Steffensen's method: Difference between revisions

There are excellent ways to solve this problem, but we will not be using those. Our purpose, after all, is to illustrate Steffensen's method: both its advantages and its drawbacks. Let us look at an advantage: to use Steffensen's method (which requires only function values, not derivatives), we do not actually have to expand that polynomial in <math>t</math>. Instead, we can evaluate <math>x(t)</math> and <math>y(t)</math> and plug the resulting numbers into the implicit equation. What is more, we do not even need to write <math>x(t)</math> and <math>y(t)</math> as polynomials, but instead can evaluate them directly from their control points, using [[wp:De_Casteljau's_algorithm|de Casteljau's algorithm]]:

fun de_casteljau

(c0 : double, (* control point coordinates (one axis) *)

c1 : double,

c2 : double,

t : double) (* the independent parameter *)

: double = (* value of x(t) or y(t) *)

let

val s = 1.0 - t

val c01 = (s * c0) + (t * c1)

val c12 = (s * c1) + (t * c2)

val c012 = (s * c01) + (t * c12)

in

c012

end

fun x_convex_left_parabola (t : double) : double =

de_casteljau (2.0, ~8.0, 2.0, t)

fun y_convex_left_parabola (t : double) : double =

de_casteljau (1.0, 2.0, 3.0, t)