Section 5.3.3.1: Quadratic Approximation formula

Given q(x) = A x2 + B x + C the stationary point exists at xm where

2 A xm + B = 0

or

xm = - B / 2 A

The stationary point is a minimum if A > 0.
Fit q(x) to three points (x1,f1), (x2,f2), (x3,f3).
Solve for A, B, C:
A x12 + B x1 + C = f1     (1)
A x22 + B x2 + C = f2     (2)
A x32 + B x3 + C = f3     (3)

(2) - (1) and (3) - (2) eliminate C:
A ( x22 - x12 ) + B (x2 - x1) = (f2 - f1)     (4)
A ( x32 - x22 ) + B (x3 - x2) = (f3 - f2)     (5)

(4) $ \times (x_3 - x_2)$ and (5) $\times (x_2 - x_1)$ :
A ( x22 - x12 ) (x3 - x2) + B (x2 - x1) (x3 - x2) = (f2 - f1) (x3 - x2)     (6)
A ( x32 - x22 ) (x2 - x1) + B (x3 - x2) (x2 - x1)= (f3 - f2) (x2 - x1)     (7)

(7) - (6) eliminates B:
A [ ( x32 - x22 ) (x2 - x1) - ( x22 - x12 ) (x3 - x2) ]= (f3 - f2) (x2 - x1) - (f2 - f1) (x3 - x2)     (8)

or, using difference of squares,
A (x3 - x2)(x2 - x1)[ ( x3 + x2 ) - ( x2 + x1 ) ]= (f3 - f2) (x2 - x1) - (f2 - f1) (x3 - x2)     (9)

Hence

\begin{displaymath}A = \frac{s_{32} - s_{21}}{x_3 - x_1} \end{displaymath}

where

\begin{displaymath}s_{ji} = \frac{f_j - f_i}{x_j - x_i} \end{displaymath}

Substitute into (4):

\begin{displaymath}\frac{s_{32} - s_{21}}{x_3 - x_1} ( x_2^2 - x_1^2 ) + B (x_2 - x_1) = (f_2 - f_1) \end{displaymath}

Thus

\begin{displaymath}B = \frac{ (f_2 - f_1)}{(x_2 - x_1)} - ( x_2 + x_1 ) \frac{s_... ..._1} = s_{21} - ( x_2 + x_1 ) \frac{s_{32} - s_{21}}{x_3 - x_1} \end{displaymath}

and

\begin{displaymath}x_m = - \frac{B}{2 A} = \frac{x_1 + x_2}{2} - \frac { s_{21}(x_3 - x_1)}{2(s_{32} - s_{21})} \end{displaymath}

Since f3 > f2 and f2 < f1, then s32 > 0 and s21 < 0. So s32 - s21 > 0 and A > 0: the quadratic has a minimum at xm. Note that xm lies between $ \frac{x_1 + x_2}{2} $ and $ \frac{x_2 + x_3}{2} $.


Return to Section 5.3.3
Return to Section 5.3 Index
Return to Section 5 Index

Course Organiser
Last modified: Tue Aug 25 12:02:52 BST