I want to separate the address of the authors with a box where I will put the abstract according to the graphic template below. Right alignment is important.
I want to achieve the exact same title page layout as the paper shown . For those that do not want to download the paper, the image of the title page is shown below:    The super-scripts indicating the authors doesn't have to be as shown in the example title page (e.g, a, b and c), other symbols are fine.  I am using the authblk package and know the general syntax such as:  \title{title of paper} \author[1]{author 1\thanks{details that goes into the footnote}} \author[1]{author 2\thanks{details that goes into the footnote}} \author[2]{author 3\thanks{details that goes into the footnote}} \affil[1]{university 1} \affil[2]{university 2} \renewcommand\Authands{ and } \maketitle  \begin{abstract} abstract goes here \end{abstract}   But I am not sure how to achieve the example title page shown. Any help would be appreciated!
Is it possible to change the line style in a tikz figure in the same \draw command?  I am looking for something like:  \documentclass[tikz]{standalone} \begin{document}     \begin{tikzpicture}         \draw[style1] (0,0) -- ++(1,0)  **CHANGE STYLE** -- ++(1,0);     \end{tikzpicture} \end{document}   The main motivation is that I don't want to loose the relative coordinate as I go on with a long \draw command. I'm trying to run this command in the terminal:  sudo apt install software-properties-common   This is the error message I get:  E: Could not get lock /var/lib/dpkg/lock-frontend - open (11: Resource temporarily unavailable)   E: Unable to acquire the dpkg frontend lock (/var/lib/dpkg/lock-frontend),     is another process using it? Is "double hashing" a password less secure than just hashing it once? Is hashing a password twice before storage any more or less secure than just hashing it once?  What I'm talking about is doing this:  $hashed_password = hash(hash($plaintext_password));   instead of just this:  $hashed_password = hash($plaintext_password);   If it is less secure, can you provide a good explanation (or a link to one)?  Also, does the hash function used make a difference?  Does it make any difference if you mix md5 and sha1 (for example) instead of repeating the same hash function?  Note 1:  When I say "double hashing" I'm talking about hashing a password twice in an attempt to make it more obscured.  I'm not talking about the .  Note 2:  I know I need to add a random salt to really make it secure.  The question is whether hashing twice with the same algorithm helps or hurts the hash. Why choose a low latency kernel over a generic or real-time kernel? A Particular Two-Variable System in a Group Suppose $a$ and $b$ are elements of a group $G$. If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.   I've been trying to prove but still inconclusive. Please prove to me.  Thanks very much for proof. Enable fractional scaling for Ubuntu 18.04 Prove that $\phi(n) \geq \sqrt{n}/2$ So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help? Increase close vote weight for gold tag badge holders sudo: /etc/sudoers is owned by uid 501, should be 0 To run programs as sudo without password, I edited /etc/sudoers with wrong content.   Then I re-edited it with Mac's default text editor(course sudo cannot be used when /etc/sudoers is wrong).  I tried to run sudo agin, and get this error:   sudo: /etc/sudoers is owned by uid 501, should be 0 sudo: no valid sudoers sources found, quitting   What should I do? I cannot remember the root password (maybe I haven't ever set root password before). I'm still new to ZFS. I've been using Nexenta but I'm thinking of switching to OpenIndiana or Solaris 11 Express. Right now, I'm at a point of considering virtualizing the ZFS server as a guest within either ESXi, Hyper-V or XenServer (I haven't decided which one yet - I'm leaning towards ESXi for VMDirectPath and FreeBSD support).   The primary reason being that it seems like I have enough resources to go around that I could easily have 1-3 other VMs running concurrently. Mostly Windows Server. Maybe a Linux/BSD VM as well. I'd like the virtualized ZFS server to host all the data for the other VMs so their data could be kept on a physically separate disks from the ZFS disks (mount as iscsi or nfs).  The server currently has an AMD Phenom II with 6 total cores (2 unlocked), 16GB RAM (maxed out) and an LSI SAS 1068E HBA with (7) 1TB SATA II disks attached (planning on RAIDZ2 with hot spare). I also have (4) 32GB SATA II SSDs attached to the motherboard. I'm hoping to mirror two of the SSDs to a boot mirror (for the virtual host), and leave the other two SSDs for ZIL and L2ARC (for the ZFS VM guest). I'm willing to add two more disks to store the VM guests and allocate all seven of the current disks as ZFS storage. Note: The motherboard does not have IOMMU support as the 880G doesn't support it, but I do have an 890FX board which does have IOMMU if it makes a huge difference.  My questions are:  1) Is it wise to do this? I don't see any obviously downside (which makes me wonder why no one else has mentioned it). I feel like I could be making a huge oversight and I'd hate to commit to this, move over all my data only to go fubar from some minute detail I missed.  2) ZFS virtual guest performance? I'm willing to take a small performance hit but I'd think if the VM guest has full disk access to the disks that at the very least, disk I/O performance will be negligible (in comparison to running ZFS non-virtualized). Can anyone speak to this from experience hosting a ZFS server as a VM guest? This question is pretty simple: I'd like to have the possibility to get the exact size of a picture in LaTeX with, for example, the macro \the.  In fact, I've founded the way \settoheight{} and so on, but it didn't work at all. What I've got doesn't match with what I can see in the image properties (I suppose that's the name in English). The result: 8pt versus 1300 pixels. It's may be a mis-use...  Edit: In the end, my aim was to find a way to fit the picture in LaTeX. If width and height were both less than the dimensions of the body, then it should just be centred. It look like this :     Require:   \usepackage{graphicx} \usepackage{xifthen} \usepackage{calc}   Code (pretty ugly, I agree):   % Centrering+fitting for image, with captation and label (not auto) \newlength\imageheight \newlength\imagewidth \newcommand{\pic}[3]{% \settoheight\imageheight{\includegraphics{pictures/#1.png}}% \settowidth\imagewidth{\includegraphics{pictures/#1.png}}% \ifthenelse{\lengthtest{\imagewidth &gt; \textwidth}}{\ifthenelse{\lengthtest{\imageheight &gt; \imagewidth}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[height=0.85\textheight]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[width=0.95\textwidth]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}}{\ifthenelse{\lengthtest{\imageheight &gt; \textheight}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics[height=0.85\textheight]{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}{\par\vspace{0.95em}% \begin{figure}     \centering\includegraphics{pictures/#1.png}     \caption{#2}     \label{#3} \end{figure}% \par\vspace{0.95em}}}} %   To use it : put your png files in a "pictures" directory, then, just simply call the picture like this : \pic{name_whitout_extension}{Captation_if_needed}{Label_if_needed}.    Edit 2 :   Thanks to Martin Scharrer's easyfig package, I've got a more or less 1,6 times faster solution to put my pictures, centred, with the perfect size.  CTAN:   VC:   Version v1.2 – 2012/05/15    This could be useful for some others, I hope. How to make a simple app from an applescript! I've got an applescript that I use to open files such as test.txt in vim, using the terminal.  I usually open a file by right-clicking it an selecting the service open in vim (this service is created by me and basically runs the applescript).  How do I build an actual program, that can be set in the "open with" option for a file?  This way, when I double click test.txt, the applescript gets run and the file gets opened in my vim in terminal.app.  I would also like to create custom icons that will be applied to all files I open using this app (using the "Change All..." button).  Where would I begin? Is this easy, because I'm just a novice -although I love programming and would spend much to to get to know all that is needed to do this.  the code is:  on open this_item      tell application "System Events"         if (count (processes whose name is "Terminal")) is 0 then             tell application "Terminal"                 activate                 do script with command "vim " &amp; (POSIX path of this_item) in front window             end tell         else             tell application "Terminal"                 activate                 tell application "System Events" to tell process "Terminal" to keystroke "t" using command down                 do script with command "vim " &amp; (POSIX path of this_item) in selected tab of the front window             end tell         end if     end tell  end open Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?   The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work. Hiding the sidebar in Ubuntu I have two questions :  First, I wanted to hide Ubuntu’s side-bar, so I set the settings in appearance to 'autohide'. The side bar did indeed disappear but when I roll over the left side, the side bar don't appear again. So, I am using a virtual box. Can it explain the fact that this bar doesn't show again when it is set to autohide.   Second, assuming I wouldn't be able to come over the problem, I pined to the dock every element of the sidebar, except the dash home. Is it possible either to pin it in the dock or to launch it from the terminal ? How to add modifiers using python script and set parameters? I've found some code that could add a modifier but none can set parameters like Octree. Subdomain versus subdirectory How do you copy the contents of an array to a std::vector in C++ without looping? Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula As far as every question I've seen concerning "what is $\sum_{k=1}^nk^p$" is always answered with "Faulhaber's formula" and that is just about the only answer.  In an attempt to make more interesting answers, I ask that this question concern the problem of "Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula for fixed $p\in\mathbb N$".  I've even checked this  without finding what I want.  Rule #1: Any method to compute the sum in question for arbitrary $p$ is good, either recursively or in some manner that is not in itself a closed form solution.  Even algorithms will suffice.  Rule #2: I don't want answers confined to "only some values of $p$".  (A good challenge I have on the side is a generalized geometric proof, as that I have not yet seen)  Exception: If your answer does not generalize to arbitrary $p$, but it still generalizes to an infinite amount of special $p$'s, that is acceptable.  Preferably, the method is to be easily applied, unique, and interesting.  To start us off, I have given my answer below and I hope you all enjoy. Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$ I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$. When I run the logistic regression, two independent variables have VIF values greater than 10 like 13 or so. Logistic regression is the one I will use to measure the overall change in the dependent variable with one incremental changes in every each independent variable. However, when I run the linear regression model version on my same data, the VIF values I get from all the independent variables are either slightly greater than 1 or around 5 or 6. Besides, all the coefficients are highly significant all below 0.01.  If so, is it safe to use my logistic regression model to find out the incremental effect of every each independent variable without dropping out any independent variables?   If my purpose is to find an incremental effect of every each independent variable on the overall dependent value in a logistic regression model, is multicolinearity an important problem to consider? Can I ignore it?   If my p-values are all less than 0.01 which is the same indicator as t-value, will I still need to worry about colinearity issue even though my VIF scores for two variables are around 13-14? Based on what you said, if p-value is safe enough, will this be still a problem?  I am also referring to the following website comment:                                                                     So, in sum, my ultimate goal is to use the final output from the logistic regression model generated from the independent variables and one binary dependent variable. If so, do you think I can ignore the multicolinearity problem?