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20

You certainly can do this with a multiple integral; but here's another approach. Imagine picking $n$ values, $X_1, \dots, X_n$ randomly, uniformly, and independently from $[0,1]$. Then the probability that $X_1\le X_2 \le \cdots \le X_n$ is $\frac{1}{n!}$ (since there are $n!$ orderings of these variables). But also this probability should equal the ...


5

We can define a sequence of functions as follows: $$V_1(x)=\int_0^x1dt=x\\ V_{n+1}(x)=\int_0^xV_n(t)dt$$ Then $V_n(1)$ will be the volume in $\Bbb R^n$. It may look a bit intimidating but really it is just $$V_n(1)=\int_0^1\int_0^{x_n}\cdots\int_0^{x_3}\int_0^{x_2}dx_1dx_2\cdots dx_{n-1}dx_n$$ Let's try to show that $V_n(x)=\frac{x^n}{n!}$ for all $n$, ...


5

Use the fact that $x^2+y^4 > 2|x|y^2$ for $(x,y) \neq (0,0)$. Then, $$\frac{\lvert x\rvert^2y^2}{x^2+y^4} \le \frac{\lvert x\rvert^2y^2}{2\lvert x\rvert y^2}=\frac{\lvert x\rvert}{2} \to 0$$ as $x \to 0$.


4

Hint: $\dfrac{x^2}{x^2+y^4} \leq 1 \Rightarrow \dfrac{x^2y^2}{x^2+y^4} \leq y^2$


4

I think the explanation comes from what is hidden, namely the range of the summation indices. Because of the anticommutativity $$ \alpha=\sum_{i<j}a_{ij}\,dx^i\wedge dx^j. $$ When you take the exterior derivative it comes out as (I switch temporarily to indices $r,s,t$ - bear with me for a moment) $$ \begin{aligned} ...


4

I assume that you have some knowledge of linear algebra. The quadratic form $x^2 + 2 \frac{1}{2} xy + y^2$ can also be written as a dot product between two vectors $$x^2 + 2 \frac{1}{2} xy + y^2 = \langle \left (\begin{array}{cc} 1 & 1/2 \\ 1/2 & 1 \end{array} \right ) \cdot \left (\begin{array}{c} x \\ y \end{array} \right ) , \left ...


4

The formulations is for any $\epsilon>0$ there is an $M>0$ such that $\sqrt{m^2+n^2}>M$ implies $|F(m,n)-L|<\epsilon$. This will guarantee that $f(m,n)$ gets arbitrarily close to $L$ regardless of how $(m,n)$ approaches $\infty$.


3

Consider all the points $(x,y)$ for which $xy=0$. Clearly, this can only happen if $x=0$ or $y=0$, so $f(x,y) = 1$ along the coordinate axes, and is $2$ everywhere else. It should also be clear that $f$ is continuous everywhere except the axes; the axes are the only place where a discontinuity occurs. What kind of set is $S$ then going to be? EDIT: In terms ...


3

Let $\mathbb Q^n \subset \mathbb R^n$ the subset. We know that "boundary = closure - interior". We know the subset is dense, i.e. closure = everything. We see easily that interior is empty, for if we take a point in the rationals, every neighborhood will intersect the irrationals. Hence "closure - interior = all - nothing = boundary = $\mathbb R^n\subset ...


3

Iterate the integral as $$\int_{-2}^2 \int_{-|y|}^{|y|} \int_{-\sqrt{y^2 - x^2}}^{\sqrt{y^2 - x^2}} |z| \, dz dx dy.$$ The inner integral is equal to $y^2 - x^2$, so the remaining computation should be fairly straightforward.


3

Why not think it geometrically?Consider any point $(x,y)\in \mathbb R^2$ such that $x^2+y^2<1$ Now what you will have to do is find a suitable radius that such that a open ball with that particular radius completely gets inside $B(0,1)$. So now try to find that radius by drawing a picture NOTE:when you take a point $(x,y)$ inside $B(0,1)$ then what ...


3

Consider an arbitrary point $p\in{\rm dom}(f)$. Since $f$ is differentiable at $p$ we have $$F(p+X)-F(p)=\bigl(X,f(p+X)-f(p)\bigr)=\bigl(X,df(p).X+r(X))\bigr)\ ,$$ where $$\lim_{X\to 0}{r(X)\over|X|}=0\ .$$ But this implies that $$\lim_{X\to 0} {F(p+X)-F(p)-\bigl(X,df(p).X\bigr)\over|X|}=\lim_{X\to 0}{\bigl(0,r(X)\bigr)\over|X|}=0\in{\mathbb R}^{n+k}\ ,$$ ...


3

You don't need to change variables. If you integrate with respect to $x$ first, the upper and lower limits will be $y$ and $0$ respectively. That will give you a $y e^{3+y^2}$ when you next integrate with respect to $y$. You can then use a substitution.


2

The justification is very simple and once you understand it you'll see how to use it elsewhere. So, given any values for $x$ and $y$, the value of $xy$ is just some number, call it $t$. Now, you are trying to compute $\sin (xy)$ which is just $\sin t$, so you can use Taylor's expansion for $\sin$ and compute $\sin (t)$ using it. Ahhhh, but $t=xy$ so you find ...


2

The first condition is: $$\sum_{n=1}^n 2\cdot (y_i - (ax_i+b))(-x_i) =0 \iff \sum_{n=1}^n 2\cdot (-y_ix_i + (ax_i+b)(x_i)) =0$$ $$ \iff \sum_{n=1}^n y_ix_i = \sum_{n=1}^n (ax_i+b)(x_i) \iff \sum_{n=1}^n y_ix_i =a \sum_{n=1}^n x_i^2 +b \sum_{n=1}^n x_i$$ $$\iff \overline{xy}=\overline{x^2}a+b\bar{x}$$ The second one is: $$\sum_{n=1}^n y_i = \sum_{n=1}^n ...


2

'Not closed' does not mean open, for example the set $[0,1)$ is neither open nor closed. And, as you suggest, sets can be both open and closed (as both $\varnothing$ and $\mathbb{R}$ are). You should use the definition of 'open' directly. Namely, for $X=\varnothing$ and $X=\mathbb{R}$, prove that for every $x \in X$ there exists $\varepsilon > 0$ such ...


2

The way to remain sane here is to note that the integral when $y < 0$ is the same as that when $y > 0$, and similarly for $x$ and $z$. Then we have $$ 2\int_0^2dy\, 2 \int_0^2 dx 2 \int_0^{\sqrt{y^2-x^2}}|z|\,dz= 8\int_0^2dy \int_0^y\sqrt{y^2 - x^2} dx $$ $$ 8\int_0^2dy \int_0^y\sqrt{y^2 - x^2} dx = 8\int_0^2 dy \left[ \frac{1}{2}\left( ...


2

You are asking to compute the volume of an $n$-dimensional pyramid. It is well known that the volume of a pyramid $P$ (in general of any cone) is: $$ V(P) = \frac{\text{volume of the base} \times \text{height}}{n}. $$ Since the base is again a $(n-1)$-dimensional pyramid, and the height is always one, you obtain the result by induction: $\frac{1}{n} ...


2

You are quite right that you need integrability of the derivative. But the answer is quite simple: on page 28 he writes: First of all, we will be interested almost exclusively in functions $f:\mathbf{R}^n\rightarrow\mathbf{R}^n$ which are $C^\infty$ (that is, each component function $f$ possesses continuous partial derivatives of all orders); sometimes we ...


2

You should set up a linear change of variable. The curve bounding the region $A$ can be written as $$(x-\frac y2)^2 + \frac 74 y^2 = x^2 - xy + \frac 14 y^2 + \frac 74 y^2 = 1$$ Thus define new coordinates $u = x - \dfrac y2$ and $v = \dfrac{\sqrt 7}{2} y$. The transformation $T : (x,y) \to (u,v)$ transforms $A$ into the unit disk $D$. The change of ...


2

(Edit: This answer does not cover the case of large holes; see the comments by coffeemath.) A hint: If $b<{a\over\sqrt{2}}$ then the intersection $B$ of the two cylinders is completely in the interior of the sphere. In this case you can proceed as follows: Do the problem with $1$ hole, then compute the volume of $B$, and use inclusion-exclusion. For ...


2

$$y = y_c + y_p = c_1 +C_2e^x - e^x + xe^x$$ $C_2e^x - e^x=(C_2-1)e^x=c_2e^x$ where $c_2=C_2-1$ So, you did nothing wrong : $$y = c_1 + c_2e^x + xe^x$$


2

The acceleration is defined as $$ \mathbf{a} = \frac{d \mathbf{v}}{dt}$$ and the unit norm vector as $$ \hat{\mathbf v} = \frac{\mathbf{v}}{|\mathbf{v}|}=\frac{\mathbf{v}} {\sqrt{\mathbf{v}\cdot \mathbf{v}}}$$ Now the task is to calculate $|d\hat{\mathbf v}/dt|.$ We have with the chain rule $$\frac{d \hat{\mathbf v}}{dt} = |\mathbf{v}|^{-1}\mathbf{a} ...


2

For each $a=(a_1,\ldots,a_n)$ define $G_a=\sum_{1\le i\le n}a_i\left(\frac{E_i}{T_E}-\frac{F_i}{T_F}\right)$. Then we have the identity $G_{u,v}=G_uG_v$. At this point $$\mathrm{Pr}(G_{u,v}\ge 0)=\mathrm{Pr}(G_u\ge 0)(G_v \ge 0)+\mathrm{Pr}(G_u\le 0)(G_v \le 0).$$ It implies that an answer to this question reduces to find the probability $\mathrm{Pr}(G_u\ge ...


1

If $f(a) \neq 0$, set $\varepsilon = \frac{|f(a)|}{2}$. Since $\varepsilon > 0$, by continuity of $f$ at $a$, there exists a $\delta > 0$ such that if $x \in B(a; \delta)$, then $|f(x) - f(a)| < \varepsilon$. Hence, for all $x\in B(a; \delta)$, $f(a) - \varepsilon < f(x) < |f(a)| + \varepsilon$, i.e., $f(a) - \frac{|f(a)|}{2} < |f(x)| < ...


1

Write $$\frac{r^2}{\sqrt{a^2 + r^2}} = \sqrt{a^2 + r^2} - \frac{a^2}{\sqrt{a^2 + r^2}}$$ These two terms are now standard integrals, equal respectively to $\displaystyle \frac{1}{2} \left( r \sqrt{a^2 + r^2} + a^2 \ln(\sqrt{a^2 + r^2} + r) \right)$ and $-a^2 \cdot \ln(\sqrt{a^2 + r^2} + r)$ Hence $$\int_0^1 \frac{r^2}{\sqrt{2 + r^2}} dr = \left[ ...


1

It appears that $I$ is a $3 \times 3$ matrix and $S$ is its determinant. Is that correct? If you knew that, it would be good to say so. The first paragraphs of the Wikipedia page have exactly the formula you are looking for.


1

In a large amount of cases, converting the function into polar coordinates works very well. For example, for $f(x_1,x_2) = \frac{x_1x_2}{\sqrt{x_1^2+x_2^2}}$ converts into $$f(r,\phi) = \frac{r^2\cos\phi\sin\phi}{r}$$ and the limit as $r\to 0$ is easy to calculate.


1

Your answer is correct, and it would be obtained more easily from the 2nd derivative test. The Hessian matrix of $f$ is $$ \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} $$ which has negative determinant; hence, the graph is saddle-like at every point.


1

If you just want to use brute force, consider $$x^2+y^2+x*y=C$$ as a quadratic equation in $y$; there are two solutions given by $$y_{\pm}=-\frac{1}{2} \left(x \pm \sqrt{4 C-3 x^2}\right)$$ To have the contour plot, both curves must be plotted and, because of the radical, choose the $x$ values in the range $$-\frac{2 \sqrt{C}}{\sqrt{3}} \leq x \leq \frac{2 ...



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