After years with LaTeX, I need to take control over the output formatting of my citations in various projects. I've referred to the  as well as the otherwise excellent  but I'm looking for a more pedagogical less manual-like approach.  The web is full of hacks for specifics, but provides no overall systematic understanding of the rules that govern how .bib-files are turned into footnotes and bibliographies with dots, italics and what information from bib-files are included where. I would like to and never again feeling the need to google stuff like "remove initials in name in from footcite".  I don't want to be given a fish; I want to learn how to be a fisherman.  Does anyone know of a good book to read?
Is there a 'biblatex in a nutshell' guide out there? I'd like if someone explained to me the essentials on how to use biblatex (what lines I have to write in my document, which files I have to have, how many times and what I have to compile), so then I can go to  to customise it further.
No Badge Progress color for Golden Badges? If $(X,d)$ is a complete metric space and $A$ is closed then show that for $x \in X$ there exists an element $a_0 \in A$ such that $d(x,A)=d(x,a_0)$ I’m struggling with getting \ifmmode to work correctly when used inside an array environment which is inside a displaymath environment.  Here is a minimal example:  \documentclass{article}  \newcommand{\signal}[1]{%   \ifmmode     1#1   \else     2$#1$   \fi }  \begin{document} \begin{displaymath}   \begin{array}{cc}     \signal{a} \signal{b} &amp; \signal{c} \signal{d} \\   \end{array} \end{displaymath} \end{document}   This example generates the following output:        Why doesn’t all invokations of \signal detect math-mode? Determine which circuits are parallel and which are in series I need to find the Equivalent Resistance across A and B. The problem is, I don't know which one's are in series and which ones are in parallel. How do I identify which ones are parallel or series? Careers: What does "allows remote" mean? When a posting is marked "allows remote", does it mean work in the office some days, work from home some days?  Or does it mean you could actually live out of town and maybe fly in as needed?  The reason I ask is I'm really interested in public facing software rather than internal/enterprise stuff, and most of that seems to happen in a few very expensive cities.  Given it's so hard to hire in places like SV/NYC, it seems like it could be a good scenario for companies if a developer had a really solid list of accomplishments, but I don't know how many companies think this way. Fast, accurate 2d collision I'm working on a 2d topdown shooter, and now need to go beyond my basic rectangle bounding box collision system.  I have large levels with many different sprites, all of which are different shapes and sizes. The textures for the sprites are all square png files with transparent backgrounds, so I also need a way to only have a collision when the player walks into the coloured part of the texture, and not the transparent background.  I plan to handle collision as follows:   Check if any sprites are in range of the player Do a rect bounding box collision test Do an accurate collision (Where I need help)   I don't mind advanced techniques, as I want to get this right with all my requirements in mind, but I'm not sure how to approach this. What techniques or even libraries to try. I know that I will probably need to create and store some kind of shape that accurately represents each sprite minus the transparent background.  I've read that per pixel is slow, so given my large levels and number of objects I  don't think that would be suitable. I've also looked at Box2d, but haven't been able to find much documentation, or any examples of how to get it up and running with SFML. Why am I getting a NoClassDefFoundError in Java? I am getting a NoClassDefFoundError when I run my Java application.  What is typically the cause of this? "The product key entered does not match any of the Windows image" I have a HP with Windows 7 installed. I want to do a clean installation of Windows 8 from a bootable DVD.   I purchased my Windows 8 32-bit core from the Microsoft store and I received my product key. I then burned the .ISO file to a blank DVD using imgburn. When I restart and change my BIOS to boot from internal CD/DVD and then boot from my internal DVD, I start the installation process.  After I pass the screen where I select the language and location, and click Install it stays there for a minute and then it says:     The product key entered does not match any of the Windows image   I did my research and learned that I need to add a PID.txt file to my source file containing the Windows 8 .ISO but the thing is: I do not know how to add files to the .ISO. The guide I researched is for a bootable USB and I am using a DVD. Annoying FrontEnd bugs in Mathematica 10 How to set identity_insert to on in SQL Server How to open the default webbrowser using java What are the differences between executing shell scripts using "source file.sh", "./file.sh", "sh file.sh", ". ./file.sh"? Change date format [YYYYMMDD] to [MM/DD/YYYY] Sweave.sty causing error in verbatim environment? Farkas/guards pursuing me glitch I was doing a quest for the companions. When I entered Jorrvaskr, Farkas pulled out his sword and began to talk to me. Every time I would try to leave the conversation he would just follow me and try to talk to me again. This lead to me paralyzing Farkas and running away.   By the time I got to the gates of Whiterun he had already gotten up, ran to me and started another conversation. I quickly made it to the gates and exited, he followed me through the gates but I whirlwind sprinted to Shadowmere then attempted to fast travel - which game me the "Can not fast travel when guards are pursuing you" message even though no guards were following me.   I knew Farkas wasn't far behind so I quickly jumped on a carriage and I traveled to the nearest hold which was Dawnstar. Now I can't leave Dawnstar because the same message appears. I have tried to pay off my debt to the Dawnstar guards but it didn't fix the problem. How can I resolve this issue? Resize multiple layers around their respective anchor points I have 50 or so layers (imported images as smart objects) that I would like to scale down, while they stay in place. Is there a way of doing this? I know After Effects has, is there a similar Scale property in photoshop? Circular duplicates in the close review queue? Proof if $a \vec v = 0$ then $a = 0$ or $\vec v = 0$ I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right:  for a vector space $V$ with an underlying field $F$:  Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$  After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$  Say $a\vec{v}=\vec{u}$. Since $a$ is non-zero, we can divide both sides by $a$.  $$\vec{v}=\frac 1 a \vec{u}$$  If $\vec{u}$ were $\vec{0}$ then by  $$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$ $v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well.   Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how.  Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple. Binding energy and mass I’ve been told that a greater binding energy means the nucleus is more tightly bound, and therefore that decreases the mass of the nucleus with respect to its nucleons when separated. But why does a tighter nucleus mean a smaller mass? Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$