01 - Existence of Interpolating Polynomials

Existence of Interpolating Polynomials

For any Polynomials interpolating data then

$$\exists p_n\in {\Pi }_n\text{ such that }{p}_n(x_i)=f_i\text{ for }i=0,1,\cdots n$$

Proof

Consider for $k=0,1,\cdots ,n$ the cardinal polynomial

$$L_{n,k}(x)=\frac{(x-x_0)\cdots (x-x_{k-1})(x-x_{k+1})\cdots (x-x_n)}{(x_k-x_0)\cdots (x_k-x_{k-1})(x_k-x_{k+1})\cdots (x_k-x_n)}\in {\Pi }_n$$

Then

$$L_{n,k}={\delta }_{ik}$$

as $L_{n,k}(x_i)=0$ for $i=0,\cdots ,k-1,k+1,\cdots ,n$ and $L_{n,k}(x_k)=1$
So $L_{n,k}(x_i)={\delta }_{ik}$

Thus define Lagrange Interpolating Polynomial

$$p_n(x)=\sum _{k=0}^n{f}_k{L}_{n,k}(x)\in {\Pi }_n$$

Hence

$$p_n(x_i)=\sum _{k=0}^n{f}_k{L}_{n,k}(x_i)=f_i\text{ for }i=0,1,\cdots ,n$$

Lagrange Interpolating Polynomial

$$p_n(x)=\sum _{k=0}^n{f}_k{L}_{n,k}(x)\in {\Pi }_n$$