Jim Lambers MAT 419/519 Summer Session Lecture 13 Notes

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1 Jim Lambers MAT 419/519 Summer Session Lecture 13 Notes These notes correspond to Section 4.1 in the text. Least Squares Fit One of the most fundamental problems in science and engineering is data fitting constructing a function that, in some sense, conforms to given data points. One type of data-fitting technique is interpolation. Interpolation techniques, of any kind, construct functions that agree exactly with the data. That is, given points (x 1, y 1 ), (x 2, y 2 ),..., (x m, y m ), interpolation yields a function f(x) such that f(x i ) = y i for i = 1, 2,..., m. However, fitting the data exactly may not be the best approach to describing the data with a function. High-degree polynomial interpolation can yield oscillatory functions that behave very differently than a smooth function from which the data is obtained. Also, it may be pointless to try to fit data exactly, for if it is obtained by previous measurements or other computations, it may be erroneous. Therefore, we consider another notion of what constitutes a best fit of given data by a function. One alternative approach to data fitting is to solve the minimax problem, which is the problem of finding a function f(x) of a given form for which max f(x i) y i 1 i n is minimized. However, this is a very difficult problem to solve. Another approach is to minimize the total absolute deviation of f(x) from the data. That is, we seek a function f(x) of a given form for which f(x i ) y i is minimized. However, we cannot apply standard minimization techniques to this function, because, like the absolute value function that it employs, it is not differentiable. This defect is overcome by considering the problem of finding f(x) of a given form for which [f(x i ) y i ] 2 is minimized. This is known as the least squares problem. We will first show how this problem is solved for the case where f(x) is a linear function of the form f(x) = a 1 x + a 0, and then generalize this solution to other types of functions. 1

2 When f(x) is linear, the least squares problem is the problem of finding constants a 0 and a 1 such that the function E(a 0, a 1 ) = (a 1 x i + a 0 y i ) 2 is minimized. In order to minimize this function of a 0 and a 1, we must compute its partial derivatives with respect to a 0 and a 1. This yields E E = 2(a 1 x i + a 0 y i ), = 2(a 1 x i + a 0 y i )x i. a 0 a 1 At a minimum, both of these partial derivatives must be equal to zero. This yields the system of linear equations ( m ) ma 0 + x i a 1 = y i, ( m ) ( m x i a 0 + These equations are called the normal equations. Using the formula for the inverse of a 2 2 matrix, [ ] 1 [ a b 1 = c d ad bc we obtain the solutions a 0 = x 2 i ) a 1 = d c x i y i. b a ], ( m ) x2 i ( m y i) ( m x i) ( m x iy i ) m m x2 i ( m x i) 2, a 1 = m m x iy i ( m x i) ( m y i) m m x2 i ( m x i) 2. Example We wish to find the linear function y = a 1 x + a 0 that best approximates the data shown in Table 1, in the least-squares sense. Using the summations x i = , x 2 i = , y i = , x i y i = , we obtain a 0 = a 1 = = = , = =

3 i x i y i Table 1: Data points (x i, y i ), for i = 1, 2,..., 10, to be fit by a linear function We conclude that the linear function that best fits this data in the least-squares sense is y = x The data, and this function, are shown in Figure 1. It is interesting to note that if we define the m 2 matrix A, the 2-vector a, and the m-vector y by 1 x 1 y 1 1 x 2 [ ] A =.., a = a0 y 2, y = a 1., 1 x m y m then a is the solution to the system of equations A T Aa = A T y. These equations are the normal equations defined earlier, written in matrix-vector form. They arise from the problem of finding the vector a such that Aa y is minimized, where, for any vector u, u is the magnitude, or length, of u. This magnitude is equivalent to the square root of the expression we originally intended to minimize, (a 1 x i + a 0 y i ) 2, 3

4 Figure 1: Data points (x i, y i ) (circles) and least-squares line (solid line) but we will see that the normal equations also characterize the solution a, an n-vector, to the more general linear least squares problem of minimizing Aa y for any matrix A that is m n, where m n, and whose columns are linearly independent. We now consider the problem of finding a polynomial of degree n that gives the best leastsquares fit. As before, let (x 1, y 1 ), (x 2, y 2 ),..., (x m, y m ) be given data points that need to be approximated by a polynomial of degree n. We assume that n < m 1, for otherwise, we can use polynomial interpolation to fit the points exactly. Let the least-squares polynomial have the form n p n (x) = a j x j. Our goal is to minimize the sum of squares of the deviations in p n (x) from each y-value, 2 n E(a) = [p n (x i ) y i ] 2 = a j x j i y i, 4 j=0 j=0

5 where a is a column vector of the unknown coefficients of p n (x), a 0 a 1 a =.. a n Differentiating this function with respect to each a k yields E n = 2 a j x j i a y i x k i, k = 0, 1,..., n. k j=0 Setting each of these partial derivatives equal to zero yields the system of equations ( n m ) a j = x k i y i, k = 0, 1,..., n. j=0 x j+k i These are the normal equations. They are a generalization of the normal equations previously defined for the linear case, where n = 1. Solving this system yields the coefficients {a j } n j=0 of the least-squares polynomial p n (x). As in the linear case, the normal equations can be written in matrix-vector form A T Aa = A T y, where 1 x 0 x 2 0 x n 0 1 x 1 x 2 1 x n a 0 y 1 1 A = 1 x 2 x 2 2 x n a 1 2, a = , y = y x m x 2 m x n a n y n m The normal equations equations can be used to compute the coefficients of any linear combination of functions {φ j (x)} n j=0 that best fits data in the least-squares sense, provided that these functions are linearly independent. In this general case, the entries of the matrix A are given by a ij = φ i (x j ), for i = 1, 2,..., m and j = 0, 1,..., n. Example We wish to find the quadratic function y = a 2 x 2 + a 1 x + a 0 that best approximates the data shown in Table 2, in the least-squares sense. By defining 1 x 1 x 2 1 y 1 1 x 2 x 2 a 0 2 A =..., a = a 1 y 2, y = a., 2 1 x 10 x 2 10 y 10 5

6 i x i y i Table 2: Data points (x i, y i ), for i = 1, 2,..., 10, to be fit by a quadratic function and solving the normal equations we obtain the coefficients A T Aa = A T y, c 0 = , c 1 = , c 2 = , and conclude that the quadratic function that best fits this data in the least-squares sense is y = x x The data, and this function, are shown in Figure 2. Least-squares fitting can also be used to fit data with functions that are not linear combinations of functions such as polynomials. Suppose we believe that given data points can best be matched to an exponential function of the form y = be ax, where the constants a and b are unknown. Taking the natural logarithm of both sides of this equation yields ln y = ln b + ax. If we define z = ln y and c = ln b, then the problem of fitting the original data points {(x i, y i )} m with an exponential function is transformed into the problem of fitting the data points {(x i, z i )} m with a linear function of the form c + ax, for unknown constants a and c. Similarly, suppose the given data is believed to approximately conform to a function of the form y = bx a, where the constants a and b are unknown. Taking the natural logarithm of both sides of this equation yields ln y = ln b + a ln x. 6

7 Figure 2: Data points (x i, y i ) (circles) and quadratic least-squares fit (solid curve) If we define z = ln y, c = ln b and w = ln x, then the problem of fitting the original data points {(x i, y i )} m with a constant times a power of x is transformed into the problem of fitting the data points {(w i, z i )} m with a linear function of the form c + aw, for unknown constants a and c. Example We wish to find the exponential function y = be ax that best approximates the data shown in Table 3, in the least-squares sense. By defining 1 x 1 z 1 1 x 2 [ ] c A =.., c = z 2, z = a., 1 x 5 z 5 where c = ln b and z i = ln y i for i = 1, 2,..., 5, and solving the normal equations we obtain the coefficients A T Ac = A T z, a = , b = e c = e = , 7

8 i x i y i Table 3: Data points (x i, y i ), for i = 1, 2,..., 5, to be fit by an exponential function and conclude that the exponential function that best fits this data in the least-squares sense is y = e x. The data, and this function, are shown in Figure 3. It can be seen from the preceding discussion and examples that the normal equations can be used to solve any problem that requires finding the vector x R n that minimizes b Ax, where b R m, m n, and A is an m n matrix with linearly independent columns, regardless of the interpretation of these columns. To see this, we define the function ϕ(x) = b Ax 2, x R n. Then, it can be shown through differentiation that ϕ(x) = 2(A T Ax A T b), H ϕ (x) = A T A. If x 0, then Ax 0 because A has linearly independent columns. It follows that x A T Ax = (Ax) Ax = Ax 2 > 0, so H ϕ (x) is positive definite on R n. This leads to the following theorem. Theorem Let A be an m n matrix with linearly independent columns, and let b R m. Then the vector x defined by x = (A T A) 1 A T b, that solves the normal equations A T Ax = A T b, is the strict global minimizer of b Ax, x R n. 8

9 Figure 3: Data points (x i, y i ) (circles) and exponential least-squares fit (solid curve) The matrix A + = (A T A) 1 A T is called the pseudo-inverse, or generalized inverse, of A. When A is a square, invertible matrix, then A + = A 1. Otherwise, A + is the matrix that, as closely as possible, serves as an inverse of A. It should be noted that the condition that A has linearly independent columns is essential, so that A T A is invertible. Exercises 1. Chapter 4, Exercise 1 2. Chapter 4, Exercise 4 3. Chapter 4, Exercise 7 4. Chapter 4, Exercise 10 9