Joy of Mathematics

Maximum of a function

July 17, 2022July 17, 2022 Tolaso

Given a real number $\alpha>0$ find the maximum of the function

$f(E) = \iiint \limits_{E} \left ( \alpha^2 - x^2 - y^2 - z^2 \right )\, \mathrm{d}V$

where $E$ is any region in space.

Solution

Let $D = \left \{ (x, y , z): x^2 + y^2 + z^2 \leq \alpha^2 \right \}$ . Then

$f(E) = f \left ( E \cap D \right ) + f \left ( E \setminus D \right )$

In the region $E \setminus D$ we have $x^2+y^2+z^2 > \alpha^2$ hence $\alpha^2-x^2-y^2-z^2 < 0$ which gives $f(E \setminus D) \leq 0$ . Thus ,

$\begin{align*} f\left ( E \right ) &\leq f \left ( E \cap D \right ) \\ &\leq f \left ( D \right ) \\ &= \iiint \limits_{D} \left ( \alpha^2 - x^2 - y^2 - z^2 \right )\, \mathrm{d}V \\ &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\alpha} \left ( \alpha^2 - \rho^2 \right ) \rho^2 \sin \varphi \, \mathrm{d} \left ( \rho, \varphi, \theta \right ) \\ &= 2 \pi \int_{0}^{\pi} \sin \varphi \, \mathrm{d} \varphi \int_{0}^{\alpha} \rho^2 \left ( \alpha^2 - \rho^2 \right )\, \mathrm{d} \rho \\ & = 4 \pi \int_{0}^{\alpha} \left ( \rho^2 \alpha^2 -\rho^4 \right )\, \mathrm{d} \rho \\ & = \frac{8 \pi \alpha^5}{15} \end{align*}$

Differentiable

December 27, 2020January 18, 2021 Tolaso

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable on $\mathbb{R}^n \setminus \{0 \}$ and continuous at $0$ . If

$\lim_{x \rightarrow 0} \frac{\partial f}{\partial x_i} (x) =0$

for $i=1, 2, \dots , n$ then prove that $f$ is differentiable at $0$ .

Solution

Let us assume without loss of generality that $f(0)=0$ . We will show that $f(x) = o \left ( \left \| x \right \|_{\infty} \right )$ as $x \rightarrow 0$ and that means that $f$ is differentiable at $x=0$ with $f'(0)=0$ .

Fix $\epsilon>0$ and choose $\delta \in \left( 0 , \frac{1}{n}\right)$ such that $0< \left \| x \right \|_{\infty} < \delta$ , $1 \leq i \leq n$ . Therefore $\left | f_{x_i} \right |<\epsilon$ . Now suppose $\left \| x \right \|_{\infty}<\delta$ . Let $a_k\; , \; 0 \leq k \leq n$ be the point that coincides with $x$ on the first $k$ coordinates and is $0$ elsewhere. Then $a_0,\dots,a_n$ is a path from $0$ to $x$ and each vector $a_{k+1}-a_k$ is parallel to one of the axes. Hence

$\begin{align*} \left | f(x) \right |&=\left| \sum_{k=0}^{n-1}(f(a_k)-f(a_{k-1}) \right|\\ &\leq \sum_{k=0}^{n-1} \left|f(a_k)-f(a_{k-1}) \right |\\ &\leq n\epsilon \left \|a_k-a_{k-1}| \right \|\\ &<n\delta \epsilon \\ &<\epsilon \end{align*}$

Convergence of series

October 28, 2020November 16, 2020 Tolaso

For $v =\langle x_1, x_2, \dots, x_n \rangle \in \mathbb{R}^n$ we define $\left \| v \right \|_p = \left ( \sum \limits_{i=1}^{n} \left | x_i \right |^p \right )^{1/p}$ and $\left \| v \right \|_{\infty} = \max \limits_{1 \leq i \leq n} \left | x_i \right |$ . For which $p$ does the series

$\sum_{p=1}^{\infty} \left ( \left \| v \right \|_p - \left \| v \right \|_{\infty} \right)$

converge?

Solution ( Robert Tauraso )

We may assume that $v$ is not the zero vector and $n>1$ otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.

(a) If the above condition is satisfied then, without loss of generality, let $x_1$ be the component of maximal absolute value and let $\displaystyle t = \frac{\max \limits_{2 \leq i \leq n} \left | x_i \right |}{\left | x_i \right |} \in [0, 1]$ . Hence , as $p \rightarrow +\infty$

$\begin{align*} 0 &\leq \left \| v \right \|_p - \left \| v \right \|_{\infty} \\ &= \left \| v \right \|_{\infty} \left ( \left ( 1 + \left ( n-1 \right )t^p \right )^{1/p} - 1 \right ) \\ &=\left \| v \right \|_{\infty} \left ( \exp \left ( \frac{\ln \left ( 1+ (n-1) t^p \right )}{p} \right ) - 1\right ) \\ &\sim \left \| v \right \|_{\infty} \left ( n-1 \right ) \frac{t^p}{p} \end{align*}$

and the given series is convergent because $\displaystyle \sum_{p=1}^{\infty} \frac{t^p}{p} < + \infty$ .

(b) If the above condition is not satisfied, then there are at least $2$ components of maximal absolute value and therefore

$\begin{align*} \left \| v \right \|_p - \left \| v \right \|_{\infty} & \geq \left \| v \right \|_{\infty} \left ( 2^{1/p} -1 \right )\\ &= \left \| v \right \|_{\infty} \left ( \exp \left ( \frac{\ln 2}{p} \right ) -1 \right ) \\ &\sim \left \| v \right \|_{\infty} \frac{\ln 2}{p} \end{align*}$

and the given series is not convergent because $\displaystyle \sum_{p=1}^{\infty} \frac{1}{p} = +\infty$ .

A double arctan integral

August 8, 2019August 9, 2019 Tolaso

Let $\Delta:\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 4x\}$ . Evaluate the integral:

$\mathcal{J} = \iint \limits_{\Delta} \arctan e^{xy}\,\mathrm{d}y\,\mathrm{d}x$

Solution

The key observation to nail the integral is that the domain of integration is symmetric with respect to the $x$ axis.

Rendered by QuickLaTeX.com

Hence , a symmetry might as well work. So,

$\begin{align*} \mathcal{J} &= \iint \limits_{\Delta} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &=\int_{0}^{4} \int_{-\sqrt{4x-x^2}}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &=\int_{0}^{4} \int_{-\sqrt{4x-x^2}}^{0} \arctan e^{xy} \, \mathrm{d}(y, x) + \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) \\ &= \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{xy} \, \mathrm{d}(y, x) + \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \arctan e^{-xy} \, \mathrm{d}(y, x) \\ &= \int_{0}^{4} \int_{0}^{\sqrt{4x-x^2}} \frac{\pi}{2} \, \mathrm{d}(y, x) \\ &= \frac{\pi}{2} \int_{0}^{4} \sqrt{4x-x^2} \, \mathrm{d}x \\ &= \frac{\pi}{2} \cdot 2 \pi \\ &= \pi^2 \end{align*}$

since

$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2} \quad \text{forall} \;\; x>0$

Recursive formula of volume of sphere

August 2, 2019 Tolaso

Let $\mathcal{V}_n(1)$ be the volume of the sphere centered at $0$ and radius $1$ in $\mathbb{R}^n$ . Prove that for $n \geq 3$ it holds that

$\mathcal{V}_n(1) = \frac{2\pi}{n} \mathcal{V}_{n-2}(1)$

Solution

The volume of the sphere in $\mathbb{R}^n$ is given by:

$\mathcal{V}_n (1) = \idotsint \limits_{x_1^2+x_2^2+\cdots+x_n^2 \leq 1} 1 \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )$

Parametrize the sphere by

$\begin{matrix} x_1 &= & r \cos \theta_1 \\ x_2 & = & r \sin \theta_1 \cos \theta_2 \\ x_3 & = & r \sin \theta_1 \sin \theta_1 \cos \theta_3 \\ \vdots \\ x_{n-1} & = & r \sin \theta_1 \sin \theta_2 \cdots \sin \theta_{n-2} \cos \theta_{n-1} \\ x_n & = & r \sin \theta_1 \sin \theta_2 \cdots \sin \theta_{n-1} \end{matrix}$

taking

$\begin{matrix} 0 \leq r \leq 1\\ 0 \leq \theta_i \leq \pi & & \text{forall} \;\; i=1, 2, \dots , n-2 \\ 0 \leq \theta_{n-1} < 2 \pi \end{matrix}$

It then follows from the Change of Variables formula that the rectangular volume element $\mathrm{d} \mathcal{V} = \mathrm{d}x_1 \; \mathrm{d}x_2 \cdots \mathrm{d}x_n$ can be written in spherical coordinates as

$\begin{align*} \mathrm{d}\mathcal{V} &= \left | \det \left ( \frac{\partial x_1}{\partial \left ( r , \theta_j \right )} \right ) \right | \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \theta_2, \dots, \theta_{n-1} \right ) \\ &= r^{n-1} \sin^{n-2} \theta_1 \sin^{n-3} \theta_2 \cdots \sin \theta_{n-2} \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \theta_2, \dots, \theta_{n-1} \right ) \end{align*}$

Thus,

$\begin{align*} \mathcal{V}_n(1) &= \idotsint \limits_{x_1^2+x_2^2+\cdots+x_n^2 \leq 1} 1 \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &\!\!\!\!=\int_{0}^{2\pi} \int_{0}^{\pi} \cdots \int_{0}^{1} r^{n-1} \sin^{n-2} \theta_1 \cdots \sin \theta_{n-2} \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \dots, \theta_{n-1} \right ) \\ &= \int_{0}^{2\pi} \, \mathrm{d} \theta_{n-1} \int_{0}^{1} r^{n-1} \, \mathrm{d}r \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \\ &= \frac{2\pi}{n} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \end{align*}$

Hence,

(1) $\begin{equation*} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta =\frac{n}{2\pi} \mathcal{V}_{n}(1) \end{equation*}$

In particular since $n-4 = (n-2)-2$ we get that:

(2) $\begin{equation*} \int_{0}^{\pi} \sin^{n-4} \mathrm{d}\theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta = \frac{n-2}{2\pi} \mathcal{V}_{n-2} (1) \end{equation*}$

Using $(2)$ as well as Wallis’ integral we are able to prove the result. Let us assume that $n$ is even, then:

$\begin{align*} \mathcal{V}_n(1) &=\frac{2\pi}{n} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-3} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-4} \theta \, \mathrm{d} \theta \cdots \\ &\quad \quad \quad \quad \quad \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \\ &=\frac{2\pi}{n}\int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-3} \theta \, \mathrm{d} \theta \cdot \frac{n-2}{2\pi} \, \mathcal{V}_{n-2} (1) \\ &= \frac{2\pi}{n} \cdot \left ( \frac{n-3}{n-2} \cdots \frac{1}{2} \cdot \pi \right ) \cdot \left ( \frac{n-4}{n-3} \cdots \frac{2}{3} \cdot 2 \right ) \cdot \frac{n-2}{2\pi} \, \mathcal{V}_{n-2} (1)\\ &= \frac{2\pi}{n} \cdot \mathcal{V}_{n-2} (1) \end{align*}$

If $n$ is odd we work similarly.