This question really has me stumped.  We define $\bar{\mathbb{Q}}$ to be the set of elements in $\mathbb{C}$ that are algebraic over $\mathbb{Q}$.  I have shown that this is a field.    Now, I'm assuming that we have some element $b \in \mathbb{C}$ that is algebraic over $\bar{\mathbb{Q}}$.  I want to show $b \in \bar{\mathbb{Q}}$.    My attempt:  Since $b$ is algebraic over $\bar{\mathbb{Q}}$, we know the extension $[\bar{\mathbb{Q}}(b) : \bar{\mathbb{Q}}] = n < \infty$.  Also, there's a polynomial $f \in \bar{\mathbb{Q}}[x]$ s.t $f(b) = 0$.  So, I've let $f = x^n + a_{n-1} x^{n-1} + ... + a_0$, where the $a_i \in \bar{\mathbb{Q}}$.  In particular, we know that these $a_i$ are algebraic over $\mathbb{Q}$.  Next, we consider the extension $[\mathbb{Q}(a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$.  Since all the $a_i$ are algebraic over $\mathbb{Q}$, this extension is finite.  Now, we consider $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$.  By Tower Law, this equals $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}(b)] \times [\mathbb{Q}(b) : \mathbb{Q}]$.  Both of these are finite extensions, so $[\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}]$ is finite.    And this is where I get stuck.  I'm not quite sure where to go from here and I just feel kinda confused.  I'm not sure if this is even taking me towards the answer, but I'm not sure what to do.  Any nudges would be great!  EDIT:  Corrected typos listed below.

The definition of algebraic number is that $\alpha$ is an algebraic number if there is a nonzero polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $p(\alpha)=0$. By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots. I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in $\mathbb{\bar{Q}}[x]$ we could find a polynomial in $\mathbb{Q}[x]$ that has the same roots?  I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts.


Let $\mathbb{Z}(p^{\infty})$ be defined by  $\mathbb{Z}(p^{\infty}) = \{ \overline{a/b} \in \mathbb{Q}/ \mathbb{Z} / a,b \in \mathbb{Z}, b=p^i$  $ with$ $ i \in \mathbb{N} \}$,  I wish show that any element in $\mathbb{Z}(p^{\infty})$ has order $p^n$ with $n \in \mathbb{N}$.  i try several ways but I have not been successful,   some help ??  thank you

In my recent study of Artinian modules I have been looking at Prüfer $p$-groups (which I will denote by $\mathbb Z_{p^\infty}$ from now on).   I am attempting to prove that $\mathbb Z_{p^\infty}$ is an Artinian $\mathbb{Z}$-module which is not finitely generated.      To do this I am trying to show that every proper non-trivial submodule of $\mathbb Z_{p^\infty}$ is equal to $\left(\frac{1}{p^n}\mathbb{Z_{}}\right)/\mathbb{Z}$ for some $n$.   I am using the following definition of $\mathbb Z_{p^\infty}$: $$\mathbb Z_{p^\infty}=\left\{x\in \mathbb{Q}/\mathbb{Z}\mid \exists n:p^nx=0 \right\}$$ I have read in several places, Atiyah & MacDonald Example 6.3 and Wikipedia to name a few, that the subgroups of $\mathbb Z_{p^\infty}$ are of this form. This would demonstrate that $\mathbb Z_{p^\infty}$ is Artinian. However, I can't seem to find anywhere a proof or example showing why all the subgroups are of this form. I would like to understand this further, as it is not immediate to me that this would be the case, but my attempts to figure out why have so far been futile. I may be thinking too much and missing a simple explanation. I have looked at starting with an arbitrary subgroup $H$ such that $H\neq 0$ and $H\neq\mathbb Z_{p^\infty}$ and seeing if I can deduce anything from that. I got this far, but it doesn't seem to be going anywhere. I have also looked at Sylow's theorems, but I'm not sure if they will be any use to me as $|\mathbb Z_{p^\infty}|=\infty$ (unless I'm wrong about this!).  I would be extremely grateful if anybody could offer a few hints here as I am slowly losing my sanity trying to figure it out!  Thanks,  Andy.


Recently many ad pages are opened in default browser in my Windows 10. But I don't know which app opens these ad pages. How can I find out the app. Is there any log or software which shows the app's name or location?

What should I do if my Windows computer seems to be infected with a virus or malware?  What are the symptoms of an infection? What should I do after noticing an infection? What can I do to get rid of it? how to prevent from infection by malware?   This question comes up frequently, and the suggested solutions are usually the same. This community wiki is an attempt to serve as the definitive, most comprehensive answer possible. Feel free to add your contributions via edits.


I want to turn a rectangle with ports in the top and bottom into a shape I can use with a node. How can I do this?    \documentclass[tikz,14pt,border=10pt]{standalone} %%%< \usepackage{verbatim} %%%> \begin{comment} :Title: abc :Tags: def :Author: ghi :Slug: jkl \end{comment} \title{test_portbox} \usepackage{textcomp} \usetikzlibrary{shapes,arrows} \begin{document} \begin{tikzpicture}[auto, line width = 0.5, node distance=2cm, >=triangle 45]  \draw (0.2,0) -- (0,0) -- (0,1) -- (0.2,1); \draw (0.4,1) -- (1,1) -- (1,0) -- (0.4,0);  \draw [line width=0.1] (0.2,0) circle (0.025) -- (0.4,0) circle (0.025); \draw [line width=0.1] (0.2,1) circle (0.025) -- (0.4,1) circle (0.025);  \end{tikzpicture} \end{document}

I am currently learning how to use TikZ/PGF to make some figures for a project report, and so I have a question:     Is there is a practical way of making a particular object into a node?   The object I have in mind is shown in the Figure below. I want the object to consist in the big circle, the two circular arcs, the label and anchors at north, south, east and west. The dotted lines are simply there to indicate that I am going to have paths going out of these points, which is why I need the anchors there.     I include the source, in the case that it is useful to anyone.  \begin{tikzpicture} \draw[help lines, gray, dotted] (-2,-2) grid (2,2);  \begin{scope}[shift={ (0,0) } ]     \begin{scope}[rotate=90]         \draw[thick] (-1,0) arc[start angle=-90, delta angle=90, %                               x radius=1, y radius=1] -- (0,1) %NW                      (0,-1) arc[start angle=180, delta angle=-90, %                                    x radius=1, y radius=1] -- (1,0); %SE         \draw[gray] (0,0) circle (1);     \end{scope} % Anchor replacements     \node(N1)       at (0,1)    {};     \node(S1)       at (0,-1)   {};     \node(E1)       at (1,0)    {};     \node(W1)       at (-1,0)   {}; % Label     \node(Crossing number) at (0,0) {1}; \end{scope}  \path[thick, gray, dotted] [ out=90,   in=270]     (N1.center) edge (0,2) [ out=180,  in=0]       (W1.center) edge (-2,0) [ out=270,  in=90]      (S1.center) edge (0,-2) [ out=360,  in=180]     (E1.center) edge (2,0); \end{tikzpicture}   PS. I have had a look at the Manual, but I could not find a description of how to do this.


I just installed ubuntu 20.04 in bootable pendrive.I wanted to use ubuntu in bootable pendrive,as my laptop has no free space in harddisk.Whenever i boot ubuntu 20.04 using bootable pendrive it shows checking file system everytime,after some time it shows me the installation window with two options (Try Ubuntu,Install Ubuntu).So i had to click on try ubuntu every time.Everytime the screen shows up and when i click Try ubuntu it once again shows the manufacturer logo and loads as new desktop each time and the files i stored in it and the update i made to the sofware is no more.I tried it out in different pendrive too. please help to solve the issue that it shows new setup everytime i turn on my laptop and the files in it and the changes i made gets erased

Persistent live drive with a partition for persistence There are some web sites describing  new tools new features in [new versions of] old tools  that can create a persistent live drive from an iso file with Ubuntu 19.10 and the corresponding Ubuntu family flavours.  Why and how does it work? How is it different from earlier methods for persistent live drives? Isn't live the default for a cloned drive?


I am currently in Canada as a tourist, however, according to the  I am only allowed to remain in Canada as a visitor for a total of 6 months. My question is, if I were to leave Canada and go to the US for a couple of days and then return to Canada will my status as a visitor be "renewed" i.e. would I be able to stay in Canada for another six months?  I should clarify that my visitor visa expires in 2024. Also just to provide some context, my son will be out of University by May 10, and we had both originally planned to return back to our country together. I have already applied for a status extension but I am afraid the processing times might take too long, hence why the question.

If a Swiss citizen comes to Canada and stays for 6 months, can they travel to USA for example, for a short period of time, and then return to Canada for another 6 months?  How long must they stay out of Canada before returning?


I have written the following code to solve the Longest SubStrings With All Unique Characters.  How can I fix this issue? , I understand what a null pointer exception is and why am I getting it , the question is what changes could be done such that I get the result  based on the question. Again , what could be changed logically in this particular piece of code to make it work as its intended , receiving a null pointer exception public class LongestSubStringWithAllUniqueChars {   static Map findUnique(char[] arr , int l , int r){     Map<Character,Integer> map = new HashMap();     for(int i=l;i<r;i++){         if(map.containsKey(arr[l])){             map.put(arr[l],map.get(arr[l])+1);         }else{             map.put(arr[l],1);         }     }     map.forEach((c,v) ->{         System.out.println(c+" "+v);     });     return map; } public static void main(String[] args) {     String str = "GEEKSFORGEEKS";     char[] arr = str.toCharArray();     int l = 0;int r=0;     int max =0;     Map map  = new HashMap<>();       while(r<arr.length){         r++;         map = findUnique(arr,l,r);         if(map.size() < r-l+1){             r++;             map.put(arr[l],(int)map.get(arr[l])-1);             while(map.size()< r-l+1){                 map.remove(arr[l],(int)map.get(arr[l])); // getting nullpointer here                 l++;             }          } else if(map.size()==r-l+1){             max = r-l+1;             r++;         }     }     System.out.println(max);  }  }

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


I would like to be able to write something like the following:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \newlength\aelengtha \newlength\aelengthb \newlength\aelengthc \begin{document}  \begin{tikzpicture}  \coordinate (Q) at (0,0);  \foreach \myn [count=\myc from=1]               in {apple,                   orange,                   this is a long line,                   short} {   \pgfextra{%%     \settowidth\aelengthb{\texttt{\myn}}     \ifdim\aelengtha<\aelengthb       \global\aelengtha=\aelengthb       \typeout{==>resetting=\the\aelengtha}     \fi}   \typeout{==>\the\aelengthb}   \node[anchor=base west,font=\ttfamily] at ($(Q)+(0,-\myc\baselineskip)$) {this \myn\ is an item}; }  \end{tikzpicture}  \end{document}   But, for some reason, no lengths are actually calculated.  So I tried to find a pgf solution and came up with the following:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \newlength\aelengtha \newlength\aelengthb \newlength\aelengthc \begin{document}  \begin{tikzpicture}  \coordinate (Q) at (0,0);  \foreach \myn [count=\myc from=1]               in {apple,                   orange,                   this is a long line,                   short} {   \pgfmathparse{width("\myn")}   \aelengthb=\pgfmathresult pt   \pgfextra{%%     \ifdim\aelengtha<\aelengthb       \global\aelengtha=\aelengthb       \typeout{==>resetting=\the\aelengtha}     \fi}   \typeout{==>\the\aelengthb}   \node[anchor=base west,font=\ttfamily] at ($(Q)+(0,-\myc\baselineskip)$) {this is \myn\ an item}; }  \end{tikzpicture} \typeout{==>\the\aelengtha}  \end{document}   But this isn't really calculating the width of the text I want to work with.  What I would like to do is write something like  \pgfmathparse{width("\texttt{\myn}")}   But that results in an error:  ! Argument of \XC@definec@lor has an extra }. <inserted text>                  \par  l.28     }  ?    I've also tried:  \pgfmathparse{width("\noexpand\texttt{\myn}")}   without any luck either.  How can I make a calculation of the true textwidth that's being formatted within the \foreach loop?  UPDATE  There's something about the tikzpicture environment that seems to dislike or prevent my code for measuring the width of text from working.  For example, everything compiles fine in the example below (sans tikzpicture)  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \newlength\aelengtha \newlength\aelengthb \newlength\aelengthc \begin{document}  \foreach \myn [count=\myc from=1]               in {apple,                   orange,                   this is a long line,                   short} {   \settowidth\aelengthb{\texttt{\myn}}%%   \ifdim\aelengtha<\aelengthb     \global\aelengtha=\aelengthb     \typeout{==>Length A=\the\aelengtha : recalculated}   \fi   \typeout{==>length B=\the\aelengthb} }  \typeout{==>length A=\the\aelengtha : Final}  \the\aelengtha  \end{document}   So, this lead me to consider writing the following:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \newlength\aelengtha \newlength\aelengthb \newlength\aelengthc \begin{document}  \begin{tikzpicture}  \coordinate (Q) at (0,0);  \foreach \myn [count=\myc from=1]               in {apple,                   orange,                   this is a long line,                   short} {   \pgfinterruptpicture     \settowidth\aelengthb{\texttt{\myn}}%%     \ifdim\aelengtha<\aelengthb       \global\aelengtha=\aelengthb       \typeout{==>Length A=\the\aelengtha : Resetting}%%     \fi    \typeout{==>Length B=\the\aelengthb}%%   \endpgfinterruptpicture   \node[anchor=base west,font=\ttfamily] at ($(Q)+(0,-\myc\baselineskip)$) {\myn}; }  \end{tikzpicture}  \end{document}   which works!  But, the tikz manual specifically says that \pgfinterruptpicture is not meant to be used in this fashion.  Apart from getting a valid working example, I would also like to understand why my original example above fails to work as expected.

I'm trying to use an \hbox to measure the dimensions of a certain text. This is working in normal text as in the following example: \setbox37=\hbox{this is a text that I want to measure} text width is \the\wd37, text heigh is \the\ht37\\ text contents: \box37  The output is correct as I expected: text width is 159.63913pt, text heigh is 6.94444pt text contents: this is a text that I want to measure However, when trying to do the same inside a TikZ environmnet, the box I'm using is not assigned the text I pass to \hbox, instead I gets empty box with zero width & height. Here is the code: \begin{tikzpicture}   \setbox37=\hbox{this is a text that I want to measure}   \node[draw] at (0,0) {text width is \the\wd37, text heigh is \the\ht37};   \node[draw] at (0,-1) {text contents is \box37}; \end{tikzpicture}  And here is the output:  What am I missing here?


Assume that a bird of mass 1.5 kg is held in a box at height 500m and this setup is placed on a weighing machine. Now, the bird starts to hover in the box  pushing the air downwards, and is at a height of around 450 m. If we measure the weight of the box, will we measure the bird's weight also. Consider air friction.  I believe that due to friction of air, the downward movement of the air may not touch the bottom of the box (the bird is hovering at around 450 m height) and is lost. Hence, we cannot measure the weight of bird.    Is this true?

Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?


i cannot install any software through software center and when i try to remove software center this appear  E: Could not get lock /var/lib/dpkg/lock - open (11: Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/), is another process using it?

I get this error when trying to use apt-get:  E: Could not get lock /var/lib/dpkg/lock - open (11 Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/) is another process using it?     How can I fix this?


I have a rich textbox where the user can select keywords from a list and they will be inserted in a different color. The user can also enter their own text. The program will be used to build SQL snippets.   For example, text entered by the user will be black, and keywords from that list will be in blue.  I've seen several examples of appending words of different color, but no examples of inserting words of different colors.  Below is what I have tried          public void InsertText(Color color, string text)     {         var startposition = _richTextBox.SelectionStart;         var endposition = startposition + text.Length;          _richTextBox.Text = _richTextBox.Text.Insert(startposition, text);         _richTextBox.Select(startposition, endposition - startposition);         _richTextBox.SelectionColor = color;         _richTextBox.Select(endposition, 0);     }   the problem with this code if text is inserted anywhere except at the end, the rest of the paragraph does not retain the color it had previously. Is there a way to insert text of a different color, while not interfering with the formatting of the rest of the text?

How can I paint in red every time I meet the letter "A" in RichTextBox?


In screen there is a kill command, which kills the current window. This command can be sent even to a detached screen session by running:  screen -S SessionName -X kill   This works to kill the current window only. To kill any other window, it appears the only way is to first switch to that window and then kill it. The select command allows switching between windows in a screen session. Typing the following while attached to a screen session will select the window named foo.  ^A:select foo   However, when not attached to the session, it appears the select command has no effect. Running the following two commands will kill the current window, rather than the window named foo (unless of course that is the active window).  screen -S SessionName -X select foo screen -S SessionName -X kill   I have verified that this occurs even if I attach and detach from the session before attempting to select a window (a known workaround for some other screen quirks).

I'm trying to run a minecraft server on my unRAID server.  The server will run in the shell, and then sit there waiting for input. To stop it, I need to type 'stop' and press enter, and then it'll save the world and gracefully exit, and I'm back in the shell. That all works if I run it via telnetting into the NAS box, but I want to run it directly on the box.   this is what I previously had as a first attempt:  #define USER_SCRIPT_LABEL Start Minecraft server #define USER_SCRIPT_DESCR Start minecraft server. needs sde2 mounted first cd /mnt/disk/sde2/MCunraid screen -d -m -S minecraft /usr/lib/java/bin/java -Xincgc -Xmx1024M -jar CraftBukkit.jar   MCunraid is the folder where I have the Craftbukkit.jar and all the world files etc. If I type that screen line in directly, the screen does setup detached and the server launches. If I execute that line from within the script it doesn't seem to set up a screen  for stopping the server, I need to 'type' in STOP and then press enter. My approach was  screen -S minecraft -X stuff "stop $(echo -ne '\r')"   to send to screen 'minecraft' the text s-t-o-p and a carriage return. But that doesn't work, even if I type it directly onto the command line. But if I 'screen -r' I can get to the screen with the server running, then type 'stop' and it shuts down properly.  The server runs well if I telnet in and do it manually, just need to run it without being connected from my remote computer.


In this sentence, is there an error?  In the name of revamping the law, investigation and trial should not be altered in a way which undermines the principles on which the judicial system was founded.  My book says that 'which' should be replaced by 'that. If yes, why is that?

When is it appropriate to use that as opposed to which with relative clauses?


I am studying the basics of statistical mechanics and Boltzmann distribution.   I tried to use the idea to find natural income distributions, through the method of maximization of probability using Lagrange multipliers, where:  average energy  => income per capita number of particles => number of people Energy levels => income ranges    The total population, income per capita and income ranges are fixed. There is a minimum and a maximum income. The interval between ranges are all equal.   When the income per capita is below the average between max and min income the curve seems realistic.   When it is equal to that average, all ranges have the same number of people, and the curve is linear.  But when it is above the average, the distribution is odd because there are many more people in the high income levels, but I don't think it is impossible. At least it is not contradictory. The problem is that in this case, one of the Lagrange multipliers (usually called $\beta$), is negative.  In statistical mechanics $\beta$ is associated to the inverse of absolute temperature. So it can not be negative. But on the other hand, there is nothing in the method of Lagrange multipliers that obliges it (as I can see).   The graph is shown below.   So, is the fact that $\beta$ must be positive a specific feature of energy configuration of particles, or has it a broader interpretation?  .

Could someone provide me with a mathematical proof of why, a system with an absolute  (such that of a spin system) is  than any system with a positive temperature (in the sense that if a negative-temperature system and a positive-temperature system come in contact, heat will flow from the negative- to the positive-temperature system).


Hello I am trying to move a group of directories with the mv command and curly braces expansion as follows:   #!/bin/bash ... group_name=$1 group=$((total/set)) group=$((group-1)) mkdir "$group_name""1"  mv dat{10.."$group"} "$group_name""1" ...   The function is used as follows:  grouping_dirs total set group_name  e.g: grouping_dirs 30 2 dir_name    And the list of directories I have are:  dat10 dat11 dat12 dat13 dat14 dat15 and so on.   But I want to move only the directories dat{10..14} to dir_name1.  However I get the following error:  mv: cannot stat 'dat{10..14}': No such file or directory  Even though the directories exists. I know that because I tried changing:  mv dat{10.."$group"} ... to mv dat{10..14} ...    And it worked. I think I am not correctly using " " within the curly braces, but I am rather new to bash scripting...So I would appreciate your insight! Thanks!

I want to use $var in a shell brace expansion with a range, in bash. Simply putting {$var1..$var2} doesn't work, so I went "lateral"...   The following works, but it's a bit kludgey.    # remove the split files echo rm foo.{$ext0..$extN} rm-segments > rm-segments source rm-segments   Is there a more "normal" way?


I know $(x)$ is a maximal ideal, but I don't know how to construct more maximal ideals.

Let $F$ be a field. Why does $F[x]$ have infinitely many irreducible elements?   For the case F has characteristic 0  Then x-a is irreducible for all a $\in F$ since x satisfies no non-trivial relations in F.  Obviously this argument fails for a finite field since there are only finitely many a to choose from.  So how may I construct irreducible polynomials in a finite field?  I figure it must involve higher powers of x, maybe $x^n-a$ ?


console.log("1,2,3".split(",").map(parseInt))   prints  [1, NaN, NaN]   why?

From the :  [1,4,9].map(Math.sqrt)   will yield:  [1,2,3]   Why then does this:  ['1','2','3'].map(parseInt)   yield this:  [1, NaN, NaN]   I have tested in Firefox 3.0.1 and Chrome 0.3 and just as a disclaimer, I know this is not cross-browser functionality (no IE).  I found out that the following will accomplish the desired effect. However, it still doesn’t explain the errant behavior of parseInt.  ['1','2','3'].map(function(i){return +i;}) // returns [1,2,3]


Neither the ISO nor the hashsum is available via HTTPS. Is there any way to verify you're not getting man-in-the-middled?

I'm writing a program that needs to download GPG keys for various distributions to do unattended net installs. To do this securely and in an automated way, I need to download the key via HTTPS, which means "gpg --recv-keys" isn't an option. Other distributions (e.g., Fedora) make their keys available on their websites via HTTPS (e.g., ), but I can't find the Ubuntu key file anywhere on ubuntu.com. Does anyone know where I can get it via HTTPS?


I see named functions exampled this way:  var clearClick = function() {     // do something }   ...and then used for binding like so:  $("#clear").bind(click, clearClick);   ...or with the "shorthand" methodology thus:  $("#clear").click(clearClick);   But why not use a more "normal" (similar to other programming languages) construct like this:  function clearClick() {     // do something }   It works the same, doesn't it? Is there any disadvantage to defining functions in this traditional way? Is the previous way just jQuerians flaunting their newfangledness?

I've recently started maintaining someone else's JavaScript code. I'm fixing bugs, adding features and also trying to tidy up the code and make it more consistent.  The previous developer used two ways of declaring functions and I can't work out if there is a reason behind it or not.  The two ways are:  var functionOne = function() {     // Some code };     function functionTwo() {     // Some code }   What are the reasons for using these two different methods and what are the pros and cons of each? Is there anything that can be done with one method that can't be done with the other?


Imagine we had an extra spatial dimension $f$ and so an overall spacetime interval: $ds^2=-dt^2+dx^2+dy^2+dz^2+df^2\tag{1}$ Now let's say that we're restricted to a 4-dimensional "plane" in this 5-dimensional space and that this plane is basically in the $t,x,y,z$ plane, except that it curves in the $f$ direction. i.e. $f=f(t,x,y,z)$ so: $df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\tag{2}$ and: $df^2=\left( \frac{\partial f}{\partial t}dt\right )^2+\left( \frac{\partial f}{\partial x}dx\right )^2+\left(\frac{\partial f}{\partial y}dy \right )^2+\left( \frac{\partial f}{\partial z}dz\right )^2+2\frac{\partial f}{\partial t}\frac{\partial f}{\partial x}dxdt+2\frac{\partial f}{\partial t}\frac{\partial f}{\partial y}dydt+2\frac{\partial f}{\partial t}\frac{\partial f}{\partial z}dzdt+2\frac{\partial f}{\partial y}\frac{\partial f}{\partial x}dxdy+2\frac{\partial f}{\partial y}\frac{\partial f}{\partial z}dydz+2\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}dxdz\tag{3}$ so the metric would be: $\begin{bmatrix}  &\eta _{00}+\frac{\partial f}{\partial t}\frac{\partial f}{\partial t}  &\frac{\partial f}{\partial t}\frac{\partial f}{\partial x}  &\frac{\partial f}{\partial t}\frac{\partial f}{\partial y}  &\frac{\partial f}{\partial t}\frac{\partial f}{\partial z} \\   &\frac{\partial f}{\partial t}\frac{\partial f}{\partial x}  &\eta _{11}+\frac{\partial f}{\partial x}\frac{\partial f}{\partial x}  &\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}  &\frac{\partial f}{\partial x}\frac{\partial f}{\partial z} \\   &\frac{\partial f}{\partial t}\frac{\partial f}{\partial y}  &\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}  &\eta _{22}+\frac{\partial f}{\partial y}\frac{\partial f}{\partial y}  &\frac{\partial f}{\partial y}\frac{\partial f}{\partial z} \\   &\frac{\partial f}{\partial t}\frac{\partial f}{\partial z}  &\frac{\partial f}{\partial x}\frac{\partial f}{\partial z}  &\frac{\partial f}{\partial y}\frac{\partial f}{\partial z}  &\eta _{33}+\frac{\partial f}{\partial z}\frac{\partial f}{\partial z} \end{bmatrix}\tag{4}$ This model is far to simple and cannot accommodate the real solutions to the Einstein Field Equations. If any of the non-diagonal terms are zero, you immediately have to choose between one of the partial derivatives of $f$ being zero and you quickly find out that you cannot write all the metric components in this way. Is there a way to do this with finitely many such $f$? What is the minimum number one would require to be sure that any given solution to the EFE can be satisfied with the appropriate $f$'s?

Does such a set of equations exist or does our universe have an intrinsic curvature that can't be explained by an embedding in a flat Minkowski space of 1 higher dimension? Even if general relativity can be explained by such equations, it still doesn't prove our universe is embedded in a flat Minkowski space of higher dimension.


What is this?   17.537s dev-sda2.device   systemd-analyze blame       18.346s dev-sda2.device      17.472s systemd-journal-flush.service      16.439s configure-printer@usb-002-004.service      12.998s dev-loop3.device      12.974s dev-loop4.device      12.941s dev-loop5.device      12.934s dev-loop1.device      12.910s dev-loop0.device      12.882s dev-loop2.device      11.276s systemd-udevd.service       9.327s configure-printer@usb-002-005.service       6.983s NetworkManager-wait-online.service       6.396s apt-daily-upgrade.service       3.819s udisks2.service       3.710s NetworkManager.service       3.630s snapd.service       3.331s apt-daily.service       3.106s accounts-daemon.service       3.046s ModemManager.service       2.951s plymouth-start.service       2.678s grub-common.service       2.364s snap-ubuntu\x2dmate\x2dwelcome-169.mount       2.342s snap-core-4917.mount      mount: /: /dev/sda2 already mounted on /.   I would like to delay it until after my computer has booted up to speed up my boot time.

I have Ubuntu 16.04 installed on my Laptop alongside windows. In systemd-analyze blame a service called 'dev-sda7.device' is taking too much time. How to resolve this problem or should i disable it ?  Result of systemd-analyze time  Startup finished in 4.207s (firmware) + 4.576s (loader) + 3.466s (kernel) + 33.899s (userspace) = 46.149s   Result of  systemd-analyze blame           16.326s dev-sda7.device          12.859s ufw.service          11.263s systemd-tmpfiles-setup-dev.service           7.935s NetworkManager-wait-online.service           3.203s keyboard-setup.service           2.736s vboxdrv.service           2.467s accounts-daemon.service           2.349s apache2.service           2.239s NetworkManager.service           2.163s ModemManager.service           1.963s lightdm.service           1.843s nmbd.service           1.749s samba-ad-dc.service           1.599s systemd-fsck@dev-disk-by\x2duuid-B053\x2dA56B.service           1.367s thermald.service           1.127s polkitd.service           1.112s systemd-journald.service           1.066s teamviewerd.service           1.007s udisks2.service            975ms apparmor.service            926ms plymouth-start.service   Result of cat /etc/fstab  # /etc/fstab: static file system information. # # Use 'blkid' to print the universally unique identifier for a # device; this may be used with UUID= as a more robust way to name devices # that works even if disks are added and removed. See fstab(5). # # <file system> <mount point>   <type>  <options>       <dump>  <pass> # / was on /dev/sda7 during installation UUID=493cc833-193e-435d-840a-b862ca367fba /               ext4    errors=remount-ro 0       1 # /boot/efi was on /dev/sda2 during installation UUID=B053-A56B  /boot/efi       vfat    umask=0077      0       1 # swap was on /dev/sda6 during installation UUID=a49f56b1-53c3-4eaf-9460-0a221e59957a none            swap    sw              0       0


For example see the following code snippet:          Map<String,String> unsafemap=new HashMap<>();         unsafemap.put("hello",null);         unsafemap.put(null, null);         unsafemap.put("world","hello");         unsafemap.put("foo","hello");         unsafemap.put("bar","hello");         unsafemap.put("john","hello");         unsafemap.put("doe","hello");          System.out.println("changing null values");                  for(Iterator<Map.Entry<String,String>> i=unsafemap.entrySet().iterator();i.hasNext();){                      Map.Entry<String,String> e=i.next();                      System.out.println("key : "+e.getKey()+" value :"+e.getValue());                     if(e.getValue() == null){                          //why is the below line not throwing ConcurrentModificationException                         unsafemap.put(e.getKey(), "no data");                          //same result, no ConcurrentModificationException thrown                         e.setValue("no data");                         }                      //throws ConcurrentModificationException                     unsafemap.put("testKey","testData");                  }                 System.out.println("---------------------------------");                 for(Map.Entry<String,String> e :unsafemap.entrySet()){                     System.out.println(e);                 }   Modifying the map during iteration always results in an exception, if not done using the iterator e.g. iterator.remove(). So obviously adding a new value during iteration is throwing the exception as expected but why is it not thrown if the value of a particular key/value pair is modified?

Say I am iterating over a Map in Java... I am unclear about what I can to that Map while in the process of iterating over it. I guess I am mostly confused by this warning in the Javadoc for the Iterator interface remove method:     [...] The behavior of an iterator is unspecified if the underlying collection is modified while the iteration is in progress in any way other than by calling this method.   I know for sure that I can invoke the remove method without any issues. But while iterating over the Map collection, can I:   Change the value associated with a key with the Map class put method (put with an existing key)? Add a new entry with the Map class put method (put with a new key)? Remove an entry with the Map class remove method?   My guess is that I can probably safely do #1 (put to an existing key) but not safely do #2 or #3.  Thanks in advance for any clarification on this.


I'm excited by the sound of Blender 2.80, but I'm reluctant to abandon 2.79 as I've spent so long learning it very slowly, (I'm 63). Can I load 2.80 without destroying 2.79? Will the blend files I made in 2.79 work in 2.80? PS: I don't expect to be able to run both at the same time.

I have Blender 2.79 on my Windows 7 system.    I want to install 2.8, but I'm completing some courses which use 2.79.  If I install 2.8, will it automatically replace 2.79?  Will they interfere with one another?  Thanks.


I want to replace a string found in a file with another string, but both have a special character (in this example it is a . char) e.g 1.0 and 2.0.  So this is the command currently used:  sed -i 's/1\.0/2\.0/g' /home/user1/file1.txt   What if 1.0 and 2.0 were saved in variables $i and $j? where i has the value 1.0 and j has the value 2.0, how can I still replace i with j?

I have a text file containing this:  0 0 -1 0 1 0 0 0 0 -1 0 0 1.5 0.0 1.0 1   and I want to replace the 1.0 to 2.0 so that it becomes:  0 0 -1 0 1 0 0 0 0 -1 0 0 1.5 0.0 2.0 1   so I use this command:  sed -i 's/'1.0'/'2.0'/g' /home/user1/file1.txt   or this:  sed -i 's/1.0/2.0/g' /home/user1/file1.txt   but this is the result that appears and I don't know how to fix this:  0 0 -2.0 2.0 0 0 0 -2.0 0 1.5 0.0 2.0 1


I'm experiencing an extremely unpleasant crackling noise whenever I try to use headphones. Even when I simply perform a sound test and "front left/right" is announced there is a great amount of crackling on the side that isn't being tested.  This has meant that my headphone jack is practically rendered useless because it is very uncomfortable to use headphones at any volume.  System: HP OMEN (15-ax003na) Card: HDA Intel PCH Chip: Realtek ALC295   Anyone else have this hardware or similar and experience this?  I've tried both of the solutions offered on  thread for 12.04 and it didn't make a difference.

After installing 12.04 Precise on my PC, I have encountered some problems with audio output which did no occur when using Lucid.   The sound is choppy and distorted in the lower tones. As I have no experience in setting/testing and doing anything with Audio Devices, I need help even to diagnose the problem.  Update  $ sudo lshw -c multimedia    *-multimedia           description: Audio device        product: Radeon X1200 Series Audio Controller        vendor: Hynix Semiconductor (Hyundai Electronics)        physical id: 5.2        bus info: pci@0000:01:05.2        version: 00        width: 64 bits        clock: 33MHz        capabilities: pm msi bus_master cap_list        configuration: driver=snd_hda_intel latency=32        resources: irq:19 memory:fdafc000-fdafffff   *-multimedia        description: Audio device        product: SBx00 Azalia (Intel HDA)        vendor: Hynix Semiconductor (Hyundai Electronics)        physical id: 14.2        bus info: pci@0000:00:14.2        version: 00        width: 64 bits        clock: 33MHz        capabilities: pm bus_master cap_list        configuration: driver=snd_hda_intel latency=32        resources: irq:16 memory:fe024000-fe027fff   Update 2  It has something to do with the volume. If the audio is quiet it is not choppy, if the sound is loud then it begins to be choppy.


Show if the following series $$ \sum_{n=2}^{\infty}\frac{(-1)^n}{(-1)^n + n} $$ are absolutely convergent, conditionally convergent or divergent. I think I succeeded in proving that it is NOT absolutely convergent. My attempt: $$ \sum_{n=2}^{\infty}\left|\frac{(-1)^n}{(-1)^n + n}\right| = \sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|} $$ Then we have $$ \lim_{n\to\infty}\frac{\frac{(-1)^n}{\left|(-1)^n + n\right|}}{\frac{1}{n}} = \lim_{n\to\infty}\frac{n}{(-1)^n+n} = 1 $$ Therefore $$ \sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|}\ \text{is convergent}\ <=> \sum_{n=2}^{\infty}\frac{1}{n}\ \text{is convergent.}\ $$ The last series are NOT convergent, so our series are NOT absolutely convergent. I'm still stuck on proving if it is conditionally convergent or divergent. intuitively, I think It's divergent, but I need a formal proof. It crossed my mind that I can show $a_n = \frac{1}{(-1)^n +n}$ is NOT monotonously decreasing so It's divergent(Leibniz). But I think Leibniz's theorem doesn't work in this direction. So I'm still stuck. Looking forward for any ideas!

How do I check the series $\sum_{n=2}^{\infty} \frac{(-1)^n }{n+(-1)^n}$ for absolute convergence/conditional convergence ?  How can I solve this problem?


Q: Is there a way to alter a method of an existing object in Python (3.6)? (By "method" I mean a function that is passed self as an argument.)    Example  Let's say I have a class Person having some very useful method SayHi():  class Person(object):     Cash = 100     def HasGoodMood(self):         return self.Cash > 10      def SayHi(self):         if self.HasGoodMood():             print('Hello!')         else:             print('Hmpf.')  >>> joe = Person() >>> joe.SayHi() Hello!   As you can see, the response of the person depends on their current mood computed by the method HasGoodMood(). A default person has good mood whenever they have more than 10$ cash on them.  I can easily create a person who does not care about the money and is happy all the time:  >>> joe.HasGoodMood = lambda: True >>> joe.SayHi() Hello! >>> joe.Cash = 0 >>> joe.SayHi() Hello!   Cool. Notice how Python knows that when using the original implementation of HasGoodMood, it passes silently self as the first argument, but if I change it to lambda: True, it calls the function with no arguments. The problem is: What if I want to change the default HasGoodMood for another function which would also accept self as a parameter?  Let's continue our example: what if I want to create a greedy Person who is only happy if they have more than 100$ on them? I would like to do something like:  >>> greedy_jack = Person() >>> greedy_jack.HasGoodMood = lambda self: self.Cash > 100 TypeError: <lambda>() missing 1 required positional argument: 'self'   Unfortunately, this does not work. Is there some other way to change a method?    Disclaimer: The above example is just for demonstration purposes. I know that I could use inheritance or keep a cash threshold as a property of the Person. But that is not the point of the question.

I've read that it is possible to add a method to an existing object (i.e., not in the class definition) in Python.   I understand that it's not always good to do so. But how might one do this?


After visiting some shady sites like tpb and clicking a link  my phone just reset on it's own and now my keyboard reset to default, I can't access the notifications dropdown, or use the return to desktop button.  I've read that it shouldn't be too big of a hassle to reflash it, which I am assuming would get rid of well-buried rootkits.  Is an official pure-android image available for Motorola phones still currently in retail? What's the best resource for doing this through linux adb?

There are several reasons to be looking for a ROM:   Upgrading or downgrading your stock ROM Looking for firmware from a different region/language Flashing a custom ROM for different features, removing bloat, etc.   But where can one find ROMs in the first place?  What sites are likely to have the ROM I need?  ; please add to the wiki answer instead of posting separate answers.  See also:


I just edited a typo in  which was rightfully closed as a duplicate. When I did, I saw that I could edit the automatically inserted text, possibly removing it. I didn't try to do so, but, should I be allowed to edit that text?  I'm going to post another question to test this.  EDIT: Yes, I was able to delete the automatically added text which linked my duplicate question to the "real" question... isn't that... not good? I guess you'd have to be pretty stupid to delete that text from your own question, but...

I've noticed the following happen a few times:   A user posts a question. The question is closed as a duplicate and the usual "Possible Duplicate" links are added at the top. The user promptly removes the duplicate links, even though the question is still closed as a duplicate. (excessive use of list items =P)   Now, I could understand this behavior if there was some dispute about whether or not the question was a duplicate, and the OP were to remove the links and post reasons why they disagree with the closing. However, I've noticed this happening without any attempt to defend the question (such as ).  In my opinion, no one should be allowed to remove the "Possible Duplicate" links as long as the question is still closed as a duplicate. The links should remain for the benefit of the community, so they can navigate to the duplicate questions and either post answers there or read them over and judge if the closing was actually fair. Once/if the question is reopened the links could be edited out if one so chooses.  I guess this is part discussion, part feature request:  Could the "Possible Duplicate" links remain un-editable for as long as the question is still closed?


When trying to publish questions on Stack Exchange sites I am seeing the follow behavior after entering the question tags and pressing Enter. The Post Question button is disabled and not clickable. If I press Enter one more time, the question is published. But the behavior is buggy and inconsistent . Sometimes it shows a popup asking if I want to discard the draft, and since I was used to pressing enter twice, I lose the content that I entered.     As per   Hitting Enter on the tag box should take you to the Review Question step instead of submitting the post. Thanks for the report!  But currently the button is disabled or the discard draft appears and content is lost.

I've noticed this twice in the last ~2 weeks or so. The "post your question" button can remain disabled for some reason even when your question is shown as "ready to publish". I.e., you've gotten to the "review" page after the initial edit. If I remember correctly, one case was on Stack Overflow, and the other was on . I'm not sure what exact scenario leads up to this. I haven't paid attention to the button until the last step when I'm ready to click it. The workaround has been to reload the page. (I copied my content for safe-keeping first.) This works, and the button then shows up as active. Afterwards, however, on both occasions, it forced me through the "are you a human" verification.  Update - this question itself did not exhibit this, nor did it do the .  Update - I'm using Chrome on Windows and am not sure what version it was when this happened.  It has  installed, but there aren't any scripts for any Stack Exchange sites.  It has  installed.  The virus scanner on the PC I probably was using was


i created custom page template where i inserted followed JQuery script.  <script type="text/javascript"> $( document ).ready(function() { console.log( "ready!" ); }); </script>   But it wont work. I get Object expected error.  Jquery is inserted on page header so it should be loaded.  Why is that not working?

I know that jQuery is loaded, because I can switch out the $ for 'jQuery' and everything behaves as expected, but this will be a messy script if I can't fix this  This script:  jQuery(document).ready(function(){     $("ul.vimeo_desc_feed li a").click(function(){         alert($(this).attr('href'));         return false;     }) });   Produces the error $ is not a function  This script:  jQuery(document).ready(function(){     jQuery("ul.vimeo_desc_feed li a").click(function(){         alert(jQuery(this).attr('href'));         return false;     }) });   works fine.


I have 2 cameras. It's working perfectly when I play the animation in preview.  The first camera works perfectly. When I press F12, it's rendering:   However, after the frame goes to camera 2, Blender shows this:    I already looked for a solution on the interned but could not find one (maybe I was not using the right keywords).

This is meant to solve this problem once and for all. What is every reason an object isn't showing up in a render or viewport in when using the cycles render engine?


Local variable oopsGlobal, gives an error when used outside of function fun1 but the other variable name defined in the same way does not give an error and is accessible outside of the function?   Am I missing something here?    function functionName() {   // variables created without Var always global scope   var name = "sudeep";   console.log("myfirst function"); }  function functionName1() {   // variables created without Var always global scope   console.log("Local variable with var:" + name); } // Declare your variable here var myGlobal = 10;  function fun1() {   // Assign 5 to oopsGlobal Here   var oopsGlobal = 5; //local scope to variable when accessed outside gives error so working fine }  functionName(); functionName1(); console.log(name); // **does not give error at all** fun1(); console.log(name); //**does not give error at all** console.log(oopsGlobal); //give error so works fine

What is the scope of variables in javascript? Do they have the same scope inside as opposed to outside a function? Or does it even matter? Also, where are the variables stored if they are defined globally?


I understand evaluation for cos($\pi$/8), but I cannot figure it out how to solve cos(2$\pi$/5).  Can anyone please show it?

I would like to find the apothem of a regular pentagon. It follows from   $$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$  But how can this be proved (geometrically or trigonometrically)?


def gcd(x, y):     (x, y) = (y, x) if x > y else (x, y)     for factor in range(x, 0, -1):         if x % factor == 0 and y % factor == 0:             return factor   What is the use of the code:  (x, y) = (y, x) if x > y else (x, y)   Isn’t it enough to get the factor through the rest of the code? I have tried to get rid of the first line and the code went wrong, but I still can’t understand it.

If Python does not have a ternary conditional operator, is it possible to simulate one using other language constructs?


I'm referring specifically to city names, though I suppose this applies to other things named after saints.  I'm looking at the Wikipedia pages for my "sources of truth".  is fairly split, with 170 instances of "Saint Paul" and 40 instances of "St. Paul".  , however, is almost always St. Louis, with only 5 instances of "Saint Louis" and 460 instances of "St. Louis".  I'm looking for any semblance of rhyme or reason here, but I can't seem to find any documentation on this.  I thought it might be official names, but it appears  is officially "City of Saint Augustine", but the text for the page only includes that 4 times (compared to 139), and none of those are in the actual text.

My first question here, so I'm not sure if this OT or not, so please bare with me.  Background: I don't live in US. I'm working with some data that I'm pulling from . Sometimes, I'm getting a city name as St. John and sometimes as Saint Ann.   So my question is, is there an accepted way on how people write a US city names which starts with Saint or is this just some random thing (or just specific to AdWords maybe)?  I'm asking this, because if for example I'll get both St. John and Saint John in the same file, should I consider them the same or should I be suspicious?


If $f$ is a function of arity $n$ and $t_i$ are terms and $u$ is any string of symbols over some specified language $\mathcal{L}$, then can $$ f t_1 t_2 \cdots t_n ~u $$ be a valid syntax of a term My understanding says that it can’t be a valid syntax because even if $u$ is some variable or constant then also it won’t be operated by $f$ because of arity. And if $u$ involves some other function $g$ then it might be against unique readability to not include the functions at left most of the string.

I sense that the following simple argument is invalid, but cannot figure out why. Base Case:No variable or constant is an initial segment of a term, since the only terms that begin with variables or constants have length 1, and so lack initial segments. Inductive Case:Suppose s is an initial segment of a term, t, where s is $f t_1 t_2 \cdots t_n$ for terms $t_i$. Then t is su, i.e., the string, s, followed by the string, u, for some non-empty string u.  But, then, t is not a term, since an n-ary function symbol followed by n terms, followed by a non-empty string is a not a term. Hence, no term is an initial segment of any other. I did not make use of the inductive hypothesis, that no ti in s is the initial segment of any term.  This is what makes me think I cheated somehow.  Assuming that I did, I wonder if someone could show me how to prove the conclusion by means of induction on the complexity of terms without assuming that the tis in t are not initial segments of any terms (i.e., by merely assuming that the tis in s are not). Any feedback would be greatly appreciated!  Thanks in advance.


I am learning blender and can't understand why edge loops "get broken" (May be a specific name for it).  Taking the picture below as example:   When I try to create an edge loops it stops on the top of the front face (marked edges).  I don't understand what make edge loops not being possible to be continued. I mean I expect the edgle loop to go all the way like the line I draw in red.  I could understand if there are n-gons or double vertices, but what's wrong in the mes below?   Just for info, the left object is just the "preview" of an array modifier used on the right object.

Alright so I've seen some postings about loop cuts and done some research. I am very new to blender and modeling overall. I am trying to loop cut this concrete barrier. Now I'm following a tutorial and kind of did my own thing in some areas. I cleaned it up following some instructions in another answer here on stack exchange.  I can't for the life of me figure out why I can loop cut around the whole bottom:   ...but not the whole middle:   I see this marked as answered by another question. That question refers to ngons and inside faces. Now once again I'm totally new to this. I looked on the inside. I don't believe I have inside faces or any ngons by the terms I've read online in my research.  Here is a wireframe picture and the inside view:     Maybe I'm missing something the answers on other posts are saying. If so and you know what it is I'd be very appreciative of what it is.


Is it possible to raise the thieves guild without doing the Nightingale questline? Because I have been doing a lot of quests for Vex and Delvin, but it seems I can't get the special quests. Is there anything I can do to change that, or do I have to deliver the lock pick thingy back? And does it matter what quests you do, if you don't have to deliver the lock pick back?

I completed the quest where you get the Skeleton Key and I don't really want to continue the chain until I've done a lot of the other stuff in the game, since the key is so useful. Brynjolf said before that I would become leader of the guild after everything was over.  Can I become the leader anyway without finishing the Nightingales quest chain? Are there any bonuses for being the leader?


I have a keyboard that I really like, but the driver () is only for Windows 95/98. I am running Windows 10.  I followed the tutorial here    by edit the registry value of the executable to ~ RUNASADMIN WIN95 but that doesn't work either.  Am I out of luck?

I was just told that PS2 Keyboards aren't compatible with windows 10.  Is this true.  I was under the impression that they should work fine?


I set up CentOS yet.   it's no network connection. eth0 is not device list.   I take a message network is unreachable while ping 192.168.0.1    How can I do reachable network?

When I run ifconfig -a, I only get lo and enp0s10 interfaces, not the classical eth0  What does enp0s10 mean? Why is there no eth0?


If I have a generic base class   class Base<T> { }   and I derive from it  class Derived : Base<SomeReallyLongConcreteTypeName>  {     // Can I use the type name T here rather than SomeReallyLongConcreteTypeName }    is it possible to use the base type generic type parameter name in the body of the derived class? In C++ I can add  using X = T;   into the base type and use X in the derived class but I don't think type aliases can do this in C# so is there another way?

I want to create an alias for a class name. The following syntax would be perfect: public class LongClassNameOrOneThatContainsVersionsOrDomainSpecificName {    ... }  public class MyName = LongClassNameOrOneThatContainsVersionOrDomainSpecificName;  but it won't compile.  Example Note This example is provided for convenience only. Don't try to solve this particular problem by suggesting changing the design of the entire system. The presence, or lack, of this example doesn't change the original question. Some existing code depends on the presence of a static class: public static class ColorScheme {    ... }  This color scheme is the Outlook 2003 color scheme. i want to introduce an Outlook 2007 color scheme, while retaining the Outlook 2003 color scheme: public static class Outlook2003ColorScheme {    ... }  public static class Outlook2007ColorScheme {    ... }  But i'm still faced with the fact that the code depends on the presence of a static class called ColorScheme. My first thought was to create a ColorScheme class that I will inherit from either Outlook2003 or Outlook2007: public static class ColorScheme : Outlook2007ColorScheme { }  but you cannot inherit from a static class. My next thought was to create the static ColorScheme class, but make Outlook2003ColorScheme and Outlook2007ColorScheme classes non-static. Then a static variable in the static ColorScheme class can point to either "true" color scheme: public static class ColorScheme {     private static CustomColorScheme = new Outlook2007ColorScheme();     ... }  private class CustomColorScheme  {     ... }  private class Outlook2008ColorScheme : CustomColorScheme  {     ... }  private class Outlook2003ColorScheme : CustomColorScheme  {    ... }  but that would require me to convert a class composed entirly of readonly static Colors into overridable properties, and then my ColorScheme class would need to have the 30 different property getters thunk down into the contained object. That's just too much typing. So my next thought was to alias the class: public static ColorScheme = Outlook2007ColorScheme;  But that doesn't compile. How can I alias a static class into another name?  Update: Can someone please add the answer "You cannot do this in C#", so I can mark that as the accepted answer. Anyone else wanting the answer to the same question will find this question, the accepted answer, and a number of workarounds that might, or might not, be useful. I just want to close this question out.


Has the database schema for Stack Overflow been made publicly available? If not, could it be?  If not the whole schema, how about a list of the tables used?  I want this as a learning resource.

Stack Exchange releases "data dumps" of all its publicly available content roughly every three months , and also makes that information queryable over the Internet at the  (SEDE) (updated weekly, on Sunday). Over time, as new features and other bits of data are introduced to Stack Exchange sites, the "anatomy" of the data dump is likely to change.


Using the zsh shell, I observe the same output when I execute ls > a.txt | wc and ls | wc. Could someone confirm my understanding regarding the same. In the first command the STDOUT is redirected to the file a.txt by usage of > operator and by making use of the pipe STDOUT (which currently points to the file) is redirected to the write end of the pipe, wc command reads from the pipe thus producing the same effect as ls | wc But my doubt over here is that since I see the same output this means the output of ls is written to both the pipe and a.txt (I checked the file). Why should the output be written to both, as in when ls is executing the file descriptor STDOUT would be either pointing to the write end of the pipe or to the file, hence I shouldn't  be getting output in both. Assuming I am in a directory with file which contains 2 files a.txt and sample.txt then  Running both the aforementioned commands gives output 2 2 17  a.txt contains   a.txt   sample.txt

Can somebody explain what is going on in this command:  echo foo >&2 | grep foo   Zsh (5.0.2) prints foo twice. One seemingly to stdout and one to stderr, since running these commands:  { echo foo >&2 | grep foo } >/dev/null   and:  { echo foo >&2 | grep foo } 2>/dev/null   both just print foo once.


The command which cd prints nothing. Same for whatis cd and whereis cd. How can I find out if cd is an alias, function, or bash built-in? I'm using Ubuntu 12.04.

I tried which cd and it didn't give a path but instead returned the exit code 1 (checked with echo $?). The coreutil cd itself is working, so the executable should be there, right? I also ran a find for cd, but there was no executable file shown. How is it implemented then?  Update:  I don't know if I should ask this in another post but since I think it is good here, I'm expanding (?) the post... So the answer was actually quite simple, there's no executable for that — because it's a builtin — But I've found some builtins (bash shell in Fedora) have the executable files! So builtin -> no executable isn't right I suppose? Maybe an answer explaining what builtins actually are (builtin commands?), which actually is the matter here, rather than focusing more on cd... Some good links posted previously indicate that builtins are not programs... so what are they? How do they work? Are they just functions or threads of the shell?


I have been using Windows 8 for quite some time now. Recently I upgraded it to Windows 10. I want to install Ubuntu as well in my computer. Could someone give me a step by step procedure I should follow to install it.

I'm absolutely new to Linux. I would like to know how to install Ubuntu alongside the pre-installed Windows 8+ OS.  Should I do it with Wubi, or through the Live USB/DVD?  What steps do I need to take to correctly install Ubuntu?


I'm a Chinese living in NYC. How do Americans call these things? Are they all named "trash can"?  And If I want to drop something, like an empty bottle, what is the right way to say that? "drop garbage"? "litter" ? "throw away garbage" or else?  Thanks!

There are so many ways to call these containers for waste. (correct me if some of them might sound weird/unnatural to use)     garbage can, trash can, rubbish can, pedal can, garbage bin, trash bin, rubbish bin, pedal bin   I'm totally confused. I know some of them might mean the same thing.   But please tell me which is which based on the following pictures by order? Please also let me know where you are from. (US, UK, Canada, Australia, etc...)  1:   2:   3:   4:   5:


Why $n\mathbb{Z} \times n\mathbb{Z}$ is not an integral domain?

Suppose that $R = S × T$ is a direct product of rings with $S$ and $T$ each having at least two elements. Prove that $R$ has zero divisors.


I have booked a Tatkal ticket from Nagpur to Pune with the Nagpur-Pune Express, but I have to board from Wardha. I will be traveling on the 9th of May 2016. How can I change the boarding station?

I have booked a ticket on a train from Mumbai to Chennai and boarding at Pune. Since my schedules have changed, I will be in Mumbai on that day. Can I board the same train in Mumbai in my own ticket and claim the seat/berth?


I cloned a row from a table, I want to remove the last cloned row by clicking the delete button.  Here is my script  $("#btnAddMoreUser").on("click",function(){     var newClass='newClass';     var $tableBody = $('#tblAddUser').find("tbody"),         $trLast = $tableBody.find("tr:last"),         $trNew = $trLast.clone().addClass(newClass);         $trLast.after($trNew);  });     $(".tr_clone_remove").on("click",function(){ //Once remove button is clicked     $(".newClass:last").remove(); //Remove field html     x--; //Decrement field counter });    Below is my fiddle

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


As I understand it, if one has a complete knowledge of the state of a quantum system (insofar as one knows the statistical distributions of all the observables associated with the state) then one can represent it as a state vector (or ket) in an associated Hilbert space, and we say that the system is in a pure state. Furthermore, given an (orthonormal) basis for the Hilbert space, one can express this state vector as a linear combination of the basis vectors. This is usual carried out through the use of an eigenbasis induced by an operator acting on the Hilbert space, representing some observable. We say then that the state is in a quantum superposition.   What confuses me is how this differs from a mixed state? I understand that the situation is at least somewhat different since a mixed state arises when we have a lack of knowledge about the state (i.e. we lack all the possible information we could, in principle, have about it, that is, the statistical distributions of all the observables associated with the state). Hence, we must consider a statistical ensemble of possible pure states that the system could be in, each having an associated probability. This is a so-called classical probability as it does not stem from the intrinsic probabilistic nature of a quantum system, but rather the fact that we lack all the knowledge that we could possibly have about the system.  Is it simply, that in the case of a pure state, even though we know the statistical distribution of the observables associated with this state, we do not, a priori, before measuring the given observable, Know which eigenstate the quantum system is in, and hence must consider it to be in a quantum superposition of the available eigenstates? (In this case, it would be a so-called quantum probability, since such an uncertainty does not arise from lack of information about the state of the system, but is intrinsic in the quantum nature of the system).  Apologies for the long-windedness of this post, I just thought I'd write evergthing down on my thoguhts about it, and hopefully someone can correct/ explain it to me.     Edit: I think that perhaps my confusion stems from how to interpret a quantum superposition of states. How should one interpret this physically? (if I have an understanding of this perhaps it'll clear up things a little).

According to , if a system has $50\%$ chance to be in state $\left|\psi_1\right>$ and $50\%$ to be in state $\left|\psi_2\right>$, then this is a mixed state.  Now, consider the state  $$\left|\Psi\right>=\frac{\left|\psi_1\right>+\left|\psi_2\right>}{\sqrt{2}},$$ which is a superposition of the states $\left|\psi_1\right>$ and $\left|\psi_2\right>$. Let $\left|\psi_i\right>$ be eigenstates of the Hamiltonian operator. Then measurements of energy will give $50\%$ chance of it being $E_1$ and $50\%$ of being $E_2$. But this then corresponds to the definition above of mixed state! However, superposition is defined to be a pure state.  So, what is the mistake here? What is the real difference between mixed state and superposition of pure states?


I am trying to develop a Python Add-In for ArcGIS 10.3 and finding debugging pretty laborious. At the moment when I make a change to my python script I have to do the following:   Close ArcMap Delete the "Python_Addins.esriaddin" file in my Default installation directory double-click the "makeaddin.py" file to create another "Python_Addins.esriaddin" file Install the add-in Restart ArcMap and start the project Test the tool   Surely, this can´t be the only way to test a python Add-In in ArcMap.

I have been developing a few python addins and I'm just wondering what is the best way to re-compile the addin anytime I edit the script.  Currently, if I make an edit to the script, it seems like I have to   Close ArcMap Go to My addin folder and double click the 'makeaddin' python file Double click the 'Python Addin File Restart ArcMap for the changes to come into effect   If I don't do these steps, then the edits aren't recognized. This gets pretty annoying and time consuming when I want to place maybe a simple print line somewhere or I do a small edit just to test a result.  Is there a better way of doing this?


How can I get the indices of a sum sign to appear besides the sum in math mode?  I know that this is default for inline mode, how can I set this for math mode locally (preferably w/o additional packages etc.)?

I'm very familiar with how symbols display differently in inline mode (with $ ... $) vs. how display math shows them (with \[ ... \] or $$ ... $$).  Two examples would be with limits  $\lim_{n\rightarrow \infty}f(x)$   vs.  \[ \lim_{n\rightarrow \infty}f(x) \]   and with sums  $\sum_{n=1}^{x} n^2$   vs.  \[ \sum_{n=1}^{x} n^2 \]   My Question is:     How do you display these things in inline mode (with all the nice inline formatting) as if they were in display mode ?   The inverse -- how to display things in display mode as if they were in inline mode -- is also of interest.


How can i prove the following inequality  $$\ \min{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)}\leq\frac{a_1+a_2}{b_1+b_2}\leq\max{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)} $$ for any real numbers $a_1,a_2$ and positive numbers $b_1,b_2$.  Can anyone give me some hint or reference for the proof of this inequality?

We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$?


I'm trying to renew a command in terms of its old definition... something similar to this:  \renewcommand{\vec}[1]{\vec{\mathbf{#1}}}   But this seems to send the interpreter in an infinite loop. How do I do this properly?

I'd like to "add to" an existing command by writing something like  \renewcommand{\somecommand}{\begin{something}\somecommand\end{something}}   but (not surprisingly) I get errors about being "too deeply nested".  Is there a way to take an existing command, add to it (by appending or prepending additional commands or text), and then have that "augmented" command replace the existing command?


I cannot find the fix for these problems (if there is one) anywhere. Basically whenever I start it up it gets stuck at the same spot on the loading screen (unless I've just reinstalled it) and it takes forever to load. If it does end up loading, nothing will load on the map. No pokestops, no gyms, it shows no pokemon nearby, and no pokemon encounters. The only time the app works is if I just reinstalled it, but it always ends up doing the same thing once I've closed the app. It also makes me log into my google account every single time. Is my game just fucked or is there a way to fix this?

Pokemon GO isn't working at all for me right now. I'm in one of the following circumstances and can't do anything:   My game is displaying a "Try Again" message I'm stuck on the Loading Screen When I sign in I get a message that the servers are experiencing problems. I see the map but nothing is loaded, and my team color is wrong The "loading" Pokeball just ends up spinning and I can't catch or interact with things When I try to catch a Pokemon, the Pokeball just sits on my screen doing nothing Gym opponents will drop to 1 HP and won't take any more damage or attack When I open a pokestop, the pokestop image and the name of the pokestop does not load.   What's causing this? Is there a way to fix it?


$ sudo apt-get install python2.7 Reading package lists... Done Building dependency tree Reading state information... Done You might want to run 'apt-get -f install' to correct these:  The following packages have unmet dependencies: python:i386 : Depends: python2.7:i386 (>= 2.7.5-1~) but it is not going to be installed               Depends: python-minimal:i386 (= 2.7.5-5ubuntu3) but it is not going to be installed               Depends: libpython-stdlib:i386 (= 2.7.5-5ubuntu3) but it is not going to be installed python2.7 :   Depends: python2.7-minimal (= 2.7.12-1~trusty1) but it is not going to be installed               Depends: libpython2.7-stdlib (= 2.7.12-1~trusty1) but it is not going to be installed  E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution).   Earlier I had used sudo apt-get autoremove 'python.*' command and now I am getting installation and dependencies issues while downloading Python 2.7.  I need to download Python 2.7.

Occasionally, when I'm installing stuff, I get an error like the following:  Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  package1 : Depends: package2 (>= 1.8) but 1.7.5-1ubuntu1 is to be installed E: Unable to correct problems, you have held broken packages.   How can I resolve this?


There is a c++ program:  # include <iostream>  using namespace std;  class base { public:     base()     {         cout<<"base"<<endl;         f();     }     virtual void f() {         cout<<"base f"<<endl;     } };  class derive: public base {     public:         derive()         {             cout<<"derive"<<endl;             f();         }     void f() {         cout<<"derive f"<<endl;     } };  int main() {     derive d;     return 1; }   and it outputs:  base base f derive derive f   I am wondering why base f appears?  I quess in base the constrctor expands to:          cout<<"base"<<endl;         this.f();   But this should point to derive so why base f is print out?

Why the following example prints "0" and what must change for it to print "1" as I expected ?  #include <iostream> struct base {    virtual const int value() const {       return 0;    }    base() {       std::cout << value() << std::endl;    }    virtual ~base() {} };  struct derived : public base {    virtual const int value() const {       return 1;    } };  int main(void) {    derived example; }


Let $\{a,b\}\subseteq \Bbb N$.  Is there a special name or notation for sets of this type, for example $\Bbb N^{2\ge}$?  Any subset size may be used, but the specific size and denoting that order does not matter is part of my interest in this notation, i.e., subsets of size $n$ of set $X$.  Particularly inspired by this question: .

How would one denote the set of all subsets of $A$ which have size $2$?  Would  $$\binom{A}{2}$$  be a good idea?


My computer has 1GB ram and a very end low graphics card, but I don't know how much power it has.  Unity is very slow on my computer, and I saw that there are many desktop environments which are much lighter and faster, even for a slow computer.  So, what's the lightest desktop environment?

For a given hardware configuration, how do I find out if Ubuntu will run on it? What considerations should I take into account when choosing an Ubuntu version and  such as:    with a lighter desktop than the usual Gnome and Unity  with the even lighter LXDE desktop   Obviously Ubuntu does not run on some processor architectures. So how do I go about choosing the right version and derivate. How can I find out the minmal system requirements?


For legal reasons I had to change my business name, and get a new site name and address (very long story).  I had to forward my old domain to this new one, but the old domain has like 200+ back-links, and the new site has zero. How does this impact SEO and what do I do in a situation like this? Do I start from scratch, or is there a way to capitalize on those back-links on the old domain name?

Suppose I have been using a domain, say www.bollywoodhatke.com for 3 years, and it has a good Alexa (hypothetical situation). It has backlinks and its has 7,000 pages indexed and 50,000 images indexed, but later there is a reason I am forced to change the domain name where the website is served.  In that case →  Would it be possible to tell Google that we are the same people with the same website, but the domain name changed?  I mean, can I somehow transfer the content to the new domain so that all the pages are indexed immediately, backlink juice is also transferred to the new domain, and all the pages indexed correctly?  I think moz.com  was semoz.com previously. They must have certainly done something.


Every time I turn my Mac on, after a while the app memory just runs out.      These are just from 10min apart, after working for a while. I need help figuring what those Installers are and why the kernel_task is taking so much memory. Those Google Chrome Helpers also look extremely unnecessary.  The only thing that 'solves' this is restarting the computer. How can I get rid of this cycle?

Mac Mini (Mid-2011), 16GB SSD, 16GB RAM, El Capitan (v10.11.6)  The Mac ran fine until I upgraded to Sierra. The latest Mac OS conflicted with most of the specialized apps I use every day, So I immediately returned to El Capitan.  Activity monitor shows that two "Installer/Installer.app" items are always running and are they major users of all of my available memory. Consequently, all my others apps become unresponsive.  I run Disk Utility, and the Installer.app show up there as .dmg files; but I cannot erase, delete, or unmount them. None of my 3rd party app can remove the installers.  My Mac Mini, which should be running super-fast (as it used to), now runs dreadfully and disgustingly slow. I a truly upset with Apple for this kind of garbage. To get their help: $$$.


Let $~f~$ be a function from $\mathbb{R}$ to $\mathbb{R}$ such that :  $~~\forall x \in \mathbb{R}~~ $  , $~~f(x) = f(2x) $   and $f$ is continuous at $~0$ .   prove that $f$ is a constant function .

I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker):  Let $f:\mathbb R\longrightarrow \mathbb R$ be a function which is continuous at $0$ and satisfies $f(x)=f(2x)\,\forall x\in\mathbb R$.  My question is how do I formally prove that for every $x<0$ $f(x)=c_1$, and for every $x>0$ $f(x)=c_2$, where $c_1,c_2\in \mathbb R$? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that $f(0)=c_1=c_2$ by the definition of continuity of $f(0)$ and thus $f(x)$ is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!


The recurrence relation for Derangement is as follows:  Let $D_n$ denote the number of derangements of a set $\{1,2,3,\ldots,n\}$. Then,  $$D_n=(n-1)D_{n-1}+(n-1)D_{n-2}.$$  Can someone give an intuitive explanation on how we come to find this result?

Give a combinatorial argument to prove that the number of derangements satisfies the following relation: $$d_n = (n − 1)(d_{n−1} + d_{n−2})$$ for $n \geq 2$.  I am able to prove this algebraically but not able to see the combinatorial example.


I'll make this short and hopefully I make sense.   Im trying to give a user ssh access, but, he/she must only be able to access /opt and the few subdirectories that live under it. However, here's where it gets tricky, I understand that if they are chrooted in /opt the other directories in there that aren't owned by root wont be accessible. That's a problem for me cause living inside there is /opt/Pentaho which you guessed it, is owned by the pentaho user and when you chown root:root that directory brings our Pentaho app to a grinding halt obviously. SO, is there a way for me to restrict an SSH user to /opt and be able to read and write files within the /opt/Pentaho directory or have I just lost my mind? Any insight would be great, thanks!

Is there a simple way to restrict an SCP/SFTP user to a directory? All methods that I've come across require me to set a chroot jail up by copying binaries, but I don't think that should be necessary.


I'm using  as a convenient way to keep my resource usage visible at all times. If you are not familiar with the product, it opens separate windows for memory, CPU, and disk usage with the taskbar button behaving as a bar graph. This looks great when the taskbar is set to "always combine, hide labels" but when labels are shown the meters are taking up valuable real estate. Is there a way to hide the labels or to force these icons to a fixed width while kepping the other open programs with full size labeled icons?

I run my taskbar buttons with the setting Combine when taskbar is full.   Q: Is there any way to force some of your pinned taskbar applications to only use the "icon" rather than the icon + label?  I am aware of the trick to force the size of the taskbar icons for all applications (detailed  and ), but there are some applications that I would like to only have the icon, and others where it is important to show icon and label together.


I need to install elmerfem and there's just no release file for focal fossa, which is the version I'm using. Is there any way to override that so that I can install releases meant for other ubuntu versions?

When updating, I get the following error message:   W: The repository 'http://ppa.launchpad.net/mc3man/trusty-media/ubuntu xenial Release' does not have a Release file.   Here, I find another statement on this error:   This recommends removing certain PPAs; and, I'm not sure if I should do that since it might mean not getting the updates that I need.    Is this what I should do?


What are common lower bounds for ${{2n} \choose {n}}$?    Edit:  I made a mistake in my original question.  It doesn't change my question but there is no reason for me to include the mistake.

How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?     Does anyone know any better elementary estimates?   Attempt.  We have $$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$ Then we have $$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$ So maybe, for the lower bound, we have $$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$ By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes $$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$ but this isn't $>\frac{2^n}{\sqrt{4n}}$.


I received an  for my  package. Certain commands in my package fail when the following code is compiled with LuaLaTeX (XeLaTeX works fine):  \documentclass{article} \usepackage{fontspec, csquotes, lilyglyphs} \MakeOuterQuote{"} \EnableQuotes  \begin{document} \wholeNote \end{document}   Upon investigation I found that there are some quotation marks in the package code, and I assume this is the culprit:  \ifluatex \def\lilyGetGlyph#1{\directlua{documentdata.fontchar "#1"}} \fi   (For context: this can be found here: )  So I assume that the quotation marks in the \directlua command are expanded to "smart quotes" before the Lua code can be executed.   Is this assumption right? Is there anything I can do about it? Or do I have to accept and document the incompatibility?

I ran into this with the package lilyglyphs but I'm not sure whose fault this error is.  When I compile the following with lualatex:  \documentclass{article} \usepackage{csquotes} \MakeOuterQuote{"} \EnableQuotes \begin{document} \directlua{print("hello")} \end{document}   I get an error saying   \directlua]:1: unexpected symbol near '\'. l.6 \directlua{print("hello")}    Now suppose that I do not directly write \directlua in my document, but I include a package that does it internally. Such as   \documentclass{article} \usepackage{fontspec, csquotes, lilyglyphs} \MakeOuterQuote{"} \EnableQuotes \begin{document} \wholeNote \end{document}   I have no knowledge why compilation fails, because the error now becomes   ! Missing number, treated as zero. <to be read again> \begingroup l.7 \end       {document}    From my user point of view I did nothing wrong, because I'm using the public API provided by the packages. The csquotes documentation specifically says:     Note that all characters are automatically checked for validity as they are allocated. This package will reject characters which are unsuitable as active quotes... In sum, the following   characters will be considered as reserved by this package:   A–Z a–z 0–9 . , ; : ! ? ' - # $ % & ^ _ ` ~ \ @ * { } [ ]    So my question is whose fault is this error?   My fault, the user's fault, because I should have known (but how?) that even though the quote is not listed as unsuitable in the csquotes manual, it actually is? The fault of the implementation or documentation of csquotes for failing to mention " as unsuitable? The fault of the implementation of the \directlua command because it doesn't prevent LaTeX catcode changes/active characters from impacting its argument? A shortcoming of TeX itself because of its global state based architecture?


I am running Ubuntu 14.04. When I run   sudo apt-get install gazebo2   apt-get complains it cannot find libsdformat(<=2.0)  apt-get doesn't provide libsdformat(<=2.0), so I have built libsdformat1.4 from the source.  How can I make apt-get find a dependency I have built myself?

Being an early adopter and using 11.04 daily builds I sometimes meet a situation when some packages require earlier versions of other packages than those I have installed. Usually newer (but refused) versions of needed packages provide all the functionality of their older versions.  If I use dpkg -i --force-depends-version to install dependent packages everything works fine, I've never seen a problem.  Instead of manually downloading and installing every package, can I just make the system pretend there is a specific package version installed when there is a newer? Or is there a --force-depends-version key equivalent for apt-get?


Why the field inside the conducting spherical shell is zero?  Professor Lewin said and I quote,"  there is NO charge inside the conductor  because any charge inside the conductor would make the electrons experience a force , the electrons will start to flow and they will kill the electric field."  But he says Electric field is zero inside the conductor and for that charge should be present to provide the electrons with force to cancel the field . Isn't it ?

I would like to clear things up: How exactly the electric field inside a conductor is zero? Let  a really "powerful" electric field be outside of it, how can the "few" charges in a conductor balance the field? From my point of view, it's the conductor surface which stops their movement. I'm obviously missing something fundamental here. Thanks in advance.


I know that there are functions in $L^2$ that are integrable but not continuous. Is there any function in $L^2$ that is not even integrable?

This exercise is in Stein's Real analysis.  Find a function such that $f$ is $L^1(\mathbb{R}^n)$, but not $L^2(\mathbb{R}^n)$ Find a function such that $f$ is $L^2$ but not $L^1$. Hint: consider $f(x)=|x|^{-a}$ on $|x|<1$ or $|x|>1$.  I was trying to use the hind to fin such functions. Let $f(x)=|x|^{-a}$ on $|x|<1$ and $f(x)=0$ otherwise. Since it is radial, $\int |f|dm_n = m_n(B(0,1))\int_0^1x^{-a}x^{n-1} dx$. Similarly $\int |f|^2dm_n = m_n(B(0,1))\int_0^1x^{-2a}x^{n-1} dx$. I could!'t find suitable $a$ to make this $f$ to be a desired function.. Help..


I am building a docker image based on Ubuntu 20.04 and one of the commands is: apt update && apt install -y python3-arrow python3-pip python3-paho-mqtt  As the installation proceeds, I am at some point requested to interactively choose the place I an in, when installing tzdata (which breaks the docker build): (...) Setting up libfakeroot:amd64 (1.24-1) ... Setting up libsasl2-modules-db:amd64 (2.1.27+dfsg-2) ... Setting up tzdata (2020d-0ubuntu0.20.04) ... debconf: unable to initialize frontend: Dialog debconf: (TERM is not set, so the dialog frontend is not usable.) debconf: falling back to frontend: Readline Configuring tzdata ------------------  Please select the geographic area in which you live. Subsequent configuration questions will narrow this down by presenting a list of cities, representing the time zones in which they are located.    1. Africa      4. Australia  7. Atlantic  10. Pacific  13. Etc   2. America     5. Arctic     8. Europe    11. SystemV   3. Antarctica  6. Asia       9. Indian    12. US Geographic area:   Is it possible to force a non-interactive installation of this dependency of one of the installed packages?

If I run:  sudo apt-get --yes install postfix    or  sudo bash -c 'yes | apt-get --yes install postfix'   an interactive prompt appears to configure postfix. I need to automate the installation of postfix (I can modify the configuration files after the install).  Is there some magic that lets me install postfix (and other package) without human intervention?


I use Lyx in Hebrew, which means regular text is written in Hebrew, but equations are of course written in English. I used to have an installation, which (I believe by accident during installation process) every time I entered any equation environment, my language was auto-set to English. Then, returning to regular text it was again auto-set to Hebrew.  This was very convenient, as it saved me the trouble of changing languages all the time, and of course there is zero use of Hebrew in math except Aleph for cardinality.  I now wonder whether there is a more conventional way to achieve this without trying to make random mutations to Lyx installation processes. Thanks!

Question  I want to write a book in Arabic (I have an Arabic keyboard, so I don't need a converter). By Arabic book I mean literally Arabic:    title, table of contents  etc    all in Arabic.   The equations, however, I want to write them in English/Latin.  Related    This link helped a lot but it was for Arabic texts in an English main document.    I already checked these two (and plenty other sites).


Say I have some code like this:     function a() {     console.log("A is all good");  }  function b() {     console.log("B throws an error");      throw new Error ("I am an error that comes from B");  }  function c() {     console.log("C is all good.");  }   function run() {     try {         a();          b();          c();      } catch(err) {         throw new Error ("An error occur in the run function.");      } }  run();      (Unfortunately Stack Overflow doesn't play very well with stack traces, so my snippet will be a little unclear).  The problem with this code is that when that error throws from run() I lose details of which function it was that cause that error to throw.   How can I include the stack trace of caught error into the thrown error?    Edit: - This question asks specifically about extending an error. While the dupe linked answer provides helpful details - it doesn't contain the specific question (Which I now have an answer for). I'd appreciate reopening the question so I can post my answer.

If I throw a JavaScript exception myself (eg, throw "AArrggg"), how can I get the stack trace (in Firebug or otherwise)?  Right now I just get the message.  edit: As many people below have posted, it is possible to get a stack trace for a JavaScript exception but I want to get a stack trace for my exceptions.  For example:  function foo() {     bar(2); } function bar(n) {     if (n < 2)         throw "Oh no! 'n' is too small!"     bar(n-1); }   When foo is called, I want to get a stack trace which includes the calls to foo, bar, bar.


I've started a long running process with nohup and redirected it's output to a log file  Example:  nohup ./bin/zkServer.sh start-foreground > /usr/local/myuser/logs/zkout.log 2>&1 &   I accidentally deleted the zkout.log file so now I can't see any output from the command.  I'm not interested in recovering the data, but I am wondering if there is a way to 'reattach' the stdout from my long running process to the file without having to kill/restart?  I tried just recreating /usr/local/myuser/logs/zkout.log but that doesn't work.  I also looked at this question:  but that is talking about recovering the content of the deleted file (which I don't care about). However, I do see the broken link in /proc/$procid/fd; could I just update that link to point to the new file?

I ran following commands in the order specified:  $ln a b $ls -i a b 523669 a 523669 b $rm -f a $ls -i b 523669 b   I concluded from this test that the command rm actually removes only the filename (a in this test) instead of the file, as the inode still exists and could be retrieved through another filename (b).  My question is, if a file is hard linked to only one filename, when rm is executed to the file, is the real file (i.e., the inode) removed completely? And if not, can a file inode be retrieved without a filename and only through the inode?


Why does  not split at every point encountered, even if run in an iterative fashion?  Following the suggestions of a thread here and several others, I made a set of points in order to split polylines at specified intervals. In order to do this, I used linear referencing to turn my lines into routes. Then, I exported the attribute table, divided each length by 5, and set points at multiples of 1/5 the length of the line. I used this as the event table to create a layer of events. So, each line has 4 equally spaced points along it, and I thus ultimately need the 5 segments separated by those points.  Curiously, though, when I try to do the second part "Split Line at Point," it seems that only one point is used on a given line at a time, and I'm not sure how it chooses the point (maybe order in its attribute table?) I could run the tool multiple times, but this doesn't seem to solve the problem either. For example, I could see how it would continue splitting segments as more segments are generated, but on e.g. the 4th run I see that some lines are still only split 3 times, even though there is a point within a line segment that was never split and existed before I ran the tool. I'm not sure how the splits are generated, but in any case, it doesn't seem to do what I want it to do, which is produce a split at "each" point.  What should I do to split lines at ALL points along the line?

I have a polyline layer and a layer of points that are on top of the polylines. I do not have the ArcGis advanced license so I can't use the Split Line at Point tool. Can someone help me with a standalone Python script that will split my polyline by my point layer? (By the way, I know that there is a tool like this in ET Geowizards, but I don't have the license for that so I can't make that tool into a standalone script).


I already checked  () but I still can't make it. I've done the final mission and chose option C...      ...(not to kill Trevor and Michael).   But now if I'm with Trevor and call Michael to hangout he accepts it, but before I pick him up I call Franklin and he's busy or not answering. It's always happening with every other combinations. Does anyone else have this problem?

Over the course of my entire game I have been trying to phone each of the protagonists in order to get this trophy. Each time I successfully 'Hang Out' with one of the characters the other one is always unreachable (busy tone on their cell) or not interested.   What is the easiest way to get all three characters together in order to get this trophy/achievement?   Secondly, there are obviously points in the story where this is not possible due to certain characters being otherwise occupied. Is there a best or easiest time to get the characters together?  Specifically:(End game spoiler)      I want to know if it is still possible to get this trophy in post game. I choose option C: Deathwish which resulted in all three characters surviving (and becoming friends again) but I still can't get them all together?


I know that there's a relation between the rank of matrix and whether it is invertible or not. What should the rank of the matrix equal to in order for the matrix to be invertible ? The other small question is what's an Nonsingular matrix ?

I am having trouble trying to prove that a matrix $A \in M_{n}(\mathbb{R})$ has inverse if and only if it's rank($A$) = n. What I was trying to do is to prove it using the fact that  $$rank(A) = \dim(\text{column-space of} \ A)$$ but I don´t really know how to continue.   Thank you for your help.


How should I prioritize my savings between enlarging my emergency fund, saving for retirement, and saving up for a down payment on a house (financial aspects of buying vs renting aside, I would prefer to own my home)?  Details:   I have no debt besides my credit cards (which I pay in full every month) and a car lease.  I already have an small emergency fund of approximately 2 months of expenses. I am married with 2 small children Neither my spouse nor I have access to a 401K plan right now, but this may change in the future   Edit:  It has been suggested that this question is either a duplicate of  or . I looked at the answers to these questions, but they don't really help:   Most of the answers to the  deal with investing strategies in and not prioritization of savings goals. Even the answers to the  don't seem to mention saving for a down payment at all (unless you assume that saving for a down payment is a perfect equivalent of paying of a mortgage early. Most answers I find about saving priorities on this site, or on the web in general place one of the highest priorities on "401K contributions up to your match", but don't explicitly give alternative priorities for when a 401K isn't available.

I am young, just started my career, and looking to begin investing so I can make my money work for me. I'm still living like I did in college (occasional purchase on myself, but most of it goes into savings) and I would like to take the extra money and invest it.   What would you suggest are the best options for me to begin? I was considering getting a personal investor [a financial advisor] for perhaps a year or so, but I've heard that they really are not worth the money. I wanted to get the input of the great minds here.