I badly in yet another uniform convergence problem

In summary, the conversation revolved around the topic of uniform convergence for series, specifically testing for uniform convergence on [0,1]. The attempt at a solution involved the idea of proving that the limit function is not continuous, but there was a mistake in assuming the value of a series. After correcting this mistake, it was determined that the proof was sound.
  • #1
end3r7
171
0
I hate, HATE uniform convergence for series... hate this, really do. I'm trying hard, but our prof gives us very difficult borderline problems.

Now, I am almost 100% sure this is not uniformly convergence at x=1, but since it's a stinking series, it's hard to figure out it's summation

Homework Statement


data[/b]
1) Test the following series for Uniform Convergence on [0,1]
[tex]
(1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}}
[/tex]

The Attempt at a Solution



I honestly have not a clue where to start. I want to try the following, and I don't know if it's true.

[tex]
\frac{1}{2} < \frac{1}{1+x^n}
[/tex]
for x in [0,1)

So [tex]
(1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} <
(1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}}
[/tex]

and for x in [0,1) [tex]
(1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} = \frac{1}{2}
[/tex]

and we know that and for x = 0 [tex]
(1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}} = 0
[/tex]

Thus the limit function is not continuous, so it can't be uniformly convergent.

Sound proof or bad proof?
 
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  • #2
The idea was good, but you assumed that

[tex]\sum_{n=1}^{\infty}x^n = \frac{1}{1-x}[/tex]

while actually this is true if the index starts at n=0 instead of n=1.
 
  • #3
quasar987 said:
The idea was good, but you assumed that

[tex]\sum_{n=1}^{\infty}x^n = \frac{1}{1-x}[/tex]

while actually this is true if the index starts at n=0 instead of n=1.

The logic should still work, however, correct? Or is there no fix?
 
  • #4
Isn't that simply going to come out to x/2 then? So it should still work correct?

(The limit function is not continuous, since it's greater than or equal to 1/2 at 1, and not 0)
 
Last edited:
  • #5
Now I can't find a fault. :)
 

1. What is uniform convergence?

Uniform convergence is a type of convergence that occurs when a sequence of functions approaches a single limit function at the same rate, regardless of the input value. In other words, the convergence is uniform across the entire domain of the function.

2. Why is uniform convergence important?

Uniform convergence is important because it allows for the simplification of calculations and the use of certain theorems and techniques that are not applicable to non-uniformly convergent sequences. It also ensures that the limit function is continuous and can accurately represent the behavior of the sequence of functions.

3. How is uniform convergence different from pointwise convergence?

In pointwise convergence, the convergence of a sequence of functions is dependent on the input value. This means that the rate of convergence may vary for different input values. In contrast, uniform convergence guarantees that the convergence is consistent across the entire domain of the function.

4. What are some common challenges in solving uniform convergence problems?

One common challenge in solving uniform convergence problems is finding the right choice of epsilon for a given function. Another challenge is determining the appropriate form of the limit function, as it may not always be obvious or easy to identify.

5. How can uniform convergence be applied in real-world situations?

Uniform convergence has many applications in various fields such as physics, engineering, and economics. For example, it can be used to approximate functions and solve differential equations, to evaluate the behavior of financial models, and to analyze the convergence of numerical methods used in computer simulations.

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