First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 104
Exercise 3.1-6: Consider a function f(x) such that f(2) = 1.5713, f(3) =
1.5719,f(5) = 1.5738, and f(6) = 1.5751. Estimate f(4) using a second degree interpolating
polynomial (interpolating the first three data points) and a third degree interpolating
polynomial (interpolating the first four data points). Round the final results to four decimal
places. Is there any advantage here in using a third degree interpolating polynomial?
Julia code for Newton interpolation
Consider the following finite difference table.
x f(x) f[x i ,x i+1 ] f[x i−1 ,x i ,x i+1 ]
x 0 y 0
y 1 −y 0
x 1 −x 0
= f[x 0 ,x 1 ]
f[x
x 1 y 1 ,x 2 ]−f[x 0 ,x 1 ]
1 x 2 −x 0
= f[x 0 ,x 1 ,x 2 ]
y 2 −y 1
x 2 −x 1
= f[x 1 ,x 2 ]
x 2 y 2
Table 3.1: Divided differences for three data points
There are 2+1=3divided differences in the table, not counting the 0th divided differences.
In general, the number of divided differences to compute is 1+... + n = n(n +1)/2.
However, to construct Newton’s form of the interpolating polynomial, we need only n divided
differences and the 0th divided difference y 0 . These numbers are displayed in red in
Table 3.1. The important observation is, even though all the divided differences have to be
computed in order to get the ones needed for Newton’s form, they do not have to be all
stored. The following Julia code is based on an efficient algorithm that goes through the
divided difference calculations recursively, and stores an array of size m = n +1 at any given
time. In the final iteration, this array has the divided differences needed for Newton’s form.
Let’s explain the idea of the algorithm using the simple example of Table 3.1. The code
creates an array a =(a 0 ,a 1 ,a 2 ) of size m = n +1, which is three in our example, and sets
a 0 = y 0 ,a 1 = y 1 ,a 2 = y 2 .
(The code actually numbers the components of the array starting at 1, not 0. We keep it 0
here so that the discussion uses the familiar divided difference formulas.)