Since recently (about a week ago), the SE app doesn't work for me any more - either or Android or on iOS.  When I open the app, I see the progress spinner for a short while, after which I get:   On Android: a blank screen and a white-on-red error message "Sorry, something went wrong trying to load teh feed. Pull to refresh to try again." On iOS: one hot question with blank screen below and a white-on-dark-blue error message "Request failed. Tap to retry."   See screenshots below from both.  I tried on WiFi and 4G - with the same results. I tried rebooting rebooting the phone - with the same results. My Android phone is Nexus 5X with Android 7.1.2. My iOS mobile is an iPhone 6 with iOS 12.1.2.  Is there a way for me to get around this? To start with, are there any logs I can look at to see what's going on?   &nbsp;&nbsp;&nbsp;
I'm currently unable to access anything via the Stack Exchange app under iOS; is this a known issue?
How is light affected by gravity? Light is clearly affected by gravity, just think about a black hole, but light supposedly has no mass and gravity only affects objects with mass.  On the other hand, if light does have mass then doesn't mass become infinitely larger the closer to the speed of light an object travels. So this would result in light have an infinite mass which is impossible.  Any explanations? Preamble: I know that 'give a reason when downvoting' is a very common request, and I assure you that I have read these posts and the discussions in them and my request is very different. This isn't about encouraging people to comment and is more about gathering metrics.  I'd like to see a feature whereby people who are downvoted get some better feedback on why they were downvoted, especially given that people rarely leave comments in practice.  So my suggestion is that when user clicks on the downvote, they are prompted to pick from a short list (No more than 5) of common reasons. It would be a popup that replaces the existing "Please consider adding a comment" box.  The user receiving the downvote would see how many people downvoted them for each reason, but not who did it. It would significantly address the issue of people being too lazy to comment or not wanting to comment (for fear of retaliation or any other reason) while helping to ensure the user gets useful feedback.  Perhaps the downvote could also be weighted based on whether or not a reason is picked.  For questions the list might look something like:   Not a question Not relevant to the subject area Trivial question with no research effort Poorly asked question Other   And for answers, this might be:   Nonconstructive or not an answer Incorrect or dangerous answer Poor explanation Doesn't answer the original question Other Mathematical notation around the world A Regular Tetrahedron is a cool Polyhedron. A regular tetrahedron has this property:  For any two of its vertices exists a third vertex, which forms a regular triangle with these 2 vertices.(It doesn't necessarily have to be a face of it).  Are there any other polyhedrons that have the same property?  I think there isn't such a polyhedron. But not sure how to prove it. I've tried proving that there isn't an irregular tetrahedron with this property, assuming that there is a pair of unequal sides.   Current answers are a little confused with the conditions of the problem. I remember being taught that the correct order of adjectives in English was something along the lines of "Opinion-Size-Age-Color-Material-Purpose."  However, it's been a long time and I'm pretty sure I've forgotten a few categories (I think there were eight or nine). Can anyone fill them in? Over time I realized a very annoying property of the Stack Overflow reputation system: Easy answers to popular question are rewarded with a lot more reputation points than actually great answers. In case anybody actually needs examples for this, here are two examples from :   gained me 420 rep, I invested 2 minutes writing it, and it only was a wild guess.   took me about an hour for the research and the math and gained me 40 rep.  Now I know that you are bound to get more rep for answers that are useful to more people, and I'd say that's fine. But the current situation is quite extreme: 30% of my reputation is from 5 of my 330 answers, and they are definitely far from being my best ones.  A final example that is not from my profile, but more extreme: Do you really think  is worth 40000 reputation?  My proposition to improve that situation would be to make the reputation gain from a single answer non-linear, maybe even have it capped at a certain point. I'm not sure on how that non-linear function should look exactly, and would leave good proposals for that to users with more experience than me. How do I make Units efficiently? I just blew almost all my money on a really cool lookin' ship. How do I regain solvency without spending a lot of time fiddling around in my inventory?  At the moment I've found a sequence of steps that'll turn 50 carbon and 100 Plutonium into ~5000 Units (by creating Suspension Fluid and then turning it into Electron Vapor). Are there better recipes or sequences of recipes I should look for? Or should I just drop the whole thing and go mine a particular mineral? How to run GUI applications remotely on a headless server? I'm new to Ubuntu - sorry for novice questions. What I have - is Ubuntu server, to which I have access only via ftp and putty. What I would like to do - is to run firefox (or any other browser) from that Ubuntu machine.  I've managed to install firefox , but when I'm trying to run something like this:  x-www-browser http://google.com   It is saying:  Error: no display specified   I don't understand - there is no display at all at Ubuntu sever - it is just computer case on rack, the only display I have - is mine.  Don't know, how to redirect this display output to my machine (if this is possible at all). Not every open set in $\mathbb{R}^n$, for $n\geq2$, is a disjoint union of countable open balls. I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help. change environment of a running process How might it be possible to alter some variable in the env of an already running process, for example through /proc/PID/environ? That "file" is read-only.  Need to change or unset the DISPLAY variable of a long-running batch job without killing it. Is it common to use "where" after "a time" as a relative? I heard the following comment of Mr. Marco Rubio about Mr. Donald Trump on CBS’s “Face the Nation” in a clip in February 29th AP Radio News:  “There is never gonna be a time where the Republican Party rallies around and says you have to get out or anyone has to get out for purposes of rallying around Donald Trump. Donald Trump is not Republican Conservative. Donald Trump is trying to pull off the biggest scam in American political history.”  I thought I misheard “There is never gonna be a time where –“ for “There is never gonna be a time when –“ at first, because I was in understanding that time should be trailed by “when” as a relative noun or adverb. An English Japanese dictionary at hand (Readers English Japanese Dictionary published by Kenkyusha) provides “where” singly as a relative to refer the ‘place.’ However, my lookup of the following two sources endorsed that I heard it right:  Here is what is never going to happen in this race. There's never going to be a time where the Republican Party rallies around and says you have to get out or anyone has to get out for purposes of rallying around Donald..  “Here is what is never going to happen,” Rubio continued. “There's never going to be a time where the Republican Party rallies around and says you have to get out or anyone has to ...   Is it common, at least not uncommon to use “where” after “time” as a relative? Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$     It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:  $\begin{matrix}  \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}  &amp;  \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} &amp; \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} &amp; \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} &amp; \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}  \\  \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}  &amp;  \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} &amp; \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} &amp; \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} &amp; \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} &amp; \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} &amp; \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} &amp; \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}  \end{matrix}$  one can notice then, that the first column vanish the fourth one :  $\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$  $\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$  and the third column vanish the fifth one :  $\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$  $\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$  $\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$  $\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$  while the second column is self-vanishing:  $\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$  $\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .  So the equality occurs. But how to generalize the proof? Setting "checked" for a checkbox with jQuery TikZ/PGF 3.0 datavisualization vs pgfplots To follow-up on a previous question , the current release of TikZ 3.0 has a flexible datavisualization engine that creates the same kind of plots as the more mature plotting package pgfplots.  What is the current recommendation for producing complete plots in LaTeX? Prove universal morphism is unique up to unique isomorphism. I'm following along Wikipedia page() on universal property, and this seems it should be trivial, but I couldn't finish the proof.  The definition I am working with:   The problem   So I want to prove that if $(A, u)$ and $(A', u')$ are universal morphism from $X$ to $F$, then there exists unique isomorphism $k: A \rightarrow A'$ such that $u' = F(k) \circ u$.   My attempt:   For any $B \in C$ and $f: X \rightarrow F(B)$, there exists unique $h: A \rightarrow B$ such that $f = h \circ u$.  Letting $B = A'$ and $f = u'$, we see that there exists unique $k: A \rightarrow A'$ such that $Fk \circ u = u'$.  This shows existence and uniqueness. We only need that $k$ is isomorphism, but this is the part I cannot prove.   Similarly, there exists unique $k': A' \rightarrow A$ such that $Fk' \circ u' = u$.  Plugging into each equation, we get  $$Fk \circ Fk' \circ u' = F(k \circ k') \circ u' = u'$$  $$Fk' \circ Fk \circ u = F(k' \circ k) \circ u = u$$  but this of course doesn't imply $F(k \circ k') = id$, so it seems useless.   Thanks for your help! Tikz: shading region bounded by several curves I know that my effort below can be improved upon considerably.  In particular, there must be a way of shading the region in a single scope, as opposed to subdividing as I have done (thus necessitating a hidden gray boundary).  Neither am I fond of sample points, but this is where my research has led me.  Your constructive criticism is appreciated.    \documentclass{article} \usepackage{tikz} \begin{document}  \begin{tikzpicture} %graph \draw[draw=gray!50!white,fill=gray!50!white]      plot[smooth,samples=100,domain=0:1] (\x,{0}) --      plot[smooth,samples=100,domain=1:0] (\x,{1}); \draw[fill=gray!50!white] plot[smooth,samples=100,domain=1:2.71828] (\x,{ln(\x)}) --      plot[smooth,samples=100,domain=2.71828:1] (\x,{1}); \draw[domain=0:4] plot (\x,{1}); \draw[samples=100,domain=.25:4] plot (\x,{ln(\x)}) node[above left]{$y=\ln x$}; %coordinate grid \draw (-.5,0)--(4.5,0) node[right]{$x$}; \draw (0,-1.5)--(0,2.5) node[above]{$y$}; \foreach \x in {1,2,3,4}     \draw (\x,2pt)--(\x,-2pt) node[below] {$\x$}; \foreach \y/\ytext in {1,2}     \draw (2pt,\y)--(-2pt,\y) node[left] {$\y$};     %labels  \node at (.75,.5) {$\mathcal{D}$}; \end{tikzpicture}  \end{document} Bash built-ins with no manual entries Newer "mainline" versions of the kernel are available from the , e.g. 3.4 and 3.5, while Ubuntu 12.04 is on 3.2.   What are the mainline kernels? Are they better than my existing kernel? What are the pros and cons of upgrading to a mainline kernel? Will I easily be able to go back? How do I reset the Windows XP Administrator password? I have forgotten the Administrator password of my Windows XP installation and locked out. Is there a way to reset it? I do have physical access to the PC. Find the MLE of the unknown parameter $\theta$ when $X_1,X_2,...,X_n$ is a sample from the distribution whose density function is: $$f_X(x) = \frac12e^{-|x-\theta|}, -\infty&lt;x&lt;\infty$$ What I did so far: $$\text{maximize }(\frac12)^ne^{-|x_1-\theta|-|x_2-\theta|-...-|x_n-\theta|} \\ \text{thus we want to minimize }|x_1-\theta|+|x_2-\theta|+...+|x_n-\theta|$$ Since $|x_i-\theta|$ is positive for all $i$, we want to solve for $\theta$ so: $$|x_1-\theta|+|x_2-\theta|+...+|x_n-\theta|=0$$ What do I do from here?