# Simplify this expression with divided differences.

The divided differences are defined as follows $$f[x_i] := f(x_i), \quad f[x_0, \ldots, x_n] := \frac{f[x_1, \ldots, x_n] - f[x_0, \ldots, x_{n - 1}]}{x_n - x_0} \quad \text{for } n \ge 2$$ For pairwise different $$x_0,x_1,x_2$$ simplify the expression $$f(x_0) + f[x_0,x_1](x - x_0) + f[x_0,x_1,x_2](x - x_0)(x - x_2) - f(x_1) - f[x_1,x_2](x - x_1) - f[x_1,x_2,x_0](x - x_1)(x - x_2)$$

I know that the $$n$$-th interpolation polynomial is given by $$P_n(x) = f[x_0] + (x - x_0)f[x_0, x_1] + \ldots + \prod_{j = 0}^{n - 1} (x - x_j) f[x_0, \ldots, x_n].$$ and hoped that the expression would maybe reduce to the second interpolation polynomial.

I am also aware of the mean value theorem for divided differences, but that didn't help either.

Then , I tried to utilise that the divided differences are invariant under permutations of the $$x_i$$ (namely $$f[x_0,x_1,x_2] = f[x_1,x_2,x_0]$$), so some simplification could be made, but I suspect that is not enough.

Lastly, I plain tried to write the divided differences in terms of $$f(x_i)$$ and $$x_j$$ but no cancelling was possible.

Note: This is one of 11 questions from a past exam for which one has 120 minutes and it only gives 2 from 40 possible points, and therefore there has to be a quick solution.

• Try to group the terms and rewrite them going from lower order divided differences to higher order. E.g., to start, consider rewriting the term $f(x_0) - f(x_1)$ in terms of $f[x_0, x_1]$ and then continue the process – VorKir Mar 2 at 19:59

Taking @VorKir's hint: We have $$f(x_0) - f(x_1) = f[x_0,x_1](x_0 - x_1)$$ and therefore $$f(x_0) + f[x_0,x_1](x - x_0) - f(x_1) = f[x_0,x_1](x - x_1).$$ Now, \begin{align} f[x_0,x_1](x - x_1) - f[x_1, x_2](x - x_1) & = (x - x_1) \left( f[x_0,x_1] - f[x_1, x_2] \right) \\ & = (x - x_1)(x_0 - x_2) f[x_0,x_1,x_2], \end{align} since $$f[x_0,x_1,x_2] = \frac{f[x_1,x_2] - f[x_0,x_1]}{x_2 - x_0}.$$ We have now reduced the expression down to $$(x - x_1)(x_0 - x_2) f[x_0,x_1,x_2] + f[x_0,x_1,x_2](x - x_0)(x - x_2) - f[x_1,x_2,x_0](x - x_1)(x - x_2).$$ We now use that divided differences are invariant under permutations of the $$x_i$$ , so we have $$f[x_0,x_1,x_2] = f[x_1,x_2,x_0]$$ and therefore the expression is equal to \begin{align} & \ f[x_0,x_1,x_2]\left((x - x_1)(x_0 - x_2) + (x - x_0)(x - x_2) -(x - x_1)(x - x_2)\right) \\ = & \ f[x_0,x_1,x_2]\left( (x - x_1)(x_0 - x_2) + (x_1 - x_0)(x - x_2) \right) \\ = & \ f[x_0,x_1,x_2](x_0 - x) (x_2 - x_1). \end{align}