I've been fiddling around with different desktop environments lately. I started with Gnome Fallback, then Gnome 3 then XFCE and now I decided to use Unity.  I installed ubuntu-desktop and launched "Ubuntu" in Lightdm but no panel or launcher appears. To make them appear I need to do CTRL+ALT+F1, login and execute:  export DISPLAY=:0 setsid unity   It then appears after I execute that but every time I login I need to do that. In addition, there are no indicators in the top right corner.  Does anyone know how I can make the indicators appear and make the launcher and panel appear when I log in?  Thanks,
When I login, nothing happens.   I am presented with my desktop wallpaper.    No Dash, no Launcher, nothing.
How can I change the icon of an application in the Unity launcher? I'm trying to change the icon of a specific application (Pidgin) in my Ubuntu 11.10 Unity launcher. I have tried searching in the options and in the Compiz settings but could locate no option for changing individual icons.  I can change the theme but that's not what I want.   I am willing to edit configuration files to get this change. It doesn't have to be a UI solution. Removing intercept from GLM for multiple factorial predictors only works for first factor in model I am running a binomial logistic regression with a logit link function in R. My response is factorial [0/1] and I have two multilevel factorial predictors - let's call them $a$ and $b$ where $a$ has 4 factor levels  $(a_1,a_2,a_3,a_4)$ and $b$ has 9 factor levels $(b_1,b_2,\dotsc,b_9)$. Therefore:  mod &lt;- glm(y ~ a + b, family = binomial(logit), data=pretend) summary(mod)   The model output would then show all the information about the model as well as the coefficients.  There is a factor level for both a and b missing (a1 and b1) from the summary output. I understand that it is constrained in the "intercept" of the model. I have read that if I want to remove the intercept term and see the estimates for those factor levels I can just add -1 to the model formula, i.e.:  mod2 &lt;- glm(y ~ a + b - 1, family=binomial(logit), data=pretend) summary(mod2)   In the new model (mod2) the intercept term is then gone and variable a's factor-level a1 is given amongst the list of coefficients. But, variable b's factor-level b1 is still missing and given that there is no intercept term anymore, how can I interpret the odds-ratio for that factor level then?  Could someone please explain to me how to get the coefficient for b1 too and why this is happening? Windows 10 Anniversary Upgrade error 0xc1900101 - 0x30018 referring to SYSPREP Fixing a code to highlight formulas and text on several lines In  I've found the following amazing code to highlight text on several lines.  The problem is that code dosen't work well with formulas : for example, fraction using the mode display are partially out of the highlighting frame.   Here is the code with one content showing the problem.  \documentclass{article}     \usepackage{soul}     \usepackage{tikz}     \usetikzlibrary{calc}     \usepackage{amsmath}     \usepackage{xcolor}      \makeatletter         \newcommand{\defhighlighter}[3][]{%             \tikzset{every highlighter/.style={color=#2, fill opacity=#3, #1}}%         }          \defhighlighter{yellow}{.5}          \newcommand{\highlight@DoHighlight}{             \fill[                 outer sep = -15pt, inner sep = 0pt,                  every highlighter, this highlighter              ](                 $(begin highlight)+(0,8pt)$) rectangle ($(end highlight)+(0,-3pt)$             );         }          \newcommand{\highlight@BeginHighlight}{             \coordinate (begin highlight) at (0,0) ;         }          \newcommand{\highlight@EndHighlight}{             \coordinate (end highlight) at (0,0) ;         }          \newdimen\highlight@previous         \newdimen\highlight@current          \DeclareRobustCommand*\highlight[1][]{%             \tikzset{this highlighter/.style={#1}}%             \SOUL@setup             %             \def\SOUL@preamble{%                 \begin{tikzpicture}[overlay, remember picture]                     \highlight@BeginHighlight                     \highlight@EndHighlight                 \end{tikzpicture}%             }%             %             \def\SOUL@postamble{%                 \begin{tikzpicture}[overlay, remember picture]                     \highlight@EndHighlight                     \highlight@DoHighlight                 \end{tikzpicture}%             }%             %             \def\SOUL@everyhyphen{%                 \discretionary{%                     \SOUL@setkern\SOUL@hyphkern                     \SOUL@sethyphenchar                     \tikz[overlay, remember picture] \highlight@EndHighlight ;%                 }{}{%                     \SOUL@setkern\SOUL@charkern                 }%             }%             %             \def\SOUL@everyexhyphen##1{%                 \SOUL@setkern\SOUL@hyphkern                 \hbox{##1}%                 \discretionary{%                     \tikz[overlay, remember picture] \highlight@EndHighlight ;%                 }{}{%                     \SOUL@setkern\SOUL@charkern                 }%             }%             %             \def\SOUL@everysyllable{%                 \begin{tikzpicture}[overlay, remember picture]                     \path let \p0 = (begin highlight), \p1 = (0,0) in \pgfextra                     \global\highlight@previous=\y0                     \global\highlight@current =\y1                     \endpgfextra (0,0) ;                     \ifdim\highlight@current &lt; \highlight@previous                         \highlight@DoHighlight                         \highlight@BeginHighlight                     \fi                 \end{tikzpicture}%                 \the\SOUL@syllable                 \tikz[overlay, remember picture] \highlight@EndHighlight ;%             }%             \SOUL@         }     \makeatother   \begin{document}  Lorem ipsum \highlight{dolor sit amet, consectetur adipis-icing elit, sed do eiusmod tempor} incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation \highlight[red]{ullamco $laboris$ nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit} in voluptate velit esse cillum dolore eu fugiat nulla pariatur.  Excepteur sint occaecat \highlight[green, draw=blue]{cupidatat non proident  $\dfrac{789}{7889}$ % The frame is not good for that formula ! suntinculpaquiofficiadeseruntmollitanimidestlaborum. Loremipsumdolorsitametconsecteturadipisicingelitseddoeiusmodtemporincididuntutlabore-etdoloremagnaaliqua.} I suppose I could write some more text here.  \end{document}     Is someone could help me to fix it? How to prove that $a \cdot b$ is not divisible by 5 for $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{99} + \frac{1}{100} = \frac{a}{b}$? How to set a command block with a command inside using the setblock command Minecraft Empty Chests Command Block I am using the following Command to spawn a new chest at the given coordinates with stuff in it at each new round:  /setblock 264 4 997 minecraft:chest 0 replace {Items:[{id:274,Count:1,Slot:0},{id:275,Count:1,Slot:1}]}   The problem with that is that when replacing the old chest all the stuff in the old chest is thrown out. And that's pretty much not what I want. I need to find a way to replace a chest which might have stuff in it without throwing those things out.  Someone else said I should replace the given position for one tick with lava, which I tried. Works in 90% of the tries (in the other 10% the items from the chest are thrown to far away - out of the spawned lava block), but the problem with that is that I can't use it in wooden areas because it might burn everything down.  So what can I do to fix that problem, because otherwise my custom made Pv.P map is not gonna work?  Thanks very much! My Mac Book Pro 13.-" doesn´t get past the grey booting screen with the apple and the spinning wheel (The loading bar appears but when it´s done nothing happens)   I have 220 GB in one single Partition (Macintosh HD) and there is all my work my music and the only copy of ALL MY PHOTOS. I´m so stupid I don´t have a backup of anything but know I have an external HD with 300GB space. I would like to repair my HDD but more important I need to save my files but I don´t know how to do it in Disk Utility Mode. I read there is a Diskwarrior but I don´t know how to install it without booting the mac. Please help!!!  When I start on Single User mode and enter /sbin/fsck -fs this appears  Executing fsck_hfs (version diskdev_cmds-557-393)  Checking Journaled HFS Plus volume.  The volume name is Macintosh HD  Checking extents overflow file.  Checking catalog file  Incorrect number of thread records  (4, 23745)  Checking multi-linked files.  Checking catalog hierarchy  Invalid volume directory count (It should be 175703 instead of 175701) Checking extended attributes file.  Incorrect number of extended attributes (It should be 340062 instead of 340064)  Checking volume bitmap.  Checking volume information.  The volume Macintosh HD could not be repaired after 3 attempts.  *****The volume was modified *****   What are the next steps if fsck fails to repair a drive? I have functionality on a cloud page that depends on the cloud page url. I know how to Requestparameter(), but how do I get the entire url using ampscript or SSJS? How to test a confirm dialog with Cucumber? Possible cardinality of power sets What cardinals can provably not occur as the cardinality of a power set? I know that $\mathbb N$ and natural numbers that are not powers of two are such cardinals. What else is out there? If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$ Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is  Try: From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ $$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$ $$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$ and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$ Could some help me how to find its upper bound, Thanks Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$    the only thing I could think of is that it can be written like this :  $$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$  is my answer right ? and are there any other method ?(I'm sure there are) How do you keep user.config settings across different assembly versions in .net? As far as I know, the variance inflation factor is not computed with pseudo-$R^{2}$ or generalized $R^{2}$ in binary outcome models (e.g. logistic regression).  Are there other measures of multi-colinearity than VIF appropriate to such models?  Why ought or ought not we think about multi-colinearity in such models? wpf combobox binding Can I use images rendered with blender for commercial use? If I make a scene and then render it with BR or Cycles, am I allowed to use that image in a commercial project? Do I need to include any copyright notice or license? How to set QGIS language other then OS Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$  But I don't know how to explain it. I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?  $$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\  \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\  \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$  I can't find the right words to explain why this is true?    Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0 \end{array} $$  By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:  $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\ \\ \arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\ \\ \text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Using Calculus, why is that true for every $x \in [-1,1]$?    Edit #2:  A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.  Thus the equations from "Edit #1" should be as follows: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ My windows directory is huge