Add function to oninput listener using JQuery vs adding it directly to the html tag
Event binding on dynamically created elements?
How to make \left, \right pairs of delimiter work over multiple lines? Database design for products with bundles of products I am building a database system for my retail business. I have set some tables which are:   Product Purchase Sales Balance   All are connected one another and are able to show my inventory level.  The problem I am having is I also sell bundles of products - that have different prices than their individual prices. Example: I sell an orange for $1, an apple for $1.2; I sell fruit package 1 (2 oranges and 2 apples) for $3.8, package 2 (4 oranges and 4 apples) for $7.  Is there a right way how to create relationship for these product bundles?  PS: I am using FileMaker Pro creating this. It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong. example of discontinuous function having direction derivative Is there a function (non piece-wise unlike below) which is discontinuous but has directional derivative at particular point? I have a manual that says the function has directional derivative at $(0,0)$ but is not continuous at $(0,0)$. $$f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^4} &amp; \text{ if } x \neq 0\\  0 &amp; \text{ if } x= 0 \end{cases}$$  Can anyone give me few examples which is not defined piece wise as above? Validate inputs that are not inside a form with jQuery Validation plugin How can I validate some inputs that are not inside a form tag?  All the samples I found have a form tag, but I want to validate some inputs that are not inside a form. Why does my iPhone power off when the battery is down to 30%? Apparently recurrences like this cannot be solved with the Master Theorem:  $T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log(n)}$  Because $n^{\log_b(a)} = n^1$ is not a polynomial multiple of $f(n) = \frac{n}{\log(n)}$ since $\frac{n}{\frac{n}{\log(n)}} = \log(n)$, which is less than $n^\epsilon$ for any positive $\epsilon$.  But when is this really even an issue?   For example why wouldn't the time complexity be $n \log n$ using case 2 of the Master Theorem? Unrolling the recursion:  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{\log_2(2^k)}$$  $$T(n) = n\sum_{k=1}^{\log_2(n)} \frac{1}{k}$$  Doesn't this suggest $\Theta(n \log n)$ even though it supposedly cannot be done with Master Theorem? This is from  .      Let $\mathcal{A}$ be a locally small category, and let $A,A' \in \mathcal{A}$ with $\operatorname{Hom}(-,A) \cong \operatorname{Hom}(-,A')$. Prove directly that $A \cong A'$.     My thoughts:      Let $\eta$ denote the natural isomoprhism between the two functors.    We have    $$H(A,A) \xrightarrow{ \eta_A} H(A,A')$$   So $\eta_A(id_A): A \rightarrow A'$ is a candidate. Now I want to construct inverse.    $$ H(A',A') \xrightarrow{\eta_{A'}} H(A',A)$$   Then the map $\eta^{-1}_{A'} (id_{A'}):A' \rightarrow A$.    So I'd like to show that the composition is identity. By the naturality condition of $\eta$, from $H_A(A) \rightarrow H_{A'}(A')$ (and similarly on other direction)    $$ \eta_A(id_A) \circ \eta^{-1}_{A'}(id_{A'}) = \eta_{A'}(\eta_{A'}^{-1}(id_{A'}))= id_{A'} $$   We deduce these two maps are inverses. Hence $A' \cong A$.    Is this correct? Or is there a neat way to see this? How are java.lang.Object's protected methods protected from subclasses? The keyword protected grants access to classes in the same package and subclasses ().  Now, every class has java.lang.Object as superclass ().  Hence I conclude that every class may access java.lang.Object's methods even if they are protected.  Take a look at the following example:   public class Testclass {   public Object getOne() throws CloneNotSupportedException {     return this.clone();   }   public Object getTwo() throws CloneNotSupportedException {     return ((Object) this).clone();   } }   While getOne() compiles fine, getTwo() gives   Testclass.java:6: clone() has protected access in java.lang.Object         return ((Object) this).clone();   I neither understand why getTwo() doesn't compile nor what's the difference (regarding the access of java.lang.Objects members) with getOne(). Without appealing to choice, can we prove that if $X$ is well-orderable, then so too is $2^X$? Without appealing to the axiom of choice, it can be shown that (Proposition:) if $X$ is well-orderable, then $2^X$ is totally-orderable.  Question: can we show the stronger result that if $X$ is well-orderable, then so too is $2^X$?  Proof of Proposition. Pick any well-ordering of $X$. Then the lexicographic order totally orders $2^X$.  More explicitly: for any two $f,g \in 2^X$, define $f &lt; g$ iff   there exists $x \in X$ such that $f(x) \neq g(x)$, and if $x \in X$ is minimal such that $f(x) \neq g(x)$, then $f(x)=0$ and $g(x)=1$.   It can be shown that $&lt;$ totally-orders $2^X$. Rounding integers to nearest multiple of 10 Given a group $ G $, how many topological/Lie group structures does $ G $ have? Can I use Named Credential merge fields as URL parameters on the endpoint? I have a named credential as follows:  Name   Test_Credential URL    https://www.somesite.com Identity Type    Named Principal Authentication Protocol    Password Authentication Username   foobar Password   mySomeSiteDeveloperKey Generate Authorization Header   FALSE Allow Merge Fields in HTTP Header    TRUE Allow Merge Fields in HTTP Body     TRUE   The service that I am trying to authenticate against uses URL parameters to pass in the developer key and username. Is there any way to use  to pass those in to the URL?   The Named Credential attributes seem not to get decoded until the request is actually sent, as debugging the endpoint, or headers if I set them per the linked doc, show the un-merged data. I can get the callout to go to the right place, and return the expected error because the auth information is not provided. Anybody know of a "show me my HTTP request" site that I could point to to get a better idea of how this works?  I've tried both of these:  req.setEndpoint('callout:Test_Credential/api/action?id={!$Credential.Password}&amp;account={!$Credential.UserName}');  req.setEndpoint('callout:Test_Credential/api/action' + '?id=' + '{!$Credential.Password}' + '&amp;account=' + '{!$Credential.UserName}');   (Edited to remove the part about $Credential vs. $Test_Credential as I was able to determine that it is the former, and see the merged creds by calling out to .) How to identify the process locking a file? Merging folders with mv? I understand that the architecture of Convolutional Neural Networks (CNN) and Feed forward (FNN) are quite different. And that CNNs use pooling and filters of shared weights over a patch of the image. I am not so clear on the core convolution operator (1):   If anyone could link me to an explanation, I have looked at the Colah blog and Nielsen's online book, and I understand what it is doing but don't understand the convolution operator.  Also, it looks quite similar to the the core FNN function is there and difference? (2).  (1) convolution operator is $a^1 = \sigma(b + w*a^0)$ which is equivalent to:  $a^1 = \sigma(b + \sum^4_{l=0}\sum^4_{l=0}w_{l,m}a_{j+l,k+m})$  (2) feed forward operation:  $a^1_i = \sigma(\sum^n_{j=1}w_{ij}x_j +b_i)$  Many thanks  Sources: functions taken from: When is a function satisfying the Cauchy-Riemann equations holomorphic? It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is continuously differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or ). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below.   Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then \begin{equation*} u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|) \end{equation*} and  \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|). \end{equation*} Multiplying the second equation by $i$ and adding the two together gives  \begin{align*} (u+iv)(z+h)-(u+iv)(z) &amp;= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\  &amp;= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|). \end{align*}  Now dividing by $h$ gives us the desired result.   Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? Convergence test of the series $\sum\sin100n$ I need to prove that $$\sum_{n=1}^{\infty} {\sin{100n}} \; \text{diverges}$$ I think the best way to do it is to show that $\lim_{n\to \infty}{\sin{100n}}\not=0$. But how do I prove it? How does ownership in fractional shares work? I've been trying to find out some more information about fractional shares, and how they work. I understand conceptually, you own x% of a share rather than x "full" shares, but I'm curious about how such a system is implemented.  As I understand it, in a "traditional" purchase of shares, the role of a broker is to act as an executor on your behalf, and does not hold any assets themselves. In the case of a fractional share, is this then not the case? For example, the broker holds the complete shares, and then just ensures that they keep an internal record of each partial owner in their system? So there would be no external record of partial ownership?  Thanks! How to make terminal transparent while still having its own background image? Is there a way to give terminal specific background image and make it 50% transparent at the same time?