ヘヴィサイドの階段関数と符号関数の関係
ヘヴィサイドの階段関数と符号関数の関係
\begin{align*} H_{c}\left(x\right) & =\frac{\sgn\left(x\right)+1}{2}+\left(c-\frac{1}{2}\right)\delta_{0,x}\\ & =\frac{1}{2}\sgn\left(x\right)-\left(c-\frac{1}{2}\right)\left|\sgn\left(x\right)\right|+c \end{align*}
\[ H_{a}\left(x\right)-H_{b}\left(x\right)=\left(a-b\right)\left(1-\left|\sgn\left(x\right)\right|\right) \]
\(\delta_{ij}\)はクロネッカーのデルタ
\(\sgn\left(x\right)\)は符号関数
(1)
\[ H_{\frac{1}{2}}\left(x\right)=\frac{\sgn\left(x\right)+1}{2} \](2)
\[ H_{0}\left(x\right)=\frac{1}{2}\left(\sgn\left(x\right)+\left|\sgn\left(x\right)\right|\right) \](3)
\[ H_{1}\left(x\right)=\frac{1}{2}\left(2+\sgn\left(x\right)-\left|\sgn\left(x\right)\right|\right) \](4)
\(c\in\mathbb{R}\)とする。\begin{align*} H_{c}\left(x\right) & =\frac{\sgn\left(x\right)+1}{2}+\left(c-\frac{1}{2}\right)\delta_{0,x}\\ & =\frac{1}{2}\sgn\left(x\right)-\left(c-\frac{1}{2}\right)\left|\sgn\left(x\right)\right|+c \end{align*}
(5)
\(a,b\in\mathbb{R}\)とする。\[ H_{a}\left(x\right)-H_{b}\left(x\right)=\left(a-b\right)\left(1-\left|\sgn\left(x\right)\right|\right) \]
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\(H\left(x\right)\)はヘヴィサイドの階段関数\(\delta_{ij}\)はクロネッカーのデルタ
\(\sgn\left(x\right)\)は符号関数
(1)
\begin{align*} H_{\frac{1}{2}}\left(x\right) & =\left(H_{\frac{1}{2}}\left(x\right)-\frac{1}{2}\right)+\frac{1}{2}\\ & =\frac{\sgn\left(x\right)}{2}+\frac{1}{2}\\ & =\frac{\sgn\left(x\right)+1}{2} \end{align*}(2)
\begin{align*} H_{0}\left(x\right) & =\begin{cases} 1 & 0<x\\ 0 & x\leq0 \end{cases}\\ & =\begin{cases} \frac{1}{2}\left(\sgn\left(x\right)+\left|\sgn\left(x\right)\right|\right) & 0<x\\ \frac{1}{2}\left(\sgn\left(x\right)+\left|\sgn\left(x\right)\right|\right) & x\leq0 \end{cases}\\ & =\frac{1}{2}\left(\sgn\left(x\right)+\left|\sgn\left(x\right)\right|\right) \end{align*}(3)
\begin{align*} H_{1}\left(x\right) & =\begin{cases} 1 & 0\leq x\\ 0 & x<0 \end{cases}\\ & =\begin{cases} \frac{1}{2}\left(\sgn\left(x\right)+1\right)+\frac{1}{2}\delta_{0,x} & 0\leq x\\ \frac{1}{2}\left(\sgn\left(x\right)+1\right)+\frac{1}{2}\delta_{0,x} & x<0 \end{cases}\\ & =\frac{1}{2}\left(\sgn\left(x\right)+1\right)+\frac{1}{2}\delta_{0,x}\\ & =\frac{1}{2}\left(\sgn\left(x\right)+1+\delta_{0,x}\right)\\ & =\frac{1}{2}\left(\sgn\left(x\right)+1+1-\left|\sgn\left(x\right)\right|\right)\\ & =\frac{1}{2}\left(2+\sgn\left(x\right)-\left|\sgn\left(x\right)\right|\right) \end{align*}(4)
\begin{align*} H_{c}\left(x\right) & =\begin{cases} 1 & 0<x\\ c & x=0\\ 0 & x<0 \end{cases}\\ & =\begin{cases} \frac{1}{2}\left(\sgn\left(x\right)+1\right)+\left(c-\frac{1}{2}\right)\delta_{0,x} & 0<x\\ \frac{1}{2}\left(\sgn\left(x\right)+1\right)+\left(c-\frac{1}{2}\right)\delta_{0,x} & x=0\\ \frac{1}{2}\left(\sgn\left(x\right)+1\right)+\left(c-\frac{1}{2}\right)\delta_{0,x} & x<0 \end{cases}\\ & =\frac{1}{2}\left(\sgn\left(x\right)+1\right)+\left(c-\frac{1}{2}\right)\delta_{0,x}\\ & =\frac{1}{2}\sgn\left(x\right)+\frac{1}{2}+\left(c-\frac{1}{2}\right)\left(1-\left|\sgn\left(x\right)\right|\right)\\ & =\frac{1}{2}\sgn\left(x\right)-\left(c-\frac{1}{2}\right)\left|\sgn\left(x\right)\right|+c \end{align*}(4)-2
\begin{align*} H_{c}\left(x\right) & =H_{\frac{1}{2}}\left(x\right)+\left(c-\frac{1}{2}\right)\delta_{0,x}\\ & =\frac{\sgn\left(x\right)+1}{2}+\left(c-\frac{1}{2}\right)\delta_{0,x}\\ & =\frac{1}{2}\sgn\left(x\right)+\frac{1}{2}+\left(c-\frac{1}{2}\right)\left(1-\left|\sgn\left(x\right)\right|\right)\\ & =\frac{1}{2}\sgn\left(x\right)-\left(c-\frac{1}{2}\right)\left|\sgn\left(x\right)\right|+c \end{align*}(5)
\begin{align*} H_{a}\left(x\right)-H_{b}\left(x\right) & =\left(\frac{\sgn\left(x\right)+1}{2}+\left(a-\frac{1}{2}\right)\delta_{0,x}\right)-\left(\frac{\sgn\left(x\right)+1}{2}+\left(b-\frac{1}{2}\right)\delta_{0,x}\right)\\ & =\left(a-b\right)\delta_{0,x}\\ & =\left(a-b\right)\left(1-\left|\sgn\left(x\right)\right|\right) \end{align*}
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ヘヴィサイドの階段関数の微分・積分と微分・積分表示
\[
\frac{dH\left(x\right)}{dx}=\delta\left(x\right)
\]
ヘヴィサイドの階段関数の２定義値と複号
\[
H\left(\pm1\right)=\frac{1\pm1}{2}
\]
ヘヴィサイドの階段関数の２定義値と関数
\[
f\left(x\right)H\left(\pm1\right)=f\left(\pm x\right)H\left(\pm1\right)
\]
ヘヴィサイドの階段関数の正数と負数の和と差
\[
H_{a}\left(x\right)+H_{b}\left(-x\right)=1+\left(a+b-1\right)\delta_{0,x}
\]