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Homework Help: Differential Equations

  1. Nov 2, 2005 #1
    Find the Fourier Series for f(x) =1 on the interval 0<= x <= pi in terms of [itex] \phi_{n} = \sin{nx} [/itex]. Bu integrating this series find a convergent series for the function g(x) =x on this interval assuming the set {sin nx} is complete
    Ok for the FOurier Coefficients
    [tex] c_{n} = \frac{\int_{0}^{\pi} f \phi \rho dx}{\int_{0}^{\pi} \phi^2} dx [/tex]
    this is how it is in my test
    rho is suppsod to be the weight
    ok for the numerator
    [tex] \int_{0}^{\pi} \sin{nx} dx = \frac{1}{n} [- \cos{nx}]_{0}^{\pi} = (-1)^n + 1 [/tex]
    for hte denominator
    [tex] \int_{0}^{\pi} (\sin{nx})^2 dx = \frac{\pi}{2} - \frac{\sin{2n \pi}}{4} =\frac{\pi}{2} [/tex]
    so the fourier series is
    [tex] \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\sin{nx}}{n} (-1^n + 1) [/tex]
    what do they mean by integrate the series? Does it mena i should integrate the argument of this sum? ANd how would one find a convergent series for the function g(x) =x??
    Please help!
    Last edited: Nov 3, 2005
  2. jcsd
  3. Nov 3, 2005 #2


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    Yep, under certain conditions you can integrate a series term by term. E.g.

    Taylor series for [itex] \frac{1}{1+x} [/itex] with [itex] \mod{x}<1 [/itex] is

    [tex] \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-... [/tex]

    and converges. Integrating it term by term one gets a convergent series for [itex] \ln\left(1+x\right) [/itex].


    BTW, where din that [itex] \frac{1}{n} [/itex] from integrating [itex] \sin nx [/itex] go?
  4. Nov 3, 2005 #3
    but how would the ln(1/1+x) relate to my problem?? DO you mena i am supposed to expand the sin or cos by taylor series?>?

    i have edited to include that 1/n of the sine

    is my FOurier Series correct? The back of my book states something else:
    [tex] \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin{2k-1}x}{2k-1} [/tex]

    im not sure how they got this...
  5. Nov 3, 2005 #4
    how does what is in the text post #3 relate to waht i got?

    It is similar in that i do have the sine part except what about the -1^n +1 part ? It appears the book has assumed n to be even ... why??
    Also why has the book gotten 2k-1 instead of n? If n is supposed to be even then why is 2k-1, an odd number, used as n in teh same argument?
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