# Derivative of a Power Series

1. Apr 8, 2012

### stupidmonkey

1. The problem statement, all variables and given/known data
http://imgur.com/FJhgN
Give this power series J(x) (leftmost in the picture), find the first and second derivative.

2. Relevant equations
You take the derivative of a power series term by term.

3. The attempt at a solution
I don't understand how to get the J''(x) in the image, which is the solution that was given. I thought that when you take the derivative of J'(x) you start from n=1 in J'(x), but then start from n=2 in J''(x). I can get the term inside the summation though.
This question is part of a problem that asks you to prove (x^2)J(x)+xJ(x)+(x^2)J(x)=0 and the proof will only work if J''(x) starts from n=1 so I don't think the answer is a typo.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 8, 2012

### Office_Shredder

Staff Emeritus
It looks like it's just re-indexing the series. Consider the following:
$$\sum_{n=1}^{\infty} x^n$$
and
$$\sum_{n=2}^{\infty} x^{n-1}$$

These are exactly the same, we just changed where the dummy variable starts, and the terms in the series had to shift to compensate

3. Apr 8, 2012

### stupidmonkey

Hmmm but I thought when you took the derivative of
Ʃ(n=1 to infinity) ((-1)^n)(x^(2n-1))(2n)/((2^(2n))(((m+1)!)^2) you get

Ʃ(n=2 to infinity) ((-1)^n)(x^(2n-2))(2n)(2n-1)/((2^(2n))((n!)^2))

Then if you change the index to n=1 you get:

Ʃ(n=1 to infinity) (((-1)^(n+1)))(2(n+1)(2n+1)(x^(2n))/((2^(2m+2)((m+1)!)^2)
which is different...
Sorry for the ugly complicated symbols, I don't know how to make it look more clean...
Btw how do you make such a large sigma symbol?

4. Apr 9, 2012

### SithsNGiggles

Sorry for not addressing the problem, but I'm a bit rusty with power series.

In the "Reply to Thread" window (the "advanced" one with formatting toolbar), on the far right of the toolbar is the sigma symbol. Clicking that will open a sort of index of symbols/formatting/etc. in LaTeX.

You can search for LaTeX commands through Google as well. I'm sure there's also a thread about this somewhere. . .
https://www.physicsforums.com/showthread.php?t=8997

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