6
$\begingroup$

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function. Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is zero. Show that $f$ is a polynomial.

$\endgroup$

1 Answer 1

15
$\begingroup$

Say that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a\in \mathbb{C}$ there is $m\in \mathbb{N}$ such that $f^{(m)}(a)=0$.

Consider $A_{n}:=\left\{z\in\mathbb{C}:f^{(n)}(z)=0\right\}$ for each $n\in \mathbb{N}$. Note that:

$f$ is polynomial iff $A_{n}$ is not countable for some $n\in \mathbb{N}$.

Indeed, if $f$ is polynomial of degree $n$, then $f^{(n+1)}(z)=0$ for all $z\in \mathbb{C}$, then $A_{n+1}=\mathbb{C}$, so, $A_{n+1}$ is not countable. Conversely, if there is $n\in \mathbb{C}$ such that $A_{n}$ is not countable, then $A_{n}$ has a limit point, then by Identity principle we have $f^{(n)}(z)=0$ for all $z\in \mathbb{C}$, so, $f$ is a polynomial of degree at most $n-1$.

Therefore, tt suffices to show that there is $n\in \mathbb{N}$ such that $A_{n}$ is not countable. Indeed, consider $\bigcup_{n\in\mathbb{N}}A_{n}$, by hypothesis for each $a\in \mathbb{C}$ there is $m\in \mathbb{N}$ such that $f^{(m)}(a)=0$, then $\mathbb{C}\subseteq \bigcup_{n\in\mathbb{N}}A_{n}$. Therefore, $\bigcup_{n\in\mathbb{N}}A_{n}$ is not countable, then there is $n\in \mathbb{N}$ such that $A_{n}$ is not countable.

$\endgroup$
2
  • 1
    $\begingroup$ @ Diego : why $A_n$ has a limit point. $\endgroup$
    – Struggler
    Jun 17, 2016 at 11:07
  • $\begingroup$ : Also i wantto knpw that if $f^{(n)}(z) = 0$ for all $z \in \mathbb C$, then $f$ is a polynomial. please tell me about two point. thank you $\endgroup$
    – Struggler
    Jun 17, 2016 at 11:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .