I've noticed something on multiple different stack sites that confuses me. For some reason, the reputation requirements for certain privileges are different.  For example, you need 50 reputation be able to comment on Stack Overflow questions, but you only need 5 on Meta Stack Exchange.  Another example, you need 1500 reputation on Stack Overflow, 500 on Meta Stack Exchange, and 1000 on Math Stack Exchange for the same privilege; create tags.  Why is that?
So, let's take Skeptics.SE for an example. You can become a trusted user when you acquire 4000 rep., not 20000  I was like, wow, so is it like this because of the lack of users on Skeptics.Se, as in, not so much as Stack Overflow or something?  Why do some sites need less reputation to be a trusted user?
jQuery child of clicked element I've got a list of links which have a click event attached to them, I need to get the ID from the child A link. So in the example below if I clicked the first list element I'd need google retuned.   I've tried '$this a' but can't quite work out the syntax.  $("ul li").click(function(event){   $("input").val($(this).html());              });     &lt;ul&gt;     &lt;li&gt;&lt;a href="http://www.google.com" id="google"&gt;Google&lt;/a&gt; &lt;/ul&gt; Quantum Field Theory from a mathematical point of view Could humans choose to establish fibre digesting colonies in our guts? Humans don't digest fibre, and nor do any animals, but some creatures have micro-organisms (bacteria, I'm guessing) in their gut which digest fibre for them.  My question is, could humans take some sort of pill and establish our own fibre-digesting colonies in our intestines?  Has this been tried?  Supposing it could be done, would you suppose it would have any negative effects on our health?  (i.e.. would those whose gut digested fibre for them be stuck on a low-fibre diet?).    Purpose of question: just wondering.  It would be cool to be able to eat bark. Incorrect plot using pgfplots (trigonometric functions like cos, sin and tan) I'm trying to plot a function with pgfplots for LaTeX. Unfortunately, pgfplots plots the function incorrectly.  Here is the LaTeX code I use:  \begin{figure}[H] \centering \begin{tikzpicture}     \begin{axis}[xlabel=Gameplays,ylabel=Rating]         \addplot+[gray,domain=1:30]         {0.5 + 0.5 * ((atan(x) * 2 / pi)^11.79)};     \end{axis} \end{tikzpicture} \caption{Omzetten van aantal gameplays tot impliciete rating} \label{fig:omzetten_impliciete_ratings} \end{figure}   Here is how the function should look like:     This is how it is plotted:    Does anyone know what I do wrong?  Thanks in advance! I had an article, but I lost it.  It showed and described a couple of C/C++ tricks that people should be careful. One of them interested me but now that I am trying to replicate it I'm not being able to put it to compile.  The concept was that it is possible to change by accident the value of a const in C/C++  It was something like this:  const int a = 3;          // I promise I won't change a const int *ptr_to_a = &amp;a; // I still promise I won't change a int *ptr; ptr = ptr_to_a;  (*ptr) = 5;               // I'm a liar; a is now 5   I wanted to show this to a friend but now I'm missing a step.  Does anyone know what's missing for it to start compiling and working?  ATM I'm getting invalid conversion from 'const int*' to 'int*' but when I read the article I tried and it worked great. Is it possible to reference an excel cell in a comment? I have an Excel balance sheet with a total at the bottom. On that cell I have a comment that says that the reason the total is X is because of what happens in J2. What happens if I insert another row before that, now my comment is wrong because all the rows jumped up a value.  Is there any way to reference a cell dynamically in an Excel commment? Difference between "does have" and "has" Possible Duplicate:                 What is the difference between does have and has? For example, compare she does have a car and she has a car. Converting a system command to python for file search and delete How do rays of light getting focused at a point create an image?  How are the information of object's shape and dimensions expressed as an image when light rays converge?  Is it understandable by solving wave equation for light? Travelling from India to canada via Dubai and London. Do I need a transit visa I am travelling to Calgary from Delhi. I have an Air India flight from Delhi to Dubai then British Airways flight from Dubai to London and Air Canada flight from London to Calgary. My query is "Do I require a transit visa." Plz help as I am travelling in four days Building the connection between cosine similarity and correlation in R Is it possible to find the origin of a virus? Is there any method with which you can determine the origin for a given piece of malware in Windows?  One of my PCs was recently infected with the  password stealing trojan. Is there any way to tell from what source this virus was downloaded from? How does commands like ls know what its stdout is?  It seems ls is operating different depending on what the target stdout is. For example if I do:  ls /home/matt/tmp    the result is:  a.txt b.txt c.txt   However if I do  ls /home/matt/tmp | cat   the result is (i.e. new line per result):  a.txt b.txt c.txt   The process is passed a file descriptor 1 for stdout right? How does it determine how to format the result? Does the file descriptor reveal information? In plain English, how does casting spells and using creature abilities work with the stack and priority? I'm a relatively new player to Magic: the Gathering. I was reading  to try to understand exactly when it's okay to cast instants (e.g. ) and use some abilities (e.g.  should be tappable sometime during other peoples' turns), but I don't seem to be making much progress: as I come to understand some things, others make less sense.  My confusion is boundless  For instance: to cast  one needs priority. Spells go on the stack, and priority gets passed at some points (I have no idea when), and effects on the stack are resolved when both players have passed priority. But when is that? I can't find any mention of when the other player gets the priority to cast an instant, except on their own turn. Furthermore,  states that during the Main Phase:     505.4. Second, the active player gets priority. Players may cast spells and activate abilities. The active player may play a land.   So players (plural) may cast spells, but only the active player has priority, so only the active player can cast an instant. The turn rules, and the rules on , never mention any point at which the other player gets priority, so when can they possibly Cancel?  In addition, you can only cast a spell when the stack is empty, so the entire stack must be able to be resolved in your turn at some point.  I need an explanation!  I could go on a bit more. There is also  which can be overloaded to "Counter each spell you don't control", which seems like it could only possibly work with an understanding of the stack.  Could someone explain, in simple English:   When can Instants be cast? When can tap abilities be used? What do Instants and abilities have to do with the Stack and Priority? Node for setting render colour, regardless of lighting I'm trying to make a material node setup to set final render colors, regardless of lighting conditions. Is such a setup possible? Many To many for Bulk records I have created two custom objects (A and B)and a junction object(AB). i.e. Both A and B have Many to Many relationship. Now I need to insert 5000 records. How can I automatically populate the relation in Junction object. DO I have to write any trigger or Do I have to do Manually mapping the records? Given an alphabet of size $n$, how many strings of length $c$ contain every single letter of the alphabet at least once?  I first attempted to use a recurrence relation to work it out:  $$ T(c) = \left\{ \begin{array}{cr} 0 &amp;\mbox{ if $c&lt;n$} \\ n! &amp;\mbox{ if $c = n$} \\ T(c-1) \cdot n \cdot c &amp;\mbox{ if $c &gt; n$} \end{array} \right. $$  As there's no strings that contain every letter if c &lt; n, and if c = n then it's just all permutations. When c > n you can take any string of size (c-1) that contains all letters (of which there are $T(c-1)$ to choose from), you choose which letter to add (of which there are $n$ choices) and there are $c$ different positions to put it. However, this gives out results that are larger than $n^c$ (the total number of strings), so it can't be right, and I realised it was because you could count some strings multiple times, as you can make them taking different inserting steps.   Then I thought about being simpler: you choose n positions in the string, put each letter of the alphabet in one of those positions, then let the rest of the string be anything:  $$ {c\choose{n}} \cdot n! \cdot n^{c-n} $$  But again this counts strings multiple times.  I've also considered using multinomial coefficients, but as we don't know how many times each letter appears in the string it seems unlikely they would be much help. I've also tried several other methods, some complicated and some simple, but none of them seem to work.  How would you go about working out a formula for this? I'm sure there's something simple that I'm missing. Proving a combinatorial identity without double counting Consider the following problem:  We have a collection of m white and n black balls. We pick k balls (k ≤ m+n) without replacement. What is the expected number of white balls in the sample?  Counting method 1: The ratio of white balls in the collection is m/(m+n). If we repeatedly sample the collection, we expect that on average, our sample has the same ratio. Therefore, the expected number of white balls in a sample of size k is km/(m+n). As a side note, I think the above argument somehow depends on the unbiasedness of  the mean estimator. Am I right?  Counting method 2: We directly compute the expected number of white balls in the sample: $$ \frac{\sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i}}{\binom{m+n}{k}} \enspace.$$  So, we reach at the identity: $$ \sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i} = \frac{km}{m+n}\binom{m+n}{k} \enspace.$$  How to prove the above result without double counting? It seems to be provable using  as well as $i\binom{m}{i} = m\binom{m-1}{i-1}$. But is there a simpler approach (which does not depend on extra identities such as Vandermonde's)? How can I suspend/hibernate from command line? How can I suspend or hibernate my laptop using command line, without installing additional software? According to preliminary estimates, the project should be completed by 2020 at a cost of about RUB 40 bn.    or      According to preliminary estimates, the project should be completed by 2020 at the cost of about RUB 40 bn.    Are both options okay, or is the first better?   We are introducing the sum to the reader, so logically it should be the first option. But there's a lot of google-hits for "at the cost of" too.