Convergence of sin(x/n) in A[0,pi]

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Homework Help Overview

The discussion revolves around the convergence of the sequence of real-valued functions fn(x) = sin(x/n) to f(x) = 0 on the interval A = [0, pi]. Participants are exploring the nature of this convergence, particularly uniform convergence as n approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the behavior of the function as n increases, with some noting that x/n approaches zero for all x in the interval. There is a question about the use of Euler's formula and whether it can be applied given the real-valued nature of the functions. Others express uncertainty about how to demonstrate the convergence effectively.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants have provided insights into the conditions for uniform convergence and the implications of the sup norm, while others are still grappling with the conceptual understanding of the problem.

Contextual Notes

There is an emphasis on the requirement to show convergence for any value of x within the specified interval, and some participants are questioning the assumptions regarding the use of complex analysis tools.

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Homework Statement



Show that the sequence real-valued functions fn(x) = sin(x/n) converges uniformly to f(x) = 0 on A=[0,pi]

Homework Equations



I don't believe there are any

The Attempt at a Solution



Well, my first thought was that if x/n = pi then sin(x/n) is obviously going to be zero. I really can't get past that, I mean it seems like stating that and that if x/n is any multiple of pi then sin(x/n) will be zero. I guess I can't see past the obvious and into how this problem might actually be challenging.
 
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You are supposed to show that fn(x) goes to zero as n goes to infinity for *any* value of x between 0 and pi.
 
Oh. Well then. You can't use euler's because the function is real valued, correct? Or, can you expand it and just leave off the imaginary part?

I can't think of another way to show that this function is going to zero.
 
Well, as n goes to infinity, x/n goes to zero for all x on [0,pi] And sin(0)=0
 
If the sup norm ( sup |fn(x)-f(x)|, over x in X) goes to zero then f_n converges uniformly to f on X. This is biconditional.

I suspect you can also show this from the definition of uniform convergence.
 

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