extract paragraph match with words using sed or awk
How to grep for text in a file and display the paragraph that has the text?
Here's a question I've been asked;  Given distinct primes $p,q,r$, show that any group $G$ of order $pqr$ is not simple.   So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only $$\dbinom{3}{2} + \dbinom{3}{1} + 1$$ of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of $n$ distinct primes.  Is there some other approach? Thanks for any insights Why is the Night's Watch so depleted at the start of ASOIAF? At the start of AGOT, the total number of the Night's Watch is said to be less than a thousand. During Aegon's conquest, the Watch numbered ten thousand men. Out of the nineteen castles on the wall, only three are manned today. Castle black itself is once said to have housed 5000.   This rapid deterioration doesn't make sense to me from a world-building perspective. The NW have continued to maintain themselves pretty well for nigh on 8000 years (if the tales can be believed that is, Sam Tarly finds no written evidence of Lord Commanders earlier than #673, which suggests that 8000 years may be an inflated 'legendary' number).   Whereas in the real world, one of the oldest and most venerable institutions, the Catholic Church, has been around for less than 2000 years and has been administered by ~200 popes. The Night's Watch has had a few rocky years, sure, but for it to maintain itself so well for so long and then deteriorate only in the last 300 years seems strange to me.   The NW has two purposes. The ostensible one of defending against the Others, and the (arguably much more real, at least at the start of AGOT) purpose of serving as a dumping ground for the 'losers' of Westeros. By 'losers', I mean the people on the losing side of various wars and political battles, as well as criminals and n-th sons of noble houses with nothing to inherit.  I am currently reading through Fire and Blood, and was struck by one chapter, where an armed revolt of members of the Faith were defeated by Maegor, and 2000 brothers were forced to take the black. Those sorts of numbers don't exist in the recent histories it seems.   One would expect that after Robert's Rebellion, there would be a surfeit of defeated Targaryen loyalists who were forced to take the black, but apart from a smattering of named characters (including Ser Alliser Thorne), there seem to be hardly any Night's watch recruits from recent wars.  Given the way the Night's watch is continually bolstered by the losers of wars and by criminals and other people who elsewise have nothing going for them, why, at the start of AGOT, is the force so depleted? Check if element found in array c++ I know that the Linux kernel supports TRIM as of version 2.6.33, so there should be TRIM support in ubuntu.  Is TRIM enabled by default or do I need to change some settings or install additional software for it to work? If so, how? How do I show a function on 2-adic units is continuous (using the 2-adic metric)?   I'd be happy to learn the general rule or definition. But in particular I need to show that $f(x)=\dfrac{3x+1}{2^{v_2(3x+1)}}$ is continuous at all odd numbers.  Since the function is isometric in $\lvert\cdot\rvert_2$, i.e. since $\lvert f(x)\rvert_2=\lvert x\rvert_2$ every orbit of the function is duplicated, multiplied by any power of $2$.  This can therefore be formulated in various ways; as a function though the odd numbers, through the dyadic rationals, but the following seems good to work with.  If we let $n$ be a positive integer and $k$ be the power of $2$ we can define:  $f(2^k(2n+1))=(3(2n+1)+1)\cdot2^k\cdot\lvert3(2n+1)+1\rvert_2$  So $f(2^k(2n+1)=(3n+2)\times2^k\times\lvert3n+2\rvert_2$  Here is what I have so far:  I think to prove continuity in the 2-adic metric I need $\lvert x_n-x\rvert_2=0\implies\lvert f(x_n)-f(x)\rvert_2=0$  I think $k$ and $n$ are independent so I think I can examine them independently.  Taking $k$ to infinity brings both $x$ and $f(x)$ to zero so that seems to satisfies the continuity requirement.  Moving on to $n$; this seems to be an exercise in proving convergence within odd integers which are in a sense a subset of the 2-adic units.  I know all Cauchy sequences in these converge to 2-adic units but not much more than that.  However I do have a little insight into this particular function.  For example if we examine the inputs $x$ which map to any given output of $f(x)$, it can fairly easily be shown that these $x$'s take the form of a set $\left\{4^mp+\dfrac{4^m-1}{3}:m,n\in\mathbb{N}\right\}$ so, at least for any given output $f(x)$ the inputs always converge to $x=\frac{-1}{3}$ as the intervals between them become large powers of $2$.  In fact the orbit of the function $a(2^kx)=2^k(4x+1)$ on variation of $x$ and holding $k$ fixed is in a sense orthogonal to $f(x)$; which is a restatement of the above except not in closed form. I'm flying to Tokyo. The flight was sold through Finnair. The first leg is operated by American Airlines. The second and final legs are operated by Finnair.  When I checked in online, it said that some airports on my journey (didn't say which one) don't allow mobile boarding passes?? Meaning I will need a physical copy.  Unfortunately I don't have easy access to a printer. Will I still be able to print all my boarding passes at the airport even though I already checked in online? Unusual placing of a second impression of a USA visa immigration stamp. Does it mean anything? My parents traveled to the USA a couple of weeks ago (Caracas - Miami). It wasn't their first time; they've probably come and gone a dozen times by now, and they've never had a single issue. This time, the immigration guy was kind of pompous and, once they told him they weren't staying at a hotel but with some friends, he began looking at them somewhat funny. Finally, he stamped their passports (normally, with entry date and a 6-month term of stay) but he put an additional stamp, one that no one they asked had seen before:    It seems to be the same usual entry stamp, but cut in half. What does it mean? Should we be concerned? "pros and cons", "black and white", "ups and downs". Always in a fixed sequence, is there a word or phrase for these? Is there a word or phrase for two nouns or adjectives joined by a conjunction (usually "and") in a fixed sequence?        alive and well           fast and furious   hat and gloves   pen and pencil   law and order   wit and wisdom   salt and pepper   bacon and eggs   horse and carriage   Adam and Eve   ladies and gentlemen   loud and clear   heads or tails   yes or no   dead or alive What kind of rhetoric is (this particular) "No one ever [...]" The President also knows that we have to stop blaming victims for these crimes.  No one ever asks the person who got robbed at gunpoint in the street -- why were you there, what were you doing, what were you wearing?  What did you say?  Did you offend someone?  We encourage people to come forward.  We don't have to explain why someone took our money.     I think the "No one ever [...]" claim is literally false, and actually obviously so. So, I understand it to mean:          The President also knows that we have to stop blaming victims for these crimes.  No one in their right mind ever asks the person who got robbed at gunpoint in the street -- why were you there, what were you doing, what were you wearing?  What did you say?  Did you offend someone?  We encourage people to come forward.  We don't have to explain why someone took our money.      Supposing that I am right, what is this kind of rhetoric called? I've used following codes and no luck  1  //Hide admin footer from admin function change_footer_admin () {     return ' '; } add_filter('admin_footer_text', 'change_footer_admin', 9999);  function change_footer_version() {     return ' '; } add_filter( 'update_footer', 'change_footer_version', 9999);   2  function wpbeginner_remove_version() { return ''; } add_filter('the_generator', 'wpbeginner_remove_version');   3  function my_footer_shh() {     remove_filter( 'update_footer', 'core_update_footer' );  } add_action( 'admin_menu', 'my_footer_shh' );   Still I see those texts and version on my footer.   So what is the accurate way to remove these? I've added these code snippets into my child theme function.php How to develop a SharePoint Designer Workflow Action for use in SharePoint (2013) online Calculating $\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$ Find the value of $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$   The first idea I had was to apply Cauchy's integral formula. But can this be done? $e^{1/z^2}$ is not holomorphic on the interior of $|z|=2$. Still, if we would apply this we would obtain:  $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz=-2\pi ie$.  Another way I think, is to apply the residue theorem. For $1-z$ has a simple zero at $z=1$ and $e^{1/1^2}\neq 0$, then the residue of $e^{1/z^2}/(1-z)$ at $z=0$ is $-e$. Again we obtain $2\pi iRes(g;0)=-2\pi ie$, where $g=\frac{e^{1/z^2}}{1-z}$.  But again, it seems to me this is not right because $g$ has two singularities, at $z=1$ and $z=0$. What we might do about $z=0$? The answer is not $-2\pi i e$? Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even ) I'm trying to prove the following Integer Partition claim :   $p$(n│even number of ODD parts) = $p$(n│distinct parts ,number of ODD parts is even) .  So I tried to prove a stronger claim :   $p$(n│number of odd parts) = $p$(n│number of distinct parts) :  Using generating functions :   DISTINCT PARTS =  $$(1+x)\cdot(1+x^2 )\cdot(1+x^3 )\cdot(1+x^4 )\cdot\cdot\cdot\cdot$$  = $$\frac {1-x^2} {1-x} \cdot \frac {1-x^4} {1-x^2} \cdot \frac {1-x^6} {1-x^3} \cdot \frac {1-x^8} {1-x^4}\cdot\cdot\cdot\cdot$$  $$ \frac {1} {1-x} \cdot \frac {1} {1-x^3} \cdot \frac {1} {1-x^5} \cdot \frac {1} {1-x^7} \cdot\cdot\cdot\cdot $$  = ODD PARTS  But I'm not sure regarding the proof , does it really prove that even number of odd parts equals distinct parts ,where number of ODD parts is even ?   Thanks I have a set of 10 tables and each has an extra column that I don't want. Specifically, these tables are descriptions of database tables exported by phpMyAdmin, and they all have a "MIME" column that I would like to remove. Is there a way to do this other than using string manipulation / find and replace?   I have tried setting the column to p{0in}, but the column title still remains. probability of getting 50 heads from tossing a coin 100 times How can I vertically align elements in a div? I have a div with two images and an h1. All of them need to be vertically aligned within the div, next to each other. One of the images needs to be absolute positioned within the div. What is the CSS needed for this to work on all common browsers? &lt;div id=&quot;header&quot;&gt;   &lt;img src=&quot;..&quot; &gt;&lt;/img&gt;   &lt;h1&gt;testing...&lt;/h1&gt;   &lt;img src=&quot;...&quot;&gt;&lt;/img&gt; &lt;/div&gt; If $n$ vectors are linearly independent, is their span $\mathbb{R}^n$? Have $n$ vectors in $\mathbb{R}^n$.  If the $n$ vectors are linearly independent, can we conclude that their span is $\mathbb{R}^n$? jQuery/JavaScript: accessing contents of an iframe I would like to manipulate the HTML inside an iframe using jQuery.  I thought I'd be able to do this by setting the context of the jQuery function to be the document of the iframe, something like:  $(function(){ //document ready     $('some selector', frames['nameOfMyIframe'].document).doStuff() });   However this doesn't seem to work. A bit of inspection shows me that the variables in frames['nameOfMyIframe'] are undefined unless I wait a while for the iframe to load. However, when the iframe loads the variables are not accessible (I get permission denied-type errors).  Does anyone know of a work-around to this? There are a lot of legendaries available in ORAS, however a lot of them will only show up/be available if certain prerequisites are met.  I know Groudon/Kyogre are available through the Story, and      Rayquaza &amp; Deoxys   through the postgame Delta episode, but what about the other available legendaries? What do I need to do to catch them? How do I run specific sudo commands without a password?