04 - Hermite Interpolation Theorem

Hermite Interpolation Theorem

Exists unique polynomial $p_{2n+1}\in {\Pi }_{2n+1}$ such that

$$p_{2n+1}(x_1)=f_i\text{ and }{p}_{2n+1}^{\prime }(x_i)=g_i$$

for $i=0,1,\cdots ,n$

Proof - Construction

Let $L_{n,k}(x)$ be the Lagrange basis polynomial then

Let

$$\begin{array}{rl}{H}_{n,k}(x)& =[L_{n,k}(x){]}^2(1-2(x-x_k)L_{n,k}^{\prime }(x_k))\\ K_{n,k}(x)& =[L_{n,k}(x){]}^2(x-x)k\end{array}$$

where

$$\begin{array}{rlrlrl}{H}_{n,k}(x_i)& ={\delta }_{ik}& & \text{ and }{H}_{n,k}^{\prime }(x_i)& & =0\\ K_{n,k}(x_i)& =0& & \text{ and }{K}_{n,k}^{\prime }(x_i)& & ={\delta }_{ik}\end{array}$$

Then

$$p_{2n+1}(x)=\sum _{k=0}^n[f_k{H}_{n,k}(x)+g_k{K}_{n,k}(x)]$$

Hence $p_{2n+1}(x)$ satisfies conditions