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almost there

Member

Registered: 2009-11-11

Posts: 21

complex power series

f is holomorphic on the complex plane. Assume that for each complex w there is at least one n such that b_n (w)=0, where b_n (w) is the coefficient of (z-w)^n in the power series expansion of f at w. Show that f is a polynomial.

I do not remember power series expansion. How is the coefficient b_n (w) found? And even though I do not remember power series expansion, wouldn't the power series expansion of a polynomial be the original polynomial?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: complex power series

How is the coefficient b_n (w) found?

And even though I do not remember power series expansion, wouldn't the power series expansion of a polynomial be the original polynomial?

About the point w = 0, sure.  But not anywhere else.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: complex power series

Consider the function on the disk.  That R^2 is uncountable means that for some n, there is a limit point in the set {p : a_n = 0 in the power series about p}.  The nth derivative of f is therefore zero an each of these points.  Since there is an infinite amount of points in a compact set, there is a limit point.  This implies that the nth derivative of f is zero on the entire disk.  Therefore the nth derivative of f is zero everywhere giving that f is a polynomial.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."