Is there command in Ubuntu that is equivalent on restarting explorer.exe in Windows?
How do I just restart the nautilus after adding a nautilus script without having to log out.
How to make a process invisible to other users? How could you launch a process and make it invisible to the top command? The process is started by a normal user (not root), and should not be visible to other normal users. Let $C$ be the class of all real-valued continuously differentiable functions $f$ on the interval $[0,1]$ with $f(0)=0$ and $f(1)=1$. How to show that $$\inf_{f\in C}\int_0^1|f'(x)-f(x)|dx=\frac{1}{e}?$$  I have been able to show that $1/e$ is a lower bound. Indeed, $$\begin{align*} \int_0^1|f'(x)-f(x)|dx &amp;= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\ &amp;\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx \\ &amp;= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx  \\ &amp;= f(1)e^{-1}-f(0)e^{0}\\ &amp;=\frac{1}{e}. \end{align*}$$ But how to show this is the infimum? Is there a function $f\in C$ such that we get $\int_0^1|f'(x)-f(x)|dx=1/e$? 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 do I install drivers for the Atheros AR8161 Ethernet controller? Is there a streaming API for JSON? Is DOM the only way to parse JSON? Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)&gt;0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$. I thought about taking the bounded set $H\cap D(0,R), R&gt;0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$. How to disable or uninstall Microsoft Edge in Windows 10 Proving that cardinality of the reals = cardinality of $[0,1]$ Show there are infinitely many primes that are equivalent to 1 mod 8. Show there are infinitely many primes that are equivalent to $1 \pmod{8}$.  Hello there! I have been trying to do this problem for a pretty long time with no avail.    I noticed that this is really similar to Euclid's proof that there are infinitely many primes.  However, I can't find a way to use that here.  Can anyone help me? How can I check if my (unlocked) phone will work with a particular country's SIM cards, for 3G and calls? I've used my phone overseas with local SIM cards a lot, but recently had a surprising experience where one local SIM card simply wouldn't work with my phone but did work first time on my partner's, even though both were similar (unlocked Android phones) and another SIM card from a different local provider did work. Looking into it, it looks like there are various different standards around, but it all gets very technical very quickly.  Is there any simple way of looking up which SIM cards in a particular country will work with a particular model of phone, for calls, texts and internet access? Some icons are correct, and some are whack. Turns out it's not just the hot questions list, but the whole icon set. Avoiding Windows 10 Creator, but not other updates Command "find -name" must be enclosed in quotes or it doesn't work. Why is that? find . -name *.rb   doesn't work on my ubuntu  I have to do  find . -name "*.rb"   to get it to work.  Why is that? Launcher Icons disappear Some Launcher icons disappear. I locked them but they have disappeared again after boot.  I have Ubuntu 12.04.2 LTS which I have installed today and updated. I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! Write a number with two decimal places SQL Server How do I troubleshoot a Windows freeze or slowness? A machine is described as being "really slow" or "freezing". How can I troubleshoot/resolve this issue? What should be the first steps to take? Energy conservation of Virtual Particles - Quantum Fluctuation? I (as a middle-school student) was wondering how virtual particles even conserve energy of the entire system? I don't mean just the particle's energy, but conservation with respect to the surroundings?  Let me explain more, Einstein proved that $E = mc^2$ and equivalently, we can say $m = \frac{E}{c^2}$ . We however know mass causes curvature in space-time (gravity well). Now that statement being completely valid we can use Quantum Fluctuation, implying energy and time can be related by the relation:  $\Delta{E}\Delta{t}\approx\frac{\hbar}{2}$ and by that simple relation we can make the following points:   As the energy increases, the time allowed to be "borrowed" decreases, before it gives it back to the universe. As time increases, the energy allowed to be "borrowed" decreases.    These are completely satisfactory points I've made, I believe.  Now, I have read that the vacuum is filled with constantly bubbling virtual particles that get created from the energy "borrowed" by the above relation and is destroyed again, which satisfies the energy conservation laws.   This seems to imply that  in the short time the virtual particle exists, they must have a very small mass which is: $m_{1} = \frac{\Delta{E}}{c^2}$ and using basic classical mechanics rather than GR (for sake of simplicity) we calculate that the gravitational force mediated by this short-lived virtual particle;  $F_{g}= G\frac{m_1 * m_x}{r^2}$   where $m_x$ is an particle\mass in question. Nevertheless since we know the force set upon an object moves the object in a direction, we can say it would increase its kinetic energy by this factor: $F_g = \frac{1}{2}mv^2$ which is equal to $G\frac{m_1 * m_x}{r^2} = \frac{1}{2}mv^2$.  Surely, this implies that every time quantum fluctuations happen, energy of real-particles\masses increases and therefore violates energy conservation, but what prevents this violation from taking place?  If virtual particles are not really present and interact with objects like I thought, how can Hawking radiation (currently a large concept in black hole physics) then be accepted in physics community? Why is my attempt at downloading Big Sur failing? I am currently using a MacBook Pro 13&quot; Mid 2018 with Touch Bar. Today macOS Big Sur was released and I tried to download it but it's getting stuck at 484.4 MB. Initially, it shows 16-17 minutes left, but as soon as it reaches 484.4 MB it's stuck and the timer goes till 10 hours! My Wi-Fi is totally stable with a speed of 100 Mbps. I tried to restart my Mac but it still gets stuck. Earlier there were similar issues in the App Store (even now they are). The issue was I was not able to download an app unless I connect to the US VPN. I even contacted support but they were not able to help out. It wasn't a big deal as a VPN could be used. But now even with VPN, it gets stuck. How to map stock ticker symbols to ISIN (International Securities Identification Number)? How can I can convert ticker symbols (which include information on the exchange) to an International Securities Identification Number (ISIN) (which contains no information on the exchange)? Also I am interested in the other way round — given an ISIN and a market identification code (MIC) how can I determine the ticker symbol? Any databases around?