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Uniform convergence

Given the sequence of functions f_n:\mathbb{R} \rightarrow \mathbb{R} where n \in \mathbb{N} such that

\displaystyle f_n(x) = \frac{n}{n^3+x^2}

Prove that

(i) the serieses \displaystyle \sum_{n=1}^{\infty} f_n and \displaystyle \sum_{n=1}^{\infty} f'_n converge uniformly to functions f, g:\mathbb{R} \rightarrow \mathbb{R} .

(ii) the functions  f, g are continuous.

(iii) f'=g.

(iv) it holds that

\displaystyle \int_{-1}^{1} f(x) \, {\rm d}x = 2 \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}} \quad, \quad \int_{-\pi}^{\pi} x^4 g(x) \, {\rm d}x = 0

Solution

(i) This is an immediate consequence of the Weierstrass M Test . We simply note that \displaystyle 0 \leq f_n(x)\leq \frac{1}{n^2} and of course

\displaystyle |f_n'(x)|=\frac{|2nx|}{(n^3+x^2)^2} \leq \frac{2}{n^{7/2}}

(ii) The uniform limit of continuous functions is continuous. This is enough for us to extract that both f and g throughout \mathbb{R}.

(iii) For a limit of differentiable functions, a sufficient condition for claiming that the derivative of the limit is the limit of the derivatives is the uniform convergence of the sequence of derivatives. Since the series for g converges uniformly, we can claim that f'=g.

(iv) For the first integral we have that

    \begin{align*} \int_{-1}^{1} f(x) \, {\rm d}x &= \int_{-1}^{1} \sum_{n=1}^{\infty} f_n(x) \, {\rm d}x \\ &= \sum_{n=1}^{\infty} \int_{-1}^{1} f_n(x) \, {\rm d}x\\ &= \sum_{n=1}^{\infty} n \int_{-1}^1\frac{{\rm d}x}{n^3+x^2}\\ &= 2\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}} \end{align*}

and for the second integral we have

    \begin{align*} \int_{-\pi}^{\pi} x^4 g(x) \, {\rm d}x &=\int_{-\pi}^{\pi} x^4 \sum_{n=1}^{\infty} f_n'(x) \, {\rm d}x \\ &=2 \int_{-\pi}^{\pi} x^4 \sum_{n=1}^{\infty} \frac{nx}{\left ( n^3 +x^2 \right )^{2}} \, {\rm d}x \\ &= 2\sum_{n=1}^{\infty} n \cancelto{0}{\int_{-\pi}^{\pi} \frac{x^5}{\left ( n^3 +x^2 \right )^{2}} \, {\rm d}x}\\ &= 0 \end{align*}

since the function \displaystyle \frac{x^5}{\left ( n^3 +x^2 \right )^{2}} is odd.

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