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Thread: Pointwise and uniform convergence

  1. #1
    Member thaopanda's Avatar
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    Pointwise and uniform convergence

    a) For each n $\displaystyle \in$ N consider $\displaystyle F_{n}$ : $\displaystyle [0, \infty) \rightarrow $ R be given by $\displaystyle F_{n}(x) := (x^n)e^{-nx}$. Show that {$\displaystyle F_{n}$}$\displaystyle _{n \in N}$ converges in a pointwise fashion to the zero function on $\displaystyle [0, \infty)$.

    b) Decide whether the convergence in part a is uniform on $\displaystyle [0, \infty)$.

    please help
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    a) For each n $\displaystyle \in$ N consider $\displaystyle F_{n}$ : $\displaystyle [0, \infty) \rightarrow $ R be given by $\displaystyle F_{n}(x) := (x^n)e^{-nx}$. Show that {$\displaystyle F_{n}$}$\displaystyle _{n \in N}$ converges in a pointwise fashion to the zero function on $\displaystyle [0, \infty)$.

    b) Decide whether the convergence in part a is uniform on $\displaystyle [0, \infty)$.

    please help
    a) Well, for all $\displaystyle x$, $\displaystyle \lim_{n\to\infty}\frac{x^n}{e^{nx}}=0$. One way to see this is that if you use L'Hopital's Rule $\displaystyle n$ times, you'll get $\displaystyle \lim_{n\to\infty}\frac{n!}{n!e^{nx}}=0$

    b) The derivative of $\displaystyle x^ne^{-nx}$ is $\displaystyle nx^{n-1}e^{-nx}(1-x)$, so the max on $\displaystyle [0,\infty)$ occurs at $\displaystyle x=1$ (for all $\displaystyle n$), and it follows that $\displaystyle x^ne^{-nx}\leq e^{-n}$, which can clearly be made less than any $\displaystyle \epsilon$. (Choosing $\displaystyle n>-\ln\epsilon$ does it.)
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