Firstly, I apologise if my thinking is completely incorrect - I am by no means a physicist (yet!), I have included layman illustrations to hopefully assist in explaining my thinking.  I am imagining the big bang to be, for lack of a better word, an explosion, thus the universe to be progressively expanding:    Therefore, to me it seems logical to suggest that there is an edge.  If this is the case, then is it reasonable to suggest that explosion + edge = 'bright edge'? Where light projects in all directions at the edge, not just outwards into the 'nothingness'.     So, this brings me to my question: why, when someone looks deeply enough 'at the edge' do they see an absence of light?   Again, apologies if I'm jumping to irrational conclusions &amp; am way off.
By looking at this picture:    The earth is near the center of the universe. I've read that the universe look the same no matter where the observer is located. It is the same distance everywhere.  So I understand that for general relativity the universe need to be homogeneous and isotropic, so it will look the same no matter where I am.  But what if I'm on one of the planets near the right or left of the image, then if I draw the same picture of the universe, but from my perspective, then I would be also located in the center? If that's not the case (I'm actually near an edge), then part of my sky would be completely dark, and all the sky that way won't be isotropic?
Yu Yuphoria bricked after installing custom recovery and boot file My YU YUPHORIA got bricked while I was working with it. I want to install custom ROM into it for that I am installing custom recovery and boot file but the file that is found on Internet was not working and that's why I am stuck with company logo. I can't go into recovery mod, neither it boot, and USB debugging mode is turned off. I tried to install fresh ROM from fastboot but after installing every driver properly command prompt is showing "waiting for the device". Please sugest some way to get rid of this problem. Best way to remove an event handler in jQuery? I have an input type="image". This acts like the cell notes in Microsoft Excel. If someone enters a number into the text box that this input-image is paired with, I setup an event handler for the input-image. Then when the user clicks the image, they get a little popup to add some notes to the data.  My problem is that when a user enters a zero into the text box, I need to disable the input-image's event handler. I have tried the following, but to no avail.  $('#myimage').click(function { return false; }); I found an asnwered question concerning this topic already here:  Copying \captionsetup[subfigure]{justification=centering} doesn't help me thought for I am using different packages.  I would like my Captions and Subcaptions to be justified as the text below. Just under my pictures. Any suggestions?  Thanks a lot  \documentclass{article} \usepackage{subfig} \usepackage[labelfont=bf, justification=justified]{caption} \usepackage[demo]{graphicx}% Remove demo option in real document  \begin{document}     \begin{figure} %   \captionsetup[subfigure]{justification=centering}     \centering       \subfloat[Subcaption one \newline including a linebreak ]   {     \includegraphics[width=0.2\textwidth]{image1}     }     \hspace{1cm}     \subfloat[Subcaption two \newline including a linebreak]     {     \includegraphics[width=0.2\textwidth]{image2}     }        \caption{Figure caption \newline with additional text right here} \end{figure}  \begin{center} This is correctly justified Text \\ including a longer linebreak than blabla. \end{center} \end{document} Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct? LaTeX Classes or Styles for Schedules and/or Calendars? Maybe TeX isn't the best for this particular problem, but any teachers out there use a particular class or package to plan lessons?  Right now I'm just using a spreadsheet, but I always find myself wishing there was a 'lessonplan.sty' or 'classschedule.sty' file out there somewhere so I could still use the LaTeX for equations, tables, lists etc.    A bit of searching yielded a few results, but there doesn't seem to be a consensus.  Anyone used any of these:        Any recommendations from anyone on best way to make calendar-based class schedules and/or lesson plans? Video editing software options? ASCII art generator Is there a way to set transparency of windows in GNOME? Sometimes ago I used KDE 3.5 I think, and really liked the transparency option for windows, which gave all my text editor a really killer look and feel, really nice/ethereal environment for coding.  I haven't managed to find such a possibility with GNOME.  Any idea if it is possible and how? custom button visibility list view There a  who has no valid parent user. It is indicated that his chat account is associated to a Maths.SE account, which does not appear to exist. The user was still able to chat. Is it normal? If it's a bug, could someone look into it?  Edit: if you browse the Google cache for users named Skullpatrol, it turns out a single series of associated accounts. The user is now named . I don't know what went wrong, maybe an account merge or something like that, but it needs someone from staff looking into it with some superpowers. I also tried to update the parent user of the chat account to this &quot;Rob&quot; profile, but the interface goes all buggy on me (works for other users, fails only for this one): How do I copy to the clipboard in JavaScript? I have the following recurrence that I've been pounding on:  $$ a(0)=1\\ a(1)=1\\ a(2)=2\\ a(2n)=a(n)+a(n+1)+n\ \ \ (\forall n&gt;1)\\ a(2n+1)=a(n)+a(n-1)+1\ \ \ (\forall n\ge 1) $$  I don't have much background in solving these things, so I've been googling around looking for different techniques. Unfortunately, I haven't been able to correctly apply them yet. So far I've tried:   Characteristic equations Generating functions Plugging it into Wolfram Alpha Telescoping Observation   I'm sure that one or more of these is the way to go, but my biggest problem right now is figuring out how to deal with the idea that there are different equations for odd and even values of $n$. I'm not necessarily looking for a solution (though it would be gratefully accepted), but any nudges in the right direction would be much appreciated.    To follow up, it turned out that the speed problem described in my comment, below was related to the Decimal implementation in Python 2.7. Switching to Python 3.3 (and Java) made everything many orders of magnitude better. "Being" or "to be"? Which is better structured?     "She loves to be herself instead of showing off"    or     "She loves being herself instead of showing off"   or     "She loves being and not appearing" Addition with NULL values How to download all files from Ubuntu One? gluten free hot cross buns Help with completing the proof: Any unbounded sequence is either infinitely large or has a convergent subsequential limit Given $\{x_n\}$ is an unbounded sequence, prove:   $$ \lim_{n\to\infty}x_n = \infty\ \  \text{or}\ \ \lim_{n_k \to \infty} = A $$   In other words any unbounded sequence either infinitely large or has a convergent subsequence.     Everything before udpate section is wrong.  Part 1. Proving $|x_n| &gt; M \implies \lim_{n\to\infty} x_n = \infty$  $\Box$ Start with the definition of an unbounded sequence: $$ \forall M &gt; 0 \in \Bbb R \ \exists N \in \Bbb N: \forall n &gt; N\implies |x_n| &gt; M $$  Suppose that: $$ \lim_{n\to\infty}x_n \ne \infty $$ That means: $$ \forall n\in\Bbb N\ \exists C &gt;0 \in\Bbb R : |x_n| &lt; C $$  Now taking $M^\prime = \max\{M, C\}$ we have that: $$ \begin{cases} \begin{align} \exists N \in \Bbb N : &amp;\forall n&gt;N \implies &amp;|x_n| &gt; M^\prime \\ &amp;\forall n \in \Bbb N: &amp;|x_n| &lt; M^\prime \end{align} \end{cases} $$  So we have arrived to a contradiction which means the assumption is wrong and: $$ \lim_{n\to\infty}x_n = \infty $$ $\Box$     Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.   Two questions is my mind:   Is part 1 correct?  How do I proceed with the second part?     I'm using the following definition. $$ \lim_{n\to\infty}x_n = +\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &gt; \epsilon \\ \lim_{n\to\infty}x_n = -\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &lt; -\epsilon \\ $$ A sequence is considered infinitely large when: $$ \lim_{n\to\infty}x_n = \infty\ \ \text{when}\ \ \lim_{n\to\infty}x_n = + \infty \ \ \text{or} \ \ \lim_{n\to\infty}x_n = - \infty $$    Update.   Looks like i've messed up a lot of things. I will try once again. As shown in comments and answers the statement holds only in case: $$ \lim_{n\to\infty} |x_n| = \infty \iff \forall \epsilon &gt; 0 \exists N \in \Bbb N: \forall n &gt; N \implies |x_n| &gt; \epsilon $$  Consider the following: $$ \lnot P = \lnot\left(\lim_{n\to\infty}|x_n| = \infty \right) \iff \exists \epsilon &gt; 0\ \forall N_1 \in \Bbb N : \exists n &gt; N_1 \land |x_n| &lt; \epsilon $$  But on the other hand it is given that $x_n$ is unbounded: $$ Q = \forall M &gt; 0\ \exists N_2 \in \Bbb N : |x_{N_2}| &gt; M $$  Construct a negative expression for boundedness: $$ \lnot P = \exists M &gt; 0\ \forall N_2 \in \Bbb N : |x_{N_2}| &lt; M $$  If $S = \lnot P \implies \lnot Q$ then $S = P \lor \lnot Q$  Let $\epsilon = M$, choose $N = \max\{N_1, N_2\}$ then both statements are true and: $$ \exists \epsilon &gt; 0\ \forall N = \max\{N_1, N_2\}: \exists n &gt; N \land |x_n| &lt; \epsilon $$  Now either $P$ is true, which would mean $\lim |x_n| = \infty$, or $\lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence. How should I deal with becoming discouraged as a graduate student? Probability expected value football problem The number of injuries at a college football game has the following probability distribution:      $$\begin{array}{|l:l|} \hline \textrm{Number of injuries} &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ \hdashline \textrm{Probability} &amp; 0.14 &amp; 0.33 &amp; 0.23 &amp; 0.17 &amp; 0.09 &amp; 0.04 \\ \hline \end{array}$$      The coach must file a report whenever there is more than one injury in any game.      a. Calculate the expected number of reports filed in the next 20 games.      b. Calculate the standard deviation of the number of reports filed in the next 20 games.      c. Calculate the expected number of football games to be played before the first game in which there are at least four injuries. I have a realy strange problem in Ubuntu 11.10. I have by accident my password set to empty. If I wan't to install software now the system asks for a password, but I haven't one anymore. If I enter my old password it sais it's wrong and if I leave it empty it also does. Now I wan't to set my password back, but the system asks for the old password before I can change it. That password doesn't exist, so I can't do anything anymore! Can someone help me please?