I am facing a problem with Firefox when logging onto my Stack Overflow account.  I click on "log in," then I choose "connect with Facebook."  But after that, nothing happens, except for the redirection to the main page.  If I try to log in again, a message displays on the top of the page telling me that I am already connected.  Notice: this problem did not exist on Chrome in my case.  This is the picture of my case:
Here's what is happening; everytime I go to Stack Overflow (stackoverflow.com, NOT meta) with Firefox, it gives me the "Welcome back" tab that pops up from the top. Whenever I click in the "click here to refresh the page" link, all it does is refresh the page without logging me back in.  Even if I try to log in using the "log in" button at the top, all it does is pop up the tab that says "Welcome name here, you're being redirected." The problem is I'm not being logged in, not to mention that it keeps refreshing the page over and over again without giving me a choice to log in with Facebook.  For some reason, I was capable of logging in to Meta Stack Overflow using Firefox. I am still unable to log in to Stack Overflow. Any ideas?
no Wi-Fi on Ubuntu 14.04 LTS Asus notebook with MEDIATEK MT7630e I'm new to Ubuntu. I installed it today, but the first thing I encounter is that I can only connect to the Internet through wifi.  Can someone help me?   iwconfig result :  eth0      no wireless extensions.  lo        no wireless extensions.   lspci:  00:00.0 Host bridge: Intel Corporation 3rd Gen Core processor DRAM     Controller (rev 09) 00:01.0 PCI bridge: Intel Corporation Xeon E3-1200 v2/3rd Gen Core processor PCI Express Root Port (rev 09) 00:02.0 VGA compatible controller: Intel Corporation 3rd Gen Core processor Graphics Controller (rev 09) 00:14.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB xHCI Host Controller (rev 04) 00:16.0 Communication controller: Intel Corporation 7 Series/C210 Series Chipset Family MEI Controller #1 (rev 04) 00:1a.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB Enhanced Host Controller #2 (rev 04) 00:1b.0 Audio device: Intel Corporation 7 Series/C210 Series Chipset Family High Definition Audio Controller (rev 04) 00:1c.0 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 1 (rev c4) 00:1c.1 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 2 (rev c4) 00:1c.3 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 4 (rev c4) 00:1d.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB Enhanced Host Controller #1 (rev 04) 00:1f.0 ISA bridge: Intel Corporation HM76 Express Chipset LPC Controller (rev 04) 00:1f.2 SATA controller: Intel Corporation 7 Series Chipset Family 6-port SATA Controller [AHCI mode] (rev 04) 00:1f.3 SMBus: Intel Corporation 7 Series/C210 Series Chipset Family SMBus Controller (rev 04) 01:00.0 3D controller: NVIDIA Corporation GF117M [GeForce 610M/710M/820M / GT 620M/625M/630M/720M] (rev a1) 03:00.0 Network controller: MEDIATEK Corp. MT7630e 802.11bgn Wireless Network Adapter 04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. Device 5289 (rev 01) 04:00.2 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller (rev 0a)   lspci -nnk | grep 0280 -A2:  03:00.0 Network controller [0280]: MEDIATEK Corp. MT7630e 802.11bgn Wireless Network Adapter [14c3:7630]     Subsystem: Foxconn International, Inc. Device [105b:e074] 04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. Device [10ec:5289] (rev 01) How do I register a custom URL protocol in Windows? How do I register a custom protocol with Windows so that when clicking a link in an email or on a web page my application is opened and the parameters from the URL are passed to it? Virtualbox Windows 10 Guest installation is freezing up in 14.04 I've installed VirtualBox using the Software Centre and I'm running a system on Ubuntu 14.04 with 8gb RAM, 64-bit dual core processor.  I want to run Windows 10, I've tried both 64 and 32 bit versions, in virtualbox. However, every time I start up windows 10 it gets to the point where the dots go round in a circle, then it freezes up and crashes back to the virtualbox manager. The only information it gives me is that the virtual machine was aborted.  I've changed the settings around a lot and nothing is working.  Any help would be appreciated.  Oh, and I've run Windows 10 technical preview on this machine before, a while ago  Thanks for any help in advance. I have this new problem since 2014 begin. I cant remotely login to this enterprise network for giving them support. Its keep showing this new error.  How to fix it please?    EDIT: More than a decade ago, I read a short story about a small group of people at the end of the world. The story was premised on some sort of mass-produced device which took ambient heat and turned it into power. The devices had been overused, and now the world was freezing.   I think the people were essentially bandits; the main character may have been a woman who lured men to her house and killed them for their body heat? There was also a strong implication that one person she had done this to was an alien who had come to warn that the devices were destroying the world.  I think the story was in a hardbound collection, and I think the collection was from the late 70's or very early 80's.  (Calling it a "short story" is just meant to distinguish it from a full novel — it may have actually been a novella or similar.) How to find the matrix of a quadratic form? I was wondering. If I have a bilinear symmetric form, it is easy to find its matrix. But, when I have a quadratic form, which is the procedure to do that?  I heard that one possibility is: If $q$ is my quadratic form, then $$f(x,y) = \frac{1}{4}q(x+y) - \frac{1}{4}q(x-y)$$ is the bilinear symmetric form associated, so the method reduces to find the matrix of $f(x,y).$   The point is that seems a little noising... How to find, in practice, the matrix of quadratic form?  Thanks in advance. Fix terminal after displaying a binary file UTF8 to/from wide char conversion in STL Is it possible to convert UTF8 string in a std::string to std::wstring and vice versa in a platform independent manner? In a Windows application I would use MultiByteToWideChar and WideCharToMultiByte. However, the code is compiled for multiple OSes and I'm limited to standard C++ library. Why aren't Christoffel symbols tensors? - asked from a fibre bundle perspective Does Quicksilver's walkman speed up with him? Can I mask an input text in a bat file? Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$ I used to have an icon on the Windows Vista taskbar showing the battery amount remaining (laptop). It has mysteriously vanished.  How can I get it back?  I followed the Windows  (see How can I hide the battery icon in the notification area? - hoping to reverse the instructions) but the Power checkbox on the Notification Area tab was greyed out and could not be checked or unchecked (it is currently unchecked). Are the floor functions of $0.999\cdots$ and 1 equal?  It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ? Let $\mathbb{S} = \{ (X,Y,Z) \in \mathbb{R}^3 :X^2 + Y^2+Z^2=1\}$ denote the Riemann sphere. The following problem is taken from Gamelin "Complex Analysis" chapter one:     What is the image of the spherical cap $ C := \{(X,Y,Z) \in \mathbb{S} : A \leq X \leq 1\}$, where $A$ is some fixed real number between $-1$ and $1$, under the stereographic projection $ \pi(X,Y,Z) = \frac{X+iY}{1-Z}$?   Here is my work so far:  ( The case $A=1$ is trivial, so assume $A &lt;1$)  We want to find $\pi(C) = \{z \in \mathbb{C} \mid \exists(X,Y,Z) \in C: \pi(X,Y,Z) = z \}= \{z \in \mathbb{C} \mid \pi^{-1}(z) \in C\}$, where the last equality holds by $\pi$ being a bijection. Write $z=x+iy$.  Here, $ \pi^{-1}(z) \in C\iff A\leq \frac{2x}{1+x^2+y^2} \leq 1 \iff (\star) :A \leq \frac{2x}{1+x^2+y^2}$, since the inequality $2x \leq 1+x^2+y^2$ holds for all $x,y \in \mathbb{R}$.  $A=0$: $( \star)$ reduces to $ 0 \leq x$, so in this case the image is $\{z \in \mathbb{C} \mid \Re(z) \geq 0 \}$.   $A &gt;0$: Since $A \neq 0$ in $(\star)$ we have,  $ \frac{2x}{1+x^2+y^2} \geq A \iff Ax^2-2x+Ay^2+A \leq 0 \iff (x-\frac{1}{A})^2 +y^2 \leq \frac{1}{A^2}-1$. Thus the image is the interior and boundary of the circle centered at $\frac{1}{A}$ with radius $\sqrt{\frac{1}{A^2}-1}$. Note $\frac{1}{A^2}-1 \geq 0$ as $A \in (0,1)$.  $A &lt;0$ : By symmetry and the above, here the image is $ \mathbb{C} / \{ $ interior and boundary of circle centered at  $  \frac{1}{A}$ with radius $\sqrt {\frac{1}{A^2}-1} \}$.  Is my working correct? Many Thanks! Add table row in jQuery In this  to a question about a series, a theorem was stated:     If $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the  of the set.   My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it? Show the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$ Show that for $n \geq 5$, the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$.     I'm not sure on how to handle this problem. I know that the commutator subgroup $G'$ will consist of products of commutators $[\sigma_{i}, \tau_{i}]$ for permutations $\sigma_{i}, \tau_{i} \in S_{n}$. However, I don't know where to start. What would be the best way to approach this? $p=4n+3$ never has a Decomposition into $2$ Squares, right? Primes of the form $p=4k+1\;$  have a unique decomposition as sum of squares $p=a^2+b^2$ with $0&lt;a&lt;b\;$, due to .  Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic residues $a^2+b^2$ with $a,b\in \Bbb{N}$ $$ a^2 \bmod 4 +b^2 \bmod 4 \le 2? $$  If so, are there alternate ways to prove it? I've already tried clicking and holding the left button of my mouse, and it just won't work! Whenever I press it, it opens the game menu instead.