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190 chapter 88.7.4 Linear Quadratic Fit (Extension)As indicated earlier, as long as the function being fitted depends linearly on theunknown parameters a i , the condition of minimum χ 2 leads to a set of simultaneouslinear equations for the a’s that can be solved on the computer using matrixtechniques. To illustrate, suppose we want to fit the quadratic polynomialg(x)=a 1 + a 2 x + a 3 x 2 (8.63)to the experimental measurements (x i ,y i ,i=1,N D ) (Figure 8.5 right). Because thisg(x) is linear in all the parameters a i , we can still make a linear fit even thoughx is raised to the second power. [However, if we tried to a fit a function of theform g(x)=(a 1 + a 2 x) exp(−a 3 x) to the data, then we would not be able to makea linear fit because one of the a’s appears in the exponent.]The best fit of this quadratic to the data is obtained by applying the minimumχ 2 condition (8.53) for M p =3 parameters and N D (still arbitrary) data points.A solution represents the maximum likelihood that the deduced parametersprovide a correct description of the data for the theoretical function g(x). Equation(8.53) leads to the three simultaneous equations for a 1 , a 2 , and a 3 :∑N Di=1∑N Di=1∑N Di=1[y i − g(x i )] ∂g(x i )σi2 =0,∂a 1[y i − g(x i )] ∂g(x i )σi2 =0,∂a 2[y i − g(x i )] ∂g(x i )σi2 =0,∂a 3∂g∂a 1=1, (8.64)∂g∂a 2= x, (8.65)∂g∂a 3= x 2 . (8.66)Note: Because the derivatives are independent of the parameters (the a’s), the adependence arises only from the term in square brackets in the sums, and becausethat term has only a linear dependence on the a’s, these equations are linearequations in the a’s.Exercise: Show that after some rearrangement, (8.64)–(8.66) can be written asSa 1 + S x a 2 + S xx a 3 = S y , (8.67)S x a 1 + S xx a 2 + S xxx a 3 = S xy ,S xx a 1 + S xxx a 2 + S xxxx a 3 = S xxy .−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 190