# Power series, why did this stay constant?

1. Dec 2, 2007

### rocomath

so we took the derivative of the series, which i understand. but why is it that when the new n position was changed, why didn't the 2n+1 change as well? and i know that the n position changed b/c if it rained 0, we would have had x to the -1 as one of our terms. but i'm confused why 2n+1 is constant.

2. Dec 2, 2007

### siddharth

Do you mean the range of summation? It doesn't matter if you start the sum from 0 or 1, because for n=0 the first term of the series on the right hand side is 0. Note that the derivative for the first term n=0 on the left hand side is d/dx(x^0) = 0.

Last edited: Dec 2, 2007
3. Dec 2, 2007

### d_leet

The first term is actually 1/2, but the point remains that this is a constant which has zero for a derivative.

4. Dec 2, 2007

### Gib Z

I think he meant that he was confused why the other n terms changed whilst the denominator did not.

Think carefully about what you actually took respect to, and what variables are held constant.

5. Dec 2, 2007

### cristo

Staff Emeritus
I don't think he's asking that, since the answer would simply be "you're differentiating wrt x, not n." I suspect his question has been answered by the above posts.

If not, then rocophysics, could you clarify what your questions actually is?

6. Dec 2, 2007

### rocomath

$$2^{n+1}$$

How come it did not become 2^(n+2) or something like that after n went from 0 to 1. Is it because we took the derivative wrt to x, and not n? If so, then my question was answered, I was just so confused.

Thanks!

Last edited: Dec 2, 2007
7. Dec 2, 2007

### cristo

Staff Emeritus
Ok, initially there's no reason to set the sum from 1 to infinity in the derivative, so we have $$\frac{d}{dx}\left(\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}x^n\right)=\sum_{n=0}^{\infty}\frac{n}{2^{n+1}}x^{n-1}$$

But, if you plug n=0 into the right hand side, you obtain 0 for all x, hence we can write the sum without the first term.

(This is pretty much what d_leet said above).