I'm looking for the asymptotic approximation to the spherical Bessel function of the first kind $j_n (\rho)$ for $\rho\gg n^2$. This is mainly to speed up calculation of a light propagation problem. There are examples for spherical Hankel functions but I've not been able to find an equivalent for $j_n (\rho)$. Does anyone know of any examples of this?
[Math] Asymptotic Approximation of Spherical Bessel Function for Large Arguments
asymptoticsbessel functions
Related Solutions
It is $$ j_n (x) = \sqrt {\frac{\pi }{{2x}}} J_{n + \frac{1}{2}} (x) \sim \frac{1}{{\sqrt {2x(2n + 1)} }}\left( {\frac{{ex}}{{2n + 1}}} \right)^{n + \frac{1}{2}} \sim \frac{1}{{2\sqrt 2 n}}\left( {\frac{{ex}}{{2n}}} \right)^n, $$ as $n\to +\infty$ (cf. http://dlmf.nist.gov/10.19.E1). You can obtain this by noting that for large $n$, the Bessel function is controlled by the leading term of its Taylor series and using Stirling's formula for the gamma function to simplify the result.
I was interested by the transcendental equation. Letting $x=ka$, for stability reasons, I prefered to change its writng and consider that we look for the zero of function $$f(x)=\left(1-x^2\right) \sin (x)-x \cos (x)$$ Since it is odd, discarding the trivial solution $x=0$, focus on the positive roots which are closer and closer to multiples of $\pi$.
Written as series around $x=k \pi$, $$f(x)=-\pi k-\pi ^2 k^2 t-\frac{3\pi k}{2} t^2+\frac{\pi ^2 k^2-4}{6} t^3+\frac{7\pi k}{24} t^4+\frac{16-\pi ^2 k^2}{120} t^5-\frac{11\pi k}{720} t^6+O\left(t^{7}\right)$$ where $t=x-k \pi$. Now, using series reversion $$x_{(k)}=\pi k-\frac{1}{\pi k}-\frac{5}{3 \pi ^3 k^3}-\frac{73}{15 \pi ^5 k^5}-\frac{1826}{105 \pi ^7 k^7}+O\left(\frac{1}{k^9}\right)$$ which seems to work quite well.
$$\left( \begin{array}{ccc} k & \text{estimation} & \text{solution} \\ 1 & 2.747869217 & 2.743707270 \\ 2 & 6.116769339 & 6.116764264 \\ 3 & 9.316615752 & 9.316615629 \\ 4 & 12.48593738 & 12.48593737 \\ 5 & 15.64386611 & 15.64386611 \end{array} \right)$$
Edit
Considering that we look for the first non trivial zero of function $$f(x)=j_n(x)-\frac{x }{n}\,j_{n-1}(x)$$ we have $$f'(x)=\sqrt{\frac{\pi }{2}}\,\,\,\frac{x^2-n(n+1) }{n \,x^{3/2}}J_{n+\frac{1}{2}}(x)$$ $$f''(x)=\sqrt{\frac{\pi }{2}}\,\,\,\frac{(n+1) \left(x^2-n(n-1) \right)J_{n+\frac{1}{2}}(x)-x \left(x^2-n (n+1)\right)J_{n+\frac{3}{2}}(x) }{n \,x^{5/2} }$$ The first derivative cancels at $$x_*=\sqrt{n(n+1)}$$ To get an estimate, expand as series to get $$x_0=x_*+\sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$ that is to say $$x_0=\sqrt{n(n+1)} +\sqrt{\sqrt{n(n+1)}\,\,\frac{ j_{n-1}\left(\sqrt{n(n+1)} \right)}{j_n\left(\sqrt{n(n+1)} \right)}-n}$$ Iterations of Newton method $$x_{k+1}=x_k+\frac{x_k \left(x_k J_{n-\frac{1}{2}}(x_k)-n J_{n+\frac{1}{2}}(x_k)\right)}{\left(x_k^2-n(n+1)\right) J_{n+\frac{1}{2}}(x_k)}$$
A few values for comparison $$\left( \begin{array}{cccc} n & x_0 & x_1 & x_2 & \text{solution} \\ 1 & 2.6691616 & 2.7445999 & 2.7437074 & 2.7437074 \\ 2 & 3.8798178 & 3.8702541 & 3.8702386 & 3.8702386 \\ 3 & 5.0331291 & 4.9740292 & 4.9734204 & 4.9734204 \\ 4 & 6.1579411 & 6.0635123 & 6.0619498 & 6.0619498 \\ 5 & 7.2649512 & 7.1428326 & 7.1402287 & 7.1402288 \\ 6 & 8.3595285 & 8.2144886 & 8.2108446 & 8.2108444 \\ 7 & 9.4448169 & 9.2801189 & 9.2754678 & 9.2754680 \\ 8 & 10.522842 & 10.340863 & 10.335248 & 10.335242 \\ 9 & 11.594999 & 11.397549 & 11.391017 & 11.391017 \\ 10 & 12.662292 & 12.450800 & 12.443395 & 12.443395 \\ 20 & 23.176232 & 22.866090 & 22.851803 & 22.851767 \\ 30 & 33.541865 & 33.167335 & 33.148172 & 33.148111 \\ 40 & 43.834852 & 43.410759 & 43.387766 & 43.387766 \\ 50 & 54.083230 & 53.618117 & 53.591940 & 53.592598 \\ 60 & 64.300908 & 63.800426 & 63.771501 & 63.771384 \\ 70 & 74.495928 & 73.964130 & 73.932777 & 73.932645 \\ 80 & 84.673409 & 84.113369 & 84.079832 & 84.079686 \\ 90 & 94.836838 & 94.250979 & 94.215451 & 94.215292 \\ 100 & 104.98871 & 104.37900 & 104.34163 & 104.34031 \\ 200 & 206.13568 & 205.34898 & 205.29815 & 205.29790 \\ 300 & 306.94240 & 306.03338 & 305.97340 & 305.97309 \\ 400 & 407.58540 & 406.57967 & 406.51253 & 406.51216 \\ 500 & 508.12878 & 507.04171 & 506.96858 & 506.96818 \\ 600 & 608.60398 & 607.44601 & 607.36768 & 607.36725 \\ 700 & 709.02905 & 707.80782 & 707.72487 & 707.72440 \\ 800 & 809.41545 & 808.13681 & 808.04967 & 807.51428 \\ 900 & 909.77091 & 908.43955 & 908.34858 & 907.81456 \\ 1000 & 1010.1010 & 1008.7207 & 1008.6262 & 1008.0934 \end{array} \right)$$
It has been numerically checked that for any $n$ (at least for $n \leq 1000$) $$f(x_0) \times f''(x_0) > 0$$ which implies, by Darboux theorem, that the solution will always be reached without any overshoot. However, since this product decreases quite fast when $n$ increases, we can expect more iterations for large $n$ (this is quite clear from the table).
I was hoping that, starting from $x_0$, Halley or Housholder methods would be leading to better $x_1$; this is not the case, the improvement being systematically quite marginal.
However, there is a way to avoid iterations (this means "almost" exact solution). Build around $x_0$ the $[1,m]$ Padé approximant which will write $$f(x) \sim \frac {f(x_0)+a^{(m)}(x-x_0) } {1+\sum_{i=1}^m b_i^{(m)} (x-x_0)^i}$$ which is at least $O\big((x-x_0)^{m+1} \big)$; all required coefficients are defined by the function and derivatives values at $x=x_0$. This gives $$x_{(m)}=x_0-\frac{f(x_0) } {a^{(m)} }$$ This has been done for this problem with $m=1,2,3$ and the results are better and better.
Best Answer
The spherical Bessel function is defined as $j_{n}(z)=\sqrt{\tfrac{1}{2}\pi/z}J_{n+\frac{1}{2}}\left(z \right).$ To get an asymptotic approximation for $j_n,$ use the Hankel expansion http://dlmf.nist.gov/10.17.E3: $$J_{\nu}(z) \sim \left(\frac{2}{\pi z}\right)^{\frac{1}{2}}\left( \cos\omega\sum_{k=0}^{\infty} (-1)^{k}\frac{a_{2k}(\nu)}{z^{2k}}-\sin\omega\sum_{k=0}^{\infty}(-1)^{k} \frac{a_{2k+1}(\nu)}{z^{2k+1}}\right),$$ with $\omega=z-\tfrac{1}{2}\nu\pi-\tfrac{1}{4}\pi,$ and $$a_{k}(\nu)=\frac{(4\nu^{2}-1^{2})(4\nu^{2}-3^{2})\cdots(4\nu^{2}-(2k-1)^{2})}{k!8^{k}} \;\cdot$$