$E=mc^2$ Does not this imply that there is no energy without mass,  or is it strictly a conversion formula?
Einstein equation  where $E$ is energy, $m$ mass, and $c$ the speed of light in vacuum. So does it mean objects without any mass does not posses energy for eg lights photons does not have mass but how can they posses energy
Using first person or third person in papers? I know that I can use something like cat test.txt | pr -w 80 to wrap lines to 80 characters  wide, but that puts a lot of space on the top and bottom of the printed lines and it does not work right on some systems  What's the best way to force a text file with long lines to be wrapped at a certain width?   Bonus points if you can keep it from breaking words. How to restore the down arrow icon on the Downloads folder in the home directory? I removed the original Downloads folder in the home directory and created a new one with the same name, but the down arrow icon on the new Download folder now disappeared. Could someone please tell me how to restore that icon?   In addition, from the image above we see "Downloads" disappears also in "Places" category in the left panel. I think it should be there originally. If after restoring the down arrow icon, the "Downloads" is still not in "Places" category, I would also like to ask you to help me put "Downloads" back to "Places". Thank you.  PS: I am not using any theme. Possible to draw an Item in a Grid with only 3 sides of a Frame? A probability theory question about independent coin tosses by two players Say Bob tosses his $n+1$ fair coins and Alice tosses her $n$ fair coins. Lets assume independent coin tosses. Now after all the $2n+1$ coin tosses one wants to know the probability that Bob has gotten more heads than Alice.    The way I thought of it is this : if Bob gets $0$ heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set $\{1,2,\dots,n+1\}$. And if Bob gets $x$ heads then the number of heads that Alice can get is anything in the set $\{0,1,2,..,x-1\}$. So\begin{align}P(\text{Bob gets more heads than Alice})&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&amp;= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]&amp; = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align}   How does one simplify this?      Apparently the answer is $\frac{1}{2}$ by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. How to use siunitx with non numerical values? I understand that something like \SI{5}{\meter\per\second} defines a certain space between the number (5) and units (m/s). What should I use if instead of a numerical quantity I want a variable: R m/s? \SI{R}{\meter\per\second} fails because it expect a number in first parameter, and R \si{\meter\per\second} or $R$ \si{\meter\per\second} show different space.  Which is the correct use?  \documentclass{standalone} \usepackage{siunitx} \sisetup{per-mode=symbol} \begin{document} \begin{tabular}{r} \SI{5}{\meter\per\second}\\ $R$ \si{\meter\per\second}\\ R \si{\meter\per\second}\\ \end{tabular} \end{document} I was curious how I would go about running multiple commands via shell script.  For example, just to try it out, I'd like to be able to connect a .desktop to a .sh that will run all the update codes so I don't have to type it out. I know how to execute a single command via a .sh, not multiple.  Any way? C# Textreader and TextWriter are the another way to read and write file respectively...      I added a text box and button I called tbUpdate and btUpdate respectively...      In the right hand column, we’ll use three TextBlocks; the first and third as fixed labels (with the values Likeability: and % respectively)...      WorkerSupportsProgress and WorkerSupportsCancellation which hold true/false values that let the backgroundworker report progress or cancel an async progress respectively...   While learning many the many different aspects of development, I have not come across a book or series where the word respectively is not used in explaining an example. Honestly I guess it's implied that we know what it means. To me I always kind of skim over it because I can usually figure out what they're saying through the surrounding context. Can anyone explain this? I'm currently using an entry-level DSLR with 18-55mm lens and having great trouble trying to creating a blurred background effect in my photos. Now matter what aperture or shutter speed I choose, I'm not able to get it.  There's a general question about , but what specific things can I do to get this effect with an entry-level lens? How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number? It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational? I found this interesting problem in :     Given a point $P$ in space, and given a piece of malleable material of constant density, how should you shape and place the material in order to create the largest possible gravitational field at $P$?   Any ideas? How can I parse integers passed to an application as command line arguments if the app is unicode?  Unicode apps have a main like this:  int _tmain(int argc, _TCHAR* argv[])   argv[?] is a wchar_t*. That means i can't use atoi. How can I convert it to an integer? Is stringstream the best option? I lost the sa password on a machine, and when I log in to the machine directly using an account in the admin group, SQL Server Management Studio will not allow me to log in using Windows authentication.  My plan was to simply log into the server, connect via Windows Authentication, and reset sa to use a new password. Since I cant' connect via Windows Authentication, this won't work.  How else can I reset the sa password? Three side-by-side tables I am using this LaTeX code to generate three tables side by side. However, it showing errors and not giving satisfactory result. Can anyone please help me to short out the problem.  \begin{table} \centering \makebox[0pt][c]{\parbox{1.2\textwidth}{% \begin{minipage}[b]{0.265\hsize}\centering \begin{tabular}{ | c | c |c | } \multicolumn{3}{c} \hline {$q=1/10$} \\ \hline T &amp; $r_h$ &amp; $\omega$ \\  \hline 0.233 &amp; 0.135 &amp; 2.38486-0.34667I  \\  0.256 &amp; 0.140 &amp; 2.38092-0.35837I \\ 0.273 &amp; 0.145 &amp; 2.38049-0.35982I \\  0.278 &amp; 0.146 &amp; 2.38007-0.36130I \\  0.281 &amp; 0.147 &amp; 2.37964-0.36282I \\  \hline 0.284 &amp; 0.761 &amp; 2.54150-2.03209I \\ 0.285 &amp; 0.765 &amp; 2.54515-2.04292I \\ 0.289 &amp; 0.800 &amp; 2.57799-2.13761I \\ 0.302 &amp; 0.900 &amp; 2.67987-2.40712I \\ 0.312 &amp; 1.000 &amp; 2.79214-2.67552I \\ \hline \end{tabular} \end{minipage} \hfill \begin{minipage}[b]{0.265\hsize}\centering \begin{tabular}{ | c | c | c | } \multicolumn{3}{c} \hline {$q=1/9$} \\ \hline T &amp; $r_h$ &amp; $\omega$ \\ \hline 0.227 &amp; 0.150 &amp; 2.34385-0.41094I  \\  0.245 &amp; 0.155 &amp; 2.34215-0.41723I \\ 0.260 &amp; 0.160 &amp; 2.34056-0.42416I \\  0.271 &amp; 0.165 &amp; 2.33908-0.43175I \\  0.276 &amp; 0.167 &amp; 2.33852-0.43497I \\  \hline 0.280 &amp; 0.720 &amp; 2.50248-1.92299I \\ 0.283 &amp; 0.750 &amp; 2.52893-2.00424I \\ 0.286 &amp; 0.780 &amp; 2.55660-2.08533I \\ 0.289 &amp; 0.800 &amp; 2.57568-2.13931I \\ 0.302 &amp; 0.900 &amp; 2.67807-2.40842I \\ \hline \end{tabular} \end{minipage} \hfill \begin{minipage}[b]{0.265\hsize}\centering \begin{tabular}{ | c | c | c |} \multicolumn{3}{c} \hline {$q=1/8$} \\ \hline T &amp; $r_h$ &amp; $\omega$ \\  \hline 0.246 &amp; 0.180 &amp; 2.29733-0.50611I  \\  0.255 &amp; 0.185 &amp; 2.29654-0.51364I \\ 0.263 &amp; 0.190 &amp; 2.29589-0.52172I \\  0.269 &amp; 0.195 &amp; 2.29534-0.53033I \\  0.273 &amp; 0.199 &amp; 2.29496-0.53758I \\  \hline 0.275 &amp; 0.675 &amp; 2.46067-1.80426I \\ 0.277 &amp; 0.700 &amp; 2.48131-1.87194I \\ 0.288 &amp; 0.800 &amp; 2.57245-2.14170I \\ 0.302 &amp; 0.900 &amp; 2.67555-2.41027I \\ 0.317 &amp; 1.000 &amp; 2.78870-2.67796I \\ \hline \end{tabular} \end{minipage}% }} \end{table} Does a constantly accelerating charged particle emit EM radiation or not? The  gives the recoil force, $\mathbf{F_{rad}}$, back on a charged particle $q$ when it emits electromagnetic radiation. It is given by:  $$\mathbf{F_{rad}} = \frac{q^2}{6\pi \epsilon_0 c^3}\mathbf{\dot{a}},$$  where $\mathbf{\dot{a}}$ is the rate of change of acceleration.  If a particle has a constant acceleration, $\mathbf{\dot{a}}=0$, then there is no reaction force acting on it. Therefore the particle does not lose energy.   Does this mean that a constantly accelerating charged particle does not emit electromagnetic radiation? Some questions about Hilbert matrix This Exercise $12$ page $27$ from Hoffman and Kunze's book Linear Algebra.     The result o Example $16$ suggests that perhaps the matrix      $$A = \begin{bmatrix} 1 &amp; \frac{1}{2} &amp; \ldots &amp; \frac{1}{n}\\  \frac{1}{2} &amp; \frac{1}{3} &amp; \ldots &amp; \frac{1}{n+1}\\ \vdots &amp; \vdots &amp;  &amp; \vdots \\ \frac{1}{n} &amp;\frac{1}{n+1} &amp;\ldots &amp;\frac{1}{2n-1} \end{bmatrix}$$      is invertible and $A^{-1}$ has integer entries. Can you prove that?   I will copy their words before that example:      Some people find it less awkward to carry along two sequences of   matrices, one describing the reduction of $A$ to the identity and the   other recording the effect of the same sequence of operations starting   from the identity.   In this , Adrián Barquero says:      If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's   Linear Algebra book then I was in the same situation you're right now   a few years ago, but I was adviced at the moment that this particular   exercise wasn't as easy as one might expect at first.      The matrix you have is called the Hilbert matrix and the question you   have was already asked a couple of times in math overflow  and   . They have excellent answers so I will just point you to them.   My question: Is it possible answer the question made by the authors without using high techniques?  I am not allowed to use determinants here.  PS: Those answers in MO don't answer the question above. I apologise if this question is unappropriated or if I was unable to understand their answers. CASE WHEN with GROUP BY I have a table that looks like this (oversimplified case of the actual case to make my point)  CREATE TABLE sales (    id int,    currency VARCHAR(5),    price int    shop_id int );   Now I need to make a query that will give me something like this table, grouping by shops yet giving sum of sales per currency My problem is when I make currency column non aggregate it fails because my query becomes non standard. When I add currency to group by then each currency creates new row, something I don't want. I have run out of option and would appreciate any help or pointer to the right direction.  I use MySQL but would appreciate if I would get the solution that works with more databases (PostGreSQL being at the top of the list after MySQL)  Query that works with undesired results  SELECT   shop_id,           (CASE WHEN currency= "GBP" THEN SUM(price) ELSE 0 END) AS POUND,           (CASE WHEN currency= "USD" THEN SUM(price) ELSE 0 END) AS USDOLLAR GROUP BY shop_id, currency   The data that should be produced by the query  shop_id     |   POUND   |   USDOLLAR     -----------------------------------------     1       |   400     |   300 -----------------------------------------     2       |   250     |   100 -----------------------------------------     3       |   400     |   100 ----------------------------------------- Why is 4 not an instance of Number? Just curious:    4 instanceof Number => false new Number(4) instanceof Number => true?   Why is this? Same with strings:    'some string' instanceof String returns false new String('some string') instanceof String => true String('some string') instanceof String also returns false ('some string').toString instanceof String also returns false   For object, array or function types the instanceof operator works as expected. I just don't know how to understand this.  [new insights]  Object.prototype.is = function() {         var test = arguments.length ? [].slice.call(arguments) : null            ,self = this.constructor;         return test ? !!(test.filter(function(a){return a === self}).length)                : (this.constructor.name ||                   (String(self).match ( /^function\s*([^\s(]+)/im )                     || [0,'ANONYMOUS_CONSTRUCTOR']) [1] ); } // usage var Newclass = function(){};  // anonymous Constructor function var Some = function Some(){}; // named Constructor function (5).is();                     //=&gt; Number 'hello world'.is();           //=&gt; String (new Newclass()).is();        //=&gt; ANONYMOUS_CONSTRUCTOR (new Some()).is();            //=&gt; Some /[a-z]/.is();                 //=&gt; RegExp '5'.is(Number);               //=&gt; false '5'.is(String);               //=&gt; true Problem from Herstein on group theory The problem is:  If $G$ is a finite group with order not divisible by 3, and $(ab)^3=a^3b^3$ for all $a,b\in G$, then show that $G$ is abelian.  I have been trying this for a long time but not been able to make any progress. The only thing that I can think of is: $$ab\cdot ab\cdot ab=aaa\cdot bbb\implies(ba)^2=a^2b^2=aabb=(\text{TPT})abba.$$ Now, how can I prove the last equality? If I write $aabb=abb^{-1}abb$, then in order for the hypothesis to be correct, $b^{-1}abb=ba\implies ab^2=b^2a$. Where am I going wrong? What should I do? Why do backslashes appear twice?