Capturing counter value in Javascript closure
JavaScript closure inside loops – simple practical example
Downloading from the main Ubuntu archive is slow even when it's not release day, how can I get apt-get to automatically use a mirror that is close to me? Automatically switch gpu graphics driver when gaming I have installed Ubuntu GNOME 15.10. I installed the nvidia tested proprietary driver and then I restarted. Afterr that I selected as primary GPU the Intel driver for normal use. However, when I enter a steam game(Portal) the graphics is bad so In order to play normal I have to switch manually to the nvidia driver again and reboot again. After the game I have to manually select as primary driver the Intel one and reboot again. It is so boring. QUESTION: How to automatically switch gpu graphics driver when I enter a game or something that needs more GPU? What is the probability that all three tosses result in the same outcome? A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used? Prove that a finite union of closed sets is also closed Let $X$ be a metric space.  If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.  I'm looking for a direct proof of this theorem.  (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.")  Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.  Here are the definitions I'm using:  Let $X$ be a metric space with distance function $d(p, q)$.  For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) &lt; r\}$.  Any $p \in X$ is a limit point of $E$ if $\forall r &gt; 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$.  Any subset $E$ of $X$ is closed if it contains all of its limit points. What are valid values for "data defined" labeling settings? I'm using QGIS 1.8. I'd like to customize label features on an existing shapefile using the new label engine and it's "data defined settings" option. I'd like to know what the attribute table field types should be for each of the settings: All 7 font options, all 2 buffer options, and all 6 position options. Should they all be numeric fields, or some text, or both? I want to create these fields in the attribute table correctly so that I can map them to the label field settings. I can't find information on the proper field types anywhere. Thanks for your thoughts. I have created a C# application which uses a SQL Server database. I have other computers connected to me and to each other in a workgroup. I have shared my C# application with others. When they open the application they get the error      A network related or instance-specific error occured while establishing a connection to SQL Server. the server was not found or was not accessible   But the application is working fine on my PC. The connection string I am using is   Data Source=ASHISHPC1\SQLEXPRESS;Initial Catalog=ACW;User ID=ash159;Password=ashish159   which is stored in a .config file.   The application is working fine on my PC. What must I do? I have enabled the TCP/IP in the server but the same error persists. Some change in connection string or something else? Putting char into a java string for each N characters How to show that when the linear transformation $\Phi =\psi \circ \phi : E \rightarrow G$ is injective, then $\phi$ is also injective? How to show that when the linear transformation $\Phi =\psi \circ \phi : E \rightarrow G$  (where $\phi : E \rightarrow F$ and $\psi : F \rightarrow G$) is (1) injective, then $\phi$ is injective and when (2) $\Phi$ is surjective, then $\psi$ is also surjective ?  I've tried to prove it by the definition of injection and surjection, but I'm not convinced myself by it.  For example, in the first case :  If $\Phi$ is injective, $\forall g \in G$, we have $0$ or $1$ $e \in E$ such as $\Phi(e)=g$.  $\Rightarrow \Phi(e) = \psi(\phi(e)) = \psi(f) =g$.   So $\forall f \in F$, we also have $0$ or $1$ $e \in E$ such as $\phi(e)=f \in F$.   $\Rightarrow \phi$ is injective.  Is this right or did I prove nothing at all ? I think it is easier to convince myself/ to understand with surjectivity but still.. How to filter IPv4 and IPv6 addresses? The total number of locks exceeds the lock table size Why isn't reflexivity redundant in the definition of equivalence relation? An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.  Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.  For any $a$ and $b$, $a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.  Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al Show that $\left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right|$. I'm trying to prove that  $$ \left| \frac{1}{x}-\frac{1}{y} \right| $$  is a metric on $X = \mathbb R^+$.  The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e.  $$ d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. $$  So, ok, I have that  $$ d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| $$  ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by $ \left| \frac{1}{x}-\frac{1}{z} \right| $, but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks. How can I prevent iPhoto/Photos/Image Capture from launching everytime I plug in my iPad/iPhone? The impression that I got from the  films was that if  was successful, most of the world was going to suffer greatly. Basically everyone would serve Voldemort as slaves or die. So what's the point of serving someone who's probably going to ultimately end up betraying and/or enslaving you? I know the  hated , but is that really enough to justify tearing down Wizard society? Why exactly do the Death Eaters serve Lord Voldemort? Fear? Is the post-Voldemort world not that bad? (I for one welcome our ) Name of component on output side cable of a DC power adapter? Does the Hunter's Mark spell chain to multiple creatures? I'm looking to play a ranger. The Hunter's Mark spell says:     If the target drops to 0 hit points before this spell ends. you can use a bonus action on a subsequent turn of yours to mark a new creature.   Does that mean I can keep marking a new enemy each time the last one dies until the hour's up? How do I install Waterfox? I tried installing Office 2007 in Ubuntu with Wine but it was throwing an error after installing and also when I tried to open a PowerPoint document and save it. Has anybody gotten Office 2007 installed without any issues? Please help me. Why doesn't the UK ask for travel bookings for a tourist visitor visa? Probability of sampling with and without replacement In sampling without replacement the probability of any fixed element in the population to be included in a random sample of size $r$ is $\frac{r}{n}$. In sampling with replacement the corresponding probability is $\left[1- \left(\frac{1}{1-n}\right)^r\right]$.  Please help me show how this is proved.