I am going to be doing a complete disassemble of a laptop to move it into a new casing and I was worried about static electricity damaging any of the components.  I’ve read about grounding yourself and I will have an anti-static wrist strap and mat. However, I was confused on where to exactly connect them to when working on a laptop. For desktops, I read about connecting the straps to the casing of the desktop while it’s plugged in with the power off. I also read about some people advising to hook up to something like a radiator or water pipe or to touch them first, or to the screw in a three prong power outlet. However, I don’t have a radiator or water pipes nearby and the screw for my power outlets are painted.    Where would I connect my straps to when working on a laptop to ensure that I’m grounded? Does both me and the laptop have to be grounded when working on it? I thought I read something about how both you and the device have to be grounded, but I may have been mistaken or misunderstood.  Is there something I can just plug into the wall (like a lamp with a three prong connection) and connect the strap to a metal part of it?

I want to do laptop maintenance and looking to ground myself with an anti-static wrist strap.   How, or where, do I ground myself to the laptop? Where can I find a grounding point in order to not make a mistake and spoil something?


I've created a views slideshow which uses a custom content type for each slide. However I want to prevent any user to access these nodes using the url itself like .../node/1. There seems to be a  from another stackoverflow post, but it's not supported by drupal 8.

I created a node which has an entity reference to select a user. Then, I created a view to show the node only to the user referenced in the node.  This works fine, but the problem is that the node can be viewed directly through its URL. I need to restrict the node to be only shown through the view and not by its URL.  How can I do this?


I'm working on the following exercise in order to practice differential calculus:     Suppose $f:\mathbb R \rightarrow \mathbb R$ is of class $C^1$. Let $F:\mathbb R^2 \rightarrow \mathbb R$ be given by:      $$F(x, y)=\left\{\begin{matrix} \frac{f(y)-f(x)}{y-x} \text{, if }x\neq y\\  f'(x)\text{, if }x=y \end{matrix}\right.$$   Show that $F$ is differentiable on $(x, y)$ whenever $x \neq y$. If there exists $f''$, show that $F$ is differentiable on $(x, x)$ for all $x$.  I have proven that $F$ is differentiable whenever $x \neq y$. Also, I already have calculated the partial derivatives of $F$ on $(x, x)$. They are:  $$\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}=\frac{f''(x)}{2}$$  So it remains to show that:  $$\lim_{(h, k)\rightarrow 0}\frac{F(x+h, x+k)-F(x, x)-\frac{f''(x)}{2}(h+k)}{\|(h, k)\|}=0$$  And this is where I am stuck. How do I compute this limit?  Edit: If i can prove that the limit above holds restricted to $h \neq k$, then I know how to prove that the limit above holds. Therefore it remains to show that:  $$\lim_{(h, k)\rightarrow 0, h \neq k}\frac{\frac{f(x+k)-f(x+h)}{k-h} -f'(x)-\frac{f''(x)}{2}(h+k)}{\|(h, k)\|}=0$$

Suppose $f: \mathbb{R} \to \mathbb{R}$ is a $C^{1}$ function. Then, define a new function $F: \mathbb{R}^{2} \to \mathbb{R}$ by: $$ F(x,y) = \begin{cases}   \displaystyle \frac{f(y) - f(x)}{y - x} &\text{ if } x \neq y \\   \displaystyle f'(x) &\text{ otherwise.} \end{cases} $$ Then, if $f''(x)$ exists, $F$ is differentiable. I can prove that $F$ is differentiable if $x \neq y$, since under these conditions $F_{x}$ and $F_{y}$ are $C^{1}$. So it's left to prove $F$ is also differentiable if $x = y$.   At first, I conjectured that, for example, $F_{x} (a,a)$ would be $f''(a)/2$, but I'm having a hard time proving it. I started using the definition $\lim_{h \to 0} (F(a+h, a) - F(a,a)) / h$ and applying the MVT found $\bar{a}$ between $a$ and $a + h$ s.t. this difference quotient is: $$ \frac{1}{h}(f'(\bar{a}) - f'(a)) $$ so I tried dividing and multiplying by $\bar{a} - a$, thinking it would be possible to prove that $\lim_{h \to 0} (\bar{a} - a)/h = 1$, but so far I've only been able to bound it above by $1$.  Is it true? Does this conjecture even makes sense? I'm lost in thinking about any other candidates for the differential in these points. Any help would be appreciated!


I was given a task to find  $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use  L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.

One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?  $$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$


I am facing a problem, when I have a historic map in WMS layer and I need to get buildings from this map (approx. 2000) to vector layer fo further analysis.   Is there any workflow how to automatically vectorize these objects from WMS (for example with ArcScan)?   I don´t want to do it all manually, because it is very time consuming.

I'm looking for tool that will be convert  WMS to vector layer.  Are they ready tool?(for example in python)


I need to include recursively class header files.    "Foo.h"  #ifndef FOO_H #define FOO_H #include "Bar.h"  class Foo { public:     Bar* barMember; }; #endif   "Bar.h"   #ifndef BAR_H  #define BAR_H  #include "Foo.h"  class Bar {  public:      Foo* fooMember;  };  #endif   In this case I am getting errors like      'class' does not name a type   Consider that in this case Foo is the main class that includes a lot of other classes as a members. But with one member I need to have bidirectional connection.  So why do I have such problems?

What are the C++ coding and file organization guidelines you suggest for people who have to deal with lots of interdependent classes spread over several source and header files?  I have this situation in my project and solving class definition related errors crossing over several header files has become quite a headache.


I am trying bring text in middle regarding image but it's not working.  I tried this but not working either    HTML

          Some Text

   CSS  .cz {      width:100%;      height: 30%;     float:left;     position:relative; }  .ca {      margin: 5px 0px 0px 30px; }  .cb {   position:absolute;     text-align: center;     vertical-align: middle;     right:0;     top:0;     margin: 20px 20px 0px 0px; }

Why won't vertical-align: middle work?  And yet, vertical-align: top does work.   span{   vertical-align: middle; }
      Doesn't work.


I was having a discussion with a friend about supersonic flights.   It was then when he proposed an idea on an interesting experiment that can be done during a supersonic flight. The experiment goes like this.  You board a supersonic jet along with your friend. The flight is cruising at Mach 1 above Amazon rainforests, or some other place with comparatively low air currents. You have somehow got the permission to do the experiment which involves getting out of the plane.  You both take all the safety measures and stand on both the ends of the aeroplane with yourself near the nose and your friend near the tail, both facing in the direction of the aeroplane's motion.  Your friend yells out something at you, but you can't hear it, because it's Mach 1.  You ask the pilot to slow down the plane, and the plane gradually decelerates. What happens now?  Your friend's sound catches up with you, and you hear the message!  Now, you yell out something at your friend.  You speed up the plane, and your friend catches up with the message.  Now, something remarkable happens when you do this. Your friend hears the message in reverse!!  Now the most interesting part:   If you keep on accelerating and decelerating the plane about the speed of sound, you can make your message to be heard multiple times by your friend!! It is weird, but is such an observation plausible? UPDATE: Most of the comments posted here are about the rewinding part. In fact, that is only a part of the question. The real question emphasizes on the idea, of receiving a signal by overtaking it.

Would you, as a stationary observer, hear a sound in reverse when the source of the sound travels with twice the speed of sound? Of course, he wouldn't hear anything at all before the airplane passed by and the sound reaches him, but what about after? Does it reverse due to the Doppler effect?


I´m writting my thesis in icelandic on Overleaf. The English references with more than one authors in the bibliography are automatically with "and" instead of icelandic "og", is it possible to use newcommand or something to change it on Overleaf? see example here:   @incollection{wals-36,   address   = {Leipzig},   author    = {Michael Daniel and Edith Moravcsik},   booktitle = {The World Atlas of Language Structures Online},   editor    = {Matthew S. Dryer and Martin Haspelmath},   publisher = {Max Planck Institute for Evolutionary Anthropology},   title     = {The Associative Plural},   url       = {https://wals.info/chapter/36},   year      = {2013} }  Daniel, M. and Moravcsik, E. (2013). The associative plural. In Dryer, M. S. and Haspelmath, M., editors, The World Atlas of Language Structures Online. Max Planck Institute for Evolutionary Anthropology, Leipzig.   Here´s beginning of the document(deleted parts which are not related to the question):  \documentclass[12pt]{article} \usepackage[icelandic]{babel} %\usepackage[utf8]{inputenc} %\usepackage[T1]{fontenc}  \usepackage{setspace} \usepackage{natbib} \usepackage{url} \usepackage{gb4e}  \bibliography{heimildir} \bibliographystyle{apalike}

How can I change the citation conjunction from english "and" to catalan "i"?  Otherwise  Author = {Kirk, D. and MacPhail, A.},   would get cited as (Kirk and MacPhail, 2002) when it should be (Kirk i MacPhail, 2002).  When there are more than 2 authors I get et al. which is fine.  I've seen  and  which do not work with apalike style.  Also, I'd rather not use biblatex if possible.


This is what my vector starts as:  vector<string> inventory; inventory.push_back("sword"); inventory.push_back("armor");   the player of this game will be able to choose two of three items with this code block:  while(paid < 2) {     cout << "choice: ";     cin >> equip;     ++paid;     inventory.insert(inventory.begin(), equip); }  if(paid == 2) {     cout << "\nYou are out of money\n"; }   then there is a character that will interact with one of the inventory items if you purchased it.   if((bazaar == "woman") || (bazaar == "WOMAN")) {     cout << "you approach the woman, and she gives you a look.";      if(...)     {      } }   in that second if statement, I want to be able to check for a string in the inventory vector, and have two separate things happen, depending on whether or not that string is in the vector somewhere.  (as a side note, this is code for a text based game played in the console window)

All I want to do is to check whether an element exists in the vector or not, so I can deal with each case.  if ( item_present )    do_this(); else    do_that();


I want to build a script getting 2 lists (same size) from the user, L1 and L2, and build a dictionary: each object from L1 is a key, and each object from L2 is the value (for same index)  I'm trying to make a combined list first, in order to use the dict() function.  here is my script. It runs, but prints nothing. I want to print just to make sure I get build the right list to apply the dict() function to...      print 'enter 2 lists with same len:'  L1=input('enter first list:') L2=input('enter second list:')  if len(L1)==len(L2) and isinstance(L1,list) and isinstance(L2, list):     D_list=[]     D_ij=[]     for i in L1:         if isinstance (i,(int, str)):             for j in L2:                 if isinstance(j, (tuple, list)):                      for i in L1:                         for j in L2:                                D_ij=[L1[i], L2[j]]                             D_list.extend([D_ij])                              print D_list  else: print 'error'

Imagine that you have the following list. keys = ['name', 'age', 'food'] values = ['Monty', 42, 'spam']  What is the simplest way to produce the following dictionary? a_dict = {'name': 'Monty', 'age': 42, 'food': 'spam'}


My code is below:  <div id="action-buttons" class="container">      <button type="button" name="MyButton" id="closeButton">close RMA</button> </div> <div> <button type="button" id="refreshButtons">refresh Buttons</button> </div>     <script type="text/javascript">             $(document).ready(function () {                  $("#refreshButtons").on('click', function () {                     $.ajax({                         url: "/renderButtons.html",                         error: function () {                             alert('Error message');                         },                         success: function (HtmlResult) {                             $("#action-buttons").html(HtmlResult);                         },                         complete: function() {                             alert('Action buttons - loaded successfully');                         }                      });                  });                   $("#closeButton").on('click', function () {                     alert('some text');                 });                });     </script>   When i click on #closeButton I see the Alert. That's correct.   But when I load the div#action-buttons content (by clicking #refreshButtons) my alert doesn't show. Why?  Why there is no message in browser console?.  "/renderButtons.html" is simply html file:  <button type="button" name="MyButton" id="closeButton">close RMA</button>

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


After upgrading from Ubuntu 18 to 20.04.1 I can no longer read .epub books with Calibre. On opening a book.epub file, the book is opened for editing and I get an error message reading "Failing to launch the worker process used for rendering the preview". The error details are: calibre, version 4.99.4 ERROR: Failed to launch worker: Failed to launch the worker process used for rendering the preview Traceback (most recent call last): File "/usr/lib/calibre/calibre/gui2/tweak_book/preview.py", line 92, in run self.worker = offload_worker(priority='low') File "/usr/lib/calibre/calibre/utils/ipc/simple_worker.py", line 260, in offload_worker listener, w = create_worker(env=env, priority=priority, cwd=cwd, func='offload') File "/usr/lib/calibre/calibre/utils/ipc/simple_worker.py", line 130, in create_worker address, listener = create_listener(auth_key) File "/usr/lib/calibre/calibre/utils/ipc/server.py", line 141, in create_listener l = LinuxListener(address=address, authkey=authkey, backlog=backlog) File "/usr/lib/calibre/calibre/utils/ipc/server.py", line 110, in init self._listener._unlink.cancel() AttributeError: 'NoneType' object has no attribute 'cancel' I've tried reinstalling Calibre but that didn't help. Any suggestions? Thanks,

Calibre stopped working recently. Tried uninstalling and reinstalling. When I open a terminal and call $calibre I get the following error message: Traceback (most recent call last):   File "/usr/bin/calibre", line 20, in <module>     sys.exit(calibre())   File "/usr/lib/calibre/calibre/gui_launch.py", line 73, in calibre     main(args)   File "/usr/lib/calibre/calibre/gui2/main.py", line 543, in main     listener = create_listener()   File "/usr/lib/calibre/calibre/gui2/main.py", line 514, in create_listener     return Listener(address=gui_socket_address())   File "/usr/lib/calibre/calibre/utils/ipc/server.py", line 110, in __init__     self._listener._unlink.cancel() AttributeError: 'NoneType' object has no attribute 'cancel'  Do I need to install some dependency that the sudo apt install didn't get? I tried installing via command line and the software installer/updater GUI. I'm on version Ubuntu 20.04.1 LTS. I made sure that my software in general is up to date also.


I am working in an application where i have to use keypress event for the data manipulation on dynamic textboxes.But the problem is that when i am trying to apply the  keypress event on the dynamic text box then  it is not working,I am giving a fiddle .  $("#DynamicTextBox").keypress(function (e) {      alert('hiii');   });   I want to apply keypress on DynamicTextBox,myPicker1,myPicker2 but it is not working


I am trying to solve the problem : Let $A,B,C,D,$ be commuting $n\times n$ matrices over the field $F$ (it is not given whether any of these matrices are invertible). Show that the determinant of the $2n\times 2n$ matrix $$\begin{bmatrix} A&B\\C&D \end{bmatrix}$$ is $\det(AD-BC)$.  I know that there is an answer given in a paper, which mentions working over $F[x]$. However, I have come up with the following idea:     Motivated by the adjoint formula for the inverse, we get that   $$\begin{bmatrix} A&B\\ C&D \end{bmatrix} \begin{bmatrix} D&-B\\ -C&A \end{bmatrix}= \begin{bmatrix} AD-BC & BA-AB\\ CD-DC&AD-BC \end{bmatrix} = \begin{bmatrix} AD-BC & 0\\ 0&AD-BC \end{bmatrix} $$   Hence   $$\det \left(\begin{bmatrix} A&B\\ C&D \end{bmatrix}\right)\det\left(\begin{bmatrix} D&-B\\ -C&A \end{bmatrix}\right)= \det(AD-BC)^2$$   But then    $$\det\left( \begin{bmatrix} A&B\\ C&D \end{bmatrix}\right)= -\det\left(\begin{bmatrix} B&A\\ D&C \end{bmatrix}\right)= -\det\left(\begin{bmatrix} B&D\\ A&C \end{bmatrix}\right)= \det\left(\begin{bmatrix} D&B\\ C&A \end{bmatrix}\right) $$   Now, thinking of $\det$ in terms of permutations, suppose for a given permutation in the $\det$ sum that we pick $m$ elements in the first $n$ rows from the $-B$ side. Then we must also pick $m$ elements from the last $n$ rows from the $-C$ side; i.e. the sign changes must cancel out. Thus,   $$\det\left(\begin{bmatrix} D&B\\ C&A \end{bmatrix}\right)=\det\left(\begin{bmatrix} D&-B\\ -C&A \end{bmatrix}\right)$$   which shows that   $$\det \left(\begin{bmatrix} A&B\\ C&D \end{bmatrix}\right)=\pm \det(AD-BC)$$   Is there a way complete the proof from here? All I need to show is that $$\det \left(\begin{bmatrix} A&B\\ C&D \end{bmatrix}\right)\quad\text{ and }\quad \det(AD-BC)$$ have the same sign.

I know that there are three important results when taking the Determinants of Block matrices  $$\begin{align}\det \begin{bmatrix} A & B \\ 0 & D  \end{bmatrix} &= \det(A) \cdot \det(D) \ \ \ \ & (1) \\ \\ \det \begin{bmatrix} A & B \\ C & D  \end{bmatrix} &\neq AD - CB & (2) \\ \\ \det \begin{bmatrix} A & B \\ C & D  \end{bmatrix} &= \det \begin{bmatrix} A & B \\ 0 & D - CA^{-1}B \end{bmatrix} \\ \\ &= \underbrace{\det(A)\cdot \det\left(D-CA^{-1}B\right)}_\text{if $A^{-1}$ exists} \\ \\ &= \underbrace{\det\left(AD-CB\right)}_\text{if $AC=CA$} & (3) \end{align}$$  Now I understand in result $(3)$, that all that row operations are being performed to bring it into the form we see in $(1)$, but I can't seem to convince myself that result $(1)$ is true in the first place.  Furthermore in result $(3)$, I understand that,  $\det(A)\cdot \det\left(D-CA^{-1}B\right) = \det\left(A(D-CA^{-1}B)\right)= \det(AD-CB)$, via the product rule for determinants I also understand that we need $A^{-1}$ to exist, for the initial row operation to reduce the matrix into an upper triangular form $U$, and I understand that we require $AC = CA$, to allow commutativity when we multiply $ACA^{-1}B$ to reduce it to $CB$.  Can someone provide proofs for results $(1)$ and $(2)$, as I can't seem to find proofs for them in any of the textbooks I have at my disposal


I wanted to open a question titled "What's the best SQLite Cocoa library?". The suggestions told me that the question might be closed for being subjective, but also suggested some similar questions. The answer I found this way was better than the answers that I found when I searched using the search box. In fact, the question I found didn't even appear in the list from the direct search.   Is there some way to use the better search without opening a question? Is there a reason they're so different?

When I search for similar questions that I am going to ask, I usually can't find anything through the search box on Stack Overflow. But when I enter my question on the "Ask Question" page, the questions suggested through the "Questions with similar titles" box are usually relevant. Do you experience the same thing?


I'm sorry if this is an obvious question, but I've tried looking everywhere and can't seem to find an answer.  Are WordPress themes automatically Child Theme-able? Or is there a process to make a theme ready for Child Themes?

One of the requests on themeforest is that themes are required to provide child theme support.  I know what child themes are, although I haven't worked a lot with them, but I would like to know is there something that my theme must have to support child themes?


I have seen many people use "anyways" in chat and comments on SE. But I think I have read in many books as "anyway".   Which is the correct word?   suggests that "anyways" is an informal language.     Definition of anyways    [1]    US, informal:  anyhow, anyway    [2]    (a) archaic:  anywise    (b) dialect:  to any degree at all

I have seen these words which looks pretty much the same in three different forms, why? what are the differences?  Which one is correct between anyway and anyways? Are they spoken English or can I write those in my essays are they formal words?     We're going to do it anyway!      Great, the service was terrible, but anyways the meal was gorgeous.      So, she's all "Anyways, let's go shopping at Target! "      I do not know any way to fix it.


Let $f : \mathbb{Q} → \mathbb{Q}$ be a ring homomorphism. Show that f is the identity.  Im having trouble with this problem. I started considering two elements in $\mathbb{Q}$ and the definition of ring homomorphism:  Let $x, y \in \mathbb{Q}$ such that $x = \frac{m}{n}, y=\frac{m'}{n'}$with $m,m',n,n'\in \mathbb{Z}$ then:  $f(\frac{m}{n}*\frac{m'}{n'}) = f(\frac{m}{n})*f(\frac{m'}{n'})$   and   $f(\frac{m}{n}+\frac{m'}{n'}) = f(\frac{mn'+m'n}{nn'})=f(\frac{m}{n})+f(\frac{m'}{n'})$   Clearly, my goal is to show that:  $f(\frac{m}{n}*\frac{m'}{n'}) =\frac{m}{n}*\frac{m'}{n'}$  or   $f(\frac{m}{n}+\frac{m'}{n'}) =\frac{m}{n}+\frac{m'}{n'}$  I've been thinking for like an hour and still achieve nothing, any ideas/hints?

If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.  I'm wondering if you can help me with this one?


I was editing a post over in the Math Stack Exchange chat. I clicked in chat once (like to click off the box) and hit backspace to go back a page. Weirdly enough, a weird symbol appeared. I could replicate this arbitrarily to produce blank messages. I searched the ascii character code and apparently it is ascii code 8 which corresponds to the "backspace character". Apparently, the box in chat interprets that as a symbol being typed instead of as an actual command to erase.  UPDATE    Cause I happen to know a bit of html, I decided to use an html interpreter thingy to try and narrow down the lines of code that are producing the bug. I might make a duplicate over on stack overflow later to see if they can help get everything broken down.  The code I am currently breaking down piece by piece (i.e. controlled erasure of lines of code) is as follows:    <!DOCTYPE html>

                     <textarea id="input"></textarea>

I typed . I believe I tab-completed @rumtscho and fully typed @MarthaF (since she doesn't tab complete). However, other people tell me that a Unicode character was inserted into the message; I can't see it:  Originally, I abandoned this bug report for a lack of reproducibility. But on another machine, I can see the character and managed to reproduce it: press tab, then press backspace. It works anywhere in a message, not just in a @-ping. (Actually, it seems that you just need to un-focus the input box; clicking elsewhere then hitting backspace works, too)  First machine is Firefox 12 (Iceweasel) on Debian GNU/Linux AMD64, testing+unstable. No user scripts. Greasemonkey is installed, but disabled in the extension manager Second machine is Firefox 10 (Iceweasel) on Debian GNU/Linux AMD64, testing+unstable. No user scripts, Greasemonkey not even installed Third machine (just tested) is Firefox 11 (Iceweasel) on Debian GNU/Linux AMD64. Greasemonkey installed and enabled, and there is a global userscript. Does not happen on Chromium 18.0.1025.151, running on machine #3.  So far, that's all the machines I've tested on, and it happens on all of them. It only happens in chat. Easiest way to produce it is to click on the page background, somewhere outside the text box, to un-focus the text box. Then press backspace. Note that not all browsers will display the character, even if its there (check  to see if yours does)


Show that if $p \equiv 3$ mod $4$ is a prime in $\mathbb{Z}$, then $p$ is a prime in $\mathbb{Z}[i]$  I think I probably have to use something related to norm to solve this problem but I can't seem to figure out anything. Can someone suggest how I can solve this problem? Thanks

So far, I know if $p$ is a rational prime, then $(1)$ if $p\equiv 3\mod4$, then $p$ is prime in $\mathbb{Z}[i]$. $(2)$ If $p\equiv1\mod4$  then $p=π_1 π_2$ where $π_1 $ and $π_2$ are conjugate, Then $π_1 $ and $π_2$ are primes in $\mathbb{Z}[i]$. $(3)$ $2=(1+i)(1-i)$, then $(1+i)$and$(1-i)$ are primes in $\mathbb{Z}[i]$.  What's are all the prime elements in Gaussian integers $\mathbb{Z}[i]$? For example, $-3$ are prime in $\mathbb{Z}[i]$, but not in the above $3$ cases.


I was watching a Silverlight tutorial video, and I came across an unfamiliar expression in the example code.  what is => ? what is its name? could you please provide me a link? I couldn't search for it because they are special characters.  code:          var ctx = new EventManagerDomainContext();         ctx.Events.Add(newEvent);         ctx.SubmitChanges((op) =>         {             if (!op.HasError)             {                 NavigateToEditEvent(newEvent.EventID);             }         }, null);

I just came over this syntax in some of the questions in this forum, but Google and any other searchengine tends to block out anything but letters and number in the search so it is impossible to search out "=>".  So can anyone tell me what it is and how it is used?


I am working on implementing an 8-bit adder abstracted with code in Java. This 8-bit adder is built from 8 full adder circuits. For those who don't know what a full adder is, it's a circuit that computes the sum of 2 bits.   My intention was8-bit  to use a for loop to add each corresponding bit of the adders 2, 8-bit inputs such that a new bit of the 8-bit result is computed each time the for loop iterates.  Would it be possible to store the new computed bit of each iteration in a variable holding the 8-bit result using bit shifting?  Here's an example to help explain what I am asking. The bold bit would be the one that is shifted into the int holding the result.     0b00001010   +     0b00001011   First Iteration (addition starting w/ LSB)     Sum: 1      Result: 0b00000001      Carry: 0   Second Iteration     Sum: 0      Result: 0b00000001      Carry: 1   Third Iteration     Sum: 1      Result: 0b00000101      Carry: 0   Fourth Iteration     Sum: 0      Result: 0b00000101      Carry: 1   Fifth Iteration     Sum: 1      Result: 0b00010101      Carry: 0   Sixth, Seventh, Eigth Iteration     Sum: 0, 0, 0 respectively      Result: 0b00010101      Carry: 0, 0, 0 respectively

byte x = -1; for(int i = 0; i < 8; i++) {     x = (byte) (x >>> 1);     System.out.println("X: " + x); }   As I understand it, java stores data in two's-complement, meaning -1 = 11111111 (according to wikipedia).    Also, from the java docs: "The bit pattern is given by the left-hand operand, and the number of positions to shift by the right-hand operand. The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension. "  Which means that >>> would shift a 0 to the left most bit every time.  So I expect this code to be  iteration: bit representation of x  0: 11111111  1: 01111111  2: 00111111  3: 00011111  ...so on  However, my output is always X: -1, meaning (I guess) that >>> is putting the sign bit in the left most position.  So I then try >>, and same result.  What's going on?  I would expect my output to be: X: -1, x: 127, x: 63, etc.


I have an old Win XP laptop which has wifi in it. I would like to convert this laptop to a wifi hotspot. I should be able to connect from both phones as well as computers. Is there any in-built mechanism to do this within XP?

I have a laptop which is connected to the internet with ADSL modem, and I would like to share the internet for any laptop connecting to wireless router from my laptop using hotspot.  This is what I want:  ┌────────────────────┐ │ ADSL modem         │ └───┬────────────────┘     │ ┌───┴───────────────────────┐ │ laptop's wired adaptor    │ │                           │ │ Hotspot software          │ │                           │ │ laptop's wireless adaptor │ └───┬───────────────────────┘     │ ┌───┴──────────────────────┐ │ wireless router(linksys) │ └───┬──────────────────────┘     │ ┌───┴─────────────────────────────┐ │ Laptop computers and cellphones │ └─────────────────────────────────┘   How can I do that?


Entire company uses Linux distributions for users and and servers solutions. Company employees, when authenticate themselves to pc and are connected do company internal network, should get authorization for various internal applications accesable over web interface, file servers and other utility services. Users can fork from home, offshore and so on, connected over vpn or work inside company physically. What are solutions for authorizing user for various application access and their rights ?

A searched for answers but have found nothing on here...  Long story short: a non-profit organization is in dire need of modernizing its infrastructure. First thing is to find an alternatives to managing user accounts on a number of Linux hosts.  We have 12 servers (both physical and virtual) and about 50 workstations. We have 500 potential users for these systems. The individual who built and maintained the systems over the years has retired. He wrote his own scripts to manage it all. It still works. No complaints there. However, a lot of the stuff is very manual and error-prone. Code is messy and after updates often needs to be tweaked. Worst part is there is little to no docs written. There are just a few ReadMe's and random notes which may or may not be relevant anymore. So maintenance has become a difficult task.   Currently accounts are managed via /etc/passwd on each system. Updates are distributed via cron scripts to correct systems as accounts are added on the "main" server. Some users have to have access to all systems (like a sysadmin account), others need access to shared servers, while others may need access to workstations or only a subset of those.  Is there a tool that can help us manage accounts that meets the following requirements?   Preferably open source (i.e. free as budget is VERY limited) mainstream (i.e. maintained) preferably has LDAP integration or could be made to interface with LDAP or AD service for user authentication (will be needed in the near future to integrate accounts with other offices) user management (adding, expiring, removing, lockout, etc) allows to manage what systems (or group of systems) each user has access to  - not all users are allowed on all systems support for user accounts that could have different homedirs and mounts available depending on what system they are logged into. For example   sysadmin logged into "main" server has main://home/sysadmin/ as  homedir and has all shared mounts  sysadmin logged into staff workstations would have nas://user/s/sysadmin as homedir(different from above) and potentially limited set of mounts, a logged in client would have his/her homedir at different location and no shared mounts.   If there is an easy management interface that would be awesome. And if this tool is cross-platform (Linux / MacOS / *nix), that will be a miracle!   I have searched the web and so have found nothing suitable. We are open to any suggestions. Thank you.  EDIT: This question has been incorrectly marked as a duplicate. The linked to answer only talks about having same homedirs on all systems, whereas we need to have different homedirs based on what system user is currently logged into(MULTIPLE homedirs). Also access needs to be granted only to some machinees not the whole lot. Mods, please understand the full extent of the problem instead of merely marking it as duplicate for points...


I've just been looking through a few man pages for a few different commands including grep and ifconfig.  I've noticed over a few pages, the content uses a strange syntax to notate what i think are quotations (back-tick followed by a single or double quote):  `text'     Why can't they use ' or " to open and close quotations?  Update  I now realise that this should be bolding out the characters instead of noting quotes. Is there any reason my system is ignoring these when formatting? I'm using OSX.

I've noticed that man pages and other documents formatted by Unix utilities often use double backticks `` followed by double single quotes '' to wrap quoted phrases instead of the double quote character ". Single quotes are similarly replaced. Why is this?  Here are a couple examples, from the man page for grep:   To find all occurrences of the pattern `.Pp' at the beginning of a line:         $ grep '^\.Pp' myfile   The apostrophes ensure the entire expression is evaluated by grep instead  of by the user's shell.  The caret `^' matches the null string at the  beginning of a line, and the `\' escapes the `.', which would otherwise match  any character.   The grep utility is compliant with the IEEE Std 1003.1-2008 (``POSIX.1'')  specification.


The comment flag dialog .  A side effect of that change was to .  In particular, the clickable regions on the new flag dialog are pretty skinny, as shown in red below:    This is frustrating, because it requires me to put my mouse in just the right place, which takes extra effort.  I wish the clickable regions were instead larger, so they included all of the text associated with the corresponding option, like this:    That would be better, because it would make it easier to select the option.  It would also match the way  on this site.  Can this change be made?

Currently, in the , when you want to select an option, you have to click either on the radio button, or on the top part of the description. This is because only the top part of the description is contained in a <label>.  This doesn’t match the behavior of the other older modals. I’d like the entire text to be clickable to select the corresponding radio button, instead of having to click on this rather thin strip.  This is a screenshot of the flag modal with the <label> being inspected. The bottom description should be part of the <label>, too, or perhaps in another <label> for the same radio button — whichever works best.    Other people requesting this feature   For older flag modals,  similar to this one has been proposed in 2014. Before I asked my question, this has been requested in this comment (which I overlooked at first):     [feature-request] Can you make the flag option's subtext clickable? (Like in all other flag dialogs) –    After my question, ’s  to the question  requested this feature.


In my document (KOMA script article) I use the listings package for code listings and the caption package for additional control over captions.  My problem is that some listings flow over multiple pages. When this occurs I would like to have something like "Listing n, continued", or something to that effect at the start of each additional page --- so a contination caption.  I know the captions package supports breaking up captions and doing the "continued" stuff, but am unsure how to get the listings package on board. Any ideas?

I have the following LaTeX code which works fine:  \documentclass[12pt,a4paper]{article} \usepackage[a5paper]{geometry}% just for the example \usepackage[utf8x]{inputenc} \usepackage{listings} \usepackage[labelfont=bf, font=small]{caption} \usepackage[framemethod=tikz]{mdframed}   \lstset{ breaklines=true, breakatwhitespace=false, xleftmargin=1em, frame=single, numbers=left, numbersep=5pt, }  \newcommand\mylstcaption{}  \surroundwithmdframed[ hidealllines=true, middleextra={   \node[anchor=west] at (O|-P)     {\lstlistingname~\thelstlisting\  (Cont.):~\mylstcaption};}, secondextra={   \node[anchor=west] at (O|-P)     {\lstlistingname~\thelstlisting\  (Cont.):~\mylstcaption};}, splittopskip=2\baselineskip ]{lstlisting}  \begin{document}  \renewcommand\mylstcaption{Example listing of code} \begin{lstlisting}[language=C, caption=\mylstcaption, label=lst:c1]  struct safe_buffer { struct list_head node;  /* original request */ void    *ptr; size_t  size; int direction;  /* safe buffer info */ struct dmabounce_pool *pool; void    *safe; dma_addr_t  safe_dma_addr; };  struct dmabounce_pool { unsigned long   size; struct dma_pool *pool; #ifdef STATS unsigned long   allocs; #endif };  struct dmabounce_device_info { struct device *dev; struct list_head safe_buffers; #ifdef STATS unsigned long total_allocs; unsigned long map_op_count; unsigned long bounce_count; int attr_res; #endif struct dmabounce_pool   small; struct dmabounce_pool   large;  rwlock_t lock;  int (*needs_bounce)(struct device *, dma_addr_t, size_t); };  #ifdef STATS static ssize_t dmabounce_show(struct device *dev, struct device_attribute *attr, char *buf) { struct dmabounce_device_info *device_info = dev->archdata.dmabounce; return sprintf(buf, "%lu %lu %lu %lu %lu %lu\n", device_info->small.allocs, device_info->large.allocs, device_info->total_allocs - device_info->small.allocs - device_info->large.allocs, device_info->total_allocs, device_info->map_op_count, device_info->bounce_count); }  static DEVICE_ATTR(dmabounce_stats, 0400, dmabounce_show, NULL); #endif  \end{lstlisting}  \end{document}   Part of the result looks like:  page 1    page 2    Now I would like to have similar looking captions (same font size etc.) which is currently not the case.


Why does mass seem to increase as you near the speed of light?

I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.     However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do relativistic correction to the mass so that it becomes[...]   We all know about the famous theory of relativity, but I couldn't quite grasp the "why" of its concepts yet. This might shed new light on what I already know about time slowing down for me if I move faster.  Why does the (relativistic) mass of an object increase when its speed approaches that of light?


How do I get the text of an item in Listbox by passing the value of that item. I don't need the selected item, only the one I specified by value.  I use jQuery.

All right, say I have this:  <select id='list'>     <option value='1'>Option A</option>     <option value='2'>Option B</option>     <option value='3'>Option C</option> </select>   What would the selector look like if I wanted to get "Option B" when I have the value '2'?  Please note that this is not asking how to get the selected text value, but just any one of them, whether selected or not, depending on the value attribute. I tried:  $("#list[value='2']").text();   But it is not working.


For the testing code as follow:  #include <stdio.h> int addTen(int x, int b[]) {   b[2] =  x + b[2];   return b[2]; } void main(void) {    int a[3] = {4,5,6};    int i = 2;    printf("%i %i %i \n", a[i], addTen(10,a), a[i]); }   Why is the output is 16, 16, 6? I know that even if the compiler processes the order from right to left like  a[i] <- addTen(10,a) <-a[i]. After calling addTen(10,a), a[i] is already 16 (not 6). So why the output is not 16, 16,16? THANKS!

Can it be assumed a evaluation order of the function parameters when calling it in C ? According to the following program, it seems that there is not a particular order when I executed it.   #include <stdio.h>  int main() {    int a[] = {1, 2, 3};    int * pa;      pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa), *(pa++),*(++pa));    /* Result: a[0] = 3  a[1] = 2    a[2] = 2 */     pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(pa),*(++pa));    /* Result: a[0] = 2  a[1] = 2     a[2] = 2 */     pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(++pa), *(pa));    /* a[0] = 2  a[1] = 2 a[2] = 1 */  }


What is the value of:  $$\prod_1^\infty \frac{p_i^2}{p_i^2 -1 }$$ Where $$p_i$$ are the prime numbers: 2, 3, ...

Let $p_n$ denote the $n$'th prime number.  How would one go about proving that infinite products like: $$\prod_{k=1}^\infty1 - \frac{1}{(p_k)^2} = \frac{6}{\pi^2}$$  or  $$\prod_{k=1}^\infty\frac{{p_k}^2}{{p_k}^2 - 1} = \frac{\pi^2}{6}$$ are correct?  Is there any way to prove it except by exhaustion?


From the documentation, there is something called action character=@ to unfold macros. But if i try to use it with this code:   \newcommand{\hierarchy}[1] {  \bracketset{action character=@}  \begin{forest} forked edges, for tree={     font=\sffamily\bfseries,     line width=1pt,     draw=linecol,     ellip=5,     align=center,     grow'=0,     edge+={color=linecol, line width=1pt, rounded corners=5pt, -Stealth},     drop shadow,     anchor=parent,     l sep'+=5pt, }, where={isodd(n_children())}{%     for n/.wrap pgfmath arg={{#1}{calign with current edge}}{int((n_children()+1)/2)} }{}, where level=2{tier=second}{}     @{#1} \end{forest} }  }   there is a error which says:     Package forest error: nodewalk stepped to the invalid node   I guess it has something to do with the second #1 expression in   where={isodd(n_children())}{%      for n/.wrap pgfmath arg={{#1}{calign with current edge}} {int((n_children()+1)/2)} }{},   Any suggestions how to handle this?  Here is an complete example document:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[edges]{forest} \usetikzlibrary{arrows.meta,shapes.geometric,shadows} \title{Tests}   \date{October 2017}   \tikzset{ellip/.append style={rounded rectangle, inner color=blue!30, outer color=blue!40}}  \newcommand{\hierarchy}[1] {  {\footnotesize  \bracketset{action character=@}  \begin{forest}     forked edges,     for tree={         font=\sffamily\bfseries,         line width=1pt,         draw=black,         ellip=5,         align=center,         grow'=0,         edge+={line width=1pt, rounded corners=5pt, -Stealth},         drop shadow,         anchor=parent,         l sep'+=5pt,     },     where={isodd(n_children())}{%         for n/.wrap pgfmath arg={{#1}{calign with current edge}}{int((n_children()+1)/2)}     }{},     where level=2{tier=second}{}     @{#1} \end{forest} } }  \begin{document}  \hierarchy{  [Test1[Test2][Test3[...]]] }  \end{document}

I've seen sometimes that people use a double pound sign (##) when defining/using arguments.  What is the difference between the normal argument, #1, and the double sign one, ##1? Are there any restrictions for its use? Can you list the good practices, if any, for this type of arguments.


Just curious how I can make a bash script run not only when I call it but all the time in the background.  I have a bash script that is supposed to look for files and place them in the right folders, and I just want it so i don't have to keep running that script manually. I just want it running in the background and continually will place the files into the right folders as more files are added.

There is a directory A whose contents are changed frequently by other people.  I have made a personal directory B where I keep all the files that have ever been in A.  Currently, I just occasionally run rsync to get the files to be backed up from A to B. However, I fear the possibility that some files will get added in A, and then removed from A before I get the chance to copy them over to B.  What is the best way to prevent this from occurring? Ideally, I'd like to have my current backup script run every time the contents of A get changed.


I have a board with various logos. I want to make them equal. So I set the equal height, e.g. 10cm.  However, this is not quite right, as logos that are long (e.g. "nice-company"), they visually dominate over logos that are more square (e.g. "box").  Do you have any smart suggestion how to make them visually equals?

A regular challenge that I face is balancing two (or more) logos. Whether they need to be equal or one needs to be given priority. Some logos however might be squares, others rectangles, circles, etc...   As a quick sample I snagged a nice high resolution photo from  and pretend its going to be an advertisement that needs to show some footwear brand logos on it. In this case I used Adidas and Saucony.  This is certainly not balanced even though they're height is the same:    Likewise this is not balanced even though they're width is the same:    So then the question becomes when are they in balance, is this balance:    Maybe a little more adjustment, is this balance:    And in these examples I'm only working with two Black Logos. Factor in color and it can get even more difficult to balance. Another issue being when there's tiny text as part of the logo.  So the question is, other than estimating / eye-balling it, what tips, methods or even tools are there to balance different sized logos?


How can I print a counter in the greek alphabet? \newcounter{nc} \setcounter{nc}{2} \arabic{nc} %print 2 \greek{nc} %I'd like to see \beta

I wish to use numbering like 1\alpha, 1\beta,... in some subequations. I know that this is obtainable by redefining the \theequation like the following  (which gives lower-case Roman numerals, 1.i, 1.ii,...)  \begin{subequations} \renewcommand{\theequation}{\theparentequation{}.\roman{equation}} \begin{align}   a1 &= b1 \\   a2 &= b2 \end{align} \end{subequations}   The problem is I don't know the equivalent of \roman that will yield the Greek alphabet.  A related question (which might subsume the above) is: can someone complete the set of commands  \arabic, \alph, \Alph, \roman, \Roman, ...


Having a Boxplot like the following, how do I read it?

I have some data in which there are 5 categorical explanatory variables (concern, breath, weath, sleep, act) and 1 continuous response variable (tto). In addition, every categorical explanatory variable is divided into 5 levels which show how strong a person feels about it. level 1 and level 5 show the perfect and worst states respectively.   I was advised to create a box plot to see the relationship between the explanatory variables and the response variable. The plot is given below. However, I do not know how to read a box plot. Can any one please help me interpret it.


First of all I know the differences between & and * for variables but not for the objects of a class  consider I have a class let's call it a "Math"  Math math; // a simple object called math Math* mathPtr = &math; // this will holds the address of math object Math& mathRef = math ; // this one is exactly equivalent to the previous one   What are the differences ?!

I know references are syntactic sugar, so code is easier to read and write. But what is the difference between a pointer variable and a reference variable?


I am trying to model a tree with curves, and I need to extrude 2 paths from 1 vertex, but when I try to do it, it separates the vertex. If I try to join it with F, it says "Can't create a segment".  Is it possible to make a branch with curves and why not?    Is there any workaround or addon? Because I need it with curves.

I want to extrude a curve in two directions.    Or join both curves into one, is it possible? I've tried copying it and then join segments, but I don't see Remove Doubles option while working with curves.


How does one go about showing that the dimension of a vector space of infinite sequences is uncountable? My methods was to try and show the existence of an uncountable, linearly independent of sequences which implies that the basis must be uncountable. I have tried the set of following sequences i.e. 1) An = n^r where r is any real number 2) set of convergent sequences. However, I cant seem to show that they are uncountably infinite dimensional. Any tips?  P.S Other solutions are welcomed but i have read other similar posts and do not understand the solutions

How do you show that the set of infinite sequences with entries in $\Bbb R$ is uncountably infinite-dimensional as a vector space over $\Bbb R$.


I'm working on a little chatbot (no spam!) that I would like to be able to read questions and answers and possibly comment if there is an easily detectable duplicate.  To avoid accidental spamming with comments of the main site, is there a test site that I could use to do this?  Also, if there is not, should I make that a proposal on Area 51?

Notes:  Answers will occasionally be moved to the  when there are too many of them. When using them to report bugs, keep an eye on the post.    This site, Meta Stack Exchange, is usually used as a "test bed" for pending changes happening to the way text is rendered on all network sites. Changes to text rendering that are going to be rolled out to the network are usually deployed here first to test them out and fix any potential glitches. Currently, there are no such changes pending, but if there were, keep in mind that text posted here may render differently from the other sites until the changes roll out to them as well.    As per Jeff's suggestion in : You can use this question as a formatting sandbox. You can:  edit  itself ( questions such as this one require only 100 reputation to edit instead of 2,000) post answers to this question (This question might be , requiring earning 10 reputation on this site to answer) post comments to this question or its answers test suspected bugs with the editor or the rendering post ponies (only when mods are asleep)   Beware that since the  in December 2010, and  added in March 2011, no syntax highlighting is applied unless the question's tags or an inline hint enable it. So, to test highlighting here in the sandbox:  , you can use GitHub-flavor fenced code blocks in your posts, specifying the language after the opening fence: ```html While not hinted otherwise: <html></html> source <b>goes</b> "here". ```  ```js var a = 3; while (not (a > 0)) {     alert("JavaScript code <b>goes</b> here."); } ```   Or:  Create code blocks in any way (using the fence notation above, the four-space indent form, or using HTML <pre><code> tags).  Save your post.  Use your browser's developer tools to edit the resulting HTML. To open developer tools, press F12, or ⌥⌘ I on Mac.  Find the <pre> element and add the attribute class="prettyprint", or change it into one of the valid syntax hints linked above, like class="lang-vb prettyprint".  Run the following in the location bar: javascript:prettyPrint(); or prettyPrint() in the console.     Answers that are considered annoying or obnoxious, or that cause breakage for users, will be deleted. This is .


Is there an adjective meaning living your life as you want to etc saying, doing, behaving  and don't care what other people think about you? It can be formal or informal  I know the slang word "based" but it's very informal. urbandictionary.com/define.php?term=Based

So, for instance, someone who doesn't occupy himself with the opinions of others and what they think of him, however who wouldn't go naked on the streets, because he completely doesn't care about what people think.


I'm having a rough time getting blender (2.79) to play nice with user preferences.  Per the documentation @ :     For locally extracted Blender distributions, the user configuration   and runtime data are kept in the same sub-directory, allowing multiple   Blender versions to run without conflict, ignoring the USER and SYSTEM   files.   However, with a locally extracted version of blender (I have both an installed and an extracted version of blender, both 2.79), it refuses to use a local preferences file, rather it is using the "global" or system-based user preferences file.  This was a big hit to my DCC station's setup, as I wasn't expecting this behavior.  I made a TON of changes in the locally extracted version, not expecting those changes to affect the installed version.  Now we've lost a highly customized base user-preferences file.  My bad...  Alas, I need to figure out how to force blender to use a local preferences file, and not modify the system based preferences.  I don't feel like hitting my self in the face with a brick.  I'm not sure if I should submit a bug report about this, as well?  I assume I've made a mistake, yet the official documentation literally states the intended behavior is for locally extracted versions to use local configurations and runtime data, to AVOID conflicts like this.  Any guidance here would be greatly appreciated!

When I used Blender, I pressed Ctrl+U (Save Startup File), the folder generated two files, startup.blend and userpref.blend. under C:\Users\hoald\AppData\Roaming\BlenderFoundation\Blender\2.78\config  Can I change this default path into the Blender folder?


I need to determine the precise sizes of all tables in schema 'OWNER'. The thing is, I need to get tables sizes that include also Indexes and LOBs, which apparently aren't stored in dba_segments.  So this simple code actually works, but it's far from giving me the exact size results:   select segment_name,segment_type,bytes/1024/1024/1024 GB  from dba_segments  where segment_type='TABLE' and owner='OWNER'   I also tried solutions provided here, but again, it's far from the realistic result :    How I know  the actual size? I use TOAD and when I click table's Stats/Size, I see the number shown there is far from above select.   Why I need that ? Because I want to get all table names and their size that exceed 500GB.

Being used to (and potentially spoiled by) MSSQL, I'm wondering how I can get at tables size in Oracle 10g. I have googled it so I'm now aware that I may not have as easy an option as sp_spaceused. Still the potential answers I got are most of the time outdated or don't work. Probably because I'm no DBA on the schema I'm working with.  Would anyone have solutions and or recommendations?


When I test my page speed for the site () it gives the score of 0 for defer parsing of JavaScript "3.6MiB of JavaScript is parsed during initial page load. Defer parsing JavaScript to reduce blocking of page rendering." Although JS minification is there. How to resolve this? Is there any plugin or something.

I couldn't add defer tag in plugin javascripts. Google developer pagespeed test suggests me to add defer tag in contact form 7 javascripts.   This is how contact form 7 includes javascript in header.  add_action( 'wp_enqueue_scripts', 'wpcf7_enqueue_scripts' );  function wpcf7_enqueue_scripts() {     // jquery.form.js originally bundled with WordPress is out of date and deprecated     // so we need to deregister it and re-register the latest one     wp_deregister_script( 'jquery-form' );     wp_register_script( 'jquery-form', wpcf7_plugin_url( 'jquery.form.js' ),         array( 'jquery' ), '2.52', true );      $in_footer = true;     if ( 'header' === WPCF7_LOAD_JS )         $in_footer = false;      wp_enqueue_script( 'contact-form-7', wpcf7_plugin_url( 'scripts.js' ),         array( 'jquery', 'jquery-form' ), WPCF7_VERSION, $in_footer );      do_action( 'wpcf7_enqueue_scripts' ); }   Now how to add defer="defer" tag in the above code?


What happens to the image of an object kept before a lens if the upper half of lens is covered by an opaque coating?  I understand that the full image will be formed and only the intensity of the image is reduced. But is the intensity uniformly reduced? That is, will the upper and lower half of image have same intensity?  (The object is kept at a distance $1.25f$ before the lens. Half the object is above the principle axis. $f$ is the focal length of lens).

Let's say there's an object in front of a convex lens. I have a piece of paper behind the lens at the right location such that it will the lens will form an image on the paper. The object is illuminating.  Now if I put another piece of paper right before the lens, and cover up half of the lens. The image, instead of only half being visible, it will actually remain the same. The only difference is that it will be dimmer.  Now, the question is what if I remove the paper at the location where the image forms, and put my eye there. What will I see? Will I see the piece of paper which is blocking half of the lens at all? Does that mean I will just "see through" the paper?  I wanted to try this out myself, but unfortunately I don't have a lens besides me.


How to prove that     $$ \sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}} $$   using the roots of $(z+1)^n-1=0$?  My rough idea is to solve  $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roots to prove the equality.

I am stuck trying to prove   $$\prod_{k=1}^{n-1} \sin {\pi k \over n} = {n \over 2^{n-1}}$$  and I'd appreciate help.   What I have done so far:  $z^n - 1 = \prod_{k=1}^n (z - \xi^k)$ where $\xi = e^{2 \pi i k\over n}$. Dividing both sides by $z-1$ we get  $$ {z^n - 1 \over z - 1} = \prod_{k=1}^{n-1} (z - \xi^k)$$  Taking the limit $z \to 1$ on both sides, $$ n = \prod_{k=1}^{n-1} (1 - \xi^k)$$  So I'm getting close to what I want to prove but unfortunately, this is where I am stuck!      How to proceed form here?


If $a_n$ is a sequence such that   $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$   and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?   I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in  to a similar question that does not require the sequence to be increasing.  $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$  This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.    So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?

I am looking for an example of a sequence which shows that, $a_{n+1} - a_n \to 0$ as $n \to \infty$ does not imply that sequence $a_n$ converges. I have a feeling that the sequence should be an oscillatory one but I am unable to think of an example.


Hello,  Does anybody know how to vertically align two subfigures with another subfigure horizontally?  I mean i need to put two small images above each other beside another longer image as shown above.  \documentclass{article} \usepackage[pdftex]{graphicx} \usepackage{subfigure}  \begin{document} \begin{figure}     \centering         \subfigure[]{\includegraphics[width=0.5\linewidth]{cat1}}         \subfigure[]{\includegraphics[width=0.5\linewidth]{cat2}}         \subfigure[]{\includegraphics[width=0.5\linewidth]{cat3}}     \label{fig:cat1} \end{figure}  \end{document}

I'm trying to set up a main figure containing three sub figures. The idea is to split them in such way that it shows a two columns figure where the second column is made out of two sub figures located in a row configuration. This might be hard to imagine, but the idea is to get something like this:    where each sub figure has its own label and caption.   My closest attempt consisted in an array/minipage configuration. This was:   \documentclass[11pt,a4paper,oneside]{article}     \usepackage{array}     \usepackage{caption}     \usepackage{subfigure}     \begin{document}         \begin{center} \begin{figure} \begin{array}{cc} \multirow{2}{*}{ \begin{minipage}[b]{0.45\linewidth} \centering \includegraphics[scale=0.6]{smiley.eps} \caption{Happy Smiley} \label{fig:minipage1} \end{minipage}}  &  \begin{minipage}[b]{0.45\linewidth} \centering \includegraphics[scale=0.3]{smiley.eps} \caption{Happy Smiley} \label{fig:minipage1} \end{minipage}\\ & \begin{minipage}[b]{0.45\linewidth} \centering \includegraphics[scale=0.3]{smiley.eps} \caption{Happy Smiley} \label{fig:minipage1} \end{minipage} \end{array} \end{figure} \end{center}     \end{document}   but the result wasn't good.    So, is this a right approach or should I try something different?   Thanks in advance.


Please do not close as a duplicate.  does not solve this problem. I want to see the power options on the GINA screen. The title basically sums it up. I want to remove the Power Options in the Start Menu, just to prevent accidental shutdowns due to muscle memory. I still want to be able to shutdown, without using a command prompt/desktop script, to shut down or restart. So I want to keep the power options on the GINA (ctrl+alt+del) screen. The Remove and prevent access to the Shut Down, Restart.. group policy removes it from both. As far as I can tell the registry entries under Computer\HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\PolicyManager\default\Start\ also remove them from both. Is there any way to remove from just Start?

I have a PC that should always be on. I recently upgraded to Windows 10 from Windows 7. In Windows 7 I changed the default action of the power button in the start menu to log off from my machine so that I could not accidentally shut it off. However in Windows 10 the start menu changed and this doesn't appear to be possible anymore?   I wonder how the shutdown option can be removed from the start menu. I cannot find any information about it but I expect this to be possible.


Can the definite integral $$\int_0^1\dfrac{\ln(1+x^2)}{1+x}dx$$ be evaluated using the technique of “differentiation under integral sign”. I don't want a complete solution, just the parameter would do.  PS: An alternative approach (preferably simple) would also be welcome as long as it doesn't involve contour integration.


I'd really love a possibility on the web where I could get feedback on my work (I do Android apps), preferably from other programmers or people engaged in UI / UX.   I don't know how the situation is for software other than apps, but I haven't found anything like that for apps on the web except for offers that a. cost money and b. use 'normal users' as their test group.   I work at a company that employs other mobile developers, usability experts, testers etc., still I would appreciate people from other cultural or technological backgrounds to have a look at my product, share their opinion.  But foremost this would be greatly beneficial to people who work on a project on their own and rarely get any feedback until it hits the market.   I think this would be something worth thinking about within the stackexchange universum.   What do you guys think about that?

Stuff like polls, recommendations based on subjective constraints, puzzles, webcomics etc. do not belong on the serious main SE sites, where professionals should be considered at work and having just a few spare minutes ("code's compiling") to answer questions, so they should not be distracted by such things. However, I'd also like to have a home for these things still using the SE engine. For the reason I mentioned before, this needs to be a separate place though. Let's call it four.[sitename-here].com.


Lemma  Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $W$ be a $T$-invariant subspace of $V$. Suppose that $v_1, \dots , v_k$ are eigenvectors of $T$ corresponding to distinct eigenvalues. Prove that if $v_1+\dots+v_k$ is in $W$, then $v_i ∈ W$ for all $i$.  Use the above lemma and prove the given result. Question  Prove that the restriction of a diagonalizable linear operator $T$ to any nontrivial $T$-invariant subspace is also diagonalizable.  I could prove the above lemma using induction. My attempt: A linear operator $T$ on a ﬁnite-dimensional vector space $V$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$. Let $\lambda_1, \lambda_2,..., \lambda_k$ are the eigenvalues of $T$, $V=E_{\lambda_1} \bigoplus E_{\lambda_1}\bigoplus ...\bigoplus E_{\lambda_k}.$ Let $W_{\lambda_j}=E_{\lambda_j}\cap W(1\leq j \leq k)$. Will we get $W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus W_{\lambda_k}?$ Claim 1:  $W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus  W_{\lambda_k}.$  $$V=E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_1}.$$ $$V\cap W=W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k}).$$ Claim 2:  $W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})=(W \cap  E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})$  $x\in (W \cap  E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})\implies$ $x=x_1+x_2+...x_k \implies x \in W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})$. How do I prove the converse? Does there exist any counter example to prove that result is false? How do I use the above lemma and prove the result?

Suppose $V$ is a vector space over $\mathbb{C}$, and $A$ is a linear transformation on $V$ which is diagonalizable. I.e. there is a basis of $V$ consisting of eigenvectors of $A$. If $W\subseteq V$ is an invariant subspace of $A$ (so $A(W)\subseteq W$), show that $A|_W$ is also diagonalizable.   I tried supposing $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_m$, with $V_i=\{v\in V: Av=\lambda_i v\}$. Then we can write $V=V_1\oplus\cdots\oplus V_m,$ but I'm not sure whether it is true that  $$W=(W\cap V_1)\oplus\cdots\oplus (W\cap V_m),.$$  If it is true, then we're done, but it may be wrong.


So I've been trying to turn this image into a 3D model for about 3 days now and I've watched a lot of 2D to 3D videos to try to find a work around but no luck. To me the issue comes from the perspective that the image is drawn in, because from all the videos I've seen they've all used side and front views.

I have a cartoon character (a lion) that is part of my business logo that I want to put into Blender and animate it somehow. Is there any way I can take a flat picture (JPEG) and transform it into a 3D image and then animate it in Blender?


I can not see my reputation points nor badges from my Android Phone on the black strip at the top of the page.    But they are displayed when the page in seen on my PC:    Is this a change in feature or is there some specific reason for this? I hope that it's not some kind of problem from my end. This question is only for Android smartphones with the Chrome browser.

This is the right edge of my topbar, starting today:    Any particular reason why I'm not seeing my total reputation and badges there? Since it's a major part of the "gamification" setup, am I correct in thinking it's a bug? Is it exclusive to Safari iOS?  Update: I highly doubt it's my viewport, since there's plenty of space for it:    If this is a viewport problem, why is there so much space for it?


If $p$ is an odd prime of the form $p=3 \pmod{4}$, then prove that   $$ \left(\frac{p-1}{2}\right)!=1 \text{ or } -1\pmod{p}. $$

I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.


I'm trying to put my code in an edit while asking a question.  I'm getting this error:     It looks like your post is mostly code please add more Detail   I checked  to improve the code but unfortunately the error is not solved.   Is there a way to directly adjust  the indent of the code so that I can paste code directly?

Error Message:  Oops! Your question couldn't be submitted because: Please add some context to explain the code sections (or check that you have not incorrectly formatted all of your question as code).  How do I fix this issue? It says that I need to explain my code more when I try to post this as google-apps-script. I have explained almost every part of the code. Also, this code was posted previously and I just added a few extra lines with explanations. What do they mean by "context" function doGet() {   var app = UiApp.createApplication();   //Create listBox   var listBox = app.createListBox();   //Add items to listBox   listBox.addItem("item 1").addItem("item 2").addItem("item 3").setName("myListBox");   //Create hidden app   var hidden = app.createHidden().setName('hidden').setId('hidden')   //Create button handler   var handler = app.createServerHandler("buttonHandler");   // pass the listbox into the handler function as a parameter and the hidden widget as well   handler.addCallbackElement(listBox).addCallbackElement(hidden);    //Create flextable   var table = app.createFlexTable().setId("myTable");    //Create flex table style attributes     table.setStyleAttribute("border-style", "solid")     table.setStyleAttribute("border-width", "1px")     table.setCellPadding(1);      //Create flex table headers   table.setWidget(0, 0, app.createLabel("Product Type")).setStyleAttribute("color", "blue");   table.setWidget(0, 5, app.createLabel("Size"));    var button = app.createButton("+", handler);    // add all widgets to the app   app.add(table).add(hidden);   app.add(listBox).add(button)   return app; }  function buttonHandler(e) {   var app = UiApp.getActiveApplication();   // get the position (is a string)   var pos = e.parameter.hidden;   // initial condition, hidden widget is empty    if(pos==null){pos='0'};   // convert to number    pos=Number(pos);   var table = app.getElementById("myTable");   // add the new item at the right place   table.insertRow(pos).insertCell( pos, 0).setText(pos, 0, e.parameter.myListBox);   // increment position   ++pos;   // save value   app.getElementById('hidden').setValue(pos);   // update app   return app; }


Problem:  I used Fedora for some time and it would start in a usual way.  But recently, it started to display this screen:    And it would display it only after reboot. If I however just turn on PC, it would load and stuck in this screen:    After reboot, I select Fedora (5.0.17-300.fc30.x86_64) 30 (Thirty) and it works.  Question:  What are these other choices displayed and how can I remove the ones that are invalid ?    EDIT:  Solution: I managed to remove old kernels by running dnf update and following  .

I have the below kernels installed on my machine:  [rajiviyer@rajiv-pc dnf]$ sudo dnf list installed kernel Installed Packages kernel.x86_64                     4.16.3-301.fc28                      @anaconda kernel.x86_64                     4.18.14-200.fc28                     @updates  [rajiviyer@rajiv-pc dnf]$    New Kernel: 4.18.14-200.fc28  Old Kernel: 4.16.3-301.fc28  Below is the list of dependencies for the Old Kernel: 4.16.3-301.fc28  [rajiviyer@rajiv-pc dnf]$ sudo dnf list installed kernel*4.16.3-301.fc28.x86_64 Installed Packages kernel.x86_64                            4.16.3-301.fc28               @anaconda kernel-core.x86_64                       4.16.3-301.fc28               @anaconda kernel-modules.x86_64                    4.16.3-301.fc28               @anaconda kernel-modules-extra.x86_64              4.16.3-301.fc28               @anaconda [rajiviyer@rajiv-pc dnf]$    I got a reference article which shows the steps to remove old kernel:   Kindly advice if I should proceed with the steps laid down in the article.  OR  Any GUI tool to remove kernel?  Post kernel removal what are post tasks need to be carried out? For e.g Updating GRUB


At First run everthing is good it's running with no error but when i try to refresh it using swiperefreshlayout boom! it crash :( i got this error.   java.lang.NullPointerException         at parse.FeedDataParser.getFeed(FeedDataParser.java:25)         at better.business.solution.feedssample.SampleGraph$1.onCompleted(SampleGraph.java:57)         at com.facebook.GraphRequest$5.run(GraphRequest.java:1243)         at android.os.Handler.handleCallback(Handler.java:615)         at android.os.Handler.dispatchMessage(Handler.java:92)         at android.os.Looper.loop(Looper.java:137)         at android.app.ActivityThread.main(ActivityThread.java:4745)         at java.lang.reflect.Method.invokeNative(Native Method)         at java.lang.reflect.Method.invoke(Method.java:511)         at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:786)         at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:553)         at dalvik.system.NativeStart.main(Native Method)   this is my code in FeedDataParser      public List<ResultFeed> getFeed(){      try {         List<ResultFeed> result = new ArrayList<ResultFeed>();         JSONObject jsonObject = mFeedJson.getJSONObject("feed");         JSONArray jsonArray = jsonObject.getJSONArray("data");         for(int ctr=0; ctr<jsonArray.length();ctr++){              ResultFeed resultFeed = new ResultFeed(jsonArray.getJSONObject(ctr));              result.add(resultFeed);         }         return result;       } catch (JSONException e) {         e.printStackTrace();     }     return null; }   and my asynctask which call everytime i refresh my app.      private List<ResultFeed> feedItemList;     private MyAdapter myAdapter;     private SwipeRefreshLayout mSwipeRefreshLayout;    public InitializeApplicationsTask(Context context,AccessToken accessToken,OnDataFullFeedListener dataFullFeedListener,List<ResultFeed> feedItemList,MyAdapter myAdapter,RecyclerView mRecyclerView,SwipeRefreshLayout mSwipeRefreshLayout) {          this.feedItemList = feedItemList;         this.myAdapter = myAdapter;         this.mSwipeRefreshLayout = mSwipeRefreshLayout;          GraphRequestFeeds graphRequestFeeds = new GraphRequestFeeds(dataFullFeedListener,context);         graphRequestFeeds.resultInBatchReq(accessToken);          Log.e("constructor",""+accessToken);      }      @Override     protected void onPreExecute() {          super.onPreExecute();         Log.e("onPre", "jac");     }      @Override     protected List<ResultFeed> doInBackground(Void... params) {           Log.e("doinBack", "jac");         Log.e("feeditemlist1",""+feedItemList);         return null;     }      @Override     protected void onPostExecute(List<ResultFeed> resultFeeds) {         super.onPostExecute(resultFeeds);            mSwipeRefreshLayout.setRefreshing(false);      }   Plsss 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?


We can create a driver object using the following ways.  WebDriver driver = new FirefoxDriver();  FirefoxDriver driver = new FirefoxDriver();   FirefoxDriver is a class that is implementing WebDriver Interface.  In the first statement we are using the interface name and in the second one we are using class name. Is there any difference between the two like we can't access some methods etc.?  I looked for this question on stackoverflow and found that first one is dynamic binding and second one is static binding but is there any real difference between the two declarations above?  If we don't have any difference why we should have two declarations?  Can someone answer this please?  Thank you.

I have seen this mentioned a few times and I am not clear on what it means. When and why would you do this?   I know what interfaces do, but the fact I am not clear on this makes me think I am missing out on using them correctly.   Is it just so if you were to do:  IInterface classRef = new ObjectWhatever()   You could use any class that implements IInterface? When would you need to do that? The only thing I can think of is if you have a method and you are unsure of what object will be passed except for it implementing IInterface. I cannot think how often you would need to do that.   Also, how could you write a method that takes in an object that implements an interface? Is that possible?


I wanted to fully uninstall QGIS (previously installed by  osgeo4w). The problem is that it does not appear on the list of programs to uninstall neither in the control panel>programs and features or in settings>apps & features. What I'm I missing?

I installed QGIS using OSGEO4W, and would love to know how to uninstall it, because when I go to "Control Panel > Programs > Uninstall" the program does not appear in the list of installed programs.   How can I do this?


I need help with this exercise: Consider the sequence $\{x_n\}$ that verifies $$\lim_{n\rightarrow \infty} x_n = +\infty$$ Now prove that $$\lim_{n\rightarrow\infty} \frac{x_1+x_2+...+x_n}n= +\infty$$  Also I need to find an example that the converse does not apply.  Thank you for all your answers.

More specifically the question is: If I have a sequence $(u_n)_{n\in\mathbb{N}} \subset \mathbb{R}$ that diverges to infinity. Then it's Cesàro summation sequence $(s_n)_{n\in\mathbb{N}}=(\frac{1}{n+1}\sum_{k=0}^n u_k)_{n\in\mathbb{N}}$ is also divergent to infinity.  This seems true but I can't prove it.


146x = 12 (mod 421) I found out that 421 is a prime number and I still did not know how to start after going over the notes I feel like it is similar to solve 7x = 1 (mod 180) if I know how to solve one of them, any tips? Thank you.

What is the best way to solve equations like the following:  $9x \equiv 33 \pmod{43}$  The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.  Is there a more efficient way ?   Help would be greatly appreciated!


Let $k$ be an algebraically closed field and let $\ Y\subset \mathbb{A}^n(k)$ be the curve given parametrically by $x=t^3, y=t^4,z=t^5$  I want to show  (i) $I(Y)$ is a prime ideal of height 2  (ii) $I(Y)$ cannot be generated by 2 elements  Since $I(Y)$ is kernel of the homomorphism sending $x\mapsto t^3,y\mapsto t^4,z\mapsto t^5$ into the integral domain $k[t]$, $I(Y)$ must be prime. And since we can consider $k\subset k[x,y,z]/I(Y)\subset k[t]$, by considering transcendence degree we have $\dim(k[x,y]/I(Y))=1$ and hence the height of $I(Y)$ equals 2.  Now I want to prove the second statement. I guess $I(Y)=\left<x^4-y^3,x^5-z^3,y^5-z^4\right>$, but I don't know how to prove it (although one inclusion is obvious), and I don't know how to use this "result" to prove (ii).  Any help or hints are appreciated, thank you!

How can I determine the height and the least number of generators of the ideal $ I=(xz-y^2,x^3-yz,z^2-x^2y) \subset K[x,y,z] $?      I tried to calculate the dimension of the vector space $I/I\mathfrak m$ with $\mathfrak m=(x,y,z)$, but I'm not able to find it.  I think the height is $2$, because the affine variety should be a curve, and also I know that the ideal is prime, so I need a chain $0\subset \mathfrak p \subset I$. Maybe $(xz-y^2)?$ Is it prime?


I'm looking for an auctioneering cms which is basically a copy of Ebay, but just more modern and open source.  I would love it if it would already have a connector to a Big Data DB like Hadoop.  Does any of you have any good ideas?

This is a general, community wiki  question to address non-specific "I need a CMS or Wiki that does x, y, and z..." questions.  If your question was closed as a duplicate of this question and you feel that the information provided here does not provide a sufficient answer, please open a discussion on .    I have a list of features that I want for my website's Content Management System (CMS) - where can I find a [free] script that includes all of them?


Okay, so I'm not entirely sure if this is the right place to ask this question but I genuinely don't know the meaning behind what pops up when I use this function of the "Disks" program. I'm currently trying to examine the HDD of my brother's PS4 because it's been having issues booting up and properly functioning after having been shut off.  I guess what I'm trying to ask about is what the categories "Normalized", "Threshold", "Worst", and "Types" mean.

How can I check the SMART status of a drive under 14.04 and beyond? I've seen  but it doesn't seem to apply under 14.04 and later.


I have 64-bit Windows 10 Pro, upgraded from Windows 7. My problem is that my windows explorer cannot find my directories if it would need to move or rename them. It can create directories, but they always end up having name "New Folder", and I get this:   If I use cmd.exe, I can move and rename my directories just fine.  Apparently explorer.exe always first creates them with name "New Folder" and then renames it and renaming fails.  I have three hard drives: 500GB SSD drive, and 3TB and 1TB HDD drives. The behaviour is identical with directories in all these three drives.

I can copy, move, create, and delete them. I can rename them using the REN command from a command prompt. I cannot rename folders through Windows Explorer anywhere on my machine, including on other drives.   When I try to do so, I receive an error titled "Rename Folder" that says "Can't find the specified file. Make sure you specify the correct path and file name." Then when I click Try Again or Cancel, I get another error titled "Rename" that says "The file or folder does not exist."  As far as I can tell, nothing else is wrong with my computer, but I'm pulling my hair out trying to fix this problem. I don't know for sure if superuser is the right place to ask this question, but feel free to redirect me.  Summary of Failed Attempts:   Booted into Safe Mode  Made a New User Account System Restore Point (two days ago, 9 days ago)  Deleting / Restoring / Modifying FolderDescriptions registry entries SFC /scannow  Removing windows update KB2984006  Scanned for viruses using Nod32  Scanned for malware using MalwareBytes   Details:  Tried rolling back to two different system restore points, both immediately before I observed the problem and one release prior to that.  Tried restoring FolderDescriptions registry entries as found here:   Booting in safe mode as suggested here:   SFC /scannow as suggested here:   Tried removing registry keys and windows update KB2984006 as suggested here:   Tried exporting folderdescriptions from other machine and importing as suggested here:   Compared my exported registry with other machine's exported registry for both:  HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Explorer\FolderDescriptions HKEY_LOCAL_MACHINE\SOFTWARE\Wow6432Node\Microsoft\Windows\CurrentVersion\Explorer\FolderDescriptions   My FolderDescriptions branch of the registry in both cases are binary equal to those exported from my wife's Windows 8.1 laptop which works fine.


I own a  which is controlled over a capacitive control panel. I would like to make a device which controls power output by checking the temperatur inside the pot: a thermostat.  First thing which came to my mind was a complex two axes apperature which points to the control element which should be driven and removes itself afterwards. This should work but then I discovered that you may drive the stove by capacitive load: I placed a coin on the control element to power the stove on, nothing happened. Then I touched it with my finger and it turned on. Then I touched it with a 10cm wire and it turned off. It seems my fingers were influencing the experiment. When holding the wire with a piece of plastic on the coin the stove does not switch!  The stove itself reacts to permanent capacitive load on any of the control elements with a self turn-off (security mechanism) which has to be avoided.    The Problem is that I am not smart enough to control the capacitive load. My last try so far was a small wire which did not switch the control and a capacitor and a transistor to temporarily enable the capacitive load. But whenever I connect a wire to the transistors base the stove already switched. My intention was to control the transistor with an Arduino Uno which finally should also house the thermistor.         – Schematic created using   This did not do because whenether I connected a wire to the base of the transistor the stove already switched.  I also tried  which should relate to  but it causes the same problem. This is my code:  void setup() {   digitalWrite(2, LOW); }  void loop() {   pinMode(2, OUTPUT); // touch   delay(100);    pinMode(2, INPUT);  // no touch   delay(2000); }   So how can I control the capacitive load to control the stove with an Arduino board?    Edit: After further experimentation with the above solution altering a GPIO pin of the Arduino Uno to be output or input I could reliably perform 'touches'... on my smartphone. The stove ran into the same problem, after the first performed action the above mentioned security mechanism pops in and powers down the unit, so I don't know how to handle that yet. Maybe there is another sensor?

I'm trying to figure out if electric linear solenoids are the best option to activate an interface element on the touch screen of my smart phone.  I have also heard of linear actuators and pneumatics but wasn't sure if those are for different cases, even though they all operate in a similar fashion.   Also, does anyone know if there is a low power solenoid that exists? The lowest power consumption I saw is 5V-8V at 1 A but I'm looking for something that is really low power.


I have 3 objects in blender. They are inside one another(think russian nesting dolls) and each one is transparent except for the inner-most object.  My problem is I cannot view all three objects through each other at the same time. I can either see the smallest object within the largest one or the second smallest object inside the largest one but never all three.

I was just testing Eevee and found that you cannot view a glass object through another glass. Is there any workaround? It works fine only If I rotate the glass window or put my camera to the back side of the scene


Suppose $Sub(G)$ is the set of all subgroups of a group $G$. What is the asymptotic of $|Sub(S_n)|$ with $n \to \infty$?  One can get the following bounds on $|Sub(S_n)|$:  $$2^{2^{\langle \frac{n}{2} \rangle}} - 1 \leq |Sub(S_n)| \leq 2^{n!}$$  Proof:  On one hand:  $|Sub(S_n)| \leq |P(S_n)| = 2^{n!}$  On the other hand:  Suppose $k = \langle \frac{n}{2} \rangle$. Then $S_n$ has a subgroup isomorphic to $Sym(\{1, … , k\} \times \{0, 1\})$. For all $A \subset \{1, … , k\} \times \{0, 1\})$ define $\sigma_A: (p, q) \mapsto (p, q + I_A(p) \mod 2)$. It is not hard to see, that $\{\sigma_A| A \subset \{1, … , k\}\} \cong C_2^{2^k}$. And it is known, that $C_2^m$ always has $2^m - 1$ subgroups isomorphic to $C_2$.  The only information on $|Sub(S_n)|$ I was able to find online was the list of exact values of $|Sym(S_n)|$ for $n \leq 18$, calculated by Derek Holt in .

Is there an efficient way to enumerate the unique subgroups of the symmetric group? Naïvely, for the symmetric group $S_n$ of order $\left | S_n \right | = n!$, there are $2^{n!}$ subsets of the group members that could potentially form a subgroup. In addition, many of these subgroups are going to be isomorphic to each other. I feel that the question has an easy answer in terms of the conjugacy classes, but I don't see how. If the answer generalizes to all finite groups, please elaborate!


What is the number of odd coefficients in the expansion of $(1+x)^n$  if $n = 3$  The coefficients for binomial expansion of $(1+x)^3$ are 1 3 3 1, out of which all are odd ,so answer is 4.  So what will be the answer in terms of $n$?

I was wondering if there's a nice general solution for the following problem:  How many numbers in the $n^\text{th}$ row of Pascal's triangle are even? How many are odd?


I'm designing this web app., but I can't reveal its name or main function, so let's see how this can be explained...  In this app., there's a level of user that has the ability to add or remove team members (employees). For this user to do that, I thought I'd make it easy and have the user just press the "+" button and fill in a little form, but to make it even more seamless, let's make that a dialog box.  Question: Is it normal for a user to write stuff in a dialog box, or are those typically only for information displays such as alerts, etc.?  Below, you can enter the link to the dialog section on material design and a screenshot of how this dialog/form is going to look (I need to add the lines below every line like any other form, but I just stopped after this question pop right into my mind).

I'm designing a web app but can't reveal its name and its main function so let's see how this could be explained.  In this app there's a type of user that have the power to Add and Remove Team Members (Employees) so for him to do that I thought well let's make this very easy, he's just going to need to press the "+" button and fill a little form BUT to make it even more seamless let's make that a Dialog box.  Question: Is it right for an user to write stuff on a Dialog box or those are only for information display as alerts, etc.  Below you can enter the link to the Dialog section on Material Design and a screenshot of how this Dialog/Form is going to look (I need to add the lines below every line like any other form but I just stopped after this question popped right into my mind).


I wish to get into a PhD program at a US/Canada university in marketing. I have five years experience in international marketing. By qualification, I am an engineer. I graduated from LSU with a masters degree in Electrical Engineering.  I have a little research experience of about 8 months, but that was in electrical engineering.  With this background, will it be possible for me to get into a PhD program in marketing at all? I know what I want to research on, and I am sure I can prepare a decent SOP. But will that be enough? I am targeting fall 2017.

Can I be admitted to graduate school in a different field from my degree?  Specific cases include:   If I've taken plenty of advanced courses in field X in the process of completing a degree in another field, can I apply to graduate school in X? What if I haven't taken many courses in X, but I have acquired a good grasp of X through self-study or working in a related field? What if I've never studied X, but I have done very well in an unrelated field?  Could I be admitted to graduate school in X on the basis of general intellectual promise, and then make up the missing background after enrollment?   Note that this question is an attempt to provide a comprehensive answer, to avoid the need for a profusion of field-specific questions on this topic (see the associated ). Please feel free to edit the question or answer to improve them.


I'm trying to install Ubuntu Gnome (on Windows 10), but once I pressed F12 key, I obtained the following picture:    Each time I selected my USB device, it brought me back in the exact same image. Yet I have well installed Ubuntu on that key .   Could anyone be able to tell me why I am not able to install Ubuntu? What do I have to do?

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


Determine if $x(t)$ is stable, where $x(t)$ is a solution to the differential equation:  $x'''+x''+4x'+4x = e^t(H(t) - H(t-3))$  and x(t) is defined as:  $x(t) = 0.1e^{-t} + 0.1e^t - 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})\times H(t-3)$  How do I approach this? Do I use eigenvalues? If so, I get 2i, - 2i and 1, is that correct and what do these values imply?  Any help is greatly appreciated!

Determine wether the differential equation is stable or not for the following conditions:  $x'''(0) = x''(0) = x'(0) = x(0) =0$  $\frac{d^3x(t)}{dt^3}+\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+4x(t) = e^t(H(t)-H(t-3))$  Now, I got that the solution to this differential equation reduces x(t) by use of Laplace Transforms, however, how do I determine whether or not it is stable?  $x(t) = 0.1e^{-t} + 0.1e^t - 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})*H(t-3)$  Now I first thought I would look at it graphically and I used Matlab software to get a plot as shown below:    And from the plot above it clearly appears to diverge (though this is not a proof).  Anyone have an idea as to an elegant proof to show the solution is unstable?  Anyone help would be greatly appreciated!


I think that this needs the use of a strong form of induction. So, I tried it with $k$ $3\le k\le n$, where $n$ is arbitrarily chosen. But I can't proceed beyond this.

Let $F_n$ be the Fibonacci sequence where  $F_0$ = 0  , $F_1$ = 1 and  $F_n$ = $F_{n-1}$ + $F_{n-2}$.                   I want to prove the following by induction. $$\sum_{k=1}^{n}F_k = F_{n+2}-1$$ Here is what I have so far.  Can anybody tell me if I am right?  Base case = n=1  $F_k$ = 1 = LHS  $F_{1+2}$ - 1 = 2 - 1 = 1 = RHS  Statement is true for n=1.  Assume statement is true for n=i.  $$\sum_{k=1}^{i}F_k = F_{i+2}-1$$  Prove statement is true for n=i+1.  $$\sum_{k=1}^{i+1}F_{k} = F_{((i+1)+2)}-1$$  $F_1$ + $F_2$ +...+ $F_i$ + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{i+2}$ - 1 + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{((i+1)+2)}-1$ = $F_{((i+1)+2)}-1$  Statement holds for n=i+1


I feel that this sentence has a missing word. Am I right?     Eating a healthy diet may take some getting used to, but it has many benefits for your health.   I think it must be     take some time getting used to

The following sentence is from a mathematical lecture note :     It takes a little bit of getting used to the idea of a function that cannot actually be evaluated at any specific point, but with some practice you will find that it will not cause any significant conceptual difficulty.   Is there anything wrong with this sentence? I guess it is supposed to be "It takes a little bit of time for getting used to the idea...". Any idea for understanding the sentence?


Have a kids alarm clock that requires 5v 500ma for power supply option (but can also run on 3x AAA batteries). I have a universal adapter that can supply 4.5 or 6v (with the correct polarity) but it rated at 300ma max.  I've tried it and it seems to work fine.  I understand that I may run into problems with the alarm clock funtionality / etc and you can't be totally sure - but is it "safe"?  Do I risk things catching fire for example?  If this is unsafe at all I will go spend the money!

Power supplies are available in a wide range of voltage and current ratings.  If I have a device that has specific voltage and current ratings, how do those relate to the power ratings I need to specify?  What if I don't know the device's specs, but am replacing a previous power supply with particular ratings?  Is it OK to go lower voltage, or should it always be higher?  What about current?  I don't want a 10 A supply to damage my 1 A device.


If $F\subseteq L\subseteq K$ are fields with $K/L$ and $L/F$ Galois, then $K/F$ is Galois?.   My intuition tells me that this is not true, but I can not find a counterexample, could someone give me one please? Thank you.

Let F be a char 0 field, K be a normal extension of F and L be a normal extension of K. Can it be proved or disproved that L is normal extension of F ?


I made a dual boot on my laptop.   I backed up all my files on an external HDD, destroyed the internal hard drive's partition table by creating a new one from an Ubuntu live USB,  Partitionned as following:   Windows (10) partition (150 GB) formatted as NTFS Ubuntu (18.04.1) partition (150 GB as well) formatted as Ext4 DATA partition (1.5 TB) formatted as NTFS for now, looking for a better choice   Installed Windows 10 - I had to delete the Windows partition in the Windows installer, then it created all it needed by selecting the empty space. Installed Ubuntu - Only customized the root mount point and GRUB location (on the Ext4 partition)   Now I'm looking to mount my data partition in both Windows and Ubuntu automatically at startup. I, of course, have disabled fast boot in Windows. I have no idea how to achieve this in Ubuntu (I already know how to assign a folder to a specific location on the hard drive in Windows).  Ideally I'd use ExFAT but I'm not sure Linux can read/write on it.  EDIT: Here's the output of sudo parted -l:  Model: ATA ST2000LX001-1RG1 (scsi) Disk /dev/sda : 2000GB Sector Size (logical/physical): 512B/4096B Partition table : gpt Disk Flags:   Number  Start   End     Size  File System  Name                          Flags  1      1049kB  524MB   523MB   ntfs                 Basic data partition          hidden, diag  2      524MB   629MB   105MB   fat32                EFI system partition          boot, esp  3      629MB   646MB   16,8MB                       Microsoft reserved partition  msftres  4      646MB   165GB   164GB   ntfs                 Basic data partition          msftdata  5      165GB   329GB   165GB   ext4  6      329GB   2000GB  1671GB  ntfs                 Basic data partition          msftdata  Thanks in advance.

I'm getting a PC with Windows and I want to dual boot using Ubuntu as my primary OS.  I want to create a Home partition to store all my files and hopefully be able to access them on Windows 7 without having to manually mount the partition or anything.  I want to know what kind of partition should I create and if I can share my Home partition to simplify matters.  Thanks