i'm trying to take the 2nd derivative of a spherical bessel function. So far i've found these recursion relations that are useful from:
$(1) j'_{n}(x) = -j_{n+1}(x) + \frac{n}{x}j_{n}(x)$
$(2) j'_{n}(x) = j_{n-1}(x) - \frac{n+1}{x}j_{n}(x)$
What i've attempted is something like this:
So starting with (1): $j'_{n}(x) = -j_{n+1}(x) + \frac{n}{x}j_{n}(x)$
I'm wondering if you can say : $j''_{n}(x) = -j'_{n+1}(x) +D[\frac{n}{x}j_{n}(x)]$ where $D[\frac{n}{x}j_{n}(x)]$ is some product rule differentiation that i'm not sure how you would apply.
If this is valid I'm also wondering if you can use (2) on the $-j'_{n+1}(x)$ term so that:
$-j'_{n+1}(x) = j_{(n+1)-1}(x) - \frac{(n+1)+1}{x}j_{n+1}(x)= j_{n}(x)-\frac{n+2}{x}j_{n+1}(x)$
and would the $D[\frac{n}{x}j_{n}(x)]$ term just be a simple product so that :
$D[\frac{n}{x}j_{n}(x)] = \frac{n}{x}j'_{n}(x) + \frac{n}{-x^2}j_{n}(x)$
Yes, you can differentiate the recursion equations. Alternatively, you know from the original equation that
$$x^2 j_n^{\prime\prime} + 2 x j_n^\prime + \left[ x^2-n(n+1) \right] j_n = 0$$
So, use this together with your recursion equation for $j_n^\prime$.