Newtonian Interpolation

It is therefore better to approximate the function f polynomially by the Newtonian term

$$f(z)=a_{0}+a_{1}(z-x_{0})+a_{2}(z-x_{1})(z-x_{0})+a_{3}(z-x_{2})(z-x_{1})(z-x_{0})+\cdots$$

with the coefficients $a_{i}$ being the divided differences

$$a_{0}=f[x_{0}] = f(x_{0}),$$

$$a_{1}=f[x_{0},x_{1}] = \frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}},$$

$$a_{k}=f[x_{0},x_{1},\ldots ,x_{k}] = \frac{f(x_{k})-a_{0}-\sum\limits _{i=1}^{k-1}a_{l}(x_{k}-x_{0})\cdots (x_{k}-x_{l-1})}{(x_{k}-x_{0})\cdots (x_{k}-x_{k-1})}$$

$$= \frac{f[x_{0},x_{1},\ldots ,x_{k-1}]-f[x_{1},x_{2},\ldots ,x_{k}]}{x_{0}-x_{k}}.$$

which are approximations of the i-th derivative of f at $x_{0}$.

In the Newtonian interpolation, to add another interpolation point it is only necessary to calculate the bottom line of the following scheme:

$$\begin{array}{lllllll} f(x_{0}) & \ddots & & & & & \\ f(x_... ...},\ldots ,x_{N}] & \cdots & f[x_{0},\ldots ,x_{N}]. \end{array}$$

In practice, this process is continued until self-consistency. Numerical problems with a small distance between the sampling points can be overcome by permuting these sampling points. A polynomial expansion is also possible for the consideration of non-hermitian operators thats eigenvalues do not lie on a real interval but inside a disk in the complex plane [2].