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Math Help - Sum of a periodic function

  1. #1
    Senior Member bkarpuz's Avatar
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    Sum of a periodic function

    Dear MHF members,

    Question. Suppose that f:\mathbb{R}\to\mathbb{R} is a continuous periodic function with a period of 2\pi. If \alpha/\pi is irrational, then for any x\in\mathbb{R} show that the sum \frac{1}{N}\sum_{n=1}^{N}f(x+n\alpha) converges to \frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)\mathrm{d}u.

    Thanks.
    bkarpuz

    Exercise 19 in Chapter 3 of W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Science/Engineering/Math, 3rd edition, 1976.
    Last edited by bkarpuz; May 17th 2011 at 11:04 AM. Reason: Remark by Tinyboss
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  2. #2
    Senior Member Tinyboss's Avatar
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    I assume f satisfies some continuity condition, else it's definitely not true. Hint: show that \{x+n\alpha\text{ mod }2\pi\}_{n\in\mathbb{N}_+} is dense in [0,2pi].
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    Question. Suppose that f:\mathbb{R}\to\mathbb{R} is a continuous periodic function with a period of 2\pi. If \alpha/\pi is irrational, then for any x\in\mathbb{R} show that the sum \frac{1}{N}\sum_{n=1}^{N}f(x+n\alpha) converges to \frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)\mathrm{d}u.

    Thanks.
    bkarpuz

    Exercise 19 in Chapter 3 of W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Science/Engineering/Math, 3rd edition, 1976.
    Tinyboss gave you an idea how to prove this. This theorem is a specific case of one of the equidistribution theorems
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