I am using ubuntu 14.04, kindly assist how to install drivers for my ATI Radeon graphics card [Mobility Radeon HD 5650/5750 / 6530M/6550M].  I read some documentation on the given site but I am not sure if I really need to do all this.
I am planning on doing a fresh install of Ubuntu and want to know what is the correct way to install ATI Catalyst Video Driver?      There are multiple valid answers for this question spanning over several versions of Ubuntu. For your convenience an index of each below:
For any prime $p > 3$, why is $p^2-1$ always divisible by 24? Have we completed Diablo 3 in 9 hours? My friend and i decided to play some Diablo 3. We played for almost 9 hours straight and killed Diablo on the hard difficult. We reached level 41.  We think 9 hours is short for a game like Diablo 3.   Have we missed something? Why are we only level 41? Is there more story? How to add CSS and JS files on node pages from theme's template.php I am trying to add JavaScript and CSS files to one node type page from my template.php file but it is not working.  Here is my template.php code:   &lt;?php // $Id: template.php,v 1.13 2010/12/14 01:04:27 dries Exp $  function mttheme_preprocess_html(&amp;$variables) {   $node = node_load(arg(1));   if ($node-&gt;type = 'sales_team_page') {     drupal_add_css('misc/jquery.fancybox-1.3.4.css');     drupal_add_js('misc/jquery.mousewheel-3.0.4.pack.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );     drupal_add_js('misc/jquery.fancybox-1.3.4.pack.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );     drupal_add_js('misc/jquery.fancybox-calls.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );   } }   Does anyone know why it is not working? There is no set of all sets.. So there are a few posts about this already, but they skip over the problem I have.  The proof of Cantor's theorem elegantly shows that if we consider $f:A\rightarrow \mathcal{P}(A)$, the set $B=\{x\in A : x\not\in f(x) \}$ by its definition cannot lie in the image of $f$, precluding it from being a bijection - hence power sets have strictly greater cardinality.  But now, I thought one (obvious) way to characterise the "set of all sets" would be $S=\mathcal{P}(S)$, as indeed, it contains all possible sets (perhaps here is the problem, but I don't immediately see how using this as a definition for the set of all sets is bad). Now consider simply taking the identity function $i:S\rightarrow S$. Then $B=\{x\in S : x\not\in x \}$ - the set of all sets which do not contain themselves. But this doesn't exist. It's exactly the set of Russell's paradox. If this isn't a set, and so is not in $S$, then the proof no longer offers an obstruction to $i$ being surjective.    Each other post I've seen on the matter uses Cantor's theorem to show such a set doesn't exist. My query is to find that something else that must be happening, since I think the proof has issues for this particular set. I'm expecting other issues to arise with the definition $S=\mathcal{P}(S)$... Are we slightly lighter during the day and slightly heavier at night, owing to the force of the Sun's gravity? Remove blank attributes from an Object in Javascript How do I remove all attributes which are undefined or null in a JavaScript object?  (Question is similar to  for Arrays) Evaluate $\lim\limits_{x \to a} \frac{x^m-a^m} {x-a}$ Do we have any experimental evidence to confirm that light is an electromagnetic wave? Or is it confirmed simply by Maxwell's equations showing a similarity in speed? For $f$ Riemann integrable prove $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$ Suppose $f$ is a Riemann integrable function on $[0,1]$. Prove that $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$   This is what I am thinking: Fix $n$. Then by Jensen's Inequality we have $$0\leq\left(\int_0^1x^nf(x)dx\right)^2 \leq \left(\int_0^1x^{2n}dx\right)\left(\int_0^1f^2(x)dx\right)=\left(\frac{1}{2n+1}\right)\left(\int_0^1f^2(x)dx\right).$$Thus, if $n\to\infty$ then $$0\leq \lim_{n\to \infty}\left(\int_0^1x^nf(x)dx\right)^2 \leq 0$$ and hence we get what we want. How correct (or incorrect) is this? Has 'Ubuntu Software' been renamed to 'Snap Store'? How can I use QGIS to create a metro style map I was writing test classes for my apex Rest Service that I had built for an Ipad app.  I noted some very interesting things while experimenting. Before doing anything, I just did a google search and found an interesting blog post by Jeff.    @RestResource(urlMapping='/v.9/member/*/results/*')  global with sharing class MemberRestSvc {  @HttpGet  global static ReturnClass doGet() {  String[] uriKeys = RestContext.request.requestURI.split('/'); // get the member name from the uri String memberName = uriKeys.get(uriKeys.size()-3);  // do awesome programming stuff here &amp; catch any exceptions try {    List&lt;Contact&gt; contacts = [Select Id From Contact where member_name__c = :memberName];   return new ReturnClass('true', 'Query executed successfully.', contacts);  } catch (Exception e) {   return new ReturnClass('false', e.getMessage(), null); }   }   global class ReturnClass {  global String success; global String message; global List&lt;Contact&gt; records;    global ReturnClass(String success, String message, List&lt;Contact&gt; records) {   this.success = success;   this.message = message;   this.records = records;  }  }  }   And here is the test class  @isTest private class Test_MemberRestSvc {  static { // setup test data   }  static testMethod void testDoGet() {  RestRequest req = new RestRequest();  RestResponse res = new RestResponse();  req.requestURI = 'https://cs9.salesforce.com/services/apexrest/v.9/member/me/results/today';   req.httpMethod = 'GET'; RestContext.request = req; RestContext.response = res;  MemberRestSvc.ReturnClass results = MemberRestSvc.doGet();  System.assertEquals('true', results.success); System.assertEquals(10, results.records.size()); System.assertEquals('Query executed successfully.', results.message);  }  }   I just commented a few lines and ran and it. I still achieved enough test coverage so my question is why do I need to simply instantiate the Rest request and response objects since I am directly calling the class?    @isTest   private class Test_MemberRestSvc {    static {     // setup test data     }  static testMethod void testDoGet() {  /*RestRequest req = new RestRequest();  RestResponse res = new RestResponse();  req.requestURI =    'https://cs9.salesforce.com/services/apexrest/v.9/member/me/results/today';   req.httpMethod = 'GET'; RestContext.request = req; RestContext.response = res; */Commented the code meant for REST apex intialization and merely calling the method!!  MemberRestSvc.ReturnClass results = MemberRestSvc.doGet();  System.assertEquals('true', results.success); System.assertEquals(10, results.records.size()); System.assertEquals('Query executed successfully.', results.message);  }  }   Can someone explain this behavior? Also what are the best practices for writing Test classes for apex Rest services? If I create a custom environment based on the equation environment, I get a spacing issue: the next line of text after my environment closes doesn't start on the left margin.  Example code:  \documentclass{article} \begin{document} \newenvironment{myenv}{\begin{equation}}{\end{equation}} \noindent $| \leftarrow$ this is the left margin. \begin{myenv} e ^{i \pi} + 1 = 0 \end{myenv} $| \leftarrow$ this isn't the left margin. \end{document}   Output:    Removing the newline after \end{myenv} solves the spacing problem, but the resulting code is not readable.  Of course, in this example I could just use the equation environment itself. My real issue involves a more complex custom environment that suffers from the same problem, and which also includes an equation environment within it. Why does the natural ring homomorphism induce a surjective group homomorphism of units? I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity? Intersection of all $p$-Sylow subgroups is normal Thats my . I loaded this to PostgreSQL DB using PostGIS shape file loader. I ran pgr_anlayzeGraph() on it and found it had a lot of isolated segments.  Next I ran pgr_nodeNetwork() to create a new table 'ways_ways'. I made the topology  for it and ran the following query in Qgis sql window.   SELECT * FROM ways_ways JOIN (SELECT seq, id1 AS node, id2 AS edge, cost FROM  pgr_astar('SELECT id, source, target, cost, x1, y1, x2, y2 FROM ways_ways',  1, 108, false, false) )AS route ON ways_ways.id = route.node;   Given below are the screenshots: Original floorplan and the prescribed route  query ouput   Clearly , I don't have a continuous output. I ran pgr_analyseGraph on ways_ways to find that it has no more isolated segments . It has 51 deadends, which is acceptable.   Did I miss any step in pgrouting? Is there anyway to see which node has id 1 and id 108 in qgis?  i ran the same query in another map(which i downloaded) and got output for all possible source and target ids. So how can i create a way(a route) that is completely noded? When installing user applications, where do "best practices" suggest they be located? Occasionally I install applications manually, rather than using apt or another package manager.   What location (/usr/, /usr/local/, /opt/, /home/, etc.) is suggested by "best practices" for the installation of user applications? Is Gauss's law wrong, or is it possible that $\int_s{\vec E} \cdot d\vec{s}=0$ does not imply $\vec E = 0$? This is a question from David J Griffith's Introduction to Electrodynamics.      A specified charge density $\sigma(\theta)=k\cos(\theta) $ is glued over the surface of a spherical shell of radius $R$. Find the resulting potential inside and outside the sphere.   The question was solved using legendre polynomials and the final answer for potential inside the sphere was : $V(r,\theta) = \frac{kr}{3\epsilon_0}cos\theta$  This final answer is confusing because The Electric field inside the sphere is coming out to be dependent on $r$ and $\theta$ whereas electric field inside a shell, no matter what the charge distribution is outside, is $zero$ from gauss's law.      my doubts:         why is the electric field inside non zero?    Can gauss's law explain this, or does it fail here?      Since solving using ordinary surface integral gave me the same result and since the. divergence inside the shell is $zero$, I concluded that legendre polynomials and gauss's law in differential form are correct. So the problem should be with the integral form of gauss's law : $\int_s{\vec E}.d\vec{s} = \frac{q}{\epsilon_0}$  The answer I got for this doubt is that  "since the charges are glued to the surface, and not evenly distributed, electric field inside need not be zero. "     This is not convincing because proof of gauss's law does not expect the charges to be free to move. presence of an exrernal force that would hold the charges in place does not change the theorem. That is    Say only a single charge $q_i$ is present outside      then $\int_s{\vec E_i}.d\vec{s} = \frac{q_{inside}}{\epsilon_0}=0$      Now if there are more charges, following any distribution, net electric field $\vec E = \vec E_1 + \vec E_2+\vec E_3+...$       So the net flux,       $\int_s{\vec E}.d\vec{s} = \int_s{\vec E_1}.d\vec{s}+\int_s{\vec E_2}.d\vec{s}+\int_s{\vec E_3}.d\vec{s}+. . .=0$   Or is it possible that $\int_s{\vec E}.d\vec{s}=0$ does not imply $\vec E = 0$? Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$  Show that $AB=0$.  My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$  then  $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't  I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you. I would like to remove selected commit log entries from a linear commit tree, so that the entries do not show in the commit log.  My commit tree looks something like:  R--A--B--C--D--E--HEAD   I would like to remove the B and C entries so that they do not show in the commit log, but changes from A to D should be preserved. Maybe by introducing a single commit, so that B and C become BC and the tree looks like.  R--A--BC--D--E--HEAD   Or, ideally, after A comes D directly. D' representing changes from A to B, B to C and C to D.  R--A--D'--E--HEAD   Is this possible? if yes, how?  This is a fairly new project so has no branches as of now, hence no merges as well.