My question:   It has two answers, but neither one answers my question. (Note, also, that both answers are from the same user) Is there any way I can get more people to see other than starting a bounty? Although I've edited it, I have not seen additional activity, nor have I seen it on the front page.

I asked a question almost as soon as I got my account. I got a few decent answers, but none of them helped solve my problem. The question has long since become buried by newer questions, and there's still no accepted answer. Unfortunately, because there are quite a few answers with votes, to a casual observer it probably looks like the question was answered and I just neglected to mark an answer as accepted, and the system does not consider it "unanswered".  Or, I asked a question that was a bit more difficult to answer. After waiting for some time, I received no answers, little or no comments, and only a few views. The question does show up in the "unanswered" list, but no one has answered my question.  In either case, what's the protocol for me to try to get this question answered?  I see several options:   Post the question a second time, possibly linking to the original, to get attention for the original question. This is encouraged on sites like Reddit.  However, it's probably not appropriate here, since we're encouraged to check for duplicates and not submit a question that's already been asked by someone else, so this probably falls under that heading. Wait for some kind of wiki-like talk page to be implemented, then post a plea on the talk page for input and hope that activity on the talk page bumps up the question. Edit the question, but do nothing further on the Stack Exchange site. If I edit my post to say that I haven't accepted any of the current answers but am looking for more, then the question might get answered eventually by people browsing through old questions to find something they can answer. Post the question on other sites, but link to my original question on Stack Exchange. People will either answer here, or they'll answer on the other site and I can post the accepted answer here myself.  This might be considered spamming, so it might be better to do this without the link.      For more information, see the "" page in the .


Nothing is so happy as to hear from your family. Nothing is so happy as hearing from your family.  Which is more natural for native speakers?

As a native speaker of English, the gerund version of this sentence sounds better:  infinitive:     When used together in chains,   extension methods are an unprecedented   tool to produce extremely concise   code.   gerund:     When used together in chains,   extension methods are an unprecedented   tool for producing extremely concise   code.   But how can I explain to someone learning English how to decide in situations like these whether to use the infinitive or the gerund?


Workflow: Passing JSON to angular scope and using ng-repeat in the View to display it.  Using following function to show dialog on click of elements of a particular class.   $(function () {         $("#dialog").dialog({             autoOpen: false,             show: {                 effect: "blind",                 duration: 1000             },             hide: {                 effect: "explode",                 duration: 1000             }         });          $(".opener").on("click", function () {             $("#dialog").dialog("open");         });     });      $(function () {         $("#imagePopup").dialog({             autoOpen: false,             show: {                 effect: "blind",                 duration: 1000             },             hide: {                 effect: "explode",                 duration: 1000             }         });   Want it to get triggered on click of a particular class.
             @*                 {{ item.CreatorID}}             	*@                              {{item.ItemName}}             	                              {{item.CreatorName}}             	                              {{item.Length}}             	                              {{item.Width}}             	                              {{item.Price}}             	                              {{item.ItemDescription}}             	                              {{item.Quantity}}

     <button class="opener">Opener</button>     <button class="opener">Opener11</button>

Sample text

   Here when I am clicking on Button Opener& Opener11 popup appears but not on the click of
 which has the same class(inside ng-repeat).   Can anyone tell me if i'm missing anything?

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


Find $\gcd(a^{2^n}+1,a^{2^m}+1)$ where $m,n$ are two distinct positive integers.   Let $\gcd(a^{2^n}+1,a^{2^m}+1)=d$. Then $d\mid a^{2^n}+1$ and $d\mid a^{2^m}+1$. So, $a^{2^n}\equiv -1 \pmod d$ and $a^{2^m}\equiv -1 \pmod d$.  Let $m>n$. Then $m-n>0$. From $a^{2^n}\equiv -1 \pmod d$ we get $(a^{2^n})^{2^{m-n}}\equiv1 \pmod d$. But $(a^{2^n})^{2^{m-n}}=a^{2^n\cdot 2^{m-n}}=a^{2m}\equiv -1 \pmod d$. Then, $1\equiv -1\pmod d$, i.e., either $d=1$ or $d=2$. How do I find the value of $a$ for $d=1$ or $d=2$?

How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k>1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ?


I'm developing a website and I want to integrate a Live Web TV.  I have this dedicated server :  Quad Core Q6600 4x 2.40+ GHz 4GO RAM 1TO Disk Space Unmetered Bandwidth 100Mbps (goes down to 10Mbps if I reach 3TO of bandwidth. Possibility to swtich to 1 Gbps if I pay more)    I think I will use Red5 (installation will be made by Re5 Team) because it's open source and free of charge. Bitrate will be about 500kbps   I predict to have about :   100 users watching my live event I will proadcast a Live for Only 2 hours daily.  Videos on-demand should be always available for recroded events.   Can my server support this kind of activity ? Or do I need more ressources? Note that I have my customers websites hosted on this server. Do I have to make a VPS plan to the WebTV and seperate ressources ? Of course if this server supports that traffic.  I really don't want to go with company who provides this kind of services because they are too expensive.

This is a  about  Related:      I have a question regarding capacity planning. Can the Server Fault community please help with the following:   What kind of server do I need to handle some number of users? How many users can a server with some specifications handle? Will some server configuration be fast enough for my use case? I'm building a social networking site: what kind of hardware do I need? How much bandwidth do I need for some project? How much bandwidth will some number of users use in some application?


I'm trying to create a small program which will display a message box if a certain program has been exited.

When I get a reference to a System.Diagnostics.Process, how can I know if a process is currently running?


What value should be used for scale while calculating slope percent in QGIS?    Some documents suggest is as 111120 and some as 365130.  I am using ESPG 4326 projection and do I need to convert this value to some other value using lat like given in this link:

I used an SRTM DEM to create a hillshade in QGIS 2.2.0. However, the resultant hillshade seems to be vertically exaggerated. Also, changing the settings for scale and z-factor does not have any effect on the resultant hillshade. I have tried both options in QGIS, Analysis-> DEM (Terrain models) and Terrain analysis -> Hillshade. Both methods gave the same results.  Is this an isolated incident, a bug, or am I missing something?


From the Rules (EXTRA ATTACK)  "Beginning at 5th level, you can attack twice, instead of once, whenever you take the Attack action on your turn."  If I have the EXTRA ATTACK feature (level 5), AND I am using two different light weapons, I'll get three attacks.  Do the rules ever explicitly stipulate that the two attacks from Extra Attack MUST come from the same weapon in this situation?  I can't find anything saying they must.  Assuming you can make one attack with each weapon, and then a third attack as a bonus action with either weapon (there is no handedness in D&D 5e) the end result would still only be 3 attacks, but might make for a more "fun" 1/2/1 rather than always being the same old 1/1/2 attack sequence.  Assuming using Extra Attack for a 1/2 weapon combo is legal (again- I see nowhere that it's not), does that mean ASIDE FROM 1/1/2:  1/2/2 is the only other legal combination (the bonus attack must be made with the opposite hand of the first attack when the action was taken).  1/2/1 is the only other legal combination (the bonus attack must be made with the opposite hand of the last attack before the bonus action).  OR since both weapons have been used to attack does the player then have the option of which weapon to use?

The fighter's  feature (PHB, p. 72) says:     Beginning at 5th level, you can attack twice, instead of once, whenever you take the Attack action on your turn.   Assuming I am holding two different weapons, and don't want to spend my bonus action on Two-Weapon Fighting, can I attack once with each weapon?


I basically want LaTeX to do following:  So everytime I do \cite{source} I get (author et al, year) out. Instead of the weird [1] or [name] tags. And if there is no author, it should do following (title, year).

I am really new to LaTeX and I need a comprehensive solution for my Thesis  I have this problem since two days and I couldn't find any solution in any other post.  I work with BibDesk on Mac  And I used Packages apacite and natbib  @article{Ha89, Author = {Peter M. Haas}, Date-Added = {2017-02-06 21:15:57 +0000}, Date-Modified = {2017-02-06 21:45:06 +0000}, Journal = {International Organization}, Lastchecked = {03/02/2010 12:54}, Number = {3}, Pages = {377-403}, Publisher = {Peter Haas}, Read = {0}, Title = {Do regimes matter? Epistemic communities and Mediterranean pollution control}, Urldate = { http://www.jstor.org/stable/2706652}, Volume = {43}, Year = {1989}}   this is a stripped version of my document:  \documentclass[11pt, oneside]{article}  \usepackage{lipsum} \usepackage[margin=1in, left=1.5in, includefoot]{geometry}   \usepackage{ragged2e} % REFERENCE PREAMBLE %\usepackage{url} \usepackage{apacite} \usepackage{natbib} %\usepackage{biblatex-apa}                                                   % Bullet preamble \renewcommand{\labelitemi}{$\bullet$} \renewcommand{\labelitemii}{$\diamond$} \renewcommand{\labelitemiii}{$\circ$} % graphics preamble \usepackage{graphicx}   %Allows you to import images \usepackage{float}      %Allows for control of float positions %HEADER AND FOOTER STUFF \usepackage{fancyhdr}  \pagestyle{fancy} \usepackage{graphicx}        \usepackage{amssymb} %Zeilenabstand 1,5 wenn aktiviert  \usepackage{setspace}\makeatletter\newcommand{\MSonehalfspacing}{\setstretch{1.44}\ifcase \@ptsize \relax\setstretch {1.448}\or\setstretch {1.399}\or\setstretch {1.433}\fi}\newcommand{\MSdoublespacing}{\setstretch {1.92}\ifcase \@ptsize \relax\setstretch {1.936}\or\setstretch {1.866}\or\setstretch{1.902}\fi}\makeatother\MSonehalfspacing %Zeilenabstand Code fertig \usepackage[hidelinks]{hyperref}   %allows for clickable reference %                       BEGIN OF DOCUMENT \citeA{source} %                       REFERENCES  \renewcommand{\sectionmark}[1]{\markright{\thesection\ #1}} \fancyhf{} \rhead{\fancyplain{}{Identifying the Regime}} \lhead{\fancyplain{}{List of References}} \cfoot{\fancyplain{}{\thepage}}  \setlength\bibitemsep{1.9\itemsep} \bibliography{/User/Desktop/Thesis/references/mybib.bib}                   \bibliographystyle{apa} \addcontentsline{toc}{section {\numberline{}References}                                                       \end{document}   I would like to have the bibliography entry include a retrieved from    I tried to deactivate hyperref – as suggested in another post – but it didn't work. I also tried to activate and deactivate natbib and apacite as well.   I would really appreciate any help...


I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:  $$\lim_{n \to \infty} nx_n=0$$  My idea is: proving this fact aims to prove that given $\epsilon >0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n >n_0$ we have $|nx_n|<\epsilon$.  Now, I know that $\sum x_n$ converges, so that given $\epsilon'>0$ there's some $k_0 \in \mathbb{N}$ such that if $k > k_0$ we have:  $$\left|\sum_{i=1}^{k}x_i - S\right|<\epsilon'$$  Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 > \cdots > x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:  $$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$  I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?  Thanks very much in advance for your help.

I'm studying for qualifying exams and ran into this problem.  Show that if $\{a_n\}$ is a nonincreasing sequence of positive real numbers such that $\sum_n a_n$ converges, then $\lim_{n \rightarrow \infty} n a_n = 0$.  Using the definition of the limit, this is equivalent to showing \begin{equation} \forall \varepsilon > 0 \; \exists n_0 \text{ such that } |n a_n| < \varepsilon \; \forall n > n_0 \end{equation} or \begin{equation} \forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{n} \; \forall n > n_0 \end{equation} Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!


Ok read this a long time ago and have forgotten the title/ author. Set in an ice world, people skate around, there are tribes of outcasts called 'cutters' who can transform people's appearance. The main character has black hair, but with red hairs that come through as he gets older. His father was an unknown evil king or something with the same hair, his Mum would pluck the boys red hairs out so he wouldn't guess who his real father was. Thanks all!

I'm trying to find a book I read long back. This book may have won a Hugo/Nebula (Trust me I've googled to death this one!).  Don't remember much of the story (that's why I want to re-read it). What I remember is that some of the powers in the book are called "Gods". And they have brains the size of the planets (or planets being their brains).  The protagonist in the climax, solves a theorem (The Great Theorem, perhaps) by remembering one of the obscure theorems (Jacuzi (?)) and incorporating it. This [Great] theorem helps spaceships helps find the shortest route from point A to B. He surprises the antagonist by reaching the vantage point quickly with the help of his new knowledge, and blasts his enemy's spaceship.  The ending of the book - as I remember it is - that the protagonist wants to become a God himself.  Please let me know the title & author of this book if you've read it.  PS #1:  BTW, It's not Asimov's The Gods Themselves :) PS #2:  It's published at the least before 1998


In my webapp I have opened a modal and in that modal I'm processing some data using with ajax in another page(ajax.php), after the processing is done in the ajax.php I do:  header("location: blah.php");   But since I'm in another page and the whole process is going on with ajax request the page doesn't do anything, when I check the ajax request via FireBug, I see the request goes red with this error:  302 Moved Temporarily error   How can I redirect in this scenario?  Thanks in advance

How can I redirect the user from one page to another using jQuery or pure JavaScript?


I am trying to set up a method to convert java.util.date typed variables into java.sql.date typed variables so they can be persisted to database.  However, the code I set up keeps returning a null pointer exception, even though I am getting values on the argument being passed into the method.  public static java.sql.Date convertUtilDateSQLDate(java.util.Date utilDate){          long utilDateTime = utilDate.getTime();          java.sql.Date sqlDate = new java.sql.Date(utilDateTime);          return sqlDate;      }   Does anyone know why this might be happening?

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?


Have a problem where mesh somehow using material in a different way. I have texture node set up. However the whole mesh is only covered with color instead of it. Color seems from a texture part. Have tried to add vertex colors but material shows as expected and the mesh seems to have another color and nothing changes it. s the texture itself is scaled to max somehow. How to cover the mesh with texture properly?  Adding this material into a cube or other basic mesh results to a good texturing with a proper placement. At least please give some clues on where to look for or what to look for?

I am creating a model of a glass and I would like to put a drawing on it (a png picture).  I know how to do it using Blender Render, but not with Cycles Render. Help :)


I own a small VPS hosted by Hetzner on which i run a small Minecraft game server. The VPS is running on Ubuntu, and the only software i installed are Java JRE and the required software to run a Minecraft game server (+ additional plugins).  I ran a tcpdump over night and here is the result:   Could anyone help me read it?  Is it regular?  First of july i received a mail from abuse@hetzner.com (my host) stating that Columbia University sent them an automated notice regarding unwanted traffic coming from my ip address.  Here is the original Columbia University message:   But it also said:     It is possible that this alert is the result of a reflection attack against your network with a spoofed origin matching Columbia   University's network.   Details are provided below.  Please take all   necessary steps to mitigate such attacks, or ignore this notice if   this traffic is spoofed.   So could that be just a random occourrence? Or is my server hacked?

This is a  about Server Security - Responding to Breach Events (Hacking)   See Also:                  Canonical Version I suspect that one or more of my servers is compromised by a hacker, virus, or other mechanism:   What are my first steps? When I arrive on site should I disconnect the server, preserve "evidence", are there other initial considerations? How do I go about getting services back online? How do I prevent the same thing from happening immediately again? Are there best practices or methodologies for learning from this incident? If I wanted to put a Incident Response Plan together, where would I start? Should this be part of my Disaster Recovery or Business Continuity Planning?   Original Version       2011.01.02 - I'm on my way into work at 9.30 p.m. on a Sunday because our server has been compromised somehow and was resulting in a    attack on our provider. The servers access to the Internet   has been shut down which means over 5-600 of our clients sites are now   down. Now this could be an FTP hack, or some weakness in code   somewhere. I'm not sure till I get there.      How can I track this down quickly? We're in for a whole lot of   litigation if I don't get the server back up ASAP. Any help is   appreciated. We are running Open SUSE 11.0.            2011.01.03 - Thanks to everyone for your help. Luckily I WASN'T the only person responsible for this server, just the nearest. We managed   to resolve this problem, although it may not apply to many others in a   different situation. I'll detail what we did.      We unplugged the server from the net. It was performing (attempting to   perform) a Denial Of Service attack on another server in Indonesia,   and the guilty party was also based there.      We firstly tried to identify where on the server this was coming from,   considering we have over 500 sites on the server, we expected to be   moonlighting for some time. However, with SSH access still, we ran a   command to find all files edited or created in the time the attacks   started. Luckily, the offending file was created over the winter   holidays which meant that not many other files were created on the   server at that time.      We were then able to identify the offending file which was inside the   uploaded images folder within a  website.      After a short cigarette break we concluded that, due to the files   location, it must have been uploaded via a file upload facility that   was inadequetly secured. After some googling, we found that there was   a security vulnerability that allowed files to be uploaded, within the   ZenCart admin panel, for a picture for a record company. (The section   that it never really even used), posting this form just uploaded any   file, it did not check the extension of the file, and didn't even   check to see if the user was logged in.      This meant that any files could be uploaded, including a PHP file for   the attack. We secured the vulnerability with ZenCart on the infected   site, and removed the offending files.      The job was done, and I was home for 2 a.m.            The Moral   - Always apply security patches for ZenCart, or any other CMS system for that matter. As when security updates are released, the whole   world is made aware of the vulnerability.   - Always do backups, and backup your backups.   - Employ or arrange for someone that will be there in times like these. To prevent anyone from relying on a panicy post on Server   Fault.


I have a micro plots map shapefile with NDVI data. I want to create a fishnet of polygons to divide the map into the micro plots. I have being able to create a fishnet with polygons but I cannot place it on top of my map.    I know my map has a -32 angle rotation. I have tried to modify the coordinates but still get the wrong placement for fishnet as you can see in the image.  I am now using ArcMap 10.5

I'm working on this project and I need a grid with a certain rotation, at about 60 degrees, for example.  Starting from that I need to make a block of 20mx50m, and the last thing is that inside each block there must be a small set of sub-blocks.   The result is this:    Of course it's been done manually, but I was thinking of a way of making it done automatically because I must do 30 blocks like this one, maybe more, and it is consuming a lot of my time.


What I know is that voltage is the electrical pressure which is required to drive the electrons in a circuit. According to that we have higher currents when there is a high voltage. Ohm's law confirms that. What I don't understand is that some books say that it is the current that is more dangerous to the human body and not the voltage. What I mean is higher currents are dangerous but high voltages aren't. But if we have a high voltage aren't we indirectly meaning that there is high current since they are directly proportional to each other.

Though voltage and current are two interdependent physical quantity, I would like to know what gives more "shock" to a person - Voltage or Current? In simple words, will it cause more "electric - shock" when the voltage is high or Current is more?


Let $A$ be a strictly diagonally dominant matrix of dimensions $n \times n$. ("Strictly diagonally dominant" means that $\left|a_{i,i}\right| > \sum\limits_{j \neq i} \left|a_{i,j}\right|$ for all $i \in \left\{1,2,\ldots,n\right\}$, where $a_{u,v}$ denotes the $\left(u,v\right)$-th entry of $A$.) Prove that $A$ is invertible.  My attempt builds on the proof of Gershgorin's circle theorem, given in the Wikipedia article  Let $\lambda$ be an eigenvalue of $A$, and scale its corresponding eigenvector $x$ so that $x_{i} = 1$ and $|x_{j}| \leq 1$ for $j \neq i$ Then $Ax = \lambda x$, and in particular $\sum_{i\neq j}a_{i,j}x_{j} = \lambda - a_{ii}$ (1) Now because $A$ is strictly diagonally dominant it holds for every $i$ that, $\sum_{j \neq i}|a_{i,j}| < |a_{i,i}|$ and since $|x_{i}| \leq 1$ the following should hold: $\sum_{i\neq j}|a_{i,j}x_{j}| \leq \sum_{j \neq i}|a_{i,j}| < |a_{i,i}|$ But here I get stuck, feel like I want to use (1) in some way to complete the proof and put $\lambda = 0$ to get a contradiction, thus proving that if $A$ is strictly diagonally dominant, it has non-zero eigenvalues which should imply invertibility.. Am I in the right direction?

I try to find a good proof for invertibility of strictly diagonally dominant matrices (defined by $|m_{ii}|>\sum_{j\ne i}|m_{ij}|$).  There is a proof of this in  but I'm wondering whether there are are better proof such as using determinant, etc to show that the matrix is non singular.


Visa for Bahrain is now available online and on arrival. Do you still need to take Bahrain visa in advance to get transit visa for KSA? Does anyone know the telephone/mobile numbers for Waseet agents? Approximately how many days does it take to get KSA transit visa?

Is it now possible for a Kuwait resident (engineer) along with family to get a transit visa to travel by road through Saudi Arabia to Bahrain? If so, which agency helps to process the transit visa in Kuwait?  Please advise how to get a by-road transit visa through Saudi Arabia.  (note: I have a multiple entry visa to Saudi Arabia, however I have to get the transit visa for my family with that.)


is it possible to configure adobe reader advanced print options from commandline when printing. I want to print the pdf document as an image. There is a 'Print as an image' option in the adobe acrobat XI reader printer advanced option.

Is there any official command line (switches) reference for the different versions of Adobe (formerly Acrobat) Reader?  I didn't find anything on .  Especially I want to:   Start Reader and open a file Open a file at a specific position (page) Close Reader (or single file)


I created two edge loops and during creation (or after) I want to move them apart. I can move them together during creation but I wanted them a certain distance from the existing edges on either side of the newly created edge loops.

I am wondering if Blender has some hidden function or anything that can make even space between edges when adding new edge loops with ctrl+R. Or even space after adding them, like if it was a mirror effect of the selected edge, so that the other edge moves the oposite way if you know what I mean.  I made a GIF from 3ds max that shows the method I am looking for:      Figured out with help from Blender discord community that I can select my 2 edges and scale with Z, but this only works if the axises are perfectly alligned. What if I got something bent with different size like a cone, the vertices of the edges would not follow the edge with this method :(  Any ideas?


My Drupal site has been compromised.   Although I applied all modules updates and update my Drupal core to 7.35, it is still compromised.  What would the next logical step to take be?

I just updated all my sites using the patch method of resolving the Drupal SA-CORE-2014-005 exploit. I just read reports that just yesterday there is someone from a Russian IP infiltrating drupal sites.     My main concerns are now:   How do I tell if my sites have been comprised? What should I search for in my apache access logs to detect if my site was a victim or not? So far what are these hackers doing to comprised sites?


I don't know we're you find it and I really need it for enchanting. I know its underground but I can't find it anywhere.

I have the 1.8.8 updated Minecraft Console edition. In the previous version, lapis lazuli, was highly prevalent, on par with red stone or something. Since the update, I have not found nearly as much. (4 blocks total) I have found 5 times that in diamonds. I am playing in the level of bedrock to + 50 above bedrock, and I haven't found hardly any. I am using the branch style of mining, which has been highly effective in finding everything but lapis. What happening? Has lapis become more rare in version 1.8.8?


Given the subspace $Y = \lbrace (x,\sin(\frac{1}{x})) : x > 0 \rbrace \subset \mathbb{R}^2$, I know that $Y$ is connected being the image of a connected of $\mathbb{R}$ (i.e. $(0,+\infty)$.), through a continuous function. I know also that that the closure of a connected is connected, so $\overline{Y}$ is connected, but I'd like to prove that $\overline{Y} = Y \cup \lbrace (0,t) : \lvert t \rvert \leq 1 \rbrace$ Are there any simple method to prove it ?

Let $S$ and $V$ be the following subsets of $\mathbb{R}^2$.  $$ S = \left\{ \ x \times \sin \frac{1}{x} \ \colon  \ 0 < x \leq 1 \ \right\}, \qquad V = \left\{ \ 0 \times y \ \colon \ -1 \leq y \leq 1 \ \right\} = \{ 0 \} \times [-1, 1]. $$ Then how to show rigorously that the closure $\overline{S}$ of $S$ is given by  $$ \overline{S} = S \cup V? $$  By rigorously, I mean using the smme kind of argument as is used in real analysis or topology, such as in (most of) the proofs in Rudin or Munkres.   Intuitively, it is clear that as $x \to 0+0$, then $\sin \frac{1}{x}$ will fluctuate its full range of $[-1, 1]$, and for any point $0 \times y$, where $y \in [-1, 1]$, there will be points of $S$ arbitrarily close to it.   But I'm unable to rigorously show this.


given two sets $A$ and $B$. How do we construct a set $C$ bijective to $B$, but disjoint to $A$?

I was recently thinking about the fact (in ZFC) that, given a (first-order) structure $A$ that embeds into another structure $B$, there is some structure $C$ isomorphic to $B$ such that the domain of $A$ is a subset of the domain of $C$. This is trivial if we can prove the following:     Given any injection from $S$ into $T$, there is some set $U$ disjoint from $S$ that bijects with $T$.   I can prove this within either ZC (using AC) or ZF (using Foundation). My ZC proof uses transfinite induction to pick $U$ out from $(P(T)×2)∖S$. My ZF proof constructs $U = \{ \{S,x\} : x∈T \}$. But I don't see a way to prove it within just Z. In one attempt, I tried to use one of $k$ disjoint copies of $T$ where $k$ is a well-ordered set that does not inject into $S$, but if every copy intersects $S$ then I seem to need a choice function on $S$ to obtain a contradiction. At least this means that I can prove it for countable $S$, but what about uncountable $S$?  I may be missing something easy, because intuitively it cannot be possible that every set that bijects with $T$ has a non-empty intersection with $S$, even in the absence of choice...


I need to use an inequality $(x+y)^p\leq x^p+y^p$ for positive $x$, $y$ and $0<p<1$ in a proof, but I'm not sure how to derive it and I wasn't able to find it anywhere. I think it might hold, because I put a number of queries with random rationals like (x+y)^(0.767809)<=x^(0.767809) + y^(0.767809) into Wolfram Alpha as well as $p=0.1$, ..., $0.9$ and it always said the inequalities were true for positive $x$ and $y$. I tried transforming it to use concavity/convexity, but got nowhere. Could somebody point me to a proof somewhere or give me a hint?

If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$    Trial: For $m=0$, $(p+q)^0=1 < 2= p^0+q^0$  and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.   So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 < m < 1$.   Please help.


Can anybody help me in creating subdoain through webmin. I want to create subdomain like test.xxxxx.com for this I tried with several reference site but no luck. exp.      My current httpd.conf look like:   <VirtualHost *:80> SSLEngine off  DocumentRoot /var/www/html/******/web DirectoryIndex index.php <Directory "/var/www/html/*****/web"> AllowOverride All Allow from All </Directory> ServerName www.******/.com ServerAlias ftp.*****.com SSLEngine off  SSLVerifyClient optional </VirtualHost>   Please help me to solve this issue.

Can someone please help me in using WEBMIN to setup virtual hosts.   My current domain www.MYDOMAINLOLFAKE.com currently functions.   Settings are as follows -  Apache - Handles the name-based server www.MYDOMAINLOLFAKE.com on all addresses Address Any Port 80 Server Name www.MYDOMAINLOLFAKE.com Document Root /var/www/html  BIND DNS Server - Master Zone MYDOMAINLOLFAKE.com ns1.mydomainlolfake.com IPHERE - works ns2.mydomainlolfake.com IPHERE - works mydomainlolfake.com IPHERE - works www.mydomainlolfake.com IPHERE -works mail.mydomainlolfake.com IPHERE - works ftp.mydomainlolfake.com IPHERE - works  What I need - something.mydomainlolfake.com -- CANT GET THIS TO WORK  What I tried - Create new virtual host Handles the name-based server something.mydomainlolfake.com on something.mydomainlolfake.com Address Any Port 81 Document Root: /var/www/vhosts/something  What happens -  I create the new VHOST and then ALL address try to go to that new Document root. I need different addresses to go to their respective folders. Can someone please give me better instructions on how to set that up using webmin?  TLDR# How do I make a something.mydomainlolfake.com subdomain work in webmin on my CENTOS 6 web server?


I just installed Ubuntu 18.04 as a second OS in my Acer notebook, (alongside with Windows 10) using 2 partitions, / and /home (sda5 and sda6). However, I found that I might have reserved less space than I'd have liked. So I was pretending to reinstall it.   Is it safe to just format both partitions from the Live USB, free some more from the Windows partition (sda3) using Windows' Disk Manager (already freed 40 GB) and create 2 new partitions with the allocated space distributed correctly? Is it possible that Grub2 will break after it?

Previously, I have installed Windows 7 on my 320 GB laptop with three partitions 173, 84 and 63 GB each. The 63 GB partition was where the Windows was installed. The rest were for file containers.  Now I changed my OS to Ubuntu 12.04 LTS. I installed Ubuntu by replacing the entire Windows 7 on the 63 GB partition. The rest of the partitions remain as an NTFS Windows partition and I can still access them both (the 173 and 84 GB partitions).  Now I want to change the two partitions of Windows into an Ubuntu format partitions plus most importantly, I want to extend the 63 GB partition to more than a 100 GB because at the moment I am running out of disk space. Whenever I try to install any application, especially using wine, it always complains for a shortage of disk space.  How do I do the extending activity before I entirely format my laptop again and lose all the important files on my partitions?


I am trying to prove  $ \sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}=\binom{n+1}{k+1} $  Whichever numbers for $k,n$ I try, the terms equal, but when I try to use induction by n, I fail to prove the induction step:  Assume the equation holds for $n$. Now by Pascal's recursion formula,  $ \binom{n+2}{k+1}=\binom{n+1}{k+1} + \binom{n+1}{k}\\   =\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}+\binom{n+1}{k}, $  by induction assumption. In order to complete the proof, I would need to show  $ (n-k+2) \binom{n-(n-k+2)}{k-1} = \binom{n+1}{k} $  but the left-hand side is zero. What am I doing wrong?  EDIT:  There were links to similar questions when my question was marked as duplicate. However, these links are now gone, so I add them here as they were useful to me:    -- hint to Vandermont's identity  -- basically the same question, but two days earlier.   (I did search, but did not found these.)

I was trying to simply an expression in an exercise related to randomized algorithms. Here is the expression which I have obtained at the end.  $$ \displaystyle\frac{\displaystyle\sum_{i=1}^{n+k-1} i \binom{n-i}{k-1}}{ \displaystyle{n \choose k}}$$  Is there any way to simplify the numerator so that the whole expression simplifies into a nice closed formula? A combinatorial approach would be greatly appreciated.


What is the difference between the way a microscope magnifies say a star and a telescope say a cell? Why is it, that the eyepiece of a telescope magnifies stronger when it is smaller and with a microscope it is the other way around?

In compound microscope, we take such an objective lens which has small focal length. While in astronomical telescope, we take such an objective lens which has large focal length.   Why don't we use same objective lens in both? The function of both the devices is to enlarge an object at infinity.


Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$ be a real polynomial such that $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$.  Show that $P$ has a root in $[0,1]$. I would like proof verification and alternative proofs.  My solution:      notice that the antiderivative of $P(x)$ is $\frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n}+\dots + \frac{a_1x^2}{2}+a_0x+C$. From here $\int_{0}^1P(x)dx=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$. If $P$ had no roots then $P$ would be a strictly positive(or negative) continuous function, so its integral would be strictly positive ( or negative), a contradiction.

Knowing that $$ \frac{a_0}{1} + \frac{a_1}{2} + \frac{a_2}{3} +\cdots + \frac{a_n}{n+1} =0$$ Prove that  $$ a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0$$ has at least one real solution.  I suspect that it's proven with the intermediate value theorem. But can't find two numbers that satisfy it.


What is the difference between "an application instance" and "instance of an application"? When we prefer the two words version and when the "of" version?

I would be very grateful for some advice on how to decide whether to use an attributive noun or the "of genitive" (periphrastic genitive).  It seems to me that an attributive noun is generally preferable, except when it leads to ambiguity, often due to the combination of several modifiers with mixed dependent relationships.  Additionally, the of-genitive, when used stand-alone, seems to impart a more formal tone, a higher register.  Simple examples (preferred first, less-preferred second):  The department secretary vs. the secretary of the department.  But:  The excellent secretary of the department vs. the excellent department secretary.  Any additional guidelines would be most welcome.


In my study i have two groups of newborns: one group with Hemodinamically significant PDA(HSPDA)labeled 1 and one group with non-hemodinamically significant PDA(nHSPDA) labeled 0.   Each of these groups has a set of tests(like blood Ph, sugar blood levels etc) and the data is ordinal(ph of 7.27, sugar blood of 100 etc). The tests are dependent on wether the PDA is significant or not.   Which test do i apply to see if there is a correlation between the resuslts of a certain test and the presence of a significant PDA? Do i just compare the test results between the groups?

I'm struggling with my lab report. I have a questionnaire which contains Likert responses (1-7) as well as questions that require only yes or no answers. I want to see if there is a correlation between higher scores on the likert scale and responses to the binary answers to the other questions.   Specifically - the likert response is about anxiety levels. The binary (yes/no) answer is about negative consequences due to alcohol use. I want to just run the one ordinal likert item with all the negative consequence responses to see if there is a relationship. Can anyone help me?     Thanks so much for the fast comments. It definitely assures me that Spearman's Rho can be used but my main problem is the fact that i am mixing binary and ordinal variables together in the one correlation matrix.  I have one ordinal variable (anxiety - 1-7 on likert scale) crossed with 13 dichotomous (0 or 1 representing yes or no) to answers ("have you ever been hospitalised due to your alcohol use?). I get statistically significant answers (<.05 and < .01 but the Correlation coefficients are tiny (i.e o.2) at best.   the hospitalisation is the DV an the anxiety levels the IV.   I just think i'm doing something wrong here. should i collapse the 13 binary variables in to one some how?   i can't see how SPSS will recognise the values between 1-7 when correlating with binary answers. i keep thinking it will just read that there is something in that box (ordinal) but not recognise the value (i.e 1-7) . but people keep saying that SPSS does in fact recognise this difference and account for it.


I currently have a few static IPs set in my router and I'm wondering if there is any way that i can access/ping/refer to them via a DNS request (i.e. instead of 192.168.0.8 use something like "RapberryPi") that way I don't have to always remember the IP address. I know that this can be done if i setup a server within the NAT but I will like to implemented at the router level. Do routers usually have a "host" file or something of that sort?

I have just had my broadband upgraded to  which means my router has had to change. Previously I was using a Thomson TG585v7 ADSL thingy provided by my ISP. One nice thing about that router was that it would act as a local DNS and allow me to find my other machines on the network by using <hostname>.lan instead of IP addresses.  I've now switched to a Linksys WRT320N ethernet router which connects to my fibre modem, and which I've updated to the latest firmware (1.0.04). I have 2 machines on the network which I regularly want to connect to each other; one is running Windows 7 and the other is running OS X Lion.  The problem I'm having is that I'm unable to connect to my Win7 machine from the OS X machine using the Win7 machine's name. However, I'm able to ping the OS X machine from my Win7 machine using the hostname.


I'm just reading through some vector calculus notes and it begins to assume knowledge of the notation $\delta_{ij}$ and $\epsilon_{ijk}$ in reference to tensors, but I am not familiar with this. It uses them as if they are constant and well-defined.  If someone could just explain this I would appreciate it.  Thank you.

I'm trying to grasp Levi-civita and Kronecker del notation to use when evaluating geophysical tensors, but I came across a few problems in the book I'm reading that have me stumped.   1) $\delta_{i\,j}\delta_{i\,j}$  2) $\delta_{i\,j}  \epsilon_{i\,j\,k}$  I have no idea how to approach evaluating these properties. Without knowing i, j, or k, how would I approach? I don't feel confident using the notation further until I can understand these properties.


I was struggling for ages on this question, and even after careful considering I cannot make sense of this with my knowledge of algebra or set theory. How can I solve this?

How many integer solutions does $xy+9(x+y)=2006$ have? Here $x$ and $y$ are both integers.  My trying:        I have tried to solve this problem. But I have no idea to solve this. Please help


So I have the following JS code.     const romanNumeral = {     1000: "M",     900: "CM",     500: "D",     400: "CD",     100: "C",     90: "XC",     50: "L",     40: "XL",     10 : "X",     9: "IX",     5: "V",     4: "IV",     1: "I"   }    for (const key in romanNumeral) {     console.log(romanNumeral[key])   }    Why on earth does the loop start to output from the bottom entry of the object? This doesn't make any sense to me. Is the FOR IN loop implementing some kind of sorting when looping through entries?

Does the "for…in" loop in Javascript loop through the hashtables/elements in the order they are declared? Is there a browser which doesn't do it in order? The object I wish to use will be declared once and will never be modified.  Suppose I have:  var myObject = { A: "Hello", B: "World" };   And I further use them in:  for (var item in myObject) alert(item + " : " + myObject[item]);   Can I expect 'A : "Hello"' to always come before 'B : "World"' in most decent browsers?


The image attached, when I try to select part of the stadium, the whole stadium is selected, how can I separate this into different parts so I can edit different parts.   Thanks

I want to make the teeth a separate child object from the mouth. I want to know the best way to do this.   I have tried going into edit mode with Tab and trying to select all the vertices but that doesn't move the teeth. Is there an easy way to select just that part and make it a separate child mesh to the rest of the mouth?


Hello stackexchange users! I am somewhat confused on $\sqrt{-1}$. I believe the answer should be no solution, or inconsistent, or at least something in that nature. Am I correct or am I most likely wrong? Any help or insight would be deeply appreciated

At school, I really struggled to understand the concept of imaginary numbers.  My teacher told us that an imaginary number is a number that has something to do with the square root of $-1$.  When I tried to calculate the square root of $-1$ on my calculator, it gave me an error.  To this day I still do not understand imaginary numbers.  It makes no sense to me at all.  Is there someone here who totally gets it and can explain it?  Why is the concept even useful?


The condition to receive flight compensation via EC261 are:  the person has:   a confirmed reservation on the flight, and arrived in time for check-in as indicated on the ticket or communication from the airline, or, if no time is so indicated, no less than 45 minutes prior to the scheduled departure time of the flight OR have been transferred from the flight for which he/she held a reservation to some other flight   In addition, if the delay is more than 5 hours, the passenger has a right to forfeit the flight and receive the ticket price back.  Does it mean that if the airline notifies about a long delay and passenger cancels the ticket, he still has to arrive in time for check-in to receive cash compensation?

Vueling is currently experiencing , with many flights delayed by several hours. Assuming I'm at the airport and my flight is certainly delayed by more than 3 hours, can I simply book another flight and claim compensation later?  According to EU's :     In addition, if you are denied boarding, your flight is cancelled or arrives more than 3 hours late on arrival at the final destination stated on your ticket, you may be entitled to compensation of EUR 250 - 600, depending on the distance of the flight.   But it's not clear if I actually have to board the flight or if simply having a purchased ticket is sufficient. I presume it would be possible to negotiate with Vueling representatives at the airport, but I'd rather not wait in the inevitable 100-person line to their counter if I'm covered by EU law.


I'm having difficulty figuring out my visa situation going through Istanbul.    I'm a US citizen.  I'll be traveling from Mumbai, and connecting through Manama, on Gulf Air before changing airlines to Turkish Airlines in Istanbul and flying to Vienna which is my final destination.    Do I need a visa with the diplomatic spat between the US and Turkey? And do I need to pick up my luggage and go through customs again to get onto my Turkish airlines flight?

I have an Indian passport and am planning to visit Germany with a valid visa. To book a flight for Germany (Mumbai – Dusseldorf), I came across some connected flights that stop at Istanbul.  Do I need to have visa for Turkey?  This is the first time I am travelling outside my country, and any information on this will be helpful.


In the case of a particle on a line, if I prepare an (although non-normalizable) eigenstate of the momentum so that the uncertainty in the measurement of the momentum vanishes, the uncertainty principle is (at least in a physicist's sense) still satisfied as the position measurement would yield a value from $-\infty$ to $\infty$ with an equal probability and the uncertainty in the outcome of the position measurement thus diverges.   But, in the case of a particle on a ring, if I prepare an eigenstate of the momentum (which will even be normalizable) so that the uncertainty in the measurement of the momentum vanishes, there seem to be some subtleties around how to think of the uncertainty principle. Naively, the eigenvalues of the position operator range only from $0$ to $2\pi$ and thus, although they are all equiprobable outcomes and the uncertainty in the outcome of the position measurement is maximal, this uncertainty (the variance) can only be finite as the range of the eigenvalues is of finite span. Thus, in this naive look, it seems that the product of the uncertainties in the measurements of the position and the momentum vanishes as the uncertainty in the measurement of the momentum vanishes and that in the measurement of the position is yet finite.   In the derivation of the uncertainty principle, we essentially only take into account the commutator of the operators which remains the same for the position and the momentum operators whether we consider a particle on a ring or a line. Moreover, we tacitly assume normalizable states which would suggest that the case of a particle on a line with a momentum eigenstate should be more subtle than the case of a perfectly normalizable momentum eigenstate of a particle on a ring. But, in fact, the roles seem reversed.   I can only think of one place from which the subtlety is arising. The uncertainty principle somehow takes all the identified positions such as $0,2\pi,4\pi,...$ as distinct and computes a divergent uncertainty as if it were simply a particle on a line. But, I don't know why it should be the case or how we could prevent it from doing it (via deriving the uncertainty relations more carefully) if it is indeed doing so.

Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?


Perhaps a basic question, if I have an un-normalised histogram, my understanding was that the integral is simply the total number of points.  If I then fit a curve over the the histogram, then surely if I integrate this curve from $-\infty$ to $+\infty$ I should get a value that is very close to the total number of points in my histogram's data set -- correct?   If the histogram and the curve is centred around $0$ are there any special considerations, for example normalising?

I asked a question  but I did a rather poor job of explaining my problem the question was poorly formed.  If I have some histogram with $N_{total}$ number of total data points, and $N_{bins}$ number of bins. If I integrate across the maximum bin and the minimum bin, then I just get back $N_{total}$, likewise if I integrate between two bins on the histogram, I simply sum the counts within those two bin points.  Now I can fit a function, $f(x)$, that describes the histogram. If I Integrate the function $f(x)$ between two points - which correspond to two bins, what I want to get back is a value which is close to the total counts in the region between the two bins.  Is there a relation that I need to use which scales with the number bins, such that the result of the integral $\int_{max\ bin}^{min\ bin}f(x)\ \ dx \approx N_{total}$  I hope this is clearer than my first attempt!


I'm in an introductory discrete mathematics course and i'm having a lot of trouble with this proof. I'm not sure if i'm heading in the right direction or not but I used the fact that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ and started trying to algebraically reduce them to eachother but haven't succeeded. Any help would be much appreciated.

I want to prove this equation, $$ \binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a} $$ I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.   (Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.


there is a way to get a value from a class in the stylesheet without having any element in the DOM with that class?? using vanilla JS so basically having something like .someclass{ color:red }  how can I get the value of color ? I can't use document.getElemnt... because there is no element in the DOM with that class yet.

I would like to return a string with all of the contents of a CSS rule, like the format you'd see in an inline style. I'd like to be able to do this without knowing what is contained in a particular rule, so I can't just pull them out by style name (like .style.width etc.)   The CSS:  .test {     width:80px;     height:50px;     background-color:#808080; }   The code so far:  function getStyle(className) {     var classes = document.styleSheets[0].rules || document.styleSheets[0].cssRules     for(var x=0;x<classes.length;x++) {         if(classes[x].selectorText==className) {             //this is where I can collect the style information, but how?         }     } } getStyle('.test')


prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$  .A little use of calculator shows that  $\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}=5.44$.Thus the inequality is indeed true. Generalising this result with $$f(x)=x-\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}<0$$ does not help as we see that $$8>\sqrt{8}+\sqrt[3]{8}+\sqrt[4]{8} \tag !$$ Repeated efforts of Bernoullis inequality have failed


i need to buy new ipod 5th gen. I keep seeing that many of them are iCloud locked. Can this lock be removed and how do I do it?

I bought a used iPhone 4.  The previous owner re-set the phone, but did not remove the device from their iCloud.  It is asking me for their iCloud information when I try to set-up the phone.  I don't have contact with the previous owner anymore.  Is there a way to remove the device from their iCloud so I can use a new one?


I'd like to use the tech for . The dynamic tile layout is really cool. I wonder how it is built. After viewing the code, I just find jquery js reference. Is it built using jquery? If so, what module it uses exactly?

What tools and technologies are used to build the Stack Exchange Network?  See also:


I've recently installed Oracle VM VirtualBox on my Windows 10 computer, and I'm trying to make an Android Virtual Machine. The problem is that when I go to start the VM, it displays the following error:  VT-x is not available (VERR_VMX_NO_VMX).  Result Code:    E_FAIL (0x80004005)   Component:  ConsoleWrap   Interface:  IConsole {872da645-4a9b-1727-bee2-5585105b9eed}   I've already checked that my CPU is compatible with VT-x, and it's activated in the BIOS (In my case, I activated VT-d because there isn't an option for VT-x, but I've read that they're the same thing. If I'm wrong please correct me). I also have disabled Hiper-V, but it didnt' work.  Do you have any ideas on what's the cause of this problem?  Edit:  I have a Gigabyte GA-H270M-DS3H motherboard with the F7 BIOS Version. I have an Intel I7 7700 (not K) CPU  Regarding , I've tried everything there, except for turning on VT-x on the BIOS, because I don't have the option, and because I thought that VT-d was the same thing. If not, how do I turn on VT-x?

Recently I started having an issue with VT-x on my Windows 10 running on HP EliteBook. I was using Oracle VirtualBox to run some VMs on it before and VT-x was enabled in BIOS and was working fine.  I started getting VT-x is not available error every time I was trying to start any of my VMs. That started happening after I installed Docker on my machine, so I tried uninstalling it and rebooting. Didn't help.  I checked BIOS settings and it says VT-x is enabled but checking with Intel(R) Processor Identification Utility shows that VT-x is not supported and Oracle VirtualBox also doesn't see it.  There were no hardware changes and the only software change was installing/uninstalling Docker.  How can this issue be fixed?


import java.util.Scanner; public class SwitchCase {      public static void main(String[] args) {          Scanner sc=new Scanner(System.in);          String ch="NULL";         do          {             System.out.println("Enter your choice:");             String str=sc.next();             switch(str)           {              case "Add":             {             System.out.println("Enter 2 nos:");             int a=sc.nextInt();             int b=sc.nextInt();             int c=a+b;             System.out.println("Adiition is:"+c);             break;             }              case "Sub":             {                 System.out.println("Enter 2 nos:");                 int a1=sc.nextInt();                 int a2=sc.nextInt();                 int a3=a1-a2;                 System.out.println("Sub is:"+a3);                 break;              }             case "Multiply":             {                 System.out.println("Enter 2 nos:");                 int a1=sc.nextInt();                 int a2=sc.nextInt();                 int a3=a1*a2;                 System.out.println("Sub is:"+a3);                 break;             }              case "Divide":             {                 System.out.println("Enter 2 nos:");                 int a1=sc.nextInt();                 int a2=sc.nextInt();                 int a3=a1/a2;                 System.out.println("Sub is:"+a3);                 break;             }             case "Fib":             {                 System.out.println("Enter the range:");                 int n=sc.nextInt();                   if (n == 0)                  {                      System.out.println("0");                  }                   else if (n == 1)                   {                        System.out.println("0 1");                  }                  else                   {                        System.out.print("0 1 ");                        int a = 0;                        int b = 1;                        for (int i = 1; i < n; i++)                        {                            int nextNumber = a + b;                            System.out.print(nextNumber + " ");                            a = b;                            b = nextNumber;                        }                              }         }     }         System.out.println("Do you want to continue? y or n");       ch=sc.next();         }           while(ch=="y");      }  }   After the while loop,when "y" is entered by user,it should ask "Do you want to continue" but the same does not happen.I debugged the code,it runs fine till "Do you want to continue?" But after that :  source not found error comes, when debugged in eclipse

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?


Let $f:(0,\infty)\to\mathbb R$ be a differentiable function for which $f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$ for every $0<x\ne1$. Then $f'(1)=0$  I have this question, and I was wondering how to switch the fact the f is differentiable for every positive x into differentiable in a chosen interval that will help me with my proof.. I tried to make quiet a general interval in $[\frac{1}{x},x]$ but I'm afraid that I can't acctualy do that, and that it's not accurate.. d. True We know $f$ is differential in $(0,\infty)$, therefore $f$ is also differentiable in the interval $[\frac{1}{x},x]$ for any $x ∈ R$ such that $0<\frac{1}{x}<1<x$. Since $f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$, by Darboux's theorem there exist a point $c ∈ R$ such that $\frac{1}{x} < c < x$ and $f'(c)=0$. Assume toward contradiction that c ≠ 1, then  we have $f'\left(c\right)\cdot f'\left(\frac{1}{c}\right)<0$. But since $f'(c)=0$, we have $0 \cdot f'\left(\frac{1}{c}\right)<0 \ \ \ ➜ \ \ \ 0 < 0$. Contradiction! Therefore $c = 1$ and $f'(1) = 0$.

Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})<0$ for every $0<x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0< \delta < 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})<0$ So by Darboux Theorem, there exists a $x_{0}<c< \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)


im doing models of interiors and one of the parts used molding around doors. is there a easy way of doing my corners other than having to get 2 frames and work inwards towards eachother?

I'm working on making a triple glazing corner sample, and I want to be able to mitre the profile at 45 degrees so I can duplicate it, rotate it and join them to make an elbow. I've muddled through making the profile, but now I'm stuck on how to get that 45 degree angle.  I've had a look around Google, perhaps I'm not phrasing my question correctly but I cant find any answers. I did try creating a cube to use as a sort of cutting form, but Blender wouldn't let me use a Boolean modifier to do it. Any thoughts?  Thank you!  What I'm trying to achieve:    What I have at the moment:


For example I have file main.css  #test { my-property-name: 10px; }   and index.html

   Can I somehow get value of my-property-name for this div from js file?

I am trying to read a custom (non-standard) CSS property, set in a stylesheet (not the inline style attribute) and get its value. Take this CSS for example:  #someElement {   foo: 'bar'; }   I have managed to get its value with the currentStyle property in IE7:  var element = document.getElementById('someElement'); var val = element.currentStyle.foo;   But currentStyle is MS-specific. So I tried getComputedStyle() in Firefox 3 and Safari 3:  var val = getComputedStyle(element,null).foo;   ...and it returns undefined. Does anyone know a cross-browser way of retreiving a custom CSS property value?  (As you might have noticed, this isn't valid CSS. But it should work as long as the value follows the correct syntax. A better property name would be "-myNameSpace-foo" or something.)


Sorry if this is a duplicate, but could we have a search bar next to the page numbers so that we can easily find the nth page? For example, there are 740 pages in Meta SE unanswered, so it would be difficult to find the 100th page without changing the URL manually. It would be great to have this additional feature, perhaps next to the 'next' button?

If there are total 500 pages, and i have to see some content on page no. 250 , i should be able to jump to that direct, using Next <> Prev is annoying


I have designed a product (small spectrometer) using FT234XD usb-to-uart bridge and it connects to any mobile phone via USB cable (90mA maximum current). My device acts as USB slave and can connect to any mobile phone which supports OTG. However, we are facing one problem. We are selling our product with lenovo tablet and user won't be able to disconnect the device and charge the tablet regularly.  I came across several links stating that most of the usb hardware in mobile phones and tablets have a mode in which mobile/tablet can act as USB master and still charge itself at the same time. So, we have tried several resistors across ID pin and ground of the usb connector but tablet either charge only or it detects the device.   Finally, I am considering one FT311D chips which may work for us but I am not sure if it will be possible with it. Can anyone help me with this problem? Should we try other things with existing hardware (FT234XD) or try this new FT311D chip?   EDIT:  My Tablet is Lenovo Tab 3 Essential Tablet. I am planning to try FT312D. Good news is that my team ran few software tests and get to know  Tab software supports Android usb accessory mode. In this mode, FT312D will be the host and I think I can charge the tab and communicate with it.  I think It will work because tab will be in slave mode and can charge itself while communicating with my device and only thing required support of accessory mode in software. Am I correct?  EDIT 2  My device has PIC24E 16-Bit microcontroller. There is no USB Protocol specific firmware in PIC24EP. PIC24EP just communicates with mobile application using string based commands. PIC sends commands over UART and FT234XD converts them to USB. Commands received from Android application is converted to commands over UART and is returned to PIC. So, I think changing USB part of my device doesn't require any change in firmware of my primary microcontroller.

A number of Android tablets are powered through the Micro B USB port but also provide USB Host support through the same port (e.g. Motorola Xoom 2, Acer A510/700). I need to be able to use USB Host support while charging, at the full power the device accepts (or at least sufficient power to keep the tablet from draining under modest use).  I have rigged up a system with splitter cables and a powered hub. This works on some devices (HP TouchPad, Archos) but not others (Google/Asus Nexus 7, Coby Kyros). And when it does work,  it slow charges (probably at USB 2's 500ma).  How can this problem be solved? Are there any solutions on the market that will solve my problem? I see there are some USB chips by Silego (their Battery Charger ID line) that may solve the problem, but I don't know of any cables, hubs, etc. that use it.


When I insert a pen drive in the USB slot of my Mac mini 2018 the Bluetooth mouse and keyboard lose connection. Anyone knows how to solve this?  I have Mojave 10.14.2 and both mouse and keyboard are Apple.

I recently switched from my beloved MacBook Pro Retina (Late 2013) i5 2.6/16/512 to MacBook Pro (2017) Non-TB i5 2.3/16/512.  I purchased what from the build-quality looked like a good USB-C-to-DisplayPort cable with 4k/60Hz support, just to run into the very problem the Logicboard repair guru Louis Rossmann had described in : As soon as you plug in the USB-C dongle, the (2.4 GHz) Wifi stops working despite full antenna signal. Unplug the USB-C connector and you are immediately back in the game.  The big question I'm asking myself now: Who is to blame? Is this a design flaw in the current generation MacBooks with improper shielding or is it a design flaw in the various dongles?  I mean the fact that even Apple’s very own USB-C Digital AV Multiport Adapter is causing these WiFi issues (as you can learn from the many horrific ; but AFAIK have been fixed through a ), show that there’s something going horribly wrong.


I am interested in how to round variables to two decimal places. In the example below, the bonus is usually a number with four decimal places. Is there any way to ensure the pay variable is always rounded to two decimal places?    pay = 200 + bonus;   Console.WriteLine(pay);

I want to do this using the Math.Round function


I would like to simplify  $\det\begin{pmatrix}1&x_1^2&x_1^3&x_1^4\\1&x_2^2&x_2^3&x_2^4\\1&x_3^2&x_3^3&x_3^4\\1&x_4^2&x_4^3&x_4^4 \end{pmatrix}$  which is not exactly like the Vandemonde matrix by the exponents, but I was wondering if there is a way to compute it in $\mathbb{Q}[x_1,x_2,x_3,x_4]$  I tried computing it like a normal matrix but after a very long computation I don't find an "easy" formula to describe it, I also tried following the general proof of the Vandermone matrix but without success. I think there is more behind it, does anyone know about this?

Consider the following  Vandermonde matrix  $$ V_n = \begin{pmatrix}   1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\   1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{pmatrix}. $$  It is well-known [] that the determinant of $V_n$ is defined by  $$ \displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right)\tag{1} $$  Let $V_{n-1}^{(i,j)}$ be a square matrix such that it is obtained by the removing  $i$th row and $j$th column of $V_n$.   My question: Is it possible to get a closed-form for the determinant of $V_{n-1}^{(i,j)}$ similar to $(1)$.   My try: If $j=n$ then $V_{n-1}^{(i,n)}$ is a Vandermonde matrix and there is a closed-form for its  determinant as $(1)$.  Thanks for any suggestions.   Edit: I think the general case of the proposed question is as follows; what is the closed-form of the determinant of the next matrix   $$ w_n = \begin{pmatrix}   1 & x_1^{i_1} & x_1^{i_2} & \cdots & x_1^{i_{n-2}} & x_1^{i_{n-1}} \\   1 & x_2^{i_1} & x_2^{i_2} & \cdots & x_2^{i_{n-2}} & x_2^{i_{n-1}} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\   1 & x_n^{i_1} & x_n^{i_2} & \cdots & x_n^{i_{n-2}} & x_n^{i_{n-1}} \end{pmatrix}. $$ where $i_t$, for $1\leq t \leq n-1$, are positive integer numbers such that  $i_1<i_2<\cdots<i_{n-2}<i_{n-1}$.


Googling for solutions to this problem seems to lead to third party solutions like , sometimes with scripting like .  I'm no stranger to either, but It seems crazy to me that there isn't a better solution.  I'm on Windows 10.  I use HDMI audio to my monitor from my AMD video card, and I use a set of headphones plugged into the back of my motherboard.  I use both often, but right now I need to  switch the default playback device to switch between them (sometimes, for reasons I don't yet know how to replicate, I even need to fully disable the device currently making sounds.    This is what I see in the volume mixer after starting my computer up with AMD HDMI as the default, and opening Chrome to watch a video:    When I switch to the Speakers device    I see that it is not assigned any applications, and I see no way to reassign applications:    I can usually hammer all of the applications over to the Speakers device by switching the default playback device, and usually vice versa.  But that sometimes that only switches some applications, and sometimes it doesn't do anything.  But that's besides the point...  How can I assign an application to a specific audio playback device?  The Volume Mixer looks like the right place, but it doesn't seem to do it unless I'm missing something.

Anyone know how I can set for an application to output its audio to a specific device in Windows? I'm dealing with applications where it's not possible to go into their settings and choose which output device you want to use.


I need to determine if an object already exists in an array in javascript.  eg (dummycode):  var carBrands = [];  var car1 = {name:'ford'}; var car2 = {name:'lexus'}; var car3 = {name:'maserati'}; var car4 = {name:'ford'};  carBrands.push(car1); carBrands.push(car2); carBrands.push(car3); carBrands.push(car4);   now the "carBrands" array contains all instances.  I'm now looking a fast solution to check if an instance of car1, car2, car3 or car4 is already in the carBrands array.  eg:  var contains =  carBrands.Contains(car1); //<--- returns bool.   car1 and car4 contain the same data but are different instances they should be     tested as not equal.  Do I have add something like a hash to the objects on creation? Or is there a faster way to do this in Javascript.  I am looking for the fastest solution here, if dirty, so it has to be ;) In my app it has to deal with around 10000 instances.  no jquery

What is the most concise and efficient way to find out if a JavaScript array contains a value? This is the only way I know to do it: function contains(a, obj) {     for (var i = 0; i < a.length; i++) {         if (a[i] === obj) {             return true;         }     }     return false; }  Is there a better and more concise way to accomplish this?


I've used my 32 Gb USB flash to try ubuntu. Now I don't want ubuntu anymore so I formated my USB flash. But size of it now 5 Mb instead of 32 Gb (likely this is because ubuntu uses two partitions and first one is very small)  In Windows I've found that the rest of the USB just "Unallocated" but I can't "Shrink".    How can I have entire flash back? "Format" doesn't help.

Out of the box, Windows will not allow you to partition a USB Thumb / Flash Drive. How can you do this?  In case anyone is wondering Why you would do this?  It's interesting in the case of a  drive where you can take a large thumbdrive (like 4 or 8 GB) and format a small amount of the drive to be used normally as a Fat32 drive and leave the remaining partition encrypted using BitLocker. This allows you to keep some non secret data on the unencrypted partition(like portable apps).


Is there a better solution than this :  2^n mod(m) {       if n=0         return 1;      r=2^(n-1)mod(m);       if 2r < m          return 2r;     if 2r > =m          return 2r-m; }

I have to compute $$2^{p-1} \mod p$$ and show by Fermat's little theorem that $p$ isn't prime. I know what the question is asking but I'm not sure how to reduce the exponent on $2^{p-1}$ to a more reasonable number. Does that make sense? Any general tips would be appreciated.


I am wanting to experiment with Ubuntu 64-bit, the latest desktop version using vmware. I found the isos I think, but I don't know which one is just the OS.

What's the difference between the server version of Ubuntu and the desktop version?


I am tring to program Game in Life in python. My output looks like this. I don't really want to use any modules (those where installations and download is needed) because my old teacher said that if he won't be able to start program he will fail you. That is why I need basic python. I would love to change colors of my "1" to green to make it easy to spot changes while program is working. I heard about colorama, but I am not sure how to make it possible to choose every "1" from my list and paint it without changes in output besides colors. Screenshot:

How can I output colored text to the terminal in Python?


There's a  (attributed to Shog9 but not verifiable by me) that     If six users without the ability to delete "recommend deletion", the post will also be automatically deleted.   If this is true, I find it worrisome. It creates a parallel vote-to-delete that users between 2000 and 20,000 rep can vote in, even if some of these users are not even trusted to vote to close.  What's more, these users are presented with the choice in a way that stongly encourages them to vote for deletion, in that the only alternatives (apart from "don't know") is "looks good" and "edit it to look good". This suggests to users that if they understand the question and the answer (and so "don't know" is not right), there is no middle ground between answers that "look good" (or can be made to do so) and answers that ought to be deleted. In effect the UI is suggesting that unless the user would upvote the answer, he should vote for deletion -- even though the vote is presented to him as a "recommendation".  It feels very backwards to encourage users to vote to delete by making it free, whereas merely downvoting an answer costs 1 rep.  At least at Math.SE we have a significant population of answers that are not good answers, but nevertheless shouldn't be deleted -- downvotes are the right way to deal with them. These answers are often short enough to trigger the low-quality filter. Their major characteristic is that they often embody the first instinctive reaction of somebody looking at the question, but are actually wrong for subtle reasons explained in comments or other answers. Having them stay visible but downvoted conveys valuable information to the reader, namely that this intuitively "obvious" answer is in fact wrong. Deleting them would entail a loss in usefulness.  If the secret parallel deletion vote is real, I think that   It should be made much more obvious to the user casting the "recommendation" that such a recommendation is in fact a vote that will cause automatic deletion without being reviewed by a moderator or high-rep user. The review interface really should provide a middle way between "looks good, I will upvote" and "burn it with fire!", such as "no action is needed" or "downvoting is good enough for this".

are pretty cool... You get a birds-eye view of activity on the site, a "dashboard" view of what's happening. Some of the individual tools haven't scaled particularly well with Stack Overflow's growth, but the concept behind them is still sound: we trust you to enough to be a bee watcher now.  ...and then there's . What a let-down! Once upon a time, this queue contained spam and offensive flags, which one might reasonably assume were important enough to put in front of the more trusted members of the site. Re-flagging them brought them that much closer to deletion, while editing them offered salvation to some hapless author. Nowadays, it's a bunch of Not an Answer flags and a smattering of assorted cruft. 10K users can't even vote to delete these; only 20K users have that privilege. Re-flagging them does nothing but increase their priority in the moderator flag queue, where they frequently outrank more pressing issues;  has . Also, it's , and the behavior has diverged far enough from that of the moderator flag queue on which it is based that it has become an active hindrance to further development of the tooling there. Worst of all, we have  that's available to anyone with the editing privilege. It even has logic built in to prioritize answers likely to be deleted for users with . It's like we gave you a car for your 10th birthday, and then replaced it with a rusty bicycle when you turned 16. It was a nice idea, but it has outlived its usefulness. Proposed changes The /tools/flagged route goes away. Period. No replacement. The rest of the 10K tools stick around as informational pages. Not an Answer flags go into /review/low-quality, . Then we beef up the Low Quality review process to make better use of more experienced reviewers and solve this whole "declined / helpful / disputed" flag debate once and for all:  Effective # of reviews required == ReviewsRequired + # of applicable flags (where ReviewsRequired is 2 on Stack Overflow, 1 everywhere else). So 1 VLQ or NAA flag means EffectiveReviewsRequired=3. LQ tasks are not dequeued until one of the following conditions is met:  Post is edited from within review. Outcome: flags are marked "helpful" (current behavior). Post accumulates 3 Delete votes (can only happen when post scores <= 0 and reviewers have >= 20K rep). Outcome: post is deleted, flags are marked "helpful". Task accumulates EffectiveReviewsRequired "Looks Good" reviews. Outcome: if the number of (Recommend)Delete reviews is >= the number of Looks Good reviews, mark flags "disputed" and raise DisputedLowQuality mod flag. Otherwise, mark flags "declined". Task accumulates 6* RecommendDelete + Delete reviews. Outcome: mark flags "helpful". If the post scores > 0 then raise DisputedLowQuality mod flag, else just delete post (current behavior).   As under the current system, flags on posts that've already completed one full review cycle without being deleted should skip /review and go directly into the mod queue.  End result  Flags get handled faster, more accurately, and with less wasted effort from 10K reviewers. Moderators are free to focus on situations that can't be handled by the community - the ! Developers are free to make enhancements to the moderator tooling without having to work around 10K user restrictions.  Questions Am I forgetting anything here? I haven't really spent much time in the 10K flag queue since back when it was filled with spam flags and therefore useful - is there a use-case that you depend on that would be lost with its removal? Post 'em below. *: on Stack Overflow, only .


I tried using __name__ == "__main__" in my python program, but didn't understand it properly. Below is my code block:  def main():   print 'hello' if __name__ == '__main__':   main()

Given the following code, what does the if __name__ == "__main__": do?  # Threading example import time, thread  def myfunction(string, sleeptime, lock, *args):     while True:         lock.acquire()         time.sleep(sleeptime)         lock.release()         time.sleep(sleeptime)  if __name__ == "__main__":     lock = thread.allocate_lock()     thread.start_new_thread(myfunction, ("Thread #: 1", 2, lock))     thread.start_new_thread(myfunction, ("Thread #: 2", 2, lock))


How can I investigate if an object appears through a camera in BGE?  Ray sensor does not suit me cause it is one line, while I am searching for the whole screen.

What I want to do here is create a system where an enemy object has some kind of "sight", as in the enemy can detect a player object when the player object comes in its field of view. I am currently thinking of the following two ways:   A spotlight that is parented to the enemy object acts as it's field of view. The enemy object casts multiple rays, with each ray being 5 degrees away from the previous one, creating an isosceles triangle of some sorts with rays between it.   The first method is problematic, and I do not know how to go about with the second method. How can I go about implementing it? Also, is there another way of doing the same that is better than the two methods listed above? If so, please do mention it.


Receiving permission denied error when writing to the file.  Any help is appreciated. Thank you

When you attempt to modify a file without having write permissions on it, you get an error:  > touch /tmp/foo && sudo chown root /tmp/foo > echo test > /tmp/foo zsh: permission denied: /tmp/foo   Sudoing doesn't help, because it runs the command as root, but the shell handles redirecting stdout and opens the file as you anyway:  > sudo echo test > /tmp/foo zsh: permission denied: /tmp/foo   Is there an easy way to redirect stdout to a file you don't have permission to write to, besides opening a shell as root and manipulating the file that way?  > sudo su # echo test > /tmp/foo


Let $G$ be a set with an associative operation defined on it.  Assuming $G$ is finite and both cancellation laws hold, show that $G$ is a group.  Is knowing that $G$ is associative under an operation and finite enough to say that it is closed? Also how can I build an identity from this?

Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.  Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.  2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.  3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.  4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have  $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.  Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.  Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right?


I have just restored a backup taken from xtrabackup (only for testing, so not critical luckily), and I am getting the following error:  InnoDB: Error: log file ./ib_logfile0 is of different size 0 125304832 bytes than specified in the .cnf file 0 262144000 bytes!   So I updated my my.cnf file to match this value:  [mysqld] . . .  innodb_log_file_size = 125304832   But I am stil getting an error:  log file ./ib_logfile0 is of different size 0 125304832 bytes than specified in the .cnf file 0 124780544 bytes!   But my my.cnf file is now set to 125304832 bytes so why does MySQL still think it is only 124780544 bytes?

So I'm fairly new to tuning InnoDB. I'm slowly changing tables (where necessary) from MyIsam to InnoDB. I've got about 100MB in innodb, so I increased the innodb_buffer_pool_size variable to 128MB:  mysql> show variables like 'innodb_buffer%'; +-------------------------+-----------+ | Variable_name           | Value     | +-------------------------+-----------+ | innodb_buffer_pool_size | 134217728 | +-------------------------+-----------+ 1 row in set (0.00 sec)   When I went to change the innodb_log_file_size value (example my.cnf on  comments to change the log file size to 25% of the buffer size. So now my my.cnf looks like this:  # innodb innodb_buffer_pool_size = 128M innodb_log_file_size = 32M   When I restart the server, I get this error:      110216  9:48:41  InnoDB: Initializing buffer pool, size = 128.0M   110216  9:48:41  InnoDB: Completed initialization of buffer pool   InnoDB: Error: log file ./ib_logfile0 is of different size 0 5242880 bytes   InnoDB: than specified in the .cnf file 0 33554432 bytes!   110216  9:48:41 [ERROR] Plugin 'InnoDB' init function returned error.   110216  9:48:41 [ERROR] Plugin 'InnoDB' registration as a STORAGE ENGINE failed.   So my question: Is it safe to delete the old log_files, or is there another method to change the innodb_log_file_size variable?


I'm a brand new user to Stack Exchange and I was wondering if I could get some guidance on how this site works and how to ask questions. I have been told by my friends I need to be clear and understandable when writing, otherwise I will get downvoted and put in "the slaughterhouse".  When I first signed up, there was a page that explained how to write good questions. However, it seemed quite vague and I'm not sure it was massively helpful. Can I have some guidance from you guys with more experience in the Stack Exchange community on the basics of how Stack Exchange works, as well as an explanation of how to write good questions on Stack Exchange?  It would be very much appreciated!

As a new user, I find it impossible to contribute to the community or to even begin contributing. I know why the reputation system is in place and I respect it, but at the same time it is too hard on new users.  It started with me wanting to upvote a question...     "Vote Up requires 15 reputation"   Just to appreciate a good question that helped me I need 15 rep? OK..., so I find out how I can get reputation points. More or less, I need to ask a question for +5 and add an answer for +10. I don't have any questions at the moment and I don't want to spam a random question, but I did happen to have a slightly different answer to the earlier question that I wanted to upvote. So I try to answer...     This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.   Now... I know this doesn't apply to every question, but at the same time, this means I'm forced to answer yet another question, and I don't want to go around looking for random questions and spam irrelevant answers...  Now I'm stuck. So I try to talk to some other users in the chat rooms about how to go about getting reputation points and how they first started. I entered into a "casual chat room".     You must have 20 reputation on Meta Stack Overflow to talk here.   Yes... I can understand why this is enforced, because we don't want new users to spam chat rooms, but all I wanted to do was ask for help. So I can't even talk to anyone about my newbie problems. So okay, I'll just settle for upvoting comments in the question that helped me. I don't see an upvote hover-over button so I can't even do that.    So now I'm stuck at 1 reputation point, unable to cross that 15 rep barrier, and even after posting this question I'll only have 6 rep points and still unable to do anything, and I'll be forced to look for questions to answer. Yes, I should be contributing to the community, but I shouldn't need to begin answering or making questions to upvote questions and answers. This only spurs more spam/irrelevant answers if anything.   Very, very, very new-user unfriendly.


Say a neutron which have magnetic moment despite carrying no charge. Isn't both the electric field and magnetic field consist of virtual photons? So how can the same virtual photons give rise to 2 different properties? Since virtual photon is the excitation of a quantum field how come the same excitation of the quantum field can have different behavior like giving particle electric charge or spin?

I continue to find it amazing that something as “bulky” and macroscopic as a static magnetic or electric field is actually a manifestation of virtual photons.  So putting on your QFT spectacles, look closely at the space near the pole of a powerful magnet – virtual photons! Now look between the plates of a charged capacitor – virtual photons again!   But if it’s all virtual photons, how do we get the difference between a magnetic and electric field?  [This question needs the tag "magnetostatics" as well].


Let $R$ be a finite Boolean ring with $1 \neq 0$. Show that $R\cong \mathbb{Z}_2\times \mathbb{Z}_2 \times \dots \times \mathbb{Z}_2$.   This is exercise 2 on p267 in the book abstract algebra of Dummit and Foote. The author gives the hint to use that if $e$ is an idempotent of a ring $R$ with $1$, then  $$R \cong Re \times R(1-e)$$  where $Re$ is a ring with identity $e$ and $R(1-e)$ a ring with identity $1-e$.  My attempt:  If $R = \{0,1\}$, then $R \cong \mathbb{Z}_2$ is trivial.  Hence, we may assume that $\{0,1\}$ is properly contained in $R$. Thus,we can pick an element $x \notin\{0,1\}$. We obtain because every element in a Boolean ring is an idempotent:  $$R \cong Rx \times R(1-x)$$  $Rx$ is a ring with identity $x \neq 0$ and $R(1-x)$ is a ring with identity $1-x \neq 0$. Moreover, $Rx$ and $R(1-e)$ have cardinality strictly smaller than the cardinality of $R$. These rings are also Boolean. Induction yields the result. $\quad \square$  Is this correct?

I'm trying to prove that every finite Boolean ring, $R$, is isomorphic to a finite number of copies of $\mathbb{Z} / 2 \mathbb{Z}$:  \begin{align} R \cong \mathbb{Z} / 2 \mathbb{Z} \; \times \cdots \; \times \; \mathbb{Z} / 2 \mathbb{Z} \end{align}  I'm aware that there are answers to this question on this website. However, almost all of the answers seem to use facts about modules and vector spaces defined over fields. This question is posed in section 7.6 in Dummit and Foote's Abstract Algebra. This section is part of the chapters which introduces the basics of ring theory (rings, homomorphisms, ideals, Chinese Remainder Theorem etc.).  I'm not even sure how to being answering this question. I know that if $R$ were an integral domain, then we would have that $R \cong \mathbb{Z} / 2 \mathbb{Z} $.   Please only give hints, considering that that the statement has to be proved using the material only included up till chapters 9 in the textbook.


I installed mysql 5.7 and removed mysql 8.0 on Ubuntu 20.04.  I used followed . But, when Ubuntu update starts, it try to uninstall MySQL 5.7 and install MySQL 8.0.  How can I prevent it?

Is it possible to install MySQL Server 5.7 on Ubuntu 19.10? If yes, how?


I need to prove that :  $$\lim_{n \to \infty} \frac{x^n}{n!}=0$$  I tried to prove it using epsilon-delta, but I still haven't managed to prove it.  I think that maybe with sandwich theorem it possible to prove it.  Any help will be appreciated.

Why is     $$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$   Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is     $$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$     This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .


I have a problem since the update of Ubuntu 17.10  I cannot create a new file from the GUI. I have to create the files from the console.      Any idea?

How can I add a right-click desktop menu entry to create a text file to Gnome 3 in Ubuntu 16.04.02?


I've install Ubuntu latest version 14 . now I Have completed with installation. but Have not created disk partition . Pleas guide me how to make Partitions after Installation

I have a 80Gb HDD with one partition for Ubuntu and want to partition it into two partitions 60G/20G.   How can I do this?


package test2;  public class NewJFrame extends javax.swing.JFrame {      private static void valueGen() {         String x = jTextField1.getText();         System.out.println(x);     }      public NewJFrame() {         initComponents();     }      private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) {         jTextField1.setText("Hello");     }      public static void main(String args[]) {         valueGen();          java.awt.EventQueue.invokeLater(new Runnable() {             public void run() {                 new NewJFrame().setVisible(true);             }         });     }      private javax.swing.JButton jButton1;     private javax.swing.JTextField jTextField1; }   I have a program as shown above.  I need to access the value of jTextField1 from the function valueGen(). But I get the error:  non-static variable cannot be referenced from static context   What should I do in order to access the value of jTextField1 from valueGen()?

The very common beginner mistake is when you try to use a class property "statically" without making an instance of that class. It leaves you with the mentioned error message:     You can either make the non static method static or make an instance of that class to use its properties.   Why? I am not asking for solutions. I would be grateful to know what is the reason behind it. The very core reason!  private java.util.List<String> someMethod(){     /* Some Code */     return someList;             }  public static void main(String[] strArgs){                // The following statement causes the error. You know why..     java.util.List<String> someList = someMethod();          }


Take any $K_{1}$ and $K_{2}$ closed sets. Define $A = \{||x-y||: x \in K_{1}, y \in K_{2}\}$. I'm asked to prove (or give a counterexample) that $A$ is closed, in general.  A classmate came up with the following counterexample:      Consider $K_{1},K_{2} \subseteq \mathbb{R}^2$ such that $K_{1} = \{(x,y) \in \mathbb{R}^2: x\geq 1, y=0\}$, i.e, the horizontal axis from $1$ on and $K_{2} = \{(x,y) \in \mathbb{R}^2: x\geq 1, y=1/x\}$. Clearly, both sets are closed. Notice that there are elements in $A$ that are arbitrarily close to $0$ but $0 \notin A$. Then, $A$ is not closed.   I could come up with some more examples, but all of them in the same spirit of the one above. I could also prove that if either $K_{1}$ or $K_{2}$ is compact, then $A$ is closed. So, a counterexample should rely on two unbounded sets, since I'm thinking about Euclidean spaces for the sake of simplicity.  In special, I was trying to find a counterexample in which $K_{1}$ and $K_{2}$ are subsets or the real numbers, but I had no success! Any ideas? Thanks a lot in advance!

Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?  (By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)  Thanks!


I have an available update for Xcode that is in the 7gb range, and when I try to update I get an alert saying there isn't enough disk space - I checked my space and I have 19gb available (of 120gb).  Is this correct, or some kind of error? It seems like I have more than enough space.  Update: I deleted some earlier device support files and I now have 24gb free space and I still don't have enough space for the update. I feel like something's amiss here, as the update is only 7gb - I'm not sure if I should keep deleting things to free up space - will it even make a difference?  Update 2: I restarted my computer with 24gb free space and when I clicked Update, the update seemed to go forth and download - however when the download completed, the button switched back to "Update". When I clicked on it again, I got the "not enough space" alert again - even though now I have 26.5gb free space. I restarted again and keep getting the same message. Truly don't know what's going on here! Here's a screenshot, as requested, of my storage window:    Update 3:  Screenshot after running df -h in terminal as per request:

I am trying to update Xcode to 8.0. My OS is 10.11.6 El Capitan. However I am getting:     We could not complete your update   There is not enough disk space available to install the product.   I have got 14.22 GB free space. How much do I need?  UPDATE: I do not know the exact figure, though I did manage to update with 17.21 GB free space later on.