Proving Sigma^infty_{n=1} r^n*cosn(t)

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The discussion centers on proving the equation Σn=1 rncos(n(t)) = (r cos(t) - r2)/(1 - 2r cos(t) + r2) for 0 < r < 1. The user is guided to express z = re^(it) in the context of the summation formula Σn=0 zn = 1/(1-z). The discussion emphasizes separating real and imaginary parts to derive the final result, highlighting the importance of correctly applying the theorem regarding summation of complex series.

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  • Complex numbers and Euler's formula (e^(it) = cos(t) + isin(t))
  • Understanding of infinite series and convergence criteria
  • Familiarity with summation formulas and their applications
  • Basic knowledge of trigonometric identities
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  • Study the derivation of the summation formula for complex series
  • Learn about separating real and imaginary parts in complex summations
  • Explore the convergence of series involving trigonometric functions
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How to prove Sigma^infty_{n=1} r^n*cosn(t) = (r cos (t) - r^2)/(1-2r*cos(t) + r^2) ??

Homework Statement


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 question 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 don't 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?

Homework 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

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
 
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laura_a said:

Homework Statement


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 question 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 don't 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?

Homework 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

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

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
 
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.

Thanks for your help.
Laura
 

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