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Fourier series coefficients: proof by induction

  1. Feb 12, 2015 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data
    Given f = a0 + sum(ancos(nx) + bnsin(nx))
    and f' = a0' + sum(an'cos(nx) + bn'sin(nx))
    The sums are over all positive integers up to n.

    show that a0' = 0, an' = nbn, bn' = -nan

    Then prove a similar formula for the coefficients of f(k) using induction.

    2. Relevant equations


    3. The attempt at a solution
    OK I showed the first part (for the first derivative coefficients) by taking the derivative of f, that was easy enough. I have to confess that the teacher's question also said HINT: (Integrate by parts) but it was much simpler to just take the derivative of f.

    I don't know if that's part of the reason i'm having trouble with the induction part or not, but it seems inefficient to integrate backward when it's such a simple thing to show these coefficients with the derivative. I have not had a lot of experience with induction, and I haven't had to do much as an engineering (as opposed to math) major.

    So induction works like this: prove it for a single case, assume something is true for some integer k, then use that assumption to show it is also true for k+1. (This is to the best of my understanding)

    After considering the problem for some time, i don't think it's too much to show that for any k, a0(k) = 0, since the derivative of 0 is 0, and I've shown that the derivative of a0 = 0.

    It's the other coefficients that trip me up, because every time the derivative is taken, the coefficients will take on a value related to its opposite (in terms of functionally odd or even, i mean) coefficient. For example an' = nbn, notice that the derivative an' relies on the value of bn.

    I guess I'm just not sure how to start. I feel like I could just do the derivative a bunch of times and look for a pattern, but I feel like that defeats the purpose of proof by induction.

    EDIT:
    Ok, I wrote out f(k), then took the derivative and got that an(k+1) = nbn(k), but this isn't actually true, because, I did the bunch of derivatives and found that the pattern is actually

    an(k+1) = nkbn(k)

    How can I show that by induction? I can see that every time the function is differentiated it's going to cause another multiplication by n because of the chain rule, I'm just not quite sure how to show it mathematically.

    EDIT2: I integrated f(k+1) but it didn't seem to really add anything new as far as a solution goes... also the HINT was to integrate by parts but looking at the expression, I don't really see anything to integrate by parts even if i wanted to.
     
    Last edited: Feb 12, 2015
  2. jcsd
  3. Feb 12, 2015 #2

    vela

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    I think your teacher wants you to calculate the Fourier coefficients for f' using the usual integral formulas for the coefficients. You'll get integrals of the form
    $$\int_0^T f' \sin nx\,dx \\ \int_0^T f'\cos nx\,dx $$ which you can evaluate by integrating by parts.
     
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