I have 7 or 8 accounts across the Stack Exchange site. When I bring the StackExchange dropdown menu down, it lists all my accounts for each forum and my reputation for each as well. Why, when I click on some of them, do I have to either "Log In" again, or refresh the page when the notification pops up the top saying Welcome back scrowler, you have been logged in. Please refresh the screen  I don't want to log in again and again, and I don't want to have to refresh the screen (call me impatient).  Request:   click on one of the SE networks in the drop down have the link automatically generate the required authentication cookies required for the site continue   Objective:   prevent the need to log in, or refresh the page after automatic login when clicking on a linked SE account
I'm looking for an answer considering both technical limitations and user experience.   Does it need a reload from a technical standpoint?
I was asked this question in an interview.  Lets say we have a correlation matrix of the form \begin{bmatrix}1&amp;0.6&amp;0.8\\0.6&amp;1&amp;\gamma\\0.8&amp;\gamma&amp;1\end{bmatrix}  I was asked to find the value of gamma, given this correlation matrix. I thought I could do something with the eigenvalues, since they should be all greater than or equal to 0.(Matrix should be positive semidefinite) - but I don't think this approach will yield the answer. I am missing a trick.  Could you please provide a hint to solve for the same? How to raise a complex number to the power of another complex number? How do I calculate the outcome of taking one complex number to the power of another, ie  $\displaystyle {(a + bi)}^{(c + di)}$? Look and you shall find (a correlation) Force.com IDE 35.0.0.201510211056 and Force.com IDE Debugger 35.0.0.201510211056 in eclipse but it doesn't show this option to run test class, is their any way i can see it because it show the Apex_Test_Result panel but running a test class option is missing. plz help What is the easiest way to export just the selected objects in Illustrator? The title says it all, I want to know what the simplest way to export the selected objects.  In Inkscape, there is a menu option called export selected which does just this.    The aforementioned option lets you export the selection as a composite image or as a batch of individual images.  Both would be nice, but in this scenario I want to export the selected objects as a single image.  I can already create a new document, and paste in the selected items.  I am looking for a way that is easier than this.  How can I save / output / export only the objects I currently have selected, to an image file, in Adobe Illustrator CS5? I'm putting some type on a path, a circle; but when I do it, the type is upside down. How can I flip it so it reads the right way up? See image below: Does $\frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}$ converge? Does the sequence $$\displaystyle \frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}$$   converge?   Attempt. Since $\Big(\frac{k}{k+1}\Big)^k \rightarrow 1/e\neq 0$ and the terms are positive, the series $\sum\limits_{k=1}^{\infty}\Big(\frac{k}{k+1}\Big)^k$ diverges to $+\infty$. I find hard to determine if $n$ or the sum goes faster to $+\infty.$   Thanks in advance. \colorbox height too long \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{xcolor} \usepackage{minted} \usepackage{parskip}  \definecolor{codegray}{gray}{0.9} \newcommand{\code}[1]{\colorbox{codegray}{\texttt{#1}}}  \begin{document}  \code{test code}\\ anotherline  \end{document}   This is working good on sharelatex.com and looks like I expect and want it:   but on my local project using this command gives me this:   any idea what could be causing this colorbox height going to the next line? Is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true? If $A$ is a $n\times m$ matrix, is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true?  I'm thinking that rank-nullity would be the simplest and easiest way to prove this, but would applying the definition of linear independence to the components of $(A^T)$ be more illustrative? I'm a bit hung up on this "necessity" issue, and not sure if it has a particular logical meaning in linear algebra. I have a problem with my touchpad; I can't scroll down with 2 fingers and I don't know why. I have searched on Google about my problem and I've found some answers but none of them don't work for me.  One solution I found was to go in System Settings -&gt; Mouse &amp; Touchpad -&gt; Touchpad and check the Two-finger scrolling, but I don't have that option.    Another solution was to try to reinstall, which also didn't work. Yet another solution was to go in Ubuntu Software Center, type synaptic and install the Synaptic Package Manager but that didn't work, either.      What can I do to be able to scroll with two fingers? Proof that every metric space is normal 407 Proxy Authentication Required When I try to install a software using Ubuntu Software center I get:  Failed to download repository information Check Your Internet connection   When I try to do a apt-get install something, I get:  407  Proxy Authentication Required   I use a proxy server that requires a user-name and a password. I have set my systems proxy manually, by plugging in the required numbers in the Networks proxy and applied it system wide. I guess the problem now is plugging in my user-name and password.  When I use INTERNET via Mozilla, it specifically asks me for my user-name and password. Consider a hypothetical situation: In which a time traveler travels in a spaceplane at close to speed of light circling the earth and I get a live telecast of inside of that plane to my television. How time dilation will effect him and what will I observe sitting next to the tv seeing him live. Considering time runs slower at speed of light, will he be looking like a slow motion video? (Or let he also watching live streaming of us at earth, what it will look to him) P.S. I'm not sure if it even makes some sense but if it does kindly explain in logical details. Sorry for the the bad english, I'm not a native speaker. What's the possibility of adding a "Did you mean ..." in the tagging system? Best way to follow a log and execute a command when some text appears in the log How can mixed content compromise an entire HTTPS session? Google Chrome is probably one of the stricter browsers out there in terms of highlighting SSL errors.  Most browsers will point out when you are viewing an HTTPS page which loads insecure images/Javascript, I understand how this is a security issue - an attacker could have modified the non-SSL content in a way which gives them access to the secure content, injected some javascript for example.  However once you have viewed an HTTPS page with mixed content, Google Chrome will continue showing you the mixed content warning (a red line through the 'https://') for as long as you stay browsing the same domain, even for subsequent pages which only include other HTTPS resources. Only once you navigate to another domain, or open a fresh tab does it tell you you're secure again.  What is the danger it's protecting you from here? How could mixed content on a previous page actually affect the security of subsequent secured pages? I understand that mixed content should be flagged up, but there must be a specific reason that Google is persisting this warning, other than "Well you still need to know about it" How to prove that $E[Y]=E[E[E[Y|X_1, X_2]]]$ ?  PS. I don't see how $E[E(Y|X_{1},X_{2})|X_{1}]=Y[Y|X_{1}]$ and $E[Y]=E[E(Y|X_{1})]$ can be used here. But it feels close. Please help, I'm stuck PPS. Thanks @Deep North. Let's consider the case when X1 is independent of X2 Obtaining a formula for prediction limits in a linear model (i.e.: prediction intervals) Let's take the following example:  set.seed(342) x1 &lt;- runif(100) x2 &lt;- runif(100) y &lt;- x1+x2 + 2*x1*x2 + rnorm(100) fit &lt;- lm(y~x1*x2)   This creates a model of y based on x1 and x2, using a OLS regression.  If we wish to predict y for a given x_vec we could simply use the formula we get from the summary(fit).  However, what if we want to predict the lower and upper predictions of y? (for a given confidence level).  How then would we build the formula? Formula label outside of formula Primary ideals in Noetherian rings For an $R$-module $M$ I have the following definition for a submodule $N\subset M$ to be $\mathfrak{p}$-primary: this is the case when $\text{Ass}(M/N) = \{\mathfrak{p}\}$, that is, $M/N$ is coprimary or $(0)$ is a primary submodule.  Now for an ideal $\mathfrak{q} \subset R$ this definition gives that $\mathfrak{q}$ is $\mathfrak{p}$-primary iff $\text{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}$. It can be shown that this is equivalent with $\mathfrak{p}^n \subset \mathfrak{q}$ for some $n\in \mathbb{N}$ and the condition \begin{align*} \forall r,s \in R: rs \in \mathfrak{q}, r\not\in \mathfrak{p}\Rightarrow s \in \mathfrak{q}. \end{align*}     Now I suspect that if $R$ is Noetherian, and if $\mathfrak{q}$ is $\mathfrak{p}$-primary, then we have $\sqrt{\mathfrak{q}} = \mathfrak{p}$. I have shown this for PID's, but does this generaly hold?    And if so, is every ideal $\mathfrak{q}\subset R$ then $\sqrt{\mathfrak{q}}$-primary?   Many thanks.