Mathematical Methods for Physics and Engineering - Matematica.NET

31.6 THE METHOD OF LEAST SQUARESy7654321001234 5xFigure 31.9 A set of data points with error bars indicating the uncertaintyσ =0.5 onthey-values. The straight line is y = ˆmx + ĉ, where ˆm and ĉ arethe least-squares estimates of the slope and intercept.L ∝ exp(−χ 2 /2) is Gaussian. From the discussions of subsections 31.3.6 and31.5.6, it follows that the ‘surfaces’ χ 2 (a) = c, where c is a constant, boundellipsoidal confidence regions for the parameters a i . The relationship between thevalue of the constant c and the confidence level is given by (31.39).◮An experiment produces the following data sample pairs (x i ,y i ):x i : 1.85 2.72 2.81 3.06 3.42 3.76 4.31 4.47 4.64 4.99y i : 2.26 3.10 3.80 4.11 4.74 4.31 5.24 4.03 5.69 6.57where the x i -values are known exactly but each y i -value is measured only to an accuracyof σ =0.5. Assuming the underlying model for the data to be a straight line y = mx + c,find the LS estimates of the slope m and intercept c and quote the standard error on eachestimate.The data are plotted in figure 31.9, together with error bars indicating the uncertainty inthe y i -values. Our model of the data is a straight line, and so we havef(x; c, m) =c + mx.In the language of (31.92), our basis functions are h 1 (x) =1andh 2 (x) =x and our modelparameters are a 1 = c and a 2 = m. From (31.93) the elements of the response matrix areR ij = h j (x i ), so that⎛ ⎞1 x 11 x 2R = ⎜⎝.⎟. ⎠ , (31.100)1 x Nwhere x i are the data values and N = 10 in our case. Further, since the standard deviationon each measurement error is σ, we have N = σ 2 I,whereI is the N × N identity matrix.Because of this simple form for N, the expression (31.98) for the LS estimates reduces toâ = σ 2 (R T R) −1 1 σ 2 RT y =(R T R) −1 R T y. (31.101)Note that we cannot expand the inverse in the last line, since R itself is not square and1275