Power series, why did this stay constant?

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.

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.

d_leet

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

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.

cristo

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?

rocomath

Sorry for the confusion. I'm confused about this part:

[tex]2^{n+1}[/tex]

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!

cristo

Ok, initially there's no reason to set the sum from 1 to infinity in the derivative, so we have [tex]\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}[/tex]

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).