Uniform Convergence of fn(x) in [0,1]?

Click For Summary

Homework Help Overview

The discussion revolves around the uniform convergence of the function fn(x) = nx^2/(1+nx) on the interval [0,1]. The original poster notes that fn(x) converges pointwise to f(x) = x as n approaches infinity but seeks clarification on whether this convergence is uniform.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the maximum of the difference |fn(x) - f(x)| to assess uniform convergence. Questions arise regarding the existence of an N that satisfies the uniform convergence criteria for a given ε.

Discussion Status

Some participants suggest that the maximum of |fn(x) - f(x)| is 1/(1+n) and discuss whether this leads to a conclusion about uniform convergence. There is an ongoing examination of how to determine N based on ε, with some guidance provided on the formulation of N.

Contextual Notes

Participants are considering the implications of their findings on the convergence behavior and the mathematical expressions involved, including the need for careful notation in their calculations.

math8
Messages
143
Reaction score
0
Let fn(x)=nx^2/1+nx ; x lies in [0,1]

Is the convergence uniform?

Since lim as n-->infinity of fn is x, I can see that fn(x) converges pointwise to f(x)= x
But I get stuck when I try to show the convergence is uniform or not.
 
Physics news on Phys.org
Since you know the limit is f(x)=x, look at |fn(x)-f(x)|. What's the maximum of that function on the interval [0,1]?
 
I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?
 
math8 said:
I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?

Can you make that conclusion? If I give you an e>0 can you find an N such that 1/(1+n)<e for all n>N? Sure you can. If you're not sure you'd better figure out how pick a corresponding N.
 
that N should be > than 1-e/e right?
 
math8 said:
that N should be > than 1-e/e right?

Sure. Use parentheses when you write something like (1-e)/e, ok? 1-e/e at first glance looks like 1-(e/e), which looks like 0.
 
Right, thanks.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
26
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
6K
  • · Replies 1 ·
Replies
1
Views
7K