# 理查德森外推法

## 推导

$D = D \left( h \right) + a h^p + b h^q + \ldots$ （1）

$D = D \left( h_2 \right) + a h_2^p + b h^q_2 + \ldots$ （2）

$\left( 1 - r \right) D = D \left( h \right) + a h^p + b h^q - r D \left(h_2 \right) - r a h^p_2 - r b h^q_2 + \ldots = D \left( h \right) - r D \left( h_2 \right) + a \underbrace{\left( h^p - r h_2^p \right)}_0 + b \left( h^q - r h_2^q \right)$
$\left( 1 - r \right) D = D \left( h \right) - r D \left( h_2 \right) + b \left( h^q - r h_2^q \right)$
$D = \frac{D \left( h \right) - r D \left( h_2 \right)}{1 - r} + \frac{b\left( h^q - r h_2^q \right)}{1 - r} = \frac{D \left( h \right) - r D\left( h_2 \right)}{1 - r} + \frac{b \left( 1 - r \frac{h_2^q}{h^q}\right) h^q}{1 - r}$
$D = \frac{D \left( h \right) - r D \left( h_2 \right)}{1 - r} + \frac{b\left( 1 - \left( \frac{h}{h_2} \right)^p \left( \frac{h_2}{h} \right)^q \right) h^q}{1 - r} = \underbrace{\frac{D \left( h \right) - r D \left(h_2 \right)}{1 - r}}_{D^{\ast} \left( h \right)} + \underbrace{\frac{b\left( 1 - \left( \frac{h_2}{h} \right)^{q - p} \right)}{1 - r}}_{b^{\ast}} h^q$

$\therefore D = D^{\ast} \left( h \right) + b^{\ast} h^q$

$D^{\ast}(h)$代替了$D(h)$，为$D$的新的数值近似。新近似相比最初形式具有更高阶的误差项，数值精度由此提高，此方法即为理查德森外推法

## 示例

$f' \left( x_n \right) = \underset{D \left( h \right)}{\underbrace{\frac{f\left( x_n + h \right) - f \left( x_n - h \right)}{2 h}}} -\underset{a}{\underbrace{\frac{f''' \left( x_n \right)}{6}}} h^2 -\underset{b}{\underbrace{\frac{f^{\left( 5 \right)} \left( x_n\right)}{120}}} h^4$

$D^{\ast} = \frac{D \left( h \right) - r D \left( h_2 \right)}{1 - r} = \frac{\frac{f \left( x_n + h \right) - f \left( x_n - h \right)}{2 h} - \frac{1}{4} \frac{f \left( x_n + 2 h \right) - f \left( x_n - 2 h \right)}{4 h}}{1 - \frac{1}{4}}$
$D^{\ast} = \frac{8 \left[ f \left( x_n + h \right) - f \left( x_n - h\right) \right] - f \left( x_n + 2 h \right) + f \left( x_n - 2 h\right)}{12 h}$

## 参考文献

• Extrapolation Methods. Theory and Practice by C. Brezinski and M. Redivo Zaglia, North-Holland, 1991.