Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

122 CHAPTER 9. LEAST SQUARES
Exercises
(9.1) How do you know that the choice of constants c i in our least squares analysis actually find a
minimum of equation 9.1, and not, say, a maximum?
(9.2) Our “linear” least squares might be better called the “affine” least squares. In this exercise
you will find the best linear function which approximates a set of data. That is, find the
function f(x) = cx which is the least squares best approximant to the given data
x x 0 x 1 . . . x n
y y 0 y 1 . . . y n
(9.3) Find the constant that best approximates, in least squares sense, the given data x, y. (Hint:
you can use equation 9.2 or equation 9.3 using a single basis function g 0 (x) = 1.) Do the x i
values affect your answer?
(9.4) Find the function f(x) = c which best approximates, in the least squares sense, the data
x 1 −2 5
y 1 −2 4
(9.5) Find the function ax + b that best approximates the data
x 0 −1 2
y 0 1 −1
(9.6) Find the function ax 2 + b that best approximates the data
x 0 1 2
y 0.3 0.1 0.5
(9.7) Find the constant c such that f(x) = ln (cx) best approximates, in the least squares sense,
the given data
x x 0 x 1 . . . x n
y y 0 y 1 . . . y n
(Hint: You cannot use basis functions and equation 9.2 to solve this. You must use the
Definition 9.1.1.) The geometric mean of the numbers a 1 , a 2 , . . . , a n is defined as ( ∏ a i ) 1/n .
How does your answer relate to the geometric mean?
(9.8) Write code to find the function ae x + be −x that best interpolates, in the least squares sense,
a set of data {(x i , y i )} i=n
i=0
. Your m-file should have header line like:
function [a,b] = lsqrexp(xys)
where xys is the (n + 1) × 2 matrix whose rows are the n + 1 tuples (x i , y i ). Use the normal
equations method. This should reduce the problem to solving a 2 × 2 linear system; let
octave/Matlab solve that linear system for you. (Hint: To find x such that Mx = b, use x =
M \ b.)
(a) What do you get when you try the following?
octave:1> xs = [-1 -0.5 0 0.5 1]’;
octave:2> ys = [1.194 0.430 0.103 0.322 1.034]’;
octave:3> xys = [xs ys];
octave:4> [a,b] = lsqrexp(xys)
(b) Try the following:
octave:5> xys = xys = [-0.00001 0.3;0 0.6;0.00001 0.7];
octave:6> [a,b] = lsqrexp(xys)
Does something unexpected happen? Why?