我希望把这些东西都缩小一点，这样它们就能放在同一页上。也许可以把两个列表放在一起，但我不知道怎么做
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}
\usepackage{mathtools}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{apacite}
\begin{document}
\begin{align*}
\ce{i &-> i + r_I, \\
r_i &-> r_I + I, \\
I + Lac &-> I \cdot Lac, \\
I \cdot Lac &-> I + Lac, \\
I + o &-> I \cdot o, \\
I \cdot o &-> I + o, \\
o + RNAP &-> o \cdot RNAP, \\
o \cdot RNAP &-> o + RNAP, \\
o \cdot RNAP &-> o + RNAP + r, \\
r &-> r + Z, \\
Lac + Z &-> Z, \\
r_I &-> \emptyset, \\
I &-> \emptyset, \\
I \cdot Lac &-> Lac, \\
r &-> \emptyset, \\
z &-> \emptyset
}
\end{align*}
where
\begin{table}[ht]
%\caption{Nonlinear Model Results} % title of Table
\centering % used for centering table
\begin{tabular}{c c} % centered columns (4 columns)
\hline\hline %inserts double horizontal lines
Species name & Symbol \\ [0.5ex] % inserts table
%heading
\hline % inserts single horizontal line
Regulatory gene & $i$ \\ [1ex] % inserting body of the table
Repressor protein & $r_I$ \\ [1ex]
Inhibitor protein & $I$ \\ [1ex]
Lactose & $Lac$ \\ [1ex]
Operator region & $o$ \\ [1ex]
RNA polymerase & $RNAP$ \\ [1ex]
mRNA & $r$ \\ [1ex]
$\beta$-galactosidase & $Z$ \\ [1.5ex] % [1ex] adds vertical space
\hline %inserts single line
\end{tabular}
\caption{Lac Operon species}
%\label{table:nonlin} % is used to refer this table in the text
\end{table}
Using SPN theory the system is represented as
\[N \, = \, (P,T,\text{Pre},\text{Post},M,h,c), \quad
P = \scriptsize{\begin{pmatrix}
i \\
r_I \\
I \\
Lac \\
I \cdot Lac \\
o \\
I \cdot o \\
RNAP \\
RNAP \cdot o \\
r \\
Z \\
\end{pmatrix}}
%, \quad
% T = \begin{pmatrix}
% \text{Enzyme-substrate association} \\
% \text{Enzyme-substrate dissociation} \\
% \text{Product formation} \\
% \end{pmatrix}
\] \\
\[ T = \scriptsize{\begin{pmatrix}
\text{Inhibitor transcription} \\
\text{Inhibitor translation} \\
\text{Lactose inhibitor binding} \\
\text{Lactose inhibitor dissociation} \\
\text{Inhibitor binding} \\
\text{Inhibitor dissociation} \\
\text{RNAP binding} \\
\text{RNAP dissociation} \\
\text{Transcription} \\
\text{Translation} \\
\text{Conversion} \\
\text{Inhibitor RNA degradation} \\
\text{Inhibitor degradation} \\
\text{Lactose inhibitor degradation} \\
\text{RNA degradation} \\
\text{z degradation} \\
\end{pmatrix}} \\ \]
\setcounter{MaxMatrixCols}{20}
\[ Pre = \scriptsize{\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}}, \quad
Post = \scriptsize{\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}} \] \\
\[M = \scriptsize{\begin{pmatrix}
1 \\
0 \\
50 \\
20 \\
0 \\
1 \\
0 \\
100 \\
0 \\
0 \\
0 \\
\end{pmatrix}}, \quad
c = \scriptsize{\begin{pmatrix}
0.02 \\
0.1 \\
0.005 \\
0.1 \\
1 \\
0.01 \\
0.1 \\
0.01 \\
0.03 \\
0.1 \\
1e-05 \\
0.01 \\
0.002 \\
0.01 \\
0.001
\end{pmatrix}}, \quad
h(x,c) = \scriptsize{\begin{pmatrix}
c_1i \\
c_2r_I \\
c_3ILac \\
c_4I \cdot Lac \\
c_5Io \\
c_6I \cdot o \\
c_7oRNAP \\
c_8o \cdot RNAP \\
c_9o \cdot RNAP \\
c_{10}r \\
c_{11}Lacz \\
c_{12}r_I \\
c_{13}I \\
c_{14}I \cdot Lac \\
c_{15}r \\
c_{16}Z \\
\end{pmatrix}}\]
\end{document}
目前，这占用了太多的空间，而且由于有大量的空白空间，这也是必然的。
答案1
您的文档中有很多错误。在数学模式下，为了使用不同的（较小的）字体；或者使用\scriptstyle或\scriptscriptstyle。永远不要\scriptsize。
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsmath}
\usepackage[version=4]{mhchem}
\usepackage{mathtools}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{apacite}
\begin{document}
\begin{align*}
\ce{i &-> i + r_I, \\
r_i &-> r_I + I, \\
I + Lac &-> I \cdot Lac, \\
I \cdot Lac &-> I + Lac, \\
I + o &-> I \cdot o, \\
I \cdot o &-> I + o, \\
o + RNAP &-> o \cdot RNAP, \\
o \cdot RNAP &-> o + RNAP, \\
o \cdot RNAP &-> o + RNAP + r, \\
r &-> r + Z, \\
Lac + Z &-> Z, \\
r_I &-> \emptyset, \\
I &-> \emptyset, \\
I \cdot Lac &-> Lac, \\
r &-> \emptyset, \\
z &-> \emptyset
}
\end{align*}
where
\begin{table}[ht]
%\caption{Nonlinear Model Results} % title of Table
\centering % used for centering table
\begin{tabular}{c c} % centered columns (4 columns)
\hline\hline %inserts double horizontal lines
Species name & Symbol \\ [0.5ex] % inserts table
%heading
\hline % inserts single horizontal line
Regulatory gene & $i$ \\ [1ex] % inserting body of the table
Repressor protein & $r_I$ \\ [1ex]
Inhibitor protein & $I$ \\ [1ex]
Lactose & $Lac$ \\ [1ex]
Operator region & $o$ \\ [1ex]
RNA polymerase & $RNAP$ \\ [1ex]
mRNA & $r$ \\ [1ex]
$\beta$-galactosidase & $Z$ \\ [1.5ex] % [1ex] adds vertical space
\hline %inserts single line
\end{tabular}
\caption{Lac Operon species}
%\label{table:nonlin} % is used to refer this table in the text
\end{table}
Using SPN theory the system is represented as
\[N \, = \, (P,T,\text{Pre},\text{Post},M,h,c), \quad
P = \scriptstyle{\begin{pmatrix}
i \\
r_I \\
I \\
Lac \\
I \cdot Lac \\
o \\
I \cdot o \\
RNAP \\
RNAP \cdot o \\
r \\
Z \\
\end{pmatrix}}
%, \quad
% T = \begin{pmatrix}
% \text{Enzyme-substrate association} \\
% \text{Enzyme-substrate dissociation} \\
% \text{Product formation} \\
% \end{pmatrix}
\] \\
\[ T = \scriptstyle{\begin{pmatrix}
\text{Inhibitor transcription} \\
\text{Inhibitor translation} \\
\text{Lactose inhibitor binding} \\
\text{Lactose inhibitor dissociation} \\
\text{Inhibitor binding} \\
\text{Inhibitor dissociation} \\
\text{RNAP binding} \\
\text{RNAP dissociation} \\
\text{Transcription} \\
\text{Translation} \\
\text{Conversion} \\
\text{Inhibitor RNA degradation} \\
\text{Inhibitor degradation} \\
\text{Lactose inhibitor degradation} \\
\text{RNA degradation} \\
\text{z degradation} \\
\end{pmatrix}} \\ \]
\setcounter{MaxMatrixCols}{20}
\[ Pre = \scriptstyle{\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}}, \quad
Post = \scriptstyle{\begin{pmatrix}
1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}} \] \\
\[M = \scriptstyle{\begin{pmatrix}
1 \\
0 \\
50 \\
20 \\
0 \\
1 \\
0 \\
100 \\
0 \\
0 \\
0 \\
\end{pmatrix}}, \quad
c = \scriptstyle{\begin{pmatrix}
0.02 \\
0.1 \\
0.005 \\
0.1 \\
1 \\
0.01 \\
0.1 \\
0.01 \\
0.03 \\
0.1 \\
1e-05 \\
0.01 \\
0.002 \\
0.01 \\
0.001
\end{pmatrix}}, \quad
h(x,c) = \scriptstyle{\begin{pmatrix}
c_1i \\
c_2r_I \\
c_3ILac \\
c_4I \cdot Lac \\
c_5Io \\
c_6I \cdot o \\
c_7oRNAP \\
c_8o \cdot RNAP \\
c_9o \cdot RNAP \\
c_{10}r \\
c_{11}Lacz \\
c_{12}r_I \\
c_{13}I \\
c_{14}I \cdot Lac \\
c_{15}r \\
c_{16}Z \\
\end{pmatrix}}\]
\end{document}