Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Given a continuous function $f$ of period $2\pi$ and $\alpha$ in $\mathbb{R}$

can you show for any real $x$ that $\lim_{ N\to \infty} \sum_{n=1}^N f(x+n\alpha) = \lim_{ N\to\infty} \sum_{n=1}^N f(x_n)$

that is, the 'average value' of $f$ diluted by $N$ is the same as the average value of the sequence $x+n\alpha$ since $\alpha/\pi$ is never irrational

where $\alpha/\pi$ is irrational

share|cite|improve this question

If $\alpha/\pi$ is rational, then your claim is false (in general, unless we got lucky), since you're sampling the same points over and over again.

Conversely, if $\alpha/\pi$ is irrational, then you're sampling random points of the function (by Weyl's equidistribution theorem), and then with some work you can probably prove that in the limit you do get the mean value of $f$.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.