ABSTRACT
Equation (4.166) can be solved for a, b, and c by Gauss elimination. A linear fit to a set of bivariate data may be inadequate. Consider the quadratic
bivariate polynomial:
I z = a + bx + cy + dx2 + el +fxy I
The sum of the squares of the deviations is given by
S(a, b, c, d, e,f) = L (Z; - a - bx; - cy; - dif - e.0 - .fx;)!i The function S(a, b, ... ,f) is a minimum when
as ~ 2 .. 2 aa = L... 2(Z; - a - bx; - cY; - dx; - ey; - .fx;)!;)(-I) = 0
as ~ 2 .. 2 af = L... 2(Z; - a - bx; - cY; - dx; - ey; - .fx;)!;)(-X;)!;) = 0
Dividing Eqs. (4.169) by 2 and rearranging yields the normal equations:
(4.167)
(4.168)
(4.169a)
(4.169f)
aN + b LX; + eLY; + dLxT + e L.0 +fLX;)!; = LZ; (4.170a) a LX; + b LXT + c LX;)!; + d LX; + e LXJIT +fLxTY; = LX;Z; (4.170b) a LY; + b LX;Y; + c L.0 + d LXTY; + eLY; +f LXJIT = LY;Z; (4.170c) a LXT + b LX; + c LXTY; + d LX; + e LxT.0 +f LX;Y; = LxTZ; (4.170d) a L.0 + b Lx;.0 + c LYT + dLxT.0 + eLY; +fLx;,YT = L.0z; (4.170e) a LX;)!; + b LxTY; + c Lx;.0 + d LX;Y; + e LX;YT +f LxT.0 = LX;)!;Z;
(4.170f)
Equation (4.170) can be solved for a to fby Gauss elimination.
