Prove that there are no analytic function satisfying this property This kind I'm kindly asking you to help me in solving this question:


  Prove that there are no analytic functions $f:\mathbb R\to \mathbb R$ such that $$f\left(\frac 1n\right)=\frac{(-1)^n}{n^2},$$ for every $n\in\mathbb N$.


Thank you very much in advance for any hint because I'm out of ideas.

Regards
In fact there is no $f$ two times differentiable in a neighborhood of $0$ such that $f(n^{-1})=\frac{(-1)^n}{n^2}$. Indeed, if $f$ is such a function, then $f(0)=0$, $f'(0)=\lim_{n\to \infty}nf(n^{-1})=\lim_{n\to\infty}\frac{(-1)^n}n=0$ and $f''(0)=\lim_{n\to\infty}n^2f(n^{—1})$ but the last limit is the limit of $(-1)^n$, which doesn't exist.
suppose such $f$ were exist, define $g(z)= f(z)- (-1)^nz^2$ which is also holomorphic, but $g(1/n)=0\forall n$ and Uniqueness theorem says that $g(z)\equiv 0$, $f(z)=(-1)^nz^2$ but which is not unique, so no such analytic function cant exist.