Convergence of Subsequences of Cosine Function in Real Analysis

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SUMMARY

The discussion focuses on proving that for any limit point L within the interval [-1, 1], there exists a subsequence of the cosine function, cos(n), that converges to L. The key approach involves demonstrating that the sequence of natural numbers n mod 2π is dense in the interval [0, 2π], leveraging the fact that π is irrational. This density ensures that values of cos(n) can be found within any ε-neighborhood of L, thus confirming the convergence of the subsequence to L. The continuity of the cosine function further supports this conclusion.

PREREQUISITES
  • Understanding of real analysis concepts, particularly limits and convergence.
  • Familiarity with the properties of the cosine function and its continuity.
  • Knowledge of density in metric spaces, specifically regarding irrational numbers.
  • Basic understanding of the pigeonhole principle and its applications in proofs.
NEXT STEPS
  • Study the proof of density of n mod 2π in [0, 2π] using irrational numbers.
  • Explore the continuity properties of trigonometric functions, particularly cosine.
  • Learn about subsequences and their convergence in real analysis.
  • Investigate applications of the pigeonhole principle in mathematical proofs.
USEFUL FOR

Students and educators in real analysis, mathematicians interested in convergence properties of sequences, and anyone studying the behavior of trigonometric functions in mathematical contexts.

kbfrob

Homework Statement


Show that for any L\in[-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\pik
 
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Think about the graph of y = cos(x) and the parts of this graph that lie between the horizontal lines
y = L + \epsilon
and
y = L - \epsilon.
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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