Fixed Point Iteration for Solving Equations

In summary, the conversation discusses using fixed point iteration to find the solution to an equation and the attempt at solving it. The correct formula for fixed point iteration is provided and the importance of starting with the correct initial value is emphasized. The conversation also mentions the conditions for a function to have a fixed point and suggests trying x0=1 and x0=2 to find the solution.
  • #1
Huumah
28
0

Homework Statement


Apply fixed point iteration to find the solution of each equation to eight correct decimal places

x3=2x+2



The Attempt at a Solution


I have tried to rewrite the equation for in every possible way to solve for one x and pluggin in my guess( have tried -2,-1,0,1,2,3,4)

and finding x1 and then x2 and plugging them all inn seperatly.

rIvNr0a.png


But my answer switches from positive values to negative values and never seems to be converging to the answer which is 1.76929235


I can understand the sample problem but I'm stuck on this problem.
 
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  • #2
Huumah said:

Homework Statement


Apply fixed point iteration to find the solution of each equation to eight correct decimal places

x3=2x+2

The Attempt at a Solution


I have tried to rewrite the equation for in every possible way to solve for one x and pluggin in my guess( have tried -2,-1,0,1,2,3,4)

and finding x1 and then x2 and plugging them all inn seperatly.

rIvNr0a.png
The formula you show above is incorrect. It should be
$$ x_{n+1} = \frac{2(x_n + 1)}{x_n^2}$$
Huumah said:
But my answer switches from positive values to negative values and never seems to be converging to the answer which is 1.76929235I can understand the sample problem but I'm stuck on this problem.

If you start with x0 = 1, what are the next three numbers you get?
If you start with x0 = 2, what are the next three numbers you get?
 
  • #3
Don't forget that there are conditions for a function to have a fixed point.

The expression I used is:

[itex]
x_{n+1}=\sqrt { \frac { 2\left ( x_n + 1 \right )}{x_n} }
[/itex]

Try with [itex]x_0 = 1[/itex] and [itex]x_0 = 2 [/itex] and let me know.

J.
 

1. What is fixed point iteration?

Fixed point iteration is a numerical method used to approximate the root of a given function. It is an iterative process where the value of the function at each iteration is used to calculate the value of the next iteration until a desired level of accuracy is achieved.

2. How does fixed point iteration work?

The fixed point iteration method uses an initial guess for the root of the function and then repeatedly applies a specific formula to find the next approximation. This process continues until the desired level of accuracy is reached or until the method fails to converge.

3. What is the convergence of fixed point iteration?

The convergence of fixed point iteration refers to how quickly the approximations approach the true root of the function. The method is said to have linear convergence if the error decreases at a constant rate, and quadratic convergence if the error decreases at a faster rate.

4. What are the advantages of using fixed point iteration?

Fixed point iteration is a relatively simple method to implement and can be used to solve a wide range of functions. It also allows for the use of initial guesses, which can improve the efficiency of finding the root compared to other methods.

5. What are the limitations of fixed point iteration?

Fixed point iteration may fail to converge if the initial guess is chosen poorly or if the function has complex roots. It also requires the function to be differentiable, which may limit its use in certain cases. Additionally, it may take a large number of iterations to reach the desired level of accuracy, making it computationally expensive.

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