Derivation

Take the result of Theorem 10 with the quartic Lagrange Polynomial:

.

Again, we can take the derivative of both sides to obtain a formula for . First, consider the Lagrange Polynomial:

> restart;

> q:=(xx,n)->product((xx-x[i]),i=0..4)/(xx-x[n]):q(xx,0);for j from 0 to 4 do qq[j]:=unapply(q(xx,j),xx); od;

> P4:=xx->sum(q(xx,k)*f(x[k])/qq[k](x[k]),k=0..4);P4(xx);

> dP4:=diff(P4(xx),xx);

or, with equidistant points such that

> x[1]:=x[0]+h;x[2]:=x[0]+2*h;x[3]:=x[0]+3*h;x[4]:=x[0]+4*h;

> dP4;

The error term can also be dealt with:

> err:=(D@@5)(f)(xi(xx))/(5!)*(xx-x[0])*(xx-x[1])*(xx-x[2])*(xx-x[3])*(xx-x[4]);

> derr:=diff(err,xx);

This error term is difficult to manipulate, but can be simplified by evaluating the expression at any of the data points

> dfx:=dP4+derr;

> dfxc:=simplify(subs(xx=x[2],dfx));

> dfxs:=simplify(subs(xx=x[0],dfx));

>

The first formula can be written in the form:

.

This represents a centered five-point formula.

The other formula is useful when all data points lie on one side of the point of interest:

.