I use memoir class and, in some situations, I use the \sidefootnote command. It works but not the right justification. I've tried with \newcommand*{\sidefootform}{\rightskip=\z@ \@plus 2em} but I've got an error message (command already in use). I've tried with \renewcommand*{\sidefootform}{\rightskip=\z@ \@plus 2em} but error message. I've tried with:  \usepackage{ragged2e} \newcommand*{\sidefootform}{\RaggedRight}   but no effect at all. Any suggestion? Thanks PM
I would like to set up memoir so that when I use \footnote{foo}, foo is displayed in the margin with the proper label and preferably break over pages.  There is an option \footnotesinmargin which gives almost exactly what I need, only justified. I want it to use ragged2e and change alignment based on even/odd page so that the margin note is always flushed towards the main text area and allows hyphenation on the side of the edge.  \documentclass[12pt, draft]{memoir} \footnotesinmargin \usepackage{polyglossia} \setmainlanguage{english} \usepackage{fontspec} \defaultfontfeatures{Ligatures=TeX} \setmainfont[Ligatures=TeX,Numbers=OldStyle]{Minion Pro} \usepackage{kantlipsum}  \begin{document} \kant[1]\footnote{\kant[1]} \kant[2]\footnote{\kant[2]} \kant[3-5] \end{document}   Possible partial work-around  I have remembered a nice thesis based upon the Classic thesis template and found the  which indeed contains a possible solution:  \usepackage{marginnote} \usepackage{ragged2e} \newcommand{\sidemark}[1]{#1.{\:}}  \newcommand{\marginelement}[2][0]{%   \marginnote{%     \strictpagechecktrue     \checkoddpage     \ifoddpage%       \RaggedRight\footnotesize%     \else%       \RaggedLeft\footnotesize%     \fi%     #2%   }[#1\onelineskip]% }%  \newcommand{\sidenote}[2][0]{%   \footnotemark%   \ignorespaces%   \marginelement[#1]{%     \sidemark{\thefootnote}%     \ignorespaces#2%   }%   \unskip% }%   Nevertheless, I would still prefer a memoir-specific solution with the re-definition of footnotes when used with the option \footnotesinmargin mainly for the sake of portability. This work-around also does not work with hyperref (there is no hypertext connection) and does not break over pages automatically.
How can I write a unit test to determine whether an object can be garbage collected? Problem:  Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow  rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS} $ is the Hilbert Schmidt Norm.  Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.  Proof of the implication $\Leftarrow $:   If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$ If $rank(A)=1$, than: $ A=\begin{pmatrix} A_{1}\\  \alpha _{2}A_{1}\\  ...\\  \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$   and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$  Proof of the implication $\Rightarrow $   We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$  On the other hand: $ A=\begin{pmatrix} A_{1}\\  A_{2}\\  ... \\ A_{n}  \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\  \left \langle A_{2},x \right \rangle\\  ...\\   \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} $$  Where $ \left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$  Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} $ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$  Please let me know if my solution makes sense. why are the best questions the ones which have been closed? I frequently come across a really useful answer on StackOverflow which has been closed because it doesn't match the Q&amp;A format.  Examples include  ,  ,   ,  and there are others.  Where should this information be exchanged if it's not on stackoverflow?  It seems the moderators are often somewhat overzealous...  ---edit  When I say best, I mean they are informative and useful for someone who's looking for an answer to them.  Even if they do become a bit of a discussion, I value other people's opinions on the relative merits of the different JPA frameworks.  I come here looking to choose one and the answer helps me.  So what if it's not a definitive answer?  ---edit  I don't care how popular a question is, nor how interesting it is.  Users are interested in stuff that's useful to them.  Plenty of those were useful (as judged by the upvotes on the answers: 'This answer is useful'), but closed because they didn't have a definitive answer, as far as I can tell.  ---edit  Seems I'm not the first person to think this: there are various similar questions to mine like this one:   In it the user lists lots of other useful questions which have been closed:                    So isn't there a place on stackoverflow for these questions?  A discussion section, maybe? How can I prevent data theft from a company?  Are there any solutions to this problem?  Is there any software that encrypts important data (.doc, .xls, etc.) and prevents it from copying?  I need a solution where employees can read/write documents but can not copy then to a USB drive or e-mail them. Beamer consists of multiple frame layouts How to build Blender as a Python module? I have seen this question now many times and every question here and on other forum points to  experimental tutorial by . Now sadly a lot of answers are invalid right now because this central link is down. I couldn't find anything by Googling so if anybody has a fork of the tutorial or a second source that would be great? The link can't be gone for too long, I saw an answer with it 4 months ago.  The only other thing I have found is , but it also references the now missing tutorial. Any ideas how to find it? Uncountable sets of Hausdorff dimension zero Let $A \subset \mathbb{R}$ be a countable set.  It is easy to see that $A$ has Hausdorff dimension $\dim_H(A) = 0$.  Do there exist uncountable sets $A \subset \mathbb{R}$ with $\dim_H(A) = 0$? I always know that the natural logarithm $\log_e$ is the inverse of the exponential function $e$, but to my surprise when reading elementary functions in complex analysis, I discover that that is not always true. $$\log(e^z)=z+2k\pi i$$  My question here is that, why is the above expression true? Where and how does the Bleach anime diverge from the manga? I've seen almost all of the Bleach anime (at least what's been released dubbed), and I'm now interested in reading the manga. However, when the manga closely follows the anime I've already seen, I tend to get bored quickly. Thus, I'd want to pick up the manga at a point right before the anime began to diverge.  At what point in the manga is this? Eclipse Optimize Imports to Include Static Imports Is there anyway to get Eclipse to automatically look for static imports? For example, now that I've finally upgraded to Junit 4, I'd like to be able to write:  assertEquals(expectedValue, actualValue);   hit Ctrl + Shift + O and have Eclipse add:  import static org.junit.Assert.assertEquals;   Maybe I'm asking too much. Under Windows 7, how can I delete a folder whose name ends with "..."? How to splice together several instances of InterpolatingFunction? Create a new cmd.exe window from within another cmd.exe prompt I'be been running 18.04 since a clean SSD install on the day of it's official release, without issues. Power on to login was seconds (max 10)  Then, I did a regular upgrade this morning:  $ sudo apt update &amp;&amp; sudo apt dist-upgrade   The packages installed/upgraded were:  Install: linux-headers-4.15.0-24:amd64 (4.15.0-24.26, automatic), linux-headers-4.15.0-24-generic:amd64 (4.15.0-24.26, automatic), linux-modules-extra-4.15.0-24-generic:amd64 (4.15.0-24.26, automatic), linux-modules-4.15.0-24-generic:amd64 (4.15.0-24.26, automatic), linux-image-4.15.0-24-generic:amd64 (4.15.0-24.26, automatic) Upgrade: gnome-control-center-data:amd64 (1:3.28.1-0ubuntu1.18.04.1, 1:3.28.1-0ubuntu1.18.04.2), linux-headers-generic:amd64 (4.15.0.23.25, 4.15.0.24.26), gnome-control-center:amd64 (1:3.28.1-0ubuntu1.18.04.1, 1:3.28.1-0ubuntu1.18.04.2), linux-image-generic:amd64 (4.15.0.23.25, 4.15.0.24.26), linux-signed-generic:amd64 (4.15.0.23.25, 4.15.0.24.26), gnome-control-center-faces:amd64 (1:3.28.1-0ubuntu1.18.04.1, 1:3.28.1-0ubuntu1.18.04.2), linux-generic:amd64 (4.15.0.23.25, 4.15.0.24.26)   I rebooted once the upgrade had completed, and noted a 2-3 minute delay on the Ubuntu loading/splash screen (prior to login) (without any progress/activity indicated on the dots).  I powered off and tried to boot again, but am getting this delay consistently now. Also close down is much slower too.  Update #1 (2018-07-03): Analysis on systemd:  $ sudo systemd-analyze blame 3min 53.073s plymouth-quit-wait.service     2min 20.699s snapd.seeded.service          49.949s snapd.service           6.186s NetworkManager-wait-online.service           1.148s dev-sda2.device           1.098s plymouth-start.service   Showing that plymouth-quit-wait.service (which I now believe is related to the Ubuntu loading/splash screen) and snapd.seeded.service were by far the longest running services to initiate. So I compared the times before the dist-upgrade and after:  $ journalctl -u plymouth-quit-wait.service --since today -- Logs begin at Fri 2018-04-27 13:01:30 BST, end at Tue 2018-07-03 12:38:05 BST. -- Jul 03 04:15:43 user-laptop systemd[1]: Starting Hold until boot process finishes up... Jul 03 04:15:46 user-laptop systemd[1]: Started Hold until boot process finishes up. -- Reboot -- Jul 03 04:21:17 user-laptop systemd[1]: Starting Hold until boot process finishes up... Jul 03 04:24:52 user-laptop systemd[1]: Started Hold until boot process finishes up.   Before the upgrade plymouth-quit-wait.service took 3 seconds. After the upgrade it took 3 minutes 35 seconds  $ journalctl -u snapd.seeded.service --since today -- Logs begin at Fri 2018-04-27 13:01:30 BST, end at Tue 2018-07-03 12:42:14 BST. -- Jul 03 04:15:43 user-laptop systemd[1]: Starting Wait until snapd is fully seeded... Jul 03 04:15:43 user-laptop systemd[1]: Started Wait until snapd is fully seeded. -- Reboot -- Jul 03 04:22:47 user-laptop systemd[1]: Starting Wait until snapd is fully seeded... Jul 03 04:24:49 user-laptop systemd[1]: Started Wait until snapd is fully seeded.   Before the upgrade snapd.seeded.service took 0 seconds. After the upgrade it took 2 minutes 2 seconds.  Update #2 (2018-07-06): This morning's boot saw the return of the delay. So I guess we're still waiting for the kernel/plymouth/snapd update.  Update #3 (2018-07-12): The problem seems to be resolved, but I didn't see any update to snap or plymouth, and I'm still running 4.15.0-24 kernel. So I'm not sure which package update fixed the problem, or if it just resolved itself somehow. Reading the bug updates on launchpad it's unclear to me what was done (or is being done) to what package/s. If anyone can clarify that would be very useful. How to test with Math.random in JavaScript? In the book , at page 27 we find the exercise asserting that       Exercise I.XXXVI.   The underlying space of a zero-dimensional scheme is discrete; if the scheme is Noetherian, it is finite.     Thoughts I thought that a zero-dimensional scheme is one such that, in every local ring $\mathscr O_{X,p}$, the only prime ideal is its unique maximal ideal $\mathfrak m_{X,p}.$ Hence, supposing that $X=\operatorname{Spec} R$ is affine, we deduce that every prime ideal is maximal, so each ponit in $X$ is closed. But how does this imply the discreteness? It only implies the Hausdorff-ness of $X$, right? Furthermore, I cannot perceive what the "Noetherianity" has to do with the finiteness claimed here. In the affine case, it seems to be claiming that there are only a finite number of prime ideals in a Noetherian ring, if its scheme is zero-dimensional? For example, $\mathbb Z$ is a Dedekind domain, hence its scheme is zero-dimensional, but it has infinitely many primes: does this contradict the statement?    Thanks in advance for any reference or hint, and point out any inappropriate point if any is presented. Edit I see why my example cannot work now: it is one-dimensinoal: don't forget the pirme $(0)$. But I am still wondering why the statement is true... How to avoid the 'not properly formatted as code' alert What's wrong with this division? I think there's a lot for me to learn about data types. Why this happens  double result = ((3/8)*100).ToString();   it gives zero .. should be 37,5 ... :( Pascal's triangle in tikz I would like to typeset the top part of Pascal's triangle. To get the triangle with the names of the binomial coefficients, i.e., {n \choose k}, I used the following code  \begin{tikzpicture} \foreach \n in {0,...,4} {   \foreach \k in {0,...,\n} {     \node at (\k-\n/2,-\n) {${\n \choose \k}$};   } } \end{tikzpicture}   The result is this   Now I want to be equally lazy and do something like this for the values of the binomial coefficients, i.e., replace {\n \choose \k} in the node label with \CalculateBinomialCoefficient{\n}{\k} where \CalculateBinomialCoefficient is a hypothetical macro that calculates the binomial coefficient. Has anyone done something like that?  The result should look like this: $f$ be an analytic function defined on $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$ I came across the following problem that says:  Let $f$ be an analytic function defined on $\mathbb{D}=\{z\in\mathbb{C}:|z|&lt;1\}$ such that the range of $f$ is contained in the set $\mathbb{C}\setminus (-\infty,0]$. Then  $f$ is necessarily a constant function. there exists an analytic function $g$ on $\mathbb{D}$ such that $g(x)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\geq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\leq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$.   I have to determine which options are correct. Can someone help in the right direction? Thanks in advance for your time.