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Homework Help: Derivative of Fourier series

  1. Oct 1, 2009 #1
    We were given in a previous question,

    s_{N}(x) = \frac{4}{\pi}\sum_{n=0}^{N-1}\frac{sin(2n+1)x}{2n+1}

    1. The problem statement, all variables and given/known data

    Show that
    s'_{N}(x) = \frac{2sin(2Nx)}{\pi sinx}, x \neq l\pi


    s'_{N}(x) = \frac{4N}{\pi}(-1)^l, x = l\pi

    where l is any integer.

    3. The attempt at a solution

    Utterly stumped on this one, I'm aware it's not *NORMAL* differentiation, how exactly do you go about differentiating a series? We've never been taught that and I'm an attentive math student.

    So I can't even start (and this is the last question), this is frustrating me...

    Last edited: Oct 1, 2009
  2. jcsd
  3. Oct 1, 2009 #2
    don't panic. all you have is a finite sum.
    the formula [tex] (f+g)' = f' + g' [/tex] does work for finite sums!
    the only problem left is finding the right trigonometric identity to use!
  4. Oct 1, 2009 #3
    But the function is w.r.t x, so my first thought was just to drop the x off. But that's wrong...

    I don't get it, what's f and what's g?

    Thanks for somewhere to start though
  5. Oct 2, 2009 #4
    dear ceasius, you are given a sequence of functions
    [tex]s_1 = \frac{4}{\pi} \sin x , s_2 =\frac{4}{\pi} (\sin x + \frac{\sin (3 x}{3})), s_3 = \frac{4}{\pi} ( \sin x + \frac{\sin (3x)}{3} + \frac{\sin(5x)}{5}),..[/tex]
    your task is to prove that
    [tex] \frac{d s_n(x)}{dx} =\frac{sin(2nx)}{\pi \sin x} ,x \ne \pi l[/tex]
    for every n. the way to do this is by using induction on n.
    the case n=1 is simple
    [tex] \frac{4}{\pi} \frac{d \sin x}{dx}=\frac{4}{\pi} \cos x[/tex]
    there is a trigonomtric formula that says
    [tex] \sin(a) \cos(b) = \frac{1}{2}(\sin(a-b) + \sin (a+b)) [/tex]
    apply this formula on [tex] \sin x \cos x [/tex], check for which x [tex]\sin x = 0[/tex] and you will get the right result. It's a good practice to try n=2
    [tex] \frac{d s_2(x)}{dx}=\frac{4}{\pi}\frac{d(\sin(x)+\frac{\sin(3x)}{3})}{dx}=\frac{4}{\pi}(\frac{d\(sin(x)}{dx}+\frac{d\sin(3x)}{3 dx})=\frac{2 \sin(2x)}{\pi \sin x}+ \frac{4}{\pi} \cos(3x) [/tex]
    now apply the trigonometric formula on [tex] \sin x \cos(3x) [/tex] and you will get the right result.
    I hope it's clear what's left to do:smile:
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