The 'last seen' date on the profile page seems to have lost its color. This makes me sad. Let me illustrate:    I really like the color because it tells me stuff without me actually having to read. Please bring back color to mSE.
I am used to see timestamp everywhere by it's color. The recent timestamps are a little bit reddish everywhere. E.g. in Meta Stack Overflow:  Last edit:        Last seen:        But in Meta Stack Exchange they have same color everywhere even for recent time:  Last edit:        Last seen:         Please make it something like this:
How to prevent ENTER keypress to submit a web form? Intuition of the meaning of homology groups Are virtual particles only a fictive tool in equations? There is no "action at a distance" in nature. Attraction of a piece of iron by a magnet, attraction between distant electric charges of opposite sign, have to be mediated by something. The  are proposed as an explanation.  If they have an observable effect, it seems like they must exist. If so, why do we call them virtual? It can't be that they have short lifetimes, as there are short-lived particles (resonances) that we don't call "virtual."  On the other hand, some people say that the so-called virtual particles only exist on paper, or that they are just a technique to help us handle our equations.  So what is the meaning of the word "virtual" here? Do these particles really exist, or not? Does voltage drop occur with zero load and zero resistance Is there an official Marvel Cinematic Universe viewing order? How do people at Hogwarts usually tell the time? Classes and lessons at Hogwarts are presumably scheduled at fixed times of day. The first class might start, for example at nine in the morning.  To be on time for one's lessons, it is therefore crucial to know the current time but we know that electric watches of course do not work at Hogwarts because of all the magical interference. Are there magic watches? Is there maybe a spell to tell the time?  How do people at Hogwarts usually tell the time? Which manga chapters correspond to the 2nd season (or "Season 3" on Netflix) of the anime, "Revival of the Commandments"? The 2nd season (labeled as "Season 3" on Netflix) of the anime, Nanatsu no Taizai: Imashime no Fukkatsu (or The Seven Deadly Sins: Revival of The Commandments), has ended.  Which manga chapters correspond to each of the anime episodes? Neat series convergence result This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution. How to tenderize large squid? My whole life I've been buying small squid (maybe 4-6" long bodies without the head and tentacles), which are very mildly chewy and tender.  All it takes is sautee them in a little OO and they are good to eat, IOW, no additional tenderizing is needed.  Yesterday I got some large squid, about 1 lb each before cleaning, probably 10"+ long without the head, probably 5-6 times as much meat as the little ones I usually get.  I remember trying to cook that before and they came out very tough and with a bit of unpleasant odor.  Is there a way to marinate and tenderize large squid and to somehow neutralize the odor? I know several questions were asked about the difference between "going to" and "will". Based on several answers (see, for instance, ,  and ), I understood that "will" is more spontaneous and "going to" is used with more planned actions.  So, it seems that everything is pretty fine. However, in , Kosmonaut has an answer in which he states: "Let's say that tomorrow you will walk your dog from 7 - 8 AM".  On the one hand, you probably planned to walk your dog long before and thus I should use "Let's say that tomorrow you're going to walk your dog from 7 - 8 am".   On the other hand, since I'm saying "let's say...", I'm deciding right now (thus, unplanned) that you will walk your dog. So, even though in this hypothetical situation you made a plan, I'm in a more spontaneous mood deciding right now that that's what you will do tomorrow, and, thus, I should use "will".  Which one (if any) of the above explanations is right? In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.  The first issue I have is that my book began with the statement that given  $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  the problem $L[y] \ = \ f$ can be re-cast in the form $L[y] \ = \ \lambda y$.  Now it could be a typo on their part but I see no justification for the way you can just do that!  More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:  Given a linear second order ode   $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  it is an exact equation if it is derivable from a differential equation of one order lower, i.e.  $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$  The equation is exact iff  $$a''(x) - b'(x) + c(x) = 0. $$  If $F$ is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.  This equation is exact iff  $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$  If you expand this out you get the Adjoint operator  $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$  If you expand $L^*$ you see that we can satisfy $L \ = \ L^*$ if $a'(x) \ = \ b(x)$ &amp; $a''(x) \ = \ b'(x)$ which then turns $L[y]$ into something of the form  $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$  Thus we seek an integratiing factor $\alpha(x)$ so that we can satisfy this &amp; the condition this will hold is that $\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}$  Then we're dealing with:  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$  But again, by what my book said they magically re-cast this problem as  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$  Then calling  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$  a Sturm-Liouville problem.  My question is, how can I make sense of everything I wrote above? How can I clean it up &amp; interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues &amp; finding full solutions - what's going on? I want to be able to look at $a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)$ &amp; know how &amp; why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness &amp; integrating factors, thanks for reading! How do I change my username? Some time ago, when I installed Ubuntu, I chose a rather stupid username for my account that I do not want to use anymore. How do I change this (including the name of my home directory, and the name in the terminal) without losing settings for applications? How do I keep permissions and my keys for various authentification (e.g. email, SSH, GPG and more)? What settings could possibly get lost if I changed my username? Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ On the uniqueness of the Riemann-Christoffel tensor According to Section 6.2, Gravitation and Cosmology by Weinberg, the Riemann-Christoffel tensor is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives. The reasoning behind the same goes like this:   In a class of frames where $\Gamma^{\lambda}_{\mu\nu} =0$, the transformation rule for $\dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\kappa}}$ involves an inhomogeneous term which is symmetric in $\mu$, $\nu$, and $\kappa$. Thus, if one is to construct a tensor which is a linear combination of the first order derivatives of the Christoffel symbol then the only way to do so is by eliminating the inhomogeneous part of the transformation and this could be done only by making the combination explicitly antisymmetric in $\mu$ and $\kappa$. Since in these frames, $R^{\lambda}_{\mu\nu\rho}$ $ = \dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\rho}} -  \dfrac{\partial \Gamma^{\lambda}_{\mu\rho}}{\partial x^{\kappa}} $, $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated using the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  I think the presented argument can only suffice to prove that $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated from the first derivatives of the Christoffel symbol and is linear in them. I can't figure out why this suffices to assert that $R^{\lambda}_{\mu\nu\rho}$  is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  Edit: As the metric is covariantly constant, $\dfrac{\partial g_{{\mu}{\nu}}}{\partial x^{\rho}} = \Gamma^{\kappa}_{{\mu}{\rho}}g_{{\kappa}{\nu}}+\Gamma^{\kappa}_{{\nu}{\rho}}g_{{\kappa}{\mu}}$.  Therefore,   $\dfrac{\partial^2 g_{{\mu}{\nu}}}{\partial x^{\xi}\partial x^{\rho}} = \bigg(\dfrac{\partial \Gamma^{\sigma}_{{\mu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\mu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\nu}}+\bigg(\dfrac{\partial \Gamma^{\sigma}_{{\nu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\nu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\mu}} + \bigg(\Gamma^{\sigma}_{{\mu}{\rho}}\Gamma^{\sigma_1}_{{\nu}{\xi}}+\Gamma^{\sigma}_{{\nu}{\rho}}\Gamma^{\sigma_1}_{{\mu}{\xi}}\bigg)g_{{\sigma}{\sigma_1}} $  Now, if I want to show that the Riemann-Christoffel tensor is the only (non-trivial and independent) tensor that can be formulated out of the linear combinations of the second order derivatives of the metric tensor then I can equivalently show that the above expansion can be expressed linearly in terms of the Riemann-Christoffel tensor. But I am stuck over how to do it. Also, I am a little bit unclear as to what a linear combination means in this context. The coefficient of the second order term can be only a scalar constant or can it be the metric or can it be the metric that is getting summed over any of the indices as well? I used the jQuery Form plugin for asynchronous form submission. For forms that contain files, it copies the form to a hidden iframe, submits it, and copies back the iframe's contents. The problem is that I can't figure out how to find what HTTP status code was returned by the server. For example, if the server returns 404, the data from the iframe will be copied as normal and treated as a regular response.  I've tried poking around in the iframe objects looking for some sort of status_code attribute, but haven't been able to find anything like that.    The $.ajax() function can't be used, because it does not support uploading files. The only way to asynchronously upload files that I know of is using the hidden iframe method. Prove that for all integers $ a $ and $ b $, if $ a $ divides $ b $, then $ a^{2} $ divides $ b^{2} $. Pluralizing abbreviations where the noun is not the last word Ifstream read function doesn't load into vector Inverse of a matrix is expressible as a polynomial? Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'. Is calling a linear-equation a linear-function, misnomer or completely wrong? From my college life, I remember many professors used to call a linear-equation a linear-function, however:  A standard definition of linear function (or linear map) is:  $$f(x+y)=f(x)+f(y),$$ $$f(\alpha x)=\alpha f(x).$$  Where as linear equation is defined as:  $$f(x)=mx+b.$$  So, linear-equation is NOT a linear-function, according to the definitions defined above.  Though, for $b=0$ the linear-equation becomes a linear-function, but it is not true in general.  Question: Is it misnomer to call a linear-equation a linear-function, or it is completely wrong to say that? And linear-equation must be considered strictly as an affine-mapping.