Keyboard layout shortcut interferes with Firefox shortcut
OS keyboard shortcuts conflict with apps keyboard shortcuts in GNOME 3
I'm looking for a way to insert continuous colour legend for a raster layer.  I managed to create discrete symbols with values along the bar, but I would like to have a continuous bar with some values on the bar, similar to the old plugin "One-Band Raster", that it doesn't work with the QGIS version higher than 1.7  I tried to set in properties->style, where I selected the  Render Type "Singleband pseudocolor" and see the "generate new color ramp" box.  I tried both "Continuous" or "Equal interval". However, in the Composer the legend always is displayed with discrete symbols.  Is there any way that can make QGIS Composer to display a nice continuous band for a raster layer? CreateService get error: ERROR_INVALID_ADDRESS (0x000001e7) Cancelling upvote on comment? As mentioned on :     For something that can be done so   easily by accident, there should be   the ability to undo mistakes.   I know that I have clicked several times on a comment up-vote without actually wanting to.  I would suggest being able to cancel the up-vote (maybe only for a couple of minutes or even less than a minute, no problem, as long as I can undo a miss-click). How to align text in TikZ nodes by baseline? I need to draw a graph where some nodes have the same y-coordinate. I want the text within the nodes on the same y-coordinate to be baseline-aligned. However, the text seems to be aligned by the bottom lines, not the baseline.   Here's what I tried to do:  \documentclass[final,t]{beamer} \mode&lt;presentation&gt; \usepackage{tikz} \usepackage{pgfplots} \tikzstyle{gate2} = [circle,fill=white,draw=black,minimum size=1cm] \begin{document}     \begin{tikzpicture}         \node[gate2] (mu_0) at (0, 0) {$\mu_0$};         \node[gate2, baseline = (mu_0.base)] (Lambda_0) at (2,  0) {$\Lambda_0  $};     \end{tikzpicture} \end{document}   What I see is that the shapes are at the same y-coordinate (as expected), but mu is higher than Lambda. How do I specify that mu and Lambda should be on the baseline? What is a field, really? Faster language/keyboard swapping I like making a habit to use the different languages I study, to keep that knowledge fresh. So I like typing in those languages when I get the chance. But swapping keyboards under OSX is just a little too cumbersome.  For example, I was just typing the name of one of my co-workers in its original 汉子 form instead of as an English transliteration. It's nice. It reminds me to pronounce her name properly, with tone, etc., which otherwise doesn't come naturally. The transliteration of her name is especially misleading. Everyone at the company is pronouncing it wrong.   But to swap into writing 汉子 I have to mouse up to the top menu bar, select a new keyboard, type, mouse back up to the top bar, and then select back into English. I know there's a keyboard shortcut, but with 5 different keyboards active in my menu, cycling through 5 to get to the one I want feels just as cumbersome.  Is there a way to attach a hotkey to a specific keyboard, or some other way to make swapping character sets lighter? How bad for TeX is omitting braces {}, even if the result is the same? Sometimes writing braces and omitting them produces the same result:  \tilde{\psi}  \mathrm{e}^{\mathrm{i} \pi}   vs  \tilde \psi   \mathrm e^{\mathrm i \pi}   Or (taken from Werner's comment below)   \frac{1}{2} vs \frac12   If that's the case, how bad for the quality of my code is omitting them? Of course, if one compiles a pair of formulas, there is no difference. But, since I ignore how TeX works, my point is: if one writes long codes with these habits,   doesn't TeX go somehow crazy because I'm not writing as it expects?   For instance, does it have any effect on how fast are long documents compiled? Any reason to keep them? Polarization vectors of massive and massless particles Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$ Calculate intersection of a line with a plane I was reading something about: “Intersection of a line with a plane”. But I don’t fully understand how to calculate it on their way.  The goal is to calculate where the plane and the line will “cross”. These items are given:   The eye is in position: O(0,0,-10)   A point of a 3D-object is P(10,20,30) The screen has the equation z=0   I have a questions about how they calculate this:    This part is clear. That is the location form the eye (vector representation).   They specify the location of the plane, there is nothing strange here (vector representation).   Now they calculate what part of the line will cross, how did they calculate this? How did they know a = 1/4? And b = 10/4?  This is everything that was given. Beta regression of proportion data including 1 and 0 How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? An awful lot of people seem to use the phrase "substitute X for Y" to mean "replace X with Y", while I've always used and understood it as "replace Y with X". This makes sense to me, given that a substitute is the replacement, not the thing which has been replaced.  I thought I might be able to figure things out from a dictionary, but now after seeing two sorts of definitions, I'm even more confused:   to put or use in the place of another to take the place of; replace   The first one suggests that the object will be the substitute (replacement), the thing put into place, while the second suggests that the object will be the thing which has been replaced.  Is one of these usages more correct in any sense? (I know this is probably a pointless question, since people will continue to speak the way they speak.) Is there perhaps a difference between American and British English?   And most importantly, is there any reliable way to tell what someone means when they ask how to substitute honey for sugar, or is the verb simply guaranteed to cause confusion? Spacing before and after section titles Tools to get a pictorial function call graph of code How can I check the integrity of a downloaded Ubuntu CD? I've just downloaded a CD image of Ubuntu. I know the installer has a check CD option, but how can I check that the image is good before burning it to CD? AutoScroll on chat exchange doesn't work in chrome 61 I updated the chrome to version 61, and the auto-scroll stopped working, it is necessary to scroll down the page manually. The problem does not occur in the previous version or in firefox.  Also, the reply icon no longer working when the message replied to is not in the current chat page. (Middle click still works and opens the link in new tab.) What do you call a person who keeps reminding others of one’s favors for them? When you fight with a person who then keeps telling you I did that I did this for you and so on. I want to tell them that it's inappropriate to remind others of the favors you did for them, especially if I appreciate those favors and try my best to return them. To mean "He is a friend of mine who I love" or "He is a friend of mine that I love", can I say "He is a friend of mine I love"? When I omit "who" or "that" in that sentence, it sounds wrong to me. I think I am not used to that structure. What do you think?  Note: I know (that) relative pronouns can be dropped like in "He is someone (that/who) I love", but in the sentence I gave, I think it is wrong to omit it because "a friend" was already modified by "of mine".  Context: I am talking about a friend, and I am saying I love him. Principal $n$th root of a complex number This is really two questions.   Is there a definition of the principal $n$th root of a complex number? I can't find it anywhere. Presumably, the usual definition is $[r\exp(i\theta)]^{1/n} = r^{1/n}\exp(i\theta/n)$ for $\theta \in [0,2\pi)$, but I have yet to see this anywhere. Is this because it has bad properties? For instance, according to this definition is it true that for all complex $z$ it holds that $(z^{1/a})^b = (z^b)^{1/a}$?