C07-1.mws

COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/


CHAPTER 7 TAYLOR and LAURENT SERIES

Section 7.1 Uniform Convergence

Throughout this text we have compared and contrasted properties of complex functions with functions whose domain and range lie entirely within the reals. There are many similarities, such as the standard differentiation formulas. However, there are also some surprises, and in this chapter we will encounter one of the hallmarks distinguishing complex functions from their real counterparts.

It is possible for a function defined on the real numbers to be differentiable everywhere and yet not be expressible as a power series (see Exercise 7.2.20 at the end of Section 7.2). In the complex case, however, things are much simpler! It turns out that if a complex function is analytic in the disk
D , its Taylor series about alpha will converge to the function at every point in this disk. Thus, analytic functions are locally nothing more than glorified polynomials.

Definition 7.1: Uniform convergence

The sequence {S[n](z)} converges uniformly to f(z) on the set T if for every 0 < epsilon , there exists a positive integer N[epsilon] (which depends only on epsilon ) such that if N <= n , then abs(S[n](z)-f(z)) < epsilon for all z*epsilon*T .

Example 7.1, Page 262. The sequence {S[n](z)} = {exp(z)+1/n} converges uniformly to the function f(z) = exp(z) on the entire complex plane because for any epsilon >0 , abs(S[n](z)-f(z)) < epsilon is satisfied for all z for n > N[epsilon] , where N[epsilon] is any integer greater than 1/epsilon . We leave the details for showing this as an exercise.

There is a useful procedure known as the Weierstrass M-test that can help determine whether an infinite series is uniformly convergent.

Theorem 7.1 (Weierstrass M-test)

Suppose the infinite series sum(u[k](z),k = 0 .. infinity) has the property that for each k, abs(u[k](z)) < M[k] for all z*epsilon*T .
If
sum(M[k],k = 0 .. infinity) converges, then sum(u[k](z),k = 0 .. infinity) converges uniformly on T.

Theorem 7.2 gives an interesting application of the Weierstrass M-test.

Theorem 7.2 Suppose the power series sum(c[k]*(z-alpha)^k,k = 0 .. infinity) has radius of convergence 0 < rho .

Then for each r , `0 < r` < rho , the series converges uniformly on the closed disk D[r](alpha) = {`z :  `*abs(z-alpha) <= r} .

Corollary 7.1 For each r , `0 < r` < 1 , the geometric series converges uniformly on the closed disk D[r](0) = {`z :  `*abs(z) <= r} .

Theorem 7.3 Suppose {S[n](z)} is a sequence of continuous functions defined on a set T containing the contour C. If {S[n](z)} converges uniformly to f(z) on the set T, then
(i) f(z) is continuous on T, and
(ii) limit(int(S[k](z),z = C .. `.`),k = infinity) = int(limit(S[k](z),k = infinity),z = C .. `.`) = int(f(z),z = C .. `.`) .

Corollary 7.2 If the series sum(c[k]*(z-alpha)^k,k = 0 .. infinity) converges uniformly to f(z) on the set T, and C is a contour contained in T, then
sum(int(c[k]*(z-alpha)^k,z = C .. `.`),k = 0 .. infinity) = int(sum(c[k]*(z-alpha)^k,k = 0 .. infinity),z = C .. `.`) = int(f(z),z = C .. `.`) .

Example 7.2, Page 266. Show that Log(1-z) = -sum(z^n/n,n = 1 .. infinity) , for all z in D : abs(z) < 1 .

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> log(1-z):
t := taylor(f(Z), Z=0, 11):
s := subs(Z=z,t):
p := z -> subs(Z=z,convert(t, polynom)):
`f(z) ` = f(z);
`f(z) ` = s;
P[10](z) = p(z);

`f(z)  ` = ln(1-z)

`f(z)  ` = series(-1*z-1/2*z^2-1/3*z^3-1/4*z^4-1/5*z^5-1/6*z^6-1/7*z^7-1/8*z^8-1/9*z^9-1/10*z^10+O(z^11),z,11)

P[10](z) = -z-1/2*z^2-1/3*z^3-1/4*z^4-1/5*z^5-1/6*z^6-1/7*z^7-1/8*z^8-1/9*z^9-1/10*z^10

Or we could use Maple's "unapply" procedure.

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> log(1-z):
s := taylor(f(z), z=0, 11):
p:=unapply(convert(taylor(f(z),z=0,11),polynom),z):
`f(z) ` = f(z);
`f(z) ` = s;
P[10](z) = p(z);

`f(z)  ` = ln(1-z)

`f(z)  ` = series(-1*z-1/2*z^2-1/3*z^3-1/4*z^4-1/5*z^5-1/6*z^6-1/7*z^7-1/8*z^8-1/9*z^9-1/10*z^10+O(z^11),z,11)

P[10](z) = -z-1/2*z^2-1/3*z^3-1/4*z^4-1/5*z^5-1/6*z^6-1/7*z^7-1/8*z^8-1/9*z^9-1/10*z^10

Sum up the terms to verify that we have things right.

> f:='f': n:='n': s:='s': S:='S': z:='z':
f := z -> log(1-z):
S10 := z -> sum(-1/n*z^n, n=1..10):
S := z -> sum(-1/n*z^n, n=1..infinity):
`f(z) ` = f(z);
s[10](z),` = `,Sum(-1/n*z^n, n=1..10) = S10(z);
s[infinity](z),` = `,Sum(-1/n*z^n, n=1..infinity) = S(z);

`f(z) ` = ln(1-z)

s[10](z), ` = `, Sum(-1/n*z^n,n = 1 .. 10) = -z-1/2*z^2-1/3*z^3-1/4*z^4-1/5*z^5-1/6*z^6-1/7*z^7-1/8*z^8-1/9*z^9-1/10*z^10

s[infinity](z), ` = `, Sum(-1/n*z^n,n = 1 .. infinity) = ln(1-z)

The real variable plot of y = f(x) and y = s[10](x) is:

> plot({f(x),S10(x)},
x=-0.999..0.999, y=-3..0.7,
title=`y=ln(1-z) and y=s10(x)`,
labels=[` x`,`y `],
tickmarks=[5,7],
view=[-1..1,-3..0.7]);

[Maple Plot]

End of Section 7.1.