How to explain the strange behaviour about Function Nothing
Using Sequence to delete a part of an expression
Can I disable pinch to show all open tabs in Safari? Prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I need to prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I am getting stuck midway through the inductive step.   Here is what I have:  $\forall n\geq 2$, where $n\in\mathbb{N}, P(n)$ is the statement "$\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$".  Base case:  $\sum\limits_{i=2}^{2}\frac{1}{(i-1)i}$ = $\frac{1}{2}$ = $\frac{(2-1)}{2}$ so $P(2)$ is true.  Inductive assumption: Assume $P(n)$ is true. Need to show $\sum\limits_{i=2}^{n+1}\frac{1}{(i-1)i}$ = $\frac{((n+1)-1)}{(n+1)}$.  $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ + $\frac{1}{n^{2}}$ = $\frac{(n-1)}{n}$ + $\frac{1}{n^{2}}$  = $\frac{n(n-1) + 1}{n^{2}}$ = $\frac{n^{2}-n+1}{n^{2}}$  I got stuck here and tried working backward from $\frac{((n+1)-1)}{(n+1)}$, which gives me:  $\frac{((n+1)-1)}{(n+1)}$ = $\frac{n}{(n+1)}$   Any suggestions on what to do next or where I may be going wrong? Showing that $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ contains multiplicative inverses Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?   I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$. It is known that a superconductor is a material with electrical resistance zero. My question is, it is exactly zero, a theoretical zero, or for practical realistic reasons it is effectively zero? What is syntax highlighting and how does it work? How to save a copy of a multiplayer world in the new 0.7+ menus? How do you save a joined game after the 0.7.1 update? The menus have all changed since 0.6.x and I can't find the option to save a copy of a joined world anymore. (My friend has got a diamond house on survival and I want to save it but it doesn't let me!) Do immature fruits perform photosynthesis? Most immature fruits are green: peppers, pine cones, plums, lots of them. I want to know if the green is from chlorophyll in the cells. Do the fruit cells perform photosynthesis? When you cover a green stem or leaf, it will turn pale and stretch. That is because the stems have little need for chlorophyll in the dark, which is why they are pale. They stretch because the auxins in the stems are not destroyed by the photons, and the stems stretch out and topple over. If a green fruit is covered, will it turn pale and stretch like that? ASP.NET Stress Testing Is there a way to test an application where you simulate a hundred different clients connecting to a IIS server and asking the same data? At the customer where our project is running they have 400 computers and they often do stress tests with all computers. I on the other hand have only got my laptop... (and a development server).  (In my case the data is asked through a WebORB Gateway). Front suspension or full suspension mountain bike When should I use # and = in ASP.NET controls? I have been using ASP.NET for years, but I can never remember when using the # and = are appropriate.  For example:  &lt;%= Grid.ClientID %&gt;   or  &lt;%# Eval("FullName")%&gt;   Can someone explain when each should be used so I can keep it straight in my mind? Is # only used in controls that support databinding? To prove Cayley-Hamilton theorem, why can't we substitute $A$ for $\lambda$ in $p(\lambda) = \det(\lambda I - A)$? Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.  We note that $p(\lambda) = \det(\lambda I - A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I - A) = \det(AI - A) = 0$? We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks Can the Eclipse Java formatter be used stand-alone Is there a way to use the formatter that comes with eclipse, outside of eclipse?  I would like to format some java files using my formatter.xml file that I have configured using eclipse.  Does anyone have any code examples that would allow me to do this?  I would also like to use this standalone, so the specific jars that are used would be nice. Baking with oven without bottom heating element My mom wants to do baking but her oven doesn't have bottom heating element. It only has an element on top, with a fan at the back.  She told me that she can't make me a good pizza or cake without it.  Is there any way around? Probability of the union of $3$ events? I need some clarification for why the probability of the union of three events is equal to the right side in the following:  $$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$  What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram?  Here on page 3, this is explained but not in enough details that I can understand it: \num{100000} prints 100 000  I want to have 1 00 000.  Question: How to set siunitx package to do the above?  Some more  \num{1000} => 1 000  \num{10000} => 10 000  \num{100000} => 1 00 000  \num{1000000} => 10 00 000  \documentclass[12pt]{article} \usepackage{siunitx} \sisetup{group-separator = {,}}  \begin{document}  \num{1000}  \num{10000}  \num{100000}  \num{1000000}  \num{10000000}  \num{100000000}  \num{1000000000}  \SI{1000000000}{m}  \end{document}   The exact rules are described on the :     The Indian Numbering System uses separators differently from the Arabic system; besides the three least significant digits of the integer part, a comma divides every two rather than every three digits. Cron Job not working in plugin I have written a plugin that will convert posts to excel format and then email the .xls to an email id. I can make it work with when function is declared in functions.php but does not work with function defined in plugin file.   if ( ! wp_next_scheduled( 'xls_func_hook1' ) ) {   wp_schedule_event( time(), 'hourly', 'xls_func_hook1' ); }  add_action( 'xls_func_hook1', 'sendxls1' ); //senxls1 is a function in functions.php if ( ! wp_next_scheduled( 'xls_func_hook2' ) ) {    wp_schedule_event( time(), 'hourly', 'xls_func_hook2' ); } add_action( 'xls_func_hook2', 'export2excel' );   The full code is at I'm trying Google's PageSpeed online service  Ironically, it's primarily highlighting Google's own services as something that needs improvement on my site  1) jQuery from Google: blocking. So I moved all javascript from &lt;head&gt; to the end of the document before &lt;/body&gt;. That helped  2) Linking to external Google Font CSS (in &lt;head&gt;): blocking. But the font is critical to the design of the page and should load before much else  3) Google Analytics: Caching is not good. (Google has set it internally to 2 hours expiration). Don't know how to change this (this is also placed at the bottom of page)  The Google Font is highlighted as a big priority to change.  How can I fix this? Where/how should I call the the font? Problem  Let $n \geq 3$ be an odd number and let $G=\{1,...,2n\}$ be a group of order $2n$. Let $\phi:G \to S_{2n}$ be the morphism defined by $\phi(g_i)(g_j)=g_ig_j$ and let $H=\phi^{-1}(A_{2n})$. Show that $H$ is a proper normal subgroup of $G$.  So basically the result of this problem is that every group of order $2n$, with $n$ odd, is not simple.   The morphism $\phi$ can be seen as the action of $G$ on itself by left multiplication. Since $A_{2n} \lhd S_{2n}$, then $H=\phi^{-1}(A_{2n}) \lhd G$. I am stuck at showing that $H \neq \{1\},G$. All I could think of is that $G$ has an element $g$ of order $2$, then $\phi(g)$ is of order $1$ or $2$, but it is easy to see that $\phi(g)=1$ if and only if $g=1$, so $ord(\phi(g))=2$. I don't know which are the elements of order two in $S_{2n}$. I would appreciate some help to solve the problem, thanks in advance. How to grep for text in a file and display the paragraph that has the text?