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13

Let's employ the weirdo substitution taught by my brother: $\sin x=\tanh t$. Doing so, one will get \begin{align} K&=\int_0^{\infty}\sqrt{\frac{(1-\tanh t)^{n-2}}{(1+\tanh t)^{n+2}}}\ln\left(\frac{1-\tanh t}{1+\tanh t}\right)\ \frac{dt}{\cosh t}\\[10pt] &=\int_0^{\infty}\sqrt{\left(\frac{\cosh t-\sinh t}{\cosh t+\sinh t}\right)^{n-2}}\frac{\cosh t}{(\...


12

Hint. One may evaluate $(1)$ with the following steps. From the geometric series evaluation $$ \sum_{n=2}^{\infty}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}=\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}},\quad 0<x<\frac{\pi}2, $$ one may write $$ I=\int_0^{\pi/2}\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}}...


5

Hint: $$|f_1(x)| \leqslant M, \\|f_2(x)| \leqslant M(x-a), \\ |f_3(x)| \leqslant \frac{1}{2!}M(x-a)^2, \\ \ldots$$


3

First we have : $$ f(x) = ((x_1)^2 +(x_2)^2)^2 - (x_1^2 + x_2^2) = x_1^4 + 2x_1^2x_2^2 + x_2^4 -x_1^2 -x_2^2 $$ Then we have $$ \frac{\partial}{\partial x_1} f(x) = 4x_1^3 + 4x_1 x_2^2 - 2x_1 = 0 $$ $$ \frac{\partial}{\partial x_2} f(x) = 4x_2^3 + 4x_1^2 x_2 - 2x_2 = 0 $$ Clearly $(0,0)$ is a solution. We can then divide the first equation by $x_1$ and the ...


3

In polar coordinates the expression is $$r\frac{\cos^2t\sin^3t}{\cos^4 t + \sin^4 t}.$$ The denominator in this fraction has a positive minimum; thus the fraction is a bounded function of $t,$ and the $r$ in front guarantees a limit of $0.$


3

Hint $$\int{\frac{1}{1+{{\sin }^{2}}t\,}}\,dt=\int{\frac{1+{{\tan }^{2}}t}{1+2{{\tan }^{2}}t\,}}\,dt=\frac{\sqrt{2}}{2}{{\tan }^{-1}}(\sqrt{2}\tan x)+c$$


2

Eric, first of all, $\omega$ is a $1$-form on the image of $g$ and $R$ is a region contained in the image of $g$, so it certainly makes sense to integrate $\omega$ over $\partial R$. Here's what I suggest: (1) Compute $g^*\omega$ (you should get $\cos(s-t)\,ds + dt$). (2) Compute $d(g^*\omega)$ and $g^*(dx\wedge dy)$. How do they compare? (3) Using the ...


2

This is a multi-variable optimization problem. The question is just a lot of text to test if you understood one thing: If you have a function $$f : \mathbb{R}^n \to \mathbb{R}$$ Then the gradient, that is $$\nabla f = \text{grad}\ f = \begin{pmatrix}\frac{\partial f}{\partial x_1} \\ \vdots \\\frac{\partial f}{\partial x_n} \end{pmatrix}$$ if evaluated at ...


2

Note that: $\rho^2 \sin \theta d\rho d\phi d\theta$ is the volume element in spherical coordinates. So the volume is the volume of a part of a sphere as noted in OP, but not a half-sphere. The limits of integrations are: $0\le \rho \le 1$ (this means that the sphere has radius $1$) and $0\le \phi \le \pi/2$,$0\le \theta \le \pi/2$ and this means that we ...


2

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$Your understanding is perfectly correct, and although it may be overly-careful for everyday use, it's a good idea to be certain what symbols mean! :) (If you can find a library copy, Jan J. Koenderink's Solid Shape is a highly worthwhile if somewhat idiosyncratic read, ...


2

The first two answers are correct. For the third question: the directional derivative is equal to $0$ if $v =(v_1,v_2)$ is perpendicular to $\nabla h$ , so: $$v\cdot \nabla h = 0 \Rightarrow \dfrac{v_1}{a^2}x+\dfrac{v_2}{b^2}y = 0 \Rightarrow y = -\left( \dfrac{b}{a} \right)^2\left(\dfrac{v_1}{v_2}\right)x\Rightarrow y=cx \quad\text{where } c=-\left( \dfrac{...


2

I suppose that we have $f(0,0)=0$. Now note that we have always $|y|\leq \sqrt{x^2+y^2}$. Hence for $(x,y)\not =(0,0)$: $$|f(x,y)-f(0,0)|=\frac{x^2}{x^2+y^4}|y|\leq 1\times \sqrt{x^2+y^2}= \sqrt{x^2+y^2}$$ This inequality is also true for $(x,y)=(0,0)$ and it is easy to finish.


2

basically what you have to do is set the function: $$ F(x,y,z,u,v)=\left(\begin{array}{c} F_1\\ F_2\\ F_3 \end{array}\right)=\left(\begin{array}{c} 2x+y+2z+u−v−1\\ xy+z−u+2v−1\\ yz+xz+u^2−v \end{array}\right) $$ Then you notice F(1,1,-1,1,1)=0. And you find the derivative with respect the variables you want to clear $$ \frac{\partial F}{\partial (x,y,z)}=\...


2

What you want is to choose $\delta = \min\{ \frac{\epsilon}{4},\frac{1}{2}\}$ and then you know the following. $$|\frac{x}{y} -1| = |\frac{x-y}{y}| \leq \frac{|x-1|+|y-1|}{|y|} < \frac{2\delta}{|y|} < 4\delta < \epsilon$$ The only thing I changed was that because we know that $\delta \leq \frac{1}{2}$ then $y > \frac{1}{2}$ or $\frac{1}{|y|} &...


1

You multiple-integrals approach works, but there is a more compact one, through: $$ \frac{1}{n^k}=\frac{1}{(k-1)!}\int_{0}^{1}t^{n-1}\left(-\log t\right)^{k-1}\,dt \tag{1}$$ that leads to: $$ (k-1)!\sum_{n=1}^{N}\frac{r^n}{n^k} = \int_{0}^{1}\sum_{n=1}^{N} r^n t^{n-1}\left(-\log t\right)^{k-1}\,dt=\color{red}{r\int_{0}^{1}\frac{1-(rt)^N}{1-rt}(-\log t)^{k-1}...


1

Perhaps an example will be illuminating: Consider $k=1$, and $n=2$, let $A=\mathbb{R}$ and let $B=\{(x,x)\in\mathbb{R}^2:x\in\mathbb{R}\}$, that is, the points of the line $y=x$. Consider $f:A\rightarrow B:x\mapsto (x,x)$. $B$ is a submanifold of $\mathbb{R}^2$ of dimension $1$, and $f$ is a diffeomorphism between $A=\mathbb{R}$ and $B$.


1

To show that $f$ is differentiable at $(0,0)$, we need to show that $$ 0 = \lim_{\Vert h \Vert \to 0}\left(\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert}\right) $$ with $\Delta f(0)=(\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0))$. In this case, the answer is no, the function is not differentiable at $(0,0)$. By your ...


1

Contour line is defined as $\{ (x,y) | f(x,y) = constant \}$. So for $e^{x+y}=C$, you get $x+y = \ln C$, which is a straight line. The answer is correct. Another strange example would be $(15x+6y)^{100}$.


1

Hint: If $f$ is a continuously-differentiable real-valued function of two variables and $$ \gamma(t) = \bigl(x_{0} + t(x - x_{0}), y_{0} + t(y - y_{0})\bigr) $$ is the constant-speed parametrization of the segment from $(x_{0}, y_{0})$ to $(x, y)$ over $[0, 1]$, then $$ f(x, y) - f(x_{0}, y_{0}) = \int_{0}^{1} \nabla f\bigl(\gamma(t)\bigr) \cdot \gamma'(t)\...


1

The Jacobian matrix of $F$ is given by $$\begin{pmatrix}-\frac{1}{2}e^{-x} & \frac{1}{2}\\ \frac{1}{3} & -\frac{1}{3}e^{-y} \end{pmatrix}$$ where the absolute value of the largest eigenvalue in absolute value (sorry for the pun) is $\leq \frac{1+e^{-x}}{2}$ by the Gershgorin circle theorem. It follows that $F$ is a contraction and has a unique fixed ...


1

For all $(x,y)\in D$ and all $(h,k)\in\mathbb{R}^2$, one has: $$\mathrm{d}_{(x,y)}F\cdot(h,k)=\left(\frac{1}{2}(k-e^{-x}h),\frac{1}{3}(h-e^{-y}k)\right).$$ Let $\|\cdot\|$ be the operator norm associated with $\|\cdot\|_1$ on $\mathbb{R}^2$, where $\|(x,y)\|_1:=|x|+|y|$. Let me remind you, that by definition, one has: $$\forall u\in\mathcal{L}(\mathbb{R}^2),\...


1

Start by convincing yourself that $$\lim_{(x, y) \to (0, 0)} \frac{x^4 + y^4}{x^2 + y^2} = 0$$ Now once this is true, consider the following: When $z$ is very small, $$\frac{g(z)}{z} \approx 2$$ This is a problem unless $\lim_{z \to 0} g(z)$ exists and is a very particular number. Think about what this number ought to be.


1

For $0\le t \le 1,$ let $$g(t) = f((-\sqrt 2,-\sqrt 2) + t(\sqrt 2,\sqrt 2)).$$ Then $g$ is differentiable on $[0,1].$ Note that $g(1)-g(0) = f(0,0) - f(-\sqrt 2,-\sqrt 2).$ By the mean value theorem, $$g(1) - g(0) = g'(c)\cdot 1 = \nabla f ((-\sqrt 2,-\sqrt 2) + c(\sqrt 2,\sqrt 2)) \cdot (\sqrt 2,\sqrt 2))$$ for some $c\in (0,1).$ Now $|\nabla f | \le 1 ...


1

It is obvious that $f$ admits its partial derivatives on the punctured plane. It thus suffices to consider the point $(0,0)$. What you say in your edit is true. The "putting in 0" problem is resolved as follows: you want to evaluate $f$ at $0$. But by definition this $0$. You are right in that $\partial f/\partial x(0)$ exists. To show the existence of $\...


1

Hint $$x=r\cos\theta\quad,\quad y=r \sin\theta$$ $$\frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial r}$$ $$-2{{\operatorname{sech}}^{2}}\,r\,{{\tanh }^{2}}r\cos \theta +{{\operatorname{sech}}^{4}}\,r\cos \theta=\cos \theta\frac{\partial v}{\partial x} +\sin \theta\...


1

Consider $f: \Bbb R^2 \times \Bbb R^3 \to \Bbb R^3$, $$f(u,v,x,y,z) =( 2x+y+2z+u-v-1, xy+z-u+2v-1, yz+xz+u^2-v)$$ We have: $$\frac{\partial f}{\partial(x,y,z)} = \begin{pmatrix} 2 & 1 & 2 \\ y & x & 1 \\ z & z & x + y \end{pmatrix} $$ So: $$\frac{\partial f}{\partial(x,y,z)}(1,1,-1,1,1) = \begin{pmatrix} 2 & 1 & 2 \\ 1 &...


1

\begin{align} & D=\{(x,y,z)|{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\le 1\,,x,y,z\ge 0\} \\ & V=\int_{0}^{1}{\int_{0}^{\sqrt{1-{{y}^{2}}}}{\int_{0}^{1-\sqrt{{{x}^{2}}+{{y}^{2}}}}{dzdydx}}}=\int_{0}^{\frac{\pi }{2}}{\int_{0}^{\frac{\pi }{2}}{\int_{0}^{1}{\rho {{\sin }^{2}}\phi \,\,d\rho }d\phi }d\theta } \\ \end{align}


1

Your function $f(x_1,x_2)$ is of the form $F(x_1,x_2) = f(g(x_1,x_2),h(x_1,x_2))$ I think. Where $g$ and $h$ are the first and second entry respectively. So the gradient that you are looking for is of the form: $J_f(x_1,x_2) = \begin{bmatrix} \frac{\partial{g(x_1,x_2)}}{\partial{x_1}} & \frac{\partial{g(x_1,x_2)}}{\partial{x_2}} \\ \frac{\partial{h(x_1,...


1

Your count seems to imply that you consider $0$ to be included in $\mathbb N$. Consider the $d$ elements as $d$ strings of ones separated by $d-1$ zeros. Each arrangement with $n$ ones corresponds to exactly one of your multiindices. Indexing these arrangements has been discussed at least once on this site; see Fast way to get a position of combination (...



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