I have produced two tables that when ran from the same LaTeX file are positioned on the same page, one under the other. However, when the two tables are called upon by using \input they appear on two seperate pages.  Here is a MWE (with the pacakges used in my paper) that produces the problem:  \documentclass[11pt,a4paper]{article} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage[stable]{footmisc} \usepackage{mathpazo} \usepackage{amsmath} \usepackage{graphicx} \usepackage{xcolor} \usepackage{microtype} \usepackage[authoryear, round]{natbib} \usepackage[colorinlistoftodos]{todonotes} \usepackage{setspace} \usepackage[colorlinks=true, citecolor=red]{hyperref} \usepackage{mdframed} \usepackage{rotating} \usepackage{lscape, longtable} \newcommand\iid{i.i.d.} \newcommand\pN{\mathcal{N}}  \begin{document} \include{Results_Tables/tablea} \include{Results_Tables/tableb} \end{document}   Whereas using this amended MWE would produce the desired output, where both tables are within one tex file:  \documentclass[11pt,a4paper]{article} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage[stable]{footmisc} \usepackage{mathpazo} \usepackage{amsmath} \usepackage{graphicx} \usepackage{xcolor} \usepackage{microtype} \usepackage[authoryear, round]{natbib} \usepackage[colorinlistoftodos]{todonotes} \usepackage{setspace} \usepackage[colorlinks=true, citecolor=red]{hyperref} \usepackage{mdframed} \usepackage{rotating} \usepackage{lscape, longtable} \newcommand\iid{i.i.d.} \newcommand\pN{\mathcal{N}}  \begin{document} \include{Results_Tables/tablea_and_b} \end{document}
There are two different commands to incorporate another file into the source of some document, \input and \include. When should I use one or the other? What are the differences between them? Are there more things like them to be aware of?
What is the Mediator in this Feynman Diagram $q\bar{q} \rightarrow gg$? Let's consider the t-channel version of the process $q\bar{q} \rightarrow gg$. What is the mediator in this case, a $q$ or a $\bar{q}$? Naively, I would have thought a $q$ (as to have one fermion line), but I am not $100 \%$ sure. How do I share internet with Windows from my Linux box? I have a Linux (Ubuntu 12.04) PC connected to the internet with a Greenpacket WiMax USB modem. I want to share the Internet connection with another computer running Windows 7 Home Premium, connected to the Linux PC over a LAN. Is this possible? How? Is the reverse possible instead (connecting the internet to the Windows computer and sharing it with Linux)? Complex numbers equation: $z^4 = -16$ How to change hair color along length of strand? At some point (not right away), I'll be wanting to make a character model which has multicolor hair.  Namely, it will be one color closer to the roots, and another color closer to the tips.  Is there any way to do this in blender?  I would be using Cycles for this. How to represent kWh usage by year against average temperature? Let $q \in \mathbb C$, $|q|=1$ and $q^n \neq 1, \forall n \in \mathbb N$. Show that $\{q^n:  n \in \mathbb N\}$ is dense in $S^1$.  My attempt: As $(q^n)$ is limited, there is subsequence $(q^{n_j})$ convergent. So given $\epsilon &gt; 0, \exists j_0 \in \mathbb N, \quad |q^{n_j} - q^{n_k}| &lt; \epsilon, \quad \forall j,k \geq j_0$. Supose that $n_j &gt; n_k$, then  $|q^{n_j} - q^{n_k}| = |q^{n_k}||q^{n_j - n_k} - 1| = |q^{n_j - n_k} - 1| &lt; \epsilon, \quad \forall j, k \geq j_0$. So, there is a subsequence of $(q^n)$ that converges to $1$.  Using this fact, how can I ensure that for all $w \in S^1$ there is a subsequence of $(q^n)$ that converges to  $w$? Prove that: $\sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}}<3,\,\forall n\in\mathbb N.$ How full is too full for mechanical hard drives? I have heard many claim that it doesn't matter how full a drive is until it starts cutting into temp and virtual memory space.  This doesn't make sense to me, given the nature of how the data is transacted on a hard drive. The inside of the platter presents less data per revolution than the outside of the drive does, by significant factors. The inside 40% of the radius of full size hard drive is used for the spindle, so only the outside 60% is used for data storage, but that still means that the inside track of a hard drive presents data 60% slower than the outside track.  By my calculation, a Hard drive that is only 10% full should perform about 2.25 times faster than a hard drive that is 90% full, assuming that the flow is constrained by other factors.  Am I wildly off base here? For all the drives I know, even the top speeds of the first 1% of the drive would be well within the bandwidth provided by a SATA 2 connection. Hello everyone interested. Here's a seemingly simple problem.  Suppose $G$ is a complete, simple graph with vertices $\{1,2,..,n\}$, $n\geq k\geq 3$.  How many cycles of length $k$ can we have on $G$?  The statement seems simple but I have trouble putting everything together.  So I say fix a vertex and start forming cycles with length k. But I get nowhere... What is the proper approach to this problem since counting is not easy? Is there a relevant theorem that I am unaware of?  Thanks for any help... Is linear algebra more “fully understood” than other maths disciplines? In , it was discussed how LA is a foundation to other branches of mathematics, be they pure or applied. One answer argued that linear problems are fully understood, and hence a natural target to reduce pretty much anything to.   Now, it's evident enough that such a linearisation, if possible, tends to make hard things easier. Find a Hilbert space hidden in your domain, obtain an orthonormal basis, and bam, any point/state can be described as a mere sequence of numbers, any mapping boils down to a matrix; we have some good theorems for existence of inverses / eigenvectors/exponential objects / etc..  So, LA sure is convenient.  OTOH, it seems unlikely that any nontrivial mathematical system could ever be said to be thoroughly understood. Can't we always find new questions within any such framework that haven't been answered yet? I'm not firm enough with G&ouml;del's incompleteness theorems to judge whether they are relevant here. The first incompleteness theorem says that discrete disciplines like number theory can't be both complete and consistent. Surely this is all the more true for e.g. topology.  Is LA for some reason exempt from such arguments, or does it for some other reason deserve to be called the best understood branch of mathematics? How to replace a character by a newline in Vim I'm trying to replace each , in the current file by a new line:  :%s/,/\n/g    But it inserts what looks like a ^@ instead of an actual newline. The file is not in DOS mode or anything.  What should I do?  If you are curious, like me, check the question  as well. Content Assist for JSF2 + CDI (weld) beans + Eclipse Helios Sole minimal element: Why not also the minimum? A minimal element (any number thereof) of a partially ordered set $S$ is an element that is not greater than any other element in $S$.  The minimum (at most one) of a partially ordered set $S$ is an element that is less than or equal to any other element of $S$.    Let's consider the power set $\mathcal P (\{x,y,z\})$ together with the binary relation $\subseteq$.  The Hasse diagram shows what element(s) we're looking for:    It's easy to see that:   $\emptyset$ is a minimal element $\emptyset$ is the minimum   Now if we remove $\emptyset$ and consider $\mathcal P (\{x,y,z\})\setminus \emptyset$ instead, we get the following:   $\{x\}$, $\{y\}$ and $\{z\}$ are minimal elements there is no minimum   (1) We know that a minimum is unique and it is always the only minimal element.  (2) And from the example above, it seems that, if a sole minimal element exists, it is always the minimum.  But I read that (2) is false. Why? This or That person Today I heard a speaker use the phrase "this person", referring to the subject-person of the story they were telling, and I heard it as (that person). Between the words “this” or “that” in context with the phrase and story-tense, which should you use. Proof that the irrational numbers are uncountable Browse an UNC path using Windows CMD without mapping it to a network drive C:\&gt; cd \\somewhere '\\somewhere' CMD does not support UNC paths as current directories.   What I usually do to get around this is to map that directory to a network drive and then I could easily access it from the command prompt.  But is there an easier way on how to get around this? In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type   $$ M=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$  where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.  Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.  Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.   Now come the questions:   I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 &amp; 0 \\ b &amp; 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?   By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix}$, $y=\begin{pmatrix} 1 &amp; 0 \\ c &amp; e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 &amp; 0 \\ b(1-e)+c(d-1) &amp; 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.   The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?   2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.   2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?  2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?  EDIT: I have found the essentially same question How to custom (split in two)? Ibidem It seems like I'm doing something wrong when using foreach variables to indicate node numbers. The two foreach loops are anchoring the paths to the east of each node. Why does this happen?    \documentclass{article}  \usepackage{tikz} \usetikzlibrary{backgrounds,fit,patterns,decorations.pathreplacing,calc}  \makeatletter \tikzset{   labelnode/.style={draw=none, fill=none, black},   latentnode/.style={draw, minimum width=5mm, shape=circle, ultra thick, black},   dagconn/.style={arrows=-&gt;, black, thick}, } \makeatother  \begin{document} \begin{tikzpicture}[scale=1]  \pgfmathsetmacro{\colgap}{1.6}  \pgfmathsetmacro{\rowgap}{1.3}   \node[labelnode] (x1) at (\colgap*0,\rowgap*1) {$X_1$};  \node[labelnode] (x2) at (\colgap*1,\rowgap*1) {$X_2$};  \node[labelnode] (x3) at (\colgap*2,\rowgap*1) {$X_3$};   \node[latentnode,blue] (x4) at (\colgap*0,\rowgap*0) {$X_4$};  \node[latentnode,red] (x5) at (\colgap*1,\rowgap*0) {$X_5$};  \node[labelnode] (x6) at (\colgap*2,\rowgap*0) {$X_6$};   \foreach \c in {1,...,2} {    \foreach \r in {1,...,2} {      \pgfmathsetmacro{\st}{(((\r-1)*3)+\c)}      \pgfmathsetmacro{\en}{\st+1}      \draw[dagconn] ([xshift=0.2cm]x\st) to ([xshift=-0.9cm]x\en);    }  }   \foreach \c in {1,...,3} {    \foreach \r in {1,...,1} {      \pgfmathsetmacro{\st}{((\r-1)*3)+\c}      \pgfmathsetmacro{\en}{\st+3}      \draw[dagconn] (x\st) to (x\en);    }  } \end{tikzpicture} \end{document} On a "1-D line", the point that minimizes the sum of the distances is the median, WHY? There is a set A which contains some discrete points (1-dimension), for example {1, 3, 37, 59}. And I want to pick one point from A which can minimize the sum of distances between this point and others.  For this "1-D line", many people say the median is the right point, but could anybody give me a concrete proof on it?