Conference rusure::math

827.0. "Fitting a parabola" by TFH::MARSHALL (hunting the snark) Sat Feb 20 1988 19:49

   I need to fit a parabola to a set of data and was given
   _Data_Analysis_for_Scientists_and_Engineers_ by Stuart L. Meyer, who 
   gives the following method of fitting a parabola:

	Determine these values:
		A = sum[x*(1/s)]
		B = sum[(1/s)]
		C = sum[y*(1/s)]
		D = sum[(x^2)*(1/s)]
		E = sum[x*y*(1/s)]
		F = sum[(y^2)*(1/s)]
		G = sum[(x^2)*y*(1/s)]
		H = sum[(x^3)*(1/s)]
 		I = sum[(x^4)*(1/s)]
	where:
		x := the position (in time) of the sample point
		y := the mean value of the sample point
		s := variance of the sample point
	create matrix AA:
                [ B A D ]
           AA = [ A D H ]
                [ D H I ]
	create vector U:
                [ C ]
	    U = [ E ]
                [ G ]
	invert matrix AA to get error matrix AA':
	    AA' = INV(AA)
	best-fit coefficient vector:
                [ A* ]
	    C = [ B* ] = U * AA'
                [ C* ]
	correlated errors from error matrix
		S(1) = SQR(AA'(1,1))
		S(2) = SQR(AA'(2,2))
		S(3) = SQR(AA'(3,3))
		to get the best fit coefficients:
		A* = C(1) � S(1)
		B* = C(2) � S(2)
		C* = C(3) � S(3)
	for the parabola:
		y = (C*)x� + (B*)x + (A*)

   This works quite well and is probably very standard, but what do I do 
   when the variance (s) = 0? Suppose I take ten samples at each 'x', and 
   'y' is the same in each case, then the variance calculates to zero.
                                                   
                  /
                 (  ___
                  ) ///
                 /

    If you only have one sample point per abscissa, then leave out the
    (1/s) factors and solve the normal equations without them.  This
    is because the best you can do is assume that the variances are all
    the same, and hence they drop out of the equations...

    You can go back after fitting the parabola and estimate the overall
    variance by the averaged discrepancy between the quadratic and the
    actual data at over all points.

    - Jim