Showing the uniform convergence of a gaussian function-like series

[dane502]

Homework Statement

Prove that the series [tex]\sum_{n=0}^\infty e^{-n^2x^2}[/tex] converges uniformly on the set [tex]\mathbb{R}\backslash\ \big] -\epsilon,\epsilon\big[[/tex] where [tex]\epsilon>0[/tex]

Homework Equations

n/a

The Attempt at a Solution

I have tried using Weierstrass M-test but I have not been able to find a suitable series.
As my topic implies I thought I could use some a series of a bell-curve like function, which I have some experience with from a probability course.
But I have a problem finding a suitable expression - let alone showing that this series convergence.

I would appreciate if someone could help getting me started (preferably without solving the entire exercise).

 

[obafgkmrns]

Try finding a function > exp(-n^2 * x^2) for every n and whose integral over R is finite.

 

[LCKurtz]

Homework Statement

Prove that the series [tex]\sum_{n=0}^\infty e^{-n^2x^2}[/tex] converges uniformly on the set [tex]\mathbb{R}\backslash\ \big] -\epsilon,\epsilon\big[[/tex] where [tex]\epsilon>0[/tex]

Homework Equations

n/a

The Attempt at a Solution

I have tried using Weierstrass M-test but I have not been able to find a suitable series.
As my topic implies I thought I could use some a series of a bell-curve like function, which I have some experience with from a probability course.
But I have a problem finding a suitable expression - let alone showing that this series convergence.

I would appreciate if someone could help getting me started (preferably without solving the entire exercise).

Well, since [itex]e^{-n^2x^2}[/itex] decreases as |x| increases you have

[tex]e^{-n^2x^2}\le e^{-n^2\epsilon^2}[/tex]

if |x| ≥ ε. Now since negative exponentials decay fast, you should be able construct a comparison with Σ1/n².

 

[dane502]

Got it! Thank you very much, LCKurtz.

For my personal interest, would someone care to comment on whether or not the sets
[tex]
\mathbb{R}\backslash\ \left\{ 0 \right\}
[/tex]

and
[tex]
\mathbb{R}\backslash\ \left] -\epsilon,\epsilon \right[
[/tex]
where [tex]\epsilon>0[/tex], are the same?

 

[LCKurtz]

Got it! Thank you very much, LCKurtz.

For my personal interest, would someone care to comment on whether or not the sets
[tex]
\mathbb{R}\backslash\ \left\{ 0 \right\}
[/tex]

and
[tex]
\mathbb{R}\backslash\ \left] -\epsilon,\epsilon \right[
[/tex]
where [tex]\epsilon>0[/tex], are the same?

No, they aren't. ε/2 is in the first but not the second.

 

[dane502]

That is a convincing argument, although I have a hard time visualizing the difference between the two sets, when ε→0..
With regards to the original topic, I have another question. We have shown that the series converges uniformly, which is all I needed to show, but i would also like to know to WHAT it converges, ie. the limit. Would someone care to comment on that?

 

[dane502]

If I may add another question to the above, are the two sets equal for ε→0?

 

[124124]

If I may add another question to the above, are the two sets equal for ε→0?

For ε→0 the sets are the same. But if only the only requirement is ε>0, then LCKurtz's argument stands. I don't know how to calculate the value of the limit, but try using Maple, Mathematica etc.

 

[dane502]

I only have maple, and it is unable to evaluate the sum. Does anybody have another idea?

 

[LCKurtz]

I only have maple, and it is unable to evaluate the sum. Does anybody have another idea?

I don't think you are likely to find a formula for the sum. Probably the best you could hope for is that it is a common enough sum that it has been given a name and its properties have been studied as, for example, Bessel functions have. Or it may be just another convergent series with a no-name sum.