Pointwise Error

If p is the interpolating polynomial through $x_0,x_1,\ldots ,x_n$then the interpolating polynomial through these points and x' is given by

$$p'(x)=p(x)+f[x_0,x_1,\ldots ,x_n,x']\prod _{i=0}^n(x-x_i),$$

where

$$f(x')=p'(x')=p(x')+f[x_0,x_1,\ldots ,x_n,x']\prod _{i=0}^n(x'-x_i).$$

Since x' is an arbitrary added point, we may drop the prime to obtain the error at the general point x, or the pointwise error, as

$$f(x)-p(x)=f[x_0,x_1,\ldots ,x_n,x]\prod _{i=0}^n(x-x_i).$$

Comparing this with our previous expression for the pointwise error, we obtain

$$f[x_0,x_1,\ldots ,x_n,x]=\frac{f^{(n+1)}(\xi )}{(n+1)!}.$$

Note that

• the special case n=0 gives

  $$f[x_0,x]=\frac{f(x_0)-f(x)}{x_0-x}=f'(\xi ),$$

  the mean value theorem,
• if f is itself a polynomial of degree not greater than n, then $f[x_0,x_1,\ldots ,x_n,x]=0$, that is, all (n+1)st order differences are zero.