This question on Multilayer intersection in QGIS was asked by "julsbreakdown" a year go. But I am a newbie and I am not really sure how it works with SQL as suggested in the responses. And I am using the same image posted by "julsbreakdown." Here is the link of the previous thread:   Using QGIS (or its plug-in) or python script, how do I get the polygon (x) which is the intersection of four polygons?
I am trying to find a way with QGIS to get the intersection area of more than 2 layers.  All the tools in QGIS take just the 2 input layers.  For example in the image below, I have 4 layers and I want to retrieve just the black area.   Furthermore, I need to retrieve the count attribute and add it to the intersected area attribute table (in this case 4)    I know there is a tool that does the trick in ArcGIS, but is there a QGIS solution?
Problem to launch java at Debian: "error while loading shared libraries: libjli.so" How to change ubuntu install from legacy to uefi Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$ Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ (other than the trivial case $p=0$).  $$p=0\colon\,\sum_{k=0}^n\binom{n}kk^0=2^n$$  I know that $\sum_{k=0}^n\binom{n}k=2^n$ and $\sum_{k=0}^nk^n=\frac{k^{n+1}-1}{k-1}$ but I am unsure of whether these would be of much use now.    Additionally, what about the similar series $\sum_{k=0}^n\binom{n}kk^n$ where $p=n$? I'm trying to show that  $$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$  Where $a_{i}$ are the roots of the function $f(x)$. I've tried to proceed by using a dummy function $g(x)$ and carrying out:  $$\int_{-\infty}^{\infty}dx\,\delta\big(f(x)\big)g(x)$$  Then making the coordinate substitution $u$ = $f(x)$ and integrating over $u$. This seems to be on the right track, but I'm unsure where the absolute value comes in in the denominator, and also why it becomes a sum.  $$\int_{-\infty}^{\infty}\frac{du}{\frac{df}{dx}}\delta(u)g\big(f^{-1}(u)\big) = \frac{g\big(f^{-1}(0)\big)}{\frac{df}{dx}\big(f^{-1}(0)\big)}$$  Can any one shed some light? Wikipedia just states the formula and doesn't actually show where it comes from. Does Stack Exchange use caching and if so, how? Can anyone help me with this?  What is the sum of this series:  $\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$  I got it after plugging $x=-1$ in a Fourier series   Thank you! When you pass a collection like list, array to another function in python, does it make a copy of it, or is it just a pointer? 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Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .  If the solution from that book is required I can post it but it is not complete as I mentioned.   Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist. Postfix SMTP banner show multiple hostnames A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.  I found the lengths of the lines $AQ$ and $AP$ in terms of $x$ and $y$ and used $AQ=2AP$ to get this equation $11x^2-40x-16=0$  Working: $AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}$  $AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}$  $AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}$  $\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0$  Using the quadratic formula I found $x=4,x=-\frac{4}{11}$  I then substituted these values into the two equations in the question to get the coordinates $(4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})$ or $(4,8),(-\frac{4}{11},\frac{136}{11})$  But the answers in the book are $(4,\frac{7}{2}),(7,5)$ or $(\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})$  I don't know where I went wrong in my method.  Edit: Found some errors in my method and calculation.  Using $x=4$ in $2y=3x-5$ gives $P(4,\frac{7}{2})$  Line with points $AP$ has equation $y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$  $Q$ lies on the same line. Therefore, the intersection between $y=\frac{1}{2}x+\frac{3}{2}$ and $y+x=12$ is the point $Q$. Working: $\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7$ Therefore, $y=-7+12=5$. Point Q has coordinates $(7,5)$.  But $x=-\frac{4}{11}$ is not the other $x$ coordinate of $P$ according to the answers in the book. I have no idea where I went wrong at this point. SQL Switch/Case in 'where' clause Using customized Coordinate System in ArcGIS Desktop? Restoring default toolbar layout in Blender 2.81 How do I fix a NoSuchMethodError? I'm getting a NoSuchMethodError error when running my Java program.  What's wrong and how do I fix it? 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