Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 11 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.