I want to add few jquery in my basic page of drupal website to call some api with parameter. Can any one explain me how can i accomplish this. $.ajax or $.post we can do right but its not happening in basic page. I wanted to triggerd that jquery when i click html submit button... *in this case i am not using hook_form actully.**  This is only for one page...basically its a drupal basic page.
Is there a location where I can drop in a custom JavaScript file in Drupal 7.x?  It just includes simple event actions for menus, and I don't want to edit the theme/module files.  Or do I have to just add the "include" hook to the theme/module?
A non-blocking read on a subprocess.PIPE in Python Possible Duplicate:          In  the following claim was mentioned:  $R/I \otimes_R M \cong M / IM$  So I tried to prove it. Can you help me finish my proof? Thanks!    We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: $$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$  $R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:  $$  (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$  Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM  \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$  so we want to show two things:   (i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R  IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here.   (ii) and that $M /IM \cong (R/I) \otimes_R M / IM $  Can you help me? Thanks. How can I completely remove a objects (e.g. particle system or groups used in particle systems) in blender using python?  The problem is this: I have a script, with which I create a scene (e.g. adding some normal objects and some group of objects with a particle system). For a research study I would like to repeat that that several times, without restarting blender. So I can create a scene, delete everything, e.g. with a script like this:  import bpy scn = bpy.context.scene for obj in scn.objects:     obj.select = True bpy.ops.object.delete()   If I rerun the script to recreate the scene with objects using particle system, the latter is not shown in the scene anymore (only the placeholders of the particle system). Now I'm not sure where exactly the problem lies, but I guess not everything is immediately deleted, running the script above. Only the startup of blender helps (but that's not what I want) or pushing the purge button. The purge button seems to start the command:  bpy.ops.outliner.orphans_purge()   But I can't get this piece of code to run in the console. It says that "the context is incorrect". Is there a way to purge the unused data by python script? Thanks a lot for any help. Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through the point $P=(12,3)$ I was solving a few problems from a textbook and I came across this one:     Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through   the point $P=(12,3)$   I could find the tangency points $$ but my solution was completely messy (I had four equations that I had to go back-and-forth with.  Is there an elegant/simple solution to that?  Tangency points:  $$P_1 = (0,3)$$ $$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$ Trouble trying to use Windows Search to search for file with a question mark in the content It seems that Windows Search is having trouble when I specify a question mark character as the content that I'm searching for within my files.  I've tried:   content:? content:"?" content:"\?" content:\? content:%3f content:"%3f" content:\3f content:"\3f" content:\00\00\00\3F   but each of these yield no results when I do indeed have files that contain ? within the scope of the search. If I instead search for something like content:happy it will indeed find all files with the phrase "happy" within those same files.  My assumption is that Windows Search is treating some characters such as the question mark as a special character likely for the purpose of wild card expressions. To test this theory I've also searched for files containing an asterisk * and indeed I have the same issue as when searching for content of question mark. What I'm wondering is whether there is any way to get a search to work where I can search for a literal expression without any type of wildcard matching or at least a way to escape the special characters within the expression.  By the way I'm using Windows 7 Enterprise. Formatting Ticks and FrameTicks labels with a function Powershell Try Catch Continue Derivative of inverse function at a point. Let $f\left(x\right)=x^{13}+3x^{9}+2x^{5}+x^{2}+x+2$ find $\left(f^{-1}\right)'\left(1\right)$. Can somebody show me the method? I am a little confused. Proof: $\|X\|_2\leq \sqrt{n}\|X\|_\infty$ (Euclidean Norm)  Proof that $\|X\|_2\leq \sqrt{n}\|X\|_\infty$, if $x \in \mathbb{R^n}$.   With that, proof $\|X\|_\infty \to 0 \implies \|X\|_2\to 0$  I'm stuck:  $\|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{n}\sqrt{x^2_1+x^2_2+...+x^2_n}$  I don't know how to use the fact: $\|X\|_\infty=\max(|x_1|,|x_2|,...,|x_n|)$ What regular expression will match valid international phone numbers? Custom cursor in WPF? I want to use an image or icon as a custom cursor in WPF app. What's the best way to do it? This occurs in the  released recently to the public. When someone posts a code block in their question or answer, the code block is cut off on the right hand side, and I can't scroll to the right to view the rest of it. How can I view these code blocks? Inequality about infinite cardinals The following is a question I'm interested in solving:    The idea is to solve it without using AC. As @Asaf Karagila pointed out in the comment section:  "Even if $\kappa$ is an infinite cardinal, you can't assume that $\kappa \geq \omega$ without using AC".  I am still trying to think this through: My reasoning being, that as $\kappa &gt; n \forall n \in \omega$, then $\kappa \geq \sup\{n:n\in \omega\}=\omega$ as $\omega$ is a limit ordinal.  Where does my reasoning fail? And also, a detailed solution would be very appreciated. Natbib puts reference data one line per item I'm using:  \documentclass[oneside,article,openbib,12pt,titlepage]{memoir} \usepackage[authoryear,round]{natbib} % Better bibliography support \bibpunct{(}{)}{,}{a}{,}{,}   in the preamble of my document and  \backmatter \renewcommand{\bibname}{References} \bibliographystyle{chicago}    \bibliography{listb,mybib}   in the body of the file. But a typical formatted reference is:     Freely, I. (1997).   A small paper.   The journal of small papers -1. to appear.   I've tried other non-natbib styles (e.g., regstud), and the reference flows and wraps as it should. So how does one get natbib to do the same?  Also, how does one get natbib to include the authors' entire first names? Remotely run script on Unix, get output locally? I need to run a (Tcl, or whatever) script on a remote Unix machine, from a (local) Windows command shell, and I want the stdout/stderr of the script to show up in the Windows CLI. The exit code would also be useful.  Is this possible using ssh (putty)? Or by any other means?  Thanks! I want to create a wave made out of particles/dots. The wave should look like this:  I tried to achieve this effect by adding an ocean modifier to a plane. Then I added a particle system to the plane but now I don't know how to go on. How can I make the particles lay on the 'ocean' and move when the ocean is moving? PS: I know there is a way to do this with a material in the Blender Render. But firstly I want to create the whole thing in Cycles and secondly I want to have particles. Why do we need sigma-algebras to define probability spaces? Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$ Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic? Is it a problem if a UK visa was issued but not used? Last year, my wife and I were issused 6-month visieor visas for the UK. She did not travel with me and, therefore, didn't use her visa. I am currently applying for a visitor visa for my daughter and myself.  Will the fact that my didn't use her visa have any negative affect on our new application?