I wanna preventDefault a dynamically added link that is not getting recognized by the on() method. However, the static link which was already on the page preventDefault()'s itself with the same function. &lt;!DOCTYPE html&gt; &lt;html&gt; &lt;head&gt;     &lt;title&gt;         Testing Phase     &lt;/title&gt; &lt;script src=&quot;https://code.jquery.com/jquery-3.6.0.js&quot; integrity=&quot;sha256-H+K7U5CnXl1h5ywQfKtSj8PCmoN9aaq30gDh27Xc0jk=&quot; crossorigin=&quot;anonymous&quot;&gt;&lt;/script&gt;     &lt;script type=&quot;text/javascript&quot;&gt;         $(document).ready(function(){             $(&quot;#txt&quot;).focus();             $('#btn').click(function(){                 var inpt = $(&quot;#txt&quot;).val();                 $(&quot;#dv&quot;).append(&quot;&lt;p class='eo'&gt;&quot;+inpt+&quot;--&gt;&lt;a href='dynamic' class=''&gt;This is dynamic link, it won't get preventDefault() with the on() method.&lt;/a&gt;&lt;/p&gt;&quot;);                 x++;             });             $('.eo').on('click', function(e){                 e.preventDefault();             });           });     &lt;/script&gt; &lt;/head&gt; &lt;body&gt; &lt;input type=&quot;text&quot; id=&quot;txt&quot; name=&quot;&quot;&gt; &lt;input type=&quot;button&quot; id=&quot;btn&quot; name=&quot;&quot; value=&quot;Add&quot;&gt; &lt;div id=&quot;dv&quot;&gt;     &lt;a href=&quot;nowhere&quot; class=&quot;eo&quot;&gt;This static link works with the on() method!&lt;/a&gt; &lt;/div&gt; &lt;/body&gt; &lt;/html&gt;
I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.
"We plan" versus "we are planning" Is there a semantic difference between the following two sentences?     In the future, we are planning to migrate our tool to the Z3 solver.         In the future, we plan to migrate our tool to the Z3 solver. We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:  Take $(\sqrt{2})^{\sqrt{2}}$  If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.  I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?   p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$. Let $G$ be a graph of minimum degree $k&gt;1$. Show that $G$ has a cycle of length at least $k+1$.  How would I show this? I understand that a cycle is a sequence of non-repeated vertices and the degree of a graph is the number of neighbors the vertex has. Getting MAC Address I need a cross platform method of determining the MAC address of a computer at run time.  For windows the 'wmi' module can be used and the only method under Linux I could find was to run ifconfig and run a regex across its output.  I don't like using a package that only works on one OS, and parsing the output of another program doesn't seem very elegant not to mention error prone.   Does anyone know a cross platform method (windows and linux) method to get the MAC address?    If not, does anyone know any more elegant methods then those I listed above? Does any test actually prove causation? $S$ be a collection of subsets of $\{1,...,100\}$ ; any two sets in $S$ has non-empty intersection , what is the maximum possible value of $|S|$? I have a variable, x, and I want to know whether it is pointing to a function or not.  I had hoped I could do something like:  &gt;&gt;&gt; isinstance(x, function)   But that gives me:  Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in ? NameError: name 'function' is not defined   The reason I picked that is because  &gt;&gt;&gt; type(x) &lt;type 'function'&gt; Did English ever have a formal version of "you"? From the top of my head, Danish "De" (practically never used), German "Sie", Chinese "您", French "vous", Spanish "usted" are a formal way of addressing someone, especially if one isn't familiar with the addressee. Did English ever have this? It sounds as though Proto-Indo-European might have had this (based on my 4 examples), but perhaps someone can enlighten me? How does centrifugal force work? When asking about , people sometimes refer to the strict aliasing rule. What are they talking about? I have an object already created in blender, a tree, or a human, or a pipe.. and I need this to be drilled with multiple holes.   The holes are painted in an image, that is, it is an image that has a repeating an images pattern (they can be circles, squares, or any form).  Here are some examples for drilling-pattern: On a systems level, I understand that as electrons are pushed into a wire, there is a net field and a net electron velocity.  And I've read that the net electron drift is slow.  But electricity travels through the wire, essentially at c, and I want to understand that mechanism.  My apologies if my question is poorly stated, I know bare bones undergraduate quantum theory circa 1990s but it doesn't explain the motion of electricity in detail.  Here's my conception, I'm hoping someone will fill in the holes, ha ha.  An electron moves into the wire.  It's got a kinetic energy.  After travelling a short distance, it spontaneously emits a photon, which hits another electron in a valence shell.  That electron then presumably does the same.  If this conception is simply wrong, please enlighten me.  The questions that arise:   Presumably at this level, electrons are acting more like waves and less like particles, but is there any classical component in the picture, ie are electrons coming in imparting other electrons with kinetic energy through repulsion, or does it not work that way? If electrons momentarily have energy, then pass it on by a photon, what determines when that photon is emitted, and what frequency it will be?  I assume that electrons in this cloud are not limited by any kind of exclusion principal, and that any frequencies are possible? Why should a photon emitted by an electron be in the direction of travel?  Conservation of momentum tells me that if an electron is moving, the photon should be emitted in that direction, slowing the electron, but could an electron emit a photon in the opposite direction?  If it did, I assume it would somehow have had to absorb energy from elsewhere? That sounds possible by analogy with quantum tunneling. What is the mechanism by which electrons propagating increase the temperature of the material?  Are they transmitting energy to the electrons in the valence shell, which tug at the nucleus, do some photons hit the nuclei directly, or is there some other way?  Presumably, electricity travels slower than light, because there is some time in each exchange, and some time when electrons are moving at sublight speeds before emitting a photon.  By how much is this slower than light, and what is the speed of each interaction? On the nilpotence of the matrix $AB-BA$ I want to assign the result of an expression (i.e., the output from a command) to a variable and then manipulate it – for example, concatenate it with a string, then echo it.  Here's what I've got: #!/bin/bash cd ~/Desktop; thefile= ls -t -U | grep -m 1 &quot;Screen Shot&quot;; echo &quot;Most recent screenshot is: &quot;$thefile;  But that outputs: Screen Shot 2011-07-03 at 1.55.43 PM.png Most recent screenshot is:   So, it looks like that isn't getting assigned to $thefile, and is being printed as it's executed. What am I missing? Meaning of $\int \phi^\dagger \hat A \psi \:\mathrm dx$ While analysing a problem in quantum Mechanics, I realized that I don't fully understand the physical meanings of certain integrals. I have been interpreting:    $\int \phi^\dagger \hat A \psi \:\mathrm dx$  as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$ when one tries to observe the observable corresponding to $\hat A$" $\int \phi^\dagger  \psi \:\mathrm dx$ as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$".   All integrals are over all space here, and all $\phi$s and $\psi$s are normalized.  Now, I realize that  interpretation of the first integral doesn't really make sense when put side by side with the second one.   For example, when $|\phi\rangle$ is an eigenstate of $\hat A$, I get two different expressions for "the square root of probability that one will get the corresponding eigenvalue of $\psi$ when one tries to observe the observable corresponding to $\hat A$". As far as I can tell, $\int \phi^\dagger \hat A \psi \:\mathrm dx \neq\int \phi^\dagger  \psi \:\mathrm dx $, even if $|\phi\rangle$ is an eigenstate of $\hat A$. I am inclined to believe that the second integral is the correct answer here (it comes naturally when you split $|\psi\rangle$ into a linear combination of basis vectors). But I am at a loss as to the interpretation of the first integral.  So, my question is, what are the physical interpretations of $\int \phi^\dagger \hat A \psi \:\mathrm dx$ and $\int \phi^\dagger  \psi \:\mathrm dx$?  (While I am familiar with the bra-ket notation, I have not used it in this question as I don't want to confuse myself further. Feel free to use it in your answer, though) Inner automorphisms form a normal subgroup of $\operatorname{Aut}(G)$ For an arbitrary group $(G,\cdot)$ let $\operatorname{Aut}(G) = \{f: G \to G \mid f \text{ is an isomorphism}\}$ be the set of all automorphisms of the group $G$. We assume that $(\operatorname{Aut}(G),\circ)$ where $\circ$ is composition of mappings is a group.  1) Prove that for arbitrary $a \in G$ there is an automorphism $p_a: G \to G;\quad p_a(x) = a^{-1}xa$.  2) Prove, that $\operatorname{Inn}(G) = \{p_a \mid a \in G\}$ is a normal subgroup of $(\operatorname{Aut}(G),\circ)$. How to stop fluid inflow? I have fluid simulation with an inflow object. I want the inflow to go for 2 seconds, then have the inflow stop (so no more fluid is added) but keep the simulation going  (so the existing fluid sloshes around).  However, I do not see any way to set start and ending values on my inflow object.  How do I stop inflow at a certain frame/time in my simulation? Prove that if the identity is written as the product of $r$ transpositions, then $r$ is an even number Theorem. If the identity is written as the product of $r$ transpositions, $id=τ_1τ_2\dots τ_r$, then $r$ is an even number. Proof. We will employ induction on $r$. A transposition cannot be the identity; hence, $r &gt; 1$. If $r = 2$, then we are done. Suppose that $r &gt; 2$. In this case the product of the last two transpositions, $τ_{r−1}τ_r$, must be one of the following cases: $(ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),$ where $a$, $b$, $c$, and $d$ are distinct. The first equation simply says that a transposition is its own inverse. If this case occurs, delete $τ_{r−1} τ_r$ from the product to obtain $id=τ_1τ_2···τ_{r−3}τ_{r−2}$. By induction $r − 2$ is even; hence, $r$ must be even. In each of the other three cases, we can replace $τ_{r−1} τ_r$ with the right-hand side of the corresponding equation to obtain a new product of $r$ transpositions for the identity. In this new product the last occurrence of $a$ will be in the next-to-the-last transposition. We can continue this process with $τ_{r−2} τ_{r−1}$ to obtain either a product of $r − 2$ transpositions or a new product of $r$ transpositions where the last occurrence of $a$ is in $τ_{r−2}$. If the identity is the product of $r − 2$ transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with $τ_{r−3}τ_{r−2}$. At some point either we will have two adjacent, identical transpositions canceling each other out or $a$ will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix $a$ in this instance. Therefore, the identity permutation must be the product of $r − 2$ transpositions and, again by our induction hypothesis, we are done. Question: That is the proof written in the book, but unfortunately after more than $10$ times reading I don't understand the last two paragraph (i.e., from &quot;In each of the other three cases&quot;). Is there another simpler proof or could someone please clarify what is going on? Thank you. Is something wrong with StackOverflow's spam filter? Recently there have been huge numbers of spam topics related to football streaming. They have all been spam flagged and deleted but there are more getting posted all the time, I just flagged another 6 or so a few minutes ago. This has been going on for around 3 days now with these topics getting constantly posted over and over again in the  tag. Simple homework question about normally distributed variables The question states:      Consider a set of random variables $X_i$, where $i=1,...n$. Each $X_i$ is   normally distributed with mean $0$ and variance $1$, i.e. $X_i$ are $\mathcal N(0,1)$.   What is the mean and the variance of the random variable $Y$, where   $Y=X_1+...+X_n$.   How do I do this?