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How can one find out the asymptotic behavior of a Bessel function? If we start from $z= 0$, we can get a Taylor series. But in physics, we have to know the asymptotic behavior of the solution at both $z=0$ and $z= \infty$. The Taylor series is not very helpful for assessing the asymptotic behavior at $z=\infty $.

Beamer
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    Using Mathematica to expand around $z=0$, I found $J_n(z)=z^n\left( \frac{1}{2^n \Gamma(1+n)} -\frac{z^2}{2^{2+n}(1+n)\Gamma(1+n)} + \dots \right)$. Around $z=\infty$, $J_n(z)=\cos (\pi/4 + n\pi/2 -z) \left( \sqrt{2/\pi z} + z^{-5/2}(40n^2-9-16n^4)/(64\sqrt{2\pi}) + \dots\right).$ – JPhy Nov 13 '14 at 09:54
  • Also, have a look at http://www.math.colostate.edu/~shipman/47/volume2a2010/Sekeljik.pdf, it seems like what you're looking for, and goes into more detail. – JPhy Nov 13 '14 at 10:20

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