Roche-Schlömilch の剰余項 - yoshitake-hのブログ

n 回微分可能な関数 $f(x)$ と $p>0,\; x_0,\;x_1$ に対し， $0<\theta <1$ が存在して，
$f(x_1) = \sum_{k=1}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x_1-x_0)^k \\ + \frac{f^{(n)}( (1-\theta )x_0+\theta x_1)}{(n-1)!}\cdot \frac{(1-\theta )^{n-p} }p (x_1-x_0)^n.$

$R_n(t) = f(x_1) - \sum_{k=0}^{n-1} \frac{f^{(k)}(t)}{k!} (x_1-t)^k$
とおくと，
$R_n(x_1)=0, \\ (R_n)'(t) = -\sum_{k=0}^{n-1} \left( \frac{f^{(k+1)}(t)}{k!}(x_1-t)^k - \frac{f^{(k)}(t)}{k!}k(x_1-t)^{k-1} \right) \\ =-\frac{f^{(n)}(t)}{(n-1)!}(x_1-t)^{n-1}.$

$g(t)=(x_1-t)^p$ とおくと， $g'(t)=-p(x_1-t)^{p-1}.$

Cauchy の平均値の定理より， $0<\theta <1$ が存在して，
$\frac{R_n(x_0)-R_n(x_1)}{g(x_0)-g(x_1)} = \frac{(R_n)'( (1-\theta )x_0+\theta x_1)}{g'((1-\theta )x_0+\theta x_1)}. \\ \therefore \quad \frac{R_n(x_0)}{(x_1-x_0)^p} = \frac{ \displaystyle -\frac{f^{(n)}( (1-\theta )x_0+\theta x_1)}{(n-1)!}(1-\theta)^{n-1}(x_1-x_0)^{n-1} }{-p(1-\theta)^{p-1}(x_1-x_0)^{p-1} }. \\ \therefore \quad R_n(x_0) = \frac{f^{(n)}( (1-\theta )x_0+\theta x_1)}{(n-1)!}\cdot \frac{(1-\theta )^{n-p} }p (x_1-x_0)^n. //$