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manged
▪ I. † manged, a. Obs. [f. mange n.1 + -ed2.] Suffering from mange.c 1410 [see mange a.].▪ II. † manged, ppl. a. Obs. [f. mang v. + -ed1.] ? Rendered stupid or helpless.1508 Kennedie Flyting w. Dunbar 546 Hangit, mangit, eddir-stangit, stryndie stultorum. 1508 Dunbar Tua mariit Wemen 118 Than mak I ... Oxford English Dictionary
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Jeremy Cockrill
Before he was elected in 2020, he manged his family's business, Fortress Windows and Doors, in North Battleford, Saskatchewan. wikipedia.org
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Ubiquiti APs on manged or unmanaged switch? I have three APs, one UAP-AC-IW and two UAP-AC-LITE. I have a Netgear Smart Managed Plus Switch, mainly for VLANs with plenty of open ports. I need to either add an unmanage...
If you use an unmanaged switch to a WAP that uses VLANs, you have no idea what will happen until you do. There is no standard for what an unmanaged switch will do with a VLAN tagged frame. Some unmanaged switches will drop the frame as damaged, some will strip off the VLAN tag and forward the frame,...
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Chain rule in conditional probability I'm currently taking a class on computer vision and I have the following problem: I should prove that, if my probability function is given as $p(a,b,c,d)$ and it can be displayed ...
In general the statement is not true. It may hold for some models; but in general the only thing you can say is that $$p(a,b,c,d) =p(a)p(b|a)p(c|a,b)p(d|a,b,c)$$ with any other permutation of the variables also valid. If you set as an assumption that $$p(a,b,c,d) =p(a)p(b)p(c|d)p(d|a,b),$$ Then, we ...
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Clarification on proof that all groups of order $< 60$ are solvable I've manged to prove that all groups of order $< 60$ are solvable, using Burnside's theorem. However, I found an alternate proof here Question about ...
Let $N$ be a maximal normal subgroup. Then $G/N$ is simple and abelian (because it is smaller than $A_5$), hence cyclic. $N$ is strictly smaller than $G$, hence we may assume by induction that itis solvable, hence $G$ is solvable.
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Asymptotics of the integral $\int_0^\pi x^n \sin(x)dx $ I am going through de bruijn's book on asymptotic methods. In the end of a chapter on Laplace's method for integrals, there is an exercise to show the following ...
For any $x\in(0,\pi)$ we have $$ \frac{\sin x}{x(\pi -x)} = \frac{1}{\pi}+K x(\pi-x),\qquad K\in\left[\frac{1}{\pi^3},\frac{4(4-\pi)}{\pi^4}\right] \tag{1}$$ hence $$ \int_{0}^{\pi}x^n\sin(x)\,dx = \frac{\pi^{n+2}}{(n+2)(n+3)}+\pi^n O\left(\frac{1}{n^3}\right)\tag{2}$$ due to $\int_{0}^{\pi}x^\alpha...
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Proving that greedy algorithm on TSP does not produce optimal solution I know that solving a TSP requires considering all possible cycles in the graph, and that a nearest neighbor greedy algorithm does not always prod...
Here is a counterexample where the optimal solution is of weight $13$ but it takes more than $100$ if we use the greedy algorithm starting from any vertex: ![enter image description here](
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Solving a congruence with high numbers and computing the Legendre symbol I am asked to compute $\left(\frac{77}{257}\right)$ specifically using Euler's criterion. I have manged to get the following: $$\left(\frac{77}{...
$77^{128}\pmod{257}$ is simple to compute through repeat-squaring: 1. $77^{2}\equiv 18\pmod{257}$ 2. $77^{2^2}\equiv 18^2 \equiv 67\pmod{257}$ 3. $77^{2^3}\equiv 67^2\equiv 120\pmod{257}$ 4. $77^{2^4}\equiv 120^2\equiv 8\pmod{257}$ 5. $77^{2^5}\equiv 8^2 \equiv 64\pmod{257}$ 6. $77^{2^6}\equiv 64^2\...
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Find the area of the surface obtained by rotating the curve about x-axis y= $(9-x^2)^.5$ $; 0<=x<=2$ I did the $\int_0^2$ and put that into the $$SA = \int_0^2 2\pi (\sqrt{9-x^2})*\sqrt{1+(-x/\sqrt{9-x^2})^2} $$ My...
\begin{align} \int_0^2 2\pi (\sqrt{9-x^2})\cdot\sqrt{1+(-x/\sqrt{9-x^2})^2}dx &=\int_0^2 2\pi (\sqrt{9-x^2})\cdot\sqrt{\dfrac{9-x^2+x^2}{9-x^2}}dx\\\ &=\int_0^2 2\pi\sqrt{9}\,dx\\\ &=6\pi\int_0^2dx\\\ &=6\pi x{\huge\vert}_0^2=12\pi \end{align}
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prove that if $\exists a\in A\space:\space stb(a)\not=\{e\}$ then the action is unfaithful let $G$ be abelian group which acts on non empty set $A$. prove that if $\exists a\in A\space:\space stb(a)\not=\\{e\\}$ th...
Suppose $G$ acts transitively on $A$, for a moment, and let $g\neq e$ be in the stabilizer of $a\in A$. For an arbitrary element $b\in A$, there exists $h\in G$ such that $b=h\cdot a$. Then $g\cdot b=g\cdot(h\cdot a)=h\cdot (g\cdot a)=h\cdot a=b$, so $g$ would also stabilize $b$. Thus $g$ stabilizes...
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Idempotence of the interior of the closure I'm reading the Complex Analysis text by Ahlfors. I'm stuck on exercise 5 on chapter 3: Prove that $\overline{\overline{\overline{{\overline{X}}^c}^c}^c}^c=\overline{{\overl...
Provided that you know the following: For any set $A$, $Int(A)\subset A \subset Cl(A)$ and for any sets $A \subset B$, $Int(A) \subset Int(B)$ and $Cl(A) \subset Cl(B)$ As you said, $Int(Cl(A))\subset Cl(Int(Cl(A)))$ and taking interiors gives $Int(Int(Cl(A))) \subset Int(Cl(Int(Cl(A))))$ so $Int(Cl...
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How to find branch cut of $\ln(\sin(z))$? I know that $\ln(z)$ is undefined when $z=0$ or $z=\inf$ , so I manged to find the critical points to be $z=\\{k*\pi,\inf*i, -\inf*i\\}$ , for every integer k, but I don't kno...
First: $$\sin(x+iy)= \sin(x)\cosh(y)+i\cos(x)\sinh(y)$$ Now it depends how you choose the branch cut of the logarithm, I'll pick it along the negative real axis. To find points on the branch cut, we must thus have: $$ \sin(x)\cosh(y)<0\\\ \cos(x)\sinh(y)=0 $$ The first condition gives $\sin(x)<0$. T...
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Legendre Symbol sum Let $p \equiv 3 \pmod 8$. Show: $\sum_{a=1}^{p-1} a\Big(\frac{a}{b}\Big)=\sum_{a=1}^{(p-1)/2} (2a-p)\Big(\frac{a}{b}\Big)$ and $\sum_{a=1}^{p-1} a\Big(\frac{a}{b}\Big)=\sum_{a=1}^{(p-1)/2} (p-4a)...
Try to reduce it to first identity by changing summation index to $a' = 2a$.
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Show that $\left\{ \omega \mid sup_{n}\{h_n(\omega)\} > c \right\} = \bigcup_{n=1}^{\infty} \left\{ \omega \mid h_n (\omega) > c \right\}$ Given a sequence of measurable functions $h_n (\omega)$ I manged to prove that...
If $\omega$ is in the union, then $h_{n_0}(\omega)>c$ for some $n_0$, hence $\sup_nh_n(\omega)\geq h_{n_0}(\omega)>c$. This shows that the right-hand side is contained in the left-hand side.
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Lorenz equations and find a minimal trapping region. Consider Lorenz's equations $x^{'}= \sigma (y-x)$ $y^{'}= (rx-y-xz)$ $z^{'}= (xy-bz)$ $\sigma, r, b>0$ are parameters of the system. The question is as follows...
We first see that the time derivative of the function $$V = rx^2+\sigma y^2+\sigma(z−2r)^2$$ is $$\dot V = -2\sigma(rx^2+y^2+bz^2-2brz)$$ We know that a trapping region exists. Because $\dot V$ goes to $-\infty$ in all directions, the region where $\dot V \ge 0$ is bounded. In fact, if $C$ is the ma...
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