I am quite absent minded, and I forget everything I want to install - especially after a restart  Is there a way that I can set another command to run (in a separate terminal, of course) when the currently running instance of apt has finished?  I expect this could be done by repeatedly trying to run the command, but I'm looking for something more like this:  WHEN apt EXIT = True; do run next-apt-command   I also expect that it could check the lock file, to see if it is available?
I have seen many softwares such as Update Manager and Synaptic Package Manager, they wait if some other program is using the /var/lib/dpkg/lock and is locked. How can we do this through the Terminal? I saw apt-get's manual but didn't find anything useful.
If Durmstrang and Beauxbatons were hidden from common knowledge, how did people from other schools go to them for the Triwizard tournament? So in GoF, they state that Beauxbatons and Durmstrang are hidden from view and don't have their locations public.     “Durmstrang’s another wizarding school?” said Harry      “Yes,” said Hermione sniffily, “and it’s got a horrible reputation.      According to An Appraisal of Magical Education in Europe, it puts a   lot of emphasis on the Dark Arts.”      “I think I’ve heard of it,” said Ron vaguely. “Where is it? What   country?”      “Well, nobody knows, do they?” said Hermione, raising her   eyebrows.      “Er — why not?” said Harry.      “There’s traditionally been a lot of rivalry between all the magic   schools. Durmstrang and Beauxbatons like to conceal their whereabouts   so nobody can steal their secrets,” said Hermione matter-offactly.   So how do students get to the other school for the Triwizard tournament? The future of meta.stackoverflow and meta.stackexchange Historically this site, , has been the place to ask questions about not just Stack Overflow, but the functioning of the entire Stack Exchange (2.0) network, including Area 51 and stackexchange.com and careers.  (Yes, there was a , but that dealt with the old legacy Stack Exchange 1.0 sites. It was deprecated a while ago, and currently redirects here.)  Now that we've gotten many, many months under our belt from the SE 1.0 days, I think it's time that we gave Stack Overflow its own, true per-site meta, and moved the higher level network meta discussions to meta.stackexchange.com.  Over time, it's begun to bug me more and more that requests for Stack Overflow, the site, kind of get buried here on meta.stackoverflow under the avalanche of network issues. Retag requests, synonym requests, things truly specific to Stack Overflow, the site, not the network. This is unfair to our vastly largest and flagship Q&amp;A site, and it's unfair to the Stack Overflow users to mix their requests in with network level concerns.  It's also odd that Stack Overflow is the only network site without a proper, dedicated per-site meta, where its meta has its own reputation system not tied to the parent site in any way.  This is mostly a quirk of history more than anything else, and I believe it is now an appropriate time to split into two sites:   , a proper per-site meta for Stack Overflow with integrated rep , a global network meta with its own reputation system   Yes, this will be painful. But I believe it is a good, necessary, and healthy step both for the future of the network and the future of Stack Overflow.  Some things to discuss:   which questions should be migrated from here to meta.so? Obviously site-specific things like the retag-requests, and perhaps questions with the  tag. Anything that is 100% wholly specific to Stack Overflow and not generalizable to the rest of the network should be moved over. This may be a pretty small, narrow list of questions, and that's OK. Should we do some meta spring cleaning in our transition? Which questions / tags should be blown away as no longer relevant, referring to ancient versions of Stack Overflow or issues that have long since ceased to exist and aren't instructive for any future visitors? Your meta reps will generally be unaffected, except insofar as we delete for spring cleaning, or move site-specific things to meta.so. We still plan to have a distinct reputation system for the new meta.se site, just like here.   I do not have a timeline for this change, that is up to Jarrod and David to decide. It may be months away. However, we feel pretty strongly that this is something that we need to do to clean things up and pave the way for the future -- so I expect it will happen in the next few months, giving us lots of time to decide how we want to do it, per the above.  Update  . What is the purpose of the word 'self'? Why don't the White Walkers take the sea route to Westeros After watching 6 seasons of Game of Thrones, we know that the only roadblock for White Walkers is "The Wall" to stop them from attacking Westeros. However, we see that King Stannis attacked wildlings through a ship route without crossing the Wall.  What is stopping the White Walkers from doing the same? They can ask wights to make Ships or rowboats. They have a lot of forests to get wood for making ships. They can simply land at the coast of the North and attack Winterfell or whatever they want. I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Biblatex ieee incorrect display of volume and issue number Whey I try to run the program Layout Editor (available on ) on Ubuntu 14.04.1 64-bit LTS I get the following output in the terminal:  $ layout layout: /usr/lib/x86_64-linux-gnu/libstdc++.so.6: version `GLIBCXX_3.4.20' not found (required by layout)   I tried the command strings on /usr/lib/x86_64-linux-gnu/libstdc++.so.6 as following and here is the input and the output:  $ strings /usr/lib/x86_64-linux-gnu/libstdc++.so.6 | grep GLIBCXX GLIBCXX_3.4 GLIBCXX_3.4.1 GLIBCXX_3.4.2 GLIBCXX_3.4.3 GLIBCXX_3.4.4 GLIBCXX_3.4.5 GLIBCXX_3.4.6 GLIBCXX_3.4.7 GLIBCXX_3.4.8 GLIBCXX_3.4.9 GLIBCXX_3.4.10 GLIBCXX_3.4.11 GLIBCXX_3.4.12 GLIBCXX_3.4.13 GLIBCXX_3.4.14 GLIBCXX_3.4.15 GLIBCXX_3.4.16 GLIBCXX_3.4.17 GLIBCXX_3.4.18 GLIBCXX_3.4.19 GLIBCXX_DEBUG_MESSAGE_LENGTH   How can I resolve this issue I somehow guess that it is related with GCC but cannot figure out how to fix the problem. Furthermore, I have tried a PPA as suggested on  and there is a similar question for an older version of Ubuntu at . However, after I ran sudo update I get numerous warnings and the list I obtained with the strings command did not have new entries after installing gcc 4.9, so I purged the PPA() with ppa-purge. Any help, suggestions on how to proceed are greatly appreciated. Normalizer of upper triangular group in ${\rm GL}(n,F)$ The following question has already appeared on mathstack:     If $B$ is the subgroup of ${\rm GL}(n,F)$ consisting of upper triangular matrices then normalizer of $B$ in ${\rm GL}(n,F)$ is $B$ itself.   I know a proof of this using Bruhat decomposition of ${\rm GL}(n,F)$.   Question: Can we prove above theorem without using Bruhat decomposition?    Why came to this question: Consider the general linear Lie algebra $L=\mathfrak{gl}(n,F)$; in it, let $T=\mathfrak{t}(n,F)$ be the upper triangular sub-algebra. Then normalizer of $T$ in $L$ is $T$ itslef, and this can be proved just by considering a very simple decomposition of ${\mathfrak gl}(n,F)$: write any element as sum of upper triangular matrix and lower triangular matrix whose diagonal is $0$.   But then for problem above, is it necessary to use Bruhat decomposition? Writing function definition in header files in C++ I have a class which has many small functions. By small functions, I mean functions that doesn't do any processing but just return a literal value. Something like:  string Foo::method() const{     return "A"; }   I have created a header file "Foo.h" and source file "Foo.cpp". But since the function is very small, I am thinking about putting it in the header file itself. I have the following questions:   Is there any performance or other issues if I put these function definition in header file? I will have many functions like this. My understanding is when the compilation is done, compiler will expand the header file and place it where it is included. Is that correct? First Question for me.  I have benefited from other answers here but have not fully resolved my problem.  Here's my problem...  The LM311 Comparator output does not drive the N channel MOSFET fully on.  The LM311 has a 110mv ref voltage on the inverting input. The non-inverting input is driven by an opto-isolator output. The Opto-isolator is driven by 1 of 4, AC 60Hz voltages.  &lt;1VAC, 9VAC, 55VAC and 115VAC.  The goal of the design is to have the FET fully turn on the 12VDC fan when there is appreciable AC voltage.  So fan fully On for the 9, 55, and 115VAC voltages.  It seems it should work because I measure voltages on the non inverting input of 1.6, 2.2 and 3VDC respectively for the 3 appreciable AC voltages.  And my ref voltage on the inverting input is 110mv.  So it "almost" works because the fan is off at the &lt;1VAC, but not fully "On" for the other AC voltages.  I believe this is because Vgs is 167mv, 6.1V, 7.7V and 9.9 Volts for the 4 AC voltages listed above.  And the fan is only getting 7, 8 and 10VDC instead of a desired 12VDC. Would really like to be able to fully turn this FET all the way on.  So Close!  Any input or guidance is appreciated.  Ps-Things I think I've done right? Flyback diode across the fan. Putting N channel FET to switch to ground on low side of fan. LM311 has Open collector output using correctly?  PSS-I realize that the pinouts for the LM311 are not right in my schematic.  I have them right in my protoboard circuit.  Its a single device.  Pin 2 is noninveritng and pin 3 in Inverting  Thanks Again.      The suggested duplicate question and answer doesn't help for the following reasons: Both diagram and schematic are"No file found" making it pretty difficult to follow the answers. I am not using any positive feedback in my circuit as I don't think I need one. I have a simple pull up resistor to my 12V supply, not a voltage divider to the power supply as is mentioned in the answer. Thanks.  Just for the record I wanted to post a more accurate schematic and thanks for the response on using a Cap to hold up the opto isolator output. I actually have a full wave input to the phototransistor but I'm sure the cap will help anyway. To answer the last question, the Pin 7 comparator output has a pullup resistor to 12v. The issue is still that Vgs is only 10v when the comparator has 3v on pin 2 and 110mv on pin 3. I would have expected the LM311 output to go much closer to 12v (Vcc) with those inputs. The intent is for the FET to go completely on for pin 2 voltage > pin 3. Removing all \n\r characters from a node XSLT? wondered if you could help me please? I have node in xml that is as followed  $LOG: 08880xbpnd $ fhdsafidsfsd df sd fsd f sd fsd  I was wondering is there anyway to make all the text go on to one line so that it then can be passsed through to a javascript function? so it would turn out like this   $LOG: 08880xbpnd $fhdsafidsfsddfsdfsdfsdfsd Don't Have A .minecraft Folder on Windows 10 PC Conjectures Involving floor(x)/piecewise continuity with regards to Integration and Differential Equations [This answer has been heavily edited in response to a long chain of comments on Eric Stucky's answer.]  I came up with these few theorems and I am curious whether or not my hypothesis are true. I don't really have a method for proving these, mind you. I'm more considering these as things I've noticed to be true. I just don't know whether they are always true.  Some of these are definitions to which I will append "def." to the end of the title. Think of these as a given. It's like how the integral is defined abstractly as area. I'm defining terms.  Jump Series Theorem def.  A function $f$ [consisting of floor] is said to have a jump series if there is a series $\Sigma=\sum_{n=1}^\infty a_nH_n$ of scaled step functions such that $f+\Sigma$ no longer has any discontinuity attributable to floor. The sequence of scaling factors $a_n$ is called the jump sequence for the function.  Definition of Implied Derivative def.  We define the evaluation of the implied derivative of a function $y$ at the point $x$ as the set of values $y^{\to}(x) = \{y | y = \lim_{h \to 0^+} \frac {y(x+h) - f(h)}{h} \lor y = \lim_{h \to 0^-} \frac {y(x+h) - f(h)}{h}\}$. It is in this sense that the implied derivative has the potential of being a multi-valued operator.  Jump Series Resolution Theorem  [The original wording is preserved; I have a formalism but it assumes that having a jump function is equivalent to being piecewise continuous, which I'm not sure is true.]   The integral of a floor-based function is floor treated as a constant minus the appropriate jump series of the integral.  Integration Continuity Theorem  The (ordinary) indefinite integral of any function $f$ having a jump series must be continuous (or fixably discontinuous) on the domain of $f$. However, this can vary for the implied indefinite integral.  Calculus Jump Theorem def.  [Again, the original wording is preserved; the meaning of this phrase has not been properly illuminated]  Any form of jumping within a function representable as a floor function can be considered as a portion of the constant of the implied integration for floor.  Jump Location Theorem  For any function $a(x)$ with a jump series, the function $\lfloor a(x)\rfloor$, has points of discontinuity precisely at those $x$ such that $a(x) - [a(x)] = 0$.  Composite Floor Function Jump Sequence Theorem One  If $a$ is a function such that $a(x) - \lfloor a(x) \rfloor = 0$ when $x - \lfloor x \rfloor = 0$, then the jump sequence of the composite function $f(x - \lfloor a\rfloor)$ is a constant value $f(0) - f(1)$.  Implied Differential Equations Conjecture  Any continuous solution to an equation constructed with implied derivatives of various orders is the same as the solution to the similarly constructed differential equation. How do I bind the tmux prefix to a Super? I find even Ctrl+b to be very emacs like but I understand the point. I'm wondering if I could bind it to a single keypress of a key I don't other wise use? namely Super_L (also known as the left windows key. for why I say Super_L start xev in a terminal and press that key) We know that wizards in the Harry Potter universe are subject to regular physical harm (e.g. witness injuries from being hit by Bludgers or just colliding with things when playing Quidditch).  Therefore, it's almost a certainty that an average wizard would be vulnerable to, say, a bunch of bullets.   There may likely be spell(s) to protect from projectiles - witness Dumbledore's  deployed when Voldemort sent a bunch of glass at him during the  at the end of Harry Potter and the Order of the Phoenix (according to , possibly an unusually powerful Protego or the Silver shield).   But:   those spells obviously take effort/energy/time to cast (not everyone's Dumbledore) AND more likely than not most Death Eaters wouldn't recognize a Muggle gun for a threat (or realize what kind of threat) till too late.   So, the guns should at best allow an over-matched OotP member (e.g. Harry or Hermione) to greatly equalize the power between any regular Death Eaters and themselves, and even perhaps help against Voldemort (who can't be killed with a bullet, but, at best, would need to expend magical energy and time into conjuring a shield, and, at worst, be stopped/interrupted/driven off).  So I'm looking for an explanation - ideally in-universe, but possibly just some statement by JKR - of why the Muggle-born OotP - who are likely very much at ease AND familiar with Muggle technology and live in the late 20th century - do not use guns. Or, for that matter, ANY Muggle technology of the late 20th century? Cue Arthur C. Clarke's "indistinguishable from magic" meme.  Even if they are on some kind of idiotic moral crusade to never kill their opponents (even the Jedi aren't this dumb), given the time-frame, non-lethal weapons would have already existed, such as rubber bullets and tasers.  NOTE: Please don't offer the "if they do it, Death Eaters would start using guns as well" theory. First, Death Eaters would have used the guns if they knew how to, anyway. Second, the guns, like any "secret weapon", could be reserved for strategic battles, say the defense of Hogwarts in Deathly Hallows.    The origin of the question stems from two things.  First, Harry's fascination with all things that are different in the magical world (e.g. moving photographs), based on his knowledge of the Muggle material world, as well as the somewhat amusing cluelessness of the "Muggle artifacts" expert, Mr. Weasley, regarding Muggle items and how they work.   Basically, from the books, it's clear that the Wizarding world wouldn't know details about guns or what they are for or how they work.  The second thing is it's clear, at least initially, that wizards are afraid of Muggles finding out about them. Hence all the efforts to cloak their society (basically, do you REALLY want every Muggle gunning for you, even with Magic)? The idea of how control over the Muggle world seems to be more indirect, by Death Eaters asserting mind-control over the government. This means there's an implicit understanding by Death Eaters that they don't stand a chance against armed Muggles even given the magical disparity.  Considering Harry and Hermione's overall adaptability and inventiveness (and brains in the latter case), it's mind-boggling that the idea of fighting Death Eaters with modern technology completely slipped their mind. I mean, it's a basic part of human myth (see Steel vs. Magic themes in Conan books, or Beowulf, or heck, A Yankee in King Arthur's Court). So it's not like the idea of using technology against magic would be impossible to come up with. The question begs itself - why not? Considering the fact that Rowling generally tried to be logical and consistent in building the HP universe, I feel like there must be some in-universe explanation for this that I just didn't notice.   (in-universe answers only unless there are some authoritative out-of-universe statements that are part of canon). Integrating MySQL with Python in Windows I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let $n = (a_ka_{k-1}\ldots a_2a_1a_0)_{10} = \displaystyle{\sum_{j=0}^{k}}a_{k-j}10^{k-j}$. The expression $$ Q_{3}^{\prime}(n) = (a_2a_1a_0)_{10} - (a_5a_4a_3)_{10} + (a_8a_7a_6)_{10} -\ldots  $$ are called alternating sum of the digits of third order of $n$. For example,  $$ Q_{3}^{\prime}(123456789) = 789-456+123=456 $$ Proposition: $7 | n \ \Leftrightarrow \ 7 | Q_{3}^{\prime}(n)$.  proof. ??  Thanks for any help. My eth0 has gone and I don't have internet and network connection I had internet connection, but when I upgraded my VGA and restarted my PC then everything for eth0 has gone.     When I used ifdown:     ifdown: interface eth0 not configured   Is there something I could do?    EDIT: Content of /etc/network/interfaces    auto lo iface lo inet loopback   When I ifconfig I don't have these lines:    inet addr:192.168.1.5  Bcast:192.168.1.255  Mask:255.255.255.0 inet6 addr: fe80::219:5bff:fe5e:a5e/64 Scope:Link   In the Terminal typed:  sudo gedit /etc/network/interfaces   and then added these lines in the bottom:  auto eth0 iface eth0 inet dhcp   then I typed:    sudo ifdown eth0 sudo ifup eth0   when I send the second command I received the below message:    No DHCPOFFERS received No working leases in persistent database - sleeping   Any help? Move post to different user I am writing a small KOMA-Script scrartcl document with multiple sections. In the table of contents, LaTeX fills the current line with dots between the end of the section name and the page number. Is there any way to customize this behavior? I would like to reduce the space between the single dots (i.e. use more dots in the same space).