proof of identity theorem of power series

We can prove the identity theorem for power seriesusing divided differences. From amongst the pointsat which the two series are equal, pick a sequence $\\{w_{k}\\}_{k=0}^{\\infty}$ which satisfies the followingthree conditions:

1.
$\\lim_{k\\to\\infty}w_{k}=z_{0}$
2.
$w_{m}=w_{n}$ if and only if $m=n$ .
3.
$w_{k}\eq z_{0}$ for all $k$ .

Let $f$ be the function determined by one power seriesand let $g$ be the function determined by the otherpower series:

	$\\displaystyle f(z)$	$\\displaystyle=\\sum_{n=0}^{\\infty}a_{n}(z-z_{0})^{n}$
	$\\displaystyle g(z)$	$\\displaystyle=\\sum_{n=0}^{\\infty}b_{n}(z-z_{0})^{n}$

Because formation of divided differences involvesfinite sums and dividing by differences of $w_{k}$ ’s(which all differ from zero by condition 2 above,so it is legitimate to divide by them), we maycarry out the formation of finite diffferenceson a term-by-term basis. Using the result aboutdivided differences of powers, we have

	$\\displaystyle\\Delta^{m}f[w_{k},\\ldots,w_{k+m}]$	$\\displaystyle=\\sum_{n=m}^{\\infty}a_{n}D_{mnk}$
	$\\displaystyle\\Delta^{m}f[w_{k},\\ldots,w_{k+m}]$	$\\displaystyle=\\sum_{n=m}^{\\infty}b_{n}D_{mnk}$

where

D_{mnk}=\\sum_{j_{0}+\\ldots j_{m}=n-m}(w_{k}-z_{0})^{j_{0}}\\cdots(w_{k+m}-z_{0}%)^{j_{m}}.

Note that $\\lim_{k\\to infty}D_{mnk}=0$ when $m>n$ , but $D_{mmk}=1$ . Sincepower series converge uniformly, we mayintechange limit and summation to conclude

	$\\displaystyle\\lim_{k\\to\\infty}\\Delta^{m}f[w_{k},\\ldots,w_{k+m}]$	$\\displaystyle=\\sum_{n=m}^{\\infty}a_{n}\\lim_{k\\to\\infty}D_{mnk}=a_{m}$
	$\\displaystyle\\lim_{k\\to\\infty}\\Delta^{m}g[w_{k},\\ldots,w_{k+m}]$	$\\displaystyle=\\sum_{n=m}^{\\infty}b_{n}\\lim_{k\\to\\infty}D_{mnk}=b_{m}.$

Since, by design, $f(w_{k})=g(w_{k})$ , we have

\\Delta^{m}f[w_{k},\\ldots,w_{k+m}]=\\Delta^{m}g[w_{k},\\ldots,w_{k+m}],

hence $a_{m}=b_{m}$ for all $m$ .