Constructor is not called in this aggregation class
Most vexing parse
Is product of two closed sets closed? If $A$ is closed in $X$ and $B$ is closed in $Y$,   does it follow that $A \times B$ is closed in $X \times Y$ ? Long-distance Mind Trick I have some troble understanding the Kernel density estimation.  If a consider the next example:  s&lt;-0 z&lt;-density(rnorm(25)) f&lt;-function(i)approx(z$x,z$y,xout=i)$y for(i in seq(min(z$x),max(z$x),length.out = 10000)) s&lt;-s+f(i) s   I expected getting that s is approximately equals to one since I am considering an approximation for the integral of the Kernel density function (that I suppose is also a density). However, I don't get the expected result. Why it doesn's sum one?  The idea behind this is that I wanted to use the Kernel as the density of my data points. I was interpolating to get my explicit function and use with other x values. Or do you know another way to get a density function estimated by fitting? How to Customize the Email template on chatter posts Bounded measurable function and integral with charcteristic function Correct plural form of a noun preceded by "zero" When using zero as a quantifier, is it correct to use the singular form on the object of the quantifier, or the plural form?  It sounds confusing when I put it that way, but what I mean is: Which is correct?   Your password expires in 0 days. Your password expires in 0 day.   Essentially I suppose I'm asking, does "singular" mean "one" or "the opposite of more than one", as zero is not "plural" in the traditional "more than one" sense?  I'm pretty sure "days" sounds correct, but I can't be sure. I have an object already created in blender, a tree, or a human, or a pipe.. and I need this to be drilled with multiple holes.   The holes are painted in an image, that is, it is an image that has a repeating an images pattern (they can be circles, squares, or any form).  Here are some examples for drilling-pattern: Firstly, this is a homework problem so please do not just give an answer away. Hints and suggestions are really all I'm looking for.  I must prove the following using mathematical induction:     For all $n\in\mathbb{Z^+}$, $1+2+2^2+2^3+\cdots+2^n=2^{n+1}-1$.   This is what I have in my proof so far:     Proof: Let $p\left(n\right)$ be $\sum_{j=0}^n 2^j=1+2+2^2+\cdots+2^n=2^{n+1}-1$. Then $P\left(0\right)$ is    \begin{align} \sum\limits_{j=0}^0 2^j=1=2^1-1.\tag{1} \end{align} Hence    $P\left(0\right)$ is true. Next, for some $k\in\mathbb{Z^+}$, assume    $P\left(k\right)$ is true. Then \begin{align}  \sum\limits_{j=0}^{k+1}2^j&amp;=\left(\sum\limits_{j=0}^k  2^j\right)+2^{k+1}\tag{2}\\ &amp;=2^{k+1}-1+2^{k+1}\tag{3} \\ &amp;=2\cdot 2^{k+1}-1\tag{4}\\ &amp;=2^{k+2}-1\tag{5}\\ &amp;=2^{\left(k+1\right)+1}-1.\tag{6} \end{align}   Therefore, $P\left(n\right)$ holds for all $n$.$\;\;\;\;\blacksquare$   Is everything right? My professor went over the methodology in class quite fast (and I had a bit too much coffee that night) and I want to make sure I have it down for tomorrow night's exam. What did Hisoka do in the Hunter Exam? In the first Hunter Exam of the series (Hisoka's second),  was introduced into the series removing another applicant's arms from his body for bumping into him and not apologizing.    What did he actually do? We know Hisoka's abilities are Bungee Gum and Texture Surprise, but they don't explain how the arms are converting into flower petals as seen above. Prove that the multiplicative inverse of $a$ modulo $m$ exists if and only if $a$ and $m$ are coprime. Prove that the multiplicative inverse of $a$ modulo $m$ exists if and only if $a$ and $m$ are coprime.  Can someone help me with this? Windows 7 task bar at autohide, make it totally disapear? My desktop look is empty so all I have is a nice wallpaper. My Windows taskbar has the "Taskbar Appearance" setting to "Auto-hide the taskbar". Although hidden, it leaves a thin top layer so you can still see it a bit, and where the pinned shortcuts are. Is it possible to totally have it hidden without a trace? This is a serious question. It really bugs m. How to create Mist in Blender I already tried several ways to create Mist in Blender, but it never worked 100%. I use Blender 2.78 and I want to create Mist like in this scene here.  What are the exact steps?     Here is what I have tried so far:  I already tried this:  but it simply did not work .. I dont know why.  (EDIT: I managed to fix it: Check out this graphic:   Until now I already used objects successfully as Mist-creators (Volume Scatter under Volume Material). The bad thing about using Objects as Mist-creator is that they dont have an endless mist range + the mist can not be bright. Normally it appears grayish result is bad..  Check it out  Please help Setting items to status-completed before they're deployed In several scenarios, including two today ( and ), we have had bugs set to  simply because the fix has been checked in. Just because the code portion has been completed does not mean the work is done, and this can be confusing for users who come across the same problem. Until then it really is  or maybe a tag that doesn't yet exist (status-checked-in?).  Particularly in times like this week, where the stability of the system is a priority, I don't think this status should be set merely because the code has been checked in - it should happen when the code is actually deployed and verified. Currently the devs are being very good about telling us that the code is done but they're not sure when it will be deployed, but I think that is because of the stability issue. I am trying to boot Ubuntu on my computer.  When I boot Ubuntu, it boots to a black screen.  How can I fix this? Table of Contents: This is a question from Stromberg related to Steinhaus' Theorem:     If $A$ is a set of positive Lebesgue measure, show that $A + A$ contains an interval.   I can't quite see how to modify the Steinhaus proof though. I have Ubuntu 17.10 with GNOME 3.  What happens: I have (for example) LibreOffice Calc open, with some data columns in it.  I highlight the columns I want to chart, then I select INSERT > CHART.  The chart child window dialog pops up on top of my data, so I want to move it to the side.  When I grab the top bar and move it, the parent (spreadsheet) window minimizes (I think it actually disappears).  Continuing the above example, if I click FINISH on the chart window, I can then see the minimized spreadsheet application, which is the window title bar plus the menu bar just below it.  This seems like really odd behaviour and certainly falls under the category of "not being useful to the user", since the whole idea of moving the child window in the first place was just to be able to better see the spreadsheet!  In other applications, for example the PDF viewer, a slightly different problem occurs.  When I try to move the child window (for example the print dialog) the parent does not minimize but instead follows the child window around.  This is less surprising than having the parent disappear entirely but equally useless in terms of allowing the user to see the information present in the parent window.  Update 2017/11/06  Pomsky's answer below fixes the problem, but leaves me wondering why the window manager is making the decision that all child windows have modal behaviour (or don't), and not the applications themselves; and further, why modal behaviour prevents moving the child screen around when it obscures useful info on the parent screen.  Neither of these seem like good design to me... Cannot select text in Safari address bar Manipulating Partial Derivatives of Inverse Function In lectures we're told:  $$\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}$$  as long as the same variables are being held constant in each partial derivative.  The course is 'applied maths', i.e non-rigorous, so don't confuse me.  But anyway, if we have:  $$\xi = x - y \qquad \eta = x$$  Then $\dfrac {\partial x} {\partial \xi} = 0$ and $\dfrac {\partial \xi} {\partial x} = 1$.  The rule presumably fails because one of the partial derivatives is $0$.  But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact $0$, but then I've worked out the partial derivative manually anyway. How to make a half cylinder? I am having a tough time building a half cylinder and have not found any tutorial showing me how to do that. The energy of an electromagnetic wave The intensity of an electromagnetic wave is only related to its amplitude $E^2$ and not its frequency. A photon has the same wavelength as the wave that's carrying it, and its energy is $h  f$.   So if a laser wave is kept at the same amplitude and the wave length is reduced, why does its intensity remain the same even though its photons now carry less energy?  Why are the intensities of electromagnetic waves so different to sound waves (and other waves travelling through a medium) which are related to $f^2  E^2$?