First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 117
z f(z) 1st diff 2nd diff 3rd diff 4th diff 5th diff
z 0 = −1.5 0.071
f ′ (z 0 )=1
−0.032−1
z 1 = −1.5 0.071
1.6+1.5 = −0.33
f[z 1 ,z 2 ]=−0.032 0.0065
−1+0.032
z 2 =1.6 -0.029
1.6+1.5
= −0.31 0.015
f ′ (z 2 )=−1 0.10 -0.005
0.0055+1
z 3 =1.6 -0.029
4.7−1.6
=0.32 -0.016
f[z 3 ,z 4 ]=0.0055 0
1−0.0055
z 4 =4.7 -0.012
4.7−1.6 =0.32
f ′ (z 4 )=1
z 5 =4.7 -0.012
Therefore, the Hermite polynomial is:
H 5 (x) =0.071 + 1(x +1.5)−0.33(x +1.5) 2 + 0.0065(x +1.5) 2 (x − 1.6)+
+ 0.015(x +1.5) 2 (x − 1.6) 2 −0.005(x +1.5) 2 (x − 1.6) 2 (x − 4.7).
Julia code for computing Hermite interpolating polynomial
In [1]: using PyPlot
The following function hdiff computes the divided differences needed for Hermite
interpolation. It is based on the function diff for computing divided differences for
Newton interpolation. The inputs to hdiff are the x-coordinates, the y-coordinates,
and the derivatives yprime.
In [2]: function hdiff(x::Array,y::Array,yprime::Array)
m=length(x) # here m is the number of data points. Note n=m-1
#and 2n+1=2m-1
l=2m
z=Array{Float64}(undef,l)
a=Array{Float64}(undef,l)
for i in 1:m
z[2i-1]=x[i]
z[2i]=x[i]