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I've got determine all entire functions $f$ such that $f(\frac{1}{n})=f(\frac{1}{n²})$ is fullfilled for all $ n\in \mathbb{N}$.

Knowing, that they are entire we can write down both terms as a taylor series around $0$: $f(1/n)=\sum_{k=0}^\infty a_k(\frac{1}{n})^k=a_0+a_1\frac{1}{n}+a_2\frac{1}{n}^2+a_3\frac{1}{n}^3+a_4\frac{1}{n}^4...$

$f(1/n^2)=\sum_{k=0}^\infty a_k(\frac{1}{n^2})^k=a_0+a_1\frac{1}{n^2}+a_2\frac{1}{n^2}^2+a_3\frac{1}{n^2}^3+a_4\frac{1}{n^2}^4...$

When they are equal, then the coefficients are $0$ for all odd numbers, that is $a_k=0$ for $k=2n+1$, That means $a_1=a_3=a_5=...=0$. Then because of the equality of both series we get $a_2=a_1=0$ which then implies $a_4=a_2=0$ and so on. So there is no entire function other then a constant function, right? Is there maybe a more simple proof for this?

Edit: My argumentation for $f(0)=0$ is wrong. So $a_0$ is the only coefficent I have.

Thesinus
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2 Answers2

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Let $g(z) = f(z^2)$. Then $g$ is also entire and in particular, holomorphic. We have $f(1/n) = g(1/n)$ for every $n$ and the sequence $1/n$ has an accumulation point. Then there is a theorem which allows us to conclude that $f=g$, i.e. $f(z) = f(z^2)$ for any $z \in \Bbb{C}$.

Now suppose $|z|<1$ and define the sequence $z_0=z,z_{n+1} = z_n^2$. Obviously $z_n \to 0$ and therefore $f(z_0) = f(z_n) \to f(0)$. This implies that $f$ is constant in the unit disk. The same theorem quoted above allows us to conclude that $f$ is constant ($f$ is equal to a constant function on a set which has an accumulation point).

Community
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Beni Bogosel
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Take the function $g(z)=f(z)-f(z^2)$

Then $x_n=\frac{1}{n}$ is a sequence of roots in $\Bbb{C}$ and has a limit point.

So from Identity theorem we have that $f(z)=f(z^2)$

Also $f(z)=f(z^2)=....=f(z^{2n}) ,\forall n \in \Bbb{N}$

If $|z|<1$ then $f(z)=f(z^{2n}) \to f(0)$ so $f(z)=f(0), \forall z \in B(0,1)$

Thus $f(z)=f(0), \forall z \in \Bbb{C}$ by identity theorem again.

Marios Gretsas
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