First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 101
Solution. We want to compute
p 2 (x) =f[x 0 ]+f[x 0 ,x 1 ](x − x 0 )+f[x 0 ,x 1 ,x 2 ](x − x 0 )(x − x 1 ).
Here are the finite differences:
x f[x] First divided difference Second divided difference
x 0 = −1 f[x 0 ]=−6
f[x 0 ,x 1 ]= f[x 1]−f[x 0 ]
x 1 −x 0
= 3
x 1 =1 f[x 1 ]=0 f[x 0 ,x 1 ,x 2 ]= f[x 1,x 2 ]−f[x 0 ,x 1 ]
x 2 −x 0
= 1
f[x 1 ,x 2 ]= f[x 2]−f[x 1 ]
x 2 −x 1
=6
x 2 =2 f[x 2 ]=6
Therefore
p 2 (x) =−6 + 3(x +1)+1(x +1)(x − 1),
which is the same polynomial we had in Example 49.
Exercise 3.1-4:
Consider the function f given in the following table.
x 1 2 4 6
f(x) 2 3 5 9
a) Construct a divided difference table for f by hand, and write the Newton form of the
interpolating polynomial using the divided differences.
b) Assume you are given a new data point for the function: x =3,y =4. Find the
new interpolating polynomial. (Hint: Think about how to update the interpolating
polynomial you found in part (a).)
c) If you were working with the Lagrange form of the interpolating polynomial instead of
the Newton form, and you were given an additional data point like in part (b), how
easy would it be (compared to what you did in part (b)) to update your interpolating
polynomial?