Proving Uniform Convergence of f_n = x/sqrt(1+nx^2) on R

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In summary, the goal is to prove that the function f_n = \frac{x}{\sqrt{1+nx^2}} converges uniformly to 0 on all real numbers. This means that for any epsilon > 0, there exists an N where n>=N implies that | f_n(x) - f(x) | < epsilon for all x in the domain. To do this, we let f(x) = lim n-> infinity f_n(x) and choose an N such that n > \frac{1}{\epsilon^2} implies | f_n(x) - 0 | < epsilon for all x. The proof also involves showing that the pointwise limit of f_n(x) is indeed 0,
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Homework Statement


Prove that [tex]f_{n} = \frac{x}{\sqrt{1+nx^2}}[/tex] is uniformly convergent to 0 on all real numbers


Homework Equations


{f_n} is said to converge uniformly on E if there is a function f:E->R such that for every epsilon >0, there is an N where n>=N implies that | f_n(x) - f(x) | < epsilon, for all x in E.


The Attempt at a Solution


Let f(x) = lim n-> infty f_n(x), and let epsilon > 0. Then it is obvious, that if n>1, that as n -> infty, the limit goes to 0, and thus we would need to show that [tex]\frac{x}{\sqrt{1+nx^2}} < epsilon[/tex]| , which happens as long as n > [tex]\frac{\frac{x}{epslion}-1}{x^2}[/tex]. So, I feel like I got it, except for the 'obvious' statement that f(x) = 0. Am I doing this right? Thanks ahead of time.
 
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i) you want to show lim n->infinity f_n(x)=0 for all x. That tells you the pointwise limit is 0. ii) your algebra is wrong, what happened to the sqrt? ii) There shouldn't be an x in your expression for choosing n. That's what 'uniform' convergence is all about.
 

1. What does it mean for a sequence of functions to be uniformly convergent on R?

Uniform convergence on R means that the sequence of functions approaches the same limit at every point on the real number line. This is different from pointwise convergence, where the limit may vary at different points.

2. How is uniform convergence on R different from uniform convergence on a subset of R?

Uniform convergence on R means that the sequence of functions approaches the same limit at every point on the real number line. However, uniform convergence on a subset of R only requires the sequence of functions to approach the same limit on that particular subset. The functions may behave differently on other parts of the real number line.

3. How can I prove that a sequence of functions is uniformly convergent on R?

To prove uniform convergence on R, you need to show that for any given epsilon (ε), there exists a natural number N such that for all x in R and for all n ≥ N, the absolute difference between the function and its limit is less than ε. This is known as the epsilon-N definition of uniform convergence.

4. What are the applications of uniform convergence on R?

Uniform convergence on R is an important concept in the study of real analysis and functional analysis. It has applications in fields such as physics, engineering, and mathematics. It is also used in the development of numerical methods for solving differential equations.

5. Can a sequence of discontinuous functions be uniformly convergent on R?

Yes, a sequence of discontinuous functions can be uniformly convergent on R. This is because uniform convergence only requires the functions to approach the same limit, regardless of their continuity. However, the limit function must still be continuous in order for the sequence to be considered uniformly convergent.

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