Using complex exponentials to prove 1+acostheta

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Homework Help Overview

The discussion revolves around proving a series involving complex exponentials and trigonometric functions, specifically the expression 1 + a cos(θ) + a² cos(2θ) + a³ cos(3θ)... and its equivalence to (1 - a cos(θ))/(1 - 2a cos(θ) + a²). The subject area includes complex analysis and trigonometry.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express the series using complex exponentials and has made some progress but encounters difficulties in setting up the summation formula. Some participants question the condition |a| < 1, which is confirmed to be true. Others explore the relationship between the terms in the series and their complex representations.

Discussion Status

Participants are actively engaging with the problem, discussing the implications of using complex exponentials and confirming assumptions about the variable a. There is a recognition that taking the real part of the sum may be necessary, and some guidance has been offered regarding the nature of the results.

Contextual Notes

There is an assumption that |a| < 1, which is critical for the convergence of the series being discussed. The original poster acknowledges a typo in their initial post, indicating an ongoing refinement of their approach.

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Homework Statement


Use complex exponentials to prove 1 + acos(theta) + a^2cos(2theta) + a^3cos(3theta)... = (1 - acos(theta))/(1 - 2acos(theta) + a^2)


Homework Equations



euler's e^itheta/2 +e^-itheta/2=2cos(2theta)

The Attempt at a Solution



a^(n)cos(ntheta) = e^nitheta = e^-nitheta

from there i got the series

(a^n(e^itheta)^n)/2

now from here I think I setup the summation formula but this is where I get stuck. Any help is greatly apprecited.
 
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Is |a| < 1?
 
Yes, a is a real constant and |a| < 1. sorry about that
 
<br /> a^n \cos n\theta = a^n\frac {e^{in\theta} + e^{-in\theta}} {2}<br /> = \frac {a^ne^{in\theta} + a^ne^{-in\theta}} {2}<br /> = \frac {p^n + q^n} {2}<br /> \\ p = (\ln a)e^{i\theta}, \ |p| &lt; 1<br /> \\ q = (\ln a)e^{-i\theta}, \ |q| &lt; 1<br />

What is the sum of p^n and q^n?
 
That makes sense. That will then give me a real and an imaginary result of which I take the real I believe. Also, I apologize for the typo on the first post.
 
Well, you can take the real part, but the sum is real anyway.
 

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