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

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In summary: She then realizes that the summation formula includes a sum, and not just a single term, and uses the given theorem to split up the real and imaginary parts. This helps her to simplify the expression and ultimately arrive at the desired equation of Sigma^infty_{n=1} r^n*cosn(t) = (r cos (t) - r^2)/(1-2r*cos(t) + r^2) for 0 < r < 1. She also notes that a similar process can be used to solve for sin as well.
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laura_a
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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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  • #2
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
 
  • #3
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
 

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

What is the formula for "Sigma^infty_{n=1} r^n*cosn(t)"?

The formula is a geometric series: r*cos(t) + r^2*cos(2t) + r^3*cos(3t) + ... + r^n*cos(nt), where n ranges from 1 to infinity.

How do you calculate the value of "Sigma^infty_{n=1} r^n*cosn(t)"?

To calculate the value, you can use the formula for the sum of a geometric series: a / (1 - r), where a is the first term (r*cos(t)) and r is the common ratio (r*cos(t)).

What is the convergence of "Sigma^infty_{n=1} r^n*cosn(t)"?

The series converges if the absolute value of the common ratio (r*cos(t)) is less than 1. In other words, if |r*cos(t)| < 1, then the series converges. Otherwise, it diverges.

Can "Sigma^infty_{n=1} r^n*cosn(t)" be used to approximate other functions?

Yes, the sum of a geometric series can be used to approximate other functions. However, the accuracy of the approximation depends on the value of r*cos(t) and the number of terms used in the series.

What are some real-life applications of "Sigma^infty_{n=1} r^n*cosn(t)"?

Some real-life applications include modeling the oscillation of a pendulum, analyzing the motion of a spring, and studying the behavior of various electrical circuits. The series can also be used in signal processing and image compression algorithms.

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