Fixed point iteration

Here we focuses on Fixed-Point Iterative Technique for solving nonlinear Equations in Numerical Analysis. It is one of the opened-iterative techniques for finding roots called Fixed-Point of Non-linear Equations.

1. Overview of Fixed Point Iteration
Numerical Analysis
Isaac Amornortey Yowetu
NIMS-Ghana
June 21, 2020

2. Background of Fixed Point Iteration
1
Background of Fixed Point Iteration
2
Steps In Solving Fixed Point Iteration
Graphical Example
Finding the derivative ofg(x)
3
Performing Iteration to nd Solution (approximate)
4
Application of Fixed-Point Iteration
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3. Background of Fixed Point Iteration
Background of Fixed Point Iteration
Fixed point iteration is a method that is used to compute a
determinate xed-point of a specic iterative function. The method
gives rise to sequences ofx0;x1;x2;x3; :::which aimed to converge at
x

.
The functionfmust be continuous in order to arrive at the
determined xed-pointx

. Also, the xed pointx

can be satised if
the given functionf(x)can be expressed asx=g(x).
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 2 / 10

4. Steps In Solving Fixed Point Iteration
Steps In Solving Fixed Point Iteration
Givenf(x) =0
1. Expressf(x) =xg(x) =0
x=g(x)
We are able to generate two functions:
y1=x (1)
y2=g(x) (2)
Thexvalue at whichx=g(x)is the solution of interest
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 3 / 10

5. Steps In Solving Fixed Point IterationGraphical Example
Graphical Example
Figure:
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 4 / 10

6. Steps In Solving Fixed Point IterationFinding the derivative ofg(x)
Take an initial guessx02[a;b]
2. Finding the derivative ofg(x)
jg
0
(x0)j<1
Condition to consider forg(x)
forn=0;1;2; :::
ifjg
0
(x0)j<1:
there will be convergence to certainxn
Hence,g(x)is a good one
else:
xn+1=g(xn)will keep diverging.
Hence, consider choosing a dierentg(x)from step (1)
end if
end for
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 5 / 10

7. Performing Iteration to nd Solution (approximate)
Performing Fixed Point Iteration
Now that we are able to get a goodg(x), we can perform our
iterations.
x1=g(x0)
x2=g(x1)
x3=g(x2)
x4=g(x3)
.
.
.
xn+1=g(xn)
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 6 / 10

8. Performing Iteration to nd Solution (approximate)
Stopping Criteria
Error Formula
Absolute Error=jxx

j
Relative Error=
jxx

j
jxj
Wherex;x

are true-value and approximated-value respectively.
These error formula is not of direct use as the true valuexis not
known.
Commonly Use Stopping Criteria
jxn+1xnj< "
jxn+1xnj
jxn+1j
< "or
jxnxn+1j
jxnj
< "
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 7 / 10

9. Application of Fixed-Point Iteration
Application of Fixed-Point Iteration
Example 1
Find a root ofxe
x
=1 onI= [0;1]using xed-point iteration.
Considering ourf(x) =xe
x
1=0
Solution
We could expressf(x)asx=e
x
y1=xandy2=g(x) =e
x
g(x) =e
x
(3)
g
0
(x) =e
x
(4)
choosex0=0:52[0;1]
jg
0
(0:5)j=j e
0:5
j=0:6065<1:Hence, g(x) is good
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10. Application of Fixed-Point Iteration
Solution Continue...
x1=g(x0) =e
0:5
=0:6065 (5)
x2=g(x1) =0:5452 (6)
x3=g(x2) =0:5797 (7)
x4=g(x3) =0:5601 (8)
x5=g(x4) =0:5649 (9)
# (10)
x100=g(x99) =e
0:5671
=0:5671 (11)
After successive iterations, the approximated value converges
x

=0:5671
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 9 / 10

11. Application of Fixed-Point Iteration
Example 2
Consider the nonlinear equationx
3
=2x+1 with a solution with the
intervalI= [1:5;2:0]using xed-point iteration with initial guess
x0=1:5, nd the approximated root.
Considering ourf(x) =x
3
2x1=0
Solution
We could expressf(x)asx
3
=2x+1()x= (2x+1)
1
3
y1=xandy2=g(x) = (2x+1)
1
3
g(x) = (2x+1)
1
3 (12)
g
0
(x) =
2
3
(2x+1)
2
3 (13)
choosex0=1:52[1:5;2:0]
jg
0
(1:5)j=j
2
3

2(1:5) +1
2
3
j=0:2646<1:Hence, g(x) is good
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 9 / 10

12. Application of Fixed-Point Iteration
Solution Continue...
x1=g(x0) =

2(1:5) +1
1
3
=1:5874 (14)
x2=g(x1) =1:6102 (15)
x3=g(x2) =1:6160 (16)
x4=g(x3) =1:6175 (17)
x5=g(x4) =1:6179 (18)
# (19)
x100=g(x99) =

2(1:6180) +1
1
3
=1:6180 (20)
After successive iterations, the approximated value converges
x

=1:6180
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 10 / 10

13. Application of Fixed-Point Iteration
End
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