ヘヴィサイドの階段関数の2定義値を引数に持つ関数の和と差
ヘヴィサイドの階段関数の2定義値を引数に持つ関数の和と差
\[ f\left(H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)=\left(f\left(0\right)+f\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(f\left(0\right)-f\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right) \]
\[ f\left(H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)=\left(f\left(0\right)+f\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(f\left(0\right)-f\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right) \]
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\(H\left(x\right)\)はヘヴィサイドの階段関数\begin{align*}
f\left(H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right) & =f\left(0\right)H\left(\mp_{1}1\right)+f\left(1\right)H\left(\pm_{1}1\right)\pm_{2}\left(f\left(0\right)H\left(\pm_{1}1\right)+f\left(-1\right)H\left(\mp_{1}1\right)\right)\\
& =f\left(0\right)H\left(\mp_{1}1\right)+f\left(\pm_{1}1\right)H\left(\pm_{1}1\right)\pm_{2}\left(f\left(0\right)H\left(\pm_{1}1\right)+f\left(\pm_{1}1\right)H\left(\mp_{1}1\right)\right)\\
& =f\left(0\right)\left(H\left(\mp_{1}1\right)\pm_{2}H\left(\pm_{1}1\right)\right)+f\left(\pm_{1}1\right)\left(H\left(\pm_{1}1\right)\pm_{2}H\left(\mp_{1}1\right)\right)\\
& =f\left(0\right)\left(H\left(\pm_{2}1\right)\mp_{1}H\left(\mp_{2}1\right)\right)+f\left(\pm_{1}1\right)\left(H\left(\pm_{2}1\right)\pm_{1}H\left(\mp_{2}1\right)\right)\\
& =\left(f\left(0\right)+f\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(f\left(0\right)-f\left(\pm_{1}1\right)\right)H\left(\mp_{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定義値を引数に持つ関数の和と差 |
URL | https://www.nomuramath.com/rj29dak3/ |
SNSボタン |
mzp関数の定義と負数の関係
\[
\mzp_{a,b}\left(x_{1},x_{2};-x\right)=-\mzp_{-b,-a}\left(-x_{2},-x_{1};x\right)
\]
ヘヴィサイドの階段関数と符号関数の積
\[
\sgn\left(x\right)H_{a}\left(x\right)=H_{0}\left(x\right)
\]
ヘヴィサイドの階段関数と単位ステップ関数の定義
\[
H_{a}\left(x\right)=\begin{cases}
0 & \left(x<0\right)\\
a & \left(x=0\right)\\
1 & \left(0<x\right)
\end{cases}
\]
ヘヴィサイドの階段関数の極限表示
\[
H_{\frac{1}{2}}\left(x\right)=\lim_{k\rightarrow\infty}\frac{1}{2}\left(1+\tanh\left(kx\right)\right)
\]