This topic has been touched on  but not really answered.  I have a dual monitor system and the colors differ wildly.  I currently live Buenos Aires where color correction hardware costs premium prices.  I do some graphic design, but don't require a pro-level calibration.  That said, I'd like my monitors to be set as close to "true color" as possible.  I've located the useful and free , but it seems to adjust the entire system internally at startup.  I could use the  to set a different ICC or ICM profile for each monitor, but the Monitor Calibration Wizard outputs its own format for profiles.
I use a dual monitor.  The colors appear entirely different on the new monitor than they do on my laptop.  When you are selecting colors etc, to be included in a website design, it totally freaks me out.  Is there no software solution that finds how the settings of one of the monitor is and puts that to the second one?  Otherwise what all settings do I need to change manually on the second monitor to make it appear as it does on the laptop?  Thanks in advance.
Smooth option sometimes produces incorrect arrow tips in PGFplots It seems that the smooth option applied to lines sometimes effects the arrow tips and produces incorrect results.  No problem with vertical lines, but any other seems to have a problem. Is this a known bug, or is there a reason why the smooth should not be applied to straight lines?  Updated: to show problem does NOT occur with TikZ, and problem also occurs with curves if smooth option is applied.  \documentclass{article} \usepackage{pgfplots} \begin{document}  \tikzstyle{MyStyle}      =[-&gt;, black, ultra thick,       ] \tikzstyle{MyStyleSmooth}=[-&gt;, black, ultra thick, smooth]  \begin{tikzpicture} \begin{axis}     % 'smooth' option seems to have no effect on lines '1' and '2'     \addplot [MyStyle      , green ] coordinates{(0,0) (0,1)} node [above] {1};     \addplot [MyStyleSmooth, blue  ] coordinates{(1,0) (1,1)} node [below] {2};     %     % But, can't seem to have the 'smooth' on lines, '4', '5' and '6'     \addplot [MyStyle      , orange] coordinates{(2,0)   (3,0)   } node [right] {3};     \addplot [MyStyleSmooth, red   ] coordinates{(4,0)   (2,0.5) } node [below] {4};     \addplot [MyStyleSmooth, red   ] coordinates{(2,1)   (3,1)   } node [left ] {5};     \addplot [MyStyleSmooth, red   ] coordinates{(3,0.8) (2,0.8) } node [right] {6};      % Similar problem with curves: "non-smooth" is ok, "smooth" is not     \addplot[MyStyle,       mark=none, domain=0:1,samples=50, blue] (x,x*x+1);     \addplot[MyStyleSmooth, mark=none, domain=0:1,samples=50, red ] (x,x*x+2);  \end{axis} \end{tikzpicture}  \begin{tikzpicture} % These are all fine     \draw [MyStyle      , green ] (0,0)   -- (0,1)    node [above] {1};     \draw [MyStyleSmooth, blue  ] (1,0)   -- (1,1)    node [above] {2};     %     \draw [MyStyle      , orange] (2,0)   -- (3,0)    node [right] {3};     \draw [MyStyleSmooth, red   ] (4,0)   -- (2,0.5)  node [below] {4};     \draw [MyStyleSmooth, red   ] (2,1)   -- (3,1)    node [right] {5};     \draw [MyStyleSmooth, red   ] (3,0.8) -- (2,0.8)  node [left ] {6}; \end{tikzpicture} \end{document} I'm a mathematician interested in abstract QFT.  I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator.  I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator.  But I don't see how this forces $T^2=-1$.  (Or maybe it doesn't force it, it merely allows it?)  Here's another version of my question.  There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$.  If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$.  So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$.  (I'm assuming Euclidean signature here.)  My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?  (I posted a  on physics.stackexchange.com last week, but there were no replies.)    EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search.  If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$.  I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator. I would like to know the best way to write test methods for controller extension classes that are used by Visualforce.  What is the minimum set code that I need to cover the basic operation of the controller without taking into account any custom functionality that I might add.  Here is my example class with the test method ready to be completed.  public with sharing class myExtension{      private final Account acct;      public myExtension(ApexPages.StandardController stdController){         this.acct = (Account)stdController.getRecord();     }      public static testMethod void testmyExtension(){      } } What is aperture, and how does it affect my photographs? Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$ I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.  Now he showed me an exercise he was having trouble with, which states that one should prove $$n^{n+1} &gt; (n+1)^n$$  for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.  Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?  Thanks for answers in advance. Oracle SQL Syntax: Quoted identifier Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great! What formula should be used to determine "hot" questions? Right now  is fairly broken -- it's a simple descending sort by views. As Joel said , it is &quot;a self-fulfilling prophecy.&quot; But this is not intentional, it's only because we haven't had time to improve it yet! As I sit down to write a better algorithm, I thought you guys might have some insight into what would make a good formula to determine the &quot;hot&quot; questions -- over the last 24 hours, the last week, the last month, and all time. Here are the variables we have to work with:  Votes [aka Score] Views Answers Answer Votes [aka Score] Whether the question has an accepted answer or not Time question was originally asked Time of last activity on question Reputation of asker Reputation(s) of answerers  Joel cited the reddit formula as an example, but I think our site is a little different than just a basic link submission site. Plus, we have more variables and data to work with. I found a  describing a few formulas: Y Combinator Hacker News  (p - 1) / (t + 2)^1.5 p = votes (points) from users t = time since submission in hours  Reddit  t = (time of entry post) - (Dec 8, 2005) x = upvotes - downvotes y = {1 if x &gt; 0, 0 if x = 0, -1 if x &lt; 0) z = {1 if x &lt; 0, otherwise x} log(z) + (y * t)/45000  Thoughts? Ideas? I don't want to overthink this; I just want the &quot;hot&quot; questions to be relevant and useful to us. So the simpler the algorithm the better! Based on my analysis of the above and the comments so far, here's the second version of what I have implemented so far. This might suck. I don't know:  (log(Qviews)*4) + ((Qanswers * Qscore)/5) + sum(Ascores) -------------------------------------------------------- ((QageInHours+1) - ((QageInHours - Qupdated)/2)) ^ 1.5  Note that accepted answers weight not at all in hotness. This is intentional, as I feel accepted answers are a fine social contract, but not a good data point for question or answer quality. This algorithm will heavily favor questions with LOTS of answers, as the sum(Ascores) are now included -- one assumes if there are lots of answers, there will be a lot more voting on the answers, too. Update: Note that this formula is what is used on the  linked from the homepage of each site. It is not the formula used to determine the  list. See . How to get the insert ID in JDBC? How do I send all information in /var/adm/message file to a remote system? I'm trying to figure out how to send all information that is normally identified in the /var/adm/message file on a Solaris system to a remote system. Is it possible to use NBT tags in commands in Minecraft: Bedrock Edition? This question was recently split into different questions for each workaround. See   meta posts for more information.   Can you use NBT tags in commands? I’ve tried commands that normally work in Java Edition like: give @s minecraft:planks 30 1 {display:{Lore:[&quot;Wooden planks&quot;]}} give @s stone_pickaxe{Unbreakable:1b} give @s item_frame{EntityTag:{Invisible:1b}} summon villager ~ ~ ~ {Offers:{Recipes:[{buy:{id:&quot;emerald&quot;,Count:5b},sell:{id:&quot;diamond_sword&quot;}}]}} summon tnt ~ ~ ~ {Fuse:20s} The supremum norm is submultiplicative Is the following proof correct:  Let $X$ be a compact Hausdorff space and $C(X)$ the space of continuous functions $f: X \to \mathbb{C}$. We can equip $C(X)$ with the (edit: sorry, semi-)norm $\lVert f \rVert_\infty = \sup_{x\in X} |f(x)|.$ I want to prove that for $f,g \in C(X)$, $\lVert fg \rVert_\infty \leq \lVert f \rVert_\infty \lVert g \rVert_\infty $. Let $z \in X$ be arbitrary then $$ \lvert f(z)g(z) \rvert \leq \lvert   \; \sup_{x} \lvert f(x) \rvert  g(z)  \;\rvert=\sup_{x} \lvert f(x) \rvert  \lvert    g(z)\rvert \leq \sup_{x} \lvert f(x) \rvert  \sup_{y} \lvert g(y) \rvert.  $$  Since $z$ was arbitrary $$ \sup_{z} f(z)g(z)\leq \sup_{x} f(x) \sup_{y} g(y) $$  that is the sought. Given a continuous random variable $X$ and the pdf find.... There is a pattern excercise:  First pattern: We have $X$ and the probability density function is usually:  $$ f(x) = \begin{cases}   {ax+bx^2} &amp;  {0\lt x \lt 1} \\   0 &amp; \text{else}   \end{cases} $$  Then we are given $E|X| = 0.6$   And usually asks to find:  a) $P\{X&gt;0.5\}$  b) $Var(X)$    Second pattern:  $$ f(x) = \begin{cases}   {cx^4} &amp;  {0\lt x \lt 2} \\   0 &amp; \text{else}   \end{cases} $$  Where c is a constant. Find:  a) Mean value of $X$  b) Variance of $X$     Third pattern (it's more like a reverse problem):  We are given $Y= \sqrt{X}$, find it's cumulative distribution function (CDF) and  it's probability density function (pdf).    Update:  The third one is transferred to a .  This is not homework. I'm self-studying and in need of some help. Thank you. sometimes when writing mathematical expression with partial derivatives, one needs to write a vertical bar afterwards and write the value at the which the partial derivatives are being computed as subscripts. See for example the bottom post here:    However, the \mid symbol doesn't produce a tall bar that corresponds to the size of the partial derivatives term. Is there a way to have a taller bar?   PS: Some other posts here mention \middle| but this doesn't compile on my computer.  Thanks in advance.... Create a table with LaTeX Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i=1,\dots,n$.    I cannot prove that such an element there exists. I was trying to prove it by induction on $n$. If $n=1$ it is true by Nakayama, so suppose it is true for $n-1$. Then for every $i$ I can find an $x_i\in M$ such that $\frac{x_i}{1}\not\in \mathfrak m_jM_{\mathfrak m_j}$ for every $j\neq i$. If $\frac{x_i}{1}\not\in \mathfrak m_iM_{\mathfrak m_i}$ for some $i$ we are done, so suppose $\frac{x_i}{1}\in\ m_iM_{m_i}$ for every $i$. I don't know how to go on, could you help me? How to report that a question should be migrated to another site? Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:  $\forall$ $\delta &gt;0$ $\exists$ $\epsilon&gt;0$ such that if $0&lt;|x-a|&lt;\delta$ then $|f(x)-L|&lt;\epsilon$  I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work. I would like to know how to find a set of coordinates on a small circle some distance from another point.   For example, lets say A = (39.73, -104.98) #Denver  B = (39.83, -106.06) #point approximately 50 nautical miles away from A along the great circle track to C  C = (40.75, -111.88) #Salt Lake City, approximately 323 nautical miles away from A  How do I get the set of orange coordinates when I don't know the angle but I do know the cross track distance?  For concreteness, lets say the outer points near B have a cross track distance of +- 30 nautical miles and the inner points have a cross track of +- 15nm.  And the points near C have a cross track of +- 25 nm.  Also I have an initial bearing from A to C of -1.34 #radians Address is invalid on local machine [on windows 8 only]