ヘヴィサイドの階段関数の２定義値の和と差
ヘヴィサイドの階段関数の２定義値の和と差
(1)
\begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =H\left(\pm_{2}1\right)\pm_{1}H\left(\pm_{2}1\right)\\ & =2H\left(\pm_{1}1\right)H\left(\pm_{2}1\right) \end{align*}(2)
\begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right) & =H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\\ & =2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{2}1 \end{align*}(3)
\begin{align*} H\left(\pm_{1}1\right)\mp_{2}H\left(\mp_{1}1\right) & =H\left(\mp_{2}1\right)\pm_{1}H\left(\pm_{2}1\right)\\ & =2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{1}\pm_{2}1 \end{align*}(4)
\begin{align*} H\left(\mp_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =H\left(\pm_{2}1\right)\mp_{1}H\left(\mp_{2}1\right)\\ & =-2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)+1 \end{align*}-
\(H\left(x\right)\)はヘヴィサイドの階段関数(1)
\begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =\frac{\pm_{1}1+1}{2}\pm_{2}\frac{\pm_{1}1+1}{2}\\ & =\frac{\pm_{2}1+1}{2}\pm_{1}\frac{\pm_{2}1+1}{2}\\ & =H\left(\pm_{2}1\right)\pm_{1}H\left(\pm_{2}1\right) \end{align*} \begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =\frac{\pm_{1}1+1}{2}\pm_{2}\frac{\pm_{1}1+1}{2}\\ & =\frac{1}{2}\left(1\pm_{1}1\right)\left(1\pm_{2}1\right)\\ & =2\frac{1\pm_{1}1}{2}\cdot\frac{1\pm_{2}1}{2}\\ & =2H\left(\pm_{1}1\right)H\left(\pm_{2}1\right) \end{align*}(2)
\begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right) & =H\left(\pm_{1}1\right)\pm_{2}\left(1-H\left(\pm_{1}1\right)\right)\\ & =H\left(\pm_{1}1\right)\mp_{2}H\left(\pm_{1}1\right)\pm_{2}1\\ & =H\left(\mp_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\pm_{2}1\\ & =\pm_{2}\left(1-H\left(\mp_{2}1\right)\right)\pm_{1}H\left(\mp_{2}1\right)\\ & =\pm_{2}H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\\ & =H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right) \end{align*} \begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right) & =H\left(\pm_{1}1\right)\pm_{2}\left(1-H\left(\pm_{1}1\right)\right)\\ & =H\left(\pm_{1}1\right)\mp_{2}H\left(\pm_{1}1\right)\pm_{2}1\\ & =2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{2}1 \end{align*}(2)-2
\begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right) & =\frac{\pm_{1}1+1}{2}\pm_{2}\frac{\mp_{1}1+1}{2}\\ & =\frac{\pm_{2}1+1}{2}\pm_{1}\frac{\mp_{2}1+1}{2}\\ & =H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right) \end{align*} \begin{align*} H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right) & =\frac{\pm_{1}1+1}{2}\pm_{2}\frac{\mp_{1}1+1}{2}\\ & =\frac{1}{2}\left(1\pm_{1}1\right)\left(1\mp_{2}1\right)\pm_{2}1\\ & =2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{2}1 \end{align*}(3)
\begin{align*} H\left(\pm_{1}1\right)\mp_{2}H\left(\mp_{1}1\right) & =\left[H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right)\right]_{\pm_{2}\rightarrow\mp_{2}}\\ & =\left[H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right]_{\pm_{2}\rightarrow\mp_{2}}\\ & =H\left(\mp_{2}1\right)\pm_{1}H\left(\pm_{2}1\right) \end{align*} \begin{align*} H\left(\pm_{1}1\right)\mp_{2}H\left(\mp_{1}1\right) & =\pm_{1}\left\{ H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right\} \\ & =\pm_{1}\left\{ 2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{2}1\right\} \\ & =2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{1}\pm_{2}1 \end{align*}(3)-2
\begin{align*} H\left(\pm_{1}1\right)\mp_{2}H\left(\mp_{1}1\right) & =\pm_{1}H\left(\pm_{1}1\right)\mp_{2}\mp_{1}H\left(\mp_{1}1\right)\\ & =\pm_{1}\left\{ H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right)\right\} \\ & =\pm_{1}\left\{ H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right\} \\ & =H\left(\mp_{2}1\right)\pm_{1}H\left(\pm_{2}1\right) \end{align*}(4)
\begin{align*} H\left(\mp_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =\left[H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right)\right]_{\pm_{1}\rightarrow\mp_{1}}\\ & =\left[H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right]_{\pm_{1}\rightarrow\mp_{1}}\\ & =H\left(\pm_{2}1\right)\mp_{1}H\left(\mp_{2}1\right) \end{align*} \begin{align*} H\left(\mp_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =\pm_{2}\left(H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right)\\ & =\pm_{2}\left(2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)\pm_{2}1\right)\\ & =-2H\left(\pm_{1}1\right)H\left(\mp_{2}1\right)+1 \end{align*}(4)-2
\begin{align*} H\left(\mp_{1}1\right)\pm_{2}H\left(\pm_{1}1\right) & =\pm_{2}\left(H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right)\right)\\ & =\pm_{2}\left(H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right)\\ & =H\left(\pm_{2}1\right)\mp_{1}H\left(\mp_{2}1\right) \end{align*}
ページ情報
| タイトル | ヘヴィサイドの階段関数の２定義値の和と差 |
| URL | https://www.nomuramath.com/nu1mccd2/ |
| SNSボタン |
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とクロネッカーのデルタの関係
\[
H_{a}\left(n\right)-H_{b}\left(n-1\right)=a\delta_{0,n}+\left(1-b\right)\delta_{1,n}
\]
ヘヴィサイドの階段関数の問題
\[
f\left(H\left(\pm_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)=\left\{ f\left(0\right)g\left(0\right)+f\left(\pm1\right)g\left(\mp1\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(0\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\right)\right\} H\left(\mp_{2}1\right)
\]
ヘヴィサイドの階段関数の極限表示
\[
H_{\frac{1}{2}}\left(x\right)=\lim_{k\rightarrow\infty}\frac{1}{2}\left(1+\tanh\left(kx\right)\right)
\]