I've tried many things but cron just won't restart pm2. I set the crontab -e with: SHELL=/bin/sh PATH=/bin:/sbin:/usr/bin:/usr/sbin  */10 * * * *  /usr/bin/node /usr/bin/pm2 restart all  And the I type: cat /var/log/cron.log  And it says: Apr  2 01:00:01 fatsecret CRON[8202]: (peteblank) CMD (/usr/bin/node /usr/bin/pm2 restart all) Apr  2 01:00:01 fatsecret CRON[8201]: (peteblank) MAIL (mailed 78 bytes of output but got status 0x004b from MTA#012) Apr  2 01:10:01 fatsecret CRON[8454]: (peteblank) CMD (/usr/bin/node /usr/bin/pm2 restart all) Apr  2 01:10:01 fatsecret CRON[8453]: (peteblank) MAIL (mailed 78 bytes of output but got status 0x004b from MTA#012)  So it looks like its restarting every 10 minutes, but its not.
This is a  about using cron &amp; crontab.   You have been directed here because the community is fairly sure that the answer to your question can be found below. If your question is not answered below then the answers will help you gather information that will help the community help you. This information should be edited into your original question.  The answer for 'Why is my crontab not working, and how can I troubleshoot it?' can be seen below.  This addresses the cron system with the crontab highlighted.
How to integrate $\int\sqrt{x^2+1}dx$ $$\int\sqrt{x^2+1}dx$$ I've been attempting this problem for days and I've made very little progress. I understand that I should substitute $x=\tan(x)$ and $dx=\sec^2(x)dx$ but beyond that I am very much lost. Changing top, bottom, left & right margins on the fly I’m trying to find a way of being able to modify all of the margins of a page whenever I want to and as many times as I want to on a single page in order to be able to put text very easily where I want on a page.   I have thought of using the geometry package and using   \newgeometry{left=6.75cm, right=1cm, top=1cm, bottom=1cm}  %…  \restoregeometry   but unfortunately this only works for whole pages, so I can only change the margins once per page (and then they are valid for the whole page without having the possibility of changing them again).   I found the following script on the internet which enables users to change margins on the fly, just as I want to. The code below works fine and enables me to change the left and right margins of my page as many times as I want to, but unfortunately, it doesn’t give me the option of changing the top and bottom margins.  % placed before \begin{document}  \newenvironment{changemargin}[2]{% \begin{list}{}{% \setlength{\leftmargin}{#1}% \setlength{\rightmargin}{#2}% }% \item[]} {\end{list}}  % And then inside the document, whenever I want to use % the command to change the margins on the current page  \begin{changemargin}{1.9875cm}{-3.7625cm} % this means the left margin increases by 1.9875cm compared % to the default left margin (4.7625cm) (so left margin=6.75cm here) % and the right margin decreases by 3.7625cm (so right margin=1cm)  % ... (text or pictures in new environment)  \end{changemargin}   For this reason, I tried to adapt the code above and came up with the code below:  % placed before \begin{document} \newenvironment{changemargin}[3]{% \begin{list}{}{% \setlength{\leftmargin}{#1}% \setlength{\rightmargin}{#2}% \setlength{\topmargin}{#3}% }% \item[]} {\end{list}}  % And then to use the command: \begin{changemargin}{1.9875cm}{-3.7625cm}{1cm} % ... \end{changemargin}   It compiles and still changes the left and right margins but the top margin remains unchanged. (I haven’t yet tried to change the bottom margin).  Can anyone help me get this code to work completely, and enable me to modify the left, right, top and bottom margins of a page? Thanks. Can regular expressions be used to match nested patterns? Is it possible to write a regular expression that matches a nested pattern that occurs an unknown number of times? For example, can a regular expression match an opening and closing brace when there are an unknown number of open/close braces nested within the outer braces?  For example:  public MyMethod() {   if (test)   {     // More { }   }    // More { } } // End   Should match:  {   if (test)   {     // More { }   }    // More { } } Why was Enterprise on a five-year mission in TOS? There is a small difference between intro texts in TOS:     Space: the final frontier. These are the voyages of the starship Enterprise. Its five-year mission: to explore strange new worlds, to seek out new life and new civilizations, to boldly go where no man has gone before.   and in TNG:     Space: the final frontier. These are the voyages of the starship Enterprise. Its continuing mission: to explore strange new worlds, to seek out new life and new civilizations, to boldly go where no one has gone before.   (emphasis mine)  It's pretty obvious that no man was changed to no one to include women. But, is there any explanation why was five-year changed to continuing? Also, why was it a five-year mission at all?  My only assumptions are:   Gene Roddenberry hoped to get funding to run the show for 5 seasons. Gene Roddenberry just made up a number without any special meaning. Gene Roddenberry was a .   Is there any official explanation, in universe or not? Fingerprint GUI not installing in Ubuntu 20.04 I'm trying to install fingerprint-gui on my Lenovo Ideapad 520 having Ubuntu 20.04 on which fingerprint isn't working for now. I tried going directly to Settings → Users → Automatic login and enabled the fingerprint login, but the error is:  Could not access any fingerprint readers Please contact your system administrator for help!  I also tried to install Fingerprint GUI: sudo apt-add-repository ppa:fingerprint/fingerprint-gui &amp;&amp; sudo apt-get update  sudo apt-get install libbsapi policykit-1-fingerprint-gui fingerprint-gui  but encountered this error: E: Package 'policykit-1-fingerprint-gui' has no installation candidate E: Package 'fingerprint-gui' has no installation candidate  What could be the problem here? A property of every real polynomial. Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$  Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.   Is this property (if true) a famous one?  Thank you for your help. Subsample of a random sample: random sample? Let's say you have a large random sample of soccer players in Europe but you are only interested in what happens in Spain. Could you reduce your sample to players in Spain and still call it a random sample (but of a different population)? If not, how would you call that subsample and which specific precautions should you take to be able to make inferences on the population of spanish soccer players?  My feeling is that using that subsample would be fine as long as it is large enough, but maybe I am missing something. I know blender can display both orthographic and perspective projection. But, when we look at a cube in blender's viewport, I'm unable to distinguish by eye, if the projection in the viewport is a one-point, a two-point or a three-point perspective. What does blender use, and is this something changeable? What is a combo update and how is it different than the regular update? On proving that a finitely generated group has a finite number of subgroups with index $n$. Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.   is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$  If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.  The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$. Read from location on console C# How to install Canon LBP2900b drivers? How to get the size of a JavaScript object? Can I create a Controller that simply returns an image asset?  I would like to route this logic through a controller, whenever a URL such as the following is requested:  www.mywebsite.com/resource/image/topbanner   The controller will look up topbanner.png and send that image directly back to the client.  I've seen examples of this where you have to create a View - I don't want to use a View. I want to do it all with just the Controller.  Is this possible? Topologist's sine curve is connected I just came across the example of the topologist's sine curve that is connected but not path-connected. The rigorous proof of the non-path-connectedness can be found .   But how can I prove that the curve is connected? To be honest, even intuitively I am not being able to see that the curve is connected. I am thinking if it is proved that the limit point of $\sin(1/x)$ as $x \to 0=0$, then it would be proved. But, why is this true? IMO, this limit doesn't exist. Intuitively also, it seems that the graph would behave crazily and not approach a particular value as a tends to $0.$  EDIT (Brett Frankel): There are a few different working definitions of the topologist's since curve. For the sake of clarity/consistency, I have copied below the definition used in the linked post: $$ y(x) = \begin{cases} \sin\left(\frac{1}{x}\right) &amp; \mbox{if $0\lt x \lt 1$,}\\\ 0 &amp; \mbox{if $x=0$,}\end{cases}$$ How to avoid bad box with long words with no hyphenation? Black screen after installation of Ubuntu 18.04 Interval preserving transformations are linear in special relativity In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:  Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,x^3\right)$ that preserves intervals, that is, such that  $$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$  is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).  Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas? Number of permutations of $n$ elements where no number $i$ is in position $i$ I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, $3,1,5,2,4$ is an acceptable permutation where $3,1,2,4,5$ is not because 5 is in position 5.  I know that the number of total permutations is $n!$. Is there a formula for how many are acceptable given the case that no position holds its own number? If $A$ is a non-square matrix with orthonormal columns, what is $A^+$? If a matrix has orthonormal columns, they must be linearly independent, so $A^+ = (A^T A)^{−1} A^T$ . Also, the fact that its columns are orthonormal gives $A^T A = I$. Therefore,  $$A^+ = (A^T A)^{−1} A^T = (I)^{-1}A^T = A^T$$  Thus, $A^+ = A^T$. Am I correct? Thank you.