- #1
gametheory
- 6
- 0
Homework Statement
Solve equations 1) and 2) for J[itex]_{p+1}[/itex](x) and J[itex]_{p-1}[/itex](x). Add and subtract these two equations to get 3) and 4).
Homework Equations
1) [itex]\frac{d}{dx}[/itex][x[itex]^{p}[/itex]J[itex]_{p}[/itex](x)] = x[itex]^{p}[/itex]J[itex]_{p-1}[/itex](x)
2) [itex]\frac{d}{dx}[/itex][x[itex]^{-p}[/itex]J[itex]_{p}[/itex](x)] = -x[itex]^{-p}[/itex]J[itex]_{p+1}[/itex](x)
3) J[itex]_{p-1}[/itex](x) + J[itex]_{p+1}[/itex](x) = [itex]\frac{2p}{x}[/itex]J[itex]_{p}[/itex](x)
4) J[itex]_{p-1}[/itex](x) - J[itex]_{p+1}[/itex](x) = 2J[itex]^{'}_{p}[/itex](x)
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[itex]_{p}[/itex] on the left of each equation or am I supposed to simply solve equation 1) as J[itex]_{p-1}[/itex](x) = x[itex]^{-p}[/itex][itex]\frac{d}{dx}[/itex][x[itex]^{p}[/itex]J[itex]_{p}[/itex](x)] and equation 2) as J[itex]_{p+1}[/itex](x) = -x[itex]^{p}[/itex][itex]\frac{d}{dx}[/itex][x[itex]^{-p}[/itex]J[itex]_{p}[/itex](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[itex]_{p+1}[/itex] into J[itex]_{p}[/itex] and I integrated on both sides and just got J[itex]_{p+1}[/itex] = J[itex]_{p+1}[/itex] 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.