Does it imply that $\int _0^\infty f_n\to \int _0^\infty f$?
Does $\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x)\,dx\,$?
I am trying to get the Linux distro "PIXEL" to install some PPAS for Wine, and I have run into a issue where it will not install because of the following error:   sudo: add-apt-repository: command not found   I am trying to install wine using the PPA add-apt-repository ppa:ubuntu-wine/ppa and I would like some assistance doing so. sed one-liner to delete everything between a pair of brackets? I am working with some text that is full of stuff between brackets [] that I don't want. Since I can delete the brackets myself, I don't need the one-liner to do that for me, but I do need a one-liner that will remove everything between them.  What about parentheses () instead of brackets? I was wondering in C++ if I have an enum can I access the value at the second index?  For example I have  enum Test{hi, bye};   if I want 'hi', can I do something like Test[0], thanks. Contour graphing algorithm I am not sure if this belongs on Mathematics Stack Exchange, but it is somewhat relevant here.       The Problem If you've installed any graphing/plotting apps on your smartphone, you will notice that the app will only take the equation in form of a single variable for 2D (e.g  $y=f(x)$) or in 2 variables for a 3D plot (e.g $z=f(x,y)$). This is problematic as many equations are tough to reduce to the form of $y=f(x)$ or $z=f(x,y)$. The plotters like this because this is the only way they can plot it in form of lines connecting multiple points. These plotters compute the $y$ value at fixed intervals of $x$ and join the point to the previous one. The brute force approach to counter this is to check if the function satisfies it at each point on the image. The problem is that this not only takes more time, but gives discontinuous graph for the function as the image cannot have infinite resolution.         The Question The only way to fix this is to identify which point comes after the next in the plot and join them together by a line. That is where I am stuck. How do you identify which point is sequentially the next one for a given function? Keep in mind that for a human, this might be very easy to do. But, for a computer, this requires a general algorithm. Any ideas? how to prove an element is non-zero in a tensor-product I was studying the following example from Atiyah &amp; MacDonald's Introduction to Commutative Algebra:   Let $x$ be the non-zero element in $N := \mathbf{Z}/ 2\mathbf{Z}$, $M := \mathbf{Z}$, and $M' := 2 \mathbf{Z}$. The element $2 \otimes x$ is zero in $M \otimes N$, but non-zero in $M' \otimes N$. I suppose this can be seen by the fact that $2 \otimes x$ generates $M' \otimes N$, and the tensor-product $M' \otimes N$ is non-zero.   However, I was wondering, what one would do if that element weren't a generator of $M' \otimes N$. So my question is:   What are the methods to prove that an element is non-zero in a tensor-product of modules?   Thanks a lot! If in a topological space only finite subsets are compact sets, is it then the discrete topological space?   Thank you. My question is how would I go about proving this?     Prove that $R$ is a local ring if and only if all elements of R that are not units form an ideal.   I understand that I need to prove both directions so:  $(\Rightarrow)$ Local ring means has a unique maximal ideal, so I want to show that this implies the elements are not units.  $(\Leftarrow)$ non unit elements of $R$ form an ideal, so if I show this is unique maximal ideal I can then conclude local ring?  Any hints would be appreciated. Computing $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x$ with residue calculus How best to include other scripts? The electroweak and strong forces seem to be completely different types of forces to gravity. The latter is geometric while the former are not (as far as I'm aware!). So why should they all be described in the same way? - Why are gravitons needed?  Alternatively, I suppose my question is why a theory of everything has to be some form of quantum gravity. Is there a reason why a theory of everything requires gravity to be quantised? Is it possible for the quantum world and gravity to be described by some overarching theory which does not require gravity to be quantised? Or is there some reason why this is not the case? How can I install the most used plugins like flash player and mp3 in Ubuntu? Why do adult insects have 6 legs? Apparently, there is an advantage to having 6 legs in the insect world. What is that advantage, if anything? Why would such an advantage exist for insects, but not for other, larger land animals?  What do those middle legs do that the front and hind legs cannot, if anything? Are they really essential, or do they just generate more muscle power? The Monty Hall problem It's clear to me by inspection that if we fix $\beta = \frac{1-\mu}{\mu} \alpha$ (thereby fixing the mean) and let $\alpha \rightarrow 0$, the Beta distribution approaches a Bernoulli($\mu$) distribution.  For example:  par(mfrow = c(1, 2),     oma = c(0, 0, 1.5, 0)) xx = seq(0, 1, length.out = 1000) mus = c(.2, .7) for (ii in 1:2) {   mu = mus[ii]   matplot(xx, sapply(10^(-1:-5), function(al)      pbeta(xx, al, (1-mu)/mu * al)),     type = 'l', lty = ii,     main = paste('Mean:', mu),     ylab = 'Cumulative Probability', xlab = 'x') } title('Beta Approaches Bernoulli', outer = TRUE)     That is, if $X \sim B(\alpha, \frac{1-\mu}{\mu} \alpha)$, its CDF $F_X(x; \alpha, \mu)$, satisfies  $$ \lim_{\alpha \rightarrow 0} F_X(x; \alpha, \mu) = 1 - \mu = F_Y(x) \quad \forall x \in (0, 1)$$  Where $Y \sim$ Bernoulli($\mu$), and the convergence is non-uniform. I tried to do a more formal proof of this but couldn't make any progress, even in simple cases (like $\mu = \frac12$). The  makes reference to the $\mu = \frac12$ case several times without proof. Perhaps I'm missing something easy about doing the integral, or some fact about the Beta/incomplete Beta function. Any help would be appreciated! I have been told [Warning: I leave this because it's what I asked and allows to understand the dialogues in the comments, but Azad, whom I thank, has pointed that the formula does not hold in general in the form it is expressed] that the angular momentum of and rigid body with respect to any point $P$ can always be expressed as $$\mathbf{L}_{P}=\mathbf{r}_{cm}\times M\mathbf{v}_{cm}+\big(\sum_im_iR_i^2\big)\boldsymbol{\omega}$$ where $\mathbf{r}_{cm}$ is the position of the centre of mass with respect to $P$, $M$ the mass of the body, $R_i$ the distance of the $i$-th point, having mass $m_i$, composing the body, and $\sum_im_iR_i^2=I$ its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity $\boldsymbol{\omega}$.  I know that the velocity $\mathbf{v}_i$ of each point $P_i$, having mass $m_i$, of a rigid body of mass $M$ can be see as the sum of a translation velocity of one of its points $C$ plus a rotation velocity around that point: $\mathbf{v}_i=\mathbf{v}_{C}+\boldsymbol{\omega}\times\overrightarrow{CP_i}$. If we chose $C$ as the centre of mass I see that $$\mathbf{L}_{cm}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{i}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{cm}+\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$$$=\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i}) $$because, if I am not wrong, $\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_C=(\sum_i m_i\overrightarrow{CP_i})\times\mathbf{v}_C=\mathbf{0}$ since $\sum_i m_i\overrightarrow{CP_i}$ is the position of the centre of mass with respect to itself, which is $\mathbf{0}$.  How can it be proved that $\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=(\sum_im_iR_i^2)\boldsymbol{\omega}$? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, $\forall t\quad I(t)= I(t_0)$, the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}=I\boldsymbol{\alpha}_{cm}$ where $\boldsymbol{\alpha}$ is the angular acceleration around the centre of mass. I heartily thank you for any answer!    Some unfruitful trials: by using the "BAC CAB identity" as suggested by Azad, whom I heartily thank, $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$, I can see that$$\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=\sum_im_i\|\overrightarrow{CP_i}\|^2\boldsymbol{\omega}-m_i(\overrightarrow{CP_i}\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$which, by decomposing every $\overrightarrow{CP_i}$ into an axial component $\mathbf{A}_i$ and a radial component $\mathbf{R}_i$, whose norms respectively are $A_i$ and $R_i$, with $R_i$ as the distance from $i$ to the axis of rotation, becomes $$\sum_im_iR_i^2\boldsymbol{\omega}+\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$but I cannot prove that $\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}=\mathbf{0}$. AutoHotkey Equivalent for OS X? Is there an equivalent product/method for AutoHotkey. For those that don't know, its a product that allows you to program your mouse movements and keyboard. This allows me to "macro" certain functions on programs instead of having to do it manually. In the old days they used to call them keyboard stuffers. But that is only half of the solution as I need a "mouse movement" stuffer as well.  Is there such a beast? Is there a noun that doesn't have a plural in any of its forms? While attempting to answer  @ProgramFox has been trying to come up with an example of a noun with no plural.   The best I've come up with is egotism and Judaism.  ()  Some of the contenders I've already shot down are air, water, and money.  (Airs, waters, and monies are all legitimate plurals.)  My question is, a) Do my two examples have actual plurals?  b) Are there any other non-ism nouns that don't have a plural? How do I get 0-padded numbers in {} (brace expansion)? I'm trying to write a simple script to retrieve mem and swap usage from a list of hosts. Currently the only way I've been able to achieve this is to write 3 seperate scripts:  for a in {1..9}; do echo "bvrprdsve00$a; $(ssh -q bvrprdsve00$a "echo \$(free -m|grep Mem|/bin/awk '{print \$4}';free -m|grep Swap|/bin/awk '{print \$4}')")"; done &gt; /tmp/svemem.txt; for a in {10..99}; do echo "bvrprdsve0$a; $(ssh -q bvrprdsve0$a "echo \$(free -m|grep Mem|/bin/awk '{print \$4}';free -m|grep Swap|/bin/awk '{print \$4}')")"; done &gt;&gt; /tmp/svemem.txt; for a in {100..218}; do echo "bvrprdsve$a; $(ssh -q bvrprdsve$a "echo \$(free -m|grep Mem|/bin/awk '{print \$4}';free -m|grep Swap|/bin/awk '{print \$4}')")"; done &gt;&gt; /tmp/svemem.txt   The reason for this is that the hostname always ends in a 3 digit number and these hosts go from 001-218 so I've needed to do a different for loop for each set (001-009, 010-099, 100-218). Is there a way in which I can do this in one script instead of joining 3 together? My husband wanted me to set up a Facebook account for him and I accidentally used our shared email address, which is set against my own account.  Since then, neither of us has been able to log in, so I can't alter any settings.  Can anyone advise how to resolve this please? I am just curious that the following format looks good to you or not? The preamble I used is just   \documentclass[12pt,a4paper]{article}  \usepackage{amsmath} \usepackage{amsthm} \usepackage{amssymb}   I don't feel it's so good because the first paragraph has no indent but the second does. If the first paragraph is much longer, than it's fine. But if not, it looks strange. Should I leave it, or you have suggestions for this?