Inverse interpolation

Given $(x_{k}, f_{k})$ from a function $f (x)$ , supose we want to find an approximation of the value of $x$ such that $f (x) = c$ , where $c$ is a given value.

We will solve the equation $x = g (c)$ where $g$ is the inverse function of $f$ . Then we will interpolate this function $g (y)$ and will evaluate it in $y = c$ , or, in other words, if we use Newton's method we will put in the first column the values $f_{j}$ and in the second one the values $x_{j}$ and proceed the same way.

Example

For example, let's suppose that we want to calculate a zero of the function $f (x) = x^{3} - 15 x + 4$ knowing that this is close to $x = 0.3$ . Then we will do quadratic interpolation, for example, of the inverse of $f (x)$ . We then first evaluate the function in three points close to $x = 0.3$ :

$x$	$0.2$	$0.3$	$0.4$
$f (x)$	$1.008$	$- 0.473$	$- 1.936$

Now we fill in the table to calculate the divided differences of Newton, but exchanging the columns, obtaining the coefficients of the interpolating polynomial:

$1.008$	$0.2$
		$- 0.0675$
$- 0.473$	$0.3$		$0.00028963$
		$- 0.0684$
$- 1.936$	$0.4$

Thus the interpolating polynomial is:

$\begin{aligned} P_{3} (y) = & 0.2 + 0.0675 \cdot (y - 1.008) + 0.00028963 \cdot (y - 1.008) (y - 0.473) \\ = & 0.2679019090 - 0.067654 y + 0.00028963 y^{2} \end{aligned}$

So an approximation of the zero of the function is:

$P_{3} (0) = 0.2679019090$