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, we can conclude that a0=b0.

Let R be chosen such that both series convergePlanetmathPlanetmath when |z-z0|<R. From the set of points at which the two power seriesMathworldPlanetmath are equal, we may choose a sequence {wk}k=0 such that

  • |wk-z0|<R/2 for all k.

  • limkwk exists and equals z0.

  • wkz0 for all k.

.

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

limkn=0an(wk-z0)n = n=0limkan(wk-z0)n=a0
limkn=0bn(wk-z0)n = n=0limkbn(wk-z0)n=b0

Because n=0an(wk-z0)n=sumn=0an(wk-z0)n for all k, this means that a0=b0.

We will now prove that an=bn for all n by an inductionMathworldPlanetmath argumentPlanetmathPlanetmath. The intial step with n=0 is, of course, the result demonstrated above. Assume that am=bm for all m less than some integer N. Then we have

n=Nan(w-z0)n=n=Nbn(w-z0)n

for all wS. Pulling out a common factor and relabelling the index, we have

(w-z0)Nn=0an+N(w-z0)n=(w-z0)Nn=0bn+N(w-z0)n.

Because z0S, the factor w-z0 will not equal zero, so we may cancel it:

n=0an+N(w-z0)n=n=0bn+N(w-z0)n

By our weaker result, we have aN=bN. Hence, by induction, we have an=bn for all n.

Title proof of identity theorem of power seriesPlanetmathPlanetmath
Canonical name ProofOfIdentityTheoremOfPowerSeries
Date of creation 2013-03-22 16:47:38
Last modified on 2013-03-22 16:47:38
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 7
Author rspuzio (6075)
Entry type Proof
Classification msc 30B10
Classification msc 40A30