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5

Your conjecture is false. Consider $Z=(X_1,-X_1,-X_1)$ where $X_1\sim N(0,1)$ and $x=(0,0,0)$. Then because $Z_2$ and $Z_3$ are of the same sign $$P(Z_1>0 \mid Z_j>0 \text{ for exactly 1 component j})=P(Z_1>0\mid Z_1>0, Z_2\leq 0, Z_3\leq 0)=1$$ while $$P(Z_1>0 \mid Z_j>0 \text{ for exactly 2 components j})=P(Z_1>0\mid Z_1\leq 0, ... 4 If you let t = (x-y)^2, the equations say$$ \eqalign{1 \pm 2 e^{t} \sqrt{t} &= 0\cr y &= e^{t}\cr}$$From the first equation,$$ 1/2 = 2 t e^{2t} $$so 2t must be one of the branches of W(1/2) where W is the Lambert W function. If you're interested in real solutions, there is only one real branch. Moreover we must have ... 4 Note that open and closed are not mutually exclusive descriptors (though you might be used to thinking that way due to learning about the closed [a,b] and the open (a,b) first). A set can be both open and closed (such as \mathbb{R}^n and \emptyset). 3 Hint: Let us make the bottom "small" while the top stays "big." Let y=x^{10}-x. 3 a) Since$$\lim_{x\to 0}f(x,0)=1\neq 0 \lim_{(x,y)\to (0,0)}f(x,y)$$it can't be continuous. b) yes, it's correct. 3 f(x,y,z) = z^2-x^2-y^2-1 is continuous from R^3 \to R, and (0,\infty) is open , thus f^{-1}(0,\infty) is open, and you are done. 3 If f(\textbf{x}) = g(r), r = |\textbf{x}|, and n \ge 3, show that$$\nabla^2 f = {{\partial^2 f}\over{\partial x_1^2}} + \dots + {{\partial^2f}\over{\partial x_n^2}} = {{n-1}\over{r}}g'(r) + g''(r)$$for \textbf{x} \neq 0. \nabla^2 is a rotationally invariant differential operator, i.e. for any orthonormal coordinate systems, (x_1, \dots, x_n) ... 3 I claim that the limit is equal to zero, if \alpha>2, and does not exist, if \alpha\le2. Assume first that \alpha>2. We can then use the AM-GM inequality, a+b\ge2\sqrt{ab}, valid for all positive numbers a,b, to give the denominator a lower bound$$ x^4+y^2\ge 2\sqrt{x^4y^2}=2x^2|y|. $$Using this we get$$ ...

3

Substitute $x=r \sin \theta$ and $y=r \cos \theta$. $$\lim_{r \to 0} 4 r^3 \cos \theta \sin^2 \theta=0$$

2

Recall that $x^2+y^2\geqslant 2xy$ for any choice of $x,y$, so that $$\left|\frac{4xy^2}{x^2+y^2}\right|\leqslant \left|\frac{4xy^2}{2xy}\right| = |2y|$$

2

hint: $0 \leq \left|\dfrac{4xy^2}{x^2+y^2}\right| \leq |4x|$

2

I think the problem here is that you have chosen a poor example. The formula $\nabla\cdot \mathbf g$ that you are trying to compute is one side of Gauss's law of gravity, $$\nabla\cdot \mathbf g = -4\pi G \rho.$$ In this formula, $\rho$ is the mass density at the point where you compute $\nabla\cdot \mathbf g$. The formula $\mathbf g = \dfrac{Gm \mathbf ... 2 There are various different conventions, and it varies depending on field and location: in some countries, people write rot for the curl of a vector field, for example. And it also depends on whether handwriting or typing (I've yet to see someone use convincing boldface in normal handwritten working!) In particular, the following can all mean a vector with ... 2 Here's an incomplete answer. In some special circumstances the guess is true, but in some others it is not. Assuming the components$Z_j$,$j=1,\ldots,n$are uncorrelated (and thus independent, since they're jointly normal) this reduces to a problem on Bernoulli random variables: Let $$Y_j = \begin{cases} 1 &\text{if }Z_j>x_j, \\ 0 & ... 2 Note that \mathbb{R}^3 explicitly introduces coordinates for three dimensional space, and coordinates are not necessary to have three dimensional affine space. You are right in observing that once we identify points by coordinates, there is an ambiguity in whether we mean a point or a vector when coordinates are given. But this scarcely means we don't ... 2 In case you are not familiar with the topological concept of continuity: Let A:=\{(x,y,z)\in\mathbb R^3:z^2-x^2-y^2>1\} and let (x,y,z)\in A be arbitrary. Then there is some natural number n such that 1+\dfrac{1}{n}<z^2-x^2-y^2. Now define \varepsilon:=\min\left\{1,\dfrac{1}{3n(|x|+|y|+|z|+1)}\right\}. It follows that if (u,v,w) is any ... 2 No, the "this means" doesn't mean that. In fact MVT is false for f:\Bbb R\to\Bbb R^2. For example, let f(t)=(\cos(t),\sin(t)), a=0, b=2\pi. Then f(b)-f(a)=0, but f'(t)=(-\sin(t),\cos(t)), so there is no t with f'(t)=0. 2 It is straightforward with partitions of unity. The collection of sets (U_x) for x \in A is an open cover of A, so there exists a smooth partition of unity (\rho_x) subordinate to this cover. Then \tilde{f} = \sum_{x \in A} \tilde{f}_x \rho_x is smooth (where \tilde{f}_x: U_x \to \mathbb{R} are the local extensions) and agrees with f on A. 1 Ok! You have the third components equal, so \sin u^1=\sin v^1. From the first and second components equality:$$(2+\cos u^1)\cos u^2=(2+\cos v^1)\cos v^2 \space\space\space\space\space\space\space(1)$$and$$(2+\cos u^1)\sin u^2=(2+\cos v^1)\sin v^2\space\space\space\space\space\space\space (2)$$Now put them into the square and add:$$(2+\cos ... 1 Use the Squeeze Theorem. Notice that$t^{2} < t^{2} + 2x^{2}$, so $$\frac{t^{2}}{t^{2} + 2x^{2}} \leq 1$$ Then we have that $$0 \leq \left| \frac{t^{2}\sin^{2}{x}}{2x^{2} + t^{2}} \right| \leq |\sin^{2}{x}|$$ and you can apply the Squeeze Theorem accordingly, noting how$|\sin^{2}{x}|$behaves as$t,x \rightarrow 0$1 Hint: Let$S = \{ (x,y)\in\mathbb R^2 \mid 0 \le y \le x \le b \}$and denote the indicator function of$S$by$\mathbb 1_S. Then, we have \begin{align*} \int_0^b \left(\int_0^x f(x,y) \; \mathrm dy \right) \mathrm dx = \int_0^b \left( \int_0^b \mathbb 1_S(x,y) f(x,y) \; \mathrm dy \right) \mathrm dx. \end{align*} Now apply Fubini's theorem on the right ... 1 This isn’t the correct generalization of the Mean Value Theorem to\mathbb R^n$-valued functions with$n>1$. Let’s look at just the$\mathbb R\to\mathbb R^n$case. You can apply the MVT to each component$f_j$of$f$individually, as you’re doing, to find a$z_j$such that$f_j(b)-f_j(a)=f_j'(z_j)(b-a)$, but the problem you run into is that there’s no ... 1 You need to consider a$f\in C^2$to user any derivative tests. Note that you are differentiating w.r.t.$t$, so careful use of chain rule yields $$\frac{d}{dt} (\nabla f(x+t))=(\frac{d}{dt}\nabla f)(x+t)\cdot \frac{d}{dt}(x+t)=\nabla^2 f(x+t)\cdot 1$$ where$1$is the unit vector in$\mathbb{R}^n$. Using the same principle for your application yields ... 1 You are justified. If you apply Stokes theorem with cylindrical polar coordinates to $$F(\mathbf{x})=P\mathbf{e}_r + \frac{Q}{r}\mathbf{e}_\theta$$ over a flat surface orientated in the positive$z$direction you can prove this. 1 Your application of Green’s Theorem is justified. You can think of$r$and$\theta$as the labels of axes in a different Cartesian plane. You have to be a little careful about$\mathcal C$and$\mathcal R$with this point of view, though—they need to be replaced by their preimages under the polar-to-Cartesian map. 1 So first you want to get rid of those absolute value signs. We have: $$z = \left\{\begin{array}{ll} x-y & \text{if } x-y \ge 0\\ y-x & \text{if } x-y \lt 0 \end{array} \right.$$ Notice that$y \le x$is the area below$y=x$and$y \ge x$the area above$y=x$, bounded on all 4 sides by the unit square of ccourse. Taking into account these domain ... 1 I assume the unit square is the region $$S = \{ (x,y) \colon 0\le x\le 1\text{ and }0\le y \le 1 \}$$ There are a couple ways to do this problem, the easiest is to do an area integral (double integral) over the region$S$. (The other way would be to treat$z = |x-y|$as a function of 3 variables and do a volume integral. Both methods amount to the same ... 1 You have $$f(x,1)-f(2x,1)=\frac{1}{x}-\frac{1}{2x}=\frac{1}{2x}$$ and $$\lim\limits_{x \to 0^+} \frac{1}{2x}=+\infty$$ Hence the function cannot be uniformly continuous. 1 Yes, this is true. Hint: The functions$f(x) = \| x\|^2$and$g_a(x) = \| x-a\|^{-2}$are smooth (in the second case only for$a \notin X\$, of course.)

1

Sets can be open, closed, both or neither. It's all about points. Consider a set A and consider points in and out of A. If every neighborhood of a point x, whether or not x is in A or not, has a point (other than x) that is part of A, we call that a limit point. Intuitively it is right up next to some point of A. If a limit point isn't in A, then it's ...

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