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Thread: Integrating the Fourier coefficient

  1. #1
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    Question Integrating the Fourier coefficient

    Consider a Fourier series of
    $$-100S(x) = \sum_{n=1}^\infty b_n \sin(n\pi x)$$
    where $S(x)$ is sigmoid function. The Fouerier coefficient is given by
    $$b_n=2 \int_0^1 S(x) \sin(n\pi x)dx$$
    Right?


    If yes, what is the solution of this integral to calculate the Fourier coefficient for the above series?
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  2. #2
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    Re: Integrating the Fourier coefficient

    Yes, the sigmoid function is an odd function so has a Fourier sine series. The sigmoid function is \frac{e^x}{e^x+ 1}. The coefficient is given by \int_0^1 \frac{e^x}{e^x+ 1}sin(n\pi x)dx. I would integrate that using sin(n\pi x)= \frac{e^{n\pi ix}- e^{-n\pi ix}}{2i} and combining the exponentials.
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  3. #3
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    Re: Integrating the Fourier coefficient

    Quote Originally Posted by HallsofIvy View Post
    Yes, the sigmoid function is an odd function so has a Fourier sine series. The sigmoid function is \frac{e^x}{e^x+ 1}. The coefficient is given by \int_0^1 \frac{e^x}{e^x+ 1}sin(n\pi x)dx. I would integrate that using sin(n\pi x)= \frac{e^{n\pi ix}- e^{-n\pi ix}}{2i} and combining the exponentials.
    Thanks for the clarification, but I lost. You mean
    \int_0^1 \frac{e^x}{e^x+ 1}\frac{e^{n\pi ix}- e^{-n\pi ix}}{2i}dx

    Could you please elaborate how to integrate it?
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