Can't vertical center h1 inside div
How can I vertically center a "div" element for all browsers using CSS?
I use youtube-dl to download videos for watching later offline. Usually, it works fine. However, due to frequent changes on the video sites, it sometimes requires a quick update. It wasn't a problem before on 13.04 but after upgrading to 13.10, I get  # youtube-dl -U It looks like you installed youtube-dl with a package manager, pip, setup.py or a tarball. Please use that to update.   I tried updating it via apt-get but there is no new package. So the app's update mechanism has been disabled and the maintainers aren't keeping up, what is the solution? How do I update it?  And why was it changed at all? It was updating a package manager version just fine in raring.   I'm already running into compatibility problems.   $ youtube-dl "http://vimeo.com/55784152" [vimeo] 55784152: Downloading webpage [vimeo] 55784152: Extracting information ERROR: Unable to extract info section; please report this issue on https://yt-dl.org/bug . Be sure to call youtube-dl with the --verbose flag and include its complete output. Make sure you are using the latest version; type  youtube-dl -U  to update.   QUICK UPDATE:  As requested,  # type youtube-dl youtube-dl is hashed (/usr/bin/youtube-dl)  # apt-cache policy youtube-dl youtube-dl:   Installed: 2013.08.29-1   Candidate: 2013.08.29-1   Version table:  *** 2013.08.29-1 0         500 http://pl.archive.ubuntu.com/ubuntu/ saucy/universe amd64 Packages         100 /var/lib/dpkg/status How can I remove myself from India's visa blacklist? Cannot redeem bounty I have about 25K of bounty vouchers but I cannot seem to redeem them. I believe I am sitting in the proper station and it has a office to redeem claimed bounties.  Bounty: 25,380 CR BOUNTY IN INDEPENDENT ERAVATE FREE  STATION: CLEVE HUB / INDEPENDENT ERAVATE NETWORK   If I am in the wrong station how do I find the information needed to travel to the correct station. Where can i see the faction running the station prior to flying there? continuity, discontinuity derivative and relation to being derivative but its partials are not continuous is $$f(x) =\begin{cases} x^2\sin(\frac{1}{x}) \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$ a continuous function specially at point x=0?  And why being derivable its derivative is not continuous at x=0 since  $$f`(x) =\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$?  I suspect this as to do with being derivable a function f $\mathbb{R}^n\rightarrow\mathbb{R}^m$ that don't imply its partial derivatives are continuous .this I don't get it so far.however partials derivatives if continuous is derivable.  Can you provide me a more general case than just one variable where f is derivable but its partials are not continuous? Does a spellcaster know when concentration ends? Does a spellcaster know when concentration ends on one of their spells?  For example, when a wizard casts invisibility, that target is invisible until they attack or after one hour. So if the character attacks and the spell ends is the wizard aware that their spell that they are concentrating on ended earlier than expected? Make a momentary switch control a toggle What are the simplest, cheapest, smallest ways to make a momentary switch produce a 2-state toggling output (latching momentary switch)?  In other words, the output is continuously low, and when you momentarily press the button/tact switch, the output changes to continuously high, and then when you press it again, it switches back to low. Installing Crystal Reports 8.5 on Windows 7 I have successfully installed Crystal Reports 8.5 on my Windows 7 desktop, but when I try to launch the application i get this:    Any experience with this?  Besides the obvious "update to as newer version", does anyone have any suggestions? Computing $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x$ with residue calculus I know there are many ways to prevent image caching (such as via META tags), as well as a few nice tricks to ensure that the current version of an image is shown with every page load (such as image.jpg?x=timestamp), but is there any way to actually clear or replace an image in the browsers cache so that neither of the methods above are necessary?  As an example, lets say there are 100 images on a page and that these images are named "01.jpg", "02.jpg", "03.jpg", etc. If image "42.jpg" is replaced, is there any way to replace it in the cache so that "42.jpg" will automatically display the new image on successive page loads? I can't use the META tag method, because I need everuthing that ISN"T replaced to remain cached, and I can't use the timestamp method, because I don't want ALL of the images to be reloaded every time the page loads.  I've racked my brain and scoured the Internet for a way to do this (preferrably via javascript), but no luck. Any suggestions? How to remove top and bottom whitespace of longtable? I am using longtable instead of tabular environment and noticed that it puts some additional whitespace at the top and at the bottom of the table. Is there some way to remove it (something less hacky than \vskip -\baselineskip before the table)?  Small piece of code for illustration:  \documentclass[12pt]{book} \usepackage{longtable}  \begin{document}  Linelineline  \begin{longtable}{l l} cell1 &amp; cell2 \end{longtable}  Linelineline  \begin{tabular}{l l} cell1 &amp; cell2 \end{tabular}  Linelineline  \end{document}   And that's what I am getting:    Common tabular environment does not have any additional margins, but longtable does. Idiom for explaining something to someone who clearly knows the topic (probably better than you) I have to prove by induction that $$\sum_{r=0}^{n}\begin{pmatrix}n \\ r \end{pmatrix}=2^k$$, using the identity $\begin{pmatrix}n \\ r-1 \end{pmatrix}+\begin{pmatrix}n \\ r \end{pmatrix}=\begin{pmatrix}n+1 \\ r \end{pmatrix}$. I'm fine with everything except the inductive step, where I only managed to get as far as $$\sum_{r=0}^{k+1}\begin{pmatrix}n \\ r \end{pmatrix}= \sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}+\begin{pmatrix}n \\ k+1 \end{pmatrix}$$ I tried to then use the identity and substitute $\begin{pmatrix}n \\ r \end{pmatrix}=\begin{pmatrix}n+1 \\ r \end{pmatrix}-\begin{pmatrix}n \\ r-1 \end{pmatrix}$ but all this did was just take me around in a circle. I also couldn't substitute $2^k$ for $\sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}$ since this didn't properly simplify to give $2^{k+1}$. However, I think that I have to obtain the expression $\sum_{r=0}^{k}\begin{pmatrix}n \\ r \end{pmatrix}$ since it simplifies to $2^{k+1}$. But I've not been able to obtain this in anyway so I'm wondering if anyone can guide in what I need to or what direction I need to take to complete the inductive step. Can someone please give me some intuition as to when to choose either SVM or LR? I want to understand the intuition behind what is the difference between the optimization criteria of learning the hyperplane of the two, where the respective aims are as follows:     SVM: Try to maximize the margin between the closest support vectors LR: Maximize the posterior class probability    Let's consider the linear feature space for both SVM and LR.  Some differences I know of already:   SVM is deterministic (but we can use Platts model for probability score) while LR is probabilistic. For the kernel space, SVM is faster (stores just support vectors) Actually snapping to pixel grid Does anyone know if there's a way to have Illustrator CC (2014) constrain control points to the pixel grid? I understand that "Align to pixel grid" makes it so that a shape will try to match the pixel grid, but the actual control points don't line up to said grid... The image I've attached illustrates my problem. How to prove that there exists a prime number between $n$ and $ n!$, for all $ n&gt; 2$?    ( gives a much better bound, but this question is about obtaining a self-contained proof.) Currently the migration path options from SO are:   Meta.SO Serverfault SuperUser Webmasters Programmers   If we are going to be limited to 5 options, shouldn't we periodically re-evaluate which sites are most appropriate?   (10k only) indicate that the top sites for the last 90 days are:   Superuser - 564 - 5% rejected ServerFault - 278 - 10% rejected Tex - 157 - 13% rejected DBA - 144 - 2% rejected Sharepoint - 127 - 1% rejected Programmers - 126 - 26% rejected CodeReview - 100 - 3% rejected Meta.SO - 81 - 18% rejected Wordpress - 75 - 14% rejected Webmasters - 58 - 8% rejected   All these migrations for sites out of the list required a flag and a moderator to process it.  This is just extra work for the mod team!  If we had the top 5 destinations by volume as the 5 options, we would have had a net savings of 160 flags to be processed.  I suspect we will actually see these numbers change quite a bit if it's easier for users to put questions in the appropriate spot without moderator intervention.  Can we have a review of the migration path list based on migration volume?  Think of the flags! I have installed GCC compiler by installing the build-essential package.  After the installation I wrote a simple C program. I tried to run it with the following command:  gcc First.c ./a.out   but I'm getting a bash: ./a.out: Permission denied message. I don't know what to do now. I've made a simple snowfall particle simulation (using gandalf3's solution: ). Is there a way to loop the animation, so the transition between last frame and first frame is not visible? Is there a setting to do it? Modified pattern does not see the pattern color option My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me.     Let $m$ and $n$ be nonnegative integers with $n\geq m$. There are $m+n$ people in line to get into a theater for which admission if $50$ cents. Of the $m+n$ people, $n$ have a $50$-cent piece and $m$ have a $\$ 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is   $$ \frac{n-m+1}{n+1}\binom{m+n}{m}. $$     I first noted that the first person to enter must be one of the $n$ with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the $n$ people with a half-dollar enters, the number of half-dollars in the register increases by $1$, and when one of the $m$ people with a bill enters, the number of half-dollars decreases by $1$ but the number of bills increases by $1$.   I tried to model this by looking at paths in $\mathbb{Z}^2$. The $x$-axis is like the number of half-dollars, and the $y$-axis is the number of bills. You start at $(0,0)$, and you can take steps forward $(1,0)$ or backwards diagonally $(-1,1)$ corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make $m+n$ moves, and I figured maybe the number of such paths is counted by $\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you.