均差

均差（Divided differences）是递归除法过程。在数值分析中，可用于计算牛顿多项式形式的多项式插值的系数。在微积分中，均差与导数一起合称差商，是对函数在一个区间内的平均变化率的测量^[1]^[2]^[3]。

均差也是一种算法，查尔斯·巴贝奇的差分机，是他在1822年发表的论文中提出的一种早期的机械计算机，在历史上意图用来计算对数表和三角函数表，它设计在其运算中使用这个算法^[4]。

给定n+1个数据点

(x_{0},y_{0}),\ldots ,(x_{n},y_{n})

定义前向均差为：

{\begin{aligned}{\mathopen {[}}y_{\nu }]&=y_{\nu },\quad \nu \in \{0,\ldots ,n\}\\{\mathopen {[}}y_{\nu },\ldots ,y_{\nu +j}]&={\frac {[y_{\nu +1},\ldots ,y_{\nu +j}]-[y_{\nu },\ldots ,y_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\quad \nu \in \{0,\ldots ,n-j\},\ j\in \{1,\ldots ,n\}\\\end{aligned}}

定义后向均差为：

{\begin{aligned}{\mathopen {[}}y_{\nu }]&=y_{\nu },\quad \nu \in \{0,\ldots ,n\}\\{\mathopen {[}}y_{\nu },\ldots ,y_{\nu -j}]&={\frac {[y_{\nu },\ldots ,y_{\nu -j+1}]-[y_{\nu -1},\ldots ,y_{\nu -j}]}{x_{\nu }-x_{\nu -j}}},\quad \nu \in \{j,\ldots ,n\},\ j\in \{1,\ldots ,n\}\\\end{aligned}}

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假定数据点给出为函数 ƒ，

(x_{0},f(x_{0})),\ldots ,(x_{n},f(x_{n}))

其均差可以写为：

{\begin{aligned}f[x_{\nu }]&=f(x_{\nu }),\qquad \nu \in \{0,\ldots ,n\}\\f[x_{\nu },\ldots ,x_{\nu +j}]&={\frac {f[x_{\nu +1},\ldots ,x_{\nu +j}]-f[x_{\nu },\ldots ,x_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\quad \nu \in \{0,\ldots ,n-j\},\ j\in \{1,\ldots ,n\}\end{aligned}}

对函数 ƒ 在节点 x₀, ..., x_n 上的均差还有其他表示法，如：

{\begin{matrix}{\mathopen {[}}x_{0},\ldots ,x_{n}]f\\{\mathopen {[}}x_{0},\ldots ,x_{n};f]\\{\mathopen {D}}[x_{0},\ldots ,x_{n}]f\\\end{matrix}}

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给定ν=0：

{\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0}\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {{\mathopen {[}}y_{1},y_{2}]-{\mathopen {[}}y_{0},y_{1}]}{x_{2}-x_{0}}}\\{\mathopen {[}}y_{0},y_{1},y_{2},y_{3}]&={\frac {{\mathopen {[}}y_{1},y_{2},y_{3}]-{\mathopen {[}}y_{0},y_{1},y_{2}]}{x_{3}-x_{0}}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n}]&={\frac {{\mathopen {[}}y_{1},y_{2},\dots ,y_{n}]-{\mathopen {[}}y_{0},y_{1},\dots ,y_{n-1}]}{x_{n}-x_{0}}}\end{aligned}}

为了使涉及的递归过程更加清楚，以列表形式展示均差的计算过程^[5]：

{\begin{matrix}x_{0}&[y_{0}]=y_{0}&&&\\&&[y_{0},y_{1}]&&\\x_{1}&[y_{1}]=y_{1}&&[y_{0},y_{1},y_{2}]&\\&&[y_{1},y_{2}]&&[y_{0},y_{1},y_{2},y_{3}]\\x_{2}&[y_{2}]=y_{2}&&[y_{1},y_{2},y_{3}]&\\&&[y_{2},y_{3}]&&\\x_{3}&[y_{3}]=y_{3}&&&\\\end{matrix}}

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用数学归纳法可证明^[6]：

{\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0}\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{0}}{x_{0}-x_{1}}}+{\frac {y_{1}}{x_{1}-x_{0}}}\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {y_{0}}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {y_{1}}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {y_{2}}{(x_{2}-x_{0})(x_{2}-x_{1})}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n}]&=\sum _{j=0}^{n}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}\\\end{aligned}}

此公式体现了均差的对称性质。^[7]故可推知：任意调换数据点次序，其值不变。^[8]

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对称性：若 $\sigma :\{0,\dots ,n\}\to \{0,\dots ,n\}$ 是一个排列则

f[x_{0},\dots ,x_{n}]=f[x_{\sigma (0)},\dots ,x_{\sigma (n)}]

线性：

{\begin{aligned}(f+g)[x_{0},\dots ,x_{n}]&=f[x_{0},\dots ,x_{n}]+g[x_{0},\dots ,x_{n}]\\(\lambda \cdot f)[x_{0},\dots ,x_{n}]&=\lambda \cdot f[x_{0},\dots ,x_{n}]\\\end{aligned}}

莱布尼茨法则（英语：General Leibniz rule）：

(f\cdot g)[x_{0},\dots ,x_{n}]=f[x_{0}]\cdot g[x_{0},\dots ,x_{n}]+f[x_{0},x_{1}]\cdot g[x_{1},\dots ,x_{n}]+\dots +f[x_{0},\dots ,x_{n}]\cdot g[x_{n}]

均差中值定理（英语：Mean value theorem (divided differences)）：

\exists \xi \in (\min\{x_{0},\dots ,x_{n}\},\max\{x_{0},\dots ,x_{n}\})\quad f[x_{0},\dots ,x_{n}]={\frac {f^{(n)}(\xi )}{n!}}

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通过对换 n 阶均差中(x₀,y₀)与(x_n-1,y_n-1)，可得到等价定义：

{\begin{aligned}{\mathopen {[}}y_{0},y_{1},\dots ,y_{n-1},y_{n}]&={\frac {{\mathopen {[}}y_{1},y_{2},\dots ,y_{n}]-{\mathopen {[}}y_{0},y_{1},\dots ,y_{n-1}]}{x_{n}-x_{0}}}\\&={\frac {{\mathopen {[}}y_{1},\dots ,y_{n-2},y_{0},y_{n}]-{\mathopen {[}}y_{n-1},y_{1},\dots ,y_{n-2},y_{0}]}{x_{n}-x_{n-1}}}\\&={\frac {{\mathopen {[}}y_{0},\dots ,y_{n-2},y_{n}]-{\mathopen {[}}y_{0},y_{1},\dots ,y_{n-1}]}{x_{n}-x_{n-1}}}\\\end{aligned}}

这个定义有着不同的计算次序：

{\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0}\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {{\mathopen {[}}y_{0},y_{2}]-{\mathopen {[}}y_{0},y_{1}]}{x_{2}-x_{1}}}\\{\mathopen {[}}y_{0},y_{1},y_{2},y_{3}]&={\frac {{\mathopen {[}}y_{0},y_{1},y_{3}]-{\mathopen {[}}y_{0},y_{1},y_{2}]}{x_{3}-x_{2}}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n}]&={\frac {{\mathopen {[}}y_{0},\dots ,y_{n-2},y_{n}]-{\mathopen {[}}y_{0},y_{1},\dots ,y_{n-1}]}{x_{n}-x_{n-1}}}\\\end{aligned}}

以列表形式展示这个定义下均差的计算过程^[9]：

{\begin{matrix}x_{0}&[y_{0}]=y_{0}&&&\\&&[y_{0},y_{1}]&&\\x_{1}&[y_{1}]=y_{1}&&[y_{0},y_{1},y_{2}]&\\&&[y_{0},y_{2}]&&[y_{0},y_{1},y_{2},y_{3}]\\x_{2}&[y_{2}]=y_{2}&&[y_{0},y_{1},y_{3}]&\\&&[y_{0},y_{3}]&&\\x_{3}&[y_{3}]=y_{3}&&&\\\end{matrix}}

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Thumb — 《自然哲学的数学原理》的第三编“宇宙体系”的引理五的图例。这里在横坐标上有6个点H,I,K,L,M,N，对应着6个值A,B,C,D,E,F，生成一个多项式函数对这6个点上有对应的6个值，计算任意点S对应的值R。牛顿给出了间距为单位值和任意值的两种情况。

牛顿插值公式，得名于伊萨克·牛顿爵士，最早发表为他在1687年出版的《自然哲学的数学原理》中第三编“宇宙体系”的引理五，此前詹姆斯·格雷果里于1670年和牛顿于1676年已经分别独立得出这个成果。一般称其为连续泰勒展开的离散对应。

使用均差的牛顿插值法为^[10]：

{\begin{aligned}N_{n}(x)&=y_{0}+(x-{x}_{0})\left([{y}_{0},{y}_{1}]+(x-{x}_{1})\left([{y}_{0},{y}_{1},{y}_{2}]+\cdots \right)\right)\\&=[y_{0}]+[{y}_{0},{y}_{1}](x-{x}_{0})+\cdots +[{y}_{0},{y}_{1},\ldots ,{y}_{n}]\prod _{k=0}^{n-1}(x-{x}_{k})\end{aligned}}

可以在计算过程中任意增添节点如点(x_n+1,y_n+1)，只需计算新增的n+1阶均差及其插值基函数，而无拉格朗日插值法需重算全部插值基函数之虞。

对均差采用展开形式^[11]：

{\begin{aligned}N_{n}(x)&=y_{0}+y_{0}{\frac {x-{x}_{0}}{x_{0}-x_{1}}}+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}+\cdots +\sum _{j=0}^{n}y_{j}{\frac {\prod _{k=0}^{n-1}(x-{x}_{k})}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}\\\end{aligned}}

以2阶均差牛顿插值为例：

{\begin{aligned}N_{2}(x)&=y_{0}\left(1+{\frac {x-{x}_{0}}{x_{0}-x_{1}}}+{\frac {(x-x_{0})(x-x_{1})}{(x_{0}-x_{1})(x_{0}-x_{2})}}\right)+y_{1}\left({\frac {x-{x}_{0}}{x_{1}-x_{0}}}+{\frac {(x-x_{0})(x-x_{1})}{(x_{1}-x_{0})(x_{1}-x_{2})}}\right)+y_{2}{\frac {(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}\\&=y_{0}{\frac {(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}}+y_{1}{\frac {(x-x_{0})(x-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}}+y_{2}{\frac {(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}}\\&=\sum _{j=0}^{2}y_{j}\prod _{\begin{smallmatrix}k=0\\k\neq j\end{smallmatrix}}^{2}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}

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当数据点呈等距分布的时候，这个特殊情况叫做“前向差分”。它们比计算一般的均差要容易。

定义

给定n+1个数据点

(x_{0},y_{0}),\ldots ,(x_{n},y_{n})

有着

x_{i}=x_{0}+ih,\quad h>0{\mbox{ , }}0\leq i\leq n

定义前向差分为：

{\begin{aligned}\triangle ^{0}y_{i}&=y_{i}\\\triangle ^{k}y_{i}&=\triangle ^{k-1}y_{i+1}-\triangle ^{k-1}y_{i},\quad 1\leq k\leq n-i\\\end{aligned}}

前向差分所对应的均差为^[12]：

f[x_{0},x_{1},\ldots ,x_{k}]={\frac {1}{k!h^{k}}}\Delta ^{(k)}f(x_{0})

例子

{\begin{matrix}y_{0}&&&\\&\triangle y_{0}&&\\y_{1}&&\triangle ^{2}y_{0}&\\&\triangle y_{1}&&\triangle ^{3}y_{0}\\y_{2}&&\triangle ^{2}y_{1}&\\&\triangle y_{2}&&\\y_{3}&&&\\\end{matrix}}

展开形式

差分的展开形式是均差展开形式的特殊情况^[13]：

{\begin{aligned}\triangle ^{k}y_{i}&=\sum _{j=0}^{k}(-1)^{k-j}{\binom {k}{j}}y_{i+j},\quad 0\leq k\leq n-i\end{aligned}}

这里的表达式

{n \choose k}={\frac {(n)_{k}}{k!}}\quad \quad (n)_{k}=n(n-1)(n-2)\cdots (n-k+1)

是二项式系数，其中的(n)_k是“下降阶乘幂”，空积(n)₀被定义为1。

插值公式

其对应的牛顿插值公式为：

{\begin{aligned}f(x)&=y_{0}+{\frac {x-x_{0}}{h}}\left(\Delta ^{1}y_{0}+{\frac {x-x_{0}-h}{2h}}\left(\Delta ^{2}y_{0}+\cdots \right)\right)\\&=y_{0}+\sum _{k=1}^{n}{\frac {\Delta ^{k}y_{0}}{k!h^{k}}}\prod _{i=0}^{n-1}(x-x_{0}-ih)\\&=y_{0}+\sum _{k=1}^{n}{\frac {\Delta ^{k}y_{0}}{k!}}\prod _{i=0}^{n-1}({\frac {x-x_{0}}{h}}-i)\\&=\sum _{k=0}^{n}{{\frac {x-x_{0}}{h}} \choose k}~\Delta ^{k}y_{0}\\\end{aligned}}

无穷级数

牛顿在1665年得出并在1671年写的《流数法》中发表了ln(1+x)的无穷级数，在1666年得出了arcsin(x)和arctan(x)的无穷级数，在1669年的《分析学》中发表了sin(x)、cos(x)、arcsin(x)和e^x的无穷级数；莱布尼茨在1673年大概也得出了sin(x)、cos(x)和arctan(x)的无穷级数。布鲁克·泰勒在1715年著作《Methodus Incrementorum Directa et Inversa》^[14]中研讨了“有限差分”方法，其中论述了他在1712年得出的泰勒定理，这个成果此前詹姆斯·格雷果里在1670年和莱布尼茨在1673年已经得出，而约翰·伯努利在1694年已经在《教师学报》发表。

他对牛顿的均差的步长取趋于0的极限，得出：

{\begin{aligned}f(x)&=f(a)+\lim _{h\to 0}\sum _{k=1}^{\infty }{\frac {\Delta _{h}^{k}[f](a)}{k!h^{k}}}\prod _{i=0}^{k-1}((x-a)-ih)\\&=f(a)+\sum _{k=1}^{\infty }{\frac {d^{k}}{dx^{k}}}f(a){\frac {(x-a)^{k}}{k!}}\\\end{aligned}}

使用普通函数记号表示幂运算， $p_{n}(x)=x^{n}$ ，有：

{\begin{aligned}p_{j}[x_{0},\dots ,x_{n}]&=0\qquad \forall j<n\\p_{n}[x_{0},\dots ,x_{n}]&=1\\p_{n+1}[x_{0},\dots ,x_{n}]&=x_{0}+\dots +x_{n}\\p_{n+m}[x_{0},\dots ,x_{n}]&=\sum _{k_{0}+\cdots +k_{n}=m}{\begin{matrix}\prod _{t=0}^{n}x_{t}^{k_{t}}\end{matrix}}\\\end{aligned}}

此中n+1元m次齐次多项式的记法同于多项式定理。

泰勒级数和任何其他的函数级数，在原理上都可以用来逼近均差。将泰勒级数表示为:

f=f(0)p_{0}+f'(0)p_{1}+{\frac {f''(0)}{2!}}p_{2}+\dots

均差的泰勒级数为：

f[x_{0},\dots ,x_{n}]=f(0)p_{0}[x_{0},\dots ,x_{n}]+f'(0)p_{1}[x_{0},\dots ,x_{n}]+\dots +{\frac {f^{(n)}(0)}{n!}}p_{n}[x_{0},\dots ,x_{n}]+\dots

前 $n$ 项消失了，因为均差的阶高于多项式的阶。可以得出均差的泰勒级数本质上开始于：

{\frac {f^{(n)}(0)}{n!}}

依据均差中值定理（英语：Mean value theorem (divided differences)），这也是均差的最简单逼近。

均差还可以表达为

f[x_{0},\ldots ,x_{n}]={\frac {1}{n!}}\int _{x_{0}}^{x_{n}}f^{(n)}(t)B_{n-1}(t)\,dt

这里的B_n-1是数据点x₀,...,x_n的n-1次B样条，而f⁽ⁿ⁾是函数f的n阶导数。这叫做均差的皮亚诺形式，而B_n-1是均差的皮亚诺核。

[1]
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Tamara Lefcourt Ruby; James Sellers; Lisa Korf; Jeremy Van Horn; Mike Munn. Kaplan AP Calculus AB & BC 2015. Kaplan Publishing. 2014: 237. ISBN 978-1-61865-686-5.
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Isaacson, Walter. The Innovators. Simon & Schuster. 2014: 20. ISBN 978-1-4767-0869-0.
[5]

${\begin{array}{lcl}{\begin{matrix}x_{0}&x_{0}^{2}&&\\&&x_{0}+x_{1}&&\\x_{1}&x_{1}^{2}&&1&\\&&x_{1}+x_{2}&&0\\x_{2}&x_{2}^{2}&&1&\\&&x_{2}+x_{3}&&&\\x_{3}&x_{3}^{2}&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n}&\\&&\sum _{i=0}^{n-1}x_{0}^{n-1-i}x_{1}^{i}\\x_{1}&x_{1}^{n}&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&\\&&{\frac {x_{1}^{n+1}-x_{1}x_{0}^{n}+x_{1}x_{0}^{n}-x_{0}^{n+1}}{x_{1}-x_{0}}}=x_{1}{\frac {x_{1}^{n}-x_{0}^{n}}{x_{1}-x_{0}}}+x_{0}^{n}=x_{1}\sum _{i=0}^{n-1}x_{0}^{n-1-i}x_{1}^{i}+x_{0}^{n}=\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}\\x_{1}&x_{1}^{n+1}&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&&\\&&\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}&\\x_{1}&x_{1}^{n+1}&&{\frac {\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}-\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}}{x_{2}-x_{0}}}={\frac {\sum _{i=0}^{n-1}x_{1}^{i}(x_{2}^{n-i}-x_{0}^{n-i})}{x_{2}-x_{0}}}=\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}\\&&\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}&\\x_{2}&x_{2}^{n+1}&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&&&\\&&\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}&&\\x_{1}&x_{1}^{n+1}&&\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}&\\&&\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}&&{\frac {\sum _{i+j+k=n-1}{x_{1}^{i}x_{2}^{j}x_{3}^{k}}-\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}}{x_{3}-x_{0}}}=\sum _{i+j+k+l=n-2}{x_{0}^{i}x_{1}^{j}x_{2}^{k}x_{3}^{l}}\\x_{2}&x_{2}^{n+1}&&\sum _{i+j+k=n-1}{x_{1}^{i}x_{2}^{j}x_{3}^{k}}&\\&&\sum _{i=0}^{n}x_{2}^{n-i}x_{3}^{i}&&\\x_{3}&x_{3}^{n+1}&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{3}&&&&\\&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}&&\\x_{1}&x_{1}^{3}&&x_{0}+x_{1}+x_{2}&&\\&&x_{1}^{1}+x_{1}x_{2}+x_{2}^{2}&&1&\\x_{2}&x_{2}^{3}&&x_{1}+x_{2}+x_{3}&&0\\&&x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}&&1&\\x_{3}&x_{3}^{3}&&x_{2}+x_{3}+x_{4}&&\\&&x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}&&&\\x_{4}&x_{4}^{3}&&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{4}&&&&&\\&&x_{0}^{3}+x_{0}^{2}x_{1}+x_{0}x_{1}^{2}+x_{1}^{3}&&&\\x_{1}&x_{1}^{4}&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}+x_{0}x_{2}+x_{1}x_{2}+x_{2}^{2}&&&\\&&x_{1}^{3}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2}+x_{2}^{3}&&x_{0}+x_{1}+x_{2}+x_{3}&\\x_{2}&x_{2}^{4}&&x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}+x_{1}x_{3}+x_{2}x_{3}+x_{3}^{2}&&1&\\&&x_{2}^{3}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}+x_{3}^{3}&&x_{1}+x_{2}+x_{3}+x_{4}&&0\\x_{3}&x_{3}^{4}&&x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}+x_{2}x_{4}+x_{3}x_{4}+x_{4}^{2}&&1&\\&&x_{3}^{3}+x_{3}^{2}x_{4}+x_{3}x_{4}^{2}+x_{4}^{3}&&x_{2}+x_{3}+x_{4}+x_{5}&&\\x_{4}&x_{4}^{4}&&x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}+x_{3}x_{5}+x_{4}x_{5}+x_{5}^{2}&&&\\&&x_{4}^{3}+x_{4}^{2}x_{5}+x_{4}x_{5}^{2}+x_{5}^{3}&&&&\\x_{5}&x_{5}^{4}&&&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{5}&&&&&&\\&&\sum _{i=0}^{4}x_{0}^{4-i}x_{1}^{i}&&&&\\x_{1}&x_{1}^{5}&&\sum _{i+j+k=3}x_{0}^{i}x_{1}^{j}x_{2}^{k}&&&&\\&&\sum _{i=0}^{4}x_{1}^{4-i}x_{2}^{i}&&\sum _{i+j+k+l=2}x_{0}^{i}x_{1}^{j}x_{2}^{k}x_{3}^{l}&&&\\x_{2}&x_{2}^{5}&&\sum _{i+j+k=3}x_{1}^{i}x_{2}^{j}x_{3}^{k}&&\sum _{i=0}^{4}x_{i}&&\\&&\sum _{i=0}^{4}x_{2}^{4-i}x_{3}^{i}&&\sum _{i+j+k+l=2}x_{1}^{i}x_{2}^{j}x_{3}^{k}x_{4}^{l}&&1&\\x_{3}&x_{3}^{5}&&\sum _{i+j+k=3}x_{2}^{i}x_{3}^{j}x_{4}^{k}&&\sum _{i=1}^{5}x_{i}&&0\\&&\sum _{i=0}^{4}x_{3}^{4-i}x_{4}^{i}&&\sum _{i+j+k+l=2}x_{2}^{i}x_{3}^{j}x_{4}^{j}x_{5}^{l}&&1&\\x_{4}&x_{4}^{5}&&\sum _{i+j+k=3}x_{3}^{i}x_{4}^{j}x_{5}^{k}&&\sum _{i=2}^{6}x_{i}&&\\&&\sum _{i=0}^{4}x_{4}^{4-i}x_{5}^{i}&&\sum _{i+j+k+l=2}x_{3}^{i}x_{4}^{j}x_{5}^{k}x_{6}^{l}&&&\\x_{5}&x_{5}^{5}&&\sum _{i+j+k=3}x_{4}^{i}x_{5}^{j}x_{6}^{k}&&&&\\&&\sum _{i=0}^{4}x_{5}^{4-i}x_{6}^{i}&&&&&\\x_{6}&x_{6}^{5}&&&&&&\\\end{matrix}}\\\end{array}}$
[6]

${\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0}\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}={\frac {y_{0}}{x_{0}-x_{1}}}+{\frac {y_{1}}{x_{1}-x_{0}}}\\&=\sum _{j=0}^{1}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{1}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {{\cfrac {y_{1}}{x_{1}-x_{2}}}+{\cfrac {y_{2}}{x_{2}-x_{1}}}-{\cfrac {y_{0}}{x_{0}-x_{1}}}-{\cfrac {y_{1}}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}\\&={\frac {y_{0}}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {y_{1}}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {y_{2}}{(x_{2}-x_{0})(x_{2}-x_{1})}}\\&=\sum _{j=0}^{2}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{2}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n}]&=\sum _{j=0}^{n}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{n}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n+1}]&={\frac {\sum _{j=1}^{n+1}{\frac {y_{j}}{\prod _{k=1,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-\sum _{j=0}^{n}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {1}{\prod _{k=1,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-{\frac {1}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {x_{j}-x_{0}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-{\frac {x_{j}-x_{n+1}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {x_{n+1}-x_{0}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {y_{n+1}}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}+{\frac {y_{0}}{\prod _{k=1}^{n+1}(x_{0}-x_{k})}}\\&=\sum _{j=0}^{n+1}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\end{aligned}}$
[7]
《数值分析及科学计算》薛毅（编）第六章第2节 Newton插值. P200.
[8]
《数值分析及科学计算》薛毅（编）第六章第2节 Newton插值. P201.
[9]

${\begin{matrix}x_{0}&x_{0}^{3}&&&&\\&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}&&\\x_{1}&x_{1}^{3}&&x_{0}+x_{1}+x_{2}&&\\&&x_{0}^{2}+x_{0}x_{2}+x_{2}^{2}&&1&\\x_{2}&x_{2}^{3}&&x_{0}+x_{1}+x_{3}&&0\\&&x_{0}^{2}+x_{0}x_{3}+x_{3}^{2}&&1&\\x_{3}&x_{3}^{3}&&x_{0}+x_{1}+x_{4}&&\\&&x_{0}^{2}+x_{0}x_{4}+x_{4}^{2}&&&\\x_{4}&x_{4}^{3}&&&&\\\end{matrix}}$
[10]
The Newton Polynomial Interpolation. [2019-04-19]. （原始内容存档于2019-04-19）.
[11]

${\begin{array}{lcl}{\begin{aligned}N_{1}(x)&=[y_{0}]+[{y}_{0},{y}_{1}](x-{x}_{0})=y_{0}+y_{0}{\frac {x-{x}_{0}}{x_{0}-x_{1}}}+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}=y_{0}(1+{\frac {x-{x}_{0}}{x_{0}-x_{1}}})+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}\\&=y_{0}{\frac {x-{x}_{1}}{x_{0}-x_{1}}}+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}=\sum _{j=0}^{1}y_{j}\prod _{k=0,k\neq j}^{1}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\{\begin{aligned}N_{n}(x)&=\sum _{j=0}^{n}y_{j}\prod _{k=0,k\neq j}^{n}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\{\begin{aligned}N_{n+1}(x)&=N_{n}(x)+[{y}_{0},{y}_{1},\ldots ,{y}_{n+1}]\prod _{k=0}^{n}(x-{x}_{k})\\&=\sum _{j=0}^{n}y_{j}\prod _{k=0,k\neq j}^{n}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}+\sum _{j=0}^{n+1}y_{j}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}\left({\frac {\prod _{k=0,k\neq j}^{n}(x-{x}_{k})}{\prod _{k=0,k\neq j}^{n}(x_{j}-x_{k})}}+{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}\left({\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x_{j}-x_{n+1})}{\prod _{k=0,k\neq j}^{n+1}(x_{j}-x_{k})}}+{\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x-x_{j})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}{\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x-x_{n+1})}{\prod _{k=0,k\neq j}^{n+1}(x_{j}-x_{k})}}+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n+1}y_{j}\prod _{k=0,k\neq j}^{n+1}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\\end{array}}$
[12]
Burden, Richard L.; Faires, J. Douglas. Numerical Analysis 9th. 2011: 129.
[13]

${\begin{aligned}\triangle ^{k}y_{i}&=\sum _{j=0}^{k}{\frac {k!}{\prod _{l=0,\,l\neq j}^{k}(j-l)}}y_{i+j},\quad 0\leq k\leq n-i\\&=\sum _{j=0}^{k}{\frac {k!}{j!(-1)^{k-j}(k-j)!}}y_{i+j},\quad 0\leq k\leq n-i\\&=\sum _{j=0}^{k}(-1)^{k-j}{\binom {k}{j}}y_{i+j},\quad 0\leq k\leq n-i\end{aligned}}$
[14]
Methodus Incrementorum Directa et Inversa（页面存档备份，存于互联网档案馆）

Louis Melville Milne-Thomson. The Calculus of Finite Differences. American Mathematical Soc. 2000. Chapter 1: Divided Differences [1933]. ISBN 978-0-8218-2107-7.
Myron B. Allen; Eli L. Isaacson. Numerical Analysis for Applied Science. John Wiley & Sons. 1998. Appendix A. ISBN 978-1-118-03027-1.
Ron Goldman. Pyramid Algorithms: A Dynamic Programming Approach to Curves and Surfaces for Geometric Modeling. Morgan Kaufmann. 2002. Chapter 4:Newton Interpolation and Difference Triangles. ISBN 978-0-08-051547-2.

[WilsonAdamson2008-1] [1]
Frank C. Wilson; Scott Adamson. Applied Calculus. Cengage Learning. 2008: 177. ISBN 0-618-61104-5.

[RubySellers2014-2] [2]
Tamara Lefcourt Ruby; James Sellers; Lisa Korf; Jeremy Van Horn; Mike Munn. Kaplan AP Calculus AB & BC 2015. Kaplan Publishing. 2014: 237. ISBN 978-1-61865-686-5.

[HungerfordShaw2008-3] [3]
Thomas Hungerford; Douglas Shaw. Contemporary Precalculus: A Graphing Approach. Cengage Learning. 2008: 211–212. ISBN 0-495-10833-2.

[Isaacson2014-4] [4]
Isaacson, Walter. The Innovators. Simon & Schuster. 2014: 20. ISBN 978-1-4767-0869-0.

[5] [5]

${\begin{array}{lcl}{\begin{matrix}x_{0}&x_{0}^{2}&&\\&&x_{0}+x_{1}&&\\x_{1}&x_{1}^{2}&&1&\\&&x_{1}+x_{2}&&0\\x_{2}&x_{2}^{2}&&1&\\&&x_{2}+x_{3}&&&\\x_{3}&x_{3}^{2}&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n}&\\&&\sum _{i=0}^{n-1}x_{0}^{n-1-i}x_{1}^{i}\\x_{1}&x_{1}^{n}&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&\\&&{\frac {x_{1}^{n+1}-x_{1}x_{0}^{n}+x_{1}x_{0}^{n}-x_{0}^{n+1}}{x_{1}-x_{0}}}=x_{1}{\frac {x_{1}^{n}-x_{0}^{n}}{x_{1}-x_{0}}}+x_{0}^{n}=x_{1}\sum _{i=0}^{n-1}x_{0}^{n-1-i}x_{1}^{i}+x_{0}^{n}=\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}\\x_{1}&x_{1}^{n+1}&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&&\\&&\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}&\\x_{1}&x_{1}^{n+1}&&{\frac {\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}-\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}}{x_{2}-x_{0}}}={\frac {\sum _{i=0}^{n-1}x_{1}^{i}(x_{2}^{n-i}-x_{0}^{n-i})}{x_{2}-x_{0}}}=\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}\\&&\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}&\\x_{2}&x_{2}^{n+1}&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{n+1}&&&\\&&\sum _{i=0}^{n}x_{0}^{n-i}x_{1}^{i}&&\\x_{1}&x_{1}^{n+1}&&\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}&\\&&\sum _{i=0}^{n}x_{1}^{n-i}x_{2}^{i}&&{\frac {\sum _{i+j+k=n-1}{x_{1}^{i}x_{2}^{j}x_{3}^{k}}-\sum _{i+j+k=n-1}{x_{0}^{i}x_{1}^{j}x_{2}^{k}}}{x_{3}-x_{0}}}=\sum _{i+j+k+l=n-2}{x_{0}^{i}x_{1}^{j}x_{2}^{k}x_{3}^{l}}\\x_{2}&x_{2}^{n+1}&&\sum _{i+j+k=n-1}{x_{1}^{i}x_{2}^{j}x_{3}^{k}}&\\&&\sum _{i=0}^{n}x_{2}^{n-i}x_{3}^{i}&&\\x_{3}&x_{3}^{n+1}&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{3}&&&&\\&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}&&\\x_{1}&x_{1}^{3}&&x_{0}+x_{1}+x_{2}&&\\&&x_{1}^{1}+x_{1}x_{2}+x_{2}^{2}&&1&\\x_{2}&x_{2}^{3}&&x_{1}+x_{2}+x_{3}&&0\\&&x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}&&1&\\x_{3}&x_{3}^{3}&&x_{2}+x_{3}+x_{4}&&\\&&x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}&&&\\x_{4}&x_{4}^{3}&&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{4}&&&&&\\&&x_{0}^{3}+x_{0}^{2}x_{1}+x_{0}x_{1}^{2}+x_{1}^{3}&&&\\x_{1}&x_{1}^{4}&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}+x_{0}x_{2}+x_{1}x_{2}+x_{2}^{2}&&&\\&&x_{1}^{3}+x_{1}^{2}x_{2}+x_{1}x_{2}^{2}+x_{2}^{3}&&x_{0}+x_{1}+x_{2}+x_{3}&\\x_{2}&x_{2}^{4}&&x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}+x_{1}x_{3}+x_{2}x_{3}+x_{3}^{2}&&1&\\&&x_{2}^{3}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}+x_{3}^{3}&&x_{1}+x_{2}+x_{3}+x_{4}&&0\\x_{3}&x_{3}^{4}&&x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}+x_{2}x_{4}+x_{3}x_{4}+x_{4}^{2}&&1&\\&&x_{3}^{3}+x_{3}^{2}x_{4}+x_{3}x_{4}^{2}+x_{4}^{3}&&x_{2}+x_{3}+x_{4}+x_{5}&&\\x_{4}&x_{4}^{4}&&x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}+x_{3}x_{5}+x_{4}x_{5}+x_{5}^{2}&&&\\&&x_{4}^{3}+x_{4}^{2}x_{5}+x_{4}x_{5}^{2}+x_{5}^{3}&&&&\\x_{5}&x_{5}^{4}&&&&&\\\end{matrix}}\\\\{\begin{matrix}x_{0}&x_{0}^{5}&&&&&&\\&&\sum _{i=0}^{4}x_{0}^{4-i}x_{1}^{i}&&&&\\x_{1}&x_{1}^{5}&&\sum _{i+j+k=3}x_{0}^{i}x_{1}^{j}x_{2}^{k}&&&&\\&&\sum _{i=0}^{4}x_{1}^{4-i}x_{2}^{i}&&\sum _{i+j+k+l=2}x_{0}^{i}x_{1}^{j}x_{2}^{k}x_{3}^{l}&&&\\x_{2}&x_{2}^{5}&&\sum _{i+j+k=3}x_{1}^{i}x_{2}^{j}x_{3}^{k}&&\sum _{i=0}^{4}x_{i}&&\\&&\sum _{i=0}^{4}x_{2}^{4-i}x_{3}^{i}&&\sum _{i+j+k+l=2}x_{1}^{i}x_{2}^{j}x_{3}^{k}x_{4}^{l}&&1&\\x_{3}&x_{3}^{5}&&\sum _{i+j+k=3}x_{2}^{i}x_{3}^{j}x_{4}^{k}&&\sum _{i=1}^{5}x_{i}&&0\\&&\sum _{i=0}^{4}x_{3}^{4-i}x_{4}^{i}&&\sum _{i+j+k+l=2}x_{2}^{i}x_{3}^{j}x_{4}^{j}x_{5}^{l}&&1&\\x_{4}&x_{4}^{5}&&\sum _{i+j+k=3}x_{3}^{i}x_{4}^{j}x_{5}^{k}&&\sum _{i=2}^{6}x_{i}&&\\&&\sum _{i=0}^{4}x_{4}^{4-i}x_{5}^{i}&&\sum _{i+j+k+l=2}x_{3}^{i}x_{4}^{j}x_{5}^{k}x_{6}^{l}&&&\\x_{5}&x_{5}^{5}&&\sum _{i+j+k=3}x_{4}^{i}x_{5}^{j}x_{6}^{k}&&&&\\&&\sum _{i=0}^{4}x_{5}^{4-i}x_{6}^{i}&&&&&\\x_{6}&x_{6}^{5}&&&&&&\\\end{matrix}}\\\end{array}}$

[6] [6]

${\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0}\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}={\frac {y_{0}}{x_{0}-x_{1}}}+{\frac {y_{1}}{x_{1}-x_{0}}}\\&=\sum _{j=0}^{1}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{1}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {{\cfrac {y_{1}}{x_{1}-x_{2}}}+{\cfrac {y_{2}}{x_{2}-x_{1}}}-{\cfrac {y_{0}}{x_{0}-x_{1}}}-{\cfrac {y_{1}}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}\\&={\frac {y_{0}}{(x_{0}-x_{1})(x_{0}-x_{2})}}+{\frac {y_{1}}{(x_{1}-x_{0})(x_{1}-x_{2})}}+{\frac {y_{2}}{(x_{2}-x_{0})(x_{2}-x_{1})}}\\&=\sum _{j=0}^{2}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{2}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n}]&=\sum _{j=0}^{n}{\frac {y_{j}}{\prod _{k=0,k\neq j}^{n}(x_{j}-x_{k})}}\\{\mathopen {[}}y_{0},y_{1},\dots ,y_{n+1}]&={\frac {\sum _{j=1}^{n+1}{\frac {y_{j}}{\prod _{k=1,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-\sum _{j=0}^{n}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {1}{\prod _{k=1,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-{\frac {1}{\prod _{k=0,\,k\neq j}^{n}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {x_{j}-x_{0}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}-{\frac {x_{j}-x_{n+1}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {{\frac {y_{n+1}}{\prod _{k=1}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}y_{j}\left({\frac {x_{n+1}-x_{0}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)-{\frac {y_{0}}{\prod _{k=1}^{n}(x_{0}-x_{k})}}}{x_{n+1}-x_{0}}}\\&={\frac {y_{n+1}}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}+\sum _{j=1}^{n}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}+{\frac {y_{0}}{\prod _{k=1}^{n+1}(x_{0}-x_{k})}}\\&=\sum _{j=0}^{n+1}{\frac {y_{j}}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\end{aligned}}$

[7] [7]
《数值分析及科学计算》薛毅（编）第六章第2节 Newton插值. P200.

[8] [8]
《数值分析及科学计算》薛毅（编）第六章第2节 Newton插值. P201.

[9] [9]

${\begin{matrix}x_{0}&x_{0}^{3}&&&&\\&&x_{0}^{2}+x_{0}x_{1}+x_{1}^{2}&&\\x_{1}&x_{1}^{3}&&x_{0}+x_{1}+x_{2}&&\\&&x_{0}^{2}+x_{0}x_{2}+x_{2}^{2}&&1&\\x_{2}&x_{2}^{3}&&x_{0}+x_{1}+x_{3}&&0\\&&x_{0}^{2}+x_{0}x_{3}+x_{3}^{2}&&1&\\x_{3}&x_{3}^{3}&&x_{0}+x_{1}+x_{4}&&\\&&x_{0}^{2}+x_{0}x_{4}+x_{4}^{2}&&&\\x_{4}&x_{4}^{3}&&&&\\\end{matrix}}$

[10] [10]
The Newton Polynomial Interpolation. [2019-04-19]. （原始内容存档于2019-04-19）.

[11] [11]

${\begin{array}{lcl}{\begin{aligned}N_{1}(x)&=[y_{0}]+[{y}_{0},{y}_{1}](x-{x}_{0})=y_{0}+y_{0}{\frac {x-{x}_{0}}{x_{0}-x_{1}}}+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}=y_{0}(1+{\frac {x-{x}_{0}}{x_{0}-x_{1}}})+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}\\&=y_{0}{\frac {x-{x}_{1}}{x_{0}-x_{1}}}+y_{1}{\frac {x-{x}_{0}}{x_{1}-x_{0}}}=\sum _{j=0}^{1}y_{j}\prod _{k=0,k\neq j}^{1}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\{\begin{aligned}N_{n}(x)&=\sum _{j=0}^{n}y_{j}\prod _{k=0,k\neq j}^{n}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\{\begin{aligned}N_{n+1}(x)&=N_{n}(x)+[{y}_{0},{y}_{1},\ldots ,{y}_{n+1}]\prod _{k=0}^{n}(x-{x}_{k})\\&=\sum _{j=0}^{n}y_{j}\prod _{k=0,k\neq j}^{n}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}+\sum _{j=0}^{n+1}y_{j}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}\left({\frac {\prod _{k=0,k\neq j}^{n}(x-{x}_{k})}{\prod _{k=0,k\neq j}^{n}(x_{j}-x_{k})}}+{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}\left({\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x_{j}-x_{n+1})}{\prod _{k=0,k\neq j}^{n+1}(x_{j}-x_{k})}}+{\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x-x_{j})}{\prod _{k=0,\,k\neq j}^{n+1}(x_{j}-x_{k})}}\right)+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n}y_{j}{\frac {\left(\prod _{k=0,k\neq j}^{n}(x-{x}_{k})\right)(x-x_{n+1})}{\prod _{k=0,k\neq j}^{n+1}(x_{j}-x_{k})}}+y_{n+1}{\frac {\prod _{k=0}^{n}(x-{x}_{k})}{\prod _{k=0}^{n}(x_{n+1}-x_{k})}}\\&=\sum _{j=0}^{n+1}y_{j}\prod _{k=0,k\neq j}^{n+1}{\frac {x-{x}_{k}}{x_{j}-x_{k}}}\\\end{aligned}}\\\end{array}}$

[12] [12]
Burden, Richard L.; Faires, J. Douglas. Numerical Analysis 9th. 2011: 129.

[13] [13]

${\begin{aligned}\triangle ^{k}y_{i}&=\sum _{j=0}^{k}{\frac {k!}{\prod _{l=0,\,l\neq j}^{k}(j-l)}}y_{i+j},\quad 0\leq k\leq n-i\\&=\sum _{j=0}^{k}{\frac {k!}{j!(-1)^{k-j}(k-j)!}}y_{i+j},\quad 0\leq k\leq n-i\\&=\sum _{j=0}^{k}(-1)^{k-j}{\binom {k}{j}}y_{i+j},\quad 0\leq k\leq n-i\end{aligned}}$

[14] [14]
Methodus Incrementorum Directa et Inversa（页面存档备份，存于互联网档案馆）

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均差

均差

定义

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展开形式

插值公式

无穷级数

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牛顿插值法

前向差分

幂函数的均差

泰勒形式

皮亚诺形式

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参见