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Divided differences
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===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>
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