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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

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1 Answer

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 $f$.

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