Here is what I do:   Search for a tag Open a discussion from list of discussions returned by search result Tap back button   This should take the view to search result but instead it returns back to tags search.  I believe this behaviour to be a bug and it is unpleasant to search for your tag every time.   App Version: 1.0.85 Device Manufacturer: asus Device Model: ASUS_Z00AD OS Version: 5.0 (WW_Z00A-WW_2.20.40.139_20151104_2652_user_rel-user-20151104)
This looks a little bit like a variation on  but specific to the tags view.  Reproduction (app version 1.0.71):   Select any site (say Meta) from the sidebar Switch to the tags view using the top bar Choose a tag, any tag ( is nice)   optionaly refine your search with more terms  Go to any question found, go back   you end on the unfiltered tags view  Go back again   lucky you, your search is here
I have a batch file which is in a directory and must be run from there as well because it updates files within this directory. This works perfectly fine, except when the user runs the batch file as administrator (required on Vista). Then the starting directory is C:\Windows\System32. Is there any way to still be able to know from which directory the batch file was run? I dont want the user to enter the directory manually. PHP generate file for download then redirect How to automatically keep user remembered in Devise There was this cool tv show I used to watch as a kid (in the 1990's I think). It was about a world where water was extremely rare and people used to fight over it, however there was this person who had this large 3D polyhedron shaped device (big enough to fit a few people in it) that could synthesise water.  Now, this device was either a time travel device or could just make water (not sure but I remember some time travel aspect as they could either remember the world the way it was where water was plentiful or could travel back to it on occasion).  Anyway, would love to know if this rings any bells?????  It was on in the UK (but maybe the US too, not sure). Is JavaScript's "new" keyword considered harmful? In another , a user pointed out that the new keyword was dangerous to use and proposed a solution to object creation that did not use new. I didn't believe that was true, mostly because I've used Prototype, Scriptaculous and other excellent JavaScript libraries, and everyone of them used the new keyword.  In spite of that, yesterday I was watching Douglas Crockford's talk at YUI theater and he said the exactly same thing, that he didn't use the new keyword anymore in his code ().  Is it 'bad' to use the new keyword? What are the advantages and disadvantages of using it? Maclaurin Expansion for $e^{e^{z}}$ at $z=0$ Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46. Attempted proof based on  and .  Given: $a,b$ elements of $G$, and $ab$ has finite order $n$.   Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$.   I must prove that $n$ is the smallest positive integer        such that $(ba)^n = e.$ Notice:  $\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &amp;= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies  \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} $ $\implies |ba| \le n$.  Now I must prove that there's no positive integer $m$ such that $m&lt;n$ and $(ba)^m = e.$ For a proof by contradiction, suppose $|ba|&lt;\color{magenta}{n}$. Then we could apply the same reasoning to find that $ |ab| ≤ |ba| &lt; \color{magenta}{|ab|}$,  which is absurd. So $|ba| = |ab| = n. \quad \blacksquare$  Questions   Whence does $ |ab| ≤ |ba| $ arise, in the second last line?   What's the intuition of the claim proved?    How do you prognosticate that $|ba|&lt;\color{magenta}{n}$ is impossible, and that you should prove it by contradiction? In a situation where say a group (or at least a plurality) of men is being addressed — for example on a sign passed by many married men — which is correct?  &quot;Remind your wife.&quot;  or  &quot;Remind your wives.&quot; Uniformly Continuous Function sending Bounded Set to Unbounded One Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and let $f: X \to Y$ be a uniformly continuous function. If $A \subset X$ is bounded, must $f(A) \subset Y$ be bounded?  It is clear to me that in metric spaces that satisfy the Heine-Borel property, such as $\mathbb{R}^{n}$, the answer to this question is yes. However, I can see no reason why this should hold for arbitrary metric spaces. Any ideas? Thanks! Prove the Probability of Two Events I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets:  $$P(A \cup B) = P(A) + P(B) - P(A \cap B) $$  Note that the function $P$ is the probability of an event happening.  From the third axiom, we are given that for any two disjoint sets, that is $ A \cap B = \emptyset$:  $$P(A \cup B) = P(A) + P(B)$$    My thought process follows from this third axiom, but I run into a problem near the end:  Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so  $$P(A \cup B) = P(A) + P(B)=P(A-B)+P(B-A)$$  From this, it appears that we are essentially taking the symmetric difference of A and B.  This, to me, seems to be equivalent to:  $$P(A-B)+P(B-A)=P(A)+P(B)-P(A \cap B)$$  Hence, by my logic:  $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$  Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them.  If the sets are disjoint, then the intersection is merely null.  If they intersect, then we are removing the "extra pieces".  I feel like this is inadequate proof despite my ability to convince myself that it is true..  Where, if anywhere, did I go wrong?  Thanks! What time zone do SO's servers operate in? What time zone do Stack Overflow's servers operate in?  I noticed that it couldn't be my time zone, or the ones before me (I'm in North American Eastern Time), because I logged into SO earlier today (around 12 pm), then I logged in just a few minutes ago (around 8 pm), then about a minute later I got the Fanatic badge (yay me!). Let $L$ and $R$ be $n \times n$ matrices. I am trying to solve the following minimization problem  $$ L^{k+1} := \arg \min_{L} \lambda \|L\|_{\ast} + \frac{1}{2\mu}\|L-R^{k}\|_{F}^{2} $$ $$ \mbox{s.t.}\ L \succeq 0 $$  where $\|\cdot\|_{\ast}$ and $\|\cdot\|_{F}$ denote the nuclear and Frobenius norms, respectively. Also, $k$ expresses the $k$th iteration when solving an optimization problem successively.  I don't know how to solve this problem. Would you tell me the way? Why do I see different versions of node and nodejs I ran the following commands to install the stable version of node  sudo npm cache clean -f sudo npm install -g n sudo npm install -g n   Now, when I run node -v, I see v10.0.0, whereas nodejs -v returns v4.2.6. As far as I'm aware, node and nodejs are the same thing. Why am I seeing this? Thanks. Hey guys I found one or two papers which use ridge regression (for basketball data). I was always told to standardize my variables if I ran a ridge regression, but I was simply told to do this because ridge was scale variant (ridge regression wasn't really part of our course, so our lecturer skimmed through it).  These papers I read didn't standardize their variables, which I found a bit surprising. They also ended up with large values of lambda (around the 2000-4000 level) via cross-validation, and I was told that this is due to not standardizing the variables.  How exactly does leaving the variable(s) unstandardised lead to high lambda values and also, what are the consequences of not standardizing the variables in general? Is it really such a big deal?  Any help is much appreciated. Is floating point math broken? How can I achieve having logical page numbers shown in acrobat reader? Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $ What's a good word for a person, or a state, when a person is constantly waiting for life to get good, but does nothing to actually make it happen I've written a song about people who constantly wait for something good to happen to them and am looking for a word to describe this state/paradox/condition.   The song is about people who expect life to get better for no reason, talk about moving to a bigger city or a different place as if that would magically make them rich, famous and happy, and keep talking about how they are going to do this or that, but never do it. I was under the impression that there's a psychological term for this state, but can't google it up, so I probably made it up.  This is essentially a song about myself, I keep talking about how I love to make music, or shoot videos, but I never actually produce anything, yet it's the only thing I can think of. When it comes to the moment to take action – I just stay at home watching tv shows.  I know it's a broad question and list of topics, but I guess I'm looking for a word that describes this state – of doing NOTHING. Here's a screen capture from a site where I can view /site-analytics:    The orange "all posts" curve should be strictly larger than the blue "questions" curve, but they are mistakenly displaying the same data. What does this depiction of a black hole in the movie Interstellar mean?