I have the following piece of code to format the excel files that I get and save them as modified.xlsx  public void FormatFile() {   FileName = Convert.ToString("test.xlsx");             SourceFileLocation = ("C:\Test");              SourceFileName = Path.Combine(SourceFileLocation, FileName);             saveLoc = Path.Combine(SourceFileLocation, "ModifiedExcel.xlsx");               var excel = new Excel.Application();             var workbook = excel.Workbooks.Open(SourceFileName);             //var sheet = (Excel.Worksheet)workbook.Worksheets.Item[1]; // 1 is the first item, this is NOT a zero-based collection             try             {                 foreach (Excel.Worksheet tempSheet in workbook.Worksheets)                 {                     if (((Excel.Worksheet)(tempSheet)).Name.Contains("Sheet1"))                     {                         if (File.Exists(saveLoc))                         {                             File.Delete(saveLoc);                         }                         tempSheet.Select();                         workbook.SaveAs(saveLoc);                     }                      System.Runtime.InteropServices.Marshal.ReleaseComObject(tempSheet);                 }                  workbook.Save();                 workbook.Close();                 excel.Quit();                  System.Runtime.InteropServices.Marshal.ReleaseComObject(workbook);                 System.Runtime.InteropServices.Marshal.ReleaseComObject(excel);             }             catch(Exception)             {                 System.Runtime.InteropServices.Marshal.ReleaseComObject(workbook);                 System.Runtime.InteropServices.Marshal.ReleaseComObject(excel);             }   I release the excel objects but when I check my task manager I can still see excel processes running. If i have run this code multiple time, that many processes are created. Am I not releasing the excel objects correctly?

I'm using the Excel interop in C# (ApplicationClass) and have placed the following code in my finally clause:  while (System.Runtime.InteropServices.Marshal.ReleaseComObject(excelSheet) != 0) { } excelSheet = null; GC.Collect(); GC.WaitForPendingFinalizers();   Although this kind of works, the Excel.exe process is still in the background even after I close Excel. It is only released once my application is manually closed.  What am I doing wrong, or is there an alternative to ensure interop objects are properly disposed of?


of the above mentioned movie has a few possible explanations for coherently explaining some of the events happening in the film, but there seems to be no explanation that fits all the events. Is it confirmed if the movie was created with a coherent storyline in the first place, keeping in mind that earlier there were shelved plans for a TV series based on the same?

I know that David Lynch's movies are all very deep and hard to grasp and that probably there is no simple answer to this question, as there are different ways to interpret the movie.   Still I would like to know, what is the main story behind Mulholland Drive and what is the most common interpretation of what it is about?  I watched it several times and read the 10 clues that Lynch gave to unlock the riddle, but I still don't think I understand the movie. Is it all a dream? A metaphor? Whats with the character switch? Does the appearance of the strange old couple has something to do with sexual abuse? What about the monster in the alley?  What the hell is going on in Mulholland Drive?


Snippet 1:  class Room(Object):       pass    Snippet 2:  class Room():       pass   What is the different between both of the code and why we write object in the above code?

What is the difference between old style and new style classes in Python?  When should I use one or the other?


Is there a way to host a SSL-tunneled (using stunnel) OpenVPN server and a regular Apache HTTPS server on the same server using HAProxy? It seems that by using stunnel, the format of the packets should all the way be the same. Is there any ways to make a difference between the two kinds of packets (which is not detectable by DPI) for the HAProxy to sort them out? I tried to use different domain names for the HTTPS server and the OpenVPN Server (though they point to the same IP address) but seems it doesn't work because OpenVPN does not use SNI.  Thanks a lot!

I'm trying to set up OpenVPN to listen on port 443, and then pass all HTTPS traffic to Apache, by using the port-share option. Relevant config snippets are: OpenVPN local ${PUBLIC_IP} port 443 port-share localhost 443  Apache with SSL Listen localhost:443  My OpenVPN client connects just fine, but when opening the HTTPS enabled page, I get errors. Firefox says:  SSL received a record that exceeded the maximum permissible length. (Error code: ssl_error_rx_record_too_long)  Curl says  curl: (35) error:140770FC:SSL routines:SSL23_GET_SERVER_HELLO:unknown protocol  The request ends up on Apache, since I see in the error logs the following messages: [Wed Oct 06 01:10:20 2010] [error] [client 127.0.0.1] Invalid method in request \x16\x03\x01 [Wed Oct 06 01:11:04 2010] [error] [client 127.0.0.1] Invalid method in request \x16\x03\x01 [Wed Oct 06 01:11:51 2010] [error] [client 127.0.0.1] Invalid method in request \x16\x03\x01  The messages entry for a HTTPS connection is Oct  6 01:13:21 ns1 openvpn[20154]: Re-using SSL/TLS context Oct  6 01:13:21 ns1 openvpn[20154]: LZO compression initialized Oct  6 01:13:21 ns1 openvpn[20154]: Control Channel MTU parms [ L:1544 D:140 EF:40 EB:0 ET:0 EL:0 ] Oct  6 01:13:21 ns1 openvpn[20154]: Data Channel MTU parms [ L:1544 D:1450 EF:44 EB:135 ET:0 EL:0 AF:3/1 ] Oct  6 01:13:21 ns1 openvpn[20154]: Local Options hash (VER=V4): 'c0103fa8' Oct  6 01:13:21 ns1 openvpn[20154]: Expected Remote Options hash (VER=V4): '69109d17' Oct  6 01:13:21 ns1 openvpn[20154]: TCP connection established with ${CLIENT_IP}:56203 Oct  6 01:13:21 ns1 openvpn[20154]: TCPv4_SERVER link local: [undef] Oct  6 01:13:21 ns1 openvpn[20154]: TCPv4_SERVER link remote: ${CLIENT_IP}:56203 Oct  6 01:13:21 ns1 openvpn[20154]: ${CLIENT_IP}:56203 Non-OpenVPN client protocol detected Oct  6 01:13:21 ns1 openvpn[20154]: TCP/UDP: Closing socket  Using httpd-2.2.3-43.el5.centos and openvpn-2.1.1-2.el5 . What should I do to make port sharing work?  Update: Using port 443 port-share localhost 10443  and Listen localhost:10443  makes no difference.  Update 2 : some command output [root@ns1 ~]# openvpn --help | grep port-share --port-share host port : When run in TCP mode, proxy incoming HTTPS sessions [root@ns1 ~]# netstat -nltp | grep 443 tcp        0      0 127.0.0.1:10443             0.0.0.0:*                   LISTEN      20088/httpd          tcp        0      0 ${PUBLIC_IP}:443             0.0.0.0:*                   LISTEN      20066/openvpn


I have been using SE sites (mainly SO) on my desktop computer and on my Samsung S4 mini mobile phone for years. It works well on of both devices. Recently I got a Samsung Galaxy Tab S2 9.7, and the SE App looks like this on it:    There are 2 columns, which is not my favorite, but okay. But the text in each entry in the columns is wrapped much more often than necessary. It does not only look ugly, it's actually hard to read this way.  Please fix that: make the actual column width adapt to the available width. And if possible, make configurable whether 1 or 2 columns are shown.  BTW: OS is LineageOS 15.1, but this should not matter.

The lines wrap so much on v1.0.85 that only a few items can be displayed.  Edit: This is on the world's most low end tablet, Azpen A727.  Screenshots of the old:    and new, showing loss of quality


Been messing around with Ubuntu, I actually installed it as my only OS but since my graphics card(AMD) kinda sucks on linux(For gaming purposes) -  I have to have a windows partition.  However, dual-booting simply won't work out for me. I currently installed a fresh copy of Windows 7.. That's fine. When I run the Ubuntu installer, it just doesn't recognize Windows 7. I'm left with wiping the entire hard disk and installing only Ubuntu, or basically nothing else(well, there is the "something else", but it doesn't recognize windows either).  What do I do? It's a problem with the new UEFI bios as far as I understand, right?  I tried booting from both USB/SD and CD/DVD with the same result, so I suppose something is up with the file structure or the UEFI/legacy booting. But before I mess with the BIOS, I'd like a pointer to what I should do exactly? I've tried shrinking my hard drive into a 200GB partition which I thought Ubuntu would recognize(and then install on), but it only recognizes all 1TB I have..? Thanks in advance!

I'm having a problem installing Ubuntu, and Kali as well. I downloaded the iso from the official site & tried installing using USB. I also tried desktop installation, but the problem is same.  It shows a single drive i.e. my total Hard Disk and if I select it, it asks me to create a new partition & shows total space as FREE.  It's with all Ubuntu I downloaded, 11.10 Server, 12.10 and 13.04. I also tried to install BackTrack's new version Kali Linux but again I got the same problem.  I hope I meet thier system requirements. Here's my system details for your ease:   OS: Windows 7 Ultimate Processor: Intel P4 2.80 GHz RAM: 3GB Hard Disk: 320GB No of Local Drives: 4 (1 Primary, 3 Logical) Free Space: 195GB (about) Motherboard: G31 Cache: 1MB Help me, please.


import java.util.*;   public class Main {      //This method is complete. Do not change it.     public static void main(String[] args) {         Random r = new Random();         r.setSeed(System.currentTimeMillis());         int n = r.nextInt(6) + 1;         List[]a = new ArrayList[1];          makeLists(a);          System.out.println(a.length);         for(int i = 0; i < a.length; i++)             System.out.println(a[i].toString());     }      private static void makeLists(List[] a) {           for (int i=0; i<a.length; i++){       List<Character> chars = new ArrayList<Character>();       chars.add(0,'A');       chars.add(1,'B');       chars.add(2,'C');       chars.add(3,'D');       chars.add(4,'E');       chars.add(5,'F');       chars.add(6,'G');       chars.add(7,'H');       chars.add(8,'I');       chars.add(9,'J');       chars.add(10,'K');       chars.add(11,'L');       chars.add(12,'M');       chars.add(13,'N');       chars.add(14,'O');       chars.add(15,'P');       chars.add(16,'Q');       chars.add(17,'R');       chars.add(18,'S');       chars.add(19,'T');       chars.add(10,'U');       chars.add(21,'V');       chars.add(22,'W');       chars.add(23,'X');       chars.add(24,'Y');       chars.add(25,'Z');        a[i] = chars;   I have to print out the linked list where data for 0 is A, 1 is A and B, when I do this above I get an error message illegal position 0. I believe that I need a nested for loop to print this, but I am not sure where to even begin.

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 want to have a large file accessible from Heroku for processing which will go a lot faster if it runs on Heroku than locally. I shared the file and got the link . It brings up a page saying it cannot preview, so it gives the option to download. That link is . But then it still gives another page saying it cannot virus scan 97MB and there is yet another button to confirm download.    If you press this final button in the browser, it will download, but you can't do it from a script because it sets a cookie or something. So how do I get a direct, direct link to a large file in Google Drive?  I found this  but it just gives the same link above, with the virus scan warning.  I don't have Amazon S3.

I am trying  but it does not work. I am not sure on how to repeat the question. It downloads a 'file is too big to be antivirus scanned' warning HTML and when I try https://drive.google.com/uc?export=download&confirm=no_antivirus&id= it still downloads that.  Edit: I tried to wget --save-cookies /tmp/cookie.txt --load-cookies /tmp/cookie.txt and repeat it, still no dice.


Find the remainder when $2^{1990}$ is divided by $1990$. I didn't get answer by Euler's generalization.

I actually do not have the basic idea on how to approach these type of questions....so please tell me a generalized method about all this too.  It came in RMO, and the question is:  What is the remainder when $2^{1990}$ is divided by $1990$ ?


import sqlite3 conn = sqlite3.connect('ProcessPower.db')  c = conn.cursor() user_input = (input("Enter a Type: "))  c.execute("SELECT * FROM Instrumentation WHERE Type = ?", user_input) rows = c.fetchall() for row in rows:     print(row) conn.close()

Using SQLite3 with Python 2.5, I'm trying to iterate through a list and pull the weight of an item from the database based on the item's name.  I tried using the "?" parameter substitution suggested to prevent SQL injections but it doesn't work. For example, when I use:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = ?", item)     self.cursor.close()   I get the error:     sqlite3.ProgrammingError: Incorrect number of bindings supplied. The current statement uses 1, and there are 8 supplied.   I believe this is somehow caused by the initial creation of the database; the module I made that actually creates the DB does have 8 bindings.  cursor.execute("""CREATE TABLE Equipment      (id INTEGER PRIMARY KEY,      name TEXT,     price INTEGER,      weight REAL,      info TEXT,      ammo_cap INTEGER,      availability_west TEXT,     availability_east TEXT)""")   However, when I use the less-secure "%s" substitution for each item name, it works just fine. Like so:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = '%s'" % item)     self.cursor.close()   I can't figure out why it thinks I have 8 bindins when I'm only calling one. How can I fix it?


just wanted to know why this program prints "25 and 7", instead of "25 and 25".  int b = 7; int c = b; b = 25;  System.out.print(b + " and " + c)   I am a little bit confused because the variable b has been replaced by the new value of 25, then why c still has its same value of 7?  Thank you in advance!

I always thought Java uses pass-by-reference. However, I've seen a couple of blog posts (for example, ) that claim that it isn't (the blog post says that Java uses pass-by-value). I don't think I understand the distinction they're making. What is the explanation?


I've heard this phrase but can't remember it--something like "[verb]ing the [noun]".  Something to describe a working on a task that began with a simple task but turned into a long string of working on requisite tasks, maybe circling back to the original.  "Last night a light bulb in my bathroom burned out and I didn't have a spare on hand, so I decided to start replacing my bulbs with LED ones, so tried to research the best ones but my Internet connection was down, so I tried to power-cycle my cable modem but couldn't see the back of it to plug it back in, so went into the bathroom to find a hand mirror but it was dark in there."  The circling back part might be called a deadlock, but that's not what I'm looking for, just the multiple-sidetrack situation.  Also not looking for "down the rabbit hole" which I think of as something different.

In some situations, trying to solve a problem uncovers another problem. Attempts to fix that uncover yet another. A few more rounds and one can be doing something seemingly completely unrelated to the main problem. Is there an expression for that?


As we all know, if you differentiate distance with reference to time, you get speed, and likewise, differentiating speed you get acceleration. However, if you keep differentiating, to the rate of change of acceleration and so forth (more emphasis on the so forth): is there any point to this? Does it have any useful application in any industrial or practical industry, as opposed to theoretical physics?  Also, what could we possibly learn about a situation by understanding the rate of the rate of the rate of change of the objects distance?

For example if the force on a particle is of the form $ \mathbf F = \mathbf F(\mathbf r, \dot{\mathbf r}, \ddot{\mathbf r}, \dddot{\mathbf r}) $, then the equation of motion would be a third order differential equation, what will require us to know the initial conditions $ \mathbf r(0), \dot{\mathbf r}(0), \ddot{\mathbf r}(0) $ in order to get the exact solution.  EDIT: As akhmeteliless mentioned the Abraham–Lorentz force is an example for such force. But, how such force is possible if the Lagrangian contains only the coordinates and their first time derivatives? Shoudn't the equations of motion be second order differential equations?


I've got a register which can either be 0x02 or 0x01 (0010 or 0001). What is the most elegant way of setting the bits in a single operation? E.g. if the register is currently 0x02, I need to turn off bit 2 and turn on bit 1.  Many thanks

How do you set, clear, and toggle a bit?


I need to upgrade Python on 30 laptops at several remote sites. These laptops have Debian 9 with Python 3.5, and the library they'll be using for the planned activity (arcade) requires Python 3.6.   I have an update process that can deploy package updates, but it needs to be able to install quickly, so compiling Python is not an option.

I'd like to install the latest Python, which is 3.6 at the time of this post. However, the repository is saying that Python 3.4.2 is the newest version.  I've tried:   $ sudo apt-get update $ sudo apt-get install python3  python3 is already the newest version.  $ python -V  Python 3.4.2   To upgrade to Python 3.6 on my Windows workstation, I simply downloaded an exe, clicked "next" a few times, and it's done. What's the proper and officially accepted procedure to install Python 3.6 on Debian Jessie?


I am having trouble doing this question and I do not know where to start. Any help is appreciated.

Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$ $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$


My graphic designer left our organization, and my Photoshop abilities less advanced. I'm having trouble figuring out how to make the letters flush with the sides of the cube.   The .psd file I have does not have a 3D layer, just the sides. I'm sure it's something simple, but it eludes me.

I'm using Photoshop CS6 and I'm trying to get the typing aligned with the box, kind of 3D like. I cannot seem to find the effect, I'm sure its simple.  Any help?


Here is my derivation. $i^2= ((-1)^{0.5})^2$ $=((-1)^2)^{0.5}$ = $1$ Is there a rule saying that 2nd line is not allowed? I have seen on a related post that sqrt(a).sqrt(b)=sqrt(a.b) is true only if a and b are above zero But i do not understand what it can mean for a and b (2 complex numbers) to be greater than zero?  Please explain this rule.  And also if the rule can be more generally stated? Note: i learnt in school that square root of x means the POSITIVE square root of x. That was true throughout all algebra before we got to complex numbers. I now understand that that referred to the "principal" root. (Root with the highest real element). Also when we define i as square root of -1, this is a strange statement actually because also, -i is also sqr rt of -1.

I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &= i \\\\\ \frac1{\sqrt{-1}} &= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\  \sqrt{\frac1{-1}} &= \frac1i \\\\  \sqrt{\frac{-1}1} &= \frac1i \\\\  \sqrt{-1} &= \frac1i \\\\  i &= \frac1i \\\\  i^2 &= 1 \\\\  -1 &= 1 \quad !!? \end{align}


Many questions on stack overflow are duplicates. These are often closed, but what would be nicer is if the question could be declared a synonym of another (by users of sufficiently high rep, or by votes).   Why do this? Well one reason duplicates come up is that there's lots of different ways of phrasing the same basic question. This means that when I type in the title for my question and it suggests related ones, it may completely miss the identical question due to it being worded completely differently. Having the ability to link different questions could go some way to alleviating this problem, and linking the good answers to the previous incarnation of the question to the newly asked question, instead of having "answers" consisting of stuff like:     Dupe:    P.S. This may well have been asked before, but I've checked all the suggested Related Questions and it doesn't look like it!

What should I do when I see a question that is a duplicate of another one?   Should I answer it? Should I downvote it? Should I comment? Should I edit the question to indicate it's a duplicate? If I can, should I (vote to) close the question? What happens when I vote to close as duplicate? Or: Why did a comment with my name on it just appear? Should I flag it for moderator attention? What about similar or related questions?


Is there a possibility to place pairs of figures on the even page and the corresponding odd page, respectively? The scenario is that I have e.g. 2 figures showing the same statistic (separating males and females). Each figure spreads over a whole page. Now, it would be very good for improved readability (for comparing males' and females' results) if this pair would be placed on corresponding even and odd pages:  |~~~~~~~~~~||~~~~~~~~~~| |Text Lorem||elitr, sed| |ipsum dolo||eirmod te | |sit amet  ||fot.      | |sed diam  ||          | |ut et doro||          | |~~~~~~~~~~||~~~~~~~~~~| |~~~~~~~~~~||~~~~~~~~~~| |-FigureA-+||-FigureA-+| |-females+-||-males+-++| |++++---+++||+++-+----+| |-++-++--++||+++-+-----| |++-++-++-+||--+-+--+-+| |~~~~~~~~~~||~~~~~~~~~~|  \documentclass[twoside=on]{scrartcl} \begin{document}     Lorem ipsum dolor sit amet, consetetur sadipscing elitr,  sed diam nonumy eirmod tempor invidunt ut labore et dolore  magna aliquyam erat, sed diam voluptua. At vero eos et accusam   et justo duo dolores et ea rebum. Stet clita kasd gubergren,   no sea takimata sanctus est Lorem ipsum dolor sit amet.       \begin{figure}         \includegraphics{statisticA-females.pdf}     \end{figure}     \begin{figure}         \includegraphics{statisticA-males.pdf}     \end{figure} \end{document}   Hence, what I need, is something like \cleardoublepage without leaving float mode for figures (without creating empty space, at least, as Long as there is text left).  Is this possible? Thanks for any suggestions!  Fabian

I would like to reproduce the following layout with two figures on one page and two others on the facing pages.    I would like the first couple to be on an even-numbered page and the second on an odd-numbered page so that all four figures can be seen together when the manuscript is printed and bound.


I really like the look and feel of ubuntu before they ditched unity. The last version to use unity was 17.04. The problem is 17.04 ended life in January 13, 2018. How do I make ubuntu 20.4 (the latest version) look like ubuntu 17.04?

I recently installed Ubuntu 18.04 and I'm not keen on the desktop environment. I want the desktop env. I used to have in Ubuntu 14.04. How can I get it?  I've attempted to use , but when I try to install the Unity's 7.2.5 version using the command:  apt install unity=7.2.5   it gives me   E: Version '7.2.5' for 'unity' was not found   For sure I need to add a PPA for this, but which one!?


While studying polynomials a question struck in my mind that  To have a polynomial $p(x)$ with integer coefficients and $a,b,c$ are three distinct integers, then is it possible to have $$p(a)=b$$$$p(b)=c$$$$p(c)=a$$

Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated.


Please help!  H/T, $50\%$-$50\%$ independent events. I pick Heads. What is the probability that after $29$ attempts of a fair coin, there's no consecutive Heads (...HH...) at all?  I would appreciate seeing how it's worked out...  Thanks a bunch.

A fair coin is tossed $10$ times. What is the probability that no two   consecutive tosses are heads?   Possibilities are (dont mind the number of terms):  $H TTTTTTH$, $HTHTHTHTHTHTHT$.  But except for those,  let $y(n)$ be the number of sequences that start with $T$  $T _$, there are two options, $T$ and $H$ so,  $y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$  Let $x(n)$ be the number of sequences that start with $H$,   $H _$ the next option is only $T$ hence,  $x(n) = y(n - 1)$   The total number of sequences $F(n)$ is:  $$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$  We are after $F(10)$,  $$F(10) = 2y(9) + y(8)$$  $$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$  $$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$  But I'm not quite sure where this will take me though.


I'm trying to open Google chrome on my Ubuntu 14.04 . But I get the following error.  command :   # google-chrome-stable Result :      abort (core dumped)  google-chrome-stable  command :   # google-chrome Result :      abort (core dumped)  google-chrome   I tried re-installing but this didn't help.

It was working properly before, suddenly it stopped working. When I tried to open by clicking on its icon, but nothing happens.  As already here are many questions related to this, but none of them are working for me. I already checked  and .  I uninstalled Chrome by following the instructions from .  sudo apt-get purge google-chrome-stable rm ~/.config/google-chrome/ -rf   Again I installed Chrome by:  sudo apt-get install google-chrome-stable   But the problem is still thesame.  When I enter google-chrome in terminal I get following error.  [12289:12323:0723/014456:ERROR:nss_util.cc(744)] Error initializing NSS with a persistent database (sql:/home/maddy/.pki/nssdb): libsoftokn3.so: cannot open shared object file: No such file or directory [12289:12323:0723/014456:ERROR:nss_util.cc(212)] Error initializing NSS without a persistent database: NSS error code: -5925 [12289:12323:0723/014456:FATAL:nss_util.cc(214)] nss_error=-5925, os_error=0


Recently, I finally got the official drivers for my NVIDIA 560 Ti card installed on Ubuntu 14.04 (hooray) However I started looking into installing Steam and I'm getting segmentation errors when I try to run the software. I tried installing 32-bit libs and it seemed like they weren't available or they were already installed. Upon further investigation, I found that a solution is to install the proprietary drivers, install steam then switch back to the other drivers. I'm not really sure what "proprietary drivers" are in all honesty. Has anyone gone through this process that could provide some insight here?  (I installed the official 64-bit driver from the NVIDIA site for my 560 Ti just for reference. And the Ubuntu version installed is 64-bit as well)  Update:  This is the error text I get when trying to run steam after installing it via the ubuntu store.  Running Steam on ubuntu 14.04 64-bit STEAM_RUNTIME is enabled automatically Installing breakpad exception handler for appid(steam)/version(1401381906_client) /home/dbrewer/.steam/steam.sh: line 755:  3943 Segmentation fault      (core dumped) $STEAM_DEBUGGER "$STEAMROOT/$PLATFORM/$STEAMEXE" "$@" mv: cannot stat ‘/home/dbrewer/.steam/registry.vdf’: No such file or directory Installing bootstrap /home/dbrewer/.steam/bootstrap.tar.xz Reset complete! Restarting Steam by request... Running Steam on ubuntu 14.04 64-bit STEAM_RUNTIME has been set by the user to: /home/dbrewer/.steam/ubuntu12_32/steam-runtime Installing breakpad exception handler for appid(steam)/version(1401381906_client) /home/dbrewer/.steam/steam.sh: line 755:  4066 Segmentation fault      (core dumped) $STEAM_DEBUGGER "$STEAMROOT/$PLATFORM/$STEAMEXE" "$@"   What I get when I run "steam --reset"  mv: cannot stat ‘/home/dbrewer/.steam/registry.vdf’: No such file or directory Installing bootstrap /home/dbrewer/.steam/bootstrap.tar.xz Reset complete!

[SOLVED] I managed to fix the issue by uninstalling the proprietary AMD drivers, rebooting, starting Steam letting it update, reinstalling AMD drivers, reboot. It seems to be working and I can now play all my Steam linux games. I just installed Ubuntu 13.10 64bit on my main PC. I also installed Steam (.deb from ). The install went without errors. On running Steam, nothing happens. I run Steam from the terminal and receive two segmentation faults, always occurring on different lines of Steam.sh when I run Steam: /home/samathy/.local/share/Steam/steam.sh: line 755: 11111 Segmentation fault   /home/samathy/.local/share/Steam/steam.sh: line 755: 11239 Segmentation fault   The rest of the output on the terminal is just dump-saving text. I can't see anything about GLX issues, or locale issues (they seem to be popular issues with steam) I'm running Ubuntu 13.10 64bit with Xubuntu (installed after the Ubuntu install.) My specs are as follows: AMD FX 6350. AMD HD7870 (With AMD proprietary beta drivers) 12GB DDR3 Samsung 840 EVO 120GB. This problem is a little weird to me because Steam runs fine on my Ubuntu 13.10 + Xubuntu-desktop, Intel Atom N270 netbook without any fiddling, I can even run games. I've googled a lot and only really found solutions to other steam issues, not my own. If anyone could help, that'd be wonderful. Thanks! edit: I should mention that I've tried doing ~$steam --reset  It does not produce errors its self, but also doesnt solve the problem either. I have also tried installing Steam from Ubuntu package manage, but that results in the same errors.


I've started using English language about 4 years ago after I moved to England. I came across this practice a few times: when people speak about their health or similar things they say this and touch a wood nearby. What does this mean and is there a historical story behind it? Appreciate if anyone could explain this.

In the US, it is common to use the idiomatic expression "knock on wood" to keep "a good thing going."  Is it the same in the UK, or do they use a different idiomatic expression?


I'm planning to buy a Nexus 5 next month, but before that I'd like to know if I will be allowed to update it to the next Android version, when it get released, that for now is being called Android L.

Android 5.0 (Lollipop) was officially announced on October 15th, 2014.  You can review the high-level changes at the  and the announcement on the official .  Of course, not all devices will get this update right away and others won't get it at all. Each manufacturer and carrier often choose to add their custom modifications which tend to delay the official release.  When will each device get Android 5.0, either officially or unofficially, if ever?  See also:


So correct if I am wrong, but these sets are:  $\mathbb Q (\sqrt2 + \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 +d \sqrt6 : a,b,c,d \in \mathbb Q \}$  $\mathbb Q (\sqrt2 , \sqrt3 )=\{a+b\sqrt2 + c\sqrt3  : a,b,c \in \mathbb Q \}$  Was trying to see if the LHS is a subset of RHS and vise versa.  Let $A \in \mathbb Q (\sqrt2 , \sqrt3 )$ with $A=a+b\sqrt2 + c\sqrt3= a+b\sqrt2 + c\sqrt3  + d \sqrt6 - d \sqrt6 = a+b\sqrt2 + c\sqrt3  + d \sqrt6 - d \sqrt3 \sqrt2 $ but i can't really do anything after this. If you factor the root $3$ or $2$, it doesn't seem to help.

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ?    $$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$   $$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace  $$  So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.  How to conclude from there?


I have read , which gave me some info but I feel I am not fully getting it. What does the execute permission actually mean on a folder? I fully understand what it means on a file, but on a folder.

In my CMS, I noticed that directories need the executable bit (+x) set for the user to open them. Why is the execute permission required to read a directory, and how do directory permissions in Linux work?


if a character is given the ability to take the dash action as a bonus action or take multiple actions on the same turn, can said character use all of them to take the dash action?  considering a wood-elf, monk-fighter multiclass with a bunch of wizard friends about to take part in an Olympic event. the wizards sneak up and caste haste, longstrider, and expedited retreat on him at the same time. the person would have a base movement of 65 feet + 10 feet for longstrider and 3 dash actions per turn. (5 with action surge) with a movement speed of 225 ft per round (37.5 ft per second) or 375 ft/round (62.5 ft/s) with action surge. other than local traffic laws is this scenario valid RAW.  addendum; if it is possible to take multiple dash actions on the same turn, can a rogue attempt to hide multiple times on a single turn?

A rogue receives a bonus Cunning Action, and one option is to Dash. My understanding is a Dash allows a player to move up to twice his speed for his turn.  If a rogue is using a regular action to Dash (hence, is already dashing) can he also use his cunning action as a bonus to Dash again — effectively moving 4 times normal speed in a turn?


I have imported a .csv file containing unique identifiers with two decimal places (e.g. 53500001.00, 53500001.01 etc.). I am joining this file to a vector file with the same identifiers. However, when I import the .csv file the identifiers ending with .00 have been reduced to integers in the attribute table and the join returns null values in those instances.  It is census data so I can't go back through and manually change this as it's hundreds of thousands of lines. Is there a way to change the precision of the attribute table similar to in excel?

I have data in an Excel file that I want to join with a shapefile in my project. The Excel file contains values like volumes. The Excel file and the shapefile I want to join it with both contain the same property names.  The problem is, when I join the two, I get too many decimals in the attribute table. I get six, but I only want it to be one.  I tested to change the format of the cells to one decimal, and then I saved the file as *csv. But, somehow I seem to loose data in the process? It should be 42 rows that contains data in the column i am working with, but I end up with only like 10?  Another thing is that my property names contains the letters Å. Ä and Ö, perhaps this complicates things?  I read the question , but it doesn't mention the problem with the data loss.  My question was tagged with , but I didn't write what version I'm using. I use QGIS 2.2.0. Would it help to upgrade to the latest version?


Is correct grammatically to say      "Jewish is the nation that Jesus came from!"   as a declarative sentence ?  I ask my question because of that's strange to me to see the end of the sentence ends with preposition (as you can understand by yourself, I'm not English native speaker...)

When I learned English at school, I was taught that I should not end a sentence with a preposition.  Is it correct to end a sentence with a preposition?     To avoid starting a sentence with a conjunction, I don't write sentences similar to "I miss you. And your kisses." Is there a similar way to avoid ending a sentence with a preposition?


Prop: Empty set is a subset of every set.   Pf:. First consider the empty set. $\emptyset \subseteq \emptyset \iff \emptyset \bigcap \emptyset = \emptyset $ and $\emptyset \bigcup \emptyset = \emptyset$. $\emptyset \bigcap \emptyset =\{x\in \!\, \emptyset  \ \text{and} \ x\in \!\, \emptyset \!\,\}=\emptyset$ and $\emptyset \bigcup \emptyset =\{x\in \!\, \emptyset  \ \text{or} \ x\in \!\, \emptyset \!\,\}=\emptyset$. Thus, $\emptyset \subseteq \emptyset$.  Now consider a non-empty set $S$. $\emptyset \subseteq S \iff \emptyset \bigcap S = \emptyset $ and $\emptyset \bigcup S = S$. $\emptyset \bigcap S =  \{x \in S \ \text{and} \  x \in \emptyset \} = \emptyset$ and $\emptyset \bigcup S =\{x\in \!\, S  \ \text{or} \ x\in \!\, \varnothing \!\,\}=S$.  Hence, $\emptyset \subseteq S$ as required $\square$. Does my proof look right?

Recently I finished my first pure mathematics course but with some intrigue about some proofs of definitions by contradiction and contrapositive but not direct proofs (the existence of infinite primes for example), I think most of them because the direct proof extends away from a first mathematics course or the proofs by contradiction/contrapositive are more didactic. The one that most bothers me in particular is the demonstration that the empty set is a subset of every set, and it is unique. I understand the uniqueness and understand the proof by contradiction:  "Suppose $\emptyset \subsetneq A$ where $A$ is a set. So it exists an element $x \in \emptyset$ such that $x \notin A$ wich is absurd because $\emptyset$ does not have any elements by definition."  but I would like to know if there exists a direct proof of this and if indeed extends from a first course. Thanks beforehand.


My application should run on a headless server, thus I cannot rely on Blender GUI, and I need a stand-alone python script.. I have been looking all over the place for an example of the creation on a particle system in pure Python API, but all the examples either assume the GUI usage or are not up to date with the API. Can someone provide a minimal snippet for the creation of a particle system in Python? Thanks, Simon

I have been using this code to add a new particle system and create a name for it... I need help to choose the particle type "HAIR" and set other parameters.. This script does not work because the [part.001] keeps changing and going up. It will only work once when creating the first name. I will need one particle system, so I can apply the code on other objects as well. It is a bit confusing. The MBall is used as a particle to cover the entire surface of an object.  import bpy  obj = bpy.context.active_object obj.modifiers.new("part", type='PARTICLE_SYSTEM')  bpy.data.particles["part"].type = 'HAIR'  bpy.data.particles["part"].count = 15000 bpy.data.particles["part"].use_advanced_hair = True bpy.data.particles["part"].render_type = 'OBJECT' bpy.data.particles["part"].instance_object = bpy.data.objects["Mball"] bpy.data.particles["part"].particle_size = 0.15 bpy.data.particles["part"].emit_from = 'FACE' bpy.data.particles["part"].distribution = 'RAND' '


If I have a [digital] photograph of an object, can I estimate it's approximate size?  I notice with interest that my camera (Nikon D3300) seems to record the approximate focal length of the lens in the Exif data. (It seems to round it to the nearest "common" nominal length.) However, I see nothing in the Exif data about the plane of focus. Presumably the camera system has that data somewhere (e.g., some lenses have it printed on the focusing scale), but I don't see it recorded anywhere.  Assuming I had the lens focal length and the lens focus distance, presumably it should be quite straight forward to estimate object size.  (In case it matters, I'm particularly interested in extremely tiny objects such as ants, grains of sand... stuff you can't exactly take a ruler to. I don't know whether that makes size estimates easier or harder.)

A lot of people have asked if we can measure the distance of an object in photo. But I was wondering if we can measure the size of an object in photo. Of course assuming we know everything from sensor size to height of the photographer etc.  Example: If I take a photo of a bird in flight with only sky as background, there would be no objects in the frame to give an idea of size of the bird. So if I could calculate the size to some approximate level, it would be really nice to know.


Is there a way or a package to guess the type of a file in Python? For example, is it a way to detect a file could be open as ascii, unicode or binary?   Thanks in advance!

Let's say you want to save a bunch of files somewhere, for instance in BLOBs. Let's say you want to dish these files out via a web page and have the client automatically open the correct application/viewer.  Assumption: The browser figures out which application/viewer to use by the mime-type (content-type?) header in the HTTP response.  Based on that assumption, in addition to the bytes of the file, you also want to save the MIME type.  How would you find the MIME type of a file? I'm currently on a Mac, but this should also work on Windows.   Does the browser add this information when posting the file to the web page?  Is there a neat python library for finding this information? A WebService or (even better) a downloadable database?


I think stackoverflow.com is great, and I would like to contribute. Usually, I find the answers to my questions are already there. I'd like to upvote the best answers and questions, but I do not have any reputation here. How do I get initial reputation?  The canonical way seems to ask a question. Which I hereby did.  Any other suggestions? I'd like to avoid stupid, pointless things (which I already failed to do by asking this question...).

On Stack Exchange, users may gain a certain level of reputation.  What does reputation do? How can a user gain or lose reputation?   See also   in the Help Center  - Maximum amount of votes a user can use in a day and reputation requirements to vote.  - How to get an exact overview of the actions that got you your reputation.  - A full list of all privileges granted at each reputation level on all sites.


I'm trying to fix a bug that occurs seldomly (a concurrentModificationException) while looping through a Set of generics Set (on the clean() method).  for (final K key : this.bean.keySet()) {     getProxy(key).clean(); }   All the answers I found suggest to use an iterator and then calling the method remove() on it but here I am not looping through a list where I need to remove an item.  There is maybe also the possibility to convert my Set to an array and lock it (using java.util.concurrent.Locks) / make it synchronous.  I am confused about the best approach to fix my problem, the solutions above don't really satisfy me as they make me lose all benefits of multithreading.

I am using a Collection (a HashMap used indirectly by the JPA, it so happens), but apparently randomly the code throws a ConcurrentModificationException. What is causing it and how do I fix this problem? By using some synchronization, perhaps?  Here is the full stack-trace:  Exception in thread "pool-1-thread-1" java.util.ConcurrentModificationException         at java.util.HashMap$HashIterator.nextEntry(Unknown Source)         at java.util.HashMap$ValueIterator.next(Unknown Source)         at org.hibernate.collection.AbstractPersistentCollection$IteratorProxy.next(AbstractPersistentCollection.java:555)         at org.hibernate.engine.Cascade.cascadeCollectionElements(Cascade.java:296)         at org.hibernate.engine.Cascade.cascadeCollection(Cascade.java:242)         at org.hibernate.engine.Cascade.cascadeAssociation(Cascade.java:219)         at org.hibernate.engine.Cascade.cascadeProperty(Cascade.java:169)         at org.hibernate.engine.Cascade.cascade(Cascade.java:130)


Based on my understanding of the “Double-Slit Quantum Eraser Experiment”, documented here: , it seems that Faster-Than-Light communication is possible.  Of course, this violates the “no-communication theorem”, so I assume it’s due to a misunderstanding on my part.  In the experiment, changing whether the polarization of the ‘p’ photon is erased or not changes whether or not interference patterns appear for the ‘s’ photon.  Suppose Alice and Bob generate many entangled photons and then travel to opposite sides of the galaxy, Alice taking the ‘p’ photons and Bob taking the ‘s’ photons.  At a predetermined time Bob releases his photons one by one into a double slit setup as in the experiment.  If he sees an interference pattern, he knows that Alice must have “erased” the polarization information on her corresponding ‘p’ photons, and if not that she has instead measured it.  Of course, even stranger situations are possible.  The fact that the ‘s’ photon in the experiment showed interference patterns when the ‘p’ photon was going to be erased, but hadn’t been yet, makes me think that photons can retrieve information from the future.  For instance, measuring the ‘s’ photon and detecting interference patterns or not, while the ‘p’ photon is bouncing around in some system, waiting to have its polarization be either erased or measured depending on the result of some future event, like an election result.  If the ‘s’ photon shows an interference pattern, then you would know the result of the electron (or know that forces beyond your control erase the information for some other reason).  I’d be very appreciative if you could clear this up for me.  My search online has not been fruitful, since the only similar questions I’ve seen just say something like “When you have entangled photons, measurements on one member of the pair cannot tell you anything about whether the other member has been measured or not.”, which seems to be contradicted by the above experiment.  Thanks!

It is often stated, particularly in popular physics articles and videos, that if one measures a particle A that is entangled with some other particle B, then this measurement will immediately affect the state of the entangled partner.  For example, if Alice and Bob share an entangled pair of electrons and Alice measures her spin in the $x$ direction, then Bob's spin will also end up spinning in that direction, and similarly if she measures in the $z$ direction. Moreover, the effect will be instantaneous, regardless of the spatial distance between the two particles, which seems at odds with special relativity.  Can I use a scheme like this to communicate faster than light?


I guess I'm stupid or what but I don't get this:  I'm using the  RemoteIPHeader X-Forwarded-For within a VirtualHost configuration in order to determine GEO location of our users within our PHP application.  But I also would like to log the Remote IP of the clients within the apache logs.  The Apache combined log format looks like this:  LogFormat "%h (%{X-Forwarded-For}i) %l %u %t \"%r\" %>s %b \"%{Referer}i\" \"%{User-Agent}i\"" combined   According to documentation:  RemoteIPTrustedProxy Directive      The RemoteIPTrustedProxy directive adds one or more addresses (or   address blocks) to trust as presenting a valid RemoteIPHeader value of   the useragent IP. Unlike the RemoteIPInternalProxy directive, any   intranet or private IP address reported by such proxies, including the   10/8, 172.16/12, 192.168/16, 169.254/16 and 127/8 blocks (or outside   of the IPv6 public 2000::/3 block) are not trusted as the useragent   IP, and are left in the RemoteIPHeader header's value.   In my case the directive looks like this:  RemoteIPTrustedProxy 172.31.0.0/16   Where I'm saying trust any PROXY forwarding from this local subnet.  BUT, when I set this Apache stops logging the remote IP because the forwarding proxy is valid now ?  So If I set:  RemoteIPTrustedProxy 10.10.10.10    The IP above doesn't exist. So the proxy servers from 172.31.0.0/16 are invalid now and therefore apache starts logging the remote IP header again?  Please, try to give me some guidance here, as obviously I don't understand this.  Thanks a lot !

I'm using Apache/2.4.27  Within the VirtualHost I'm forwarding the remote client IP header from the Loadbalancer with:  RemoteIPHeader X-Forwarded-For   Which is needed by the application served by that Virtualhost.  This is the log format within the main httpd.conf context.  LogFormat "%h (%{X-Forwarded-For}i) %l %u %t \"%r\" %>s %b \"%{Referer}i\" \"%{User-Agent}i\"" combined   When I have the RemoteIPHeader X-Forwarded-For listed within the virtualhost, Apache stops writting the remote-client IP into logs.  When I remove it from the VirtualHost, the remote-client IP starts appearing again within the logs.  Any ideas why the RemoteIPHeader X-Forwarded-For doesn't play with (%{X-Forwarded-For}i) from the LogFormat ?  Thanks !


When you're looking at a question that has a "migrated from..." banner (like ), and then somebody edits a post, spawns one of those yellow boxes that prompt you to click to reload that post, and you click on that box, the "migrated from..." banner vanishes.   This bug appears to have been fixed for other types of banners (see ).

I've come across a small issue which occurs when a question has been closed but the question itself or one of its answers are being edited.  Steps to reproduce:   Visit a closed question. (even closed years ago)   In a different tab or browser, edit the question or one of its answers. "an edit has been made to this post; click to load" message will appear above the edited post in the first tab/browser, click it. The .question-status div becomes hidden, which includes the closed notice below a question and, for a duplicate-closed question, the list of duplicates above the question body. This gives the appearance that the question is no longer closed (until you refresh).   Some screenshots:                I know it isn't common/important but it did confuse me when looking at a particularly bad question - I thought it had been reopened.   Is this an intentional feature? If so, what was the thinking behind it?    Another example  Before    After


A Tikz picture has four nodes and each node has a new tikz picture, an inheritance problem coming. I have four Tikz pictures where each is inside a node and each node has size of 4cm times 4cm (square) but this size is inherited to other nodes where node size only ~1mm as shown in Bad Demo below with big black circles. So  How can I overwrite Tikz node settings in the inner nodes?  Bad Demo Four tikz pictures jammed together bad    Four tikz pictures with larger node but node style inherited, how to overwrite it?   where my goal is to have each node of size 4cm and each single tikz picture inside each node -- by this I try to make my tikz pictures more accessible and visually pleasing.  MWE \documentclass[english]{article} \usepackage{tikz} \usepackage{pgfplots}  \begin{document} \begin{tikzpicture}  \node[minimum size=4cm](q1)[label=(a) $x_1+x_2$]{ \begin{tikzpicture}[x=2cm, y=2cm] \coordinate [label=$x_1$] (1) at (0,0); \coordinate [label=$x_2$] (2) at (1,0); \node[draw=none,circle,inner sep=1pt,fill=black] (11) at (1){}; \node[circle,inner sep=1pt,fill=black] (22) at (2){}; \end{tikzpicture} };   \node(q2)[right =of q1,label=(b) $x_1+x_1x_2+x_2x_3$]{ \begin{tikzpicture}[x=2cm, y=2cm] \coordinate [label=$x_1$] (1) at (0,0); \coordinate [label=$x_1x_2$] (2) at (1,0); \coordinate [label=$x_2x_3$] (3) at (2,0); \draw (1)--(2)--(3); \node [circle,fill=black,inner sep=1pt] (11) at (1){}; \node [circle,fill=black,inner sep=1pt] (22) at (2){}; \node [circle,fill=black,inner sep=1pt] (33) at (3){}; \end{tikzpicture} };   \node(q3)[below =of q1,label=(c) $x_1x_2x_3+x_1+x_2+x_3$]{ \begin{tikzpicture}[x=2cm, y=2cm] \coordinate [label=left:$x_1$] (1) at (0,0); \coordinate [label=right:$x_2$] (2) at (2,0); \coordinate [label=below:$x_3$] (3) at (1,-2); \coordinate [label=right:$x_1x_2x_3$] (123) at (1,-1); \draw (1)--(123)--(2); \draw (3)--(123); \end{tikzpicture} };   \node(q4)[right =of q3,label=(d) $x_1x_2x_3+x_1x_2+x_2x_3+x_2x_3$]{ \begin{tikzpicture}[x=2cm, y=2cm] \coordinate [label=left:$x_1$] (1) at (0,0); \coordinate [label=right:$x_2$] (2) at (2,0); \coordinate [label=below:$x_3$] (3) at (1,-2); \coordinate [label=right:$x_1x_2x_3$] (123) at (1,-0.8); \draw (1)--(123)--(2)--(1); \draw (3)--(123); \draw (1)--(3); \draw (3)--(2); \end{tikzpicture} }; \end{tikzpicture} \end{document}  where [minimum size=4cm] creates the two big circles, its removal returns the jammed pictures in the beginning. Trials  Trial 1: scope is suggested to avoid the overwriting but LaTeX Error: Environment scope is undefined, why?

update i discovered lately  where Nickolay Kolev uses  \tikz inside a tikzpicture environment. A lot of users think that is a wrong practice and I agree that is very strange but sometimes very useful. I agree too that it's preferable to avoid this kind of situation. I use this possibility in  pgfornament.  I don't understand why if I invert the two lines of the tikzpicture (the wrapper) I get different results. overlay I think, resets the bounding box but why inside the second tikzpicture, the overlay resests the general bounding box. When I use a scope, the problem disappears ?   update I suppose that tikz's experts think it's preferable to use a scope and the question is stupid ...  Or in the pgfmanual, I read      All graphic options are local to the {tikzpicture} to which they   apply.   The picture in red has no bounding box.  \documentclass[11pt]{scrartcl} \usepackage{tikz}     \begin{document}  baseline\begin{tikzpicture}[baseline=(current bounding box.north west)]   \tikz[overlay] \path (0,0) --  (1,1) node  {\color{blue}$\bullet$};  \draw[help lines,blue!20] (0,0) grid (2,2) ; \end{tikzpicture}baseline% %  \begin{tikzpicture}[baseline=(current bounding box.south west)]    \draw[help lines,red!20] (0,0) grid (2,2) ;   \tikz[overlay] \path (0,0) --  (1,1) node  {\color{red}$\bullet$};  \end{tikzpicture}baseline     \end{document}   With pgfinterruptboundingbox     Picture in green. The bounding box is correct. The first bounding box is protected.  The solution is to place the line with overlay inside a pgfinterruptboundingbox environment so the problem is linked to the current bounding box.    baseline\begin{tikzpicture}[baseline=(current bounding box.south west)]       \draw[help lines,green!50] (0,0) grid (2,2) ;   \begin{pgfinterruptboundingbox}    \tikz[overlay] \path (0,0) --  (1,1) node  {\color{green}$\bullet$};    \end{pgfinterruptboundingbox} \end{tikzpicture}baseline       No problem with a scope   \documentclass[11pt]{scrartcl} \usepackage{tikz}  \begin{document} baseline\begin{tikzpicture}[baseline=(current bounding box.north west)]   \tikz[overlay] \path (0,0) --  (1,1) node  {\color{blue}$\bullet$};  \draw[help lines,blue!20] (0,0) grid (2,2) ; \end{tikzpicture}baseline  \begin{tikzpicture}[baseline=(current bounding box.south west)]    \draw[help lines,red!20] (0,0) grid (2,2) ;  \begin{scope} [overlay] \path (0,0) --  (1,1) node  {\color{red}$\bullet$}; \end{scope}   \end{tikzpicture}baseline  \end{document}    ( What is the real differences between a scope and a tikzpicture ?)  update The question is perhaps too general, better is How tikz performs overlay ?


I have on my webserver apache2 and php:    But if I write this code on my website:  <?php phpinfo(); ?>   I don't get the output you should get. The site just displays this :  <?php phpinfo(); ?>   Any solution?

I have just reinstalled Apache and PHP5 but when I visit my doc root it just downloads a "download" file. And when I try documentroot/index.php it just downloads the PHP file.  I have PHP installed but Apache doesn't seem to want to execute it. People suggested adding a 'Load Module' line to my httpd.conf but I don't have that file.


I am looking into buying a LENOVO RD330 SERVER with a ThinkServer RAID 500 Adapter II RAID controller. I am having problems finding out if all of the hardware that I am wanting will be compatible with Ubuntu. I have check the "Certified Hardware" list but the list is very limited. Both the computer and the RAID controller say they support RedHat and Suse but obviously Ubuntu isn't mentioned. I have talked with my vendor and they also not certain if this hardware will work. I have also Google around and the RAID controller appears to be an LSI chipset. But again I haven't been able to find any definitive information saying that this will work.  Any suggestions on how I can find out if the hardware will work?

I'm thinking about installing Ubuntu Desktop, but I don't know what flavor is the best for my system. What are the minimum and recommended hardware requirements? What kind of CPU? How much memory? Should I have Hardware Acceleration? What flavor should I use?  This is an attempt of a canonical answer. My answers have the "official  requirements", the recommended are a mix of official sources and opinion based (along with the answer it's told the source). You can comment or edit if you feel that the information is obsolete or incomplete. It is a good rule of thumb that any system capable of running Windows Vista, 7, 8, x86 (Intel) OS X will almost always be a lot faster with any Ubuntu flavor even if they are lower-spec than described below.


I have asked this question on Grammarly.com too, but didn't get any answers.  English is not my first language and I'm trying to get better at it therefore I can't give much input or put my "research" here. Here is the sentence:     Max had an opportunity to play the role of interpreter in an international film festival organised here.   Please explain if any article is needed or not.

From the movie  starring Leonardo DiCaprio who plays the role of the young King Louis XIV. d'Artagnan is responsible for security. Here's a short conversation between the two:     King Louis XIV: Perhaps l haven't met a woman with a heart like my own until recently.   d'Artagnan: Would that be Christine?   King Louis XIV: It is good that you watch me, but I fear you watch me too closely.   d'Artagnan: Did you send Raoul to the war so that you could be free to pursue her?   King Louis XIV: Do you question my honor?   d'Artagnan: No, I do not question your honor. It's you I care about.   King Louis XIV: Raoul will return soon. You have my word.   d'Artagnan: Thank you, Your Majesty.   King Louis XIV: I am a young king, but I am king.   d'Artagnan: Then be a good king, Your Majesty.   Can anyone explain why there is no indefinite article in front of the second "king"?


I need to use internet, but without being able to be called. So basically just block all incoming calls. I used to enable flight mode, but doing so, I'm not able to connect to the internet.  Is there an alternative?

It's kinda disturbing when people keep calling me during work or meetings, but to me it's okay if I receive short messages or Whatsapp message because I can reply them few minutes later in a short break. I'm new Android user and wonder if my phone could do that for me.


Hi I am creating pdf on my windows 8 machine. I am using texlive 2019. While giving command pdflatex test.tex I am getting this error.  I have no clue why this error is happening.  This is pdfTeX, Version 3.14159265-2.6-1.40.20 (TeX Live 2019/W32TeX) (preloaded format=pdflatex)  restricted \write18 enabled. entering extended mode (./test.tex LaTeX2e <2018-12-01> (c:/texlive/2019/texmf-dist/tex/latex/base/article.cls Document Class: article 2018/09/03 v1.4i Standard LaTeX document class (c:/texlive/2019/texmf-dist/tex/latex/base/size10.clo)) No file test.aux. [1{c:/texlive/2019/texmf-dist/fonts/map/pdftex/updmap/pdftex.map}] (./test.aux)  )<c:/texlive/2019/texmf-dist/fonts/type1/public/amsfonts/cm/cmr10.pfb> Output written on test.pdf (1 page, 9062 bytes). Transcript written on test.log.   Please guide me is there any package I have to install or any settings I need to make.

I've installed a backported version of TL2012 on Ubuntu 12.04, and every time I start pdftex, xetex, luatex, pdflatex, xelatex, or lualatex, I get the statement restricted \write18 enabled. entering extended mode. I don't recall seeing this in Miktex so I wonder whether I did something wrong during the installation. Interestingly, I don't get the message when I run tex.  First of all, what does it mean? Second, is it a normal message for Tex Live (2012)? Finally, if it is an error, how can I fix it?


I'm planning to move a Large Database from SQL Server 2012 to 2008/R2. I tried the scripting thing, it works on a small-medium type of database but not on Large database w/c is about a hundred MB. Any tips on how can I make this? Or Is it recommended to do this for a Large Databases?

What are my options if I need to move a database from SQL Server 2012(32bit) to SQL Server 2005(64bit)?  I know I cannot:   restore a backup of the database on SQL Server 2005 detach & attach   I know I can:   use import data wizard, and I tried it on one database, but it only moves data, and even that was troublesome as I needed to do a lot of work creating temp tables to maintain identity columns, recreating all FKs, indexes etc.    Is there an easier option?


From my readings I found that both do the same job by collecting crash reports to send it to Canonical.    So why there is two daemon processes doing the same Job? If there differences between them, so what are these differences?

On one of my machines I have a process running called "whoopsie". I'm running 12.04 server and never specifically installed anything with this name.  Google seems to imply that it has something to with error logs but I'm not finding too much information. The fact that I didn't manually install it and the 3 other servers I checked did in fact have no such running process OR executable made me a bit confused.  Does anyone know what the "whoopsie" process is?  Does anyone know what packages might have installed it? The server is quite plain, it has a LAMP stack, Samba and print servers and the Nagios NRPE plugin, nothing more installed, just standing there being a nice backup-server.  Some more info:  $ whoopsie -h Usage:   whoopsie [OPTION...]  Help Options:   -h, --help           Show help options  Application Options:   -f, --foreground     Run in the foreground   and  USER       PID %CPU %MEM    VSZ   RSS TTY      STAT START   TIME COMMAND   whoopsie   913  0.0  0.4  24448  2092 ?        Ssl  May07   0:00 whoopsie   and  $ sudo cat /etc/passwd | grep whoop whoopsie:x:107:118::/nonexistent:/bin/false


I'm having a little trouble with this one. So far I've found that $$g(k) = \sqrt{\frac{2}{\pi}}\frac{\sin{k}}{k}$$ when $f(x) = 1, |x| < 1$ and $0$ elsewhere. I've also found the convolution $(f*f)(x)$ is $$\frac{1}{\sqrt{2\pi}}\begin{cases}0&x\le -2\\\\\int_{-1}^{x+1}(1)\,dx=(x+2)&-2<x<0\\\\\int_{x-1}^1(1)\,dx=2-x&0\le x<2\\\\0&x\ge 2\end{cases}$$ And the Fourier transform of this is $$\frac{2}{\pi}\frac{\sin^2{k}}{k^2}$$ I'm sure that from here the answer is quite simple, however I'm having a hard time figuring out the last step. Any help is appreciated.

How would I integrate  $$\int_{-\infty}^{+\infty} \frac{\sin^{2}(x)}{x^{2}}\,dx$$  using Fourier Transform methods, i.e. using Parseval's Theorem ?   How would I then use that to calculate:  $$\int_{-\infty}^{+\infty} \frac{\sin^{4}(x)}{x^{4}}\,dx$$?


So the way I was taught to decompose into partial fractions was something like this:  Say I begin with a fraction like $\frac{1}{x^2-1}$ that I want to decompose that into two fractions with linear denominators.  I would do $$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1},$$  which then gives $1=A(x+1)+B(x-1)$.  Then to solve quickly, you can set $x=1$, and solve for $A$, then $x=-1$ and solve for $B$.  My question is, why can you let $x=1$ and $x=-1$?  Aren't those excluded values?  Isn't the equation $$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$$ only valid for all $x$ other than $1$ and $-1$?

Let’s suppose we want to decompose $\frac {9}{9-x^2}$ by partial fractions. I was taught that we proceed by writing $$ \frac{9}{9-x^2} = \frac{a}{3+x} + \frac{b}{3-x}, $$ where $a$ and $b$ are unknown constants.  Clearing the fractions, we get $$(3-x)a + (3+x)b = 9.$$ To find $a$, we assume that $x=-3$ and solve the resulting equation; and similarly, to find $b$, we assume that $x=3$ and solve.  But when we assume it we are diving by $0$ in the original fraction which isn't possible!  I am confused.  How does this procedure give the correct result?


I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:   $13^{100} \bmod 7$ $7^{100} \bmod 13$   I've also heard of this formula:  $$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n $$  But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)  Any hints or pointers in the right direction would be great.


a little while ago was everything working perfectly. But now, when i have even just basic cube in the scene, then when i want to add texture or materieal, or even if i wont add any of them to the cube, when i go to Rendered viewport shading, all i see is this (just a grey cube with nothing on it, i have set the color tu blue but nothing):  And when i render it it shows just checkered cube like this:  I know its probably gonna be something dumb, but i just cant figure out what the problem is. Thanks for help.

I don't know why but now, when i go in material preview or rendered (in solid mode it works fine), my object is entirely gray, or invisible idk, because it has the same color of the background. I've been working for days on this project without having this issue, it suddenly showed up. I tried opening other old projects and suddenly they all do the same so i guess it is some blender program setting that applies for all the projects, but idk how it happened cuz i didn't change anything :/, anyone got any tips?


Let $A$ and $B$ are matrices on field $F$ of size $m \times n$ and $n \times m$ respectively. Can we claim, that nonzero eigenvalues of matrices AB and BA are the same and each of them has the same multiplicity for this matrices if:  a) $F = \mathbb{C}$  b) $F$ — arbitrary field?

Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?   I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?


My wife is a Indian passport holder and Permanent Resident of Canada. She is traveling from Canada to India via Amsterdam (Delta flight) and will have 1 hour 20 minutes layover at Amsterdam.  Will she require a transit visa for Amsterdam?

I found many related questions on this site but I am still not sure about the rules. How can I decide if I need a visa to transit? Schengen members as of May 2021 are as follows:  Austria Belgium Czech Republic Denmark (excluding Greenland and the Faroe Islands - but an open border with the Schengen Area is maintained) Estonia Finland France (excluding overseas departments and collectivities) Germany Greece Hungary Iceland Italy Latvia Liechtenstein Lithuania Luxembourg Malta Netherlands (excluding Aruba, Curaçao, Sint Maarten and the Caribbean Netherlands) Norway (excluding Svalbard) Poland Portugal Slovakia Slovenia Spain (except Ceuta and Melilla) Sweden Switzerland


Consider the following two examples:    First:  var x = (function(){  return 786;  }());   Second:  var y = function(){  return 786;  }();   I know that we can't use function(){  return 786;  }(); directly as an anoomymus self calling function but now since I am assigining it to variable x I can use this function. So,     Is there any difference in result of the behaviour of variable x in First and Second method?

Possible Duplicate:          This is something I haven't quite figured out yet, but I have been using function(){}() just because my VIM syntax highlight screws up if I add the parenthesis, although I've seen (function(){})() around many times, maybe its an IE thing?  edit:  var singleton = function() {     // code }();  var singleton = (function() {     // code })();


One central observation is that the vast majority of Robo-Advisors base their investment approach on ETFs. About 70% of the examined providers even operate on an “ETF-only” approach, to maximize cost-efficiency for highly diversified yet small portfolios. ()  To me this seems like a really good deal for the company offering the robo advisor and a really bad deal for the customer. After all, choosing a handful of index based funds in various geographic regions is not a complicated operation and will reflect the overall market performance quite well while maintaining a good diversity. Is there any good reason to pay a fee of typically 0.5-1% for a robo advisor?

I recently become interested in wealthfront, and plan to make the first real investment in my life. However, as I read more about mutual funds and ETFs, I realize that wealthfront is actually built upon on vanguard's mutual funds and ETFs.  The fees of wealthfront is higher than vanguard's mutual funds and ETF. Then I get a question: why pay more if I can buy from vanguard directly.  I find one answer to that wealthfront is more tax efficient. But this seems a minor reason to me. The other answer is that the minimum investment of wealthfront is smaller. I actually don't care about that.   Then what's the real advantage of wealthfront over vanguard?


I have tried tried to open my older qgis files made on a mac on my new PC.  Wien(2.8) only seems to allow .qgs files so all my folders appears empty when I access them from QGIS on the PC.

I made a project in my personal computer (mac OS Xlion) and now I would like to modify it on my work computer (windows 7), Is it possible?


can someone provide any Idea on below requriement: my requirment is : We received  JSON data ( complex have Tree structured more than 300 values) , Based on JSON data , need to generate Response . For example : workbench --> RestExplorer - >  / services/apex/apexRest/Id = 000000000 Then i need to generate Response as Below : caseInfo -- PartyId : 00000 -- ProductType : ABc --Product : abc Arrangement --arrangemetn Number :9999 --isseleccted :Yes --totalnumberrecords : 20 How can i achive this functionality ...Pls provide any sample data or any references Thanks

This is a canonical question and answer developed by the community to help address common questions. If you've been directed here, or your question has been closed as a duplicate, please look through the resources here and use them to shape more specific questions. To browse all canonical questions and answers, navigate to the  tag.  This question and answer is owned by the community! If you can improve it and make it more useful, you're welcome to edit it.  I have a specific JSON structure in mind. I need to either deserialize this structure that's coming in from a web service, or I need to serialize data into this structure to transmit it to another system. How should I get started on this project in Apex?


I  sometimes wonder where to ask a particular question (on Ubuntu? Unix and Linux? Superuser?) because all of the sites will have great people to help. This is not a common case for me but I already had it happening.   It may be, though, that a question goes unanswered for a while and would be ,  all in all, a better fit for another site.   Is it acceptable, then, to migrate it ? Or move (delete and repost)? Or copy?

When asking questions related to SharePoint, I generally post them on . But sometimes I don't receive any responses to my question, so I delete the question from SharePoint Stack Exchange and repost it on Stack Overflow.  Is this behavior correct, or will I be penalized for it if this happens frequently?


If $R$ is a reduced Noetherian ring and $U \subset R$ is a multiplicatively closed set, how do we get $K(R[U^{-1}]) = K(R)[U^{-1}]$?   asks the same question I have but I don't see how the accepted answer answers the OP's question.  I got up to the point where we use the Chinese remainder theorem just like in the post, but do not see where to go from here. Can someone further elaborate the answer given? More specifically:  How do we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$? How are we concluding $K(R[U^{-1}])=K(R)[U^{-1}]$?

Let $R$ be any ring. The total quotient ring of $R$, denoted $K(R)$, is $R$ localized at the set of non zero-divisors. I am trying to solve Exercise 3.15 in Eisenbud's Commutative Algebra:  a) Let $R$ be a Noetherian reduced ring and let $U$ be a multiplicative subset. Then $K(R[U^{-1}])=K(R)[U^{-1}]$. b) If $R$ is a Noetherian ring, then $K(R[U^{-1}])=K(K(R)[U^{-1}])$.  For a), I was able to show that all prime ideals in $K(R)$ are maximal. Further, there are finitely many of them. Let these ideals be denoted $P_1,\ldots ,P_k$. By the Chinese Remainder Theorem, we have that $$K(R)=\prod_{i=1}^k K(R/P_i)$$ How should I proceed? Also, can I have a hint to start part b)? Should I use the universal property of the localization somehow?


The Borel-Cantelli Lemma: Let $\{E_k\}_{k=1}^\infty$ be a countable collection of measurable sets for which $\sum_{k=1}^\infty m(E_k) < \infty$. Then almost all $x \in \mathbb R$ belong to at most finitely many of the $E_k$'s.   Here is the proof from Royden's book.  For each n, the the countable subadditivity of measure, $$m(\bigcup_{k=n}^\infty E_k) \leq \sum_{k=n}^\infty m(E_k) < \infty.$$  Hence, by the countinuity of measure, $$m(\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)) = \lim_{n \to \infty} m(\bigcup_{k=n}^\infty E_k) \leq \lim_{n \to \infty} \sum_{k=n}^\infty m(E_k) = 0.$$  Therefor almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ and therefore belong to at most finitely many $E_k$'s.   I don't understand the last sentence. How can we conclude that if almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ then they belong to at most finitely many $E_k$'s. If $x \in \bigcap_{n=1}^\infty (\bigcup_{k=2n}^\infty E_k)$, it may fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$, but it still belongs to infinitely many $E_k$'s.

The Borel-Cantelli Lemma in  seems to be a sort of "corollary" of the .  It says:  "Let $(E_k)_{k=1}^{\infty}$ be a countable collection of measurable sets for which $\sum_{k=1}^{\infty} m(E_k) < \infty$. Then almost all x $\epsilon \ \mathbb{R}$ belong to at most finitely many of the $E_k$'s"  I don't get three things:   "Then almost all x $\in \ \mathbb{R}$ belong to at most finitely many of the $E_k$'s"   What does that mean? Since the Borel-Cantelli Lemma comes after the "almost everywhere" definition, I guess: $\exists \ Z \ \subset \mathbb{R}$ s.t. $m(Z) = 0$ and $\forall x \ \in \ \mathbb{R} \setminus Z$, x belongs to $(E_{j}, E_{j+1}, E_{l}, ..., E_{m})$.   In the proof, one can see the $m(\limsup \ E_k) = 0$ part, stated as $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$. For some reason, $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$ implies that "Therefore almost all x $\in \ \mathbb{R}$ fail to belong to $\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]$ and therefore belong to at most finitely many $E_k$'s"   Okay, why? What is the contradiction if almost all $x \epsilon \ \mathbb{R}$ belong to all the $E_k$'s? Or a countably infinite subcollection?   What is the significance of this? Why not just state the Borel-Cantelli Lemma as "Let $(E_k)_{k=1}^{\infty}$ be a countable collection of measurable sets for which $\sum_{k=1}^{\infty} m(E_k) < \infty$. $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$"?


I received my Schengen visa today from the French Embassy. I wrote in my application that I will stay for 14 days. They gave me 1 month (valid from 23-10-18 until 21-11-18) but on the side it says duration of stay 14 days. Can I stay for only 14 days?  Now I regret that I did not make it for 20 days. I had no idea that they would just allow me 14 days. I thought that if it's valid for one month I could stay for one month. Also, is it okay if I change my mind and stay only 3 days in Paris and go to Rome for 5 days?

I'm from Tunisia and I got a Schengen visa from the Austrian Embassy for  28 days, the “duration of stay”, and the number of entries is 1. But the validity of the visa is from 02/08/2015 to 13/09/2015.  Do I have to stay only 28 days, for example from 02/08/2015 to 29/08/2015 or can I stay there until 12/09/2015 ?  And if I have got to stay only 28 days, then can I choose the days of stay, that's mean can I stay in Austria for example from 09/08/2015 to 05/09/2015?


I've been searching for hours to find a regex that does this for me, all the ones I've found either require dashes or limit something else that I don't need.  Like this one:      ^(?([0-9]{3}))?[-. ]?([0-9]{3})[-. ]?([0-9]{4})$  Basically I want to allow these types of input for phone number, these are all valid:  +000 111111111  +00011111111111  0022 111111111111  0022111111111  +333-4444-5555-6666  000-7878787-0000-4587  Note that the number of digits is not limited, I only want the validation to filter out empty inputs and the alphabet. Also filter out all other characters except a maximum of 4 dashes, max. 4 spaces and an optional single plus sign.  Is this possible through preg_match or not?  Any help is appreciated, thanks!

I'm trying to put together a comprehensive regex to validate phone numbers. Ideally it would handle international formats, but it must handle US formats, including the following:   1-234-567-8901 1-234-567-8901 x1234 1-234-567-8901 ext1234 1 (234) 567-8901 1.234.567.8901 1/234/567/8901 12345678901   I'll answer with my current attempt, but I'm hoping somebody has something better and/or more elegant.


I would expect the following command to delete all files with extension .x or .z, but instead it seems to only delete the first file listed:  mkdir foo touch foo/{a.x,b,c.z} find foo -type f -name '*.x' -o -name '*.z'  # correctly lists a.x and c.z find foo -type f -name '*.x' -o -name '*.z' -exec rm -f {} \+       # only deletes c.z   Why is this happening?

When I'm using  find . -type f -name "*.htm*" -o -name "*.js*" -o -name "*.txt"  it finds all the types of file. But when I add -exec at the end:  find . -type f -name "*.htm*" -o -name "*.js*" -o -name "*.txt" -exec sh -c 'echo "$0"' {} \;  it seems it only prints .txt files. What am I doing wrong?  Note: using MINGW (Git Bash)


Is mathematics a science? I have long considered this to be open to debate, but my professor said that he once heard the quote,   "Mathematics is not a science. It is a language   that is free of contradictions, used by all sciences."    It seems to be that most non-mathematicians do not think math is a science, while most mathematicians do.  What do you guys think?

Possible Duplicate:          I would like to know if mathematics is considered a science? I've searched the internet and asked many people for insight to no avail.  I thought science was based on observation, experimentation and falsification.  I don't understand how mathematics can satisfy all the above.  Especially, when there are some branches of mathematics that need not actually correspond to the physical world.  Maybe I have a naive definition or unsophisticated understanding of both mathematics and science.    Thank you in advance.  Sincerely, Maxwell


Quote from here:      Allocate drive space      Use the checkboxes to choose whether you'd like to Install Ubuntu   alongside another operating system, delete your existing operating   system and replace it with Ubuntu, or — if you're an advanced user —   choose the 'Something else' option.   Question: does choosing the second option "Replace Windows 7 with Ubuntu" delete my drive & drive data also, or does it only delete the C drive?

I am a newbie in linux. I wanted to install ubuntu to my desktop. I currently have two partitions, Windows 8 in installed on one, and i have some important files in another partition. While installing ubuntu, if I chose "Replace Windows 8 with Ubuntu", do I lose the other partition too?   And are there other ways to leave the other partition as it is, and install ubuntu in the partition  where windows was installed.


My question is that when I slide my home screen, my background wallpaper also slides. Not giving me a complete view of full wallpaper on a particular screen.  How to fix it?

I just changed my background on my home screen on my Galaxy S4 and every time I scroll to a different page the picture moves. How do I get this to stop?


The speed of an electron is very slow  while going through a conductor(like 1mm per second),but the current passes instantaneous.why?

How is it possible for current to flow so fast when charge flows so slowly?  We know electrons travel very slowly while charge travels at ~the speed of light.


For instance, can someone please explain to me the purpose of the asterisk in line 2 below?   m = Basemap(projection='merc', llcrnrlon=lon.min(), urcrnrlon=lon.max(), llcrnrlat=lat.min(), urcrnrlat=lat.max()) x, y = m(*np.meshgrid(lat,lon))

Does * have a special meaning in Python as it does in C? I saw a function like this in the Python Cookbook:  def get(self, *a, **kw)   Would you please explain it to me or point out where I can find an answer (Google interprets the * as wild card character and thus I cannot find a satisfactory answer).


Sometimes you ask a question on Stack Overflow, or Server Fault, and I'm sure it will happen on Super User, that lends itself to multiple correct responses, like a list. Is there anything to handle such a thing? The situation I am thinking about is when the response would end up being a list and multiple people would add input, but there is more than one correct answer.   Is the proper thing to do for this type of question just not to award anyone a correct answer or award the correct answer the best answer? What happens if you award the correct answer and someone comes along latter with a better answer?  Example: This is oversimplified to make the point but hopefully you get the idea.  Q: What is a good place to find information on programming?  A1: Stack Overflow (stackoverflow.com)  A2: MSDN (msdn.microsoft.com)  A3: Channel 9 (channel9.msdn.com)  Now obviously this question has a best answer, but the others are good too. I kind of had a hard time thinking of this example, but in the past this has bothered me. Maybe it really doesn't occur often enough to be an issue.

There are absolutely loads of payment gateways out there, and I'd like a comprehensive list of them. There's already a question asking for the "best" ones,  and another asking for ones suitable for a charity .  Would such a question be on topic? And would it need to be CW? The obvious way to do it is to have one CW answer and edit all other answers into it. I don't know whether the question itself should be CW, or just the main answer.


Question:  $$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$  What we did:  we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this: $\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1}  $, for which we know how to solve the first and second integral but not the third...  Thanks

I have to evaluate:  $$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ I can't get the right answer! So please help me out!


According to the Rob Hyndman's , "section 5.3 - Selecting predictors", when selecting predictors in a regression model:      Adjusted $R^2$       Computer output for regression will always give the $R^2$    value, discussed in Section 5/1. However, it is not a good measure of the predictive ability of a model. Imagine a model which produces forecasts that are exactly 20% of the actual values. In that case, the $R^2$ value would be 1 (indicating perfect correlation), but the forecasts are not very close to the actual values.      In addition, $R^2$ does not allow for "degrees of freedom''. Adding any variable tends to increase the value of $R^2$ even if that variable is irrelevant. For these reasons, forecasters should not use $R^2$ to determine whether a model will give good predictions.      An equivalent idea is to select the model which gives the minimum sum of squared errors (SSE), given by      $SSE = \sum e_i^2$  , (with $e_i = y_i - \hat{y_i}$ )       Minimizing the SSE is equivalent to maximizing $R^2$ and will always choose the model with the most variables, and so is not a valid way of selecting predictors.   Now based on :   $R^2 = 1 - \frac{SSE}{SST}$ , with $SST= \sum (y_i - \bar{y})^2$   My questions:    I can't understand, based on the the definitions of $R^2$ and $SSE$, why $R^2$ would always increase when the number of variables in the model increases? and why does the $SSE$ would also decrease? Especially if this happens as he says even when the variable is irrelevant?  Does this relation between SSE and number of variables hold only for linear models?  We use the $SSE$ for training ML models in general all the time, but from the above quoted reasons, that would be a bad idea. Is this conclusion (maximizing $R^2$/minimizing $SSE$ is not a good measure of the predictive ability of a model) specific to forecasting problems (i.e. when using regression models specifically for forecasting purposes) or does it apply to any type of regression model? And if so, why is the use of $SSE$ so relevant if it is such a bad idea?

In my econometric textbook(Introductory Econometrics) covering OLS, the author write, "SSR must fall when another explanatory variable is added."  Why is it?


Can you use the Healing Extra Restorative to heal loss of Toughness (-x Toughness save) due to save vs damage?     Restorative: Your Healing effect can restore character points removed by Weaken effects with the appropriate descriptors, such as injury, disease, or poison. You restore points equal to your rank to the affected trait(s).

Why does "Healing" effect only apply to status conditions like daze or hindered? Why doesn't heal loss of Toughness defense if toughness the equivalent of the games HP?


I want to draw some knots for my master's dissertation, but I dont want to do that by including pictures of them. Can I do this on latex? If this is possible how? If not, are there any apps which can help me with it?

I was wondering if anyone could suggest software/packages to create nice knot diagrams (hopefully with a link to images they have made in the past).  I have used xy-pic recently but am mainly interested in hearing about other options.


The package system is broken  If you are using third party repositories then disable them, since they are a common source of problems. Now run the following command in a terminal: apt-get install -f  The following packages have unmet dependencies.  libegl1-mesa : Depends: libgl1-mesa-dri (= 10.5~git1411191810.9460cd~gd~t) but 10.4~git1411171930.920f87~gd~t is installed  libgbm1 : Depends: libgl1-mesa-dri (= 10.5~git1411191810.9460cd~gd~t) but 10.4~git1411171930.920f87~gd~t is installed  libgl1-mesa-dri : Breaks: libgl1-mesa-dri:i386 (!= 10.4~git1411171930.920f87~gd~t) but 10.5~git1411191810.9460cd~gd~t is installed  libgl1-mesa-dri:i386 : Breaks: libgl1-mesa-dri (!= 10.5~git1411191810.9460cd~gd~t) but 10.4~git1411171930.920f87~gd~t is installed  libgl1-mesa-glx : Depends: libgl1-mesa-dri (= 10.5~git1411191810.9460cd~gd~t) but 10.4~git1411171930.920f87~gd~t is installed E: Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages. E: Unable to correct dependencies   I'm on Ubuntu 14.04. I was using the update drivers from , so the first thing I did was remove that PPA, but this doesn't seem to have helped.  EDIT: I've followed these instructions, but to no avail:      Running these commands achieves nothing:  sudo apt-get clean sudo apt-get -f install sudo apt-get update   I don't have synaptic installed and when I try to install it, or remove packages using sudo apt-get remove --dry-run I just get more errors about unmet dependencies. I've been using Ubuntu for years and have always been able to resolve problems like this in the past, but something is eluding me here.

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


Let $M$ be an infinite set.       Proposition 1:   For any $n \in \mathbb{N}$, there exists an injection from $\{1, \cdots, n\}$ to $M$.       (1) Since $M \neq \emptyset$, there exists $x \in M$. Define $f(1)$ as $f(1) := x$. $f$ is an injection from $\{1\}$ to $M$. (2) Suppose that there exists an injection $f$ from $\{1, \cdots, n\}$ to $M$. Since $M$ is an infinite set, $M - \{f(1), \cdots, f(n)\}$ is not an empty set. So, there exists $x \in M - \{f(1), \cdots, f(n)\}$. Define $g(1), \cdots, g(n+1)$ as $g(1) := f(1), \cdots, g(n):=f(n)$ and $g(n+1) := x$. Obviously, $g$ is an injection from $\{1, \cdots, n+1\}$ to $M$.    Let $n_1$ be an arbitrary natural number. If I wanna calculate $h(n_1)$, then I get an injection $g$ from $\{1, \cdots, n_1\}$ to $M$ by Proposition 1. And I return $g(n_1)$ as the value of $h(n_1)$. And I store the pairs $(1, g(1)), \cdots, (n_1, g(n_1))$ to my database.    If I wanna calculate $h(n_2)$ for $n_2 \leq n_1$, then I search my database and I get the value $g(n_2)$ from my database and I return $g(n_2)$ as the value of $h(n_2)$.  If I wanna calculate $h(n_3)$ for $n_3 > n_1$, then I add the pairs $(n_1 + 1, g(n_1+1)), \cdots, (n_3, g(n_3))$ to my database by Proposition 1 and I return $g(n_3)$ as the value of $h(n_3)$.    I can calculate $h(n)$ for any $n \in \mathbb{N}$.    From above, we get an injection $h : \mathbb{N} \to M$.    Why is my proof wrong?  By the way. Suppose that a man wanna know if I have an injection $h : \mathbb{N} \to M$ or not. Then how can the man know if I have  an injection $h : \mathbb{N} \to M$ or not?

Reading the comment to  the user makes the argument that one can pick a dense subset of a separable vector spaces without relying the axiom of countable choices. Since the comment got a lot of up votes, I guess it's correct, but to me this seems very strange. My rule of thumb was/is the following: We can get by without a choice function if   There is a canonical choice at each iteration. The proof doesn't actually require the iteration process to terminate.   None of these seem to be satisfied in the present case, unless there is a canonical choice for the dense subset, or the set is dense at some finite iteration, so my question is:   When can I tell if an induction arguments relies on the existence of a countable choice function?


I keep getting the error in the title when I try to run this code. I thought that the problem was to do with the variables not being integers, but that didn't fix the problem. Here is the code: shoeSize=int(input("Please enter your shoe size.")) ageOne=int(input("Please enter your age."))  if ageOne >= 18:     print("Your membership fee is £" & shoeSize*10) elif ageOne <= 65:         print("Your membership fee is £" & shoeSize*5) else:         print ("You cannot join!")

I want to do something like String.Format("[{0}, {1}, {2}]", 1, 2, 3) which returns:  [1, 2, 3]   How do I do this in Python?


I would like to make a Norwegian index for a book. Is there a MWE  which can help me?

I am making a songbook, using the songs package. The book is in norwegian, and thus we need the norwegian letters æ,ø,å to be indexed after z. As far as i understand, this is only possible with the xindy package. Creating the index is no problem, but I want to customize how it looks. As pr. now, it looks like this:     Using makeindex, I was able to make it look correct   Here is the code for makeindex  \usepackage[makeindex]{imakeidx} % Normal LaTeX indexing \makeindex[columns=1, title=Sangregister,program=makeindex,intoc=true,options=-s idxconf.ist] % Properties of index   This is for xindy  \usepackage[texindy]{imakeidx}  \def\xindylangopt{-M lang/norwegian/latin1-test } \makeindex[columns=1,name=master,title =Sangregister,intoc=true,options=\xindylangopt]   The layout-options, is in the file idxconf.ist, which works on makeindex looks like this  headings_flag 1 heading_prefix "{\\bfseries " heading_suffix "}\\nopagebreak\n" delim_0 " \\dotfill " delim_1 " \\dotfill " delim_2 " \\dotfill "   Is it possible to use the idxconf.ist file with xindy? If not, how do I customize xindy to look like the last picture.  Here is a minimum working example:  xindy:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[norsk]{babel} \usepackage[texindy]{imakeidx} \def\xindylangopt{-M lang/norwegian/latin1-test} \makeindex[columns=1,name=xindytest,title =Sangregister,  options=\xindylangopt] \begin{document} start \index{a}\index{æ}\index{ø}\index{å} end blabla \printindex \end{document}   makeindex, which looks correct, but æøå is treated as symbols.  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[norsk]{babel} \usepackage[makeindex]{imakeidx} \makeindex[columns=1, title=Sangregister,program=makeindex,intoc=true,options=-s idxconf.ist] % Egenskaper til indeks \begin{document} start \index{a}\index{æ}\index{ø}\index{å} end blabla \printindex \end{document}


I have the following set of lines in my code      RestRequest req = RestContext.request;     RestResponse res = RestContext.response;     AreaRequestWrapper areaRequestRecord = (AreaRequestWrapper)System.JSON.deserialize(requestString, AreaRequestWrapper.class);             if(areaRequestRecord.OpportunityId == null || areaRequestRecord.OpportunityId == '')             {                 res.statusCode = 401;                  res.responseBody = Blob.valueOf(res.statusCode+' : Mandatory field OpportunityId was not sent in the Request JSON');             }   My issue is that, when I am writing the test class for the above code, the the code is covered till line   res.statusCode = 401;   But the responseBody line does not get covered.

This is a canonical question and answer developed by the community to help address common questions. If you've been directed here, or your question has been closed as a duplicate, please look through the resources here and use them to shape more specific questions. To browse all canonical questions and answers, including more unit test resources, navigate to the  tag.  This canonical question is intended to address several classes of common questions by providing a quick summary and links to comprehensive resources:   How is code coverage obtained and calculated? How do I increase my coverage?  Why can't I cover these lines? Why isn't the body of my loop or conditional statement covered?


Today someone accidentally edited one of my answers (he thought he was editing his own).  He quickly noticed his error and reverted it, but it left me thinking:  What should I have done if he hadn't noticed his error?

So... from a while back, I posted a question dealing with implementing a certain feature within an iPad app (no specifics, see  for question), and I noticed that there was an approved edit by a user named Zephyer who posted something about Windows 7.  After a face-palm, I re-edited my question to bring it back in line with my query.  So, my question then is: What the heck do we do with the occasional stupid/irrelevant edit?  I know it doesn't happen often, but there's got to be something other than just re-editing I can do!    EDIT: Excellent suggestions.  To clarify two things though:   This post is quite old.  I was a new user at the time, unaware of the rollback feature if it even existed at the time of it's posting. I just wanted to know if there was some way to get the attention of the editor.  It doesn't need to be a downvote or a flag, just a message, or an indication.


This person I am referring to is Martha Wayles who was born in 1748 and died in 1782. I've been trying to wrap my brain around the x cousin x times removed thing but it is still confusing me.   The red circle is me. The pink circle is the cousin in question. And the blue one is one I'd like to know how we are connected as well.

How do I figure out if someone is a first cousin, second cousin, cousin once-removed, etc?


On page 157 of Schwartz's QFT book, He write that “$\sigma_i$ do not change under rotations”.  If so, changes in $\psi$ and $B$ cancels, so we can get that $(\vec{\sigma} \cdot \vec B)\psi$ is rotationally invariant.  But why Pauli matrices are the same in any frame? Any hint or reference would be helpful!

From my understanding, gamma matrices transforms under Lorentz transformation $\Lambda$ as \begin{equation} \gamma^{\mu} \rightarrow S[\Lambda]\gamma^{\mu}S[\Lambda]^{-1} = \Lambda^{\mu}_{\nu}\gamma^{\nu} \end{equation} Where $S[\Lambda]$ is the corresponding Lorentz transformation in bispinor representation. So my question is: When we change from on frame to another, are we allowed to write $\gamma'^{\mu} = \Lambda^{\mu}_{\nu} \gamma^{\nu}$ where $\gamma'^{\mu}$ is the transformed version of $\gamma^{\mu}$? If yes, then when we write down $\gamma^{\mu}$ explicitly (in some representation) as we do in any standard QFT textbook like \begin{equation} \gamma^{\mu} = \begin{pmatrix} 0 & \sigma^{\mu} \\  \bar{\sigma}^{\mu} & 0 \end{pmatrix} \end{equation} do we assume any specific frame of reference? If so, which frame? Because if I apply Lorentz transformation such as a boost along $x$-direction I will have \begin{equation} \gamma'^{0} = \cosh(\eta)\gamma^0 + \sinh(\eta)\gamma^1 = \begin{pmatrix}  0 & \cosh(\eta) + \sigma^1 \sinh(\eta) \\ \cosh(\eta) - \sigma^1\sinh(\eta) & 0 \end{pmatrix} \end{equation} and similarly for $\gamma^i$. I understand that at the end any choice of reference frame does not matter because that's the point of relativistic theory like QFT. The term like $\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$ in the theory will remain invariance under Lorentz transformation. But just like how we have the momentum shell condition $p^{\mu}p_{\mu} = -m^2$ in all frames but $p^{\mu}$ itself will change from one frame to the other and in particle rest frame we have $p^{\mu} = (m,0,0,0)$ it seems to me that by writing down $\gamma^{\mu}$ explicitly as above we are picking a specific frame. Could someone please clarify this to me?


I have a client with two CMS (Hubspot and WordPress) on two different subdomains.  WordPress: www.domain.tld Hubspot: www2.domain.tld  The WordPress environment is used for the main website and the Hubspot is only used for the blog (yeah, I know, it's strange). The client wants to keep the Hubspot environment, but I'm not sure about the subdomain actually used (www2).  The www2 isn't used like a mirror.  I want to know if a subdomain like www2 affects the SEO.  Thx,

If I have example.com, and blog.example.com, do search engines see those as one site or two? Will using subdomains hurt SEO? I have some technical reasons why I can't use example.com/blog.