I'm trying to find all instances of the word "eNews" on a page and replace it with;   &lt;span style="text-transform: lowercase;"&gt;e&lt;/span&gt;News    I found a few other answers on here but they were all directed towards looking in specific classes. I tried using the all elements selector to find the word but the replace I've written just replaces the entire page:  $('*:contains("enews")').each(function () {     $(this).html('&lt;span style="text-transform: lowercase;"&gt;e&lt;/span&gt;News') });   Here is a fiddle of the sort of page set up with the word spread about the page:
I basically need to highlight a particular word in a block of text. For example, pretend I wanted to highlight the word &quot;dolor&quot; in this text: &lt;p&gt;     Lorem ipsum dolor sit amet, consectetuer adipiscing elit. &lt;/p&gt; &lt;p&gt;     Quisque bibendum sem ut lacus. Integer dolor ullamcorper libero.     Aliquam rhoncus eros at augue. Suspendisse vitae mauris. &lt;/p&gt;  How do I convert the above to something like this: &lt;p&gt;     Lorem ipsum &lt;span class=&quot;myClass&quot;&gt;dolor&lt;/span&gt; sit amet, consectetuer adipiscing elit. &lt;/p&gt; &lt;p&gt;     Quisque bibendum sem ut lacus. Integer &lt;span class=&quot;myClass&quot;&gt;dolor&lt;/span&gt; ullamcorper     libero. Aliquam rhoncus eros at augue. Suspendisse vitae mauris. &lt;/p&gt;  Is this possible with jQuery? Edit: As  , this is quite possible without jQuery - but I was hoping there might be a special method of jQuery which would let you do selectors on the text itself. I'm already using jQuery heavily on this site, so keeping everything wrapped up in jQuery would make things perhaps a bit more tidy.
LWC to display salesforce objects as a picklist onclick I have a requirement where I need to display list of salesforce objects as a picklist when a radio button is clicked.  Being new to LWC development, I could not figure out what is the right approach to do so. I have come across  from Salesforce articles where we can display radio buttons but as per my scenario, i need to display list of all salesforce objects as picklist when a radio button option is selected.  Can anyone please suggest the right approach to do so. I need to add a coloured line before my caption; this is what I used:  \usepackage[labelfont=bf]{caption}  \DeclareCaptionFormat{myformat}{\hrulefill #1  #2 #3} \captionsetup[figure]{format=myformat}   but the figure turns out to be line this:    and not a horinzontal line which what I want  2) I want the line to be coloured RGB [152 204 204] with thickness about the same as \top rule. Can anyone show me how to do this? Creating Two Executable Jars Using maven-assembly-plugin A google search gives number of methods to hide ALL tabs in chrome , but I want to hide 1 particular tab / change the display name of that tab.   Reason: I have an online music playing website open, which keeps showing tracks in the tab name and I just want to remove it from view of my work. Need rtl8814au driver for kernel 5.3 on Ubuntu 19.10 remove divs and spans from view output In my page.tpl.php I'm doing this to display one of my view:  print views_embed_view('teaserview', 'default');   That gives me the following html output:  &lt;span thmr="thmr_10"&gt; &lt;div class="view view-teaserview view-id-teaserview view-display-id-default view-dom-id-e051ba535ce9aac6cd9a0d564b3dbb29 contextual-links-region"&gt;     &lt;div class="contextual-links-wrapper"&gt;         &lt;span thmr="thmr_16"&gt;         &lt;ul class="contextual-links"&gt;             &lt;li class="views-ui-edit first last"&gt;&lt;span thmr="thmr_17"&gt;&lt;a href="/kbi/admin/structure/views/view/teaserview/edit/default?destination=node/3"&gt;Edit view&lt;/a&gt;&lt;/span&gt;&lt;/li&gt;         &lt;/ul&gt;         &lt;/span&gt;     &lt;/div&gt;     &lt;div class="view-content"&gt;         &lt;span thmr="thmr_15"&gt;         &lt;div class="views-row views-row-1 views-row-odd views-row-first views-row-last"&gt;             &lt;span thmr="thmr_14"&gt;             &lt;span thmr="thmr_11"&gt;&lt;span thmr="thmr_12"&gt;&lt;span thmr="thmr_13"&gt;&lt;img src="http://127.0.0.1/kbi/sites/default/files/teaser_2.jpg" width="1029" height="308" alt="" /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;         &lt;/div&gt;         &lt;/span&gt;     &lt;/div&gt; &lt;/div&gt; &lt;/span&gt;   But the only thing that I really need is  &lt;img src="http://127.0.0.1/kbi/sites/default/files/teaser_2.jpg" width="1029" height="308" alt="" /&gt;   Is there a possibility to modify the views_embed_view call so that I get only the needed data?   There would be the possibility to modify all the views .tpl.php files but I hope there's an easier way to do this.   I also read about the semantic views module but I didn't understand how to configure my view with it so that i get only the needed markup as output. it didn't seem to have any effect. Hill Cipher known plaintext attack I know a plaintext - ciphertext couple of length 6 for a hill cipher where its key is a [3x3] matrix.  Based on what I've read and learned, to attack and crack keys of [n x n], if we know a plaintext - ciphertext duo of length $n^2$ then we have our set of n equations with n variables, and this is generally solvable. However in my case there is only length of $6$ instead of $n^2 = 9$ .   My question is, how will I solve this problem and find the key? Or, since this is an obvious homework question, what should be my way of thinking? I cannot think of anything else but matrix multiplications and inverses but they do not help me at all. How can I add a newline in the text of a label in WPF such as the following?  &lt;Label&gt;Lorem    ipsum&lt;/Label&gt; Prove that $J^+AJ$ has positive eigenvalues Let $J \in \mathbb{R}$ be a $m \times n$ matrix of rank $n$ and a $A \in \mathbb{R}$ be a $m \times m$ diagonal positive definite matrix. Denote by $J^+$ a  of $J$.   As was shown  $J^+AJ$ is not positive definite in general.   However, I was wondering whether one can prove or disprove that $J^+AJ$ has real and positive eigenvalues. When writing (a blog post, script, etc..) what is the proper way to indicate two or more instances of a single letter? For instance, in Monty Python's Bookshop Sketch:     C: I wonder if you might have a copy of "Rarnaby Budge"?      P: No, as I say, we're right out of Edmund Wells!      C: No, not Edmund Wells - Charles Dikkens.      P: (pause - eagerly) Charles Dickens??      C: Yes.      P: (excitedly) You mean "Barnaby Rudge"!      C: No, "Rarnaby Budge" by Charles Dikkens.  That's Dikkens with two Ks, the      well-known Dutch author.      P: (slight pause) No, well we don't have "Rarnaby Budge" by Charles Dikkens      with two Ks, the well-known Dutch author, and perhaps to save time I      should add that we don't have "Karnaby Fudge" by Darles Chickens, or      "Farmer of Sludge" by Marles Pickens, or even "Stickwick Stapers" by Farles      Wickens with four M's and a silent Q!!!!!  Why don't you try W. H. Smith's?      C: Ah did, They sent me here.   I had always believed that plural never uses an apostrophe before the 's' (it's only used for possession), but I have rarely seen in written material the format "four Ms". (On a side note, whoever wrote  also used "two Ks".)  On a side note, and perhaps this should be a separate question, if a Compact Disc is a CD, then two Compact Discs would be two CDs right? (I see "CD's" written everywhere) Adjust sound notification per chat room At the top right of a chat room, just above the room title, there is a speaker icon. Clicking on this icon brings up the settings for "Sound notifications":    Currently, a change to this setting affects all chat rooms. Selecting "visible room" means that an audio notification is given for any activity in any room that is currently open in a browser tab (regardless of whether that is the active tab).  It would be useful to be able to adjust audio notification settings for individual rooms. For example, The Nineteenth Byte is quite a busy room, so I'd like to be notified only when mentioned. The Cornell Box is a much quieter room, and I'd like to be notified any time there is activity regardless of whether I am mentioned. Currently I can set all rooms to notify me of any activity, or I can set all rooms to notify me only when mentioned, but I can't choose a different setting for each room.  This would also be particularly useful when setting up a room for a very specific purpose, such as discussion of a particular question. Retrieving Maxwell's equations from the minimum action principle I'm currently working at the start of Alexei Tsvelik's book Quantum Field Theory in Condensed Matter Physics. I'm kinda stumped on a few essential steps.  Starting with the action:  $$S = \int dt \int d^3 x \mathcal{L}_{\mathrm{EM}} = \int dt \int d^3 x \left( -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + j_\mu A^{\mu}\right).$$  In the absence of any sources $j_\mu = 0$ and $c=1$, which after pulling in the negative sign implies  $$S = \frac{1}{2}\int dt \int d^3 x  \,\,\left(E^2 - B^2\right)$$ The next step is to take $\vec{E} = -\vec{\nabla} \phi + \partial_t \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$. Letting $\phi = 0$ and directly inserting these into the above yields:  $$S = \frac{1}{2} \int dt \int d^3 x  \,\left( (\partial_t \vec{A})^2 - (\vec{\nabla} \times \vec{A})^2 \right) $$  Now considering $\vec{A} = \vec{A}_0 + \delta \vec{A}$ as some minimal change in our vector potential, with $\vec{A}_0$ gives a minimum for the action and putting this into the above expression we have: $$\delta S = \int dt \int d^3 x \left(\partial_0 \vec{A}_0 \partial_0 \delta \vec{A} + \left(\vec{\nabla} \times \vec{A}_0\right)\cdot\left(\vec{\nabla} \times \delta \vec{A} \right) \right)+ \mathcal{O}(A^2)\,\,\,\,\,\,\, (1)$$ where I have the term  $$\mathcal{O}(A^2) = \int dt \int d^3 x \left( (\partial_0 \vec{A}_0)^2 + (\partial_0 \delta \vec{A})^2 - (\vec{\nabla}\times\vec{A}_0)^2-(\vec{\nabla}\times \delta\vec{A})^2\right)$$  The next step is where I lose our author and go from "Oh ok I get this" to "completely clueless". He rewrites $\delta S$ as:  $$\delta S =  \int dt \int d^3 (\delta x) \vec{A} \mathcal{F(\vec{A}_0)}+ \mathcal{O}(A^2)$$  where $\mathcal{F} = \frac{\delta S}{\delta A}$. What is the purpose of $x\to \delta x$ in the above expression? Is the functional derivative $\mathcal{F}$ a vector quantity? He has it bolded with a bolded $\vec{A}$ in the above expression. One last question takes the above expression further. We assume $\delta \vec{A}$ vanishes at infinity, as it should, and integration by parts yields:  $$\delta S =  -\int dt \int d^3 x \left( \partial^2_0 \vec{A}_0 - (\vec{\nabla} \times \vec{\nabla})\times \vec{A}_0  \right)\delta \vec{A}\,\,\,\,\,\,\,\, (2)$$ How exactly did we go from (1) to (2)? Thanks I appreciate your time! Why does Newton's Third Law actually work? My father explained to me how rockets work and he told me that Newton's Third Law of motion worked here. I asked him why it works and he didn't answer. I have wasted over a week thinking about this problem and now I am giving up. Can anyone explain why Newton's Third Law works? For reference, Newton's third law:  To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. QTimer can only be used with threads started with QThread EM algorithm gaussian mixtures- derivation I'm trying to make sense of a derivation I'm following from the lecture notes of Stanford's ML course.  Specifically the notes are here:   I'm particular I'm interested in the step where they are optimizing the lower bound of the log-likelihood by finding the derivative of that function w.r.t the mean (top of page 7). I'm a bit stuck on going from line 1 to line 2 in that sequence.  To copy here (using abbreviated notation for space):  $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\log\frac{\alpha_j \exp\psi_{j}^{(i)}\phi_j}{w_j^{(i)}}  \qquad(1) $$ $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\psi_{j}^{(i)}\qquad(2) $$ where $\psi_{j}^{(i)} =-\frac{1}{2}(x^{(i)}-\mu_j)^T\Sigma^{-1}_{j}(x^{(i)}-\mu_j)$ and $\alpha_j=\frac{1}{(2\pi)^{n/2}|\Sigma_j|^{1/2}}$.  The quantity $\mu_s$ is the mean (of a multivariate gaussian distribution) for class $s$.    In short, I'm not sure how they get from (1) to (2).  From just changing the log of (1) I see:  $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\log \left(\alpha_j \exp\psi_{j}^{(i)}\phi_j\right) - w_j^{(i)}\log w_j^{(i)} \qquad(1) $$  To get to their form (2) my impression is that they take pre-emptively discard terms that would contain derivatives of $w_j^{(i)}$ (w.r.t. $\mu_s$).  At least if you assume that much, you get to (2).  However, as they show earlier in that PDF,  $$ w_j^{(i)} = p(z^{(i)}=j | x^{(i)}) =\frac{p(x^{(i)}|z^{(i)}=j)p(z^{(i)}=j) }{\sum_j p(x^{(i)},z^{(i)}=j) } $$ i.e. this quantity is the posterior distribution of the class labels.  Intuitively, it seems like $\partial w_j^{(i)}/ \partial \mu_s$ should be non-zero.  After all, changing the location of the mean $\mu_s$ would change your posterior probability for a particular class given the same data.  (Right?).  Just from a mechanical point of view, $\partial w_j^{(i)}/\partial \mu_j$ would seemingly have some non-zero terms since $p(x^{(i)}|z^{(i)}=j) \sim \mathcal{N}(\mu_j, \Sigma_j)$.  If I work through the algebra, I get  $$ \frac{\partial w_j^{(i)}}{\partial \mu_j} = w_j^{(i)}(1-w_j^{(i)})\frac{\partial \psi^{(i)}_{j}}{\partial \mu_j} $$  What am I missing here?  If I take the log-likelihood (without even thinking about EM algorithm),    $$ \ell = \sum_{i=1}^{m}\log \sum_{j=1}^{k}p(x^{(i)}|z^{(i)}=j)p(z^{(i)}=j) $$ and differentiate w.r.t $\mu_s$, I correctly get  $$ \mu_s^* = \frac{\sum_{i=1}^{m}w_s^{(i)}x^{(i)}}{\sum_{i=1}^{m}w_s^{(i)}} $$  so I'm comfortable with the eventual result, just not the intermediate steps. When is the limit of a sum equal to the sum of limits? I was trying to solve a problem and got stuck at the following step:  Suppose ${n \to \infty}$ .  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$  Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.  Now we have  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$  As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...  $$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$  ...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:  $= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$  and the results we get are not the same.  Where did I go wrong? Shell Script for logging into a ssh server The Setup: Let's say we want to study a Euclidean $\mathrm{CFT}_2$ on $\mathbb R^2$ with coordinates $\sigma^1$ and $\sigma^2$ and metric   $ds^2 = (d\sigma^1)^2+(d\sigma^2)^2$.    It seems to me that in the usual discussion (e.g. di Francesco, Ginsparg, Polchinski), one proceeds to consider an analytic continuation of the CFT to $\mathbb C^2$ with coordinates $z^1, z^2$ and complex metric  $ds^2 = (dz^1)^2+(dz^2)^2$  and then, one performs the coordinate transformation $z = z^1+iz^2$ and $\bar z = z^1-iz^2$.  In this way the coordinates $z$ and $\bar z$ can be considered "independent" because they are coordinates on a complex two-dimensional manifold.  Also, in these coordinates the metric becomes  $ds^2 = dz\,d\bar z$  and it becomes clear that conformal mappings consist of mappings: $(z, \bar z)\to (f(z), g(\bar z))$.  My confusion is this: Since our original theory was on $\mathbb R^2$, books say that when we do calculations, we should consider the physical theory as living on the copy of $\mathbb R^2$ embedded in $\mathbb C^2$ given by the condition $\bar z = z^*$.  But consider the mapping $(z, \bar z)\to (z^2, \bar z)$.  This is a conformal mapping on $\mathbb C^2$, but it does not map the surface $\bar z = z^*$ to itself; for example the point $(z, \bar z)=(2,2)$ gets mapped to the point $(z^2, \bar z) =(4,2)$ and $2$ is clearly not equal to $4^*$.  In particular, it seems to me that analytic continuation to a CFT on $\mathbb C^2$ enlarges the set of mappings one can have, so what relevance does it really have to the original CFT on $\mathbb R^2$? How to create UEFI-only bootable USB live media? Algorithms for computing multiplicative order modulo $n$