Computing a Geometric Power Series with Cosine Terms

Click For Summary

Homework Help Overview

The problem involves computing the infinite series \(\sum_{n=0}^{\infty} p^n \cos(3nx)\) under the condition that \(|p| < 1\), where \(p\) is a real number. The context is centered around series convergence and manipulation of trigonometric functions within power series.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster considers whether the series could be approached as a telescoping series but expresses uncertainty about its nature. Another participant suggests rearranging the cosine term using exponential forms to facilitate the summation, questioning the validity of this approach.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the series. Some guidance has been offered regarding the use of exponential forms, but there is no explicit consensus on the best approach yet.

Contextual Notes

There is a mention of the condition \(|p| < 1\) which is crucial for the convergence of the series, and the original poster reflects on the nature of the series, indicating some confusion about its classification.

quasar_4
Messages
273
Reaction score
0

Homework Statement



Compute [tex]\sum_{n=0}^{\infty} p^n cos(3nx)[/tex] for [tex]\abs{p} \textless 1[/tex], where [tex]p \in \mathbb{R}[/tex].

Homework Equations





The Attempt at a Solution



I was thinking that maybe this could be approached as a telescoping series, but I'm not really sure if it is. Would that be the most expedient approach? Clearly it isn't geometric, and I'm not sure how to find the sum of a general power series.
 
Physics news on Phys.org
No one? Someone has to know. There are a lot of clever people on this forum.
 
What if I rearrange it using the fact that cos(3nx) = 1/2(exp(3nix)-exp(-3inx)). Could I then write [tex]\sum_{n=0}^{\infty} (pe^{3xi})^n + \sum_{n=0}^\infty (pe^{-3ix})^n[/tex] and try to work from there? (ie, is that valid?)
 
Ok, I was being silly, as usual - it is actually geometric after all.

Thank you to me for figuring out this problem.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K