Mathematics of Physics and Engineering

(^210) Power Series
EXERCISE 4.3.7.c Find the radius of convergence of the power series
7.2n+^1 (dnV

E(5+(_ln«

fc>l

Let us now state and prove the main result of this section.

Theorem 4.3.4 A function f = f{z) is analytic at the point ZQ if and
only if there exists a power series Ylk>oak(z ~ z°)k vy^^1 some radius of
convergence R > 0 so that f(z) = ^2k>0 ak{z — zo)k for all \z — zo\ < R.

Proof. Step 1. Let us show that an analytic function can be written as a
convergent power series. Recall that, by definition, / = /(z) is analytic at
ZQ if and only if / has a derivative in some neighborhood of ZQ. Therefore,
there exists an R > 0 so that / is analytic in the open disk GR = {z :
\z — zo\ < R}. Take z G GR and find a number p so that \z — zo| < P < R-
Then the circle Cp with center at ZQ and radius p encloses z and lies entirely
in GR (draw a picture). By the Cauchy Integral Formula,

Next, we note that \z — ZQ\ < |£ — ZQ\ = p for C, € Cp and use the formula
for the geometric series to write

(z - Z 0 )

We now plug the result into (4.3.5) and find

z 0 )k dC (4.3.7)

Finally, let us assume for the moment that we can integrate term-by-term
in (4.3.7), that is, first do the integration and then, summation; this is
certainly true for sums with a finite number of terms, but must be justified
for (4.3.7), and we will do the justification later. Then the term-by-term
integration results in the equality

/(z) = J>fe(z - z 0 )fe, (4.3.8)

fc>0