I need to evaluate $\sum_{n = -\infty}^{\infty} J_0(\alpha n) z^{-n}$ in closed form, where $z$ is complex variable and $J_0()$ is the zeroth order Bessel function of the first kind. How do I evaluate this summation?
How to evaluate $\sum J_0(\alpha n) z^{-n}$ in closed form?
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sequences-and-series
special-functions
power-series
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1Explanation. In "signal processing" they say "Z transform" but mathematicians don't use that terminology. The question may be translated to:evaluate $$\sum_{n=-\infty}^{\infty}J_0(\alpha n)z^{-n}$$ is closed form. – 2012-03-24
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0Since $J_0(x)$ is asymptotically proportional to $x^{-1/2} \cos x$, wouldn't the series diverge for all $z$? – 2012-03-28
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0@StevenStadnicki Since $Z = r e^{j\theta}$, for large enough value of $r$ the sum $\sum |J_0(\alpha n) r^{-n}|$converges? – 2012-03-28
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0@sauravrt But you're summing over positive _and_ negative $n$, so if the terms go to $0$ on one side (e.g., if $|r|< 1$ then the terms go to $0$ for $n\rightarrow\infty$) then they grow unboundedly on the other (e.g., $n\rightarrow -\infty$). The harmonic term tweaks this behavior slightly, but not enough to get any kind of convergence. – 2012-03-28
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This is not a full answer, but maybe it will get you somewhere.
You can prove (or maybe already know) that in fact $J_0(-z)=J_0(z)$. So we can split up the sum as $$ \underset{n=-\infty}{\overset{\infty}{\sum}}J_0(an)z^{-n}=\underset{n=-\infty}{\overset{-1}{\sum}}J_0(an)z^{-n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n} = \underset{n=1}{\overset{\infty}{\sum}}J_0(an)z^{n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n}$$ $$ = J_0(0)+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n}) = 1+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n})$$
I am not sure if this will lead to something useful, but it may simplify things somewhat.