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On uniform convergence of sequence

  1. Oct 5, 2005 #1
    Suppose (f_n} is a sequence of functions where f_n(x) = x / (1 + n^2 x^2).

    I am finding the pointwise limit of the sequence of {f_n'(x)} on the interval
    (-oo, + oo)...in which {f_n'(x)} is the sequence of functions obtained from the derivative of x / (1 + n^2 x^2) and I am trying to find out if this {f_n'(x)}
    converges uniformly or not.

    SOlution:

    Well, the derivative of x / (1 + n^2 x^2) is :

    (1 - n^2 x^2) / ( 1 + n^2 x^2 )^2.

    Now,
    when x = 0, lim (n -> +oo) f_n'(0) = (1 - n^2 0^2) / ( 1 + n^2 0^2 )^2 = 1

    When x is not 0,

    lim (n -> +oo) f_n'(x) = (1 - n^2 x^2) / ( 1 + n^2 x^2 )^2

    lim (n -> +oo) f_n'(x) = (1 - n^2 x^2) / ( 1 + 2n^2 x^2 + n^4 x^4 )

    Multiplying both numerator and denominator by 1/n^4 yields

    lim (n -> +oo) f_n'(x) = ( 0 - 0) / ( 0 + 0 + x^4 ) = 0 / x^4 = 0.
    Since for a large value of x, I can make my n larger than x.

    Hence, f(x) = 0 when x = 0 and f(x) = 1, when x is not 0.

    Since the function f(x) is not continuous on the interval (-oo, +oo), then
    {f_n'(x)} does not converge uniformly on that interval.



    Have I done my solution correctly?
     
  2. jcsd
  3. Jan 3, 2010 #2
    It looks correct.
     
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