TeX - LaTeX Asked by Fel on April 24, 2021
I want to write something on the left side of the cases brackets like shown in the picture. The code I am using:
begin{equation}
begin{split}
& E(T,alpha) =
&
begin{cases}
1 & text{$alpha < alpha_{vi}$}
textrm{$E_0(T)$} & textrm{$alpha$ = $alpha_{vi}$}
text{$E_0(T) + (alpha-alpha_{vi})(E_infty(T)-E_{0}(T))$} & text{$alpha_{vi} < alpha < 1$}
text{$E_infty(T)$} & text{$alpha = 1$}
end{cases}
end{split}
label{eq:chile}
end{equation}
One possibility is:
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
begin{split}
& E(T,alpha) =
&
begin{matrix}
textsf{I} [0.5ex]
textsf{II} [0.5ex]
textsf{III} [0.5ex]
textsf{IV}
end{matrix}quad
begin{cases}
1 & text{$alpha < alpha_{vi}$}
textrm{$E_0(T)$} & textrm{$alpha$ = $alpha_{vi}$}
text{$E_0(T) + (alpha-alpha_{vi})(E_infty(T)-E_{0}(T))$} & text{$alpha_{vi} < alpha < 1$}
text{$E_infty(T)$} & text{$alpha = 1$}
end{cases}
end{split}
end{equation}
end{document}
Answered by Zarko on April 24, 2021
A possibility with {NiceArray}
of nicematrix
. The output will be correct even if you change the height or the depth or the rows of array.
documentclass{article}
usepackage{nicematrix}
begin{document}
begin{equation}
begin{split}
& E(T,alpha) =
&
begin{NiceArray}{>{color{red}}llc}
text{I} & 1 & alpha < alpha_{vi}
text{II} & E_0(T) & alpha$ = $alpha_{vi}
text{III} & E_0(T) + (alpha-alpha_{vi})(E_infty(T)-E_{0}(T)) & alpha_{vi} < alpha < 1
text{IV} & E_infty(T) & alpha = 1
CodeAfter
SubMatrix{{}{1-2}{4-2}{.}
end{NiceArray}
end{split}
label{eq:chile}
end{equation}
end{document}
You need several compilations (because nicematrix
uses PGF/Tikz under the hood).
Answered by F. Pantigny on April 24, 2021
Still another solution, without cases, replaced with blkarray
and the bigdelim
package:
documentclass{article}
usepackage{amsmath}
usepackage{blkarray}
usepackage{bigdelim}
usepackage[svgnames]{xcolor}
begin{document}
begin{equation}
begin{split}
& E(T,alpha) =
&
begin{blockarray}{>{color{IndianRed}mathsf{Roman{BAenumi}}}l@{;}c@{}ll}
& ldelim{{4}{*}& 1 & alpha < alpha_{vi}
& & E_0(T) & alpha = alpha_{vi}
& & E_0(T) + (alpha-alpha_{vi})(E_infty(T)-E_{0}(T)) & alpha_{vi} < alpha < 1
& & E_infty(T) & alpha = 1
end{blockarray}
end{split}
end{equation}
end{document}
Answered by Bernard on April 24, 2021
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