New answers tagged

2

If it's continuous, it's bounded (image of the compact set $[-\pi,\pi]$ is a compact hence bounded set), so $0\le |f(x)|^2\le M$. Then by Lebesgue's DCT or Fatou's lemma or whatever you want, $$\int_{-\pi}^\pi |f(x)|^2\,dx\le \int_{-\pi}^\pi M\,dx = 2\pi M < \infty$$ which is exactly what it means to be in $L^2([-\pi,\pi])$


0

The expression you want to keep small is $$ \frac{1}{a+x}-\frac{1}{a+b} =\frac{b-x}{(a+b)(a+x)} $$ and you can assume $x>0$ by taking $|x-b|<b$. Then $a+x>a$ and so $1/(a+x)<1/a$. Therefore $$ \left|\frac{1}{a+x}-\frac{1}{a+b}\right|= \left|\frac{b-x}{(a+b)(a+x)}\right|< |b-x|\frac{1}{a(a+b)} $$ So, if $$ |b-x|<\varepsilon a(a+b) $$ you ...


2

The given ODE is a linear homogeneous equation of second order. This implies that the set of solutions is a two-dimensional complex vector space. If the equation $a\lambda^2+b\lambda +c=0$ has two different solutions $\lambda_1$, $\lambda_2\in{\mathbb C}$ then the general solution is given by $$y(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}\ .$$ If the ...


1

For a very trivial example, let $X$ be any set, and let $d$ be the discrete metric on $X$; $d(x,y)=0$ if $x=y$, and $d(x,y)=1$ otherwise. Clearly this $d$ is an ultrametric, and it generates the discrete topology. For a much more interesting example, let $D=\{0,1\}$ with the discrete topology, and let $X=D^{\Bbb N}$, the Cartesian product of countably ...


-1

Use Laplace transform: $$ay''(t)+by'(t)+cy(t)=0\Longleftrightarrow$$ $$\mathcal{L}_t\left[ay''(t)+by'(t)+cy(t)\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$ $$a\left[s^2y(s)-sy(0)-y'(0)\right]+b\left[sy(s)-y(0)\right]+cy(s)=0\Longleftrightarrow$$ $$as^2y(s)-asy(0)-ay'(0)+bsy(s)-by(0)+cy(s)=0\Longleftrightarrow$$ ...


1

If $a = 0$ and $b = 0$, then we have $c \, y(t) = 0$, which is not even an ODE. If $c \neq 0$, then the solution is $y (t) = 0$. If $c = 0$, every function from $\mathbb R$ to $\mathbb C$ is a solution. If $a = 0$ and $b \neq 0$, then we have a 1st order ODE $$\dot y + \left(\frac{c}{b}\right) y = 0$$ whose solution is $y (t) = \beta \, \exp\left(- ...


1

Let's rewrite the integral a bit as $$ \int_{D_1(0)}\left|\nabla f\right|^2\;dv = \int_{D_1(0)}\nabla f\cdot\nabla f\;dv. $$ I note the unit disk centered at the origin as $D_1(0)$. Okay, so that integrand should look familiar. I've seen in before in the product rule for the divergence of a scalar times a vector: $$ \nabla\cdot(f\mathbf{v}) = \nabla ...


1

One can prove that for a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ such that $\lVert f(\mathbf{x}) \rVert \leq \lVert \mathbf{x} \rVert^2$, for $\mathbf{x}$ in some open set containing $\mathbf{0}$ , $f$ is differentiable at $\mathbf{0}$ and $\mathbb{D}f(\mathbf{0})=\mathbf{0}$. Then considering that $r^2\leq r, 0\leq r\leq 1$, the result follows for ...


1

From (1) we have $x_2 = \frac14 x_1' - \frac12 x_1$, and differentiating yields \begin{align} x_1'' &= 2x_1' + 4x_2'\\ &= 2x_1'+4(3x_1-2x_2)\\ &= 2x_1' + 12x_1 - 2x_1'+4x_1,\\ \end{align} so we may instead consider the second-order ODE $$x_1''-16x_1=0. \tag 3$$ Suppose $$x_1(t) = \sum_{n=0}^\infty a_n t^n.$$ Then differentiating and substituting ...


0

Well, it's easy to see that if $\int_{|x|=r} k(x) \, dS(x) = 0$ for every $r > 0 $, then there exists $\delta_r \in \mathbb{S}^{n-1} _r = \{ x \in \mathbb{R}^n \, : \, |x| = r\} $ such that $k(\delta_r)=0$. Firstly I start noticing that under the conditions given (i.e. $|h| < |x|/2$) we have $$|k(x + h) - k(x)| \le \frac{ |h|^{\alpha}}{|x|^{n+\alpha}} ...


0

First observe that any rational number in $\mathbb Q$ has a unique factorization into prime numbers (allowing negative exponents of course). Next, let $p$ be a prime and for any $\frac{a}{b}\in\mathbb Q$ define $v_p\left(\frac{a}{b}\right)$ to be the exponent of the prime $p$ in the factorization of $\frac{a}{b}$, and $v_p(0)=\infty$. Now let ...


1

I suggest to start by drawing: draw the bounding box $[0,1)\times (-1,1)$, place your pencil at $0,0$, and try to trace several functions. The following are observations and hints, no logical proofs. A first idea is that, since you want a continuous function, a monotonous one might be troublesome in mapping a semi-open interval like $[0,1)$ to an open one ...


0

To me you only demonstrated that $\underline{\mu}(\bigcup S_i) \geq \sum\underline{\mu}(S_i)$. Now you need to show that $\underline{\mu}(\bigcup S_i) \leq \sum\underline{\mu}(S_i)$ and for that you need the semifinite property of $\mu$ for the case where $\exists A \subset \bigcup S_i, A \in {\cal M}$ and $\mu(A) = \infty$ exactly like Weltschmerz was ...


2

3 $\Rightarrow$ 2 Let $\varepsilon >0$. Let $L$ be such that $\forall x,y ||A(x-y)|| \le L \cdot||x-y||$. In particular, if you take $y=0$, $\forall x ||A(x)|| \le L ||x||$. Then, if $\eta = \varepsilon / L$, $||x|| < \eta \Rightarrow ||A(x)|| \leq \varepsilon$, so that $A$ is continuous in $0$. 2 $\Rightarrow$ 1 If $A$ is continuous at $x_{0}$, then ...


2

For $d_1$ consider $f_n(x) = [\sin(nx)]/n, n =1,2,\dots $ Then $f_n \to 0$ in the $1$ norm, but $d_1(f_n) = 1$ for all $n.$


0

In the Wolfram Alpha definition, it is required that the domain be path-connected to begin with. I guess it depends on which definition you're using; if only your condition (on loop shrinking) were imposed, then $3$ would be simply connected.


3

For $d_1$ you can refer to the comment, and even try to find a counter-example by reading the proof for $d_2$ and see where the proof can't be applied for $d_1$ and find a counter-example from here. For the norm $||.||_2$, you can show the continuity by noticing that : $$|d_2(f)-d_2(g)|=|f'(0)-g'(0)|\le ||f'-g'||_{\infty} \le ||f'-g'||_{\infty} + ...


0

Let $L_1, L_2$ be different limits of subsequences. Let $\epsilon = {1 \over 3} |L_1-L_2|$. Then $B(L_1,\epsilon)$, $B(L_2,\epsilon)$each contain an infinite number of points of the sequence. In particular, the sequence cannot be Cauchy.


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


1

Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. We have \begin{align} ...


1

$\forall \epsilon>0, \exists \delta>0$ such that $|x - b|< \delta \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$ $|\frac 1{a+x} - \frac 1{a+b}| = |\frac {b-x}{(a+x)(a+b)}| < \frac {\delta}{|(a+x)(a+b)|}$ If you can prove that $|(a+x)(a +b)|> 0$ then you are done. (which is pretty much what you told us, I just need to do certain ...


1

I think we should find a way to keep $|x-b|$ on the LHS of the inequality. For a fixed small $\epsilon>0$, we want to find $\delta>0$ such that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ when $|x-b|<\delta$. Note that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon \Leftrightarrow \frac{|x-b|}{|a+x||a+b|}<\epsilon \Leftrightarrow ...


0

I don't know what you mean by "matrix" linear regression, and your question isn't all that clear. However, suppose you're doing multiple linear regression with $K$ predictors (including the constant predictor) and $n$ cases. Suppose the errors (not to be confused with the (observable) residuals) all are independent and each is distributed as ...


1

Hint: The density $f(a)=\frac12e^{-|a|}$ belongs to the Laplace distribution. The Laplace distribution arises when you subtract two iid exponential(1) variables. Second hint: If $U$ has uniform distribution on $[0,1]$ then $X:=-\ln (U)$ has exponential(1) distribution.


1

Given $u,v$ with $\|u\|\ge\|v\|\ge r>0$ we have $\|u-v\|^2\ge \left\|\frac{\|v\|}{\|u\|}u-v\right\|^2$. This follows from $$ \begin{align}\left\langle u-\frac{\|v\|}{\|u\|}u,\frac{\|v\|}{\|u\|}u-v\right\rangle&=\|v\|\|u\|-\langle u,v\rangle-\|v\|^2+\frac{\|v\|}{\|u\|}\langle u,v\rangle\\&=\left(1-\frac{\|v\|}{\|u\|}\right)(\|v\|\|u\|-\langle ...


0

I outlines of my argument for key question 1) For 3) it is clear that the $ \gamma_j (t) = - i \lambda_j$ are also continuous real functions where $ i ^ 2 = -1 $. My remarks 1) The first point is that the concept of continuous functions is local. So all my reasoning is locally at one point. 2) The fact that $ \lambda $ are defined and real ($ A (t) $ ...


2

The claim is false unless we add the facepalm condition that $V$ is not empty. So assuming $V\ne 0$, we can pick $v\in V$, let $R=\|a-v\|$, and as suggested consider $W:=V\cap \overline{ B(a;R)}$. This set is closed and bounded, hence compact. Therefore, the continuous function $W\to \Bbb R$, $x\mapsto d(a,x)$ attains its minimum at some point $b\in W$. Then ...


1

$$ h(x)=f(x)\frac{d}{dx}\,(\log f(x))=f(x)\,\frac{f'(x)}{f(x)}=f'(x). $$ Since $f$ is logarithmically convex on $(-\infty,0)$ it is also convex. This means that $f'$ is increasing on $(-\infty,0)$. Similarly, $f'$ is decreasing on $(0,\infty)$. This implies that $f'$ attains a maximum at $x=0$. However $f'$ could be constant on an interval around $x=0$. This ...


1

Yes, your argument is correct. Except for a slight notational nitpick: You ought to say that e.g. $A$ is countable, not that $|A|$ is countable. The notation $|A|$ means the cardinality of $A$, but it is the set itself that is "countable".


1

The set $X$ is the unit sphere in $\Bbb{R}^4$. Any two distinct $x,y\in\Bbb{R}^4$ are contained in some plane $Y\subset\Bbb{R}^4$, and hence $x,y\in X\cap Y$. But $X\cap Y$ is then a circle in $Y$, which is of course path connected. So $x$ and $y$ are connected by a path in $X\cap Y\subset X$, hence $X$ is path connected.


1

The manipulations I am about to describe are valid unless $a_3 = a_1 = 0$, but if this is the case, the first equation is just a quadratic and the continuity of the location of the roots with respect to the coefficients follows in the same way. Solve the first equation for $\sqrt{f}$. Square both sides. Substitute the quadratic for $f$. Multiply both ...


1

For continuity away from the origin, consider points $z = re^{i \theta}$ and $z_0 = r_0e^{i \theta_0}.$ Suppose $|z - z_0| < \delta.$ In this case, $z$ is in a disk of radius $\delta$ with center $z_0$. Using some geometry we find $$|\theta - \theta_0| \leqslant \arcsin \frac{\delta}{r_0},$$ and $$| \sin \theta - \sin \theta_0| = 2 \left|\sin ...


0

Using Euler's identity you can write a direct formula for $f$, which makes continuity obvious where polar coordinates are unique. Geometrically speaking, you function works like this: draw a line $L_z$ through your given point $z$ and the origin, and then project the intersection of $L_z$ and the unit circle onto the y-axis. In particular, $f$ is ...


0

For discontinuity at $0$, approach $0$ along the lines $\theta = \pi/3$ and $\theta = \pi/4$. For continuity, let $\varepsilon > 0$ and $x = r_0(\cos \theta_0 + \mathrm{i} \sin \theta_0)$ be given. Can you find a $\delta$ (depending on $x$ and $\varepsilon$) such that for any $y = r(\cos \theta + \mathrm{i} \sin \theta)\in \Bbb{C}$ with $|x-y| < ...


1

Use the implicit function theorem to show that the derivative exists. As for computing it: we will have a very nice derivative if $A'(t)$ commutes with $A(t)$. In particular, we find that $$ \frac d{dt}\sqrt{A(t)} = \frac 12[A(t)]^{-1/2}A'(t) $$ this can be confirmed via the power series for $x \mapsto \sqrt{x}$ centered at $x = 1$, an applying the ...


2

The key is to first bound $x$ around $2$. To that end, we first take $|x-2|<1$ so that $1<x<3$. Then, since $3<x+2<5$, we have $$\begin{align} |(4x^2-1)-15|&=4|x-2||x+2|\\\\ &\le 20|x-2|\\\\ &<\epsilon \end{align}$$ whenever $0<|x-2|<\delta=\min\left(1,\frac{\epsilon}{20}\right)$. And we are done!


0

Since I like having things approach zero, let $x = 2+y$, so $x \to 2$ is the same as $y \to 0$. Then $\begin{array}\\ 4x^2-1 -15 &=4(2+y)^2-16\\ &=4(4+4y+y^2)-16\\ &=16+16y+4y^2-16\\ &=16y+4y^2\\ &=4y(4+y)\\ & \to 0 \text{ as } y \to 0 \end{array} $


1

For $a \implies b$, consider a subset $X \subseteq \Bbb R^n$, and the collection of functions $$ f_1(\mathbf x) = x_1, f_2(\mathbf x) = x_2,\dots, f_n(\mathbf x) = x_{n} $$ Note that $\mathcal A$ contains the constant function $$ g(x) = (f_1)^0(f_2)^0 \cdots (f_n)^0 $$ moreover, $\mathcal A$ separates points since for any two elements $\mathbf a = ...


1

Let $a_{n_k}$ be such that $a_{n_k} \to \limsup_n a_n $ then $f(\limsup_n a_n) = \lim_n f(a_{n_k}) \le \limsup_n f(a_n)$.


1

Since $u(x,y),v(x,y),a$, and $b$ are all real numbers, you can actually compute $|f(z)-L|$ using the real and imaginary parts: \begin{align*} |f(z)-L|=|u(x,y)+iv(x,y)-(a+ib)| &= |u(x,y)-a+i(v(x,y)-b)|\\ &= \sqrt{(u(x,y)-a)^2 + (v(x,y)-b)^2}. \end{align*} Now by dropping the $(v(x,y)-b)^2$ or the $(u(x,y)-a)^2$, you get that the above quantity is ...


0

I just stumbled across this post after I was trying this problem. Part one of martins incorrect but a small change can fix it. We dont know that there exists a finite open cover, ${\cup_{k=1}^n I_k}$ of $E$ such that if $O=\cup_{k=1}^n I_k$ then $m(O\setminus E) < \epsilon$ since the definintion of lebesgue measure uses countable coverings and finite ...


1

A set $X$ to be closed or not, depends of where it is refered to. (Or better yet: where it's neighbourhood is defined; but a neighbourhood is always a subset of the superset.) Example 1: Consider $\Bbb{R}$ the set of real numbers, and the set of rational numbers $\Bbb{Q}$. $\Bbb{Q}$ is closed in itself, but not in ${\Bbb{R}}$. More formally: Let ...


1

A possible integration strategy could be derived from putting $$ \eqalign{ & e^{\,i\,n\,\theta } - e^{\,i\,\left( {n - 1} \right)\,\theta } = e^{\,i\,\left( {n - {1 \over 2} + {1 \over 2}} \right)\,\theta } - e^{\,i\,\left( {n - {1 \over 2} - {1 \over 2}} \right)\,\theta } = \cr & = e^{\,i\,\left( {n - {1 \over 2}} \right)\,\theta } ...


2

Noting that by the equation$$p^2 + q^2 + 1 = z^{-2}$$the equations for $\dot{p}$ and $\dot{q}$ can be simplified. Furthermore, the equation for $z$ is independent from the rest, which means that we can solve for $z$ first, then for $p$ and $q$. Geometrically, our equation looks like an eikonal equation since it can be written as$$\left|\nabla u\right|^2 = ...


0

Let $$f(x)=\frac{x^{32}}{x^{32}-1}\cdot \left( \frac{2}{\delta(1-x^2)} \right)^{16} ,$$ then since $0<x<1$ one deduce easily that $x^{32}-1$ is negative, thus on $ \left( 0,1 \right)$ our function is negative since all factors are positve with the exception of $\frac{1}{x^{32}-1}$. Please check the computations using the rules for derivatives and ...


1

No, of course not. Consider piecewise linear functions. The local Lipshitz constant is the slope of the lines. If this changes with a jump, the Lipshitz constant will do so, too.


0

I think what the problem is trying to say is that if the point $(x,f(x))\in B$, then $(rx,rf(x))\in rB$, so we can parameterize $rB$ as $$\left(u,rf\left(\frac ur\right)\right)$$ For $u\in[0,r]$. Then the perimeter of the region bounded by $u=0$, $y=0$, $u=r$, and $y=rf\left(\frac ur\right)$ is $$\begin{align}P(r)&=rf(0)+r+rf\left(\frac ...


2

It seems to me that $$\lim_{\theta\rightarrow\pi^-}\left(e^{in\theta}-e^{i(n-1)\theta}\right)=\lim_{\theta\rightarrow\pi^-}e^{in\theta}(1-e^{-i\theta})=2e^{in\pi}\ne0$$ So the integral diverges because $$\lim_{\theta\rightarrow\pi^-}\frac{|\sin\theta|}{(\pi-\theta)}=1$$ So the integrand looks like $$\frac{2e^{in\theta}}{\pi-\theta}$$ As ...


2

$\phi$ will be continuous. Indeed, fix any $u\in C_{[0,1],\mathbb{R}}$, and let $\delta > 0$ such that $\delta(2\lVert u\rVert_\infty + \delta) \leq \varepsilon$. Fix any $\varepsilon > 0$. If $v$ is such that $\lVert u - v\rVert_\infty \leq \delta$, then $$\begin{align} \lvert \phi(u)-\phi(v) \rvert &= \left\lvert \int_{[0,1]} u^2-v^2\right\rvert ...


0

Hint $$\lim_{z \to z_0} f(z) = c \Longleftrightarrow \lim_{z \to z_0} p(f(z)) = p(c)$$ $$\lim_{z \to z_0} f(z) = \infty \Longleftrightarrow \lim_{z \to z_0} p(f(z)) = \infty$$ then use the caracterisation of singularities thanks to limits.



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