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Math Help - Proving Uniform Convergence with epsilon proof.

  1. #1
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    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.
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    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.
    Thanks from mathguy25
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    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.
    Last edited by mathguy25; March 3rd 2014 at 05:23 PM.
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    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
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    Re: Proving Uniform Convergence with epsilon proof.

    Quote Originally Posted by Hartlw View Post
    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.
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    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.
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    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.
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    Re: Proving Uniform Convergence with epsilon proof.

    Quote Originally Posted by mathguy25 View Post
    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?
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    Re: Proving Uniform Convergence with epsilon proof.

    N > 2 ln (1/e) where e is epsilon
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    Re: Proving Uniform Convergence with epsilon proof.

    Quote Originally Posted by mathguy25 View Post
    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)
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