I'm not familiar with how these devices work. I'm looking to duplicate an ID card for ease of use and this is all the information I can find on it:  MIFARE ISO 14443A (13.54 MHz)  So, is there a simple cloner or duplicator that supports this? If not, is there an arduino device or something else that can do it?  Thanks.
I have got a NFC key to enter in my room and i want to make a copy of this key in case of loss. I read the nfc tag key with my nexus 5 with Broadcom BCM20793M NFC controller. The phone reads the tag and retrivies all the infos but when i try to copy the tag it gives an "Error reading the tag" message. The tag is a  NXP MIFARE Classic. I read that it is not compatible with the newest chips so i wanna buy an Arduino NFC reader to read and copy the tag.  Might it works?    Below all the infos retrieved from the nfc key:
How do I do simple user input in python? Problems with pie charts There seems to be in increasing discussion about pie charts.   The main arguments against it seem to be:   Area is perceived with less power than length. Pie charts have very low data-point-to-pixel ratio   However, I think they can be somehow useful when portraying proportions. I agree to use a table in most cases but when you are writing a business report and you've just included hundreds of tables why not having a pie chart?  I'm curious about what the community thinks about this topic. Further references are welcome.  I include a couple of links:         In order to conclude this question I decided to build an example of pie-chart vs waffle-chart. Usage of Some: Does "some" require a plural or singular verb? What should I consider when choosing continuous lighting vs. strobes for studio photography? $SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom$(GL_n(\mathbb R), A)$, $A$ is an abelian group. Try to prove the following proposition.     $SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom($GL_n(\mathbb R), A)$, where $A$ is an abelian group.   My first attempt is to show that $SL_n(\mathbb R)$ is in form of $ABA^{-1}B^{-1}$, then it's clear that $ABA^{-1}B^{-1}$ is in $ker(f)$. However, there exists an elementary row reduction matrix, i.e. $E_{ij}(m)$ adding m times $j$-th row to $i$-th row. Then $E_{ij}(m)\in SL_n(\mathbb R)$. Now, it is possible that $E_{ij}(m)$ is not in $ker(f)$. Then I try to find any relation between $\mathbb R^{\times}$ and $A$, but, except knowing they are both abelian groups, I cannot figure out what else connects those two group. Could you give me some hints? Google chrome is reading characters backwards Relationship between Partial Harmonic Sum and Logarithm. How do I show that if there are functions $f,g$ such that$$ f(g(x)+g(y))=bx+cy $$holds for all $b,c\in\mathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?  Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$ y^2+z^2=f(x,g(y-x)+g(z-x)). $$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems? I know $[0,1]$ denotes the interval between 0 and 1 with the boundary. And $(0,1)$ denotes the interval between 0 and 1 without the boundary. Today, I encounter some expressions as $]0,1]$, $]-\infty,+\infty]$. What does $]0,1]$ mean? Thanks. Error message "Strict standards: Only variables should be passed by reference" Perfect set without rationals Interaction term positive while components are negative? I am confused about the interpretation of my interaction term , I have been through similar questions but I am still not sure about the interpretation that I should give to my output .  So this is my model , Y  stands for earnings management , X1 = CSR , X2=CEO duality , as you can see both of the coefficients are negative CSR reduces earnings management manipulation and Duality in my sample reduces em as well  which is in line with the literature and confirm my hypothesis , but to my surprise my interaction term is positive and significant does that mean that CSR combined with duality increase EM ? Or that CEO duality moderates the effect of CSR on EM knowing that in papers that I've read, they only use&quot; moderate &quot;when the interaction term and the components are all negative . Could you please help me ? I am supposed to comment my results but I don&quot;t understand this . Thank you How align the qedsymbol after of a big formula in display mode? Description of the problem My problem is this. For the proof of a math theorem i use the proof environment of the package amsthm. It adds a \qedsymbol in the last line of the proof, but if the proof ends with a displayed formula the qedsymbol appears in the next line leaving an unwanted blank space. The documentation of  teaches to fix this problem placing the command \qedhere after of the displayed formula but inside of the appropriate environment. This works well if the formula is of one line like \[ax^2+bx+c=0.\qedhere \]  but if the proof ends in a matrix, or cases with a &quot;great height&quot; the qedsymbol is fixed in the baseline at right and it this looks higher than expected (for my taste). The result is showed in this MWE \documentclass{article} \usepackage[width=7cm]{geometry} % Only for reducing space in this MWE  \usepackage{amsmath} % For `bmatrix` \usepackage{amsthm} % For `proof` environment \begin{document} \begin{proof} And this proof ends with \[ \begin{bmatrix}     1 &amp; 0 \\     0 &amp; 1 \\     0 &amp; 0 \end{bmatrix}.\qedhere \] \end{proof} \noindent Text \end{document}  that produces  What I want I'd like something like  with the qedsymbol aligned with the bottom of the matrix. Is this possible? Proving that expression is equivalent to the definition of derivative Let $f$ be differentiable at $x=a$.   Prove that if $x_n \to a^+$ and $y_n \to a^-$ then:  $$\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).$$  Every option that I think about seems to my very trivial, so I believe that I am doing something wrong.  Both numerator and denominator approach zero as $n\to\infty$ as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“$\frac{0}{0}$”).   Any direction? There is a way to use my Galaxy tab s like a wacom table, on linux? A is similar to $A^k$, then each eigenvalue of $A$ is a root of unity Let $A \in \mathbb{C}(n,n)$ and $k \geq 2$ be an integer such that $$A \sim A^k$$. Show that if $A$ is non-singular then each eigenvalue of $A$ is a root of unity.   Attempt: Since $A \sim A^k$, $$PA = A^kP$$ where $P$ is an invertible matrix. Since $A$ is invertible, $0$ cannot be an eigenvalue of $A$. Suppose $$Av = \lambda v \quad v \neq 0$$ then $$PAv = \lambda Pv$$ $$\therefore A^k(Pv) = \lambda (Pv) $$ but this implies that $Pv$ is an eigenvector of $A$. But the eigenvalues of $A^k$ are $\lambda^k$ $$\therefore \lambda^k=\lambda$$ which gives the conclusion required.   My questions is: Is the logic correct? If so, Am I missing any details? If not, then how could I approach this?   Thanks! How do I remove a property from a JavaScript object? Say I create an object as follows: let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot;,   &quot;regex&quot;: &quot;^http://.*&quot; };  How should I remove the property regex to end up with new myObject as follows? let myObject = {   &quot;ircEvent&quot;: &quot;PRIVMSG&quot;,   &quot;method&quot;: &quot;newURI&quot; }; What to write in visa application form and what documents to submit as an unregistered freelancer? There are many visa applicants who are working as an unregistered freelancer (photographer, construction contractor, fashion designer, journalist etc ). Some of these freelancers make really handsome monthly income and don't really need to work as a full time employee. Most of them receive variable amount of payments (cash, cheques, demand drafts etc) When it comes to filling a visa application form (USA, UK, Schengen etc), there is always an employment section in all visa form, eg for Schengen visa employment form :    This is a very important factor to make an application really strong.  Questions:   1 Unregistered freelancers can just simply write their profession? what documents to produce along with the application?  2 Bank statements with variable methods of payments could convince the visa officer that these are all legitimate payments? Which Windows 7 log file contains device connection/disconnection information? Windows 7 frequently plays the usb/new device connection/disconnection sound. It happens both when the computer is idle and under load. All usb device connections have been checked (though I can't rule out a bad cable yet). I thought a good way to track down the issue would be to wait for the sound and then check the log file for the latest connected/disconnected device. I just need to know which log file has this information.  Edit: I'm going to leave the answer accepted, however the issue persists. I get that sound seemingly randomly throughout the day. Anyone else have any ideas? How to solve ~(P → Q) : P & ~Q by natural deduction? I'm trying to solve the following by natural deduction:  ~(P → Q) : P &amp; ~Q  It's a trivial problem if identities are used, as can be seen by the following:   {1}      1.   ~(P → Q)                    Prem. {1}      2.   ~(~P ∨ Q)                   1 Mat. Impl. {1}      3.   ~~P & ~Q                    2 De Morgan's {1}      4.   P & ~Q                      3 DNE   Starting from step 2 of the previous proof, it's also fairly straightforward to reach a conclusion using natural deduction; i.e. starting from ~(~P ∨ Q), which is an identity of ~(P → Q):  ~(~P ∨ Q) :  P &amp; ~Q   {1}      1.   ~(~P ∨ Q)                   Prem. {2}      2.   ~P                          Assum. {2}      3.   ~P ∨ Q                      2 ∨I {1,2}    4.   ~(~P ∨ Q) & (~P ∨ Q)        1,3 &I {1}      5.   P                           2,4 RAA {6}      6.   Q                           Assum. {6}      7.   ~P ∨ Q                      6 ∨I {1,6}    8.   ~(~P ∨ Q) & (~P ∨ Q)        1,7 &I {1}      9.   ~Q                          6,8 RAA {1}      10.  P & ~Q                      5,9 &I   I also found that it's fairly easy to conclude ~Q:   {1}      1.   ~(P → Q)                    Prem. {2}      2.   Q                           Assum. {3}      3.   P                           Assum. {2,3}    4.   P & Q                       2,3 &I {2,3}    5.   Q                           4 &E {2}      6.   P → Q                       3,5 CP {1,2}    7.   ~(P → Q) & (P → Q)          1,6 &I {1}      8.   ~Q                          2,7 RAA ... {1,2}    12.  P & ~Q   In spite of how easy it looks, I can't figure out how to do it. I believe it can be done in 12 steps. Thanks!