5: Let p > 0. Let [tex]f_n (x) = x n^p e^{-nx}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Question on uniform convergence

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**