Adding texture on a material
How to place a .png texture with transparency over a material
Variable with colon dash ${VAR:-} Can an observer B move in such a way that his clock will run faster than that of a fixed observer A? My understanding of Special and General Relativity comes from books which attempt to explain them to non-experts in these fields.   Let A and B be observers, which at some time t(0), are together at some location P fixed to the earth's surface. A and B are provided with "clocks", which at time t(0), are identical lumps of the same radioactive element-that has a very long half-life. A and A's "clock" remain fixed to the earth's surface. B and B's "clock" leave A and stay together, but eventually return to A after some interval of time T (as measured by A's "clock"). If, during this time T, B's velocity relative to A is sufficiently close to the velocity of light or B remains close to a sufficiently massive body, it seems that time can "pass more slowly" for B than for A. So that when B rejoins A, B's "clock" will contain more of the radioactive element than A's "clock"   My question is: Are there any scenarios that B together with B"s "clock" can undergo (during the time T) which will cause time to "pass more quickly" for B than for A? So that when B rejoins A, B's "clock" will contain less of the radioactive element than A's "clock". If there are no such scenarios, is there any fact or law in Relativity theory which rules them out? No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. Suppose I installed ubuntu with the user usr1 then I added usr2. I then create a group grp1 where this 2 users are added. A folder is owned by usr1 and has group grp1. But when I login as usr2 I cannot write. The folder has permissions 755...  usr2@kmserver:/media/data/www$ groups usr1 usr1 : grp1 ... usr2@kmserver:/media/data/www$ groups usr2 usr2 : grp1 ... usr2@kmserver:/media/data$ ls -l total 20 drwx------ 2 root root 16384 2011-05-12 10:53 lost+found drwxrwxr-x 2 grp1   grp1    4096 2011-05-12 15:15 www   Whats wrong? How To Store Files In An EXE Alright, so I'm working on programming my own installer in C#, and what I'd like to do is something along the lines of put the files in the .exe, so I can do  File.Copy(file, filedir);  Or, if this isn't possible, is there another way of doing what I am attempting to do? What should my Objective-C singleton look like? My singleton accessor method is usually some variant of:  static MyClass *gInstance = NULL;  + (MyClass *)instance {     @synchronized(self)     {         if (gInstance == NULL)             gInstance = [[self alloc] init];     }      return(gInstance); }   What could I be doing to improve this? "the police conspiracy" vs. "the police's conspiracy" Set of branch points isn't discrete, but branch points are isolated? I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related: Is there a standard pronunciation for gender-neutral -@ or -x More and more I see, especially in activist communities, Spanish-derived words ending in x or @ in order to neuter the gendering inherent in the original language. For example: latinx (or latin@) as opposed to latina and latino. How, though, are these endings pronounced? Please provide evidence of how this orthography is most commonly pronounced. Note 1: I recognize this is in an English Language site, rather than a Spanish language site, but this phenomenon is very much happening in English, albeit with Spanish loan words. Therefore the question is relevant to non-Spanish-speaking English speakers. Note 2: A question like this can lead to politically charged or, at least, sarcastic comments. Please refrain from bringing in your biases for or against. Note 3: This question is not seeking opinions as to how some people pronounce the suffixes. Answers should address evidence of a developing orthodoxy in pronunciation. Proof Verification: Converse of Intermediate Value Theorem A function $f$ is increasing on $A$ if $f(x)\leq f(y)$ for all $x&lt;y$ in $A$. Show that the Intermediate Value Theorem does have a converse if we assume $f$ is increasing on $[a,b]$.  The converse of IVT along with the additional hypothesis would look like this: Lef $f:[a,b] \rightarrow \mathbb{R}$ be an increasing function on $[a,b]$ which satisfies: if $L$ is a real number such that $f(a)&lt;L&lt;f(b)$ (or $f(a)&gt;L&gt;f(b)$), then there exists a point $c\in (a,b)$ where $f(c)=L.$ If the preceding conditions are met, then $f$ is continuous on $[a,b]$.  The proof (if it's correct) I'm going to show is only for the $c \in (a,b)$, but I think for end points ($a$ or $b$), the method should be similar.  We want to show that given $c \in (a,b)$, for all $\epsilon&gt;0$, there exists a $\delta&gt;0$ such that $|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon$.  Since $f$ is increasing, we know that $f(a)\leq f(c)$. If $f(c)-\epsilon&lt;f(a)$, then set $x_1=a$, if $f(c)-\epsilon\geq f(a)$, then we know by hypothesis that a $x_1&lt;c$ exists such that $f(x_1)=f(c)-\epsilon.$ In either case, we have for $x \in (x_1,c]$ $$ f(c)-\epsilon \leq f(x_1) \leq f(x) \leq f(c) .$$  In a similar way, since we know that $f(c)\leq f(b)$, we can deduce that a $x_2&gt;c$ exists such that, for $x \in [c,x_2)$ $$f(c)\leq f(x)\leq f(x_2)\leq f(c)+\epsilon .$$  Pick $\delta = \min \{c-x_1,x_2-c \}$, we then have $$|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon .$$  Is this correct? In the proof that I saw, they originally used $\frac{\epsilon}{2}$, instead of just $\epsilon$, but I think this one works just as well, what do you guys think? Antecedent Precedence? We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$? Why did Voldemort take the wand of one of his most loyal servants? In Harry Potter and the Deathly Hallows, chapter one The Dark Lord Ascending, Voldemort believes he must use a wand other than his own in order to kill Harry, when Harry is moved from Privet Drive. He demands Lucius Malfoy's wand, and gets it.   My question is, why did Voldemort take the wand of one of his most loyal servants, rendering Lucius Malfoy useless if a battle was to occur?  If I remember correctly, he had Ollivander at his disposal, and he was the one who told him to use another wand in first place. Why not ask Ollivander to create another wand, or just use ones that he had ready to sell in his shop at Diagon Alley? He could even have taken Ollivander's own wand. If my memories are correct, a wand's importance was compared to that of a limb. MediaElement not playing audio from stream WP7 Show that $\lim\limits_{y\downarrow 0} y\mathbb{E}[\frac{1}{X};X>y]=0$. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a nonnegative random defined on this space. For any $A\in \mathcal{F}$, let $\mathbb{E}[X;A]:=\mathbb{E}[X\mathbb{1}_{A}]$, where $\mathbb{1}_{A}$ denotes the indicator random variable on $A$. Without assuming $\mathbb{E}\left[\frac{1}{X}\right]&lt;\infty$, show that  \begin{eqnarray} \lim\limits_{y\downarrow 0} ~~y~\mathbb{E}\left[\frac{1}{X};X&gt;y\right]=0. \end{eqnarray} Any initial ideas will be greatly appreciated. How do I take a screenshot on my Mac? What's the easiest way to take a screenshot of my Mac?  I've used Grab.app, but it's a bit annoying just to get a quick screenshot. Is there a faster way? Default currency symbol I'm looking for a way to get the Default currency symbol (in a none-multi-currency org) of the salesforce org locale (to have it displayed on a visualforce page in a field that can't be &lt;apex:inputField&gt; or &lt;apex:outputField&gt;) preferably query-able from apex (via the database, a describe or whatever) and then again the actual symbol, meaning € for Euro, $ for dollar, and so on...  There must surely be some sort of mapping hidden in the code somewhere salesforce does this via the input/outputfields if bound to an sobject of type currency  Can anyone point me in the right direction? InputStream returning NullPointerException (Java) Why are my render times 30+ minutes, but other people's are ~3 minutes? I am currently using Sheep It! to render my blend (built on v2.68c) on multiple computers, and I noticed something rather strange. According to the log on-site, the frames rendered from other people's blends were completing in 3 to 4 minutes. This was on an 4 core i5. My main rig (with an 8 core i7) was completing frames in no less than half an hour. How is this possible?  To clarify, these people's frames are very complex, often with very detailed textures. I have 2 objects in my blend. Below are the options I changed for my blend:   Aspect Ratio: 1000px x 1000px Output: PNG Clamp Indirect: 0.98 Render Sampling: 128 Min Light Bounces: 0 Max Light Bounces: 4 Cycles Render   Is there something in there that I should change to make it go faster? I can't really comprehend how it's going to quickly.  EDIT: To answer what @Duarte asked...   System Specs: i7-920, 3GB DDR3, GT 630, Windows 10 I have two circles, extended to form a sort of coin with a star on it. It is not complex at all. I have Firefox and Steam running in the background, but not actively. I've been away from the computer for 8 hours and noticed this behavior. (I am running Blender from Steam.) I made sure when I began rendering that the blend was set for CPU rendering.          And to answer what @pycoder asked, yes, I am using Cycles Render. Thermodynamics of supercooled water