First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 107
end
pr=1.0
for j in 1:(m-1)
pr=pr*(z-x[j])
sum=sum+a[j+1]*pr
end
return sum
Out[5]: newton (generic function with 1 method)
Let’s verify the code by computing p 3 (1.761) of Example 57:
In [6]: newton([1.765,1.760,1.755,1.750],[0.92256,0.92137,0.92021,0.91906]
,1.761)
Out[6]: 0.92160496
Exercise 3.1-7: This problem discusses inverse interpolation which gives another
method to find the root of a function. Let f be a continuous function on [a, b] with one
root p in the interval. Also assume f has an inverse. Let x 0 ,x 1 , ..., x n be n +1 distinct
numbers in [a, b] with f(x i )=y i ,i =0, 1, ..., n. Construct an interpolating polynomial P n
for f −1 (x), by taking your data points as (y i ,x i ),i=0, 1, ..., n. Observe that f −1 (0) = p, the
root we are trying to find. Then, approximate the root p, by evaluating the interpolating
polynomial for f −1 at 0, i.e., P n (0) ≈ p. Using this method, and the following data, find an
approximation to the solution of log x =0.
x 0.4 0.8 1.2 1.6
log x -0.92 -0.22 0.18 0.47
3.2 High degree polynomial interpolation
Suppose we approximate f(x) using its polynomial interpolant p n (x) obtained from (n +1)
data points. We then increase the number of data points, and update p n (x) accordingly. The
central question we want to discuss is the following: as the number of nodes (data points)