Convergence of Subsequences of Cosine Function in Real Analysis

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Homework Help Overview

The discussion revolves around the convergence of subsequences of the cosine function, specifically showing that for any limit L within the interval [-1, 1], there exists a subsequence of cos(n) that converges to L.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the idea of finding a subsequence that converges to a specific limit L, with one participant suggesting the use of the graph of the cosine function to identify values within a certain range. Another participant questions the assurance of finding a natural number n within that range.

Discussion Status

The discussion is ongoing, with participants sharing thoughts on the density of n mod 2π in the interval [0, 2π] and its implications for continuity of the cosine function. Some participants express uncertainty about the proof of density and its connection to the problem at hand.

Contextual Notes

There is a mention of the irrationality of π as a key factor in demonstrating density, and the pigeonhole principle is referenced as part of the reasoning process. Participants are navigating through assumptions related to natural numbers and the behavior of the cosine function.

kbfrob

Homework Statement


Show that for any L[tex]\in[/tex][-1,1] there exists a subsequence of cos(n) such that that subsequence converges to L

The Attempt at a Solution


I have no idea.
I suppose that the ultimate goal would be to find a subsequence nk so that nk converges to x, where x = cos-1(L) + 2[tex]\pi[/tex]k
 
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Think about the graph of y = cos(x) and the parts of this graph that lie between the horizontal lines
[tex]y = L + \epsilon[/tex]
and
[tex]y = L - \epsilon[/tex].
Some of the values of cos(n) will lie in this band.
 
how can you guarantee that there is a value of n in that band? n is a natural number so I'm not sure how you can say that it is in there.
 
The easiest way is to show that n mod 2*pi is dense in [0,2pi]. Then use that cos is continuous. To show the density use that pi is irrational. If r is irrational then n*r mod 1 is dense in [0,1]. I KNOW this is true. For some reason the proof doesn't stick in my head. I've looked it up more than once. Do I have to do it again, or can you figure it out? I know it involves the pigeonhole principle.
 

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