How could I typeset a formula like this:
It should respect typographic rules, but it must look very similar.
You could use the aligned environment (requires the amsmath package), e.g.
\left. \begin{aligned}
A &= T \\
B &= F
\end{aligned} \right\} \implies A \vee B = T \vee F = T
(Assuming you meant to use logical disjunction and logical implication)
\[
\left.\begin{array}{l}
A = T \\ B = F
\end{array}\right\}
\implies A \lor B = T \lor F = T
\]
This code should do what you want:
\[
\left.
\begin{array}{c}
A = T\\
B = F
\end{array}
\right\}
\Rightarrow A \vee B = T \vee F = T
\]
Related
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}
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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}
I am a beginner in latex.I have the following piece of latex code. The code is working fine but I want that all the equality operators of each equation should be aligned. How can it be done?
\begin{enumerate}
\item[Commutative] $a+b = b+a$
\item[Associative] $a+(b+c)=(a+b)+c$
\item[Distributive] $a(b+c)=ab+ac$
\end{enumerate}
I'm unsure if this is possible inside the enumerate environment. An easy alternative is to use a tabular environment instead. In the example below the left and right sides of the equation are contained in two separate columns, with an = appearing between them.
\documentclass[12pt]{article}
\begin{document}
\begin{tabular}{l r#{$=$}l}
Commutative & $a+b$ & $b+a$ \\
Associative & $a+(b+c)$ & $(a+b)+c$ \\
Distributive & $a(b+c)$ & $ab+ac$ \\
\end{tabular}
\end{document}
Using \eqmakebox[<tag>][<align>] (from eqparbox) you can have all elements under the same <tag> be placed in a box of maximum width, together with individual <align>ment as needed. Below I has \eqmakebox[LHS][r] to ensure all elements tagged LHS is right-aligned. The result is alignment around the =.
\documentclass{article}
\usepackage{eqparbox}
\begin{document}
\begin{enumerate}
\item[Commutative] $ a + b = b + a $
\item[Associative] $a + (b + c) = (a + b) + c$
\item[Distributive] $ a(b + c) = ab + ac $
\end{enumerate}
\begin{enumerate}
\item[Commutative] $ \eqmakebox[LHS][r]{$a + b$} = b + a $
\item[Associative] $\eqmakebox[LHS][r]{$a + (b + c)$} = (a + b) + c$
\item[Distributive] $ \eqmakebox[LHS][r]{$a(b + c)$} = ab + ac $
\end{enumerate}
\end{document}
Alternatively you can measure the widest element yourself:
\newlength{\widestelement}
\settowidth{\widestelement}{$a + (b + c)$}
and then use
\begin{enumerate}
\item[Commutative] $ \makebox[\widestelement][r]{$a + b$} = b + a $
\item[Associative] $\makebox[\widestelement][r]{$a + (b + c)$} = (a + b) + c$
\item[Distributive] $ \makebox[\widestelement][r]{$a(b + c)$} = ab + ac $
\end{enumerate}
I am new to R markdown. I have a simple question about how to remove the automated equation numbering in the output PDF. Use the following codes for example:
\begin{align}
\sum_{i = 1}^{n} | (f + g )(x_i) - (f + g)(x_{i-1}) | &= \sum_{i = 1}^{n} | \{ f(x_i) + g(x_i) \} - \{ f(x_{i - 1} + g(x_{i-1}) \} |
\\ &\leq \sum_{i = 1}^{n} | f(x_i) - f(x_{i - 1} | + \sum_{i = 1}^{n} | g(x_i) - g(x_{i - 1})|
\\ &\leq V(f, P) + V(g, P)
\end{align}
It will automatically label the three lines of equations with (1), (2), (3). How do I suppress the numbering?
Thanks for your time!
I was actually looking up how to add numbering... so perfect info swap! Try:
$$
\begin{aligned}
\sum_{i = 1}^{n} | (f + g )(x_i) - (f + g)(x_{i-1}) | &= \sum_{i = 1}^{n} | \{ f(x_i) + g(x_i) \} - \{ f(x_{i - 1} + g(x_{i-1}) \} |
\\ &\leq \sum_{i = 1}^{n} | f(x_i) - f(x_{i - 1} | + \sum_{i = 1}^{n} | g(x_i) - g(x_{i - 1})|
\\ &\leq V(f, P) + V(g, P)
\end{aligned}
$$
This is more of a LaTeX thing than an RMarkdown thing: from my understanding, the $$ $$ isn't numbered, and the aligned environment is intended for use within other environments in order to display a single equation.
So, within the $$ $$ environment, you can use the aligned environment to display an equation, but it won't be numbered.
I missed the response from Werner earlier, but using align* works too. I guess the difference is the nuance in the initial intention.
(Also btw, this means if you use aligned within align, you get a block equation with only one number.)
I have a simple alignment question for Latex tables. At
the moment it looks as follows:
\begin{center}
\begin{tabular}{| c | l | l | }
\hline
\tt {a} & $a = b + c + d + e + f + g + $ \\
& $ e + f + g + h + i + j$ \\ \hline
\end{tabular}
\end{center}
The problem is, that the output looks as follows:
a = b + c + ...
e + f + ....
However, I would like to have it looks as
a = b + c + ...
e + f + ...
Anyone an idea how I could do that in a table?
Thanks,
Klaus
Please use the align environment for multiline equations.
You can add a \phantom{a = } to indicate the should-be-existing spaces.
& $\phantom{a = }e + f + g + h + i + j$ \\ \hline
As KennyTM pointed out, you shouldn't typeset multiline equations using tables like this. But if you must do it, you could do it like this:
\begin{center}
\begin{tabular}{| c | l #{} l | }
\hline
\tt {a} & $a =\;$ & $b + c + d + e + f + g + $ \\
& & $e + f + g + h + i + j$ \\ \hline
\end{tabular}
\end{center}
The usual way is to make an additional column for the a = part; right-aligning it, and removing spacing with the next column for aesthetics:
\begin{tabular}{ c r #{} l } % you have one superfluous l
\tt {a} & $a =$ & $b + c + d + e + f + g +$ \\
& & $e + f + g + h + i + j$ \\
\end{tabular}
Maybe you will need an explicit space after the = sign, so that it's properly spaced with the b.
Another solution would be to have a multiline equation in a single cell of the table, but that amounts to the same (you will need an array environment or something similar to wrap the left part).