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Messages - Yiyi Cheng

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MAT334--Lectures & Home Assignments / Re: 2.5 Q31
« on: November 13, 2018, 10:59:00 PM »
This would be my answer to this question, feel free to ask me any question ;)
$\sum_{-\infty}^{\infty}{J_n(u)z^n}= G(z;u)$
$G(z;u)= e^{(\frac{u}{2})(z-\frac{1}{z})}= G(-\frac{1}{z};u)$
$=\sum_{-\infty}^{\infty} J_n(u)\frac{(-1)^n}{z^n}$
$=\sum_{-\infty}^{\infty} J_{-n}(u)(-1)^n z^n$
Therefore, we get $J_{-n}= (-1)^n J_n(u)$

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