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Simfish
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5: Let p > 0. Let [tex]f_n (x) = x n^p e^{-nx}[/tex]
(i) For what values of p is[tex] f_n (x) -> 0[/tex] uniformly on [0,1]?
(ii) For what values of p is it true that [tex]lim_{n->\infty} \int_0^1 f_n (x) dx = 0[/tex]
For (i), we know that uniform convergence implies that the convergence to 0 depends ONLY on the value of N, so it should converge to 0 for all x. Now, we can make our function arbitrarily small for any p, if p < 1,since then for n -> infinity, both functions converge to 0.
Let's try (i) first.
We want to find the maximum value of [tex] f_n (x)[/tex]. So...
[tex]
\begin{align}
\left[ f_n (x) \right]' = \left[- x n^{p+1} e^{-nx} + n^p e^{-nx}\right] = 0 \\
\left[ e^{nx} n^{p} (-nx + 1) \right] = 0 \\
-nx + 1 = 0 \\
nx = 1 \\
x = 1/n,\\
\end{align}
[/tex]
plug back into equation. the value of the maximum is... [tex]
\left[ n^{p-1} * \frac{1}{e} \right]
[/tex]
which can only imply uniform convergence for all x for p [tex]<[/tex] 1.
But what of (ii)? Integration by parts.
[tex]
x n^p \left[\frac{-x e^{-nx}}{n} - \frac{e^{-nx}}{n^2} \right]_{0}^{1} = \left[\frac{-e^{-n}}{n} - \frac{e^{-n}}{n^2} + \frac{1}{n^2} \right] x n^p
[/tex]
...which would imply convergence of the integral for all p < 2. Am I wrong anywhere? Someone else said that p > 0, but this relation seems to be completely relevant for all negative values of p as well.
(i) For what values of p is[tex] f_n (x) -> 0[/tex] uniformly on [0,1]?
(ii) For what values of p is it true that [tex]lim_{n->\infty} \int_0^1 f_n (x) dx = 0[/tex]
For (i), we know that uniform convergence implies that the convergence to 0 depends ONLY on the value of N, so it should converge to 0 for all x. Now, we can make our function arbitrarily small for any p, if p < 1,since then for n -> infinity, both functions converge to 0.
Let's try (i) first.
We want to find the maximum value of [tex] f_n (x)[/tex]. So...
[tex]
\begin{align}
\left[ f_n (x) \right]' = \left[- x n^{p+1} e^{-nx} + n^p e^{-nx}\right] = 0 \\
\left[ e^{nx} n^{p} (-nx + 1) \right] = 0 \\
-nx + 1 = 0 \\
nx = 1 \\
x = 1/n,\\
\end{align}
[/tex]
plug back into equation. the value of the maximum is... [tex]
\left[ n^{p-1} * \frac{1}{e} \right]
[/tex]
which can only imply uniform convergence for all x for p [tex]<[/tex] 1.
But what of (ii)? Integration by parts.
[tex]
x n^p \left[\frac{-x e^{-nx}}{n} - \frac{e^{-nx}}{n^2} \right]_{0}^{1} = \left[\frac{-e^{-n}}{n} - \frac{e^{-n}}{n^2} + \frac{1}{n^2} \right] x n^p
[/tex]
...which would imply convergence of the integral for all p < 2. Am I wrong anywhere? Someone else said that p > 0, but this relation seems to be completely relevant for all negative values of p as well.
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