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Consider the dynamical system defined by the iteration of the map $$f(x) = \frac{x}{2} + \frac{2}{x}.$$ Prove that for all initial condition $x_0 \in [2,\infty)$, we have: $$(1) |f^{n}(x_0) - 2| \leq \frac{1}{2^n} |x_0 - 2|, \forall n \geq 0$$

Proof: Suppose (1) is true for certain $n>0$. $$|f^{n+1}(x_0) - 2| = |x_{n+1} - 2| = |\frac{x_n}{2} + \frac{2}{x_n} - 2| \leq |\frac{x_n}{2} - 2| + |\frac{2}{x_n}| = |\frac{1}{2}(x_n - 4)| + |\frac{2}{x_n}| = \frac{1}{2}|x_n - 4| + |\frac{2}{x_n}| \leq \frac{1}{2}|x_n - 2| + \frac{1}{2}|2 - 4| + |\frac{2}{x_n}| \leq (\frac{1}{2})(\frac{1}{2^n})|x_0 - 2| + \frac{1}{2}|2 - 4| + |\frac{2}{x_n}| = (\frac{1}{2^{n+1}})|x_0 - 2| + 1 + |\frac{2}{x_n}|$$

The problem is that I don't know if I made some mistakes throughout my calculations or if it is possible to get rid of: $1 + |\frac{2}{x_n}|$.

Albelaski
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2 Answers2

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There is no mistake in the calculation. But due to the application of the triangle inequality your estimates are not good enough to prove the desired conclusion.

If you write $$ x_{n+1} - 2 = \frac {x_n}2 + \frac 2{x_n} - 2 = \frac{(x_n-2)^2}{2x_n} $$ instead then you can conclude that $$ x_n \ge 2 \text{ for all } n \ge 0 $$ which then implies $$ 0 \le x_{n+1} - 2 \le \frac 12 (x_{n} - 2) $$ and finally $$ | x_{n+1} - 2| \le \frac{1}{2^n} |x_0 - 2| \, . $$

Martin R
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Another way to prove this inequality.

First, notice that, for all $x \geq 2$, $$f'(x) = \frac{1}{2} - \frac{2}{x^2} \geq 0$$

so $f$ increases on $[2, +\infty)$. Because $f(2)=2$, you have $f([2, +\infty)) \subset [2, +\infty)$. By induction, it is easy to show that $(x_n)$ is well defined and that $x_n \in [2, +\infty)$ for each $n$.

Now, let's apply mean value theorem to $f$, between, $2$ and $x_n$. Because $f'(x) \leq \frac{1}{2}$ for all $x \geq 2$, you get $$|f(x_n)-f(2)| \leq \frac{1}{2}|x_n - 2|$$

i.e. $$|x_{n+1}-2| \leq \frac{1}{2} |x_n - 2|$$

From this inequality, a short induction shows that $$|x_n - 2| \leq \frac{1}{2^n}|x_0-2|$$

TheSilverDoe
  • 29,720