Question

Can someone help me with this power series expansion?

Answer 1

I am trying to evaluate this integral
\[\Phi=-2R\mu G\int^\pi_0\frac{d\theta}{\left(r^2+R^2-2Rr\cos\theta\right)^{1/2}}\]

Answer 2

letting \(x=R/r\) I have\[\Phi=-2R\mu x\int\frac{d\theta}{\left(1+x^2-2x\cos\theta\right)^{1/2}}\]

Answer 3

I guess I'm waiting for the power series to come into play.

Answer 4

now my textbook wants me to expand this is a power series making certain to keep all terms of order \(x^2\). How do I do this? the next step shows\[\Phi=-2\mu Gx\int^{\pi}_0\left[\left(1-\frac{x^2}{2}+x\cos\theta\right)+ \frac{3}{8}\left(x^2-2x\cos\theta\right)^2+\dots\right]d\theta\]

Answer 5

I get lost at this step

Answer 6

Wait what? I'm lost too. So it's trying to expand this as a function of x or theta? I was hoping you'd be able to use a trig identity to get rid of the bottom.

Maybe if I had more context, what is this? You're trying to do something in physics so perhaps it's an approximation to something that I could understand better, but it looks like the law of cosines essentially.

Answer 7

I am trying to find the potential of a field of a thin ring.

Answer 8

just uploading the page

Answer 9

Alright thanks, this looks fun overall, hopefully we can figure it out haha.

Answer 10

so I understand what happened up to the power series expansion. I am unsure of what math happened there. The rest is straightforwards as well

Answer 11

I guess it comes down to me really just not wanting to take the derivative of these functions to see if this is truly just a substitution of the power series.

Answer 12

Probably should just ask your professor about this, since this seems fairly odd to just sort of bust out the power series like this almost magically. I assume that this is really just a power series of (1+x^2-2xcos(theta))^(-1/2) and when you use the far-field approximation that essentially means that x^3 and higher terms become negligible, so we get rid of them to make the approximation simpler.

Answer 13

alright well thanks for trying!