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TiKz 图片内的绘图

tag-icon • tag-icon tikz-pgf
TiKz 图片内的绘图

我有一张 tikz 图片，想包含一条具有以下形式的曲线：

1/(x^2+1)

我可以绘制正域的函数，例如[1,4]：

\draw[domain=1:4,smooth,variable=\x] plot ({\x},{1/(\x^2+1)});

我希望能够绘制[-4,4]范围内的函数。

答案1

解决方案是替换\x^2为\x*\x

它看起来像这样：

\begin{tikzpicture}
\fill[pattern=north west lines,opacity=.3] (-4,-1.11803*3) -- plot [domain=-4:4] ({\x},{-3/(\x*\x+2)-1.11803*3}) -- (4,-1.11803*3) -- cycle;

\draw[<-]    (1.5,{-3/(1.5*1.5+2)-1.11803*3}) -- (2,-4.5)   node[right] {$U(x)$}; 
\draw[<-]    (-1.5,-3.6) -- (-2,-4.5)   node[left] {$\displaystyle\int_{-\infty}^{\infty}\!\!\!\!\mbox{d}x \:U(x)$}; 
\draw[red,thick,domain=-4:4,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});
\draw[red,dotted,thick,domain=-4:-5,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});
\draw[red,dotted,thick,domain=4:5,smooth,samples=200,variable=\x] plot ({\x},{-3/(\x*\x+2)-1.11803*3});

\draw (0,0) circle (1.5cm);
\draw[->]    (0,0) -- (0,-1.5) node[midway,left] {$R$}; 
\draw[->]    (0,0) -- (1,-1.11803) node[midway,right] {$R$}; 
\draw[<->]    (1,-1.11803) -- (1*3,-1.11803*3) node[midway,right] {$R$}; 
\draw[<->]    (0,-1.5) -- (0,-1.11803*3) node[midway,right] {$z$}; 
\draw[-latex]    (-5,-1.11803*3) -- (5,-1.11803*3); 
\draw[|->|]    (0,-1.11803*3-0.2) -- (3,-1.11803*3 -0.2) node[midway,below] {$x$};
\end{tikzpicture}

结果是： 在此处输入图片描述

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