Can I use the names or logos of real football teams in my game?
Legal issues around using real players names and team emblems in an open source game
How can I create launchers on my desktop? How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it? Bernhard Elsner, alias MathOMan, posted this exercise in plane Geometry,  on January $29$th, $2010.$  "Let $\mathcal{C}$ be a circle, $A,B$ two distinct points on $\mathcal{C}$ and $M$ be the midpoint of the chord $[AB]$. Take two other chords,$[PQ]$ and $[SR]$, that pass through $M$. Let $C$ (resp. $D$) be the intersection of $[AB]$ with $[PS]$ (resp. $[RQ]$). Prove that $M$ is the midpoint of the chord $[CD]$."    To prove it I've written the following (failed) argument, in the  of this post (translation of mine):    The figure is symmetric with respect to $M$: $\overline{AM}=\overline{MB}$, $\overline{PM}=\overline{MU}$, $\overline{RM}=\overline{MW}$, $\overline{QM}=\overline{MV}$, $\overline{QR}=\overline{VW}$, $\overline{SM}=\overline{MT}$. From $\dfrac{\overline{SC}}{\overline{DT}}=\dfrac{\overline{CM}}{\overline{MD}}=1$ follows that $\overline{CM}=\overline{MD}$.  Here is an extract of the author's reply (translation of mine):   "It is not clear that  $\overline{QM}=\overline{MV}$. Is the point $V$ on the line $(QM)$ defined by this equality or is $V$ defined as the intersection point of the lines $(QM)$ and $(WC)$? Why are both definitions to give the same point?  Let $C^{\prime }$ be the intersection of $(SP)$ and $(VW)$, and $D^{\prime }$ the intersection of $(TU)$ and $(QR)$. (...)  One still has to show that $C^{\prime }=C$ and $D^{\prime }=D$."   I have agreed with these objections.  Until now no proof has been posted. The author considers that the "proof is not quite simple".  Q. What is the theorem this exercise refers to? Or how does one prove it? Intuition behind measurable random variables and $\sigma$-algebra I've been trying to understand $\sigma$-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture.  I'll try to explain the counter-intuition I get with the classical example of : the probability space $\Omega = \{ HH, HT, TH, TT \}$ and a r.v. $X(\omega)$ equal to the number of heads.  At times $0$, $1$ and $2$ the available information is represented using $\sigma$-algebras $\mathcal{F}_0=\{\emptyset,\Omega\}$, $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}$ and $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}$.  One can notice that $X(\omega)$ is not measurable with respect to $\mathcal{F}_0$ and $\mathcal{F}_1$, because $X^{-1}((\frac{3}{2}; +\infty))=\{HH\}$. To me it is quite surprising: intuitively $X$ makes perfect sense at all times. In particular it has an expected value at time $0$, which I interpret as that the probability and value of all outcomes $\{\omega\}$ can be computed. How do you think of a non-measurable function?  Here's another way of expressing the same confusion. The most natural choice of $\sigma$-algebra in a finite discrete case is $\mathcal{F}=2^\Omega$, and it is implicitly used in all elementary probability problems. However, this choice of $\mathcal{F}$ does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "$\sigma$-algebra is known information" make sense only in conditioning? Why is it convenient then? What's the incentive to flag duplicates? What is a "locked" post? A post can be locked.   What does it mean to lock a post? Who can lock a post? When should a post be locked? How can one have a mini table of contents for the current chapter appear just after the chapter name listing all sections of that chapter? Creating mask file from shapefile using ArcGIS for Desktop? I have a shapefile of Lakes over my study area. Each polygon has attribute information of each lake, name, shape etc. Actually I want to eliminate all lakes from analysis. In other word, I want lakes to have 0 value and the area that out of boundary of lake to have their orginal pixel values. when I do subset using this shapefile, the output is not I want because my shapefile deoesn't have any value outer boundary of lake.   How can I build a proper mask file (inner part of lake has 0 value, outer is 1) using this shapefile in ArcGIS Desktop? When is R squared negative? My understanding is that $R^2$ cannot be negative as it is the square of R. However I ran a simple linear regression in SPSS with a single independent variable and a dependent variable. My SPSS output give me a negative value for $R^2$. If I was to calculate this by hand from R then $R^2$ would be  positive. What has SPSS done to calculate this as negative?   R=-.395 R squared =-.156 B (un-standardized)=-1261.611   Code I've used:  DATASET ACTIVATE DataSet1.  REGRESSION /MISSING LISTWISE /STATISTICS COEFF OUTS R ANOVA             /CRITERIA=PIN(.05) POUT(.10) /NOORIGIN             /DEPENDENT valueP /METHOD=ENTER ageP   I get a negative value. Can anyone explain what this means? NFS: mount.nfs: Protocol not supported How to make a standalone document with one equation? I was surprised to find that standalone package can not compile a document with an equation:  \documentclass{standalone} \begin{document} \begin{equation} % not working! F(V, T) = E(V) + D(T) \end{equation} \end{document}   Is there an option or some workaround that allows this? The current error is something like:   ! Missing $ inserted.   in line 4. (The idea of course is that the including (main) document can just \input the standalone without knowing if is going to be an equation or something else) If the Universe is expanding, how is that galaxies run into each other? As the title says. If everything is moving outward, thrown by the Big Bang, how could two Galaxies end up with velocity vectors that run them into each other? evaluate the limit as $x$ approaches $0$ of $(\cos x-1)/x$ without l'Hopital's rule I know with l'Hopital's rule it becomes $-\sin(x)$ which has the limit $0$. However, I have been wondering how to evaluate this limit without l'Hopital's rule. Multiple object instances represented by two slightly shifted rectangles In a software diagram, I'd like to be able to represent multiple instances of an object by two rectangles with one of them slightly shifted. I recently learned about the pic style so I'm using it with two nodes to achieve what I want.  The following example does work, but I'd like to know if I can do the same without using the fill=white. My hack kind of assumes the entire picture background is white and that may not always hold. I'd like to be able to crop the unwanted picture part out.  \documentclass[crop,border=3pt]{standalone}  \usepackage{pgf,tikz} \usetikzlibrary{positioning}  \begin{document} \begin{tikzpicture}[     every node/.style={font=\sffamily\small},     block/.style={         rectangle, draw, align=center, minimum width=2cm,         minimum height=1.4cm, font=\sffamily\small     },     multinode/.pic={         \node [block] at (0.1,0.1) {};         \node [block, \tikzpictextoptions, fill=white] at (0,0) {\tikzpictext};% &lt;-- FILL WHITE     } ]     \node [block] (STG) at (2,5) {single\\instance};     \pic [pic text={multi\\instance}] at (5,5) {multinode}; \end{tikzpicture} \end{document}   The following images show first one version with the fill white hack, and then one version without it. Proving that $x_n\to L$ implies $|x_n|\to |L|$, and what about the converse? Problem 3. Show that for a sequence $(x_n)$ the following are true:   (i) $\lim x_n=0$ if and only if $\lim |x_n|=0$.   (ii) $\lim x_n=L$ implies $\lim |x_n|=|L|$. Is the converse true? Prove or give a counterexample.    (i) is already done, easy.  I'm halfway done with (ii), I split it into three cases:  $L = 0$, $L &gt; 0$ and $L &lt; 0$.  For $L = 0$ I just refer to part (i).  For $L &gt; 0$, we know that $|x_n - L| &lt; \epsilon$   if lim $|x_n|$ = $|L|$, then we must have  $||x_n| -|L|| &lt; \epsilon$  but by the reverse triangle equality, $||x_n| - |L|| &lt; |x_n - L|$ so clearly $||x_n|-|L|| &lt; \epsilon$, thus  lim $x_n = L$ $\implies$ lim $|x_n| = |L|$ for $L &gt; 0$  for some reason I'm confused as to part three, $L&lt;0$. I'm having trouble seeing the difference between $L&gt;0$ and $L&lt;0$, and i'm starting to think that splitting it up like that isn't necessary at all. MathJax isn't currently working on the Android app, which makes it totally useless for the 'technical' sites like  or .  For example, in the Android app:    Same post in a (mobile) browser:     Please add this functionality! What is the pronunciation of the possessive words that already end in s? Possible Duplicate:           My name is Greg — this is Greg 's post.  If my name ended with an 's', I am aware of the proper apostrophe usage (James&nbsp;→ James'), but how should this be pronounced?  Phonetically, am I Jameses best friend or James best friend? What part of speech is handsome in the sentence:     The bridegroom looked so handsome. If I'm floating in space and I turn on a flashlight, will I accelerate? Photons have no mass but they can push things, as evidenced by .   Can photons push the source which is emitting them? If yes, will a more intense flashlight accelerate me more? Does the wavelength of the light matter? Is this practical for space propulsion? Doesn't it defy the law of momentum conservation?  Note: As John Rennie mentioned, all in all the wavelength doesn't matter, but for a more accurate answer regarding that, see the comments in  .  Related Wikipedia articles: , Definite article, repetition in lists Which would be correct?        The fridge, the shelves, the walls, the floor, and John himself, were all covered in foam.   The fridge, shelves, walls, floor, and John himself, were all covered in foam.      The the can be applied to all the nouns, so is there a need to repeat it? Where can I send feature requests? Both for Ubuntu as a whole and individual (supported) apps?