# 微分演算子の交換則

$f\left(D\right)g\left(D\right)=g\left(D\right)f\left(D\right)$

が成り立つ．すなわち，交換則が成り立つ．

## ■具体的な解説

$f\left(D\right)=D+{C}_{1}$　･･････(1)

$g\left(D\right)=D+{C}_{2}$　･･････(2)

とする．

$\left\{f\left(D\right)g\left(D\right)\right\}y=f\left(D\right)\left\{g\left(D\right)y\right\}$

(微分演算子の積を参照)

$=\left(D+{C}_{1}\right)\left\{g\left(D\right)y\right\}$

$=D\left\{g\left(D\right)y\right\}+{C}_{1}\left\{g\left(D\right)y\right\}$

(微分演算子の和を参照)

$=D\left\{\left(D+{C}_{2}\right)y\right\}+{C}_{1}\left\{\left(D+{C}_{2}\right)y\right\}$

$=\left({D}^{2}+{C}_{2}D\right)y+\left({C}_{1}D+{C}_{1}{C}_{2}\right)y$

$={D}^{2}y+{C}_{2}Dy+{C}_{1}Dy+{C}_{1}{C}_{2}y$

$={D}^{2}y+{C}_{1}Dy+{C}_{2}Dy+{C}_{1}{C}_{2}y$

$=\left({D}^{2}+{C}_{1}D\right)y+\left({C}_{2}D+{C}_{1}{C}_{2}\right)y$

$=D\left\{\left(D+{C}_{1}\right)y\right\}+{C}_{2}\left\{\left(D+{C}_{1}\right)y\right\}$

$=D\left\{f\left(D\right)y\right\}+{C}_{1}\left\{f\left(D\right)y\right\}$

$=\left(D+{C}_{2}\right)\left\{f\left(D\right)y\right\}$

$=g\left(D\right)\left\{f\left(D\right)y\right\}$

$=\left\{g\left(D\right)f\left(D\right)\right\}y$

よって

$f\left(D\right)g\left(D\right)=g\left(D\right)f\left(D\right)$

が成り立つ．

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