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Chapter 8: Problem 19

Show that the sequence \(\left(x^{2} e^{-n x}\right)\) converges uniformly on \([0, \infty)\).

Short Answer

Expert verified
The sequence \(\left(x^{2} e^{-n x}\right)\) converges uniformly on [0, 鈭�).

Step by step solution

01

Find the derivative

Firstly, differentiate the function \(f(x) = x^{2} e^{-n x}\) using the product and chain rules. The derivative will be \(f'(x) = 2x e^{-n x} - n x^{2} e^{-n x}\).
02

Find the critical points

Now, find the values of x where the derivative equals 0. This can be done by taking \(f'(x) = 0\) and solving the equation for x, yielding \(x = 0\) and \(x = 2/n\). These are the critical points of the function.
03

Find the maximum of the function

Next, to find the maximum of the function \(f(x)\) on the interval [0, 鈭�), we evaluate the function at the critical points and at infinity, and pick the largest value. We get \(f(0) = 0\), \(f(2/n) = 4e^{-2}/n\), and we can see that \(f(x)\) approaches 0 as \(x\) approaches infinity by L'Hopital's rule.
04

Show that the sequence converges to 0

From the maximum value \(f(2/n) = 4e^{-2}/n\), we see that this approaches 0 as \(n\) approaches infinity, which shows that the sequence \(x^{2} e^{-n x}\) converges to 0 as \(n\) goes to infinity.
05

Demonstrate uniform convergence

The distance between \(f_n(x) = x^{2} e^{-n x}\) and the limit function 0 is less than or equal to \(4e^{-2}/n\), and so for every 蔚 > 0 there exists an \(N\) such that if \(n > N > 4e^{-2}/蔚\), then \(|f_n(x) - 0| < 蔚\) for all \(x\) and all \(n > N\). Thus, by definition, we can conclude the sequence converges uniformly on [0, 鈭�).

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Most popular questions from this chapter

Evaluate \(\lim \left(e^{-n x}\right)\) for \(x \in \mathbb{R}, x \geq 0\)

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