# Entire functions that satisfy certain equation.

Problem

Find all the entire functions that satisfy $$n^2f(\dfrac{1}{n})^3+f(\dfrac{1}{n})=0 \space \space \forall \space n \in \mathbb N$$

My idea was to define a function $$g(z)=z^2f(z)^3+f(z)$$

Since $f$ is an entire function and $g$ is composition of entire functions, then $g$ is entire as well. The function $g$ vanishes on the sequence $\{\dfrac{1}{n}\}_{n \in \mathbb N}$ and $\dfrac{1}{n} \to 0 \in \mathbb C$, then $g$ is identically zero (if a function $f$ defined on a region $\Omega$ vanishes on a sequence of different points with a limit point in $\Omega$, then $f$ is identically zero).

We have $$0=g=z^2f(z) ^3+f(z),$$ from here we obtain $$f(z)=-z^2f(z)^3$$

I have the feeling that I have not answered the complete solution, but I don't know what else I can say about $f$. I would appreciate some help to complete my solution or suggestions of another answer.

• Your functional equation needs a negative exponent. – Adam Hughes Jul 3 '14 at 17:09
• Oops, you're right, then $g$ is not entire, but it is holomorphic in a region $\Omega$. – user156441 Jul 3 '14 at 17:12
• The functional equation for $f$ implies $f(0)=0$, so the Taylor series for $f(z)^3$ has lowest power at least 3, canceling out the $z^{-2}$. – Adam Hughes Jul 3 '14 at 17:17

Clearly $f(z)=0$ is a solution, so assume not. Then the functional equation is:

$$z^{-2}f(z)^3+f(z)=0$$

satisfied on some infinite subsequence of $\{{1\over n}:n\in\mathbb{N}\}$ where $f(z)\ne 0$, which is possible since $f\not\equiv 0$.

As a preliminary note, this is an analytic function because the original functional equation implies $f(0)=0$ so that the Taylor series for $f$ has lowest term at least degree 1, so when we cube it the lowest power is at least 3, so the pole ostensibly introduced by the $z^{-2}$ term is seen to be removed, so that the function is analytic.

Then we see that $f(z)^2=-z^2$ along this subsequence, which implies $f(z)=\pm iz$ since that is a set with an accumulation point in the domain $\mathbb{C}$.

• I have one sort of stupid doubt: you have the functional equation $z^{-2}f(z)^3+f(z)=0$, as you are discarding the case $f=0$ (the zero function), you arrived to $f(z)^2=-z^2$; to arrive to that equation, in one step you must have divided by $f(z)$ on both sides, but $f(z)=0$ for some points (for example $\{\dfrac{1}{n}\}$) so my concrete question is: is it ok in general to divide by a function that vanishes at least at one point? – user156441 Jul 3 '14 at 17:21
• @user156441 I only do the division on the infinite subsequence on which $f(z)\ne 0$, that still has an accumulation point, so it works just fine. In particular I'm not really "dividing" I'm using that $ab=0\iff a=0\text{ or }b=0$ so that $f(z)(z^{-2}f(z)^2+1)=0$ and along the subsequence, since $f(z)\ne 0$ we must have the other option, i.e. $f(z)^2=-z^2$. – Adam Hughes Jul 3 '14 at 17:22
• I've updated my answer to include all the clarifications you asked for so that people don't have to search through the comments. – Adam Hughes Jul 3 '14 at 17:29
• Thanks, Igot this this, but have another doubt: what you get from the functional equation is $f(z)=iz$ or $f(z)=-iz$ for those points in the sequence, but what about the rest of $\mathbb C$? how could one conclude that $f(z)$ must be $f(z)=iz$ or $f(z)=-iz$ for all $z$ (I am discarding the zero function here)? Are you using the identity theorem? – user156441 Jul 3 '14 at 17:29
• It's an entire function, agreement along a sequence with an accumulation point is all you need, just like you tried to do with your original approach. I am indeed using the identity theorem. – Adam Hughes Jul 3 '14 at 17:30

To make your original attempt work,, rewrite the given condition as $$f(\frac 1n)^3+ \frac1{n^2}f(\frac1n)=0$$ and let $g(z)=f(z)^3+z^2f(z)$. As before, $g(\frac1n)=0$ and hence $g(z)=0$. Then $$0=f(z)+z^2f(z)= f(z)(f(z)-iz)(f(z)+iz).$$ At least one factor must be $=0$ for a converging sequence, hence is identically zero. We conclude that $f(z)=0$, $f(z)=iz$, $f(z)=-iz$ are the only three solutions.