.
Compactness in Metric Spaces
129
Chapter 30.
Ascoli's Theorem
133
Partial Solutions to Exercises
137
Greek Letters
147
Index
149
Preface
Our lives and the universe barely work, but that's OK; it's amazing and great that they work at all. I think it has something to do with math, and especially real analysis, the theory behind calculus, which barely works. Did
you know that there are functions that are not the integral of their derivatives, and that a function can be increasing and have a negative derivative? But if you're a little careful you can get calculus to work. You'll see.
The theory is hard, subtle. After Newton and Leibniz invented the calculus in the late 1600s, it took puzzled mathematicians two hundred years,
until the latter 1800s, to get the theory straight. The powerful modern approach using open and closed sets came only in the 1900s. Like many others, I found real analysis the hardest of the math major requirements; it took me half the semester to catch on. So don't worry: just keep at it, be patient, and have fun. This text is designed for students. It presents the theoretical intellectual breakthroughs which made calculus rigorous, but always with the student in mind. If a shortcut or some more advanced comments without proof provide better illumination, we take the shortcut and make the comments. The result is a complete course on real analysis that fits comfortably in one semester.
vii
Preface
This text developed with my one-semester undergraduate analysis course at Williams College. I would like to thank my colleagues and students, especially Ed Burger, Tom Garrity, Kris Tapp, and Nasser Al-Sabah `05, and my editors Ed Dunne and Tom Costa. -Frank Morgan
Department of
Mathematics: and Statistics
Williams College Williamstown, Massachusetts
www.williams.edu/Mathematics/fmorgan
[email protected] Part I
Real Numbers and Limits
Chapter 1
Numbers and Logic
1.1. Numbers. Calculus and real analysis begin with numbers: The natural numbers
N = {1, 2, 3.... I. The integers
Z = {... , -3, -2, -1, 0, 1, 2, 3, ... } (Z stands for the German word Zahl for number). The rationals Q _ {p/q in lowest terms: p E Z, q E N} _ {repeating or terminating decimals}. (Q stands for quotients). The reals I[8 = {all decimals}
with the understanding that .999 . = 1, etc. Reals which are not rational are called irrational. Thus the set of irrationals is the complement of the set of rationals, and we write
{irrationals} _ Q6 = IR - Q.
1.2. Intervals in R. For a < b, define intervals
[a,b]={xEIR:a<x20=1.
n n = nl/n = (elnn)1/n = e(Inn)/n
-
e° = 1.
(The exponent (Inn)/n --40 because Inn N), we will be taking x close to some fixed target p, taking Ix - pl small. To say how small, we'll use the Greek letter before epsilon e, namely delta b, and require that I x - PI < S.
4.1. Definitions. We say limx_,p f (x) = a ("the limit as x approaches p of f (x) equals a") if, given e > 0, there exists a b > 0 such that
0 < Ix-pI <s= If(x) -al <E. Of course we have to assume that x lies in the domain E of f, and hence we need to assume that p is an accumulation point of E. Notice that by requiring that 0 < Ix - p1, we do not allow x to equal p. The limit depends only on values f (x) at nearby points, not on f (p). If it happens that f (x) = f (p), or p is not an accumulation point of E, then f is called continuous at p. Examples of continuous functions include powers of x, sin x, cos x, ex, and combinations of continuous functions such as ex cos x
x2+1 (as long as the denominator is never 0).
4.2. Example. Consider the function f (x) = sin -1, defined on E = lib-{0}, graphed in Figure 4.1. Then limx-,o f (x) does not exist (Exercise 4). 21
4. Functions and Limits
22
-0.8
-0.6
-0.4
Al
0.2
10.4
0.6
0.8
s
Figure 4.1. For f (x) = sin 1, limy-o f (x) does not exist.
-
4.3. Example. Consider the function f (x) = x sin -1, defined on E = II8 {O}, graphed in Figure 4.2. Then limx.o f (x) = 0 (Exercise 3). If we define f (0) = 0, then f will be continuous.
4.4. Example. Consider the function f (x) on IR which is 1 on nationals and 0 on irrationals. Then lim f (x) never exists, and f is continuous nowhere. This function is called the characteristic function of the rationals and written
f =XQ, using the Greek letter chi.
4.5. Proposition. If
f (x) exists, then it is unique.
Proof. The proof is just like the proof that the limit of a sequence is unique. Suppose that lim,;,P f (x) has two distinct values a, b. Choose S > 0 such
that if 0 < Ix - pI < 5, then
If(x)-aI < Ib-aI/2
and
If(x)-bI < Ib-aI/2.
Since for limits we always assume that p is an accumulation point of the domain, some xo satisfies 0 < Ixo - pI < S. Then lb - al < If - al + I f(xo) - bI < Ib - al/2 + lb - al/2 = Ib - al,
4.6. Proposition
23
Figure 4.2. For f (x) = x sin z , lima--.o f (x) = 0.
the desired contradiction.
4.6. Proposition. Suppose that limp f (x) = a and limx,p g(x) = b. Then
(1) lim,p cf (x) = ca, (2) limx,r(f + g) (x) = a + b, (3) limx,p (f g) (x) = ab, and (4) limx,p (f /g) (x) = a/b if b
0.
Proof. The proof is similar to the proof of 3.6; see Exercises 5 and 6.
4. Functions and Limits
24
Exercises 4 1. Give an example of a function f : R --+ Il8 which is continuous except at the integers. 2. Give an example of a function f : JR - JR which is continuous only at 0.
3. Prove that for the function f (x) of Example 4.3, lim.,,o f (x) = 0.
4. Prove that for the function f (x) of Example 4.2, lim.,;,o f (x) does not equal 0. 5. Prove 4.6(2).
6. Prove 4.6(3).
7. Prove that the sum of two continuous functions is continuous. 8. Prove that the product of two continuous functions is continuous.
9. Give an example of a function which is continuous only at the integers. Graph it. 10. Give an example of two functions, both discontinuous at 0, whose sum is continuous at 0. Give an example of two functions, both discontinuous at 0, whose product is continuous at 0.
11. Consider the function f : JR -+ JR which is 0 at irrationals and 1/q at a rational p/q (in lowest terms with q positive). Where is f continuous?
Part II
Topology
Chapter 5
Open and Closed Sets
Although sequences played the crucial role in the rigorous understanding of calculus developed in the 1800s, over the next fifty years mathematicians sharpened the theory using the concepts of open and closed sets. If it took
mathematicians fifty years to get used to these more abstract ideas, you shouldn't worry if it takes you a couple of weeks. It's actually quite fun if you take your time and think about lots of examples.
5.1. Definition. A point p in RT is a boundary point of a set S in IRT if every ball about p meets both S and its complement So. The set of boundary points of S is called the boundary of S and written 8S. (That funny symbol, sometimes called del, is not a Greek letter. It is used for partial differentiation too.) It follows immediately that a set S and its complement SC have the same boundary. For example, consider a ball about a point a of radius r > 0:
B(a,r) _ {Ix - al < r} (see Figures 5.1 and 5.2). Its boundary is the sphere:
8B(a, r) =fix - al = r}. In this case the boundary of S is part of S. In R2, we sometimes call the ball a disc and we usually call its boundary a circle rather than a sphere. In II81, the ball B(a, r) is just the interval [a - r, a + r] and its boundary is just the two endpoints. 27
5. Open and Closed Sets
28
3D B(a,r)
Figure 5.1. The boundary of the ball B(a, r) is the sphere. The closed ball includes its boundary.
2D B(a,r)
1 D B(a,r)
a r Figure 5.2. The boundary of a 2D ball or "disc" is a circle. The boundary of a 1D ball or interval is two points.
As another example, consider all of R' with one point, the origin, removed:
S=R--{0}. Its boundary is the single point {0}. In this case, the boundary of S is not part of S. The boundary of the rationals Q is all of R. The boundary of R is the empty set.
5.2. Definition. A set S in R' is open if it contains none of its boundary points. A set S in R7z is closed if it contains all of its boundary points. It follows immediately that a set S is open if and only if its complement So is closed.
Many a set contains just part of the boundary and hence is neither open nor closed. "Not closed" does not mean "open." These terms are not opposites!
5.3 Proposition
29
CLOSED
OPEN
NEITHER
Figure 5.3. Some sets are closed or open but most are neither.
For example, the ball B (a, r) is closed. If you remove its boundary, the resulting set is open and is called an open ball, for which unfortunately we have no special symbol. R' - {0} is open. The rationals Q are neither open nor closed. The reals R are both open and closed. See Figure 5.3 for a few more examples. Proposition 5.3 gives nice direct characterizations of open and closed. A set is open if it includes a ball about every point; of course the ball has to get smaller as you get close to the boundary. A set is closed if it includes all accumulation points. See Figure 5.4.
5.3. Proposition. A set S in R'ti is open if and only if about every point of S there is a ball completely contained in S. A set S is closed if and only if it contains all of its accumulation points.
Proof. Suppose that about some point p of S there is no ball completely contained in S. Then every ball about p contains a point of So as well as
5. Open and Closed Sets
30
Figure 5.4. An open set includes a ball about every point. A closed set includes all of its accumulation points.
the point p in S. Consequently, p is a boundary point of S, and S is not open. Conversely, suppose that S is not open. Then some point p of S is a boundary point, and there is no ball about p contained in S. Suppose that an accumulation point p of S is not contained in S. Then every ball about p contains a point of So (namely, p) and a point of S (because p is an accumulation point of S). Therefore, p is a boundary point not contained in S, so that S is not closed. Conversely, suppose that S is not closed. Then So contains some boundary point p of S, which is an accumulation point of S, so S does not contain all of its accumulation points.
5.4. Proposition. Any union of open sets is open. A finite intersection of open sets is open. Any intersection of closed sets is closed. A finite union of closed sets is closed.
Proof. To prove the first statement, suppose that x belongs to the union of open sets Ua. Then x belongs to some Up. By Proposition 5.3, some ball about x is contained in Up, and hence in the union of all the Ua. Therefore, by Proposition 5.3 again, the union is open. To prove the second statement, let U be the intersection of finitely many open sets U2. Let p belong to U. Since p belongs to Ui, there is a ball B(p, ri) contained in Ui. Let r = min{ri}. Then B(p, r) is contained in every Ui and hence in U. Therefore U is open. Let C be an intersection of closed sets Ca. Then CC is the union of the
open sets CC, and hence open. Therefore C is closed. Similarly let C be a union of finitely many closed sets Ci. Then CC is the intersection of the open sets CP, and hence open. Therefore C is closed.
5.5. Definitions (see Figure 5.5). The interior of a set S, denoted int S or S, is S - 8S. The closure of S, denoted cl S or S, is S U ,9S. An isolated point of S is the only point of S in some ball about it.
Exercises 5
31
.
P
Figure 5.5. A set, its interior, its closure, and an isolated point p.
5.6. Proposition. The interior of S is the largest open set contained in S, and the closure of S is the smallest closed set containing S. Proof. Exercises 11 and 12.
5.7. Topology. In 1l? or in more general spaces, the collection of open sets is called the topology, which determines, as we'll soon see, continuity, compactness, connectedness, and other "topological" properties.
Exercises 5 1. Say whether the following subsets of T1 are open, closed, neither, or both. Give reasons.
a. [0,1); b. 7L;
_c. {xEIIB: sinx>0}, d. U= 2[1/n,1) 2. Show by example that the intersection of infinitely many open sets need not be open.
3. Show by example that the union of infinitely many closed sets need not be closed.
4. Is all of IR the only open set containing Q? Prove your answer correct.
5. If S = [0,1), what are 8S, S, and S?
32
5. Open and Closed Sets
6. If S = 7G, what are 8S, S, and S?
7. If S = B(0,1) in lR2, what are 8S, S, and S? 8. Give an example of open sets U1 D U2 D n UZ is closed and nonempty.
such that the intersection
9. Give an example of nonempty closed sets C1 D C2 D intersection n Cz is empty.
such that the
10. Prove that every point of S is either an interior point or a boundary point.
11. Prove that the interior of S is the largest open set contained in S. (First prove that it is open.) 12. Prove that the closure of S is the smallest closed set containing S. 13. Prove that a boundary point of S is either an isolated point or an accumulation point.
14. Prove or give a counterexample: two disjoint sets cannot each be contained in the other's boundary. 15. A subset So of a set S is dense in S if every ball about every point of S contains a point of So. Are the rationals dense in the reals? Are rationals with powers of 2 in the denominator dense in the reals? Are the points with rational coordinates dense in RI?
Chapter 6
Continuous Functions
Continuous functions are the bread and butter of calculus. We'll now give three equivalent definitions of continuous functions. The first, our original definition, uses the idea of limit. The second uses the concept of a convergent sequence. The third, modern definition uses the concept of open set. It is the shortest and the most abstract, and it takes a little while to get used to.
6.1. Proposition. Let f be a function from Rn to R. The following are equivalent definitions of what it means for f to be continuous everywhere:
(1) For every point p, given E > 0, there exists S > 0, such that
Ix - pI < S=> If(x)-f(p)I <e. (2) For every point p, for every sequence xn -4p, f (xn) -p f (p). (3) For every open set U in R, f -l U is open.
(2) because if all points x near p have Proof. (1) 4* (2). Easily (1) values f(x) near f(p), then certainly the xn for n large will have values f (xn) near f (p). Conversely, suppose that (1) fails. This means that for some E > 0, no small 5 bound on Ix-pI, for example S = 1/n, guarantees that If (X) - f (p) I < E. Thus there must be some sequence x, with Ixn -PI < 1/n and with If (xn) - f (p) I > e, so that (2) fails.
(1) = (3). Let p E f -1U. Then f (p) E U and hence some small ball B(f (p), E) C U. By (1), choose S > 0 such that
Ix - pI < S= If(x)-f(p)I <e, and hence
Ix-pI IR satisfies f (0) = 0 and I f'(x) I < M. Prove that I f (x) I < MIx1. Apply this to the function f (x) = sinx. 3. Discuss the logical chain of reasoning from compactness of [a, b] to Corollary 14.5.
4. Show that the Cantor function is a continuous map of the Cantor set onto [0, 1], solving part of Exercise 13.6.
Define a map f from C into [0, 1] as follows. Given a point in the Cantor set, represent it by a base three decimal without is in it, such as 5.
.0222022002... , change the 2s to is to get something like .0111011001, and then interpret it as a base two decimal in [0,1]. Is f continuous? surjective?
Is f related to the Cantor function?
Chapter 15
The Riemann Integral
This chapter defines the standard, Riemann integral of a function on a bounded interval [a, b] in 11 and shows that the process works for every continuous function f on [a, b], using the fact that a continuous function on a compact set is uniformly continuous.
15.1. The Riemann integral. The Riemann integral fQ f (x) dx of a function f over an interval [a, b] represents the area under the graph. The area may be approximated as in Figure 15.1 by chopping the interval up into narrow subintervals of perhaps variable thickness Ax, approximating each subarea by a skinny rectangle of height f (x), thickness Ax, and area AA = f (x) .x, and adding them up:
A :: E f (x) Ax. This approximating sum is called the Riemann sum. To get the exact area,
we take the limit as the maximum thickness goes to 0, and call this the Riemann integral: fb
f (x) dx = lim E f (x) Ox. The limit must be independent of the choice of subintervals and of the choice
of where we evaluate f (x) in ;each subinterval. If the limit exists, we say that f is integrable on [a, b]. If f (x) is a constant c, then f is integrable and fb
f (x) dx = c(b - a). 65
15. The Riemann Integral
66
Figure 15.1. The area under the graph of f may be approximated by the sum of areas of skinny rectangles of height f (x) and thickness Ox.
Indeed, every Riemann sum
f(x)Ax=c>i x=c(b-a). This corresponds to the fact that the area of a rectangle of height c and width b - a is c(b - a).
15.2. Nonintegrable functions. One function which is not integrable on [0, 1] is the characteristic function XQ of the rationals, which equals 1 on the rationals and 0 off the rationals. Indeed, no matter how small Ax, there .are Riemann sums equal to one, obtained by always choosing to evaluate XQ(x) at a rational, and there are Riemann sums equal to zero, obtained by always choosing to evaluate XQ(x) at an irrational. Another function which is not integrable on [0, 1] is the function f (x) _
1// because no matter how small Ax, there are Riemann sums arbitrarily large, obtained by choosing to evaluate f (x) at a point very close to 0 on the first subinterval. In general, an integrable function must be bounded (Exercise 3). Fortunately, every continuous function is integrable:
15.3. Theorem. Every continuous function is integrable on [a, b].
Proof. By Exercise 8.6, it suffices to show that any sequence of Riemann sums with Ax - 0 is Cauchy. By Theorem 11.2, the continuous function f on the compact set [a, b] is uniformly continuous: given e > 0, there exists 5 > 0 such that (1)
JAXI b by
I
ra
b
f (x) dx = -
Jb
f (x) dx.
Finally, instead of fa f (x) dx, we sometimes just write fa f
.
15.5. Proposition. Suppose that f and g are integrable on the relevant intervals and that C is a constant. a.
faCf =Cfa f.
b. faf+g=f f+fag c. fa f = fa f + fbC f .
d. faf HfaIfI e. If f 0, by Corollary 15.6, rb b+tb f(x) Iy_
kob f (x)
dx
Ab
ix
b Obi f (x)
If Ab < 0, the sign of both the numerator and the denominator change, the As Ab -p 0, the fraction remains unchanged, and the inequalities still left-hand side and the right-hand approach f (b); hence so does the fraction. Therefore hold..
d db
Jb+Ob f(x) b
J f (x) dx = bmo
dx
Ob
- f (b)
as desired.
Proof of II. Note that by I,
db CF(b) d
ab
f (x) dx = F'(b) - f (b) = f (b) - f (b) = 0.
By Corollary 14.5 to the Mean Value Theorem, b
F(b)-1. f(x)dx=C. Plugging in b = a yields
F(a) = C. Therefore b
Ja
f (x) dx = F(b) - F(a).
16.2. Remark. Why don't we just define fa f (x) dx as F(b)-F(a)? First of all, we may not know of any antiderivative F. Second of all, how could we expect that to have anything to do with area or other applications? The only sound approach is to define area somehow (the Riemann integral) and then figure out an easy way to compute it.
16.3. Remark. There are amazing generalizations of the Fundamental Theorem to functions on domains in R' and more general domains, which go by names such as Green's Theorem, Gauss's Theorem or the Divergence Theorem, and Stokes's Theorem.
Exercises 16
73
-
Exercises 16 1. Use the Fundamental Theorem to compute that fo x2 dx = 1/3. Is this easier than direct computation from the definition? 2. Compute db fQ x2 dx two ways, first using 16.1(11), then using 16.1(1).
3. Let F(x) = fox
e_t2
dt. Compute F'(x) and F'(0).
4. What is da .1 a f (x) dx?
Chapter 17
Sequences of Functions
What does it mean for a sequence of functions to converge to some limit function? There is an easy definition, just looking at one point at a time.
,17.1. Definition (pointwise convergence). A sequence of functions fn converges to f pointwise on some domain E if for every x E E, the sequence of numbers ,,,(x) converges to f (x); i.e., for every x in E, given e > 0, there is an N, such that
n > N = I fn (x) - f (x) < E.
For example, if f, (x) = x/n, then fn --> f , where f (x) = 0, on R. As a second example, if fn (x) = xn, as in Figure 17.1, then on [0, 1] fn - f, where
for0<x 0, there is an N, such that
n>N=Ifn(x)-f(x)I<E for all x in E.
75
17. Sequences of Functions
76
x Figure 17.1. The continuous functions to a discontinuous function.
x" converge pointwise
The only difference in the two definitions is that in the second, N appears
before x, and hence the same N must work for all x, whereas in the first definition, N is allowed to depend on x. It is amazing that such a seemingly small change can be so important. Uniform convergence is just what you need to get continuous limits: 17.3. Theorem. A uniform limit of continuous functions is continuous. Proof. The idea of the proof is that by uniform convergence, we can handle all points near any particular point p by looking at one fn, that is uniformly near f. Given e > 0, choose N such that for all x in the domain D,
Ifn(x)-f(x)I <E. Since fN+l is continuous, given a point p in D, we can choose S > 0 such
that Ix - pi < S= IfN+1(x)-fN+1(p)I <e.
17.5. Theorem
77
Then if Ix - pI < S, f uniformly, then f 2 -p f 2 uniformly. First give a counterexample. Second add a simple hypothesis and prove the revised statement. Prove or give a counterexample: If nonvanishing (never 0) functions f,,, -p f uniformly, then 1/fn -> 11f uniformly.
4.
5. Consider continuous functions from [0, 1] to IR
0
0, every f,, I < M. Then
J
lim f," = lim j fn,.
Proof. Apply Lebesgue's Dominated Convergence Theorem, with g(x) _ . M. Since the domain is bounded, M has finite integral. For switching the order of integration in a double integral, the following theorem is very useful.
18.3. Fubini's Theorem. For a double integral, it is OK to switch the order of integration if either order yields a finite answer when the integrand is replaced by its absolute value.
18.4. Caveat on measurability. Technically, Theorems 18.1-18.3 have an additional hypothesis, that the integrands be "measurable." Measurable functions include continuous functions, piecewise continuous functions, perhaps altered on countable sets or sets of measure 0, and much more. They include all functions that you have ever heard of and all functions that can arise in the real world. Indeed, there is a theory of mathematics (without the Axiom of Choice) in which all functions are measurable. So this is not something to worry about. We give one more very useful theorem for switching integration with respect to one variable x with differentiation with respect to another variable t.
18.5. Leibniz's Rule. Suppose that J (t) f (x) t) dx exists, that a(t), b(t), and f (x, t) are all continuously differentiable with respect to t, and that there
is a function g(x) > at (x, t) with r J(t) g(x, t) dx bounded. Then d dt
(I
f
b(t)
Jab(t)
f (x, t) dx =
°Q(t)
C7 f
at
We allow a = -oo or b = +oo.
(x,
t) dx + f (b(t), t) Y(t)
- f (a (t), t) d (t).
Exercises 18
83
Exercises 18 fi x2-(S,nnx)/ndx. Justify.
1. Compute
z
2. Compute limn, f 0O '+(-')'e 1"x dx. Justify. 3. Compute limn,,,,, hl z dx. Justify. 4. Compute fy, 0 fx l0 xy cos xy2 dx dy. Justify. Answer: 3/400.
5. Compute fy 0 f2 Y xe-y2/y dx dy. Justify. Answer: 1- 1/e. 2t
6. Compute At fy 1 S'nx dx. Justify. Answer: 2t (sin 4t - sin t) for t 0 0, 2 for t = 0.
7. Compute d f1 i Answer:
-e3_1
ti
e-X2t
dx at t = 1. Justify. -.17478.
8. Compute y f °OZ+y at y = 0. Justify. Answer: -1/4. 9. Give a counterexample to the following converse to Lebesgue's Dominated
Convergence Theorem: If f,,, -> f and f f,,, --+ f f (all finite), then there exists g > If,, I .with f g < oo.
Chapter 19
Infinite Series E an
Infinite sums or series turn out to be very useful and important because in the limit you can actually attain an apparently unattainable value or function by adding on tiny corrections forever. In theory, they are no harder
than sequences, because to see if a series converges you just look at the sequence of subtotals or partial sums. For example, we say that the infinite series 1
1
1
1 24816+...
converges to 1 because the sequence of partial sums 1
3
7
15
2' 4' 8' 16' converges to 1.
19.1. Definition. The series
°_ a,.,, converges to L if given E > 0 there
is an No such that
N>No=* >an-L <E. n=1
Otherwise the series diverges (perhaps to +oo, perhaps to -oo, or otherwise "by oscillation"). Of course if a series converges, the terms must approach 0 (Exercise 12). The converse fails. Just because the terms approach 0, the series need not converge. For example, see the harmonic series below.
19.2. Proposition. Suppose that E°O_1 an converges to A and E°°_1 b,,, converges to B. Then EO°_1 can converges to cA and E°°_1
converges
to A+B. 85
19..Infinite Series E an
86
Proof. This proposition follows immediately from similar facts about sequences, as you'll show in Exercise 12. We now summarize some of the famous convergence tests from calculus.
19.3. Geometric series. A geometric series (obtained by repeatedly multiplying the initial term ao by a constant ratio r) (1)
E aorn = ao + aor + aor2
+ aor3 +...
,
converges to j if Irl < 1 and diverges if Irl > 1 (and ao 0). For example, the series at the beginning of Chapter 19, a geometric series with initial term ao = 1/2 and ratio r = 1/2, converges to 111/2 = 1.
Proof. If Irl > 1 (and ao # 0), the terms do not approach 0, so the series diverges. Suppose Irl < 1. Let Sn denote the subtotal Sn = ao + aor + aor2 + aor3 + ... + aorn;
rSn = (1 - r)Sn = ao Sn =
+ aorn + aorn+1;
aor + aor2 + aor3 +
- aorn+l;
ao(1 - rn+1)
1-r
Asn -oo,Sn -> 1Or' 19.4. p-test. The p-series
converges if p > 1 and diverges if p < 1.
For example, taking p = 1, the harmonic series 1
1
1
1
1
n=1+2+3+4+...
diverges, despite the fact that the terms approach 0. As a second example, taking p = 2, 1
n2
1
1
1
1 22+22+32+42+...
converges, although we do not yet know what it converges to. We omit the proof of the p-test, which uses comparison with integrals.
19.5. Comparison test. If Ian < bn and E bn converges to /3, then E an converges, and its limit a < ,Q. For example, 1+n converges by comparison with verges by the p-test.
n , which con-
19.6. Alternating series
87
Proof. It suffices to show that the sequence of subtotals, which is obviously bounded by,6, is Cauchy. For No < M < N, N
M
1: an
N
N
M
N
- E a, = E an : E Ian1 < E bn =
n=1
n=1
n=M+1
n=M+1
n=M+1
M bn - E bn
n=1
n=1
which is small because the sequence of subtotals of > bn is Cauchy.
Remark. The first million terms do not affect whether or not a series converges, although of course they do affect what it converges to. For example, 2 ( 1'0 n ) n converges by comparison with En=2 (2) n = 1 because for nOO=glo n large, 1 glo 10 n < 2. The limit, however, is huge; just the single term when I
)1010 = 1010000000000 n - 1010 is (100 to
19.6. Alternating series. If the terms of a series E an alternate in sign and I an l decrease with limit 0,
jail
?1a21>la31...-*0,
then the series converges.
For example, the alternating harmonic series (_1)n+l = 1 1 1 1
n
1
2+3
4
converges (to In 2).
Proof. We may assume that the first term is positive. The subtotals after an odd number of terms are decreasing and positive, and hence converge to a limit Ll. The subtotals after an even number of terms are increasing and bounded above by the first term, and hence converge to a limit L2. Since their differences approach 0, Ll = L2 is the limit of the sequence of subtotals.
19. Infinite Series E an
88
Exercises 19 For exercises 1-11, prove whether or not the series converges. If you can, give the limit.
1 10+loo+1000 2.
65+ 2512
25+
4. En=1 n 1/n 5.
6.
°°
1
Li°°
1
:n=1 nfnn=0
7. E00 n=1 51nn n
8. En'=2 In n . Why did we start the series with n = 2 instead of n = 1? (-1)n
9. E°° n=2 nl/n
10. Fn3 (n2+n-i)(Inn) . 11. EO° n=1
6n2 89n+73 n 4-213n
12. Prove that if a series converges, the terms approach 0. 13. Prove Proposition 19.2.
Chapter 20
Absolute Convergence
20.1. Definitions. A series E a,,, converges absolutely if the series of absolute values > janI converges. Otherwise it converges conditionally (or diverges)-
For example, the alternating harmonic series converges conditionally. It follows from the Comparison Test 19.4 that absolute convergence implies convergence.
A rearrangement of a series has exactly the same terms, but not necessarily in the same order. You might expect that order should not matter, and it does not if the series converges absolutely (Prop. 20.2), but it does matter if the series converges merely conditionally (Prop. 20.3). Here is an example:
2_2+2_2+33+33+33+44+44+44+44+... 1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
with partial sums 1
1
1
1
1
1
1
1
1
0, 2 , 0, 3 , 0, 3 , 0, 3 , 0, 4 , 0, 4 , 0, 4 , 0, 4 , 0, .. .
2,
converges to 0, but the rearrangement
2+222+3+3+3333+4+4+4+4+4444+... 1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
with partial sums 1
2' 1'
1
2,
0'
1
2
3, 3,
1'
2
1
3, 3,
0'
1
4'
1 3 2' 4' 1'
3
4,
1
1 2' 4' 0'
.
.
diverges by oscillation.
20.2. Proposition. If a series converges absolutely, then all rearrangements converge to the same limit. 89
20. Absolute Convergence
90
Proof. Let > an denote the original series, which converges absolutely to some limit L, and let E bn denote the rearrangement. Given E > 0, choose N1 such that if M, N > N1, then N
(1)
N
Ean-L <e/2.
E lanI,<E/2 and n=M
n=1
Choose N2 such that the first N2 terms of E bn include the first N1 terms
of T an. Let N > N2. Choose N3 such that the first N3 terms of E a, include the first N terms of E bn. Then N
N
Ebn-L n=1
N3
b, n=1
N3
a, + n=1
1: an - L n=1
Since N3 > N1, by (1) the last expression is at most e/2. Consider the middle expression. All of the terms of the first sum are included in the second sum, and the first N1 terms of the second sum are included in the first sum. Hence by (1) the middle expression is at most e/2. We conclude
that N
Ebn-L <E, n=1
which means that the rearrangement converges to the same limit L.
20.3. Proposition. Suppose that a series > an converges conditionally. Then its terms may be rearranged to converge to any given limit, or to diverge to ±oo, or to diverge by oscillation.
Proof sketch. First we claim that the amount of positive stuff, the sum of the positive terms, must diverge to +oo, and that the amount of negative stuff must diverge to -oo. If both converged, the series would converge absolutely. If just one converged, the series would diverge. Hence both must diverge. Second, note that because the series converges, the terms approach 0. Our rearrangements will keep the positive terms in the same order and the negative terms in the same order. To get big (positive) limits, we'll front
load the positive terms. To get small (negative) limits, we'll front load the negative terms. To get a rearrangement with prescribed limit L, take positive terms until you first pass L heading right. Then take negative terms until you first pass L heading left. Then take positive terms until you pass L heading right. Then take negative terms until you pass L heading left. Since there is an infinite amount of positive stuff and of negative stuff, you can continue
20.5. Root test
91
forever. Since the terms go to 0, the amount by which you overshoot goes to 0. Hence the rearrangement converges to L. To get a rearrangement which diverges to +oo, take positive terms until you pass 1, then a negative term, then positive terms until you pass 2, then a negative term, then positive terms until you pass 3, and so on. To get a rearrangement which diverges to -oo, take negative terms until you pass -1, then a positive term, then negative terms until you pass -2, then a positive term, then negative terms until you pass -3, and so on. To get a rearrangement which diverges by oscillation, take positive terms until you pass 1, then negative terms until you pass -1, then positive terms until you pass 2, then negative terms until you pass -2, and so on.
20.4. Ratio test. Given a series E an, let p = limIfI pQanl (Greek letter rho). If p < 1, then the series converges absolutely. > 1, then the series diverges. If p = 1 or the limit does not exist, the test fails; the series could converge absolutely, converge conditionally, or diverge. For example, consider the series 00
n=O
1
1
1
1
1
1
1
1
1
1
1
0!+1!+2!+3!+4!+...-1+1+2+6+24+...
n!
To get to the nth term from the previous term, you multiply by 1/n, so the limiting ratio p = 0. Consequently the series converges absolutely (to e).
Proof of Ratio test. Exercise 10. Although I don't usually teach the following test in calculus, it turns out to be of theoretical importance in real analysis.
20.5. Root test. Given a series E an, let p = lim sup n a,, . If p < 1, then the series converges absolutely. If p > 1, then the series diverges. If p = 1, the test fails; the series could converge absolutely, converge conditionally, or diverge.
Proof. If p < 1, to give us some room to work, let p < a < 1 (v is the a, i.e., Ian1 < am, so that Greek letter sigma). Then for all large n, " the series converges absolutely by comparison with the geometric series. If p > 1, for all large n, IanI > 1, and the series diverges because the terms do not approach 0.
20. Absolute Convergence
92
Exercises 20 1. What are the possible values of rearrangements of the following series: a.
F°°
1 .
n=1 2n 1
b. 1:0 1 00
C. Ln=1
(an)n .
(-1)n . rn-
I
d. En= 1 ,' 1
Prove whether the following series 2-9 converge absolutely, converge conditionally, or diverge. Give the limit if you can.
i
2. E0n=0 5n 3.
00 n=0
4. E00 n=0
(-1)n 5n 5n I,
5. E00 o (-5 )n
i
6. E00 n=0 nn
00 (-1)n . 7. En=0 771/5
(-1)n 00 8. En=0 1+1/n'
9. V'00
n=0
nsinn en
10. Prove the Ratio test 20.4. Hint: Use the proof of the root test as a model.
Chapter 21
Power Series
Power series are an important kind of series of functions rather than just numbers. We begin with consideration of general series of functions.
21.1. Definitions. A series E fn of functions converges (pointwise or uniformly) if the sequence of partial sums converges (pointwise or uniformly). The series E fn converges absolutely if > IfnI converges absolutely. There is a nice test due to Weierstrass for uniform convergence, essentially by comparison with a series of constants.
21.2. Weierstrass M-test. Suppose Ifn I < Mn where each Mn is a positive number and E Mn converges. Then E fn converges uniformly. Proof. For each x, E fn(x) converges by comparison with E Mn to some f (x) To prove the convergence uniform, note that 00 00 N N fn(x)
- > fn(x)
fn(x) - f(x)
?1m
n=1
n=1
n=N+1
E Mn,
n=N+1
which is small for N large, independent of x. For example, comparison with
Sly converges uniformly on R by Weierstrass M-test . By Theorem 17.3, the limit is continuous.
21.3. Definition. A power series is a series E anxn of multiples of powers of x, such as 00 xn=1+x+-x2+6x3+...
n! 93
21. Power Series
94
which we will identify as ex in Chapter 25. A function defined by a convergent power series is called real analytic.
Of course a power series is more likely to converge if x is small. You might expect a power series to converge absolutely for x small, converge conditionally for x medium, and diverge for x large. The truth is even a bit simpler.
21.4. Radius of convergence. A power series E anxn has a radius of convergence 0 < R < oo, such that (1) for IxI < R, the series converges absolutely;
(2) for IxI > R, the series diverges.
For x = ±R, the series might converge absolutely, converge conditionally, or diverge.
On every interval [-Ro, Ro] C (-R, R), the series converges uniformly and the series may be integrated term by term. Note that this result admits the possibilities that the series always converges (the case R = oo) or never converges unless x = 0 (the case R = 0).
Proof. To prove (1) and (2), it suffices to show that if the series converges at xo, then it converges absolutely for IxI < Ixol. Since it converges at xo, the terms anx0 converge to 0; in particular, I anx0 j < C. Therefore lanxnl < CIx/xoln,
and E anxn converges absolutely by comparison with the geometric series >CIx/xo1n. Next we prove uniform convergence on [-Ro, Ro]. For some leeway, choose R1 between Ro and R. Since the series converges for x = Rl; the terms anRl converge to 0; in particular, I anRl I < C. Hence for all IxI < Ro,
Ianxn) < C(Ro/Rl)".
To apply the Weierstrass M-test, let Mn = C(Ro/Rl)n; then E Mn is a convergent, geometric series. By Weierstrass, E anxn converges uniformly on [-Ro, Ro]. In particular, for [a, b] C (-R, R), by Theorem 17.5, the integral of the series equals the limit of the integrals of the partial sums, which of course equals the limit of the partial sums of the integrals of the terms, which equals the series of the integrals of the terms.
21.10 Taylor's formula
95
For example,
[x++++j
1/2'
x2
J
x3
x4
11/2 o
1
1 1 1 2+222+233+244+...
21.5. Hadamard formula. The radius of convergence R of a power series E anxn is given by the formula 1
lim sup I an I
1/n'
under the agreement that 1/0 = oo and 1/oo = 0.
Proof. Using the Root test (20.5), we have p = limsup IanxnI1/n = lxl R Hence by the Root test, the series converges absolutely if x1 < 1, i.e., if x < R, and diverges if x R > 1, i.e., if lx > R. Therefore, RRis the radius of convergence.
21.6. Corollary. Given a power series E anxn, the "derived series" E nanxn-1 has the same radius of convergence.
Proof. Multiplication by x (which does not depend on n) of course yields a series with the same radius of convergence: > nanxn. Since lim n1/n = 1, by Hadamard's theorem the radius of convergence is the same as for anxn.
21.7. Differentiation of power series. Inside the radius of convergence, a power series may be differentiated term by term:
(ax7) = > nanxn,-1
Proof. On an interval [-Ro, Ro] C (-R, R), E anxn converges uniformly to some f (x), and the derived series > nanxn-1 converges .uniformly to some g(x). Since g can be integrated term by term, f xRo g = f (x) - f (-Ro). By the Fundamental Theorem of Calculus, g(x) = f'(x), as desired.
21.8. Corollary. A real-analytic function is infinitely differentiable (C°°). 21.9. Corollary. Inside the radius of convergence, you can antidifferentiate a power series term by term.
21.10. Taylor's formula. If f (x) _ > anxn, then an denotes the.nth derivative.
where f (n)
21. Power Series
96
Proof. f(x) = ao + aix + a2x2 + a3x3 + , f(0) = ao; f'(0) = 1a1; f'(x) = a1 + 2a2x + 3a3x2+--- , f"(x) = 2a2 + 3.2a3x + , f"(0) = 2a2; f-'(0) 3!a3+--- , f'11(x) = = 3! a3; and so on.
Caveat. Given a C°° function f , Taylor's formula yields an associated Taylor Series. Even if the domain of f is all of IR, the Taylor Series could have radius of convergence 0, and even if the series converges, it need not converge to the original function f. In a more advanced course, you will see examples of this strange behavior.
21.11. Power series about c
0. Power series j anx' are best when c, we could use E an (x - c)". There are only x is near 0 (x ti 0). For x minor changes to the theory. The interval of convergence now goes from c - R to c + R, and Taylor's formula becomes an =
(1)
f (n) (c) n1
For example, f (x) = 1/x has no power series about 0 because it is not even defined there. However, in terms of u = x - 1, (2)
1 = X
1 1+u
= 1 - u + u2 - u3 + u4 -...
(for Jul < 1)
=1-(x-1)+(x-1)2-(x-1)3+(x-1)4-...
(for Ix-11 °°_1 . a. Use the Ratio test to show that the radius of convergence R = 1. b. Use the Hadamard formula to show that R = 1.
It follows that the series converges uniformly on every interval [-Ro, Ro] C (-1, 1). Use the Weierstrass M-test to show that it actually
c.
converges uniformly on [-1, 1].
Exercises 21
97
3. Continuing Exercise 2, consider g(x) = x f'(x).
E'
a. Compute that g(x) = 1 b. TRUE/FALSE. By Corollary 21.6, g(x) has the same radius of convergence R = 1. c. Use the Ratio test and Hadamard formula to show that R = 1. d. It follows that the series for g(x) converges absolutely for jxj < 1. At the endpoints of the interval of convergence 1 and -1, does the series converge absolutely, converge conditionally, or diverge?
4. Find the Taylor Series and its radius of convergence for
a. f(x) = eS; b. f(x) = x2.
5. Leibniz's formula for 7r.
a. Justify that for lx i < 1, 1 1
1+x2
-x2+x4-x6+x8-...
b. Justify that for I xI < 1, x5 x3 tan lx=x-3 + - 7 -} 5
x7
x9
-...
9
(Note that it does not suffice to show.that this is the Taylor series for tan -1 x; see the Caveat after 21.10.) c. Assuming that part b also holds for x = 1 (as it does), deduce Leibniz's famous formula for 7r: 4
4
4
4
4 1-3+5-7+9-..
6. Check Taylor's formula 21.11(1) for the power series 21.11(2).
7. Find a series for 1/x in powers of u = x - 2 by noting that
1' x
1
2+u
_
1
1
2 1+u/2
and using the formula for a geometric series. Check Taylor's formula 21.11(1).
Chapter ,22
Fourier Series
22.1. Sines and cosines. The functions sinnx and cosnx on [-7r, 7r] are beautiful oscillations representing pure tones. A majestic orchestral chord, represented by a much more complicated function, is made up of many such pure tones. The remarkable underlying mathematical fact is that every smooth function on [-7r, 7r] can be decomposed as an infinite series in terms of sins and cosines, called its Fourier series. This is in strong contrast to the fact that only real-analytic functions are given by Taylor series in powers of x. Apparently, for decomposition, sines and cosines work much better than powers of x. There is good reason for studying sines and cosines so much, starting in high school. 22.2. Theorem (Fourier series). Every continuous, piecewise differentiable function f (x) on [-ir, -7r] is given by a Fourier series 00 (1)
f(x) = Ao + E (An cos nx + Bn sinnx),
where 1
Ao = 1(2)
An = Bn
1
ff
J
f (x) dx, cos nx dx,
fsinnxdx,
except that at the endpoints the series converges to the average of f (-ir) and f(70 99
22. Fourier Series
100
Outside [-ir, -7r], the Fourier series just keeps repeating every 27r, so it doesn't always equal f, unless f is also 21r periodic. The proof of Theorem 22.2 is just beyond the scope of this text, although Theorem 25.1 at least proves that the series converges if f is smooth; it takes more work to prove that it converges to f.
22.3. The coefficients. The above formulas 22.2(2) for the Fourier coefficients An, and Bn, due to Euler, deserve comment. First note that A0 is just the average value of f (x), about which f varies or oscillates. The other formulas are best understood by comparison with formulas for the coefficients of a vector
v = vlel + v2e2 + ... +. vNeN
(1)
in RN, expressed as a linear combination of standard basis vectors en. Such basis vectors are orthonormal, that is, em en = 0, except that em em = 1. It follows immediately that the coefficients vn satisfy (2)
by just taking the dot product of each side of (1) with en. Analogously, on. the space of continuous functions on [-7r, 7r], one can define a dot product by 1
fg=
(3)
7r
f
7r
f (x) g(x) dx, 7r
integrating all products of values instead of just summing products of components. We want to write everything in terms of the functions cos nx and sin nx, which turn out to be orthonormal functions for this dot product (see Exercise 1). The formulas for the Fourier coefficients 22.2(2) just say that each Fourier coefficient is obtained by dotting the function f (x) with the associated orthonormal function, just like the corresponding formula (2) for vectors.
22.4. Example. For example, consider the piecewise smooth function 0
if-ir<x (1/n2) = 7r2/6.
1
1
1
22. Fourier Series
102
22.6. Perspective. Since a Fourier series is determined by the coefficients, Fourier series provide a way to encode a function by two sequences A, B,, of numbers with physical significance corresponding to "frequencies." For a violin string, the predominant frequencies are nice multiples of each other, called harmonics, and produce a rich, harmonious sound. In many applications, it is often the case that certain frequencies. dominate, and the function can be well approximated by finitely many terms. Such mathematics can provide a way to encode or compress complicated data, sounds, or images. JPEG is exactly such a method for efficient storage of images. Dolby stereo filters out hissing noise by avoiding the frequencies responsible for such noise. The next chapter (23) will show how Fourier series can be used to solve important differential equations.
22.7. The sine series. If f (x) is an odd function, which means that f (-x) = -f (x), it follows immediately (Exercise 5) from the formulas 22.2(2) for the coefficients that every A,, vanishes (is 0), so that the Fourier series consists entirely of sines.
Every function f (x) on (0, 7r) can be extended to an odd function on (-7r, ir), whose sine series holds for the original function f (x) on (0, 7r): 00 (1)
f (x) _
B,, sin nx, ti.=1
where (2)
Bn = 2 7r
Jo
' f (x) sin nx dx.
The new factor of 2 comes because we are integrating only from 0 to 7r; we wouldn't need it if we integrated the extension from -ir to it.
22.8. General intervals. Fourier series (22.2) immediately generalize from [-7r, 7r] to [-L, L], just by replacing x with irx/L: 00
(1)
f (x) = Ao +
(A,, cos n=1
nix L
+ B,, sin
n7rx) L
,
22.8. General intervals
103
where 1
Ao = 2L (2)
A, = 2L Bn
1
= L L-2
L
ff(x)dx,
f
L L
f
L
f (x) Cos nLx dx, L
nirx f (x) sin L dx ,
104
22. Fourier Series
Exercises 22 1. Verify that the functions cos x and sin x are orthonormal on [-7r, Tr] for the dot product 22.3(3). 2. What is the Fourier series for f (x) = sin x on [-7r, 7r] ?
3. Compute the Fourier series for Example 22.4 using 22.2(2).
Hint: Use integration by parts and notice that cos n7r + 1 is 0 if n is odd and 2 if n is even.
4. Show that the Fourier series for f (x) = x (-ir < x < -7r) is given by
x=2
(isinx- l2sin2x+3sin3x-
Taking x = 7r/2, obtain Leibniz's amazing formula for 7r: 4
4
4
4
1
3
5
7
5. Show that if f (x) is an odd function, which means that f (-x) = -f (x), then the coefficients A,, vanish (are 0), and you get a Fourier series of sines. Similarly if f (x) is an even function, which means that f (-x) = f (x), then the coefficients B,,, vanish, and you get a Fourier series of cosines.
6. Show that the Fourier series for the function
for -7r < x 0.
25.3. Proposition. exp(-b) =
ex(b)
In particular, exp(-1) = 1/e, and
exp(x) is positive for all x.
Proof. This is the part of the development where you might think you would have trouble. Dividing 1 by an infinite series is quite a long division
problem. But it turns out that there is a slick proof. Notice that by the product rule dx
(exp(x) exp(-x)) = exp(x) exp(-x) - exp(x) exp(-x) = 0.
Hence exp(x) exp(-x) is a constant function: exp(x) exp(-x) = C. Plugging in x = 0, we see that C = 1. Hence exp(-x) = 1/ exp(x), as desired.
25.4. Proposition. exp(a + b) = exp(a) exp(b). Proof. Since by the quotient rule d exp(a + x) exp(a + x) exp(x) - exp(a + x) exp(x) dx
exp(x)
-
exp2(x)
- 0,
exp(a+x) = Cexp(x). Plugging in x = 0 yields C = exp(a), as desired.
25.5. Corollary. exp(n) = en; exp(p/q) = 9 eP. Proof. By Proposition 25.4, exp(n) = exp(1) exp(1) ... exp(1) = e'. Likewise, (exp(p/q))q = exp(p) = eP, which implies that exp(p/q) = eP/q. If we now define e' as exp(r) for all real numbers, this definition will agree with any others by continuity.
25.6. Corollary. (1) exp(x) is strictly increasing. (2) For every n, there is an en, > 0, such that for x > 0, exp(x) > en,x". (3) Given C > 0 and n, for x large, exp(x) > Cx".
Proof. Conclusion (1) follows because the derivative exp(x) is positive. Conclusion (2) follows from the defining series, with e,, = 1/n!. In particular, given C > 0, for x > 0, exp(x) > (1/(n + 1)!)xn+1, which is greater than Cxt once x > C(n + 1)!.
Exercises 25
113
25.7. The natural logarithm. It follows from Corollary 25.6(1) that exp(x) is as bijective map from R onto the positive reals. The inverse function from the positive reals onto R is called the natural logarithm function, written In x in calculus books and log x in more advanced mathematics. Its properties, such as (log x)' = 11x, follow from the properties of exp(x). See Exercise 4.
25.8. Complex exponentials. Although we have dealt only with real series, all of our results hold for series of complex numbers z = x + iy, including the definition and properties of exp(x). Here i2 = -1, i3 = -i,
i4 = 1.....
Exercises 25 1. Development of the sine and cosine functions. Define x2
cosx=
2
+
x4
x6
4! x5
6!
sinx=x - x33 + 5 -
x7 --
+
x8 8!
-,
+x99 - .
a. Check that the radii of convergence are oo, so that these functions are defined for all x.
b. Show that cos(-x) = cosx and that sin(-x) sin(x). c. Show that (sin x)' = cosx and that (cosx)' sin x. 2 d. Prove that sin x + cost x = 1. Hint: First show the derivative 0, so that sin 2 x + cost x = C. Then use
x=0tofind C. Note that this implies that I sin x < 1 and I cos x < 1. 2. By plugging into the series for ex, verify Euler's identity ez0
= cos 0 + i sin 0.
3. Use Euler's identity and the fact that ei(a+b) = eiaeib to deduce that sin(d + b) = sin a cos b + cos a sin b,
cos(a + b) = cos a cos b - sin a sin b.
25. The Exponential Function
114
4. Since the exponential function has a nonvanishing (never 0) continuous derivative, it follows from the Inverse Function Theorem (which did not quite make it into this book) that its inverse, log x, is continuously differentiable.
a. If y = log x, then x = ey. Differentiate implicitly to deduce that (log x)' = 1/x. b. Prove that log cx = log c + log x. 5. Obtain a power series for logx in powers of x - 1 by integrating 21.11(2). Check Taylor's formula (21.11(1)) for this series. 6.
Prove that e = limn. + (1 + 1/n)'. (This is sometimes used as the
definition of e.)
Hint: It suffices to show that
log(1 + 1/n)n = n log(1 + 1/n) -p 1. By the definition of the derivative, log(1 + Ox) log' lim g( 1) = Ix->O Ox Therefore, taking Ox = 1/n, nlog(1 + 1/n) -* log'(1) = 1.
Chapter 26
Volumes of n-Balls and the Gamma Function
There is a wonderful formula (26.5) for the volume of the ball B(0, r) in I[8": IT-/2
vol B (0, R) =
(n/2)
Rn.
For example, the "volume" or area of a ball in R2 is (222), r2 = 'irr2. The volume of a ball in R4 is 2ir2r4. However, according to this formula, the "volume" or length of a ball in R1, which should be 2r, is (%2), r. What is (1/2)! ? For our formula to work, we would need (1/2)! = -vfir/2.
Fortunately, there is a nice extension of (x-1)! from integers to real numbers greater than 1, called the Gamma Function r(x).
26.1. Definition. For x > 1, define the Gamma Function r(x) by
r (x) =
f
et1 dt.
This integral converges because et grows faster than any power of t.
26.2. Proposition. r(D; + 1) = xr(x).
Proof. Using integration by parts, f u dv = uv - f v du, with u = t', dv = e-tdt, du f= xtx-1, v = -e-t, 00
00
r(x + 1) =
J0
e-ttX dt = -e-ttx1o 00 +
J0
xe-ttx-1 dt = 0 + xr(x).
a
115
26. Volumes of n-Balls and the Gamma Function
116
26.3. Proposition. For every nonnegative integer n, F(n + 1) = n!.
-
We can use this relationship to extend the definition of the factorial function to every nonnegative real number by x! = F(x + 1).
Proof by induction. First we note that for n = 0,
I'0+1 =I'1 =
e-tdt=-e-t°°- 0- (-1) =1=0!.
Second, assuming the result for smaller values of n, we see by Proposition 26.2 that F(n + 1) = n r (n) = n(n - 1)! = n!.
26.4. Proposition. F(1/2) _ V'7r. Proof. By making the substitution t = s2, dt = 2s ds, we see that
F(1/2) =
f
e-tt-1/2 dt =
es2s ds = J 00 eds. J z
z
0
Now comes the famous trick-multiplying r(1/2) by itself and switching to polar coordinates with dA = r dr dB:
r(1/2) I'(1/2) = 00
jO27r
=
:;;:
:1::
=
oo
=oJr=o
1
z
e-T r dr dO = 21r --e-T 2
°°
z
= 7r. Jn
Therefore 1'(1/2) = V?-r.
In particular, (1/2)! = 1'(3/2) = (1/2)1'(1/2) = //2, as we hoped. Note too that our formula gives for the volume of a ball in R3: X3/2
x.3/2
4
vo1B(0,r) =
(3/2)!r3 the correct, familiar formula.
26.5. Proposition. The volume of a ball in I18' is given by rn/2
volB(0,R) = (n/2)[Rn Proof. We haven't defined volume; we will assume here that it can be computed as in calculus, by integrating over slices. We've already checked this formula for n = 1 and n = 2. We'll prove it by induction, assuming the result for n - 2. (By jumping two dimensions at a time, we can use a much
26.7. Stirling's Approximation
117
simpler polar coordinates proof.) We view the nD ball {x2 + x2 + x2 +
+
x2 < R2} as a 2D ball {r2 = xi + x2 < R2} of (n - 2)D balls of radius s satisfying s2 x3 + + xn = R2 - r2 and volume (by induction) ?r(n,-2)/2
n-2 = ((n/2) - 1)!' n-2 7r(n/2)-1
((n - 2)/2)!
s
Hence,
(n/2)-1 vol B (0, R) =
Ifsn-2 dA
((n/2) - 1)t 7r(n/2)-1
((n/2) - 1)! ((n/2) - 1)!
j 27r
R
(R2
- r2)(n.-2)/2r dr dO
=0 r=0 1 (R2 - r2)f/2
-2
R
n/2
1 Rn
, (n/2)-1
_ ((/2) n
2r
- 1) 2/2 n = (n/2) Rn
0
7rn/2
27r
.
26.6. Corollary. The volume of the (n - 1)D sphere in 1R is given by vol S(0, R) = n
7r
n/2
(n/2)
Rn
For example, the circumference of a circle in R2 is given by 2R = 27rR.
Proof. The volume of the sphere is just the derivative of the volume of the ball (the greater the surface area, the faster the volume of the ball increases).
26.7. Stirling's Approximation. The Gamma function may be used to derive an excellent asymptotic approximation to n! n!
27rn(n/e)n.
(Technically this asymptotic symbol - means that as n -> oo, 1.) It follows that e n monh -+ mone -' money).
Exercises 27 Is each of the following a metric on the space of continuous functions from [0, 1] to R? If not, explain why not. 1.
a. p(f, g) = If (1/2) - g(1/2) I b. p(f, g) = sup{If'(x) - g'(x)I}.
c. p(f, 9) = fo I f (x) - 9(x) I dx.
d. P(f 9) = f o I f (x) - 9(x) I2 dx. e. p(f, g) _
How many words can you find in B ("pat" ,1)? How many elements (counting words and gibberish) are there in B ("pat", 1)? 2.
Chapter 28
Analysis on Metric Spaces
Analysis on metric spaces is very similar to analysis in R. Just replace the distance Ix - yj with the more general p(x, y).
28.1. Definition. A sequence an in a metric space X converges to a limit
L (an - L) if, given e > 0, there is some N such that whenever n > N, p(an,L) < E.
In C[0,1], fn -f f under the sup metric therefore means that fn -p f uniformly. So fn(x) = x/n converges to f (x).= 0, but fn(x) = xn does not converge (uniformly) to f (x) = 0. Most of the results on sequences of Chapter 3 remain true in metric spaces. The exception is Proposition 3.6 because in a general metric space there is no addition, multiplication, or division. 28.2. Definition. A sequence an in a metric space X is bounded if all the terms lie in some ball. A sequence an in a metric space X is Cauchy if, given e > 0, there is some N such that whenever m, n > N, p(a,n, an) < E. Every convergent sequence is Cauchy (Exercise 3), but whether every Cauchy sequence converges depends on the metric space. It is true in the reals. It fails in the space of rationals, since 1, 1.4, 1.41, 1.414, ..., which converges to
in IR, does not converge in Q, even though it is still Cauchy.
The problem is that
is missing-the space is somehow not "complete."
28.3. Definition. A metric space X is complete if every Cauchy sequence converges (to a point of X). 125
28. Analysis on Metric Spaces
126
is complete, but Q is not complete (Exercises 16 and 17). C [0, 1] is complete (Exercise 18), essentially because a uniform limit of continuous functions is continuous and therefore is not missing from the space. Our word space W is complete because every point is isolated (no other points within distance 1/2) and hence all Cauchy sequences are eventually constant-the only way'for words to be very close together is for them to be identical. If a metric space X is not complete, one can form the "completion" by adding limits of all Cauchy sequences. Instead of defining the reals via decimals, some texts define the reals as the completion of the rationals. You have to be careful because different Cauchy sequences correspond to the same real number. 11
Topology on a metric space X is very similar to topology in R.
28.4. Definitions. A point p in a metric space X is a boundary point of a set S if every ball
B(p,r) = {x E X : p(x,p) < r} about p meets both S and So. The set of boundary points of S is called the boundary of S and written 8S. A set S is open if it contains none of its boundary points. A set S in Rn is closed if it contains all of its boundary points. For example, a ball B(p, r) is closed; if you remove the boundary, the resulting set is open and is called an open ball. The interior of a set S, denoted int S or S, is S - S. The closure of S, denoted cl S or S, is S U 8S. An isolated point of S is the only point of S in some ball about it. Every ball about an accumulation point of S contains infinitely many points of S. All of the results of Chapters 5-7 continue to hold in metric spaces.
Exercises 28
127
Exercises 28 1. In the space C[O, 1], does the sequence f, converge or diverge (under the sup metric)? If it converges, what is the limit? a. f,,(x) = x2 /n; b. fn(x) = x2n
2. Suppose that a sequence a,, converges in a metric space. Prove that the limit is unique and that the sequence is bounded. 3. Prove that a convergent sequence in a metric space is Cauchy. 4. Prove that a set S in a metric space is open if and only if the complement So is closed.
5. Prove that a set S in a metric space is open if and only if about every point of S there is a ball completely contained in S. 6. Prove that a set S in a metric space is closed if and only if it contains all of its accumulation points. 7. Say whether the set is open, closed, neither, or both in the'space C[O, 1].
a. If (1/2) < 6}; b. {f(1/2) < 6};
c. B(0,1).
8. Prove the following statements in a metric space. Any union of open sets is open. A finite intersection of open sets is open. Any intersection of closed sets is closed. A finite union of closed sets is closed.
9. Prove that in the word space W every singleton is open, and hence every set is open and closed.
10. If S = {0 < f (1/2) < 1} C C[0,1], what are 8S, S, and S? 11. Prove that the interior of S is the largest open set contained in S. 12. Prove that the closure of S is the smallest closed set containing S.
128
28. Analysis on Metric Spaces
13. Prove that a is an accumulation point of S if and only if a is the limit of a sequence of other points in S. 14. Prove that a boundary point of S is either an isolated point or an accumulation point.
15. Prove that every point of S is either an isolated point or an accumulation point. (Why does this not imply Exercise 14?) 16. Show that Q is not complete.
17. Prove that R is complete. 18. Prove that C[O, 1] is complete.
19. Do Proposition 6.1 and proof generalize to real-valued functions on a metric space?
Chapter 29
Compactness in Metric Spaces
This is the moment when metric space topology takes a nasty turn. In a general metric space, the three characterizations of compactness of Theorem 9.2 are no longer equivalent: a closed and bounded set need not satisfy the other two. First we will give two natural examples of this phenomenon. Then Theorem 29.3 explains that the other equivalences continue to hold.
29.1. Example. Let ]R°° denote the space of all real vectors (X1, x2, X3.... ),
with infinitely many components, within finite distance of the origin:
R' _{(mil, x2, X3, ...) : X1
X2
X 3+-'' < oo}
with metric
1/2
P(x, y)
y'')2)
This space is a lot like R'. The unit ball B(0,1) is closed and bounded, but unlike R", the unit ball B(0,1) admits sequences with no convergence subsequences. For example, let
el = (1,0,0,0,...), e2 = (0,1,0,0,...), e3 = (0, 0,1, 0, ... )
and so on. This sequence has no convergent subsequence because it has no apart. The problem Cauchy subsequence because all terms are distance is that R°° is infinite dimensional. 129
29. Compactness in Metric Spaces
130
29.2. Example. Similarly, consider the unit ball B(0, 1) in C[0,1]. It is closed and bounded, but admits sequences (of functions) with no (uniformly) convergent subsequences. (Recall that convergence under the sup metric of
C[0,1] means uniform convergence.) For example, let fn (x) = x7L, as in Figure 17.1. Then any subsequence converges pointwise to
f(X)-
] 0 for0<x 0, for every x E K, B(x, e) is contained in some U. Otherwise, let xn be a sequence of points in K such that B(xn,1/n) is contained in no Ua. By hypothesis (1), some subsequence converges to a point a in K. Some B(a, r) is contained in some U. Hence, once p(xn, a) < r/2, B(xn, r/2) is contained in Ua, a contradiction once 1/n < r/2.
Exercises 29
131
Now as long as possible choose disjoint balls of radius e/2 centered at points yn in K. This cannot go on forever to produce an infinite sequence because such a sequence yn could not have a convergent subsequence because it is not Cauchy. Hence, after choosing some yN, for every y in K, B(y, e/2)
intersects some B(yn, s/2). Hence, the finite collection {B(yn, e)} of balls twice as large covers K. Choose Uan containing B(y,,, E). Then {Uan} is the desired finite subcover of K.
Remark. All of the results of Chapters 10-12 continue to hold in metric spaces. See for example Exercise 29.3.
Exercises 29 1. Prove that 29.3(1) = 29.3(3). 2. Here is a trivial "artificial" example of a closed and bounded set which is not compact. Let X be the integers with metric p(m, n) = 1, except that p(n, n) = 0. Check that p is a metric. Show that X is closed and bounded, but not compact. 3.
Prove that in a metric space, a continuous real-valued function on a
nonempty compact set attains a maximum. (Cf. Corollary 10.2.) . 4. Do Theorem 11.2 and its proof generalize to a metric space?
Chapter 30
Ascoli's Theorem
To solve problems, one needs compactness in order to extract from a sequence of approximate solutions an exact solution in the limit. For sophisticated problems, the desired solution is not just a number, but rather a function, perhaps describing an economically ideal schedule of production or the most efficient shape of an airplane wing. Unfortunately, as we have seen, even the closed unit ball in the space C[O, 1] of real-valued continuous functions on the unit interval under the sup metric is not compact. Fortunately, Ascoli's Theorem provides useful compact spaces of functions. The main hypothesis is a uniform kind of continuity called equicontinuity. See if you can spot the difference from uniform continuity:
30.1. Definition. A set of real-valued functions f on a domain D C R is equicontinuous if, given e > 0, there is a 5 > 0 such that
y - xI < b= If(y)-f(x)I
<E.
The difference from uniform continuity is that b is independent not only of x but also of f : the same b works for all f. Note how equicontinuity fails for our previous Example 29.2 (fn (x) _ x"'). As n approaches infinity, xn increases rapidly near 1, and for given e > 0, S approaches 0; S is not independent of which In we consider.
30.2. Ascoli's Theorem. Let .F be a closed, bounded, equicontinuous subset of C[0,1]. Then .T is compact. 133
30. Ascoli's Theorem
134
Proof. Let f,,, be a sequence of functions in F. Note that for any- fixed point p, f,,,(p) is a bounded sequence of real numbers and hence has a convergent subsequence. Similarly, if q is another point, we can take a subsubsequence converging at q as well as at p. Indeed, given any finite number of points, it is easy to find a subsequence converging at those points. Begin by taking a subsequence Si that converges at 0, 1/2, and 1, i.e., at all multiples of 1/2, to some limit values h(0), h(1/2), h(1). By throwing away the beginning of the subsequence, we may assume that all values are within 1/2 of the limit values at those three points. Next take a subsequence S2 of Sl that converges at all multiples x of 1/4 to some limit values h(x), such that all values are within 1/4 of the limit values at those five points. In general, take a subsequence SN of SN_1 that converges at all multiples x of 1/2N to some limit values h(x), such that all values are within 1/2N of the limit values at those 2N + 1 points. Now we are ready to construct the desired subsequence of f,,,. As the first term, gl, choose the first term of S1. For 92, choose a later term of S2. In general, for gN, choose a still later term of SN. We claim that the sequence g,, is uniformly Cauchy. Given e > 0, by equicontinuity choose b > 0 such that
Iy - xI < s = Ign.(y) - g.(x)I < E/3.
(1)
Choose N such that 1/2N < b and 2/2N < e/3.
(2)
Suppose m, n > N and let x be any point. Choose k/2N such that ix - k/2NI < 1/2N < b.
(3)
Then I gm.(x) - gn(x) I
I gm(x) - gm.(k/2N)I + I gm(k/2N) - gn(k/2N) I + I gn(k/2N) - gn (x)I
< e/3 + 2/2N + e/3 < e/3 + e/3 + e/3 = E.
(The estimates on the first and third terms follow from (1) and (3). The estimate on the second term follows from the construction of the gn and We have shown that the sequence gn is uniformly Cauchy. In particular, the gn converge pointwise to some limit h. Since the sequence is uniformly Cauchy, the convergence is uniform, and the limit h is continuous. Thus the sequence gn, a subsequence of the original sequence f, converges in the
Exercises 30
135
space C[O, 1] (under the sup metric). Since .F is closed, the limit h lies in T. Therefore F is compact.
30.3. The Ascoli-Arzela Theorem. Arzela generalized Ascoli's Theorem from a bounded interval to any compact metric space. Much work in analysis has centered on finding nice spaces of functions with good compactness properties.
Exercises 30 Give an example. of an equicontinuous sequence of functions in C[0,1] which has no convergent subsequence. 1.
2. Give an example of a bounded sequence of functions in C[O, 1] which has no convergent subsequence. 3.
(Saroj Bhattarai and Brian Simanek). Prove the converse of Ascoli's
Theorem.
Hint: For the hard part, equicontinuity, use proof by contradiction: assume that for some e > 0 there are points in [0, 1] and functions in f such that xn - yn I < 1/n but I f (xn) - f (ym) I >- E. 4. Prove that the set of "Lipschitz" functions in C[O,1] satisfying
If(x)-f(y)I 0, choose N so that for some m > N, lam - L I < s/2 and for all m, n > N, Ian - an I < s/2. Then for all n > N,
Ian - LI < Ian - aml + Iam - LI < e/2+e/2=e. 8.7. 1, +oo, -oo, 1, 2, 1.
Chapter 9. 9.1. Let S and T be compact subsets of ][8n. Since S and T are both closed, S fl T is closed. Since S and T are both bounded, S fl T is bounded. Therefore S fl T is compact.
9.14. Immediately, from the definition, sup S > s for all sin S. If a < sup S, then there is a point of S larger than a, and hence a point of S larger than a, so that a fails to satisfy a > s for all s in S.
Chapter 10. 10.5. Since g is continuous, g(K) is compact by Theorem 10.1. Likewise since f is continuous, f (g (K)) = (fog) (K) is compact. Of course, you need to assume that g(K) is contained in the domain of f.
Chapter 11. 11.3. Suppose that f and g are uniformly continuous. Since f is uniformly continuous, given e > 0, we can choose Si > 0 such that (*)
Iyi - x1 I < S1
I f (yi) - f (xi) I < e.
Since g is uniformly continuous, we can now choose 5 > 0 such that whenever
I y - xI < S, then I g(y) - g(x) I < Si, which in turn implies that I f (g (y)) f (g(x)) I < e, by (*) with yi = g(y) and xl = g(x). Thus Iy - XI < S =
I(f 0 g) (Y) -.(f 0 g)(x)I < s+
Partial Solutions to Exercises
141
and f o g is uniformly continuous.
Chapter 12. 12.1. The disjoint open sets U1 = (-oo, 1/2) and U2 = (1/2, oo) separate the integers into two nonempty pieces.
12.7. Suppose that f takes on two different values, yl and y2. Choose an irrational number y3 between yl and y2. Then the open sets U1 = (-oo, y3) and U2 = (y3i oo) separate f (118) into two nonempty pieces. This contradicts
Theorem 12.4, which says that the continuous image of a connected set is connected.
12.8. By Proposition 12.2, Theorem 12.4, and Proposition 12.3, it must be an interval. By Theorems 10.1 and 9.2, it must be a closed interval.
12.11. First suppose that S is totally disconnected. By definition, S has at least two points. If S contained an interval, a separation of two points of the interval. would prove the interval disconnected, a contradiction. Conversely, suppose that S has two points but no interval. Let p1i p2 be distinct points of S. Since S does not contain an interval, there is a point p3 between pl and p2 not in S. Then the open sets U1 = (-oo,p3) and U2 = (p3i oo) provide the required separation to show that S is totally disconnected.
Chapter 13. 13.1. Let a be a point of the Cantor set. Let a,,, be one of the endpoints of the interval of S,, containing a. (If one of the endpoints is a, choose the other one.) Then a,, is in C and a is the limit of the sequence a,, because the length of the intervals goes to 0.
Chapter 14. 14.3. Since [a, b] is compact (Theorem 9.2), every continuous image is compact (Theorem 10.1), and hence every continuous function has a maximum (Corollary 10.2). This fact is used in the proof of Rolle's Theorem (14.3), to find a place where the derivative vanishes. Rolle's Theorem leads immediately to the Mean Value Theorem (14.4) and finally to Corollary 14.5,
Partial Solutions to Exercises
142
which says that on an interval where f' is always 0, f is constant. This final result will be a key ingredient in the proof of the Fundamental Theorem of Calculus (16.1).
Chapter 15. 15.1. [6, 10].
15.4. Let f be a nonnegative function with Riemann integral equal to A. Chop the interval up into identical subintervals with Ax small enough to guarantee that every Riemann sum is less than A + 1. Since every contribution is nonnegative, each subinterval contributes at most A + 1 to the Riemann sum. Therefore, on every subinterval f is bounded above by (A + 1)/Ax. 15.6. An unbounded interval cannot be chopped up into finitely many small intervals.
15.7. Yes. There will always be just one subinterval on which the chosen f (x) could be 0 or 1, and as the subinterval width shrinks, the effect becomes negligible. Similarly, finitely many discontinuities are OK.
Chapter 16. 16.4. -f (a).
Chapter 17. 17.3. A simple hypothesis is that f be bounded. 17.8. Given E > 0, choose S such that
Ix - yl < S=> If(x)-f(y)l <E/3. Second, choose M such that 1/M < S and C/M < E/3. Third, choose N such that
n>N
fn(k/M) - f (k/M) I < E/3 (k=0,1,2,-.-,M)-
Partial Solutions to Exercises
143
Now suppose that n > N. Given x, choose k such that Ix - k/MI < 1/M, and, hence, by the Lipschitz condition
I fn(x) - fn(k/M)I < C/M < e/3. Now
Ifn(x) -.f(x)I < fn(x) -.f.(klM)I +Ifn(k/M) -.f(k/M)I +If(k/M) -.f(x)I < C/M + e/3 +,-/3 < e/3 + e/3 + s/3 = E.
Chapter 18. 18.4. Compute by switching the order of integration. Justify by Fubini's Theorem because the integral of the absolute value is at most the integral of (ir/3)(10)(1) = 10ir/3, which integral is (10)(7r/3)(107r/3) < oo. 18.8. In justifying the use of Leibniz's Rule, note that for y near 0, 8
1
ay x2 +Y2
2y (x2 + y2)2
whose integral even from 1 to oo is finite.
Chapter 19. 19.1. Converges to 1/9.
Chapter 20. 20.1. 1, -1/3, [-oo, +oo] ,+00.
Chapter 21. 21.1b. TRUE.
N = p(an, Ll) < e/2 and p(an, L2) < e/2. Then p(Li, L2)
p(Li, an) + p(an, L2) < e/2 + e/2 = s = p(L1, L2),
Partial Solutions to Exercises
146
a contradiction. To show that the sequence is bounded, choose N such that
n>N=p(an,L) 0, choose N such that m, n > N implies that I f. (x) - fn (x) I < e/2 for all x. Now given any x, choose m > N (depending on x) such that Ifm(x) - f (x) I < e/2. It follows that for n > N, I fn (x) - f (x) I < e, uniform convergence. Finally, as a uniform limit of continuous functions, f is continuous. 28.19. yes.
Chapter 29. 29.3. The proofs of Theorem 10.1 and Corollary 10.2 hold in metric spaces.
Chapter 30. No solutions.
Greek Letters a
A alpha
Q B beta ry
r gamma
b A delta
v
N nu
E xi
o 0 omicron H pi
7r
e
E
epsilon
p P rho
S
Z
zeta
o
B
H eta O theta
t
I
77
iota
n K kappa A A lambda
µ M mu
E sigma
r T tau
v T upsilon cp
-D
phi
X X chi
V) ' psi w
S2
omega
147
Index composition of function,. 3a
L1 correspondence, 7
abbreviatiotu. G absolute convergence. 89 accumulation point. 1& 126 Al-Sabah, Nasser. viii alternating series. t32 Armstrong, Zan. 411
Ascoli's Theorem. 133. 1.3
ball B(n, r), 27 Bernoulli, Daniel, 10.i. 109 Bhattarai, Saroj, 135 bijectivu, L
Bohnhorst. Kristin, 13 13olzano-N'eierstrass. 3K, IL 4 K 1:311 boundary. ')9 126 bounded, 15. 38. 41 66. 12 i
Burger. Ed. iv. viii ('[0. 1 122. 11Et.. 121. 128, 1311 133. L3a calculus. 511
22. at
closed sets, 28. 29 1211
closed sets, unions and intersections, 3Q closure, 30, 126 comparison test. 81i complement. 3. 1i. 22 complete. 125
converge absolutely. tom1, 13
converge conditionally. 81 converge polmwrrx:, Z5 converge uniformly, 75 125 converse, ti Corvetti. Candice. dfi Costa. Tom. viii countable. 9 countable subcover, 12 countable union. 111 dense. 32 derivative. 61
differentiable. lil
Cartesian pro..luct. 111 Cauchy sequence. 20. 39 48 66_6 125. 131
compact. 9.1.. 45. 1291
Constantine, Alexandra. 145 contains, ti continuous. 21 continuous functions. 33 cuntrap ositive, 5 converge, l_t. 85. 43. 125
difference. G
Cantor function. 63 Cantor set, 53 at Cantor, G(org. U
characteristic function circle, 22
conditional convergence. ts9 connected. .19
differentiation of power series, 915 dimension, a5 disc. 27L disconnected, totally, 50 distance. 3. L21
diverge. L4. gel it)
domain. Z Dominated Convergence Theorem. at
Dunne, Ed. viii email address, viii English. 5
149
Index
150
equicontinuous, 133 Euler's identity. 113 exponential function, 111
logarithm, 113 logic, 5 lowersemicontinuous, 135
Fourier coefficients, 100 Fourier series, 99. 102, 109 fractals, 53 Fubini's Theorem, 82 functions, 1 Fundamental Theorem of Calculus. 71
Mandelbrot, Benoit, 56
Gamma Function r(x). U.S Garrity, Tom. viii
metric, Euclidean, 4 metric, sup, 122, 125 metric, taxicab, 121 minimum, 42. 46, fil Murphy, Erin, iv
geometric series, 8fi Gibbs phenomenon, LID greatest lower bound, 44 Greek letters. L47
Hadamard formula, 95 harmonic series. 86. 87 Heine- Borel, 41, 45, 134 image, 2 implication, 5 infimum, 44 infinite sets, 9 injective, Z integers, 3 integrable, 65, 61 interior, 30. 126 Intermediate Value Theorem, 5Q
maximum, 42, 44-46, 111 Mean Value Theorem, 62. 63, 72 measurability, 82 measure, 54 Menger sponge, 5Si metric space, 119, 121, 125 129.
natural numbers, 3 Newton, vii, 60. 106 nonintegrable functions. fib one-to-one, 7 onto, 7 open cover, iv, 41, 134 open sets, vii, 28. 29, 126
open sets, unions and intersections, 3Q "or". 5
p-test, 86 power series, 93
intersect, 6
radius of convergence, 94
intersections, 30, 43
range, 7 rates of growth, 1H ratio test, 91 rationals, 3. 22 real analytic, 94. 95 real numbers, 1, 3 rearrangement, 89 relatively open, 34 resonance, 111 Riemann integral, 65 Riemann sum, 65 root test, 95
interval, 27. 49 intervals, 3
Inverse Function Theorem, 114 inverse image, 7, 8
irrationals, 3 4. 8 isolated point, 311, 126 JPEG. 1112
least upper bound, 44 Lebesgue integral. 81 I.ebesgue's Dominated Convergence Theorem, 81 Leibniz, vii, &fl Leibniz's formula for ^.r, 97, 144
self-similar, 55 sequence, 13, 125 sequence of functions, 75, 125
Ieibniz's Rule, 82 length, 8 lunit. 1 13, 14, 18. 125
series, 85
limit of function, 2.1 limit, unique, 15
shortest paths, 135 Sierpinski's carpet, 56 Sil a, Cesar, iv Simanek, Brian, 135 sine and cosine functions, 113
liminf, 39 lim sup, 39
Lipschitz constant. 79 Lipschitz function, 79, 1.35
series Y(1/n2). 101 sets, 6
sphere, 2,7
Index
springs. 106
Stirling's Approximation to n!, Lli Stokes's 'T'heorem, 12 strings. 105 subsequence, 32 subset. b
supremum, 3:3 surjective. 1 switch limit and integral, 77, f1.1
switch order of integration. 82 Tacoma Narrows suspension bridge, LU Tapp, Kris. viii Taylor's formula, 95, 911
topolo y. 25, 3! 126 triangle inequality, 4, 121
uncountable, I1 51 uniform continuity, 3i unions, .30, 13 volumes of 71-balls. 11
Voss, It. F., wave equation. 1116
webpage. iv. viii Weicrstrays Al-test. 9,1 words W, 122. 126, 121
151
