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Math Help - Fourier Series Limit proof

  1. #1
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    Fourier Series Limit proof

    I'm working on a problem at the moment and I just can't figure it out, here goes:

     \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx = f(0)

    with l > 0 and f(x) differentiable on the interval [-l,l]

    Things that come to mind are Riemann-Lebesgue Lemma -- from Wolfram MathWorld Riemann's lemma which states that any limit of n to infinity of an integral of a function on -l to l of the function times sin(nx) or cos(nx) tends to 0.
    And ofcourse that \int_0^{\infty}\frac{sin(x)}{x}dx = \frac{\pi}{2}

    Does anyone have any ideas on how I can proceed?
    Last edited by TiRune; May 28th 2009 at 11:20 AM.
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  2. #2
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    Clarity?

    Substitute y=\frac{L}{\pi}x, so dx=\frac{\pi}{L}dy, and sub g(x)=\frac{f(x)}{x}

    Then <br /> <br />
\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx <br />
becomes <br /> <br />
 \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy<br />

    So by the Riemann-Lebesgue Lemma, the LHS is zero, so f(0)=0

    *What exactly is it you have to prove? Are you trying to find a function f(x) that satisfies the equation, or are you supposed to prove the equality holds for all f(x)?
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  3. #3
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    We have to prove that the limit over that integral is equal to f(0) (not zero, but the value f(0) for any f differentiable above -l,l).

    Although in the case x is not 0 this proves that the LHS goes to 0, but not that it goes to f(0)
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  4. #4
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Media_Man View Post
    Substitute y=\frac{L}{\pi}x, so dx=\frac{\pi}{L}dy, and sub g(x)=\frac{f(x)}{x}

    Then <br /> <br />
\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx <br />
becomes <br /> <br />
 \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy<br />

    So by the Riemann-Lebesgue Lemma, the LHS is zero, so f(0)=0

    *What exactly is it you have to prove? Are you trying to find a function f(x) that satisfies the equation, or are you supposed to prove the equality holds for all f(x)?
    So you basically "transformed" the LHS so that you could use the Riemann-Lebesgue Lemma showing that it is  0 . But this does not mean its  f(0) .
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  5. #5
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Media_Man View Post
    Substitute y=\frac{L}{\pi}x, so dx=\frac{\pi}{L}dy, and sub g(x)=\frac{f(x)}{x}

    Then <br /> <br />
\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx <br />
becomes <br /> <br />
 \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy<br />

    So by the Riemann-Lebesgue Lemma, the LHS is zero, so f(0)=0

    *What exactly is it you have to prove? Are you trying to find a function f(x) that satisfies the equation, or are you supposed to prove the equality holds for all f(x)?
    Shouldn't it be  \frac{L}{\pi^2} \lim_{n \to \infty} \int_{-L^2/\pi}^{L^2/\pi} g(y) \sin \left (\frac{n \pi y}{L} \right) \ dy ?
    Last edited by Sampras; May 28th 2009 at 05:10 PM.
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  6. #6
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    Blooper, and yes, f(0)=0

    Quote Originally Posted by Media_Man View Post
    Substitute y=\frac{L}{\pi}x, so dx=\frac{\pi}{L}dy, and sub g(x)=\frac{f(x)}{x}
    This was a typo. Sub y=\frac{\pi}{L}x and dx=\frac{\pi}{L}dy, in order to change the limits of integration from [-L,L] to [-\pi,\pi] in accordance with the R-L lemma. Thanks for catching that blooper, Sampras.

    But this does not mean its f(0)=0.
    Yes it does. Completing the RHS of the substitution, <br /> <br />
\frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy=f(0)<br />

    Regardless of what f is, it needs to sit on the origin. Here is a proof: If \int_{-a}^a h(x)dx=0 for all a\in\mathbb{R}, then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that h(-x)=h(x) but \lim_{n\rightarrow\infty} h(x)-h(-x) does need to be zero.

    Consider this counterexample: f(x)=x+1, so f(0)\neq0 Theorem: \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l \frac{x+1}x sin(nx) dx\neq0

    Proof: \lim_{n\rightarrow\infty} \frac{x+1}x sin(nx) - \frac{-x+1}{-x} sin(-nx)= \lim_{n\rightarrow\infty} \frac{\sin(nx)}x [(x+1)-(1-x)]= \lim_{n\rightarrow\infty} 2\sin(nx)\neq 0

    *I'm not sure how rigorous this is, but because the \frac{x+1}x sin(nx) is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.
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  7. #7
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Media_Man View Post
    This was a typo. Sub y=\frac{\pi}{L}x and dx=\frac{\pi}{L}dy, in order to change the limits of integration from [-L,L] to [-\pi,\pi] in accordance with the R-L lemma. Thanks for catching that blooper, Sampras.



    Yes it does. Completing the RHS of the substitution, <br /> <br />
\frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy=f(0)<br />

    Regardless of what f is, it needs to sit on the origin. Here is a proof: If \int_{-a}^a h(x)dx=0 for all a\in\mathbb{R}, then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that h(-x)=h(x) but \lim_{n\rightarrow\infty} h(x)-h(-x) does need to be zero.

    Consider this counterexample: f(x)=x+1, so f(0)\neq0 Theorem: \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l \frac{x+1}x sin(nx) dx\neq0

    Proof: \lim_{n\rightarrow\infty} \frac{x+1}x sin(nx) - \frac{-x+1}{-x} sin(-nx)= \lim_{n\rightarrow\infty} \frac{\sin(nx)}x [(x+1)-(1-x)]= \lim_{n\rightarrow\infty} 2\sin(nx)\neq 0

    *I'm not sure how rigorous this is, but because the \frac{x+1}x sin(nx) is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.
     dx = \frac{L}{\pi} dy .
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  8. #8
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    Gahh

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  9. #9
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    I totally get it now, thanks a lot both of you! XD
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