Showing a function converges uniformly

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In summary: So basically, the larger N, the closer the function gets to the max value of N, making it impossible to get closer to 0 uniformly. In summary, the conversation discusses the issue of proving that a series converges uniformly and the confusion surrounding the definition and example provided in the textbook. The main point is that although the series has pointwise convergence to 0, it is not uniformly convergent due to the fact that the larger N is, the closer the function gets to the maximum value of N, making it impossible to get closer to 0 uniformly.
  • #1
trap101
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let fn(x) = n/(1+n2x2) - (n-1)/(1+(n-1)2x2) in the interval 0<x< L

I am trying to show that this series converges uniformly.

I have solved that the sum of the series from n=1 to n = N is:

N/(1+N2x)

now by definition a series converges uniformly if:

max (a≤x≤b) |f(x) - Sn(x)| ---> 0 as N-->∞

my issue is that in the solution example they provided in the textbook they said the series does not uniformly converge because:

max(0,L) 1/(1+N2x2) = N

how did they get 1/(1+N2x2) from the definition of uniform convergence? What is it that I am not interpreting right to get a solution?
 
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  • #2
trap101 said:
let fn(x) = n/(1+n2x2) - (n-1)/(1+(n-1)2x2) in the interval 0<x< L

I am trying to show that this series converges uniformly.

On what interval? Are you sure you stated the interval correctly? Was it [a,L) for any a between 0 and L and you restated it (0,L)?

I have solved that the sum of the series from n=1 to n = N is:

N/(1+N2x)

now by definition a series converges uniformly if:

max (a≤x≤b) |f(x) - Sn(x)| ---> 0 as N-->∞

my issue is that in the solution example they provided in the textbook they said the series does not uniformly converge because:

max(0,L) 1/(1+N2x2) = N

That isn't true on (0,L). It's true on [0,L]. On (0,L) is is a sup, not a max.

how did they get 1/(1+N2x2) from the definition of uniform convergence? What is it that I am not interpreting right to get a solution?

I don't know where those last two lines came from. And I would like to see the exact statement of the problem. I don't believe it is true as you have stated it.
 
  • #3
LCKurtz said:
On what interval? Are you sure you stated the interval correctly? Was it [a,L) for any a between 0 and L and you restated it (0,L)?



That isn't true on (0,L). It's true on [0,L]. On (0,L) is is a sup, not a max.



I don't know where those last two lines came from. And I would like to see the exact statement of the problem. I don't believe it is true as you have stated it.




You were right in terms of the interval supposed to be closed on [0,L] as well as the exact definition of uniform convergence I just didn't communicate it correctly. If I polished up on my latex it would be easier for me to restate. I attached a jpeg of the question and statement that is confusing me so you can see the exact question.
 

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  • #4
That last equation$$
\max_{(0,L)}\frac 1 {1+N^2x^2}=N$$obviously has typos and should read$$
\max_{[0,L]}\frac N {1+N^2x^2}=N$$
 
  • #5
LCKurtz said:
That last equation$$
\max_{(0,L)}\frac 1 {1+N^2x^2}=N$$obviously has typos and should read$$
\max_{[0,L]}\frac N {1+N^2x^2}=N$$



I'm still having issues with how they got N, because are we not evaluating the difference between f(x) and the sum of the series? I don't see how that difference translates into just N.
 
  • #6
OK, maybe they meant to write$$
\sup_{(0,L)}\frac N {1+N^2x^2}=N$$The point is that although you have pointwise convergence to ##0## it is not uniform because the larger N is the bigger the sum is near zero. You can't uniformly bound the sum near 0 no matter how large N is.
 
  • #7
LCKurtz said:
OK, maybe they meant to write$$
\sup_{(0,L)}\frac N {1+N^2x^2}=N$$The point is that although you have pointwise convergence to ##0## it is not uniform because the larger N is the bigger the sum is near zero. You can't uniformly bound the sum near 0 no matter how large N is.


Ahhhh, now I see. Thanks.
 

FAQ: Showing a function converges uniformly

What is the definition of uniform convergence?

Uniform convergence is a property of a sequence of functions, where the distance between the values of the functions at any given point approaches zero as the number of terms in the sequence increases. In other words, the functions become "closer" to each other as the sequence progresses.

How do you determine if a function converges uniformly?

To determine if a function converges uniformly, you must first show that it converges pointwise. Then, you must show that the convergence is independent of the choice of x, meaning that the distance between the values of the function at any given point must approach zero uniformly as the number of terms in the sequence increases.

What is the difference between pointwise and uniform convergence?

Pointwise convergence refers to the behavior of a sequence of functions at a specific point x. It means that as the number of terms in the sequence increases, the values of the functions at x will get closer and closer to the limit. Uniform convergence, on the other hand, means that the convergence is independent of the choice of x, and the distance between the values of the functions at any given point approaches zero as the number of terms increases.

What are some common techniques for proving uniform convergence?

One common technique for proving uniform convergence is the Cauchy criterion, which states that a sequence of functions converges uniformly if and only if the sequence of partial sums satisfies the Cauchy criterion. Another technique is the Weierstrass M-test, which involves finding a convergent series that bounds the sequence of functions.

Why is uniform convergence important in mathematics?

Uniform convergence is important in mathematics because it allows us to make stronger conclusions about the behavior of a sequence of functions. For example, if a sequence of functions converges uniformly, we can switch the order of integration and limits, which can be useful in solving certain problems. It also plays a crucial role in the study of series, as uniform convergence is necessary for term-by-term differentiation and integration.

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