Page 255 - Advanced engineering mathematics

7.8 Least Squares Vectors and Data Fitting 235

Finally,
⎛ ⎞
3

−112 9
t
A B = ⎝ −2 ⎠ = .
−242 0
7
The auxiliary lsv system is

6 10 9
X = .
10 24 0
This has a unique solution, which is the unique least squares vector for the system:
−1
6 10 9 12/22 −5/22 9 108/22
∗ .
X = = =
10 24 0 −5/22 3/22 0 −45/22
We will apply least squares vectors to the problem of drawing a straight line that is, in some
sense, a best fit to a set of given data points in the plane. We can see the idea by looking at an
example. Suppose (perhaps by experiment or observation) we have data points

(0,−5.5),(1,−2.7),(2,−0.8),(3,1.2),(5,4.7),
which we will label (x j , y j ) (from left to right) for j = 1,2,3,4,5. We want to draw a straight
line y = ax + b that is a “best fit” to these points. For each of the observed points (x j , y j ), think
of ax j + b as an approximation to y j ,so
ax 1 + b ≈ y 1 ,
ax 2 + b ≈ y 2 ,
.
.
.
ax 5 + b ≈ y 5 .

Consider the system
⎛ ⎞ ⎛ ⎞
10 −5.5
−2.7
11
⎜ ⎟ ⎜ ⎟
⎜ ⎟ b ⎜ ⎟
⎟
⎜
⎜ 12 ⎟ = −0.8 .
⎜ ⎟ a ⎜ ⎟
⎝ 13 ⎠ ⎝ 1.2 ⎠
15 4.7
This has the form AX = B with A defined so that row j of the matrix product AX is ax j + b,
and this is set equal to the column matrix B listing the given y j ’s. Of course, ax j + b is only
approximately equal to y j . We want a line that “best approximates” these points, so we obtain a
∗
and b by solving for a least squares vector X for this system.
Once we decide on this approach, the rest is arithmetic. Compute

5 11
t
A A = ,
11 39
and

39/74 −11/74
t −1
(A A) = .
−11/74 5/74
The unique least squares vector is

∗ t −1 t −5.0229729···
X = (A A) A B = .
2.001351351···

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October 14, 2010 14:23 THM/NEIL Page-235 27410_07_ch07_p187-246