Question 8.2: Compute the third divided difference using the recursion rel...
Compute the third divided difference using the recursion relationship Eq. (8.2.28).
f[x_n, x_{n−1}, … , x_1, x_0] = \frac{f[x_n, x_{n−1}, … , x_2, x_1] − f[x_{n−1}, … , x_1, x_0]}{x_n − x_0} (8.2.28)
The recursion relationship gives us
f[x_3, x_2, x_1, x_0] = \frac{f[x_3, x_2, x_1] − f[x_2, x_1, x_0]}{x_3 − x_0} (8.2.33)
We then apply the recursion relationship to each term on the right-hand side,
f[x_3, x_2, x_1] = \frac{f[x_3, x_2] − f[x_2, x_1]}{x_3 − x_1} (8.2.34)
f[x_2, x_1, x_0] = \frac{f[x_2, x_1] − f[x_1, x_0]}{x_2 − x_0} (8.2.35)
Finally, we need to apply the recursion relationship again to the terms on the right-hand side to get the remaining three divided differences,
f[x_3, x_2] = \frac{f(x_3) − f(x_2)}{x_3 − x_2} (8.2.36)
f[x_2, x_1] = \frac{f(x_2) − f(x_1)}{x_2 − x_1} (8.2.37)
f[x_1, x_0] = \frac{f(x_1) − f(x_0)}{x_1 − x_0} (8.2.38)
Having gotten to the end of the recursions, we now work our way upward to generate the desired result. We first substitute the first divided differences into the second divided differences,
f[x_3, x_2, x_1] = \frac{f(x_3) − f(x_2)}{(x_3 − x_2)(x_3 − x_1)}− \frac{f(x_2) − f(x_1)}{(x_2 − x_1)(x_3 − x_1)} (8.2.39)
f[x_2, x_1, x_0] =\frac{ f(x_2) − f(x_1)}{(x_2 − x_1)(x_2 − x_0)}− \frac{f(x_1) − f(x_0)}{(x_1 − x_0)(x_2 − x_0)} (8.2.40)
The latter results are then substituted into Eq. (8.2.33) to get the final result. In the general case, where the positions x_i are not specified, the end result can be a bit of a mess. Let’s hold off on any further algebra here until Example 8.6.
b_1 = f[x_1, x_0] (8.2.23)