Just out of curiosity. Maybe there are some technical reasons why the drivers are not up to date.  14.04 64b In additional drivers I have 331, but at the nvidia website they have 340. I'm coming back now to ubuntu, but before, lets say 6 months or 1 year ago was the same. The repositories were always behind.
Why are packages in the official Ubuntu repositories older than the latest (upstream) versions from Debian Sid, PPAs, the authors, etc.?
What are the shortcomings of the Mean Absolute Percentage Error (MAPE)? The  () is a common accuracy or error measure for time series or other predictions,  $$ \text{MAPE} = \frac{100}{n}\sum_{t=1}^n\frac{|A_t-F_t|}{A_t}\%,$$  where $A_t$ are actuals and $F_t$ corresponding forecasts or predictions.  The MAPE is a percentage, so we can easily compare it between series, and people can easily understand and interpret percentages.  However, I hear that the MAPE has drawbacks. I'd like to understand these drawbacks better so I can make an informed decision about whether to use the MAPE or some alternative like the MSE (), the MAE () or the MASE (). Motley fool says you can make $15,978 more per year with Social Security. Is this for real? The Motley fool has an ad campaign that says you can make $15,978 more per year on Social Security by doing 17 minutes of work using a legal loophole.  I don't want to link to their ad (although, you can find it by searching ).  What are they talking about here?  Is there any truth to this?  If someone could explain it to me without me having to watch the entire ad video or subscribe to their newsletter, I'd appreciate it. At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result) filling an occluded plane with the smallest number of rectangles I've got a specific problem which I'll try to describe as clearly as possible.   I have a defined rectangular region on a cartesian plane, and within that region there are other given rectangular sub-regions that are described in terms of their 4 vertices, ie {(x1, y1), (x1, y2), (x2, y1), (x2, y2)}, so that these regions form 'occlusions' on the plane. These regions don't overlap, but they can form more complex polygons when different-sized rectangles adjacent to each other appear joined.  here's an illustration:   I am interested in the space between these shapes, and how to define the space in the same way the occlusions are defined, that is as a set of rectangles. In particular I want the definition optimised so that the space is described using the smallest possible number of rectangles. For example, an incomplete rendering might look like this:   Can anyone suggest a way forward with this? How can I have the original set of vertices (describing the black rectangles) generate the 'complementary' set of vertices (red rectangles) such that the number of red rectangles is minimal?  I suspect it's a variation of a 'packing problem', but I have a feeling it might be fairly simple... Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases. I've been wondering, what is the difference between seek and search? When should one be preferred over the other? Why does Beamer not change the itemize symbol when using \setbeamertemplate? This code:  \documentclass[10pt]{beamer} \setbeamertemplate{itemize item}[circle] \setbeamertemplate{itemize subitem}[triangle] \setbeamertemplate{itemize subsubitem}[square]   \begin{document} \begin{frame} \begin{itemize} \item first level     \begin{itemize}     \item second level         \begin{itemize}         \item third level         \end{itemize}     \end{itemize} \end{itemize} \end{frame} \end{document}    produces this output:    I expected the symbol of the first level item to be a circle, not a square. Is there an error in my code? The log file was too long to include in the question, so I . Proving Pascal's identity So I came across Pascal's identity: Prove that for any fixed $r\geq 1$, and all $n\geq r$, $$ \binom{n+1}{r}=\binom{n}{r}+\binom{n}{r-1}. $$  I know you can use basic algebra or even an inductive proof to prove this identity, but that seems really cumbersome. I was wondering if anyone had a "cleaner" or more elegant way of proving it. For example, I think the following would be a decent combinatorial proof.   Proof: Let $S$ be a set with $n+1$ elements, and consider some fixed $x\in S$. There are $\binom{n+1}{r}\;\; r$-subsets of $S$--count them according to whether or not they contain $x$: there are $\binom{n}{r}$ not containing $x$, (each formed by choosing $r$ of the remaining $n$ elements in $S\setminus\{x\}$), and there are $\binom{n}{r-1}\;\; r$-sets containing $x$, (each formed by selecting an additional $r-1$ elements in $S\setminus\{x\}$).   Is that right? Are there any other efficient ways of doing it? Filter "Hot Network Questions" by excluding sites I was going to make a filter for the &quot;&quot; by excluding some sites from it.  I noticed  which talks about &quot;negative filtering&quot; but does not explain how one can do it or if it is possible.  There does not seem to be a way to do this, though.  On the filter page, my only options are:  So how can I do &quot;All sites except [list]&quot;? It would also be nice if this could include similar functionality with the meta checkbox.  Something like &quot;All main &amp; meta sites except...&quot; Perfect set without rational numbers Sorry if this problem is repeated.  Is there a nonempty perfect set in $\mathbb{R}^1$ which contains no rational number?  Proof sketch: This set must be uncountable because any nonempty perfect set in $\mathbb{R}^k$ is uncountable.   We know that Cantor set (C) is perfect set on a line. If we take $E=C+\sqrt{2}=\{x+\sqrt{2}: x\in C\}$. Will this set be an answer to above problem? I am trying to identify a sci-fi movie from the 80's or 90's, with the following parameters:  Setting:    Scenes of a red desert planet   Culture:    Either people roam fairly solo and try to acquire resources to survive, or you live in a weird society where if a stranger is captured he needs to play in a game of fight to the death.  Mobility:      One foil character has a giant battle-bot-looking tank as his form of transportation through the desert. Protagonist has a car that has a plastic half bubble on top to look out from.   Robot:      The protagonist runs around in the beginning of the film with a female robot/android whose face is melted off after it has acid touch its face.(This scene happens next to the protagonist's rover)  Love interest:      Somehow protagonist comes across female protagonist in desert. She smells really bad and is hungry. They sleep outside, and he tells her to sleep away from him. He uses a thermal blanket(the first I ever saw in my life). She wakes up next to him, smelling horrible. He bathes her in some water hole they find(as if water were not scarce and they could just waste water.)   Pupa monsters:      Scene where they are in a building and they are trying to get out. They come across a room that is white, in which multiple giant, white, snot, pupa are hanging securely from the ceiling. Through curiosity, they disturb them, and now they have to get out of that room fast.  Boss Antagonist:      Pervert bionic man, that is actually attached to a giant robotic arm, as only form of movement. He is in black and has a pale bald head as his most human looking part. Symbol Table from Apex Class In an earlier question  I asked in general about the symbol table for an Apex class.  Using the Tooling API wrapper from  I am able to get part way there. But am unGodly confused.  If I go to the developer console, query editor, check the "Use Tooling API" box and run this query:  Select Id, Name, symbolTable FROM ApexClass Where Name Like '%NewCaseListController%'   I get back an ID, Name and [object Object] for the symbol table, meaning there is a symbol table object there.   But if I try the same query in Apex code in the execute anonymous window I get the error:   System.CalloutException: Web service callout failed: Unable to parse callout response. Apex type not found for element location   If I do this in Apex I can get a result back, and according to the code for the class the result should be of type ToolingAPIWSDL.QueryResult. The QueryResult should contain a member consisting of a list of sObject_x records sObject_x[], unless I am mis-reading what that means.  ToolingAPIWSDL.QueryResult result = tooley.query('select Id, Name from ApexClass WHERE Name Like \'NewCaseListController\'');   I can take the result and apparently put the sObject_x object in it like this:  List&lt;ToolingAPIWSDL.sObject_x&gt; objList = new List&lt;ToolingAPIWSDL.sObject_x&gt;(); for(Integer i = 0; i&lt;result.records.size(); i++){     objList.add(result.records[i]); }   At least that does not throw an error and I can write:     System.debug('objList: ' + objList[0]);  And get a result, but how do I access the symbol table? Or more to the point, how can I get a list of methods and variables? I am assuming the symbolTable is the only way? The staircase paradox, or why $\pi\ne4$ What is wrong with this proof?    Is $\pi=4?$ $\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$? I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.          $\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $,  what is the     nearest value of $y^2 - y$?          $\textbf {(1) } 1 \qquad \textbf {(2) } \sqrt5 \qquad \textbf {(3) } 4 \qquad \textbf {(4) } 2\sqrt5$      I worked the problem and got $$y^2 - y = 5+ \sqrt{5-\sqrt{5+\sqrt{5-...}}} -\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} \\ = 5+ \sqrt{5-y} - y = 5-y + \sqrt{5-y}$$ What do I do next? Table and figure side-by-side with independent captions Looking to put a table and a figure side by side so that the table and figure have their own captions. A crude sketch    ------------              -----------------     TABLE    | a |  b |                     FIGURE   | c |  d |  ------------              Table 2: Caption           -----------------                            Figure 7: Caption   The closest I've gotten is:  \begin{table}% \centering \parbox{0.4\textwidth}{ \begin{footnotesize} \begin{tabular}{| l | r |} \hline some &amp; table\\ \end{tabular} \end{footnotesize} \caption{Table} \label{tab:table} } \qquad \begin{minipage}[c]{0.53\textwidth}% \centering     \includegraphics[width=1\textwidth]{awesome} \caption{Figure} \label{fig:figure} \end{minipage} \end{table}   But that gives two Table X: Caption captions. I didn't find anything in subfig.  Help? How to install Microsoft fonts on Ubuntu 16.10? Calculate limit with L'Hospital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.   I have done the following:    It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.   So, we can use L'Hospital's rule:  \begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&amp;=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &amp;=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &amp;=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ &amp; =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ &amp; =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}   We calculate the two limits separately   $\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :    \begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1  \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq  2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}    How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ? How can I find $f(0)$, $f'(0)$, and $f'(x)$ given that  $f(x+y)=f(x)+f(y)+x^2y+xy^2$  and  $\lim_{x\to0}\frac{f(x)}{x}=1$. How can I clip an image via a bezier-path? On my Windows machine, my main hard drive has the letter C: and the name "Local disk".   To list the drive letters in Java on Windows, the File object has the static listRoots() method. But I can't find a way to acquire the drive names (as opposed to the drive letters) on Windows.  Has anyone tried this before?