How to prove that an entire function f, which is representable in power series with at least one coefficient is 0, is a polynomial?
Entire functions representable in power series
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complex-analysis
functional-analysis
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5Welcome to Math.SE. Your proposition is false as stated: $\exp x -1$ is entire, can be represented with the power series $ x+ x^2/2! + x^3/3! + \cdots $ which has a 0 coefficient, yet is not a polynomial. Are you sure this is the correct version of the problem you intended to ask? – 2012-09-16
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2@Ragib: Presumably the question is: if $f$ is entire and at each point $z$ there is *some* derivative $n = n(z)$ such that $f^{(n)}(z) = 0$ then $f$ must be a polynomial (this is an easy consequence of the fact that a non-constant entire function can have at most countably many zeros). This statement is in fact true (but much harder to prove) even for $C^\infty$-functions on intervals of $\mathbb{R}$, see [this MO thread](http://mathoverflow.net/questions/34059/). – 2012-09-16
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1[This question](http://math.stackexchange.com/q/132304) might be (the contrapositive of) what is intended -- ignore the part on Casorati-Weierstrass. – 2012-09-16
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0@t.b. : How can then I prove that for an entire polynomial function the set Dn is uncountable? – 2012-09-16
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0@abby: If $p$ is a polynomial of degree $n$ then $p^{(n+1)}(z) \equiv 0$, so $D_{n+1} = \mathbb{C}$. – 2012-09-16
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0@t.b.: Thanks for helping, I realized the answer once wrote my previous comment :). – 2012-09-16
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0I see. This happens often :) So, does the other thread address your intended question? – 2012-09-16