New answers tagged

0

Nonsingularity of $Df(x)$ follows from the general theory of matrices. If a matrix $A$ satisfies $\|A\|<1$, then $I-A$ is invertible. Its inverse is give by Neumann series $$(I-A)^{-1}=\sum_{n=0}^\infty A^n.$$


0

Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$ (polar coordinates), hence $$x^2+y^2=r^2\cos^2(\theta)+r^2\sin^2(\theta)=r^2$$Now,$$\lim_{(x,y)\to(0,0)}\frac{\ln(1-x^2-y^2)}{x^2+y^2}=\lim_{r\to 0}\frac{\ln(1-r^2)}{r^2}\underset{\text{L'Hopital}}{=}\lim_{r\to 0}\frac{-\frac{2r}{1-r^2}}{2r}=-\lim_{r\to 0}\frac{1}{1-r^2}=-1$$


1

Hint: Use the squeeze theorem and the inequality $$\frac{-x}{1-x} \leqslant \ln(1-x) \leqslant -x$$


0

I think your formula switched $f$ and $g$, I found: $(g\circ f)''(x)[u,v]=g''(f(x)[f'(x)[u], f'(x)[v]]+g'(f(x))\circ f''(x)[u,v]$. Write $f=(f^1,...,f^k)$ and plugging in $e_i=u, e_j=v$ gives: $\frac{\partial (g\circ f)}{\partial x_i\partial x_j}=\frac{\partial f^h}{\partial x_i}\frac{\partial f^\ell}{\partial x_j}\frac{\partial^2 g}{\partial x_h\partial ...


0

Here are a couple of suggestions. The rank theorem (also called the constant rank theorem) says that in a neighborhood of each point of $\mathbb R^d$, you can choose smooth coordinates $(u^1,\dots,u^d)$ for $\mathbb R^d$ and $(v^1,\dots,v^k)$ for $\mathbb R^k$ in which $\xi$ has the form $\xi(u^1,\dots,u^d) = (u^1,\dots,u^k)$. By means of a partition of ...


0

The points of intersection satisfies the equation : $$x^2+y^2+z^2-2=x^2+y^2-z$$ so we get $$z^2+z-2=0$$ it has two solutions $$z=1$$ or $$z=-2$$ which is exclused because $$x^2+y^2>=0$$ so $z=1$ and $x^2+y^2=1$ gives the parametrization $c(t)=(cost,sint,1)$.


0

From this 3D graph you can see that the boundary of the constrained region has two parts: the bottom of the paraboloid $x^2+y^2=z$ for $0\le z\le 1$, and the cap of the sphere $x^2+y^2+z^2=2$ for $1\le z\le \sqrt 2$. How did I get those limits for $z$? Equate the right-hand sides of the equations $x^2+y^2=2-z^2$ and $x^2+y^2=2$ to get $2-z^2=2$ which has ...


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Simply take $g_i (x)=x_i$ and $f(x)=x_n$ to get a counterexample. Here, $x =(x_1,\dots,x_n)\in \Bbb{R}^n$.


2

If we write $r:=\vert\vert(x,y)-(x_0,y_0)\vert\vert$, differentiability of $f$ is equivalent to the existence of $(a,b)$ such that $f(x,y)=f(x_0,y_0)+a(x-x_0)+b(y-y_0)+o(r)$. Squaring this gives $f(x,y)^2=f(x_0,y_0)^2+2(a(x-x_0)+b(y-y_0))f(x_0,y_0)+(a(x-x_0)+b(y-y_0))^2+o(r)$ (noting that $(f(x_0,y_0)+a(x-x_0)+b(y-y_0))\times o(r)$ is $o(r)$. By switching ...


0

Of course it is a straightforward consequence of the chain rule. For an elementary solution, you might want to remark that $f^2-g^2=(f-g)(f+g)$. Now choose $f$ and $g$ properly and let $(x,y) \to (x_0,y_0)$.


2

You are correct until partial difference of $f(\mathbf{x})$. Next, the vector $\mathbf{x} - \nabla f(\mathbf{x})$ is replaced for $\mathbf{y}$. Therefore, $$ \begin{align} \mathbf{y} &=\mathbf{x}-\nabla f(\mathbf{x}) \\ &=\mathbf{x}-(Q\mathbf{x}-\mathbf{b}) \\ &=(E-Q)\mathbf{x}+\mathbf{b} \end{align} $$ Then, $E$ is an identical matrix. So, ...


1

The average of $10$, $99$, and $101$ is $\dfrac{10+99+101} 3 = 70$, and that is not the same as the average of the smallest and largest of these numbers, which is $\dfrac{10+101} 2 = 55.5$. That is because $99$ is a lot bigger than the average of the smallest and largest ones. It is not generally true that the average of the numbers in a list is the same ...


1

Your second approach is wrong. Consider for example this situation: The company has a $99.9\%$ probability of having $0$ daily profit and a probability of $0.01\%$ of having a profit of one million. Now the average profit will be $0.999 \cdot 0 + 0.001 \cdot 10^6 = 10000$. But if you simply take the average of the maximal and the minimal possible profit, ...


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Official Web Assign answer If we position the hyperboloid on coordinate axes so that it is centered at the origin with axis the $z$-axis then its equation is given by $$\left(\frac xa\right)^2+\left(\frac yb\right)^2-\left(\frac zc\right)^2=1.$$ Horizontal traces in $z=k$ are $$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1+\left(\frac ...


1

Integrate wrt to $x$ to get that $$\frac{\partial u}{\partial y} = z(y)$$ Integrate wrt to $x$ $$u = \int z(y)dy + g(x) = f(y) + g(x)$$ for two functions $f,g$. Now add the boundary conditions to find $$u(x, x^3) = f(x^3 ) + g(x) = \sin x^6$$ $\frac{\partial u}{\partial x} (x,x^3) = g'(x) = 0 \implies g(x) = B \in \mathbb R$ hence $f(x^3) = \sin x^6 ...


1

Here is a simple proof using index notation and BAC-CAB identity. $$\begin{align} \nabla \times \left( {{\bf{A}} \times {\bf{B}}} \right) &= {{\bf{e}}_i}{\partial _i} \times \left( {{A_j}{{\bf{e}}_j} \times {B_k}{{\bf{e}}_k}} \right)\\ &= {\partial _i}\left( {{A_j}{B_k}} \right){{\bf{e}}_i} \times \left( {{{\bf{e}}_j} \times {{\bf{e}}_k}} \right)\\ ...


0

Hint In polar coordinates, which we denote by $(\,\cdot\,,\,\cdot\,)_p$, we see that $$f(x, y) = (e^x \cos y, e^x \sin y) = (e^x, y)_p .$$


1

Either you note that $f(z)=e^z$ and know (or show) that $e^z$ takes all complex values except zero, or you solve explicitly the equations $e^x\cos y=a$ and $e^x\sin y=b$: These equations give $e^x=\sqrt{a^2+b^2}$ and since the real exponential $e^x$ never vanishes you get $a\ne0$ or $b\ne0$. In particular, $x=\log\sqrt{a^2+b^2}$. If $a=0$, then you can ...


1

The notes that you are following should be corrected by erasing the initial fraction: $$ \frac{dE}{dt}(t) = \int_\Omega (u_tu_{tt} + c(x)^2 \nabla u \cdot \nabla u_t + q(x)uu_t)\, dx. $$ In particular, the derivative of $(u_t)^2$ is obtained exactly as you wrote. As for the middle term, you have instead $$ \frac{\partial}{\partial t}|\nabla u|^2 = ...


0

There is a result stating that a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is differentiable at a point $(x,y)$ if its partial derivatives exist in some neighbourhood of $(x,y)$ and are continuous at $(x,y)$ - e.g. Theorem 2.3.4 of Multidimensional Real Analysis I by Kolk & Duistermaat. So you need only calculate the partial derivatives of $f$ ...


1


0

The sine/cosine functions map $\mathbb{R}$ to the closed interval $[-1,1]$. (THIS IS WRONG: Moreover, the embedding $\mathbb{R} \to \mathbb{R}^2, x \mapsto (x, 0)$ satisfies both assertions.)


3

The function isn't bounded above since $$f(x,0)=2x^4-3x^2+2\rightarrow\infty$$ Since the function is even in both variables we can work only with $\;x,y\ge0\;$ , and $$f(x,y)\xrightarrow[(x,y)\to(0,0)]{} 1+1=2$$ But for any point on the circle $\;x^2+y^2=1\;$ or on the ellipse $\;2x^2+y^2=1\;$ the value of the function is $\;1\;$ : $\;f(x,y)= 1$ Yet ...


1

The best way to view this in my opinion is by looking at the actual ODE you get when you solve the characteristic equation. This ODE is what you get along the path $x=x_0+\frac{3}{2}t$. For each $x_0$ you have $v(t)=u(t,x_0+\frac{3}{2}t)$ with $v(0)=f(x_0)$. It satisfies $\frac{dv}{dt}=\frac{1}{2} \cos(x(t))=\frac{1}{2} \cos(x_0+\frac{3}{2}t)$. The solution ...


0

In spherical coordinates: $$ D=\{(\rho,\theta,\phi)\;|\;0\le \theta \le 2\pi, 0\le \rho \le a, 0\le \phi \le \pi/6\} $$ Careful, $\rho$ is always positive, and $a$ is the radius of the sphere. $\tan \phi = 1/\sqrt{3}$ will give you the right angle, as you suggested. In cylindrical coordinates: $$ D=\{(r,\theta,z) \;|\;0\le \theta \le 2\pi, 0\le r \le a/2, ...


0

First, the result isn't an scalar, is a (row) vector. Second, lousy notation. The usual formula for the chain rule is $$D(h\circ g)(x) = Dh(g(x))Dg(x)$$ where the product in the RHS is the matrix product. In your case $\nabla^T h(g(x))$ (i.e., $Dh(g(x))$) is a row vector. See https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant. EDIT: example with ...


2

$$3u_x+2u_t=\cos(x)$$ HINT : The equations characteristics are : $$\frac{dx}{3}=\frac{dt}{2}=\frac{du}{\cos(x)}$$ First characteristic : $\frac{dx}{3}=\frac{du}{\cos(x)} \quad\to\quad 3du-\cos(x)dx=0 \quad\to\quad u-\frac{1}{3}\sin(x)=c_1$ Second characteristic : $\frac{dx}{3}-\frac{dt}{2}=0 \quad\to\quad 2x-3t=c_2$ General solution on implicite form, ...


1

We are given the vector $\vec F=\frac{\vec r}{r^2}=\hat r$, where $\hat r$ is the radial unit vector. The divergence of $\vec F$ for $0<a\le r\le b$ is $$\begin{align}\nabla \cdot \vec F&=\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r}+\frac{1}{r\sin \theta}\frac{\partial (\sin \theta F_{\theta})}{\partial \theta}+\frac{1}{r\sin ...


1

For the first equation, complete the square in $z$... that is, convert the equation to the form $x^2 + y^2 + (z - q)^2 = r^2$. That will give you a sphere centered at $\{0,0,q\}$ of radius $r$. You must express $q$ and $r$ in terms of your $a$. For the second equation, note that the region is rotationally symmetric about the $z$ axis because the $x$-$y$ ...


1

Or does everyone really just refer to points and vectors interchangeably in the Euclidean space $\Bbb R^n$? Yes. Everyone does. Okay, hyperbolic statements like that one are almost guaranteed to not be true. But still, it is close to true. Practically no one balks at the concept. However, it is also not completely true that they are used ...


0

Putting $x=0$ and solving the two equations we obtain a point which satisfies both equations i.e.$(0,-1,3)$. Therefore this point would lie on the line of intersection of the two planes. Let equation of line be $$ \frac{x-0}{l}=\frac{y+1}{m}=\frac{z-3}{n}$$. Angle between this line and both the planes will be $0$. Therefore, $3l-m+n=0$ $m+n=0$ Solving ...


0

We know that the two planes hit at an intersection, and thus their intersection should be orthogonal to the "facing" of said planes. Hence, you can find the (orthogonal) vector of intersection by taking the cross product the two normal vectors of these planes, since the cross product of two vectors produces a vector that is perpendicular to both of the ...


1

One way is to recognize that the intersection must satisfy the equation for both planes, and must therefore satisfy their sum: $$ (3x-y+z)+(y+z) = 4+2 $$ $$ 3x+2z = 6 $$ You can then let $x = t$, and then $3t+2z = 6$, whence we get $z = 3-\frac{3}{2}t$. You can then rewrite your first equation as $$ y = 3x+z-4 $$ to obtain an expression for $y$ in ...


0

This page has a good explanation of the technique and way to think about the multiple boundaries. This page goes into more detail about why the technique works.


1

A first integration with respect to $x$ is clearly easier. I suggest you to apply Fubini theorem and reverse the order of integrals. $$\int_{0}^{1}\int_{x}^{1} y^2 \sin(2\pi \frac{x}{y})dydx = \int_{0}^{1} y^2\left( \int_{0}^{y} \sin(2\pi \frac{x}{y})dx \right)dy.$$ And $$\int_{0}^{y} \sin(2\pi \frac{x}{y}) dx = \frac{y}{2\pi} [-\cos(2\pi \frac{x}{y})]_0^y ...


0

The limit of the numerator is $1$. The limit of the denominator is $0$. Are you sure that your numerator isn't supposed to be $e^{x+y}-x-y-1$? That would be more interesting.


3

Along $y = -x$ for example: $$\lim_{(x,y)\rightarrow(0,0)}\frac{ e^{x+y} - x - y}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{2}}\cdot{\lim_{x\rightarrow 0}\frac{1}{|x|}} \rightarrow \infty$$


1

A change to polar coordinates will suffice: $$ \lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)} - r(\sin\theta+\cos\theta)}{r}=\lim_{r\to 0}[\frac{ e^{r(\sin\theta+\cos\theta)}}{r} - (\sin\theta+\cos\theta)] $$ Which does not exist since $\lim_{r\to 0}\frac{ e^{r(\sin\theta+\cos\theta)}}{r}$ does not exist for any $\theta$.


3

You can do linear algebra in terms of $n$-tuples $(x_1,x_2,\ldots, x_n)$ and matrices $\bigl[a_{ik}\bigr]$, or in terms of "abstract" vectors ${\bf x}$ and linear maps $A$. Similarly with derivatives in a multivariate setting: You can consider component functions $f_i$ $(1\leq i\leq m)$ and their partial derivatives $f_{i.k}$, as defined in $(25)$, or you ...


0

As already answered by SAUVIK, if $f:E\longrightarrow F$, $$Df(x)\in L(E,F)\text{ and } Df:E\longrightarrow L(E,F).$$ So, $$D^2 f:E\longrightarrow L(E,L(E,F)).$$ Now, the space $L(E,L(E,F))$ can be identified of the space of bilinear functions form $E$ to $F$ via the isomorphism $$g\to\hat g,\qquad\hat g(x,y) = (g(x))(y).$$ The trick can be obviously ...


0

From what I've experienced, we use separation of variables when we face linear PDEs and we can safely use them when we're faced with problems that have well-defined coordinates. For example, in the heat, wave and Laplace's equation, we have the first one depending on space, the second one depending on space and time, and the last one again on space. We can ...


0

$ x = y^2 - z^2 + 6 = 81(\cos^2 t - \sin^2 t) + 6 $ The parametrization is $ (81(\cos^2 t - \sin^2 t) + 6, 9\cos t, 9\sin t) $


1

The mistake you seem to be making is that you misunderstood what $\Sigma$ is. It is the cilinder without the two 'caps' at the ends (it is what you would get is you took a piece of A4 paper and folded two sides together). This means that the boundary of $\partial \Omega$ are the two circles at the ends of the cilinder. While checking the rest of your ...


0

Yes, its possible to derivate G(x,y). The easiest way, in my opinion, is, first, by using the properties of the exponential. e^(x-z) = e^z / e^z, where z=n-y, and the you do the derivate, because the integral only depends on t.


0

If we have a 3-place function $f(y,y',x)$ it is common notation to use $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial y'}$ and$\frac{\partial f}{\partial x}$ to mean respectively the first, second and third partial derivatives. In the example you provide, it is taking the first and second derivatives wrt $\alpha$ (the third does not appear ...


1

I hope I have a possible solution. Le us assume that, for all the components $J_i$ of $\boldsymbol{J}$, there is a function $f_i\in C^4(\mathring{A})$ such that $\nabla^2 f_i=J_i$, with $\bar{V}\subset \mathring{A}$ and $\boldsymbol{x}\in \mathring{V}$ and there exists a $\delta$ such that, for all $\epsilon\le \delta$, $\epsilon >0$, the region ...


0

The double integral can be split into several integrals involving only trigonometric functions. But these integrals cannot be expressed in terms of a finite number of standard functions. So, it is doubtful that a closed form could be obtained to express the double integral :


3

Let $g(x,y)=(x^3,y)$ and $h(x,y)=(\sin x\cosh y,\cos x\sinh y)$, so $f=h\circ g$. Then $g(0,0)=(0,0)$ and $g$ is invertible, and $h$ is locally invertible at $(0,0)$ since $\displaystyle\frac{\partial(h_1,h_2)}{\partial(x,y)}=\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(\sinh^2y+1)+\sin^2x\sinh^2y=\sinh^2y+\cos^2x$, so ...


0

Hint: In three dimensions you would check if the vector field has a non-zero curl. Checking $F = (x f(u), x g(u), 0)$ we get $$ \DeclareMathOperator{curl}{curl} $$ \begin{align} \curl F &= (0, 0, \partial_x(x g(u)) - \partial_y(x f(u))) \\ &= (0, 0, g(u) + x g'(u)y - x f'(u) x) \\ &= (0, 0, g(u) + u g'(u) - x^2 f'(u)) \end{align} So we need $g$ ...


1

Let's think about how this works for a curve in $\Bbb R^3$. We show that a curve $\gamma$ has torsion identically $0$ if and only if $\gamma$ lies in an affine plane. Namely: If there is a constant unit vector $\mathbf C$ so that $\mathbf C\cdot(\gamma-\gamma_0) = 0$, then we differentiate a few times to find that $\mathbf e_1\cdot\mathbf C = \mathbf ...



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