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Consider Newton's method on finding the roots of $x^3-x=0$, how to show that $x_n$ converges to $1$ for any $x_0>1/\sqrt{3}$?

My attempt: The Newton's method says $x_{k+1}=x_k-\frac{x_k^3-x_k}{3x_k^2-1}$. I tried to prove $|x_{k+1}-1|/|x_k-1|$ is some positive number less than 1. Thus it is equivalent to show$|\frac{x_k-\frac{x_k^3-x_k}{3x_k^2-1}-1}{x_K-1}|<1$ when $x>1/\sqrt{3}$. But I have difficulty proving this.

Could someone kindly help? Thanks so much!

Tony
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  • It's easier imo to show that $x_k$ is monotone and bounded. Then convergence follows from the monotone convergence theorem. To get you started: write $x_{k+1} = \frac{2x_k^3}{3x_k^2-1}$ and study the function $f(x) = \frac{2x^3}{3x^2-1}$. Show that $f(x) < x$ for $x > 1$ and use this to show that $x_{k}$ is decreasing (to cover the case $1/\sqrt{3} < x_0 < 1$ note that this implies $x_1>1$). – Winther Jan 12 '16 at 01:35
  • @Winther Thanks! I have solved this question. – Tony Jan 12 '16 at 03:34
  • Great. It might be a good idea to write up an answer then. You can also accept it yourself so that this quesiton gets marked as solved. – Winther Jan 12 '16 at 03:36

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