Uniform Intervals

We take $x_i=x_0+ih,~i=0,1,\ldots$, where h is the interval width. The divided differences can be simplified, since we no longer need to show the interval width explicitly. For this case we define the forward difference by

$$\Delta f(x)=f(x+h)-f(x)\mbox{ or } \Delta f(x_i)=f(x_{i+1})-f(x_i).$$

The connection between this and the equivalent divided difference is given by

$$f[x_0,x_1]=\frac{f(x_1)-f(x_0)}{x_1-x_0}=\frac{\Delta f(x_0)}{h},$$

or $\Delta f(x_0)=hf[x_0,x_1]$. We now regard $\Delta$ as the forward difference operator, which operates on f(x_i) to give f(x_i+1)-f(x_i). Thus

$$\begin{array} {lcl}\Delta ^2f(x_i) & = & \Delta (\Delta f(x_... ...+1})-f(x_i))\\ & = & f(x_{i+2})-2f(x_{i+1})+f(x_i).\end{array}$$

It is easy to check that $\Delta$ is a linear operator, that is, it satisfies $\Delta (\lambda f(x)+\mu g(x))=\lambda \Delta f(x)+\mu \Delta g(x)$, which allows simpler manipulations.

Proceeding slightly differently, we obtain

$$\begin{array} {lcl}\Delta ^2f(x_i) & = & f(x_{i+2})-f(x_{i+1... ...hf[x_i,x_{i+1}]\\ & = & h^2f[x_i,x_{i+1},x_{i+2}]. \end{array}$$

You will use induction in an exercise to prove the results

$$\begin{array} {lcl}\Delta ^nf(x_i) & = & \sum _{k=0}^n(-1)^{... ...}]\\ & = & h^nf^{(n)}(\xi ),\quad x_i<\xi <x_{i+n}.\end{array}$$

Newton's divided Difference Interpolating Polynomial now becomes

$$\begin{array} {lcl}p(x) & = & f(x_0)+(x-x_0)f[x_0,x_1]+(x-x_... ...ts +\frac{s(s-1)\cdots (s-n+1)}{n!}\Delta ^nf(x_0),\end{array}$$

which we met earlier as Newton's Forward Difference Formula. The differences are computed in a tabular form similar to that for divided differences, but rather simpler.

We mention in passing the central difference operator, $\delta$,which is defined by

$$\delta f(x)=f(x+\frac{1}{2}h)-f(x-\frac{1}{2}h),$$

which is more symmetric than the forward difference operator, but has the drawback of including intermediate tabular values. We establish its connection as follows:

$$\begin{array} {lcl}\delta f(x+\frac{1}{2}h) & = & f(x+h)-f(x... ... & = & f(x+h)-2f(x)+f(x-h)\\ & = & \Delta ^2f(x-h).\end{array}$$

In general,

$$\begin{array} {lcl}\delta ^{2n}f(x_i) & = & \Delta ^{2n}f(x_... ...f(x_i+\frac{1}{2}h) & = & \Delta ^{2n+1}f(x_{i-n}).\end{array}$$

There is also a backward difference operator, $\nabla$, which is defined by

$$\nabla f(x)=f(x)-f(x-h),$$

but we shall do nothing with it.