# 制御と振動の数学/Laplace 変換/有理関数の原像/別法

${\displaystyle F(s)\sqsubset f(t)}$ ならば ${\displaystyle {\frac {dF(s)}{ds}}\sqsubset -tf(t)}$

${\displaystyle F(s)=\int _{0}^{\infty }f(t)e^{-st}dt}$

${\displaystyle s}$ で項別微分すると，

${\displaystyle {\frac {dF(s)}{ds}}=\int _{0}^{\infty }(-t)f(t)e^{-st}dt}$

となる． ${\displaystyle \diamondsuit }$

よって補題2.3

${\displaystyle {\frac {d(FG)}{ds}}={\frac {dF}{ds}}G+F{\frac {dG}{ds}}}$

の原像であることが分かる[1]

${\displaystyle T(f+g)=Tf+Tg}$
${\displaystyle T(f\circ g)=Tf\circ g+f\circ Tg}$

なる形式を有するとき，${\displaystyle T}$ は微分演算と考えてよいのである．

この公式 3を用いて公式 2を導け．

(1)

${\displaystyle {\frac {d}{ds}}{\frac {1}{(s^{2}+\beta ^{2})^{n-1}}}=-2(n-1){\frac {s}{(s^{2}+\beta ^{2})^{n}}}}$

を念頭において，

${\displaystyle {\frac {1}{(s^{2}+\beta ^{2})^{n-1}}}\sqsubset f_{n-1}}$

${\displaystyle -2(n-1){\frac {s}{(s^{2}+\beta ^{2})^{n}}}\sqsubset (-t)f_{n-1}}$

${\displaystyle {\frac {s}{(s^{2}+\beta ^{2})^{n}}}\sqsubset {\frac {t}{2(n-1)}}f_{n-1}}$

(2)

${\displaystyle {\frac {d^{2}}{ds^{2}}}{\frac {1}{(s^{2}+\beta ^{2})^{n-1}}}={\frac {d}{ds}}\left(-{\frac {2(n-1)s}{(s^{2}+\beta ^{2})^{n}}}\right)}$
${\displaystyle ={\frac {-2(n-1)}{(s^{2}+\beta ^{2})^{n}}}+{\frac {4n(n-1)s^{2}}{(s^{2}+\beta ^{2})^{n+1}}}}$
${\displaystyle ={\frac {-2(n-1)}{(s^{2}+\beta ^{2})^{n}}}+4n(n-1)\left({\frac {(s^{2}+\beta ^{2})-\beta ^{2}}{(s^{2}+\beta ^{2})^{n+1}}}\right)}$
${\displaystyle ={\frac {-2(n-1)}{(s^{2}+\beta ^{2})^{n}}}+{\frac {4n(n-1)}{(s^{2}+\beta ^{2})^{n}}}-{\frac {4\beta ^{2}n(n-1)}{(s^{2}+\beta ^{2})^{n+1}}}}$
${\displaystyle ={\frac {4n^{2}-4n-2n+2}{(s^{2}+\beta ^{2})^{n}}}-{\frac {4\beta ^{2}n(n-1)}{(s^{2}+\beta ^{2})^{n+1}}}}$
${\displaystyle ={\frac {2(2n-1)(n-1)}{(s^{2}+\beta ^{2})^{n}}}-{\frac {4\beta ^{2}n(n-1)}{(s^{2}+\beta ^{2})^{n+1}}}}$

この原像は，

${\displaystyle t^{2}f_{n-1}=2(2n-1)(n-1)f_{n}-4\beta ^{2}n(n-1)f_{n+1}}$
${\displaystyle 4\beta ^{2}n(n-1)f_{n+1}=2(2n-1)(n-1)f_{n}-t^{2}f_{n-1}}$
${\displaystyle f_{n+1}={\frac {1}{4\beta ^{2}}}\left\{{\frac {2(2n-1)}{n}}f_{n}-{\frac {t^{2}}{n(n-1)}}f_{n-1}\right\}}$
${\displaystyle ={\frac {1}{2\beta ^{2}}}\left\{{\frac {2n-1}{n}}f_{n}-{\frac {t^{2}}{2n(n-1)}}f_{n-1}\right\}}$

これが求める結果である． ${\displaystyle \diamondsuit }$

1. ^
${\displaystyle t(f*g)=(tf)*g+f*(tg)}$
より
${\displaystyle (-t)(f*g)=\{(-t)f\}*g+f*\{(-t)g\}}$