How many closed polygons will $n$ coplanar lines, all intersecting each other (no three concurrent), form on the plane?  Can we find a recursion formula for this?? Is there a way out?
Show that $n$ lines separate the plane into  $\frac{n^2+n+2}{2}$ regions if no two of these lines are parallel and no three pass through a common point.   I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0.  But I really have no idea how to begin the inductive step.  How do we know what k+1 we're supposed to arrive at?  Thanks!
We use unit length Quaternion to represent rotations. Following is a general rotation matrix obtained ${\begin{bmatrix}m_{00} &amp; m_{01}&amp;m_{02} \\ m_{10} &amp; m_{11}&amp;m_{12}\\  m_{20} &amp; m_{21}&amp;m_{22}\end{bmatrix}}_{3\times 3}\tag 1 $.   How do I accurately calculate quaternion $q = q_1i+q_2j+q_3k+q_4$for this matrix?Means   how can we write $q_i$s interms of given $m_{ij}$  accurately? What is the percentage increase to any value when initial value is 0? This is a very simple maths problem, which doesn’t seems simple to me when I just put some extra thoughts on it. So need to borrow some extra brains for this.   Question: If cost of coffee in my office was Dollar 0 till yesterday and today it is Dollar 2.5 then what is the percentage increase in the cost of coffee?  Initially it seemed like 2.5% increase, simply by distributing number from 0 to 100, but this is not the case.   Some suggested it is infinitely more than the previous value, because we are calculating percentage by dividing by zero. But what about this if the cost of coffee was $0.25? Isn’t still infinitely more! Something is wrong here.   PS: Yes, coffee were free in our office but not anymore, and we are still recovering from this shocker. 😥 Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$  So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$ Sponsor Gallery, Page, or Something Like That? I am trying to pass a member function within a class to a function that takes a member function class pointer.  The problem I am having is that I am not sure how to properly do this within the class using the this pointer.  Does anyone have suggestions?  Here is a copy of the class that is passing the member function:  class testMenu : public MenuScreen{ public:  bool draw;  MenuButton&lt;testMenu&gt; x;  testMenu():MenuScreen("testMenu"){     x.SetButton(100,100,TEXT("buttonNormal.png"),TEXT("buttonHover.png"),TEXT("buttonPressed.png"),100,40,&amp;this-&gt;test2);      draw = false; }  void test2(){     draw = true; } };   The function x.SetButton(...) is contained in another class, where "object" is a template.  void SetButton(int xPos, int yPos, LPCWSTR normalFilePath, LPCWSTR hoverFilePath, LPCWSTR pressedFilePath, int Width, int Height, void (object::*ButtonFunc)()) {      BUTTON::SetButton(xPos, yPos, normalFilePath, hoverFilePath, pressedFilePath, Width, Height);      this-&gt;ButtonFunc = &amp;ButtonFunc; }   If anyone has any advice on how I can properly send this function so that I can use it later. Disable automatic activation of Gnome Shell activities on mouse over I want to write a function in Python that returns different fixed values based on the value of an input index.    In other languages I would use a switch or case statement, but Python does not appear to have a switch statement.  What are the recommended Python solutions in this scenario? Loading PyQt5 modules from QGIS3/PyQGIS I am trying to talk to the PyQGIS API for use in a 3rd party application using the offical .dmg image provided on the QGIS homepage.  After setting PYTHONPATH to /Applications/QGIS3.4.app/Contents/Resources/python and initializing everything as stated in , I am facing the following error when running from PyQt5.QtGui import or from qgis import *:    ImportError: dlopen(/Applications/QGIS3.4.app/Contents/Resources/python/PyQt5/QtGui.so, 2): Library not loaded: @executable_path/../Frameworks/QtGui.framework/QtGui   Referenced from: /Applications/QGIS3.4.app/Contents/Resources/python/PyQt5/QtGui.so   Reason: image not found   Looks like I need to set some linkers or similar?   Side facts: - I have no problems when using the homebrew version of QGIS). - I am on macOS 10.14.6 (Mojave) remote ssh tcpdump command, through a gateway, to be opened, in realtime, in Wireshark, on local machine I (Host A - Windows based using plink) am trying to connect to a remote linux server (Host C), through a gateway linux server (Host B), and run this tcpdump on Host C to be piped all the way back to Host A (through Host B I assume), and opened, in real time, with Wireshark on Host A:  Below is what I saw as the command when not using a gateway ("direct remote fetch").  So, how would I make this work going through my gateway host B scenario above?  Host A$ plink.exe -ssh -pw abc123 root@Host C "tcpdump -s0 -U -n -w - -i eth0 not port 22" | "C:\Program Files\Wireshark\Wireshark.exe" -k -i - Photoelectric effect intensity I understand the PE effect quite well but I'm failing to understand one thing. Intensity is the amount of energy per second incident to a given area.   So can you can increase the intensity by either increasing the electrical energy or making the area smaller? By increasing the electrical energy, what exactly is happening?  I assume that by increasing intensity you are increasing the number of photons incident to an area of the metal thus increasing the rate of photoelectric emission. But why doesn't increasing intensity via more electrical power lead to greater energy of the electrons, if by increasing intensity you are increasing the energy of the incident waves?  I understand the results but I'm trying to understand why the intensity works the way it is thanks Outcast boy meets girl underwater How do I detect the direction of 2D rectangular object collisions? Interaction with the site on the app should count towards Enthusiast/Fanatic If I do any of the below, which would normally constitute "activity" on any Stack Exchange site (including mobile):   View my profile Vote on any question or comment Visit one or more questions   ...then Stack Exchange considers me to have "visited" the site.  If I do any of the below on the mobile app:   Vote on any question or comment Visit one or more questions   ...then Stack Exchange does not consider me to have visited the site.  If one wants the app to have feature parity with the mobile site, then shouldn't my participation within the app also count towards the Enthusiast/Fanatic badge? Listing only directories in UNIX creating a db schema I am trying to create a schema page in LaTeX. It shall look like this.   How can I create schema code which look like this image? There should be also a click to go function. When the user clicks on the user_schema link he should be redirected to that page.  What classes I can use for this? Any suggestions would be appreciated.    \documentclass{article} \usepackage[utf8]{inputenc}  \begin{document} \maketitle  \section{User schema} \mySchema{Users} {     _id: ObjectId - Id do registro     name: String - Nome do usuário     date: Date - Data de Cadastro     ren_date: Date - Data de renovação do Cadastro     email: String - Lorem ipsum dolorsitamet, consecteturadipisicingelit, sed do eiusmodtemporincididunt ut labore et dolore magna aliqua. Ut enim ad minimveniam, exercitationullamcolaborisnisi ut aliquipexeacommodoconsequat. Duisautereprehenderit in voluptatevelit esse cillumdolore eu fugiatnullapariatur. Occaecatcupidatat non proident, sunt in culpa quiofficiadeseruntmollit.     access_log: Array [          {              date: String - Data da ocorrência.             user: ObjectId - Usuário que executou.             action: String - Ação Executada.         }         {              date: String - Data da ocorrência.             user: ObjectId - Usuário que executou.             action: String - Ação Executada.         }     ]     \link{group}{group_id}: ObjectId - Id do grupo. } \end{document}   Here is the image how should it appear: Getting different answers when integrating using different techniques Question:  Is it possible to get multiple correct results when evaluating an indefinite integral?  If I use two different techniques to evaluate an integral, and I get two different answers, have I necessarily done something wrong?    Often, an indefinite integral can be evaluated using different techniques.  For example, an integrand might be simplified via partial fractions or other algebraic techniques before integration, or it might be amenable to a clever substitution.  These techniques give different results.  For example, looking over a few other questions on MSE:      From : evaluate $$ \int x(x^2+2)^4\,\mathrm{d}x. $$    Via the substitution $u = x^2+2$, this becomes $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + \frac{32}{5} + C.     $$ However, multiplying out the polynomial and integrating using the power rule gives $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + C    $$     From :  evaluate   $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x. $$    Simplifying the integrand using partial fractions then integrating gives $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{-2}{(x+1)} - \ln|x+1| + C.    $$ Via integration by parts, we get $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{x-1}{(x+1)} - \ln|x+1| + C.    $$     From : evaluate   $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x. $$    Using the substitution $u = \sec(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\sec^2(\pi x)}{4\pi} + C.    $$ Using the substitution $u = \tan(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\tan^2(\pi x)}{4\pi} + C.    $$ $M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Why doesn't mass of bob affect time period? The gravitation formula says $$F = \frac{G m_1 m_2}{r^2} \, ,$$ so if the mass of a bob increases then the torque on it should also increase because the force increased. So, it should go faster and thus the oscillation period should be decrease.  My physics book says that period is only affected by effective length and $g$. Why doesn't mass of bob affect the period? The center of the dihedral group How to prove that the center of the dihedral group $D_{2n}$ is $\{1,r^{n}\}$ and the center of $D_{2n-1}$ is $\{1\}$?  I don't know how to prove it in this general case. Particle energy in general relativity? Assume that a particle is moving with four-velocity $u^\mu$ through a spacetime with metric $g_{\mu\nu}$ ($-+++$ signature).  Let us also assume that there exists a time-like vector field $t^\mu=(1,0,0,0)$.  Is it generally true that the energy $E$ of a particle is given by the following expression:  \begin{eqnarray*} E &amp;=&amp; -mt_\mu u^\mu\\ &amp;=&amp; -mg_{\mu\nu} t^\nu u^\mu\\ &amp;=&amp; -mg_{00}u^0\\ &amp;=&amp; -mg_{00}\frac{dt}{d\tau}? \end{eqnarray*}  In the special case that $t^\mu$ is a Killing vector and $u^\mu$ is tangent to a geodesic then $E$ is a constant of motion of the particle. But the energy of a particle need not remain constant in the general case - is that true?  If $E=-mt_\mu u^\mu$ is indeed the particle mass/energy then does it act as a source of spacetime curvature given that it depends on the $00$-component of the spacetime metric ifself? I suppose that is not a problem.