How do I generate LaTeX code faster? I am a new member on stack exchange and I find it very tedious to write codes for mathematical equations in order to ask and answer questions. Is there a shorter and quicker way (with help from some websites and applications) to do that for free?
Slow math typing I have been using LaTeX for quite some time, but I find the math writing to be a bit frustrating. It seems to take me a relatively long time to code-write long math expressions, and sometimes it's hard to see where { start and end.  I am interested in writing as fast as possible.  Is code-typing each symbol really the fastest way? Is there any good program (in the spirit of Word's MathType) that enables producing math expressions by button clicking, that you recommend?
I'm working on a fairly big document that has a lot of graphs of functions- so far it has around 200, by the end it will have around 1500. I use pgfplots for each graphic, and the functions are pretty basic: exponential, rational, polynomial, and trigonometric.  Once I am happy with each graph, I don't necessarily need to see it every time I compile, and would like to minimize compilation time for future drafts. Naturally, the document is already chopped up into chapter files, but each chapter contains around 100 figures.  Having studied the pgfplots documentation, I haven't found a draft mode, but I'm interested in knowing if one could be created.    I'm envisaging something similar to the draft mode of the graphicx package, in which graphics are not included, and blank boxes are put in their place.  Some things I've considered:   the  documentclass; but with about 1500 figures, wouldn't this mean over-modulization?  tikzexternalize, but this still requires an initial 'monster' compilation to get the images externalized before the draft option could be invoked   Here's a MWE to play with  \documentclass{article} \usepackage{pgfplots}  \begin{document}  \begin{tikzpicture}     \begin{axis}         \addplot {x^2};     \end{axis} \end{tikzpicture}  \end{document} Why does the speed of light in vacuum have no uncertainty? I could understand that the definition of a second wouldn't have an uncertainty when related to the transition of the Cs atom, so it doesn't have an error because it's an absolute reference and we measure other stuff using the physical definition of a second, like atomic clocks do.  But why doesn't the speed of light have uncertainty? Isn't the speed of light something that's measured physically?  Check out that . Mirrored object not symmetrical I am trying to mirror an object and the original is symmetrical along all axis, but when I attempt to mirror across the y axis, the mirrored side is tilted to a different angle. How to properly install Microsoft Office in Ubuntu 14.04 with wine? I have been trying to install it many times using the Wine version avaliable from the Sofware Center. Is there a proper way to configure it for allowing the installtion? I have association bonus on superuser.com. After that I've started a bounty worst of 50 rep and earned more than 10 rep. Now I have 89 rep (38 of which is earned at superuser which is more than 10) but still not able to answer protected questions.  So, currently new user with 10 rep can answer protected question and I with 89 (38) rep can't.   How does it work now: after registration on new Stack Exchange site you get 101 rep Association Bonus and are able to answer protected questions only when you have 111+ rep on that site. How should it work: after registration on new stackexchange site you get 101 rep Association Bonus and are able to answer protected questions only when you have earned 10+ rep on that site. Can I change a user's keyboard input? I have HTML code like this :  &lt;div&gt;        &lt;a&gt;Link A1&lt;/a&gt;        &lt;a&gt;Link A2&lt;/a&gt;        &lt;a&gt;Link A3&lt;/a&gt; &lt;/div&gt;  &lt;div&gt;        &lt;a&gt;Link B1&lt;/a&gt;        &lt;a&gt;Link B2&lt;/a&gt;        &lt;a&gt;Link B3&lt;/a&gt; &lt;/div&gt;   When user clicks a link from above HTML, I want to get the jQuery object of the corresponding &lt;a&gt; element, and then manipulate its sibling. I can't think of any way other than creating an ID for each &lt;a&gt; element, and passing that ID to an onclick event handler. I really don't want to use IDs.   Any suggestions? Modern Linux commands that parallel uuencode/uudecode functionality I need to get binary files into text format.  Years ago, I used uuencode/uudecode, and it was easy:  uuencode file.xls &gt; file.txt # Delete file.xls # Do whatever with file.txt # Now I need file.xls back uudecode file.txt &gt; file.xls   But uuencode/uudecode is not on my system, and as I look around, the yum/rpm process is ugly--this gives me the clue that maybe no one uses it any more and instead there is something more modern to do this.  I looked at xxd and hexdump but couldn't get them to do it.  Could just be dumb user issues.  What can I use on Linux to do this?  Would like it to be something that tends to be in the native install so I don't need to yum/rpm it if possible.  Just for your own understanding, what I'm specifically trying to do now is source control an *.xls file but the corporate tool doesn't allow *.xls in.  I respect that constraint normally, but in my context, an *.xls is a configuration table for my code so It is logically like source code, but the corporate SVN trigger is physical--if it's *.xls, it gets rejected, so I'll encode it and source control the *.txt file.  Thanks! Note: this post is different to , which just cites part from somewhere else, without any further analysis and thinking. Furthermore, none of the answers to that post answers my question - the situations where &quot;will (going to) be doing&quot; is more appropriate.   I will be participating in IELTS test next Tuesday afternoon.  In an online class, the teacher says the following at the beginning of the lesson.  I will be using NotePod (pointed out by red rectangle) to write relevant points we talk about today.   I guess &quot;will be using&quot; is more appropriate than &quot;will use&quot; there, right?  A  explains  This &quot;tense&quot; is used for reporting a scheduled activity with special focus on its timing— as if one is standing in the future and experiencing the flow of the time. The focus is on the temporary, ongoing, or repetitive nature of the activity.   I understand the grammar rules of those verb tenses and the difference between this group  I am going to drive home tomorrow.   I am going to be driving home tomorrow.  The former is just convey my activity in the near future while the latter probably makes an implication that &quot;don't disturb me as I am driving in that time&quot; However, in the following situations, I don't see the difference between 2 verb tenses.  Welcome back to this class. Today's lesson is for the IELTS students who are going to be taking the IELTS test. it's for everybody to be honest with you. this lesson is useful for anybody who needs to do some writing.   who are going to take the IELTS test ... What really allows airplanes to fly? What aerodynamic effects actually contribute to producing the lift on an airplane?  I know there's a common belief that lift comes from the Bernoulli effect, where air moving over the wings is at reduced pressure because it's forced to travel further than air flowing under the wings. But I also know that this is wrong, or at best a minor contribution to the actual lift. The thing is, none of the many sources I've seen that discredit the Bernoulli effect explain what's actually going on, so I'm left wondering. Why do airplanes actually fly? Is this something that can be explained or summarized at a level appropriate for someone who isn't trained in fluid dynamics?  (Links to further reading for more detail would also be much appreciated) Finite choice without AC Can anyone explain how we choose one sock from each of finitely many pairs without the axiom of choice?  I mean : To choose one sock from each of infinitely many pairs of socks requires the Axiom of Choice, but for shoes the Axiom is not needed.  The idea is that the two socks in a pair are identical in appearance, and so we must make an arbitrary choice if we wish to choose one of them. For shoes, we can use an explicit algorithm -- e.g., "always choose the left shoe." Why does Russell's statement mention infinitely many pairs? Well, if we only have finitely many pairs of socks, then AC is not needed -- we can choose one member of each pair using the definition of "nonempty," and we can repeat an operation finitely many times using the rules of formal logic (not discussed here). Gravity in 2d space and inverse linear law In our three-dimensional universe, gravity obeys the inverse square law. In a four-dimensional universe, gravity would be expected to obey the inverse cube law et cetera.  In a two-dimensional universe, one would similarly expect gravity to obey an inverse linear law. I've seen it claimed that this is not actually true, when you work it out according to general relativity there would actually be no gravity at all in such a universe. Is this true? If so, is there a layman's explanation of why? Is it related to the fact that gravitational potential wells would be infinitely deep?  Would electrostatic forces follow the same law as gravity?  If there existed infinitely long strings in our universe, would they have the same gravitational effect as point particles in a 2d universe? Can I use my Windows 7 license in a dual-boot configuration? Possible Duplicate:          I've purchased a retail license of Windows 7 Ultimate for my machine at home.  I'm thinking of creating a dual-boot install:   64-bit install for work apps (need the RAM for virtual machines and RAMDisk) 32-bit install for entertainment apps (Media Centre, games, etc)   How does retail licensing work in such situations? Can I activate the same license on both installations?  Edit:  I would like to add that I want to have separate partitions for work and play. Pass values from parent to child component on click lightning I have this below code where i am trying to first search an account name and on hit button it displays the result set.  On the button click i am also loading an child component which is displaying the id of the record.  I need the child component should get displayed once i click on the record.  How can i achieve this -   Parent Component -   &lt;aura:component controller="SimpleAccountController"&gt;     &lt;aura:attribute name="account" type="account[]"/&gt;     &lt;ui:outputText aura:id="outName" value="" class="text"/&gt;     &lt;ui:inputText aura:id="name" label="Enter Name:" placeholder="Your Name" /&gt;     &lt;ui:button aura:id="button" buttonTitle="Click to see what you put into the field" class="button" label="Click me" press="{!c.getInput}"/&gt;     &lt;aura:iteration var="acc" items="{!v.account}"&gt;             &lt;a onclick="{!c.accountselected}" data-recId="{!acc.Id}"&gt;         &lt;p&gt;{!acc.Name} : {!acc.Type} : {!acc.Industry} : {!acc.Rating}&lt;/p&gt;             &lt;/a&gt;         &lt;c:HelloWorkdChildCmp selectedaccount="{!acc.Id}" /&gt;     &lt;/aura:iteration&gt; &lt;/aura:component&gt;   COntroller -   ({     getInput : function(cmp, evt) {         var myName = cmp.find("name").get("v.value");         var myText = cmp.find("outName");         var greet = "Hi, " + myName;         myText.set("v.value", greet);         var action = cmp.get("c.seaAcc");         action.setParams({ "s" : myName });         action.setCallback(this, function(response) {             var state = response.getState();             if (state === "SUCCESS") {                 cmp.set("v.account", response.getReturnValue());                 //alert("From server: " + response.getReturnValue());             }         });         $A.enqueueAction(action);     },      accountselected : function(cmp, evt) {        console.log(evt.currentTarget.getAttribute("data-recId")) ;     }  })   Child component -   &lt;aura:component &gt;     &lt;aura:attribute name="selectedaccount" type="String"/&gt;     &lt;div&gt;Id: {!v.selectedaccount}&lt;/div&gt; &lt;/aura:component&gt; The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group, I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question. My Question is :   Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$     So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:           Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??       I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?  Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you. Hamiltonian Cycle Problem At the moment I'm trying to prove the statement:     $K_n$ is an edge disjoint union of Hamiltonian cycles when $n$ is odd.   ($K_n$ is the complete graph with $n$ vertices)  So far, I think I've come up with a proof. We know the total number of edges in $K_n$ is $n(n-1)/2$ (or $n \choose 2$) and we can split our graph into individual Hamiltonian cycles of degree 2.   We also know that for n vertices all having degree 2, there must consequently be $n$ edges. Thus we write $n(n-1)/2 = n + n + ... n$ (here I'm just splitting $K_n$'s edges into some number of distinct Hamiltonian paths) and the deduction that $n$ must be odd follows easily.  However, the assumption I made - that we can always split $K_n$ into Hamiltonian paths of degree 2 if $K_n$ can be written as a disjoint union described above - I'm having trouble proving. I've only relied on trying different values for $n$ trials and it hasn't faltered yet. So, I'm asking:     If it is true, how do you prove that if $K_n$ can be split into distinct Hamiltonian cycles, it can be split in such a way that each Hamiltonian cycle is of degree 2? How to make a cyclorama I can create a collection of pictures giving a 360-degree view from a certain spot, just taking a bunch of photographs from the same spot but in a different direction.  However, what software could I use (on Windows) to stitch these images together to get one big image?  I prefer something that would stitch them together without showing any seams. I keep on seeing a fair number of questions (such as ) which get marked as answered within 5-10 minutes of being posted. Sometimes the questions like this are insanely popular and actually see some proper discussion, however often I see questions have few views and only 1 answer. The trouble with these questions is that very few people have scrutinised the answer for correctness, or possibly a better solution. I always make a point of not accepting my answers until a day or so has passed — would it not be a good idea to put a short (1 hour or so) time limit on accepting answers on new questions with few views or votes, or perhaps just an additional confirmation dialog, something along the lines of:  Your question is new and not many people have had a chance to look at it yet.  Are you sure you don't want to wait to see if anyone else has a better suggestion? if $(x_0,y_0)$ is local extrema in $ax^2 + by^2 + cxy + dx + ey + l$ then its global too. An exercise on a book asks me to prove that if a point $(x_0,y_0)$ is a local extrema for the function $$f(x,y) = ax^2 + by^2 + cxy + dx + ey + l$$ then it's also a global extrema.  The exercise asks me to consider that $g(t) = f(x_0+ht,y_0+kt)$ is a parabola, but I couldn't understand how it helps. Maybe since it's a parabola, I know that this derivative has to be $0$ in some point, and also that this function $g(t)$ has a global extrema point, but I couldn't derive anything from this. I also tried to find the derivatives, equal them to $0$ and solve the system. I got:  $$x_0 = -\frac{2 b d - c e}{4 a b - c^2}, y_0 = -\frac{   c d - 2 a e}{-4 a b + c^2}$$  But when I try to find   $$f\left(-\frac{2 b d - c e}{4 a b - c^2},-\frac{   c d - 2 a e}{-4 a b + c^2}\right)$$ so I can try to prove:  $$f(x,y)-f(x_0,y_0)$$  is either greater or less $0$ depending on the case (so I can do $f(x,y) \le f(x_0,y_0)$ and $f(x,y) \ge f(x_0,y_0)$)  things get hairy. How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}&lt;\binom{2n}{n}&lt;\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ &gt; 1 ?     Does anyone know any better elementary estimates?   Attempt.  We have $$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$ Then we have $$\left(1+\frac{k}{2(n-k)}\right)&gt;\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$ So maybe, for the lower bound, we have $$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}&gt;\frac{2^n}{\sqrt{4n}}.$$ By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes $$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$ but this isn't $&gt;\frac{2^n}{\sqrt{4n}}$.