剰余演算の定数倍
剰余演算の定数倍
\(\beta\ne0\)とする。
\(\left\lfloor z\right\rfloor \)は床関数
\(\sgn\left(z\right)\)は符号関数
\(\beta\ne0\)とする。
(1)
\[ \frac{1}{\gamma}\mod\left(\alpha,\beta\right)=\mod\left(\frac{\alpha}{\gamma},\frac{\beta}{\gamma}\right) \](2)
\[ \frac{1}{\delta}\mod\left(\alpha,\beta,\gamma\right)=\mod\left(\frac{\alpha}{\delta},\frac{\beta}{\delta},\frac{\gamma}{\delta}\right) \](3)
\[ \mod\left(\alpha,\beta,\gamma\right)=\sgn\left(\beta\right)\mod\left(\alpha\sgn\left(\beta\right),\left|\beta\right|,\gamma\sgn\left(\beta\right)\right) \]-
\(\mod\left(\alpha,\beta\right)\)は剰余演算\(\left\lfloor z\right\rfloor \)は床関数
\(\sgn\left(z\right)\)は符号関数
(1)
\begin{align*} \frac{1}{\gamma}\mod\left(\alpha,\beta\right) & =\frac{\beta}{\gamma}\left(\frac{\alpha}{\beta}-\left\lfloor \frac{\alpha}{\beta}\right\rfloor \right)\\ & =\frac{\beta}{\gamma}\left(\frac{\alpha/\gamma}{\beta/\gamma}-\left\lfloor \frac{\alpha/\gamma}{\beta/\gamma}\right\rfloor \right)\\ & =\mod\left(\frac{\alpha}{\gamma},\frac{\beta}{\gamma}\right) \end{align*}(2)
\begin{align*} \frac{1}{\delta}\mod\left(\alpha,\beta,\gamma\right) & =\frac{1}{d}\left(\mod\left(\alpha-\gamma,\beta\right)+\gamma\right)\\ & =\mod\left(\frac{\alpha-\gamma}{\delta},\frac{\beta}{\delta}\right)+\frac{\gamma}{\delta}\\ & =\mod\left(\frac{\alpha}{\delta},\frac{\beta}{\delta},\frac{\gamma}{\delta}\right) \end{align*}(3)
\begin{align*} \mod\left(\alpha,\beta,\gamma\right) & =\sgn\left(\beta\right)\mod\left(\frac{\alpha}{\sgn\left(\beta\right)},\left|\beta\right|,\frac{\gamma}{\sgn\left(\beta\right)}\right)\\ & =\sgn\left(\beta\right)\mod\left(\alpha\sgn\left(\beta\right),\left|\beta\right|,\gamma\sgn\left(\beta\right)\right) \end{align*}ページ情報
タイトル | 剰余演算の定数倍 |
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剰余演算の引数
\[
\mod\left(\alpha,\beta,\gamma\right):=\mod\left(\alpha-\gamma,\beta\right)+\gamma
\]
剰余の剰余
\[
\mod\left(\mod\left(\alpha,n\beta\right),\beta\right)=\mod\left(\alpha,\beta\right)
\]
剰余演算と床関数・天井関数の関係
\[
\alpha=\beta\left\lfloor \frac{\alpha-\gamma}{\beta}\right\rfloor +\mod\left(\alpha,\beta,\gamma\right)
\]
複素数と複素共役の実数での剰余演算
\[
\mod\left(\alpha,1\right)=\mod\left(\Re\alpha,1\right)+i\mod\left(\Im\alpha,1\right)
\]