Can someone please suggest how to solve this problem. The above solution is not correct because apparently it assumes that the potential along the x=0 and y=0 planes is constant but it is not. It is only constant at the boundries around the circle as described above.
Here is the hint that we...
cazlab,
thanks for you suggestion but I think I HAVE TO solve it with conformal map transformations.
Can someone else please suggest a solution or a comment on this exisiting solution?
Sorry I had a mistake for the n=0 term. I took it to be 1/2 instead of 1. here is the corrected version.
The integral now becomes:
\int\frac{2\cos{x}-1}{5-4\cos{x}}dx=\tan^{-1}(3\tan\frac{x}{2})-\frac{x}{2}
Again using mathematica
SammyS,
I think your suggestion with the S'(x) was genius. Here is what I got for final result. Can you please check me?
I'm stuck now with finding an analytical solution to:
\int\frac{3}{10-8\cos{x}}dx but I solved in mathematica and it gives:
\tan^{-1}(3\tan\frac{x}{2})
Any...
If it is not of too much trouble can you please expand on your answer a little further.
Do you think there is a way to convert this into.
\sum_{n=1}^{\infty}f(x)g(x) =\sum_{n=1}^{\infty}f(x) *[]= \frac{2sin(x)}{5-4cos(x)} *[]
and consider f(x)=\frac{sin(nx)}{2^n}. Finding the [] then would...
Homework Statement
An infinite hollow conducting cylinder of unit radius is cut into four equal parts by planes x=0, y=0. The segmments in the first and third quadrant are maintained at potentials +V_{0} and -V_{0} respectively, and the segments in the second and fourth quadrant are maintained...
Please also find the attached document for full derivation so far. I am stuck with the \frac{1}{n} term. Can someone please suggest a solution or help.
Thank you so much!
HallsofIvy,
I am sorry if I was a little misleading. The problem is simply to find the sum. The second part of the problem is actually to plot the answer that we find as a function of x and the sum of the first five terms for n=1,2,3,4,5 in the interval -5<x<5.
I think the idea is to compare...
By the way the result given above by expressing \sin(nx)=\frac{\exp(inx)-\exp(-inx)}{2i} Then I applied the geometric infinite series for the
\frac{\exp(ix)}{2} to the n-th power and \frac{\exp(-ix)}{2} to the n-th power.
Doing some algebra we get that
\sum_{n=1}^{\infty}...
Find the convergent sum and find the sum of first five terms
\sum_{n=1}^{\infty} \frac{sin(nx)}{2^nn}
from 1 to infinity.
I have found so far that:
\sum_{n=1}^{\infty} \frac{sin(nx)}{2^n} = \frac{2sin(x)}{5-4cos(x)}
I am not sure how to consider the \frac{1}{n} term.
Can someone please help?
No. They're my questions, the answers of which I think are an attempt to better answer the "Why or Why Not" part of the problem.
It makes sense that the initial momentum before collision would be zero:
p_initial = mv/(1-v^2/c^2) \hat{x} + mv/(1-v^2/c^2) \hat{-x} = 0
p_final of the photon may...
1. Problem Statement What is the wavelength of each of the two photons produced when a particle pair of a positron and electron accelerated at 25GeV collide? Can the net result of this annihilation be only ONE single photon? Why or why Not?
Homework Equations
Etotal(initial) =...
I understand! r_{\perp} = rsin{\theta} Then you get \int^{\pi}_{0} sin^3{\theta} d\theta ,
which is just \frac{4}{3} using sin^3{\theta} = sin{\theta}(1-cos^2{\theta}) identity and it works out to be I = \frac{2}{5} MR^2. Thanks!