Error Function for an Hermite Polynomial

Our goal is to demonstrate that, for the Hermite polynomial

H_{2n+1}(x)=\sum_{j=0}^{n}f(x_{j})H_{j}(x)+\sum_{j=0}^{n}f'(x_{j})\hat{H}_{j}(x)

where

H_{j}(x)=[1-2(x-x_{j})L'_{j}(x_{j})]L_{j}^{2}(x)

\hat{H}_{j}(x)=(x-x_{j})L_{j}^{2}(x)

the error function is given by the equation

f(x)-H_{2n+1}(x)=\frac{f^{\left(2n+2\right)}\left(\eta\left(x\right)\right)}{\left(2n+2\right)!}\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}

where

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

H_{2n+1}(x)+\lambda\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}

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

\phi\left(x\right)=f\left(x\right)-\left(H_{2n+1}(x)+\lambda\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}\right)=f\left(x\right)-H_{2n+1}(x)-\lambda\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}

Since \hat{x} is an interpolating point, \phi\left(\hat{x}\right)=0. Substituting this into the above and solving for \lambda, we have

\lambda=\frac{f\left(\hat{x}\right)-H_{2n+1}(\hat{x})}{\overset{n}{\underset{j=0}{\prod}}\left(\hat{x}-x_{j}\right)^{2}}

For the other interpolating points, we know that

f\left(x_{j}\right)-H_{2n+1}(x_{j})=0

and, since the Hermite polynomial also interpolates at the first derivative,
f'\left(x_{j}\right)-H'_{2n+1}(x_{j})=0

and finally, obviously,
\overset{n}{\underset{j=0}{\prod}}\left(x_{j}-x_{j}\right)^{2}=0

we can say

\phi\left(x_{j}\right)=f\left(x_{j}\right)-H_{2n+1}(x_{j})-\lambda\overset{n}{\underset{j=0}{\prod}}\left(x_{j}-x_{j}\right)^{2}=0

and

\phi\left(\hat{x}\right)=f\left(\hat{x}\right)-H_{2n+1}(\hat{x})-\lambda\overset{n}{\underset{j=0}{\prod}}\left(\hat{x}-x_{j}\right)^{2}=0

It’s also possible to say that

\phi'\left(x_{j}\right)=f'\left(x_{j}\right)-H'_{2n+1}(x_{j})=0

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
\vdots
\phi^{\left(2n+2\right)}\Longrightarrow 1 zero

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

\phi^{\left(2n+2\right)}\left(\eta\left(x\right)\right)=0

where \eta\left(x\right) is the value where the zero exists.

At this derivative, from our previous considerations,

\phi^{\left(2n+2\right)}\left(x\right)=f^{\left(2n+2\right)}\left(x\right)-H_{2n+1}^{\left(2n+2\right)}(x)-\lambda\left(\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}\right)^{\left(2n+2\right)}

It is fair to say that, because of the degree of the polynomial,

H_{2n+1}^{\left(2n+2\right)}(x)\equiv0

The last term could be quite complex to differentiate, but let us
consider the following:

\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}=x^{2n+2}+r\left(x\right)

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

\left(\overset{n}{\underset{j=0}{\prod}}\left(x-x_{j}\right)^{2}\right)^{\left(2n+2\right)}=\left(2n+2\right)!

Substituting,

\phi^{\left(2n+2\right)}\left(x\right)=f^{\left(2n+2\right)}\left(x\right)-\lambda\left(2n+2\right)!

Solving,

\lambda=\frac{f^{\left(2n+2\right)}\left(x\right)-\phi^{\left(2n+2\right)}\left(x\right)}{\left(2n+2\right)!}

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

\lambda=\frac{f^{\left(2n+2\right)}\left(\eta\left(x\right)\right)}{\left(2n+2\right)!}

Recalling

\phi\left(\hat{x}\right)=f\left(\hat{x}\right)-H_{2n+1}(\hat{x})-\lambda\overset{n}{\underset{j=0}{\prod}}\left(\hat{x}-x_{j}\right)^{2}=0

or

f\left(\hat{x}\right)-H_{2n+1}(\hat{x})=\lambda\overset{n}{\underset{j=0}{\prod}}\left(\hat{x}-x_{j}\right)^{2}

we can substitute and achieve our original goal

f\left(\hat{x}\right)-H_{2n+1}(\hat{x})=\frac{f^{\left(2n+2\right)}\left(\eta\left(x\right)\right)}{\left(2n+2\right)!}\overset{n}{\underset{j=0}{\prod}}\left(\hat{x}-x_{j}\right)^{2}

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