Hermite Interpolation

Section 4.4 Hermite Interpolation

In the previous section, we studied polynomial interpolation in which a polynomial passes through the points \((x_{0},f_{0}), (x_{1},f_{1}),\ldots, (x_{n},f_{n})\text{.}\) In this section, we consider the polynomial also satisfies the derivative conditions \((x_{0},f'_{0})\text{,}\) \((x_{1},f'_{1})\text{,}\) \(\ldots\text{,}\) \((x_{n},f'_{n})\text{.}\) This is called Hermite Interpolation.

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Consider an example in which a polynomial \(P(x)\) needs to satisfy the points \((0,0)\) and \((2,0)\) as well having a slope of 1 at \(x=0\) and a slope of \(-1\) at \(x=2\text{.}\) Visually, this is:

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Figure 4.4.1.

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where the lines in the plot are the tangent lines to the polynomial. This polynomial must be at most order 3 (with 4 unknowns) to satisfy the 4 conditions. A crude way to proceed would be to write:

\begin{equation*} \begin{aligned}P(x)\amp = a_{0} + a_{1}x + a_{2} x^{2} + a_{3} x^{3} \\ \text{and its derivative}\qquad\qquad \amp \\ P'(x)\amp = a_{1} + 2a_{2} x + 3a_{3} x^{2}\end{aligned} \end{equation*}

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Applying \(P(0)=0\) and \(P'(0)=1\) leads to \(a_{0}=0\) and \(a_{1}=1\text{.}\) If we set \(P(2)=0\) and \(P'(2) = -1\) we get:

\begin{equation*} \begin{aligned}P(2)\amp = (2) + a_{2} (2)^{2} + a_{3} (2)^{3} \\ P'(2)\amp = 1 + 2 a_{2} (2) + 3 a_{3} (2)^{2}\end{aligned} \end{equation*}

and solve for \(a_{2}\) and \(a_{3}\) (using some linear algebra) results in \(a_{2}=-1/2\) and \(a_{3}=0\text{,}\) thus the polynomial is:

\begin{equation*} \begin{aligned}P(x)\amp = x -\frac{1}{2}x^{2}\end{aligned} \end{equation*}

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In short, Hermite polynomials exist and are unique much like Lagrange polynomials. The proofs are similar to that shown in Section 4.2 however a bit more complicated in that the form of the interpolation is more complicated due to the requirement that the derivative needs to match. See Burden and Faires, Burden:2004fk for more information about this.

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Subsection 4.4.1 Hermite Interpolation Implementation

The form presented above is not a simple way to find the polynomial coefficients as the degree of the polynomial gets large. Instead, we proceed in a way which closely follows that of Newton Divided Differences.

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The idea behind using Newton Divided Differences in Hermite interpolation is as follows. Let \(z_{2i}=x_{i}\) and \(z_{2i+1}=x_{i}\) for \(i=0,1,\ldots, n\text{.}\) However some of the difference formulas will have the form:

\begin{equation*} \begin{aligned}f[z_{0},z_{1}]=f[x_{0},x_{0}]\amp = \frac{f[x_{0}]-f[x_{0}]}{x_{0}-x_{0}}\end{aligned} \end{equation*}

which has the form \(0/0\text{.}\) When this situation arises, let \(f[x_{i},x_{i}]=f'(x_{i})\text{.}\)

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The Newton Divide Difference table for two points \(x_{0}\) and \(x_{1}\) (or the four points \(z_{0},z_{1},z_{2},z_{3}\)).

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		First divided	Second divided
\(z\)	\(f(z)\)	difference	difference

\(z_{0}=x_{0}\)	\(f[z_{0}]=f(x_{0})\)
		\(f[z_{0},z_{1}]=f[x_{0},x_{0}]=f'(x_{0})\)
\(z_{1}=x_{0}\)	\(f[z_{1}]=f(x_{0})\)		\(\displaystyle f[z_{0},z_{1},z_{2}] = \frac{f[z_{1},z_{2}]-f[z_{0},z_{1}]}{z_{2}-z_{0}}\)
		\(\displaystyle f[z_{1},z_{2}] = \frac{f[z_{2}]-f[z_{1}]}{z_{2}-z_{1}}\)
\(z_{2} = x_{1}\)	\(f[z_{2}]=f(x_{1})\)		\(\displaystyle f[z_{1},z_{2},z_{3}] = \frac{f[z_{2},z_{3}]-f[z_{1},z_{2}]}{z_{3}-z_{1}}\)
		\(\displaystyle f[z_{2},z_{3}] = f[x_{1},x_{1}]=f'(x_{1})\)
\(x_{3} = x_{1}\)	\(f[z_{3}]=f(x_{1})\)

If additional points are needed, the divided difference table can be continued in a similar manner.

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Example 4.4.2.

Use the Newton Divided Difference table to produce the Hermite Polynomial that satisfies \(P(0)=0, P(2)=0, P'(0)=1\) and \(P'(2)=-1\text{.}\)

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First let \(z_{0}=z_{1}=x_{0}=0\) and \(z_{2}=z_{3}=x_{1}=2\text{.}\)

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\(z\)	\(f(z)\)	1st D.D.	2nd D.D.	3rd D.D.

\(0\)	0
		\(\boxed{1}\)
\(0\)	\(0\)		\(-1/2\)
		\(0\)		\(0\)
\(2\)	\(0\)		\(1/2\)
		\(\boxed{-1}\)
\(2\)	\(0\)

where the two boxed values are the derivative values given in the problem.

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And as we saw in Newton divided difference, we produce a polynomial based on the top diagonal elements of the table. The only difference is recall that since there are two \(z=0\) terms, there will be a \((z-0)^{2}\) term as well. The Hermite Polynomial will be:

\begin{equation*} H(z) = 0 + 1 \cdot z - \frac{1}{2}z^{2} + 0 \cdot z^{2} (z-2) \end{equation*}

or writing it back in terms of \(x\text{:}\)

\begin{equation*} H(x) = x - \frac{1}{2}x^{2} \end{equation*}

which is the same as seen above.

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The next example applies this technique to finding a Hermite polynomial to \(\sin x\) on \([0,\pi]\text{.}\)

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Example 4.4.3.

Find the Hermite Polynomial that satisfies both the function value and derivative of \(f(x)=\sin x\) for \(x_{0}=0, x_{1}=\pi/2\) and \(x_{2}=\pi\text{.}\)

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Solution.

We will first produce a Divided Difference table:

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\(z\)	\(f(z)\)	1st D.D.	2nd D.D.	3rd D.D.	4th D.D.	5th D.D.

0	0
		\(\boxed{1}\)
0	0		\(\dfrac{4-2\pi}{\pi^2}\)
		\(2/\pi\)		\(\dfrac{4\pi-16}{\pi^3}\)
\(\pi/2\)	1		\(-\dfrac{4}{\pi^2}\)		\(\dfrac{16-4\pi}{\pi^4}\)
		\(\boxed{0}\)		0		0
\(\pi/2\)	1		\(-\dfrac{4}{\pi^2}\)		\(\dfrac{16-4\pi}{\pi^4}\)
		\(-2/\pi\)		\(\dfrac{16-4\pi}{\pi^3}\)
\(\pi\)	0		\(\dfrac{2\pi-4}{\pi^2}\)
		\(\boxed{-1}\)
\(\pi\)	0

And then read the Hermite polynomial off the top diagonal row:

\begin{equation*} \begin{aligned}H(x)\amp = 0 + 1 \cdot x + \frac{4-2\pi}{\pi^{2}}\cdot x^{2} + \frac{4\pi-16}{\pi^{3}}\cdot x^{2} (x-\pi/2) \\\amp \qquad + \frac{16-4\pi}{\pi^{4}}\cdot x^{2} (x-\pi/2)^{2}\end{aligned} \end{equation*}

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A plot both the polynomial and the sine function is:

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Figure 4.4.4.

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and this shows that to the visual eye, they appear the same. If we plot the absolute value of the difference:

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Figure 4.4.5.

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you can see that the error seems to be bounded by 0.003. There are error bounds for Hermite polynomials, however we will not cover them here.

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