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Derivative of Fourier series

  1. Sep 25, 2014 #1

    etf

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    Let's say I have Fourier series of some function, [tex]f(t)[/tex], [tex]f(t)=\frac{a0}{2}+\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})[/tex], where [tex]a[/tex] and [tex]b[/tex] are lower and upper boundary of function, [tex]a0=\frac{2}{b-a}\int_{a}^{b}f(t)dt[/tex], [tex]an=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt[/tex], bn=[tex]\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt[/tex]. My question is, can I find derivative of Fourier series on this way:
    [tex]\frac{d}{dt}\sum_{n=1}^{\infty}(an\cos{\frac{2n\pi t}{b-a}}+bn\sin{\frac{2n\pi t}{b-a}})=\frac{d}{dt}(\sum_{n=1}^{\infty}an\cos{\frac{2n\pi t}{b-a}}+\sum_{n=1}^{\infty}bn\sin{\frac{2n\pi t}{b-a}})=[/tex]
     
    Last edited: Sep 25, 2014
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  3. Sep 25, 2014 #2

    etf

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    [tex]=\sum_{n=1}^{\infty}\frac{d}{dt}(an\cos{\frac{2n\pi t}{b-a}})+\sum_{n=1}^{\infty}\frac{d}{dt}(bn\sin{\frac{2n\pi t}{b-a}})[/tex]
    ?
    Are there some limitations when I can do this and when I can't do this?
    PS. I had to write this in second message since for some reason my latex code can't fit in one message :)
     
  4. Sep 25, 2014 #3

    mathman

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    I think I know what you meant - your equation is an identity (I don't think you meant it). I presume your question is can you differentiate term by term to get the derivative of a Fourier series. The answer is no different from the same question for any infinite series.

    Google the question and you will get plenty of answers.
     
  5. Sep 25, 2014 #4

    HallsofIvy

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    As long as a sum of functions is uniformly convergent, then you can differentiate the sum "term by term". Fourier series (and Taylor's series) are always uniformly convergent.
     
  6. Sep 25, 2014 #5
    I think the Fourier series of a discontinuous function is only pointwise convergent, not uniformly convergent.
     
  7. Sep 25, 2014 #6

    jbunniii

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    Indeed, the partial sums of a Fourier series are continuous, and the uniform limit of a sequence of continuous functions is continuous. So no discontinuous function can have a uniformly convergent Fourier series.

    Moreover, uniform convergence of a sequence of differentiable functions ##f_n## to some function ##f## is not sufficient to ensure that ##f## is differentiable. If the sequence of derivatives ##f_n'## converges uniformly, then it is sufficient, and ##\lim f_n' = f'##.
     
    Last edited: Sep 25, 2014
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