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Math Help - Periodic functions - prove that p is a rational number?

  1. #1
    Super Member fardeen_gen's Avatar
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    Periodic functions - prove that p is a rational number?

    If the function given by f(x) = \sin x + \cos px is periodic, prove that p is a rational number.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    If the function given by f(x) = \sin x + \cos px is periodic, prove that p is a rational number.
    suppose T \neq 0 is a period of f. then \sin(x+T)+\cos(px+pT)=\sin x + \cos px. call this (1). put x=0 in (1) to get: \sin T + \cos pT = 1. call this one (2). now differentiate (1) twice to get:

    \sin(x+T)+p^2 \cos(px+pT)=\sin x + p^2\cos px, and put x=0 to get: \sin T + p^2 \cos pT=p^2. call this (3). now (2) and (3) give us: (p^2-1)(\cos pT - 1)=0. therefore either p^2=1 and so

    p=\pm 1 \in \mathbb{Q}, or \cos pT = 1 and thus by (2): \sin T=0. hence pT=2k \pi and T=\ell \pi, for some integers k and \ell \neq 0. thus: p=\frac{2k}{\ell} \in \mathbb{Q}.
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  3. #3
    Senior Member pankaj's Avatar
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    Period of \sin x=2\pi

    Period of \cos px=\frac{2\pi}{p}

    Period of f(x)=L.C.M\left\{2\pi,\frac{2\pi}{p}\right\}= 2\pi L.C.M\left\{1,\frac{1}{p}\right\}.

    If f(x) is to be periodic then the L.C.M must exist.This L.C.M will exist only if p is a rational number and 2\pi L.C.M\left\{1,\frac{1}{p}\right\}also may not be the fundamental period.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by pankaj View Post
    If f(x) is to be periodic then the L.C.M must exist.
    How do you know that the period of the sum must be a period of both?
    Assuming this, of course, the problem is easy; but you should supply proofs of your statements!

    See my other thread.
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  5. #5
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Bruno J. View Post
    How do you know that the period of the sum must be a period of both?
    Assuming this, of course, the problem is easy; but you should supply proofs of your statements!

    See my other thread.

    I had been trying to find a proof of such a statement myself but could only come up with examples
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