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Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion

$$f(z) = \sum_{n = 0}^{\infty}c_n(z-z_0)^n$$

is equal to $0$. Prove that $f$ is a polynomial.

The problem hints to use the fact that $c_nn! = f^{(n)}(z_0)$ and use a countability argument.

My attempt at a solution

The only thing I can think of in this case is that the coefficients $c_n$ of the above Taylor expansion are defined as:

$$c_n = \frac{f^{(n)}(z_0)}{n!}$$

The only way one of these $c_n$ could be zero is if the derivative of order $n$ vanishes everywhere. Thus, this would mean that $f$ is a polynomial of order $k < n$.

Does this suffice as a proof? How would I use a countability argument to prove this instead? Thanks.

r123454321
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  • No, that is not a proof. You are not given that there is a fixed $n$ such that $f^{(n)}(a)=0$ for all values of $a$; you are given that for each $a$ there is an $n$ (depending on $a$) such that $f^{(n)}(a)=0$. (By the way, it would be less confusing if you'd stick with either $z_0$ or $a$ instead of switching back and forth.) – bof Sep 24 '14 at 05:25

2 Answers2

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No, that's not true. It's quite possible for some derivative of $f$ to be $0$ at some points $a$ without vanishing everywhere. What you have to use is that for every $a$ there is such an $n$.

Hints:

  1. The zeros of a non-constant analytic function form a discrete set, i.e. have no limit point in the domain of the function.
  2. A discrete subset of $\mathbb C$ is countable.

BTW: there is a much harder version of the problem that replaces $\mathbb C$ by a real interval and replaces analytic by $C^\infty$.

Robert Israel
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  • Okay, thanks -- so we know that $f$ has countably many zeros -- does this imply that the function can be approximated at any $z_0$ by a Taylor expansion of countable degree? Thus, there exists an $n$ such that for every $z_0$, $f^{(n)}(z_0($ vanishes everywhere? – r123454321 Sep 24 '14 at 22:03
  • I think you're confused. If $f$ is analytic on all of $\mathbb C$, it can be approximated anywhere by a Taylor series centred anywhere else. But that has nothing to do with having countably many zeros, and your "Thus," doesn't follow. – Robert Israel Sep 25 '14 at 00:00
  • Further hint: if $f$ is not a polynomial, each $f^{(n)}$ has countably many zeros. What can you say about the union of all these countable sets? – Robert Israel Sep 25 '14 at 00:01
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If none of the derivatives are exactly zero, the set of $a \in \mathbb{C}$ such that there exists $n \in \mathbb{N}$ such that $f^{(n)}(a)=0$, i.e. $\cup_{n=0}^\infty (f^{(n)})^{-1}(0)$ is countable. This uses that an analytic function has only countably many zeroes, the quickest way I can see to prove this is the fact it has only finitely many in any compact set (or you get a limit point) and $\mathbb{C}$ is a countable union of compact sets (e.g. positive integer radius balls around 0)

Matt Rigby
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  • Sorry -- I don't quite follow. – r123454321 Sep 24 '14 at 22:37
  • We want to show that for every $z_0$, there is such an $n$ such that $f^{(n)}(z_0)$ vanishes everywhere. If this is true, then $f$ is a polynomial of order $k < n$, and we are done. We know a discrete subset (i.e. $f$'s zeroes) of $\mathbb{C}$ is countable, so we know $f$ has countably many zeroes. I'm not following how what you said fits into this, or in what direction you're proceeding? Thanks. – r123454321 Sep 24 '14 at 22:40
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    A countable union of countable sets is countable, so if no derivative is identically zero, as each derivative is analytic (hence it's set of zeroes is countable), the set of points such that some derivative vanishes is countable. But the hypothesis says that every $z_0 \in \mathbb{C}$ has some derivative vanishing at $z_0$; so the set of points where some derivative is zero is uncountable, hence some derivative is exactly zero, implying $f$ is a polynomial. – Matt Rigby Sep 24 '14 at 22:46