# Tag Info

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 ODEx_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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