Convergent Sum of Sine Series with Added Term for Continuity

[Hells_Kitchen]

Find the convergent sum and find the sum of first five terms
<br /> \sum_{n=1}^{\infty} \frac{sin(nx)}{2^nn}<br /> from 1 to infinity.

I have found so far that:
<br /> \sum_{n=1}^{\infty} \frac{sin(nx)}{2^n} = \frac{2sin(x)}{5-4cos(x)}<br /> I am not sure how to consider the \frac{1}{n} term.

Can someone please help?

 

[HallsofIvy]

What exactly does "find the convergent sum" mean? If you mean just "find the sum", it would seem very strange that you are also asked to find the sum of the first 5 terms. You can show that the series does converge by the "comparison test".

 

[Hells_Kitchen]

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
<br /> <br /> \sum_{n=1}^{\infty} \frac{sin(nx)}{2^n} = \frac{2sin(x)}{5-4cos(x)}<br /> <br />

 

[Hells_Kitchen]

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 how the overall sum compares to the sum of the first five terms. I am guessing that after five terms there should not be very much change because of the 1/n term and 2^n term for large n the f(nx) should be approximately zero.

What do you think?

 

[Hells_Kitchen]

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!

 

[SammyS]

The 1/n factor you left out of each term will make all the difference in the world.

I suggest expressing sin(nx) as a Taylor series, using perhaps k for an index, and switching the order of summation.

 

[Hells_Kitchen]

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 considerf(x)=\frac{sin(nx)}{2^n}. Finding the [] then would solve the problem ...

 

[SammyS]

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 considerf(x)=\frac{2sin(x)}{5-4cos(x)}. Finding the [] then would solve the problem ...

Sure, I can expand that. In fact I was in the process of doing that when a new idea occurred to me.

New idea:
Evaluate S'(x) in a manner similar to what you did in the pdf in the link you posted. Taking the derivative w.r.t x will get rid of the 1/n.

You get a sum similar to the one you evaluated except it has cos(nx) rather than sin(nx).

(I'll continue to LaTeX-ify details for my previous suggestion - just in case, but I like the idea in this post much better.)

 

[SammyS]

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 considerf(x)=\frac{sin(nx)}{2^n}. Finding the [] then would solve the problem ...

Expanded version:

\sum_{n=1}^{\infty} \frac{sin(nx)}{2^nn} = \sum_{n=1}^{\infty}\left(\frac{1}{n\,2^n}\,\sum_{k=0}^{\infty}\frac{(-1)^k\,n^{2k+1}}{(2k+1)!}\,x^{2k+1}\right)

=\sum_{k=0}^{\infty}\left(\frac{(-1)^k\,x^{2k+1}}{(2k+1)!}\, \sum_{n=1}^{\infty}\frac{n^{2k+1}}{n\,2^n}\,\right)

=\sum_{k=0}^{\infty}\left(\frac{(-1)^k\,x^{2k+1}}{(2k+1)!}\, \sum_{n=1}^{\infty}\frac{n^{2k}}{2^n}\,\right)​

But, I don't think the sum over n is all that easy to evaluate..

 

[Hells_Kitchen]

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 suggestions how to handle this animal to solve it analytically?

Anyway thank's again for the great suggestion!

 

[Hells_Kitchen]

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]

Looks very nice.

Looking at the initial form for S(x) (in your Original Post), it's clear that S(0) = 0, so the constant of integration is zero and all is well!

 

[SammyS]

OPPS! There is a problem with the solution: S(x)=\tan^{-1}(3\tan\frac{x}{2})-\frac{x}{2}

When x is an odd integer multiple of π, S(x) is discontinuous because of arctan(3tan(x/2)), which is periodic and has jump discontinuities at odd multiples of π.

This can be fixed by replacing the trailing x/2 with arctan(tan(x/2)), since it is being subtracted and it has exactly the same jump discontinuities.

If you look at the graph of S'(x) it's clear that its integral should be continuous and periodic.

 

[SammyS]

It's too late to edit the previous post.

To make S(x) continuous at x = (2n+1)π, for n an integer, define:
S(x)=0, if x=(2n+1)π,
S(x)=tan^-1(3tan(x/2)) - tan^-1(tan(x/2)), otherwise.