Last Updated on September 23, 2026 by Maged kamel
Fixed-point iteration.
The objective of our lecture is to understand the following points: What does fixed-point iteration mean? What is the linear-approximation Newton method for root finding? We get x1 usingg fixed-point iteration; if we plug in x1 again, e get x2.
We substitute to get x3, and we repeat the process until the resulting x is the same for successive steps.
If you wish to review the PDF data used in the illustration, please continue reading.
Solved problems accompany all these methods. If we start with x0, then we get an Expression for X1. We substitute our first choice of the Value of x.
Normally, we write the function as y = f(x), but if we want Expression for x, we can put x on the left-hand side and rearrange our equation to obtain an Expression f(x)=0, arranged as follows.
We put X on the left-hand side, so it becomes g(x) in the Expression. The new form will be x_i = g (x_i). We can get a Value for x starting with X =a
Solved example 1 using fixed-point iteration.
Solve the following equation x^3 + 5x = 20 numerically. Answer 3 decimal places.
Start with X0 = 2. Sometimes in the example, the author gives us a starting point, then we rearrange the equation to become as follows:
1-We choose to put x^3 on the left-hand side, so we move 5x to the right with a negative sign.
2- The equation will become xx^3 = 20- 5x; then, for the X Value, we take the cubic root of the equation. So X is the 3rd root of (20-5*x) we call it g(x).
How to get the x1 Value by fixed-point iteration?
3- Our starting Value of x we call it x0=2, this is a Value that he has given us and substituted in the equation, then we get a Value of x1=2.154, the calculation can be viewed from the next equation.
How to get the x2 Value by fixed-point iteration?
For the next point, we call it x2.
4-If we plug that Value here, we getx2;2; then if we substitute tx2x2 Vvalue iitwill become the third root of (20 -10.77) =2.098.
How to get x3 Value by fixed-point iteration?
5- Again, we put theValuee of 2.098 in the Expression of g(x); x3 = we get X3, which is 2.119, and use the Value of X3. We substitute; we get X4; the Value is 111.
We proceed the same with x4; we plug it here. We get X5, X6, X7, and X8. After 7 iterations, the right-hand side is close to 20 for x7 = 2.113.
Table of x values based on the fixed-point iteration method.
The Table shows the different values obtained from the fixed-point iteration; please refer to the following slide image.
These are two graphs; the upper one shows the f(x) function and its intersection with the x-axis. The root is between 2.1 and 2.11 for the function x^3 + 5x = 20.
Using fixed-point iteration creates a new function called g(x); the graph is shown. The intersection of g(x) with the function y=x will give the root Value, which is x7=2.113
Solved example-2 using fixed-point iteration.
The solved example-2. It is required to find the root for x^4-x-10=0; the same procedure that we have adopted for the previous example will be followed. Create a graph of g(x) = (10 + x)^4, with x0 = 4. Plug in to get the Value of x1.
The slide image shows the Table of points (x, g(x)) from x = 4 to x = 1.8555. We are looking for the intersection point between t(x) and y = x, or simply when we plug in a certain Value of x and get the same Value. The point coordinate is (1.85555, 1.8555), obtained after 6 iterations.
The curve is drawn with the x and y axes shown. The zero point will be very close to 1.8555. The slide image shows a Table of x values from x=4 to x=0.50 and the corresponding f(x) values.
We are looking for the intersection point between 1.75 and 2.00. The previous Value of 1.8555, obtained from the fixed-point iteration, will give a y Value close to 0.
This is a Link to download the PDF used for the illustration of this Post.
The next Post is: What is the Linear Approximation method?
A very important source is Holistic Numerical Methods.