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0

You might try $f(t,x) = t^{-1/2} x^2$ for $t > 0$, $f(0,x) = 0$.


0

Using John Hughes' comment, you get $\displaystyle\int_0^{\pi}\int_0^{\pi}\sin(x+y)dydx=\int_0^{\pi}\bigg[-\cos(x+y)\bigg]_{y=0}^{y=\pi}\;dx=\int_0^{\pi}\big(-\cos(x+\pi)-(-\cos x)\big)dx$ $\hspace{.3 in}\displaystyle=\int_0^{\pi}2\cos x \;dx=2\big[\sin x\big]_0^{\pi}=0$ $\;\;\;$(using that $\cos(x+\pi)=-\cos x$)


5

Since $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ we have, for $x\in[0,2]$, $$-\frac{1}{4}\le x^2-x \le 2$$ Then \begin{align*} \int_0^2e^{-\frac{1}{4}}\,dx&\le\int_0^2e^{x^2-x}\,dx\le\int_0^2e^2\,dx\\ 2e^{-\frac{1}{4}}&\le\int_0^2e^{x^2-x}\,dx\le 2e^2 \end{align*}


0

The first statement is true both ways. Specifically, suppose $(X, ||\cdot||)$ is a normed linear space. Then the norm $||\cdot ||$ is induced by an inner product iff the parallelogram law holds in $(X,||\cdot||)$. For the second statement, this is not true. Call the condition $d(x,y)=d(x+a,y+a)$ translation invariance, and the condition $d(x,y)=d(ax,ay)$ ...


1

Just remember that $$\lim_{\text{a number}\to0} \frac{\sin(\text{that number})}{\text{that very same number}} = 1$$


1

for $f(x,y)=\sin xy^2$, it is $f_x(0,y)=y^2 cos 0y^2=y^2$


2

Provided you are aware of continuity of polynomials in $\mathbb R^n$ you can plug $t:=xy^2$ The limit then becomes $\lim_{t \rightarrow 0} \dfrac {\sin t}{t}y^2$


1

Hint: \begin{align*} \frac{\sin xy^2}{x}=y^2\frac{\sin xy^2}{xy^2} \end{align*} now take the limit!


4

The simplest (and shortest) is with equivalents: $$\sin ax\sim_0 ax,\enspace\text{hence}\quad \frac{\sin ax}x\sim_0\frac{ax}x=a.$$


4

For fixed $y\ne 0$, change the variable to $u=xy^2$. This transforms your limit to $$ \lim_{u\to 0}\frac{\sin u}{uy^{-2}} = \frac{1}{y^{-2}}\lim_{u\to 0}\frac{\sin u}{u} = y^2 \lim_{u\to 0}\frac{\sin u}{u}$$ Alternatively, if you know the standard calculus toolkit, you can brute-force it using l'Hospital, which amounts to recognizing the limit as the ...


1

Since you are dealing with a moment of inertia, the claim just follows from the Huygens-Steiner/parallel axis theorem: the absolute minimum is attained by the centroid of $\{a_1,\ldots,a_m\}$. On the other hand, your argument is perfectly fine, but you do not need to compute any Hessian matrix, since the moment of inertia is a convex function as a sum of ...


1

For a strictly increasing function, only the first derivative must be strictly positive, not all of them.


1

The second sub-function $xy^2$ clearly has all partial derivatives everywhere. The first sub-function $\dfrac{\sin(xy^2)}{y}$ has partial derivatives everywhere it is defined, namely $y\ne 0$: since that sub-function is used only where $y>0$ it clearly has all partial derivatives in that domain. So the only possible problem is the boundary between those ...


1

Hint: $f$ is $C^\infty$ in both $\{y>0\}$ and $\{y<0\},$ so $f$ has partial derivatives of all orders in those domains. So all you have to check is the whether $\partial f \partial x, \partial f \partial y$ exist at points of the $x$-axis.


1

Your computation of $x_0$ is correct, except that you forgot to divide through $m$ in the end. ($x_0$ is the center of gravity of the $a_k$.) The argument that $\lim_{|x|\to\infty} f(x)=\infty$ is correct. Then you can, for example, argue that for a sufficiently large radius $r$, the function $f$ restricted to the closed ball of radius $r$ around $0$ has a ...


1

By the squeeze theorem, your inequality does show that the limit exists and is zero, when used along with the fact that $0 \le \frac{x+y+\sin xy}{x^2+y^2+\sin^2 (xy)}$ when $x \ge 1, y \ge 1$.


2

HINT: rewrite your exprssion in the form $$\frac{\frac{x+y}{x^2+y^2}+\frac{\sin(xy)}{x^2+y^2}}{1+\frac{\sin(xy)^2}{x^2+y^2}}$$


1

Apply the variation of parameters to obtain the particular solution. The particular solution will be represented by $$ y_p(x)=g_1(x)y_{1}+g_2(x)y_{2} $$ where $$ g_{1}(x)=-\int \frac{f(x)y_2(x)}{W(x)}dx,~g_2(x)=\int\frac{f(x)y_1(x)}{W(x)}dx $$ for $f(x)=\frac{2}{e^x+1}$. The Wronskian of $y_1$ and $y_2$ is given by $$ W(x)=\begin{pmatrix} e^{-x} & e^x ...


5

It should be fairly easy to follow the path outlined by John Hughes' comment, but even quicker to see that the integral must be $0$ due to symmetry. Along the diagonal $x+y=\pi$, we have $\sin(x+y)=0$. A point above the diagonal has a negative value that corresponds exactly to the positive value at the point you get by reflecting across the diagonal. (And ...


1

I think it would be best to send the authors an email and ask them, since they are the only ones who knows by certain. Nevertheless, let me make a guess: I think that $$ u\wedge M=\min(u,M) $$ and that $$ u\vee M=\max (u,M). $$ I'm not certain of what they want to do, but I guess that they are cutting the sequence $\phi_n$ off at the values $\pm M$ (in ...


4

Swap the order of integration: $$\int_{-2}^2\left(\int_{0}^{\sqrt{4-x^2}}y\,dy\right)\,dx$$ But $\int_{0}^{z}y\,dy = \frac{z^2}{2}$, so the above is: $$\int_{-2}^2 \frac{4-x^2}{2}\,dx$$ Which is easy to do. Without swapping: $$\int_{0}^2\left(\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}y\,dx\right)\,dy$$ With the inner integral equal to $2y\sqrt{4-y^2}$. We ...


2

In polar coordinates, $r$ is going from $0$ to $2$ and $\theta$ is going from $0$ to $\pi$.


4

$s \exp(n-1)\log\xi(s)$ is a strictly increasing function of real $s$, hence it follows that all its $k-th$ derivatives are nonnegative. How does that follow? $\log(x)$ is strictly increasing and its second derivative is $-\frac{1}{x^2}$.


2

To show that $F$ is not conservative on $\mathbb{R}^2 \setminus \{ 0 \}$, compute the curve integral of $F$ along the unit circle. It will turn out to be non-zero, which shows $F$ can't be conservative. For b), the domain $\Omega$ is simply connected, so the fact that $\operatorname{rot}(F) = 0$ on $\Omega$ is enough to guarantee that $F$ is convervative ...


1

Your partial sum is $H_n-H_{\left\lceil\frac{n}{2}\right\rceil}$, where $H_n$ is the $n^{\rm{th}}$ harmonic number Since $H_n \approx \log n + \gamma,$ this will approach $\log n - \log {\left\lceil\frac{n}{2}\right\rceil}=\log 2$


0

Following Fedja's hint, let $\phi$ be a nonnegative test function supported in $(1,2)$ such that $\phi=1$ on $(5/4,7/4)$. For $b>0$, let $\phi_{b}(x):=\phi(bx)$. Observe that $$\int_{\mathbb{R}}\phi_{b}(x)e^{1/x}dx=b^{-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx,\qquad\forall b>0$$ Suppose that there is a distribution $u\in\mathcal{D}'(\mathbb{R})$ of order ...


0

It is an easy consequence of U.B.P. applied to the normed space $X^*$ and the set of operators $S:={J(x_n); n\in N}$. Indeed, for every $f \in X^*$ there is a $C_f$ such that $|y(f)|\leq C_f$ for all $y\in S$. Applying the U.B.P we have that there is $C$ so that $\Vert y\Vert\leq C$ for all $y\in S$. But every y of S is of the form $y=J(x_n)$, and thus ...


1

$$\left| \int_a^b F(s,x_k(s)) - \int_a^b F(s,x(s)) \right| \le\int_a^b\left|F(s,x_k(s))-F(s,x(s))\right|\,dx$$ As, $F$ is uniformly continuous so, $\left|F(s,x_k(s))-F(s,x(s))\right|<\epsilon/(b-a)$ , whenever $|x_k(s)-x(s)|<\delta$. Can you continue ?


3

The Cauchy-Kowalevski Theorem, together with Lewy's example, provides some food for thought. The CKT is the main result about local existence of a solution of an analytic first order system of PDEs. Roughly speaking, if all the coefficients, the force and the boundary datum are analytic at some point the PDE admits a local solution which is $C^{\infty}$ and ...


3

zhw points out a nice property of analytic functions: If $f,g$ are analytic on $(a,b)$ and $f(x_n)=g(x_n)$ for a sequence of distinct points converging to some some $x_0\in(a,b)$, then $f(x)=g(x)$ for all $x\in(a,b)$. This becomes false if we loosen the restriction of analyticity, as can be seen by considering the functions ...


1

(A) is false: On $[0,1]$ take $f(x)= x^2-1/2$ and $P= \{0,1\}.$ Then $M(P,f) = 1/2, \int_0^1f = -1/6, \sup |f| = 1/2.$ So the left side equals $2/3,$ the right side equals $1/2.$ (C ) is true: We have $$M(P,f) - \int_a^b f = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(M_k - f) = \sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}(f(c_k) - f(t))\ dt$$ $$ \implies |M(P,f) - ...


0

Say we have a continuous function $f :\Bbb{Q}\to \Bbb{Q}$ for which we know $f(a)<0<f(b)$ for fixed rational numbers $a<b$. Now set $$ M:=\{x\mid a\leq x\leq b \text{ and } f(x)<0\}. $$ Clearly,$M$ is bounded above. Suppose that $M$ has a least upper bound $\alpha$. It is not hard to see that this entails $f(\alpha)=0$ (if you don't believe ...


1

The set $$\left\{\sum_{k=1}^n\frac1{k!}:n\in\Bbb N\right\}$$ is contained in the rationals, and has upper bound. It is not difficult to show that $3$ is an upper bound. But in the reals, the series converges to $e$, which is not rational. For a relatively easy proof, see here.


3

$\sup S$ is shorthand for the least upper bound of $S$ in some set $T\supset S$ with respect to an order $\leq$ on $T$. However, we often omit $T$ and $\leq$ when they are obvious from the context. Consider the case when $T=\mathbb{Q}$ and $\leq$ is the usual order. Since $\sup\left\{ x\in\mathbb{Q}:x^{2}<2\right\} $ has an upper bound but no least upper ...


1

Hint: Note that $f(x) = 0 \iff x = 0$ $\|f(x)\| = \|x\|^3$ $f(x)/\|f(x)\| = x/\|x\|$ (for $x \neq 0$) That is enough to deduce injectivity.


1

$y=x\Vert x \Vert^2$ implies $\Vert y\Vert=\Vert x\Vert^3$. $g(y):=y / \Vert y\Vert^\frac{2}{3}$ is the inverse and is defined on $y\neq o$ So you have $f^{-1}(y)=g(y)$, if $y\neq 0$ and $0$ otherwise.


3

$ x \to -x $ is decreasing , $ x \to x^3 $ is increasing, but $x \to x^3-x $ is neither so W can't be a subspace of $V$.


1

This is correct for one direction- i.e. you have shown that if $L=\lim \sup_{k\to\infty} a_k$ the (i) and (ii) hold. You now only need to show the opposite. It turns out that (i) implies $L\leq\lim \sup_{k\to\infty} a_k$ and (ii) implies $L\geq \lim \sup_{k\to\infty} a_k$, so you can do these separately. Aim to prove these statements indirectly e.g. the ...


1

By considering the $y$-derivative, the $y$-coordinate of the gradient is zero iff $y=x$. The $x$-coordinate is then zero iff $y=0$, so the only gradient-zero point is $(0,0)$. To determine whether this is a minimum or a maximum (or otherwise), we can examine a small perturbation near $(0, 0)$. That is, for small-in-modulus $x, y$, is $f$ positive, negative, ...


0

=> Y'Z(X+X')+XZ(Y+Y') //BY DISTRIBUTIVE AND ASSOCIATIVE PROPERTY => Y'Z+XZ //X+X'=1 AND Y+Y'=1 USING BOOLEAN TABLE PROOF => Z(Y'+X) //BY DISTRIBUTIVE PROPERTY


0

The answer is yes. We want to prove that for every $\epsilon > 0$ there is $\delta > 0$ such that if $0 < r < \delta$ we have $$\frac{1}{\text{area}(\partial B(0,r))} \Big|\int_{\partial B(0,r)} (f(x)-f(0))\,dS(x)\Big| \le \epsilon.$$ So let $\epsilon > 0$ be given. By assumption $f$ is continuous, hence there is $\delta$ such that ...


0

You are on the right track with your two cases. First note that $$\left|X\setminus\{x\}\right|=\left|\{i\in\mathbb{N}:i<n-1\}\right|<\infty.$$ Since we have a function, $g$, from a finite set to another finite set of the same cardinality, we only need to show that $g$ is injective or surjective. To show that $g$ is surjective let ...


18

For example, $$ \lambda_n := e - \sum_{k=0}^n\frac{1}{k!} $$ are all irrational numbers, but their limit is zero.


10

You are implicitly claiming that the set of irrational numbers is closed in $\mathbb R$, which it is not. Indeed its closure is $\mathbb R$ (it is a dense subset), i.e. every real number is the limit of a sequence of irrational numbers.


23

You can have a rational limit of a sequence of all irrational numbers. Consider for example $\sqrt{2{\sqrt{2{\sqrt{2\ldots}}}}} = 2$


1

Hint: $F = \nabla f$ for some $f: \mathbb R^3 \to \mathbb R$ means that $F$ is a conservative field. A vector function is a conservative field $\iff$ it has a $0$ curl.


2

Since the square root portions in both the numerator and denominator are the same, I would think about factoring out the numerator. Note that: $$2 \sqrt{9n^2 + 20n + 10} - 6n - 10 = 2(\sqrt{9n^2 + 20n + 10} - 3n - 5) $$ After you separate your numerator appropriately, then the answer should be more obvious. $$\lim_{n \rightarrow ...


1

$\displaystyle \lim_{n \rightarrow +\infty}\frac{2\sqrt{9n^2+20n+10}-6n-5}{\sqrt{9n^2+20n+10}-3n-5}\;,$ Now Let $\displaystyle n=\frac{1}{y}$ So when $\lim_{n\rightarrow \infty}\;,$ Then $\lim_{y\rightarrow 0}$ So Limit $\displaystyle \lim_{y\rightarrow 0}\frac{2\sqrt{10y^2+20y+9}-5y-6}{\sqrt{10y^2+20y+9}-5y-3} = \lim_{y\rightarrow 0}\frac{2\cdot ...


3

Hint: Did you try to get rid of the square root in the denominator? Remember: $$\frac{A}{\sqrt{B} + C} = \frac{A(\sqrt{B}-C)}{B-C^2}$$



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