First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 116
Then the Hermite polynomial can be written as:
H 2n+1 (x) =f[z 0 ]+f[z 0 ,z 1 ](x − z 0 )+f[z 0 ,z 1 ,z 2 ](x − z 0 )(x − z 1 )+
...+ f[z 0 ,z 1 , ..., z 2n+1 ](x − z 0 ) ···(x − z 2n )
=f[z 0 ]+
2n+1
∑
i=1
f[z 0 , ...., z i ](x − z 0 )(x − z 1 ) ···(x − z i−1 ).
There is a little problem with some of the first divided differences above: they are undefined!
Observe that
f[z 0 ,z 1 ]=f[x 0 ,x 0 ]= f(x 0) − f(x 0 )
x 0 − x 0
or, in general,
f[z 2i ,z 2i+1 ]=f[x i ,x i ]= f(x i) − f(x i )
x i − x i
for i =0, ..., n.
From Theorem 58, weknowf[x 0 , ..., x n ]= f (n) (ξ)
for some ξ between the min and max
n!
of x 0 , ..., x n . From a classical result by Hermite & Gennochi (see Atkinson [3], page 144),
divided differences are continuous functions of their variables x 0 , ..., x n . This implies we can
take the limit of the above result as x i → x 0 for all i, which results in
f[x 0 , ..., x 0 ]= f (n) (x 0 )
.
n!
Therefore in the Hermite polynomial coefficient calculations, we will put
f[z 2i ,z 2i+1 ]=f[x i ,x i ]=f ′ (x i )=y ′ i
for i =0, 1, ..., n.
Example 63. Let’s compute the Hermite polynomial of Example 62. The data is:
i x i y i y i
′
0 -1.5 0.071 1
1 1.6 -0.029 -1
2 4.7 -0.012 1
Here n =2, and 2n +1=5, so the Hermite polynomial is
H 5 (x) =f[z 0 ]+
5∑
f[z 0 , ..., z i ](x − z 0 ) ···(x − z i−1 ).
i=1
The divided differences are: