Are the deleted users automatically renamed to user&lt;random number>, e.g. user354353, or user240249 or are they actually naming their accounts like this?
It's been more than a month since I have started using . It has these users with user names in the format of "user-random number". I have observed them in some of the stack activities, most of the time asking questions.   Does the moderators create them for their own purpose. If yes what purpose do they serve?   I mean a user is free to have any user name. But it's just a hunch that such users are actually moderators. I may be wrong; just need some clarification.
LaTeX Classes or Styles for Schedules and/or Calendars? C# and USB HID Devices What defines a PC's consciousness of self, memories and knowledge? Reversing the $T(n) = \frac{n(n+1)}{2}$ formula Can someone reverse this formula? Sorry if it's too basic, I'm an old guy and my math is rusty :-D  $s =\frac{n(n+1)}{2}$  If it ain't clear, by reverse I mean obtaining the value of n by providing s  Thanks Is there a case insensitive jQuery :contains selector? Is there a case insensitive version of the  jQuery selector or should I do the work manually by looping over all elements and comparing their .text() to my string? Is there an comparison of the various mapping libraries for Android? I am currently looking at the various libraries that I can use to show a map in my custom Android App. I have found several like Mapsforge, OSMDroid, ArcGIS runtime for Android, Google Maps for Android and so on. I'm sure that I am missing a few.   Is there a comparison chart somewhere about the differences in functionalities and pros and cons of each one?   If there isn't one, we could make this question community wiki, to serve alongside  &amp; . Every time I try to install "WINE" VIA the Software Center on Ubuntu 12.10, I get this error:  The following packages have unmet dependencies:  wine1.4: PreDepends: dpkg (&gt;= 1.15.7.2~) but 1.16.7ubuntu6 is to be installed          Depends: libc6 (&gt;= 2.14) but 2.15-0ubuntu20 is to be installed          Depends: wine1.4-amd64 (= 1.4.1-0ubuntu1) but 1.4.1-0ubuntu1 is to be installed          Depends: wine1.4-i386 (= 1.4.1-0ubuntu1) but it is not going to be installed   When I try installing it VIA the terminal, this is what I get:  Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  wine1.4 : Depends: wine1.4-i386 (= 1.4.1-0ubuntu1) but it is not installable            Recommends: gnome-exe-thumbnailer but it is not going to be installed or                        kde-runtime but it is not going to be installed            Recommends: ttf-droid            Recommends: ttf-mscorefonts-installer but it is not going to be installed            Recommends: ttf-umefont but it is not going to be installed            Recommends: ttf-unfonts-core but it is not going to be installed            Recommends: winbind but it is not going to be installed            Recommends: winetricks but it is not going to be installed E: Unable to correct problems, you have held broken packages.   how do I fix this? Point vs Pixel: What is the difference? What is the difference between points and pixels? Why do clocks measure arc-length? How to integrate $ \int x^n e^x dx$? QCompleter Custom Completion Rules counting proof for $\sum_{k=0}^{n-1} 2^{k} = 2^n - 1$ I was hoping someone could help me come up with a combinatorial proof, one with an easy to understand 'story' for $\sum_{k=0}^{n-1} 2^{k} = 2^n - 1$. For instance I understand that $2^n$ is the number of possibilities when creating subsets of size $n$. For example, if you're given a choice of any topping on a pizza, and you can only have said topping once (decide to include or exclude) then you have $2^n$ choices. But I'm not sure how the $-1 $ would be incorporated, or why the LHS would be $1+2+4+8+...+2^{n-1}$ If $T^n$ is $q$-contractive, $T$ exactly has one fixed point Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? I am creating a type of adventure map. Once you enter the map, you step on some pressure plates that are supposed to seamlessly close the wall behind you. For some reason this command does not work when in a command block or just in text. Here is the command:  /fill -17 4 -109 -18 5 109 minecraft:light_gray_concrete   The output it gives me is "That position is not loaded". I am on a singleplayer superflat world in 1.14.3. Any help would be very appreciated! In Obi-Wan's Order 66 warning transmission why didn't he tell the surviving Jedi about the Sith, including Palpatine's true identity? If you've seen the premiere episode of Rebels, you will have seen Obi-Wan's warning message that he broadcast to Jedi throughout the galaxy in RotS. In this message he speaks cryptically &amp; warns of a "dark shadow" falling &amp; to avoid the temple &amp; await a "new hope".   Wouldn't it have been useful to inform the Jedi that their mortal enemy the Sith had returned? And specifically that Palpatine himself was a Sith? You'd assume that the majority of the Jedi who were not in the loop of the Council would be unaware of the existence of the Sith. Yoda mentions a few times that they should keep it under wraps. I think now would've been the time to reveal the truth. The more info the surviving Jedi have the better. Who knows if the likes of Kanan knows anything about the Sith's return! I find Obi-Wan's omission of this information bizarre. Running two independent instances of Chrome I use Chrome as my main browser.  I also spend a lot of time on a particular webapp that my company uses.  I'd like to keep those two things separate.  I can run the webapp in a separate window - but I'd like to go further than that.  I'd like to have it be identified by a different task in the Win 7 taskbar; ideally a different icon; and ideally a different title in Process Monitor.  How can I run a second, independent instance of Chrome, with a different title and icon, that won't merge with my main Chrome browser. How can I check if my (unlocked) phone will work with a particular country's SIM cards, for 3G and calls? I've used my phone overseas with local SIM cards a lot, but recently had a surprising experience where one local SIM card simply wouldn't work with my phone but did work first time on my partner's, even though both were similar (unlocked Android phones) and another SIM card from a different local provider did work. Looking into it, it looks like there are various different standards around, but it all gets very technical very quickly.  Is there any simple way of looking up which SIM cards in a particular country will work with a particular model of phone, for calls, texts and internet access? Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.     Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.   This is what I tried.  Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$. This is a well-defined bilinear map.  By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.  This map is surjective.  So all we need to show is that the kernel of $F$ is $0$.  But I am unable to do this. Can I run 64 bit apps in 32 bit Linux? There are some proprietary apps that are being provided only in 64-bit version. Not that it's a very good idea to change it, but for fun. According to  post, there are still some problems even after changing entries in /etc/passwd, /etc/shadow, and /etc/sudoers. Any suggestions?