Hermite's Interpolation Method

Topics: Interpolation

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Let $y=f(x)$ be a function whose exact definition we don’t know. All we know is various values of $f(x)$, like so:

${\mathbf{x}}_{\mathbf{i}}$	${\mathbf{y}}_{\mathbf{i}}=\mathbf{f}({\mathbf{x}}_{\mathbf{i}})$
$x_0$	$y_0$
$x_1$	$y_1$
$x_2$	$y_2$
$x_3$	$y_3$
$⋮$	$⋮$
$x_n$	$y_n$

Just like the tables we use for Newton’s divided differences method, and unlike the ones used for Newton’s finite differences method, the spacing between every $x_i$ does not need to be constant.

We will be able to interpolate $y_k=f(x_k)$ for any $x_k$ within the constraints of the table by using a method known as Hermite’s interpolation method.

$z_i$ §

Just like Newton’s divided and finite differences method, we will use a table to aid us. This time around, though, instead of using $x_i$ as the main “measure”, we will use $z_i$.

For every $i=0,1,\dots ,n$, let $z_{2i}$ be:

$$z_{2i}=z_{2i+1}=x_i$$

In a Table

More simply, in a table:

${\mathbf{z}}_{\mathbf{i}}$	$\mathbf{f}({\mathbf{z}}_{\mathbf{i}})$
$z_0=x_0$	$f(z_0)=f(x_0)$
$z_1=x_0$	$f(z_1)=f(x_0)$
$z_2=x_1$	$f(z_2)=f(x_1)$
$z_3=x_1$	$f(z_3)=f(x_1)$
$z_4=x_2$	$f(z_4)=f(x_2)$
$z_5=x_2$	$f(z_5)=f(x_2)$
$⋮$	$⋮$
$⋮$	$⋮$
$z_{2n}=x_n$	$f(z_{2n})=f(x_n)$
$z_{2n+1}=x_n$	$f(z_{2n+1})=f(x_n)$

Differences §

We’ll also use differences, just like Newton’s finite and divided differences methods.

In fact, most of the differences will be just like the ones used in Newton’s divided differences method.

The only exceptions will be in the even-numbered first differences (if we start counting from $0$):

$$\begin{array}{rl}f[z_0,z_1]& =f^{\prime }(x_0)\\ f[z_2,z_3]& =f^{\prime }(x_1)\\ f[z_4,z_5]& =f^{\prime }(x_2)\\ ⋮& \\ f[z_{2i},z_{2i+1}]& =f^{\prime }(x_i)\end{array}$$

How to find the derivative?

Let’s say we’re calculating all of these differences from scratch. If we don’t know the function yet, how can we derivate it?

Easy. We obtain an approximating polynomial first by using one of the many other interpolation methods we know, then we derivate it.

The rest are the familiar divided differences. For rest of the first differences:

$$\begin{array}{rl}f[z_1,z_2]& =\frac{f(z_2)-f(z_1)}{z_2-z_1}\\ f[z_3,z_4]& =\frac{f(z_4)-f(z_3)}{z_4-z_3}\\ ⋮& \\ f[z_i,z_{i+1}]& =\frac{f(z_{i+1})-f(z_i)}{z_{i+1}-z_i}\end{array}$$

In general, for the $n$th difference:

$$f[z_i,z_{i+1},z_{i+2},\dots ,z_{i+n}]=\frac{f[z_{i+1},z_{i+2},\dots ,z_{i+n}]-f[z_i,z_{i+1},\dots ,z_{i+n-1}]}{z_{i+n}-z_i}$$

Differences Table

Similarly to Newton’s finite and divided differences method, we can build a differences table to better visualise all these differences:

${\mathbf{z}}_{\mathbf{i}}$	$\mathbf{f}({\mathbf{z}}_{\mathbf{i}})$	$\mathbf{f}[{\mathbf{z}}_{\mathbf{i}},{\mathbf{z}}_{\mathbf{i}+1}]$	$\mathbf{f}[{\mathbf{z}}_{\mathbf{i}},{\mathbf{z}}_{\mathbf{i}+1},{\mathbf{z}}_{\mathbf{i}+2}]$	$\mathbf{f}[{\mathbf{z}}_{\mathbf{i}},{\mathbf{z}}_{\mathbf{i}+1},{\mathbf{z}}_{\mathbf{i}+2},{\mathbf{z}}_{\mathbf{i}+3}]$
$z_0=x_0$	$f(z_0)=f(x_0)$
		$f[z_0,z_1]=f^{\prime }(x_0)$
$z_1=z_0$	$f(z_1)=f(z_0)$		$f[z_0,z_1,z_2]$
		$f[z_1,z_2]=\frac{f(z_2)-f(z_1)}{z_2-z_1}$		$f[z_0,z_1,z_2,z_3]$
$z_2=z_1$	$f(z_2)=f(z_1)$		$f[z_1,z_2,z_3]$
		$f[z_2,z_3]=f^{\prime }(x_1)$		$f[z_1,z_2,z_3,z_4]$
$z_3=z_1$	$f(z_3)=f(z_2)$		$f[z_2,z_3,z_4]$
		$f[z_3,z_4]=\frac{f(z_4)-f(z_3)}{z_4-z_3}$		$f[z_2,z_3,z_4,z_5]$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$
$⋮$	$⋮$	$⋮$	$⋮$	$f[z_{2n-2},z_{2n-1},z_{2n},z_{2n+1}]$
$z_{2n}=z_n$	$f(z_{2n})=f(z_n)$	$⋮$	$f[z_{2n-1},z_{2n},z_{2n+1}]$
		$f[z_{2n},z_{2n+1}]=f^{\prime }(x_n)$
$z_{2n+1}=x_n$	$f(z_{2n+1})=f(x_n)$

Interpolation Formula §

Given an $x_k$ contained within the constraints of the table (i.e. $x_0\le x_k\le x_n$), we can use these differences to interpolate its $f(x_k)=H_{2n+1}(x_k)$ with the following formula that is used to obtain an interpolating polynomial.

We’ll choose a specific $z_i$ in our table as an anchor value; we’ll set this value to be $z_0$.

$$\begin{array}{rl}& \\ H_{2n+1}(x_k)=& f(z_0)+(x_k-z_0)f[z_0,z_1]+(x_k-z_1)(x_k-z_0)f[z_0,z_1,z_2]\\ & +(x_k-z_2)(x_k-z_1)(x_k-z_0)f[z_0,z_1,z_2,z_3]+\dots \\ & +(x_k-z_{k-1})\dots (x_k-z_0)+[z_0,\dots ,z_k]\end{array}$$

Indeed, this formula is equivalent to Newton’s finite and divided differences’ forwards interpolation formulas. We can come up with a backwards interpolation formula for Hermite’s method, but notice we can just flip our table upside down and then interpolate with this formula to get the same result.