The post  shows that watch is the utility that is useful for the invoking a command at fixed interval repeatedly.  Now, I have very long commands, and have used aliases to group them logically to get a quicker output, like  $ alias c1='grep "checking for" file1.log' $ alias c2='grep "validated" file1.log' $ (echo "Checking: "`c1`) &amp;&amp; (echo "Validated: "`c2`)   The output of 3rd command is like  Checking: 100 Validated: 80   It is a long running process, I need to check the status of this process, to get the mentioned counts. But invoking the above with watch gives error      $ watch '(echo "Checking: "c1) &amp;&amp; (echo "Validated: "c2)'   c1 command not found   I can put the entire command in there and remove the aliases, but is there any other work-around to get the aliases working with watch command?  Note: Did quickly go through the man page for watch, but didn't find any reference to alias in specific.
If a run the watch command containing an alias, it will not expand the alias. I have tried both with single quote and double quotes, in fact given the following alias:  # alias ll alias ll='ls -l --color=tty'   The following command will fail  # watch ll sh: ll: command not found   Shouldn't command line expansion work in this case?
Inserting an image in plain TeX I used the OzTeX implementation of TeX, also I used to typeset my docs in plain TeX, not LaTeX for me, so I had no problem to insert an image using the OzTeX, now I'm back to TeX docs, am using the TeXWorks frontend, but I can't insert an .eps image file, it seems the \includegraphics macro its only for LaTeX, so I can't use it, any idea on how to insert an image when using plain TeX? What is the difference between Ubuntu and its derivatives? What different Ubuntu derivatives are there? And what is special with them? How to "shuffle" an array? Intersection Number of $B = Y^2 - X^3 + X$ and $F = (X^2 + Y^2)^3 - 4X^2Y^2$ using the fact $I(P,F \cap B) = ord_P^B(F) $. Why does the reputation need to be publicly displayed? Define a mathematical set General Information about Eigenvalues for an 3x3 symmetric matrix How do we find general information about the eigenvalues of an arbitrary 3x3 symmetric matrix without resorting to explicitly computing the solutions to the cubic characteristic equation?:  \begin{bmatrix}     a   &amp; b &amp; c  \\     b    &amp; d &amp; e \\     c &amp; e &amp;f \end{bmatrix}  (In this case the sum and product of the eigenvalues, and the sum of the reciprocals of the eigenvalues.) Since diagonalization preserves the trace and determinant of matrices, would it be best to just find the corresponding diagonal matrix and read off the sum and product of the eigenvalues from there? The problem is I can't think of a way to diagonalize a matrix without explicitly computing the eigenvalues. Complex analysis : If $z =re^{i\theta}$, then prove that $|e^{iz}| =e^{-r\sin\theta}$ Problem :   If $z =re^{i\theta}$, then prove that $|e^{iz} | =e^{-r\sin\theta}$   My working :   $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$   $\Rightarrow iz = ir(\cos\theta +\sin\theta) $ = $-r\sin\theta +ir\cos\theta $  $\Rightarrow e^{iz} =e^{(-r\sin\theta +ir\cos\theta)} = e^{-\sin\theta} e^{ir\cos\theta} $   $\Rightarrow |e^{iz}| = |e^{-r\sin\theta}||e^{ri\cos\theta}|$   Now please guide how to proceed further to get the result... thanks. How do I prove that $$ a^3+b^3+c^3−3abc = 1 $$ if   $ a = 1+\frac{x^3}{3!}+\frac{x^6}{6!}......$ , $b=x+\frac{x^4}{4!}+\frac{x^7}{7!}....$ , $c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}......$ ?  I have tried the following :-  I discovered that $a+b+c=e^x$ and that $a=\frac{d(b)}{dx}$, $b=\frac{d(c)}{dx}$, $c=\frac{d(a)}{dx}$ . But I am unable to proceed further with these information. Partial Sums of Geometric Series This may be a simple question, but I was slightly confused.  I was looking at the second line $S_n(x)=1-x^{n+1}/(1-x)$.  I was confused how they derived this.  I know the infinite sum of a geometric series is $1/(1-x)$.  I just can't figure out how the partial sums, $S_n(x)$, have $1-x^{n+1}$ on the numerator.  How was this derived?  Thank you.     Example 5.20.   The geometric series   $$     \sum_{n=0}^\infty x^n   = 1 + x + x^2 + x^3 + \dotsb $$   has partial sums   $$     S_n(x)   = \sum_{k=0}^n x^k   = \frac{1 - x^{n+1}}{1 - x} \cdotp $$   Thus, $S_n(x) \to 1/(1-x)$ as $n \to \infty$ if $|x| &lt; 1$ and diverges if $|x| \geq 1$, meaning that   $$     \sum_{n=0}^\infty x^n   = \frac{1}{1-x}   \qquad   \text{pointwise on $(-1,1)$}. $$   (Original image .) Where $\phi(n)$ is the Euler phi function, how do you find all $n$ such that $\phi(n)|n$? Can future execution start before the synchronous execution is completed Is there a way for @future annotated method to start execution before the synchronous request which invokes it has completed.  In other words, do I have some guarantees that the timing will look like this:  -start sync---call future---end sync---execute future-&gt;   instead of like this:  -start sync---call future---execute future---end sync-&gt;   I couldn't find the execution order guarantees in the documentation, although  states:     The reason why sObjects can’t be passed as arguments to future methods   is because the sObject might change between the time you call the   method and the time it executes. In this case, the future method will   get the old sObject values and might overwrite them.  To work with   sObjects that already exist in the database, pass the sObject ID   instead (or collection of IDs) and use the ID to perform a query for   the most up-to-date record.   This can be read as saying that the future is not queued at the end of the transaction, otherwise Salesforce could guarantee that the sObject sent is the most recent version. And since it's queued during the transaction, it might imply that it also might execute and complete while the synchronous transaction is still running.  On the other hand,  implies that futures are only fired after the transaction completes:     Invoking Database.rollback not only rolls back inserts, but it goes   much, much further.   [...]   It also rolls back:      Calls to @future methods. Any methods called within the rolled back part of your script will not be invoked (also why @future doesn't start until your script completes). Installing mysql-workbench in Ubuntu 19.04 The error looks like this:  Reading package lists... Done Building dependency tree        Reading state information... Done Package mysql-workbench is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source E: Package 'mysql-workbench' has no installation candidate   How do I install it? I'm trying to make a quicklist on the Workspace Switcher launcher that allows for quick switching to a certain workspace. (My friend sometimes uses my computer and just doesn't enjoy keyboard shortcuts, so I want to help him out.)  So is there a command one could type in a terminal that would force a certain workspace to be focused on? Yes, the plot is pretty weird. There is this "star" (whose incarnation is a woman dressed in white) and some random guy who falls in love with her. There's the arc of "if you cross a specific border you'll die" which applies to our lady "star". Well, I also remember that they have some kind of a tubular case, and when one opens it, lightning files out.  I watched the film somewhere around 2010, it's in English and when I try to remember it, I get the vibe of a mediocre Hollywood fantasy romance. Proving that $\lim_{x \to 0} \frac{f(x)}{|x|}=1$ implies $f$ is not differentiable at $0$. $$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$. I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient? Assume toward contradiction that $f$ is differentiable at $0$. Then by definition we know that $f$ is continuous at $0$. Therefore $\lim _{x\to 0}f\left(x\right)=f\left(0\right)$ We know that $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$ Therefore we can conclude $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$ But we can also rewrite this as $\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$ Therefore $f(0) = 0$. Since $f$ is differentiable at $0$ we have $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$ Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$ $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$ Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$. Therefore $f$ is not differentiable at 0. I was trying to create an AnyDice function to model the Great Weapon Fighting fighting style (which lets you reroll 1s and 2s), but I couldn't get it to work on any arbitrary dice.  I've found this one:  function: reroll R:n under N:n {    if R &lt; N { result: d12 } else {result: R} } output [reroll 1d12 under 3] named "greataxe weapon fighting"   And it works fine. But I don't know how to make the function generic so i don't need to change the d12 every time i want a different dice to reroll.  I've tried  function: reroll R:n under N:n {    if R &lt; N { result: d{1..R} } else {result: R} } output [reroll 1d12 under 3] named "greataxe weapon fighting"   but it is not giving the right probabilities. Maybe if I could fetch the die size inside the function... Modifying .sty files? Yesterday I did some changes in the layout of my TeX document. I couldn't accomplish the changes until I opened the .sty files and modified some of their code. I am sure this is not the best idea. For one reason, after the packages got an update, I will have to modify the file again (and again and again).  Another way would be to create my own version of that .sty file and call that instead of the original one, but this might not be the best idea either. (again: at least the updates speak against it)  How can I accomplish the following changes by adding code to a local file (e.g. my .tex preamble) instead of changing global files (like the package .sty files)?  I wanted to change the following lines in ledmac.sty:  \def\printlines#1|#2|#3|#4|#5|#6|#7|{\begingroup   \setprintlines{#1}{#2}{#3}{#4}{#5}{#6}%   \ifl@d@pnum #1\fullstop\fi   \ifledplinenum \linenumrep{#2}\else \symplinenum\fi     %% #2 into: \arabtxt{\n{#2}}   \ifl@d@ssub \fullstop \sublinenumrep{#3}\fi   \ifl@d@dash \endashchar\fi   \ifl@d@pnum #4\fullstop\fi   \ifl@d@elin \linenumrep{#5}\fi   \ifl@d@esl \ifl@d@elin \fullstop\fi \sublinenumrep{#6}\fi \endgroup}   The second thing is a line in ledpar.sty:  \newcommand*{\l@dlinenumR}{%   \numlabfont\linenumrepR{\line@numR}\Rlineflag%       %% numlabfont into: numlabfontR   \ifsublines@     \ifnum\subline@num&gt;\z@       \unskip\fullstop\sublinenumrepR{\subline@numR}%     \fi   \fi}   (In the latter case, just adding \renewcommand with the modified code did not work.) My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here? Peppers evolved chemicals like capsaicin to discourage animals from eating them, and that chemical itself can cause internal damage if consumed in large quantities, though this is usually only a concern for the hottest of peppers (which people do eat).   Still, why is it that humans seem to be the only animal that often "like" spicy food?