The asymptotic expansion for the Bessel function $J_n(x)$ with integer $n$ is known and given at https://dlmf.nist.gov/10.17. Is there a way to find the asymptotic expansion for $1/J_n(x)$? I'm not sure how given the expansion for $J_n(x)$ is not a pure power series.
Thanks a lot!
Best Answer
It is problematic to take the reciprocal of the standard compound asymptotic expansion of $J_\nu (z)$. Instead, we can use the following representation in terms of the modulus and phase functions: $$ J_\nu (z) = M_\nu (z)\cos (\theta _\nu (z)) = \sqrt {\frac{2}{{\pi z\,\theta '_\nu (z)}}} \cos (\theta _\nu (z)). $$ Then, if $J_\nu(z)\neq 0$, $$ \frac{1}{{J_\nu (z)}} = \sqrt {\frac{{\pi z\,\theta '_\nu (z)}}{2}} \sec (\theta _\nu (z)). $$ The phase function and its derivative possess the asymptotic expansions $$ \theta _\nu (z) \sim z - \left( {\tfrac{1}{2}\nu + \tfrac{1}{4}} \right)\pi + \frac{{4\nu ^2 - 1}}{{2(4z)}} + \frac{{(4\nu ^2 - 1)(4\nu ^2 - 25)}}{{6(4z)^3 }} + \ldots $$ and $$ z\,\theta '_\nu (z) \sim z - \frac{{(4\nu ^2 - 1)}}{{2 (4z) }} - \frac{{(4\nu ^2 - 1)(4\nu ^2 - 25)}}{{2(4z)^3 }} + \ldots , $$ as $z\to\infty$ in the sector $|\arg z|\le \pi-\delta<\pi$, provided $\nu=o(z)$.