5.4 General Linear Least-Squares linear, non-linear, and linearizable equations the general linear equation notational simplification and matrix least-squares solution errors in the coefficients an example with Gaussian peaks alternative matrix formalism an example from multi-component spectrophotometry 5.4 : 1/17

What is a Linear Equation? With respect to curve fitting, a linear equation is defined as one that is linear in the unknown parameters. In the following examples let a represent the unknown parameters, k represent known constants, x represent the independent, error-free variable, and y represent the dependent random variable. The randomness of y follows a normal pdf. Linear Equations Non-Linear Equations 5.4 : 2/17

coefficient transformation Linearizable Equations Some equations can be made linear in their coefficients by a mathematical transformation. non-linear coefficient linear coefficient coefficient transformation N.A. Two important restrictions: the pdf for z may not be normal, voiding the least-squares analysis. CLT comes to the rescue for large numbers of measurements, though. the pdf for the transformed coefficients may not be normal, voiding the use of many hypothesis tests. 5.4 : 3/17

The General Linear Equation Consider a general equation linear in its n+1 coefficients, where the fi(xi) are functions of x that contain only known constants. For the straight line equation, f0(x) = 1 and f1(x) = x. There are N data pairs, (xi,yi). The method of maximum likelihood produces one equation for each of the n+1 unknown equation coefficients. 5.4 : 4/17

Notational Simplification Define two new variables to replace the summation operations. br is an (n+1) column vector, where 0  r  n. ar,c is an (n+1)(n+1) matrix, where 0  r  n and 0  c  n. For all sums, 1  i  N. With this change in variables, the set of equations becomes, which can be further simplified using matrix notation, where m = n+1 is the number of rows (and the number of columns). 5.4 : 5/17

Matrix Least-Squares Solution The matrix equation is solved by pre-multiplying both sides by the inverse alpha matrix, a-1, and remembering that a-1a = 1. The matrix least-squares solution: the N, x,y-pairs of data are collected the n+1 functions, f(x), are evaluated for each of the x-values the b terms are computed and arranged into a column vector the a terms are computed and arranged into a matrix the inverse alpha matrix, a-1, is computed the beta vector is left-multiplied by the inverse alpha matrix to obtain the n+1 equation coefficients in the form of a column vector, a 5.4 : 6/17

Errors in the Coefficients The a-1 matrix is called the variance/covariance matrix. The diagonal elements are the variances of the corresponding coefficients. In the common case of an unweighted least-squares (all si = s), the above expression is modified to, where the variance is estimated in the usual manner. Remember, n+1 coefficients have already been computed from the N data values, so the degrees of freedom are N-(n+1). 5.4 : 7/17

Error Matrix for y = a0 + a1x For the straight line equation, f0(x) = 1 and f1(x) = x. For an unweighted least-squares the a matrix is given by the following. The a-1 variance/covariance matrix yields the expressions derived on slide 5.3-7. where 5.4 : 8/17

Covariance The covariance of two random variables is given by the following. For statistically independent random variables, <xy> = mxmy, and cov(x,y) = 0. The least-squares solution on the previous slide showed that the slope and intercept are only statistically independent when the xaxis values are centered about zero. When the deviation in the slope is positive, the deviation in the intercept will be negative. Except for centered data, the two random variables, a0 and a1 are NOT statistically independent - the error in one of them influences the error in the other. 5.4 : 9/17

Example Matrix Least-Squares Consider a situation where the measured data are some combination of three known Gaussian functions. The measured data are shown at the right. The goal is to obtain the least-squares solution to, y = a0f0(t) + a1f1(t) + a2f2(t) 5.4 : 10/17

Matrix Solution 5.4 : 11/17

Alternative Matrix Formalism Suppose that the linear components of the equation to be fit are available as data vectors and not functions. An excellent example would be a multi-component determination using Beer's Law. Al =el,0C0 + el,1C1 + el,2C2 With this method the spectrum of each component, el,i, is known, but doesn't have a functional form. The collected data and known molar absorptivities might be arranged in a table as follows. l A el,0 el,1 el,2 400 0.000 10,342 105 405 0.001 2 10,281 108 ●●● 650 536 5.4 : 12/17

Matrix Algebra Let A(N) be a column vector of absorption values, where N is the number of wavelengths. Let e(N,m) be a matrix of e-values, where m = n+1 is the number of components. And, let C(m) be a column vector of the m concentrations. This cannot be solved directly using the inverse matrix since e is not square. Instead, pre-multiply both sides by the transpose of e. Note that the product, eTe, is square, mm, and has an inverse. Finally, pre-multiply both sides by the inverse, (eTe)-1. The term in brackets is the variance/covariance matrix. 5.4 : 13/17

Multi-Component Spectrophotometry Consider the analysis of a solution containing p-biphenyl (B), p-terphenyl (T), and p-quaterphenyl (Q). For the three compounds, molar absorptivities, e, are given in the table and spectra in the graph. l B T Q 240 13,900 5,350 2,600 250 15,500 13,100 4,960 260 10,400 24,000 10,900 270 4,440 32,900 21,300 280 2,590 33,300 32,600 290 24,800 40,200 300 12,100 38,500 310 3,760 28,300 320 400 14,400 330 5,200 5.4 : 14/17

Data, Vectors and Matrices Absorbance for the mixture was measured at 10 wavelengths. (240,0.610)(250,0.886) (260,1.095)(270,1.248) (280,1.310)(290,1.188) (300,0.822)(310,0.501) (320,0.310)(330,0.109) e is a 103 matrix formed by the right three columns on the last slide. 5.4 : 15/17

Solving for Concentrations Compute the inverse alpha matrix. Compute the concentrations. Plot the regression points, where 5.4 : 16/17

Concentration Errors Compute the standard error of the fit. Compute the coefficient standard deviations using the diagonal elements of the inverse alpha matrix. 95% Confidence limits with 10-3 = 7 degrees of freedom: 5.4 : 17/17