When I boot (Ubuntu 14.04) Grub does not load the OS. If I list Advanced Options, grub lists kernels up to 3.13.0-100-generic, however uname -r outputs 3.13.0-96-generic.  I have installed grub Customizer, and it also only sees   linux-image-extra-3.13.0-32-generic         linux-image-extra-3.13.0-92-generic         linux-image-extra-3.13.0-93-generic         linux-image-extra-3.13.0-96-generic       Any idea what is going on?
I went into Live CD and activated Terminal.  Then, I typed gksu nautilus to gain temporary access to my root directory. Once the Manager window opened, I clicked on File System and navigated to the /etc/default/grub directory. I then double clicked on the Grub file to make changes to the TIMEOUT and TIMEOUT_QUIET.  I saved the file after I made the changes. Then I went back to the terminal mode and typed:  gksu update-grub   The error message I got was:     Can not find a device for / (is /dev mounted?)   I'm not sure what to do at this point. How can I run update-grub to update the changes? Thanks in advance for your help. Please be very detailed and specific with your response as I am totally new to this environment.
A problem regarding to Schur's lemma Let $\rho: G \to GL_n(\mathbb{C})$ be an irreducible representation, and $g\in Z(G)$. Show $\rho(g)$ is a scalar multiple of the identity matrix $I$.   I think I have it, here is my solution:  Since $\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})$ and $\rho$ is irreducible, consider a nonzero eigenvalue of $\rho(g)$, say $\lambda$, we have $\rho(g) -\lambda $ is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to $\lambda$ are in the kernel).  But I didn't use the condition that  $g\in Z(G)$, so are there something wrong with my solution? Why couldn't Ellie's experience be corroborated by simply re-using the machine? find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.  Edit: This problem is taken verbatim from Exercise 2.7.11 on page 68 of Abbott's Understanding analysis. In QGIS, I want to make polygons from a set of intersecting lines. Looking at the screenshot here, I want chop the purple line (it's a closed line, not a polygon) where it crosses the green and blue lines. The result hopefully would be a polygon where I have hatched in red. I have quite a few to do.    I have tried the QGIS tool previously called Polygonizer (see ), but now called Polygonize. This is a tool that, given a layer containing a set of crossing lines arranged essentially like a # symbol, should create a polygon from the enclosed space in the middle.  (In QGIS 2.01, Polygonize is in the top menu bar: go to Processing -> Commander, then type "Polygonize" on the command line, select "Processing algorithm: Polygonize" from the list that appears, and press Enter.)   At the momemt Polygonize doesn't work properly. Is there another way to do what I want to do?   For example, wouldn't it be great if the QGIS 'Split Feature' command was able to work like MapInfo's 'Polyline Split', where you select the polygons to split, then select an already-drawn polyline to split by, and voila. But Split Feature seems not to work that way.    UPDATE: Detail of problems with running Polygonize.  I have 64-bit QGIS 2.01 installed to Windows 7 from the installer package, QGIS-OSGeo4W-2.0.1-3-Setup-x86_64.exe.  The routine processed an output from my shapefile, but did not create a polygon from the region enclosed by intersections between the blue, green and purple lines. (Note: in the actual input file, all these lines are on the same layer; the screenshot just colours them differently for illustrative purposes). What I got instead was a simple conversion of the purple outline to a polygon, as though the program completely ignored the blue and green lines.  When I node-edited the closed purple line, to open it so that I had a set of open crossing lines like in the # symbol, and then ran Polygonize again, nothing happened. No crash, but no visible polygon in the output file.    When I used Nick Hopton's own  data, I got a crash; see screenshot: When overriding equals in Java, why does it not work to use a parameter other than Object? I ran into an interesting behavior recently.  It seems that if I override .equals() to take a parameter other than Object, it doesn't get called.  Can anyone explain to me why this is happening?  It seems to violate my understanding of polymorphism in OOP, but maybe I'm missing something.  Here's much simpler code that shows what I'm seeing:  public class MyClass {   private int x;   public MyClass(int n) { x = n; }   public boolean equals(Object o) { return false; }   public boolean equals(MyClass mc) { return x == mc.x; }   public static void main(String[] args) {     List&lt;MyClass&gt; list = new ArrayList&lt;MyClass&gt;();     list.add(new MyClass(3));     System.out.println("Contains 3? " + list.contains(new MyClass(3)));   } }   When this is run, it prints "Contains 3? false".  It looks like the equals(Object) function is called, even though there is another that would work.  By contrast, if I write equals like this the code works as expected:  public boolean equals(Object o) {   if(!(o instanceof MyClass))     return false;   MyClass mc = (MyClass)o;   return x == mc.x; }   Why isn't it figuring out which version of the function to call based on the type of the parameter? Wouldn't the thermodynamic cost of creating alternate universes make the Many Worlds interpretation implausible? What is the difference between "source x", ". x" and "./x" in Bash? I want to copy/modify/delete a Minecraft world. Where can I find my saves folder? Was Thanos really committed or just a hypocrite? The title is extremely vague, because this question contains HEAVY spoilers for Avengers: Infinity War, specifically the very last scenes of the movie. I can't reasonably put the entire question in spoiler tags, so consider this your warning.  At the end of the movie, Thanos pretty much wins. He gets all the Infinity Stones, Gamora and Vision have been brutally murdered, and the rest of the cast completely beaten. Nobody could stop him. So he executes his plan.  As far as I know (unless I missed something), his plan is to eradicate exactly half of all mortals in the universe, and from his own words, in a fair lottery (random chance), without taking into account a person's wealth, social status, or anything else. Pure chance decides whether you are sacrified for the greater good or live through this greater good (according to him).  During the movie, Thanos and one of his underling (Ebony Maw) say themselves that being sacrified is an honor and should be appreciated. As you would expect, most people do not appreciate being sacrified.  At the end of the movie, Thanos uses all 6 stones to do a finger snap, deleting half the universe.  The question is: Did he exclude himself from the targets? Was there a 50% chance that Thanos himself would disappear right there?  As far as I know, this is not explicitly answered in the movie itself. I am looking for similar situations in the comics (or in the movie if I happened to miss something) where you could know, or have an educated guess, as to Thanos being the kind of person who follows through to the end, even if it means dying, or the kind of person who gets scared of death when it comes down to it, and comes off as an hypocrite.  Some people also mentioned his "deal" with Strange would make him spare Tony from the snap too. I'm also wondering about that, if there is a link between the two, but it's not the focus of the question.  The reason I'm wondering is that Thanos seems to me like a very extreme utilitarist. Even though he might be wrong about the value of his plan (Do the remaining half really become happy?), in-universe it seemed successful at least once (Gamora's home planet), so his reasoning is not bad, even if the premise might be horribly flawed. Excluding himself from the snap would make it into a straight villain, while considering himself equal to the rest would make him into a wannabe hero with very very evil methods. This makes a huge difference in the interpretation of the movie and his character. Below, the  of second-order logic is used.    My question is about a second-order analogue of $ZFC$ other than the usual "second-order $ZFC$." Rather than define the latter, I'll just state that its (set-sized) models are exactly the $V_\kappa$s with $\kappa$ strongly inaccessible (or if you prefer, the uncountable Grothendieck universes).  Specifically, I'm interested in the second-order theory consisting of the usual (first-order) axioms of Pairing, Extensionality, Union, Choice, Infinity, and Powerset, and the Separation and Replacement schemes ranging over second-order formulas. I'll call this "$ZFC_2^{scheme}$."    Specifically, for each formula $\varphi(x, y_1,...,y_k)$ of second-order logic with only the displayed variables, each of which is first-order, we have a corresponding Separation and Replacement instance. We do not allow second-order parameters here, so formulas with second-order free variables don't yield Separation/Replacement instances.   The passage from $ZFC$ to $ZFC_2^{scheme}$ has a nice uniformity: it's an example of a more general operation $\mathcal{ZFC}$ which spits out a $ZFC$-analogue given a logic at least as strong as first-order logic.    This is the reason for the parameter restriction above: I want $ZFC_2^{scheme}$ to look just like $\mathcal{ZFC}(\mathcal{L})$ for any other logic $\mathcal{L}$, and not take into account the particular feature that we now have a new kind of variable at our disposal - and in particular, take as "syntax-free" an approach to logics as we can get away with. (This is a choice of arguable value, but it's the one I'm making for now.)   It's easy to show that every model of $ZFC_2^{scheme}$ is well-founded, so to understand its models we only need to look at transitive sets. A natural hope at this point is that $ZFC_2^{scheme}$ is just second-order $ZFC$ in disguise, that is, that for a transitive set $M$ we have $M\models ZFC_2^{scheme}$ iff $M=V_\kappa$ for some strongly inaccessible $\kappa$. However, this isn't at all obvious to me (although it is easy to show the right-to-left direction).     Question. What are the (set-sized) models of $ZFC_2^{scheme}$?   (Note by contrast that the "arithmetic analogues" $PA_2^{scheme}$ and second-order $PA$ are equivalent since each pins down $\mathbb{N}$ up to isomorphism - although that leads to .)  (I'm happy to drop Choice if that would help.)  EDIT: As Hanul Jeon pointed out below,  shows that consistently $ZFC_2^{scheme}$ (there called "$ZFC_2^{def}$") has countable models, so we have a partial answer. However, I don't see at the moment a way to get an outright $ZFC$ proof that $ZFC_2^{scheme}$ is strictly weaker than second-order $ZFC$. Is there a way in javascript to detect if the unload event is caused via a refresh, the back button, or closing the browser? Deleting Individual SE Accounts Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a entire function. Suppose that there are $M$, $r&gt;0$ and $n\in \mathbb{N}$ such that $\left|f(z)\right|&lt;M\left|z\right|^n$ for all $z \in \mathbb{C}$ with $\left|z\right|\geq r$.  Show that $f$  polynomial of degree at most $n$.  Remark: I tried to follow the proof of Liuville's Theorem but I have complications with the condition  $\left|z\right|\geq r$.. How to define a recurrence relation in Maple without the command proc? I made a script that should notify me when there's a new chapter of manga that I'm reading. I used the command notify-send to do this. The program works when I am trying to run it in terminal. The notification is showing. However, when I placed this in my crontab, the notification doesn't show. I'm pretty sure that the program is running since I made it to create a file for me. The file was created, but the notification didn't show.  Here's my script  #!/bin/bash    #One Piece Manga reminder     #I created a file named .newop that contains the latest chapter.     let new=$(cat ~/.newop)     wget --read-timeout=30 -t20 -O .opreminder.txt http://www.mangareader.net/103/one-piece.html  if (( $(cat .opreminder.txt | grep "One Piece $new" | wc -l) &gt;=1 ))     then         (( new+=1 ))         echo $new         echo $new &gt; ~/.newop         notify-send "A new chapter of One Piece was released."     else         notify-send "No new chapter for One Piece."         notify-send "The latest chapter is still $new."     fi         exit   And here's what I wrote in my crontab  0,15,30,45 12-23 * * 3   /home/jchester/bin/opreminder.sh How to I stop a launchd script from running on wake I have a simple launchd script that plays a sound every 30 minutes and it runs out of the user library ~/Library . It doesn't run if my machine is asleep, but does run the last event missed when the machine wakes. I think this is the expected behavior. My question is, how do I stop this happening. If I missed the half-hour event, I don't want it to be run at 20 minutes past the hour.  Here's the core of my script. It calls a Python script. I tried to put a block in the python script to check that I am within a few minutes of each 30 minutes, but that doesn't seem to work on wake. (Is it missing the system clock somehow?)  &lt;key&gt;ProgramArguments&lt;/key&gt; &lt;array&gt;     &lt;string&gt;/usr/local/bin/python3&lt;/string&gt;     &lt;string&gt;/Users/pheon/Documents/playsound.py&lt;/string&gt; &lt;/array&gt;  &lt;key&gt;StartCalendarInterval&lt;/key&gt; &lt;array&gt;     &lt;dict&gt;         &lt;key&gt;Minute&lt;/key&gt;         &lt;integer&gt;0&lt;/integer&gt;     &lt;dict&gt;         &lt;key&gt;Minute&lt;/key&gt;         &lt;integer&gt;30&lt;/integer&gt;     &lt;/dict&gt; &lt;/array&gt;   Here is a snippet from the python code which checks the time before playing the sound.  time0 = datetime.datetime.now() if (time0.minute % 30) &lt; 2:     subprocess.run(["/Users/pheon/bin/afplay-vol.sh", "1", bell],check=True) How to disable Ctrl+Shift+U? Ctrl+Shift+U is a combined shortcut in Eclipse. But when I press it, it shows up a u, just like input with an underline. I guess this shortcut has been declared by Ubuntu, so I can't use it. I used to solve this problem by typing Caps Lock first. Is there any better method? I would like to set a maximum execution time for sql queries like set_time_limit() in php. How can I do ? A guide to moderating chat Chat is an offshoot of the main site, and using it is  during their time here.  Chat is our place for:        Real-time collaboration   Meeting fellow members of the community   Less structured, casual (but still roughly on-topic) conversation      Our chat system exists to serve the main site; users of each site can gather and commune with each other.  When in chat, we expect all users to abide by our  policy, which means that we expect community members to treat each other with respect.    While it is a place where the community can go to interact on a more casual level, chat still needs to be moderated by the community and there are tools in place to help with the moderation. This raises a few questions, that are addressed here: Inspired by a , here's a list of the 'etymology' for the  hats. For pictures and a list of criteria, see  instead.  The list is a community wiki; feel free to edit it if you have more information about a hat. If you like to do some research, most hats have been featured in earlier editions, so you might be able to find more information in old Meta Stack Exchange questions.