Here are the Code examples of of this chapter. You can compile them online right on this web page by pressing the Typeset / Compile button. You can also edit them for testing, and compile again.
For a better view with the online compiler, I sometimes use \documentclass[border=10pt]{standalone} instead of \documentclass{article}. Instead of having a big letter/A4 page, the standalone class crops the paper to see just the visible text without an empty rest of a page.
Any question about a code example? Post it on LaTeX.org, I will answer. As forum admin I read every single question there. (profile link).
\documentclass{article}\pagestyle{empty}\begin{document}\section*{The golden ratio}The symbol for the golden ratio is the Greekletter \( \varphi \). Its value is the positive solutionof \( x^2 - x - 1 = 0 \).It can be calculated to:\[\varphi = \frac{1 + \sqrt{5}}{2} = 1.618 \ldots\]\end{document}
\documentclass{article}\pagestyle{empty}\begin{document}\[x^{n_1} \neq x^n_1\]\[{x^2}^3 = x^{2^3}\]\end{document}
\documentclass{article}\pagestyle{empty}\usepackage{mathtools}\mathtoolsset{showonlyrefs}\begin{document}\begin{gather}y = x^2 + 1 \label{xx}\\z = 0\end{gather}See \eqref{xx}.\end{document}
\documentclass{article}\usepackage{dsfont}\begin{document}\[\lim_{n\to\infty} \sup_{x\in\mathds{R}} f_n(x^2)< n \Big(\sum_{x\in\mathds{R}, n\in\mathds{N}}\big| f_n(x^2) \big| \Big)\]\end{document}
\documentclass{article}\usepackage{amsmath}\usepackage{amsthm}\usepackage{thmtools}\newtheorem{thm}{Theorem}\newtheorem{lem}[thm]{Lemma}\theoremstyle{definition}\newtheorem{dfn}[thm]{Definition}\theoremstyle{remark}\newtheorem*{note}{Note}\pagestyle{empty}\begin{document}\listoftheorems\newpage\begin{dfn}The longest side of a triangle with a right angleis called the \emph{hypotenuse}.\end{dfn}\begin{note}The other sides are called \emph{catheti},or \emph{legs}.\end{note}\begin{thm}[Pythagoras]\label{pythagoras}In any right triangle, the square of the hypotenuseequals the sum of the squares of the other sides.\end{thm}\begin{proof}The proof has been given in Euclid's Elements,Book 1, Proposition 47. Refer to it for details.The converse is also true, see lemma \ref{converse}.\end{proof}\begin{lem}\label{converse}For any three positive numbers \(x\), \(y\),and \(z\) with \(x^2 + y^2 = z^2\), there is atriangle with side lengths \(x\), \(y\) and \(z\).Such triangle has a right angle, and the hypotenusehas the length \(z\).\end{lem}\begin{note}This is the converse of theorem \ref{pythagoras}.\end{note}\end{document}
\documentclass[12pt]{beamer}\usefonttheme[onlymath]{serif}\usepackage{breqn}\begin{document}\begin{frame}\begin{dmath*}\left( \frac{f}{g} \right)^\prime(x)= \lim_{h \rightarrow 0}\left( \frac{1}{ g(x+h) g(x) } \right)\left[\frac{ f(x+h) - f(x) }{h} g(x)-\frac{ g(x+h) - g(x) }{h} f(x)\right]= \frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g^2(x)}\end{dmath*}\end{frame}\end{document}
\documentclass{article}\usepackage{dsfont}\usepackage{mathtools}\begin{document}\[\adjustlimits\lim_{n\to\infty} \sup_{x\in\mathds{R}}f_n(\cramped{x^2})< n \Big( \smashoperator{\sum_{x\in\mathds{R},n\in\mathds{N}}}\big\lvert f_n(\cramped{x^2}) \big\rvert \Big)\]\end{document}
\\documentclass[border=10pt]{standalone}\usepackage{tikz-cd}\usepackage{amsmath}\DeclareMathOperator{\im}{im}\pagestyle{empty}\begin{document}\begin{tikzcd}G \arrow[r, "\varphi"]\arrow[d, twoheadrightarrow, "\pi" left] & \im \varphi \\G/\ker \varphi \arrow[ru, hook, dashed, "\tilde{\varphi}" below] & \\\end{tikzcd}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{tikz-cd}\usepackage{amsmath}\DeclareMathOperator{\im}{im}\pagestyle{empty}\begin{document}\begin{tikzcd}G \arrow[r, "\varphi"]\arrow[d, "\pi" left] & \im \varphi \\G/\ker \varphi \arrow[ru, "\tilde{\varphi}" below]\end{tikzcd}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{amsmath}\usepackage{tikz}\usetikzlibrary{fit}\newcommand{\overlay}[2][]{\tikz[overlay,remember picture, #1]{#2}}\tikzset{highlighted/.style = { draw, thick, rectangle,rounded corners, inner sep = 0pt,fill = red!15, fill opacity = 0.5}}\newcommand{\highlight}[1]{%\overlay{\node [fit = (left.north west) (right.south east),highlighted] (#1) {}; }}\newcommand{\flag}[2]{\overlay[baseline=(#1.base)]{\node (#1) {$#2$};}}\begin{document}\[M = \begin{pmatrix}\flag{left}{1} & 2 & 3 & 4 & 5 \\6 & 7 & 8 & 9 & 10 \\11 & \flag{before}{12} & \flag{right}{13} & 14 & 15 \\16 & 17 & 18 & 19 & 20\end{pmatrix}\highlight{N}\qquadM^T = \begin{pmatrix}\flag{left}{1} & 6 & 11 & 16 \\2 & 7 & \flag{after}{12} & 17 \\3 & 8 & \flag{right}{13} & 18 \\4 & 9 & 14 & 19 \\5 & 10 & 15 & 20\end{pmatrix}\highlight{NT}\]\overlay{\draw[->, thick, red, dotted] (before) -- (after);\draw[->, thick, red, dashed] (N) -- (NT)node [pos=0.68, above] {Transpose};\node[above of = N ] { $N$ };\node[above of = NT] { $N^T$ };}\end{document}
\documentclass{article}\usepackage{amsmath}\usepackage{amsthm}\newtheorem{thm}{Theorem}\newtheorem{lem}{Lemma}\theoremstyle{definition}\newtheorem{dfn}{Definition}\theoremstyle{remark}\newtheorem*{note}{Note}\begin{document}\begin{dfn}The longest side of a triangle with a right angleis called the \emph{hypotenuse}.\end{dfn}\begin{note}The other sides are called \emph{catheti},or \emph{legs}.\end{note}\begin{thm}[Pythagoras]\label{pythagoras}In any right triangle, the square of the hypotenuseequals the sum of the squares of the other sides.\end{thm}\begin{proof}The proof has been given in Euclid's Elements,Book 1, Proposition 47. Refer to it for details.The converse is also true, see lemma \ref{converse}.\end{proof}\begin{lem}\label{converse}For any three positive numbers \(x\), \(y\),and \(z\) with \(x^2 + y^2 = z^2\), there is atriangle with side lengths \(x\), \(y\) and \(z\).Such triangle has a right angle, and the hypotenusehas the length \(z\).\end{lem}\begin{note}This is the converse of theorem \ref{pythagoras}.\end{note}\end{document}
\documentclass{article}\usepackage{amsmath}\usepackage{amsthm}\usepackage{amsthm}\newtheorem{thm}{Theorem}\newtheorem{lem}[thm]{Lemma}\theoremstyle{definition}\newtheorem{dfn}[thm]{Definition}\theoremstyle{remark}\newtheorem*{note}{Note}\begin{document}\begin{dfn}The longest side of a triangle with a right angleis called the \emph{hypotenuse}.\end{dfn}\begin{note}The other sides are called \emph{catheti},or \emph{legs}.\end{note}\begin{thm}[Pythagoras]\label{pythagoras}In any right triangle, the square of the hypotenuseequals the sum of the squares of the other sides.\end{thm}\begin{proof}The proof has been given in Euclid's Elements,Book 1, Proposition 47. Refer to it for details.The converse is also true, see lemma \ref{converse}.\end{proof}\begin{lem}\label{converse}For any three positive numbers \(x\), \(y\),and \(z\) with \(x^2 + y^2 = z^2\), there is atriangle with side lengths \(x\), \(y\) and \(z\).Such triangle has a right angle, and the hypotenusehas the length \(z\).\end{lem}\begin{note}This is the converse of theorem \ref{pythagoras}.\end{note}\end{document}
\documentclass{article}\usepackage{amsmath}\usepackage[amsmath,amsthm,thmmarks]{ntheorem}\newtheorem{thm}{Theorem}\newtheorem{lem}{Lemma}\theoremstyle{definition}\newtheorem{dfn}{Definition}\theoremstyle{remark}\newtheorem{note}{Note}\begin{document}\begin{dfn}The longest side of a triangle with a right angleis called the \emph{hypotenuse}.\end{dfn}\begin{note}The other sides are called \emph{catheti},or \emph{legs}.\end{note}\begin{thm}[Pythagoras]\label{pythagoras}In any right triangle, the square of the hypotenuseequals the sum of the squares of the other sides.\end{thm}\begin{proof}The proof has been given in Euclid's Elements,Book 1, Proposition 47. Refer to it for details.The converse is also true, see lemma \ref{converse}.\end{proof}\begin{lem}\label{converse}For any three positive numbers \(x\), \(y\),and \(z\) with \(x^2 + y^2 = z^2\), there is atriangle with side lengths \(x\), \(y\) and \(z\).Such triangle has a right angle, and the hypotenusehas the length \(z\).\end{lem}\begin{note}This is the converse of theorem \ref{pythagoras}.\end{note}\end{document}
\documentclass{article}\usepackage{amsmath}\usepackage{amsthm}\usepackage{thmtools}\usepackage{xcolor}\declaretheorem[shaded={bgcolor=red!15}]{Theorem}\declaretheorem[thmbox=L]{Definition}\begin{document}\begin{Definition}The longest side of a triangle with a right angleis called the \emph{hypotenuse}.\end{Definition}\begin{Theorem}[Pythagoras]\label{pythagoras}In any right triangle, the square of the hypotenuseequals the sum of the squares of the other sides.\end{Theorem}\end{document}
\documentclass{article}\usepackage{tikz}\usetikzlibrary{matrix}\usepackage{amsmath}\DeclareMathOperator{\im}{im}\begin{document}\begin{tikzpicture}\matrix (m)[matrix of math nodes,row sep = 3em,column sep = 4em]{G & \im \varphi \\G/\ker \varphi & \\};\path[-stealth](m-1-1) edge node [left] {$\pi$} (m-2-1)(m-1-1.east |- m-1-2)edge node [above] {$\varphi$} (m-1-2)(m-2-1) edge node [below] {$\tilde{\varphi}$}(m-1-2);\end{tikzpicture}\end{document}
\documentclass{article}\usepackage{tikz}\usetikzlibrary{matrix,arrows.meta}\usepackage{amsmath}\DeclareMathOperator{\im}{im}\begin{document}\begin{tikzpicture}\matrix (m)[matrix of math nodes,row sep = 3em,column sep = 4em]{G & \im \varphi \\G/\ker \varphi & \\};\path(m-1-1) edge [->>] node [left] {$\pi$} (m-2-1)(m-1-1.east |- m-1-2)edge [->] node [above] {$\varphi$} (m-1-2)(m-2-1.east) edge [{Hooks[right,length=0.8ex]}->,dashed] node [below] {$\tilde{\varphi}$} (m-1-2);\end{tikzpicture}\end{document}
\documentclass{article}\usepackage{tikz}\usetikzlibrary{matrix,calc}\begin{document}\begin{tikzpicture}[-stealth,label/.style = { font=\footnotesize }]\matrix (m)[matrix of math nodes,row sep = 4em,column sep = 4em]{A_0 & A_1 & A_2 & A_3 & A_4 \\B_0 & B_1 & B_2 & B_3 & B_4 \\};\foreach \i in {1,...,4} {\pathlet \n1 = { int(\i+1) } in(m-1-\i) edge node [above, label] {$f_\i$} (m-1-\n1)(m-2-\i) edge node [below, label] {$f^\prime_\i$} (m-2-\n1)(m-1-\i) edge node [left, label] {$g_\i$} (m-2-\i);}\path (m-1-5) edge node [left, label] {$g_5$} (m-2-5);\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\begin{document}\begin{tikzpicture}\begin{axis} [axis lines=center]\addplot [domain=-3:3, smooth, thick] { x^3 - 5*x };\end{axis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\begin{document}\begin{tikzpicture}\begin{axis} [grid, xtick = {-360,-270,...,360}]\addplot [domain=-360:360, samples=100, thick] { sin(x) };\end{axis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\usepgfplotslibrary{polar}\begin{document}\begin{tikzpicture}\begin{polaraxis}[hide axis]\addplot[domain=0:180,smooth] {sin(x)};\end{polaraxis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\usepgfplotslibrary{polar}\begin{document}\begin{tikzpicture}\begin{polaraxis}\addplot[domain=0:360,samples=300] {sin(6*x)};\end{polaraxis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\begin{document}\begin{tikzpicture}\begin{axis} [title = {$f(x,y) = \sin(x)\sin(y)$},xtick = {0,90,...,360},ytick = {90,180,...,360},xlabel = $x$, ylabel = $y$,ticklabel style = {font = \scriptsize},grid]\addplot3 [surf, domain=0:360, samples=60]{ sin(x)*sin(y) };\end{axis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{pgfplots}\usepgfplotslibrary{shift}\begin{document}\begin{tikzpicture}\begin{axis} [shift3d]\addplot3 [surf, colormap/hot2, domain = -2:2, samples = 50]{ x/exp(x^2+y^2) };\end{axis}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{tkz-euclide}\begin{document}\begin{tikzpicture}\tkzDefPoint(0,0){A}\tkzDefPoint(4,1){B}\tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D}\tkzDrawPolygon(A,B,C)\tkzDrawPoints(A,B,C,D)\tkzLabelPoints[below left](A)\tkzLabelPoints(B,D)\tkzLabelPoint[above](C){$C$}\tkzDrawCircle[dotted](A,B)\tkzDrawCircle[dotted](B,A)\tkzCompass[color=red, very thick](A,C)\tkzCompass[color=red, very thick](B,C)\tkzCompass[color=red, very thick](A,D)\tkzCompass[color=red, very thick](B,D)\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{tkz-euclide}\begin{document}\begin{tikzpicture}\tkzDefPoint(0,0){A}\tkzDefPoint(4,1){B}\tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D}\tkzDrawPolygon(A,B,C)\tkzDrawPoints(A,B,C,D)\tkzLabelPoints[below left](A)\tkzLabelPoints(B,D)\tkzLabelPoint[above](C){$C$}\tkzDrawCircle[dotted](A,B)\tkzDrawCircle[dotted](B,A)\tkzCompass[color=red, very thick](A,C)\tkzCompass[color=red, very thick](B,C)\tkzCompass[color=red, very thick](A,D)\tkzCompass[color=red, very thick](B,D)\tkzDrawArc[fill=blue!10,thick](A,B)(C)\tkzDrawArc[fill=blue!10,thick](B,C)(A)\tkzDrawArc[fill=blue!10,thick](C,A)(B)\tkzInterLC(A,B)(B,A)\tkzGetPoints{F}{E}\tkzDrawPoints(E)\tkzLabelPoints(E)\tkzDrawPolygon(A,E,D)\tkzMarkAngles[fill=yellow,opacity=0.5](D,A,E A,E,D)\tkzMarkRightAngle[size=0.65,fill=red,opacity=0.5](A,D,E)\tkzLabelAngle[pos=0.7](D,A,E){$\alpha$}\tkzLabelAngle[pos=0.8](A,E,D){$\beta$}\tkzLabelAngle[pos=0.5,xshift=-1.4mm](A,D,D){$90^\circ$}\tkzLabelSegment[below=0.6cm,align=center,font=\small](A,B){Reuleaux\\triangle}\tkzLabelSegment[above right,sloped,font=\small](A,E){hypotenuse}\tkzLabelSegment[below,sloped,font=\small](D,E){opposite}\tkzLabelSegment[below,sloped,font=\small](A,D){adjacent}\tkzLabelSegment[below right=4cm,font=\small](A,E){Thales circle}\end{tikzpicture}\end{document}
\documentclass[border=10pt]{standalone}\usepackage{tkz-euclide}\begin{document}\begin{tikzpicture}\tkzDefPoints{0/0/A, 5/0/B, 1/4/C}\tkzDefCircle[in](A,B,C)\tkzGetPoint{M}\tkzGetLength{r}\tkzDefCircle[circum](A,B,C)\tkzGetPoint{N}\tkzGetLength{R}\tkzDefPointBy[projection=onto A--B](M)\tkzGetPoint{a}\tkzDefPointBy[projection=onto B--C](M)\tkzGetPoint{b}\tkzDefPointBy[projection=onto A--C](M)\tkzGetPoint{c}\tkzDrawCircle[R](M,\r pt)\tkzDrawCircle[R](N,\R pt)\tkzDrawPolygon[very thick](A,B,C)\tkzDrawLines[dotted](N,A N,B N,C)\tkzDrawLines[dashed](M,a M,b M,c)\tkzMarkRightAngles(M,a,B M,b,C M,c,C)\tkzDrawPoints(A,B,C,M,N,a,b,c)\tkzLabelPoints[below left](A,M,a,c)\tkzLabelPoints[below right](B)\tkzLabelPoints[above](C,b)\tkzLabelPoints[below](N)\end{tikzpicture}\end{document}
\documentclass{article}\usepackage{tikz}\pagestyle{empty}\begin{document}\noindentIn a right-angled triangle the two shortest sidesgot widths of 3 and 7, respectively. Then, thelongest side has a width of \pgfmathparse{min(3^2 + 7^2)}\pgfmathresult.The smallest angle is about \pgfmathparse{atan(3/7)}\pgfmathprintnumber[precision=2]{\pgfmathresult} degrees.\end{document}
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