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**almost there****Member**- Registered: 2009-11-11
- Posts: 21

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?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"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..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"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..."

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