Interpolation of Hermite

The Hermite polynomial is the one that interpolates a set of points and the value of their derivatives in any points we want. That is, let's suppose that we have $(x_{k}, f_{k})$ and $(x_{k}, f_{k}^{'})$ .

Then we construct the same table as in Newton's method, placing $x_{k}$ in the first column and writing twice the same point if we know the value of the derivative at this point; in the second column the values of $f$ corresponding to the $x$ of the same line. Namely if we know the value of $f$ in $x_{0}$ and of its derivative we will write $x_{0}$ twice and next to them we will write $f_{0}$ . For example,

$x_{0}$	$f_{0}$
$x_{0}$	$f_{0}$
$x_{1}$	$f_{1}$
$x_{1}$	$f_{1}$

From here on we proceed the same way, but with the difference that we have to define $f [x_{i}, x_{i}] = f_{i}^{'}$ , the value of the derivative in $x_{i}$ .

$x_{0}$	$f_{0}$
		$f_{0}^{'}$
$x_{0}$	$f_{0}$		$f [x_{0}, x_{0}, x_{1}]$
		$f [x_{0}, x_{1}]$		$f [x_{0}, x_{0}, x_{1}, x_{1}]$
$x_{1}$	$f_{1}$		$f [x_{0}, x_{1}, x_{1}]$
		$f_{1}^{'}$
$x_{1}$	$f_{1}$

Therefore, if we have $n + 1$ values of the function and $n + 1$ values of the derivatives, the Hermite polynomial will have a $2 n + 1$ degree .

Let's consider an example:

Example

Let's suppose that we want to calculate $f (\frac{1}{8})$ where $f (x) = \tan (π x)$ from Hermite interpolation in $0, \frac{1}{4}$ .

To obtain the result, we draw a table as in Newton's interpolation but repeating every point which derivative we know. This is:

$0$	$0$
		$f^{'} (0) = π$
$0$	$0$		$\frac{4 - π}{\frac{1}{4} - 0} = 16 - 4 π$
		$\frac{1 - 0}{\frac{1}{4} - 0} = 4$		$\frac{8 π - 16 - 16 + 4 π}{\frac{1}{4} - 0} = 148 π - 128$
$\frac{1}{4}$	$1$		$\frac{2 π - 4}{\frac{1}{4} - 0} = 8 π - 16$
		$f^{'} (\frac{1}{4}) = 2 π$
$\frac{1}{4}$	$1$

Proceeding as with Newton's interpolation, we get: $P_{3} (x) = π x + (16 - 4 π) x^{2} + (48 π - 128) x^{2} (x - \frac{1}{4})$

Now, $\tan (\frac{π}{8}) \approx P_{3} (\frac{1}{8}) = 0.4018 \dots$