I want to start Zookeeper daemon after Ubuntu server 16.04 booting (not after logging) under user named zookeeper. So I changed the file /etc/rc.local like the following:  #!/bin/sh -e # # rc.local # # This script is executed at the end of each multiuser runlevel. # Make sure that the script will "exit 0" on success or any other # value on error. # # In order to enable or disable this script just change the execution # bits. # # By default this script does nothing.  echo 'never'; defrag_file_pathname  su -c '$ZOOKEEPER_HOME/bin/zkServer.sh start' zookeeper &amp;  exit 0   , adding the line su -c '$ZOOKEEPER_HOME/bin/zkServer.sh start' zookeeper &amp; before exit 0. But the process is not started after restarting!  What's wrong here?  details: The zookeeper user is in sudo group and has password.  details: When I run command su -c '$ZOOKEEPER_HOME/bin/zkServer.sh start' zookeeper &amp; in terminal, it needs password to run.
I would like to run a script from the main ubuntu shell as a different user that has no password.  I have full sudo privileges, so I tried this:  sudo su -c "Your command right here" -s /bin/sh otheruser   Then I have to enter my password, but I am not sure if that script is now really running under that user.  How can I confirm that the script is really running under that user now?
Exponent and abelian groups Let $S = \{ax + by: x,y \in \mathbb{Z}\}$  and let $e &gt; 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$  I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.  I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this. Search favorite questions by tag Favorite questions on Stack Overflow I feel is highly unused feature of all the features provided on the site. (This again is my personal opinion)  I have favorited around 90 question till date and it gets very very very difficult if I want to search for a question which I remember I have favorited, which defeats the entire purpose of one favouriting a question, so here is my small suggestion.  Favorites by tags  Screen 1        Screen 2        Please share your opinions on this. Why are Grothendieck's and Hartshorne's definitions of quasi-coherence equivalent? Hartshorne's Algebraic Geometry defines an $\mathcal O_X$-module $\mathscr F$ to be quasi-coherent if there is an open affine cover $(U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}$ of $X$ such that each $\mathscr F\rvert_{U_i}$ is isomorphic to the sheaf of modules $\widetilde{M_i}$ associated to some $A_i$-module $M_i$. It is later proved that this is equivalent to the statement that for any open affine $U=\operatorname{Spec} A\subseteq X$, there exists an $A$-module $M$ with $\mathscr F\rvert_U\cong\widetilde M$.  On the other hand, Grothendieck's Éléments de Géométrie Algébrique defines $\mathscr F$ to be quasi-coherent if every point $x\in X$ admits an open neighborhood $U\subseteq X$ such that $\mathscr F\rvert_U$ is isomorphic to the cokernel of a morphism of free $\mathcal O_U$-modules.  Now I wonder how those definitions are equal. I tried proving it by myself, but both directions seem quite nontrivial to me... Any hints? Example of a Subgroup That Is Not Normal Can you kindly provide an example of a subgroup that is not normal?  I have been told many times that, for coset multiplication to be defined, the subgroup must be normal.  I have seen the proof and examples of quotient group multiplications.  Now, I am trying to find out where the process will break down if the subgroup is not normal.  Suppose that $x \in a_1H = a_2H, y \in b_1H = b_2H$, where $H$ is not normal. So $x = a_1h_1 = a_2h_2$ for some $h_1, h_2 \in H$, and $y = b_1h_1^* = b_2h_2^*$ for some $h_1, h_2 \in H$.  $xy = (a_1b_1)(h_1h_1^*) = (a_2b_2)(h_2h_2^*)$.  However, both $h_1h_1^*, h_2h_2^*$ are still elements of $H$.  So I don't see how $a_1b_1H \neq a_2b_2H$. Cancelled cut and paste, what happens? Is there a recursive version of the dict.get() built-in? How do I report a bug? From official :     Due to a head injury, Cyclops is unable to shut off his optic blasts   at will and must therefore wear ruby quartz lenses to block the beams.   From the same page:     Cyclops possesses the mutant ability to project a beam of heatless   ruby-colored concussive force from his eyes, which act as   inter-dimensional apertures between this universe and another.   When his eyes are always blocked by ruby-colored beam of concussive force, how does he see? How can I take a site offline using Drush? I accidently ran  sudo chmod 755 -R /   instead of  sudo chmod 755 -R ./   I stopped it after few seconds, but now there is some problems such as  sudo: must be setuid root   How can I revert permissions back? Solving "mv: Argument list too long"? I have a folder with more than a million files that needs sorting, but I can't really do anything because mv outputs this message all the time  -bash: /bin/mv: Argument list too long   I'm using this command to move extension-less files:  mv -- !(*.jpg|*.png|*.bmp) targetdir/ Save Illustrator as PDF with transparent background What is JavaScript's highest integer value that a number can go to without losing precision? Is this defined by the language? Is there a defined maximum? Is it different in different browsers? If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$ I'm trying to prove the following:     If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n&lt;\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2&lt;\infty$, then $\sum_{n=1}^\infty a_n^2 &lt;\infty$.   The context here is functional analysis homework, in the subject of Hilbert spaces.  Here's what I've thought:  Let $f=(a_n)&gt;0$. Then the problem reads: if $\int f\overline{g}&lt;\infty$ for all $g&gt;0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.  I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle &lt;\infty$ for all $g&gt;0,g\in \ell^2$ then $f\in \ell^2$.  I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P  From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$  but it's not useful here. How do I avoid the "S to Skip" message on boot? Tourist to F1 change of status within 60 days I have a PhD place at a US university starting Aug 21 but want to visit a different university for casual research for 6 weeks from July 1. This is longer than the 30 day grace period for the F1. Can I arrive on the visa waiver program or tourist visa in July and then just stay on for the PhD? I guess I might need to change status. Will this work?  Please don’t suggest exiting and reentering, that’s too expensive.  (British masters student in UK) How to check if a string "StartsWith" another string? Is there a way to force a line break inside a math mode text?  My example:  \[   \underbrace{....}_\text{Some long text that should be multiline} \]   Trying  \text{Some long text that\\ should be multiline}   Didn't do the trick. Is there a third Holmes sibling?