Although in $\mathsf{Set}$, canonical projections from a product are surjective and canonical injections to a coproduct are in fact injections, there seems to be nothing forcing this to be the case elsewhere, and indeed Wikipedia indicates without proof that they needn't be epic/monic. Unfortunately, I'm a rank beginner and have not yet knowingly read about a category less rich in monomorphisms and epimorphisms than $\mathsf{Set}$ (though Aluffi has already introduced a couple examples that are richer in them), so I don't know where to try to find a counterexample. Any hints?  To clarify: I am seeking canonical injections to coproducts that are not monomorphisms and/or canonical projections from products that are not epimorphisms.

Are all  $\Pi A_\alpha \stackrel{\pi_i}\longrightarrow A_\alpha$ projection maps epic, given that $\Pi A_\alpha$ be the product of $A_\alpha$s? Of course, assuming that the product exists.


I am working on website index page and my code is working fine on my pc. But on 25% inner div moving down . Here is snapshot:    and on 25% screen:    want it on the center and even i use position: fixed but nothing works. However i change structure but nothing work.    .txg{      position: absolute;         top: 40%;         left: 50%;         -moz-transform: translateX(-50%) translateY(-50%);         -webkit-transform: translateX(-50%) translateY(-50%);         transform: translateX(-50%) translateY(-50%);     }     .homelogo {      text-align: center;      line-height: 1em;      font-family: helvetica, arial, sans-serif;      font-weight: bold;      background: linear-gradient(to right, #7db9e8 50%,#1e5799 50%);      -webkit-background-clip: text;      -webkit-text-fill-color: transparent;      font-size: 50px;     }     .teg{      position: absolute;         top: 50%;         left: 48%;         -moz-transform: translateX(-50%) translateY(-48%);         -webkit-transform: translateX(-50%) translateY(-48%);         transform: translateX(-50%) translateY(-48%);     }     .teg ul{      list-style: none;     }     .teg li{      display: inline-block;      background-color: #07ABD8;      padding:15px;      border-radius: 8px;      font-size: 15px;      text-transform: uppercase;      font-family: helvetica, arial, sans-serif;     }     .teg li a{      text-decoration: none;      color: white;     }
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I tried to prove this question but without a success: Let $K_1 \text{and } K_2$ be definable sets, prove that $K_1∪K_2$ is definable.  What I tried to do is to assume: $K_1=\text{Ass}(X)=\{v\mid v\models X\}$ and $K_2=\text{Ass}(Y)=\{v\mid v\models Y\}$ and to say that $K1\cup K2=\text{Ass}(X\cap Y)$. But I couldn't prove that: $K_1\cup K_2\supseteq\text{Ass}(X\cap Y)$.  I would like to get help with this proof

I tried to prove this question but without a success: Let $K_1$ and $K_2$ be definable sets, prove that $K_1\cup K_2$ is definable.  What I tried to do is to assume: $K_1=Ass(X)=\{ v|v \vDash X \}$ and $K_2=Ass(Y)=\{ v|v \vDash Y \}$ and to say that $K_1\cup K_2=Ass(X \cap Y)$. but I couldn't prove that: $K_1\cup K_2 \supseteq Ass(X \cap Y)$ .   I would like to get help with this proof


$\sum _{k=1}^{n}\mathrm{sin}kz=\frac{\mathrm{sin}\frac{n+1}{{2}}\cdot \mathrm{sin}\frac{nz}{2}}{\mathrm{sin}\frac{z}{2}}$ Proof for $z\neq 0$

I have to calculate the sum of this series $$ \sum_{k=1}^n\sin{k\theta} $$ I tried solving it like this $$ \sum_{k=1}^n\sin{k\theta} = \operatorname{Im}\sum_{k=1}^ne^{(i\theta)k} $$ I recognized it as a geometric sum, but to get the sum I first had to check $$ \left|e^{i\theta}\right| = 1 $$ which means that the geometric sum is divergent. Apparently that is not the right answer, but I can't find the spot where I made the mistake. Could you help me finding it?


So I accidentally run in terminal command export PATH="/usr/bin/php/bin" which override my defaults and none of the commands aren't responding.  MacBook-Pro-Kil:home $ sudo -bash: sudo: command not found   Is it possible somehow restore it? My system is 10.9.5

I was trying to add something to $PATH and it went totally wrong.  I now can't run any commands such as ls.  I've looked at this   and used the following lines:  PATH=/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/X11/bin PATH=$PATH:~/bin   These lines fix the problem temorarily; however, when I restart terminal it seems to forget these changes.  How do I permanently reset my $PATH?  I'm running the most recent version of Mountain Lion.


I apologize for creating a duplicated question. I'm sure this has been asked before but I simply could not apply the other fixes to mine due to my lack of understanding and knowledge.  So I have a class called Patients.cs and within that, all of a patient's data is stored. I have added a field for the birthdate with the following line   public DateTime birthdate { get; set; }   I am now creating a front-end form that will allow users to input information into the database through the webform. The field responsible for inserting the birthdate is as follows   <%--Birthdate--%>         Birthdate
           <asp:TextBox  class="form-control" ID="inputBirthDate" runat="server" placeholder="day/month/year"></asp:TextBox>
   For the back-end of the form page, I've created a new method that submits the data upon submitting the form and it is as follows  public void AddPatient() {      Patient newPatient = new Patient()     {         firstName = inputFirstName.Text,         middleName = inputMiddleName.Text,         lastName = inputLastName.Text,         gender = inputGender.SelectedValue,         birthdate = inputBirthDate.Text,         age = ,   The first few lines of the method / function seem to work fine, but when it comes to the birthdate, I get the following error Cannot implicitly covert type 'string' to 'System.DateTime'. I am unsure as how to handle this and then how to be able to display a patients age based on the birthdate they have put.

Given a DateTime representing a person's birthday, how do I calculate their age in years?


How do I find sums like these?--  $$S=\displaystyle\sum_{k=0}^{39} \dbinom{200}{5k}$$  that is, when there is a summation of binomial coefficients, but with jumps of some terms..?


I try to make an array with days of the month but I get NullPointerException at build.  This is my source code:      Calendar gCal = new GregorianCalendar(2016, Calendar.JANUARY, 1);     int currentDay = gCal.get(Calendar.DAY_OF_WEEK);     lows = (GridView)findViewById(R.id.MonthView);     gCal.add(Calendar.DAY_OF_YEAR, Calendar.SUNDAY - currentDay);     for(int i=0;i<=34;i++){         lunas[i] = String.valueOf(gCal.get(Calendar.DAY_OF_MONTH));         gCal.add(Calendar.DAY_OF_YEAR, 1);     }   and this is my logCat error:  01-22 14:11:40.221: W/dalvikvm(19656): threadid=1: thread exiting with uncaught exception (group=0x41604898) 01-22 14:11:40.221: E/AndroidRuntime(19656): FATAL EXCEPTION: main 01-22 14:11:40.221: E/AndroidRuntime(19656): java.lang.RuntimeException: Unable to start activity ComponentInfo{ro.thich.programdetura/ro.thich.programdetura.MainActivity}: java.lang.NullPointerException 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2221) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2271) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread.access$600(ActivityThread.java:144) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1263) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.os.Handler.dispatchMessage(Handler.java:99) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.os.Looper.loop(Looper.java:137) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread.main(ActivityThread.java:5169) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at java.lang.reflect.Method.invokeNative(Native Method) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at java.lang.reflect.Method.invoke(Method.java:525) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:737) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:553) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at de.robv.android.xposed.XposedBridge.main(XposedBridge.java:132) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at dalvik.system.NativeStart.main(Native Method) 01-22 14:11:40.221: E/AndroidRuntime(19656): Caused by: java.lang.NullPointerException 01-22 14:11:40.221: E/AndroidRuntime(19656):    at ro.thich.programdetura.MainActivity.onCreate(MainActivity.java:48) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.Activity.performCreate(Activity.java:5133) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1087) 01-22 14:11:40.221: E/AndroidRuntime(19656):    at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2185) 01-22 14:11:40.221: E/AndroidRuntime(19656):    ... 12 more   and this is line 48 of my MainActivity.java :  lunas[i] = String.valueOf(gCal.get(Calendar.DAY_OF_MONTH));   I don't understand where the problem is. Can you help me?

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


I have been using this site for a year now. I've asked many questions and also answered many times.  Today one of my questions was put on hold by some experienced users. This is fine, but I was not notified of this in the notification panel. Only when I checked in on hold question did I learn that it was put on hold.  Why was there no notification in my inbox after it was put on hold?

I think it would really help if you get notified that a question of yours was closed.  Especially for the multi-site power user, usually (at least for me) a question is asked and then left alone for while. I rely on the network-wide inbox to tell me when someone has commented or answered, but if I don't receive any messages then I tend to assume that the question wasn't answered or even noticed. It's only when I check the question a day or two later to see what I can add do I find out it was closed.  This would be especially helpful for duplicates, where the answers you need are already there. A message to the effect of, "Closed as duplicate of question" is just like notifying me that a new answer has been added!  Unfortunately, question closures can remain unnoticed by their askers until months have passed. By then, the questions may either be too old to migrate to the correct site, or may , depending on the specific closing reason. We should notify users of these question closures so that they can improve them and learn from their mistakes.  Could closed questions get added to the inbox?


Last night I discovered what looks like an unusual file named "~" (tilde) in the root of my "Macintosh HD" directory on my MacBook.    Finder reports the file is 2.2MB in size, of kind "TextEdit.app Document", and was created on 22nd January 2014 (so about a month ago).  I haven't noticed any operational issues with OS X or it's applications.   I haven't yet touched the file but would obviously like to remove it.  I am very conscious that '~' is a special alias to the user home directory and thus I would appreciate any recommendations in removing the file in a safe manner that isn't likely to nuke my home directory or screw up the operation of applications or OS X. I don't know whether to delete or rename and whether to use Finder or the command line.   It's entirely possible that I created the file by accident since I've recently been drafting and running various shell scripts that use the ~, user home alias.  MacBook Pro 15" (Retina, Mid 2012) Yosemite


How to install Canon LBP2900C on Lubuntu? What I did using terminal is sudo dpkg -i [name of file inside debian] , and another installation terminal code. But it is throwing error saying dependency problems - leaving unconfigured Errors were encountered while processing: cndrvcups-common

I cannot configure my Canon lbp2900b printer in Ubuntu 20.04. I have a dual boot of Windows 10 and Ubuntu 20.04, and it is working fine in Windows 10. When I turn on the printer in Ubuntu it is showing Configuring printer please wait... and after some time it is showing Printer added Canon lbp2900, but when I print a test page it's showing Printing test page and then it is showing Printing completed but there is no response from the printer. I installed the driver for linux_64 from the official Canon website. Results of lpinfo -v : file cups-brf:/ serial serial:/dev/ttyS0?baud=115200 network https network beh direct ccp direct hp network ipps network http network ipp network socket network lpd direct hpfax  and the command systemctl list-units "ippusbxd" | grep service shows nothing.


I was watching a video on propeller gyroscopic effect, please take a look at this 1 minute segment (up to 6:06):    Alternatively, here is a summarizing image from the above video:    Could you please explain mathematically why the person says that, when a force is applied at one point on the gyroscope, its effect gets translated by 90 degrees in the direction in which the gyroscope is turning? Thanks a lot!

I understand how torque mathematically causes a change to the direction of angular momentum, thus precessing the .  However, the direction, either clockwise or counterclockwise, of this precession seems to me a bit arbitrary beyond mathematical definitions. (What if torque were defined by a "left hand rule", for example?).  What's the fundamental physical reason that the precession of a gyroscope would be directed in a certain direction?


I need to write a long paragraph in my table, but the text goes out of the margins.. Here is my code:  \begin{tabular}{l l}         \textbf{P1.1} & aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\\         \textbf{P4.1} &  bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\\  \end{tabular}   How can I correct this?

Why isn't the table adjusting the text to make line breaks when necessary? Can I make them manually? What could be done to fix these margins? Edit: MWE below: \documentclass[a4paper,12pt,spanish]{article} \usepackage{multirow} \usepackage[utf8]{inputenc} \usepackage{geometry} \usepackage[spanish]{babel} \usepackage{graphicx} \begin{document} \begin{center} \begin{tabular}{|r|ccccc|} \hline    & Descripción & Narración & Diálogo & Exposición & Argumentación \\ \hline   Técnico-científico &  &  &  & Habitual & \\   Jurídico-administrativo &  &  &  &  & \\   Humanístico &  &  &  & Estudio & Ensayo\\   Periodístico &  &  &  & Noticia & Art. opinión\\   Literario & Descripción & Novela & Teatro & \multicolumn{2}{c|}{Lengua literaria}\\ \hline \end{tabular} \end{center} \end{document}  Edit 2: Problem remains when changing tabular with tabu. Aditionally, it gives me a non fatal error when compiling: Overfull \hbox (57.79063 too wide) (page 2), lines 112-123. Since this table appears in the page 2 and the code lines match too, it probably wants to tell me something, but I don't know what it means.


Show, if $f:X->\mathbb R$ is continuous at point $a \in X$, then it is bounded in a neighborhood of the point $a$. That is, it exists $\delta>0$ such that $U_{\delta}=X \cap (a-\delta,a+\delta)$ or $f(U_{\delta})$ is bounded. Help me, please

I have this question :  Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.  I want to know if my proof is valid :  If continuous in $x_0$ then :  $$\lim_{x \to x_0} f(x)= f(x_0)$$  Therefore :  All $\epsilon>0$ exists $\delta>0$ so any $x$ that implies $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$.  Therefore for $\epsilon=1$ there is $\delta>0$ so any $x$ that implies $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<1$.  $|x-x_0|<\delta \rightarrow x_0-\delta<x<x_0+\delta$.  Lets choose $x_1,x_2$ that implies $x_0-\delta<x_1<x_2<x_0+\delta$  Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there.


In "The Two Towers", in the chapter "The White Rider", Gandalf describes his battle with the Balrog. He mentions being underground, deeper than even where the dwarves went.     'We fought far under the living earth, where time is not counted. Ever he clutched me, and ever I hewed him, till at last he fled into dark tunnels. They were not made by Durin's folk, Gimli son of Glóin. Far, far below the deepest delvings of the Dwarves, the world is gnawed at by nameless things. Even Sauron knows them not. They are older than he. Now I have walked there, but I will bring no report to darken the light of day. In that despair my enemy was my only hope, and I pursued him, clutching at his heel. Thus he brought me at last to the secret ways of Khazad-dûm: too well he knew them all. Ever up now we went, until we came to the Endless Stair.'   Clearly Gandalf does not want to describe this place any further (he describes this place as a "despair", which implies it's not something he'd readily mention again), so I doubt he brings it up again later in the book, or in Return of the King (although I haven't read past "The White Rider" yet, and I certainly wouldn't remember something this obscure from the last time I read these books years ago).  Is this place mentioned anywhere else in any of Tolkien's other works, such as the Silmarillion? Any and all further information on this place would be appreciated, as well as the "nameless things" that Gandalf mentions.  The very similar question  identifies that it is some kind of lake, but doesn't add any further information on it besides what's in the text I've quoted. Specifically, I'm looking for any further information outside of this chapter.  I've just found  about the creatures, which I wanted to update this question with, since they are related.

I was wondering; what is the name of the place Gandalf falls to when the Balrog drags him down, after the fight in Moria?   What information from the Legendarium is there available about this place?


For a submitted paper, what does the status "Evaluating Recommendation" mean and how long does it last?

What steps does a manuscript typically go through from submission to publication (or rejection) in a typical journal? How are these steps referred to, in particular by editorial systems, and how long do they each typically take?  Note that this question is about the typical situation and hence not about:   Journals with an atypical workflow, e.g. those that allow for an instantaneous reviewer–author interaction. Exceptional steps or rare occurrences such as withdrawal or .   This is a canonical question on this topic as per . Due to its nature, it is rather broad and not exemplary for a regular question on this site. Please feel free to improve this question.


There are no rationals $r, s$ such that $\sqrt{3} = r + s\sqrt{2}$. Our professor gave us lengthy and a bit of a complex proof. I tried to construct my own, more simple proof. Could you please verificate it ?  Suppose there are $r,s$ such that$\sqrt{3} = r + s\sqrt{2}$. Then $\sqrt{3}-r= s\sqrt{2} \to \frac{\sqrt{3}}{\sqrt{2}} - \frac{r}{\sqrt{2}} = s$. $\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} = \frac{1}{2} * \sqrt{6}$  Clearly this number is irrational $\frac{r}{\sqrt{2}} = \frac{r\sqrt{2}}{2} = \frac{r}{2} * \sqrt{2}$ Again this is irrational. The only way to substract irrationals and receive an integer is if they are equal. Clearly, this is impossible in our case. Hence, this is contradiction.  Actually could you please explain why difference of irrationals is equal to irrational or $0$. UPD: I now realize my proof was wrong. I constructed a new one. $\sqrt{3} - r = s\sqrt{2}$. Squaring this equation we get $3-2r\sqrt{3} + r^2 = 2s^2 \to 3-2s^2+r^2=2r\sqrt{3}$ Clearly, $2r\in Q $ and so RHS is irrational. Suppose $r=0$. Then  $3-2s^2=0 \to s=\sqrt{\frac{3}{2}}$ Contradiction, since s is rational. Suppose $r\neq 0$ Then there is contradiction since LHS is rational and RHS is irrational. Therefore there does not exist such rational r, s.

Possible Duplicate:          Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?


I am using selenium webdriver with python. I am not able to find a way in python by which I can switch the webdriver to a new browser.  My web application have  multiple links and buttons  Clicking on any of  them leads to a new browser window being opened.    for clicking on buttons and link I have used below code   driver.find_element_by_css_selector("input[type=\"submit\"]").click() driver.find_element_by_id("btn1").click() driver.find_element_by_css_selector("#linksPlace > #OIT> a").click()   For selecting subsequent browser windows, I have used below code  for handle in driver.window_handles:     print handle     driver.switch_to_window(handle)     driver.get_window_position(handle)     driver.save_screenshot('E:\Selenium scripts\demo.png')   However  I am not able to navigate to that new browser window. I have used below command as well, but things didn't work       driver.switch_to_window(driver.window_handles[-1])   Is there any way by which I can navigate to new browser window? I have seen that there are functions which exist in java and c# to do the same, but struggled so far in python.

I have clicked an image in a slide show and it generates a new window. Once the window opens I need to bring focus on new window and do some stuff. I have attached my code below, but it is not successful.  @Test public void testCreateAccount() throws Exception{  //OPEN manorama online  url driver.navigate().to("http://www.manoramaonline.com/cgi-bin/MMOnline.dll/portal/ep/home.do?tabId=0&BV_ID=@@@");  // to click a link inside a slide show  while (!driver         .findElement(                 By.xpath("/html/body/div[2]/div[2]/div[4]/div/div/div/div/center/div/div[2]/span/div/a/img"))         .isDisplayed()) { } driver.findElement(         By.xpath("/html/body/div[2]/div[2]/div[4]/div/div/div/div/center/div/div[2]/span/div/a/img"))         .click();    String newwindowTitle = driver.getTitle();  switchWindow(newwindowTitle);  }  protected static boolean switchWindow(String title) throws IOException {      String currentWindow = driver.getWindowHandle();     Set<String> availableWindows = driver.getWindowHandles();     if (!availableWindows.isEmpty()) {     for (String windowId : availableWindows) {     if (driver.switchTo().window(windowId).getTitle().equals(title)) {     return true;     } else {         driver.switchTo().window(currentWindow);     }}}      return false;        }


So I downloaded Ubuntu as a dual boot but I now change my mind and would like  to delete the dual boot and install it as my primary OS how would I go about doing that?

I want to keep Ubuntu but remove Windows.


Does the special unitary Lie group $SU(5)$ contains $SU(3) \times SU(2) \times U(1)$ as a subgroup?  Can one show which of the following embeddings are possible rigorously: $$SU(5) \supset SU(3) \times SU(2) \times U(1)?$$ $$SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_2}?$$ $$SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3}?$$ $$SU(5) \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3 \times \mathbb{Z}_2}?$$  Here $\mathbb{Z}_n$ means the cyclic group of order $n$.  it looks to me that the first one is impossible, can we prove it is a no go?  (p.s. $SO(10)$ and $SU(5)$ are different groups - how can the two questions be duplicate to each other, without knowing the way of embedding?)

Let $G$ be $SO(10)$ or $Spin(10)$.  Does either of them $G=SO(10)$ or $G=Spin(10)$ contain  $SU(3) \times SU(2) \times U(1)$ as a subgroup?  Can one show which of the following embeddings are possible rigorously: $$G\supset SU(3) \times SU(2) \times U(1)?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_2}?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3}?$$ $$G \supset \frac{SU(3) \times SU(2) \times U(1)}{\mathbb{Z}_3 \times \mathbb{Z}_2}?$$  Here $\mathbb{Z}_n$ means the cyclic group of order $n$.  it looks to me that the first one is impossible, can we prove it is a no go? for $G=SO(10)$ or $G=Spin(10)$?


I am writing a program that is supposed to take 10 test scores from students and their names and put that information into a list inside of a list. I saw a problem with the fact that I was repetitively .append -ing things to my 'info' list and was getting duplicate data. However, when I try to fix it the program keeps returning either empty lists or lists with only the second set of name and test scores. I do not know why this is happening, any help at all would be greatly appreciated.  I have tried:   del info[:] info.clear() for i in range len(info):    info.pop()   w  testinfo = [] score = 0 testnum = 0 name = '' info = [] info2 = [] name = input('Enter a student name') while name != '0':     info.append(name)     for i in range(0, 10):         testnum = testnum+1         print('Enter a score for test', testnum)         score = int(input())         info.append(score)     testnum = testnum-10     testinfo.append(info)     name = input('Enter a student name')     del info[:] print(testinfo)   Expected result:[[student1name,1testscore1,1testscore2,etc.],[student2name,2testscore1,2testscore2,etc.]]  Actual result: [[], []] or [[student2name,2testscore1,2testscore2,etc.], [student2name,2testscore1,2testscore2,etc.]]

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


I'm building an automation script in JavaScript to pass variables into an input form into a child window. Every time I go to return the element ID from the child window, the JS console tells me it     cannot find the property 'onload' of an undefined or null reference.   My first thought was I wasn't calling the window properly. But I've checked, both URLs are from the same domain, so it wouldn't violate the same origin principle. I'm just completely stuck and hoping someone could help.  I can't give any more detail than what I have in the code because I am working on a proprietary system.   var WinName = "";  function getWindowName() {   var d = new Date();   d = d.toUTCString();   d = d.split(' ');    for (var i = 0; i < d.length; i++) {      d[i]= d[i].replace(/\W/,'');     d[i]= d[i].replace(":",'');    }   WinName = d.join('');   return WinName; }  var b = getWindowName(); var a = window.open([URL], b).focus(); a.onload = document.getElementById([ElementID]);'

I've opened a new window with window.open() and I want to use the reference from the window.open() call to then write content to the new window.  I've tried copying HTML from the old window to the new window by using myWindow.document.body.innerHTML = oldWindowDiv.innerHTML; but that's doesn't work.  Any ideas?


I'm writing a document and want to include an appendix of pictures at the end. Each picture has a caption associated with it, and a label. However, each time I try to compile the code, LaTeX automatically creates a page break between the section heading and the first figure. I know that use of \includegraphics can create a page break, but even if I use the figure environment without a picture as below, LaTeX starts a new page between the section heading and caption. Any ideas how I can stop this from happening?  \section{Appendix A} \begin{figure}     \centering     \caption*{BlaBlaBla}     \label{pic1} \end{figure}

How to influence the position of float environments like figure and table in LaTeX?   This is a general question and should collect useful answers for all users. I hope we can use this as a reference


Basically I can't find Skype or Spotify in the right click menu, where I can find all other applications I've installed.  I am looking for some way to be able to start them (even if it is through a search).   Note: I know I can start them from the console, but I am wondering if there is another way

After sudo apt-get install awesome awesome-extra and logging into it (with no previous configuration that I know of), there's no menu entry for Firefox. All my other stuff is there - Mutt, Meld, jEdit, BOINC, etc., only Firefox is missing. I have no ~/.config/awesome/ directory, so there should be no override in place.


Subj. The expiration time is about 1 hour or so. This is very inconvenient:   If I leave a query open, after some time it becomes inoperative - impossible to execute and the editor glitches I have to login every time - which I don't have to at other SE sites

I'm pretty sure that since a month or so,  seems much more eager to get rid of my session and log me out in that process. I don't mind logging in a couple of times per day, if that helps to reduce resource consumption, but the user experience isn't all that great when it happens. Here is why: I'm most often in the middle of writing a query for an MSE question, for example . So I have the edit window open and am creating a new revision of my query. When I click "run," I do notice the button gets grayed-out, as it normally does, but I never get any result back. There is also no error message. To recover from this I have to copy the SQL text I have typed, open a new tab and load SEDE, log in, switch back to the stalled tab, refresh, paste the SQL text, run query. Repeat the process after 10 minutes. If the intended behavior is to log users out of their session on such short notice (I recall sessions lived for 24 hours), can there then be a slightly better handling in the browser of that state, so it is clear you're logged out once you click the run button? I prefer, however, to have the longer allowed sessions back. I'm experiencing this behavior mostly in Chrome on Win 10.


Show that $6^{(k+1)+1} + 7^{2(k+1)-1}$ is a multiple of $43$ knowing that $6^{k+1} + 7^{2k-1}$ is a multiple of 43.  I have written it as $6^{k+1} 6^1 + 7^{2k-1} 7^2$ but idk what to do after.

Use mathematical induction to prove that 6(n+2) + 7(2n+1) is divisible by 43 for n >= 1.  So start with n = 1:  6(1+2) + 7(2(1)+1) = 63 + 73 = 559 -> 559/43 = 13. So n=1 is divisible  Let P(k): 6(k+2)+7(2k+1) , where k>=1  Show that P(k+1): 6((k+1)+2) + 7(2(k+1)+1) is true  = 6(k+1+2) + 7(2k+2+1)  = 6(k+3) + 7(2k+3)  I'm unsure where to go from here, I've tried several directions after this but have got nowhere. I don't know how to get the 43 or 559 out the front.  Any help would be great


Can we get that info the same way we do for FTP since HTTP is a plain text protocol?

Recently I was thinking about how websites like gmail and amazon use HTTPS during the login process when accessing your account. This makes sense, obviously, since you're typing in your account username and password and you would want that to be secure. However, on Facebook, among countless other websites, their logins are done with simple HTTP. Doesn't that mean that my login name and password are completely unencrypted? Which, even worse, means that all those people who login to their facebooks (or similar sites) at a wifi hotspot in public are susceptible to anyone getting their credentials using a simple packet sniffer (or something similar)? Is it really that easy? Or am I misunderstanding internet security?  I'm a software engineer working on some web related stuff, and although at the current time I'm not too involved with the security aspect of our software, I knew I should probably know the answer to this question, since it's extremely fundamental to website security.  Thanks!


I got Firefox Quantum and also I have uGet but I can't download because I don't have flashGot(I can't add to Firefox because it says that is not compatible with Firefox Quantum). Please help me.  Thank you

I want to use Internet Download Manager for downloading on Firefox. According to this  it can be done with the extension called Flashgot. But Flashgot does not work on new Firefox Quantum. However, I found  that can be used to interrupt browser's downloading jobs by redirecting these jobs to an external download manager like Internet Download Manager (IDM). But I don't know how to integrate IDM with this new extension. I wrote IDM's location (/.wine/drive_c/Program Files (x86)/Internet Download Manager/IDMan.exe) in the box that says "Path to the executable" but when I try to download something, an error is appearing.     The preference page of the extension        The error that is appearing while trying to download files     How can I make it work? Thanks beforehand!


This sequence appears Cauchy. But how to rigorously approach this problem.

Suppose that for the sequence $\{x_n\}$ there exists $0 < C < 1$ such that  $$\left|x_{n+1} - x_n\right| < C\cdot \left|x_n - x_{n-1}\right|.$$  Prove that ${x_n}$ is Cauchy.


is it currently possible to enter sweden for a non EU / EEA citizen with a UK work permit visa? this is for a short holiday. edit: the type of visa is a work permit / residence visa, not a temporary visa as referenced in the linked question

I'm a UK passport holder; my wife isn't. We don't live in UK. If she gets a UK visit visa, does this entitle her to travel to the Schengen area or do I need to get her a separate Schengen visa?


I want to be able to download specific file from a magnet/.torrent (not all the files) using a cli client for my headless server. I know that GUI clients usually have this option but was looking for a command line tool. Though  answers on the options available, it doesn't cater to this specific question

Yesterday one of my friends asked me if there is some torrent client that runs on the command-line.  I had no answer for him, but I think that you have some.


I start saying that I know this question has been already proposed a thousand times on this forum BUT I couldn't find a direct answer so I try asking on my own.  What's the intuitive meaning of this definition of measurable set?  $\cdot$ Let $X$ be a set and $\lambda^*$ an outer measure. Then $X$ is said to be measurable if for each $B\in X \lambda^*(B)=\lambda^*(B\cap X) + \lambda^*(B\cap X^C)$  Or equivalently if  $\lambda_*(X)=\lambda^*(X)$.  I feel OK with the second definition because it looks kind of similar to the intuition in Riemann Jordan "measure", but why are the two equivalent?  Thanks.

This week I saw the definition of a measurable set for an outer measure.     Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.   This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition.   The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.     Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"?


I have the Steam version of ME2 and am able to log into the Cerberus Network without issue.  However, whenever I start up the game the "Cerberus laptop" spins for a few seconds and then says:     Communication with the Mass Effect 2 server was interrupted or has timed out   After this I have to log in again, which works perfectly so I don't think it's actually a communication issue.  It's getting pretty annoying to have to do this every time I play — is there any way to fix it so I'll just stay signed in?

Whenever I start up Mass Effect 2 and it tries to auto-login to my EA Online Account, I see this screen.    Then, I have to manually log in, and I see this screen.    Empty login screen. Every time. So I have to input my email address and password again. Every time. Once I do this manually, it works. However, the next time I close and reopen the game, I have to go through this entire process again. The auto-login isn't working for me.  Is there any way to get Mass Effect 2 to remember my new login information?


Most of us are aware of the famous "Basel Problem":  $$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$  I remember reading an elegant proof for this using complex numbers to help find the value of the sum.  I tried finding it again to no avail.  Does anyone know a complex number proof for the solution of the Basel Problem?

As I have heard people did not trust Euler when he first discovered the formula (solution of the ) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs.   I believe many of you know some nice proofs of this, can you please share it with us?


Input:  df=pd.DataFrame({'text':['value 123* 333','122* 666','722 888*']}) print(df)              text 0  value 123* 333 1        122* 666 2        722 888*   I need to extract from df['text'] only numeric values, but withou *label my code:  df.text.str.extract(r'([0-9]+|[0-9]+\.[0-9]+)')   But with this code, values with the * char on the right are returned.  Expected output:  text 333 666 722

I am tired of always trying to guess, if I should escape special characters like '()[]{}|' etc. when using many implementations of regexps.  It is different with, for example, Python, sed, grep, awk, Perl, rename, Apache, find and so on. Is there any rule set which tells when I should, and when I should not, escape special characters? Does it depend on the regexp type, like PCRE, POSIX or extended regexps?


$a^{\dagger}$ and $a$ are ladder operators:  $a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle  \,\,\text{and}\,\,  a|n\rangle= \sqrt{n}|n-1\rangle $  Is the state $|n\rangle$ an eigenfunction of $a\dagger$ ?

How can I find the eigenstates of creation and annihilation operator in QM?   My attempt:  Such eigenstate will obey: $$ a^{\dagger} |\psi \rangle = \alpha |\psi \rangle. $$  We can expand $|\psi \rangle$ in terms of the quantum SHM eigenstates: $|\psi \rangle = \sum_{n=0}^{\infty} c_n |n\rangle $.  Knowing the action of the creation operation of quantum SHM eigenmodes ($a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle)$:  $$ a^{\dagger} |\psi \rangle = a^{\dagger} \sum_{n=0}^{\infty} c_n |n\rangle = \sum_{n=0}^{\infty} c_n \sqrt{n+1}|n+1\rangle$$  from which the state $|0\rangle$ is now missing, so it will never be equal to the RHS of the first expression, $\alpha |\psi\rangle = \sum_{n=0}^{\infty} \alpha c_n |n\rangle $.


Here are two javascript snippets. i can't understand thedifferent between the two. it seems both allow newing up instances.  var Test1 = (function () {             function Test1(_x, _y) {                 this.x = _x;                 this.y = _y; console.log("test1");             }             return Test1;         }()); var Test2 = function () {             function Test2(_x, _y) {                 this.x = _x;                 this.y = _y; console.log("test2");             }             return Test2;         }();  new Test1(1,2);  new Test2(2,4);   Edit: it looks the same to me BUT, i was poking around with TypeScript and it generated the classes like Type1. there must be a reason they chose to go with that. after all the shortest code is better?  Thanks

I'm new to both JavaScript and . In YUI library examples, you can find many uses of this construct:  (function() {     var Dom = YAHOO.util.Dom,     Event = YAHOO.util.Event,     layout = null,         ... })();   I think the last couple of parentheses are to execute the function just after the declaration.  ... But what about the previous set of parentheses surrounding the function declaration?  I think it is a matter of scope; that's to hide inside variables to outside functions and possibly global objects. Is it? More generally, what are the mechanics of those parentheses?


My laptop wifi is broken, so I bought a usb wifi card TL-WN823N(rtl8192cu module). (I know linux but I'm really new to linux drivers things.)  Since I'm using "14.04.2 LTS, Trusty Tahr", I need to install the driver myself. I downloaded the driver from , compiled and installed it. During that I met some compilation error and solved according to . And reboot.  Now the wlan1 seems DISABLED .  cipher@Joy:~$ sudo lshw -c network   ...   *-network DISABLED       description: Wireless interface       product: Wireless 7260       vendor: Intel Corporation       physical id: 0       bus info: pci@0000:03:00.0       logical name: wlan0       version: 6b       serial: 28:b2:bd:a3:76:7c       width: 64 bits       clock: 33MHz       capabilities: pm msi pciexpress bus_master cap_list ethernet physical wireless       configuration: broadcast=yes driver=iwlwifi driverversion=3.16.0-30-generic firmware=25.228.9.0 latency=0 link=no multicast=yes wireless=IEEE 802.11bgn       resources: irq:65 memory:f1400000-f1401fff   *-network DISABLED                  # This is the usb wifi card.    description: Wireless interface    physical id: 3    bus info: usb@2:1    logical name: wlan1    serial: 24:24:20:4e:49:43    capabilities: ethernet physical wireless    configuration: broadcast=yes driver=rtl8192cu multicast=yes wireless=unassociated   When I tried to get it up manually:  cipher@Joy:~$ sudo ifconfig wlan1 up SIOCSIFFLAGS: Operation not permitted   What are the proper reasons to get this? How can I get it up?  Here are some info about the current state:  Output of lsmod:  cipher@Joy:~$ lsmod | grep 8192 8192cu                527333  0   module depends:  cipher@Joy:~$ grep "8192cu" /lib/modules/3.16.0-30-generic/modules.dep kernel/drivers/net/wireless/rtlwifi/rtl8192cu/rtl8192cu.ko: kernel/drivers/net/wireless/rtlwifi/rtl_usb.ko kernel/drivers/net/wireless/rtlwifi/rtlwifi.ko kernel/drivers/net/wireless/rtlwifi/rtl8192c/rtl8192c-common.ko kernel/net/mac80211/mac80211.ko kernel/net/wireless/cfg80211.ko updates/dkms/8192cu.ko:   Some output of lspci:  cipher@Joy:~$ lspci 02:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. RTS5227 PCI Express Card Reader (rev 01)   Output of rfkill:  cipher@Joy:~$ sudo rfkill list all 0: phy0: Wireless LAN      Soft blocked: yes      Hard blocked: no   I think it's the device of the system wifi card. The usb card seems missing from rfkill.  Some output of iwconfig:  cipher@Joy:~$ iwconfig wlan1     unassociated  Nickname:"<WIFI@REALTEK>"           Mode:Auto  Frequency=2.412 GHz  Access Point: Not-Associated              Sensitivity:0/0             Retry:off   RTS thr:off   Fragment thr:off           Power Management:off           Link Quality:0  Signal level:0  Noise level:0           Rx invalid nwid:0  Rx invalid crypt:0  Rx invalid frag:0           Tx excessive retries:0  Invalid misc:0   Missed beacon:0   system modules:  cipher@Joy:~$ cat /etc/modules lp rtc # rtl8192cu 8192cu   blacklist about rtl8192cu:  cipher@Joy:/etc/modprobe.d$ grep 8192 *  blacklist-8192cu.conf:blacklist rtl8192cu blacklist-8192cu.conf:blacklist rtl8192c_common blacklist-native-rtl8192.conf:## This file ships with the rtl8192-fixes DKMS module. blacklist-native-rtl8192.conf:blacklist rtl8192cu blacklist-native-rtl8192.conf:blacklist rtl8192c_common   I've tried most posts about installing wireless drivers from google, and they brought me to the current situation.  I think I'm almost get the usb wifi card work, but how to get this continue: get the wlan1 up ?  Any experts can give any hints? Thanks in advance.

I have purchased a TP-LINK TL-WN823N (EU) V2.0 device, which is a USB device that should work as a Network Interface Card, allowing me to connect my computer to wireless networks. Unfortunately, I am unable to find any wireless networks using this device using either wicd or network-manager (no networks appear in the list of networks supplied by either application).  A CD came with the device claiming "Linux support". The CD contained a URL to a corrupted archive file for downloading installation resources. I try to mask my immense frustration.  Here is the additional dmesg output I receive when I connect the device to my computer (similar content is present when I boot the computer with the device connected):  [17096.009912] usb 2-1.2: new high-speed USB device number 4 using ehci-pci [17096.103501] usb 2-1.2: New USB device found, idVendor=2357, idProduct=0109 [17096.103513] usb 2-1.2: New USB device strings: Mfr=1, Product=2, SerialNumber=3 [17096.103527] usb 2-1.2: Product: 802.11n NIC [17096.103529] usb 2-1.2: Manufacturer: Realtek [17096.103530] usb 2-1.2: SerialNumber: 00e04c000001   This is my lsusb, where the offending device corresponds to Bus 002 Device 004:  Bus 002 Device 003: ID 0a5c:5800 Broadcom Corp. BCM5880 Secure Applications Processor Bus 002 Device 004: ID 2357:0109   Bus 002 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 001 Device 004: ID 0c45:6433 Microdia  Bus 001 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub   As lsusb -v:   It seems the device is working to me, but no new interface is found in 'ifconfig':  docker0   Link encap:Ethernet  HWaddr 02:42:6e:bf:92:16             inet addr:172.17.0.1  Bcast:0.0.0.0  Mask:255.255.0.0           UP BROADCAST MULTICAST  MTU:1500  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0            RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)  enp0s31f6 Link encap:Ethernet  HWaddr 30:5a:3a:e1:96:b1             UP BROADCAST MULTICAST  MTU:1500  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)           Interrupt:16 Memory:df200000-df220000   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:412 errors:0 dropped:0 overruns:0 frame:0           TX packets:412 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1           RX bytes:31876 (31.8 KB)  TX bytes:31876 (31.8 KB)   Running 'lshw -class network' as root contains only information about the 'enp0s31f6' interface.  Additionally for sanity, my 'uname -a':  Linux home-z170a-ubuntu 4.4.0-22-generic #39=Ubuntu SMP Thu May 5 16:53:32 UTC 2016 x86_64 x86_64 x86_64 GNU/Linux   I am using Ubuntu 16.04.  My questions are:   Why are no networks present in the list for wicd or network-manager? Why is there no additional network interface created when I plug in the USB device? Is it possible to create this network interface?   Thanks for reading.


This question is pretty simple: are there any feats for D&D 3.5 that allow a non-barbarian character to enter a rage mode? It doesn't matter if the feat is a third party resource. I also don't care about prerequisites.  I am after feats specifically, and not interested in non-feat options.

I'm building a druid character for an upcoming E6 game, and I'm wondering if I can get myself rage along with 6 levels of shapeshift variant druid.  Unfortunately, shapeshift is obligatory, which means no druidic avenger. Are there any feats/races/templates/whatever which could grant rage? In case it is important, flaws are allowed, splatbooks are generally allowed and we are starting on 2nd level.  For those not knowing what E6 is, here is a . TLDR version: as dnd up to 6th level, one feat every 5kxp thereafter, level adjusted races get lesser point buy but no level adjustment.


I try to render some shapes with EEVEE bloom effect and transparent background. I understood that the rendered file output had to be OpenEXR instead of PNG format in order to keep the effect and the transparency. Nevertheless, when opening the OpenEXR files into the VSE, and adding a background clip, there are always some black contour artifacts around the rendered shapes. How could I get rid of that and have it done properly from the beginning ? Blender version used is 2.90. Note : I tried to tick the use alpha box in the compositor. It works better but the colors are changed.

was solved by saving to EXR, however when loading that EXR into the VSE over a background a dark border appears around the transparent text that cuts through the glare and bloom. I'm using Blender 2.92.0. EEVEE with bloom Render layers -> Glare -> Composite (with "UseAlpha" checked)


which one is correct? Could we consider both somehow correct?   Using idioms and expressions is appreciated.   Or  Using idioms and expressions are appreciated.   Thanks a lot.

Making adult decisions are/is really not fun.     What is the proper answer and why? My gut says "is" but I can't explain why.


I have a month field which stores the value of month as 1,2,3 etc. How can I convert this to Jan, Feb, March etc.  Following is my controller code:   public List<OpportunityData> getOpportunityData() { Integer regnInt = (regn!=null) ? Integer.valueOf(regn) : -1;   AggregateResult[] result = [SELECT Month__c, Sum(Rental_Revenue_Actual__c) Reve, Sum(Rental_Revenue_Goal__c) Gol FROM User_Stats__C where Year__c = '2013' and Region__c =: regnInt and District__c =: dist and Sales_Rep__c != null and Rental_Revenue_Actual__c != null and Rental_Revenue_Actual__c != 0 and Rental_Revenue_Goal__c != null and Rental_Revenue_Goal__c != 0 GROUP BY Month__c ORDER BY Month__c];  List<OpportunityData> oppsData = new List<OpportunityData>(); for (AggregateResult a : result) {     OpportunityData opp = new OpportunityData((String)a.get('Month__c'),                                               (Double)a.get('Reve'),                                               (Double)a.get('Gol'));         oppsData.add(opp); } oppsData.sort(); return oppsData; } public class OpportunityData implements Comparable { public Integer compareTo(Object o) {     return month-((OpportunityData)o).month; } public Integer month { get; set; } public Double Revenue { get; set; } public Double Goal { get; set; } public Double Attnmnt { get; set; }         public OpportunityData(String mon, Double Reve, Double Gol) {     month = integer.valueof(mon);     Revenue = Reve/1000;     Goal = Gol/1000;     Attnmnt = Reve / Gol * 100;              } public String getMonth() {  return DateTime.newInstanceGMT(System.Date.newInstance(2000, month, 1), Time.newInstance(0, 0, 0, 0)).formatGMT('MMM'); }   VF page:     <apex:chart height="350" width="700" data="{!OpportunityData}" background="#FFFFFF">     <apex:legend position="right"/>     <apex:axis type="Numeric" position="left" fields="Revenue,Goal" title="Revenue (In $ '000)" minimum="0"/>     <apex:axis type="Category" position="bottom" fields="Month" title="Month of the Year">         <apex:chartLabel rotate="315"/>     </apex:axis>     <apex:barSeries title="Actual, Goal" orientation="vertical" axis="left" xField="Month" yField="Revenue,Goal" colorset="#32CD32,#1E90FF" groupGutter="10">         <apex:chartTips height="20" width="120"/>     </apex:barSeries>     <apex:axis type="Numeric" position="right" fields="Attnmnt" title="Attainment (%)" grid="true" minimum="0"/>    <apex:lineSeries title="Attnmnt(%)" axis="right" xField="Month" yField="Attnmnt" fill="false" strokeColor="#FFD700" strokeWidth="2" Smooth="3" markerType="circle" markerSize="3" markerFill="#808080"/> </apex:chart>   Please Help.

In a visual force page, I can take a date and do this...  <apex:outputText value="{0,date, MMMM d,  yyyy}">     <apex:param value="{!relatedTo.my_date__c}"/> </apex:outputText>   I want to be able to do that in Apex code.  So I want a method which takes a date and returns a certain format as a string.  I check the API:   and sorry if I am stupid but not so obvious how to do this?  Many thanks.


I am trying to write a query on SQL Server to merge data items in 1 record  Input table:  sales_ref_no    Description 001 Hello 001 Hi 002 Dear 002 All 002 Please 003 Thanks   Output table:  sales_ref_no    Description 001 Hello | Hi 002 Dear | All | Please 003 Thanks   The description under the same sales_ref_no is accumulated under the same record using a | delimiter  Can anyone help?  Thanks!

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


Here comes my case:  I conducted an experiment with roughly the following design:  30 participants, each with a unique id, were asked to rank using a likert scale how much they liked images of forests. All the participants ranked the first 8 images and then the following 5 images were randomly drawn from a pool of 15 images. Therefore, in total each participant viewed 13 images of forests, but not all images were viewed by each participant. As the responses are ordinal I have gone with a cumulative link mixed effects model to preserve the structure of the data.   And here comes when I need your expertise. So far, I believe that the random terms of my model should take into account that:  Participants (id)  Image (id)   However, Im confused about how to incorporate the random effects as each participants views some but not all of the same images. Thus-far I have come to the conclusion and coded it as:   clmm(likert_Rating ~  Experience +  X.4 + X.3 + (1|part_id)+(1|Plot_ID), data=TotalF)  However, no matter how many models I try, I am never sure about what terms would be reasonable to include as randoms.  I would really appreciate if some of you could point me in the right direction as I am struggling to decide.  Many thanks for any help.

Here comes my case:  I conducted an experiment with roughly the following design:  30 participants, each with a unique id, were asked to rank using a likert scale how much they liked images of forests. All the participants ranked the first 8 images and then the following 5 images were randomly drawn from a pool of 15 images. Therefore, in total each participant viewed 13 images of forests, but not all images were viewed by each participant. As the responses are ordinal I have gone with a cumulative link mixed effects model to preserve the structure of the data.  And here comes when I need your expertise. So far, I believe that the random terms of my model should take into account that:  Participants (id) Image (id)  However, Im confused about how to incorporate the random effects as each participants views some but not all of the same images. Thus-far I have come to the conclusion and coded in R package ordinal it as:  clmm(likert_Rating ~ Experience + X.4 + X.3 + (1 | part_id) + (1 | Plot_ID),       data = TotalF)   However, no matter how many models I try, I am never sure about how to include the randoms effects using this design.  I would really appreciate if some of you could point me in the right direction as I am struggling to decide how to include the random effects.


Let's say you leave your terminal open and a co-worker comes along and types:  alias    exit='echo nope' alias unalias='echo nope' alias    type='echo nope' alias builtin='echo nope' alias   alias='echo nope'   How could you run one of these commands anyway?

I am using tcsh. bash and zsh and other suggestions won't help here.  I have several aliases that are named the same thing as another command, so if I did unalias it, typing the same thing would now do something different.  Most of the time I want the aliased command, which is why I have them. However, sometimes I want the unaliased command.  Without actually unaliasing and redefining the command, is there a simple way to tell tcsh to use the unaliased command instead?  For example, vi is aliased to vim, but sometimes I want to just use vi. cd is aliased to change my window title, but sometimes I want to leave it alone.  Obviously I could type /usr/bin/vi but since cd is a shell built-in command, there is no equivalent. Is there a general solution?


I'd like to create a custom exception with a constructor taking some context and building its message itself. In Java, that would look like (see )  public MyException(String message, IContext context){     super(message + " " + context.someData); }   In C# however, the compiler tells me that base() is not valid in this context.  public MyException(IContext context) {     base(String.Format("Context is: {0}", context)); }   What's wrong?

If I inherit from a base class and want to pass something from the constructor of the inherited class to the constructor of the base class, how do I do that?  For example, if I inherit from the Exception class I want to do something like this:  class MyExceptionClass : Exception {      public MyExceptionClass(string message, string extraInfo)      {          //This is where it's all falling apart          base(message);      } }   Basically what I want is to be able to pass the string message to the base Exception class.


The test itself lies in attempting to apply this new update to a card belonging to Judas, whom is a legitimate user of the system.   Is Judas considered the subject or the object? I'm considering Judas to be the object and hence applying whom after the comma. Is this correct?  I read about whom vs who at , but I'm not sure if I'm applying it correctly to this case. I've also tried the whom+him test by asking Whom does the card belong to? with the answer It belongs to him.

I can never figure out whether I should use who and whom. Most people use who for both colloquially, but some people say this is not correct.  What’s the rule for using who and whom correctly?


I've got a tcpdump file.  I want to remove certain packets. They are easy to spot because they have   0x0000: a6a6 a6a6 a6a6 a6a6 ........   as the first line of data.  The full packet looks (mutatis mutandis) like:  16:24:46.132673 00:35:1a:71:91:37 (oui Unknown) > 05:05:05:c5:05:05 (oui Unknown), ethertype Unknown (0x88be), length 60:      0x0000:  a6a6 a6a6 a6a6 a6a6 0101 2400 00b4 b83e     0x0010:  931f 00be 7c80 545a 0202 0000 0000 00a5     0x0020:  a5a5 a500 0000 0000 0000 0000 0000   So what I'd like to say is:  "Remove lines that contain this pattern, and the line before the pattern, and the two lines after it".  Sed or awk or grep would be good, e.g.  $ sed -i 'magic' file   Any ideas what goes in place of magic?

Consider a text file with the following entries:  aaa bbb ccc ddd eee fff ggg hhh iii   Given a pattern (e.g. fff), I would like to grep the file above to get in the output:  all_lines except (pattern_matching_lines  U (B lines_before) U (A lines_after))   For example, if B = 2 and A = 1, the output with pattern = fff should be:  aaa bbb ccc hhh iii   How can I do this with grep or other command line tools?    Note, when I try:  grep -v 'fff'  -A1 -B2 file.txt   I don't get what I want. I instead get:  aaa bbb ccc ddd eee fff -- -- fff ggg hhh iii


What is the method finding the closed form of $\displaystyle\sum_{k=1}^{n}\frac{2k-1}{2^k}$?

What is the sum of $\sum\limits_{i=1}^{n}ip^i$ and does it matter, for finite n, if $|p|>1$ or $|p|<1$ ?   Edition :  Why can I integrate take sum and then take the derivative ? I think that kind of trick is not always allowed.  ps. I've tried this approach but I made mistake when taking derivative, so I've asked, mayby I should use some program (or on-line app) for symbolic computation.


Is there a way to use the limit definition to prove that the limit as $x$ approaches 0 of $\dfrac{\sin x} x = 1$? I get the definition (sort of), but it doesn't seem to have any practical applications in terms of extension of knowledge, just raw verification and mathematical rigor.   Thanks.  EDIT: John: it was a typo. Sorry for the confusion, the limit of sin(x)/x as x approaches infinity seems much more obvious. I am more interested as x approaches 0. My question really comes down to a much simpler concept: it seems like proofs often reveal to us new information through a series of logically consequent statements. However, I can't tell what the new information gained by epsilon delta proofs are, and I was wondering if there was some new mathematical idea that was revealed by this method.

How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.  This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.


Ubuntu is being updated, usually with UBUNTU BASE on a weekly basis, and has been for some time.  When the INSTALL button is pressed, the software goes thru the update process.  PROBLEM:  Ubuntu does NOT request a password to authorize the update.  Isn't this a protocol violation?  Doesn't this risk entry to viruses, trojans, and the like?

Oneiric update manager no longer asks for a password to update already installed software.  This is a change in behavior from previous versions.  Is there a way to revert this behavior and get the Update Manager in Oneiric to ask for a password again?  Thanks!


I have this code I am reviewing:             var phrases = db2                 .Query<Phrase>("SELECT C.Id As CategoryId" +                               " FROM Category AS C" +                               " JOIN Phrase AS P ON C.Id =          P.CategoryId");   phrases has a type of List<Phrase>  Now I would like to find the first element of the list. But with my IDE when I hover over "phrases."  on another line all I see is the option to get a .count .add .addRange .find and similar.   Can someone explain how I can get the first element?

I often want to grab the first element of an IEnumerable<T> in .net, and I haven't found a nice way to do it.  The best I've come up with is:  foreach(Elem e in enumerable) {   // do something with e   break; }   Yuck!  So, is there a nice way to do this?


This is an extension question of .  They have given a plenty nice way to fold a A4 sheet in 3 equal parts.   What is the Mathematics behind the foldings? Can we use the same way for $A5,A6,\ldots$?

This is something that always annoys me when putting an A4 letter in a oblong envelope: one has to estimate where to put the creases when folding the letter. I normally start from the bottom and on eye estimate where to fold. Then I turn the letter over and fold bottom to top. Most of the time ending up with three different areas. There must be a way to do this exactly, without using any tools (ruler, etc.).


I'm using Desktop 10.2 and I tried with this script to add new text element "road 90"  to a lot of mxd's, but Something wrong with the code:  import arcpy from arcpy import env    env.workspace = r"F:\Projects\Nadav\road90\gis" for mxdname in arcpy.ListFiles("*.mxd"):     print mxdname     mxd = arcpy.mapping.MapDocument(r"F:\Projects\Nadav\road90\gis\\" + mxdname)     for TextElement in arcpy.mapping.ListLayoutElements(mxd, "TEXT_ELEMENT"):         TextElementName = "road 90"         if TextElementName == "TextElementName":             true             print 'text'     mxd.save() del mxd

From what I can tell python at ArcGIS Desktop 10.0 cannot create new  nor . Yet, 10.1 seems to support the ability to at least create a clone, of a existing , but not create one from scratch.  Can someone confirm this?   Does anyone know of any workaround to create a textElement from scratch?


Signing a tribute.  Which way is correct?  Liz, Fred, and Nicole  or   Liz, Fred and Nicole

Should I put a comma before the last item in a list?     I would like crackers, cheese and some soda.   I would like crackers, cheese, and some soda.


I am looking for a tool to flip the rows and columns of a tablular table.  For example  \begin{tabular}{lll}                                                                                1 & 2 & 3 \\        4 & 5 & 6 \end{tabular}   should become  \begin{tabular}{ll}        1 & 4 \\        2 & 5 \\        3 & 6 \end{tabular}   This is possible using . I would however prefer a simpler solution with less overhead.  Ideally   it should run on Linux, however an online solution would be acceptable too if there is one, and whatever possible in cells is conserved (comments for instance)

I would like to be able to transpose a tabular(x) table like in the following example. It is not necessary to be completely automated, but I don't want to change the data order, which is a painful and stupid. By "not completely automated" I mean that e.g. the headers can be done seperately etc.  Let's see the MWE:  \documentclass{article}  \usepackage{booktabs} \def\Midrule{\midrule[\heavyrulewidth]}  \begin{document}  \centering  \begin{tabular}{*6c}\toprule \bfseries Name & Alice & Bob & Chuck & Dave & Eve\\\Midrule \bfseries Sex & Female & Male & Male & Male & Female\\\midrule \bfseries Age & 18 & 19 & 20 & 21 & 22\\\bottomrule \end{tabular}  \bigskip  Which I want to look like  \bigskip  \begin{tabular}{*3c}\toprule \bfseries Name & \bfseries Sex & \bfseries Age\\\Midrule Alice & Female & 18\\\midrule Bob & Male & 19\\\midrule Chuck & Male & 20\\\midrule Dave & Male & 21\\\midrule Eve & Female & 22\\\bottomrule \end{tabular}   \end{document}     Note: I know that , but I would like to get it without expl3 (mostly because people would might use my code quite often have an old LaTeX installed and I cannot be convincing them to change this).


I have a content type with a number of fields. But one of the fields in the edit form must be seen by administer and people with "Reviewer" role. How can i do it?

What I need to do  I have some user account fields that should be administrator-only (read/write/edit).  For example, the IP address used to register.  So, I want to restrict access to administrators.  Current state of field permissions in Drupal 8 (as of Feb 19, 2016)   In Drupal 7, I used the  module.  However,  There is a , but this has been going on for seven years and is marked as "postponed."


W.K.T, any user(with more than 2k rep) can review 20 suggested edits per day.  Today I saw the newly appointed moderator  reviewed more than 20.    Is it possible for moderators to review more than the daily limited count?  If yes, can any moderators can do it at any time?  I didn't see this before.

Today a moderator :    But the daily limit is 40. Is it just moderators who can do this? And why are they allowed to review 20 more?  I have checked , but it does not mention anything about reviewing more tasks.  Or is Bill special ;).


I am writing some equations in latex with   \begin{dmath} equation \end{dmath}   By default the equations are centered. I want to align the equations to start from the left of the page. How can I do it?

I wanted to left align a set of equations. I've tried the solution provided in  . It worked. But, I wanted to have a single equation number for the entire flalign in order to refer to it later. Like in case of equation+split. While using equation+align, I am able to get it. But, if I use equation+flalign the equations are still being centered. Is there a way to do it?  Example using equation+align,  \begin{equation}      \begin{align}       \mbox{Minimize } & x_1+x_2+x_3 \\       \mbox{Subject to} & \\        & x_1+x_2 \leq 10 \\        & x_2+x_3 \leq 8 \\        & x_1+x_3 \leq 5       \end{align} \end{equation}   I would like to do this while the equations are left aligned.  It will be even better if the equations can be spaced a little (for example, 1 cm) from the left margin instead  of starting from the left margin.  Thanks in advance


I want to write a query similar to this:  Select * from table where EMPLID = 1 OR EMPLID = 2 OR EMPLID = 3 OR EMPLID = 4;   Is there any easy way to break this down or do I have to use OR for each case?

Is there an elegant way to do this:  SELECT Cols from MyTable WHERE  zip = 90210 OR zip = 23310 OR zip = 74245 OR zip = 77427 OR zip = 18817 OR zip = 94566 OR zip = 34533 OR zip = 96322 OR zip = 34566 OR zip = 52214 OR zip = 73455 OR zip = 52675 OR zip = 54724 OR zip = 98566 OR zip = 92344 OR zip = 90432 OR zip = 91532 OR ...   (zip codes in this post are ficticious and bear no resemblance to actual zip codes living or dead)


I bought a ibm thinkpad t430used and it came with Windows 10.  Since my SSD is only 120GB I wanted to remove windows, and reallocate that current 72GB that Windows 10 is using to my ubuntu install.  How can I do that?  My partitions currently look like:   Can someone please help? Thanks.

I want to keep Ubuntu but remove Windows.


The question and my solution, where do I get from here? We can start with a

I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes.  The problem is:   Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.     Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$. Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$. Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.   I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces).  Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$).  I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$.  Lastly $\lambda q \in Int \; D$ implies $\lambda <1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.    Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$.  The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.    My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed.  Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem.


I'm into Astrophotography lately.   I'm using a Canon 550D with kit-lens.   1) I found there was a displacement of stars (obviously) in the picture that I stacked. How to solve this as I don't use a Star-tracker/telescope  2) Also, is it necessary to use  while taking pictures of Milkyway? Is there a good tutorial to follow?  Also, what will the correct exposure settings? I read somewhere about 500/focal-length calculations that I couldn't comprehend.   Thanks for your time.

I want to take pictures of  or  or , but I don't have an equatorial tracking mount, and don't want to spend that much money on one.  How can I get clear, low-noise shots of things in the night sky without noticeable motion blur, without spending too much money? Is it possible to build a DIY tracking mount? Are there other techniques or devices that could help?


I was mathematically shown 1 = 2 by a function that states the following  $$x^2-x^2 = x^2-x^2 $$   $$x(x-x)=(x-x)(x+x)$$  dividing by $(x-x)$ we get...  $$x=x+x$$     $$ x=2x$$      $$1=2$$  I can see that mathematically he was right, but for sure that I was missing something as it doesn't make mathematical sense

I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &=& xy \nonumber \\ x^2 - y^2 &=& xy - y^2 \nonumber \\ (x + y)(x - y) &=& y(x-y) \nonumber \\ x + y &=& y \nonumber \\ 2y &=& y \nonumber \\ 2 &=& 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &=& y \\ x &=& y - y \\ x &=& y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close?


We use vinegar to prevent fruit from turning black.When we don't have vinegar what alternative is used to prevent fruits from turning black.

I love to use Apples in my Desserts. They look really delicious when I just make them, but by the time they are consumed as dessert, the apples have all turned brown. and suddenly the dessert doesn't look all that nice anymore. How do I prevent this?  UPDATE: I remember reading somewhere that dipping apples in lemon juice prevents the oxidation which turns them brown, but this effects the flavor of my recipe. What else can I do?


As the question states, show that the property exhibited can only be satisfied by a logarithmic function i.e no other family of functions can satisfy the above property.

What are some examples of continuous (on a certain interval) real or complex functions where $f(ab)=f(a)+f(b)$ (like $\ln x$?)


I mean, how do they work, generally? How do they receive the link to a video stream itself (not just the page containing a Flash player)?  I did a search on the web but couldn't find anything useful (all links point to such services, but none of them tell how they are actually implemented).

Sites were you put in a youtube url, then it goes and gives you an mp4 or mp3 of the video.  How does that work?   Viewing the source of a youtube page gives me:   <video class="video-stream html5-main-video" x-webkit-airplay="allow" data-youtube-id="q9ioaNWaqtw" src="blob:http%3A//www.youtube.com/9a45c6e4-df42-4e8b-9fb3-43a78cccebf3" style="width: 640px; height: 360px; left: 0px; top: 0px; -webkit-transform: none;"></video>   Just going to the src gives you nothing.  As an answer, oliver says you can just "simply download them directly". How do those sites do that?  I'm planning to make one. What pieces of information, specifically, should I be targeting?  Is there any sort of open source youtube downloader I can take a look at?


i have this paragraph : "some texte  calculate((( a + b )/(c - d)) / 0.2) then some other text calculate((( a + b )/( c - d)) / 0.3)"  so i need a regex that can extract each calculate(.....) expression separately. i appreciate if anyone could help me solve this issue. Thanks a lot 🙏

I need a regular expression to select all the text between two outer brackets.  Example: some text(text here(possible text)text(possible text(more text)))end text  Result: (text here(possible text)text(possible text(more text)))


In the desktop view, when we browse a site which we are not a member of, we are shown a list of sites and the reputation and badges shown on it when we click on the hamburger.      This doesn't happen on mobile web.    Joined communities are not shown on mobile web. They show just like when we are logged out. It is difficult back to communities which we are members of.  So, add a list of joined communities list on mobile web like it is on desktop site.

If you don’t have an account for some site, you can still see a list of your communities via top bar menu of this site, e.g. on 3D Printing:        But this works only for full web, in mobile web «sign up or log in» is shown:        When you have an account then the list of your communities is shown even for mobile web:


I was writing a  on another site where saying "Done." was a sufficient post. In order to meet the character limit, I typed the sentence:     Sometimes five characters is really enough.   After looking at it, I changed it to:     Sometimes five characters are really enough.   Somehow, as a native speaker, the first one still feels better to me, although I can't figure out why that is or if my feeling is misleading me.  Is my initial pass on the sentence grammatically correct? Does it mean something different than the second take?

You don't have to call me again to confirm about our meeting - just   this one e-mail from you and my reply to it is well enough.   Should it be "is" or "are" in this case?


I found this on old CPU motherboard circuit and find its interesting.  What is the purpose of that pattern? is it only for aesthetic or is there something else? I only seen the same pattern on bluetooth module rf antenna.

I've seen many boads with weird zig-zag shaped tracks just like the one below. Why are they designed like that?


I need to prove the identity $$\sum_{k = 0}^{n} k^2\binom{n}{k} = n(n - 1)2^{n - 2} + n2^{n - 1}$$ in two ways : in algebraic way and combinatorial way.  About the algebraic way - I tried to find a formula to be placed in the identity that could help me solve the problem but without success. Combinatorial way - I couldn't find a "story" that is proof of identity.

I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$


I would like to know what type of nouns car exhaust fumes, car accident, school day, air quality... are because, many times, I write such nouns with apostrophes, like car's exhaust fumes, air's quality, and car's engine, then find out they are not written this way.  For example:     (bedroom, football, sunglasses...) = Compound nouns      (high-spirited, long-lasting, never-ending...) = compound adjectives       (car exhaust fumes, car accident, school day, air quality...) = ?   I know my question might sound silly but please bear with me.

I am firmly convinced he is innocent and his accusation is the result of the police conspiracy.     I am firmly convinced he is innocent and his accusation is the result of the police's conspiracy.    Which sentence do you think is a better choice?


Booming Blade says the following:  If the target willingly moves 5 feet or more before (the start of your next turn), the target takes 1d8 thunder damage, and the spell ends.  Is “willingly” an actual game term? Or is it just meant to be taken for the obvious meaning, that being compelled to move magically, or being shoved doesn’t count? What are some other examples of “Willingly” being used in D&D?

So this is something I just found out about that I find odd:   Opportunity attacks—trigger off of movement that uses the creatures movement (action, move action, bonus action, reaction.)  Sentinel—Whenever a creature within 5 ft of you makes an attack at a creature other than you.   Booming blade—If the target willingly moves.   Mental Prison spell—if it moves, attacks through it, or reaches any part of body out of it.   So the “willingly” is different than other rulings for how effects related to movement works. However what exactly fits willing?   Cause fear—nope magic is having them move. Dissonant whispers—same thing. They have to immediately move.   So the creature has to want to move completely of its own accord.   Crown of Madness—Humanoid creature you choose must make a wisdom saving throw. On a fail it is charmed by you. (Next is flavor text so skipping.) While charmed in this way the target must make an attack on a creature other than itself that you mentally choose on each of its turns before moving.  It can act normally if you choose no creature or if none are within reach.   Crown of madness isn't making them move but doesnt specifically say willing.  And if the ruling for willing is movement not caused by magic what about magic causing situations where you take damage if you don't move. Would a creature shimmering with evocation energy be spared because he moved out of the magical blender that is cloud of daggers? If no, then what is the go-to rule for willing movement that I should be using?  In short, what movement counts as willing, and where can it be found? Is moving out of obvious danger caused by magic, willing? Is moving away from an ally so you don't hit them, willing?   If there isn't any actual ruling on willing, that's fine; I'll go back to running it with my interpretation, so that it has same wording as AOO. If there is an actual 5e term or explanation for willing (that isn't Sage Advice), please let me know. I just don't see why it was necessary to put a term in the spell that isn't used anywhere else. I understand it could have been overlooked since it was in a splatbook or has just been ignored since most people probably assume that there is "Forced " and "Willing" movement, with the willing being defined as: any movement from an action, move action, bonus action, or reaction. In same way AOO work.


I have seen a lot of posts but couldn't find a simple answer that summarizes everything.  I am new so I am still confused.  Thanks :)

I need to configure proxy. There wasn't apt.conf file in etc/apt/ so I've created it using nano.   But I don't understand what I need to write there. I've written  http_proxy = http://lgn:pwd@192.168.1.254:8080/ there and then restarted networking. But when I try to apt-get update - it doesn't work.   Actually I don't understand what port and ip I should write in apt.conf file(there was example with 192.168.1.254 ip and 8080 port so I decided to use them).


Why is 0.5 rounded up? It is exactly halfway between 0 and 1, so it is not closer to 1...nor is it closer to 0. It is exactly halfway between them.  The same principle applies to other things like 11.5 or 23.5. The value is exactly halfway between integers. I don't see any objective round-direction in these cases.  You can say the same thing for something like 6.45 if rounding to the nearest 10th. 0.45 is no closer to 0.50 than it is to 0.40. It's exactly halfway between them.  Why are they rounded up? Is it just because we need an arbitrary rule for all cases?  Edit: I read everything on the duplicate question, but still feel left hanging. If the answer is, it's arbitrary, it just seems so unsatisfying. Is there a way to prove if something in math is arbitrary or not?

As an example, the number 15, rounded to the nearest tens, rounds to 20. I understand it's arbitrary, as 10 and 20 are equidistant from 15, I just wonder if there's any discernible logic behind the convention of rounding up. Even something like, 'It just feels more natural', would probably satisfy me. Just had to ask :)


How can I find a generator of $\mathbb{F}_{743}[x]/(x^2+1)$ ?  (x-5) ??? Well it has something to do with primes that do not divide the order of 743 which is equal 2*7*53 ....?  Does one of you know how to do this?

The task is to show that $x^2 + 1$ is irreducible over $\mathbb{F}_{743}$ and then find a generator of  $\mathbb{F}_{743}[x]/(x^2+1)$.  This question has already been asked but not really answered    I've shown that $x^2+1$ is irreducible, since $ 743 \equiv 3\pmod{4}$ and therefore $-1$ is not a quadratic residue mod 743.  But I don't understand how to find a generator of $\mathbb{F}_{743}[x]/(x^2+1)$. I assume I need to look for an element of order $743^2-1$, but is there an obvious way to do this, or is it just done by trying?


I am getting following error while trying to start VM instance The zone 'projects/chrome-parity-nnnnnn/zones/asia-south1-c' does not have enough resources available to fulfill the request. Try a different zone, or try again later. It is going on since last 7 days. I am able to create and run instances in other zones. I need to get this instance running as this is tied up with a Static IP which cannot be moved across zones. Any help will be highly appreciated.

This is a  about a temporary shortage of available resources in Google Cloud services.   I'm a Brazilian user of Google Cloud's Compute Engine for a while and I managed to use it fine for quite some months. Unfortunately, in the last one month or so, I've been facing difficulties starting my instances (around 9 AM) due to the problem "the zone does not have enough resources available to fulfill the request".    At the start, it happened once or twice and then I was able to start my instance. But now it's almost impossible to start it even after dozens of attempts! Following the message's suggestion, I tried to create other instances in southamerica-east1-a nd c (mine was in b) only to find out the same problem happening there. The impression I get from this is that Google has saturated their physical capacities in South America and are doing nothing to increase it.  So what can I do about it? I can't move my server to the USA or something because it's ping sensitive (stock market applications) and, as someone still using the free trial period, I certainly won't start paying for GC if Google may handle me a product I can't use and will take weeks or months to fix.


I am trying to search and retrieve a fact from the Drools Working Memory using a FactHandle but unable to do so. I am using the kieSession.getFactHandle(object) method in order to retrieve the fact, but it returns NullPointerException every time. Please correct me if I am implementing the FactHandle the wrong way. Below is my implementation: Person person = new Person().setName("David").setAge(33); FactHandle factHandle = kieSession.getFactHandle(person); kieSession.update(factHandle, person); // NullPointerException

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?


Can anybody tell me how to redirect the heat from going up to coming out the front? The wall and mantel gets so hot that we can't touch it, so we can't use our fireplace.

When I moved into this house, the unvented gas fireplace had no mantel or hood. It worked OK, although I thought the tile above the fireplace (and the sheetrock behind it) got rather hot. When I burned it for long periods, I would take the picture we have hanging above it down. A couple of years ago I bought a wood mantel, painted it white and hung it above the fireplace. Now the mantel gets hot. I doubt it would burst into flames, but its too hot to touch and would probably peel the paint off if left on too long.   I figured I needed a fireplace hood. Off to ebay and went and bought a hood. It sticks out 4.5 inches. I found a bargain on a brass one for $30 and thought it looked fine but my wife informed me it was ugly and looked like something from the 70's. Once again, I learned I have no taste.    Turns out the hood didn't help much, the tile above the fireplace and the mantel still gets hot. Not only that, it blocks some vents at the top of the fireplace which the instructions say is a no-no.   It appeared to me the heat needs to be directed up and away. I had some aluminum flashing laying around so I made a prototype shield. My idea was just to test it and if it works make one out of sheet metal as soon as I find a shop with a sheer and a brake. The good news is that it works great. The bad news is that its ugly.      Ok, the questions:   Has anyone else had an experience with the mantel getting so hot? Has anyone seen a mantel heat shield like this I can buy? If I stick with the homemade route - any ideas on how I can make it look better. (I do plan to paint it black and maybe cut it back some) Any other creative ideas of what i can do for this problem? Maybe put a shield directly under the mantel?


By accepting an answer you're obviously already saying that it was useful. Should an additional up-vote be reserved for answers that are particularly useful, or do people always tend to use the two in unison (or the contrary)?

Is there ever a case where you would want to accept an answer, but not upvote it?  I can't think of a single situation where this would be the case.. So would it not make sense to have the "Accept" button upvote the answer too?


I only wanted to restore my iPhone 7plus, so I downloaded iOS 10.3.2 and 10.3.1 from a trusted site. I have made backup and all is okay, then when I attempt to restore my iPhone using the downloaded version, at the end of restoration process, my iPhone stuck on connect to iTunes!  I have connected the iPhone to my computer and it shows a message: There's a problem with iPhone which needed to be updated or restored!! I tried to use update with the both iOS versions and it says iPhone has been updated successfully. However, it is still stuck on connect to iTunes?  What's wrong? How do I proceed from here?

My iPhone is stuck on Connect to iTunes Screen (support.apple.com/iphone/restore iTunes logo and up arrow with lightning cable) I have Tried :-   Connecting it to iTunes and Updating, It updates Successfully.But on the same screen again   Restoring, It restores successfully.But on the same screen again  Tried Recovery Mode Screen Method and DFU method Tried Closing and Restarting iTunes Tried Deleting 'iPhone Software Update' folder in iTunes and Re-Downloding Updates    Nothing Works... Any Help will be appreciated


The following code works just fine, but it shouldn't ! When I click the Button1, the object is destroyed first, and then its Value is used and I don't receive any Access Violation or something... Even more, the multiply operation gives the correct result, that proves that Obj1 is not destroyed ! But then again, this is not true either, because when I close the program it does'n report any memory leakage. I'm very confused.  unit Unit1;  interface  uses   Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,   Dialogs, StdCtrls;  type   MyObj = class(TObject)    Value: Cardinal;   end;    TForm1 = class(TForm)     Button1: TButton;     procedure FormCreate(Sender: TObject);     procedure Button1Click(Sender: TObject);   public    Obj1:MyObj;   end;  var   Form1: TForm1;  implementation  {$R *.dfm}  procedure TForm1.Button1Click(Sender: TObject); begin  Obj1.Free;  Obj1.Value:=Obj1.Value * 5;  Caption:=IntToStr(Obj1.Value); end;  procedure TForm1.FormCreate(Sender: TObject); begin  ReportMemoryLeaksOnShutdown:=true;  Obj1:=MyObj.Create;  Obj1.Value:=10; end;  end.

In Delphi, why does the Assigned() function still return True after I call the destructor?  The below example code will write "sl is still assigned" to the console.  However, I can call FreeAndNil(sl); and it won't be assigned.  I've been programming in Delphi for a while, but this never made sense to me.  Can someone explain?  program Project1; {$APPTYPE CONSOLE} uses SysUtils, Classes;  var   sl : TStringList;  begin   sl := TStringList.Create;   sl.Free;   if Assigned(sl) then     WriteLn('sl is still assigned')   else     WriteLn('sl is not assigned'); end.   I tried comparing the VCL operations... FreeAndNil is short and sweet and makes sense:  procedure FreeAndNil(var Obj); var   P: TObject; begin   P := TObject(Obj);   TObject(Obj) := nil;  // clear the reference before destroying the object   P.Free; end;   But TObject.Free is in mysterious assembler, which I don't understand:  procedure TObject.Free; asm         TEST    EAX,EAX         JE      @@exit         MOV     ECX,[EAX]         MOV     DL,1         CALL    dword ptr [ECX].vmtDestroy @@exit: end;


When installing the operating system, I indicated the point to use commercial drivers. Now I want to completely abandon the commercial drivers in favor of the free ones. How can i do this?

I know I can check the currently installed non-free software with the following command (assuming aptitude is FOSS):  aptitude search '?installed (?section(restricted) | ?section(multiverse))'   And I can remove the "restricted" and "multiverse" sources (and not add any PPA's) to keep that list empty. I'm assuming from this point, whenever I apt install something, that something is either in the allowed sources or it is not found.  Is this enough? Does this make sure all software I'm running (in Ubuntu) is open-source?


Suppose I have a project that uses SVN for version control and includes .tex files.  Collaborators may wish to typeset/compile these .tex files into .pdf files in order to view how they look.  This has a side effect of creating the .aux, .log, .gz, and of course the .pdf file in the directory of the .tex file.  Since these are redundant, is it possible to tell SVN to ignore those file extensions when running a command like "svn status?"  That way the repo only has .tex files in it and does not waste space on those additional files.    I found this question, which I believe asks the more or less the same thing as mine:  (1)   My issue with it, is that it seems like a lot of work for a little feature.  Basically if SVN could just be told  "hey, these files are actually semantically irrelevant/vacuous/redundant for the project.  You don't need to add them or put them under version control!," that would be great.

How do I ignore files in Subversion?  Also, how do I find files which are not under version control?


Prove $\det(A)$ is $(a+(n-1)b)(a-b)^{n-1}$ where $A$ is $n \times n$ matrix with $a$'s on diagonal and all other elements $b$, off diagonal.

I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}$$  where $0 < b < a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated.


Please give us the ability to delete our data and disappear from this site. You scare me. Edit: It's not enough to simply delete my account. The anonymous scrubbing of names is good, but I want any data stored in the SE database to be removed. My name, in particular, if they received it as metadata via their OpenID or arbitration. Edit #2: I did not ask for the license on my questions and answers. I asked for the policy on preventing SE employees from talking about me to the media, and whether or not I can be secure in my personal information that I did not disclose publicly. Edit 3: I'm not convinced that I'm safe here according to what I've read from CMs. I'll be joining the others who are leaving. Be well. I saw that StackExchange's Sara Chipps went to the press and implied that a woman who uses her real name on here is a bigot, even when her account of events states otherwise... I never gave my consent to the site to slander me to the media. And SE can obtain my real name by the arbitration opt-out, or the SSO functionality of the site. I saw that  that I will be compelled to use certain pronouns even if I am being neutral with they/them, you, and OP, and that I could be banned for declining to engage. I am not comfortable with the two things in combination with each other. Over and over, it was requested if we could remain neutral and to focus on the Q&A aspect on the site. Silence, disengaging, sticking to gender-neutral words... It's not okay. I'm unable to follow these terms of service. As someone on the autism spectrum, every single word I choose is specially chosen and orchestrated to piece together my thought. It has to be structured, logical, coherent. It's already hard enough to perceive social norms. I've been telling everyone for a decade to come to this site, that it has a Be Nice Policy that is simplistic and elegant. I felt like myself. I felt like I could use this site. But now I can mess up. I will mess up. I can't remember all the different types of pronouns. I don't know how to use them. I don't know why the rules are so draconian and are telling me that I must have a positive, active engagement with people's gender identities when I can't even do that about people's favourite food. I can't comprehend how neutrality and silence can be bannable. I've tried to think about it. I've given it serious thought and Cesar's blatant insistence on policing my own thoughts, even to the point of policing my human right to be silent, is creepy and terrifying. And I can't even control if a staff member will slander me to the media. This is not okay.

If I want to copy a method from Stack Overflow, I usually put the name of the author and the link in a comment above it. I'm just wondering if there's any license in effect for all the code put in answers and questions as sample.