# How to prove Sigma^infty_{n=1} r^n*cosn(t) = (r cos (t) - r^2)/(1-2r*cos(t) + r^2) ?

1. May 10, 2007

### laura_a

How to prove Sigma^infty_{n=1} r^n*cosn(t) = (r cos (t) - r^2)/(1-2r*cos(t) + r^2) ??

1. The problem statement, all variables and given/known data
This question has two parts and I need help with the first part (I think) because then I should be able to make a start on the second part. The exact quesiton is (where t = theta)

"Write z=re^(it), where 0 < r < 1, in the summation formula (which it says was derived in the example on the previous page, I'll put that formula below) and then with the aid of the theorem (4) which i'll also put below show that

Sigma^infty_{n=1} r^n*cosn(t) = (r cos (t) - r^2)/(1-2r*cos(t) + r^2)

when 0 < r < 1

And a similar one for sin as well, but I dont need to be spoon fed everything, I can work it out just need help on the first part with the z=re^(it) because I am sure that is going to help me with the cos thing?

2. Relevant equations

Summation formula

Sigma^{infty}_{n=0} z^n = (1/(1-z))

Theorem (4)

I think it means

Sigma^{infty}_{n=0} (x_n + iy_n) = Sigma^{infty}_{n=0} x_n + i Sigma^{infty}_{n=0} y_n

3. The attempt at a solution

The first part of the question asks to write z=re^(it) into the summation formula

Sigma^{infty}_{n=0} z^n = (1/(1-z))

so I fugure all I have to do is sub it in.. but I think I need to use the sub that e^(it) = cos(t) + isin(t)

So then I have z = r(cos(t) + isin(t))

so according to summation formula

z^n = (r(cos(t) + isin(t)))^n = 1/(1-r(cos(t) + isin(t))

And the other formula looks as though I can split those two up into real and imaginary, but as you may be able to tell I'm not really sure about this topic at all, can anyone help me get any further? And maybe let me know if I'm even on the right planet on the above attempt.

Thanks heaps

2. May 10, 2007

### HallsofIvy

Staff Emeritus
No, the summation formula does NOT say that! It says that
$$\sum_{n=0}^\infty z^n= \frac{1}{1- r(cos t+ i sin t)}$$
don't forget the sum on the left! Of course, you should know that $z^n= r^n (cos nt+ i sin nt)$.

3. May 10, 2007

### laura_a

Thanks for that, In the answer its has r^2's and 2r. I just don't see how I can get that, What is the next step. Is it something to do with seperating the real and imaginary parts? And do I have r^n(cosnt + isinnt) in the summation formula? There is no "n" in the answer which is one of the reasons why I'm totally confused.