我有以下代码：
\documentclass[tikz,border=5mm]{standalone}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage{xfp}
\usetikzlibrary{math}
\usepackage{etoolbox}
\usepackage{fp}
\usetikzlibrary{fixedpointarithmetic}
\newcommand{\ar}[1]{\fpeval{round(#1,4)}}
\begin{document}
\tikzmath{
\p=8;%número de pontos fornecidos
\q=3;%número de parâmetros a serem ajustados
\X{1}=33;
\X{2}=31;
\X{3}=29;
\X{4}=27;
\X{5}=25;
\X{6}=23;
\X{7}=21;
\X{8}=19;
%
\Y{1}=13232;
\Y{2}=11686;
\Y{3}=10236;
\Y{4}=8882;
\Y{5}=7624;
\Y{6}=6462;
\Y{7}=5396;
\Y{8}=4426;
%
function phi1(\x){%Funções da base
\y = 1;
return \y;
};
function phi2(\x){%Funções da base
\y = \x;
return \y;
};
%
function phi3(\x){%Funções da base
\y = \x^2;
return \y;
};
%
function Phi(\t,\betaum,\betadois,\betatres) {%Função linear dos parâmetros
\y = \betaum*phi1(\t) + \betadois*phi2(\t) + \betatres*phi3(\t);
return \y;
};
%
function G(\x) {
real \u; real \v;
\u = phi1(\x);
\v = phi2(\x);
\w = phi3(\x);
return {\u,\v,\w};
};
real \Matriz; int \i; int \j;
for \i in {1,...,\p}{
for \j in {1,...,\q}{
\Matriz{\i,\j}={G(\X{\i})}[\fpeval{\j-1}];
};
};
}
\tikzmath{
int \i;
int \j;
int \k;
real \m;
%
real \soma; real \somab; real \D;
\n=\q;
for \i in {1,...,\q}{
for \j in {1,...,\q}{
\soma = 0;
for \k in {1,...,\p}{
\soma = \soma + \Matriz{\k,\i}*\Matriz{\k,\j};
};
\A{\i,\j} = \soma;
};
};
for \i in {1,...,\q}{
\somab=0;
for \k in {1,...,\p}{
\somab = \somab + \Matriz{\k,\i}*\Y{\k};
};
\b{\i}=\somab;
};
%
for \i in {1,...,\n}{
for \j in {1,...,\n}{
\Alinha{\i,\j} = \A{\i,\j};
};
\blinha{\i}=\b{\i};
};
for \k in {1,...,\fpeval{\n-1}}{
for \i in {\fpeval{\k+1},...,\n}{
\m=\Alinha{\i,\k}/\Alinha{\k,\k};
\blinha{\i}=\blinha{\i}-(\m)*\blinha{\k};
\Alinha{\i,\k}=0.0;
\M{\i,\k} = \m;
for \j in {\fpeval{\k+1},...,\n}{
\Alinha{\i,\j}=\Alinha{\i,\j}-\m*\Alinha{\k,\j};
};
};
for \i in {1,...,\n}{
for \j in {1,...,\n}{
\L{\i,\j,\k} = \Alinha{\i,\j};
};
\N{\i,\k} = \blinha{\i};
};
};
}
\tikzmath{
real \s;
int \i;
int \j;
real \x;
\x{\n}=\blinha{\n}/\Alinha{\n,\n};
for \i in {\fpeval{\n-1},...,1}{
\s=0;
for \j in {\fpeval{\i+1},...,\n}{
\s=\s+\Alinha{\i,\j}*\x{\j};
};
\x{\i}=(\blinha{\i}-\s)/\Alinha{\i,\i};
};
}
\tikzmath{
int \i;
int \j;
real \ybarra; real \SSres; real \SStot;
\ybarra=0;
for \i in {1,...,\p}{
\ybarra = \ybarra + \Y{\i};
};
\ybarra=\ybarra/\p;
\SSres=0;
for \i in {1,...,\p}{
\SSres = \SSres + (\Y{\i} - Phi(\X{\i},\x{1},\x{2},\x{3}))^2;
};
\SStot=0;
for \i in {1,...,\p}{
\SStot = \SStot + (\Y{\i} - \ybarra)^2;
};
\R = 1 - (\SSres/\SStot);
}
\begin{tikzpicture}[scale=1]
\node[above] (2) at (0,-\i) {\parbox{14cm}{
Queremos ajustar os pontos \foreach \i in {1,...,\p}{(\X{\i},\Y{\i}), } ao modelo $y=\phi(x)=\beta_1 \phi_1(x) + \beta_2 \phi_2(x) + \beta_3 \phi_3(x)$ onde $\phi_1(x) = 1$, $\phi_2(x) = x$ e $\phi_3(x) = x^2$.\\
Vamos resolver o seguinte sistema de equações normais $\mathbf{A}^T \mathbf{A} \mathbf{x}=\mathbf{A}^T \mathbf{y}$, onde:\\
$\mathbf{A} = \left[
\begin{array}{rrr}
\Matriz{1,1} & \Matriz{1,2}\\
\Matriz{2,1} & \Matriz{2,2}\\
\Matriz{3,1} & \Matriz{3,2}\\
\Matriz{4,1} & \Matriz{4,2}\\
\Matriz{5,1} & \Matriz{5,2}
\end{array}
\right], \quad \mathbf{x} = \left[
\begin{array}{r}
\beta_1 \\
\beta_2 \\
\beta_3
\end{array}
\right] \quad \text{e} \quad \mathbf{y} = \left[
\begin{array}{r}
\Y{1} \\
\Y{2} \\
\Y{3} \\
\Y{4} \\
\Y{5}
\end{array}
\right]$\\[0.5cm]
resulta no seguinte sistema $\q \times \q$:\\[0.5cm]
$
\left\{
\begin{array}{lcr}
(\ar{\A{1,1}})\beta_1 + (\ar{\A{1,2}})\beta_2 + (\ar{\A{1,3}})\beta_3 & = & \ar{\b{1}} \\
(\ar{\A{2,1}})\beta_1 + (\ar{\A{2,2}})\beta_2 + (\ar{\A{2,3}})\beta_3 & = & \ar{\b{2}} \\
(\ar{\A{3,1}})\beta_1 + (\ar{\A{3,2}})\beta_2 + (\ar{\A{3,3}})\beta_3 & = & \ar{\b{3}} \\
\end{array}
\right.
$
\\
Matriz a ser escalonada:\\
$\mathbf{M} =
\left[
\begin{array}{rrr|r}
\ar{\A{1,1}} & \ar{\A{1,2}} & \ar{\A{1,3}} & \ar{\b{1}} \\
\ar{\A{2,1}} & \ar{\A{2,2}} & \ar{\A{2,3}} & \ar{\b{2}} \\
\ar{\A{3,1}} & \ar{\A{3,2}} & \ar{\A{3,3}} & \ar{\b{3}}
\end{array}
\right]
$
}};
\end{tikzpicture}
\foreach \k in {1,...,\fpeval{\n-1}}{
\begin{tikzpicture}[scale=1]
\node (Te) at (0,0) {
\parbox{14cm}{Fazendo o escalonamento no passo $k=\k$ teremos:\\[0.5cm]}};
\ifnum\k=1
\node (5) at (0,-2) {
\parbox{14cm}{Matriz atualizada:\\[0.5cm]
$
\left[
\begin{array}{rrr|r}
\ar{\A{1,1}} & \ar{\A{1,2}} & \ar{\A{1,3}} & \ar{\b{1}}\\
\ar{\A{2,1}} & \ar{\A{2,2}} & \ar{\A{2,3}} & \ar{\b{2}}\\
\ar{\A{3,1}} & \ar{\A{3,2}} & \ar{\A{3,3}} & \ar{\b{3}}
\end{array}
\right]
\Longrightarrow
\left[
\begin{array}{rrr|r}
\ar{\L{1,1,\k}} & \ar{\L{1,2,\k}} & \ar{\L{1,3,\k}} & \ar{\N{1,\k}}\\
\ar{\L{2,1,\k}} & \ar{\L{2,2,\k}} & \ar{\L{2,3,\k}} & \ar{\N{2,\k}}\\
\ar{\L{3,1,\k}} & \ar{\L{3,2,\k}} & \ar{\L{3,3,\k}} & \ar{\N{3,\k}}
\end{array}
\right]
$\\[0.5cm]
com as seguintes operações:\\[0.5cm]
}};
\foreach \i in {\fpeval{\k+1},...,\n}{
\node at (0,{-(\i-\k+1)-1.5}) {\parbox{14cm}{$m_{\k \i} = \fpeval{round(\M{\i,\k},4)}$ e $L_{\i} \leftarrow L_{\i} - (\fpeval{round(\M{\i,\k},4)}) L_{\k}$}};
}
\else
\node (5) at (0,-2) {
\parbox{14cm}{Matriz atualizada:\\[0.5cm]
$
\left[
\begin{array}{rrr|r}
\ar{\L{1,1,\fpeval{\k-1}}} & \ar{\L{1,2,\fpeval{\k-1}}} & \ar{\L{1,3,\fpeval{\k-1}}} & \ar{\N{1,\fpeval{\k-1}}}\\
\ar{\L{2,1,\fpeval{\k-1}}} & \ar{\L{2,2,\fpeval{\k-1}}} & \ar{\L{2,3,\fpeval{\k-1}}} & \ar{\N{2,\fpeval{\k-1}}}\\
\ar{\L{3,1,\fpeval{\k-1}}} & \ar{\L{3,2,\fpeval{\k-1}}} & \ar{\L{3,3,\fpeval{\k-1}}} & \ar{\N{3,\fpeval{\k-1}}}\\
\end{array}
\right]
\Longrightarrow
\left[
\begin{array}{rrr|r}
\ar{\L{1,1,\k}} & \ar{\L{1,2,\k}} & \ar{\L{1,3,\k}} & \ar{\N{1,\k}}\\
\ar{\L{2,1,\k}} & \ar{\L{2,2,\k}} & \ar{\L{2,3,\k}} & \ar{\N{2,\k}}\\
\ar{\L{3,1,\k}} & \ar{\L{3,2,\k}} & \ar{\L{3,3,\k}} & \ar{\N{3,\k}}\\
\end{array}
\right]
$ \\[0.5cm]
com as seguintes operações:\\[0.5cm]
}};
\foreach \i in {\fpeval{\k+1},...,\n}{
\node at (0,{-(\i-\k+1)-1.5}) {\parbox{14cm}{$m_{\k \i} = \ar{\M{\i,\k}}$ e $L_{\i} \leftarrow L_{\i} - (\ar{\M{\i,\k}}) L_{\k}$}};
}
\fi
\end{tikzpicture}
}
\begin{tikzpicture}
\node at (0,\n) {\parbox{14cm}{
Assim, a solução do sistema linear é calculada por substituição retroativa do seguinte modo:\\[0.5cm]
$\beta_\n = \ar{\blinha{\n}}/(\ar{\Alinha{\n,\n}}) = \ar{\x{\n}}$
}};
\foreach \i in {\fpeval{\n-1},...,1}{
\node at (0,{\i-0.5}) {\parbox{14cm}{
$\beta_\i = \left( \ar{\blinha{\i}} - \sum_{j=\fpeval{\i+1}}^{\n} a_{\i j} \cdot x_j \right)/(\ar{\Alinha{\i,\i}}) = \ar{\x{\i}}$
}};
}
\node at (0,{-\n-1}) {\parbox{14cm}{
De maneira que a solução do sistema de equações normais é:\\
$\beta^* = \left[
\begin{array}{rrrr}
\ar{\x{1}} & \ar{\x{2}} & \ar{\x{3}}
\end{array} \right]
$\\
ou seja, o modelo ajustado é\\
$\boxed{y= \phi(x) = \foreach \i in {1,...,\fpeval{\q-1}}{(\ar{\x{\i}})\phi_{\i}(x)+}(\ar{\x{\q}})\phi_{\q}(x)}$\\
Com $r^2$ dado por:\\
$\displaystyle r^2 = 1 - \frac{\text{SSres}}{\text{SStot}} = \ar{\R}$\\
onde $\displaystyle \text{SSres} = \sum_{i=1}^\p{} (y_i - \phi(x_i))^2$ e $\displaystyle \text{SStot} = \sum_{i=1}^{\p} (y_i - \bar{y})^2$
e com a soma dos residuos igual a:\\
$\text{SSres} = \ar{\SSres}$.
}};
\end{tikzpicture}
\end{document}
但是，编译器给出以下错误：dimension too large。
对于较小的数字，代码运行完美。
我怎样才能解决这个问题？