ヘヴィサイドの階段関数の問題
ヘヴィサイドの階段関数の問題
(1)
\[ f\left(H\left(\pm_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)=\left\{ f\left(0\right)g\left(0\right)+f\left(\pm1\right)g\left(\mp1\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(0\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\right)\right\} H\left(\mp_{2}1\right) \](2)
\begin{align*} f\left(H\left(\pm_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right) & =\left\{ f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(0\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(0\right)\right\} H\left(\mp_{2}1\right) \end{align*}(3)
\begin{align*} f\left(H\left(\pm_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right) & =\left(f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(0\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(0\right)\right\} H\left(\mp_{2}1\right) \end{align*}-
\(H\left(x\right)\)はヘヴィサイドの階段関数(1)
\begin{align*} 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[h\left(H\left(\pm_{1}1\right)\right)\pm_{2}h\left(-H\left(\mp_{1}1\right)\right)\right]_{h\left(x\right)=f\left(x\right)g\left(-x\right)}\\ & =\left[\left(h\left(0\right)+h\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(h\left(0\right)-h\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right)\right]_{h\left(x\right)=f\left(x\right)g\left(-x\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) \end{align*}(2)
\begin{align*} f\left(H\left(\pm_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right) & =f\left(H\left(\pm_{1}1\right)\right)g\left(\mp_{1}1+H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(\mp_{1}1-H\left(\mp_{1}1\right)\right)\\ & =\left[h\left(H\left(\pm_{1}1\right)\right)\pm_{2}h\left(-H\left(\mp_{1}1\right)\right)\right]_{h\left(x\right)=f\left(x\right)g\left(\mp_{1}1+x\right)}\\ & =\left[\left(h\left(0\right)+h\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(h\left(0\right)-h\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right)\right]_{h\left(x\right)=f\left(x\right)g\left(\mp_{1}1+x\right)}\\ & =\left\{ f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(\mp_{1}1\pm_{1}1\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\pm_{1}1\right)\right\} H\left(\mp_{2}1\right)\\ & =\left\{ f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(0\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(0\right)\right\} H\left(\mp_{2}1\right) \end{align*}(3)
\begin{align*} f\left(H\left(\pm_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right) & =f\left(H\left(\pm_{1}1\right)\right)g\left(\mp_{1}1+H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(\mp_{1}1-H\left(\mp_{1}1\right)\right)\\ & =\left[h\left(H\left(\pm_{1}1\right)\right)\pm_{2}h\left(-H\left(\mp_{1}1\right)\right)\right]_{h\left(x\right)=f\left(x\right)g\left(\mp1+x\right)}\\ & =\left[\left(h\left(0\right)+h\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(h\left(0\right)-h\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right)\right]_{h\left(x\right)=f\left(x\right)g\left(\mp_{1}1+x\right)}\\ & =\left(f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(\mp_{1}1\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\pm_{1}1\right)\right\} H\left(\mp_{2}1\right)\\ & =\left(f\left(0\right)g\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)g\left(0\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(\mp_{1}1\right)-f\left(\pm_{1}1\right)g\left(0\right)\right\} 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問題 |
| URL | https://www.nomuramath.com/h41ao97m/ |
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ヘヴィサイドの階段関数の２定義値の和と差
\[
H\left(\pm_{1}1\right)\pm_{2}H\left(\pm_{1}1\right)=H\left(\pm_{2}1\right)\pm_{1}H\left(\pm_{2}1\right)
\]
ヘヴィサイドの階段関数の極限表示
\[
H_{\frac{1}{2}}\left(x\right)=\lim_{k\rightarrow\infty}\frac{1}{2}\left(1+\tanh\left(kx\right)\right)
\]
ヘヴィサイドの階段関数の負数・和・差
\[
H_{a}\left(-x\right)=-H_{a}\left(x\right)+1+\left(2a-1\right)\delta_{0,x}
\]
ヘヴィサイドの階段関数同士の変換
\[
H_{a}\left(x\right)=H_{b}\left(x\right)+\left(a-b\right)\delta_{0,x}
\]