Product of all divisors
How to prove $ \prod_{d|n} d= n^{\frac{\tau (n)}{2}}$
Beamer: how to justify the whole text? How can I justify the text in the whole document in beamer without using \usepackage{ragged2e} and \justifying at the beginning of each frame?  \documentclass[8pt,handout,t]{beamer} \usepackage{pifont} %used inside itemize environment \item[\ding{226}] \usepackage{lipsum} \begin{document}  \begin{frame} \begin{columns}[T]  \begin{column}{0.3\textwidth} \includegraphics[width=\columnwidth]{example-image-1x1} \end{column} \begin{column}{0.65\textwidth} \begin{itemize} \item[\ding{226}] \lipsum[2] \item[\ding{226}] \lipsum[3] \end{itemize} \end{column}  \end{columns} \end{frame}  \begin{frame}  \lipsum[3]  \begin{itemize}    \item[\ding{226}] \lipsum[4]  \end{itemize} \end{frame} \begin{frame}   \begin{columns}[t] \begin{column}{0.4\textwidth} \lipsum[3] \end{column} \begin{column}{0.4\textwidth} \lipsum[5] \end{column} \end{columns} \end{frame} \end{document}     Edit: I had already checked  out but it does not work for both the columns and the itemize environment. Is it possible to play FIFA 2015, 4v4 on 2 Consoles? I've the following situation: My friends and I, we're 8 in total, want to play Fifa 15 4v4. I know that the Playstation 4 just supports 4 controllers and thats why we think about a 2 console + 2 TV's setup, so 4 players per console.   So my question is: Is it possible or is there a game mode available were we could play such a setup?  Update: After I got the first answer I've to add that I know that the only way to play on 2 consoles would be through the psn online features. So we did not plan to play on Lan. We just want to sit in the same room with 2 consoles and play against each other over psn. Is it supposed to be a HTML or an HTML I've seen many people who say:     This is a HTML page.   Yet I've also seen many people who say:     This is an HTML page.   Are both usages equally correct?  Or, which is the grammatically correct one?       Possible Duplicates: Too many of your edits were rejected, try again in 7 days First time I've got the following message on SO:     Too many of your edits were rejected,   try again in 7 days.   Although, I have made ~50 accepted edits this day.  I'm sure that should outweigh the ~5 rejected edits, right?    My suggestion:  bool allowEdit(USER u) {     if(u.rejected * 5 &gt; u.accepted)         return(false);     else         return(true); }   Think of it this way: if I do a 1000 accepted edits, and mess up on 5 [rejected], does that mean I should get banned for 7 days?    Hmmm... There seems to be a bug; the message only appears when clicking edit from review.  But otherwise, I can't edit , even when going to it from the homepage. Should my first trip be to the country which issued my Schengen Visa? How to prevent freehub body damage? I feel like this is poor design. I have two wheelsets that are less than a year old which have damaged freehub bodys due to cogs digging in to the splines on the freehub. It's so bad that I can't remove the cassette spacer without filing down the splines. I had to use  pliers and chain whip to remove the cogs that were embedded in the freehub when trying to remove the cassette.  Most of the damage is on the center of the cassette as the largest 3 cogs are held together by carrier. I'm using a shimano 105 cassette and  Novatec disc hub and Velocity ATB convertible disc hubs (which were somewhat damaged in the same fashion, but not as bad as the novatec).  I'm using this bike for cyclocross, offroad exploring, light singletrack use, Rail trails and some road riding.    How can I prevent this damage to the freehub from happening? Can I travel to Canada with valid visa on old/expired passport? I recently renewed my passport. I also have an expired passport which has my Canada visa (visitor's) valid for another three years. I have already used my Canadian visa which is multiple entry. I am planning to visit Canada this summer.  Since my valid Canadian visa is in my old/expired passport, do I need to renew/transfer my Canada visa or can I simply travel with valid visa in old passport? Of Half-elven and mortal life How can I prevent system clipboard image data being pasted into a WPF RichTextBox I have the table below and i want it to read through columns A-B and then sum the values that match their names in columns C-D.  So that in a column elsewhere of a list of names i have the sum of their values from this table. For example Ed features in both columns with a value of 1 and 3, so would return a value of 4. Struggling to get this to work nicely. My actual table consists of 4 columns names and two sets of 4 columns with numbers.    A       B       C   D Steve   Jeff    4   1 James   Calum   1   1 Alex    Andy    2   1 Peter   Ed      0   3 Owen    Harry   1   2 Jeff    Steve   4   1 James   Calum   1   1 Andy    Alex    2   1 Peter   Owen    0   3 Ed      Harry   1   2 "ERROR: Extra }, or forgotten \right." in align* enviroment Why do I get a "ERROR: Extra }, or forgotten \right." in the following align* environment?  \begin{align*}     \int\left[\left(\der{E_z}{y}-\der{E_y}{z}-\mu\der{\alpha}{t}\right)\alpha&amp;+\left(\der{E_x}{z}-\der{E_x}{x}-\mu\der{\beta}{t}\right)\beta\\     &amp;+\left(\der{E_y}{x}-\der{E_x}{y}-\mu\der{\gamma}{t}\right)\gamma\left.\vphantom{\left(\der{E_y}{x}}\right]d\omega=0.     \end{align*}   thanks JSON encode MySQL results How do I use the json_encode() function with MySQL query results? Do I need to iterate through the rows or can I just apply it to the entire results object? In contemporary American English usage, I come across sentences like:   I ain’t got no money.  Ain’t no man like him.    Saying ain’t no sounds incorrect to me because it is a double negative.  What is the origin of this particular construction? I wasn’t able to find anything online related to the origins of this incorrect usage. Add an "I don't understand" option for review audits Constant Amortized Time What is meant by "Constant Amortized Time" when talking about time complexity of an algorithm? If $x_1=5$, $x_{n+1}=x_n^2-2$, find $\lim x_{n+1}/(x_1\cdots x_n)$ Quick way to subtract white outline from knotwork pattern in Illustrator? I am working with a pretty complex traced AI that I received, and I want to basically make it so that all the visually black stuff is shapes, and all the white outlines (invisible here) are removed. Most of the spaces when a line crosses over another is created by a white outline.   I know about expanding appearance and the pathfinder tool, but because there are so many lines that are crisscrossing, I haven't had an easy way of minusing the fronts away. I could expand appearance and sort every object carefully, but I feel like there may be a quicker way of doing this.  If I put it in Photoshop as a smart object in 'screen' mode, it looks right.. I could put that back in Illy and trace it, but I don't want to complicate the paths. This is a summary of the excerpt:  Given that we have:  *$10^9$ people  *each person has a chance of 0.01 of going to a hotel  *we have $10^5$ hotels, each hotel holds 100 people  *Thus, $P(2 \text{ persons meet on the same hotel in any given day}) = 0.01 \cdot 0.01 \cdot 10^{-5}$  *and $P(2\text{ persons on the same hotel in $2$ day}) = P(2 \text{ persons on the same hotel in any given days})^2 = 10^{-18}$  the book "Mining of Massive Dataset" said that the expected number of pairs who stay in the same hotel for 2 days is:  $P(2 \text{ persons on the same hotel in $2$ days}) \cdot ( \text{number of pairs of people} ) \cdot (\text{number of pairs of days})$  this is the part where I got stuck. Can't understand how it come to this derivation. Can anybody explain it in more elaborate detail.  You can find the content on page 6 of this link:  (it is freely available) I have this question :  Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.  I want to know if my proof is valid :  If continuous in $x_0$ then :  $$\lim_{x \to x_0} f(x)= f(x_0)$$  Therefore :  All $\epsilon&gt;0$ exists $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;\epsilon$.  Therefore for $\epsilon=1$ there is $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;1$.  $|x-x_0|&lt;\delta \rightarrow x_0-\delta&lt;x&lt;x_0+\delta$.  Lets choose $x_1,x_2$ that implies $x_0-\delta&lt;x_1&lt;x_2&lt;x_0+\delta$  Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there. Can you explain the "Axiom of choice" in simple terms?