# Trying to prove an identity

1. Dec 5, 2003

### nille40

Hi everyone!
I'm trying to prove an identity, and it's driving me insane.

Let $$J_p(x) = \sum_{n=0}^{\infty} \left(-1\right)^n\frac{x^{2n+p}}{2^{2n+p}n!(n+p)!}$$

Show that
$$\frac{d}{dx}(x^{-p}J_p(x)) = x^{-p}J_{p+1}(x)$$

I get the left part to
$$\sum_{n=0}^\infty(-1)^n\frac{x^{2n-1}}{2^{2n+p-1}(n-1)!(n+p)!}$$

And the right part to
$$\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2^{2n+p+1}n!(n+p+1)!}$$

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

2. Dec 5, 2003

### HallsofIvy

Staff Emeritus
You might notice that the lower index on
$$\sum_{n=0}^\infty(-1)^n\frac{x^{2n-1}}{2^{2n+p-1}(n-1)!(n+p)!}$$
is incorrect. The first term will be n= 1, not n=0.

Try changing the index from n to j= n-1 so that the sum will be
from j=0 to infinity.

If j= n-1 then n= j+1 so, for example, the power of x, 2n-1, becomes 2(j+1)-1= 2j+1. Make that change for every n in the formula.

Of course, since n (or j) is a "dummy" variable, you could then just replace "j" with "n" or change the n in the second sum you have with j in order to compare them.

3. Dec 8, 2003

### nille40

That's it!!
I had a hard time realizing that that you could just change the index. But I realize that "infinity - 1" is, obviously, infity. Hince the delay...

Thanks for helping me!
Nille

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