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Math Help - Fourier Transform & Regularity

  1. #1
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    Fourier Transform & Regularity

    Here is a few transform that I'm trying to calculate:

    1. Calculate the fourier coefficient of e^x on [-\pi, \pi] and thus determine the sum \sum_{k=-\infty}^{\infty} 1/(1+k^2). I get F_k = (-1)^k (e^{\pi}-e^{-\pi})(1+ki) /(1+k^2). I'm not sure if this is correct and can't see how using any identities (e.g. Parseval) I can calculate the sum.

    2. Suppose that f \in \{C(\mathbb{R}) \cap L^1 (\mathbb{R})\} and p \in \mathbb{N} . Show that if |k|^p F_k is integrable then f has p continuous bounded derivatives.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by davidmccormick View Post
    Here is a few transform that I'm trying to calculate:

    1. Calculate the fourier coefficient of e^x on [-\pi, \pi] and thus determine the sum \sum_{k=-\infty}^{\infty} 1/(1+k^2). I get F_k = (-1)^k (e^{\pi}-e^{-\pi})(1+ki) /(1+k^2). I'm not sure if this is correct and can't see how using any identities (e.g. Parseval) I can calculate the sum.
    The Fourier series converges to the function at points where the function is continuous. At points where there is a jump discontinuity, it converges to the midpoint of the jump. So you know that e^x = \frac1{2\pi}\sum_{k=-\infty}^\infty \frac{(-1)^k (e^{\pi}-e^{-\pi})(1+ki)}{1+k^2}e^{ikx} when |x|<\pi. But what happens when x=\pi?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Proceeding as in...

    http://www.mathhelpforum.com/math-help/advanced-applied-math/141470-fourier-series.html

    ... we obtain...

    a_{0} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot dx = \frac{2}{\pi}\cdot \sinh \pi

    a_{n} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot \cos nx \cdot dx = (-1)^{n} \frac{2\cdot \sinh \pi}{\pi\cdot (1+n^{2})}

    b_{n} = \frac{1}{\pi} \int_{-\pi}^{+ \pi} e^{x}\cdot \sin nx \cdot dx = (-1)^{n-1} \frac{2\cdot n\cdot \sinh \pi}{\pi\cdot (1+n^{2})} (1)

    ... so that for -\pi < x < \pi is...

    e^{x} = \frac{\sinh \pi}{\pi} \cdot \{1 + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1 + n^{2}}\cdot (\cos nx - n\cdot \sin nx)\} (2)

    Now if we remember that the series (2) for x= \pm \pi converges to \frac{e^{\pi} + e^{- \pi}}{2} = \cosh \pi we obtain...

    1 + 2 \sum_{n=1}^{\infty} \frac{1}{1 + n^{2}} = \sum_{n= -\infty}^{+\infty} \frac{1}{1+n^{2}} = \frac{\pi}{\tanh \pi} = 3.153348094937... (3)

    ... and from (3) ...

    \sum_{n=1}^{\infty} \frac{1}{1 + n^{2}} = \frac{\pi}{2\cdot \tanh \pi} - \frac{1}{2} = 1.076674047468... (4)

    Kind regards

    \chi \sigma
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