Proving Uniform Convergence with epsilon proof.

Prove that f_{n}(x) = n^{2}x^{2}e^{-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 n^{2}x^{2}e^{-nx} < n^{2}x^{2}e^{-n}. I can't seem to eliminate this x^{2}. Any hints? Thanks.

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.

Re: Proving Uniform Convergence with epsilon proof.

f_{n}(x) < e^{2ln(n)} = n^{2}does not appear to help I need this less than epsilon and n^{2} is unbounded above. This does give me an idea though.

f_{n}(x) = n^{2}x^{2}e^{-nx} = e^{2ln(n)+2ln(x)-nx} = e^{2ln(n) + 2ln(x) - x + (1-n)x} = e^{2ln(n) + (2 ln(x) - x) + (1-n)x} < e^{2ln(n)+(1-n)x }=n^{2}/e^{(n-1)x} < n^{2}/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.

Re: Proving Uniform Convergence with epsilon proof.

Ref romsek post:

You finally get to f_{n}(x) < e^{2ln(n)} < ε which is impossible, as pointed out by mathguy, who still has to find n>N

On the other hand:

expand e^{nx} and divide top and bottom by n^{2}x^{2}

f_{n}(x) = 1/(1/nx +1/2 + nx/6 + n^{2}x^{2}/12 + …..}

1/nx +1/2 >1/2

nx/6 > n/6

n^{2}x^{2}/12 > n^{2}/12

.

.

fn(x) < 1/(1/2 + n/6 + ….) < 1/(n/6) < ε

n > 6/ε

Re: Proving Uniform Convergence with epsilon proof.

Quote:

Originally Posted by

**Hartlw** Ref romsek post:

You finally get to f_{n}(x) < e^{2ln(n)} < ε which is impossible, as pointed out by mathguy, who still has to find n>N

On the other hand:

expand e^{nx} and divide top and bottom by n^{2}x^{2}

f_{n}(x) = 1/(1/nx +1/2 + nx/6 + n^{2}x^{2}/12 + …..}

1/nx +1/2 >1/2

nx/6 > n/6

n^{2}x^{2}/12 > n^{2}/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.

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.

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.

Re: Proving Uniform Convergence with epsilon proof.

Quote:

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?

Re: Proving Uniform Convergence with epsilon proof.

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

Re: Proving Uniform Convergence with epsilon proof.

Quote:

Originally Posted by

**mathguy25** N > 2 ln (1/e) where e is epsilon

how did you get that?

e^{n-1}/n^{2} > 1/ε gives (n-1)-2ln(n)>ln(1/ε)