Fixed Point Iteration for Solving Equations

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The discussion focuses on applying fixed point iteration to solve the equation x^3 = 2x + 2 to eight decimal places. A user struggles with convergence, trying various initial guesses but consistently obtaining values that oscillate between positive and negative. Another participant points out an incorrect formula and provides a corrected iteration formula, suggesting starting values of x0 = 1 and x0 = 2 for further calculations. The importance of ensuring the function meets fixed point conditions is also highlighted. The conversation emphasizes the need for proper formulation and initial guesses in fixed point iteration methods.
Huumah
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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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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?
 
Don't forget that there are conditions for a function to have a fixed point.

The expression I used is:

<br /> x_{n+1}=\sqrt { \frac { 2\left ( x_n + 1 \right )}{x_n} }<br />

Try with x_0 = 1 and x_0 = 2 and let me know.

J.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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