Uniform convergence of series of functions

In summary, the conversation discusses the uniform convergence of the series \sum \frac{x^n}{1 + x^n} on the interval [0,1). It is determined that the series is not uniformly convergent, and various methods are discussed to prove this, such as using the Weierstrass M-Test and the Cauchy criterion. Ultimately, it is found that the set \sup \{|g_k(x)| : x \in [0,\, 1)\} is equal to \frac{1}{2}, which proves the series is not uniformly convergent.
  • #1
fmam3
82
0

Homework Statement


Is the series [tex]\sum \frac{x^n}{1 + x^n}[/tex] uniformly convergent on [tex][0,1)[/tex] ?

Homework Equations


The Attempt at a Solution


I'm guessing that it is NOT uniformly convergent but I don't know how to show it. But I would note that it is quite easy to show that the series is uniformly convergent on [tex][0,a][/tex] for [tex]\forall a, 0 < a < 1[/tex] (i.e. straightforward application of the Weierstrass M-Test).

I'm aware of the following fact is useful to check whether a series of functions is "not" uniformly convergent. We know that if [tex]\sum g_k[/tex] is uniformly convergent on [tex]S \subseteq \mathbb{R}[/tex], then [tex]\lim_{n \to \infty} \sup \{|g_k(x)| : x \in S\} = 0[/tex].

Hence motivated by this fact, a good trick would be to use derivatives to find the maximum. But in this case, it doesn't work so well. For convienence, set [tex]g_n(x) = \frac{x^n}{1 + x^n}[/tex] and in this case we can drop the absolute value since [tex]x \in [0,1)[/tex]. And if my differentation is correct, we should have [tex]g_n'(x) = \frac{nx^{n-1}}{(1 + x^n)^2}[/tex] and set this to zero to find the maximum / minimum and we see that [tex]g_n'(x) = 0[/tex] implies [tex]x = 0[/tex] --- and I'm very certain that [tex]x = 0[/tex], so that [tex]g_n(x) = 0[/tex], is not the maximum of [tex]g_n[/tex], and in fact it is clear that it is the minimum. Thus, the above attempt to find the supremum doesn't work so well.

I also thought about using the Cauchy criterion for uniform convergence, but there does not seem to be any "convenient" indicies for me to choose to show that it fails the criterion.Any help is greatly appreciated!
 
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  • #2
Given k, what is the set:

[tex]\sup \{|g_k(x)| : x \in [0,\, 1)\}[/tex]

Find out what that set is, then it's easy to compute its sup, and then computing the lim of those sups is going to be easy, and your proof will be done.
 
  • #3
AKG said:
Given k, what is the set:

[tex]\sup \{|g_k(x)| : x \in [0,\, 1)\}[/tex]

Find out what that set is, then it's easy to compute its sup, and then computing the lim of those sups is going to be easy, and your proof will be done.

Thanks for the reply. I thought a little bit more on the problem yesterday and I think I made it way too complicated in my previous attempt.

It's simply as follows. The number is [tex]\sup \{|\frac{x^n}{1 + x^n}| : x \in [0,1)\} = \frac{1}{2}[/tex] and this is seen by setting [tex]x = 1[/tex]. And of course, if that is equal to 1/2, then its limit is still 1/2 and thus, this series cannot be uniformly convergent.
 

1. What is uniform convergence of series of functions?

Uniform convergence of series of functions is a type of convergence that measures how closely a sequence of functions approaches a limit function. In this case, the convergence is uniform if the difference between the function and the limit function approaches zero as the independent variable approaches infinity.

2. How is uniform convergence different from pointwise convergence?

Uniform convergence differs from pointwise convergence in that it requires the convergence of a sequence of functions to be uniform over the entire domain, rather than just at individual points. This means that the difference between the functions and the limit function must approach zero at the same rate for all points in the domain.

3. What are the conditions for uniform convergence of series of functions?

The conditions for uniform convergence of series of functions are that the sequence of functions must be continuous, the limit function must exist and be continuous, and the convergence of the series must be uniform. Additionally, the rate of convergence must be the same for all points in the domain.

4. How is uniform convergence related to the continuity of the limit function?

The continuity of the limit function is a necessary condition for uniform convergence. This means that if the limit function is not continuous, then the convergence of the series cannot be uniform. However, the converse is not always true, as there are cases where the limit function is continuous but the convergence of the series is not uniform.

5. What are some real-world applications of uniform convergence of series of functions?

Uniform convergence of series of functions has many applications in various fields, including physics, engineering, and economics. It is used to approximate functions and solve differential equations, as well as to analyze the convergence and stability of numerical methods. It is also used in optimization problems and in studying the behavior of physical systems.

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