# Proving Recursion relations for Bessel Functions

1. May 5, 2012

### gametheory

1. The problem statement, all variables and given/known data
Solve equations 1) and 2) for J$_{p+1}$(x) and J$_{p-1}$(x). Add and subtract these two equations to get 3) and 4).

2. Relevant equations
1) $\frac{d}{dx}$[x$^{p}$J$_{p}$(x)] = x$^{p}$J$_{p-1}$(x)
2) $\frac{d}{dx}$[x$^{-p}$J$_{p}$(x)] = -x$^{-p}$J$_{p+1}$(x)
3) J$_{p-1}$(x) + J$_{p+1}$(x) = $\frac{2p}{x}$J$_{p}$(x)
4) J$_{p-1}$(x) - J$_{p+1}$(x) = 2J$^{'}_{p}$(x)

3. The attempt at a solution
My main problem is I'm not really sure what the question is asking me to do in the first part. Am I supposed to plug p+1 and p-1 into J$_{p}$ on the left of each equation or am I supposed to simply solve equation 1) as J$_{p-1}$(x) = x$^{-p}$$\frac{d}{dx}$[x$^{p}$J$_{p}$(x)] and equation 2) as J$_{p+1}$(x) = -x$^{p}$$\frac{d}{dx}$[x$^{-p}$J$_{p}$(x)]? I tried this way and then differentiated the series and got two infinite series I didn't know what to do with.

Next, I tried to substitute J$_{p+1}$ into J$_{p}$ and I integrated on both sides and just got J$_{p+1}$ = J$_{p+1}$ after rearranging everything.

I feel like this isn't an overly difficult problem, but I just have no idea what direction to take with it.

2. May 5, 2012

### Steely Dan

Yes, it's saying the latter, but I don't know what you mean about getting infinite series; if you expand the derivative on the RHS in each case, you can just use the product rule to get one term with the Bessel function and one term with its derivative (and adding these or subtracting should give the two results).

3. May 5, 2012

### gametheory

Nevermind, you're right, I got it. All the other problems had us manipulate the bessel function as an infinite series and I was just used to doing this. Forgot to try basic rules from calculus...haha thanks