Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

1 Answer 1

up vote 7 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.

share|improve this answer
3  
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 at 18:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.