Problem with installing Nvidia driver 319 I have Ubuntu ver. 12.04 and Nvidia ver. 304. After i updated my driver to 319 which is recommended in Additional Driver, I got stuck in the booting screen with something like: AppArmor etc. (black screen with yellow code) so i had to reinstall my OS.   Please could you help me how to install it correctly without error, because I am new with Linux.  Thank you very much!
After installing NVIDIA driver, X isn't showing I recently installed the NVIDIA 319 driver from the Jockey window, and now X won't load! I'm stuck with a blinking cursor. I used to have the 304 drivers installed.  How do I fix this?
I want to have a vector with a superscript:  \vec{o}^t   But this gives very bad results: the t is overlapping with the vector arrow. How do I fix this (and why does this give bad results in the first place) ? Embed data in a C++ program Is it true that  $$\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dx$$  Just making sure.   If not, how about:  $$\int_a^b f(x) dx = (f(b)-f(a))b - \int_{f(a)}^{f(b)}f^{-1}(x)dx$$   I'm having a hard time concentrating right now, and I'm trying to figure out how to get the area under a curve when the function is inverted. An example:    O  O?   There is a triple letter tile to the right of the bottom O, indicated with a ?.  If I place an X on it, making the words OX (vertical) and OX (horizontal), do I score 50 points (scoring the tile twice) or 34 (scoring it once)? How to apply a regex on a column fetched from DB I am using IBM DB2.  So, I am retrieving a column named SSN from a table. I have to apply the following regex on that column:  ([1-57-8][0-9]{2} |0([1-9][0-9] |[0-9][1-9]) |6([0-57-9][0-9] |[0-9][0-57-9])) ([1-9][0-9]|[0-9][1-9])([1-9]\d{3}|\d[1-9]\d{2}|\d{2}[1-9]\d|\d{3}[1-9])   Someone suggested to me that I have to write a function/procedure in the query to validate that the SSN picked from the db follows this regex else it should not be stored in the result set of the query.  I don't know how to write a function or a procedure in SQL, can you please help me out here? I thought this would be really simple, but it's presenting some difficulties. If I have std::string name = &quot;John&quot;; int age = 21;  How do I combine them to get a single string &quot;John21&quot;? Drafts for edits get lost on leaving the Android app When editing in the Stack Exchange Android app and navigating away to get some reference URLs the text is gone when I go back into the app.  Seems like the edit isn't kept in memory. Can you please save them in memory? (drafts) If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$ Is there an IDictionary implementation that, on missing key, returns the default value instead of throwing? The indexer into Dictionary throws an exception if the key is missing.  Is there an implementation of IDictionary that instead will return default(T)? I know about the TryGetValue() method, but that's impossible to use with LINQ. Would this efficiently do what I need?: myDict.FirstOrDefault(a =&gt; a.Key == someKeyKalue);  I don't think it will as I think it will iterate the keys instead of using a Hash lookup. removeGroup() in QGIS 3 I am updating a plugin from QGIS 2 to 3.  I need to iterate through the groups (the names are dynamic) and remove them with their respective child layers.  I need to get a list of groups and then iterate through the groups and delete them.  The original code is as follows:  legend = self.iface.legendInterface() group_names = legend.groups() rev = self.groups[:] rev.reverse() ...     for i, group in enumerate(rev):         legend.removeGroup(group)   I replaced self.iface.legendInterface() with QgsProject.instance().layerTreeRoot() but group_names = legend.groups() fails with the message 'QgsLayerTree' object has no attribute 'groups'   What is the proper container of the groups and how do I access the collection; and, once found, how do I delete those groups?  A search for information on how to remove groups from a layer tree in QGIS 3 has proven fruitless. How does the standard error work? I have been looking into the inner-workings of the standard error recently, and I found myself unable to understand how it works. My understanding of the standard error is that it is the standard deviation of the distribution of sample means. My questions are:  • how do we know the standard error is the standard deviation of the sample means when we usually take just a single sample?  • why doesnt the equation to calculate the standard error mirror the standard deviation equation for a single sample? ‘So the boy ... the boy must die?’ asked Snape, quite calmly.   ‘And Voldemort himself must do it, Severus. That is essential.’   Deathly Hallows - page 551 - UK Hardcover - chapter 33, The Prince's Tale   Dumbledore tells Snape it is 'essential' that Voldemort be the one to kill Harry, I'm assuming in order for the piece of Voldemort's soul in Harry to be properly killed as well.  But why?   Hermione destroyed the Hufflepuff cup; Ron destroyed the Slytherin locket; Neville killed Nagini; Harry destroyed the diadem and Tom Riddle's diary. So the Horcruxes were not immune to destruction at the hand of someone other than Voldemort. And as it ended up, Harry himself wasn't even a Horcrux, but rather just an unknowing host to a parasitic bit of Voldemort's soul.   Why was it 'essential' that Voldemort be the one to kill Harry in order for the piece of Voldemort's soul to die? Are there 'stack' sites for Mac or Windows? In educational material for the Russian language and linguistic literature the need to put accents on Cyrillic vowels often arises. However when I tried to do this with luaLaTeX the accents where misplaced.  \documentclass[11pt]{article} \usepackage[utf8]{inputenc} \usepackage{fontspec} \setmainfont{Linux Libertine:style=Regular} \begin{document}  Е́сли вы изу́чите  \'{Е}сли вы из\'{у}чите  \'{E}сли вы из\'{y}чите % Here "E" and "y" are Latin glyphs, but this workaround does not                         % work with all Cyrillic letters e.g. Я, Ю \end{document}     The last line would looks nice, but unfortunately these are Latin glyphs.  Is this a problem with the font used (Linux Libertine) or with LaTeX? Like most Python developers, I typically keep a console window open with the Python interpreter running to test commands, dir() stuff, help() stuff, etc.  Like any console, after a while the visible backlog of past commands and prints gets to be cluttered, and sometimes confusing when re-running the same command several times.  I'm wondering if, and how, to clear the Python interpreter console.  I've heard about doing a system call and either calling cls on Windows or clear on Linux, but I was hoping there was something I could command the interpreter itself to do.  Note: I'm running on Windows, so Ctrl+L doesn't work. How To Set Default Wallpaper As Login Screen Wallpaper [Automatically]? Prove existence of a real root. Problem: If   $$C_0+\frac{C_1}{2}+\cdots + \frac{C_{n-1}}{n}+\frac{C_n}{n+1} =0,$$  where $C_0,...,C_n$ are real constants, prove that the equation  $$C_0+C_1x+\cdots +C_{n-1}x^{n-1}+C_nx^n=0$$  has at least one real root between $0$ and $1$.  Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 4. My intent is to show that a composition of bijections is also a bijection by showing the existence of an inverse.  But my approach requires the associativity of function composition.  Let $f:  X \rightarrow Y, g:  Y \rightarrow Z, h:  Z \rightarrow W$ be functions.  $((f \circ g) \circ h)(x) = h((f \circ g)(x)) = h(g(f(x)))$, and $(f \circ (g \circ h))(x) = (g \circ h)(f(x)) = h(g(f(x)))$.  However, I am having problems in justifying that the two compositions, $(f \circ g) \circ h$ and $f \circ (g \circ h)$, have the same domain and range.  When I consulted ProofWiki, whose link is at the bottom, I got even more confused.  Specifically, for $(f \circ g) \circ h = f \circ (g \circ h)$ to be defined, ProofWiki requires that dom$g =$ codom$f$ and dom$h =$ codom$g$.  First of all, I think that it should be dom$g =$ range$f$ ....  Moreover, as you can see in the example below, you actually have to adjust domains and ranges of $f, g, h$ for the requirement to hold true.  Let $f: \mathbb R \rightarrow \mathbb R$ be $f(x) = 2x$, $g: \mathbb R^+ \rightarrow \mathbb R$ be $g(y) = ln(y)$, $h: \mathbb R \rightarrow \mathbb R$ be $h(z) = z - 10$.   Then $((f \circ g) \circ h)(x) = ln(2x) - 10 = (f \circ (g \circ h))(x)$, with dom$((f \circ g) \circ h) = \mathbb R^+$ = dom$(f \circ (g \circ h))$.  As a result, we need to set dom$f = \mathbb R^+$, range$f = \mathbb R^+$; dom$g$, range$g$, dom$h$, and range$h$ remain the same.  Am I allowed to do that?  This adjustment implies that when we say dom$f = X$, $f$ must be defined for all elements in $X$, but $X$ may not be the entire set of elements for which $f$ is defined. Decompose vector fields on product manifolds So, I know that tangent bundle of a product manifold $M \times N$ splits in a sum $$ T_{(x,y)}(M \times N) = T_xM \oplus T_yN, $$ so that is obvious that the sum $X \oplus Y$ of smooth vector fields $X \in \mathcal{T}(M)$ and $Y \in \mathcal{T}(N)$ is a smooth vector field of $M \times N$. I've been told that, though not every vector field in $\mathcal{T}(M \times N)$ is a sum, locally one can always find one such decomposition, which will in turn be unique due the fact the sum is a direct one.   How can I show that this decomposition exists locally? More than that, if $X = X_1 + X_2$ is the decomposition, is there a way the express the coordinate functions of $X_1$ and $X_2$ in terms of those of $X$?   First I thought about taking two frames that locally spam $TM$ and $TN$ and write down $X$ using them, but then the coordinate functions are of the form $X^i: M \times N \to \mathbb R$, and the vector field components in each subspace are not exactly fields of $M$ and $N$ because their coordinate functions don't have the right domains. Is there another better way to see this decomposition holds locally? Consider a text file with the following entries:  aaa bbb ccc ddd eee fff ggg hhh iii   Given a pattern (e.g. fff), I would like to grep the file above to get in the output:  all_lines except (pattern_matching_lines  U (B lines_before) U (A lines_after))   For example, if B = 2 and A = 1, the output with pattern = fff should be:  aaa bbb ccc hhh iii   How can I do this with grep or other command line tools?    Note, when I try:  grep -v 'fff'  -A1 -B2 file.txt   I don't get what I want. I instead get:  aaa bbb ccc ddd eee fff -- -- fff ggg hhh iii