Hot to get rid of strange object borders
Shadows along edges of mesh in 3D View
Look and you shall find (a correlation) I have several hundred measurements. Now, I am considering utilizing some kind of software to correlate every measure with every measure. This means that there are thousands of correlations. Among these there should (statistically) be a high correlation, even if the data is completely random (each measure has only about 100 datapoints).  When I find a correlation, how do I include the information about how hard I looked for a correlation, into it?   I am not at a high level in statistics, so please bear with me. A contractive mapping on $M$ is a function $f$ from the metric space $(M,d)$ into itself satisfying $$d(f(x),f(y))&lt;d(x,y)$$ whenever $x,y\in M$ with $x\ne y$. Prove that if $f$ is a contractive mapping on a compact metric space $M$, there exists a unique point $x\in M$ with $f(x)=x$.   Consider the function $g(x)=d(f(x),x)$. Note that   \begin{align*}d(f(x),x)-d(f(y),y)&amp;\leq (d(x,y)+d(y,f(x))-(d(y,f(x)-d(f(x),f(y)) \\ &amp;=d(x,y)+d(f(x),f(y)) \\ &amp;&lt;2d(x,y)  \end{align*}  And similarly for $d(f(y),y)-d(f(x),x)$, so we have $|d(f(x),x)-d(f(y),y)|&lt;2d(x,y)$. This means $g(x)$ is continuous.  So $g(x)$ is a real-valued continuous function on a compact space, so must be bounded and has a minimum $c$, i.e. there exists $x_0$ such that $d(f(x_0),x_0)$ equals the minimum $c$. What then? Why is rendered viewport faster than rendering individual frames? When I'm using rendered viewport it initially takes a while before rendering starts, but once it does, I can freely move around and rendering a new view happens almost instantaneously. When on the other hand I'm rendering a movie, for each frame Blender takes a significant amount of time before rendering even starts, usually doing the following:  Synchronizing object, Updating images, Updating meshes, Updating Scene BVH, Packing BVH,  Why does it do this? And is it possible to disable this to increase movie rendering speed? When I play Black Ops 2, it seems that my xbox controller is inverted. Example: if I press up, my character looks down and vice versa. It was not like this before. Does anyone know how to reset the controller back to its original settings? How to prevent Command-I/Command-Shift-I from opening Mac Mail when in browser? When in a browser (Chrome or Safari), the following keyboard shortcut will always bring up a window that says "Welcome To Mail":  Command+Shift+I  I would like to use this shortcut for another purpose.  Is there any way to disable this shortcut? Book about classical mechanics In any Pythagorean triplet at least one of them is divisible by $2$, $3$ and $5$. Show that if $x$, $y$, $z$ are integers such that $x^2 + y^2 = z^2$, then at least one of them is divisible by $2$, at least one is divisible by $3$, and at least one is divisible by $5$.  I know that $x,y,z$ are sides of a right-angled triangles. If I start putting values, then I can show, but have no clue about mathematical approach.  How do I solve it mathematically, using modular arithmetic? How can I find the cheapest prepaid SIM card for a given country? How can adding a 2nd IV make the 1st IV significant? I have what is probably a simple question, but it is baffling me right now, so I am hoping you can help me out.  I have a least squares regression model, with one independent variable and one dependent variable. The relationship is not significant. Now I add a second independent variable. Now the relationship between the first independent variable and the dependent variable becomes significant.  How does this work? This is probably demonstrating some issue with my understanding, but to me, but I do not see how adding this second independent variable can make the first significant. Can magnetic fields be "blocked"? For example, in the game, TitanFall, a robot stop bullets with ( presumably ) a magnetic shield.  I wish to calculate the magnetic force required to stop a bullet within a few microseconds.  But the problem I have is that my entire electronic system or the vehicle ( most likely even a building ) will be subjected to massive fields most likely destroying it as well.  Does any know of a way to "block" or divert the field around my vehicle? Is it even possible? My knowledge in magnetic fields are quite limited at this time. Irreducibility of $X^n - 2$ over $\mathbf Z[i]$. I was “playing” with Eisenstein's criterion and the polynomials $X^n-p$ ($p$ being a prime number).  Over $\mathbf Z$, the classical Eisenstein's criterion directly applies and shows that $X^n - p$ is an irreducible element of $\mathbf Z[X]$. The question was “what happens over $\mathbf Z[i]$?”  We know that in this new ring, a prime number $p$ has one of three different fates:   if $p \equiv -1\ (\mathrm{mod}\, 4)$, $p$ still is an irreducible element of $\mathbf Z[i]$, or equivalently $(p) \subset \mathbf Z[i]$ still is prime. (Unless I'm mistaken, number theorists say that $p$ is inert). Eisenstein's criterion still applies and the polynomial is irreducible over $\mathbf Z[i]$. if $p \equiv 1\ (\mathrm{mod}\, 4)$, $p$ can be written as a sum of two squares: $p = n^2 + m^2$ and is therefore reducible on $\mathbf Z[i]$: $p = (n+im)(n-im)$. These two factors aren't associated (the units in $\mathbf Z[i]$ being $\pm 1$ and $\pm i$) so $(p)$ splits: $(p) = (n+im)(n-im)$ is the product of two different prime ideals. One can still use Eisenstein's criterion with either of these prime ideals, and $X^n - p$ is irreducible over $\mathbf Z[i]$. if $p = 2$, the oddest of all primes, one still has a factorisation $p = (1+i)(1-i)$ but this time, $1+i$ and $1-i$ are associated. In NT slang, $(2)$ is ramified: $(2) = (1+i)(1-i) = (1+i)^2$ and one cannot directly use Eisenstein's criterion.   So here's my question: how do we prove (or disprove) the irreducibility of $X^n - 2$ over $\mathbf Z[i]$? You can start fire by focusing the sunlight using the magnifying glass.   I searched the web whether you can do the same using moonlight. And found  and  - the first two in Google search results.  What I found is the thermodynamics argument: you cannot heat anything to a higher temperature using black body radiation than the black body itself, and Moon isn't hot enough.  It may be true, but my gut feelings protest... The larger your aperture is, the more light you collect, also you have better focus because the airy disk is smaller. So if you have a really huge lens with a really short focus (to keep Moon's picture small), or in the extreme case you build a Dyson-sphere around the Moon (letting a small hole to the let the sunlight enter), and focusing all reflected light into a point it should be more than enough to ingnite a piece of paper isn't it?   I'm confused. So can you start fires using the Moon? C++, array of objects without What is the difference between [0-9]+ and [0-9]++? Node.js on multi-core machines Sum of points and rank according to points Start ubuntu without a desktop environment but start an X application I want to do a small computer just for internet browsing. I dont want to load a desktop environment to keep RAM, CPU and GPU ussage as low as possible. The computer is going to run on a ARM CPU. I dont care with Distro I use, I thought about good old plain Debian. Converting a .lasd (Lidar) coordinate system? How can one change a coordinate system for a .lasd (Lidar) dataset?  The file is currently in UTM_Zone_15N and I am trying to convert it to IL_East. Galois group of the product of two fields being the product of Galois groups? I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory.    Theorem Let $E_1$ and $E_2$ be Galois field extensions of $F$ with trivial intersection ($E_1 \cap E_2 = F$), then $E_1 E_2$ has Galois group $\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.  Proof   $E_1$ and $E_2$ are splitting fields (over $F$) of some polynomials $f_1$ and $f_2$ so $E_1 E_2$ is the splitting field of $f_1 f_2$ over $F$ and therefore it is Galois and we are justified using the "Gal" notation. Let $\sigma_1 \in \text{Gal}(E_1/F)$ and $\sigma_2 \in \text{Gal}(E_2/F)$ then by the fundamental theorem of Galois theory we can lift these both into elements of $\text{Gal}(E_1 E_2/F)$. The claim is that $\sigma_1 \sigma_2 = \sigma_2 \sigma_1$, to see this consider an element $\alpha \in E_1$, we have both $\sigma_2 \alpha = \alpha$ and $\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha$ because it is invariant with respect to $\sigma_2$, but we can combine these equalities to get $\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha$. Both groups $\text{Gal}(E_i/F)$ ($i = 1,2$) are included in $\text{Gal}(E_1 E_2/F)$ and the degrees say there are no more elements, furthermore since they commute we can conclude $\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.     So hopefully that theorem is, if not correct, fixable.. and can be used to prove that $\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ has Galois group $C_2^m$ ($m \le n$) over $\mathbb Q$? What is the name for moments, a parameter that is calculated