proof of identity theorem of power series

We start by proving a more modest result. Namely, we show that,under the hypotheses of the theorem we are trying to prove, wecan conclude that $a_0=b_0$.

Let $R$ be chosen such that both series converge when .From the set of points at which the two power series are equal, we maychoose a sequence ${\{w_k\}}_{k=0}^{\mathrm{\infty }}$ such that

• for all $k$.

• ${lim}_{k\to \mathrm{\infty }}{w}_k$ exists and equals $z_0$.

• $w_k\ne z_0$ for all $k$.

.

Since power series converge uniformly, we may interchange thelimit with the summation.

$\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{(w_k-z_0)}^n$$=$${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\underset{k\to \mathrm{\infty }}{lim}{a}_n{(w_k-z_0)}^n=a_0$

$\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{b}_n{(w_k-z_0)}^n$$=$${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\underset{k\to \mathrm{\infty }}{lim}{b}_n{(w_k-z_0)}^n=b_0$

Because ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n{(w_k-z_0)}^n=su{m}_{n=0}^{\mathrm{\infty }}{a}_n{(w_k-z_0)}^n$ for all $k$,this means that $a_0=b_0$.

We will now prove that $a_n=b_n$ for all $n$ byan induction argument. The intial step with $n=0$is, of course, the result demonstrated above.Assume that $a_m=b_m$ for all $m$ less thansome integer $N$. Then we have

$$\sum _{n=N}^{\mathrm{\infty }}{a}_n{(w-z_0)}^n=\sum _{n=N}^{\mathrm{\infty }}{b}_n{(w-z_0)}^n$$

for all $w\in S$. Pulling out a commonfactor and relabelling the index, we have

$${(w-z_0)}^N\sum _{n=0}^{\mathrm{\infty }}{a}_{n+N}{(w-z_0)}^n={(w-z_0)}^N\sum _{n=0}^{\mathrm{\infty }}{b}_{n+N}{(w-z_0)}^n.$$

Because $z_0\notin S$, the factor $w-z_0$ willnot equal zero, so we may cancel it:

$$\sum _{n=0}^{\mathrm{\infty }}{a}_{n+N}{(w-z_0)}^n=\sum _{n=0}^{\mathrm{\infty }}{b}_{n+N}{(w-z_0)}^n$$

By our weaker result, we have $a_N=b_N$.Hence, by induction, we have $a_n=b_n$ for all $n$.