# 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, 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

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

The sources for this are not copious; more information can be found here.

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