Mathematics of Physics and Engineering

212 Power Series

The value of \z — zo|/|z — C| is constant for all £ on Cp and is less than one,
which allows you to make the sum ^2k>N arbitrarily small, (c) Show that the
n-th derivative g^ of g(z) = 2fc>oafc(z — z°)k ea;ss inside the disk of
convergence and satisfies g^(zo) = n)an. Then use (4-3.8) and (4-3.9) to
give the rigorous proof of (4-2.16) on page 201. Hint: part (a) of this exercise
shows that you can differentiate the power series term-by-term as many times as
you want.
Corollary 4.1 Uniqueness of power series. (a) Assume that the
two power series Y^T=o ak(z ~ zo)k, Sfclo bk(z — zo)k converge in the some
neighborhood of ZQ and Y?kLoak(z ~ zo)k = J2T=o^k(z — zo)k for a^ z *n
that neighborhood. Then a^ = b^ for all k > 0.
(b) If f is analytic at ZQ, then

f(z) = f(zo) + f; ^-^- (z - z 0 )k (4.3.11)

fc=i

for all z in some neighborhood of ZQ. Therefore, if a function has a power
series representation at a point ZQ, this power series is necessarily (4-3.11).
EXERCISE 4.3.9;A Prove both parts of Corollary 4-1-
As in the ordinary calculus, the series in (4.3.11) is called the Taylor
series for / at ZQ\ when z§ = 0, we also call it the Maclaurin series.
It appears, though, that the original idea to represent real functions by a
power series belongs to neither Taylor nor Maclaurin and can be traced back
to Newton. The English mathematician BROOK TAYLOR (1685-1731) and
the Scottish mathematician COLIN MACLAURIN (1698-1746) were the first
to make this idea clear enough and spread it around, and thus got their
names attached to this power series representation, even in the complex
domain.
The methods for finding the power series expansion of a particular com-
plex function are the same as for the real functions. With these methods,
you never compute the derivatives of the function. One of the key relations
is the sum of the geometric series:

r^ = E

2

*- w

<:L

• (

4

3

12

)

fc>o

FOR EXAMPLE, let us find the Taylor series for f(z) = 1/z^2 at point z 0 = 1-

We have f(z) = -g'(z), where g{z) = 1/z. Now, 1/z - 1/(1 + (z -

1)) = Efe>o(_1)fc(z ~^1 )fc> where we used (4-3.12) with -(z - 1) instead