Biology Reference
In-Depth Information
In general, when the data set contains n data points, the matrices P and
Y* will be:
2
3
dG
ð
r
0
;
X
1
Þ
4
5
dr
2
4
3
5
ð
Þ
Y
1
G
r
0
;
X
1
dG
ð
r
0
;
X
2
Þ
Y
2
G
ð
r
0
X
2
Þ
;
dr
.
and Y
¼
.
P
¼
:
(8-29)
Y
n
G
ð
r
0
X
n
Þ
;
dG
ð
r
0
;
X
n
Þ
dr
To solve Eq. (8-28) for
e
, we multiply both sides by the transposed
matrix P
T
:
P
T
P
P
T
Y
:
e ¼
(8-30)
Now, P
T
P is a square matrix, and, if it is invertible, we can solve
Eq. (8-30) and obtain
Þ
1
P
T
P
P
T
Y
Þ:
e ¼ð
ð
(8-31)
Because
r
0
.We
call this improved guess r
1
, use it in place of r
0
in Eq. (8-26), and then
iterate. Schematically, the process can be represented as
e ¼
r - r
0
, the next guess is calculated from r
¼ e þ
e þ
guess
)
better guess.
The process terminates when the calculated value for better guess is the
same as guess; that is, when
e ¼
0. We have then found the answer
for the parameter r.
The Gauss-Newton method just described is not based upon minimizing
the SSR defined in Eq. (8-22), so how is it a least-squares procedure?
The answer is found in Eq. (8-31). When
0, we have (P
T
P)
-1
(P
T
Y*)
¼
0. Because (P
T
P)
-1
cannot be zero, as the matrix (P
T
P) was inverted, it
must be that P
T
Y*
e ¼
0. For the matrices P
T
and Y* defined in
Eq. (8-29), we then have for r
¼
¼
answer, the product
X
dG
ð
r
;
X
i
Þ
P
T
Y
¼
½
Y
i
G
ð
r
X
i
Þ
¼
0
:
;
dr
i
On the other hand, differentiating Eq. (8-22) gives
dSSR
ð
r
Þ
¼
dr
2
P
i
½
dG
ð
r
X
i
Þ
;
Thus, when r is such that P
T
Y*
Y
i
G
ð
r
X
i
Þ
:
¼
0,
;
dr
we will also have
dSSR
ð
r
Þ
¼
0. This shows that we have found
dr
the least-squares estimate for the parameter r.
