var gh = new List&lt;SomeClass&gt;&gt;();  Let's say    string SomeClass=&quot;Car&quot;;  How can I intialize this list using only name I am trying to do something like gh = new &lt;List&lt;Car&gt;&gt;();
The title is kind of obscure. What I want to know is if this is possible:  string typeName = &lt;read type name from somwhere&gt;; Type myType = Type.GetType(typeName);  MyGenericClass&lt;myType&gt; myGenericClass = new MyGenericClass&lt;myType&gt;();   Obviously, MyGenericClass is described as:  public class MyGenericClass&lt;T&gt;   Right now, the compiler complains that 'The type or namespace 'myType' could not be found." There has got to be a way to do this.
Comma with 'who' in this specific sentence Can you shed light into this?  The coach came to the defence of the player who has been sacked from the team following his differences with the captain.  OR  The coach came to the defence of the player, who has been sacked from the team following his differences with the captain.  Isn't the subordinate clause that begins with 'who', non-restrictive? I was editing my mesh and I wanted to only see a part of my mesh, so I pressed alt+B to only show a selected area. However, I did ctrl+B and created a red rectangle on my screen (I don't know what it does). Does anybody know how to get it off? Could the universe's antimatter be hiding in black holes? As far as we know almost all the mass in the universe is matter, not antimatter. There are three scenarios:   The universe started off with more matter than antimatter and the small amount of antimatter has mostly been converted to energy via collisions with matter The universe started off with equal amounts of matter and antimatter and some unknown process caused the situation we see today The universe started off with less matter than antimatter and somehow the ratio reversed (I've never heard this discussed as it seems so unlikely)   We reason that mass is matter because if there were large amounts of antimatter they should produce antimatter galaxies, which would give off huge amounts of energy when they interact with matter galaxies. This has never been observed so we conclude that all galaxies, therefore all mass in the universe is matter.  As I understand it there are two main types of black holes, stellar remnants (the collapsed remnants of dead stars with more than two solar masses) and super-massive black holes at the centre of many galaxies, including one thought to be at the centre of our own galaxy.  Stellar black holes are obviously originally composed of matter, for the same reason that all galaxies are composed of matter.  We don't know how super-massive black holes formed. Is there enough mass in them to contain all the antimatter in the universe if matter and antimatter were originally created in equal quantities? I realise this shifts the unknown mechanism from "why was more matter created than antimatter" to "why is antimatter more likely to form super-massive black holes" but is this hypothetical scenario possible? Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}), \forall a,b\in\mathbb{Q}$ Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then   $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}),  \forall a,b\in\mathbb{Q}$   It has something to do with $\sqrt{a} = -\sqrt{b}\implies a^2 = b^2$  The inclusion $\mathbb{Q}(\sqrt{a}+\sqrt{b})\subseteq \mathbb{Q}(\sqrt{a},\sqrt{b})$ is obvious  Now take $x\in \mathbb{Q}(\sqrt{a},\sqrt{b})$, that is: $x = c + d\sqrt{a} + e\sqrt{b} + f\sqrt{ab}$. I need to show that this is in $\mathbb{Q}(\sqrt{a}+\sqrt{b})$, that is, $x = g + h\sqrt{a} + i\sqrt{b}$, but I can't get that $\sqrt{ab}$ removed I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.  Now he showed me an exercise he was having trouble with, which states that one should prove $$n^{n+1} &gt; (n+1)^n$$  for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.  Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?  Thanks for answers in advance. Whats the difference between Unix, Linux and Ubuntu? What is Unix, Linux, Ubuntu?  Are they the same thing or different versions or completely different stuff?  This has always made me confused, any explanation would be great. This is an exercise in some textbooks.     Let $E$ be an algebraic extension of $F$. Suppose $R$ is ring that contains $F$ and is contained in $E$. Prove that $R$ is a field.   The trouble is really with the inverse of $r$, where $r\in R$. How to prove that $r^{-1}\in R$, in apparent lack of a characterization of $R$.  It occurred to me to use the smallest field containing $R$ ($R$ is easily shown to be an integral domain), that's the field of quotients, and proving that it's $R$ itself, but I don't really know how to proceed.  A not-too-weak, not-too-strong hint will be much appreciated.   Beware $ $ Readers seeking only hints should  beware that there is now a complete answer. What is the correct pronunciation of the word "solder"? I have been listening to a podcast where the host pronounces the word solder as "sodder" or "sod-der", even "saw-der". Same thing happened when the lecturer of one of my EE classes pronounced the word solder. As for me I pronounce the word solder as "sole-der" with a distinct L and a longish O. I'm trying to define \conjunction by means of \textbullet instead of \bullet, since the latter looks too big in logical formulas.  How can I make the spacing around \conjunction to behave like the spacing around \bullet when there is an opening parenthesis just after it?  Compare (2) and (2'):     \documentclass[12pt]{article} \usepackage[paperwidth=10cm,paperheight=7cm,margin=1em]{geometry} \usepackage{amsmath} \usepackage{amssymb} \usepackage{ebgaramond-maths}  \DeclareMathOperator\conjunction{\raisebox{-0.25ex}{\textbullet}}  \begin{document}\begin{align} \intertext {Good spacing but bad size:} (\exists x(\forall y(\phi(y)\equiv y\!=\!x)&amp;\bullet Gx)\supset\psi)\\ \exists  x(\forall y(\phi(y)\equiv y\!=\!x)&amp;\bullet (Gx\supset\psi))\\ \intertext {Good size and spacing:} (\exists x(\forall y(\phi(y)\equiv y\!=\!x)&amp;\conjunction Gx)\supset\psi)\\ \intertext {Good size but bad spacing:} \exists  x(\forall y(\phi(y)\equiv y\!=\!x)&amp;\conjunction (Gx\supset\psi)) \end{align}\end{document}   This won't do:  \DeclareMathOperator\conjunction{\raisebox{-0.25ex}{\textbullet\ }}   because it adds an unwanted extra space to (1'). Should one create slides similar to those that one uses in a good powerpoint presentation? Or are there things that a poster should include that a powerpoint should not include? (and vice versa) How do I go about porting Ubuntu Touch to different devices? I would like to port Ubuntu Touch to a device that is currently not supported. What are the requirements for doing this and where do I begin? My keyboard is typing by itself, could there be a ghost in it? Add links to official status pages in the read-only mode status message When the site is in read-only mode, I get a message like this:     This site is currently in read-only mode.   Would it be possible to update that message to provide links to these resources on SE's status:   : Official SE tweets on SE's status. : Official blog posts on SE's status. If you are reading this, you may have been affected by one of our post quality bans. If so, it is important that you  to understand what has happened and how to regain the permissions that you lost. While trying to ask a question, one could see:  We are no longer accepting questions from this account. See  to learn more.  Likewise, for answers:  We are no longer accepting answers from this account. See  to learn more.   Why am I getting this message? Are deleted posts taken into account too? Is a question/answer ban the same as a suspension? How do I avoid getting a question ban? How long do I have to wait before I can post again? What can I do to release the ban? How can I reactivate my account? Can I simply create a new account? I'll just ask somewhere else on the SE network, and they'll migrate my question to the correct site! Will a ban on one Stack Exchange site affect my standing on other sites in the network? Does this apply to meta sites too? My account is in good standing. Why am I still blocked? Salesforce lightning get current users username How I would go about getting the current users username or first and last name?  I have been looking at this for the last couple of hours - I was just wondering how I would go about getting their name and then displaying it on the page.  For example the current user is 'John Smith' and in the lightning component I want to output John Smith's name like:     Welcome to our community John Smith! I have an Optimus Elite, and it has a minimum brightness setting. In certain apps, I can reduce the brightness below the safe minimum, but when I leave the app, the brightness is increased to the system-wide minimum.  Using , I can permanently set the brightness below the safe amount, and it will stay even after going to other apps. Or using , I can accomplish almost the same thing even without needing root (it displays a translucent 'filter' over the whole screen, reducing brightness/contrast but not actually changing the backlighting power amount).  What I want to know is where the minimum screen brightness is stored. I have a rooted phone with CyanogenMod 9, so I have permissions to edit anything and could change a config or sqlite entry, but I don't know where this value is stored. What I would like to do is use the status bar brightness slider to be able to slide to the actual minimum of the screen, rather than the OS-controlled minimum. Having an app is useful, but having the functionality without the app seems to be within reach, if only I knew where the OS is getting it's safe minimum setting.  Where is this "safe minimum brightness" setting stored? If I can change it, I can remove my apps and use native controls. BASH script to set environment variables not working I've written the following script to set some environment variables when needed.  #!/bin/sh export BASE=/home/develop/trees echo $BASE export PATH=$PATH:$BASE echo $PATH   Below the command and the results I can see on my terminal: the script runs, but the variables are not set at the end.  ~$: ./script.sh /home/develop/trees /bin:......:/home/develop/trees ~$: echo $BASE  ~$:    What's wrong? Thanks in advance. Mirko skype 4.3 cannot connect to account on ubuntu 14.04 LTS I have being using Skype for a long time on my Ubuntu 14.04 LTS. From this morning on, Skype simple cannot connect to my account anymore.  That is my Skype version:     Skype 4.3.0.37   © 2014 Skype and/or Microsoft   My Ubuntu system:  LSB Version:    core-2.0-amd64:core-2.0-noarch:core-3.0-amd64:core-3.0-noarch:core-3.1-amd64:core-3.1-noarch:core-3.2-amd64:core-3.2-noarch:core-4.0-amd64:core-4.0-noarch:core-4.1-amd64:core-4.1-noarch:cxx-3.0-amd64:cxx-3.0-noarch:cxx-3.1-amd64:cxx-3.1-noarch:cxx-3.2-amd64:cxx-3.2-noarch:cxx-4.0-amd64:cxx-4.0-noarch:cxx-4.1-amd64:cxx-4.1-noarch:desktop-3.1-amd64:desktop-3.1-noarch:desktop-3.2-amd64:desktop-3.2-noarch:desktop-4.0-amd64:desktop-4.0-noarch:desktop-4.1-amd64:desktop-4.1-noarch:graphics-2.0-amd64:graphics-2.0-noarch:graphics-3.0-amd64:graphics-3.0-noarch:graphics-3.1-amd64:graphics-3.1-noarch:graphics-3.2-amd64:graphics-3.2-noarch:graphics-4.0-amd64:graphics-4.0-noarch:graphics-4.1-amd64:graphics-4.1-noarch:languages-3.2-amd64:languages-3.2-noarch:languages-4.0-amd64:languages-4.0-noarch:languages-4.1-amd64:languages-4.1-noarch:multimedia-3.2-amd64:multimedia-3.2-noarch:multimedia-4.0-amd64:multimedia-4.0-noarch:multimedia-4.1-amd64:multimedia-4.1-noarch:printing-3.2-amd64:printing-3.2-noarch:printing-4.0-amd64:printing-4.0-noarch:printing-4.1-amd64:printing-4.1-noarch:qt4-3.1-amd64:qt4-3.1-noarch:security-4.0-amd64:security-4.0-noarch:security-4.1-amd64:security-4.1-noarch Distributor ID: Ubuntu Description:    Ubuntu 14.04.3 LTS Release:    14.04 Codename:   trusty   Any suggestion why or how to fix it?  Thanks in advance. How to force Windows 7 to ask for a "fresh" IP address from DHCP server? I'm troubleshooting a certain issue with my DHCP configuration, and need my Windows machine to ask for a "fresh" IP address, so I can see which address DHCP server gives by default.   When I do ipconfig /release followed by ipconfig /renew, Windows "proposes" its old IP address to the DHCP server (just checked with Wireshark, initial "DHCP Discover" message has Option-50 (requested IP address) with Windows machine's old IP).  Tried disabling/enabling network adapter. Same behavior.  Question: How can I force Windows to just ask for a new IP address, without proposing its old IP address. Calculation of $\int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx$, where $n\in \mathbb{N}$ Compute the definite integral      $$ \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx $$      where $n\in \mathbb{N}$.   My Attempt:  Using $\cos (x) = \frac{e^{ix}+e^{-ix}}{2}$, we get  $$ \begin{align} \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx&amp;=\int_{0}^{\pi/2} \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\left(\frac{e^{inx}+e^{-inx}}{2}\right)\,dx\\ &amp;= \frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^n\cdot e^{inx}\,dx\right\}\\ &amp;=\frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2}\left(e^{2ix}+1\right)^n\,dx\right\} \end{align} $$  Letting $z=e^{4ix}$ gives us  $$ \begin{align} e^{4ix}dx &amp;= \frac{1}{4i}dz\\ dx &amp;= \frac{dz}{4iz} \end{align} $$  So the integral becomes  $$\frac{1}{4\cdot 2^n}\mathrm{Re}\left\{\int_{C}\left(\sqrt{z}+1\right)\cdot \frac{dz}{iz}\right\}$$  How can I complete the solution from here?