I am using the arduino with DHT22 temperature, humidity sensor, but i don't want to constantly power the sensor, so I would like to have a NPN 2N2222 transistor to control when to the give the power.   but i am a bit puzzled how to wire them together, which how the collector and emmitter should be wired to DHT22 sensor

Before I ask my question, I should say that I am very new to working with electronics and I may not quite know how to describe my problem.  I am trying to use a pin on my msp430 microcontroller I have as a replacement for a pushbutton on a device. The msp430 is 3.3v and the device I am switching is 5 volts. I was under the impression I could use a 2n3904 transistor in place of the button by applying current by turning on the pin on the msp430 which would let current go from the collector to the emmiter, and hopefully "push" the button.  This is my current setup    However, turning the pin on does not activate the circuit. When I connect the jumper for the onboard LED, I can verify the pin is working. also, when I connect the 5v from the collector to the base using a wire, I can activate the transistor.  What other information am I missing to solve this? Thanks in advance  EDIT: After reading the comments, adding a 4.6k resistor between the base and IO pin and connecting my 3v ground and 5v ground allowed me to control the transistor without any noticeable problems. Thank you!


I just installed ubuntu 14.04.3 (LTS). My laptop can dual-boot Windows 10 & Ubuntu. But I have a problem of using wireless network. When I open the terminal and type      Code:           iwconfig      returns the following          eth0 no wireless extensions.          lo no wireless extensions.      I can't find   wlan0 IEEE 802.11 blah blah blah...  Also when I type      Code:           sudo wpa_cli      returns the following    As far as I know, in normal circumstances, wireless interface information should be seen. Just like below picture:    I think my ubuntu can't detect wireless LAN Card.(But when I use windows, My PC can connect to wifi and can use the web)  I need your help. Thank you. Before complete question, I show my PC information.            First OS : Windows 10          Second OS : Ubuntu 14.04.3 (LTS) Desktop Version                 windows and ubuntu dual-boot               Ubuntu Kernel Version : 4.3.3-040303-generic          My Network Adapter                          Qualcomm Atheros QCA61x4 Wireless Network Adapter                  Realtek PCle GBE Family Controller                  .......................................................................  when I type     Code:           ifconfig      returns the following          eth0   Link encap:Ethernet  HWaddr 24:f5:aa:d7:f2:1d               inet addr:165.132.117.69  Bcast:165.132.119.255  Mask:255.255.252.0               inet6 addr: fe80::26f5:aaff:fed7:f21d/64 Scope:Link               UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1               RX packets:315020 errors:0 dropped:396 overruns:0 frame:0               TX packets:157677 errors:7 dropped:0 overruns:0 carrier:0               collisions:0 txqueuelen:1000                RX bytes:323556810 (323.5 MB)  TX bytes:14659107 (14.6 MB)          lo      Link encap:Local Loopback               inet addr:127.0.0.1  Mask:255.0.0.0               inet6 addr: ::1/128 Scope:Host               UP LOOPBACK RUNNING  MTU:65536  Metric:1               RX packets:4694 errors:0 dropped:0 overruns:0 frame:0               TX packets:4694 errors:0 dropped:0 overruns:0 carrier:0               collisions:0 txqueuelen:0                RX bytes:459292 (459.2 KB)  TX bytes:459292 (459.2 KB)

I just installed Ubuntu 14.04.3 LTS after installing windows 10 in my new VN7-571G and the wireless is not working. It never had: even during the installation it was not recognized. I run a script to gather the data , which is . lspci -knn | grep Net -A2 output 03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:003e] (rev 20)         Subsystem: Lite-On Communications Inc Device [11ad:0804]  The OS behaves like it has no wifi card. I do not see wifi settings in network settings I'm not a linux power user, so I do not know what to do to solve the issue. What should I do to make it work?


I overstayed in the USA on 2 occasions. The first time I stayed for 10 months, the second time, 3 years.  10 years have passed, my ban will be over. The next step is to arrange an interview for a non-immigrant visa. I plan to be completely honest about everything.  The problem is, I’m not married and I don’t own any property at home here, to prove that I have ties at home to stop me overstaying again. I do work part-time as a caretaker for a school and own my own online business. I care for my mother, who is quite dependent on me (she will be traveling with me).  I want to visit for a holiday. I’m wondering whether I am wasting my time even thinking about going to an interview. I’m a very nervous person at the best of times, never mind and intense situation like that. It would really devastate me if I was declined.  Has anybody had a similar experience and got a visa approved?

When I first went to the USA it was before fingerprinting, so I was not fingerprinted. I overstayed for 10 months. I came home and then went back to the US and got back in, and this time I was fingerprinted. I overstayed for 3 and a half years. When I left the US, I had a new passport, as if I'd lost my passport in the States.  I'm attempting to go back now for a visit. It has been 10 years since I left the States (my 10-year ban would be over). I'm told I have to apply for a non-immigrant visa, but the chances of getting approval are slim, due to the fact that I have overstayed my welcome on 2 occasions.  I'm from Northern Ireland. I can have two passports: a UK passport or an Irish passport. If I change my passport and attempt to enter, would I be denied? I honestly have no intentions of overstaying again, but I’d love to go back and see some old friends.


In this sentence:     Five birds landed on the branch.   Is the word “five” an adjective? Why or why not?

My wife and I were discussing whether it is allowable to put an adjective in front of "of them". For instance, I could say "I want 5 cats" and "I want 5 of them". However, while it sounds perfectly normal to say "I want 5 new cats", "I want 5 new of them" doesn't roll of the tongue very smoothly.  If it is incorrect to place an adjective in front of "of them"?


I want to make the default Windows search to index and search files in network too. Is there any way or tool from which I can do it?

I was able to do this in Windows XP Search and using something like Copernicus.  I don't see any way to do this with Windows 7 (other than making it available offline which means that all that data will be duplicated on my local drive.  What I've tried : I read elsewhere in the Windows 7 Help that you need to make sure the networked drive is indexed by it's O/S. So this network drive is on our Server (XP Pro, SP3) so I've installed Windows Search 4.0 and set it to indexing that drive.  I have another XP machine which is already indexed and Windows 7 still wouldn't let me add that drive to  a Library (to get it indexed) because it's not indexed. So I'm not hopeful that the above solution will work.


I've a weird bug with my username on stack exchange and overflow. When I post a comment or replay a question my user isn't DataHearth but :  a25bedc5-3d09-41b8-82fb-ea6c353d75ae  There's a screenshot          App Version: 1.0.95 Device Manufacturer: Xiaomi Device Model: MI 5s Plus OS Version: 7.0 (V9.5.2.0.NBGMIFA)

Earlier today I asked my first question on this site. So Meta SE was added to my accounts. But this account is shown three times:        One of them also has a badge while the other two don't.   The network profile doesn't show MSE at all:        Also I have got this long wierd user name and  I am not able to change it. In the edit- display name this long name is not there.My user name on other sites is shown there.   After all this happened Meta SE questions are not getting loaded in my SE app.    Edit: the issue of name is solved now. I changed the profile picture and saved changes and it worked.


How do I convert the GROUP_CONCAT row result as a separate columns for this query? SELECT Email, Address, CityName, StateName, PostalCode, GROUP_CONCAT(DISTINCT     CONCAT(       PP.PrefKey     )   )     FROM Profile P      LEFT JOIN ProfileAddress PA ON  P.ProfileID = PA.ProfileID      LEFT JOIN ProfilePrefs PP   ON PP.ProfileID = PA.ProfileID     WHERE P.ProfileID = 6;   DESIRED OUTPUT OF QUERY RESULT email | address | cityname | statename | postalcode | availability | certifications | commute .... ProfilePrefs  table

Using MySQL, I can do something like:  SELECT hobbies FROM peoples_hobbies WHERE person_id = 5;   My Output:  shopping fishing coding   but instead I just want 1 row, 1 col:  Expected Output:  shopping, fishing, coding   The reason is that I'm selecting multiple values from multiple tables, and after all the joins I've got a lot more rows than I'd like.  I've looked for a function on  and it doesn't look like the CONCAT or CONCAT_WS functions accept result sets.  So does anyone here know how to do this?


I am a permanent Resident of Canada and I am trying to understand the requirements of getting Schengen tourist visa.  I have not decided my exact dates of travel. I want to visit 6 to 7 countries. My appointment is with German Consulate for Schengen visa. All I know is I want to travel during the month of September or October for 2 weeks, depending on the best deal I get from travel agencies. How can I give them my itinerary or flight ticket details and accommodation details when I have still not decided on my exact dates?  I am also worried that, if I give hypothetical dates, they would issue a visa for just those 15 days and, later, if I cannot find any good deal for that period, it will be a waste.

My husband has a Chinese passport and I am a U.S. citizen; I don't need a visa to travel to Schengen countries. I am helping him to apply for a visa.  Our plans are to go to Paris, Italy and Spain, but we haven't nailed down definite dates; the approximate dates are August 18 to September 11. We made an appointment with Spain consulate to apply for the visa. We need to show air flight reservations, I could book it online without any problems.    Since we don't have definite dates, what dates should he put down on the application?  I know that, once they approve you, they give you 90 days, so does it matter what dates I put down on the visa application?   Our appointment at the Spanish Consulate is in a few days.


How can I change the type of text in beamer as shown in the image.

I've checked the  that can be used through beamer class. I've successfully used most of the fonts, but when I tried to include the Bradhley Hand font, it did not work (it shows ordinary characters)?! How can I use it?  For example, when I use the following :  \documentclass[serif]{beamer}  \usepackage{mathpazo}  \renewcommand{\rmdefault}{ibh}  \begin{document} \begin{frame} $\frac{ \sqrt{ \mu(i)^{ \frac{3}{2}} (i^{2} -1) } } { \sqrt[3]{\rho(i) - 2} + \sqrt[3]{\rho(i) - 1} } $ \end{frame} \end{document}   it gives:


Can someone help me understand the cross product a little better, for me it makes more sense for the new vector to be somewhere between the original vectors and closer to the bigger one but that would be just adding to one another right? how does crossing say for example vector $\vec A \times \vec B =\begin{bmatrix} 1 \\ 0 \\ 0 \\\end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\\end{bmatrix}= \vec C$?

I was wondering if anyone could give me the intuition behind the cross product of two vectors $\textbf{a}$ and $\textbf{b}$. Why does their cross product $\textbf{n} = \textbf{a} \times \textbf{b}$  give me a vector which is perpendicular to a plane?  I know I can just check this by using dot product but I'm not totally satisfied with "it just works" answer =)  Thank you for any help! =)


I just posted this comment:  and the source code is:  it works!   It seems that nothing visible here but I used this character: [alt + 0173], I knew this case is not common, but the JavaScript part can be improved to count readable characters only.  The HTML entity is: &shy; for that hidden character, its called

A follow-up to , and an extension of .   shows that it is possible to circumvent minimum character requirements using null bytes.  Can we exclude such non-printing characters as ASCII NULL and other control characters from counting toward minimum character requirements?  Edit: Users have demonstrated that , unmatched or invalid HTML tags, and other content that is not displayed after Markdown parsing can be used to bypass such requirements as well.  Can we cover this case as well, except for links?


after sudo apt-get install git:  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:  git : Depends: liberror-perl but it is not installable E: Unable to correct problems, you have held broken packages.   after sudo apt-get install liberror-perl:  Package liberror-perl is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source  E: Package 'liberror-perl' has no installation candidate

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?


Is there a difference between doing a shutdown via Ubuntu's GUI versus doing a "sudo shutdown now" from the terminal? I ask this as I notice that my machines take longer to shutdown if I shutdown via the GUI.

I was trying to shutdown my PC through SSH. I executed  sudo shutdown now  The user was logged-out and Ubuntu starting turning off, but it froze at the last screen with the Ubuntu logo and the loading dots. Any idea where the problem might be?  Also, what's the difference between shutdown and halt? What other similar commands are there?


I have been coding a basic method to take an array of strings, find the largest length and return a string array containing on values that are equal to the highest length of the array. I keep getting a null pointer exception and i am not sure why. The code is:  String[] allLongestStrings(String[] inputArray) {     int compare = 0;     int k = 0;     String[] use = new String[20];     for (int i = 0; i < inputArray.length; i++) {         if (inputArray[i].length() > compare)             compare = inputArray[i].length();      }      for (int j = 0; j < 20; j++) {         if (inputArray[j].length() - compare == 0) {             use[k] = inputArray[j];             k++;         }      }     return use;  }

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?


Which of the following is correct or better? A) Here's £5. B) Here are £5. In case both should be correct, could you explain why? Thanks a lot in advance.

Possible Duplicate:          We usually say "10 pounds", but for a single bill we say "10 Pound note" and not "10 pound(s) note". And when we have a lot of notes we say again "10 Pound notes". Why this disparity?


Yesterday, I had an exchange with my colleague about a black hole. His theory was that black holes were merely an imaginary construction that is purported to have some mathematical meaning. Is what he said true? If not, how do we know about black holes' existence and what properties do we know about them?

I'm not very knowledgeable about physics generally, but know that nothing can escape a black hole's gravitational pull, not even light (making them nearly invisible?).   My question is: What has been obtained from direct observation of a black hole to prove their existence? Please note that I'm not questioning the existence of black holes, but am just curious as to how we have come to realize it.


After thinking about ways to minimize my app size I wondered..  public int; public boolean;   does one take up more memory than the other?   if instead they are:  public int = 0; public boolean = false;   any difference there?  What about using an integer instead of a boolean( 1 & 2 ). Does that somehow take up less memory / space?

Can any one tell the bit size of boolean in Java?


i have an ubuntu-server box to which i connect via ssh.  now when i start a service (in this case teamcity) i would like to keep it running after the ssh session ends.  right now the service stops as soon as i disconnect.  i am starting the service using:  sudo service teamcity start   and the service script looks like this:  #! /bin/sh # /etc/init.d/teamcity #  # Carry out specific functions when asked to by the system case "$1" in   start)     echo "Starting script teamcity "     /home/m/TeamCity/bin/runAll.sh start     ;;   stop)     echo "Stopping script teamcity"     /home/m/TeamCity/bin/runAll.sh stop     ;;   *)     echo "Usage: /etc/init.d/teamcity {start|stop}"     exit 1     ;; esac  exit 0

I sometimes have long running processes that I want to kick off before going home, so I create a SSH session to the server to start the process, but then I want to close my laptop and go home and later, after dinner, I want to check on the process that I started before leaving work. How can I do that with SSH? My understanding is that if you break your SSH connection you will also break your login session on the server, therefore killing the long running process.


Mathematically a leading zero does not have any effect on the number value i.e. 10 = 010 = 0...010. However my java code is unexpectedly giving strange result by decreasing the value of the number on putting any leading zero. package com.start;  public class Basics {          public static void main(String[] args) {         int a = 10;         int b = 010;         int c = 00000010;         System.out.println("a = "+a);         System.out.println("b = "+b);         System.out.println("c = "+c);     }  }

This prints 83  System.out.println(0123)   However this prints 123  System.out.println(123)   Why does it work that way?


Is  $ \mathbb{Q}( \sqrt[3] 2)$ is  a normal extension of $\mathbb{Q}?$ My attempt : I think yes By definition  of normal extension : $L/K$ is a normal extension if and only  if  for every irreducible  polynomial  $f \in K[x]$  with at least  one root in $L$ Now  i take $f(x)=x^3-2$ then $f(x)$ has one root i,e $x^3=2 \implies x=\sqrt[3] 2$ So here we have atleast one root in $\mathbb{Q}$

In a question related to introductory Galois theory, I was asked to given an example of a tower of fields $F \subset K \subset E$ such that $E$ is a splitting field for some polynomial $f(x) \in F[x]$, but that $K$ need not be a splitting field. The example that comes to mind for me is $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\omega_3, \sqrt[3]{2})$.  So I know that the minimal polynomial for $\mathbb Q(\omega_3, \sqrt[3]{2})$ over $\mathbb Q$ is $x^3-2$. This polynomial does not split in the intermediate field $\mathbb Q(\sqrt[3]{2})$.  I wanted to make sure that I'm justifying that this intermediate field is not a splitting field correctly. Now I may be confused about the books terminology. When we say something is not a splitting field, we always have to refer to a specific polynomial correct? I've read a term called a normal extension which I think refers to one in which every polynomial splits, but my book doesn't mention these.   If splitting field means with respect to the polynomial $x^3-2$, then I've already demonstrated that $\mathbb Q(\sqrt[3]{2})$ is not a splitting field. Also, it couldn't be a normal extension either because that polynomial doesn't split.  Does it sound like I'm understanding this correctly or are there some subtleties I might be missing?


file1.cc  void func() {   std::cout << "test\n"; }   file2.cc  ... std::string str = GET_FILE("file1.cc"); ...   So, what I need is that str has the content of file1.cc as string, in compile time, NOT execution time.  How to implement this GET_FILE macro?

Is there a way to include an entire text file as a string in a C program at compile-time?  something like:   file.txt:  This is a little text file  main.c:  #include <stdio.h> int main(void) {    #blackmagicinclude("file.txt", content)    /*    equiv: char[] content = "This is\na little\ntext file";    */    printf("%s", content); }    obtaining a little program that prints on stdout "This is a little text file"  At the moment I used an hackish python script, but it's butt-ugly and limited to only one variable name, can you tell me another way to do it?


I have just installed KDE , set up my preferred environment (with desktop cube effects color scheme, set the splash screen to "minimalist" and also installed ksnapshot.) Installation was done with the terminal,  sudo apt-get install kde-desktop   Everything seemed to be working fine (actually pretty stellar.) Then I did a reboot to make sure all my changes could settle-in properly. When I got back to the login screen, which is the KDM login screen I selected in the setup process, however, there seems to be no user account for me (or anyone else for that matter.) If I click "switch user," my only option is something that says, "Unused (:0, vt8)" which is already selected. If I type in my previous username and password, it won't recognize it and acts as though I've typed in error.  While I cannot login, I do seem to have other options. The blue down-arrow gives me options of which desktop to use.      options are    "Defualt,"  "KDE Plasma Workspace,"  "KDE Plasma Workspace (failsafe session),"  "Ubuntu," and  "Failsafe."   The red "shutdown" button on the screen gives me the options,    "Switch User,"  "Restart X Server,"  "remote Login,"  "Console Login," and  "Shutdown."    Would someone please help me get it functioning again?It seems this has happened before.    This person closed their thread before there were any responses, so I wasn't able to post that I'm having the same issue. The "long-story short" version of this is, KDM won't accept (or seem to recognize the existence of) my username or password, or any other, and I cannot switch to another login manager to try that because that would require that it accept my credentials first. If it helps to know, before my reboot and subsequently getting stuck at KDM, LightDM logged me into KDE just fine.

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


We have an ellipse $x=a\cos(\theta)$ and $y=b\sin(\theta)$. We will find the area by adding sectors $\frac12r^2\;d\theta$. Now,  $$ \begin{align} \int_0^{2\pi}\frac12r^2\;d\theta&=\frac12\int_0^{2\pi}\left(a^2\cos^2\theta+b^2\sin^2\theta\right)\;d\theta\\ &=\frac12\int_0^{2\pi}\left(\frac{a^2+b^2}2+\frac{a^2-b^2}2\cos2\theta\right)\;d\theta\\ &=\left.\frac12\left(\frac{a^2+b^2}2\theta+\frac{a^2-b^2}4\sin2\theta\right)\right|_0^{2\pi}\\ &=\frac\pi2\left(a^2+b^2\right) \end{align} $$  Why is it not $\pi a b$?

I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating   $$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$  So we should be able to find the area of the ellipse by   $$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$  $$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$  $$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$  $$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$  $$=\pi\frac{a^2+b^2}{2}.$$  First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as .  What am I doing wrong?


Why is $S_{\ast}\left(X,A\right)$ free? it is the quotient of two free groups $S_{\ast}\left(X\right)$ & $S_{\ast}\left(A\right)$

Suppose you have a topological space $X$ and a subspace $A$. Their chain complexes are made up of free abelian groups $C_n(X)$ and $C_n(A)$ are the free abelian groups on the $n$-simplexes on $X$ and $A$, respectively.   If we take the quotient complex we get another chain $$ \cdots\to C_2(X)/C_2(A)\to C_1(X)/C_1(A)\to C_0(X)/C_0(A)\to\cdots $$ of abelian groups.   Are these quotients still free abelian groups though? I know in general that the quotient of a free abelian need not be free, but is it any different in this case however?


Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$   $n^3-1=(n-1)(n^2+n+1)$. If  $n^3-1$ is prime then $n-1=1$ or $n=2$ so $n^3-1=7$. But I need a concrete proof of it, not from intuition.

I discovered  which claims that "$7$ is the only prime followed by a cube".  I find this statement rather surprising.  Is this true?  Where might I find a proof that shows this?  In my searching, I found , which is similar but the answers seem focused on squares next to cubes.  Any ideas?


Can I use a linear polarizing filter with a DSLR camera in manual mode?

When I bought my polarizer, a friend told me that I should get a circular one, because the linear ones can mess with the autofocus. Is this true? What should each be used for?


I have the following tikzpicture:  \documentclass{article}  \usepackage{parskip} \usepackage{amssymb} \usepackage[fleqn]{amsmath} \usepackage{tikz} \usepackage{relsize}  \begin{document}      \begin{center}     \begin{tikzpicture}[scale=0.9]       % % % % % % % % left % % % % % % % %     \draw (-11.7,0.5) node {$\mathlarger{\mathlarger{\mathlarger{f=}}}$};      % small rectangles     \draw (-11,0) rectangle (-9,1) node[pos=0.5] {\large$\pi_{S_1}(I)$};     \draw (-8.6,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};     \draw (-8.2,0) rectangle (-6.2,1) node[pos=0.5] {\large$\pi_{S_2}(I)$};     \draw (-5.8,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};      \draw (-4.7,0.5) node {\huge$.\hspace{10pt}.\hspace{10pt}.$};      \draw (-3.6,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};      \draw (-3.2,0) rectangle (-1.2,1) node[pos=0.5] {\large$\pi_{S_n}(I)$};       % big rectangle     \draw (-9.5,3) rectangle (-2.7,4) node[pos=0.5] {\large\large\large$I$};      % arrows     \draw[<-] (-8.5,2.8) -- (-10,1.2);     \draw[<-] (-6.5,2.8) -- (-7,1.2);     \draw[<-] (-3.7,2.8) -- (-2.2,1.2);     \draw (-10,2) node {$\mathlarger{\mathlarger{\beta^{-1}}}$};       % % % % % % % % right % % % % % % % %     \draw (1.2,0.5) node {$\mathlarger{\mathlarger{\mathlarger{f\,'=}}}$};      % small rectangles     \draw (1.9,0) rectangle (3.9,1) node[pos=0.5] {\large$\pi_{S\,'_1}(I\,')$};     \draw (4.3,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};     \draw (4.7,0) rectangle (6.7,1) node[pos=0.5] {\large$\pi_{S\,'_2}(I\,')$};     \draw (7.1,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};      \draw (8.2,0.5) node {\huge$.\hspace{10pt}.\hspace{10pt}.$};      \draw (9.3,0.5) node {$\mathlarger{\mathlarger{\mathlarger{\bowtie}}}$};      \draw (9.7,0) rectangle (11.7,1) node[pos=0.5] {\large$\pi_{S\,'_n}(I\,')$};       % big rectangle     \draw (3.4,3) rectangle (10.2,4) node[pos=0.5] {\large\large\large$I\,'$};      % arrows     \draw[->] (4.4,2.8) -- (2.9,1.2);     \draw[->] (6.4,2.8) -- (5.9,1.2);     \draw[->] (9.2,2.8) -- (10.7,1.2);     \draw (2.9,2) node {$\mathlarger{\mathlarger{\beta}}$};      \end{tikzpicture}     \end{center}  \end{document}   The above produces this:    As you probably noticed, this is not what I intended when I used the \begin{center}\end{center} tags...  Is there a way to force the picture to be centered regardless of the left margin? In other words, how do I tell the picture to ignore the margin and to center the whole thing properly?  Thanks in advance!

I have a figure that is wider than the \textwidth of my document. (I don't want to change the \textwidth.) As a result, the leftmost part of the figure is flush with the left margin, while the rightmost part spills into the right margin too much. I'd like to center the figure, so that it will exceed the \textwidth equally on both sides.


Let $f(x)=a_0 +a_1x+\dots +a_nx^m \in\mathbb Z[x]$.  Suppose $\exists p$ prime: $p\mid a_1$, $p\mid a_2$, $\dots$, $p\mid a_n$, but $p^2 \nmid a_n$ and $p\nmid a_0$.   Prove $f(x)$ is irreducible in $\mathbb{Q}[x]$.  Note that this is not Eisentein's criterion, but it is very close.    "Proof"  (by contradiction)    We assume $f(x)=a_0 +a_1x+\dots +a_nx^m \in\mathbb Z[x]$, $\exists p$ prime: $p\mid a_1$, $p\mid a_2$, $\dots$, $p\mid a_n$, but $p^2 \nmid a_n$ and $p\nmid a_0$,  and $f(x)$ is reducible in $\mathbb{Q}[x]$.  if $f(x) $ is red. over $\mathbb{Q}$ we know it will be red in $\mathbb{Z}[x]$ so $\exists g(x),h(x)\in\mathbb Z[x]$ s.t. $$ f(x)=g(x)b(x) \text{ where } 1 \leq \deg g(x),\deg h(x) <n$$  Say $g(x)=b_rx^n+\dots+ b_0$ and $h(x)=c_sx^s+\dots +c_0$ now, $$                    p\nmid a_0 \equiv \neg (p\mid c_0 t_0)\equiv \neg  (p\mid c_0 \vee p\mid t_0)\equiv p\nmid c_0 \wedge p\nmid t_0                          $$   $$p^2 \nmid a_n\equiv p^2 \nmid c_s t_r \equiv p^2 \nmid c_s \wedge p^2 \nmid c_r$$  also, $p\mid a_n$ so $p\neq 0,\pm1 $ so $p\mid c_r r_s $  $\vdots$  Contradiction somewhere  $\vdots$

I know the 'Eisenstein's criterion'. I know that it states "($p\nmid a_{n}$), [$p\mid a_i$ for ($0\le i\le n-1$)], ($p^2\nmid a_0$)". I know regular way, but I hope to second pf way.  Let $f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}$,  deg $f(x)\ge1$. For some prime $p$, ($p\nmid a_0$) and $p\mid a_i$ for ($1\le i \le n$), ($p^2\nmid a_n$). Then $f(x)$ is irreducible over $\Bbb Q$.  My Proof. Suppose that $f(x)$ is reducible over $\Bbb Q$. Then $f(x)=g(x)h(x)$ for some $g(x),h(x) \in \Bbb Z[x]$ (by Gauss lemma), where $g(x)=b_{0}+b_{1}x+\cdots +b_{r}x^{r}$,  $\deg g(x)\ge1$, $h(x)=c_{0}+c_{1}x+\cdots +c_{s}x^{s}$,  $\deg h(x)\ge1$. So, $$f(x)=g(x)h(x)=(b_{0}+b_{1}x+\cdots +b_{r}x^{r})(c_{0}+c_{1}x+\cdots +c_{s}x^{s})=b_0c_0+\cdots+b_r c_sx^{r+s}.$$ Then $a_0=b_0 c_0$ and $a_n=b_r c_s$. Since ($p\nmid a_0$) and $p\mid a_i$ for ($1\le i \le n$), ($p^2\nmid a_n$), ($p\nmid b_0 c_0$) =>  ($p\nmid b_0$) and ($p\nmid c_0$). ($p^2\nmid b_r c_s$) => either ($p\nmid b_r$) or ($p\nmid c_s$) Say $p\mid b_r$ and ($p\nmid c_s$). Let $t$ be the largest integer s.t. ($p\nmid b_t$) and $p\mid b_i$ for ($ 0\lt t \lt i$). Why largest integer? I thinks set $t$ is smallest integer... then $a_t=(b_0 c_t +\cdots +b_{t-1} c_1)+b_t c_0$. Since $p\mid a_t$, then $p\mid b_t c_0$ -Contradiction- $p\mid b_0 c_t +\cdots +b_{t-1} c_1$ is true? Why? Set ($p\nmid b_t$) and $p\mid b_i$ for ($ 0\lt t \lt i$). It means that ($p\mid b_{t+1}$, $p\mid b_{t+2}$, $p\mid b_{t+3}$, $\dots$). I don't know it... plz help me!


After installing Kali Linux live (debian) on my USB drive. on running the display was inverted same as the mouse I have tried some basic commands like xrandr but not available.

I installed kali linux OS on my laptop.  The installation went well but when I booted in kali linux the screen was backwards. I signed in as root and the screen flipped back but the mouse is reverse (x & y axis).  When I move my mouse left it moves right when I move it down it moves up. If I want to press a button at the upper right I need to press it at the down left.  The screen is acting backwordes (x & y axis)but I see the graphics straight and the mouse is acting backwordes.  I installed kali linux again with a fresh installation. I have the same problem as before.  I installed kali linux again with a fresh installation. I have the same problem as before.   I searched the web And I don't find information on a problem like this


I am an absolute beginner in Ubuntu and I am  trying to install developer tools. Can someone guide me I have tried the following:

After upgrading from 10.04 to 12.04 I am trying to install different packages.  For instance ia32-libs and skype (4.0).  When trying to install these, I am getting the 'Unable to correct problems, you have held broken packages' error message.   Output of commands:  sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   After running this:  sudo dpkg --configure -a foo@foo:~$ sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.


I have applied for a green card and travel document which is in process.  Now I want to travel domestically in the US. Is it possible to travel with the expired visa? I attended my interview for the green card and I did not receive my travel document yet. So is it possible to travel domestically?  Can anyone advise me on this?

I am an international student and authorized to stay in US for 2 years on my I-20 form. My visa expired on June this year and I need to fly to some states outside Georgia to visit colleges. Can I fly within the US?  If I can, what kind of document will the airport check when I depart? Will they notice my expired visa and turn me in to Homeland Security? In other words, will I get into trouble?


I want to run a function that will create a file with the name of a variable and write in it the content of that variable.  list_A = 'one two three'.split()  def write_list(list_to_write):      flux = open('%s.list' %list_to_write, 'w')     flux.write('\n'.join(list_to_write))     flux.close()  write_list (list_A)   This creates a file called ['one', 'two', 'three'].list   I would like it to create a file called liste_A.list instead.  PS: I found many similar posts, but none that seemed to address my issue.

Say I have a variable named choice it is equal to 2. How would I access the name of the variable? Something equivalent to  In [53]: namestr(choice) Out[53]: 'choice'   for use in making a dictionary. There's a good way to do this and I'm just missing it.  EDIT:  The reason to do this is thus. I am running some data analysis stuff where I call the program with multiple parameters that I would like to tweak, or not tweak, at runtime. I read in the parameters I used in the last run from a .config file formated as  filename no_sig_resonance.dat  mass_peak 700  choice 1,2,3   When prompted for values, the previously used is displayed and an empty string input will use the previously used value.  My question comes about because when it comes to writing the dictionary that these values have been scanned into. If a parameter is needed I run get_param which accesses the file and finds the parameter.   I think I will avoid the problem all together by reading the .config file once and producing a dictionary from that. I avoided that originally for... reasons I no longer remember. Perfect situation to update my code!


So today I was asked a question which I found very simple yet I honestly couldn't explain (embarrassingly).   One of my mum's friends who knows I'm a scientist asked me "If I turn the light off in a room why does the light instantly disappear"  To phrase it up in a more scientific way :  If we took a box of light and sealed up the entrance to the box such that the photons can't escape, how come (since the photons are still inside the box) it appears dark?   My thoughts were to do with the source of the radiation being removed, but I have no idea why photons bouncing about inside the box do not light up the inside? Intensity maybe?   I guess its sort of like black body radiation approximated by a cavity with radiation inside and a pin small hole following Planck's model, but in this case we just close the hole rather than measure any emitted radiation.   Any ideas?   Edit Thank you all very much for your answers! I understand now. Hypothetically speaking if the walls were of a material that did not permit the adsorption of a certain frequency of the light would it just bounce around forever? (I actually have another "simple" question to ask, but I will ask it in a separate part later this evening. Thanks again :)

I am failing to explain why light won't remain inside the wooden box in the following situation. I considered a wooden box closed from all the sides, with a bulb inside it. If we switch on the bulb, light will be emitted from the bulb inside the box, if we switch off the bulb, I assumed the light to disappear, which I practically assume to be true. If I am wrong here, please explain. So, practically I assumed light to disappear if we switch off the bulb, after being switched on for a while. I got a question here, why light has to disappear the moment we switch off the bulb?      Thinking about the above situation, I considered the case of bullet being fired into the wooden face from the inside of the same box. No doubt, bullet smashes the wooden face into pieces and comes out. Here bullet has high speed and even has mass, so that, it disappears out of the wooden box. I considered the photons now, I thought photons have high speed much greater than bullet, but they have very less mass (I am not sure with rest or relativistic), so I thought that, photons should remain into the wooden box after colliding the wooden face. If they would had, I thought that there should have been light inside the box even after the bulb was switched off, but it is not. Is it that photons are piercing into the voids between the atoms with high speed to get out of wooden box?      If we say photons are getting out of the wooden box piercing into the voids, why won't it be in total internal reflection phenomenon. If we assume light to be incident at the interface from the denser medium to the rarer medium, at an angle greater than critical angle, photons should even then pierce into the voids. But, my teacher has taught me that, in total internal reflection there would be no refraction. By this even my assumption of photon getting out of the box from the voids also falls down. As radiation is a form of energy, it can't be lost with out being utilised or else it can't be lost without getting emitted out of the box so as to avoid violation of conservation of energy. Is here, the light absorbed inside the box by the medium? I don't know whether I have misunderstood here or whether  we could explain from wave nature of light, so as to why light disappears from the box?


My question might seem very dumb but here goes. Say we have y = f(x). If we have g = dy/dx, we can also say that gdx = dy, and this works. Why does it work? I know that dy/dx isn't a fraction but what underlying properties are being implicitly used when we change g = dy/dx into gdx = dy? An example, to derive one of the laws of motion we can start with a = dv/dt, then say that adt = dv and by integration of both sides we conclude that at = v-u <=> v = u + at

In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?


I was trying to run the Windows diagnostic that ships with Sony laptops when it "froze" and I've lost my Linux 12.04 partition. I had VM's and software configured that I can't afford to lose. Please help me return to the state where I was before I tried running that Windows tool.  I am running a Lunux 13.10 LiveCD.   When I run the command: sudo gpart /dev/hda, I get the following output:  *** Fatal error: open(/dev/hda): No such file or directory.   However, when I run the command sudo fdisk -l, I get the following output:  Disk /dev/sda: 640.1 GB, 640135028736 bytes 255 heads, 63 sectors/track, 77825 cylinders, total 1250263728 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x2d85c4ea     Device Boot      Start         End      Blocks   Id  System /dev/sda1            2048    28887039    14442496   27  Hidden NTFS WinRE /dev/sda2        28887040    29091839      102400    7  HPFS/NTFS/exFAT /dev/sda3        29091840  1250260991   610584576    7  HPFS/NTFS/exFAT ubuntu@ubuntu:~$ gpart /dev/hda   How should I proceed?

I was installing ubuntu 12.04 on my system. There were 4 partitions on my system and I selected one of the four partition for the installation and chose the option of resizing the partition. Initially my partition was of size 100+GB and i created another partition out of it of size 15GB (EXT4). Now the moment I changed this partition structure my original partition got lost along with its data and I am left with 50GB partition and 50GB unallocated free space.   Now the data that I have lost is meant a lot to me and I want to recover that data. So is there any way I can recover it? And I haven't checked "format" option while resizing the partition.


This is my code:  SELECT b.nameBook, a.firstName, a.lastName FROM books b     INNER JOIN books_author ba ON b.idBook = ba.BooksId     INNER JOIN author a ON a.idAuthor = ba.AuthorId     ORDER BY b.nameBook;     I don't want to repeat the books name (in this case, "Fisica 11"). I want it . How can I do it? Sorry if this is a dumb question, i'm a beginner.  Also, I wanted to ask if there is any difference between the code above and the one below and which one is better to use:  SELECT nameBook, lastName, firstName      FROM books b, author a, books_author ba      WHERE ba.AuthorId = a.idAuthor          AND ba.BooksId = b.idBook   Both of them give the same result.  I tried GROUP BY but  is what happens.  I could easily fix this by using php but I want to use SQL. Thank you.

Is there any efficiency difference in an explicit vs implicit inner join? For example:  SELECT * FROM table a INNER JOIN table b ON a.id = b.id;   vs.  SELECT a.*, b.* FROM table a, table b WHERE a.id = b.id;


I wish to achieve following. There may be many variant of this.

I have a sentence like this "पठ् + अ + ति" and I want to add some notes like shown in the attached image. Is there a package for such things or I need to work with Tikz.   Any reference material or help would be appreciated.     ============Edited question based on Solution given by Gonzolo =========== Here is MWE and the output I am getting.   \documentclass{article} \usepackage{fontspec,xltxtra,xunicode} \usepackage{tikz} \usetikzlibrary{shapes.callouts} \setmainfont[Script=Devanagari,AutoFakeBold=3.5]{Sanskrit 2003} \definecolor{myblue}{RGB}{122,175,234}  \newcommand\tikzmark[1]{%   \tikz[remember picture,overlay]\node (#1) {};} \newcommand\MyNote[3][]{% \begin{tikzpicture}[remember picture,overlay] \node[rectangle callout, fill=myblue,text width=2cm,align=center,font=\sffamily\bfseries\color{white},#1] at #2 {#3};  \end{tikzpicture}% }  \begin{document}  पठ् \tikzmark{a}\qquad + अ + \tikzmark{b} ति ।  \MyNote[callout relative pointer={(0,1)}]{([yshift=-1.5cm]a)}{क्रिया को धारण करता है वह धातु है ।} \MyNote[callout relative pointer={(0,1)}]{([yshift=-1.5cm]b)}{"अ \& ति" का कर्ता अर्थ को कह रहा है}  \end{document}


Here are my steps:  $PAP^{-1}PBP^{-1}=PABP^{-1}=AB$, since both are diaglonalizale, the equation is established, is it correct?

I'm studying Linear Algebra. I saw an example of a pair of 2 by 2 or n by n diagonalizable matrices, the product of which is not diagonalizable. Is there a similar example when I replace the condition "diagonalizable" by "invertible and diagonalizable"?


How do i make material like this lampshade? Let light through, but not be transparent.

I'm having trouble recreating this material in cycles:      It's a solid glass bulb with a LED inside and copper threaded fabric which gives it a copper colored glow.  I'm not sure how to achieve a similar hue / glow and IOR.  Any pointers into the right direction would be greatly appreciated!  Blendfile:


Disk Utility.app displays my internal HDD "APPLE HDD" to be entirely taken up by "Macintosh HD" at 759.65 GB. Seeing as it’s a 1 TB hard drive, that’s a little disconcerting.   I also can’t reclaim the space using the partition tool, because there’s no partition to remove.  How do I fix this? Do I need to do a fresh install or reformat or something?  EDIT: Found the answer to it

Using Boot Camp Assistant, I started to create a 50 GB partition, but it hung on partitioning. After waiting about 30 minutes without the progress bar moving, I quit the Boot Camp Assistant, thinking I would be able to fix the partition either by using the Boot Camp Assistant or the Disk Utility. However, the partition doesn't appear in either. Instead, my drive has just become 50 GB smaller. Looking for other solutions, people were generally able to find the partition by using Terminal application commands, however I haven't even been able to do that. (Notice in Disk Utility, it says Macintosh HD is 441.65GB, but "Capacity" is 500.28GB):    I tried booting in Recovery mode and doing First Aid in the Disk Utility. This resulted in the same "All's Fine," but still haven't gotten my 50GB back. I spent all day clearing 100GB to make room for a Windows partition, and now in the Finder I'm down to 43GB :/.  Any help would be greatly appreciated, thank you!


In my English homework, a student gets asked if she would put "Any news your end" in a formal or informal letter. My class should translate it, but in my native language it means something like "Some news you want to tell before/while dying?" and thats makes no sense to me.

Did they come to you and immediately make you feel a relationship was   being offered with the Russians that you’d like to develop? Or did you   seek it from your end?   I found the phrase in this article:


The text of one of my posts' titles was completely changed. From "Asking the wrong questions vs. asking at the wrong site" to "Why we don't accept shopping questions". (The latter could be a little different because I changed the title back and can't find the edits)  The meaning and purpose of the post was destroyed. I find this practice from a few users irresponsible, demeaning and immoral. While I did post my disappointment about the reasons one my posts was closed and the topic was related to shopping, I don't find a reason to change a stand alone post to correlate with something I previously mentioned. The post below was all about asking the right question. It had nothing to do with shopping.  What do you think of titles being completely replaced with no what the poster had intended? Regardless of some editor thinks.

I'm getting pretty close to the reputation where I can start to edit others' posts, and I'm wondering what the etiquette for when (and when not) to do so. I figure that I should edit in the following circumstances:  Grammatical or spelling errors. Clarification where the meaning is not changed. For instance if the title does not describe the question very well. Where the user has made a very minor mistake that doesn't justify a full post to clarify. To add related resources or links that will help someone answer the post, or provide context  Before the commenting feature was added, I noticed that some moderators edit posts in order to insert a reply or extension to the post. I feel this isn't good etiquette (but I could be wrong) because I don't know what to do with their changes when I come to re-edit or respond. I also feel that I should be very careful not to change the meaning of any question or answer, even if that meaning is misguided. What other rules would be best practice?  Signatures and fluff are discouraged and are fair game to be removed.  See, for example, this post by  and this post by  and also see .  Also see Jeff Atwood's response in the comments to


I'm not really sure how to approach the problem, since I'm not really sure I understand the mechanisms why it is true aside from putting some numbers in and see that it works.  Qualitatively, I'd try a proof by contradiction, trying to show that $\gcd(a-bc, b) \neq \gcd(a,b)$ and use some prime attribute to contradict the assumption.

How can I prove that $\gcd(a, b) = \gcd (a, b + ma)$?  I have tried this:  let $g = \gcd(a, b)$, then $g \mid a$ and $g \mid b$. This means that $g \mid ax+by$.  I don't know what to do next.  Thanks.


I run an organisation that cleans data from donated laptops and reinstalls minimal operating systems and office for poor schools in rural India.  I plan to boot and wipe the laptops I receive, but I don't know what to do after that to install Ubuntu. Which release/variant/version should I burn onto CD/DVD?  In terms of software, would like to install at least OpenOffice on the laptops.

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


I am new to c++ and I was using code blocks . I found that main() function is not taking return type other than int and void . I tried other data type to return the value and the returned the value of the same data type used . It was giving me an error when i use return type other than void and int . I would really appreciate it if you could explain it to me.

What is the correct (most efficient) way to define the main() function in C and C++ — int main() or void main() — and why? And how about the arguments? If int main() then return 1 or return 0?  There are numerous duplicates of this question, including:       — For C++, with a very good answer indeed.     Related:


So I'm trying to create the face of a character model by starting around the eyes and mouth but when I try and join the edges to connect it creates this weird crease. I want to smooth it out but I can't seem to get rid of it. What I'm doing is pressing F with the selected vertices to create a new face to connect and this happens.  Any help to getting rid of it is appreciated, thanks!

I am relatively new to Blender so please excuse me if this is rather amateur!   The bottom of my torus has poor surface quality, I have scaled the vertices down for a smaller (variable) tubular radius all the way around the torus and at the bottom the surface quality is awful, with inaccurate shading. Any ideas as to what is wrong and/or how I could fix this?  Thanks, itsanate


Am updating my website Backlinks and content from last 2 month but page rank is not changing,how often does Google re-calculate the page rank?

As you may know Google updates PageRank roughly 4 times a year, out of interest and not for ranking purposes I often check when the next incoming update is. I do this by searching various sites and finding out the history, Well since I'm a regular on Pro Webmasters I thought I'd ask the question and provide the answer for myself and others webmasters alike.   So the question is simple, what is the last updates and when we can expect another Google Page Rank (PR) Update.   UPDATE: Still no Google Pagerank update yet - very late.


I have watched some of the lectures given by Brian Greene and Lawrence Krauss about the observation conundrum encountered during the double slit experiment. Can anybody answer me this question?  Is there a difference in results between    case 1: observation by a sentient creature (e.g. human or cat), or  case 2: the unmonitored recording of events (to be looked at later).   It seems to me that if there is a difference, then it is clearly the presence of something conscious/sentient that is making that difference, not just the recording of the events.  It seems also to me that in either case, since electrons are everywhere, and observation of their behaviour makes a difference, there is a serious consequence - scientific method falls victim to this ambiguity in behaviour for the simple reason that it involves observation, so this must undoubtedly affect the physical rules.  Perhaps I should surmise that we are entering an era of understanding where our sentience has to be considered part of the physical reality, and though this is an old idea, most of our top physicists are still saying they are long way from getting a handle on this fact. So:    when they decide that sentience is part of the universe, does that mean "intelligent design" is again on the agenda?   So, can anyone answer, is it case 1 or case 2 above?

My question is not about (pseudo) philosophical debate; it concerns mathematical operations and experimental facts.  What is an observer? What are the conditions required to be qualified of observer, both mathematically and experimentally?


I am looking for a general rule for all possible integers $k\in(a,b)$ such that $k$ is can be expressed as a linear combination of $a$ and $b$.   I understand that k must be a multiple of $\gcd(a,b)$, since any linear combination of $a$ and $b$ will be a multiple of $\gcd(a,b)$. Is there a general formula for $k$ in terms of $a,b$?  Thanks in advance!

I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?   I tried to play around with it: $x = (c - by)/a$, hence $a|(c - by)$.     $a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?  Thanks, Chan


I have subscribed to mailing lists using Gmail's .  The subscribed email address is of the format hikingfan+mailinglists@gmail.com and I can receive emails on the same address. But when I reply to an email it is sent without the sub-address tag i.e. hikingfan@gmail.com.  The sent message is being held until the list moderator can review it for approval since the list considers it as a post by a non-member to a members-only list.  Since hikingfan+mailinglists@gmail.com is already a member of the list, is there any way to send emails with the sub-addressing tag?

I know that I can sign up to a service using user+foo@example.com, the question I have is how can I send an email from this same address?


I find my self jumping between those the tags Java and Android very often while looking for new questions.  I know that I can filter by tag1 AND tag2 like this:   That has the disadvantage that I do not get the  questions that have nothing to do with , but I am also very interested in those.  So I would like to do it filter by tag1 OR tag2 to also get the pure java + pure android questions!  For the case something like this is already possible, please kindly explain how it should be done. I tried to find such an OR filter but could not find it and at least for me would be very useful! Thx!

Is there a way to search questions that are tagged with either tag1 or tag2?  Actually, I would like the general filter,      ([tag1] AND [tag2], ...) OR ([tag3]   AND [tag4], ...) OR ...   And even     [tag1] AND ([tag2] OR [tag3])   I need that because I'm looking for interesting questions and I have to search for each tag separately.


I read this book probably close to 20 years ago (and even then I don't think the book was new). It might have even have been a short series.  It's in the future and the descendants of some spacefaring people have ended up on a planet and a small group of them have "magic powers." These powers are granted by symbiotic crystals that are tempered with various elements, e.g. fire/water. Depending on what type of crystal they bind with, it gives them control over the related element. I think this process of binding to the crystals happens in the spaceship that brought the original people to the planet.  I think that the main character binds with a crystal that wasn't tempered by any element and thus can control all of them?

I read this book a long time ago, but I'm guessing it was written sometime in either the 80's or early 90's. It was a rather long/large book.  Plot summary: Set in a fantasy land of wizards and magic, some kids (maybe) are recruited to learn magic. In the process they find out that wizards are created by being placed in a crashed spaceship that bonds them with crystals that imbue the powers. One of them is able to bond with many/multiple crystals to become very powerful.


What sort of Rice do Chinese Restaurants use? I use good quality Thai Rice at home in my rice cooker, I have tried all sorts of brands of rice and I have tried mixing different rice together, but, I have never been able to achieve the same taste of rice that I have had in Chinese Restaurants. Any advice?

I love the Chinese food we get from takeaway shops, and often to save on the cost we will cook our own rice at home. This is good, but the rice never seems to have the same texture as takeaway rice - it's less sticky, and much harder to eat with chopsticks. I usually use plain long grain rice as that's what I have in the cupboard. Does anyone know what kind they would use in a Chinese restaurant or takeaway?  Also, out of interest, what kind is used in authentic Chinese cooking? (I am well aware that this is vastly different from what we eat as "Chinese food" in the West!)


I want to be able to reference a captionless figure (e.g. \ref{figure1}). Based on , it appears that labeling a figure (e.g. \label{figure1}) actually labels the caption. How do I reference my captionless figure? I tried adding an empty caption (e.g. \caption{}) but this create an undesired Figure 1.

From what I understand, in order to refer to a figure in the text and be able to access it via clicking, I use \label{}, and then in the text I use \ref{}, with the same name inside.  What I would like is for the figure to not have the caption appear, but to keep the numbering. So that in the text, there is still "1" that I can click to jump to the figure, but for there not to be "Figure 1" or "Figure 1: etc". If I try including \caption*{} (as I've seen on other answers), the numbering goes away and is replaced with "??" which is not what I want (although it still goes the picture when I click it).  I think getting rid of the caption package and associated uses of \caption{} would do the trick normally, but within the same document later on, I am planning on cross-referencing to figures in a document that uses the captions and the caption package quite heavily, so I figure I need to be able to have the caption package here.  Example code below:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{caption} \usepackage{graphicx} \usepackage{hyperref}  \begin{document}  \section{Introduction}  Let's test this \ref{Fig:test}  \begin{figure}     \centering     \includegraphics[width=\textwidth]{example-image-a}     \caption{}     \label{Fig:test} \end{figure}  \end{document}   Any help would be appreciated.


I am an Indian citizen with F-1 status travelling from the U.S. to India. I am going to have a layover of 12 hours in Paris while on my way to India and another layover of 6 hours in Amsterdam while returning from India. Do I need any visas for sightseeing Paris and Amsterdam during the respective layovers?

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


I imported a mixamo character in blender 2.83. Its armature scale is set to 0.01.I want to apply this scale (Ctrl+A).  But whenever I apply the scale, the armature and mesh changes it location and the animation becomes all messed up.

Basically what I want to do is scale a skeleton down. The reason for this is that I'm ripping a model from another game and I want to use it in Unity. When the model is ripped from the other game, it is way too big for unity and has to be scaled down in order for it to even show up on the screen.  This is my model. It has a walking animation.    Now I want to scale the model down from it's original scale to 0.2 by 0.2 by 0.2, I scale the skeleton down so that the mesh goes with it.   Now I'll have to apply the scaling, so that it's the smaller scale is the default one, instead of the big one.   As soon as I apply it, the model moves away from it's original location, and the animation is all messed up, it goes much faster to it's walking destination because he's smaller.  How do I avoid the skeleton, and most importantly, the animation from messing all up?


How to solve the problem? (The existing system is Windows 7, 64-bit.)

I've been trying to install 12.04 for a while now, I'm running on Windows 7 and I already made the partition (it's dynamic). I've tried booting from a disc and also the Windows installer. I haven't tried used any repair-boot programs or anything. Every time I tried to load from either, I get an error:  command=C:\windows\sysnative\bcdedit.exe /set {03f9077f-661f-11e1-845a-d08b36406c3d} device partition=U;  retval=1  stderr=An error has occured setting the element data. The request isn't supported  stdout=   Then after that, it makes a log file with information that I don't understand.  If it is simply due to the fact of it being a dynamic partition, someone please help. If it is more extensive, anyone with details on possible solutions or definite ones please express them.


I'm using Fedora 17.  I was backing up some files to a USB key.  I was foolish and used the Move command instead of copy, and before the directory I was moving completed I accidentally yanked the usb key from the port.  The USB key is now empty, and the directory I moved is no longer on my hard drive.  Does anyone know if I can recover from this?  Is there a temporary directory that houses moved files?  Thanks!

I accidentally deleted a file from my laptop. I'm using Fedora. Is it possible to recover the file?


i am trying to  move a small div along a big div in the y-direction according to how much i scrolled down the page.but i've found that using setTimeout() and setInterval() gives two completely different results.actually setInterval() hanged by browser several times .what is the basic difference between the two function??  <!DOCTYPE html> <head> <title>creat a dynamic div</title> <style> #mydiv{ border:2px solid green; } </style> </head> <body> <script> var i=0; var elem1=document.createElement("div"); var atts1=document.createAttribute("style");  atts1.value="width:200px;height:3200px;border:1px solid black;background-color:orange;"; elem1.setAttributeNode(atts1); document.body.appendChild(elem1);  var elem2=document.createElement("div"); var atts2=document.createAttribute("style"); var atts22=document.createAttribute("id"); atts22.value="mydiv"; atts2.value="width:200px;height:300px;background-color:red;position:absolute;top:0px;left:300px;"; elem2.setAttributeNode(atts2); elem2.setAttributeNode(atts22); document.body.appendChild(elem2);  function moveIt(){ var a=window.pageYOffset;   if(i > (a+30)){  clearTimeout(p);  }else{ elem2.style.top=i+"px"; i=i+1; } var p=setTimeout(moveIt,200); }  window.onscroll=moveIt;  </script> </body> <html>

As far as I can tell, these two pieces of javascript behave the same way:  Option A:  function myTimeoutFunction() {     doStuff();     setTimeout(myTimeoutFunction, 1000); }  myTimeoutFunction();   Option B:  function myTimeoutFunction() {     doStuff(); }  myTimeoutFunction(); setInterval(myTimeoutFunction, 1000);   Is there any difference between using  and ?


There are many answers as to how to show item one item at a time in Beamer  But these answers do not address the case where I want to hide the first item as well (start with a completely blank slate). The first item is always displayed when you use the following code. \begin{itemize}[<+->]  ... \end{itemize}  I tried to put it in a \only<> block. However, this causes the itemize to display two items simultaneously. Is there anyway to address this problem?

In Beamer, \begin{itemize}[<+->] causes items to appear one-by-one. However, item 1 is visible starting from the first slide itself. Is there a way to insert a pause before the first item too?  of course I can manually type  \item<2-> first \item<3-> second ...   but is there a shortcut way to specify this?


I tried to install Ubuntu in VirtualBox on my laptop (Windows 10 Home). I selected the ISO file I downloaded: ubuntu-16.04.1-desktop-i386.iso, and started it up from VirtualBox. But after starting, the interface became blurred and showed macroblocking like this :    I wondered if my ISO file goes wrong, but after using winMd5Sum it shows that the MD5 is the right one. I cannot find relevant messages. Could anyone tell me how to fix the problem?

Inside a VirtualBox VM, I have installed myself an Ubuntu 16.04.1 LTS Server installation. However, when I installed the ubuntu-desktop package and rebooted, this happened:    Exact command I used: sudo apt-get update; sudo apt-get install ubuntu-desktop ..and then a reboot command.  This just seems like a corrupted top-left portion of the screen, as my cursor can appear in view (but corrupted):    This is a purely graphical issue because I can use the keyboard to successfully progress to a desktop:    Note: On startup it shows correct graphics for a split-second then reverts to glitch.    Any help on this?


I have a receipt issued by an Italian post office for my first residence permit, as well as a foreign passport with expired D visa.   Can I enter Italy with this?

I have a residence permit (permesso di soggiorno) for Italy. I am currently in the Netherlands for an exchange. My permit expired on the 30th of November and I applied for a new one but I will get the card in January. I have the documents supporting that I have applied for the renewal of the permit. In the middle I want to travel to Italy for Christmas and come back to the Netherlands.    Would this be a problem?


I've got an Ubuntu 14 VM (holding backups) with two very large disks. One disk (the primary) is locally in the data center. The other is on media physically separate (another building). My hope is to easily take a snapshot of the primary file system and then update the secondary with that data.  If I were using ZFS, this would be quite easy. I'm not sure if it is nearly so easy with LVM/ext4 though.  I'm familiar with how to take a snapshot of the file system with LVM. (). However, everything I find on google (including this link) seems to suggest mounting the snapshot and then backing up the file contents. I'd prefer to duplicate the file system itself.  The question  is very similar. Though in my case, I'd like to duplicate to a second LVM partition rather than to a file on a remote server.

I have a number of Xen virtual machines running on a number of Linux servers. These VMs store their disk images in Linux LVM volumes with device names along the lines of /dev/xenVG/SERVER001OS and so on. I'd like to take regular backups of those disk images so I can restore the VMs in case we need to (the LVM devices are already mirrored with DRBD between two physical machines each, I'm just being extra paranoid here).  How do I go about this? Oviously the first step is to snapshot the LVM device, but how do I then transfer data to a backup server in the most efficient manner possible? I could simply copy the whole device, something along the lines of:  dd if=/dev/xenVG/SERVER001OS | ssh administrator@backupserver "dd of=/mnt/largeDisk/SERVER001OS.img"   ...but that would take a lot of bandwidth. Is there an rsync-like tool for synching contents of whole disk blocks between remote servers? Something like:  rsync /dev/xenVG/SERVER001OS backupServer:/mnt/largeDisk/SERVER001OS.img   If I understand rsync's man page correctly, the above command won't actually work (will it?), but it shows what I'm aiming for. I understand the --devices rsync option is to copy devices themselves, not the contents of those devices. Making a local copy of the VM image before syncing it with the remote server isn't an option as there isn't the disk space.  Is there a handy utility that can synch between block devices and a backup file on a remote server? I can write one if I have to, but an existing solution would be better. Have I missed an rsync option that does this for me?


I've made a chest that gives you random items. It works, but now I want that is spawns a chest wit random items.   This works:  /give @p minecraft:chest 1 0 {BlockEntityTag:{LootTable:"mctools:chests/loot_table"}}   And this does not:  /setblock ~1 ~ ~ chest 1 0 {BlockEntityTag:{LootTable:"mctools:chests/loot_table"}}   What's wrong? Or it is not possible?

I tried the custom loot tables here:                                (with his website:  ).  I successfully obtained my chest with the custom loot using:   /give @p minecraft:chest 1 0 {BlockEntityTag:{LootTable:"mctools:chests/loot_table"}}   The problem is that I want to summon the chest instead of giving it to the player. I tried the command:   /setblock ~1 ~ ~ chest 0 replace {BlockEntityTag:{LootTable:"mctools:chests/loot_table"}}   The chest did place, but when I opened it, the chest was always empty. Why can't I retrieve my loot this way? And if possible, how can this be done?


Is there any way to lock a shell script from being executed twice?  I need to execute a script that transfers some files from one server to another. The script has to be executed by cron once a day. The connection between both servers should be fast, but may be slower at some times, meaning that the transfer may take more than 24 hours time. Hence, if the script is still running after 24 hours, cron should not execute it.  I thought of creating a file named myscript.lock at the start of the script and deleting it again at the end of the script execution. When the script is started it first checks if the file exists and if it does, then the script stops. This way I can guarantee only one execution. But, if the server breaks down during the script execution, the file will remain and not be deleted. Hence, every time the script is started it will think that it is already running.  Is there any locking functionality offered by the OS I could use for this matter?

I have multiple threads running from different terminal windows in ubuntu trying to write to the same file.   I would like to lock the file via terminal while writing so that the other thread(s) has to wait to write till the lock is released. How do I do this?


If I rememeber well, it must be an animated series frome the mid 90's, possibly the early 2000's. I'm mostly sure it was broadcasted on either France 3, canal J or cartoon network  Part of the plot was finding a secret stardrive hidden in the center of some planet, which when activated with a certain key put out gigantic thruster to be able to move the planet fom harm.  My memory is relly scarse on this but the series might had some part character or background in 3d instead of "hand drawn"

I remember seeing this series back when I was in the early grades of primary school (prep to 3) so this would be between 1996 and 1999  from what I remember it was set in space and the cast was made up from people from these different worlds, one of the characters is from a fire attributed world and I get the impression he was small framed (I think).  at one point in the series this entity appears called the Beast Planet appears, the one scene I do remember is that this Beast Planet attacks the fire planet, the front of the beast planet opens up with this claw and grabs the fire planet from bellow and then the fire planet is slammed into the beast planet dooming the fire planet but I think the beast planet was unharmed.  I also seem to remember that the cast discover the other planets have engines which allows the remaining planets to escape from the Beast Planet rather than them being sitting ducks.  I remember that this beast planet, despite being called a beast is actually more machine. I began remembering this series after playing Galaxy Angel where I saw the  and it reminds me of this beast planet


Suppose we have an Inertial frame S, all the clocks in this frame are synchronized. Now suppose, two seperate events occur at two different place A and B in that reference frame. Now, the events are said to be simultaneous if the clocks at the respective places registers the same time for each event(i.e two observers at the respective places read the same time of occurrence for each event). Now, suppose we have an observer at some different place C & the distance from A to C & that from B to C are not equal. Now my question is that: will the observer at C also claim about these two events to be simultaneous?  If "No" is the answer, then we have observers in one single inertial frame, disagreeing about the simultinity of two events. See, there is no relative velocity between the observers, all they are in a single inertial reference frame at different places(I'm strictly talking about a single inertial frame. No relative velocity between the observers, nothing!) How this can be possible that the observers in a single reference frame are not in agreement about the simultinity of two seperate events though all the clocks in this frame are synchronized from the beginning?

I am having a bit of confusion, let us say there are two observers observing two simultaneous lightning strikes with a finite distance between them. Now one observer is located such that he is in between the two lightning strikes, while the other is near to one strike and away from the other. In this scenario, the observer situated in between will perceive the lightning strikes to be simultaneous while the other will perceive them differently, since he is close to one and far from the other. Does this mean the Principle of Relativity holds even without actual motion?


I'm trying to validate a phone number in javascript with  the following regex  var phoneExpression = /^\(?([0-9]{9})\)?$/;   this expression works with xxxxxxxxx number but i need  a regex for  xxxxxxxxx , +xx xxxxxxxxx and +x xxxxxxxxx and all i've tried fails.

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.


Hypothetically speaking, let's say there is no trading volume during 1 year for a given stock, does that mean the stock price will remain the same regardless of how the company is performing?  Let me emphasize the word "Hypothetically", because I know it is unrealistic. I just would like to know if Supply and Demand is the only rule applied when it comes to stock price.

Let's assume there's this random company (XYZ)--that few are aware of--is selling stocks and they're doing really well. If their stock price is at $50 and no one is buying, will it still go up?  Do stock prices only go up when people are buying?


Say I want to have a command that applies to everyone on a server except 2 people. Is there a target selector for that?

I've asked this question towards a developer for Minecraft and I thought I'd ask here too since they might not see the question.  How do I select all but two types of entities in Minecraft?  For example, something like: /say @e[type=![Player,Item],r=50] (to clarify I also tried type=Item,Player, type=Item,type=Player etc)  Should select any entity that does not match players or items, so things like Creepers, Ghasts, XPOrbs, PrimedTNT and so on within a radius of 50 blocks and print them to the chat.  Instead it ignores one of the flags, and works with the other instead.


I am trying to share a report chart in a Customer Account Portal; I have checked multiple items but still no luck when viewing the portal as a Guest User:  I have created a Sharing Rule for the object; granting access to the Site Guest User The report is in a Folder shared with the Public Group: "All Customer Portal Users" I enabled "report options for external users" and the "Partner Super User Access" in the Community Settings I've checked in the Guest Profile>Object Settings>Reports but the only option is "Tab Settings", which is on I also noticed, in the "Sharing Settings>Other Settings", the tick box "Secure guest user record access" is enable and I am not able to deselect it, as it's greyed out. I tried to do a workaround creating a Public Group and including the Guest User into the group, but help Article "Guest User Security Policies and Timelines" clearly says Guest Users cannot be part of Public Groups  Am I missing something? The report is still not visible when switching to a Guest User View but it is visible as a Authenticated User View.

Is it possible, as a non logged user, to see report and dashboard in community when it is open to non logged user? And  for analytic dashboards?


I am a new Ubuntu user. Last night I installed Ubuntu 16.04 to dual boot with Windows 10, but now I can't see Windows 10 in the boot menu - although all of the Windows system files and folders are shown in my laptop.   When I boot, there are only three options:   1. Ubuntu   2. Advanced options for Ubuntu   3. System setup    I tried to install Ubuntu again, but after some stage it says:   Erase all and install Ubuntu 16.04 LTS   Install Ubuntu 16.04 LTS alongside Ubuntu 16.04 LTS..., etc.   Now what can I do?   How can I get back into Windows 10?    After getting into Windows 10, what should I do next time to properly dual boot Ubuntu with Windows 10?

I installed Windows on my computer, followed by an installation of Ubuntu. However, now I'm unable to boot into my Windows install.  What can I do to fix this?


I need to get an hr element in between 2 buttons so that it is vertically centered in between them.  This is my codepen, I've been struggling to find a decent solution. Could anybody please take a quick look?  Right now they are below and above.    <button class="left">Test</button> <hr /> <button class="right">Test</button>

I know there is a hr (horizontal rule) in html, but I don't believe there is a vr (vertical rule).  Am I wrong and if not, why isn't there a vertical rule?


Prove the direct product of rings never form a ring.  And be the direct product $A\times B$: $+$: $(x,y) + (z,t) = (x+z,y+t)$ $*$: $(x,y) * (z,t) = (x*z, y*t)$ $x,z\in A$. If we define $A$ and $B$ both commutative rings, with units and inverses itself (So $A$ and $B$ are fields). Now, $A\times B$ is commutative, it is a ring, there is a unity element. And since $A$ and $B$ have inverses, the inverse of each element in $A\times B$ $(x,y)$ is just $(x^{-1},y^{-1})$, being $x^{-1}$ the inverse of $x$ in $A$, and so to $y$ in $B$. So $$(x,y) * (x^{-1},y^{-1}) = (e_{a},e_{b})$$ Defining the unity of $A\times B$ as $(e_{a},e_{b})$ I can't see the error in my writings.

This question is from the book "Abstract Algebra" by Dummit and Foote.(Exercise: 7.6.4):  Problem: Prove that if $R$ and $S$ are any two non-zero rings then $R \times S$ is never a field.  My problem with this question is that it doesn't feel right. If $R$ and $S$ are rings then $R \times S$ is also so. (The addition and multiplication on $R \times S$ is defined as $(r_1,s_1)+(r_2,s_2)=(r_1+r_2,s_1+s_2)$ and $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$ respectively where $(r_1,s_1),(r_2,s_2) \in R \times S$.) So proving that $R \times S$ is a field reduces to proving commutativity of multiplication, existence of multiplicative indentity and existence of multiplicative inverse for each element of it. Now, if the two rings are fields themselves then multiplication is commutative, the identy is $(1_R,1_S)$ and inverse of an element $(r,s)$ is $(r^{-1},s^{-1})$. So $R \times S$ is also a field. Am I right? Please help.  Update: The question comes from a section about the Chinese Remainder Theorem. Is it possible to prove this fact using Chinese Remainder Theorem?


I have the octopus kings crown and the monkey wizards staff and pink enchanted gloves with 423 health.

I already have the pink and red leather gloves, the armor, the crown, and the monkey's staff (and they're all enchanted). Regardless of how many health potions I use, I always die. Help!


I'm using Android 2.3 and uninstalled dropbox. Then I wanted it back, but dropbox 3.01, the only official version available, is meant to work only from Android 4 on. Where can I find an old, trustable version of the dropbox 2.x that works for my phone?

Is is possible to download older versions of Android applications?   Is there some kind of online archive for all versions of Android apps? Like how you can find older versions of some Windows app.


I was recently playing Terraria on my laptop when I saw that this was not valid housing. Why? I might be stupid and there could just be a simple answer but do any of you know?

I read on the wiki that people will move in under the following conditions:   You do something to please them An empty house is available with a light source, table, and chair (also with walls).   I have a house which is about 7x8, then a roof space above that; about half the area. I could put furniture up there too, but will anyone move in?


Suppose $a_1 = 2$, and   $$ a_{n+1} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right) $$   if $n \ge 1$.   Is the sequence convergent or divergent?   If it's convergent, find   $$\lim_{n\rightarrow\infty}{a_n}$$   Usually, what's the approach we use for recursive sequence like this? A friend of mine told me that any recursive sequence is divergent, but after I checked on calculator, it seems that the series is approaching one particular value.

Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?


Lemma#3: If a is relatively prime to n, then multiplication by a permutes the elements  1,2,3,....,n-1. I didn't understand what permutes means here. If I consider $\mathbb{Z_n}$ then any element a relatively prime to n has the same order as 1 and also it generates $\mathbb{Z_n}$. Thats the maximum I know. Please I need help to understand this.Thank you.

The problem statement is: for $a,b$ coprime, prove that there exists positive integers $x,y$ such that $ax-by=1$.  The question is form Arthur Egnels' problem solving text, and from the chapter on the pigeonhole principle.    My question: His proof begins by generating a list $a,\ldots,a(b-1)$, where he then points out that each element in this list(which proceeds sequentially) doesn't have remainder $0\pmod{b}$.  He then shows that we arrive at a contradiction if we assume that we also don't get remainder $1$ in this list $\bmod b$.  To show the contradiction in this last statement, he first states that we would have positive integers $p,q $ where $ 0<p<q<b$ so that $pa\equiv qa \pmod{b}$.  He then goes on to point out that since we $a$ and $b$ are coprime, we then have that $b| q-p$.  I understand the last part here, but don't see why we get $pa\equiv qa \pmod{b}$.  Thanks in advance


So I have string like this:  "UFFKTEWKW"   And I need to convert it to a list or tuple like this:  ("UFF", "KTE", "WKW")   So every 3 letters from the string goes to separate element of list or tuple.  I can't use split() here because string doesn't have any delimiters. I don't wanna make some dummy for cycle for it. And I think there should be simple solution for it.

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


I need to center align a dynamic text using a pre-defined point (top and left). If I use the following code, the dynamic text will left align the point.  CSS   .outer {         display: block;         position: relative;         width: 100%;         max-width: 640px;         height: 480px;         background-color: #EEE;         background-size: contain;         background-repeat: no-repeat;         background-position: center center; } .inner {         display: block;         position: absolute;         min-width: 100px;         min-height: 50px;         padding: 20px; } .inner--3 {         left: 45%;         top: 200px;         text-align: center; }   HTML

             Peter Wyatt

   I want to achieve Figure 1 !    I tried everything and still it gives me Figure 2 as the output! Is there anyway I can achieve this using CSS?

Best way to center a <div> element on a page both vertically and horizontally?  I know that margin-left: auto; margin-right: auto; will center on the horizontal, but what is the best way to do it vertically, too?


A 400mm f4.0 lens is supposed to have a maximum aperture diameter of 10cm. Looking at such a lens they don't seem big enough to fit that kind of aperture including all the mechanics for control inside them. Do modern lenses have a way of focusing light into a smaller aperture to behave like a bigger one or are they really always the size that the formula (fstop = focal length / aperture diameter) suggests?

I understand that the entrance pupil is the image of the aperture as seen from the front of the lens. However, is the size of the entrance pupil the same as the size of the effective aperture? In other words, if I measure the size of the entrance pupil by looking at it, will that be equivalent to the focal length of the lens divided by the f-number (with the focus set to infinity)? I know that sometimes this is hard to estimate because of the shape of the aperture blades, but I was wondering whether pupil and aperture can be quite different (perhaps at very low magnification).


A sphere that has the radius of 1 has a right angle square based pyramid inside it with the maximum volume possible. What is the ratio between the edge of the pyramids base compared to the pyramids height? How do I find out the edge and the height, explanations please. Thank you.

What is the maximum volume of a pyramid inscribed with a sphere that has the radius 1? Explanations also please.


My current state is        and I would like to change this appearance of cites with a single number to three characters juxtaposed with two numbers, each of which denotes authors and years of a paper, as follows:        I wonder how to do this, and any comment would be appreciated!

I am in need of a particular citation style, it is different from the ones offered by \cite, \citep or \citet.   This is how i would like the marks to look like    and this is how the bibliography should be.   I am not referring to the particular \bibliographystyle but to the fact that there are personalized names instead of numbers. I am not sure of it, but I believe that these marks were handcrafted.  At present I am using BibTex, but if needed i am ready to move to other tools.  This is an excerpt of my .bib   @article{maietti2009minimalist,   title={A minimalist two-level foundation for constructive mathematics},   author={Maietti, Maria Emilia},   journal={Annals of Pure and Applied Logic},   volume={160},   number={3},   pages={319--354},   year={2009},   publisher={Elsevier} } @article{coquand1998realizability,   title={A realizability interpretation of Martin-L{\"o}f’s type theory},   author={Coquand, Catarina},   journal={Twenty-Five Years of Constructive Type Theory},   year={1998} }


There is a game going around my school that plays with the following rules:   The game is played by a single player -The game is played using a deck of italian gaming card ( $40$ cards, $4$ suit, every suit has cards going from the value of $1$ to the value of $10$ ) The player has to draw cards continuously. Every time the player draws a card he has to say a value starting from $1$ and increasing the value every draw. After saying $3$ the value becomes $1$ again. If the player draws a card with the same value as the value said before drawing (ex:  the player says "one" and draws an ace) the game stops and the player loses. If the player can't draw anymore cards the game stops and the player wins.     What I want to calculate is the probability of the player winning. Unfortunately it seems I can' t get to answer.  So, what is the probability for a player A to win?

Ok, game is as follow, with spanish cards (you can do it with poker cards using the As as a 1)  You shuffle, put the deck face bottom, and start turning the cards one by one, saying a number each time you turn a card around ---> 1, 2, 3; 1, 2, 3; etc. If when you say 1 a 1 comes out, you lose, same with 2 and 3. If you finish the deck without losing, you win.  I know some basics of probabilities, but is there a way to calculate the probability of winning the game, given a random shuffled deck?


just start learn web dev ..   i created two files     - html file     - js file  the js file i wrote code:  function func1() {     alert("alert message");      //document.writeln("document write line");      //console.log("this is log message");   }   The html file code:  <html> <head>     <script type="text/javascript" src="basicFile.js"/> </head>  <body>     <script>         func1();     </script> </body>     Why the alert does not popup ?  also the doc.writeln does not work.   When i write the js func1 directly on the html - its work fine

What is the reason browsers do not correctly recognize:  <script src="foobar.js" /> <!-- self-closing script element -->   Only this is recognized:  <script src="foobar.js"></script>   Does this break the concept of XHTML support?  Note: This statement is correct at least for all IE (6-8 beta 2).


I have object and there are some elements i don't want them (ex: 'id'), there is any way to exclude this elements from object in type script.

Say I create an object as follows: let myObject = {   "ircEvent": "PRIVMSG",   "method": "newURI",   "regex": "^http://.*" };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   "ircEvent": "PRIVMSG",   "method": "newURI" };


I have a method in java which returns the updated lines of a flat file as a "String". I  am receiving the String as a bunch of lines, what i want to do is to separate the String line by line. How can i do that in java?????

I'm trying to split text in a JTextArea using a regex to split the String by \n However, this does not work and I also tried by \r\n|\r|n and many other combination of regexes. Code:  public void insertUpdate(DocumentEvent e) {     String split[], docStr = null;     Document textAreaDoc = (Document)e.getDocument();      try {         docStr = textAreaDoc.getText(textAreaDoc.getStartPosition().getOffset(), textAreaDoc.getEndPosition().getOffset());     } catch (BadLocationException e1) {         // TODO Auto-generated catch block         e1.printStackTrace();     }      split = docStr.split("\\n"); }


In this code the result is failure sending mail as message box shown,      please guide me where   i did mistake.             i tried it in many ways i didn't get. Thanks in advance   private void button2_Click(object sender, EventArgs e)   {    try     {       MailMessage msg = new MailMessage();      msg.To.Add("someone@gmail.com");      MailAddress address = new MailAddress("someone@gmail.com ");      msg.From = address;      SmtpClient client = new SmtpClient();      client.DeliveryMethod = SmtpDeliveryMethod.Network;      client.EnableSsl = true;      client.Host = "smtp.gmail.com";      client.Port = 25;      NetworkCredential credentials = new NetworkCredential("someone@gmail.com", "password");      client.UseDefaultCredentials = false;      client.Credentials = credentials;        client.Send(msg);      } catch (Exception ex) {       MessageBox.Show(ex.Message);    }   }     <smtp deliveryMethod="Network">    <network host="smtp.gmail.com" port="25" userName="someone@gmail.com" password="password" />  </smtp>

Instead of relying on my host to send an email, I was thinking of sending the email messages using my Gmail account. The emails are personalized emails to the bands I play on my show.   Is it possible to do it?


I would like to test the exception trowed by this piece of code:  private void delete(final File file, final String fileName) {     boolean deleted = file.delete();     if (deleted) {         log.info("File {} was deleted", fileName);     } else {         throw new RuntimeException("The file exists but could not be deleted");     } }   My first idea was create a temporary file that can't be deleted in my unit test. For sure, I would like to control this behavior to delete this file after the test. But I supposed that this is not possible, it's contradictory. Set the file to "read only" not works for this case.   So, any ideas?   I could create a new class and pass the responsibility to delete the file and return the boolean, so I could mock it and test my exception. But I would like to explore another possibilities using Junit and Java before do this.

I have a Java process that opens a file using a FileReader. How can I prevent another (Java) process from opening this file, or at least notify that second process that the file is already opened? Does this automatically make the second process get an exception if the file is open (which solves my problem) or do I have to explicitly open it in the first process with some sort of flag or argument? To clarify: I have a Java app that lists a folder and opens each file in the listing for processing it. It processes each file after the other. The processing of each file consists of reading it and doing some calculations based on the contents and it takes about 2 minutes. I also have another Java app that does the same thing but instead writes on the file. What I want is to be able to run these apps at the same time so the scenario goes like this. ReadApp lists the folder and finds files A, B, C. It opens file A and starts the reading. WriteApp lists the folder and finds files A, B, C. It opens file A, sees that is is open (by an exception or whatever way) and goes to file B. ReadApp finishes file A and continues to B. It sees that it is open and continues to C. It is crucial that WriteApp doesn't write while ReadApp is reading the same file or vice versa. They are different processes.


I want to suggest Stack Exchange network to a friend who is visually impaired. He lost his vision a couple of years ago to Glaucoma and is slowly finding his feet on the web using Screen Reader Software.   I just want to know if SE network sites are accessible using screen readers.   I tried to find the answer using search but I only managed to get results from StackOverflow and UserExperience SE and not any information if SE sites were themselves accessible or not. Nor did "Questions that may already have your answer" suggestion retrieve relevant results.  Please excuse me if this question is indeed a duplicate and it would be really nice if someone could point me to previous discussions on this?   Thanks a lot.

There are some plans to put more information on  which is a Stack Exchange site. I am concerned it may not be as readable to everyone as it could be on a plain wiki.  Is there any information on how to use Stack Exchange with a screen reader?


As I understand it, $i^i$ returns more than one real number. Why does this happen? When I raise $i$ to $i$, I'm raising a specific number to itself. So why do I get multiple values of this, as if it was, for lack of a better word, some sort of a "function"?  EDIT: My question wasn't "why is $i^i$ real". I know it's real. Furthermore, I know how to calculate $i^i$. I was just looking for intuition as to why this number has multiple values.

Possible Duplicate:          My calculator (as well as WolframAlpha) gives me the approximation:  $$0.2078795763507619085469...$$  But I don't understand how exponentiating two purely imaginary constructs yields a real (albeit irrational) number. When I do $i^{i+1}$ it gives me an imaginary number as well as $(i+1)^i$. So why does $i^i$ fall into that precise point where it is real and no longer imaginary? What is happening? I understand that exponentiation is not repeated multiplication, and it wouldn't make sense to multiply $i$ by itself $i$ times (because it would only yield $i$, $-i$, $1$, or $-1$). So what are we doing behind the scenes to get such a number?