Page 237 - Matrix Analysis & Applied Linear Algebra

232 Chapter 4 Vector Spaces

Solution: Determine the parabola f(t)= α 0 + α 1 t + α 2 t 2 that best fits the
observed data in the least squares sense. Then estimate where the missile will
land by determining the roots of f (i.e., determine where the parabola crosses
the horizontal axis). As it stands, the problem will involve numbers having rela-
tively large magnitudes in conjunction with relatively small ones. Consequently,
it is better to first scale the data by considering one unit to be 1000 miles. If

1 0 0 0
   
 
 1 .25 .0625  α 0  .008 
A =  1 .5 .25  , x =  α 1  , and b =  .015  ,




1 .75 .5625 α 2 .019
   
1 1 1 .02
and if ε = Ax − b, then the object is to find a least squares solution x that
minimizes
5
2 T T
ε = ε ε =(Ax − b) (Ax − b).
i
i=1
We know that such a least squares solution is given by the solution to the system
T
T
of normal equations A Ax = A b, which in this case is
     
5 2.5 1.875 α 0 .062
2.5 1.875 1.5625 = .04375 .
   α 1   
1.875 1.5625 1.3828125 α 2 .0349375
The solution (rounded to four significant digits) is
 −4 
−2.286 × 10
x =  3.983 × 10 −2  ,
−1.943 × 10 −2

and the least squares parabola is
2
f(t)= −.0002286 + .03983t − .01943t .

To estimate where the missile will land, determine where this parabola crosses
the horizontal axis by applying the quadratic formula to find the roots of f(t)
to be t = .005755 and t =2.044. Therefore, we estimate that the missile will
land 2044 miles down range. The sum of the squares of the errors associated with
the least squares solution is

5
2 T T −7
ε = ε ε =(Ax − b) (Ax − b)=4.571 × 10 .
i
i=1