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where
A
(σ) is the derivative of
A
(σ). Then
−
A
(σ
r
)
A
(σ)
σ
r
−
σ
A
(σ)
≈
.
(8.69)
Now consider the problem
A
(σ
r
)
A
(σ)
σ
r
−
σ
−
A
(
σ
)
x
r
σ
r
−
σ
.
A
(σ)
x
r
A
(σ)
u
=
≈
x
r
=−
(8.70)
Thus, the solution to this problem is
x
r
σ
−
σ
r
.
u
=
(8.71)
A more complicated problem is
A
(σ)
b
,
A
(σ)
v
=
(8.72)
where
b
is the linear combination of all the eigenvectors,
i
γ
i
x
i
.
b
=
(8.73)
We can verify that
γ
i
x
i
σ
−
σ
i
v
=
(8.74)
i
is the solution of this more complicated problem. We then have
A
(σ)
i
(σ
−
σ
i
)
A
(σ)
γ
i
x
i
σ
−
σ
i
=
i
γ
i
[
A
(σ
i
)
+
+···
]
x
i
A
(σ)
v
=
σ
−
σ
i
=
A
(σ)
i
γ
i
x
i
=
A
(σ)
b
,
(8.75)
completing the verification. This suggests that if we have an initial value of the
eigenfrequency, σ
r
, the solution
v
will enhance the corresponding eigenvector
x
r
above all others. The iterative scheme then becomes
A
(σ
r
)
x
r
.
A
(σ
r
)
x
r
+
1
=
(8.76)
After satisfactory convergence to the eigenvector has been obtained, an improved
estimate of the eigenfrequency can be found, satisfying the equation
A
(σ
r
+
1
)
x
r
+
1
=
0.
(8.77)
Then
−
σ
r
)
A
(σ
r
)
A
(σ
r
+
1
)
=
A
(σ
r
)
+
(σ
r
+
1
+···
(8.78)
and
A
(σ
r
)
+···
x
r
+
1
−
σ
r
)
A
(σ
r
)
+
(σ
r
+
1
=
0.
(8.79)
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