Steffensen's method: Difference between revisions
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That is easier said than done.
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)
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