Problem 5. The least squares approximation method taught in class is for discrete data. There is a corresponding method for approximating data given as a continuous function. Consider approximating f(x) by a quadratic polynomial f(x) = a1 + a2x + a3x 2 on the interval 0 x 1. Do so by choosing a1, a2 and a3 to minimize the root-mean-square error E(a1, a2, a3) = Z 1 0 [a1 + a2x + a3x 2 f(x)]2 dx. Derive the linear system satisfied by the optimum choices of a1, a2, a3. What is its relation to the Hilbert matrix in Problem 2?

Problem 2. In this problem we consider the question of whether a small value of the residual kAz bk means that z is a good approximation to the solution x of the linear system Ax = b. We showed in class that, kx zk kxk kAkkA 1 k kAz bk kbk . which implies that if the condition number kAkkA1k of A is small, a small relative residual implies a small relative error in the solution. We now show computationally what can happen if the condition number is large. A standard example of a matrix that is ill-conditioned is the Hilbert matrix H, with entries (Hij ) = 1/(i+j1). For n = 8, 12, 16 (where H is of dimension nn), use Matlab to solve the linear system of equations Hx = b, where b is the vector Hy and y is the vector with yi = 1/ n, i = 1 . . . n. Clearly, the true solution is given by x = y, and we let z denote the approximation obtained by Matlab. Then calculate for each value of n the following quantities: (i) the relative error kx zk/kxk, (ii) the relative residual kHz bk2/kbk, (iii) the condition number kHkkH1k, and (iv) the product of the quantities in (ii) and (iii). Arrange all these numbers in a table. The Matlab commands 1 norm and cond can be used to compute the norm and condition numbers, respectively. When vectors are input, Matlab writes them as row vectors. To convert y to a column vector, write it as y 0 . To solve the linear system Hz = b in Matlab, type z = H \ b. An example of a Matlab loop is given below; the semicolon keeps Matlab from writing unwanted output to the screen. To avoid potential problems, type clear before running a new value of n. Example of a Matlab Loop: for i=1:10 y(i) = 1/sqrt(10); end