牛顿多项式

給定包含${\displaystyle k+1}$個數據點的集合${\displaystyle (x_{0},y_{0}),\ldots ,(x_{k},y_{k})}$。

如果對於${\displaystyle \forall i,j\in \left\{0,...,k\right\},i\neq j}$，滿足${\displaystyle x_{i}\neq x_{j}}$，那麼應用牛頓插值公式所得到的牛頓插值多項式為

${\displaystyle N(x):=\sum _{j=0}^{k}a_{j}n_{j}(x)}$

其中每個${\displaystyle n_{j}(x)}$為牛頓基本多項式（或稱插值基函數），其表達式為

${\displaystyle n_{j}(x):=\prod _{i=0}^{j-1}(x-x_{i})}$

其中${\displaystyle j>0}$，並且${\displaystyle n_{0}(x)\equiv 1}$。

係數${\displaystyle a_{j}:=[y_{0},\ldots ,y_{j}]}$，而${\displaystyle [y_{0},\ldots ,y_{j}]}$表示差商。

差商表（高階差商是兩個低一階差商的差商）

	${\displaystyle 0}$階差商	${\displaystyle 1}$階差商	${\displaystyle 2}$階差商	${\displaystyle 3}$階差商	${\displaystyle \ldots }$	${\displaystyle k-1}$階差商
${\displaystyle x_{0}}$	${\displaystyle f[x_{0}]}$
${\displaystyle x_{1}}$	${\displaystyle f[x_{1}]}$	${\displaystyle f[x_{0},x_{1}]}$
${\displaystyle x_{2}}$	${\displaystyle f[x_{2}]}$	${\displaystyle f[x_{1},x_{2}]}$	${\displaystyle f[x_{0},x_{1},x_{2}]}$
${\displaystyle x_{3}}$	${\displaystyle f[x_{3}]}$	${\displaystyle f[x_{2},x_{3}]}$	${\displaystyle f[x_{1},x_{2},x_{3}]}$	${\displaystyle f[x_{0},x_{1},x_{2},x_{3}]}$
${\displaystyle \ldots }$	${\displaystyle \ldots }$	${\displaystyle \ldots }$	${\displaystyle \ldots }$	${\displaystyle \ldots }$	${\displaystyle \ldots }$
${\displaystyle x_{k}}$	${\displaystyle f[x_{k}]}$	${\displaystyle f[x_{k-1},x_{k}]}$	${\displaystyle f[x_{k-2},x_{k-1},x_{k}]}$	${\displaystyle f[x_{k-3},x_{k-2},x_{k-1},x_{k}]}$	${\displaystyle \ldots }$	${\displaystyle f[x_{0},\ldots ,x_{k}]}$

因此，牛頓多項式可以寫作：

${\displaystyle N(x)=[y_{0}]+[y_{0},y_{1}](x-x_{0})+\cdots +[y_{0},\ldots ,y_{k}](x-x_{0})(x-x_{1})\cdots (x-x_{k-1})}$

• 插值
• 多項式插值
• 拉格朗日插值法