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Math Help - Sum of periodic functions

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Sum of periodic functions

    This theorem was inspired to me by this thread. I have found a proof of it but I'll post it later; I want to see if somebody can find another one!

    Let m(x), n(x) be continuous real-valued functions having periods p,q. Show that m(x)+n(x) is periodic if and only if p,q are linearly dependent over \mathbb{Q}.
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    Super Member malaygoel's Avatar
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    Quote Originally Posted by Bruno J. View Post
    This theorem was inspired to me by this thread. I have found a proof of it but I'll post it later; I want to see if somebody can find another one!

    Let m(x), n(x) be continuous real-valued functions having periods p,q. Show that m(x)+n(x) is periodic if and only if p,q are linearly dependent over \mathbb{Q}.
    If m(x) + n(x) has period T, then
    T=ap
    T=bq
    where a,b are natural numbers.

    hence, ap=bq
    or,
    p=\frac{b}{a}q

    now, what is meant by linear independence here?
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    MHF Contributor Bruno J.'s Avatar
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    How can you be sure that the period of the sum is a multiple of both p,q? You need to show that this is true; it's the hard part of the problem.

    If x_1,...,x_n are linearly independent over K, then whenever k_1x_1+...+k_nx_n=0 with k_1,...,k_n \in K we must have k_1=...=k_n=0. If they are linearly dependent then they are not linearly independent. In your post, p-\frac{b}{a}q=0 is a linear dependence relation over the rationals.

    But you have to show that we must have T=ap=bq for some integers a,b.
    Last edited by Bruno J.; June 29th 2009 at 09:44 AM.
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    For a,b integers:

    m(x + ap) + n(x + bq) = m(x) + n(x) (1)

    Therefore ap=bq defines all periods of the sum function.

    If the sum is periodic, p = b/a * q, where b/a obviously belongs to Q.

    Alas c1*p + c2*q = (c1 * b/a + c2) * q, for c1=-a/b and c2=1 (both Q) it is 0, that is if the sum is periodic p,q are linearly dependent over Q (2)

    If p,q are linearly dependent over Q, p = -c2/c1 * q for c1,c2 rational numbers, then expressing c1,c2 as c1N/c1D and c2N/c2D, where c1N,c1D,c2N,c2D are integers, we get c1N*c2D*p = -c2N*c1D*q.

    Taking (1) and making a=c1N*c2D and b=-c2N*c1D, ab=pq, so If p,q are linearly independent over Q, the sum function is periodic (3).

    m(x) + n(x) is periodic If(3) and only if(2) p,q are linearly independent over Q.

    (Sorry for the ugly proof)
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    MHF Contributor Bruno J.'s Avatar
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    For a,b integers:

    m(x + ap) + n(x + bq) = m(x) + n(x) (1)
    Ok so far.

    Therefore ap=bq defines all periods of the sum function.
    How so? If T is the period of m(x)+n(x), then m(x)+n(x)=m(x+T)+n(x+T); but that does NOT imply m(x)=m(x+T) and n(x)=n(x+T). You are making a huge leap here.

    If you do not use the continuity of m,n your proof is certainly flawed because we can construct non-continuous m,n whose sum is periodic, but whose periods are linearly independent over \mathbb{Q}.
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    Ok so far.

    How so? If T is the period of m(x)+n(x), then m(x)+n(x)=m(x+T)+n(x+T); but that does NOT imply m(x)=m(x+T) and n(x)=n(x+T). You are making a huge leap here.

    If you do not use the continuity of m,n your proof is certainly flawed because we can construct non-continuous m,n whose sum is periodic, but whose periods are linearly independent over \mathbb{Q}.
    Ups I'll try it later...
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Bump. Nobody has solved this one yet... last call! I'll give the solution soon.
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