Unable to learn third relentless force
Can't learn the third Word of Power from Wulfgar
Possible Duplicate:          This is something I haven't quite figured out yet, but I have been using function(){}() just because my VIM syntax highlight screws up if I add the parenthesis, although I've seen (function(){})() around many times, maybe its an IE thing?  edit:  var singleton = function() {     // code }();  var singleton = (function() {     // code })(); Motivation for Napier's Logarithms In the wikipedia article on logarithms, I am clueless about the approach and motivation for the following computations done by Napier (and the mysterious appearance of Euler's number) in this . Please help me understand:       By repeated subtractions Napier   calculated $10^7(1 − 10^{−7})^L$ for L   ranging from 1 to 100. The result for   $L=100$ is approximately $0.99999 = 1  − 10^{-5}$. Napier then calculated the   products of these numbers with $10^7(1  − 10^{−5})^L$ for L from 1 to 50, and   did similarly with $0.9995 ≈ (1 −  10^{−5})^{20}$ and $0.99 ≈ 0.995^{20}$.   These computations, which occupied 20   years, allowed him to give, for any   number ''N'' from 5 to 10 million, the   number ''L'' that solves the equation      $$N=10^7 {(1-10^{-7})}^L$$      In modern notation, the relation to   natural logarithms is:      $$L = \log_{(1-10^{-7})} \!\left(  \frac{N}{10^7} \right) \approx 10^7  \log_{ \frac{1}{e}} \!\left(  \frac{N}{10^7} \right) = -10^7 \log_e  \!\left( \frac{N}{10^7} \right)$$      where the very close approximation   corresponds to the observation that      $${(1-10^{-7})}^{10^7} \approx  \frac{1}{e}. $$   I understand the computations, but I don't know what's going on. sed fails with "unknown option to `s'" error How do I do this? I created a user without a home directory and now I want to create one. Not just a folder called new_user, I want the default home directory with all it's folders and hidden files, etc. How to initialize HashSet values by construction? Kendo Grid & Rich text Editor of Sharepoint 2013 Pagination not working with custom loop I've got a custom loop that I'm using to display some Real Estate listings that will be available within 60 days. I'm calling it with the following function:   &lt;?php  $sixtydays = date('Y/m/d', strtotime('+60 days')); $paged = (get_query_var('paged')) ? get_query_var('paged') : 1; $query = new PostsOrderedByMetaQuery(array(   'post_type' =&gt; array('post', 'real-estate'),   'meta_key' =&gt; 'Time Available',   'meta_compare' =&gt; '&lt;=',   'meta_value' =&gt; $sixtydays,   'paged' =&gt; $paged,   'orderby_meta_key' =&gt; 'Price',   'orderby_order'    =&gt; 'ASC' )); ?&gt; &lt;?php while ($query-&gt;have_posts()) : $query-&gt;the_post(); ?&gt;   While the loop works great, I can't get it to paginate. It shows the first 10 (my default) posts but doesn't show the pagination. The only way to display all posts is to show them on one page by adding 'posts_per_page' =&gt; -1, I have similar loops on other pages that have no problem paginating. The only difference with this one is that there are two meta keys that are filtering the posts.   I'm using WP Page Navi for this and the rest of my pages. I'm closing the loop and adding the pagination using the following code:   &lt;?php endwhile; // End the loop. Whew. ?&gt; &lt;?php wp_pagenavi(); ?&gt; &lt;?php wp_reset_query(); ?&gt;   How can I go about fixing this? In the following is it better to use a possessive noun with an apostrophe or an attributive noun without an apostrophe?     The following list details the assumptions that have been made in   conjunction with the Dell Company’s staff.   Or      The following list details the assumptions that have been made in   conjunction with the Dell Company staff.   Assume Company is the name of a company, such as Dell. What do we call someone who's in between -a friend and an enemy! He's a friend of mine   The message is clear. He's my friend. He wishes all good for me and helps me whenever I need one.      He's an enemy of mine   The message is clear again. He's my enemy. He wishes all bad for me and never helps me whenever I need!     He's ...... of mine!   The message is clear. He's neither friend nor enemy. He's in between. At times, he may help me but then I cannot always rely on him. He's not a friend of mine.   Other times, he does really bad to me but maybe, there was no any other go for him. He's not an enemy of mine! :) Over the years I have slowly developed a  that validates MOST email addresses correctly, assuming they don't use an IP address as the server part.  I use it in several PHP programs, and it works most of the time.  However, from time to time I get contacted by someone that is having trouble with a site that uses it, and I end up having to make some adjustment (most recently I realized that I wasn't allowing 4-character TLDs).  What is the best regular expression you have or have seen for validating emails?  I've seen several solutions that use functions that use several shorter expressions, but I'd rather have one long complex expression in a simple function instead of several short expression in a more complex function. Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ? How to decide which glm family to use? I have fish density data that I am trying to compare between several different collection techniques, the data has lots of zeros, and the histogram looks vaugley appropriate for a poisson distribution except that, as densities, it is not integer data. I am relatively new to GLMs and have spent the last several days looking online for how to tell which distribution to use but have failed utterly in finding any resources that help make this decision. A sample histogram of the data looks like the following:   I have no idea how to go about deciding on the appropriate family to use for the GLM. If anyone has any advice or could give me a resource I should check out, that would be fantastic. W.K.T, any user(with more than 2k rep) can review 20 suggested edits per day.  Today I saw the newly appointed moderator  reviewed more than 20.    Is it possible for moderators to review more than the daily limited count?  If yes, can any moderators can do it at any time?  I didn't see this before. I am travelling form Delhi (DEL) to San Francisco (SFO), I have to make a connection flight in Vienna that would take me to Frankfurt (FRA). My flight to SFO is from FRA, I have one ticket for entire journey and the flights are operated by Lufthansa or its subsidiaries.  Do I need a transit visa? My flight arrives at Vienna international airport and leaves from the same, and for second leg it arrives at Frankfurt international airport.  I asked this question as an addendum to my  but was advised to make a new post.  I hold Indian passport and valid US visa. Volume form on $(n-1)$-sphere $S^{n-1}$ Let $\omega$ the (n-1) form on $\mathbb{R}^n$ $$\omega=\sum_{j=1}^{n}(-1)^{j-1}x_{j}dx_{1}\wedge\cdots\wedge \hat{dx_{j}}\wedge\cdots dx_{n}$$ Show that the restriction of $\omega$ to $S^{n-1}$ in precisely the volume for this sphere.  What I did was: $\omega$ never vanish on the sphere, because, defining $\sigma\in \Omega^{n-1}(S)$ for $$\sigma_{p}(v_{1},...,v_{n-1})=det(p,v_{1},...,v_{n-1})$$ and $i:S^{n-1}\rightarrow \mathbb{R}^{n}$ the inclusion function, then $\omega=i^{\ast}(\sigma)$ then $\omega\not=0$ and therfore is a volume form.   How proof that $\omega$ is the volume form?   The first thing that comes to mind is show that $\int_{S^{n-1}}\omega=Vol(S^{n-1})$ but I have serious problems with the definition, I think that is to much.   How see that $\omega$ is invariant on $\mathbb{R}^{n}$ under action of $O(n)$ Is it bad to marinate meat for too long? From my experience it is always best to marinate meat for a decent amount of time, so the meat can "soak" up the marinade and be more flavourful.  However, I came across this recipe here:   In the directions, it says: Add pork and marinade for a minimum of 6 hours or overnight, but no longer than 12 hours.  I was wondering, why does it say "no longer than 12 hours"? Isn't it good to let meat marinate for as long as possible?  Thanks. I am looking to implement a biplot for principal component analysis (PCA) in JavaScript. My question is, how do I determine the coordinates of the arrows from the $U,V,D$ output of the singular vector decomposition (SVD) of the data matrix?  Here is an example biplot produced by R:  biplot(prcomp(iris[,1:4]))     I tried looking it up in the  but it's not very useful. Or correct. Not sure which. Prove the following equality using mathematical induction: Prove the following equality using mathematical induction: For any integer $n \ge 1$  $$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$  I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get   \begin{align} \sum_{i=1}^{1} \frac{1}{1(1+1)} &amp;= \frac{1}{1+1} \\ \sum_{i=1}^{1} \frac{1}{2} &amp;= \frac{1}{2} \end{align}  As for the inductive step, I inputted $n +1$ but it did not work for me because honestly I did not know what to do with the $i$ and what I should do next.  How would I go about doing the next step and an explanation for $i$ would be helpful. Let $a$, $b$, and $c$ be complex numbers with $a\neq0$. Show that the solutions of $az^2+bz+c=0$ are $z_1,z_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, just as they are in the case when $a$, $b$, and $c$ are real numbers. Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?