29.3.1 Solution to Second Problem: polynomial passing through k points

We can find solutions to the second problem by making use of two observations: 

1. Suppose we can find a solution, f_j (x) for the case in which f(x_j) is 1 and all the other f(x_i) are 0.

Then we can add up appropriate multiples of such f_j 's to get the solution:

We can do something similar in the first problem 

2. It is easy to find the f _j (x). The statement that it is zero at x_i means it has a factor (x-x_i) in it. We therefore find  that the f_j are given by

This process and these formulae are called Lagrange interpolation.

29.3.2 Solution to First Problem: filling a gap between two curves smoothly

To solve  the first problem in this way we must find g_b0 , g_b1 , g_c0 , and g_c1 that obey

g_b0(b) = 1, g_b0'(b) = 0,  g_b0(c) = 0, g_b0'(c) = 0,

g_b1(b) = 0, g_b1'(b) = 1,  g_b1(c) = 0, g_b1'(c) = 0,  

g_c0(b) = 0, g_c0'(b) = 0,  g_c0(c) = 1, g_c0'(c) = 0,  

g_c1(b) = 0, g_c1'(b) = 0, g_c1(c) = 0, g_c1'(c) = 1.

We find by inspection that:

g_b1(x) = ((x - c) / (b-c))²(x - b)

and

g_c1(x) = ((x - b) / (c - b))²(x - c).

The function ((x - c) / (b - c))² vanishes at c, has 0 derivative at c, and is 1 at b, but its derivative at b is 2 / (b - c).

We therefore obtain:

g_b0(x) = ((x - c) / (b - c))² - 2g_b1(x) / (b - c),

and similarly:

g_c0(x) = ((x - b) / (c - b))² - 2g_c1(x) / (c - b)

The general solution is then given by

f(b)g_b0(x) + f '(b)g_b1(x) + f(c)g_c0(x) + f '(c)g_c1(x).