Well, as the title says, I guess:   Start off with [foo]. Use either the keyboard shortcut or editor button to wrap the [foo] in a hyperlink. ... will result in  rather than      This is due to the system blinding/ naively escaping all []'s within the link, without checking whether it is contained with a code block or not.  Weirdly however, the syntax:  (i.e. [[foo]][1]) seems to work regardless, so I'm not sure this escaping of [] is even necessary?  This rendering bug happens both within the preview window, and within the actual post when submitted.
The link button in the post editor incorrectly escapes square brackets in inline code.  To reproduce:   Write the following MarkDown in the editor  text text `function[]` text text  Select `function[]`, press the link button (), and enter some link the result will be [`function\[\]`][1] which renders as function\[\]   This comes up very often in   when I'm linking to the documentation because the usual way to indicate that a symbol is a function in Mathematica is adding [] after it.  This is a minor thing, as it's easy to manually remove the backslashes, but it's somewhat annoying so I thought I'd mention it.    Edit by Mr.Wizard It is September 2013 and this still hasn't been fixed.  While "adding [] after it" is not presently the custom on Mathematica.SE this is still a bug and still needs to be fixed.
Proof of for-all / there-exists "negation"? Proving the measure of an increasing sequence of measurable sets is the limit of the measures Where can I find the source code for the Ubuntu Kernel? I'd like to build my own copy of the Ubuntu kernel. Where can I get the source code? Analysis's output vs analysis output vs output of the analysis US Customs when coming through Canada I have an upcoming trip where I will be flying to and from Copenhagen, through Toronto.  I notice that my departure and return gate in the Philadelphia airport is not in the international terminal.  Where will I clear customs? Blender Nodes in Editor doesn't show preview Image I have a small problem with my blender lately. I want to make some compositing with my compositing nodes but the problem is, blender won't show me the preview Image. I have a Screenshot for you guys.    The preview Image itself is enabled, but it just won't show.  Thanks for your help in advance!  -Marten I flagged a post today, typed out a fairly thorough explanation, clicked the button... and got told that I'd already flagged this post (for something else, which I'd forgotten about) so this flag wasn't allowed.  (This happens if the first flag is still active.)  The dialogue with my explanation was gone at this point, so if I wanted to communicate with the moderators about that, I was going to have to start over.  (In case you're wondering how this could happen, one flag was related to the post itself and one was related to the user -- there being no way to flag a user directly, you have to hang it off of a post.  But I've also seen this with cases where the first flag hangs around unhandled for a week or two and I've forgotten about it.)  Since the system is going to detect that I've already flagged a post and reject a second flag, can it do that after I click the "flag" link instead of waiting until the end of the process?  (Or gray out the link so I can't click it in the first place?)  2015-12-18: someone suggested to me in chat today that this had been implemented, so I tested again.  The behavior has not changed: I flagged a post using "other", then flagged it again (also "other") before the first flag was handled, and the flag was rejected only after I'd typed the message.  (I did this on a site where I'm a moderator so I could then clean up after myself.) Let $G$ be a matrix Lie group and let $H$ be a Lie subgroup of $G$. My understanding is that all $g \in G$ are matrices, since $G$ is a subgroup of the general linear group. Then aren’t all elements $h \in H$ matrices, since $h \in G$?  From what I can gather $H$ is a matrix Lie group if $H$ is a closed subset of $G$. But if $H$ is not a closed subset of $G$, why is it not a matrix Lie group? Is the reasoning in the first paragraph wrong? I have the following code that shows either a bug or a misunderstanding on my part.  I sent the same list, but modified over an ObjectOutputStream.  Once as [0] and other as [1].  But when I read it, I get [0] twice.  I think this is caused by the fact that I am sending over the same object and ObjectOutputStream must be caching them somehow.  Is this work as it should, or should I file a bug?   import java.io.*; import java.net.*; import java.util.*;  public class OOS {      public static void main(String[] args) throws Exception {         Thread t1 = new Thread(new Runnable() {             public void run() {                 try {                     ServerSocket ss = new ServerSocket(12344);                     Socket s= ss.accept();                      ObjectOutputStream oos = new ObjectOutputStream(s.getOutputStream());                     List same = new ArrayList();                     same.add(0);                     oos.writeObject(same);                     same.clear();                     same.add(1);                     oos.writeObject(same);                  } catch(Exception e) {                     e.printStackTrace();                 }             }         });         t1.start();          Socket s = new Socket("localhost", 12344);         ObjectInputStream ois = new ObjectInputStream(s.getInputStream());          // outputs [0] as expected         System.out.println(ois.readObject());          // outputs [0], but expected [1]         System.out.println(ois.readObject());         System.exit(0);     } } So Black Holes Actually Merge! In 1/5th of a Second - How? I've read a lot of conflicting answers in these forums. However, today saw the awesome announcement of gravitational waves. Two black holes merged:   Not only that, they merged FAST. In 1/5th of a second revolving around each other 250 times a second. The entire event was quicker than a heartbeat. Moreover, we observed this happening as distant outsiders. So now we can say for sure:   Objects approaching the event horizon DO NOT appear to slow down Black holes CAN merge in a finite (and quick) amount of time And all this is wrt a frame of reference far, far away   To quote the NYTimes article:  One of them was 36 times as massive as the sun, the other 29. As they approached the end, at half the speed of light, they were circling each other 250 times a second.  And then the ringing stopped as the two holes coalesced into a single black hole, a trapdoor in space with the equivalent mass of 62 suns. All in a fifth of a second, Earth time.  However, everything I've read so far has let me to believe that an outside observer should never be able to measure the collision happening in a finite time. So what exactly is happening here? I must have read at lest 5 different versions of this so far everywhere in these forums over the past several years. How to get the environment variables of a subprocess after it finishes running? Product of elements of a finite abelian group Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists). Outcast boy meets girl underwater Embedding ordinals in $\mathbb{Q}$ Can Xdebug be triggered by Drush commands? I'm running into an issue with my Features modules and installation profile when installing the site with drush si. There are a number of probably related issues in the Features issue queue, but none with good solutions. The interesting part (mentioned in some of the issues) is that it works fine when installing the site through the browser, but fails when installing with drush si.  I have an idea of what's wrong, but need to confirm it. The only way I can think of confirming it is to step through the code with a debugger (Xdebug is what I have set up). So, is it possible to trigger Xdebug from a command run with Drush? Specifically, can I get Xdebug to kick in when I run drush si? Touchpad stopped working 20.04 How are Stack Exchange sites officially branded? How are sites in the Stack Exchange network, including Stack Overflow, officially branded? How are their names abbreviated? What conventions should be used in the FAQ? Hibernate: comparing current & previous record I want to compare the current value of an in-memory Hibernate entity with the value in the database:  HibernateSession sess = HibernateSessionFactory.getSession(); MyEntity newEntity = (MyEntity)sess.load(MyEntity.class, id); newEntity.setProperty("new value"); MyEntity oldEntity = (MyEntity)sess.load(MyEntity.class, id); // CODEBLOCK#1 evaluate differences between newEntity and oldEntity sess.update(newEntity);       In CODEBLOCK#1 I get that newEntity.getProperty()="new value" AND oldEntity.getProperty()="new value" (while I expected oldEntity.getProperty()="old value", of course). In fact the two objects are exactly the same in memory.  I messed around with HibernateSessionFactory.getSession().evict(newEntity) and attempted to set oldEntity=null to get rid of it (I need it only for the comparison):  HibernateSession sess = HibernateSessionFactory.getSession(); MyEntity newEntity = (MyEntity)sess.load(MyEntity.class, id); newEntity.setProperty("new value"); HibernateSessionFactory.getSession().evict(newEntity); MyEntity oldEntity = (MyEntity)sess.load(MyEntity.class, id); // CODEBLOCK#1 evaluate differences between newEntity and oldEntity oldEntity = null; sess.update(newEntity);   and now the two entities are distinct, but of course I get the dreaded org.hibernate.NonUniqueObjectException: a different object with the same identifier value was already associated with the session.  Any idea?  EDIT: I tried the double session strategy; I modified my HibernateSessionFactory to implement a map of session and then...  Session session1 = HibernateSessionFactory.getSession(SessionKeys.DEFAULT); Session session2 = HibernateSessionFactory.getSession(SessionKeys.ALTERNATE); Entity newEntity = (Entity)entity; newEntity.setNote("edited note"); Entity oldEntity = (Entity)session1.load(Entity.class, id);  System.out.println("NEW:" + newEntity.getNote()); System.out.println("OLD: " + oldEntity.getNote()); // HANGS HERE!!!  HibernateSessionFactory.closeSession(SessionKeys.ALTERNATE);   My unit test hangs while attempting to print the oldEntity note... :-( Sound of a next scene starts one moment earlier than video I saw this several times in different movies, but I don't understand what is this supposed to mean. The main idea is when the scene is about to change, for one second or so we still see the picture from the old scene, but the sound is already coming from from the upcoming scene. Why do they do this? What's the purpose of this technique? This code is part of a larger code that generates an Arithmetic Quiz, although the code I have show below just saves the score that the user got on the test and their name. When saving the code it checks if the user has taken the quiz before. If the user has taken the quiz before it saves it as 'Older Score' or 'Oldest Score', however if the user has not taken the quiz before it generates a new row for them. Additionally it works out the average score and highest score for all users it has the data for. For my task I'm supposed to make the code as short as possible.  name = ""+first_name+" "+last_name rows_data = [] class_file_names = ['Class 1.csv', 'Class 2.csv', 'Class 3.csv'] scan = open(class_file_names[int(math_class)-1], 'r+') rows = csv.reader(scan) for row in rows:     rows_data.append(row) scan.close() class_file_names=['Class 1.csv', 'Class 2.csv', 'Class 3.csv'] csvfile=open(class_file_names[int(math_class)-1], 'w', newline="") fieldnames = ['Student Name', 'Oldest Score', 'Older Score', 'Latest Score', 'Average', 'Highest'] writer = csv.DictWriter(csvfile, fieldnames=fieldnames) writer.writeheader() nameFound=False for x in rows_data[1:]:     if x[0] == name:         x[3]=x[2]         x[2]=x[1]         x[1]=str(score)         x[4]=(int(x[1])+int(x[2])+int(x[3]))/3         x[5]=max(x[1],x[2],x[3])         nameFound=True     writer.writerow({'Student Name': x[0], 'Oldest Score': x[3], 'Older Score': x[2], 'Latest Score': x[1], 'Average': x[4], 'Highest' : x[5]})  if nameFound==False:     writer.writerow({'Student Name': name, 'Latest Score': score, 'Older Score': '0', 'Oldest Score': '0', 'Average': score, 'Highest': score}) csvfile.close() input("Test complete! Press anything to close.")