- #1
nille40
- 34
- 0
Hi everyone!
I'm trying to prove an identity, and it's driving me insane.
Let [tex]J_p(x) = \sum_{n=0}^{\infty} \left(-1\right)^n\frac{x^{2n+p}}{2^{2n+p}n!(n+p)!}[/tex]
Show that
[tex]\frac{d}{dx}(x^{-p}J_p(x)) = x^{-p}J_{p+1}(x)[/tex]
I get the left part to
[tex]\sum_{n=0}^\infty(-1)^n\frac{x^{2n-1}}{2^{2n+p-1}(n-1)!(n+p)!}[/tex]
And the right part to
[tex]\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2^{2n+p+1}n!(n+p+1)!}[/tex]
This is incorrect, since they are not equal. Please, please help me! I can post my calculations if that would help.
Thanks in advance,
Nille
I'm trying to prove an identity, and it's driving me insane.
Let [tex]J_p(x) = \sum_{n=0}^{\infty} \left(-1\right)^n\frac{x^{2n+p}}{2^{2n+p}n!(n+p)!}[/tex]
Show that
[tex]\frac{d}{dx}(x^{-p}J_p(x)) = x^{-p}J_{p+1}(x)[/tex]
I get the left part to
[tex]\sum_{n=0}^\infty(-1)^n\frac{x^{2n-1}}{2^{2n+p-1}(n-1)!(n+p)!}[/tex]
And the right part to
[tex]\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2^{2n+p+1}n!(n+p+1)!}[/tex]
This is incorrect, since they are not equal. Please, please help me! I can post my calculations if that would help.
Thanks in advance,
Nille