COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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CHAPTER 7 TAYLOR and LAURENT SERIES

Section 7.2 Taylor series Representations

In Section 4.4 we saw that functions defined by power series have derivatives of all orders (Theorem 4.16). In Section 6.5 we saw that analytic functions also have derivatives of all orders (Corollary 6.2). It seems natural, therefore, that there would be some connection between analytic functions and power series. As you might guess, the connection exists via the Taylor and Maclaurin series of analytic functions.

Definition 7.2: Taylor series

If i s analytic at , the series is called the Taylor series for centered around . When the center is , the series is called the Maclaurin series for .

Theorem 7.4 (Taylor's Theorem)

Suppose is analytic in a domain G, and that is contained in G. Then the Taylor series converges to for all z in ; that is

for all .

Furthermore, this representation is valid in the largest disk with center t hat is contained in G, and the convergence is uniform on any closed subdisk for .

Corollary 7.3 Suppose is analytic in the domain G that contains the point . Let be a singular point of minimum distance to . If , then
(i) the Taylor series converges to on all of , and
(ii) if , the Taylor series does not converge to on all of .

Example 7.3, Page 272. Show that , is valid for in : .

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1-z)^2:
t := taylor(f(Z), Z=0, 10):
s := subs(Z=z,t):
p := z -> subs(Z=z,convert(t, polynom)):
`f(z) ` = f(z);
`f(z) ` = s;
P[9](z) = p(z);

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

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1-z)^2:
s := taylor(f(z), z=0, 10):
p:=unapply(convert(taylor(f(z),z=0,10),polynom),z):
`f(z) ` = f(z);
`f(z) ` = s;
P[9](z) = p(z);

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

> f:='f': n:='n': S:='S': z:='z':
f := z -> 1/(1-z)^2:
S9 := z -> sum((n+1)*z^n, n=0..9):
S := z -> sum((n+1)*z^n, n=0..infinity):
`f(z) ` = f(z);
s[9](z),` = `,Sum((n+1)*z^n, n=0..9) = S9(z);
s[infinity](z),` = `,Sum((n+1)*z^n, n=0..infinity) = S(z);

Example 7.4, Page 273. (a) Show that , is valid for in : .

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1-z^2):
t := taylor(f(Z), Z=0, 20):
s := subs(Z=z,t):
p := z -> subs(Z=z,convert(t, polynom)):
`f(z) ` = f(z);
`f(z) ` = s;
P[18](z) = p(z);

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

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1-z^2):
s := taylor(f(z), z=0, 20):
p:=unapply(convert(taylor(f(z),z=0,20),polynom),z):
`f(z) ` = f(z);
`f(z) ` = s;
P[18](z) = p(z);

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

> f:='f': n:='n': s:='s': S:='S': z:='z':
f := z -> 1/(1-z^2):
S9 := z -> sum(z^(2*n), n=0..9):
S := z -> sum(z^(2*n), n=0..infinity):
`f(z) ` = f(z);
s[18](z),` = `,Sum(z^(2*n), n=0..9) = S9(z);
s[infinity](z),` = `,Sum(z^(2*n), n=0..infinity) = S(z);

(b) Show that , is valid for in : .

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1+z^2):
t := taylor(f(Z), Z=0, 20):
s := subs(Z=z,t):
p := z -> subs(Z=z,convert(t, polynom)):
`f(z) ` = f(z);
`f(z) ` = s;
P[18](z) = p(z);

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

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> 1/(1+z^2):
s := taylor(f(z), z=0, 20):
p:=unapply(convert(taylor(f(z),z=0,20),polynom),z):
`f(z) ` = f(z);
`f(z) ` = s;
P[18](z) = p(z);

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

> f:='f': n:='n': s:='s': S:='S': z:='z':
f := z -> 1/(1+z^2):
S9 := z -> sum((-1)^n * z^(2*n), n=0..9):
S := z -> sum((-1)^n * z^(2*n), n=0..infinity):
`f(z) ` = f(z);
s[18](z),` = `,Sum((-1)^n * z^(2*n), n=0..9) = S9(z);
s[infinity](z),` = `,Sum((-1)^n * z^(2*n), n=0..infinity) = S(z);

Theorem 7.5 (Uniqueness of power series) Suppose that in some disk we have

= . Then for .

Example 7.5, Page 274. Find the Maclaurin series of .

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> sin(z)^3:
t := taylor(f(Z), Z=0, 14):
s := subs(Z=z,t):
p := z -> subs(Z=z,convert(t, polynom)):
`f(z) ` = f(z);
`f(z) ` = s;
P[13](z) = p(z);

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

> An:='An': Bn:='Bn': f:='f': n:='n': p:='p': q:='q': s:='s': z:='z':
f := z -> sin(z)^3:
An := (-1)^n * 3/(4*(2*n+1)!):
Bn := - (-1)^n * 3 * 9^n/(4*(2*n+1)!):
`f(z) ` = f(z);
p13 := sum(An* z^(2*n+1), n=0..6):
q13 := sum(Bn* z^(2*n+1), n=0..6):
Sum(An* z^(2*n+1), n=0..6) = p13;
Sum(Bn* z^(2*n+1), n=0..6) = q13;
`Partial sum `, s[13](z) =
sum(An* z^(2*n+1), n=0..6) +
sum(Bn* z^(2*n+1), n=0..6);
s[infinity](z),` = `,
Sum(An* z^(2*n+1)+Bn* z^(2*n+1), n=0..infinity) =
sum(An* z^(2*n+1), n=0..infinity) +
sum(Bn* z^(2*n+1), n=0..infinity);

Theorem 7.6 Let and have the power series representations
for and for .
If , and is any complex constant, then
(i) for ,
(ii) for , and
(iii) for , where .

Identity (iii) is known as the Cauchy product of the series for and .

End of Section 7.2.