Prove sum identities: r</=>1, θ, e^iθ

[DieCommie]

Homework Statement

Prove the following identities:
\sum_{n=0}^{\infty} r^n \cos(n\theta) = \frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}

\sum_{n=0}^{\infty} r^n \sin(n\theta) = \frac{r\sin(\theta)}{1-2r\cos(\theta)+r^2}

Homework Equations

\cos(n\theta) = \frac{1}{2}(e^(in\theta)+e^(-in\theta))

The Attempt at a Solution

I get it to this point...

\frac{1}{2}( \sum_{n=0}^{\infty}(re^(i\theta))^n+\sum_{n=0}^{\infty}(re^-(i\theta))^n

But I don't know what to do next! r could be less than or greater than one. I need to do the general case where r could be either...

Also, I am not sure that the 'relevent equation' is really relevent

Any help/tips would be greatly appreciated, Thx!

 

[Dick]

You are now looking at a pair of geometric series. The next step should be easy. I wouldn't worry about the case r>1. In general it should diverge except for special angles.

 

[MaGG]

Wouldn't you just use the following rule for an infinite geometric series:

\sum\limits_{k = 1}^\infty {ar^{k} = \frac{a}{{1 - r}}}

 

[DieCommie]

Yes, but only if R<1. The problem conspicuously leaves out that stipulation, which is where my confusion set it.

I went ahead and solved it assuming R<1 and ignoring the other case... I hope that's good enough for some credit.

 

[MaGG]

Did you take r=re^{i\theta} and a=1? Now you've gotten me wanting to solve this problem, lol...

 

[DieCommie]

I took re^{i\theta} = r(cos\theta + isin\theta) and a=1. I have it completed assuming r<1. We will look at later.

 

[Dick]

If r=1 then the series is going to be, at best, only conditionally convergent (r>1, not even that except for some silly special cases like theta=pi). I would really only worry about r inside of the radius of convergence of the series. You can't 'prove the identities' if the sum of the series doesn't exist.