In (1) there is a property of spherical Bessel functions, which's derivation I can not find in the literature.

${\mathsf{j}_{n}^{2}}\left(z\right)+{\mathsf{y}_{n}^{2}}\left(z\right)=\sum_{k=% 0}^{n}\frac{s_{k}(n+\frac{1}{2})}{z^{2k+2}},$ where $s_{k}(n+\tfrac{1}{2})=\frac{(2k)!(n+k)!}{2^{2k}(k!)^{2}(n-k)!}.$

I tried to obtain it by myself with the use of 10.49.1

$a_{k}(n+\tfrac{1}{2})=\begin{cases}\dfrac{(n+k)!}{2^{k}k!(n-k)!},& k=0,1,\dotsc% ,n,\\ 0,&k=n+1,n+2,\dotsc.\end{cases}$

and 10.49.6


but I haven't managed with sums rearrangement. Can someone tell me where to find the derivation?


From the definition 10.47.10, it follows that $$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$ So, by the expansions 10.49.6 and 10.49.7, \begin{split} \mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) &= \sum_{k=0}^n I^{k-n-1} \frac{a_k(n+\frac{1}{2})}{z^{k+1}}\cdot \sum_{l=0}^n (-I)^{l-n-1} \frac{a_l(n+\frac{1}{2})}{z^{l+1}} \\ &= \sum_{s=0}^{2n} \frac{(-I)^s}{z^{s+2}} \sum_{k=\max\{0,s-n\}}^{\min\{n,s\}} (-1)^k a_k(n+\frac{1}{2})a_{s-k}(n+\frac{1}{2}). \end{split} From the definition 10.49.1, the inner term can be restated as $$(-1)^k\frac{s!}{2^s} \binom{s}{k}\binom{n+k}{s}\binom{n+s-k}{s}$$ and it naturally nullifies when $k$ is outside the summation range. Furthermore, if $s$ is odd then the terms for $k=k'$ and $k=s-k'$ cancel each other. So, we can set $s=2t$ and obtain $$ \mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = \sum_{t=0}^{n} \frac{(-1)^t (2t)!}{z^{2t+2}2^{2t}} \sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t}. $$

It remains to show that $$\sum_{k\geq 0} (-1)^k \binom{2t}{k}\binom{n+k}{2t}\binom{n+2t-k}{2t} = (-1)^t\frac{(n+t)!}{t!^2(n-t)!},$$ which at very least can be done with the WZ method. But perhaps it's just a consequence from something well-known.

P.S. This identity has a neat representation in terms of hypergeometric functions, which I posted in a follow-up question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.