# Thread: Proving Uniform Convergence with epsilon proof.

1. ## Proving Uniform Convergence with epsilon proof.

Prove that fn(x) = n2x2e-nx converges uniformly on [1, inf). I have to use epsilon proof.

I have found the point-wise limit f(x) = 0. I let e > 0. So far I have bounded n2x2e-nx < n2x2e-n. I can't seem to eliminate this x2. Any hints? Thanks.

2. ## Re: Proving Uniform Convergence with epsilon proof.

\large \begin{align*} &f_n(x)=(nx)^2 e^{-nx}\\ \\ &f_n(x)=e^{2\ln(nx)}e^{-nx} =\\ \\&e^{2\ln(n)}e^{2\ln(x)-nx} \\ \\ &\mbox{Note that }2\ln(x)-nx<0, \forall n>0, \forall x > 0\mbox{ so } \\ \\ &e^{2 \ln(x)-nx}<1\mbox{ and thus } \\ \\ &f_n(x)<e^{2\ln(n)} \end{align*}

You do need to prove the assertion $2\ln(x)-nx<0, \forall n, \forall x \geq 0$

I leave that to you.

3. ## Re: Proving Uniform Convergence with epsilon proof.

fn(x) < e2ln(n) = n2does not appear to help I need this less than epsilon and n2 is unbounded above. This does give me an idea though.

fn(x) = n2x2e-nx = e2ln(n)+2ln(x)-nx = e2ln(n) + 2ln(x) - x + (1-n)x = e2ln(n) + (2 ln(x) - x) + (1-n)x < e2ln(n)+(1-n)x =n2/e(n-1)x < n2/e(n-1) ---> 0 as n ---> INF

I prove that 2 ln x - x < 0 by letting g(x) = x - 2 ln x. Then g'(x) = 1 - 2/x > 0 when x > 2 and g'(x) < 0 when 0 < x < 2. Thus, g(x) > g(2) = 2 - 2 ln 2 > ln e^2 - ln 2^2 > 0 since e > 2. Thus, g(x) > 0 which implies 2 ln x - x < 0.

4. ## Re: Proving Uniform Convergence with epsilon proof.

Ref romsek post:

You finally get to fn(x) < e2ln(n) < ε which is impossible, as pointed out by mathguy, who still has to find n>N

On the other hand:

expand enx and divide top and bottom by n2x2

fn(x) = 1/(1/nx +1/2 + nx/6 + n2x2/12 + …..}

1/nx +1/2 >1/2
nx/6 > n/6
n2x2/12 > n2/12
.
.
fn(x) < 1/(1/2 + n/6 + ….) < 1/(n/6) < ε

n > 6/ε

5. ## Re: Proving Uniform Convergence with epsilon proof.

Originally Posted by Hartlw
Ref romsek post:

You finally get to fn(x) < e2ln(n) < ε which is impossible, as pointed out by mathguy, who still has to find n>N

On the other hand:

expand enx and divide top and bottom by n2x2

fn(x) = 1/(1/nx +1/2 + nx/6 + n2x2/12 + …..}

1/nx +1/2 >1/2
nx/6 > n/6
n2x2/12 > n2/12
.
.
fn(x) < 1/(1/2 + n/6 + ….) < 1/(n/6) < ε

n > 6/ε
I'm glad you were able to solve it. Sometimes I get a bit overwhelmed trying to answer 15 questions a day and don't always get it right.

6. ## Re: Proving Uniform Convergence with epsilon proof.

I understand. I was thinking the same thing and almost mentioned it out of sympathy, except for past experience.

7. ## Re: Proving Uniform Convergence with epsilon proof.

I do have an N > 0 though. It's the one used in lim (n->INF) n^2/(e^(n-1)) = INF.

8. ## Re: Proving Uniform Convergence with epsilon proof.

Originally Posted by mathguy25
I do have an N > 0 though. It's the one used in lim (n->INF) n^2/(e^(n-1)) = INF.
So what is N?

9. ## Re: Proving Uniform Convergence with epsilon proof.

N > 2 ln (1/e) where e is epsilon

10. ## Re: Proving Uniform Convergence with epsilon proof.

Originally Posted by mathguy25
N > 2 ln (1/e) where e is epsilon
how did you get that?

en-1/n2 > 1/ε gives (n-1)-2ln(n)>ln(1/ε)