I had a meeting at my Company's HQ office in Belgium from July 1 to July 15 my embassy's interview was at June 28 I had an accommodating and flight ticket from 1-7 July and of course it has been cancelled finally i've got a refusal from the embassy with code 2 (the purpose and conditions of stay plans have not been justified) Do you think i've got refused because of intervention between interview date and visit date ? And what am i suppose to do now especially we have another office meeting at Sep-15 do i have to appeal with the new ticket and accommodation date  15-30 SEP or i have to reapply one more time *i don't got any stamp on my passport  Please advise

Schengen refusal formulae can be difficult to understand sometimes.    When an applicant receives a refusal with "Justification for the purpose and conditions of the intended stay was not provided", what does it mean?


Does anybody know the best way to get a procedurally generated texture (made by using the node system in EEVEE) to work inside of Unity? I have an object that I made and textured in Blender but cannot get the textures to show in Unity. I have exported the file as both an .fbx and .glTF and Unity shows the texture for neither.

I have a sphere having a planet material MADE WITH NODES, and I want to convert this material into an image texture; because this material is made with nodes and only works in Blender, but I want to use in another software. This is my sphere material :   For Example : This is Jupiter :  And this is the texture :  In this case I used a downloaded image for Jupiter to be the material but I want the opposite, I want to change the material to image texture Like The shape of Jupiter texture. Is there anyway in Blender that can achieve my goal ? This is the file with my sphere material .  Textures Needed :  Albedo / Color / Diffuse Normal Map Roughness Glossines  I am using normal nodes and I am not useing the new Principled-shader


Just thinking out loud here and hoping for a second opinion.  I just got a RPi3 that I am loving. I was looking at using it as a remotely accessible WOL server. The PC I am trying to reach is dual-booted with Windows 10 and Ubuntu 17.10 in which I have configured VNC access to each. I got to thinking that even if I configure the WOL server and I send the magic packet, I still won't be able to choose what OS will be loaded (because AFAIK the magic packet isn't capable of relaying this sort of information).  So, Option 1) Write a python script that can change the grub configuration file to specify the preferred default boot option. Assuming the Ubuntu drive/filesystem is accessible from Windows, a python script would be written accordingly. More easily, a different python script would be made to run in Ubuntu (to make it easier than plainly editing the grub conf file).     Option 2) Not sure if this is even possible but what if I hosted a symlinked (via some intermediate cloud service) grub conf file on the RPi? This way, I wouldn't have to worry about changing the grub conf file from Windows OR Ubuntu, I could just do it on the RPi. What I'm not sure about here is when the change made on the RPi will be accessible to grub in relation to a boot sequence.   I'd appreciate any feedback regarding a more efficient way of accomplishing the above, issues/feasibility regarding the above, miscellaneous ideas, etc. Thanks!

I have a dual boot setup:   Ubuntu 16.04 LTS Windows   Currently when I want to reboot from Ubuntu into Windows I use grub-reboot with the appropriate number as argument. This works well.  However sometimes Windows needs a reboot so a certain program can install or update and I manually have to select the right grub boot menu entry. Is there a similar way (from Windows UI) to tell grub which entry to boot?  I suppose the grub-reboot command passes the argument to a file which is in turn read by grub upon reboot.  edit (concerning my selected answer): I'm currently looking into mounting the ext4 drive that contains /boot and scripting the edit. this will possibly take quite a while since i'm doing this in my free time beside my 40h/week non-tech job. ^^


Documentation for a piece of software will often include a tour of its features. Sometimes this is called "Feature Tour", and sometimes "Features Tour". From querying web search engines, it would seem that "Feature Tour" is twice as frequent, both being widely used. Can someone expand on the correctness of both, the subtle difference in meaning, and/or reasons to prefer either form?

My friend said to me one day: "We have a mice problem at UNI". Is "a mice problem" grammatical as opposed to "a mouse problem"?


I'm using lstlisting to insert a code in my text. Fine. But I'm writing in PT_BR and want to change the word "Listing" to "Listagem" in my list (list caption from english to portuguese) so my code listing will become  Listagem 1. Texto em português   instead  Listing 1. Texto em português   Now using:  \usepackage{sbc-template}   \usepackage{graphicx,url} \usepackage[brazil]{babel}    \usepackage[utf8]{inputenc}    \usepackage[square,authoryear]{natbib} \usepackage{blindtext} \usepackage{enumitem} \usepackage{listings} \usepackage{color}  \lstset{   frame=tb,   language=Java,   aboveskip=3mm,   belowskip=3mm,   showstringspaces=false,   columns=flexible,   basicstyle={\small\ttfamily},   numbers=left,   numbersep=5pt,   numberstyle=\tiny\color{gray},   keywordstyle=\color{blue},   commentstyle=\color{dkgreen},   stringstyle=\color{mauve},   breaklines=true,   breakatwhitespace=true,   tabsize=3 }   and the code in the text:  \begin{lstlisting}[language=bash,caption={Texto em português},label=mylabel]  #some code  \end{lstlisting}   How can I do this?

I have some listings that contain pseudo-code of algorithms. And I use this: \lstset{caption={Descriptive Caption Text},label=DescriptiveLabel} to add captions to them.   My problem is that the caption at the created file shows Listing 1: Descriptive Caption Text. I would like it to write something like Algorithm 1: Descriptive Caption Text.  How can this be done?


We have a website where we showcase statistics of various reports of students. There is one report for which there are many columns i.e. 13 in total and every column is important to show. However it is not fitting different screens with different height and width. An example of the same report I am attaching below, please open the image in another tab for better viewing purpose:    Tabular way definitely doesn't look good and it is hard to see the report unless one zooms in. Is there a better UX way to showcase all these data at one place which will also look good in different screen sizes?

I have a similar problem to what is mentioned in . But to make things a little more difficult, the table has approximately 1000 rows and 64 columns. I thought of splitting up the table but the users insist on viewing all the columns at once. Hence I have come up with one option where the first two to three columns with name, ID are set to a freeze state (like in excel) and rest of the columns scroll under these columns; i.e. the table gets a horizontal scroll and to solve the row issue i have given a pagination. I have recommended that the table will be displayed only if there are 100 rows, if there are more than 100 rows the user will get an export to excel/import from excel option.  Is there a better way to handle this?


I've created a public (not gallery, or private (mod-only)) room with the intent of talking to a user with low site reputation (less than 20). I know that a user needs 20 rep to chat but I noticed a button to "grant user explicit write access" which comes with the description "Explicit write access -  Even when this room is read-only or their reputation is too low, these users will be able to talk in this room."  I go to grant the user write access...   ...but when I click "change"...  ...I get a nice red error message about needing 20 rep to talk.   So how is this function supposed to work? If they need 20 rep to talk the description should not read "reputation too low".

It was nice to discover the option for a room owner to add any user to a room, even if their rep is too low.     ...so I notified the user in question, and as instructed on the next screen, I had them got to chat.stackoverflow.com so that I'd be able to add them.    ...but when they did their part, and I did my part, I got the message:    Obviously both frustrated, the user's next logical question was:  "So I just have ask a couple questions and get someone to upvote them, right?"    Needless to say, we don't want to encourage new users to post questions for the sole purpose of bypassing a frustrating, misdocumented rule.  I realize this will likely quickly get closed since it has at least one  - which hasn't been touched in 5½ years.  However it would be nice if the issue was revisited:  Ideally, it would be cool if RO's did have the option of deciding who had access to the room they "own", but short of that, is it possible to remove the 6 "misleading" words from that page?


I belive I have accepted some updates yesterday and now my laptop doesn't reach the login screen. It keeps loading and showing the Ubuntu logo, but when I press on F7 key it shows me the following errors:    Edit1: I followed this solution  the Black/purple screen after you boot Ubuntu for the first time section and i basically got the same issue, it tried to start few jobs, login services and network manager etc... but fails in all of them, i belive the network manager services is the one causing all of this, however, i switched on windows and also tried ubuntu from my usb and internet works just fine, so its not a hardware issue  Link to original image:

I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents:


I was invited to play Rocket League, but it shows a warning:    What does that mean? Is it safe to accept the invitation? What will happen?  It happens only for one person. All others work fine.

I am trying to join my friend's Garry's Mod server and Steam is telling me that Garry's Mod is trying to launch with optional parameters. I don't know why, and he is telling me that he does not know.   The code it is trying to run is:    +connect p2p:(a long string of numbers is here).   What does this mean exactly? Should I be concerned? As of now, I have not connected yet.   Update: I tried to join a friend's Borderlands 2 game and it gave me a similar thing.  It tried to run:  +connect_lobby (a long string of numbers)   Is Steam just displaying something it normally doesn't?  Update: It has stopped loading them up with optional parameters, or at least showing me that it has been.


In the below code,  function Stack() {     this.stac = new Array();     this.pop = function () {         return this.stac.pop();     }     this.push = function (item) {         this.stac.push(item);     } }   How to decide whether push and pop are instance members of Stack class (shown above) or members of Stack.prototype?

What's the difference between  var A = function () {     this.x = function () {         //do something     }; };   and  var A = function () { }; A.prototype.x = function () {     //do something };


The context of my problem is the construction of test problems for the $k$-means algorithm and variants of it. The problems should have known optima and are thus constructed from a basic building block which is a regular  $n\times n$ grid of data points filling the unit square. For $n=2$ the point set would be {(0,0),(0,1),(1,0),(1,1)} for $n=3$ the point set would be {(0,0),(0,0.5),(0,1),(0.5,0),(0.5,0.5),(0.5,1),(1,0),(1,0.5),(1,1)} and so on. I like to know the sum of squared distances of all grid points for a given $n$ to the center of gravity which is the point (0.5,0.5). The naive formula is the following: $$\mbox{SSE}(n) = \sum_{i=0}^{n-1}\sum_{j=0}^{n-1} (0.5-\frac{i}{n-1})^2+(0.5-\frac{j}{n-1})^2$$ E.g. $$\mbox{SSE}(2)=2$$ Does a closed-form solution for $\mbox{SSE}(n)$ exist (and how does it look)? Addendum: Following the remark of @Ian I think one can rewrite the sum as $$\mbox{SSE}(n)=2n \sum_{i=0}^{n-1} (0.5-\frac{i}{n-1})^2$$ which is basically a sum of powers of 2 as described in . I should be able to figure out the final formula from here. Addendum 2: And here is the formula: $$\mbox{SSE}(n)=\frac{n^3+n^2}{6(n-1)}$$

Bernoulli stated sum of series of powers as:      I had a doubt in the given formula in the picture! What if $n < p$  i.e. $1^4 + 2^4 + 3^4$  here $n = 3$ and $p = 4$ so outer summation runs ($i = 1$ to $p$) from $1$ to $4$. And we have to find $(n-i)!$ [factorial]. Won't it lead to factorial of negative number when $i > n$ ?


while I'm doing some research about the meaning of "hey yourself" I found that it means just "greet back", but does it mean something else like "don't speak this way"?

I've heard this phrase in a couple of movies, it was being used like this:     -Hey, John!      -Well, hey yourself, Mike!   Sounds pretty simple, but my question is about how appropriate is it to say that in a conversation. It sounds a little "offensive" to me, and in the movies it was being used with a negative meaning (e.g. greeting a person who you don't really like to see/talk with).   So does it really have a "negative" meaning, or am I wrong?


With the authoryear style the when using hyperref only the year is "clickable" but not the author. What should I change so to have a sigle clickable area author+year?  \documentclass{article} \usepackage{hyperref} \usepackage[style=authoryear]{biblatex} \bibliography{dummybib} \begin{document} \cite{dummyentry} \end{document}

I am using biblatex with the authoryear styles.  So far I've been able to find everything I need easily in the excellent documentation, but I'm stuck on one thing.   With my current settings (below) citations only have the year hyperlinked to the bibliography entry. I think prior to using biblatex I would have the whole citation (authors + year) as a reference - I think this is easier to click etc. when browsing.  Is there an easy way to get this (or does it involve creating a new style).  \usepackage[style=authoryear,             bibstyle=authoryear,             citestyle=authoryear,             natbib=true,             hyperref=true,             backref=true,             abbreviate=true]{biblatex}


Whenever I install steam it won't run, nothing happens when opened. I had Lubuntu but couldn't get steam working to save my life, so I installed Ubuntu that has the exact same problem....  I've been looking all over the place for answers. The most common answer i've found and tried is:  mv ~/.local/share/Steam/ubuntu12_32/steam-runtime/i386/lib/i386-linux-gnu/libgcc_s.so.1{,.disable} mv ~/.local/share/Steam/ubuntu12_32/steam-runtime/i386/usr/lib/i386-linux-gnu/libs   or along those lines (a few variations). However whenever I do this and try to run steam it says: Updating Steam, and reinstalls what I've just removed and then does nothing again. And stopping the update doesn't seem to help.

After a botched upgrade from Ubuntu 14.04 to Ubuntu 16.04 I had to wipe my system and start all over. This time I decided to go with a much less CPU intensive desktop environment and installed Lubuntu 16.04-1.  My problem: when I run Steam from the menu shortcut it acts like it is doing nothing. However it does show up in the task manager. If I just type steam in the terminal I get the following.  michael@michael-HP-Pavilion-dm1-Notebook-PC:~$ steam Running Steam on ubuntu 16.04 64-bit STEAM_RUNTIME is enabled automatically grep: symbol lookup error: grep: undefined symbol: pcre_jit_stack_alloc grep: symbol lookup error: grep: undefined symbol: pcre_jit_stack_alloc /bin/bash: /home/michael/.local/share/Steam/ubuntu12_32/steam-runtime/amd64/lib/x86_64-linux-gnu/libtinfo.so.5: no version information available (required by /bin/bash) grep: symbol lookup error: grep: undefined symbol: pcre_jit_stack_alloc awk: /home/michael/.local/share/Steam/ubuntu12_32/steam-runtime/amd64/lib/x86_64-linux-gnu/libtinfo.so.5: no version information available (required by /lib/x86_64-linux-gnu/libreadline.so.6) Installing breakpad exception handler for appid(steam)/version(1468023329) libGL error: unable to load driver: r600_dri.so libGL error: driver pointer missing libGL error: failed to load driver: r600 libGL error: unable to load driver: swrast_dri.so libGL error: failed to load driver: swrast   Then nothing happens including steam doesn't show up in the task manager. Now I did find the command.  LD_PRELOAD='/usr/$LIB/libstdc++.so.6' DISPLAY=:0 steam   This does run Steam but it is not a real solution in my eyes. How can I fix it?


I'm trying to learn about the Singleton design pattern and I found two different ways of only creating 1 instance.  public class Singleton {    private static Singleton instance; // attributes omitted    private Singleton() {     // omissions    }  public static Singleton instance() {     if (instance == null) {         instance = new Singleton();     }     return instance; } // other methods omitted }     public class Singleton {    private static int bound = 1;    public Singleton() {     if (bound == 0) {         throw new RuntimeException(         "Singleton: No more objects must be created");     }     bound--; } }   Which is preferred to use and why? Are they equally as good?

What is an efficient way to implement a singleton pattern in Java?


I don't know if it is a right place to ask, but why isn't a certain low level programming site for embedded systems in the Stack Exchange network?

I have an idea or subject that should have a Q&A, and want to propose a new site. How can I do this?  Related:


My task is to find formula for generating function of sequence $a_0, a_1...$ defined with following recurence  $a_0=1$ and $a_n=\sum_{i=0}^{n-1} (n-i)a_i$.   I rewrote the expression  $a_n=\sum_{i=0}^{n-1} (n-i)a_i=n+(n-1)a_1+(n-2)a_2+...+a_{n-1}$  and I counted few members of sequence $a_1=1,a_2=3,a_3=8,a_4=21$ etc.  There are two answears below-wchich one i correct? Is there any way how to do some backward examination to show that it gives correct solution?


I am trying to find the Jacobian matrix of $x^T A x$. Where $A \in \mathbb R^{n \times n}$ (and $A$ is symmetric) and $x\in \mathbb R^n$.  I have tried to write it in summation notation:  $$x^T A x=\sum_{i=1}^n \sum_{j=1}^n x_ia_{ij}x_j:=y$$  Since $x^TAx=y$ is a polynomial, the corresponding Jacobian matrix will look something like this:  $$J_y=\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & ........ & \frac{\partial f_1}{\partial x_n} \end{pmatrix}$$  So the $k$-th component of my Jacobian matrix is:  $$ \frac{\partial y}{\partial x_k}=\frac{\partial }{\partial x_k} \left(\sum_{i=1}^n \sum_{j=1}^n x_ia_{ij}x_j \right) $$  I am stuck here. I have tried separating this into cases where $k=i \not=j$ but I am not really getting anywhere. Is there a way to simplify this further? ( I am not necessarily looking for a complete solution but rather a few hints that will enable me to figure it out by myself. Thanks.)

It's stated that the gradient of:  $$\frac{1}{2}x^TAx - b^Tx +c$$  is   $$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$  How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?


In D&D 5e, variant humans receive an additional skill upon creation. Does this skill have to  belong to the same class as the character or can it be from any class? For example, if I made a variant human barbarian could I choose the skill acrobatics? Or, would it have to be an additional one from the barbarian skill list?

When creating a variant human you get an extra skill to be proficient in. Does that skill have to be from your class skill list, or can it be any skill available to any class? For instance, if I'm playing a variant human barbarian, can I take acrobatics as my extra skill?


==>> What are the most useful Unix shell "one-liners" that you use? <<==  Here are a couple of mine to get the conversation started:  1. Repeat the previous command with a substring replacement:  Syntax:  ^before^after^   Example:  You entered:  git clonr https://unbelievablylongurl.org/projectdirectory/evenmoreprojects/project.git   Use this:  ^clonr^clone^   And your command will be re-run with the replaced substring:  git clone http://unbelievablylongurl.org/projectdirectory/evenmoreprojects/project.git   2. Repeat the previous command with an additional string/character  Shell Idiom:  !!   Example:  You ran a command that requires superuser privileges on your box.  Re-run the command without retyping it like this:  sudo !!

One tip or trick per answer.  My favorite is   open .   Opens the folder you're currently browsing in Finder. You can also pass URLs, images, documents or else to open.  If you specify a program name with -a you can pass the URL, image, document or folder to that program instead, e.g. open -a Preview image.png, overriding the default program set for the filetype.  Please don't post duplicates. Search in the question like this: inquestion:this ls -l  Mac OS X specific answers only.


On Ubuntu 16.04.3 i installed the lubuntu-desktop via:  apt-get install lubuntu-session   But when I login into lubuntu-desktop session I cannot find a way to install the icon/indicator that allows me to switch the language from English into Greek. All the research I did are saying to use xkbmap but on "normal" lubuntu disto there is an option to select the language.  The problem is that there is no indicator.

I have recently upgraded my PC from Lubuntu 14.10 to 15.10. Before the upgrade my keyboard layout matched my Danish keyboard. But after the upgrade I probably have a standard English/US layout.   Preferences/Language Support, doesn't let me configure the keyboard layout. Preferences/Keyboard and Mouse, only let me configure stroking delay and similar.    I don't know about iBus and fcitx, as far as the tooltip infomation tells me, its for more complex languages such as Chinese.  I don't have a US icon in the taskbar, no keyboard and/or language icon at all.  Things I have tried:   Running the following in a terminal works, but only until the next reboot:  setxkbmap -layout dk  I got the following parameter in the file /etc/default/keyboard:  XKBLAYOUT="dk"  Installing and running the app Lxkeymap changes the keyboard to Danish when I run it, but rebooting will change the layout back to US.   I don't want anything fancy, I just want to set my keyboard layout to Danish. How can I do that?


As we all know L[::-1] reverses L, no matter L is a list or a string  >>> l=[1,2,3] >>> l[::-1] [3, 2, 1] >>> s 'abc' >>> s[::-1] 'cba' >>>    But I wonder what the two colons mean. I know it goes from back to front using '-1' as step length, but can you explain in more depth? Thanks :D

I need a good explanation (references are a plus) on Python's slice notation.   To me, this notation needs a bit of picking up.   It looks extremely powerful, but I haven't quite got my head around it.


I made a live usb with ubuntu on it. I can test ubuntu with it but if I want to install it onto the SAME usb stick it is not possible. How can I make a REAL installation of ubuntu onto an usb stick without requiring another usb stick? Is it possible to install ubuntu on the same usb stick you are running the live version on? If not can I somehow run it in a virtual machine with access to the usb stick and install it from there (I tried Linux live usb creator, but it didnt work). Or can I dual boot and then install on the usb stick and delete the linux partition afterwards? If there are other/better options please let me know.  Thank you for your help

If you are hurrying to reply, System → Administration → StartUp Disk Creator -- no, that's not what I'm talking about.   I want to try Ubuntu 11.04's Unity without touching my existing Ubuntu install.  To do this, I need to install the nVidia drivers first (sigh).  To do this, I need changes to persist a reboot.  To do this, I need to really install Ubuntu on a USB key.  How do you do that?    What I tried   I tried to make a USB key from , then boot from it, then choose "Install Ubuntu." The installer refused to install to the installation media itself. I tried, from my installed copy of Ubuntu:  sudo kvm /dev/sdb --cdrom .cache/testdrive/iso/ubuntu_natty-desktop-i386.iso   ...but the installer didn't detect the disk properly.


Once an tag synonym is created, it can be in error, e.g., grape --> animal. How does one undo this?

What are tag synonyms? How can tag synonyms be distinguished from "real" tags? How are new tag synonyms created? Who can create tag synonyms? How can I delete/reverse/undo bad tag synonyms? What are merged tags?


I created a list of 3 empty lists: >>> l = [[]]*3 >>> l [[], [], []]  then added an item in the second one: >>> l[1].append("j")  but it added it on every lists: >>> l [['j'], ['j'], ['j']]  why?

I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3  The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]    Then I changed one of the innermost values: my_list[0][0] = 5  Now my list looks like this: [[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]    which is not what I wanted or expected. Can someone please explain what's going on, and how to get around it?


pid=$(pgrep 'engrampa') #Get the PID of the engrampa processes . killpid=$(echo $pid | head -1) #Get only the first line of the $pid variable and put into a new variable called $killpid. kill $killpid   I want to maintain only the first line of the variable $pid.  Let's say I have 3 instances of the engrampa process open.  When I run the commands above step by step at terminal, I get exactly what I want: 2590    When I run those exaclty commands in a script I get this result: 2590 18425 18449    Why is that happening?

I'm trying to store multiple lines in a bash variable, but it doesn't seem to work.  For example, if I list /bin one file per line and store it in $LS, then I pass $LS as stdin to wc, it always returns 1:  $ ls -1 /bin | wc -l 134 $ LS=$(ls -1 /bin); wc -l <<< $LS 1   If I try to output to screen, I get various results: echo prints all the lines on a single line, while printf prints only the first line:  #!/bin/bash LS=$(ls -1 /bin) echo $LS printf $LS   So, a bash variable can contain multiple lines?


For $n \ge 1,$ and positive integers $a,b,$ show the following:      $\rm (a)$ If $\gcd(a,b)=1,$ then $\gcd(a^n,b^n)=1.$      $\rm (b)$ The relation $a^n\,|\,b^n$ implies that $a\,|\,b.$   This is a problem from Elementary Number Theory by David M. Burton. I was able to solve part $\rm(a)$ by expanding $(ax+by)^{2n-1}$ using binomial theorem and factoring. To solve part $\rm(b)$, I used Bezout's theorem along with the following hint given in the book:     Put $d=\gcd(a,b)$ and write $a=rd,b=sd,$ where $\gcd(r,s) = 1.$ By part $\mathrm{(a)},\, \gcd(r^n,s^n)=1$. Show that $r=1$, whence $a=d.$    Before seeing the hint, I tried using mathematical induction as follows.     Let $S$ denote the set of all positive integers $n$ for which $a^n\,|\,b^n\implies a\,|\,b.$ We observe that when $n=1,$ $a\,|\,b\implies a\,|\,b.$ This means that $1\in S.$       Next, assume that $k\in S$ so that $a^k\,|\,b^k\implies a\,|\,b.$ Since $a^k\,|\,b^k,$ we can write $b^k$ as $$b^k=ma^k\tag1$$ for some integer $m$, and $b$ as $b=ra$ for some integer $r$.       Multiplying $(1)$ by $b$ yields   $$\begin{align*} b^k\cdot b&=ma^k\cdot b,\\ \text{or,}\quad \quad b^{k+1}&=ma^k\cdot ra,\\ \text{or,}\quad\quad b^{k+1}&=mra^{k+1},\\ \text{or,}\quad\quad b^{k+1}&=ca^{k+a},\tag{where $c=mr$}\\ \text{i.e.,}\!\quad\quad a^{k+1}\,&|\,b^{k+1}, \end{align*}$$ which is precisely the case for $n=k+1,$ putting $k+1\in S$ whenever $k\in S$. According to the first principle of finite induction, $S$ must be the set of all positive integers.      $``\,\blacksquare"$   Is my proof correct?

I want to prove by induction that $a^n \mid b^n$ implies that $a \mid b$ holds for all integers $n\geq 1$.  Clearly for $n=1$ this is true, since if $a \mid b$, then $a \mid b$.  Suppose this is true for some $n = k$. Then $a^k \mid b^k$, so $a|b$.  $a^k \mid b^k$ means there exists some integer $m$ such that $b^k = ma^k$, and $a \mid b$ means there exists an integer $r$ such that $b=ra$.  Then we can multiply both sides of the above equality by any constant, namely $b$.   Then $b \cdot b^k = bma^k$  so $b^{k+1} = (ra)ma^k = rma^{k+1}$  which means that $a^{k+1} \mid b^{k+1}$ completing the induction step, and thus proving that for all integers $n \geq1$ the original statement holds.  Are there any logical holes in the proof?


Nucleus mass should be equal to sum of mass of protons and neutrons, but after binding of these particles, its mass is reduced. Why? Due to reduced mass of protons or neurons? Then what's the purpose of calculated mass?

As a school student, I have wondered while studying mass defect the following mysterious problem  My assumption   Just like a car's mass is constituted by each part of it(i.e total mass of car will be equal to mass of body+engine etc..), I assume that the mass of nucleus is the sum total of the mass of the particles the nucleus contains, i.e  mass of nucleus =mass of protons +mass of neutrons I assume all the protons, neutrons  are exactly the same, they have (each kind) exactly same mass I assume the charge of proton is distributed uniformly.   Details    A quote from Wikipedia:      It is observed experimentally that the mass of the nucleus is smaller   than the number of nucleons each counted with a mass of 1 a.m.u.. This   difference is called mass excess.   The mysterious problem I encounter  There is no change in the number of neutrons or protons in the nucleus of the atom. And my idea was as the mass of nucleus is the total mass of protons and neutrons, if the mass of nucleus is to change what I can think about the possibilities are   Either the number of neutrons or number of protons should change In order to fit the 2nd assumption, some part of the mass of every neutron and every proton will go (probably as energy) so that all the protons are same and all the neutrons too are same.   The 1st possible explanation is obviously wrong as we don't see any change in the number.  The 2nd possible explanation is just 'like a reason for the sake of giving a reason' logically difficult to convince myself. If that were true there should have been many types of protons and different types of neutrons (both classified according to their mass), which again is un-intuitive.  The question   What's the possible explanation to the question from where the mass is being lost? Either one of my assumption is wrong (but isn't my assumption just logical?) or there should be some other reason how the mass can vary (may be we should better define what is the mass of the nucleus). In that case, another question , If my assumption 1 is false What constitute the mass of a nucleus?  Isn't there something like a standard proton or a standard neutron (By standard I mean a fixed mass fixed charge etc) By my third assumption, If the proton loses some mass shouldn't it lose some of it's charge. And suppose we forget E=mc^2 can't we say the difference in the potential energy of electron-proton (as it changes when the charge changes) is an explanation of how the nucleus gets binding energy.


I have a machine running ubuntu server with nextcloud. At first I installed ubuntu server just for learning purposes, but the server worked just perfectly. Now I want to encrypt my full disc and the only way I know how to do it is in the middle of the OS instalation process. Is there a way to encrypt my entire hard drive through terminal commands? If so, how?

I'm running 13.10 Saucy. If I didn't enable disk encryption during installation, is there any way to enable it post facto?    I found , which says that encryption has to happen at install time, but it's also referring to Fedora. I can easily boot into a live disk if there's any way to do it from there.


How can I connect an extra HID device (for example, an extra mouse and keyboard unifier) at VM startup for Windows 10 guest? I'm using VMware Workstation 15.0.3 on Windows 10 host machine.  I can connect non-HID USB devices like USB pen automatically to VM and VM remembers it at every start and don't connect it to host machine.  However, I cannot find an easy solution for a HID device. PC have its own mouse and keyboard. Extra HID devices will be used for guest only.  How can I do it ? Thanks.

I'm using VM Ware (specifically the player) and i'd like to attach a multiple mouses and have one for each VM (same idea with keyboard but i havent tried keyboards yet). I'd like several people using their own vm. I cant seem to figure out how to lock the mouse into the VM. If i plug in a mouse windows uses both of them and i need to click onto an instance of vmware to get the mouse working there.  I tried clicking the USB area at the bottom of the player window and i got the error message     Cannot connect "BRAND USB Optical Mouse" to this virtual machine. The host requires this device for input.   Is it possible to dedicate a mouse/keyboard to a VM instance? I'm willing to use another VM software (but preferably free)


Need to check time execution on client side    var start = new Date().getSeconds(); // code... var end = new Date().getSeconds(); let time = end - start console.log(time);     Result: 0;  I suppose it's a part of second, for example 0.05. How can I get it?

I need to get execution time in milliseconds.  I originally asked this question back in 2008. The accepted answer then was to use  However, we can all agree now  that using the standard  API is more appropriate. I am therefore changing the accepted answer to this one.


I am trying to stick two faces to each other. These two faces are of the same size, but do not have the same rotation. I tried to do it manually and I of course experimented with the magnet tool, but it did not work well till now. Is there a possibility to mark two faces like "that they belong together" and stick perfectly to each other like magnets?  Thank you, Huxi

How would I make the face of an object parallel to the ground? I could just eyeball it and use the rotate tool, but it probably wouldn't be perfectly flat. Is there a way I could possibly snap the face of the object I want level to the top of a block (making it parallel with the block) and then delete the block?


I have successfully made a connection to my access database on my local computer but when I upload my config.php file which contains the connection string on my web server and when the insert command runs it displays a blank page without making entries to the database.   $conn = new COM ("ADODB.Connection")   or die("Cannot start ADO");   $connStr = "PROVIDER=Microsoft.Jet.OLEDB.4.0;Data Source=/onlineadmission.mdb;Jet OLEDB:Database Password=;";   $conn->open($connStr);   It works well on my local computer but not when uploaded. Where is the problem?

How can we connect a PHP script to MS Access (.mdb) file?  I tried by including following PHP code:  $db_path = $_SERVER['DOCUMENT_ROOT'] . '\WebUpdate\\' . $file_name . '.mdb'; $cfg_dsn = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" . $db_path; $odbcconnect = odbc_connect($cfg_dsn, '', '');   But it failed and I received following error message:   Warning: odbc_connect() [function.odbc-connect]: SQL error: [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified, SQL state IM002 in SQLConnect in C:\web\WebUpdate\index.php on line 41


I met a problem when I use python to process my data, I want a three dimensional array, the first dimension is fixed at 5, the elements in second dimension are added dynamically, it`s type is array, the third dimension is how many number in the array, it is fixed at 4. code snippet is as follows:  three_d_array = [[]] * 5 # [[], [], [], [], []] three_d_array[0] = [1,2,3,4] # [[1,2,3,4],[],[],[],[]] three_d_array = [[]] * 5 # [[], [], [], [], []] three_d_array[0].append([1,2,3,4]) # [[[1, 2, 3, 4]], [[1, 2, 3, 4]], [[1, 2, 3, 4]], [[1, 2, 3, 4]], [[1, 2, 3, 4]]]   I know *5 will generate one copy but five references, but why '=' can break this relationship? '=' must assign new memory address for this array, I wonder it assign five copies for three_d_array[1:5] or just one for three_d_array[0]?

I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3  The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]    Then I changed one of the innermost values: my_list[0][0] = 5  Now my list looks like this: [[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]    which is not what I wanted or expected. Can someone please explain what's going on, and how to get around it?


Say I’ve got a list ls = [1,2,3,4,5,6]  And I’d like to get this result: res = [[1,2],[3,4],[5,6]]  What’s the cleanest way to go about this?

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.  I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.  I was looking for something useful in itertools but I couldn't find anything obviously useful. Might've missed it, though.  Related question:


Show that if f is entire and one-to-one, then it must be of the form AZ+B, with A not equal to zero.  I am editing my question, since there are duplicates on this forum to the question of why an entire and one-to-one function must be of the form AZ+B, with A non-zero.  I am currently stuck at f(z)=AZ for A non-zero, from using Liouville's Theorem on g=z/f(z).  I'd like to show that f(z) cannot also take the form AZ^2, AZ^3, and so on...  I think that is done by using the Fundamental Theorem of Algebra and saying that an nth degree polynomial in z (with non-zero coefficient, A) has exactly n roots.  But I'm thinking about the situation when all of the roots are at one location, so that we have only one distinct root with multiplicity = n.  Then this doesn't rule out the case that f(z) is one-to-one.    What can I do to show the polynomial must only be of degree 1?  (I've seen some derivative arguments now, including @JohnHuges ' argument below, where f' is not zero, but I don't understand this argument and why we can conclude from this that f is not one-to-one...)  Thanks in advance,

Can we prove that every entire one-to-one function is linear?


I use this Enum for available user roles types: public enum Role implements GrantedAuthority {   ROLE_ADMIN, ROLE_CLIENT;    public String getAuthority() {     return name();   } }  Full code:  But when I try to convert the list using this code: claims.put("auth", roles.stream()                         .map(s -> new SimpleGrantedAuthority(s.getAuthority()))                         .filter(Objects::nonNull)                         .collect(Collectors.toList()) );  I always get NPE when I try to get s.getAuthority(). Do you know how I can fix this issue?

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'm trying to limit the access for users to change the value of text format dropdown but I can't seem to find the correct dropdown to alter. Here's what I have so far but it seems to disable access to the field itself and not the text format dropdown.  function hook_field_widget_WIDGET_TYPE_form_alter(&$element, &$form_state, $context) {   if ($element['#paragraph_type'] == 'my_paragraph_bundle') {     $element['subform']['field_text_long']['widget'][0]['#format']['#access'] = FALSE;   } }

/**  * Implements hook_form_alter().  */ function MODULE_NAME_form_alter(   &$form,   \Drupal\Core\Form\FormStateInterface $form_state,   $form_id ) {    if ($form_id == 'node_article_form') {     $form["body"]["widget"][0]['#allowed_formats'] = ['full_html']; // Your default format     $form["body"]["widget"][0]['#after_build'][] = 'MODULE_NAME_hide_format';   } }  function MODULE_NAME_hide_format(   &$element,   FormStateInterface $form_state ) {   if (isset($element["format"]["format"])) {     $element["format"]["format"]["#access"] = FALSE;   }    return $element; }


I'm fairly new to programming and trying to connect to a DB using HTML, I'm using Wildfly, Apache NB and Derby in this case, hosted locally, the thing is that the browser gives me the following error after I try to log in with the credentials that I have created in the DB: [Login screen  java.lang.NullPointerException: Cannot invoke "java.sql.Connection.prepareStatement(String)" because the return value of "model.Conexion.getConexion()" is null So I went to NB Conexion.java to check the getter but I cannot seem to find what is the issue  here is the code. Also, I have tried this with different versions of JDK, JRE more specifically u144 and u271 tried with GlassFish5, Payara and Wildfly and still no luck. I would really appreciate any help on this as my semester depends a lot on this package model;  import java.lang.reflect.InvocationTargetException; import java.sql.Connection; import java.sql.DriverManager; import java.sql.SQLException; import java.util.logging.Level; import java.util.logging.Logger;  public class Conexion {      private static Conexion conexion;     private static final String BDURL= "jdbc:derby://localhost:1527/proyectohtml;user=ROOT;password=1234";     private static Connection conn;      private Conexion(){         try {             Class.forName("org.apache.derby.jdbc.ClientDriver").getDeclaredConstructor().newInstance();             conn=DriverManager.getConnection(BDURL);         } catch (ClassNotFoundException | NoSuchMethodException | SecurityException | InstantiationException | IllegalAccessException | IllegalArgumentException | InvocationTargetException | SQLException ex) {             Logger.getLogger(Conexion.class.getName()).log(Level.SEVERE, null, ex);         }         }      public static synchronized Connection getConexion() {         if (conexion==null) {             conexion=new Conexion();         }         return conn;     } }


This is related to . That post contains (supposedly) working example using the names of many fonts, and each font name contains a space. Names include: "CMU Serif", "EB Garamond", "GFS Artemisia".  The post states that the fonts are all part of TeXLive 2013. I have a TeXLive updated 2013 distribution. I copied the code exactly and attempted to xelatex it. But it returned an error stating that a font cannot have a space in its name. How did the previous poster manage to run code in which all font names contain spaces?

I have an updated TeX Live distribution. Following  I copied the code exactly: \documentclass{article} \usepackage{fontspec} \defaultfontfeatures{Ligatures=TeX}           \newcommand\test[1]{%   #1\\{\fontspec{#1}Βίβλος γενέσεως Ἰησοῦ Χριστοῦ υἱοῦ Δαυὶδ υἱοῦ Ἀβραάμ.}   \par\medskip}      \setlength{\parindent}{0pt}      \begin{document}      \test{CMU Serif} \test{EB Garamond} \test{GFS Artemisia} \test{GFS Baskerville} \test{GFS Bodoni} \test{GFS Complutum} \test{GFS Didot} \test{GFS Olga} \test{GFS Porson} \test{GFS Solomos} \test{Junicode} \test{Linux Libertine O} \test{Old Standard} \end{document}  into c.tex and then did xelatex c. This returned the following error:  /usr/local/texlive/2013/texmf-dist/tex/latex/tipa/t3cmr.fd)kpathsea:make_tex: Invalid fontname `CMU Serif', contains ' ' fontspec error: "font-not-found"  l.14 \test{CMU Serif}  Thus its first objection is to a space in a font name. How can that linked example work with a whole list of fonts with spaces in their names? And it also appears that no font whose name contains the strings "Serif" and "CMU" is part of TeX Live 2013:  find /usr/local/texlive -type f | egrep -i  serif | egrep -i cmu  What am I missing?


It's all in the title, but I'll spell it out:  Let's say my primary Gmail address is bob@gmail.com, and my alias is bob@bobscrabshack.com.  I have my personal mail server forwarding mail addressed to bob@bobscrabshack.com to bob@gmail.com and I'm using my SMTP server to send mail via bob@bobscrabhshack.com using Gmail.  Someone sends me an email: Crab party inquiry to bob@bobscrabshack.com and I check it in Gmail. I hit reply. I want the "From:" to be automatically set to bob@bobscrabshack.com but by default, it is set to bob@gmail.com. I normally want to send mail and reply as bob@gmail.com except when the mail is addressed to bob@bobscrabshack.com. Is there a way to do this?

In short: In Gmail, is there a way to let the "send from" address change automatically depending on which of my accounts it was sent to?   In long:  I recently joined a team (of developers) and they provided me with an email address on their domain (for if I have business with a client or something).    Anything sent to me@team.com is forwarded by the team.com mail server I'm using Gmails "send mail as" functionality to send email through smtp.team.com. I receive emails and I can send them, so that works I have my default "send as" address setup as me@Gmail.com, which I would like to keep since I send 90% of my mail through there.   Now I'm wondering how to do the following, if at all possible:   when I receive an email at me@team.com (as opposed to me@Gmail.com), and I reply to it, my "send as" address will change automatically to me@team.com I can do it manually, but I know I'm going to forget sooner or later, confusing the hell out of people since the team lets me use a pseudonym.   I hope my question makes sense, and I'm sure it's easily answered by a "no, not possible" or "yes, option X in tab Y".


/tried commands with sudo/ apt-get update & upgrade apt-get install git apt-get -f install git bash: aptitude: command not found

I have problem with missing dependencies, when I try to install some software by apt-get and also from downloaded .deb files.  I am using Debian 10. After installation I removed a lot of software which I no needed eg. Deluge, some terminals (I have only LXTerminal now), some tools from Libre Office Packet etc.  Now, when I try to install Atom, TeamViewer, or some RDP tools, every time I got information about missing unresolved dependencies, common it is libgconf, libcurl etc.  How can I resolve this issue? Now I can not install a lot of must have software, and when I try to apt-get install this packets, I got error, that no libs with this name were found.  Here is my sources.list. Maybe it will be some tip to fix this problem:  deb http://security.debian.org/debian-security/ buster/updates non-free contrib main deb-src  deb-src http://security.debian.org/debian-security/ buster/updates non-free contrib main deb-src   deb http://deb.debian.org/debian/ buster-backports main contrib non-free  deb-src http://deb.debian.org/debian/ buster-backports non-free contrib main   deb http://ftp.de.debian.org/debian stretch main contrib non-free deb-src http://ftp.de.debian.org/debian stretch main contrib non-free   Stretch repos were added for test purposes, but when I comment them, it looks the same.


Is it made in Photoshop or illustrator? I tried many way but can't reach this smooth level

I've been trying with brushes but it just doesn't get close to this. Has anybody been able to accomplish this effect? how did you do it?  I'm using photoshop cs6


I use Ubuntu 14.04 LTS.  When I attempt to install updates. I receive a message that says, "The volume boot has only 0 bytes disk space remaining".  How do I clear disk space in the boot folder?  This has not occurred since I installed 14.04 LTS last year.

My /boot partition is nearly full and I get a warning every time I reboot my system. I already deleted old kernel packages (linux-headers...), actually I did that to install a newer kernel version that came with the automatic updates.  After installing that new version, the partition is nearly full again. So what else can I delete? Are there some other files associated to the old kernel images?  Here is a list of files that are on my /boot partition:  :~$ ls /boot/ abi-2.6.31-21-generic         lost+found abi-2.6.32-25-generic         memtest86+.bin abi-2.6.38-10-generic         memtest86+_multiboot.bin abi-2.6.38-11-generic         System.map-2.6.31-21-generic abi-2.6.38-12-generic         System.map-2.6.32-25-generic abi-2.6.38-8-generic          System.map-2.6.38-10-generic abi-3.0.0-12-generic          System.map-2.6.38-11-generic abi-3.0.0-13-generic          System.map-2.6.38-12-generic abi-3.0.0-14-generic          System.map-2.6.38-8-generic boot                          System.map-3.0.0-12-generic config-2.6.31-21-generic      System.map-3.0.0-13-generic config-2.6.32-25-generic      System.map-3.0.0-14-generic config-2.6.38-10-generic      vmcoreinfo-2.6.31-21-generic config-2.6.38-11-generic      vmcoreinfo-2.6.32-25-generic config-2.6.38-12-generic      vmcoreinfo-2.6.38-10-generic config-2.6.38-8-generic       vmcoreinfo-2.6.38-11-generic config-3.0.0-12-generic       vmcoreinfo-2.6.38-12-generic config-3.0.0-13-generic       vmcoreinfo-2.6.38-8-generic config-3.0.0-14-generic       vmcoreinfo-3.0.0-12-generic extlinux                      vmcoreinfo-3.0.0-13-generic grub                          vmcoreinfo-3.0.0-14-generic initrd.img-2.6.31-21-generic  vmlinuz-2.6.31-21-generic initrd.img-2.6.32-25-generic  vmlinuz-2.6.32-25-generic initrd.img-2.6.38-10-generic  vmlinuz-2.6.38-10-generic initrd.img-2.6.38-11-generic  vmlinuz-2.6.38-11-generic initrd.img-2.6.38-12-generic  vmlinuz-2.6.38-12-generic initrd.img-2.6.38-8-generic   vmlinuz-2.6.38-8-generic initrd.img-3.0.0-12-generic   vmlinuz-3.0.0-12-generic initrd.img-3.0.0-13-generic   vmlinuz-3.0.0-13-generic initrd.img-3.0.0-14-generic   vmlinuz-3.0.0-14-generic   Currently, I'm using the 3.0.0-14-generic kernel.


I imported an OSM file into a Postgres DB using the Osmosis tool. Now, I need to extract some data but I can't find any schema description (I'm really a newbie in GIS).    Where can I find it?   Do you know if there is a tutorial to learn how to query an Osmosis Postgres DB?  I need to extract the names of all streets available within a given bounding box.   How can I do that?

I want to create a query to select all ways and their nodes that exist within a bounding box using postGIS. The bounding box shall includes all details as osmosis "--bounding-box" command will retrieves.   Is there any way to do that?


I found this expression curt and powerful. But I also understand it's offensive to someone, and somehow, has a racist implication.   Is there a similar curt and powerful expression I can use without offending anybody?

I'm looking for expressions that are equivalent to      "calling a spade a spade."    In some environments and verbal contexts, this phrase needs to be avoided -- please take my word for it.    Thank you.    I shouldn't have used the gerund in my question.  Where I need to use this sort of expression is in a discussion I'm losing patience with, and I want to say, effectively, let's not pussyfoot around with euphemisms. I'm going to tell it like it is, and Let's not mince words here, both work quite well.  Thank you all.  In response to @Pharap,     Do you have evidence that someone has taken offence at this before or   is this merely preemptive?   I want to avoid problems.  I live in a small town where everybody has their sensitivities.  I was once in a hiring committee where a person objected to someone calling one of the candidates a "dark horse."  I believe I read that some town in California decided not to use the term "manhole covers" any more because of the "man" syllable.


Given the following Interfaces:  interface IEntity {     int Id{get;} }   interface IPerson : IEntity {     string Name{get;}      int Age{get;} }  interface ITeacher : IPerson  {     string StaffId{get;} }  interface IStudent : IPerson  {     string StudentId{get;}     string Courses{get;} }  interface IRepository {     T Get<T>(int id) where T : IEntity }   I have the following classes in my namespace  public class EntityBase() : IEntity {     int Id{get;set;} } public class Teacher : EntityBase, ITeacher{} public class Sudent : EntityBase, IStudent{}   Currently I am implementing this IRepository as follows:  class Repository: IRepository {     IDataContext Context{get;set;}      T Get<T>(int id) where T : EntityBase     {         if(typeof(T) == typeof(Teacher))             return Context.Get<ITeacher>(id);         if(typeof(T) == typeof(Sudent))             return Context.Get<ISudent>(id);         throw new Exception("Unknown Interface " + typeof(T).Name);     } }   Is there a betterway of implementing this? Given that our Context has no knowledge of our data types (Teacher, Student), just its interfaces (ITeacher, IStudent).  Can something like this work?  class Repository: IRepository {     T Get<T>(int id) where T : EntityBase     {         var MyInterface = FindInterface<T>();         return Context.Get<MyInterface>(id);     } }

What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime?  Consider the following sample code - inside the Example() method, what's the most concise way to invoke GenericMethod<T>() using the Type stored in the myType variable?  public class Sample {     public void Example(string typeName)     {         Type myType = FindType(typeName);          // What goes here to call GenericMethod<T>()?         GenericMethod<myType>(); // This doesn't work          // What changes to call StaticMethod<T>()?         Sample.StaticMethod<myType>(); // This also doesn't work     }      public void GenericMethod<T>()     {         // ...     }      public static void StaticMethod<T>()     {         //...     } }


It's two years later (than ).  Ubuntu is at 16.10 and we are about to celebrate 2017.  (And there is also a ZFS wrench in the Canonical gears now.)  How much longer will ext4 remain the default file system? What will succeed it and when?

Btrfs seems to have a set of useful features such us snapshots and is default in Opensuse and Suse.Fedora plans to move to it. Does Ubuntu plan to move to it? If not, why ?


So, I'm trying to remove an object from an ArrayList. When I do it and print the ArrayList again, the object is still there.       else if (cmd==4){         System.out.println("Enter the product name: ");         String nameToRemove = input.next();         for (Product x : inventoryList){            if (nameToRemove == x.getName()){               inventoryList.remove(x);            }         }      }   Why won't the remove() method do anything when I implement it?

I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?


Prove that $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$ SOURCE :   It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for the inequality. Any hint would be really helpful. Thanks in advance ! ^__^

Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$   Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.   This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.


What happens if the visa availability is longer than the availability of the passport?  In my case, a Romanian with a Romanian passport (5 years availability) who wants to get a tourist visa for United States which usually is for 10 years. A fact that might be important, Romania is not in the   Will the visa be considered expired when my passport will expire or it could be used later too (with a new passport, of course).

I have a valid B1/B2 US visa expiring about a year from now. The passport it's in expired a few years ago, and I recently got a new one.   The passport number does not match, but the national ID number, mentioned in both, does (it used to be the same as your national ID, but not anymore)  Can I travel to the US carrying both passports? Or is it asking for trouble? (I read a few years ago that it was possible, but that might have changed)  Should I just get a new visa to avoid potential complications?


I'm going through the Dungeon Master's Guide, and I see that DCs should be based on the "average person".  What I'm having a hard time with is that if someone has a +5 modifier to their History checks (quite knowledgable), and the task is "hard" (DC 20), then they still only have a 1/4th chance of success.    It seems to me the chances don't quite add up.   Do I have this correct?  How can I (or should I?) express that a person who doesn't know History has no chance of passing a History check, but a historian should probably have a better than 1/4th chance.    I am a player now, but wish to be a DM in the future.  I guess my bigger issue is the difference between an average person and someone who is well-trained.  Someone with a 0 in Arcana has a 25% chance of success against a "medium" (DC 15) task. Where as someone well trained in magic (+5 modifier to Arcana checks) has only a 50% chance. That seems quite out of whack to me.

Bounded accuracy is a major feature of the new system. It promises that bonuses to rolls won't grow too much throughout a character's career. And indeed, a first-level fighter would have an apex skill of about +5 (Strength +3, proficiency +2), whereas a twentieth-level fighter would have the same skill at at most +11 (Strength +5, proficiency +6). Supposedly, such bounded accuracy allows characters to meaningfully interact with the same threats for most of their career, if they so choose. (.)  This wasn't always the case. In 3e, everything was modeled by the rules, and a level 20 character meeting a level 1 character would not require any change to the way either are represented by the system. Such interaction was unlikely to be meaningful mechanically, as level 20 characters would outclass level 1 characters thoroughly. Thus, level 20 characters in 3e were both in mechanics and flavor significantly more powerful in every respect.  In 4e, only the immediate vicinity of PCs was ever modeled, and there was no such thing as "level 1" people near them by the time they'd hit level 30. If they were to be modeled at all, "low-level" creatures would be represented as minions of higher level (For more on this particular reasoning see , though it's not that important to the question at hand). There were no conclusions of relative power level to be drawn from the mechanics, even though the numbers kept on growing, but flavor of becoming demi-gods at high levels preserved.  In 5e, it appears that the approach has moved back to modeling the whole world, as the implementation of flat math indicates. But what about the flavor? Is the world's greatest athlete at +11 really only capable of winning about 3/4ths of the time against a village strongman at +5? If that is not the case or the intention, how is this apparent disconnect reconciled? Is comparing them in such a way actually valid?  If it is, indeed, the intent of the system (please provide some quotes if possible) that even at high levels PCs do not straddle like colossi over mere mortals, it is a serious departure from the past two editions I'm familiar with. As a qualitative change, it would impact the way I treat high-level characters and the role they play in the world, much more so than different class features or spellcasting methods of a new edition.  Does the bounded accuracy approach mean that max-level characters are no longer demigods compared to starting characters in everything other than killing?


I have no down voted questions.  Of a total of 8 questions I have 7 up votes and 0 down votes.  I have a 51 reputation which is very low but however it is one nonetheless.  And yet I am excluded and I'd like to know why please?  I have contacted the Stack Overflow people and get no reply.  If I cannot improve on my already up voted questions, what other measures can I take to be allowed back in?

If you are reading this, you may have been affected by one of our post quality bans. If so, it is important that you  to understand what has happened and how to regain the permissions that you lost. While trying to ask a question, one could see:  We are no longer accepting questions from this account. See  to learn more.  Likewise, for answers:  We are no longer accepting answers from this account. See  to learn more.   Why am I getting this message? Are deleted posts taken into account too? Is a question/answer ban the same as a suspension? How do I avoid getting a question ban? How long do I have to wait before I can post again? What can I do to release the ban? How can I reactivate my account? Can I simply create a new account? I'll just ask somewhere else on the SE network, and they'll migrate my question to the correct site! Will a ban on one Stack Exchange site affect my standing on other sites in the network? Does this apply to meta sites too? My account is in good standing. Why am I still blocked?


What is the remainder of $18!$ divided by $437$?    I'm getting a little confused in the solution. It uses Wilson's theorem    Wilson's Theorem: If $p$ is prime then $(p-1)!\equiv-1(\text{mod } p)$    So it first factors $437$ into primes. So $437 = 19 \cdot 23$. Then from Wilson's theorem notes that $18!\equiv-1(\text{mod } 19)$ so we're part way there, but also says $22\equiv22!(\text{mod }23)$ by Wilson's theorem (really don't know how they got this from $22!\equiv-1(\text{mod }23)$.  Also I'm confused how solving this leads to finding the remainder for $18!$ divided by $437$? I understand getting $18!$ from $19$ but not the $23$ part.

Suppose that $p$ is a prime. Suppose further that $h$ and $k$ are non-negative integers such that $h + k = p − 1$.   I want to prove that $h!k! + (−1)^h \equiv 0 \pmod{p}$  My first thought is that by Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$, and $h!k!$ divides $(p-1)!$ (definition of a binomial). Where would I go from here?


I recently posted a  that is still unanswered. I'm trying to put several articles into a publication. In each article the authorlist is defined using authblk. So I have:  Article 1:  \begin{document} \title{title1} \author[*]{First} \author[**]{Second} \affil[*]{FirstAffil} \affil[**]{SecondAffil} \maketitle Text \end{document}   Article 2:  \begin{document} \title{title2} \author[*]{Third} \affil[*]{ThirdAffil} \maketitle Text \end{document}   Main file:  \documentclass[a4paper]{article} \usepackage{import} \usepackage{authblk}  \begin{document} \subimport{Folder1/}{article1} \subimport{Folder2/}{article2} \end{document}   My problem was that with the second \maketitle (the one corresponding to article 2), ALL the authors appeared.   I solved this problem inserting the following code between the two \subimportcommands:  \makeatletter \renewcommand{\AB@authlist}{} \renewcommand{\AB@affilist}{} \makeatother   In this way the list of authors is reseted.  However, now in the second article the author is PRECEEDED by a comma, so I get something like this:  title1  First* and Second**  *FirstAffil  **Second Affil  Text  title2  , Third*  *ThirdAfiil  Text  I guess that there is some counter associated with the author list that is not reseted, so \maketitle assumes  that the Third author is in fact the third, and not the first one.  Does someone know how can reset this counter (if it exists) or how can I solve this problem anyway?

I'm trying to put a few articles together into a small publication. Each article has been sent to me in .tex format, so I want to include them into one file, for the whole publication. My code is something like this:       \documentclass[a4paper]{article}     \usepackage{newclude}     \usepackage{docmute,subfiles}     \begin{document}     \input{article1}     \newpage     \setcounter{section}{0}     \input{article2}     \end{document}   Each article uses authblk package, so in article1, I have:      \author[1]{A}     \author[2]{B}     \affil[1]{AffilA}     \affil[2]{AffilB}   And in article2:      \author[1]{C}     \affil[1]{AffilC}   The problem is that including the files as above, the authblk package adds all the the authors, so after compilation, all authors (A,B and C) appears in the \maketitle of article2. I guess I need to reset the author list, but I can't find out how. Any ideas?


I've installed Ubuntu 15.10 on a new Dell Precision 5510 from a Live USB stick. I used the "Try without install" mode, and from there installed Ubuntu to the hard drive. During the install, the touchpad worked. However, in the newly installed system, the Synaptic touchpad isn't registering anything.   I've done a bit of digging, and I can confirm the following:   The kernel sees the touchpad. xinput list lists the touchpad. evtest does see any events from the touch pad.   When I run xinput list, I get the following:  Virtual core pointer                        id=2    [master pointer  (3)] ⎜   ↳ Virtual core XTEST pointer                id=4    [slave  pointer  (2)] ⎜   ↳ A4Tech USB Mouse                          id=11   [slave  pointer  (2)] ⎜   ↳ ELAN Touchscreen                          id=13   [slave  pointer  (2)] ⎜   ↳ SynPS/2 Synaptics TouchPad                id=15   [slave  pointer  (2)] ⎜   ↳ DLL06E5:01 06CB:7A13 UNKNOWN              id=17   [slave  pointer  (2)]   Now doing further digging, I've established that id=15 and id=17 are both trying to hog the touchpad, and as such, it's not working.  I can see that the touchpad is recognised in Settings > Mouse and Touchpad, and that options like disabling the touchpad, etc. are also present, and the touchpad is "enabled".  Is there any way to solve this? Can I just disable the UNKNOWN touchpad? I'm not a linux expert, so might need pointers on basics. Any help is appreciated.

I just got my laptop in the mail and I installed Ubuntu 14.04 LTS. It all went smooth until it booted up and I noticed my touch pad doesn’t work at all. It works perfectly on Windows and when I use a wireless mouse it works as well.  Here's my xinput:  ⎡ Virtual core pointer                          id=2    [master pointer  (3)]   ⎜   ↳ Virtual core XTEST pointer                    id=4    [slave  pointer  (2)]   ⎜   ↳ 2.4GHz RF  KEYBOARD  AND  MOUSE               id=11   [slave  pointer  (2)]   ⎜   ↳ DLL06AB:00 06CB:78F1                          id=13   [slave  pointer  (2)]   ⎜   ↳ SynPS/2 Synaptics TouchPad                    id=15   [slave  pointer  (2)]   ⎣ Virtual core keyboard                         id=3    [master keyboard (2)]       ↳ Virtual core XTEST keyboard                   id=5    [slave  keyboard (3)]       ↳ Power Button                                  id=6    [slave  keyboard (3)]       ↳ Video Bus                                     id=7    [slave  keyboard (3)]       ↳ Power Button                                  id=8    [slave  keyboard (3)]       ↳ Sleep Button                                  id=9    [slave  keyboard (3)]       ↳ 2.4GHz RF  KEYBOARD  AND  MOUSE               id=10   [slave  keyboard (3)]       ↳ Integrated_Webcam_HD                          id=12   [slave  keyboard (3)]       ↳ AT Translated Set 2 keyboard                  id=14   [slave  keyboard (3)]       ↳ Dell WMI hotkeys                              id=16   [slave  keyboard (3)]     I would love to have my track pad working and I've done some googling and browsing of this site and nothing I've tried has worked. Any input would be great. Let me know if there’s any other information that can help remedy this. I'm pretty new to Linux. Thank you very much.  Edit: I've updated the kernel to 3.19 and checked my Additional Drivers which had nothing for me. Neither of which worked.


Here is a simple query and it's results using SQL Server:  SELECT Notes FROM DailyTaskHours WHERE Notes IS NOT NULL     AND Notes <> ''     AND NonScrumStoryId = 2264     I would like one row with one column returned. How do I accomplish this?

I'm trying to migrate a MySQL-based app over to Microsoft SQL Server 2005 (not by choice, but that's life).  In the original app, we used almost entirely ANSI-SQL compliant statements, with one significant exception -- we used MySQL's group_concat function fairly frequently.  group_concat, by the way, does this: given a table of, say, employee names and projects...  SELECT empName, projID FROM project_members;   returns:  ANDY   |  A100 ANDY   |  B391 ANDY   |  X010 TOM    |  A100 TOM    |  A510   ... and here's what you get with group_concat:  SELECT      empName, group_concat(projID SEPARATOR ' / ')  FROM      project_members  GROUP BY      empName;   returns:  ANDY   |  A100 / B391 / X010 TOM    |  A100 / A510   So what I'd like to know is: Is it possible to write, say, a user-defined function in SQL Server which emulates the functionality of group_concat?   I have almost no experience using UDFs, stored procedures, or anything like that, just straight-up SQL, so please err on the side of too much explanation :)


Let $x_n=1+\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.......+\frac{1}{\sqrt n}-2\sqrt{n+1}$   I got to know form comment that $x_n$ is increasing sequence. is there any way to prove limit of sequence.   Any help will be appreciated.

Calculate below limit $$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$


$G$ has no subgroup with index 2. How can I show that all subgroups with index 3 are normal? (The index of the subgroup is the order of the quotient set.)

Prove that if a group is containing no subgroup of index 2 then any subgroup of index 3 is normal.    Thank you.


I am helping to correct some errors in a game manual not originally written in english. I've come across a bit that I'm not sure what to do with.  The game consists of several 90-minute matches.  My question is whether this is correct or if the line should read:  The game consists of several 90-minutes matches.  I feel like the original version is correct, though I only base this off of the fact that is "sounds" right to me. I've done a bit of searching and found a blog post about something similar . It doesn't provide a definitive answer though.

Why are year, pound and mile in the singular form in the phrases below?   five-year-old children  20 pound note 10 mile run   Is that because they're acting as adjectives, which are always invariable in English?  Is it incorrect to say...   five-years-old children? 20 pounds note? 10 miles run?


I'm trying to understand how a frequentist would approach statements like "having observed X, the probability of Y happening is Z". For concreteness, let's say we conducted $n_1$ identical, independent Bernoulli trials, and observed $n_1$ successes and 0 failures. We want to attach a probability to the statement "the next $n_2$ trials will also all succeed". The Bayesian approach is straightforward: choose a prior for the Bernoulli parameter $p$, let's say a Beta distribution $\mathrm{Beta}(\alpha, \beta)$, incorporate the likelihood of the first $n_1$ observations to obtain a posterior, then compute the expected value of the probability of $n_2$ additional successes according to the posterior. If I'm not mistaken, the result is: $$\text{Prob}(\text{$n_2$ additional successes}) = \frac{B(\alpha+n_1+n_2, \beta)}{B(\alpha+n_1,\beta)}$$ I'm less sure about how a frequentist would look at the problem. Here is what I think. We decide a priori that we are going to make our claims according to the following decision tree:  If there is any failure in $n_1$ trials, then make no claim.  If there are no failures in $n_1$ trials, then claim that there will be no failures in the next $n_2$ trials.   The probability that this decision tree will lead to make an incorrect claim is: $$\text{Prob}(\text{incorrect claim}) = p^{n_1}(1-p^{n_2})$$ and this probability is maximized if the Bernoulli parameter is $$p_{\text{worst case}} = \left( \frac{n_1}{n_1+n_2}\right)^{\frac{1}{n_2}}$$ So the probability of making an incorrect claim, in the worst case scenario, is: $$\text{Prob}(\text{incorrect claim, worst case}) = \frac{n_2}{n_1+n_2}\left(\frac{n_1}{n_1+n_2}\right)^{\frac{n_1}{n_2}}$$ And this would lead me to conclude that, from a frequentist point of view: $$\text{Prob}(\text{$n_2$ additional successes}) =  1 - \frac{n_2}{n_1+n_2}\left(\frac{n_1}{n_1+n_2}\right)^{\frac{n_1}{n_2}}$$ Here the meaning of the probability is that if the experiment is repeated many times, for every time that the decision tree leads to claim there will be $n_2$ additional successes, the claim will wrong at most this fraction of the time, regardless of the true value of $p$. I am suspicious of this approach mainly because it appears to be less conservative than the Bayesian, for any reasonable choice of the prior.

I have a couple of questions for prediction and tolerance intervals.  Let's agree on the definition of the tolerance intervals first: We are given a confidence level, say 90%, the percentage of the population to capture, say 99%, and a sample size, say 20. The probability distribution is known, say normal for convenience. Now, given the above three numbers (90%, 99% and 20) and the fact that the underlying distribution is normal, we can compute the tolerance number $k$. Given a sample $(x_1,x_2,\ldots,x_{20})$ with mean $\bar{x}$ and standard deviation $s$, the tolerance interval is $\bar{x}\pm ks$. If this tolerance interval captures 99% of the population, then the sample $(x_1,x_2,\ldots,x_{20})$ is called a success and the requirement is that 90% of the samples are successes.   Comment: 90% is the a priori probability for a sample to be a success. 99% is the conditional probability that a future observation will be in the tolerance interval, given that the sample is a success.  My questions:  Can we see prediction intervals as tolerance intervals? Looking on the web I got conflicting answers on this, not to mention that nobody really defined the prediction intervals carefully. So, if you have a precise definition of the prediction interval (or a reference), I would appreciate it.   What I understood is that a 99% prediction interval for instance, does not capture 99% of all future values for all samples. This would be the same as a tolerance interval that captures 99% of the population with 100% probability.   In the definitions I found for a 90% prediction interval, 90% is the a priori probability given a sample, say $(x_1,x_2,\ldots,x_{20})$ (size is fixed) and a single future observation $y$, that $y$ will be in the prediction interval. So, it seems that both the sample and the future value are both given at the same time, in contrast to the tolerance interval, where the sample is given and with a certain probability it is a success, and under the condition that the sample is a success, a future value is given and with a certain probability falls into the tolerance interval. I am not sure if the above definition of the prediction interval is right or not, but it seems counterintuitive (at least).   Any help?


I want to add 30 days to current date and get in a format.  Im confused to add a 30 to a date and get new.  Its pure JS Solution needed.  Format : June 10, 2017

How to add days to current Date using JavaScript. Does JavaScript have a built in function like .Net's AddDay?


Let $R$ be commutative ring with identity that contains a field $K$ as a subring. If$ $R is a finite dimensional vector space over the field $K$, prove that every prime ideal in $R$ will be maximal.   My idea was to prove the integral domain $R/p$ (if $p$ is a prime ideal) was a field as for any ideal $P$, $R/P$ will be a field iff $P$ is maximal in $R$. But how can I use the fact that $R$ is finite dimensional over $K$? I don't understand.

Let $R$ be a commutative ring containing a field $k$ as sub-ring. Assume that $R$ is a finite dimensional $k$-vector space. Show that every prime ideal of $R$ is maximal.   My approach:  Let $I$ be an prime ideal of $R$, as $R$ is a finite dimensional vector space, assume $I= \text{span}(\beta_1, \beta_2...\beta_m$). Assume $I$ is not a maximal ideal and $I \subset J$, where $J$ is a proper ideal. Since $J$ is maximal, $R/J$ is a field.  Say, we choose $r_1$, $r_2\in$ $R$, such that $(r_1+J)(r_2+J)=1+J$, since $R/J$ is a field.  Now say $r_1\in I$, so, $r_1=\sum_{i=0}^m a_i\beta_i$, where all $a_i$'s are from field $k$.   As $r_i \in I$, $r_1r_2 \in I$, so $r_1r_2=\sum_{i=0}^m b_i\beta_i$. Where all $b_i$'s are from field $k$.  We have chosen $r_1, r_2$ in such a way $(r_1+J)(r_2+J)= r_1r_2+j_1=1+j_2$.  As $I \subset J$, $J$ has a basis of ($\beta_1, \beta_2...\beta_n$), where ofcourse $n>m$.  Say $j_1 = \sum_{i=0}^n c_i\beta_i$ and $j_2=\sum_{i=0}^n d_i\beta_i$.  Now, $$r_1r_2+j_1=1+j_2\Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i=1+\sum_{i=0}^n d_i\beta_i  $$  $$ \Rightarrow \sum_{i=0}^m b_i\beta_i +\sum_{i=0}^n c_i\beta_i -\sum_{i=0}^n d_i\beta_i=1 $$ $$ \Rightarrow 1 \in span(\beta_1, \beta_2... \beta_n)$$ So the multiplicative unity $1 \in J$. Hence, J is not a proper ideal of R, rather $J=R$. Hence $I$ is maximal.  So here we need not to use the property that $I$ is a prime ideal. Therefore I think either I'm missing something, or this method is completely wrong. Please help.


I have a custom list with Employee Name, Approval Status, Created Date columns in it.I am using SharePoint 2010 version. I want to display the list items depending on one of the column "Approval Status" when equals to Approved in a page. In page I have a set up of text box , date filter and a button as shown in figure   when i type in the employee name and select the dates and click on the filter button it should get the results based on the column value of approval status equals approved below.I am trying to implement this using jquery can someone help me in this scenario. thanks

I am working on SharePoint 2010, I have a custom list with columns Start date , End date, Created by, Approval Status. I have created a page where I am looking to filter the list and show the result depending on the approval status which is a column in custom list as shown in figure . Can some one help me out on how to approach for this.


I have a question about \text{}.  For convenience, I define many new commands, for example \def\im{\text{Im}}, \def\coker{\text{coker}}, etc.. I expect my new commands work like \ker, and really work that way in usual situation. But, in italic font enviroments(thm, prop, etc..), \im and \coker typed in italic font. I want they type in normal font like \ker.  Is there any way to solve this problem.

Mathematical operators, such as function names, should be set in roman type, not italics. Latex already has commands for some operators, including \max, \min, and \log.  How can I define additional such commands?


In "Harry Potter and Goblet of Fire" Harry, Ron and Hermione go through countless books in the library to find out how Harry could breathe underwater. How come that the Bubble-Head Charm wasn't mentioned anywhere?

In Harry Potter and the Goblet of Fire, Harry finds out several weeks before the second task of the Triwizard Tournament that he would be required to survive underwater for an hour. Harry, Hermione, and Ron all proceed to spend hours and hours researching a solution to this problem, even engaging the librarian's help:  So Harry, thinking that he would soon have had enough of the library to last him a lifetime, buried himself once more among the dusty volumes, looking for any spell that might enable a human to survive without oxygen. However, though he, Ron and Hermione searched through their lunchtimes, evening and whole weekends -- though Harry asked Professor McGonagall for a note of permission to use the Restricted Section, and even asked the irritable, vulture-like librarian, Madam Pince, for help -- they found nothing whatsoever that would enable Harry to spend an hour underwater and live to tell the tale. Goblet of Fire -- Bloomsbury -- chapter 26, The Second Task  Hermione, in particular, has a charm in mind as the most probable solution:  "I think your best chance is some sort of charm." Hermione -- Goblet of Fire -- Bloomsbury -- chapter 26, The Second Task  Despite many hours spent searching, and despite Hermione's formidable researching skills, they are unable to find any solution. However, in the second task itself, both Cedric Diggory and Fleur Delacour use a Bubble-Head Charm to survive underwater:  Miss Fleur Delacour, though she demonstrated excellent use of the Bubble-Head Charm, was attacked by Grindylows as she approach her goal, and failed to retrieve her hostage. ... Mr Cedric Diggory, who also used the Bubble-Head Charm, was first to return with his hostage, though he returned one minute outside the time limit of an hour. Goblet of Fire -- Bloomsbury -- chapter 26, The Second Task  The Bubble-Head Charm doesn't appear to be particularly difficult to perform, as it is in widespread usage among students by Harry's fifth year:  Dungbombs and Stink Pellets were dropped so frequently in the corridors that it became the new fashion for students to perform Bubble-Head Charms on themselves before leaving lessons, which ensured them a supply of fresh air, even though it gave them all the peculiar appearance of wearing upside-down goldfish bowls on their heads. Order of the Phoenix -- Bloomsbury -- chapter 30, Grawp  Why did Harry, Hermione, and Ron all fail to discover the Bubble-Head Charm in advance of the second task, despite spending weeks specifically researching a charm to survive underwater, and despite both Cedric and Fleur learning that charm?


I have a loglog plot where the ordinate only covers less than a magnitude. pgfplots will display the plot like this:    It is hard to make out the actual time it takes, it must be somewhere between one and a tenth of a second. Is there some way to label the ordinate with numbers between 0.1 and 1 like in a regular plot?

I want to change format of x or y number in green box of below graph .


I need to install OS X Yosemite onto my laptop, but can't find it in App Store.

Sometimes it's hard to search for older versions of macOS & OS X  on the App Store.  This is an attempt to provide the canonical QA for this issue, as per the Meta QA -  I expect it to be modified over time to hopefully become a complete information resource.  How can I get any older version of a Mac OS to be able to run as a VM, or to reinstall on my Mac?


If you set up a button, a piston and a redstone dust in this composition...    The piston will be activated while the button is on (and will retract when off).    However, I can't find the rationale why this gadget works. I guess this is some kind of BUD mechanism, but still don't understand what route the piston is powered through.  As far as I know, if I remove the dust below the button, it'll no longer trigger the piston.    Nor when the dust is powered otherly.

I've lately become hooked into building computers in Minecraft and today I decided to build a giant hard drive. I've built HDDs before, but not at the scale I'm currently attempting, and I anticipated problems. As I was wiring some pistons, they activated and pushed forward oddly. As far as I can tell, there's nothing powering them.    I thought at first that it might be the redstone torches (top left of above pic) but they aren't powered.   Here's a top view of the same pistons. I also thought the line of redstone above  them might be the problem (middle of above pic), but when I destroy it, the pistons stay extended.  Is there a reason why my pistons are extending? If so, is there a way to get them to power (when I want them to) without disturbing the pistons above?


I want to write the following equation:  For that I wrote a code: \RequirePackage{fix-cm} \documentclass[twocolumn]{svjour3}    % \smartqed  % flush right qed marks, e.g. at end of proof % \usepackage{graphicx} \graphicspath{ {./images/} } \usepackage[caption=false,font=footnotesize]{subfig} \usepackage{mathtools} \usepackage{nccmath} \usepackage{booktabs} \usepackage{float} \usepackage{multirow} \usepackage{makecell} \usepackage[caption=false,font=footnotesize]{subfig} \usepackage{varwidth} \usepackage{placeins} \usepackage{textcomp} \usepackage{natbib} \setcitestyle{aysep={}}  \bibliographystyle{spbasic} \begin{document}  \begin{align*} w_{i} =  \Bigg\{1;\hspace{0.25cm} if\hspace{0.25cm} 0.25N \leq i \leq 0.75N \\                             0.5;\hspace{0.25cm} otherwise\\ \end{align*}  \end{document}  The output:

Can any one suggess me the code to write this equation in LaTeX.


I have been trying to install the Ubuntu 16_04 version on my laptop (Dell Inspiron 5577). I also tried installing the 18_04 version.  Both of them possess different errors:   The 16_04 version while installing through a USB drive, stops as soon as the Ubuntu logo pops up at the beginning of the installation In the 18_04 version, the installation is successful however while booting, after we select Ubuntu, the screen goes black and nothing else happened   The laptop has got a SSD of 128GB, and by default it was MBR partitioned. The above two issues occurred while it was MBR partitioned.  Later I deleted the partitions and installed the Windows 10 using a USB drive which had Windows 10 GPT partitioned. Hence, now my laptop has GPT partitioning. Windows 10 installation was successful after GPT partitioning in the SSD.  I again tried installing both the versions of Ubuntu using the alongside Windows installation option. Now, the Ubuntu 18_04 is installed on the 1TB HDD but has the same above error (screen goes black and nothing else happened). I tried installing 16_04 version which was unsuccessful with the same error again (Ubuntu logo pops up at the beginning of the installation)!  The error reads as "(sdc) no cache ...." and some ACPI error.   PC Configuration:  SSD: 128 GB HDD: 1TB Ram: 8GB Graphics Card: Nvidia GTX 1050 4GB Processor: Intel i5 7th Gen   Can somebody please help me out?

I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents:


Please, I can't log into my Ubuntu account. I have tried logging in using the right password. But can gain access to my account.

I'm working on a Ubuntu system, and my client has completely forgotten his administrative password. He doesn't even remember entering one; however it is there.  I've tried the suggestions on the website, and I have been unsuccessful in deleting the password so that I can download applets required for running some files. Is there a solution?


I'm trying to create a Graph data structure and sort it using Topological Sorting. My code is as below:  public class Node<T> {     // Private member-variables     private T data;     private NodeList<T> neighbors = null;      public Node() { }     public Node(T data) : this(data, null) { }     public Node(T data, NodeList<T> neighbors)     {         this.data = data;         this.neighbors = neighbors;     }      public T Value     {         get         {             return data;         }         set         {             data = value;         }     }      protected NodeList<T> Neighbors     {         get         {             return neighbors;         }         set         {             neighbors = value;         }     } }  public class NodeList<T> : List<Node<T>> {     public NodeList() : base() { }      public Node<T> FindByValue(T value)     {         // search the list for the value         foreach (Node<T> node in this)             if (node.Value.Equals(value))                 return node;          // if we reached here, we didn't find a matching node         return null;     } }  public class GraphNode<T> : Node<T> {     private List<int> costs = null;      public GraphNode() : base()     {     }      public GraphNode(T value) : base(value)     {     }     public GraphNode(T value, NodeList<T> neighbors) : base(value, neighbors)     {     }      new public NodeList<T> Neighbors     {         get         {             if (base.Neighbors == null)                 base.Neighbors = new NodeList<T>();             return base.Neighbors;         }     }      public List<int> Costs     {         get         {             if (costs == null)                 costs = new List<int>();             return costs;         }     } }  public class Graph<T> {      private NodeList<T> nodeSet;      public List<GraphNode<T>> Sort(Graph<T> graph, Func<GraphNode<T>, IEnumerable<GraphNode<T>>> neighBours)     {         var sorted = new List<GraphNode<T>>();         var visited = new Dictionary<GraphNode<T>, bool>();          foreach (var item in graph.Nodes)         {             var _tempItem = item as GraphNode<T>;             Visit(_tempItem, neighBours, sorted, visited);         }          return sorted;     }      public void Visit(GraphNode<T> item, Func<GraphNode<T>, IEnumerable<GraphNode<T>>> neighBours, List<GraphNode<T>> sorted, Dictionary<GraphNode<T>, bool> visited)     {         bool inProcess = false;         var alreadyVisited = visited.TryGetValue(item, out inProcess);          if (alreadyVisited)         {             if (inProcess)             {                 throw new Exception("Cyclic dependency found.");             }         }         else         {             visited[item] = true;              var dependencies = neighBours(item);             if (dependencies != null)             {                 foreach (var dependency in dependencies)                 {                     Visit(dependency, neighBours, sorted, visited);                 }             }              visited[item] = false;             sorted.Add(item);         }     } }   The call to Sort method is made as below:     List<GraphNode<string>> lstSorted = _graph.Sort(_graph, x => x.Neighbors);   But I'm getting below error in the call for x.Neighbors:  Cannot convert lambda expression to intended delegate type because some of the return types in the block are not implicitly convertible to the delegate return type  Can someone please tell me how to fix this?

Can someone explain to me why in .NET 2.0 if I have an interface, IPackable and a class that implements that interface OrderItem, when I have a method that takes in a List<IPackable>, passing in a list of List<OrderItem> does not work?    Does anyone know how I could accomplish this functionality?  Code:  public interface IPackable {         double Weight{ get; } }  public class OrderItem : IPackable   public List<IShipMethod> GetForShipWeight(List<IPackable> packages) {    double totalWeight = 0;    foreach (IPackable package in packages) {         totalWeight += package.Weight;    } }   The following code does not work.  List<OrderItem> orderItems = new List<OrderItem>(); List<IShipMethod> shipMethods = GetForShipWeight(orderItems);


I have an array of objects, modeling some videogames: let games = [     {title: 'Desperados 3', score: 74, genre: ['Tactics', 'Real Time']},     {title: 'Streets of Rage', score: 71, genre: ['Fighting', 'Side Scroller']},     {title: 'Gears Tactics', score: 71, genre: ['Tactics', 'Real Time']},     {title: 'XCOM: Chimera Squad', score: 59, genre: ['Tactics', 'Turn Based']},     {title: 'Iratus Lord of The Dead', score: 67, genre: ['Tactics', 'Side Scroller']},     {title: 'DooM Eternal', score: 63, genre: ['Shooter', 'First Person']},     {title: 'Ghost Of Tsushima', score: 83, genre: ['Action', '3rd Person']},     {title: 'The Last Of Us Part 2', score: 52, genre: ['Shooter', '3rd Person']} ]  If I want to sort it by games' scores it's very easy: const gamesSortedByRating = games.sort((a, b) => b.score - a.score);  But I've found out, that in order to sort it by games titles, it cannot be (or, at least I wasn't able to) do it this way: const gamesSortedByTitle = games.sort((a, b) => a.title - b.title);  I had to write a compare function to do that: function compare(a, b) {     const titleA = a.title.toLowerCase();     const titleB = b.title.toLowerCase();        let comparison = 0;      if (titleA > titleB) {       comparison = 1;     } else if (titleA < titleB) {       comparison = -1;     }      return comparison; }  The question is is there any way to do that in a more simple way? Like the score one liner I've shown above.

I have a list of objects I wish to sort based on a field attr of type string. I tried using -  list.sort(function (a, b) {     return a.attr - b.attr })   but found that - doesn't appear to work with strings in JavaScript. How can I sort a list of objects based on an attribute with type string?


So in my story a girl is hitchhiking east to live with her boyfriend. One of the drivers she's hitching a ride with is asking her why she's going east.  "Willow decided she didn’t want to tell him that she was going to live with her dorky boyfriend, so she decided on something more ___ / she decided to give a more __ answer" 'Moving,' she answered tersely.  I'm thinking "general" or "vague", but I don't know if they're exactly describing what I'm looking for. I'm not 100% sure a phrase would work here, but I could always rework the sentence if it fits the best. I need an adjective.

I'm looking for a single word synonym of evasion, fudging or dodging the question that does not imply deceit in the answer (apart from possible awareness of the respondent that they are not actually answering the question, and perhaps an intent to divert the questioner to other topics).  Answers could be verbose or blunt non-sequiturs.  This is similar but not identical to "What do you call a response which does not address the question?" There answers have other connotations, such as a word or phrase for an inappropriate response (such as a "non sequitur"). This is a new question, which differs particularly in that I'm looking for a word that does not imply an intention to deceive.


Show that $A$ satisfies all the axioms to be a commutative ring with unity. Indicate the zero element, the unity, and the negative of an arbitrary a.   A is the set $ℤ$ of the integers, with the following "addition" $⊕$ and "multiplication" $⊙$ :   $$a \oplus b = a + b - 1$$ $$a ⊙ b = ab - (a+b) + 2$$  In regards to this question, I'm mostly struggling with how to apply these axioms with the given operations. For example, to satisfy axiom 2, does the following approach make sense? $Let\ a, b, c ∈ A$ $Then:$ $a = pq - (p + q) + 2$ $b = st - (s + t) + 2$ $c = xy - (x + y) + 2$ $And:$ $(a ⊙ b) ⊙ c = ([pq - (p + q) + 2] · [(s + t) + 2]) · [(x + y) + 2]$ $a ⊙ (b ⊙ c) = [(p + q) + 2] · ([st - (s + t) + 2] · [xy - (x + y) + 2])$

Let R be a commutative ring with identity. Define a new operation $\circ$ on $R$ by for any $a,b \in R$  $$a \circ b=a+b-ab$$  a) Prove that $\circ$ is associative  b) Prove that R is a field iff the set $\{r \in R: r \ne 1\}$ is an abelian group with respect to the operation $\circ$  Starting with a) we must prove   $a \circ ( b \circ c)= (a \circ b) \circ c$  Left side: $a \circ ( b \circ c)=a \circ (b+c-bc)=a+b+c-bc-a(b+c-bc)=a+b+c-bc-ab-ac+abc$  Right side: $(a \circ b) \circ c =(a+b-ab) a \circ c=a+b-ab+c-(a+b-ab)c=a+b-ab+c-ac-bc+abc$ and we have thus proved that $\circ$ is associative  As for part b), I am a little confused on what is being asked exactly. Is it asking to go through ALL the properties of a field, closure, associativity, commutativity, identity, inverse, distributive, showing that all hold except for when $r \ne 1$?


I have a huge data in my table so I want to fetch it in parts. So I am trying to run select top 1000 from table_name  It's giving me the top 1000 data but what will be the query if I want the next 1000 data. It's just returning me the first 1000 data.

What is the best way (performance wise) to paginate results in SQL Server 2000, 2005, 2008, 2012 if you also want to get the total number of results (before paginating)?


It is famous by Ramanujan that $1+2+3+...=\frac{-1}{12}$. But what is the value of $1+1+1=....$?  I was thinking that $2=1+1$ and $3=1+1+1$ and so on...So does this mean that $1+1+1+...=\frac{-1}{12}$.  It means that when you have infinity you can do anything and everything. Please explain this

$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.  Why should analytically continuing to $\zeta(-1)$ give the right answer?


is redirecting to the old 2012 site:   Shouldn't it redirect to ?

Visiting  redirects to . This domain should be reused, and should redirect to . Why did I think this would happen? The use of this domain was announced in the original email to moderators, which read in relevant part:  This event will run from 16 December 2013 to 3 January 2014. Users will be able to see their entire hat collection on . That site will also have a landing page, explaining the rules and other details of the event.


This sentence is from  about Michael K. Williams from "The Wire".     The scar led to his being offered roles as a thug in music videos.   As I see it, the word being is either a synonym for person, or it is an auxilliary verb to create Passive Voice. But in the case of Passive Voice, personally to me the word him would sound more correct than his: "led to him being offered". Can you explain this structure?

The rule says that we can use a possessive adjective or an object pronoun before a gerund. Is there a rule  that says when to use each or are they interchangeable? Some say that it's wrong to use an object pronoun before a gerund.


At lunch, my coworkers and I were discussing The Force Awakens, and started questioning Chewbacca's age.  A quick check of  and it says he is possibly about 230 years old as of Episode VII.  Following the  link on that page, Wookieepedia claims that they live to be 400 years old.  Do we know from canon what the average life expectancy of a Wookiee is?

How old is Chewbacca, anyway? Do Wookiees get gray fur when they are old?