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 of g(x)
3 Performing Iteration to find Solution (approximate)
4 Application of Fixed-Point Iteration
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 2 / 10

3. Background of Fixed Point Iteration
Background of Fixed Point Iteration
Fixed point iteration is a method that is used to compute a
determinate fixed-point of a specific iterative function. The method
gives rise to sequences of x0, x1, x2, x3, ... which aimed to converge at
x∗
.
The function f must be continuous in order to arrive at the
determined fixed-point x∗
. Also, the fixed point x∗
can be satisfied if
the given function f (x) can be expressed as x = 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
Given f (x) = 0
1. Express f (x) = x − g(x) = 0
x = g(x)
We are able to generate two functions:
y1 = x (1)
y2 = g(x) (2)
The x − value at which x = 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 Iteration Graphical Example
Graphical Example
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 4 / 10

6. Steps In Solving Fixed Point Iteration Finding the derivative of g(x)
Take an initial guess x0 ∈ [a, b]
2. Finding the derivative of g(x)
|g (x0)| < 1
Condition to consider for g(x)
for n = 0, 1, 2, ...
if |g (x0)| < 1:
there will be convergence to certain xn
Hence, g(x) is a good one
else:
xn+1 = g(xn) will keep diverging.
Hence, consider choosing a different g(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 find Solution (approximate)
Performing Fixed Point Iteration
Now that we are able to get a good g(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 find Solution (approximate)
Stopping Criteria
Error Formula
Absolute Error = |x − x∗
|
Relative Error =
|x − x∗
|
|x|
Where x, x∗
are true-value and approximated-value respectively.
These error formula is not of direct use as the true value x is not
known.
Commonly Use Stopping Criteria
|xn+1 − xn| < ε
|xn+1−xn|
|xn+1|
< ε or |xn−xn+1|
|xn|
< ε
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 of xex
= 1 on I = [0, 1] using fixed-point iteration.
Considering our f (x) = xex
− 1 = 0
Solution
We could express f (x) as x = e−x
y1 = x and y2 = g(x) = e−x
g(x) = e−x
(3)
g (x) = −e−x
(4)
choose x0 = 0.5 ∈ [0, 1]
|g (0.5)| = | − e−0.5
| = 0.6065 < 1. Hence, g(x) is good
Isaac Amornortey Yowetu (NIMS-Ghana) Overview of Fixed Point Iteration June 21, 2020 8 / 10

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 equation x3
= 2x + 1 with a solution with the
interval I = [1.5, 2.0] using fixed-point iteration with initial guess
x0 = 1.5, find the approximated root.
Considering our f (x) = x3
− 2x − 1 = 0
Solution
We could express f (x) as x3
= 2x + 1 ⇐⇒ x = (2x + 1)
1
3
y1 = x and y2 = g(x) = (2x + 1)
1
3
g(x) = (2x + 1)
1
3 (12)
g (x) =
2
3
(2x + 1)
−2
3 (13)
choose x0 = 1.5 ∈ [1.5, 2.0]
|g (1.5)| = |2
3
2(1.5) + 1
−2
3
| = 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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