Getting columns and inserting into array [Bash]
Getting certain columns in bash
induction with 2 recursive sequnces I'm having trouble solving this problem. I have relation for two sequences of natural numbers.   $$a(n)+\sqrt3\cdot b(n)=\left(1+\sqrt3\right)^n$$  and I have to prove that recursions:  $$\begin{align*} &amp;a(n+1)=a(n)+3b(n)\\ &amp;b(n+1)=a(n)+b(n) \end{align*}$$  will satisfy relation mentioned above. I tried solve it using induction but I failed. Can anybody help with some other ideas? Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?  Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field. Proof. We need to show that any nonzero $r \in R$ has an inverse. Consider $\{1, r, r^2, \cdots\}$. This is an infinite set of elements of $R$, and by hypothesis $R$ is finite dimensional as an $F$-vector space, so this set is linearly dependent. Hence $\sum_{i=0}^n{c_i r^i} = 0$ for some $n$ and some $c_i \in F$ not all zero.  It then goes on to show, given the above, that we can derive an inverse for $r$. However, if I consider examples like $r = 2 \in Q[\sqrt{2}]$, $r = \sqrt{2} \in Q[\sqrt{2}]$ or $r = 2 \in Q[X]/{&lt;X^2&gt;}$, the set $\{1, r, r^2, ...\}$ doesn't look linearly dependent to me. I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something? [Edit] Well yes, I am. Somehow I had managed to discount the possibility of any $c_i$ being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding. While deploying code in vsCode from sandbox to production I get the below error      sf:INVALID_OPERATION: INVALID_OPERATION: testLevel of NoTestRun cannot   be used in production organizations   As per the error, I need to change the value of testLevel but I am not able to find where and how can I change it to RunLocalTests  Could someone please help me to change the value of testLevel in vsCode Can you nest html forms? How to make Table of Contents manually I'm trying to make table of contents manually, I mean I set the page of each of sections one by one. Here's my target:  and I'm using book class here.  Here's the code that I found   \documentclass[oneside]{book} \renewcommand\contentsname{CONTENTS} \renewcommand\listtablename{LIST OF TABLES} \renewcommand\listfigurename{LIST OF FIGURES} \addtocontents{toc}{{\bfseries Contents\hfill Page No.\bigskip\par}} \addtocontents{lot}{{\bfseries No.\hfill Title\hfill\hfill Page No.\bigskip\par}} \addtocontents{lof}{{\bfseries No.\hfill Title\hfill\hfill Page No.\bigskip\par}}  \usepackage[titles]{tocloft} \usepackage{xcolor,xpatch} \renewcommand\cftbeforechapskip{0ex} \renewcommand\cftchapfont{\mdseries} \renewcommand\cftchappagefont{\mdseries} \renewcommand\cftchappresnum{Chapter~} \renewcommand\cftchapaftersnum{. } \newlength\tocindent \settowidth\tocindent{\cftchapfont\cftchappresnum9\cftchapaftersnum} \edef\cftchapnumwidth{\the\tocindent} \edef\cftsecindent{\the\tocindent} \advance\tocindent2.3em \edef\cftsubsecindent{\the\tocindent} \advance\tocindent3.2em \edef\cftsubsubsecindent{\the\tocindent} \renewcommand\cftsecdotsep{\cftnodots} \renewcommand\cftsubsecdotsep{\cftnodots} \renewcommand\cftsubsubsecdotsep{\cftnodots} \newcommand\tocmainmatter {\renewcommand\cftchappagefont{\color{white}}% } \xapptocmd\mainmatter{\addtocontents{toc}{\protect\tocmainmatter}}{}{}   \newif\ifschaptertoc \makeatletter \renewcommand\@makeschapterhead[1]% {{\parindent \z@ \centering     \normalfont     \interlinepenalty\@M     \Large \bfseries  #1\par\nobreak     \vskip 20\p@ }% \@mkboth{\MakeUppercase{#1}}{\MakeUppercase{#1}}% \ifschaptertoc \addcontentsline{toc}{chapter}{#1}% \fi } \makeatother  \usepackage{blindtext}  \begin{document} \frontmatter  \tableofcontents \schaptertoctrue  \chapter*{ACKNOWLEDGEMENTS} \blindtext  \chapter*{ABSTRACT} \blindtext  \listoftables  \listoffigures  \chapter*{ABBREVIATIONS} \Blindtext  \chapter*{NOTATIONS} \Blindtext  \mainmatter  \chapter{INTRODUCTION} \blindtext  \section{General Background} \blindtext  \section{Objective} \blindtext \subsection{Objective 1} \blindtext \subsection{Objective 2} \blindtext \subsection{Objective 3} \blindtext  \section{Scheme} \blindtext \end{document} How to make a ISO file out of a folder containing Windows setup files I copied the Windows 8 CD to my HDD. I want to convert that folder to an ISO and use that to create a bootable USB.  Please tell me how I can make an ISO image out of that folder containing files.  This is my folder which I want to convert to a ISO file: Finding the derivative of $x\uparrow\uparrow n$ I am trying to find a general derivative for the function: $f(x)=x^{x^{x^{...^{x}}}}$however to do that I must find $f^{\prime }$ and $f^{\prime \prime}$...etc. I am now trying to write down a general expression for $f^{\prime \prime}$ and I have stumble opon the series in the picture below, I wonder, do anyone have an idea of what sort of expantion it migth be ? I have tried breaking it down into three different sums, but I wasn't very successful!  EDIT  So since this post has gotten some attention and the title has been change I will tell you what I am on about, although I already have a post just for the question in the title from before. I am as I said earlier trying to find a general derivative for the function $f(x)=x^{x^{x^{...^{x}}}}$ where $n$ is an arbitrary number, natural number and greater or equal to 2. I have already made success when I tried to find the first derivative, and I have also found some interesting pattern emerge when the derivatives as well as n incresses. But my question here was if anyone knew what this expation might be or look similar to be, since I have yet to be lernt series etc. I did not know that the question needed such background since it's straight a forward question. LaTeX template for typesetting a novel I'd like to use LaTeX to typeset a novel. Could you recommend me any template?  What I mean is a template (some .tex file) where I can include my chapterN.tex files (mainly plain text) and get a pretty good design with zero work :) What is the function of the RNA primer in DNA replication? MySQL Date Between plot(density(rexp(100))   Obviously all density to the left of zero represents bias.  I'm looking to summarize some data for non-statisticians, and I want to avoid questions about why non-negative data has density to the left of zero.  The plots are for randomization checking; I want to show the distributions of variables by treatment and control groups. The distributions are often exponential-ish.  Histograms are tricky for various reasons.    A quick google search gives me work by statisticians on non-negative kernels, e.g.:  .    But has any of it been implemented in R?  Of implemented methods, are any of them "best" in some way for descriptive statistics?  EDIT:  even if the from command can solve my current problem, it'd be nice to know whether anyone has implemented kernels based on literature on non-negative density estimation Z-axis issue with the floating review info panel, permits various elements to flow over it When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail What is the "canary thread" and why is it starving? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? High Voltage Powerlines - clarification of energy loss Why $\gcd(qb+r,b)=\gcd(b,r)$? Given: $a = qb + r$. Then it holds that $\gcd(a,b)=\gcd(b,r)$. That doesn't sound logical to me. Why is this so?    Addendum by LePressentiment on 11/29/2013: (in the interest of  and averting a duplicate)  What's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally. Can I secretly kill every guard in a town? Is it possible to kill all of the guards in a town?  Ideally, I'd like to do this secretly so the entire town doesn't see my deeds and put a bounty on my head.  Edit: I was questing in Markarth and was told by a guard to back off and stop being nosy. I didn't and when I went to the Shrine of Talos I was set upon by a bunch of guards.  Now I have a huge bounty and I had to sneak/kill my way out of the city.  Every time I killed a group of guards the bounty was reduced by $1000, but I still have a $30,000 bounty.  I haven't even been through 1/3 of the city yet and know there has to be a lot of quests, so what do I do?  The only solution seems to kill every single guard (and inhabitant). When should you push variables in quotes in boolean conditions? Are the following two boolean expressions the same?  if [ -n $1 ] ; then  if [ -n "$1" ] ; then   If not - When should you put a variable in quotes? People often use sans, the French word for ‘without’ in English.   What I don’t get, is when should that switch be made. When should I say sans when I really want to say without. Are there any general rules regarding usage, or is it kind of a person-to-person basis/feeling of flowerness.