我有以下内容代码先前由 Alenanno 发布我想给出二次公式的动画证明。我的问题是：
- 我想改进代码，这样当我与 相乘时
$4a$，$4a$和$bx$从公式 2 得出，依此类推，$4aax^2$和$4ac.$ - 然后我想将
$a$和$\textcolor{red}{a}$合并，这样它就变成了$a^2$。怎么做？
\documentclass[tikz, margin=10pt]{standalone}
\newcommand\basicstuff{
\path (-9,-5) rectangle (9,5);
}
\tikzset{
bas/.style={text width=6cm}
}
\begin{document}\Huge
\foreach \x in {0,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\end{tikzpicture}
}
%
\foreach \x [
count=\xx starting from 0,
evaluate=\x as \opac using (\x/10),
evaluate=\x as \y using (3-\x)
] in {0,.25,...,2}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas, opacity=\opac] at (0,\y) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
%\node[bas, opacity=\opac] at (3,\y) {$\textcolor{green}{bx}$};
\end{tikzpicture}
}
%
\foreach \x in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\end{tikzpicture}
}
%---------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas, opacity=\opac] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas, opacity=\opac] at (4.4,0.88) {$)$};
\node[bas, opacity=\opac] at (-1.2,.88) {$4a\cdot$};
\node[bas, opacity=\opac] at (6,.88) {$\cdot4a$};
\end{tikzpicture}
}
%
%
%-------------------------------------------------------------------------------------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas, opacity=\opac] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} \,$};
\end{tikzpicture}
}
%--------------------------------------------------------------------------------------------------------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas, opacity=\opac] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\end{tikzpicture}
}
%--------------------------------------------------
%--------------------------------------------------------------------------------------------------------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas, opacity=\opac] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\end{tikzpicture}
}
%-------------------------------------------------------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\node[bas, opacity=\opac] at (6.5,-1.2) {$ = 0 $};
\end{tikzpicture}
}
%-------------------------------------------------------------------------
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10,10,10,10,9,8,...,0}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\node[bas] at (6.5,-1.2) {$ = 0 $};
%\node[bas, opacity=\opac] at (-0.9,-1.) {$ 4 \textcolor{red}{a^2x^2}$};
\end{tikzpicture}
}
\end{document}
答案1
对于第一个问题，您可以使用相同的基本思想在改变不透明度的同时逐渐改变距离。至少，我是这么认为的。我不完全理解为什么事物在等式的右边较低，即为什么 4ac 低于 4abx，而 4abx 低于 4aax^2，所以也许我没有正确理解这是如何工作的。
\documentclass[border=10pt,tikz,multi]{standalone}
% code from Alenanno's answer at http://tex.stackexchange.com/a/305603/
\newcommand\basicstuff{
\path (-9,-5) rectangle (9,5);
}
\tikzset{%
bas/.style={text width=6cm}
}
\begin{document}\Huge
\foreach \x in {0,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\end{tikzpicture}
}
\foreach \x [
count=\xx starting from 0,
evaluate=\x as \opac using (\x/10),
evaluate=\x as \y using (3-\x)
] in {0,.25,...,2}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas, opacity=\opac] at (0,\y) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
%\node[bas, opacity=\opac] at (3,\y) {$\textcolor{green}{bx}$};
\end{tikzpicture}
}
\foreach \x in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas, opacity=\opac] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas, opacity=\opac] at (4.4,0.88) {$)$};
\node[bas, opacity=\opac] at (-1.2,.88) {$4a\cdot$};
\node[bas, opacity=\opac] at (6,.88) {$\cdot4a$};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas, opacity=\opac] at (-0.9,{1-\x/5}) {$4a \textcolor{red}{ax^2} \,$};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas, opacity=\opac] at (1.5,{.9-\x/5}) {$+ \,4a \textcolor{green}{bx}$};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas, opacity=\opac] at (4.2,{.8-\x/5}) {$+ \,4a \textcolor{blue}{c}$};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\node[bas, opacity=\opac] at (6.5,{.8-\x/5}) {$ = 0 $};
\end{tikzpicture}
}
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10,10,10,10,9,8,...,0}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas] at (-0.9,-1.) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\node[bas] at (6.5,-1.2) {$ = 0 $};
%\node[bas, opacity=\opac] at (-0.9,-1.) {$ 4 \textcolor{red}{a^2x^2}$};
\end{tikzpicture}
}
\end{document}
对于第二个问题，你不能简单地使用类似下面的方法吗？
\foreach \x[ evaluate=\x as \opac using (\x/10),] in {1,...,10}{
\begin{tikzpicture}
\basicstuff
\node[bas] at (0,3) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (0,1) {$\textcolor{red}{ax^2} + \textcolor{green}{bx} + \textcolor{blue}{c}=0$};
\node[bas] at (-0.2,0.88) {$( $}; %\hspace{4.4cm} \right) \hspace {1.cm}
\node[bas] at (4.4,0.88) {$)$};
\node[bas] at (-1.2,.88) {$4a\cdot$};
\node[bas] at (6,.88) {$\cdot4a$};
\node[bas, opacity={1-\opac}] at (-0.9,-1) {$4a \textcolor{red}{ax^2} $};
\node[bas] at (1.5,-1.1) {$+ \,4a \textcolor{green}{bx}$};
\node[bas] at (4.2,-1.2) {$+ \,4a \textcolor{blue}{c}$};
\node[bas] at (6.5,-1.2) {$ = 0 $};
\node[bas, opacity=\opac] at (-0.9,-1) {$ 4 \textcolor{red}{a^2x^2}$};
\end{tikzpicture}
}
