TITLE: A question regarding power series expansion of an entire function
QUESTION [8 upvotes]: Let $f$ be an entire function and let for each $a\in \mathbb R$, there exists at least one coefficient $c_n$ in $f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n$, which is zero. Then 

$f^{(n)}(0)=0$ for infinitely many $n\geq 0$
$f^{(n)}(0)=0$ for every $n\geq 0$
$f^{(2n+1)}(0)=0$ for every $n\geq 0$
There exists $k\geq 0$ such that $f^{(n)}(0)=0$ for all $n\geq k$

We know that $c_n=\frac{f^{(n)}(a)}{n!}$ for all $n\in \{0,1,2\ldots\}$. Thus for $a=0$, $c_n=\frac{f^{(n)}(0)}{n!}$. By hypothersis, atleast one $c_n=0$. After that I could not do anything. Please help!

REPLY [5 votes]: Let $A_n =\{z\in \mathbb{C} : f^{(n)} (z) =0 \}.$ By assumption $$\mathbb{R} =\bigcup_{n\in\mathbb{N}\cup \{0\}} A_n $$ and since $\mathbb{R} $ is uncountable at least one of the sets $A_n $ is uncoutable. Let $$k=\min\{ n\in\mathbb{N}\cup \{0\} :\mbox{card}(A_n ) >\aleph_0 \}$$ then the set $A_k$ is uncountable and therefore it has an accumulation point. Hence the $f^{(k)} (x) $ is identically zero on $\mathbb{C}$ and therefore $f$ is an polynomial with degree less or equal to $k.$
So 1. and 4. are true.