# analytic function on $\mathbb{C}$ such that $f\Bigl(\frac{1}{2n}\Bigr)=f\Bigl(\frac{1}{2n+1}\Bigr)=\frac{1}{2n}$

I was working out some problems from a entrance test paper, and this problem looked a bit difficult.

• Does there exist an analytic function on $\mathbb{C}$ such that $$f\biggl(\frac{1}{2n}\biggr)=f\biggl(\frac{1}{2n+1}\biggr)=\frac{1}{2n} \quad \ \text{for all} \ n\geq 1?$$

I did apply $f$ repeatedly to get some more insights about the problem, but couldn't get anything from it. So how do i solve this one?

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Use the fact that an analytic function is determined by its values on any non-discrete set. – Chris Eagle Jan 11 '11 at 10:40
A similar solution: any f(z) analytic, agreeing with g(z)=z on a non-discrete set, would have to be equal to g(z)=z. But f(1/(2n+1)) is not 1/(2n+1) – gary Jun 25 '11 at 19:59

Suppose there is such an $f(z)$. Let $g(z) = f(z) - z$. If $f(z)$ were analytic, then $g(z)$ would also be analytic. But $g(z)$ has a limiting sequence of zeros (in particular, $g(1/2n) = 0$), and the zeros of any nonconstant analytic function are discrete. Therefore $g(z) = 0$, which is not the case; contradiction. So there is no such analytic function.