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Suppose that $f$ is an entire function, and that in every power series

$f(z)=\sum_{n=0}^{\infty} c_{n}(z-a)^n$ at least one coefficient is 0. Prove that $f$ is a polnomial.

Hint: $n!c_{n}=f^{(n)}(a)$

Actually, this is a Rudin's book's exercise.

I tried Cauchy inequality, and Liouville's theorem ( for $g(z)=\sum_{n=m}^{\infty} c_{n}(z-a)^n$ is bounded) but failed.

I really want to solve that, but I don't have any idea. I need your help.

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Consider the set $A_n=\{z\in \mathbb{C} : f^{(n)}(z)=0\}$ , prove that for sufficiently large $n$, $A_n$ is uncountable, and hence for this $n$ the function $f^{(n)}(z)$ vanishes. – Ajat Adriansyah Sep 15 '12 at 17:34
up vote 9 down vote accepted

From the hint, we see that for every $c \in \mathbf C$, some higher derivative of $f$ vanishes at $c$. Let $Z_n = \{z \in \mathbf C : f^{(n)}(z)=0\}$. If no derivative of $f$ vanishes identically, then each $Z_n$ is a closed, nowhere dense subset of $\mathbf C$. But $\mathbf C = \bigcup_{n\geq 0} Z_n$ is then impossible by the Baire Category Theorem. Thus some derivative of $f$ vanishes identically, which implies $f$ is a polynomial.

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This is a perfectly fine solution, but if the second sentence was "If no derivative of $f$ vanishes identically, then each $Z_n$ is at most countable", the Baire category would not be needed. – user31373 Sep 15 '12 at 23:28
@LVK, you are right, Baire's theorem is not required. – Bruno Joyal Sep 15 '12 at 23:42
can anybody explain why $Z_n$ is closed? – Topology Oct 26 '14 at 18:35
@Topology it's the inverse image of the closed set $\{0\}$ under the continuous function $f^{(n)}$. – Callus Jul 1 at 13:46

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