Disable global fog from camera
How do I disable fog on specific camera/s
Prove $$\int _0^\infty f^2(x) dx \leq \frac{2}{3}\cdot  \max_{x \in \mathbb R^+} f(x) \cdot  \int _0^\infty f(x) dx$$ for $f(x) \geq 0$ and convex.   I know via Holder's we can get without the constant $\frac{2}{3}$ but that should also be true for $f$ not convex. I tried some approaches using convexity but didn't get the desired result.   A hint or reference would also help. Also, it would be interesting to know when equality holds. How to add and sync folders outside SkyDrive folder? I have folders with documents all over my hard drive, some which I would like to include in SkyDrive without moving them into the SkyDrive folder.  E.g. my SkyDrive folder is in C:\Users\Name\SkyDrive, but I have some files in C:\Users\Name\SyncMePlz that I also would like to be synced with SkyDrive so that I can reach them from other devices.  How can this be achieved? What does Use an Object cover? Mean Time to Failure (MTTF): When disk manufacturers post this, how should you interpret their numbers? Mean Time to Failure (MTTF) is usually given in terms of hours, and by doing some calculations, it seems that a disk should fail only after a good number of years have gone by.  It seems that disks need repair more often than that. Does anyone know why this is so?  I figured that there is something fishy about this metric. Am interpreting something wrong here? i know that|$S_n$|=$n!$ and |$A_n$|= $\frac{n!}{2}$ but i'm not sure where to go from there. Thanks. If an ideal contains the multiplicative identity, then it is the whole ring I have to prove that if $I\subseteq R$ is an ideal, and $1\in I$, then $I=R\,$. So I know $I\subseteq R$ is an ideal if $a,b\in I$ implies $a+b\in I$, and if $a\in I$, $r\in R$, then $ra\in I$. I'm finding it hard to put this into words. Since $1\in I$ and $I\subseteq R$ is an ideal, then let $a=1$ and $r=1$ so $1\cdot 1\in R\,?$ After the upgrade of salesforce to Spring 15, I am getting this error "Error: You have exceeded the maximum number (5) of Apex scheduled jobs." when scheduling more than 5 batches. I have this issue in several environment. I used to more than 20 batches scheduled at once.   On the link below, it says that I can schedule up to 100 batches but I am being restricted to 5 only.   I can't have more than 5 jobs here. How exactly does one “control for other variables”? Here is the article that motivated this question:   I liked this article, and it nicely demonstrates the concept of “controlling for other variables” (IQ, career, income, age, etc) in order to best isolate the true relationship between just the 2 variables in question.    Can you explain to me how you actually control for variables on a typical data set?      E.g., if you have 2 people with the same impatience level and BMI, but different incomes, how do you treat these data?  Do you categorize them into different subgroups that do have similar income, patience, and BMI?  But, eventually there are dozens of variables to control for (IQ, career, income, age, etc)  How do you then aggregate these (potentially) 100’s of subgroups?  In fact, I have a feeling this approach is barking up the wrong tree, now that I’ve verbalized it.  Thanks for shedding any light on something I've meant to get to the bottom of for a few years now...! Ruby Style: How to check whether a nested hash element exists Can not move polygons but only vertices of an object by bmesh! Are threads implemented as processes on Linux? I'm going through , Advanced Linux Programming by Mark Mitchell, Jeffrey Oldham, and Alex Samuel. It's from 2001, so a bit old. But I find it quite good anyhow.  However, I got to a point when it diverges from what my Linux produces in the shell output. On page 92 (116 in the viewer), the chapter 4.5 GNU/Linux Thread Implementation begins with the paragraph containing this statement:     The implementation of POSIX threads on GNU/Linux differs from the   thread implementation on many other UNIX-like systems in an important   way: on GNU/Linux, threads are implemented as processes.   This seems like a key point and is later illustrated with a C code. The output in the book is:  main thread pid is 14608 child thread pid is 14610   And in my Ubuntu 16.04 it is:  main thread pid is 3615 child thread pid is 3615   ps output supports this.  I guess something must have changed between 2001 and now.  The next subchapter on the next page, 4.5.1 Signal Handling, builds up on the previous statement:     The behavior of the interaction between signals and threads varies   from one UNIX-like system to another. In GNU/Linux, the behavior is   dictated by the fact that threads are implemented as processes.   And it looks like this will be even more important later on in the book. Could someone explain what's going on here?  I've seen this one , but it doesn't help much. I'm confused.  This is the C code:  #include &lt;pthread.h&gt; #include &lt;stdio.h&gt; #include &lt;unistd.h&gt;  void* thread_function (void* arg) {     fprintf (stderr, "child thread pid is %d\n", (int) getpid ());     /* Spin forever. */     while (1);     return NULL; }  int main () {     pthread_t thread;     fprintf (stderr, "main thread pid is %d\n", (int) getpid ());     pthread_create (&amp;thread, NULL, &amp;thread_function, NULL);     /* Spin forever. */     while (1);     return 0; } Bring a window to the front in WPF Why AUC =1 even classifier has misclassified half of the samples? I am using a classifier which returns probabilities. To calculate AUC, I am using pROC R-package. The output probabilities from classifier are:  probs=c(0.9865780, 0.9996340, 0.9516880, 0.9337157, 0.9778576, 0.8140116, 0.8971550, 0.8967585, 0.6322902, 0.7497237)   probs shows probability of being in class '1'. As shown, the classifier has classified all of samples in class '1'.  True label vector is:  truel=c(1, 1, 1, 1, 1, 0, 0, 0, 0, 0)   As shown, classifier has misclassified 5 samples. But, AUC is:  pROC::auc(truel, probs) Area under the curve: 1   Could you please explain to me why it happens? First I tried to use integration: $$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$ $$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$ But I could not express it as a riemann integral. Now I am thinking about sandwich theorem.  $$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$ Since $\frac{a}{t+1}&gt;\frac{a}{t+2}&gt;\frac{a}{t+1}&gt;\cdots&gt;\frac{a}{n}$ $$\frac{a^n}{n!}&lt;\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$ since $\frac{a}{t+1}&lt;1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$ Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$ And by using sandwich theorem, $y=0$. Is this correct? Integral $\int^{ \pi /2}_{0} \ln (\sin x)\ dx$ $$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$  The answer is $- \frac{\pi}{2} \ln 2$.  I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$  But I didn't get the answer with it. Solve simple trig equation Taking Seats on a Plane This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless  Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their "proper" seat, or if that seat is taken, a random seat instead.  Question: What is the probability that the last person that boards will end up in his/her proper seat.  Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc... I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.   From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$  So we can do something like this:  $$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$   we can then factor all of the common terms out on the LHS.  $$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side  $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$  I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).   If anybody can provide a hint as to how to proceed I would appreciate it. Using local OpenStreetMap (*.OSM) map with Openlayers 3? What do I need to load a .OSM map in a client side javascript app? (Imagine an offline webapp). How to add a list of vectors (lists) to a directed graph as an edge weight?