How to write text that takes up no space? How does one write text such that whatever comes next is placed as if the text had not been written?  Edit: More specifically, I would like to place text in the top-left corner of a tikzpicture by writing the text right before the figure and anchoring the figure as if the text had not been written. Here is an attempt I made that does not work:  \documentclass{article} \usepackage{tikz} \begin{document}  \newcommand{\imagetop}[1]{\vtop{\null\hbox{#1}}}  \makebox(0,0){a} \imagetop{%     \begin{tikzpicture}         \draw (0, 0) rectangle (1, 1);      \end{tikzpicture} } \makebox(0,0){b} \imagetop{%     \begin{tikzpicture}         \draw (0, 0) rectangle (1, 1);      \end{tikzpicture} }  \end{document}
What are the different kinds of boxes in (La)TeX? I've seen hbox, vbox, mbox, parbox and others used in various code samples. What are the different kinds of box in (La)Tex? Which are suitable for use in LaTeX, as opposed to plain TeX?
Where can I get a Thunderbolt hub or splitter? Apple has just announced the new MacBook Airs, Mac mini, and Thunderbolt display. One of the features of the Thunderbolt display that caught my eye was its ability to daisy-chain two displays together from one Thunderbolt port (on a MacBook Pro).  I already have two 24" monitors and I don't feel like paying for two £900 displays. I worked out that all I need is a Thunderbolt hub/splitter so that I can plug the two cables that emerge into mini DisplayPort-to-DVI adapters.  Does anyone know of any Thunderbolt hubs or splitters? Or does anyone know of an alternative method? Press space to continue How do I stop a bash script until a user has pressed Space?  I would like to have the question in my script     Press space to continue or CTRL+C to exit   and then the script should stop and wait until Space is pressed. I've been trying to wrap my head around the basic concepts of category theory, and I thought I would attempt to illustrate what I understand with the category of sets, probably the easiest example. Particularly, I've been trying to prove that $id_A$ (the identity morphism on $A$, for all $A \in Obj(\mathbf{Set})$) is $1_A \colon A \rightarrow A, x \mapsto x$.  This is a very intuitive and reasonable statement, and it's trivial to prove that $1_A$ is indeed an identity morphism on $A$, and I suppose uniqueness of $id_A$ can be demonstrated analogously to uniqueness of the identity element in a monoid (considering the subcategory which has $A$ as its only object, and endofunctions on $A$ as its only morphisms).  In this manner, it is not hard to prove that the proposition in the title is true, but this demonstration requires to make an assumption or guess as to what could $id_A$ be. Specifically, the scheme of the proof is: assume $id_A$ = $1_A$, see that it works with the definition of an identity morphism, show that the identity morphism is unique, and, in conclusion, $id_A$ can only be $1_A$. What I'm looking for, nonetheless, is a somehow more direct proof, that doesn't assume $id_A$ = $1_A$ at the start. I want to place myself in a state of little or no knowledge about sets and functions, and under this assumption, why would I assume $id_A$ = $1_A$ at first? Why not try with $id_\mathbb{Z}$ = $f \colon \mathbb{Z} \to \mathbb{Z}, x \mapsto x^2 + 1$, for example? It wouldn't work, but I don't have any reason to think that $1_\mathbb{Z}$ is a better guess for $id_\mathbb{Z}$.  I suppose that the proof for which I'm asking would work for categories of sets with additional structure, and probably for posets as well, although I'm not clear as to what modifications it would require to work.  Thanks. How should a user's death be handled? How to give an image or video a TV screen effect? I thought it would be interesting to create a .blend that filters any regular video or image with vintage television look.   Edit: Although this question initially arose from my desire to create a vintage look, it has since been pointed out in a comment that all displays use arrays of RGB emitters, so it's not really an issue of vintage or not. How vintage it will end up looking is probably more a matter of the shape, size, and density of the emitters.  Reference images:       Upon examining the structure of screens, the most common seems to be rows of vertical RGB bars:    Here is another type which I believe is common in computer displays - circular dots in a hexagonal grid:    I have been thinking about how to make a screen in Blender that takes its input from an image or image sequence and for each frame converts the RGB value for a pixel (or region) and uses that to illuminate the red green and blue phosphors in the correct ratios so that when viewed from a distance an image is distinguishable.  Do you think this will require python scripting? I made some initial attempts modeling an array of phosphors then using a Light Path node and a Separate RGB node in Cycles but I haven't gotten very far. I would love to hear what you all can suggest.  Thanks! How to add numbering to reference list but still use authoryear in natbib Is there an easy way to add numbers to the reference list generated by \usepackage[authoryear]{natbib} with whatever style file (using plainnat for ease)? Specifically, I use \cite and \citenum from natbib in order to have some references show up in the text as author-year and others just appear as numbers. Switching authoryear out for numbers in the natbib options generates a reference list with numbered references, but also causes the references in the text to all appear as numbers, which I don't want.  \documentclass[12pt]{article} % use larger type; default would be 10pt \usepackage[authoryear]{natbib} \begin{document}  \cite{A} said this, but other research \citenum{B,C} says otherwise.  \bibliographystyle{plainnat} \bibliography{MalariaBib,MST} \end{document} permission error when installing ttf-mscorefonts-installer/Can't drop privileges as file couldn't be accessed by user '_apt' C++ code file extension? What is the difference between .cc and .cpp I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired. Solve $[2^{(2^{403})}] = [a]$ in $\mathbb Z_{23}$ where $0 \le a < 23$. Solve $[2^{(2^{403})}] = [a]$ in $\mathbb Z_{23}$ where $0 \le a &lt; 23$.  I've tried to combine corollaries of Fermat's little theorem and various other methods to solve this, but have always been stuck. Problem on Extremal Graphs: three independent paths The problem states that:     A simple (or strict) graph with $n$ vertices and $m&gt;\dfrac{3(n-1)}{2}$ edges, has two vertex   joined by three independent paths.   Any hints, ideas or useful results, to attack this problem? I'm using the beamer miniframes theme (with subsection=false) and the compress option. Currently the navigation shows the dots for each subsection side-by-side. I would prefer if there would be a horizontal space (e.g. with the width of a dot) between the dots of each subsection. Is this possible? Right now it looks like:  Section 1 oooooo  And I would prefer it to look like:  Section 1 ooo ooo  MWE: \documentclass[compress]{beamer}   \useoutertheme[subsection=false]{miniframes}   \begin{document}   \section{Sec 1}   \subsection{}   \begin{frame}[t]     hi   \end{frame} \subsection{}   \begin{frame}[t]     hi   \end{frame} \end{document}` Suppose that an entire function $f(z)$ satisfies $\left|f(z)\right|\leq k\left|z\right|^n$ for sufficiently large $\left|z\right|$, where $n\in\mathbb{Z^+}$ and $k&gt;0$ is constant. Show that $f$ is a polynomial of degree at most $n$. What does the size of the font translate to exactly? I have a 40 pixel font that I'm trying to translate into custom images. I'm trying to match the size of the image exactly, but I'm having some problems getting exactly the right size. Basically, what it comes down to is this. When a font says it is X pixels, what does this physically mean? I want to append some data to a layer in my PostGIS DB using a shapefile, but when I use the Import layer/file tool in QGIS, it doesn't allow me to choose the append option.  Why is this and how do I get it to allow me to append? What should the standard spelling be - British or US? I just saw someone edit the title of  question to  the spelling from favourite (The British spelling) to favorite (The US-English spelling).  Does SOFU have an accepted standard on language and spelling? Which is it? Using the properties of real numbers, verify commutativity on complex numbers. I have just started self-studying Linear Algebra after a 3 year hiatus from mathematics (in an extended neuroscience foray), and I want to ensure that I am sufficiently verifying commutativity of complex numbers by using the properties of real numbers.  My attempts...  (1) Commutativity of Addition:   Verify that $w + z = z + w$ for all $w, z \in \mathbb{C}$.  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{16mm} = (a + c) + (bi + di) \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{19mm} = (c + a) + (di + bi) \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{13mm} = c + (a + di) + bi \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = c + (di + a) + bi \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{17mm} = (c + di) + (a + bi) \hspace{11mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{13mm} = z + w \hspace{34mm}\text{by definition of }\mathbb{C}$$  $$\blacksquare$$  This seems a bit verbose to me - is it acceptable to combine steps? For example, rather than:  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ can I instead write: $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{48mm} = a + (c + bi) + di \hspace{10mm}\text{by associativity and commutativity in }\mathbb{R}$$?  (2) Commutativity of Multiplication:  Verify that $wz = zw$ for all $w, z \in \mathbb{C}$.  $$\hspace{-14mm}wz = (a + bi)(c + di) \hspace{24mm}\text{by definition of }\mathbb{C}$$ $$\hspace{4mm} = a \cdot c + a \cdot di + bi \cdot c + bi \cdot di \hspace{10mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{9mm} = a \cdot c + bi \cdot c + a \cdot di + bi \cdot di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{35mm} = c \cdot a + c \cdot bi + di \cdot a + di \cdot bi \hspace{10mm}\text{by multiplicative commutativity in }\mathbb{R}$$ $$\hspace{1mm} = c(a + bi) + di(a + bi) \hspace{20mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{5mm} = (c + di)(a + bi)\hspace{34mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{4mm} = zw \hspace{55mm}\text{by distribution in }\mathbb{R}$$ $$\blacksquare$$  This also seems a bit wonky to me. Any and all suggestions on clearing up my verification, reasoning, etc. would be greatly appreciated! how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $? It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...  Let consider sum no to infinity, but to n  $$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$ $$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$ but $$ n = ln e^n $$ so it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$  So the limit of it is $-\infty$ Have I done this well or I missed sth? How to prove $\lim\limits_{n \to \infty} (1+\frac1n)^n = e$? Is it possible to set up and watch a game between just computer AI opponents? No human players, just AI and have the ability to scroll around and see what the AIs are doing with the turn management being also automated, so the AIs just end turn when they are done. Oh, and all of this hopefully without any sort of fog of war.  This way, I can see how the AI manages resources and units and maybe learn a thing or two as a civ newb.