Our goal is to demonstrate that, for the Hermite polynomial
the error function is given by the equation
Let us begin by considering a point where , i.e., it is not equal to any of the points on which the interpolant was developed. Since our objective is to determine the error between and , because by definition the two are the same at the interpolating points , it would be pointless (sorry!) to use one of the interpolation points for .
Now we build a polynomial of degree to describe the error function . This function would interpolate at all and additionally for . This function yields zero error to itself at as an interpolating point. However, by comparing this polynomial at with , we can establish the degree of error. Let us write this polynomial as
The constant is intended to make the interpolant precise at . Let us now state the error of this new interpolant as
Since is an interpolating point, . Substituting this into the above and solving for , we have
For the other interpolating points, we know that
and, since the Hermite polynomial also interpolates at the first derivative,
and finally, obviously,
we can say
It’s also possible to say that
From this we can determine that has at least zeroes (all of the points plus the point ) in . Likewise we can say that has at least (all of the points ) zeroes in .
At this point we observe the following:
…Rolle’s Theorem states that a continuous curve that intersects the -axis in two distinct points and , and has a slope at every point for which , must have slope zero at one or more of these latter points. (Tierney, J.A. Calculus and Analytic Geometry. Boston: Allyn and Bacon, 1972, p. 128.)
There is thus at least one zero for each interval; since there are intervals, we can say from this that has at least zeroes. However, also has zeroes as an interpolant, so has a total of zeroes.
Successive differentiation will yield the following
From this we can conclude that, for the one zero of the final derivative
where is the value where the zero exists.
At this derivative, from our previous considerations,
It is fair to say that, because of the degree of the polynomial,
The last term could be quite complex to differentiate, but let us
consider the following:
where is a polynomial. Taking the derivative, disappears and we are left with
At the point , ,
we can substitute and achieve our original goal