Uniform Convergence of Series on (0,1)

Click For Summary

Homework Help Overview

The discussion revolves around the uniform convergence of the series \(\sum_{n=0}^{\infty} x^n(1-x)\) on the interval (0,1). Participants are exploring the conditions under which the series converges uniformly versus pointwise.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish pointwise convergence to 1 and then tests for uniform convergence using the supremum of the difference from 1. Some participants question the validity of the original poster's reasoning regarding uniform convergence and suggest the use of the Weierstrass M-test.

Discussion Status

Participants are actively discussing the application of the Weierstrass M-test to determine uniform convergence. There is a recognition that the choice of \(M_n\) is crucial, with some suggesting \(M_n = 1\) as a valid approach, which leads to a conclusion of uniform convergence.

Contextual Notes

There is an ongoing examination of the definitions and conditions for uniform convergence versus pointwise convergence, with specific attention to the independence of \(M_n\) from \(x\) in the context of the M-test.

latentcorpse
Messages
1,411
Reaction score
0
Does [itex]\sum_{n=0}^{\infty} x^n(1-x)[/itex] converge uniformly on [itex](0,1)[/itex]?

[itex]S_n(x)=\sum_{k=0}^{\infty} x^k(1-x)=\frac{1-x^{n+1}}{1-x}(1-x)=1-x^{n+1}} \rightarrow 1[/itex] as [itex]n \rightarrow \infty[/itex]
so we get pointwise convergence to 1

now we test for uniform convergence

[itex]d_{\infty}(S_n,1) = sup |S_n(x)-1|= sup x^{n+1}[/itex] now remembering that x is in (0,1), i said that this should tend to 0 as n goes to infinity giving uniform convergence but the answers say that it doesn't go to 0 and so convergence is not uniform, merely pointwise - why is this?
 
Physics news on Phys.org


Hint: this problem is screaming Weierstrass M-test.
 


we can find a sequence of positive integers [itex]M_n[/itex] such that [itex]|f_n(x) \leq M_n \forall x \in (0,1)[/itex] so let [itex]M_n=x^n[/itex] then since [itex]\sum M_n[/itex] converges, [itex]\sum f_n(x)[/itex] converges by Weierstrass M Test?

why is it screaming M test?
 


It is screaming M test because that is what you have to use to answer the question.

You can't use xn for Mn because Mn is independent of x. Try again.
 


just [itex]M_n=1 \forall n[/itex] then? this obviously converges and so we would get uniform convergence...?
 


Yes, Mn = 1 works and so by the M-test, you have uniform convergence on (0,1).
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K