# show uniform convergence of sequence of functions

• Apr 18th 2011, 07:06 PM
mremwo
show uniform convergence of sequence of functions
The exercise is to show that the sequence $\displaystyle x^2$$\displaystyle e^{-nx} converges uniformly on \displaystyle [0, \infty) I know that the sequence converges to 0 because \displaystyle x^2$$\displaystyle e^{-nx}$ = $\displaystyle x^2$($\displaystyle e^{-x}$)^n and $\displaystyle e^{-x}$ is always less than 1.

It makes sense that it converges uniformly since $\displaystyle e^{-nx}$ will go to 0 as $\displaystyle n \rightarrow \infty$, not depending on what x is, but I don't know how to show this using the definition of uniform convergence. Help is appreciated. Thank you!
• Apr 18th 2011, 08:39 PM
TKHunny
Spoken roughly, once it get's close, it never wanders off again. You must demonstrate, since you seem confident this converges to f(x) = 0 on the interval, that for some value of n, it is very close to zero on the entire interval AND that it never gets any worse than that.

For your sequence, the maximum deviation is always at x = 2/n. Does that do anything for us? Maximum deviation is 4/[(e^2)*(n^2)].
• Apr 19th 2011, 12:39 PM
mremwo
So the sequence is farthest from 0 at x=2/n, or the maximum of each function in the sequence is at x=2/n? So if the maximum deviation of the functions in the sequence is 4/[(e^2)*(n^2)] for each n, I need to make sure my natural number K makes it so that if n>=k then |sequence-0|<= 4/[(e^2)*(n^2)] < E ??
• Apr 19th 2011, 12:48 PM
Drexel28
Quote:

Originally Posted by mremwo
So the sequence is farthest from 0 at x=2/n, or the maximum of each function in the sequence is at x=2/n? So if the maximum deviation of the functions in the sequence is 4/[(e^2)*(n^2)] for each n, I need to make sure my natural number K makes it so that if n>=k then |sequence-0|<= 4/[(e^2)*(n^2)] < E ??

Merely note that x^2e^{-nx}= x^2/e^{nx}=x^2/{1+nx+n^2x^2/2+...}<= x^2/{1+n^2x^2} and I think it's easy to see the result from there. Namely, by Dini's theorem it's evident that since this converges pointwise to 0 on [0,1] that it converges uniformly there and on (1,∞) one has that x^2/e^{nx}<=x^2/{1+nx^2}<= 1/n
• Apr 19th 2011, 06:39 PM
chisigma
Quote:

Originally Posted by mremwo
The exercise is to show that the sequence $\displaystyle x^2$$\displaystyle e^{-nx} converges uniformly on \displaystyle [0, \infty) I know that the sequence converges to 0 because \displaystyle x^2$$\displaystyle e^{-nx}$ = $\displaystyle x^2$($\displaystyle e^{-x}$)^n and $\displaystyle e^{-x}$ is always less than 1.

It makes sense that it converges uniformly since $\displaystyle e^{-nx}$ will go to 0 as $\displaystyle n \rightarrow \infty$, not depending on what x is, but I don't know how to show this using the definition of uniform convergence. Help is appreciated. Thank you!

Each http://quicklatex.com/cache3/ql_b17e...000d399_l3.png is greater or equal to 0 in the whole interval and has a maximum at http://quicklatex.com/cache3/ql_1484...2851bd6_l3.png
where is http://quicklatex.com/cache3/ql_88e4...f17f4c9_l3.png. If You take an http://quicklatex.com/cache3/ql_9c44...75be18a_l3.png then http://quicklatex.com/cache3/ql_3a15...8805d89_l3.png in the whole interval it will be http://quicklatex.com/cache3/ql_a4cb...de624db_l3.png...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$