Fixed point iteration
Numerical Analysis
Math 465/565
1
Monday, September 9, 13
Fixed Point Iteration
Suppose we wanted to solve :
f (x) = cos(x)
x=0
or
cos(x) = x
We might consider a iteration of this type :
xk+1 = cos(xk )
Try this : Enter any number into your calculator and
repeatedly hit the cos function. You will see the value
converge to the solution of cos(x) = x. This value x is
the fixed point of the function cos(x).
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Fixed point iteration
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Fixed point iteration
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Fixed point iteration
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Convergence of fixed point iteration
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Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
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Monday, September 9, 13
Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
|g(↵)
g( )| 
|↵
|,
0
<1
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Monday, September 9, 13
Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
|g(↵)
g( )| 
|↵
|,
0
<1
for all values ↵ and in the interval [a, b]. Suppose the
initial guess x0 satisfies
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Monday, September 9, 13
Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
|g(↵)
g( )| 
|↵
|,
0
<1
for all values ↵ and in the interval [a, b]. Suppose the
initial guess x0 satisfies
|x0
g(x0 )|  |1
|
b
a
2
⌘ |1
|⇢
6
Monday, September 9, 13
Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
|g(↵)
g( )| 
|↵
|,
0
<1
for all values ↵ and in the interval [a, b]. Suppose the
initial guess x0 satisfies
|x0
g(x0 )|  |1
|
b
a
2
⌘ |1
|⇢
Initial guess is “close
enough” to a root.
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Monday, September 9, 13
Convergence of fixed point iteration
Let g(x) satisfy the Lipschitz condition
|g(↵)
g( )| 
|↵
|,
0
<1
for all values ↵ and in the interval [a, b]. Suppose the
initial guess x0 satisfies
|x0
g(x0 )|  |1
|
b
a
2
⌘ |1
Then
1. All of the iterates will be in [a, b],
|⇢
Initial guess is “close
enough” to a root.
2. The iterates will converge to a fixed point x̄, and
3. The point x̄ is the only root in [a, b].
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Convergence of the fixed point method
1. Proof by induction : Since x1 = g(x0 ), we have that
|x1
x0 |  (1
Assume that for each xk , we have |xk
|xk+1
)⇢  ⇢
x0 |  ⇢. We have that
xk | = |g(xk )
g(xk
1 )|
Use the Lipschitz condition recursively, and our requirement on x0 to get
|xk+1
xk |  |xk
... 
k
xk
|x1
2
1| 
k
x0 | 
|xk
(1
1
xk
)⇢
2|
 ...
Then show that
|xk+1
x0 |  (
= (1
⇢
Monday, September 9, 13
k
+
k 1
k+1
+ . . . + 1)(1
)⇢
)⇢
7
Convergence of fixed point method
2. (existence) First, we have to prove that the sequence
xk is a Cauchy sequence. This allows us to show that
the sequence has a limit in our interval [a, b], i.e.
lim xk = x̄,
k!1
x̄ 2 [a, b]
In fact, we have that
|xk
x̄| 
k
⇢
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Monday, September 9, 13
Convergence of the fixed point method
3. (uniqueness) Suppose we have a second root
[a, b]. Then assuming |↵
|=
6 0, we have
|↵
| = |g(↵)
g( )|  |↵
| < |↵
in
|
This contradiction implies that ↵ = .
Corallary : If |g 0 (x)| 
|g(x0 )
< 1, and
x0 |  (1
)⇢
then the conclusion of the theorem still holds.
Proof : Use the Mean Value Theorem.
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Fixed point iteration
To reformulate a root finding problem of the type f (x) =
0 into a fixed point problem, we can define g(x) =
f (x) + x.
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Monday, September 9, 13
Fixed point iteration
In general, a fixed point iteration can be used to solve
equations of the form :
To reformulate a root finding problem of the type f (x) =
0 into a fixed point problem, we can define g(x) =
f (x) + x.
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Monday, September 9, 13
Fixed point iteration
In general, a fixed point iteration can be used to solve
equations of the form :
g(x) = x
To reformulate a root finding problem of the type f (x) =
0 into a fixed point problem, we can define g(x) =
f (x) + x.
10
Monday, September 9, 13
Fixed point iteration
In general, a fixed point iteration can be used to solve
equations of the form :
g(x) = x
To reformulate a root finding problem of the type f (x) =
0 into a fixed point problem, we can define g(x) =
f (x) + x.
Then
g(x) = x
,
f (x) + x = x
,
f (x) = 0
10
Monday, September 9, 13
Fixed point iteration
In general, a fixed point iteration can be used to solve
equations of the form :
g(x) = x
To reformulate a root finding problem of the type f (x) =
0 into a fixed point problem, we can define g(x) =
f (x) + x.
Then
g(x) = x
,
f (x) + x = x
,
f (x) = 0
There are often several choices one can make to
reformulate the problem. There is no guarantee that
any of them will give a convergent method, however.
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Example
Use fixed point iteration to find the root of the
polynomial
f (x) = x3 + x2
3x
3
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Monday, September 9, 13