How to use Latex to create a notation style like this image. or any other suggestions ?
\documentclass{article}
% See http://tex.stackexchange.com/questions/112576/math-mode-in-tabular-without-having-to-use-everywhere
\usepackage{amstext}
\usepackage{array}
\usepackage{amssymb}
\newcolumntype{L}{>{$}l<{$}}
\begin{document}
\textbf{Typing rules for F$_1$}
\begin{tabular}{LLL}
\hline
\text{\footnotesize(Env $\varnothing$)} &
\text{\footnotesize(Env $x$)}\\
\frac{}{\varnothing \vdash \diamond} &
\frac{E \vdash A \quad x \notin dom(E)}{E, x:A \vdash \diamond} \\
&&\\
\text{\footnotesize(Type Const} &
\text{\footnotesize(Type Arrow)}\\
\frac{E \vdash \diamond}{E \vdash K} &
\frac{E \vdash A \quad E \vdash B}{E \vdash A \rightarrow B} \\
&&\\
\text{\footnotesize(Val $x$)} &
\text{\footnotesize(Val Fun)} &
\text{\footnotesize(Val Appl)} \\
\frac{E\vdash \diamond}{E \vdash x :E(x)} &
\frac{E,x:A\vdash b:B}{E \vdash \lambda(x:A)b :A \rightarrow B} &
\frac{E \vdash b:A \rightarrow B \quad E \vdash a : A}{E \vdash b(a) : B} \\
\hline
\end{tabular}
\end{document}
Related
Hello I'm trying to use cells in latex.
image of my cell here
How do you put the "Func2" and the "-" centered in the cell?
The code I'm using is:
\begin{tabularx}{0.8\textwidth} {
| >{\raggedright\arraybackslash}X
| >{\centering\arraybackslash}X
| >{\raggedleft\arraybackslash}X | }
\hline
\multicolumn{3}{|c|}{2D - $L = 64$} \\
\hline
Speedup(x) & Func1 & Func2 \\
\hline
Func2 & 35.4 & - \\
\hline
Func3 & 322.8 & 9.1 \\
\hline
\end{tabularx}
You could use the tabularray package. This makes it easy to change the alignment for individual cells:
\documentclass{article}
\usepackage{tabularray}
\begin{document}
\begin{tblr}{
width={0.8\textwidth},
colspec={|X|X[halign=c]|X[halign=r]|}
}
\hline
\SetCell[c=3]{halign=c} 2D - $L = 64$ \\
\hline
Speedup(x) & Func1 & \SetCell{halign=c} Func2 \\
\hline
Func2 & 35.4 & \SetCell{halign=c} - \\
\hline
Func3 & 322.8 & 9.1 \\
\hline
\end{tblr}
\end{document}
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\noindent
A =
\bigg[
\begin{smallmatrix}
1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0 \\
0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0 \\
0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0 \\
0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0 \\
0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0 \\
0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0 \\
0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0 \\
0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1 \\
\end{smallmatrix}
\bigg]
\noindent
You can use the tabularray package with its math library:
\documentclass{article}
\usepackage{amsmath}
\usepackage{tabularray}
\UseTblrLibrary{amsmath}
\begin{document}
\[
A =
\begin{+bmatrix}
1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0 \\
0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0 \\
0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0 \\
0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0 \\
0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0 \\
0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0 \\
0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0 \\
0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1 \\
\end{+bmatrix}
\]
\end{document}
I am trying to typeset the following equation in the align-evoirement
\begin{align}
t_2' &= t_2 + \frac{L/C} \\
t_1' &= t_1 + \frac{L + v\Delta t\cos\theta}{c} \\
t_2' - t_1' &= (t_2-t_1) + \frac{L-L-v\Delta t\cos\theta}{C} \\
\Delta t' &= \Delta t - \frac{v\Delta t \cos \theta}{c} \\
\Delta t' &= \Delta t \left(1-\frac{v\cos\theta}{c}\right) \\
\frac{1}{\nu} &= \frac{1}{\nu_0 \sqrt{1-\frac{v^2}{c^2}}}~\left(1-\frac{v\cos \theta {c}\right) \\
\nu &= \nu_0\frac{ \sqrt{1-\dfrac{v^2}{c^2}}}{1-\dfrac{v\cos \theta }{c}}
\end{align}
But when I tried this I got the following message:
[45]
! Missing } inserted.
<inserted text>
}
l.938 \end{align}
?
I copied the equations in Matcha (without the &), where it was perfectly working... I tried some things, but those did not seem to work...
Does anybody know what I did wrong?
At two occasions you write \frac{..} without the mandatary second argument. You must write \frac{...}{...} instead.
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
t_2' &= t_2 + \frac{L}{C} \\
t_1' &= t_1 + \frac{L + v\Delta t\cos\theta}{c} \\
t_2' - t_1' &= (t_2-t_1) + \frac{L-L-v\Delta t\cos\theta}{C} \\
\Delta t' &= \Delta t - \frac{v\Delta t \cos \theta}{c} \\
\Delta t' &= \Delta t \left(1-\frac{v\cos\theta}{c}\right) \\
\frac{1}{\nu} &= \frac{1}{\nu_0 \sqrt{1-\frac{v^2}{c^2}}}~\left(1-\frac{v\cos \theta}{c}\right) \\
\nu &= \nu_0\frac{ \sqrt{1-\dfrac{v^2}{c^2}}}{1-\dfrac{v\cos \theta }{c}}
\end{align}
\end{document}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{proof}
\usepackage[left=0.60in, right=0.60in, top=0.65in, bottom=0.65in, footskip=0.1in]{geometry}
\title{Peergrade 1}
\author{anonymous}
\date{September 2020}
\begin{document}
\maketitle
\section{Exercise 1}
\begin{tiny}
\begin{table}[h]
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
A & B & C & A -\textgreater B & B -\textgreater $\sim$C & A -\textgreater $\sim$C & C -\textgreater $\sim$A & (A -\textgreater $\sim$C) /\textbackslash (C -\textgreater $\sim$A) & (B -\textgreater $\sim$C) -\textgreater ((A -\textgreater $\sim$C) /\textbackslash (C -\textgreater $\sim$A)) & Prop 1 \\ \hline
T & T & T & T & F & F & F & F & T & T \\ \hline
T & T & F & T & T & T & T & T & T & T \\ \hline
T & F & T & F & F & F & F & F & T & T \\ \hline
T & F & F & F & T & T & T & T & T & T \\ \hline
F & T & T & T & F & T & T & T & T & T \\ \hline
F & T & F & T & T & T & T & T & T & T \\ \hline
F & F & T & T & F & T & T & T & T & T \\ \hline
F & F & F & T & T & T & T & T & T & T \\ \hline
\end{tabular}
\end{table}
\end{tiny}
\end{document}
Can anyone help me so that the table fits in LaTeX? It almost fits, but half of the last column part goes outiside the paper. I have already tried to add /small and /footnotesize, but it doesnt work.
Please have a look at http://betterposters.blogspot.de/2012/08/the-data-prison.html how to design nice looking tables
You could use a tabularx and let latex adjust the size:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{proof}
\usepackage[left=0.60in, right=0.60in, top=0.65in, bottom=0.65in, footskip=0.1in]{geometry}
\usepackage{tabularx}
\title{Peergrade 1}
\author{anonymous}
\date{September 2020}
\begin{document}
\maketitle
\section{Exercise 1}
\begin{tiny}
\begin{table}[h]
\begin{tabularx}{\linewidth}{|l|l|l|l|l|l|l|l|X|l|}
\hline
A & B & C & A -\textgreater B & B -\textgreater $\sim$C & A -\textgreater $\sim$C & C -\textgreater $\sim$A & (A -\textgreater $\sim$C) /\textbackslash (C -\textgreater $\sim$A) & (B -\textgreater $\sim$C) -\textgreater ((A -\textgreater $\sim$C) /\textbackslash (C -\textgreater $\sim$A)) & Prop 1 \\ \hline
T & T & T & T & F & F & F & F & T & T \\ \hline
T & T & F & T & T & T & T & T & T & T \\ \hline
T & F & T & F & F & F & F & F & T & T \\ \hline
T & F & F & F & T & T & T & T & T & T \\ \hline
F & T & T & T & F & T & T & T & T & T \\ \hline
F & T & F & T & T & T & T & T & T & T \\ \hline
F & F & T & T & F & T & T & T & T & T \\ \hline
F & F & F & T & T & T & T & T & T & T \\ \hline
\end{tabularx}
\end{table}
\end{tiny}
\end{document}
Just complementing the answer, if the table is very large, it's possible to rotate the sheet by using pdflscape:
\usepackage{pdflscape}
...
\begin{landscape}
(Your Table)
\end{landscape}
My solution:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{proof}
\usepackage[left=0.60in, right=0.60in, top=0.65in, bottom=0.65in, footskip=0.1in]{geometry}
\title{Peergrade 1}
\author{anonymous}
\date{September 2020}
\begin{document}
\maketitle
\section{Exercise 1}
%\begin{tiny}
\begin{table}[h]
\begin{tabular}{|l|l|l|l|l|l|l|p{20mm}|p{22mm}|l|}
\hline
A & B & C & A -\textgreater B & B -\textgreater $\sim$C %
& A -\textgreater $\sim$C & C -\textgreater $\sim$A %
& (A -\textgreater $\sim$C) \par /\textbackslash (C -\textgreater $\sim$A) %
& (B -\textgreater $\sim$C) \par -\textgreater ((A -\textgreater $\sim$C) \par %
/\textbackslash (C -\textgreater $\sim$A)) & Prop 1 \\ \hline
T & T & T & T & F & F & F & F & T & T \\ \hline
T & T & F & T & T & T & T & T & T & T \\ \hline
T & F & T & F & F & F & F & F & T & T \\ \hline
T & F & F & F & T & T & T & T & T & T \\ \hline
F & T & T & T & F & T & T & T & T & T \\ \hline
F & T & F & T & T & T & T & T & T & T \\ \hline
F & F & T & T & F & T & T & T & T & T \\ \hline
F & F & F & T & T & T & T & T & T & T \\ \hline
\end{tabular}
\end{table}
%\end{tiny}
\end{document}
I have only edited the alignment of the 8th and 9th column (p{20mm}|p{22mm}) and added three \par in the first cell in such columns (out of dollar signs, you're typing text in the cells).
Anyway the solution with tabularx by #samcarter is very good.
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I am a beginner in latex. I have the following piece of latex code. The code is working fine but I wish all the equality operators and the "if kflag=n" of each equation be aligned and written in one equation box with one equation counting number. How can it be done?
\begin{equation} %kflag=0
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=0
\end{equation}
\begin{equation} %kflag=1
\left \{
\begin{array}{rl}
T =& \alpha\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{\alpha\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=1
\end{equation}
\begin{equation} %kflag=2
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}+T_{max}-\\
&(1-D)\sigma_{max,0}\times e\times\frac{\Delta_{max}}{\delta_n}+\\
&10\times \sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& 11\times\frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=2
\end{equation}
\begin{equation} %kflag=3
\left \{
\begin{array}{rl}
T =& T_{max}+K\times(\Delta_n-\Delta_{max})\\
K =& \frac{(1-D)\sigma_{max,0}*e}{\delta_n}\\
\end{array}
\right.
\quad \text{if} \quad kflag=3
\end{equation}
Right now the equations look like this
Here's an option using a multitude of nested structures - equation for the numbering, aligned for the horizontal alignment of structures and dcases (or cases) for the left-braced content.
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\newcommand{\Ddn}{\frac{\Delta_n}{\delta_n}}
\newcommand{\smz}{\sigma_{\mathrm{max}, 0}}
\begin{equation}
\begin{aligned}
&\begin{dcases}
\phantom{K}\mathllap{T} = (1 - D) \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 0 \\ % k-flag = 0
&\begin{dcases}
\phantom{K}\mathllap{T} = \alpha \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{\alpha \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 1 \\ % k-flag = 1
&\begin{dcases}
\phantom{K}\mathllap{T} = \begin{aligned}[t]
&(1 - D) \smz \times e^{1 - \Ddn} \times \Ddn + T_{\mathrm{max}} \\
&{} - (1 - D) \smz \times e \times \frac{\Delta_{\mathrm{max}}}{\delta_n} \\
&{} + 10 \times \smz \times e^{1 - \Ddn} \times \Ddn
\end{aligned} \\
K = 11 \times \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 2 \\ % k-flag = 2
&\begin{dcases}
\phantom{K}\mathllap{T} = T_{\mathrm{max}} + K \times (\Delta_n - \Delta_{\mathrm{max}}) \\
K = \frac{(1 - D) \smz \times e}{\delta_n}
\end{dcases} & \text{if $k$-flag} = 3
\end{aligned}
\end{equation}
\end{document}