All that empty space would be a great place to put the network-wide rep and badges, as appears in the flair.    This would fill the space, in addition to making it more in line with the regular site-specific profile.
Looking at your network profile, there is no number anywhere in your entire profile that has a sum of all your reputation across the network except for the flair page, which isn't always accurate because it's cached, and isn't exact because it rounds. Yet, there's so much empty space on the profile page.  My simple idea is to list the user's total reputation and badge counts below their Gravatar image like on any normal profile page across the network (keeping the rule of only accounts with over 200 reputation). It would make it incredibly easy to find and I wouldn't have to go search for myself on Area 51 to figure it out.
How to submit a form when the return key is pressed? Can someone please tell me how to submit an HTML form when the return key is pressed and if there are no buttons in the form? The submit button is not there. I am using a custom div instead of that. A Sequence converges to a limit $L \in \mathbb R$ if, and only if, an $N$-th tail of the sequences converges to $L$ for some $N \in \mathbb N$ Two parties are trying to resolve some dispute that involves interest of both. Sometimes it's hard for them to reach an agreement on a fair basis. This is typically when another party who has no interest involved and no bias towards either party comes into play in order to settle the dispute between the two conflicting parties. The term "third party" seems being used quite often. I would also call it a "middle-standing" party or we could say this party is neutral.  These terms seem NOT formal enough to me though. I'd like to see if anybody can come up with a better one. regular expressions / pattern matching on ubuntu command line Inequality ${n \choose k} \leq \left(\frac{en}{ k}\right)^k$ This is from page 3 of  ().  Copying the relevant segment:     Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that   $$ {n \choose k} \leq \left(\frac{en}{ k}\right)^k$$   I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation? Video grid with vstack and hstack What word describes random unfitting accusations I really dont know how to describe this. Im having a discussion with someone, I was sharing information that I found on a topic. Randomly im called out as being rude when nowhere was there evidence of rudeness exchanged. This person seems to have a history of claiming people have a problem or are being rude out of the random blue.   Is there a word to describe these random accusations that dont connect to anything? I want to use input from a user as a regex pattern for a search over some text.  It works, but how I can handle cases where user puts characters that have meaning in regex? For example, the user wants to search for Word (s): regex engine will take the (s) as a group.  I want it to treat it like a string &quot;(s)&quot; .  I can run replace on user input and replace the ( with \( and the ) with \) but the problem is I will need to do replace for every possible regex symbol. Do you know some better way ? How are qubits better than classical bit if they collapse to a classical state after measurement? Classical computers store information in bits, which can either be $0$ or $1$, but, in a quantum computer, the qubit can store $0$, $1$ or a state that is the superposition of these two states. Now, when we make a "measurement" (to determine the state of qubit), it changes the state of the qubit, collapsing it from its superposition of $0$ and $1$ to the specific state consistent with the measurement result.  If from a measurement of a qubit we obtain only a single bit of information ($0$ or $1$, because the qubit collapses to either of the states), then, how is a qubit better than a classical bit? Since ultimately we are only able to get either $0$ or $1$ from either classical or quantum bit.  If you can explain with an example, it may be helpful. Send Email notification if task is not completed 7 days prior to the due date Select Input Texture file name as parameter/node Ubuntu 18.10 gnome extensions stop working after hibernation After upgrading to 18.10 I am experiencing very strange behaviour of gnome extensions. After my laptop goes into hibernation they all switch off and enabling them back on  doesn't help: nothing changes and when I refresh the page they are disabled again.   The only way to turn them back on is to reinstall them. Simple problem with Manipulate and Plot Is there a word for people who always act too friendly in order to curry favor with their acquaintances? Is there a single noun or adjective for a person who is always flattering friends and acquaintances, not only to be liked and accepted but, at other times, to curry favor with them as well ?  I'm not looking for regionalisms but a word that will be understood in any English speaking country. Understanding NSString comparison If I said, &quot;I will run, not if it is snowing, if I my shoes are comfortable.&quot; Does the &quot;if my shoes are comfortable&quot; modify the &quot;I will run&quot; clause? I agree the sentence would be clearer with the addition of &quot;and&quot; in place of the comma. However, without the &quot;and&quot; is it still clear that the &quot;if my shoes are comfortable&quot; clause modifies the &quot;I will run&quot; clause? Using Georeference information stored in tfwx (from ArcGIS) in QGIS? I have georeferenced some images (ArcGIS -> Update Georeferencing) which creates a *.tfwx file with the new parameters for georeferencing.   I renamed it to *.tfw but QGIS is not able to read the georeferencing information.   Does anybody has any useful tips how to fix this? If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective. I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof. How to (properly) remove the parentheses around the year in authoryear style? (v3) I have a question that has been answered a few times, but does not work for me (and also not for another commenter in a previously asked question).   I want to remove the parenthesis around the yead in the bibliography with biblatex style authoryear, and ideally also add a comma and space before it. I have looked at the following questions:        I use biber 2.10 and TexLive 2017/W32Tex.  This MWE should work according to the question v2 above, but does not work for me, i.e. the output is still Author, Anne (2001a). Alpha. The output I need is Author, Anne, 2001a. Alpha.  \documentclass{article}  \usepackage[style=authoryear,dashed=false]{biblatex} \usepackage{xpatch,filecontents}  \begin{filecontents}{\jobname.bib} @misc{A01,author={Author, Anne},year={2001},title={Alpha}} @misc{A02,author={Author, Anne},year={2001},title={Beta}} \end{filecontents} \addbibresource{\jobname.bib}  \xpatchbibmacro{date+extrayear}{%   \printtext[parens]% }{%   \setunit*{\addperiod\space}%   \printtext% }{}{}   \begin{document} \nocite{*} \printbibliography \end{document} Many-worlds: how often is the split how many are the universes? (And how do you model this mathematically.)