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Thread: Fourier Transform

  1. #1
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    Fourier Transform

    As part of my final year project, I'm using the identity $\displaystyle cos(z sin \theta ) = J_{0} (z) + 2 \sum _{k=1}^{oo} J_{2k} (z) cos (2k \theta )$ where J is a Bessel function of the first kind.

    Instead of just quoting this, it would be quite nice to actually prove it, so I was wondering if anyone could direct me in how to do so?
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  2. #2
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    Quote Originally Posted by whatisthisfor View Post
    As part of my final year project, I'm using the identity $\displaystyle cos(z sin \theta ) = J_{0} (z) + 2 \sum _{k=1}^{oo} J_{2k} (z) cos (2k \theta )$ where J is a Bessel function of the first kind.

    Instead of just quoting this, it would be quite nice to actually prove it, so I was wondering if anyone could direct me in how to do so?
    I'm using this definition, which is probably yours (?): $\displaystyle J_n(z)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-in\theta} e^{-i z\sin \theta} d\theta$.

    Note that $\displaystyle J_n(z)$ is the $\displaystyle n$-th complex Fourier coefficient of the function $\displaystyle \theta\in(-\pi,\pi)\mapsto e^{-i z \sin\theta}$ (extended periodically). The general theory says that its Fourier series converges: $\displaystyle \sum_{n\in\mathbb{Z}} J_n(z) e^{in \theta} = e^{-i z\sin\theta}$ for all $\displaystyle z\in(-\pi,\pi)$. Take the real part of both sides, and notice that $\displaystyle J_{-n}(z)=(-1)^nJ_n(z)$ so that terms with odd indices cancel, while terms with even (non-zero) indices count twice (this gives the 2 in your formula). That's it.
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