Editing Divided differences (section)

==Alternative characterizations==

===Expanded form===
<math display="block">
\begin{align}
f[x_0] &= f(x_0) \\
f[x_0,x_1] &= \frac{f(x_0)}{(x_0-x_1)} + \frac{f(x_1)}{(x_1-x_0)} \\
f[x_0,x_1,x_2] &= \frac{f(x_0)}{(x_0-x_1)\cdot(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)\cdot(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)\cdot(x_2-x_1)} \\
f[x_0,x_1,x_2,x_3] &= \frac{f(x_0)}{(x_0-x_1)\cdot(x_0-x_2)\cdot(x_0-x_3)} + \frac{f(x_1)}{(x_1-x_0)\cdot(x_1-x_2)\cdot(x_1-x_3)} +\\
&\quad\quad \frac{f(x_2)}{(x_2-x_0)\cdot(x_2-x_1)\cdot(x_2-x_3)} + \frac{f(x_3)}{(x_3-x_0)\cdot(x_3-x_1)\cdot(x_3-x_2)} \\
f[x_0,\dots,x_n] &=
\sum_{j=0}^{n} \frac{f(x_j)}{\prod_{k\in\{0,\dots,n\}\setminus\{j\}} (x_j-x_k)}
\end{align}
</math>

With the help of the [[polynomial function]] <math>\omega(\xi) = (\xi-x_0) \cdots (\xi-x_n)</math> this can be written as
<math display="block">
f[x_0,\dots,x_n] = \sum_{j=0}^{n} \frac{f(x_j)}{\omega'(x_j)}.
</math>

===Peano form===
If <math>x_0<x_1<\cdots<x_n</math> and <math>n\geq 1</math>, the divided differences can be expressed as<ref>{{Cite book |last=Skof |first=Fulvia |url=https://books.google.com/books?id=ulUM2GagwacC&dq=This+is+called+the+Peano+form+of+the+divided+differences&pg=PA41 |title=Giuseppe Peano between Mathematics and Logic: Proceeding of the International Conference in honour of Giuseppe Peano on the 150th anniversary of his birth and the centennial of the Formulario Mathematico Torino (Italy) October 2-3, 2008 |date=2011-04-30 |publisher=Springer Science & Business Media |isbn=978-88-470-1836-5 |pages=40 |language=en}}</ref>
<math display="block">f[x_0,\ldots,x_n] = \frac{1}{(n-1)!} \int_{x_0}^{x_n} f^{(n)}(t)\;B_{n-1}(t) \, dt</math>
where <math>f^{(n)}</math> is the <math>n</math>-th [[derivative]] of the function <math>f</math> and <math>B_{n-1}</math> is a certain [[B-spline]] of degree <math>n-1</math> for the data points <math>x_0,\dots,x_n</math>, given by the formula
<math display="block">B_{n-1}(t) = \sum_{k=0}^n \frac{(\max(0,x_k-t))^{n-1}}{\omega'(x_k)}</math>

This is a consequence of the [[Peano kernel theorem]]; it is called the ''Peano form'' of the divided differences and <math>B_{n-1}</math> is the ''Peano kernel'' for the divided differences, all named after [[Giuseppe Peano]].

===Forward and backward differences===
{{details|Finite difference}}

When the data points are equidistantly distributed we get the special case called '''forward differences'''. They are easier to calculate than the more general divided differences.

Given ''n''+1 data points
<math display="block">(x_0, y_0), \ldots, (x_n, y_n)</math>
with
<math display="block">x_{k} = x_0 + k h,\ \text{ for } \ k=0,\ldots,n \text{ and fixed } h>0</math>
the forward differences are defined as
<math display="block">\begin{align}
\Delta^{(0)} y_k &:= y_k,\qquad k=0,\ldots,n \\
\Delta^{(j)}y_k &:= \Delta^{(j-1)}y_{k+1} - \Delta^{(j-1)}y_k,\qquad k=0,\ldots,n-j,\ j=1,\dots,n.
\end{align}</math>whereas the backward differences are defined as:
<math display="block">\begin{align}
\nabla^{(0)} y_k &:= y_k,\qquad k=0,\ldots,n \\
\nabla^{(j)}y_k &:= \nabla^{(j-1)}y_{k} - \nabla^{(j-1)}y_{k-1},\qquad k=0,\ldots,n-j,\ j=1,\dots,n.
\end{align}</math>
Thus the forward difference table is written as:<math display="block">
\begin{matrix}
y_0 &               &                   &                  \\
    & \Delta y_0 &                   &                  \\
y_1 &               & \Delta^2 y_0 &                  \\
    & \Delta y_1 &                   & \Delta^3 y_0\\
y_2 &               & \Delta^2 y_1 &                  \\
    & \Delta y_2 &                   &                  \\
y_3 &               &                   &                  \\
\end{matrix}
</math>whereas the backwards difference table is written as:<math display="block">
\begin{matrix}
y_0 &               &                   &                  \\
    & \nabla y_1 &                   &                  \\
y_1 &               & \nabla^2 y_2 &                  \\
    & \nabla y_2 &                   & \nabla^3 y_3\\
y_2 &               & \nabla^2 y_3 &                  \\
    & \nabla y_3 &                   &                  \\
y_3 &               &                   &                  \\
\end{matrix} 
</math>

The relationship between divided differences and forward differences is<ref>{{cite book|last1=Burden|first1=Richard L.| last2=Faires|first2=J. Douglas | title=Numerical Analysis |url=https://archive.org/details/numericalanalysi00rlbu |url-access=limited | date=2011|page=[https://archive.org/details/numericalanalysi00rlbu/page/n146 129]|publisher=Cengage Learning | isbn=9780538733519| edition=9th}}</ref>
<math display="block">[y_j, y_{j+1}, \ldots , y_{j+k}] = \frac{1}{k!h^k}\Delta^{(k)}y_j,    </math>whereas for backward differences:{{Citation needed|date=December 2023}}<math display="block">[{y}_{j}, y_{j-1},\ldots,{y}_{j-k}] = \frac{1}{k!h^k}\nabla^{(k)}y_j.       </math>