I just discovered a strange behavior of the CSS syntax highlighting feature which you can see in the following code example:  .first_rule {    color: red; } .second_rule:hover {    color: blue; }   .first_rule will have a plain black color, but .second_rule will be highlighted in blue because it is interpreted a keyword. It seems as if .first_rule would also become a keyword, as soon as I append a pseudo class (or anything else beginning with a  colon) to it. This lets the user assume that there is a difference between the two kinds of identifiers, but indeed there isn't.  Is this worth the effort to fix it, in order to keep StackOverflow the most user friendy page on the web? :-)  PS: Sorry, StackOverflow Meta uses a very low contrast for syntax highlighting. Please use Firebug, if you can't see this properly or temporarily paste it to the real StackOverflow.

I noticed that sometimes my code gets highlighted in different colors when rendered.   What is syntax highlighting? How does it work? Why isn't my code being highlighted correctly? How do I report a bug or request a new language? How do I use syntax highlighting? What languages are currently available on Stack Exchange?


Let $\{x_n\}$ be a sequence such that $\forall m,n\in\mathbb{N}$:   $$0\leqslant x_{m+n}\leqslant x_m+x_n$$  Prove that $\lim_{n\to\infty}\frac{x_n}{n}$ exists.  From the relation, we can show that  $$0\leq x_n\leq x_{n-1}+x_1\leq x_{n-2}+2x_1\leq\dots\leq nx_1$$  Hence  $$0\leq\frac{x_n}{n}\leq x_1$$  for all $n$. This means this sequence is bounded, so it has a convergent subsequence $\left\{\frac{x_{n_k}}{n_k}\right\}$ with  $$\lim_{k\to\infty}\frac{x_{n_k}}{n_k}=c$$  for some $c$. I want to show that the $\lim_{n\to\infty}\frac{x_n}{n}=c$, and I was stuck.

If the sequence $\{x_n\}$ satisfies the property that $0 \leq x_{m+n} \leq x_n + x_m$ for all $n$, $m \in \mathbb{N}$ , show that the limit of the sequence $\left\{\frac{x_n}{n}\right\}_n$ exists. Provide an example of such a sequence.  I can show that it is bounded, but I don't know how to show that it is monotone increasing (I think it is monotone increasing, not sure).


I could swear I once saw somewhere in under duke.edu a pdf that lists undergrad institutions across the world with a way to interpret their grades qualitatively. I can't find it anymore. Is there any other document that tells for various institutions how to interpret their grades?

I'm trying to figure out the equivalence between grade schemes in different countries.  According to some Google results:   If you are a U.S. citizen, you need a GPA of 3.2+ (it varies, some say 3.2 and some others say 3.6, etc) to apply for a position in the UK (job or university) where the prerequisite is to have a minimum grade of 2:1. In UK 2:1 is earned if you have a 60-69% of the points. I've read that 2:1 is the same as 67%. In Spain we use a grading scheme of 10 points. So according to (2), for have a UK 2:1 you need to have 6-6.9 points. If you are Spanish and have 6.7, when converting it to GPA the result is 1.5.   However, if you take the above points and treat them as a math equation, the final result will be GPA 1.5 = GPA 3.2, which doesn't make any sense.  GPA 3.2 = UK 2:1 UK 2:1 = Spain 6.7 Spain 6.7 = GPA 1.5   Replacing now...  UK 2:1 = GPA 1.5     GPA 3.2 = GPA 1.5   My question is if someone has a better explanation on converting grades obtained in different countries.


I have a section containing something similar to:  \begin{align*}    ...some long content... \end{align*}   The content is long and should be split in two pages, however, the whole block is moved to the second page leaving an empty space.  Why is this happening?

The report I'm writting features a lot of long, multiple lines equations. To reduce visual polution I've been using an aligned environment nested inside equation. This way (unlike just using an align environment) I get only a single label for all the lines.  The problem is that so many long equations often need to be broken into two pages, otherwise I get lots of blank vertical space in the document. However the aligned environment does not allow the \displaybreak command.  Is there an alternative way to achieve a pagebreak inside an aligned equation?  Here's a sample ofthe equations I'm using  \begin{equation}     \begin{aligned}     ...\\     ...\\  %I need to pagebreak here     ...\\ \end{aligned}\end{equation}


Many websites (including this one) seem to think it's acceptable to assign their own custom functionality to system hotkey combinations.  For example, this text box I'm typing in right now has custom "undo" functionality which overrides the system default Ctrl+Z with its own (crappy) implementation.   How do I prevent JavaScript from intercepting some or all of my keys so I can regain control of my browser?

Many websites, especially everything involving rich text editing (this site is guilty as well), steal keyboard shortcuts normally used to control Firefox and make them do something else instead. It is totally infuriating when I press something like Cmd-number, Cmd-L, Cmd-T, or Cmd-K and it doesn't do what I want it to. Can I make it stop?  Actually, it would probably be for the best if I could forbid stealing of all Cmd-* shortcuts. I've never seen them used for anything useful. Is it possible?


Is there any way to build an enclosure that protects its contents from magnetic fields more effectively than simply keeping sources of magnetic fields away by the same distance the box would?

I need to shield my device from magnetic interference, including earth magnetic field (if you move device around, it might be enough to cause slight currents i guess) and magnetic field caused by power nets, wires with large currents e.t.c .  I know this could be achieved by making case out of , but it seems that I can't find it anywhere (especially in small quantities).  So, are the other, easier ways to shield magnetic field?  Will multiple sheets of steel work for example?


Convert the following to Cartesian (rectangular) form and provide a graph.  $$e^{i7\pi /2}$$  The problem comes after a long series of similar problems. However, the noticeable difference with this one is that it is 1.75x the value of $2\pi$. I'm supposed to understand something here, but I'm not sure what.  My guess: I would say that there is something to be learned about how there is a principle range of $theta$ values, $(\pi,\pi]$, but then after that everything is just a $+2\pi$ or $-2\pi$ multiple of the principle angle. Am I on the right track?

I need to express $z = 4e^{-i\pi/3}$ in the form of $x + yi$ and represent it on the Argand diagram.  I think that $4 = \sqrt{x^{2} + y^{2}}$ and that $\tan (\pi/3) = y/x$ but I haven't been able to do anything useful with this information...  Is this solvable via simultaneous equations?  Thanks!


Is it possible to determine exactly how many times a question has been viewed on any Stack Exchange network site?  After a page has been viewed for around a thousand times, all I get to see is view count rounded to the nearest thousand (or million).  Is there a way to get an exact count of how many times a question has been viewed?

I’ve noticed that recently view counters no longer show the exact amount of views over 1.000, like they used to.  Is that to save space or what?


I noticed that the same bug marked as fixed (status-complete)  still exists when using a mobile device:  When I'm not logged-in, and click the "Show N more comments" link, then:   the hidden coments are show the "Show N more comments" link stays visible when I click the link again, I get a small yellow popup with the message "Commenting requires an account with 50 reputation"   I reproduced this bug on an android 4.3 device with both the latest Chrome and the standard android browser.

This might be related to  but the add comment and show x more comment buttons do not work on mobile when not logged in  Images for reference:     I have so far tested this in Safari and Chrome on my iPhone


The total number of distinct $x\in[0,1]$ for which $ \int\limits_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}  = 2x -1$ is ________. My approach is as follow $y = \int\limits_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}  - 2x + 1$ $y' = \frac{{{x^2}}}{{1 + {x^4}}} - 2$ $y' = \frac{1}{{{x^2} + \frac{1}{{{x^2}}}}} - 2$ ${x^2} + \frac{1}{{{x^2}}} \ge 2\left( {AM \ge GM} \right)$ Hence $y'<0$ decreasing $y\left( 0 \right) = \int\limits_0^0 {\frac{{{t^2}}}{{1 + {t^4}}}dt}  - 2 \times 0 + 1 = 1$ The official answer is $1$ but how we will find whether $y(1)<0$

Total number of distinct $0\leq x\leq 1$ for which $\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt = 2x-1$   $\bf{My\; Try::}$ Given $\displaystyle \underbrace{\int_{0}^{x}\frac{t^2}{1+t^4}dt}_{\geq 0} = 2x-1\;,$ Where $0\leq x\leq 1$  So we get $\displaystyle x\geq \frac{1}{2}\;\;\;,$ Now  $\displaystyle \frac{d}{dx}\left\{\int_{0}^{x}\frac{t^2}{1+t^4}dt\right\} =\frac{d}{dx}(2x-1)$  So $$\frac{x^2}{1+x^4}=2\Rightarrow \underbrace{x^2+\frac{1}{x^2}}_{\geq 2} = \frac{1}{2}$$  So no real values of $x$  Is my solution is Right, If not then how can i solve it some short way, Thanks


Does there exists a non constant analytic map $f:A\to B$ . Where $A=\{z\in \mathbb C~:~ |z|\neq 0\}$ and $B=\{z \in \mathbb C ~:~ |z|>1\}$. I am unable to construct one

Does there is non constant analytic function from  $\{z\in\mathbb{C}:z\neq 0\}$ to $\{z\in\mathbb{C}:|z|>1\}?$ According to me there is no such  non constant analytic function because if there is any such function say $f,$ then $f$ can have  either a pole or essential singularity at $z=0$. In the case of pole Picard's theorem of meromoprphic function will work and in the case of essential singularity we know that image of  any neighbourhood of essential singularity  is dense in $\mathbb{C}$, so in both of the cases we get a contradiction. So no such non constant analytic function. Am i right? Please suggest me. Thanks.


What is a "treatment firm"?  Further, propensity score matching is a more robust method of matching that quantitative accounting researchers  frequently use (Rosenbaum & Rubin, 1983). With this technique, the researcher first estimates a logit or probit regression  with the treatment effect as the dependent variable (i.e., S&P 500 membership, continuing with our earlier example) and  independent variables that are known determinants of the treatment (possibly economic performance, firm size, age, industry, etc. in our example). From this first-stage regression, the researcher obtains propensity scores. 6 The researcher  then matches each treated observation to a non-treated observation based on the distance between their propensity scores  i.e., those observations with the closest propensity scores are matched). 7 Hoi et al. (2013) utilize propensity score matching in their study that links irresponsible CSR activities with tax aggressiveness. They first generate predicted propensity scores     for each firm-year using a model of the determinants of a firm engaging in a high level of negative, irresponsible CSR activity.  Then, they match a treatment firm (one with a high level of negative CSR activities) to a non-treatment firm (one with a low level of, or no, negative CSR activities) by matching observations with the closest propensity scores. Hoi et al. (2013) then retest the relation between negative CSR activities and tax aggressiveness using this matched sample. These matching techniques help mitigate the researcher’s potential failure to consider certain firm characteristics that differ between    treatment and non-treatment firms (Dhaliwal et al., 2012).

what are the meanings of "treated observation" and "non-treated observation" in this context ?  Further, propensity score matching is a more robust method of matching that quantitative accounting researchers frequently use (Rosenbaum & Rubin, 1983). With this technique, the researcher first estimates a logit or probit regression with the treatment effect as the dependent variable (i.e., S&P 500 membership, continuing with our earlier example) and independent variables that are known determinants of the treatment (possibly economic performance, firm size, age, industry, etc. in our example). From this first-stage regression, the researcher obtains propensity scores. 6 The researcher then matches each treated observation to a non-treated observation based on the distance between their propensity scores (i.e., those observations with the closest propensity scores are matched). 7 Hoi et al. (2013) utilize propensity score matching in their study that links irresponsible CSR activities with tax aggressiveness. They first generate predicted propensity scores for each firm-year using a model of the determinants of a firm engaging in a high level of negative, irresponsible CSR activity. Then, they match a treatment firm (one with a high level of negative CSR activities) to a non-treatment firm (one with a low level of, or no, negative CSR activities) by matching observations with the closest propensity scores. Hoi et al. (2013) then re- test the relation between negative CSR activities and tax aggressiveness using this matched sample. These matching techniques help mitigate the researcher’s potential failure to consider certain firm characteristics that differ between treatment and non-treatment firms (Dhaliwal et al., 2012).


me (US citizen) flying into JFK from my home in Germany with my German girlfriend, and my three teenage kids from an earlier marriage (dual citizens). can Martina, the German girlfriend, pass through the U.S. line with us, or does she need to go through on her own? Yes, she's a big girl and can handle it, but if it's not necessary then I want to stay as one group. Thanks for your insight!

Me and a friend will be travelling to the USA, Since we have a short layover in JFK airport, I was wondering if we have to use the non-US passport line, or if we can use the US passport line. (just looking for the shortest line)  I have a US Passport and my friend has a German passport. I also live in Europe right now.


I'm hoping someone knows how to solve this issue. I'm on windows 10, the latest version of blender, rendering a scene in cycles, and in the viewport everything looks good. But when I hit f12 to render, about half the objects are missing. The camera icon in the outliner is active for all objects, the missing objects are all in different collections (some are in collections with objects that don't disappear in the final render) , and the ray visibility in the options panel is active for each object. The issue persists in Eevee as well. Clearing animation data (why was that even there in the first place? i didnt create any animations) and applying location data solved the issue for a couple of the objects, but not all of them. Not sure why that would work on some but not all. Any ideas? Dont delete my post like last time for not having "enough details" I gave literally every detail could.  Im trying to get help here I can't share screenshots or the file unfortunately, as this is for an NDA work project. What's going on? Thanks in advance!

I'm hoping someone knows how to solve this issue. I'm rendering a scene in cycles, and in the viewport everything looks good. But when I hit f12 to render, about half the objects are missing. The camera icon in the outliner is active for all objects, the missing objects are all in different collections (some are in collections with objects that don't disappear in the final render) , and the ray visibility in the options panel is active for each object. The issue also happens in Eevee. I can't share screenshots or the file unfortunately, as this is for an NDA work project. What's going on? Thanks in advance! I gave all the details I can, stop closing this post, jesus. Im trying to get help here. EDIT: Okay, so clearing animation data (why was that even there in the first place? i didnt create any animations) and applying location data solved the issue for a couple of the objects, but not all of them. Not sure why that would work on some but not all. Any ideas?


I'm sure this is a FAQ, but I can't find any clear answer. From what I understand, IPv6 addresses should be allocated in /64 blocks. Many articles are warning against using anything less, which might break some protocol expectations.  Does this mean that in IPv6 a /64 block should be treated as a single IPv4 address?  Can someone give me examples of practical problems I'd be facing if I went ahead and built a local network inside the /64 block?

This is a  about IPv6 Subnetting.      Related:               I know a lot about , and as I prepare to (deploy|work on) an IPv6 network I need to know how much of this knowledge is transferable and what I still need to learn. IPv6 seems at first glance to be much more complex than IPv4. So I would like to know:   IPv6 is 128 bits, so why is /64 the smallest recommended subnet for hosts? Related to this:   Why is it recommended to use /127 for point to point links between routers, and why was it recommended against in the past? Should I change existing router links to use /127? Why would virtual machines be provisioned with less than a /64 worth of addresses? Are there other situations in which I would use a subnet smaller than /64?  Can I map directly from IPv4 subnets to IPv6 subnets? For instance, does an IPv4 /24 correspond directly to an IPv6 /56 or /120? My interfaces have several IPv6 addresses. Must the subnet be the same for all of them? Why do I sometimes see a % rather than a / in an IPv6 address and what does it mean? Am I wasting too many subnets? Aren't we just going to run out again? In what other major ways is IPv6 subnetting different from IPv4 subnetting?


I have only Ubuntu 16.04 (no Partitions) running on my MacBook (My model is A1181) but I would like to install macOS, replacing Ubuntu. How should I go about doing this?

I recently experienced a problem with my macOS Sierra that forced me to reinstall it. However the internet connection was bad, so I ended up installing Ubuntu 17 from USB instead.  I’ve now reformatted my hard disk and torrented a macOS Sierra installer. How can I burn it onto a USB stick?


I need to add to add the following characteristics to this form but I'm stuck. any help would b greatly appreciated. Thanks      One or more word characters   Exactly one at-sign  One or more word characters  Exactly one period  Two or more characters that are a-z, A-Z, 0-9, period, or hyphen  <!DOCTYPE html>   <html>   <head>   <script>     function validateForm()     {       var x=document.forms["myForm"]["email"].value;       var atpos=x.indexOf("@");       var dotpos=x.lastIndexOf(".");       if (atpos<1 || dotpos<atpos+2 || dotpos+2>=x.length)       {          alert("Not a valid e-mail address");          return false;       }     }   </script> </head>  <body>   <form name="myForm" action="demo_form.asp" onsubmit="return validateForm();" method="post">     Email: <input type="text" name="email">     <input type="submit" value="Submit">   </form> </body>

Is there a regular expression to validate an email address in JavaScript?


I have some code. In main function i push six element in BST. when i look to debugger, i see that variable  size = 5, but variable root = null. Why variable root not change.  package Search;  public class BST<Key extends Comparable<Key>, Val> { private class Node{     Key key;     Val val;     Node left;     Node right;     Node prev;     Node(Key k, Val v){         key = k;         val = v;     } } public void push(Key k, Val v){     push(root,k,v); } private void push(Node x, Key k, Val v){     if(x == null){         x = new Node(k,v);         size++;         return;     }     int cmp = x.key.compareTo(k);     if(cmp > 0)         push(x.left,k,v);     else if(cmp < 0)         push(x.right,k,v);     else         x.val = v;  } Node root = null; int size = 0; public static void main(String args[]){     BST<String,Integer> bst = new BST<String, Integer>();     bst.push("c",1);     bst.push("b",2);     bst.push("d",3);     bst.push("a",4);     bst.push("e",5);     bst.push("c",6); }   }

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


I'm reading a proof in a linear algebra book. It mentions $$p(x) -p(c)= (x - c) h(x),$$ where $c$ is a constant, and $p(x)$ and $h(x)$ are polynomials.  Can we always factor $p(x) - p(c)$ in this way?   Please give a proof.

Let $F$ be a field. $(x-c)|(p(x)-p(c))$.   So if $x-c|p(x)=q(x)-\frac{r}{x-c}$ so for $x-c|p(x)-p(c)=s(x-c)-\frac{r}{x-c}$. Is that correct? Or is there a different way to go about proving this? I have proved that $x-c|p(x)$ has a remainder of $p(c)$ so It wuld make sense that $p(x)-p(c)$ would be $x-c$.


so in this code : #include <iostream> using namespace std; int main() { int phone_number;   cout<< "Phone number: "; cin>>phone_number;  }  I want when the user input a letter (not numbers) he gets a message like " please enter a valid phone number" and keep looping untill he enter a numbers

My brother recently started learning C++. He told me a problem he encountered while trying to validate input in a simple program. He had a text menu where the user entered an integer choice, if they entered an invalid choice, they would be asked to enter it again (do while loop). However, if the user entered a string instead of an int, the code would break. I read various questions on stackoverflow and told him to rewrite his code along the lines of:    #include<iostream> using namespace std;  int main() {     int a;     do     {     cout<<"\nEnter a number:"     cin>>a;         if(cin.fail())         {             //Clear the fail state.             cin.clear();             //Ignore the rest of the wrong user input, till the end of the line.             cin.ignore(std::numeric_limits<std::streamsize>::max(),\                                                     '\n');         }     }while(true);     return 0; }   While this worked ok, I also tried a few other ideas: 1. Using a try catch block. It didn't work. I think this is because an exception is not raised due to bad input. 2. I tried if(! cin){//Do Something} which didn't work either. I haven't yet figured this one out. 3. Thirdly, I tried inputting a fixed length string and then parsing it. I would use atoi(). Is this standards compliant and portable? Should I write my own parsing function? 4. If write a class that uses cin, but dynamically does this kind of error detection, perhaps by determining the type of the input variable at runtime, would it have too much overhead? Is it even possible?  I would like to know what is the best way to do this kind of checking, what are the best practices?  I would like to add that while I am not new to writing C++ code, I am new to writing good standards compliant code. I am trying to unlearn bad practices and learn the right ones. I would be much obliged if answerers give a detailed explanation.  EDIT: I see that litb has answered one of my previous edits. I'll post that code here for reference.  #include<iostream> using namespace std;  int main() {     int a;     bool inputCompletionFlag = true;     do     {     cout<<"\nEnter a number:"     cin>>a;         if(cin.fail())         {             //Clear the fail state.             cin.clear();             //Ignore the rest of the wrong user input, till the end of the line.             cin.ignore(std::numeric_limits<std::streamsize>::max(),\                                                     '\n');         }         else         {             inputCompletionFlag = false;         }     }while(!inputCompletionFlag);     return 0; }   This code fails on input like "1asdsdf". I didn't know how to fix it but litb has posted a great answer. :)


Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $  Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13  = \bigg( ( 4^{2n} \mod 13) \cdot ( 4 \mod 13 ) \bigg) \mod 13. $ But still I don't know how to simplify the first term in the large bracket.  Any help/suggestions?

How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?


I would like to retrieve answers to old questions. Is there any way to search "only" on the questions I posted?

What is the syntax to use for searching the text of the questions that I've posted? i.e. Limit the search to just the questions that I've asked and just the Title/Subject/Question part of it....


I have the Nikkor 18-55mm f/3.5-5.6g vrii lens using it on the D3300. I have been trying to create some bokeh effect using this lens (I know it isn't a great lens for bokeh). The thing I was wondering was which setting will give me good bokeh.  1). Zooming at 18mm and keeping aperture the widest? (Sacrificing the max zoom for the lens) 2). Zooming at 55mm, this eventually wouldn't have widest aperture.  I know that keeping the aperture the widest and max zooming the lens combined gives you great bokeh. I'd like to know what I could do in my case to achieve the best possible blur.

I'm currently using an entry-level DSLR with 18-55mm lens and having great trouble trying to creating a blurred background effect in my photos. Now matter what aperture or shutter speed I choose, I'm not able to get it.  There's a general question about , but what specific things can I do to get this effect with an entry-level lens?


I have what I consider to be an above average computer. Specs Gforce GTX 780ti. 16GB Memory Intel i7 - 4790k 4.00GHz. However although it would appear to be fast when rendering the blenchmark test render, turning in a respectable 1 min 12 secs. When it came to rendering my first ever tutorial off the Blender Guru site. (The one with the Doughnuts..you may have seen it). With samples set to around 128 blender reports it would take over 4 hours to render out. Surely this can't be right. So it's out of desperation that I'm turning to anyone out there who can help me out as I haven't got a clue why this should be the case. Would therefor appreciate any help on the matter please.  Thanks Foxtrot

I am currently using Sheep It! to render my blend (built on v2.68c) on multiple computers, and I noticed something rather strange. According to the log on-site, the frames rendered from other people's blends were completing in 3 to 4 minutes. This was on an 4 core i5. My main rig (with an 8 core i7) was completing frames in no less than half an hour. How is this possible?  To clarify, these people's frames are very complex, often with very detailed textures. I have 2 objects in my blend. Below are the options I changed for my blend:   Aspect Ratio: 1000px x 1000px Output: PNG Clamp Indirect: 0.98 Render Sampling: 128 Min Light Bounces: 0 Max Light Bounces: 4 Cycles Render   Is there something in there that I should change to make it go faster? I can't really comprehend how it's going to quickly.  EDIT: To answer what @Duarte asked...   System Specs: i7-920, 3GB DDR3, GT 630, Windows 10 I have two circles, extended to form a sort of coin with a star on it. It is not complex at all. I have Firefox and Steam running in the background, but not actively. I've been away from the computer for 8 hours and noticed this behavior. (I am running Blender from Steam.) I made sure when I began rendering that the blend was set for CPU rendering.          And to answer what @pycoder asked, yes, I am using Cycles Render.


The developer has coded this but it seems like a not so clean implementation also it does not work if the user enters many colons. I would like to use regex but is it possible for regex to check for two or more colons and replace with just two? renameString = renameString.Replace(":::::::", "::"); renameString = renameString.Replace("::::::", "::"); renameString = renameString.Replace(":::::", "::"); renameString = renameString.Replace("::::", "::"); renameString = renameString.Replace(":::", "::");  Does anyone have any suggestions on a better way that this can be done? For reference here's the full code for that method: if (renameString.StartsWith(":") || renameString.EndsWith(":")) {     AcceptText.IsEnabled = false;     IsValidDeckName = false; } else if (renameString.Contains("::")) {     renameString = renameString.Replace(":::::::", "::");     renameString = renameString.Replace("::::::", "::");     renameString = renameString.Replace(":::::", "::");     renameString = renameString.Replace("::::", "::");     renameString = renameString.Replace(":::", "::");     IsValidDeckName = true;     AcceptText.IsEnabled = true; } else {     IsValidDeckName = true;     AcceptText.IsEnabled = true;  }  Any suggestions on if this can be improved would be appreciated.

How can I replace multiple spaces in a string with only one space in C#?  Example:  1 2 3  4    5   would be:  1 2 3 4 5


I'm looking for a word that describes an attribute Y, where you could say  The good thing about X is Y, the problem with X is Y  note that it's not a neutral attribute, it's simultaneously good and bad.  is there such a word in the English language?   preferably I'd like to be able to say something along the lines of:  Y is a ______ of X; it's good because of blah, but it also causes bleh problems

Take this scenario:     a tool has a specific feature that could be regarded as an invaluable benefit for. However, this feature, if used in opposite direction may be counterproductive. In other words, the strength point may be the weakness point.   For example, feature X of a security mechanism could be leveraged to break that if used maliciously. So, what's the word/phrase to describe this feature? I think it should be something other than Achilles heel.


Someone tried to make a helpful edit to one of my posts. (A community wiki post which specifically states "please add definitions as needed").   I got a notification that a suggested edit was pending, but by the time I responded, it had been rejected with two incorrect reasons.  (The edit neither "changed the meaning" nor was it incorrect.)  I know I can go ahead and edit the post to include the denied contribution, but is there a way to challenge the rejection?  The original editor deserves credit for doing something useful,and the rejectors could use a mild scolding, for rejecting something they didn't understand.

A nice user suggested  to . The edit, though very good and desirable, was rejected. I do not understand the reasons, and I did not find a way to contact those people who rejected it. What can I do to re-approve the edit?


void main() { printf("%d\n",sizeof('1')); }   Output : 4  void main() { char a='1'; printf("%d\n",sizeof(a)); }   Output : 1  Can somebody say why it is different ?

In C++, sizeof('a') == sizeof(char) == 1. This makes intuitive sense, since 'a' is a character literal, and sizeof(char) == 1 as defined by the standard.  In C however, sizeof('a') == sizeof(int). That is, it appears that C character literals are actually integers. Does anyone know why? I can find plenty of mentions of this C quirk but no explanation for why it exists.


$$\begin{align} \int \sec x \, dx  &= \int \cos x \left(  \frac{1}{\cos^2x} \right)  \, dx \\ &= \int \cos x \left(  \frac{1}{1-\sin^2x} \right) \, dx \\ & = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx  \end{align}$$  I am stuck in here. Any help to integrate secant?

The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$ is to multiply by $\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$. I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \,\mathrm d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$?


I'm running Windows 10 Professional, and I'd like to install the latest Insider build. I'm not on an Insider build right now.  Due to reasons I don't discuss here, recently I moved my bootloader to an external USB hard drive with Windows 7 and no bootloader using EasyBCD. (It didn't work... :) ) Since then, I moved it back, even recreated it a few times, but my Windows still believes it is on an USB disc.   This prevents me from installing Insider builds, as I get an error about "Windows can't be installed on an USB drive via the installer." (Free translation of mine.) How do I reset the original state and make Windows belive the correct thing, that it's installed on my C: SSD?

I have a Windows to Go (W2G) Windows 10 Home installation. The Anniversary Update was released a few days ago. However, I cannot install it because "You can't install Windows on a USB flash drive using Setup."  The fix that I've found everywhere is to change the Registry key HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\PortableOperatingSystem from 0 to 1.  However, a W2G installation IS a PortableOperatingSystem by definition. Is it safe to just change the key in this case? Do I have some other alternative (that is not reinstalling from scratch, of course) to update this system?


I have remote Ubuntu server where I deployed django project and set it (project) up by logging system. I can access server only by shell. Each log record of project save in directory like HTML file.  Opening and researching log records using vim isn't convinient for me because of it I want to open these files by browser.  Can I pass command to server to open file by my device browser.

On Windows I'm using FileZilla to upload/download files on my WordPress website, via sftp, while using ssh key.   Now Linux is my primary operating system. For the last month FileZilla client was working fine on Ubuntu 16.04, but now it crashes whenever I start it. I search on internet and found that it is happening with a lot of users. Unfortunately the FileZilla support just have one answer if we ask them:     Ubuntu 16.04 is old OS, use 17.10 or Debian latest version.   I don't want to use 17.10. So I am looking for FileZilla client alternative for Linux, which I can use as sftp client with ssh key.   I found gFTP (which is an old program and I'm not sure how much secure is it?), I also found that I can use Ubuntu file manager to connect to my server. Is that a good idea to connect to server while using Nautilus or Caja?   Actually I don't want to put my username / password in dialogue box which appear when I click File > Connect to server, instead I want to use my .ppk key. How I can add that in Nautilus or Caja?


Write a method that returns a new array by eliminating the duplicate values in the array using this header:   public static int[] eliminateDuplicates (int[] list)   all I have so far is my main, but what I wanted to do for the other method was use a for loop to check values at each space in my array and print them if they did not equal any of the other entries.  Working on it as we speak!  My output is this: The distinct numbers are: [I@4554617c  first of all this is 11 characters and at max I should be printing 10.  import java.util.Scanner;  public class EliminateDuplicates    {       public static void main(String [] Args)      {         Scanner input = new Scanner(System.in);          System.out.println("Please enter ten whole numbers: ");         int[] tenNumbers = new int[10];         for (int i=0; i<10; i++)         {            tenNumbers[i] = input.nextInt();         }                      System.out.println("The distinct numbers are: " + eliminateDuplicates(tenNumbers));       }    public static int[] eliminateDuplicates (int[] list)      {         int count = 0;          for (int i = 0; i > list.length; i++)            {             for (int j = i + 1; j < list.length; j++)               {                   if(list[i] == list[j])                     {                        list[j] = -1;                     }                                      }                  }           for (int i = 0; i < list.length; i++)            {               if(list[i] != -1)                  {                     count++;                  }             }         int[] array2 = new int[count];          int newCount = 0;         for (int i = 0; i < list.length; i++)            {               if(list[i] != -1)                  {                     array2[newCount] = list[i];                  }                              }                return array2;      }        }

In Java, arrays don't override toString(), so if you try to print one directly, you get the className + '@' + the hex of the  of the array, as defined by Object.toString():  int[] intArray = new int[] {1, 2, 3, 4, 5}; System.out.println(intArray);     // prints something like '[I@3343c8b3'   But usually, we'd actually want something more like [1, 2, 3, 4, 5]. What's the simplest way of doing that? Here are some example inputs and outputs:  // Array of primitives: int[] intArray = new int[] {1, 2, 3, 4, 5}; //output: [1, 2, 3, 4, 5]  // Array of object references: String[] strArray = new String[] {"John", "Mary", "Bob"}; //output: [John, Mary, Bob]


Whats the difference between mail and mailx?   I'm trying to troubleshoot a problem where I can send mail from server A with this echo $MESSAGE | mail -s "$SUBJECT" -r $FROM $RECIPIENTS command but executing the same command  from server B throws me this error (Both servers are RHEL)  mail: invalid option -- r Usage: mail [-iInv] [-s subject] [-c cc-addr] [-b bcc-addr] to-addr ...             [-- sendmail-options ...]        mail [-iInNv] -f [name]        mail [-iInNv] [-u user]   Now... going through mail manpages to ty to understand whats happening, man mail gives me this:  In Server A   MAILX(1)                         User Commands                        MAILX(1)  NAME        mailx - send and receive Internet mail  SYNOPSIS        mailx [-BDdEFintv~] [-s subject] [-a attachment ] [-c cc-addr] [-b bcc-addr] [-r from-addr] [-h hops]               [-A account] [-S variable[=value]] to-addr . . .        mailx [-BDdeEHiInNRv~] [-T name] [-A account] [-S variable[=value]] -f [name]        mailx [-BDdeEinNRv~] [-A account] [-S variable[=value]] [-u user]   But in server B   MAIL(1)                   BSD General Commands Manual                  MAIL(1)  NAME      mail - send and receive mail  SYNOPSIS      mail [-iInv] [-s subject] [-c cc-addr] [-b bcc-addr] to-addr... [-- sendmail-options...]      mail [-iInNv] -f [name]      mail [-iInNv] [-u user]   Now... if I try man mailx in server B I get:  MAILX(P)                   POSIX Programmerâs Manual                  MAILX(P)  PROLOG        This manual page is part of the POSIX Programmerâs Manual.  The Linux implementation of this interface may differ        (consult the corresponding Linux manual page for details of Linux behavior), or the interface may not  be  imple-        mented on Linux.  NAME        mailx - process messages  SYNOPSIS    Send Mode               mailx [-s subject] address...     Receive Mode               mailx -e                mailx [-HiNn][-F][-u user]                mailx -f[-HiNn][-F][file]   Server B has a different version of mail than Server A? Mailx and Mail are different things? I would like to use the -r option I use in server A but in server B and I don't really know whats happening.

What is this business about mail and Mail? [On a CentOS box] Why mail does not work but won't return an error either while Mail works fine with the exact same syntax?  Mail is basically a link to mail, isn't it?  Sorry, this was a mixup, apparently (for historical reasons?!) there is a mail and a Mail  Nowadays it looks like they both point to the same thing, on my machine:  sr-linux> ls -l `which mail` -rwxr-xr-x    1 root     mail        66492 Jun 24  2001 /bin/mail sr-linux> ls -l `which Mail` lrwxrwxrwx    1 root     root           14 Nov 19  2001 /usr/bin/Mail -> ../../bin/mail


I upgraded my system with a Geforce 1060 three weeks ago and after I installed the necessary  everything was running fine. But after couple of days Ubuntu(16.04 LTS 64bit) booted with the lowest possible resolution and stucked in a login loop.  I switched to the command line and went through the logs but strangely, could not find anything suspicious. I googled and went through all the suggestion you find when you google for , but nothing solved my problem.  Eventually I reinstalled the NVIDIA driver and got the GUI running again. But now, every 4 to 6 days, I have got to go through the same process, Ubuntu stucks in login and I have to install the NVIDIA driver again.  Does anybody have similar problems?

My Ubuntu is stuck in a login loop when trying to enter my desktop. When I login, the screen gets black and soon after that the login screen comes back.   I've read that the problem might be caused by an error depending on the graphics, here's my graphics card: ATI Radeon 7670M


   MyDiv1 should contain the DOM-Part of div id="a" and MyDiv2 the DOM part of id="b". So that I don't need arguments to get it, like the "this()" or "self()" function / object.  btw. a function to get the child "script" tag dom part of the javascript code inside the script tag.  EDIT: a more detailed example.  //---Global loaded script in HEAD--- function LoadAsyncScript(){     //.....     var TScriptData = xmlhttp.responseText;     var TAppID = G_UNIQUEID;     var TScript = document.createElement('script');     TScript.setAttribute('type', 'text/javascript');     TScript.innerHTML = TScriptData;     document.getElementById('div_main').insertAdjacentHTML('beforeend', '

');     var TApp = document.getElementById('_' + TAppID + '_app');     TApp.dataset['appid'] = TAppID;     TApp.appendChild(TScript);     //..... };  function CloseApp(V_AppID){     var TMyAppDiv = document.getElementById('_' + V_AppID + '_app');     TMyAppDiv.removeChild(TMyAppDiv); }; //---END OF Global loaded script in HEAD---   //Async loaded and "appendChild" script part function OnClick_Callback(){     var TMyDiv = GetMyDivOfThisFunction();     CloseApp(TMyDiv.dataget['appid']); }; //......   How I can get this work?  Thx so much.

How can I reference the script element that loaded the javascript that is currently running?  Here's the situation. I have a "master" script being loaded high in the page, first thing under the HEAD tag.  <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <script type="text/javascript" src="scripts.js"></script>   There is a script in "scripts.js" which needs to be able to do on-demand loading of other scripts. The normal method doesn't quite work for me because I need to add new scripts without referencing the HEAD tag, because the HEAD element hasn't finished rendering:  document.getElementsByTagName('head')[0].appendChild(v);   What I want to do is reference the script element that loaded the current script so that I can then append my new dynamically loaded script tags into the DOM after it.  <script type="text/javascript" src="scripts.js"></script> loaded by scripts.js--><script type="text/javascript" src="new_script1.js"></script> loaded by scripts.js --><script type="text/javascript" src="new_script2.js"></script>


That's a question I found in a french list of exercises online. I'm preparing myself for an exam, I think solving this question would really help. Thank you  The tip given by the exercise is to analyse the 7-Sylow.  I can't go any farther than $n_7=1$ or $8$ or $36$ or $288$.  Then I should analyse for each case.  I found similar questions online but couldn't quite understand the reasoning behind. They always show that the number of the Sylows can't be different than 1, which should show that the 7-Sylow is normal. The problem is that I can't understand how the usually get to show that $n_7=1$.  Would you please help me solve it or send maybe some papers about it or solved similar exercises?

The year 2016 is coming up, so it makes me wonder: how do you prove that there is no simple group of order 2016? (without invoking the 10,000 page theorem on the classification of simple groups)


What does it do and what are its applications? I am currently testing the subtitle of my beamer presentation for its string length to decide about line spacings. Could it be simpler using \empty?  If I just try \empty{word}, it seems that nothing happens. Of course, I do not expect to get a string length, but I thought that I could check the title for being empty or not.  This is how I add space by checking beamer's subtitle:  \usepackage{xstring} \StrLen{\insertsubtitle}[\subtitlelen]  \ifnum\subtitlelen>0{     \vskip0.2cm }\fi   I try to clarify my question as requested in the comments: I thought that I could check for \empty. \StrLen is easy to understand because you can output that value as text. However, this does not work for \empty so I wondered how one would use it to accomplish the same task.

I want to elaborate a command that would print differently simple differential dx and multidimensional differential d^3x. For this purpose I invented the command \dif with one optional argument. It is defined as follows:  \newcommand{\dif}[1][]{%     \def\@tmp@a{#1}%     \ifx\@tmp@a\@empty         [b1]\mathrm{d}     \else         [b2]\mathrm{d}^{#1}{} \!     \fi }   Here [b1] and [b2] are added for debugging. Branch [b1] is intended for the case when the command is called without any argument, e.g. \dif{x}. Second branch [b2] should be taken in case where, e.g., the command \dif[3]{x} is used to print d^3x. The definition, shown above, seems to work fine. However, if \@empty is substituted with \empty first branch is always ignored.   The command \empty is explained in The TeXBook, so that initially I tried it for this code and only occasionally reverted to \@empty having remembered a post at this site where it was said that the  \end command of Plain TeX is redefined to \@@end in LaTeX.     Can somebody explain what is the difference between \empty and   \@empty in LaTeX and, perhaps, suggest a better solution for my   problem?


If $f$ is a   mapping of a complete metric space $(X, d)$ into itself and $f^N$(composite $f$ for $N$ times) is a contraction   mapping for some positive integer $N$, then $f$ has precisely one fixed point. (Banach fixed point theorem is applicable)   I tried to show that $f$ is also a contraction function. I considered the sequence $x,f(x),f^2 (x),... $, but then fail pathetically in showing that it's cauchy. Please tell me if my direction is correct, any new ideas are appreciated. Thanks.

Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q<1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q<1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q <1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me?


I am looking for a full explanation for how to compute $\pi_1(S^1\times S^1 \setminus \{x_1, \ldots, x_n\})$ where each $x_i\in S^1\times S^1$. I know that this has been asked on SE before, but the answers are always brief sketches.  I think that the final answer comes out to the free product $\ast_{i=1}^{n+1}\mathbb Z_i$ where each $\mathbb Z_i=\mathbb Z$, but I have little idea as to how we arrive at this.  I would really appreciate it if someone could help me understand the full process of computing this fundamental group. It would also be great if the explanation could stick to relatively basic concepts (i.e., basic properties of free groups, Van-Kampen, etc.).  UPDATE (this is not a duplicate):   I realize that my question was too vague. Here is some more direction.   The answer should come out to a bouquet of $n+1$ circles, I believe. My intuitive ideas is as follows. The torus with one point removed deformation retracts onto $\mathbb Z \ast \mathbb Z$. Now, my justification for this is a little fuzzy, though. I can see that $S^1\times I$ deformation retracts onto $S^1$ where $I=[0,1]$, so that $\pi_1(S^1\times I)\cong \mathbb Z$, but I can't draw an explicit connection between this fact and the first deformation retraction. Any help seeing the explicit argument here is appreciated.   Moreover, I don't see in general how removing more points continues to add to the bouquet of circles. Would we need an induction argument? If so, how would it go?  I  hope this clarifies my question.

Question 5.33 from "Topology and its Applications"  by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints?


Some time ago I started self studying physics and I did quite well, despite my limited knowledge of maths (some basic algebra and geometry). Now I would like to understand calculus, but I don't know what books I should use. Clearly, first I need to revisit the basic and intermediate notions of algebra.  For this task I tried using this book:     but I found it to be somehow simplistic (problems are too easy, the theorems are not demonstrated, etc.)  What textbooks do you think are appropriate for someone with very basic knowledge of elementary algebra who wishes to self study calculus for use in physics?

I have the summer to prepare myself for a high school calculus-based physics course; however, I will have to study calculus by myself. I've completed a course in precalculus and I'm beginning to read .   I am looking for a textbook that has mathematical rigor (i.e. not memorizing formulas), yet also provides adequate foundations for a physics course. From similar stack exchange questions, my top candidates are  and . Which one of these textbooks, if any, would be the best choice for my situation?


One of our Windows 2003 servers has a fairly small C-partition, which is approaching full. I've had a quick look round using  as to where the space has gone, and a fair bit of it seems to be taken up with C:\WINDOWS\SoftwareDistribution\Download  I think this folder is used as a temporary storage for Windows Update, but as all the pending Windows updates have now been installed, is it OK to delete the contents of this folder?

I need to free space on Windows XP SP3.  Q1) What system folders can be safely delete in case I don't need to have restore points etc?  Q2) Can I delete content from C:\WINDOWS\SoftwareDistribution\Download


Every site uses a different share icon here are some examples:  Which icon is best for UX and most obvious to the user that it is a share icon?

Now days there are multiple viable share icons available, but they are mostly mobile OS specific (iOS, WP, Android...). I wonder which icon should I use for the web that most of the users would recognize?    Image source:   Has anyone done any A/B tests on how users respond to different ones?   Its a simple album page where you can get links for each album/image and then share it with a friend, nothing special, just popup on click of an icon.


I have created a function that has the capability to copy the specific data in each table cells but my problem is that it cannot be accessed after the page is loaded when I put it in $(window).load(function() {}); and if I also put it in $(document).ready(function () {});, some of the table cells is not having the result I want. I'm able to solve my problem by creating the function in both window.load and document.ready but it's not what I want since it's a bit space consuming and not a good coding practice. How can I achieve a result somewhat like this.  $(document).ready(function () {    myCreatedFunction(); }); $(window).load(function() {    myCreatedFunction(); }); //and somewhere in the page in which can be accessed by both methods  var myCreatedFunction = function(){     //some other actions }

I've recently started maintaining someone else's JavaScript code. I'm fixing bugs, adding features and also trying to tidy up the code and make it more consistent.  The previous developer used two ways of declaring functions and I can't work out if there is a reason behind it or not.  The two ways are:  var functionOne = function() {     // Some code };     function functionTwo() {     // Some code }   What are the reasons for using these two different methods and what are the pros and cons of each? Is there anything that can be done with one method that can't be done with the other?


I'm curious what is available at the 'Census Block' level,  the finest scale.  I found this dataset:        It contains HOUSING10 and POP10,  but I would like to get Vacant vs. Occupied Housing and income if possible.      In spite of the fact that it says this is answered elsewhere, it is not.   Only # of houses and total population are addressed in the other post.   The question here is...  What addtional attributes are available at the 'census block' level?  How do you obtain these additional attributes?

I would like to download 2010 Census Block Data for several towns and I can't seem to find it on American Factfinder.   Is it there? And if not, does anyone know where I could find it?


my figures in latex must have caption like this: FIGURE 1. figure name which command gives me this capital figure name?   \documentclass[11pt,a4paper]{article} \usepackage{graphicx} \usepackage{caption}     \begin{figure} \centering \caption{figure name} \end{figure}   I need to have FIGURE1.figure name instead of FIGURE 1: figure name how can i fix this?

How can I modify the caption name of a figure? For example I have  \caption{This is a figure.}   and by default the caption appears as     Figure 1: This is a figure.   However I want     Fig.1 - This is a figure.


I have a hard drive that is just for my cloud system, I need to be able to limit the folder size and prevent all the users from going over that size,

I have a directory on my system which is used for a specific reason by applications and users, but I don't want its size to be allowed to exceed 2GB, is there a way of setting up some sort of limit which just doesn't allow the file size to exceed that or any other amount I decide to set for it in the future?  When the size limit is exceeded it should undo the last change (though there should be an option to have it so that it just stops the operation and doesn't care if half a file was copied and left there) and then display a warning to the user.   I am running Ubuntu GNOME 16.10 with GNOME 3.22.


I need to run a command to uninstall forefront on some of our workstations and am having a bit of trouble formatting the command.  Here's what needs to happen (in multiple commands)  xcopy \\serverpath\Installer.exe C:\Windows Install.exe /u /s Del C:\Windows\Install.exe  I'm using psexec with a computer list but keep running into errors. This is what my command looks like as of right now.   psexec @ComputerList.txt -u domain\administrator cmd /c (xcopy "\\NetworkPath\Forefront Software\Install.exe" "C:\Windows" && Install.exe /u /s && del C:\Windows\Install.exe)   I think the spaces in the path name and quotes are messing things up...  EDIT:  Tried running it with a script with no luck. It looks like scepinstall.exe just hangs on the remote machine. I also tried using the exe that can be found in C:\Windows\ccmsetup\  Thanks for the help

I am try to use multiple command with psexec tools but i am only remotely entered in that computer but it won't be perform another command   psexec -i \\192.168.30.135 -u username -p password cmd & del abc.exe   by using this command line parameter i am able to only entered in that machine remotely but it can't perform an del command to delete that file can any one having solution about this please give me.


Which one is correct? If both were correct, in what cases they must be used?  -In the doorway -At the doorway.   For example: “He lingered in/at the doorway in hopes to cross paths with me”  Which one should I use?

I am really confused.  Which preposition is correct?     She is in/at the park.   They are in/at the park.   I am in/at the park.   Should I use in or at in these sentences?


(I am on Windows 7 by the way)  Here's the log:  ---- Minecraft Crash Report ---- // My bad.  Time: 4/18/20 10:58 AM Description: Initializing game  org.lwjgl.LWJGLException: Pixel format not accelerated     at org.lwjgl.opengl.WindowsPeerInfo.nChoosePixelFormat(Native Method)     at org.lwjgl.opengl.WindowsPeerInfo.choosePixelFormat(WindowsPeerInfo.java:52)     at org.lwjgl.opengl.WindowsDisplay.createWindow(WindowsDisplay.java:247)     at org.lwjgl.opengl.Display.createWindow(Display.java:306)     at org.lwjgl.opengl.Display.create(Display.java:848)     at org.lwjgl.opengl.Display.create(Display.java:757)     at org.lwjgl.opengl.Display.create(Display.java:739)     at bib.at(SourceFile:635)     at bib.aq(SourceFile:458)     at bib.a(SourceFile:404)     at net.minecraft.client.main.Main.main(SourceFile:123)   A detailed walkthrough of the error, its code path and all known details is as follows: ---------------------------------------------------------------------------------------  -- Head -- Thread: Client thread Stacktrace:     at org.lwjgl.opengl.WindowsPeerInfo.nChoosePixelFormat(Native Method)     at org.lwjgl.opengl.WindowsPeerInfo.choosePixelFormat(WindowsPeerInfo.java:52)     at org.lwjgl.opengl.WindowsDisplay.createWindow(WindowsDisplay.java:247)     at org.lwjgl.opengl.Display.createWindow(Display.java:306)     at org.lwjgl.opengl.Display.create(Display.java:848)     at org.lwjgl.opengl.Display.create(Display.java:757)     at org.lwjgl.opengl.Display.create(Display.java:739)     at bib.at(SourceFile:635)     at bib.aq(SourceFile:458)  -- Initialization -- Details: Stacktrace:     at bib.a(SourceFile:404)     at net.minecraft.client.main.Main.main(SourceFile:123)  -- System Details -- Details:     Minecraft Version: 1.12.2     Operating System: Windows 7 (amd64) version 6.1     Java Version: 1.8.0_51, Oracle Corporation     Java VM Version: Java HotSpot(TM) 64-Bit Server VM (mixed mode), Oracle Corporation     Memory: 43481408 bytes (41 MB) / 134217728 bytes (128 MB) up to 2147483648 bytes (2048 MB)     JVM Flags: 8 total; -XX:HeapDumpPath=MojangTricksIntelDriversForPerformance_javaw.exe_minecraft.exe.heapdump -Xmx2G -XX:+UnlockExperimentalVMOptions -XX:+UseG1GC -XX:G1NewSizePercent=20 -XX:G1ReservePercent=20 -XX:MaxGCPauseMillis=50 -XX:G1HeapRegionSize=32M     IntCache: cache: 0, tcache: 0, allocated: 0, tallocated: 0     Launched Version: 1.12.2     LWJGL: 2.9.4     OpenGL: ~~ERROR~~ RuntimeException: No OpenGL context found in the current thread.     GL Caps:      Using VBOs: Yes     Is Modded: Probably not. Jar signature remains and client brand is untouched.     Type: Client (map_client.txt)     Resource Packs:      Current Language: ~~ERROR~~ NullPointerException: null     Profiler Position: N/A (disabled)     CPU: <unknown>

Similar to  on the bugs in Minecraft, I would like a complete list of what error codes exist in Minecraft and how they occur, as well as how they can be resolved.


I recently inherited a poorly-managed SQL server.  One of the databases is in rough shape. Here are the details:   The .mdf file is a modest 369MB. The .ldf file is a bloated 29.7GB. The recovery model is set to FULL. The data file is backed up nightly without issue, and is 365MB. The transaction log has NEVER been backed up.   I've restored the nightly backup to the same SQL server and poked around in it to make sure data looks to be intact. When I restore that nightly .bak backup file it recreates the .mdf and .ldf files at 369MB and 29.7GB, respectively.  Being decent with SQL, but not what I would call an expert by any means, I have a few follow up questions on this situation that I am hoping to get feedback on:   How is a 369MB .bak file restoring both a 369MB .mdf file and a 29.7GB .ldf file? Any suggestions on what to do regarding the recovery model and bloated .ldf file? We are only concerned about doing backups once nightly - we don't need to do differential backups at this time - but it's something I will tackle later.

This one seems to be a common question in most forums and all over the web, it is asked here in many formats that typically sound like this:  In SQL Server -  What are some reasons the transaction log grows so large? Why is my log file so big? What are some ways to prevent this problem from occurring? What do I do when I get myself on track with the underlying cause and want to put my transaction log file to a healthy size?


I have a blend file that I've worked with for a long time, testing and experimenting with things like rigify and almost everything else. I'm trying to clean it up. By removing a lot of mocap actions, it's almost half in size, from 8 to 4.7 Mb.  I though the mesh might use space, but when I delete all my meshes I went from 4.7 Mb to 4.1Mb. The only important thing left now is an armature and it's drivers.   I ended up deleting everything I could, and it's still 2.3 Mb, compared to about 500k for a new empty file.  So what's using all this space? Is there a way to list all data blocks and how much disk space it's using?  The reason I want to clean this up is that I'm preparing a group for export, and don't want slow performance due to crap in the file.

I made a huge project, wanted to reduce file size from 1.3GB to below 1GB. Compressing did not help. I removed objects from scene. Also textures. I then realized, the file size even was the same when I delete ALL objects. Why?  Download of sample:   Before deleting all: 1.2GB, after: 1.1GB.


I (the room owner) was going to remove stars from an old chat message that no longer appeared in the sidebar. However, I can't find any moderation options either in the full list of starred messages or in the transcript. I couldn't load the chat back to that point either.  How can I moderate an old message? Is there some way to get the same moderation options in the transcript, instead of only in the real-time chat section?

Currently, the only way to cancel stars in a chat transcript is when the chat-message is visible in the chatroom's sidebar.    However, in a fast-moving, multi-star chatroom, the window of opportunity to cancel can be short, and rapid changes in starring can quickly make this impossible to do in a timely way.  We use stars as markers in moderator chat events (where the Community managers disseminate important information to the moderator teams) as well as for town halls during elections.  Feature request: Allow mods/room owners to cancel stars from the chat message menu box, including in the transcript.


I use three different computers and sign in to each one with the same Google Chrome account, but the bookmarks do not seem to be updating.  For example I add a bookmark when signed in on one pc, then open chrome on one of the other pc and it is not there, is there a way to force it to update the bookmarks and histrory etc.

I've been using Chrome's  which seems to work well.  Is there a way to force Chrome to do the sync as opposed to waiting for the alloted 'sync time'?    Update: Seems to sync when I restart Chrome, so that works.  Would be nice to have a "sync button" equivalent..


Suppose $R$ is a commutative ring with $1$. Further we have an isomorphism $R^m \cong R^n$. Then does that imply $m=n$?  If I consider $R=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots $, then I think $R^m=R^n$ as both $R^m$ and $R^n$ has countably infinite copies of $\mathbb{Z}$. Hence $R^m \cong R^n$ for any natural numbers $m \neq n$?

If $R$ is a commutative ring with $1\ne 0$ and $R^m \cong R^n$ as direct sums of $R$-modules, then $m=n$.   There is a hint to this problem (I don't know how to use it): Reduce this to the case of a field.  I noticed that there was . But as a beginner of commutative algebra, I don't know if my question is closely related to that one. Thanks in advance!


I am on Minecraft in creative mode on xbox one and am lost in nether. I have tried for hours to find my portal and can't. All the responses I see tell you how to get out if you are on a computer and not in xbox one. Anybody have a way to help me get out?

No, actually I didn't, but that was a close call and I'm really curious how would one go about that one.  On my third or fourth visit to the Nether, I still didn't realize the risk, and didn't place marks that would help me get back to the origin (currently I like to use flint and iron to create big flaming arrows towards the exit, visible from afar.) I wandered past one hill, and then I couldn't locate the place I came from. I wisened up and started placing marks in form of torches, towers of netherrack with glowstones on top etc. The bunch of wood I had in my stash has proven invaluable as I ran out of steaks, and had to craft a bowl and start making mushroom soup. I realized the fastest way back would be to simply die, but since I had no clue where I was, that would mean losing my quite precious equipment, which cost me a lot of time and effort to gather and enchant, for that trip specifically. I also knew I had traveled rather far from the portal before starting placing marks, so the chances of finding it again by going in any random direction were rather slim.  By sheer luck I found one of the caverns I had explored before, and found my way back, but that was really a close call and I'd prefer to know any hints what to do in Nether if you're already lost (no Obsidian to build a new portal, not willing to die and forgot the direction to the exit.)


I would like to measure the CPU usage within a Java application, for example the   System.out.println("...")    of HelloWorld application. Any idea? I do not want to measure the CPU usage of the entire application, only the methods within the application.  Is there a way to correlate the the total execution time of the application with the CPU usage of the entire application?

For a particular segment of Java code, I'd like to measure:   Execution time (most likely thread execution time) Memory usage CPU load (specifically attributable to the code segment)   I'm a relative Java novice and am not familiar with how this might be achieved. I've been referred to , however I'm not sure how that might be used, and JMX looks a bit 'heavy' for what I'm looking to do.  Ideally I'd like some measurement class that can be told what I would like to measure, with the option of calling a start() method prior to a code segment and a stop() method after. Relevant metrics would be logged to a file I specify.  For example:  import com.example.metricLogger;  metricLogger logger = new metricLogger();  logger.setLogPath(pathToLogFile); logger.monitor(executionTime); logger.monitor(memoryUsage); logger.monitor(cpuLoad);  logger.start();  /* Code to be measured */  logger.stop();   Is there any standard/common/conventional way of achieving this in Java?  Such measurements are for one-off performance comparisons, and so I'm not looking for any in-production long-term monitoring processes.  I'm more than happy to be referred to tutorials or external examples and don't expect a full answer here. That said, if anything as simple as the above can be achieved a realistic example would go down really well.


I have fragment A:  @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {      view = inflater.inflate(R.layout.fragment_main, container, false);      CustomKeyboard customKeyboard = new CustomKeyBoard(getActivity());       etAge = (EditText) getActivity().findViewById(R.id.etAge);      customKeyboard.actionEt(etAge);      return view; }   And I have class B:      public class CustomKeyboard {  private Context context;      public CustomKeyboard (Context context) {         this.context = context;     }      private void hideKeyboard(View view) {         InputMethodManager inputMethodManager = (InputMethodManager) context.getSystemService(Activity.INPUT_METHOD_SERVICE);         inputMethodManager.hideSoftInputFromWindow(view.getWindowToken(), 0);     }                 public void actionEt(Edittext edittext){                     editText.setOnFocusChangeListener((view, hasFocus) -> {                        if(!hasFocus) {                              hideKeyboard(view);                         } else {                           //........                         }                     });                 }     }   The Exception:                                                   Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'void android.widget.EditText.setOnFocusChangeListener(android.view.View$OnFocusChangeListener)' on a null object reference  What is the reason of the problem? Hope for some help.  EDIT: NOW THE APP RUNS. But the Edittexts dont hide.

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?


On my flagging history page, I'm told that I have X moderator attention flags, of which Y were marked helpful, Z were disputed, and W were declined, and the same for spam, offensive, and comment flags.  My  is to allow viewing of flags by type and how they were handled, similar to how the activity or votes tab of the user profile page can be broken down into sections.  I think this is a good idea, because if people can review their flags easily, users whose flags are declined often will find it easier to figure out why they were declined, so that they can learn from the experience.

Here is my flagging summary on SU:   Your flagging history   32 posts flagged as requiring moderator attention    2 are waiting for review   3 were marked valid   1 were marked invalid   (older flags were not recorded valid/invalid)    2 posts marked as spam    It is too complicated to look for invalid flags. Would it be possible to make filters on the top of the page (like on favorites for example) with:  By date, By type (off-topic, spam, dupes), By status (accepted, pending, rejected)  It would make it easier for reviewers to check their flagging and eventually know what to improve on.  Or is there another easy way to audit this?


I am trying to update our SQL database from a JFrame panel in Netbeans. Currently, I am trying to use a button in from the "registernewStudent" panel, where the user can assing 4 text fields, where the first (Student ID) and the last (room) should be an integer and the middle 2 are strings. However, I am getting an errormessage saying "Exception in thread "AWT-EventQueue-0" java.lang.NullPointerException"  Below are the source code for the button.  This is our first real project, so we'd appreciate any help we could get.   private void btnNyREActionPerformed(java.awt.event.ActionEvent evt) {                                             try {         String talStrang1 = txtElevID.getText();         int tal1 = Integer.parseInt(talStrang1);          String Strang2 = txtElevFN.getText();         String Strang3 = txtElevEN.getText();         String talStrang4 = txtElevSS.getText();          int tal2 = Integer.parseInt(talStrang4);          database.insert("insert into Elev values( tal1, Strang2, Strang3,tal2)");      } catch (InfException ex) {         Logger.getLogger(NyRegElev.class.getName()).log(Level.SEVERE, null, ex);     }  }


Why is it that only the particle with negative energy is absorbed into the black hole on the event horizon during Hawking Radiation? What discriminates between the particle or antiparticle?

I was reading Brian Greene's "Hidden Reality" and came to the part about Hawking Radiation. Quantum jitters that occur near the event horizon of a black hole, which create both positive-energy particles and negative-energy particles, produce the so-called Hawking radiation. However, I do not understand why only the negative-energy particles are being absorbed into the black hole, while the positive-energy particles shoot outward. Shouldn't there be 50/50 chance that each type of particle is being absorbed by the black hole? Also, the book mentions that a negative-energy particle would appear to an observer inside the black hole as positive. Why?


Do heroes of different classes get affected by Lucio's Soundwave in different ways? Will a tank be pushed back less than a lightweight character?

When playing against Lúcio, sometimes it seems I only get knocked back a little bit, other times it seem like I go flying. I haven't noticed much difference when playing, but I don't really notice a difference in how far I knock them away from me, however I try not to get too close to people in order to stay safe. Does the distance away from the target effect the distance I knock the target back when using alternate fire with sonic amplifier?


How to install kernel 4.14 on Kubuntu 19.10?   Why: Because my hardware works with that kernel.  What I tried:  dan@dan-pc:~$ apt-cache search linux-generic  linux-generic - Complete Generic Linux kernel and headers linux-image-unsigned-5.3.0-18-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-18-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-18-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-19-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-22-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-23-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-24-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-26-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-image-unsigned-5.3.0-29-generic - Linux kernel image for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-19-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-22-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-23-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-24-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-26-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-5.3.0-29-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-19-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-22-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-23-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-24-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-26-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP linux-modules-extra-5.3.0-29-generic - Linux kernel extra modules for version 5.3.0 on 64 bit x86 SMP   My problem is with an Amd r9 390 xfx to enable vulkan.  This is what I did on Manjaro Kde to make it work  Solved it by using this  1  Manjaro kde 18  Works for now on kerenl 4.14.170-1 by adding this using nano with sudo: sudo nano /etc/modprobe.d/amdgpu.conf  Put this in the file  options amdgpu si_support=0 options amdgpu cik_support=1  sudo nano /etc/modprobe.d/radeon.conf Put this in the file  options radeon si_support=0 options radeon cik_support=0  Rebuild the initramfs: mkinitcpio -P  But if i use the kernel 5.something  My screen will not boot.  After a lot of trial and error made my gpu recognize amdgpu on the most recent kernel 5.3.0-40-generic using this:  In /etc/default/grub  I changed this line  GRUB_CMDLINE_LINUX_DEFAULT="quiet splash"   Too  GRUB_CMDLINE_LINUX_DEFAULT="radeon.cik_support=0 radeon.si_support=0 amdgpu.si_support=1 amdgpu.cik_support=1 amdgpu.dc=1 amdgpu.dpm=1"   Also I have a script that forces power to gpu  !bin/bash  xset -dpms  I called amd-power.sh and i made it to autostart on system boot.  Make sure that is the first thing in autostart.  It should work for me it's working.  Finally amd r9 390 can be used with vulkan on ubuntu 19.10.  Thnaks for all the help guys  Thanks

Currently I'm running Linux-3.0 and I want to update it to Linux-3.3.1 the latest stable kernel release as mentioned at . Can I update to 3.3.1 in Ubuntu without any risk of crashes? I'm updating my kernel regularly as provided by the Update Manager..Currently I have Linux-3.0.0.17..Can I update?  NOTE:  The latest kernel version any user is supposed to use in Ubuntu is updated automatically via the Update Manager, so no action is normally required by user regarding kernel upgrades. What the question author is referring to is mainline kernel, see:


According to University Physics  Every possible motion of a rigid body can be represented as a combination of translational motion of cm and rotation about centre of mass. This is true even if centre of mass is accelerating  How can the above statement be proved. If a rigourous proof can be given thats good but only an intuitive explanation about this fact and how is it true for ANY motion and even if centre of mass is not having constant velocity would be sufficient ?

in its strong form says:     The most general rigid body displacement can be produced by a translation along a line (called its screw axis) followed (or preceded) by a rotation about that same line.   Now,  simply says that any rigid body displacement can be decomposed into a rotation plus translation. This is easy to visualize. But what Chasles' Theorem says is something much stronger.   Unfortunately, I am just not able to visualize it. Perhaps, I am comprehending it wrong. I mean how is it possible to have the axis of rotation and translation the same (or parallel) for the most general displacement.   I mean, think of this case:     A body is given a finite rotation about the X-axis and then a finite translation about the Z-axis. How can we find that "screw" axis along which both of them can be described?


Three proper ladies are traveling on a train. Each turns red within one second if they become aware of dirt on their face. They are too proper to tell the other if they have dirt, and there are no mirrors. The conductor tells them "At least one of you has dirt on your face". Within 3 seconds, all three ladies turn red.  How many have dirt on their face? What is the reasoning of each of the ladies for believing they have dirt?  Assumptions: The conductor is telling the truth...the solution is either 1, 2 or 3 ladies have dirt on their face. (The answer can't be zero).

Text below copied from       The Blue-Eyed Islander problem is one of my favorites. You can read   about it  on Terry Tao's website, along with some discussion.   I'll copy the problem here as well.      There is an island upon which a tribe resides. The tribe consists of   1000 people, with various eye colours. Yet, their religion forbids   them to know their own eye color, or even to discuss the topic; thus,   each resident can (and does) see the eye colors of all other   residents, but has no way of discovering his or her own (there are no   reflective surfaces). If a tribesperson does discover his or her own   eye color, then their religion compels them to commit ritual suicide   at noon the following day in the village square for all to witness.   All the tribespeople are highly logical and devout, and they all know   that each other is also highly logical and devout (and they all know   that they all know that each other is highly logical and devout, and   so forth).      [For the purposes of this logic puzzle, "highly logical" means that   any conclusion that can logically deduced from the information and   observations available to an islander, will automatically be known to   that islander.]      Of the 1000 islanders, it turns out that 100 of them have blue eyes   and 900 of them have brown eyes, although the islanders are not   initially aware of these statistics (each of them can of course only   see 999 of the 1000 tribespeople).      One day, a blue-eyed foreigner visits to the island and wins the   complete trust of the tribe.      One evening, he addresses the entire tribe to thank them for their   hospitality.      However, not knowing the customs, the foreigner makes the mistake of   mentioning eye color in his address, remarking “how unusual it is to   see another blue-eyed person like myself in this region of the world”.      What effect, if anything, does this faux pas have on the tribe?   The possible options are   Argument 1. The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).   Argument 2. 100 days after the address, all the blue eyed people commit suicide. This is proven as a special case of  Proposition. Suppose that the tribe had $n$ blue-eyed people for some positive integer $n$. Then $n$ days after the traveller’s address, all $n$ blue-eyed people commit suicide.  Proof: We induct on $n$. When $n=1$, the single blue-eyed person realizes that the traveler is referring to him or her, and thus commits suicide on the next day. Now suppose inductively that $n$ is larger than $1$. Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be $n-1$ blue-eyed people on this island, and so they will all commit suicide $n-1$ days after the traveler’s address”. But when $n-1$ days pass, none of the blue-eyed people do so (because at that stage they have no evidence that they themselves are blue-eyed). After nobody commits suicide on the $(n-1)^{st}$ day, each of the blue eyed people then realizes that they themselves must have blue eyes, and will then commit suicide on the $n^{th}$ day.     It seems like no-one has found a suitable answer to this puzzle, which seems to be, "which argument is valid?"   My question is...  Is there no solution to this puzzle?


At the Battle of Scarif, a pilot shouts out      More fighters! Form up at six-five!                                    as imperial fighters are dispatched.  What is 65? Is that a formation or some sort of coordinates?

K-2SO beats up the troopers easily. Up in the sky, the bombers are preparing to attack. Just as they start the attack run, TIE fighters come in from the shield and swarm the rebel fleet.  JON VANDER (GOLD LEADER): Red Leader, this is Gold Leader, we’re starting our attack run on the  shield gate. Keep it tight and watch out for those towers. GOLD PILOT: More fighters. Form up, at 65.  What does "65" mean?


To prove $$\sin{A}+\sin{B}-\cos{C}\le\frac32$$ Given $A+B+C=\pi$ and $A,B,C>0$  I have managed to convert LHS to $$1-4\cos{\frac C2}\sin{\frac{A+C-B}2}\cos{\frac{A-B-C}2}$$ but that clearly isn't very helpful  One other conversion was $$\sin A+\sin B+\cos A\cos B-\sin A\sin B$$ but not sure how to proceed.  Any hints will be appreciated

If $A,B,C$ are the angle of a triangle, then show that $\sin A+\sin B-\cos C\le \dfrac32$ I tried substituting $C=180^\circ-(A+B)$ and got stuck. I also tried using the formula $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$ without any progress. Please help!


An epimorphism is called split if it has a section (a right inverse).  An epimorphism is called effective if it has a kernel pair and is the coequalizer of its kernel pair.  The ncatlab implies  that every split epimorphism is effective, but I can't find a proof anywhere, nor have I had success in coming up with a proof or counterexample.  There are two parts to this. First, is it true that every split epimorphism has a kernel pair? I haven't managed to come up with an obvious construction of its kernel pair; what strikes me as the natural choice doesn't seem to work. (My inclination would be to try, if $f$ is a split epi with section $s$, the pair $(s \circ f, 1)$, because I think $f$ is definitely the coequalizer of this - but that doesn't seem to work.)  Second: if every split epimorphism does have a kernel pair, then is it further true that this kernel pair must have a coequalizer?

In , during the second part of the proof of Proposition 1.8, one finds the statement that (background at the end) the morphism $U\times_ST\rightarrow U$ is an effective epimorphism, and the reason given is that because it has a section over $U$.    My Question:     Why is the morphism $U\times_ST\rightarrow U$ an effective epimorphism if it has a section?   I know why it has a section; it arises from the two $S$-morphisms $U\rightarrow U$ and $U\rightarrow T$. But I don't know why this implies that it is an effective epimorphism. While it is easy to show that it is epimorphic by the existence of a section, I found no clues as to how to show the effectivity. I think this should come from some simple arguments, but cannot figure it out.  Thanks in advance for help or references.    Some backgrounds:  We are consider $U\overset{v}\rightarrow T\overset{u}{\rightarrow}S$ where $uv$ is a universal effective epimorphism and $u$ is quarrable, which means the fibre product with every other morphism $S'\rightarrow S$ exists. And we want to show that $u$ is a universal effective epimorphism.  And a morphism $u:U\rightarrow V$ is an effective epimorphism if the diagram $$V\leftarrow U\leftleftarrows U\times_VU$$ is exact.


I have the following question, from Isaacs' Algebra book.  Suppose $F\le E$ is a finite-degree normal field extension, and that $K$ and $L$ are intermediate subfields (between $F$ and $E$).  If $E$ is separable over $K$ and $L$, prove $E$ is separable over $K\cap L$.  I've tried to prove this by recognizing that $E$ is Galois over $K$ and $L$, and then showing $E$ is Galois over $K\cap L$.  Here is my attempt:  Let $M$ be the fixed field of $Gal(E/K\cap L)$, so that $K\cap L\subset M$. Now every element of $Gal(E/K)$ fixes $K\cap L$, and the same is true of $Gal(E/L)$. So $Gal(E/K)$ and $Gal(E/L)$ are subgroups of $Gal(E/K\cap L)$. By the fundamental theorem of Galois theory, $K$ (as the fixed field of the subgroup of $Gal(E/K)$) and $L$ (as the fixed field of the subgroup of $Gal(E/L)$) are fields containing $M$.  Then $M\subset K\cap L$, so $M=K\cap L$ and thus $E$ is Galois over $K\cap L$.  Can someone let me know if this proof works? Or if there is a different way to solve this exercise?

Let $E\mathop{/}F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $F\subseteq K\subseteq E$ and $F\subseteq L\subseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $K\cap L$.   So I was thinking that if $f=\min_{K\cap L}(\alpha)$. Then $\min_{K}(\alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.   Thanks!


what is $I(2n) =\int \tan^{2n}x\mathrm{d}x$  Thanks

(a) Use integration by parts to express $$I_{2n} = \int\tan^{2n}x dx$$  in terms of $I_{2n − 2}$  (b) Hence show that $$ \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{1}{\tan^{4}x}dx =\frac{\pi}{4} − \cfrac{2}{3}$$  For part (a), I've tried turning $\tan^{2n}x$ into $\tan^{2n-2}x\tan^{2}x$ and then using integration by parts, but the answer is messy and i can't simplify it in terms of $I_{2n -2}$.  For part (b), I don't know where to start from after using the solution below for part (a).  Thanks


i am just trying to find a way to run an extra command when opening an app (Firefox), but i am failing miserably...   The original question is .  I tried to add my command after the app command as in the following line:  Exec=firefox %u & /bin/monitor_process.sh firefox   But this results in the monitor_process.sh being opened by FF, showing an "open file with..." dialogue box.  I tried ; & && but to no avail...  So, any idea how to handle multiple commands in .desktop files?  Also,  Said:     that also won't work. Exec= has to have exactly one statement, no ; is allowed.

I'm running Gnome on Ubuntu 15.4.  I want to execute two or more actions at once, but let's try with two:   sudo pm-suspend gnome-screensaver-command -l   pm-suspend has been added and allowed to be executed without providing passwords by specifying:  %sudo ALL = (root) NOPASSWD: /usr/sbin/pm-suspend`   , and  have no references if having multiple Exec is possible, but unfortunately:  [Desktop Entry] _Name=Sleep and logout _GenericName=Sleep and logout  #Exec=gnome-screensaver-command -l & sudo pm-suspend # does not work  #Exec=gnome-screensaver-command -l #Exec=sudo pm-suspend                                # does not work either  Terminal=false Type=Application Categories=System;Security;   I'm open to any other suggestions how to achieve given functionality except:   make a script and run it as sudo or similar - I consider solutions basing on this insecure, let's not delve into that matter   use gnome power switch or whatever that is called (top right corner) - mine does not have an option to suspend   also, while I'd be happy to have that what is stated above, I'd be even more happy to have an answers that say how to launch multiple different commands (without script/compiling own binary etc), in case I'd like to bundle and automate some other actions.


I am trying to delete a file on an Ubuntu 12.04 system for which I have root access. However, I am getting this error:  root@CRST-Web:/home/real_property/html# rm -r -f results.php rm: cannot remove `results.php': Read-only file system   df /home/real_property/html output:  Filesystem                     1K-blocks     Used Available Use% Mounted on /dev/mapper/CRST--Web--vg-root 121407596 65468436  49748876  57% /   I tried to remount the partition rw but it failed:  # mount / -o remount,rw mount: cannot remount block device /dev/mapper/CRST--Web--vg-root read-write, is write-protected

chmod 777  -R  /mnt   rm -rf /mnt/*    rm: cannot remove 'omitted': Read-only file system   rm: cannot remove 'omitted': Read-only file system     Please show me how I can do this?


Can someone help me to prove there are infinitely many solutions to the Diophantine equation: $$x^2 − 3y^2 = 1$$ using the method of ascent. We can do this by showing how, given one solution $(u, v)$, we can compute another solution $(w, z)$ that is larger is some suitable sense. Then my proof will involve finding a pair of formulas, something like: $w = x + y$ and $z = x − y$. However I tried these formulas and they don't work. So I asked my teacher and she said that there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3.

I have seen the Pell's equation wiki page but I need to prove this from scratch without mentioning any formula. I have also seen multiple answers on this site but the answers tend to skip over and assume formulas.  This is what people do -> $x^2 - 3y^2 = 1$ has solution 1, 0.  Next they say we "observe" that $$(x',y') = (2x + 3y, x + 2y)$$ is also a solution.  What I'm asking is how they get this value. Can someone help?


Suppose I'm writing a foo class which at some point calls  LoadClassWithOptions{book}   I would like to be able to pass either oneside or twoside to the book class based on various other considerations. Is there a way to explicitly set the option prior to calling LoadClassWithOptions or do I have to use LoadClass and pass the appropriate option explicitly?

Can anyone tell me the difference between those two commands?  Edit: Thanks for the answers, but I didn't get the clue yet. Here's an example:  test.tex  \documentclass[a4paper,oneside]{mytestclass} ...   mytestclass.cls  \NeedsTeXFormat{LaTeX2e}  \ProvidesClass{testclass}[2011/01/30 v0.1 testclass] \LoadClass{book}   or  \NeedsTeXFormat{LaTeX2e}  \ProvidesClass{testclass}[2011/01/30 v0.1 testclass] \LoadClassWithOptions{book}   In both examples the options a4paper and oneside are passed to book. So, I don't see any difference in the result.


What exactly is the relationship between $\star$-products in phase-space quantum mechanics, i.e.  $$ (f \star g) (x,p) = f(x,p) e^{\frac{i \hbar}{2} ( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \cdot \overrightarrow{\partial_x})} g(x,p) ~,$$ and $\star$-products in $2n$-dimensional non-commutative spaces, i.e.,  $$ (f \star g) (X) = \left . e^{\frac{i\hbar}{2} \theta^{\mu\nu} \frac{\partial}{\partial X^\mu}\frac{\partial}{\partial Y^\nu}} f(X) g(Y)\right \vert_{Y \to X}  ~~,$$ where $\theta^{\mu\nu}$ is a constant anti-symmetric tensor?  To be a bit more specific, is there a way to reinterpret/reformulate the phase-space non-commutativity as spatial non-commutative geometry? Or something of that sort.

Can someone please explain me what is a ?  Also, how does putting $$X_a(\psi) ~=~ x_a\star\psi$$ realise $$[X_a,X_b]=i\theta_{ab}{\bf 1}?$$   Ref:


Suppose i have a string ci='SKDKFJKFLKG' and I want to store it in a matrix.I am using the following code to store it in a matrix:  def chunks(l, n):      for i in range(0, len(l), n):         yield l[i:i+n] pprint.pprint(list(chunks(ci, 1)))   Output:  ['S', 'K', 'D', 'K', 'F', 'J', 'K', 'F', 'L', 'K', 'G']   What should i do if i want only 4 characters in each row of the matrix?

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


I have a below Java Class.  public class MyClass {  private final String id;  public MyClass(final String id) {     this.id= id; }  public String getId() {     return id; }  @Override public int hashCode() {     final int prime = 31;     int result = 1;     result = prime * result             + ((id == null) ? 0 : id.hashCode());     return result; }  @Override public boolean equals(final Object obj) {     if (obj == null)         return false;     if (getClass() != obj.getClass())         return false;     final MyClass other = (MyClass) obj;     if (id== null) {         if (other.getId != null)             return false;     } else if (!id.equals(other.getId))         return false;     return true; }  }   Now I am creating 2 objects of this class.  MyClass obj1 = new Object("ABCD"); MyClass obj2 = new Object("ABCD");   As per my class definition these 2 objects are equal. But in heap I believe 2 different objects would be created.  If I want to prove although both the objects are equal but still they are different how can I do it?

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


I am trying to join attributes by location using processing (I have many so I need to do them by batch)   processing.runalg('qgis:joinattributesbylocation', firstlayer, secondlayer,u'equals',0,0,'',1,layerOutput)   The first future are integers then the outputs became all negative numbers. i.e. 4120034510 becomes -174932786  Could some one help me how to avoid this?  ( may be an easiest way to change first column to string , or better ideas)

I run the function "Join attributes by location" in Qgis 2.2 between a point vector layer and a polygon vector layer. When I open the resulting table of contents of point layer there is a column of numbers (fields from 10 to 13) that fail, because there is the same value in every row.   In "input" a list of unique numeric values (example: 100000000001, 100000000002, 100000000003...), in output results the same value in every row (example: -12345678, -12345678, -12345678...).


I'm using the latest iTunes 12.7 and iPhone software (iOS 10.3.3). Any suggestions?

I just updated itunes to 12.7.0.166 before making a backup of my iphone 5. To my surprise I can't find an apps section when configuring the phone in itunes, nor can I find the apps section in the itunes library itself.   I'm confused. Have they removed that section in the update? If so, then that would make my backups useless -- there is no way to back up my apps. Transferring purchases from the phone does nothing, it just finishes immediately.  I could not find an itunes changelog.   Do I need to downgrade itunes to make a decent restore of my phone nowadays?  Update: To clarify, I have apps that are no longer available for download. There is no other option than itunes.


I'm trying to install the proprietary AMD drivers through Software Sources. I tick the box for fglrx-updates, and it says applying changes and fills 10% of a progress bar. It then sites there for hours (I left the computer unattended).   Trying to get this process to finish, I found that the Ubuntu software center progress tab display a message reading:      Applying Changes: Waiting for required medium   What in the world does that mean and how can I get it to move on?

I have a laptop with ATI Radeon 6770M HD Hybrid graphics card. In Ubuntu 12.04, I installed the fglrx driver through "additional drivers" and it worked. (I can even switch GPUs). But in the new Ubuntu 12.10, after installing, Unity won't load.   Only the mouse and the wallpaper. If I initialize the settings sudo aticonfig --initial then after rebooting it gives a warning saying I'm in low graphics mode! How to fix this?  Earlier I used software source to install the drivers. But when using the terminal, I got this warning :   update-alternatives: warning: forcing reinstallation of alternative /usr/lib/fglrx/ld.so.conf because link group x86_64-linux-gnu_gl_conf is broken   Update : Filed a bug report in launchpad :


I don't like the Adobe download manager installing. I want to just download the program and install it. I have several systems so is it the same for each one?

I can't find Flash Player download link which will give me the full EXE/MSI file, not the tiny downloader which downloads data from the internet.  What's the proper download site for this?