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Suppose that $x_0$ is sufficiently close to $3$. Which of the following iteration $x_{n+1}$ = $g(x_n)$ will converges to the fixed point $x = 3$ ?

  1. $x_{n+1} = -16 +6x_n +\dfrac{3}{x_n}$

  2. $x_{n+1} = \sqrt{3+2x_n}$

  3. $x_{n+1} = \dfrac{3}{x_n - 2}$

  4. $x_{n+1} = \dfrac{x_n ^2 - 2}{2}$

I am confused what iteration method should be used

Amzoti
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user120386
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2 Answers2

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Hints

They provided the iteration for each choice of $g(x)$.

For each $g(x)$, you want to test the condition:

$$|g'(r)|, ~ r = 3$$

  • If $|g'(r)| < 1$, converges.
  • If $|g'(r)| > 1$, diverges.

You will find that number $2.$ converges to the root, and the others diverge or find a different root.

You should then verify these results numerically.

For example, for $2.$, choose $x_0 = 4$, we get the iterations:

  • $x_0 = 4$
  • $x_1 = 3.31662479036$
  • $x_2 = 3.10374766705$
  • $\ldots$
  • Converges to the root $r = 3$
Amzoti
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A little addition to Amzoti's answer: Of course you will have to check whether the iteration is actually a contraction in an interval around your point of interest. The easiest way for that is (here) to look at the derivative of the function $g$. However, to remain purely on paper you can verify your findings in another way:
If the mapping is a contraction, then it has a unique fixed point $x^*$ and the recursive sequence $x_{t+1}=g(x_{t})$ converges towards that point. Take a limit on both sides of the equality $\Rightarrow x^*= g(x^*)$ and solve for $x*$ to check whether it is indeed your intended fixed point.

Nox
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