Uniform convergens and continuity on R

In summary, the conversation discusses the convergence and continuity of the function f(x) = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2} on the real numbers. It is shown that the series converges uniformly and that the function is continuous on \mathbb{R}. The conversation also includes a question about proving the continuity of the function using integrals.
  • #1
Hummingbird25
86
0
Hello people, I'm tasked with showing the following:

given the series [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}[/tex]

(1) show that it converges Uniformly [tex]f_n(x) :\mathbb{R} \rightarrow \mathbb{R}[/tex].

(2) Next show the function

[tex]f(x) = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2} [/tex]

is continious on [tex]\mathbb{R}[/tex]

(1) Suppose [tex]f_n = \frac{1}{x^2 + n^2}[/tex],

Then [tex]f_n[/tex] is uniformly convergens if

[tex]sup _{x \in \mathbb{R}} |f_n(x) - x|[/tex]. Now

[tex]_{sup} _{x \in \mathbb{R}} |f_n(x) - x| = |\frac{1}{x^2 - n^2} - x| =
_{sup} _{x \in \mathbb{R}} \frac{1}{x^2 - n^2}[/tex]

The deriative of f_n(x) is non-negative on [tex]\mathbb{R}[/tex], so its increasing and is hence maximumized at [tex]x = \mathbb{R}[/tex]. So the supremum is [tex]1/n^2[/tex]. This does tend to zero as [tex]n \rightarrow \infty[/tex]. So therefore it converge Uniformly.

Am I on the right track here? If yes any hints on how to prove the continuety ?

I know that its something to do with:

\integral_{1} ^{\infty} 1/x^2 + n^2 dx = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}

Sincerely Hummingbird25
 
Last edited:
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  • #2
Hummingbird25 said:
given the series [tex]\sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2}[/tex]

(1) show that it converges Uniformly [tex]f_n(x) :\mathbb{R} \rightarrow \mathbb{R}[/tex].

you've not defined f_n

(2) Next show the function

[tex]f(x) = \sum_{n=1} ^{\infty} \frac{1}{x^2 + n^2} [/tex]

is continious on [tex]\mathbb{R}[/tex]

(1) Suppose [tex]f_n = \frac{1}{x^2 + n^2}[/tex],

Then [tex]f_n[/tex] is uniformly convergens if

[tex]sup _{x \in \mathbb{R}} |f_n(x) - x|[/tex].


You start an 'if' statement and then do not complete it with a condition about anything.
 
  • #3
Hi :)

f_n: \mathbb{R} \rightarrow \mathb{R}

f_n(x) should have said f_n

There is one sup to much sorry.

matt grime said:
you've not defined f_n

You start an 'if' statement and then do not complete it with a condition about anything.

Sincerely

Hummingbird25
 
Last edited:

FAQ: Uniform convergens and continuity on R

What is uniform convergence on R?

Uniform convergence on R is a type of convergence in mathematics, specifically in the field of real analysis. It refers to a sequence of functions that converge uniformly on a given set of real numbers. This means that for any given epsilon, there exists an N such that the difference between the function and its limit is less than epsilon for all x in the set.

How is uniform convergence different from pointwise convergence?

Pointwise convergence means that for a given x, the sequence of functions converges to the limit as n approaches infinity. However, this convergence can vary at different points. In contrast, uniform convergence guarantees that the convergence is the same for all x in the given set.

What is the significance of uniform convergence?

Uniform convergence is important in analysis because it allows for the interchange of limits and integrals. This means that if a sequence of functions is uniformly convergent, then the limit of the integral of the sequence is equal to the integral of the limit of the sequence.

How can I determine if a sequence of functions is uniformly convergent on R?

One way to determine if a sequence of functions is uniformly convergent on R is to use the Weierstrass M-test. This test states that if there exists a sequence of numbers M_n such that the absolute value of each function in the sequence is less than or equal to M_n for all x in the given set, and if the series of M_n converges, then the sequence of functions is uniformly convergent on R.

Can a sequence of continuous functions be uniformly convergent on R?

Yes, a sequence of continuous functions can be uniformly convergent on R. This is because uniform convergence does not guarantee the continuity of the limiting function, only that the convergence is uniform. However, if the sequence of functions is uniformly convergent and each function in the sequence is continuous, then the limit of the sequence will also be continuous.

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