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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On matrices

Let A, B be two 3 \times 3 matrices with real entries. Prove that

A - \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} = ABA

provided all the inverses appearing on the left hand side exist.

(Vojtech Jarnik / 2nd Category/ 2015)

Solution

Let A, B be elements of an arbitrary associative algebra with unit. Then:

\begin{aligned} \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\ &=\left ( A^{-1} \left ( \left ( B^{-1} - A \right ) +A \right )\left ( B^{-1} -A \right )^{-1} \right )^{-1} \\ &= \left ( A^{-1} B^{-1} \left ( B^{-1} -A \right )^{-1} \right )^{-1}\\ &= \left ( B^{-1}-A \right ) BA \\ &= A - ABA \end{aligned}

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A multiple sum

Evaluate

\displaystyle \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k \left ( n_1+n_2+\cdots+n_k+1 \right )}

(Vojtech Jarnik / 2nd Category/ 2011)

Solution

We are applying the classical trick, thus:

\begin{aligned} \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k \left ( n_1+n_2+\cdots+n_k+1 \right )} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k} \int_{0}^{1}x^{n_1+n_2+\cdots+n_k} \, {\rm d}x\\ &= \int_{0}^{1} \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k} x^{n_1+n_2+\cdots+n_k} \, {\rm d}x \\ &= \int_{0}^{1} \left ( -\ln (1-x) \right )^k \, {\rm d}x\\ &= k! \end{aligned}

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Riemann integrability

Let f:[0, 1] \rightarrow \mathbb{R} be defined as:

f(x)= \left\{\begin{matrix} 0&, &x \in [0,1]\cap \left ( \mathbb{R} \setminus \mathbb{Q} \right ) \\ x_n &, &x=q_n \in [0,1] \cap \mathbb{Q} \\ \end{matrix}\right.

where x_n is a sequence such that \lim x_n =0 and 0 \leq x_n \leq 1 and q_n be an enumeration of the rationals of the interval [0, 1]. Prove that f is Riemann integrable and that \bigintsss_0^1 f(x)\, {\rm d}x=0.

Solution

We define

f_n = \left\{\begin{matrix} x_n &, & x \in \{ q_1, \dots, q_n\}\\ 0 & , & x \notin \{ q_1, \dots, q_n \} \end{matrix}\right.

Apparently f_n are Riemann integrable and \bigintsss_0^1 f_n(x) \, {\rm d}x =0. Also f_n \rightarrow f uniformly because f_n, f coincide except of some points q_m (m>n). Hence , since \lim x_n=0 , we have that:

0\leq \left | f(x)-f_n(x) \right |\leq \sup \left \{ x_m \mid m>n \right \} \xrightarrow{n \rightarrow +\infty}0

Thefore f is Riemann integrable and thus:

    \begin{align*} \int_{0}^{1} f(x) \, {\rm d}x &=\int_{0}^{1} \lim_{n \rightarrow +\infty} f_n (x) \, {\rm d}x \\ &= \lim_{n \rightarrow +\infty} \int_{0}^{1} f_n (x) \, {\rm d}x\\ &= \lim_{n \rightarrow +\infty}0 \\ &= 0 \end{align*}

The exercise can also be found in mathematica.gr

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Is it a conservative field?

(i) Let \mathbb{D} \subset \mathbb{R}^2 be the unit disk and \partial \mathbb{D}  be its positive oriented boundary. Evaluate the line integral

\displaystyle \mathcal{J} = \ointctrclockwise \limits_{\partial \mathbb{D}} (x-y^3, x^3-y^2)\, {\rm d}(x, y)

(ii) Can you deduce if the function

f(x, y) =(x-y^3, x^3-y^2)

is a conservative field using the above question?

Solution

(i) We are invoking Green’s theorem. Thus:

\begin{aligned} \ointctrclockwise \limits_{\partial \mathbb{D}} \left ( x-y^3, x^3-y^2 \right ) {\rm d}\left ( x, y \right ) &=\iint \limits_{\mathbb{D}} \left ( \frac{\partial }{\partial x}Q(x, y) - \frac{\partial }{\partial y} P(x, y)\right )\, {\rm d}(x, y) \\ &= \iint \limits_{\mathbb{D}} \left ( 3x^2 + 3y^2 \right )\, {\rm d}(x, y)\\ &= 3\int_{0}^{1}\int_{0}^{2\pi} \, {\rm d} \theta \, {\rm d} \rho\\ &= 6\pi \neq 0 \end{aligned}

(ii) If f was a conservative field then then the line integral over all closed curves would have to be 0. But in the previous question we found one closed curve whose line integral is not 0. Thus f  is not a conservative field.

The exercise can also be found in the Jom Forum here.

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