I meet a lot of people regularly who have the knowledge and interest to participate in various SE communities however, many a time they do not know of SE, or of the existence of any site beyond Stack Overflow.  So, in such cases, what would be a good  that would make them log on to Stack Exchange and explore the Network?
What's the best blog post/video/tweet/whatever to explain how the Stack Exchange system works to regular people? I am trying to gauge interest for a new Stack Exchange site, but I want to explain it to people as best (and preferably) as visually as I can. Any YouTube clips breaking it down?  anyone?  Note: I am fully aware of , but mine is about Stack Exchange, not Stack Overflow.  Most people think I'm describing traditional forum software, which they know sucks. Is there some way to describe the magic of how the design of the Stack Exchange system actually produces useful results very quickly? I can describe generally that it is a Q&amp;A site that users vote on the answers, but Stack Overflow feels like so much more. Am I perhaps thinking too big about this?
I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this . How to find normal and lognormal moments, given partial information? $Y=\ln(X)$. $X$ is lognormal and $Y$ is normal. If all I know is the arithmetic mean of $Y$ and the standard deviation of $X$. What is the formula to calculate the arithmetic mean of $X$ and the standard deviation of $Y$? Do Squibs have to be pure-blood? If a Muggle was born of a wizard/witch and a Muggle, is that person a Squib, or is that reserved to people born from pure-blood families?    (Related: ) I am thinking of buying a digital SLR camera. What things should I be looking for? Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the   maximum and minimum values of $2x + 3y + z$ .       How can I able to solve the problem? Thanks for your help. Purpose: Construct a 95% confidence interval for $\theta$.   A common strategy is to construct a confidence interval for $\log(\theta)$ and then exponentiate.  Why is it valid?   My concern is that $E(\log(\theta)) \neq \log(E(\theta))$ and $Var(\log(\theta)) \neq \log(Var(\theta))$...   Do people use this back transformation because it has proven to work fine? What about other transformations (e.g. logit)? Most instructions for equipment and software make a habit of shortening sentences by dropping words of certain parts of speech - usually definite and indefinite articles.  For example:     PULL (THE) LEVER TO OPERATE      CARS MAY BE TOWED WHEN (THEY ARE) PARKED ILLEGALLY      PLEASE USE (THE) OTHER DOOR   Is there a name for this custom? And what is the best way to describe it? Overdispersion and Underdispersion in Negative Binomial/Poisson Regression I was performing a Poisson regression in SAS and found that the Pearson chi-squared value divided by the degrees of freedom was around 5, indicating significant overdispersion. So, I fit a negative binomial model with proc genmod and found the Pearson chi-squared value divided by the degrees of freedom is 0.80. Is this now considered to be underdispersed? If so, how does one go about handling this? I have read a lot about overdispersion and believe I know how to handle this but information on how to handle or determine if there is underdispersion is scant. Can anyone assist?  Thanks. proving logical equivalence $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$ I am currently working through Velleman's book How To Prove It and was asked to prove the following  $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$  This is my work thus far  $(P \to Q) \wedge (Q \to P)$  $(\neg P \vee Q) \wedge (\neg Q \vee P)$ (since $(P \to Q) \equiv (\neg P \vee Q)$)  $\neg[\neg(\neg P \vee Q) \vee \neg (\neg Q \vee P)]$ (Demorgan's Law)  $\neg [(P \wedge \neg Q) \vee (Q \wedge \neg P)]$ (Demorgan's Law)  At this point I am little unsure how to proceed.   Here are a few things I've tried or considered thus far:  I thought that I could perhaps switch some of the terms in step 3 using the law of associativity however the $\neg$ on the outside of the two terms prevents me from doing so (constructing a truth table for $\neg (\neg P \vee Q) \vee (\neg Q \vee P)$ and $\neg (\neg P \vee \neg Q) \vee \neg (P \vee Q)$ for sanity purposes)  I can't seem to apply the law of distribution since the corresponding terms are negated.  Applying demorgans law to one of the terms individually on step 2 or 3 doesnt seem to get me very far either.  Did I perhaps skip something? Am I even on the right track? Any help is appreciated UIDatePicker show only month and day I'd like to have a UIDatePicker where the user can pick a month and a day but not the year. I'm aware that leap years have an extra day, so for simplicity, let's throw that day away.  Is there any way to remove the year column or have 2 reels with month/day that act like the UIDatePicker (grey out days that don't exist in the selected month)? Contraction mapping in the context of $f(x_n)=x_{n+1}$. Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$   I tried to prove it by induction  but I got a contradiction. According to HTML specs, the select tag in HTML doesn't have a readonly attribute, only a disabled attribute. So if you want to keep the user from changing the dropdown, you have to use disabled.  The only problem is that disabled HTML form inputs don't get included in the POST / GET data.  What's the best way to emulate the readonly attribute for a select tag, and still get the POST data? What causes atoms to have their specific colors? It has been said that RSA uses a modulus product of two prime numbers for security reasons. But does RSA even work correctly if we allow composite integers instead?  I think that the answer is "NO".  Define “work correctly” as:   allows encryption and decryption to be carried out as in normal RSA, yielding the original message (message padding and unpadding can remain unspecified, or absent as in texbook RSA); a security level can be given in terms of the key generation process (which may have alternate requirements to testing that $p$ and $q$ are prime). Chat messages should use consistent icon in iOS app It'd be nice if the chat notifications in the feed used the default Stack Exchange icon like the ones in the inbox do*:     Since items in the feed are grouped so that multiple items from the same site , the site icon helps delimit grouped items a bit less subtlely than the grey divider lines, and chat replies' lack of one causes them to sort of blend in to the previous group.  *Ideally it'd use the icon of the site the room was associated with, but I don't think that information is readily available Ubuntu stuck/freeze on install I've been trying to replace my current Windows 10 installation with Ubuntu 16.04. My computer is a Lenovo Thinkpad T470. I've formatted my 15GB USB-dongle. Downloaded the software Rufus, and burned the ISO onto the USB. I've tried the settings "MBR Partition for BIOS or UEFI", "MBR Partition for UEFI" and "GPT partition for UEFI". I've tried using memtest86+ to test if I have any memory errors, I have not. I've tried disabling Secure boot, and Fast startup. I've tried using other software than Rufus, and I've tried entering the "nomodeset" in the boot options, as well "acpi=off", nothing works.   When I boot from my USB. I choose "Install Ubuntu" from the grub menu. And then select my language in the installer. It is from this point on, it gets problematic. Sometimes when I try to connect to my WIFI, it connects successfully, but when I press the "continue" button, the installer freezes and nothing happens. Sometimes I get past this point, but it then freezes later on in the installation progress - typically right when I press the "Install now" button.   It should also be mentioned that, if I press the "Quit" button in the install. I get to a black screen with a lot of text saying something like squashfs error.   Can someone please help me? When I tried booting memtest86+ it only worked when I changed my boot mode from UEFI to Legacy. Could this be it? Are there any downsides to installing an operating system in legacy?   If anyone has any other suggestions, it would be greatly appreciated!  Thank you :-) What are the ways to position things absolutely on the page? I know that there is tikz, but also eso-pic and textpos.  I'd like to see a list of possible ways (not necessarily packages) for absolute positioning of things (which I believe amounts to "boxes") on the page, together with their advantages and disadvantages (for instance: tikz's current page node is very easy to use, but usually requires two LaTeX runs, and overhead from loading tikz if you don't need it for any other things is considerable).  Both LaTeX and ConTeXt solutions are welcome $n$ choose $k$ where $n$ is less than $k$ I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, etc.  In general how do we compute $_nC_k$ when $n$ is less than $k$? Do we still use the formula $\frac{n!}{(n-k)!k!}$ and use the gamma function to compute the negative factorial?  Thanks! proof that function is constant I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker):  Let $f:\mathbb R\longrightarrow \mathbb R$ be a function which is continuous at $0$ and satisfies $f(x)=f(2x)\,\forall x\in\mathbb R$.  My question is how do I formally prove that for every $x&lt;0$ $f(x)=c_1$, and for every $x&gt;0$ $f(x)=c_2$, where $c_1,c_2\in \mathbb R$? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that $f(0)=c_1=c_2$ by the definition of continuity of $f(0)$ and thus $f(x)$ is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!