Why is it that many packages in apt-get are outdated? For example, I would like to get the latest implementation of the clojure programming language onto my machine. However, running the command apt-cache search clojure I get the following:  clojure - a Lisp dialect for the JVM clojure-contrib - user contributed set of libraries for clojure clojure1.2 - Lisp dialect for the JVM clojure1.3 - Lisp dialect for the JVM leiningen - simple build system for Clojure libclojure-maven-plugin-java - Clojure plugin for Maven libclucy-clojure - Clojure interface to the Lucene search engine libjava-jdbc-clojure - Clojure interface to sql databases via jdbc librobert-hooke-clojure - Function wrapper library for Clojure libtools-logging-clojure - Logging macros for Clojure   All the clojure implementations are old (clojure is version 1.1). The newest release of the clojure programming language is 1.5.  Why is it that packages are not up to date? Do I need to reconfigure my apt-get program?
Why are packages in the official Ubuntu repositories older than the latest (upstream) versions from Debian Sid, PPAs, the authors, etc.?
Why isn’t singular ‘they’ used with 3Sg verb forms? There are many (duplicate) questions about the acceptance, popularity and history of singular they (and their, them and themself) around here, it even got a tag of its own. If I didn’t miss anything, however, the proper verb form hasn’t been questioned yet.  As we all know, English third person singular pronouns (it, she, he and one, +body), names (Alice, Bob, …) and nouns (student, teacher, …) demand the +(e)s suffix be added to the finite verb form in simple present, where some auxiliary verbs have a “special” form (is &lt; *bes / *ares and has &lt; *haves). All other subjects don’t, including plural third person pronoun they.  When the plural you replaced thou (with thee, thy / thine), the other marked verb form that had remained in English – i.e. suffix +(e)st or +t – vanished, too. The first and second persons only ever appear as pronouns (I, we; you), not names or nouns, so there was no strong inclination to keep the verbal inflection.  The second person precedent would suggest that singular they be used with uninflected verb forms which is how it’s usually encountered in the wild. Assuming that they someday replaces he and she (and maybe it) it would lead to disagreement with the words the pronoun stands in for:        Alice         goes to her   place  by herself.   Bob           goes to his   place  by himself.   Alice and Bob go   to their place  by themselves. – (common)   Alice and Bob go   to their places by themselves. – (separate)                  She  goes to her   place  by herself.   He   goes to his   place  by himself.   They go   to their place  by themselves.   They go   to their places by themselves.                  They ?goes to their place  by themself.   ditto   They  go   to their place  by themselves.   They  go   to their places by themselves.                  They ?go to their place  by ?themselves.   ditto   They  go to their place  by  themselves.   They  go to their places by  themselves.      So why doesn’t singular they afford s on present-tense verbs like all other third person singular subjects do?   I learned about singular they only long after I had been taught English as a second language in school. That’s why it’s still a conscious decision to use it and hence I could easily adapt to use s forms with it, but would that sound and look funny / strange / wrong to native speakers? Wilson and Fermat's Theorem Split a plane into smaller plane objects I just started using Blender and I'm trying to cut a plane in equal squares and have each square be its own separate object. My main goal is create the plane and add a displace and decimate modifier to create a low poly terrain. By cutting the plane in equal squares I'll be able to tile them in a 3D maze game I'm working on. If anyone know a better way to approach this it would also be greatly appreciated. How to delete a file with a path too long to be deleted I was working with some computer vision source code I found and didn't realize that it creates a cache file that has a VERY LONG name and now I can't delete the 2 of them. I tried to rename it to a shorter name but I can't do anything to the file, I also tried to restart my computer and see if it would just go away. The error message says:  Destination Path Too Long. The file name(s) would be too long for the destination folder. You can shorten file name and try again, or try a location that has a shorter path.   How can I delete the files in this case? How do I shuffle cards for a card game? I am trying to develop an card game for Android Can anyone suggest me how to write code for effectively shuffling the playing cards? How is canonicity of derivative works determined for Star Wars? Why is $\frac{1}{\frac{1}{0}}$ undefined? Is three hours enough for an international-domestic transfer at Delhi? My International flight will be reaching New Delhi IGI at 2:45 AM and I am planning to book a domestic flight at 5:45 AM.  Is three hours sufficient for me to move from International to Domestic?  I believe there must be a shuttle service inside the airport to transfer me from International to Domestic but still I don't know if Customs time at IGI could cause any delay or not. "Try Ubuntu" option is not dislayed Antiresonance of multiple parallel decoupling capacitors: use same value or multiple values? My quesiton: Is it better practice to put same-valued capacitors in parallel of capacitors of different values to decouple the high-frequency noise caused by digital ICs? Background Digital IC need a decoupling capacitor close to their supply pins to ensure a stable voltage during power transients and to deal with noise (mostly to prevent noise generated by the IC to affect neighboring circuitry). It seems sensible to place a bulk capacitor (say 10-100uF), to act as an energy reservoir, and several smaller capacitors to deal with higher frequencies. The reason to place several small capacitors instead of just one is to deal with their Equivalent Series Inductance (ESL), which in practice, causes them to behave like an LC circuit. The effect of anti-resonance Yet, here is where best design practices and electronic myth seem to get mixed up and confusing to me. Most electronic engineers I have met like placing several decoupling capacitors of different values in parallel (with the smaller capacitors closer to the IC). The logic behind it is that the each capacitor takes care of a different noise frequency as depicted in Figure 1.  Figure 1: Impedance over frequency of three different value capacitors in parallel (cyan) vs their individual contribution (brown, blue, red). Image taken from . Note the small anti-resonance peak. It seams no major trouble , and the overall behavior of the three different capacitors in parallel is vastly superior to their individual decoupling capabilities. However, I have read in  by [Henry W. Ott] that placing capacitors of different values may cause a much greater antiresonance-peak which can be very harmful for our designs (see Figure 2). In fact, it amplifies any noise that falls into the anti-resoance frequency range, which is corroborated by .  Figure 2: from Electromagnetic Compatibility Engineering, by Henry W. Ott, section 11.4.4. The 15nH inductance makes reference to the capacitors ESL. Show that $\mathbb{Q}(\zeta_n)$ is Galois over $\mathbb{Q}$ and $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}))\cong \mathbb{Z_n}^*$ Question 1: For $n \in \mathbb{N}$ explain why $\mathbb{Q}(\zeta_n)$ is Galois over $\mathbb{Q}$   We must show that it is normal and separable.   $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is normal since it is a splitting field for $X^n-1 \in\mathbb{Q}[X]$ (I think... Is this correct?).  It is separable because it is of characteristic $0$  Question 2: $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}))\cong \mathbb{Z_n}^*$  Is this because the primitive nth root of unity $\zeta_n$ acts as a generator for a cyclic group? I am unable to fill out a decent proof on this.   Thanks for your help Finding plane equation given two lines Determine an equation of the plane containing the lines $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-5}{6}$; $r=\lt1,-1,5\gt+t\lt1,1,-3\gt$.   I calculated the cross product between the directional vector of both lines to find the normal vector $n$, but when I looked for an intersection point $r_0$ to apply the formula: $\lt r-r_0\gt \bullet \ \ n$, I did not find any.  Can I use the point $\lt1,-1,5\gt$ given in the line $r$? Or is it not possible to find an equation of the plane containing two lines that do not intersect? I have this, in D1:  =sum(D3:D)   I want D1 to always be the sum of the rest of column D (D2 is a header)  I sometimes move a row, from eg 8 to 3, and that causes D1 to change to sum(d4:d).    I want D1 to stay as my sum of the column. There is a hint to show that if there were points in $\mathbb{Q}$, then there would exist $r,s, t \in \mathbb{Z}$ with $\gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$ , and then show that any prime dividing $s$ is a quadratic residue modulo $17$.  I'm stuck on the first bit - if you suppose $\left(\frac{a}{b}, \frac{c}{d}\right)$ is a point on the curve in $\mathbb{Q}$ (so $a, b, c, d \in \mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square? When will my phone get the Android 2.3 update (Gingerbread)? In other words, why do we need to connect the battery positive to the negative to get electron flow? As far as I know, voltage difference is what drives current flow.  From what I understand, there's a surplus of negative charge (electrons) in the positive end of a battery (weird I know, but I guess they do it for mathematical reasons). Between the positive and negative ends of the battery is some kind of wall that prevents the electrons from diffusing, so they have to go the long way (through a wire to the other end of the battery) to diffuse and reach the receiver atoms/molecules.  So why not just have the surplus of electrons on one end go to ground to equalize (e.g. 5 volts to 0 volts should produce current)? Why does it have to be the other end of the battery? I'm sure there's atoms/molecules in the ground that could accept the extra electrons (receivers atoms/molecules for the donors in the battery). How does a new user begin using Stack Overflow/Stack Exchange? As a new user, I find it impossible to contribute to the community or to even begin contributing. I know why the reputation system is in place and I respect it, but at the same time it is too hard on new users.  It started with me wanting to upvote a question...     "Vote Up requires 15 reputation"   Just to appreciate a good question that helped me I need 15 rep? OK..., so I find out how I can get reputation points. More or less, I need to ask a question for +5 and add an answer for +10. I don't have any questions at the moment and I don't want to spam a random question, but I did happen to have a slightly different answer to the earlier question that I wanted to upvote. So I try to answer...     This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.   Now... I know this doesn't apply to every question, but at the same time, this means I'm forced to answer yet another question, and I don't want to go around looking for random questions and spam irrelevant answers...  Now I'm stuck. So I try to talk to some other users in the chat rooms about how to go about getting reputation points and how they first started. I entered into a "casual chat room".     You must have 20 reputation on Meta Stack Overflow to talk here.   Yes... I can understand why this is enforced, because we don't want new users to spam chat rooms, but all I wanted to do was ask for help. So I can't even talk to anyone about my newbie problems. So okay, I'll just settle for upvoting comments in the question that helped me. I don't see an upvote hover-over button so I can't even do that.    So now I'm stuck at 1 reputation point, unable to cross that 15 rep barrier, and even after posting this question I'll only have 6 rep points and still unable to do anything, and I'll be forced to look for questions to answer. Yes, I should be contributing to the community, but I shouldn't need to begin answering or making questions to upvote questions and answers. This only spurs more spam/irrelevant answers if anything.   Very, very, very new-user unfriendly. Showing that $|N \cap Z(G)| > 1$ for normal subgroups of p-groups I have a finite $p$-group $G$ and a normal subgroup $N$ which is not the trivial subgroup. I am asked to show that $|N \cap Z(G)| &gt; 1$.  There has been a similar question on MSE here:  Unfortunately, I cannot comment on other user's questions yet, so I made my own question.  I know that the trick is to use conjugation but I can't follow the logic. Can someone elaborate? Especially on the last part of the top answer. How can I change the language for dictation on the fly? With English not being my first language, I usually need support for at least two languages when using features like auto-correct and dictation.  Is it possible to toggle the language which is used for dictation via a shortcut?  As of now, I always have to manually navigate through the menus in the System Preferences which if far from convenient. I've tried to see if there are any pre-defined actions in Automator, but found none. Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.