ヘヴィサイドの階段関数の２定義値と関数
ヘヴィサイドの階段関数の２定義値と関数
(1)
\[ f\left(x\right)H\left(\pm1\right)=f\left(\pm x\right)H\left(\pm1\right) \](2)
\[ H\left(\pm1\right)=\pm H\left(\pm1\right) \](3)
\[ f\left(\pm_{1}H\left(\pm_{2}1\right)\right)=f\left(0\right)H\left(\mp_{2}1\right)+f\left(\pm_{1}1\right)H\left(\pm_{2}1\right) \]-
\(H\left(x\right)\)はヘヴィサイドの階段関数(1)
\begin{align*} f\left(x\right)H\left(\pm1\right) & =\begin{cases} f\left(x\right) & \pm1\rightarrow+1\\ 0 & \pm1\rightarrow-1 \end{cases}\\ & =\begin{cases} f\left(\pm x\right) & \pm1\rightarrow+1\\ 0 & \pm1\rightarrow-1 \end{cases}\\ & =f\left(\pm x\right)H\left(\pm1\right) \end{align*}(2)
\begin{align*} H\left(\pm1\right) & =\left[f\left(x\right)H\left(\pm1\right)\right]_{f\left(x\right)=x\;,\;x=1}\\ & =\left[f\left(\pm x\right)H\left(\pm1\right)\right]_{f\left(x\right)=x\;,\;x=1}\\ & =\pm H\left(\pm1\right) \end{align*}(2)-2
\begin{align*} H\left(\pm1\right) & =\frac{1\pm1}{2}\\ & =\pm\left(\frac{1\pm1}{2}\right)\\ & =\pm H\left(\pm1\right) \end{align*}(3)
\begin{align*} f\left(\pm_{1}H\left(\pm_{2}1\right)\right) & =\begin{cases} f\left(\pm_{1}1\right) & \pm_{2}1\rightarrow+1\\ f\left(0\right) & \pm_{2}1\rightarrow-1 \end{cases}\\ & =f\left(0\right)H\left(\mp_{2}1\right)+f\left(\pm_{1}1\right)H\left(\pm_{2}1\right) \end{align*}
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| タイトル | ヘヴィサイドの階段関数の２定義値と関数 |
| URL | https://www.nomuramath.com/h01ij6zb/ |
| SNSボタン |
ヘヴィサイドの階段関数の複素積分表示
\[
H_{\frac{1}{2}}\left(x\right)=\frac{1}{2\pi i}\lim_{\epsilon\rightarrow0+}\int_{-\infty}^{\infty}\frac{1}{z-i\epsilon}e^{ixz}dz
\]
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と絶対値・符号関数
\[
H_{a}\left(\left|c\right|x\right)=H_{a}\left(x\right)
\]
ヘヴィサイドの階段関数の問題
\[
f\left(H\left(\pm_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)=\left\{ f\left(0\right)g\left(0\right)+f\left(\pm1\right)g\left(\mp1\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(0\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\right)\right\} H\left(\mp_{2}1\right)
\]