Sol. of Non-linear Systems
1Numerical Solutions of Nonlinear Systems of Equations
NTNU
Tsung-Min Hwang November 30, 2003
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
21 – Fixed Point method . . . 3 2 – Newton’s Method . . . 4 3 – Quasi-Newton’s Method . . . 8
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
31 – Fixed Point method
Definition 1 A function
G
fromD ⊂ R
n intoR
n has a fixed point atp ∈ D
ifG(p) = p
.Theorem 1 (Contraction Mapping Theorem) Let
D = {(x
1, · · · , x
n)
T; a
i≤ x
i≤ b
i, ∀ i = 1, . . . , n} ⊂ R
n. SupposeG : D → R
nis a continuous function with
G(x) ∈ D
wheneverx ∈ D
. ThenG
has a fixed point inD
.Suppose, in addition,
G
has continuous partial derivatives and a constantα < 1
exists with∂g
i(x)
∂x
j≤ α
n ,
wheneverx ∈ D,
for
j = 1, . . . , n
andi = 1, . . . , n
. Then, for anyx
(0)∈ D
,x
(k)= G(x
(k−1)),
for eachk ≥ 1
converges to the unique fixed point
p ∈ D
andk x
(k)− p k
∞≤ α
k1 − α k x
(1)− x
(0)k
∞.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
31 – Fixed Point method
Definition 1 A function
G
fromD ⊂ R
n intoR
n has a fixed point atp ∈ D
ifG(p) = p
.Theorem 1 (Contraction Mapping Theorem) Let
D = {(x
1, · · · , x
n)
T; a
i≤ x
i≤ b
i, ∀ i = 1, . . . , n} ⊂ R
n. SupposeG : D → R
nis a continuous function with
G(x) ∈ D
wheneverx ∈ D
. ThenG
has a fixed point inD
.Suppose, in addition,
G
has continuous partial derivatives and a constantα < 1
exists with∂g
i(x)
∂x
j≤ α
n ,
wheneverx ∈ D,
for
j = 1, . . . , n
andi = 1, . . . , n
. Then, for anyx
(0)∈ D
,x
(k)= G(x
(k−1)),
for eachk ≥ 1
converges to the unique fixed point
p ∈ D
andk x
(k)− p k
∞≤ α
k1 − α k x
(1)− x
(0)k
∞.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
42 – Newton’s Method
First consider solving the following system of nonlinear equations:
f
1(x
1, x
2) = 0, f
2(x
1, x
2) = 0.
Suppose
(x
(k)1, x
(k)2)
is an approximation to the solution of the system above, and we try to computeh
(k)1 andh
(k)2 such that(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
satisfies the system. By the Taylor’s theorem for two variables,0 = f
1(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
1(x
(k)1, x
(k)2) + h
(k)1∂f
1∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
1∂x
2(x
(k)1, x
(k)2) 0 = f
2(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
2(x
(k)1, x
(k)2) + h
(k)1∂f
2∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
2∂x
2(x
(k)1, x
(k)2)
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
42 – Newton’s Method
First consider solving the following system of nonlinear equations:
f
1(x
1, x
2) = 0, f
2(x
1, x
2) = 0.
Suppose
(x
(k)1, x
(k)2)
is an approximation to the solution of the system above, and we try to computeh
(k)1 andh
(k)2 such that(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
satisfies the system.By the Taylor’s theorem for two variables,
0 = f
1(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
1(x
(k)1, x
(k)2) + h
(k)1∂f
1∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
1∂x
2(x
(k)1, x
(k)2) 0 = f
2(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
2(x
(k)1, x
(k)2) + h
(k)1∂f
2∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
2∂x
2(x
(k)1, x
(k)2)
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
42 – Newton’s Method
First consider solving the following system of nonlinear equations:
f
1(x
1, x
2) = 0, f
2(x
1, x
2) = 0.
Suppose
(x
(k)1, x
(k)2)
is an approximation to the solution of the system above, and we try to computeh
(k)1 andh
(k)2 such that(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
satisfies the system. By the Taylor’s theorem for two variables,0 = f
1(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
1(x
(k)1, x
(k)2) + h
(k)1∂f
1∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
1∂x
2(x
(k)1, x
(k)2) 0 = f
2(x
(k)1+ h
(k)1, x
(k)2+ h
(k)2)
≈ f
2(x
(k)1, x
(k)2) + h
(k)1∂f
2∂x
1(x
(k)1, x
(k)2) + h
(k)2∂f
2∂x
2(x
(k)1, x
(k)2)
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
5Put this in matrix form
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
h
(k)1h
(k)2
+
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
≈
0 0
.
The matrix
J(x
(k)1, x
(k)2) ≡
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
is called the Jacobian matrix. Set
h
(k)1 andh
(k)2 be the solution of the linear systemJ(x
(k)1, x
(k)2)
h
(k)1h
(k)2
= −
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
then
x
(k+1)1x
(k+1)2
=
x
(k)1x
(k)2
+
h
(k)1h
(k)2
is expected to be a better approximation. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
5Put this in matrix form
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
h
(k)1h
(k)2
+
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
≈
0 0
.
The matrix
J(x
(k)1, x
(k)2) ≡
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
is called the Jacobian matrix.
Set
h
(k)1 andh
(k)2 be the solution of the linear systemJ(x
(k)1, x
(k)2)
h
(k)1h
(k)2
= −
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
then
x
(k+1)1x
(k+1)2
=
x
(k)1x
(k)2
+
h
(k)1h
(k)2
is expected to be a better approximation. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
5Put this in matrix form
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
h
(k)1h
(k)2
+
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
≈
0 0
.
The matrix
J(x
(k)1, x
(k)2) ≡
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
is called the Jacobian matrix. Set
h
(k)1 andh
(k)2 be the solution of the linear systemJ(x
(k)1, x
(k)2)
h
(k)1h
(k)2
= −
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
then
x
(k+1)1x
(k+1)2
=
x
(k)1x
(k)2
+
h
(k)1h
(k)2
is expected to be a better approximation. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
5Put this in matrix form
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
h
(k)1h
(k)2
+
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
≈
0 0
.
The matrix
J(x
(k)1, x
(k)2) ≡
∂f1
∂x1
(x
(k)1, x
(k)2)
∂x∂f12
(x
(k)1, x
(k)2)
∂f2
∂x1
(x
(k)1, x
(k)2)
∂x∂f22
(x
(k)1, x
(k)2)
is called the Jacobian matrix. Set
h
(k)1 andh
(k)2 be the solution of the linear systemJ(x
(k)1, x
(k)2)
h
(k)1h
(k)2
= −
f
1(x
(k)1, x
(k)2) f
2(x
(k)1, x
(k)2)
then
x
(k+1)1x
(k+1)2
=
x
(k)1x
(k)2
+
h
(k)1h
(k)2
is expected to be a better approximation.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
6In general, we solve the system of
n
nonlinear equationsf
i(x
1, · · · , x
n) = 0
,i = 1, . . . , n
.Let
x = h
x
1x
2· · · x
ni
Tand
F (x) = h
f
1(x) f
2(x) · · · f
n(x) i
T.
The problem can be formulated as solving
F (x) = 0, F : R
n→ R
n.
Let
J(x)
, where the(i, j)
entry is ∂x∂fij
(x)
, be then × n
Jacobian matrix. Then the Newton’s iteration is defined asx
(k+1)= x
(k)+ h
(k),
where
h
(k)∈ R
n is the solution of the linear systemJ(x
(k))h
(k)= −F (x
(k)).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
6In general, we solve the system of
n
nonlinear equationsf
i(x
1, · · · , x
n) = 0
,i = 1, . . . , n
. Letx = h
x
1x
2· · · x
ni
Tand
F (x) = h
f
1(x) f
2(x) · · · f
n(x) i
T.
The problem can be formulated as solving
F (x) = 0, F : R
n→ R
n.
Let
J(x)
, where the(i, j)
entry is ∂x∂fij
(x)
, be then × n
Jacobian matrix. Then the Newton’s iteration is defined asx
(k+1)= x
(k)+ h
(k),
where
h
(k)∈ R
n is the solution of the linear systemJ(x
(k))h
(k)= −F (x
(k)).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
6In general, we solve the system of
n
nonlinear equationsf
i(x
1, · · · , x
n) = 0
,i = 1, . . . , n
. Letx = h
x
1x
2· · · x
ni
Tand
F (x) = h
f
1(x) f
2(x) · · · f
n(x) i
T.
The problem can be formulated as solving
F (x) = 0, F : R
n→ R
n.
Let
J(x)
, where the(i, j)
entry is ∂x∂fij
(x)
, be then × n
Jacobian matrix. Then the Newton’s iteration is defined asx
(k+1)= x
(k)+ h
(k),
where
h
(k)∈ R
n is the solution of the linear systemJ(x
(k))h
(k)= −F (x
(k)).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
6In general, we solve the system of
n
nonlinear equationsf
i(x
1, · · · , x
n) = 0
,i = 1, . . . , n
. Letx = h
x
1x
2· · · x
ni
Tand
F (x) = h
f
1(x) f
2(x) · · · f
n(x) i
T.
The problem can be formulated as solving
F (x) = 0, F : R
n→ R
n.
Let
J(x)
, where the(i, j)
entry is ∂x∂fij
(x)
, be then × n
Jacobian matrix.Then the Newton’s iteration is defined as
x
(k+1)= x
(k)+ h
(k),
where
h
(k)∈ R
n is the solution of the linear systemJ(x
(k))h
(k)= −F (x
(k)).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
6In general, we solve the system of
n
nonlinear equationsf
i(x
1, · · · , x
n) = 0
,i = 1, . . . , n
. Letx = h
x
1x
2· · · x
ni
Tand
F (x) = h
f
1(x) f
2(x) · · · f
n(x) i
T.
The problem can be formulated as solving
F (x) = 0, F : R
n→ R
n.
Let
J(x)
, where the(i, j)
entry is ∂x∂fij
(x)
, be then × n
Jacobian matrix. Then the Newton’s iteration is defined asx
(k+1)= x
(k)+ h
(k),
where
h
(k)∈ R
n is the solution of the linear systemJ(x
(k))h
(k)= −F (x
(k)).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
7Algorithm 1 (Newton’s Method for Systems) Given a function
F : R
n→ R
n, an initial guessx
(0) to the zero ofF
, and stop criteriaM
,δ
, andε
, this algorithm performs the Newton’s iteration to approximate one root ofF
.Set
k = 0
andh
(−1)= e
1.while
(k < M )
and(k h
(k−1)k≥ δ)
and(k F (x
(k)) k≥ ε
doCalculate
J(x
(k)) = [∂F
i(x
(k))/∂x
j]
.Solve the
n × n
linear systemJ(x
(k))h
(k)= −F (x
(k))
.Set
x
(k+1)= x
(k)+ h
(k) andk = k + 1
.end while
Output ( “Convergent
x
(k)”) or (“Maximum number of iterations exceeded”)Remark 1
(i) quadratic convergence if the starting point is near the exact solution point in terms of vector norm.
(ii) At each iteration, a Jacobian matrix has to be evaluated and an
n × n
linear system involving this matrix must be solved.Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
7Algorithm 1 (Newton’s Method for Systems) Given a function
F : R
n→ R
n, an initial guessx
(0) to the zero ofF
, and stop criteriaM
,δ
, andε
, this algorithm performs the Newton’s iteration to approximate one root ofF
.Set
k = 0
andh
(−1)= e
1.while
(k < M )
and(k h
(k−1)k≥ δ)
and(k F (x
(k)) k≥ ε
doCalculate
J(x
(k)) = [∂F
i(x
(k))/∂x
j]
.Solve the
n × n
linear systemJ(x
(k))h
(k)= −F (x
(k))
.Set
x
(k+1)= x
(k)+ h
(k) andk = k + 1
.end while
Output ( “Convergent
x
(k)”) or (“Maximum number of iterations exceeded”) Remark 1(i) quadratic convergence if the starting point is near the exact solution point in terms of vector norm.
(ii) At each iteration, a Jacobian matrix has to be evaluated and an
n × n
linear system involving this matrix must be solved.Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
83 – Quasi-Newton’s Method
Next we will extend secant method to solving system of nonlinear equations.
Recall that in one dimensional case, one uses the linear model
`
k(x) = f (x
k) + a
k(x − x
k)
to approximate the function
f (x)
atx
k. That is,`
k(x
k) = f (x
k)
for anya
k∈ R
. If wefurther require that
`
0(x
k) = f
0(x
k)
, thena
k= f
0(x
k)
. The zero of`
k(x)
is used togive a new approximate for the zero of
f (x)
, that is,x
k+1= x
k− 1
f
0(x
k) f (x
k)
which yields Newton’s method. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
83 – Quasi-Newton’s Method
Next we will extend secant method to solving system of nonlinear equations. Recall that in one dimensional case, one uses the linear model
`
k(x) = f (x
k) + a
k(x − x
k)
to approximate the function
f (x)
atx
k.That is,
`
k(x
k) = f (x
k)
for anya
k∈ R
. If wefurther require that
`
0(x
k) = f
0(x
k)
, thena
k= f
0(x
k)
. The zero of`
k(x)
is used togive a new approximate for the zero of
f (x)
, that is,x
k+1= x
k− 1
f
0(x
k) f (x
k)
which yields Newton’s method. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
83 – Quasi-Newton’s Method
Next we will extend secant method to solving system of nonlinear equations. Recall that in one dimensional case, one uses the linear model
`
k(x) = f (x
k) + a
k(x − x
k)
to approximate the function
f (x)
atx
k. That is,`
k(x
k) = f (x
k)
for anya
k∈ R
.If we further require that
`
0(x
k) = f
0(x
k)
, thena
k= f
0(x
k)
. The zero of`
k(x)
is used togive a new approximate for the zero of
f (x)
, that is,x
k+1= x
k− 1
f
0(x
k) f (x
k)
which yields Newton’s method. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
83 – Quasi-Newton’s Method
Next we will extend secant method to solving system of nonlinear equations. Recall that in one dimensional case, one uses the linear model
`
k(x) = f (x
k) + a
k(x − x
k)
to approximate the function
f (x)
atx
k. That is,`
k(x
k) = f (x
k)
for anya
k∈ R
. If wefurther require that
`
0(x
k) = f
0(x
k)
, thena
k= f
0(x
k)
.The zero of
`
k(x)
is used togive a new approximate for the zero of
f (x)
, that is,x
k+1= x
k− 1
f
0(x
k) f (x
k)
which yields Newton’s method. Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
83 – Quasi-Newton’s Method
Next we will extend secant method to solving system of nonlinear equations. Recall that in one dimensional case, one uses the linear model
`
k(x) = f (x
k) + a
k(x − x
k)
to approximate the function
f (x)
atx
k. That is,`
k(x
k) = f (x
k)
for anya
k∈ R
. If wefurther require that
`
0(x
k) = f
0(x
k)
, thena
k= f
0(x
k)
. The zero of`
k(x)
is used togive a new approximate for the zero of
f (x)
, that is,x
k+1= x
k− 1
f
0(x
k) f (x
k)
which yields Newton’s method.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
9If
f
0(x
k)
is not available, one instead asks the linear model to satisfy`
k(x
k) = f (x
k)
and`
k(x
k−1) = f (x
k−1).
In doing this, the identity
f (x
k−1) = `
k(x
k−1) = f (x
k) + a
k(x
k−1− x
k)
gives
a
k= f (x
k) − f (x
k−1) x
k− x
k−1.
Solving
`
k(x) = 0
yields the secant iterationx
k+1= x
k− x
k− x
k−1f (x
k) − f (x
k−1) f (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
9If
f
0(x
k)
is not available, one instead asks the linear model to satisfy`
k(x
k) = f (x
k)
and`
k(x
k−1) = f (x
k−1).
In doing this, the identity
f (x
k−1) = `
k(x
k−1) = f (x
k) + a
k(x
k−1− x
k)
gives
a
k= f (x
k) − f (x
k−1) x
k− x
k−1.
Solving
`
k(x) = 0
yields the secant iterationx
k+1= x
k− x
k− x
k−1f (x
k) − f (x
k−1) f (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
9If
f
0(x
k)
is not available, one instead asks the linear model to satisfy`
k(x
k) = f (x
k)
and`
k(x
k−1) = f (x
k−1).
In doing this, the identity
f (x
k−1) = `
k(x
k−1) = f (x
k) + a
k(x
k−1− x
k)
gives
a
k= f (x
k) − f (x
k−1) x
k− x
k−1.
Solving
`
k(x) = 0
yields the secant iterationx
k+1= x
k− x
k− x
k−1f (x
k) − f (x
k−1) f (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
10In multiple dimension, the analogue affine model becomes
M
k(x) = F (x
k) + A
k(x − x
k),
where
x, x
k∈ R
n andA
k∈ R
n×n, and satisfiesM
k(x
k) = F (x
k),
for any
A
k.The zero of
M
k(x)
is then used to give a new approximate for the zero ofF (x)
, that is,x
k+1= x
k− A
−k 1F (x
k).
The Newton’s method chooses
A
k= F
0(x
k) ≡ J(x
k) =
the Jacobian matrix.
and yields the iteration
x
k+1= x
k− (F
0(x
k))
−1F (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
10In multiple dimension, the analogue affine model becomes
M
k(x) = F (x
k) + A
k(x − x
k),
where
x, x
k∈ R
n andA
k∈ R
n×n, and satisfiesM
k(x
k) = F (x
k),
for any
A
k. The zero ofM
k(x)
is then used to give a new approximate for the zero ofF (x)
, that is,x
k+1= x
k− A
−k 1F (x
k).
The Newton’s method chooses
A
k= F
0(x
k) ≡ J(x
k) =
the Jacobian matrix.
and yields the iteration
x
k+1= x
k− (F
0(x
k))
−1F (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
10In multiple dimension, the analogue affine model becomes
M
k(x) = F (x
k) + A
k(x − x
k),
where
x, x
k∈ R
n andA
k∈ R
n×n, and satisfiesM
k(x
k) = F (x
k),
for any
A
k. The zero ofM
k(x)
is then used to give a new approximate for the zero ofF (x)
, that is,x
k+1= x
k− A
−k 1F (x
k).
The Newton’s method chooses
A
k= F
0(x
k) ≡ J(x
k) =
the Jacobian matrix.
and yields the iteration
x
k+1= x
k− (F
0(x
k))
−1F (x
k).
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
11When the Jacobian matrix
J(x
k) ≡ F
0(x
k)
is not available, one can requireM
k(x
k−1) = F (x
k−1).
Then
F (x
k−1) = M
k(x
k−1) = F (x
k) + A
k(x
k−1− x
k),
which gives
A
k(x
k− x
k−1) = F (x
k) − F (x
k−1)
and this is the so-called secant equation. Let
h
k= x
k− x
k−1 andy
k= F (x
k) − F (x
k−1).
The secant equation becomes
A
kh
k= y
k.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
11When the Jacobian matrix
J(x
k) ≡ F
0(x
k)
is not available, one can requireM
k(x
k−1) = F (x
k−1).
Then
F (x
k−1) = M
k(x
k−1) = F (x
k) + A
k(x
k−1− x
k),
which gives
A
k(x
k− x
k−1) = F (x
k) − F (x
k−1)
and this is the so-called secant equation.
Let
h
k= x
k− x
k−1 andy
k= F (x
k) − F (x
k−1).
The secant equation becomes
A
kh
k= y
k.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
11When the Jacobian matrix
J(x
k) ≡ F
0(x
k)
is not available, one can requireM
k(x
k−1) = F (x
k−1).
Then
F (x
k−1) = M
k(x
k−1) = F (x
k) + A
k(x
k−1− x
k),
which gives
A
k(x
k− x
k−1) = F (x
k) − F (x
k−1)
and this is the so-called secant equation. Let
h
k= x
k− x
k−1 andy
k= F (x
k) − F (x
k−1).
The secant equation becomes
A
kh
k= y
k.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
Sol. of Non-linear Systems
11When the Jacobian matrix
J(x
k) ≡ F
0(x
k)
is not available, one can requireM
k(x
k−1) = F (x
k−1).
Then
F (x
k−1) = M
k(x
k−1) = F (x
k) + A
k(x
k−1− x
k),
which gives
A
k(x
k− x
k−1) = F (x
k) − F (x
k−1)
and this is the so-called secant equation. Let
h
k= x
k− x
k−1 andy
k= F (x
k) − F (x
k−1).
The secant equation becomes
A
kh
k= y
k.
Department of Mathematics – NTNU Tsung-Min Hwang November 30, 2003
