vertical centering div without defined height and content
How can I vertically center a "div" element for all browsers using CSS?
In how many inequivalent ways can 8 people be seated at a round table? How many different arrangements are there of eight people seated at a round table, where two arrangements are considered the same if one can be obtained from the other by a rotation? Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$ $$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$   I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$  enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$  But I can find how to continue from here How do you make clothes move along with model how do you make clothes move along with your model? Do you rig the model first and then parent it to the body of the character then click cloth simulation? I am a little bit confused with the process. Thanks for the help. Quote unquote (or end-quote), unseparated by the actual quotation In spoken English, people often say "quote-unquote" (or "quote-endquote") to indicate that part of what they are saying is a quotation (scare or otherwise). Sometimes the quoted material will go between quote and unquote, and occasionaly it will precede the quote-unquote. Often, though, "quote unquote" introduces the quoted material.  Why does quote unquote precede the quotation?  It would seem that quote opens the quoted text and that unquote closes it, but then why would we close the quotation before actually providing it. It would be the equivalent of writing your quotation marks together, "", rather than around the quoted text.  What are the origins of this construction?  Three years ago ChrisR asked  on whether it is correct to use quote, unquote. The answers did not deal with origins, but one commenter suggested that quote-unquote is a corruption of quote-on-quote, but he provided no evidence, nor any real explanation. How can I work with customize completion? I want to add some new auto-completion but I don't know How the wizard works. It's here: User>Customize Completion I don't know what should be writen in the box. I want to learn to use biper and biblatex in Lyx. This is how I would do it in LaTeX traditionally with biblatex and biber  \documentclass[12pt,a4paper,oneside,pdftex]{report} \usepackage[backend=biber]{biblatex} \addbibresource{minCuts.bib} ... \printbibliography \end{document}   The error is apparently originating from the \bibliographystyle command and biblatex package getting mixed, by . But I have no \bibliographystyle command in Preamble so they should not be mixed unless Lyx is somehow importing the \bibliographystyle.   How to resolve the biblioraphystyle error in Lyx with Biber and Biblatex?    Settings by the instructions  and Error        Document > Settings > Bibliography > Biber                  In Document > Settings > LaTeX Preamble                   Insert > List / TOC > BibTeX Bibliography...                        Package biblatex Error : '\bibliographystyle' invalid. Confused about complex numbers I am confused about something: \begin{eqnarray}     (e^{2 i \pi})^{0.5} = (e^{2 i \pi \cdot 0.5})= e^{i \pi}=-1 \end{eqnarray} but \begin{eqnarray}      e^{2 i \pi}=1~ and~ 1^{0.5}=1  \end{eqnarray} where is my mistake??  thanks Apt-get stuck at 0% [Working] A proof by strong induction that $a_n\le3^n$ where $a_n=a_{n-1}+a_{n-2}+a_{n-3}$ I am not sure whether this is right. Can anyone verify, whether this proof is valid? Thanks!        Define a sequence $\{a_n\}_{n\ge0}$ as follows:   $$a_0=1,\qquad,a_1=3,\qquad,a_2=9,\qquad,a_n=a_{n-1}+a_{n-2}+a_{n-3}\text{ for }n\ge3.$$   Prove that for any positive integer $n$, $a_n\le3^n$.        Proof. Let $P(n)$: $a_n\le 3^n$, $n$ is a non-negative integer.      (i) Base case:         Consider when $n=0$. LHS$=a_0=1$, RHS=$3^0=1$ $\therefore$ LHS$\le$RHS, then $P(0)$ holds.   Consider when $n=1$. LHS$=a_1=3$, RHS=$3^1=3$ $\therefore$ LHS$\le$RHS, then $P(1)$ holds.   Consider when $n=1$. LHS$=a_2=9$, RHS=$3^2=9$ $\therefore$ LHS$\le$RHS, then $P(2)$ holds.         (ii) Inductive case:   Assume $P(i)$ is true for $0 \le i \le k$, $k\ge2$.      (iii) Inductive conclusion: Consider $n=k+1$.   $$ \begin{align*} \mathrm{LHS}=a_{k+1}&amp;=a_k+a_{k-1}+a_{k-2} \qquad\text{(by definition, since $k\ge 2$)}\\                     &amp;\le 3^k+3^{k-1}+3^{k-2} \qquad\text{(by Induction Hypothesis)}\\                     &amp;=3^k(3^{-1}+3^{-2}+3^{-3})\\                     &amp;=3^k\left(\frac13+\frac19+\frac1{27}\right)\\                     &amp;=\frac{13}{27} 3^k\\                     &amp;\le 3^{k+1} = \mathrm{RHS} \end{align*} $$   Therefore, by the Principle of Strong Induction, $P(n)$ is true for all non-negative positive integers $n$. Is there word for this, other than know-it-all? Just wondering, if there's a word for:  One who makes it appear as if they're an expert on a subject while there are also signals (subtly or blatantly noticeable) that this person doesn't know very much on the subject or at least as much as they're acting like they do on the subject; pretending to have much knowledge on a subject matter.  I found pedantic but I don't suppose it's the right word. There are many cases in maths where doing the same thing over and over again will bring you closer to a value. Derivation of polynomials is the striking example: no matter what you start with, you end up with 0. In a series expansion, the longer you continue the sequence the closer you get to the real value.  This concept applies in areas other than math (Wikipedia's everything-leads-to-philosophy phenomenon, for example). It's a little unrelated but it just proves the concept.  Hashing algorithms are just special math. Following logically, hash entropy supposedly decreases the more you hash a value. So is there a point at which any input value will give the same hash as any other?  (Let me just specify that we're not re-injecting any new data every time we hash, as if we're trying to increase the security of the hash by salting and re-hashing. It's strictly a matter of md5(md5(md5(... if that's the algorithm we're using.  I can see some problems with this already. Algorithms like md5 are designed to change their output radically when the input changes only by one bit of information. So even if a common hash for two values appears along the way, it itself will start changing. But then, does the fact that the output appears to be very different actually mean the possibilities are still many? Possible Duplicate:          The  describes the expansion of the $n$th power of the sum $(a+b)$:  $$(a+b)^n = \sum_{k = 0}^n {n\choose k}a^kb^{n-k}$$  In calculus, there is a generalization of the product rule called , which describes the expansion of the $n$th derivative of the product of two functions $f$ and $g$:  $$(fg)^{(n)} = \sum_{k = 0}^n {n\choose k}f^{(k)}g^{(n-k)}$$  It's not difficult to prove using the product rule and induction, what I'm really trying understand here is why the formula is true. Is there a way to prove it using the binomial formula? How are these two formulas related? Is there a combinatorial proof for Leibniz's rule like there is for the ?   I understand this question is a bit vague, but I hope you'll agree that these two formulas look so similar it just feels like there should be some relation between the two. Of course this is not necessarily true, but I'm hoping it's not just a crazy coincidence. Extracting a regex matched with 'sed' without printing the surrounding characters To all the 'sed' doctors out there:  How can you get 'sed' to extract a regular expression it has matched in a  line?   In other words words, I want just the string corresponding to the regular  expression with all the non-matching characters from the containing line stripped away.  I tried using the back-reference feature like below  regular expression to be isolated           gets `inserted`                here                     |                v    sed -n 's/.*\( \).*/\1/p    this works for some expressions like   sed -n 's/.*\(CONFIG_[a-zA-Z0-9_]*\).*/\1/p    which neatly extracts all macro names starting with 'CONFIG_ ....' ( found in some '*.h' file ) and prints them all out line by line             CONFIG_AT91_GPIO           CONFIG_DRIVER_AT91EMAC                    .                    .              CONFIG_USB_ATMEL           CONFIG_USB_OHCI_NEW                    .                  e.t.c.    BUT the above breaks down for something like    sed -n 's/.*\([0-9][0-9]*\).*/\1/p    this always returns single digits like                    7                  9                  .                  .                    6   rather than extracting a contiguous number field such as.                 8908078               89670890                    .                  .                    .                23019                     .                e.t.c.     P.S.: I would be grateful to feedback on how this is achieved in 'sed'.        I know how to do this with 'grep' and 'awk'       I would like to find out if my - albeit limited - understanding of        'sed' has holes in it and if there is way to do this in 'sed' which I       have simply overlooked. Boy goes insane during space travel I'm trying to remember the name of a short story I read about 25 years ago or so. I'd also like to know of any collections it appeared in.   The details I recall involve a family who are embarking on a hyperspace journey. To undertake this journey you have to be put to sleep. One of the children pretends to be asleep, and when the jump is complete the child has gone insane. From the child's point of view the jump lasted for an eternity while in reality it was only a few seconds. Show that the set of all polynomials whose constant term is a multiple of 3 is an ideal of the ring $Z[x]$.   Attempt: Suppose then we need to show   if $p(x), q(x) \in I$ then $p(x) + q(x) \in I$  and if $r(x) \in Z[x]$ , then $p(x)r(x) \in I$ and $r(x)p(x) \in I$.  And $0 \in Z$, then $0 \in I$.  Then let   $I =$ {$a_nx^n + a_{n-1}x^{n-1} + ...+a_1x + a_0$ | $a_i \in Z$, $a_0\in 3Z$}  $p(x)  = a_nx^n + a_{n-1}x^{n-1} + ...+a_1x + 3a$  and $q(x) = b_mx^n + b_{m-1}x^{n-1} + ...+b_1x + 3b$  Here we do not assume that $p(x) $ and $q(x)$ have the same degree.  Then $p(x) + q(x)  = [a_nx^n + a_{n-1}x^{n-1} + ...+a_1x + 3a]  +  [b_mx^n + b_{m-1}x^{n-1} + ...+b_1x + 3b] = 3(a + b) + (a_1 + b_1)x + ....+ (a_n + b_m)x^n$.  so $p(x) + q(x) \in I$  And $0 \in Z$ is also in $I$, since for $p(x) = 0x^n + ....+ 3(0) = 0$.  And let $r(x)= c_nx^n + ...+3c  \in Z[x] $.  Then $r(x)p(x) = 3(3ac) + (3b_1 + 3a_1) + ....+ a_nb_mx^{2n}$  So $r(x)p(x) \in I$.  Can someone please help? I am not really sure how to prove this. I am getting confused with the underscrips Any help or better approach would really help. Thanks in advance! Creating rounded corners using CSS How can I create rounded corners using CSS? Just watched the 'How Big is the Universe?' TV program and it indicated that the universe is believed to be infinite in size. They used an approach involving measuring the internal angles of triangles. Or something like that anyway.  But hang on, the Universe has a known starting time and size followed by a finite time expanding at a finite speed. So surely, by definition, it cannot be infinite? I can accept it is honking big but by definition it cannot be infinite. Are they misleading me? sum of the residue at all poles of the rational function p/q is 0. We are given two polynomials p, q with deg(q)>1+deg(p). Show that the sum of the residue at all poles of the rational function p/q is 0.   How do I go about solving this? Show that $\int_{0}^{\infty}|\frac{\sin x}{x}|\,dx=\infty $ Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .  Here is what i have tried :  I found out that for all $x&gt;0$ , $x-\frac{x^3}{6} &lt; \sin x &lt;x $  . So $1-\frac{x^2}{6}&lt;\frac{ \sin x}{x}&lt;1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}&lt;\int_{0}^{M}|\frac{ \sin x}{x}|&lt; M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved .  Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you . Short harvard citation? I need to cite the following way:  The first 3 Letters of the author + the year.  Example:     Kemper 2009 --> [Kem09]   But I don't have any clue how to do that!