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Math Help - Pointwise and uniform convergence

  1. #1
    Super Member TheChaz's Avatar
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    Cool Pointwise and uniform convergence

    From my exam this morning:
    Show that f_n(x) = nx(1-x)^n converges to f == 0 pointwise on [0, 1], but not uniformly.

    There was some hint about  \frac{n}{n + 1}^n going to 1/e or something like that.

    I tried to let E > 0, using the definitions and whatnot, but it didn't seem to go anywhere! It was MUCH easier on the last test, when f_n was cos(x/n)...

    Anywho... if you've got a solution, great. Otherwise, I'll post the answer in September after I've seen the prof again
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  2. #2
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    Pointwise convergence. As n increases, what happens? More and more points are included as not very far from f(x) = 0. In fact, there isn't a point that won't be included eventually.

    Uniform Convergence. The problem is over on the left, near x = 0. A little elementary calculus shows the first derivative having a zero at x = 1/(1+n). This certainly moves toward x = 0. What is the value of the function at x = 1/(1+n)? Whatever it is, it's limit needs to be zero if we are to establish Uniform Convergence

    It is \left(\frac{n}{n+1}\right)^{n+1}.

    Use your favorite logarithm technique to prove to what this converges.

    Rewriting as \frac{ln\left(\frac{n}{n+1}\right)}{\frac{1}{n+1}} may not be obvious, but this invites l'Hospital.

    A single application (and a little algebra) produces -(1 + (1/n)). Well, that's clear enough. So, the limit of the logarithm is -1. Where does that leave us?
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  3. #3
    Super Member TheChaz's Avatar
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    Thanks for the reply. The uniform convergence argument is surely what we were expected to produce. There WAS a hint about finding f'...

    I'm still not convinced about pointwise here. Do the function values approach 0 because of the n in the exponent?
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  4. #4
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    Define g(y)=yx(1-x)^y=\frac{yx}{e^{-y\ln(1-x)}} and take  y\rightarrow \infty (regard x as a constant).
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  5. #5
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Jose27 View Post
    Define g(y)=yx(1-x)^y=\frac{yx}{e^{-y\ln(1-x)}} and take  y\rightarrow \infty (regard x as a constant).
    Ok, that makes sense. Thanks.

    I just woke up from a dream in which I got 100% on this exam...
    THAT didn't happen!
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