I read the "" article by Thomas C. Erren and Philip E. Bourne. It gives me the general idea of how to prepare a poster.  However, I wonder whether I should add references to the poster as many parts have strong (advanced) knowledge acquired from other works? If yes, where is recommended for references in a portrait A0 poster. Thanks.
After reading this question about  and asking myself two questions about  and , I realize that these activities are for showing what you found, make the audience be attracted. If they need further explanation, they should read your paper, which has successfully defensed and proved that your work is validated. One important thing when you attempt to attract people is concentrate on the important, which means unnecessary things need to be trimmed. I think the reference is one of those. So, should we put the references in slides/poster when we already have them in paper?
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: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ What's the meaning of this present perfect sentence? What is the best regular expression to check if a string is a valid URL? I currently have a masters degree in an engineering field and I don't currently have any PhD degrees (I may study for a doctorate degree in future). It happens that when I go to conferences or scientific occasions, even in the emails I receive from people; they use a doctor prefix for my name; but I do not have a PhD degree.  What is a good etiquette to deal with such thing? Should I directly tell them that I do not have a doctorate degree? Or should I neglect such a mistake by them and don't do anything?  I think it is unethical not to correct such thing, because the person may still think that I have a PhD degree; and such wrong usage of title should be corrected some way. Once a person used doctor title for my name in an email and I just added a line in my reply to his email, stating that I have a masters degree. Compute $e^{e^z}$, where z is a complex number I am self-studying the MAT 67 linear algebra course and I got stuck at a problem in Chapter 2:     Compute the real and imaginary parts of $e^{e^z}$ for $z$ $\in$ $\Bbb C$.   I tried to expand it using Euler's formula but it seemed endless. Any help? Apply a string replacement on the citation as returned by \cite I have biblatex with options  \usepackage[backend=bibtex, style=authoryear]{biblatex}   The citations made with \cite look like this: Smith, 2019 or Smith and Baker, 2019 and so on. I would like to define a command that automatically replaces the occurrence of my name with my initials. Namely I need  \cite{SmithsPaper}  % --&gt; Smith, 2019 \cite{MyPaper}      % --&gt; Manny, 2019 %%% \newcite{SmithsPaper}  % --&gt; Smith, 2019 \newcite{MyPaper}      % --&gt; MC, 2019   I have tried with the package xstring to do something like  \newcommand{\newcite}[1]{%   \begingroup   \edef\@temp{\cite{#1}}%   \StrSubstitute{\@temp}{Manny}{MC}   \endgroup }   The problem is that I can't figure out how to tell LaTeX to expand \cite{#1}. As a result the citation comes out without the replacement. If I change \cite{#1} by simply #1 and I call \newcite{bla Manny bla} it gives bla MC bla as expected.  It shouldn't be relevant but I'm using this within the beamer documentclass. If I just render the default cube it will look like this:   It is rendered with color depth 16. I can see there are artefacts consisting of visible "line strokes" on some of the cube faces.  The bands seems to not go darker and darker, but alternates between light and dark, or is my eye fooled?  Why?  Edit  Here comes a gradient created by gimp without banding, using the same computer: How to quickly and easily create a grid with Adobe Photoshop? Suppose I need to draw a grid in Photoshop, each square being the same size, and a certain amount of pixels tall and wide, and the grid lines having a certain pixel width. Like this:     Currently I do this by drawing one square, copy+pasting+moving+merging it a few times to create one row, copy/pasting that a few times to create more rows above each other.  It works, but feels very clumsy, I'm sure there ought to be a more efficient way to do this?  What are some efficient methods for creating a grid of equally sized squares? How do I call a Windows shell command using VB6? What is the correct way to get a pgfkey value? How to deal with very long lines that do not contain spaces? CSV text files loading error / crash in QGIS 2.18.8 I got a problem when I try to load tab-separated text files in the "Create a layer from the delimited text file" tab (that one with de comma symbol). After normally selecting the X and Y fields and clicking OK, the reference coordinate system (SRC) selector screen appears, then the program closes by itself. I had uploaded more than a hundred of this type of file by the same procedure in the previous version of QGIS (2.14 Essen) and have no problem, but since I updated to version 2.18.8 (Las Palmas) I can not load any files or .CSV type.   Do anyone knows what's happening? Am I doing anything wrong? Where do I find the current C or C++ standard documents? Arbitrary intersection of closed, connected subsets of a compact space connected? Let $(B_i)_{i\in I}$ be an indexed family of closed, connected sets in a compact space X. Suppose $I$ is ordered, sucht that $i &lt; j \implies B_i \supset B_j$.  Is $B = \bigcap_i B_i$ necessarily connected?  I can prove it, if I assume $X$ to be Hausdorff as well: If $B$ is not connected, then there are two disjoint, closed, nonempty sets $C$, $D$ in $B$, such that $C \cup D = B$. Now these sets are also closed in $X$, hence by normality there exist open disjoint neighborhoods $U$, $V$ of $C$ and $D$, respectively.  Then for all $i$: $B_i \cap U^c \cap V^c \ne \emptyset$, since $B$ is contained in $B_i$ and $B_i$ is connected. Thus we must also have  $$ B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset $$  by compactness and the fact that the $B_i$ satisfy the finite intersection property. This is a contradiction to the choice of $U$ and $V$.  I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated!  Thanks,  S. L. Editing capabilities document in GeoServer I've got several layers in GeoServer, one of which I'm trying to use as a WMS in desktop programs (Manifold 8). I've set it up correctly, but whenever I bring it into Manifold, it projects the layer into EPSG:4326, which unfortunately can't work with my map - I need it in EPSG:3857. The layer in question is wrapped inside a more overarching layer definition, which contains all the layers - and coordinate systems that allow them all to be used flexibly. However, I want to put this layer on its own in the capabilities document, and only allow it to use EPSG:3857. How, in GeoServer, would I edit, or override, a capabilities document? Spoilers for the end of the second film      Towards the end of the fight against Ego, we see that the bomb Groot set off blows up Ego's core. Shortly thereafter, it cuts to a scene where Quill is shown apparently losing his powers.   Given that the movie builds up this finale so much, I'm left wondering:      Did Quill lose his powers or merely the means to access them after Ego's apparent death?   Answers from the film or comics are very much welcome. Why does my Nikon D3100 give the error "Lens not attached" with a Vivitar 200 telephoto lens? I have a geotiff raster image that has a coordinate system with longitudes from 0 to 360. The horizontal center of the image is 180 longitude. See image below:    I want to transform it to EPSG:4326 SRS with -180 180 longtitude range.  And I want the center of the image to be at Greenwich meridian (0). I guess this srs is very widely used. I expect the result look like this:    So I use a gdalwarp command to reproject:  gdalwarp -s_srs '+proj=latlong +datum=WGS84 +pm=180dW' -t_srs EPSG:4326 test_col.tif test_4326.tif   But I only get a tiff with bigger dimensions (more pixels) and EPSG:4326 metadata. The image itself looks the same, as the initial one. But I expect it to swap the hemispheres.  How do I gdalwarp an image to be strictly -180 180 EPSG:4326 with the center in 0 longitude?  This is gdalinfo of my initial file:  Origin = (-0.102272598067084,89.946211604095552) Pixel Size = (0.204545196134167,-0.204423208191126) Metadata:   AREA_OR_POINT=Area Image Structure Metadata:   INTERLEAVE=BAND Corner Coordinates: Upper Left  (  -0.1022726,  89.9462116) (  0d 6' 8.18"W, 89d56'46.36"N) Lower Left  (  -0.1022726, -89.9462116) (  0d 6' 8.18"W, 89d56'46.36"S) Upper Right (     359.897,      89.946) (359d53'50.18"E, 89d56'46.36"N) Lower Right (     359.897,     -89.946) (359d53'50.18"E, 89d56'46.36"S) Center      ( 179.8975000,  -0.0000000) (179d53'51.00"E,  0d 0' 0.00"S)   This is gdalinfo after gdalwarp     Origin = (-180.102727401932952,89.946211604095552) Pixel Size = (0.091397622896436,-0.091420837939082) Metadata:   AREA_OR_POINT=Area Image Structure Metadata:   INTERLEAVE=PIXEL Corner Coordinates: Upper Left  (-180.1027274,  89.9462116) (180d 6' 9.82"W, 89d56'46.36"N) Lower Left  (-180.1027274, -89.9699975) (180d 6' 9.82"W, 89d58'11.99"S) Upper Right ( 179.8211116,  89.9462116) (179d49'16.00"E, 89d56'46.36"N) Lower Right ( 179.8211116, -89.9699975) (179d49'16.00"E, 89d58'11.99"S) Center      (  -0.1408079,  -0.0118929) (  0d 8'26.91"W,  0d 0'42.81"S) The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied How to find integer solutions to $M^2=5N^2+2N+1$?