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Convergence in mean square (or L^2) sense

  1. Nov 9, 2009 #1
    Convergence in "mean square" (or L^2) sense

    1. The problem statement, all variables and given/known data
    This is an example from a textbook (with solutions) in which I am feeling confused.
    Let fn(x) = [n/(1+n2x2)] - (n-1)/[1+(n-1)2x2] in the interval 0<x<L. This series telescopes so that
    N
    ∑ fn(x) = N/(1+N2x2)
    n=1

    L
    ∫ [∑ fn(x)]2 dx =
    0
    L
    ∫ N2/(1+N2x2)2 dx =
    0
    NL
    ∫ N/(1+y2)2 dy (let y=Nx)
    0
    This last line -> +∞ as N->∞
    Since it does not converge to 0, the series does NOT converge in the mean-square (or L2) sense to f(x)=0.

    2. Relevant equations/concepts
    Convergence in mean square/L2 sense

    3. The attempt at a solution
    N/A

    (i) Now I don't understand why we have to use the change of variable y=Nx. What is the point of doing this?


    (ii) Also, WHY as N->∞,
    NL
    ∫ N/(1+y2)2 dy -> +∞ ?
    0

    Can someone please explain?
    Thank you!
     
  2. jcsd
  3. Nov 10, 2009 #2
    Re: Convergence in "mean square" (or L^2) sense

    Why is it a good idea to bring N into the limits of integration? After the change of variable, N appears both in the upper limit of integration AND in the integrand...
     
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