Shaping line segment end points
How do I flatten this vector logo leaving only the visible black paths?
Bluetooth headset never gives me sound options "Planar" graphs on Möbius strips May someone help me? I am trying to use induction to prove that the formula for finding the $n$-th term of the Fibonacci sequence is:   $$F_n=\frac{1}{\sqrt{5}}⋅\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}⋅\left(\frac{1-\sqrt{5}}{2}\right)^n.$$  I tried to put $n=1$ into the equation and prove that if $n=1$ works then $n=2$ works and it should work for any number, but it didn't work. I need to prove that this formula gives the $n$th Fibonacci number. I am comfortable with the different sizes of infinities and Cantor's "diagonal argument" to prove that the set of all subsets of an infinite set has cardinality strictly greater than the set itself. So if we have a set $\Omega$ and $|\Omega| = \aleph_i$, then (assuming continuum hypothesis) the cardinality of $2^{\Omega}$ is $|2^{\Omega}| = \aleph_{i+1} &gt; \aleph_i$ and we have $\aleph_{i+1} = 2^{\aleph_i}$.  I am fine with these argument.  What I don't understand is why should there be a smallest $\infty$? Is there a proof (or) an axiom that there exists a smallest $\infty$ and that this is what we address as "countably infinite"? or to rephrase the question "why can't I find a set whose power set gives us $\mathbb{N}$"?  The reason why I am asking this question is in "some sense" $\aleph_i = \log_2(\aleph_{i+1})$. I do not completely understand why this process should stop when $\aleph_i = \aleph_0$.  (Though coming to think about it I can feel that if I take an infinite set with $\log_2 \aleph_0$ elements I can still put it in one-to-one correspondence with the Natural number set. So Is $\log_2 \aleph_0 = \aleph_0$ (or) am I just confused? If $n \rightarrow \infty$, then $2^n \rightarrow \infty$ faster while $\log (n) \rightarrow \infty$ slower and $\log (\log (n)) \rightarrow \infty$ even slower and $\log(\log(\log (n))) \rightarrow \infty$ even "more" slower and so on).  Is there a clean (and relatively elementary) way to explain this to me?  (I am totally fine if you direct me to some paper (or) webpage. I tried googling but could not find an answer to my question) My bibliography (made with biblatex) has all items close to each other. I would like to introduce some space between them, e.g. half a line.  How can I do this? What rules say that summoning undead is evil, and what rules say why? How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers? I have seen many examples of how to render sprites from a spritesheet but I havent grasped why it is the most common way of dealing with sprites in 2d games.  I have started out with 2d sprite rendering in the few demo applications I've made by dealing with each animation frame for any given sprite type as its own texture - and this collection of textures is stored in a dictionary. This seems to work for me, and suits my workflow pretty well, as I tend to make my animations as gif/mng files and then extract the frames to individual pngs.  Is there a noticeable performance advantage to rendering from a single sheet rather than from individual textures? With modern hardware that is capable of drawing millions of polygons to the screen a hundred times a second, does it even matter for my 2d games which just deal with a few dozen 50x100px rectangles?  The implementation details of loading a texture into graphics memory and displaying it in XNA seems pretty abstracted. All I know is that textures are bound to the graphics device when they are loaded, then during the game loop, the textures get rendered in batches. So it's not clear to me whether my choice affects performance.  I suspect that there are some very good reasons most 2d game developers seem to be using them, I just don't understand why. I played around with it for a while, but I simply can't figure it out.  I made a tank that fires missiles, and when the missiles hit the walls, I want them to bounce off, but I want them to bounce off to the right angle.  Right now I haven't got any obstacles, the missiles just bounce off when they get outside the viewportRectangle I made.  Is the solution I'm looking for quite advanced?  Is there a relativly simple way to do it? I have a graph. This is my code.    \documentclass[10pt]{article}     \usepackage{pgf,tikz}     \usetikzlibrary{arrows,patterns}     \pagestyle{empty}      \definecolor{ttqqcc}{rgb}{0.2,0,0.8}     \definecolor{ffqqtt}{rgb}{1,0,0.2}     \definecolor{ttqqff}{rgb}{0.2,0,1}     \definecolor{uququq}{rgb}{0.25,0.25,0.25}      \begin{document}      \begin{tikzpicture}[line cap=round,line join=round,&gt;=triangle 45,x=1.0cm,y=1.0cm]      \draw[-&gt;] (-3.63,0) -- (3.67,0);     \foreach \x in {-3,-2,-1,1,2,3}       \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt)          node[below] {\footnotesize $\x$};     \draw[color=black] (3.48,0.07) node [anchor=south west] { $x$};      \draw[-&gt;] (0,-1.93) -- (0,5.57);     \foreach \y in {-1,1,2,3,4,5}       \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt)          node[left] {\footnotesize $\y$};     \draw[color=black] (0.06,5.19) node [anchor=west] { $y$};      \draw (0pt,-10pt) node[right] {\footnotesize $0$};      \clip(-3.63,-1.93) rectangle (3.67,5.57);     \draw[pattern color=ffqqtt,pattern=north east lines,fill opacity=0.1, smooth,samples=50,domain=0:1.0]      plot(\x,{\x^4-4*\x^2+3}) -- (1,0) -- (0,0) -- cycle;     \draw[pattern color=ttqqcc,pattern=crosshatch,fill opacity=0.1, smooth,samples=50,domain=1.0:1.7320508075688772]      plot(\x,{\x^4-4*\x^2+3}) -- (1.73,0) -- (1,0) -- cycle;      \begin{scope}[xscale=-1]     \draw[pattern color=ffqqtt,pattern=north east lines,fill opacity=0.1, smooth,samples=50,domain=0:1.0]        plot(\x,{\x^4-4*\x^2+3}) -- (1,0) -- (0,0) -- cycle;     \draw[pattern color=ttqqcc,pattern=crosshatch,fill opacity=0.1, smooth,samples=50,domain=1.0:1.7320508075688772]        plot(\x,{\x^4-4*\x^2+3}) -- (1.73,0) -- (1,0) -- cycle;     \end{scope}      \draw[smooth,samples=100,domain=-3.6342419080068153:3.665758091993186] plot(\x,{(\x)^4-4*(\x)^2+3});      \fill [color=uququq] (-1.73,0) circle (1.5pt);      \fill [color=uququq] (-1,0) circle (1.5pt);      \fill [color=uququq] (1,0) circle (1.5pt);      \fill [color=uququq] (1.73,0) circle (1.5pt);      \end{tikzpicture}     \end{document}   When I see the DVI output DVI, I have    And if I look at the PDF file, I have How can time dilation be symmetric? Suppose we have two twins travelling away from each other, each twin moving at some speed $v$:    Twin $A$ observes twin $B$’s time to be dilated so his clock runs faster than twin $B$’s clock. But twin $B$ observes twin $A$’s time to be dilated so his clock runs faster than twin $A$’s clock. Each twin thinks their clock is running faster. How can this be? Isn’t this a paradox? Does installing and using Wine open up your Linux platform to Windows viruses? Author information column wise I am using authblk package and I want to achieve following output using this package. Why is there such variation in PhD length internationally? I'm an undergrad in Australia, and at this stage I want to go on to do further study in maths (or maybe physics/CS, still have a few years till then).  Should I do so, I imagine that I will want to complete a PhD overseas. Possibly at a university in the US or maybe Oxbridge in the UK.  In Australia, PhDs typically take 3 or 4 years, and I would be eligible to go straight into a PhD upon completion of my undergraduate degree (it is research focused and includes an honours year, see ).  But, much of what I have found for overseas institutions has been in the range of 5-7 years, or maybe a 3-4 year PhD with a few years doing a master's degree beforehand.  Does this indicate that the level of rigour and amount of work produced is significantly different for various PhD programs, or is it the case that some programs are more relaxed and simply spread the same amount of work over a longer period of time? Using thematic layer for segmentation in eCognition? I have a thematic layer, i want to perform segmentation and classification only within the thematic layer,not in a whole image. how can i do it? I tried to segment the image after viewing the thematic layer in view setting and the thematic layer usage was enabled, but it was segmenting the whole image. I'm making an app with a custom list in multiple languages.  Is there an easy way so that I only have to create the site columns, content type and list once? So I don't have to create them for each language?  My app will be in English, French and Dutch.  My lists elements.xml looks like this  &lt;?xml version="1.0" encoding="utf-8"?&gt; &lt;Elements xmlns="http://schemas.microsoft.com/sharepoint/"&gt;     &lt;!-- Do not change the value of the Name attribute below. If it does not match the folder name of the List project item, an error will occur when the project is run. --&gt;     &lt;ListTemplate         Name="List1"         Type="10000"         BaseType="0"         OnQuickLaunch="TRUE"         SecurityBits="11"         Sequence="410"         DisplayName="List1"         Description="My List Definition"         Image="/_layouts/15/images/itgen.png"/&gt;   &lt;Field       Name="Adres"       ID="{2a32d9eb-ad3b-4dbb-abba-66dd01ca7c6e}"       DisplayName="$Resources:Adres"       Type="Text"       Required="TRUE" /&gt;   &lt;Field       Name="Naampje"       ID="{ac47be45-2f43-4d16-9a06-162c85945ec9}"       DisplayName="$Resources:Naampje"       Type="Text"       Required="FALSE" /&gt;   &lt;Field     Name="Nickname"     ID="{6B0A2CD7-A7F9-41ca-B932-F3BEBB603793}"     DisplayName="$Resources:Bijnaam"     Type="Text"     Required="FALSE" /&gt;  &lt;/Elements&gt;   Well if i change my language in Sharepoint (Office 365) it's not changing the titles from dutch to the other languages i have added. I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.   In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:  $$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$  Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$  This is a convergent geometric series as $r = \frac{2}{3}\lt1$  now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?  Thanks If votes works differently on meta sites how does that of accepting answers work? In the faq of meta sites you can read something like the following:     Voting here works a bit differently   from other Stack Exchange sites. On   Meta Stack Overflow, voting is used to   express agreement or disagreement, not   to point out a lack of quality or   helpfulness. Please don't be concerned   if you receive downvotes – members of   the community may simply disagree with   your bug, feature request, support   issue, or the nature of the   discussion.   My question is: If votes works differently on meta sites how does that of accepting answers work?  Take for instance a question on a meta site that is meant for the community to discuss some issue. What does it mean if the question is accepted? Accepting an answer is up to the author of the question so it cannot be a sign of what answer the community concludes from the discussion. It could mean that the discussion is over but that should be up to the community and not the author of the question. Is it so that accepting an answer for a discussion question on a meta site does not have a particular function? Classifying whether a collection of objects is a set Is the convex hull of closed set in $\mathbb R^{n}$ is closed? Is convex hull of closed set in $\mathbb R^{n}$ closed?