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.
Best Answer
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.