Uniform Convergence: Series Homework Help

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Homework Help Overview

The discussion revolves around demonstrating the uniform convergence of a series involving cosine functions and an alternating series component. The series is expressed as \(\frac{4b}{\pi} \sum^{\infty}_{n=1} \frac{1-(-1)^{n}}{n^{2}} \cos(nt) \cos(nx)\) for fixed \(t\). Participants are exploring the convergence properties of this series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the series, with some suggesting it may be an alternating series and others questioning the role of the constant \(b\). There are attempts to apply the Weierstrass M test and to analyze the absolute convergence of the series. Some participants express confusion over the terms of the series and their contributions to convergence.

Discussion Status

The discussion is active, with various approaches being explored, including the Weierstrass M test and comparisons to known convergent series. Participants are engaging with each other's ideas, and while there is no explicit consensus, several lines of reasoning are being developed.

Contextual Notes

There is some ambiguity regarding the constant \(b\) and its relevance to the convergence analysis. Additionally, participants are navigating the implications of the series' terms being zero for even \(n\) and the focus on odd \(n\) terms.

Somefantastik
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Homework Statement


Show uniform convergence

[tex]\frac{4b}{\pi} \sum^{\inf}_{n=1} \frac{1-(-1)^{n}}{n^{2}}cos(nt)cos(nx)[/tex]

for fixed t

Homework Equations


The Attempt at a Solution



[tex]\left| cos(nt) \right| \leq 1[/tex]

[tex]\left| cos(nx) \right| \leq 1[/tex]

[tex]lim \left|\frac{1 - (-1)^{n}}{n^{2}}\right| \ = \ 0[/tex]

What's next?
 
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Well it's an alternating series isn't it? So... see if it's conditionally convergent and absolutely convergent
 
I'd like to use the Weierstrauss M test. Can someone help me in that direction?
 
What's b? It seems like there is a lot of stuff they added here to try to confuse you. Let's ignore the constant 4b/pi for now. Then for any n, for all x the absolute value of the nth term is less than 2/n^2. We know that the series 2/n^2 converges, so this series also converges.
 
Somefantastik said:
I'd like to use the Weierstrauss M test. Can someone help me in that direction?

Then do so. For each n you want a M_n such the |f_n|<=M_n (where f_n is the nth function in your series) and the series of the M_n converges. It's really pretty straightforward. You've already handled the cos parts. Any suggestions?
 
This is not an alternative series! The terms with n even are all zeros and terms with n odd are 2/n^2 * cos ( ) * cos (). So take its absolute value and compare it with the series sum(2/n^2, n=1..infty) (I'm ignoring the constant outside the summation. Hence the series is absolutely convergent!
 
Ok thanks everybody! I think I have gleaned enough to get a good answer.
 

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