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In general, a function $f(\cdot)$ defined at points $x_1,x_2,\dots, x_k$, the $(k − 1)$th-order divided difference is defined by the recurrence relation:

$$ f[x_1,x_2,\dots...x_k]=\frac{f[x_2,\dots...x_{k-1},x_k]-f[x_1,x_2,\dots...x_{k-1}]}{x_k-x_1}, $$

with $f[x]=f(x)$. Is there a way to express analytically the $(k − 1)$th-order divided difference in terms only of divided differences of order 1?

Thanks in advance!

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Start from the expanded form of the divided differences (which uses $\omega(\xi) = (\xi - x_1) \cdots (\xi - x_k)$): $$\begin{aligned} f[x_1,\ldots,x_k] &= \sum_{i=1}^k \frac{f(x_i)}{\omega'(x_i)} \\ &= f(x_1)\sum_{i=1}^k \frac{1}{\omega'(x_i)} + \sum_{i=1}^k \frac{f(x_i)-f(x_1)}{\omega'(x_i)} \\ &= \sum_{i=1}^k \frac{\sum_{j=2}^i (x_j-x_{j-1}) f[x_{j-1},x_j]}{\omega'(x_i)} . \end{aligned}$$ The term proportional to $f(x_1)$ vanishes because it is multiplied by $1[x_1,\ldots,x_k] = 0$.

The linked Wikipedia article doesn't give a direct citation for the "expanded form" of divided differences, but it could probably be found in one of the references of the article.

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