Weneed to show that
x
k
∈
I
.
Assume
x
0
∈
I ,

x
0

x
*
 ≤
then

x
1

x
*

=

x
0

x

2
1
2
f
00
(
ξ
0
)
f
0
(
x
0
)
≤ 
x
0

x
*

x
0

x
*

M
( )
≤ 
x
0

x
*

M
( )
<

x
0

x
*
 ≤
Now assume that
x
k
∈
I
then

x
k
+1

x
*

=

x
k

x

2
1
2
f
00
(
ξ
k
)
f
0
(
x
k
)
≤
2
M
( )
<
⇒
x
k
+1
∈
I
So now

x
k
+1

x
*

≤

x
k

x
*

2
M
( )
=
≤ 
x
k

x
*

M
( )
.
.
.
≤

x

0
*

(
M
( )
k
+1
→
0
, k
→ ∞
, x
k
→
x
*
as
k
→ ∞
The need for a good initial guess
x
0
for Newton’s method should be
emphasized. In practice, this is obtained with another method, like bisection.
12.6
The Secant Method
Sometimes it could be computationally expensive or not possible to evaluate
the derivative of
f
.
The following method, known as the secant method,
12.6.
THE SECANT METHOD
199
replaces the derivative by the secant:
x
k
+1
=
x
k

f
(
x
k
)
f
(
x
k
)

f
(
x
k

1
)
x
k

x
k

1
,
k
= 1
,
2
, . . .
(12.18)
Note that since
f
(
x
*
) = 0
x
k
+1

x
*
=
x
k

x
*

f
(
x
k
)

f
(
x
*
)
f
(
x
k
)

f
(
x
k

1
)
x
k

x
k

1
,
=
x
k

x
*

f
(
x
k
)

f
(
x
*
)
f
[
x
k
, x
k

1
]
=
(
x
k

x
*
)
1

f
(
x
k
)

f
(
x
*
)
x
k

x
*
f
[
x
k
, x
k

1
]
!
=
(
x
k

x
*
)
1

f
[
x
k
, x
*
]
f
[
x
k
, x
k

1
]
=
(
x
k

x
*
)
f
[
x
k
, x
k

1
]

f
[
x
k
, x
*
]
f
[
x
k
, x
k

1
]
=
(
x
k

x
*
)(
x
k

1

x
*
)
f
[
x
k
,x
k

1
]

f
[
x
k
,x
*
]
x
k

1

x
*
f
[
x
k
, x
k

1
]
=
(
x
k

x
*
)(
x
k

1

x
*
)
f
[
x
k

1
, x
k
, x
*
]
f
[
x
k
, x
k

1
]
If
x
k
→
x
*
, then
f
[
x
k

1
,x
k
,x
*
]
f
[
x
k
,x
k

1
]
→
1
2
f
00
(
x
*
)
f
0
(
x
*
)
and lim
k
→∞
x
k
+1

x
*
x
k

x
*
= 0, i.e.
the
sequence generated by the secant method would converge faster than linear.
Defining
e
k
=

x
k

x
*

, the calculation above suggests
e
k
+1
≈
ce
k
e
k

1
(12.19)
Let’s try to determine the rate of convergence of the secant method. Starting
with the ansatz
e
k
≈
Ae
p
k

1
or equivalently
e
k

1
=
(
1
A
e
k
)
1
/p
we have
e
k
+1
≈
ce
k
e
k

1
≈
ce
k
1
A
e
k
1
p
,
200
CHAPTER 12.
NONLINEAR EQUATIONS
which implies
A
1+
1
p
c
≈
e
1

p
+
1
p
k
.
(12.20)
Since the left hand side is a constant we must have 1

p
+
1
p
= 0 which gives
p
=
1
±
√
5
2
, thus
p
=
1 +
√
5
2
=
≈
1
.
61803
(12.21)
gives the rate of convergence of the secant method. It is better than linear,
but worse than quadratic. Sufficient conditions for local convergence are as
in Newton’s method.
12.7
Fixed Point Iteration
Newton’s method is a particular example of a functional iteration of the form
x
k
+1
=
g
(
x
k
)
,
k
= 0
,
1
, . . .
with the particular choice of
g
(
x
) =
x

f
(
x
)
f
0
(
x
)
. Clearly, if
x
*
is a zero of
f
then
x
*
is a fixed point of
g
, i.e.
g
(
x
*
) =
x
*
. We will look at fixed point iterations
as a tool for solving
f
(
x
) = 0.
Example 33.
Suppose we want to solve
x

e

1
= 0
in
[0
,
1]
. Then if we
take
g
(
x
) =
e

x
, a fixed point of
g
corresponds to a zero of
f
.
Definition 22.
Let
g
is defined in an interval
[
a, b
]
.
We say that
g
is a
contraction or a contractive map if there is a constant
L
with
0
≤
L <
1
such that

g
(
x
)

g
(
y
)
 ≤
L

x

y

,
for all
x, y
∈
[
a, b
]
.
(12.22)
If
x
*
is a fixed point of
g
in [
a, b
] then

x
k

x
*

=

g
(
x
k

1
)

g
(
x
*
)

≤
L

x
k

1

x
*

≤
L
2

x
k

2

x
*
≤
· · ·
≤
L
k

x
0

x
*
 →
0
,
as
k
→ ∞
12.7.
FIXED POINT ITERATION
201