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Error Function for an Hermite Polynomial

Our goal is to demonstrate that, for the Hermite polynomial





the error function is given by the equation



f\in C^{2n+2}[a,b]

Let us begin by considering a point \hat{x}\in[a,b] where \hat{x}\neq x_{j}, 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 f(x) and H_{2n+1}(x), because by definition the two are the same at the interpolating points x_{j}, it would be pointless (sorry!) to use one of the interpolation points for \hat{x}.

Now we build a polynomial of degree 2n+2 to describe the error function f(x)-H_{2n+1}(x). This function would interpolate at all x_{j} and additionally \hat{x} for H_{2n+1}(x). This function yields zero error to itself at \hat{x} as an interpolating point. However, by comparing this polynomial at \hat{x} with f(x)-H_{2n+1}(x), we can establish the degree of error. Let us write this polynomial as


The constant \lambda is intended to make the interpolant precise at \hat{x}. Let us now state the error of this new interpolant as


Since \hat{x} is an interpolating point, \phi\left(\hat{x}\right)=0. Substituting this into the above and solving for \lambda, 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 \phi\left(x\right) has at least n+2 zeroes (all of the points x_{j} plus the point \hat{x}) in \left[a,b\right]. Likewise we can say that \phi'\left(x\right) has at least n+1 (all of the points x_{j}) zeroes in \left[a,b\right].

At this point we observe the following:

…Rolle’s Theorem states that a continuous curve that intersects the x-axis in two distinct points A\left(a,0\right) and B\left(b,0\right), and has a slope at every point \left(x,y\right) for which a<x<b, 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 n+1 intervals, we can say from this that \phi'\left(x\right) has at least n+1 zeroes. However, \phi'\left(x\right) also has n+1 zeroes as an interpolant, so \phi'\left(x\right) has a total of 2n+2 zeroes.

Successive differentiation will yield the following

\phi'\left(x\right)\Longrightarrow 2n+2 zeroes
\phi''\left(x\right)\Longrightarrow 2n+1 zeroes
\phi'''\left(x\right)\Longrightarrow 2n zeroes
\phi^{\left(2n+2\right)}\Longrightarrow 1 zero

From this we can conclude that, for the one zero of the final derivative


where \eta\left(x\right) 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 r\left(x\right) is a polynomial. Taking the 2n+2 derivative, r\left(x\right) disappears and we are left with






At the point \eta\left(x\right), \phi^{\left(2n+2\right)}\left(\eta\left(x\right)\right)=0,
and now






we can substitute and achieve our original goal



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