Bessel Recurrence relations

Question

Really short question: I am following the steps laid out by Arfken and Brown in Mathematical Methods for Physicists to derive the first two recurrence formulae for Bessel Functions. Using the generating function $$ g(x,t) = \exp\left[\frac{x}{2}(t-t^{-1})\right] = \sum_n J_n(x) t^n $$ I understand that I have to get the t and x derivatives and then compare the powers of t. But when I do the x derivative, I end up with the recurring relation: $$ t \sum J_n t^n - t^{-1} \sum J_n t^n = 2 \sum J'_n t^n\\ \implies J_{n+1} - J_{n-1} = 2\, J'_n $$ The book (and elsewhere on the internet) the left side is flipped, i.e.: $$ J_{n-1} - J_{n+1} = 2\, J'_n. $$ So, what's wrong here?

Answer 1

You have derived $$ \sum J_n t^{n+1} - \sum J_n t^{n-1} = 2 \sum J'_n t^n $$ (with appropriate indices and special cases on the boundary). If we just glance at that equation, we might think it's telling us that $J_{n+1}-J_{n-1}=2J'_n$. However, when we have two equal power series, what we should do is look at the coefficients of $t^n$ on both sides and set them equal to each other.

The coefficient of $t^n$ in $\sum J_n t^{n+1}$ is equal to $J_{n-1}$. (Do you agree?) The coefficient of $t^n$ in $\sum J_n t^{n-1}$ is equal to $J_{n+1}$. Therefore the coefficients of $t^n$ on the two sides are equal: $$ J_{n-1} - J_{n+1} = 2J'_n. $$