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How to prove that an entire function f, which is representable in power series with at least one coefficient is 0, is a polynomial?

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    I see. This happens often :) So, does the other thread address your intended question?2012-09-16

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Define $F_n:=\{z\in \Bbb C, f^{(n)}(z)=0\}$. Since for each $n$, $f^{(n)}$ is holomorphic it's in particular continuous, hence $F_n$ is closed. Since we can write at each $z_0$, $f(z)=\sum_{k=0}^{+\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$, the hypothesis implies that $\bigcup_{n\geq 0}F_n=\Bbb C$. As $\Bbb C$ is complete, by Baire's categories theorem, one of the $F_n$ has a non empty interior, say $F_N$. Then $f^{(N)}(z)=0$ for all $z\in B(z_0,r)$, for some $z_0\in \Bbb C$ and some $r>0$. As $B(z_0,r)$ is not discrete and $\Bbb C$ is connected, we have $f^{(N)}\equiv 0$, hence $f$ is a polynomial.

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    +1, although I think appealing to Baire is a bit of an overkill: if $f$ is not a polynomial, none of the $f^{(n)}$ is constant, hence each $F_n$ is countable, so $\bigcup_{n = 0}^\infty F_n \subsetneqq \mathbb{C}$.2012-09-16