- Fixed point iteration.
- Fixed point iteration video.
- Solved example-1 using fixed-point iteration.
- How to get x1 value by fixed-point iteration?
- How to get x2 value by fixed-point iteration?
- How to get x3 value by fixed-point iteration?
- The table indicates the different values based on the fixed-point iteration.
- Solved example-2 using fixed-point iteration.

## Fixed point iteration.

The objective of our lecture is to understand the following points, what is the meaning of fixed-point iteration? What is the linear approximation newton method of root finding? We get x_{1}, using fixed-point iteration, if we plug in x_{1} again we get X_{ 2}.

We substitute we get X_{ 3}, so we will repeat the process until the result of X obtained is the same for successive steps.

### Fixed point iteration video.

If you wish to watch the video on U-tube, here is the link.

If you wish to review the pdf data used in the illustration, please continue reading.

All these methods are accompanied by solved problems. If we start x_{0, }then we get an expression for X_{1}. We substitute our first choice of the value of x.

Normally we write the function as Y is an f(x), but if we want to get an expression for x. We can put x on the left-hand side and readjust our equation by getting an expression f(x)=0, to be arranged as follows.

We put X on the left-hand side so it will become in the expression = g(x). The new form will be x_{i}=g(x). We can get a value to x starting by X =a

### Solved example-1 using fixed-point iteration.

Solve numerically the following equation X^3+5x=20. Give the answer to 3 decimal places.

Start with X_{0} = 2. sometimes in the example, the author is giving us a starting point then we are rearranging the equation to become as follows:

1-We choose to let X ^3 on the left-hand side, so we are sending 5x with a negative sign.

2- The equation will become x^3=20-5x, then for the X value, we take the third root of the equation. So X is the 3rd root of (20-5*x) we call it g(x).

### How to get x_{1} value by fixed-point iteration?

3- Our starting value of x we call it x_{0}=2, this is a value that he has given us and substitute in the equation, then we get a value of x_{1}=2.154, the calculation can be viewed from the next equation.

### How to get x_{2} value by fixed-point iteration?

For the next point, we call it x_{2}.

4-If we plug that value here we get X_{2} then if we substitute the X 2 value will become the third root of (20 -10.77) =2.098.

### How to get x_{3} value by fixed-point iteration?

5- Again we put the value of 2.098 in the expression of g(x), x_{3}=we get, X_{3} which is 2.119, and use the value of X_{3}. We substitute we get X_{4} the value is 2.111.

We proceed the same with x_{4}, we plug it here. We get X_{5} and X_{6} and X_{7} and X_{8}. After 7 iterations, the right-hand side is close to a value of 20 for x_{7}=2.113.

### The table indicates the different values based on the fixed-point iteration.

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 the fixed point iteration created a new function which is called g(x), the graph is shown. The intersection of g(x) with the function y=x, will give the root value, which is x_{7}=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 g(x)=(10+x)^4, the initial point given is x_{0}=4. Plug in to get the value of x_{1}.

The slide image shows the table of points of x from x=4 till x=1.8555 and the corresponding value of g(x). We are looking for the intersection point between this g(x) and y=x, or simply when we plug in a certain value of x we get the same value in y. The point coordinate is (1.85555,1.8555), which is obtained from 6 iterations.

The curve is drawn with the shown x and Y-axis. The zero point will be very close to 1.8555. The slide image shows the table of points of x from x=4 till x=0.50 and the corresponding value of f(x).

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 the Newton-raphson method which is another iteration method.