Question 8.1: Show that Eq. (8.2.27) reduces to Eq. (8.2.20).
Show that Eq. (8.2.27) reduces to Eq. (8.2.20).
f[x_i, x_j, x_k] = \frac{f[x_i, x_j] − f[x_j, x_k]}{x_i − x_k} (8.2.27)
b_2= \left[\frac{f(x_2) − f(x_1)}{x_2 − x_1}−\frac{ f(x_1) − f(x_0)}{x_1 − x_0}\right] \frac{1}{x_2 − x_0} (8.2.20)
Let x_i = x_2, x_j = x_1, and x_k = x_0. Then we have
f[x_2, x_1, x_0] = \frac{f[x_2, x_1] − f[x_1, x_0]}{x_2 − x_0} (8.2.29)
The first divided differences are computed from Eq. (8.2.26),
f[x_2, x_1] = \frac{f(x_2) − f(x_1)}{x_2 − x_1} (8.2.30)
f[x_1, x_0] =\frac{ f(x_1) − f(x_0)}{x_1 − x_0} (8.2.31)
Substituting the latter into Eq. (8.2.29) gives
f[x_2, x_1, x_0] = \left[\frac{f(x_2) − f(x_1)}{x_2 − x_1}− \frac{f(x_1) − f(x_0)}{x_1 − x_0}\right] \frac{ 1}{x_2 − x_0} (8.2.32)
which is indeed Eq. (8.2.20) for b_2.