Convergence of Sequences in [0,1]

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Homework Help Overview

The discussion revolves around determining the convergence of specific sequences defined on the interval [0,1]. The sequences in question are: (i) \( s_n(x) = n^2x^2(1 - \cos(\frac{1}{nx})) \) for \( x \neq 0 \) and \( s_n(0) = 0 \); (ii) \( s_n(x) = \frac{nx}{x+n} \); and (iii) \( s_n(x) = n \sin(\frac{x}{n}) \). Participants are exploring both pointwise and uniform convergence for these sequences.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the pointwise and uniform convergence of the sequences, with some expressing concerns about the straightforward nature of the solutions. There is an exploration of the limits of the sequences as \( n \) approaches infinity, with specific attention to the behavior at the endpoints of the interval.

Discussion Status

Some participants have confirmed the conclusions regarding convergence but question the reasoning behind uniform convergence. Others suggest that the differentiability of the sequences and the boundedness of their derivatives may provide insight into uniform convergence. The discussion is ongoing, with participants considering various theorems and approaches to solidify their understanding.

Contextual Notes

Participants are navigating the complexities of convergence definitions, particularly in relation to uniform convergence, and are referencing theorems related to the behavior of derivatives and compactness of the interval [0,1]. There is a focus on ensuring that the reasoning aligns with established mathematical principles.

  • #31
Dick said:
Well, yes. Just interchanging the order just interchanges increasing/decreasing and signs. It doesn't make any real difference to the problem. But ##|s_n(x) - s(x)|## is still increasing as a function of x on [0,1]. Not decreasing. It's an absolute value. I'm still not sure you are clear on what's important and what is not.

Oh so It doesn't make any difference at all. I see now.

It's just |a-b| = |b-a| looked wrong to me for a moment ( Can't even explain why). Now that that's clear I think I'm good.
 
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  • #32
Zondrina said:
Oh so It doesn't make any difference at all. I see now.

It's just |a-b| = |b-a| looked wrong to me for a moment ( Can't even explain why). Now that that's clear I think I'm good.

That is great!
 

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