Sometimes all I want to say is Thank you., but I need '5' more letters 'to go'. I've seen people write "4more" or "kdjgodfjg" or other things to get around this. Can we short the comment minimum to be 10 so we don't have to do this to simply say thank you?
I understand that there is an upper limit. But why do you have to type at least 15 characters to post a comment? It is easy to work around it anyway (like adding dots or other useless things).
How to get this blur texture effect in photoshop? I'm trying to get an image to look similar to this photo:    I'd guess they took a picture of the monitor with a regular camera. But I'm more interested to know what effects / filters they used to get the blur and texture. Any suggestions would be appreciated Deriving input based on other parameters in Python script tool? I have a folder and it contains two feature classes, e.g., a.shp and b.shp. The script tool created has two parameters. The first one is the model location, and the second one needs a feature class (a.shp), as shown in the below picture.  What I really want is every time when I set up the model location for the first input, the second input will automatically find the a.shp as the input rather than the user needs to choose a.shp from that folder. I noticed that the dependency is probably what I really need, but there is no code example. Any suggestions?   I am using ArcGIS 10.3.1 Advanced License. I just would like to know if a question or an answer saved draft is visible to someone else, apart from the OP.  Will a copy of a saved draft remain visible even after it has been deleted? Who has access to it? I've had an issue for a while now which makes my Alt + Tab not work and function correctly.  I can't:   Click on it It displays items which are supposed to be hidden (?) It doesn't use the Aero theme   I've tried the following:   Rebooting Closing dwm.exe Updating Windows.   If there is a need for OS: I'm running Win 8.1 x64.  Example: $f \in {\mathscr R[a,b]} \implies f $ has infinitely many points of continuity. Centering polygon on specific coordinates using QGIS I need to place a set of polygons onto a set of points, so that each polygon is centered on its corresponding point. How do I do that? In MapInfo I can just manually edit the coordinates of the center, but there does not seem to be an equivalent in QGIS. Is ∅ equivalent to {∅}? I think they are, but I am not sure? If anyone could clarify, that would be great. Thank you! Heisenberg Uncertainty Principle scientific proof states that:    $$\sigma(x)\sigma(  p_x )\ge \frac {\hbar}{2}.$$  What is the scientific proof of this principle? Vertical component of moving weight at a 45 degree angle Here's an easier one. I use the leg press machine at the gym so I don't have to worrying about hurting myself while lifting heavier weight.  The weight glides on a track that looks to be 45 degrees.  What's the equation to figure out how much weight I would be able to squat normally.  IE the vertical component of moving 400lbs at a 45 degree angle. What is the most elegant proof of the Pythagorean theorem? What are routed events and how it's different from normal events On the commutative property of multiplication (domain of integers, possibly reals) $ab = ba$  This is, inherently, true. Some texts drop it like an axiom without any justification. But I'm a bit curious where it stems from or basically why/how it works. If anyone could enlighten me a bit further, I'd be most grateful. Peano axioms? From set theory? Help.  What bugs me with this definition is it's real world "application":  Let's say I have a power output of 5 W ( $kg\cdot m^2 \cdot s^{-3}$) and I want 30% of that power output (to have some units and natural context).  Naturally, 30% is 30 1/100 (% is the unit). For simplicity, let's express that as:  $30$%$ = 30/100 = 3/10 = 3d$ where $d = 1/10$, a simpler unit (because %W would look... Wrong?).  So 30% of power is then $5 W \cdot 3d$, and we can force it to "make sense" if we associate d with W or basically scale the unit of power by d (which is intuitively understood as being 10 times smaller then the output of 1 W) $dW$. And we want to scale these 5 units of $dW$ three times.  $5 dW \cdot 3$  But the commutative property says it is the same as scaling 3 units of $dW$ 5 times or:  $3 dW \cdot 5$  And the result is the same. This is the bit that hurts my head, the fact it is the same. I try to interpret the same as, for example 50 J of work, it's either applying 50 N over 1 m of distance or 1 N over 50 m of distance.    Down here is additional stuff I think I have (you don't have to read it if you know how to answer immediately). I don't claim it's correct, if anyone parses this information, feel free to correct me.  $m\sum\limits_{i=1}^{n}{1} = n\sum\limits_{i=1}^{m}{1}$  This is how I've been trying to break it down, as it is repeated addition. It is repeated $n$ times and multiplied by $m$, and reverse on the right side. There is $x = m - n$, so when I multiply $n$ $m$ times, each repeated addition of $n$ lacks exactly $x$ to $m$.  $mn = nm$  $x = m-n$  $m(m-x) = (m-x)m$  $m^2 - mx = m^2 - mx$  $ true $  And yes, sadly, I realise that I am using the very property I'm trying to prove ($xm$ shifted as $mx$ on the right-hand side.)  And it even calls in the distributive property of multiplication over addition which I derive (for myself, informally) from the very nature of multiplication and the way we "process" numbers:  For example, number $55$ times $2$  $55 = 5 \cdot 10^1 + 5 \cdot 10^0 = 5 \cdot 10 + 5 \cdot 1$  $(5 \cdot 10 + 5 \cdot 1) \cdot 2$  This is the basis of our positional notation which appends digits of varying orders of magnitude or units together in a way it makes "sense". Each one is b times bigger than the one to the right. Basic stuff. It is why I expressed 10 and 1 explicitly, those are what I consider units in this case.  Now, from the definition of multiplication, which is at its heart just repeated addition, it is truly the same if you add together $2$ $50$ times and then again $5$ times or "all at once" (figuratively, we are always doing the former mentally) $2$ $55$ times.  That added together gives 110, which is true.  To me, the distributive property (in case of integer multiplication) is a repercussion of the very definition of multiplication which is repeated addition (evading strict concepts). I am not certain whether this is a good way to look at it. Reading Chapter 2 of Koblitz's Introduction to Elliptic Curves and Modular Forms, I got stuck on the following question. I would like to proceed my reading, so I would appreciate any hint to this. I have no background in Algebraic Number Theory, but if someone explains in that language, I would like to see it and learn it as much as I can at this moment. References will also be appreciated!  Let $q$ be a power of prime $p$ such that $q \equiv 1 \mod 4$ and $p \nmid 2n$.  Question. Let $\chi_{4} : \mathbb{F}_{q}^{*} \rightarrow S^{1}$ be the character of order $4$, show that $\chi_{4}(-4) = 1$.  The hint in the book says we may show that $\chi_{4}(4) = \chi_{4}(-1) = 1$ for $q \equiv 1 \mod 8$ and $\chi_{4}(4) = \chi_{4}(-1) = -1$ for $q \equiv 5 \mod 8$.  If $q \equiv 1 \mod 8$, then $\chi_{4}(4) = \chi_{4}(2)^{2} = \chi_{2}(2) = \pm 1$, where $\chi_{j}$ denotes the character of order $j$. So I think $2$ must be quadratic residue in this case, but I cannot figure out why. Given (p ∧ q), use the Fitch system to prove (q ∨ r) I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.  Premises: p ∧ q  Goal: q ∨ r  Would be an assumption (p ∧ q => q ∨ r) in first step a correct one?   Then, q ∨ r Implication Elimination: 2, 1  Why is it not correct?  Thank you very much for any help! Probability of a painted cube being reassembled into itself Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$  Without using calculus. There are a few ways I've tried solving this:  $1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$  But it's apparent that nothing's useful here.  $2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$  So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$  We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$  But this inequality doesn't hold.   $3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$  I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative. How to execute file I'm editing in Vi(m) Clock time is off on dual boot Dual boot system Windows XP Pro and Ubuntu 12.04.  I have the bios set for the correct time and Ubuntu set for US Eastern time.  Ubuntu will boot up and the time will be off by -4 hours.  If I correct the time in Ubuntu then when I boot up in Windows XP the time will be off by + 4 hours.  This is a fresh install of 12.04. I did not have this problem before reinstalling Ubuntu. Error :: You must put some 'source' URIs in your sources.list I'm trying to install PIL a python module on ubuntu   I first need to install a package called python-imaging using the following command    sudo apt-get build-dep python-imaging   When I try and do this I get the error   E: You must put some 'source' URIs in your sources.list   I saw  question whihc shows how to fix via the GUI but I'm using the command line. Does anyone know how I can fix this error..?    UPDATE: Based on the below answers I've updated my package list but I now get this error. Not sure if it is related, if not I'll start a new question as I can't find reference to the error anywhere.   E: Build-Depends dependency for python-imaging cannot be satisfied because candidate version of package python-all-dev can't satisfy version requirements Reduction Integration of $I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$ How do I derive this using reduction formula?  $$I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$$  I got $\sqrt{x^2+a^2}(x^n) - \frac{1}{n-1}(x^{n-1})\sqrt{x^2+a^2} + \frac{1}{n-1}(I_n) $ . I'm not too sure if this is correct. I would appreciate any form of help or correction to my answer. Thank you.