0
$\begingroup$

In class we looked at the following exercise:

Let $a \ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as

$$z_{n+1} = \frac{1}{2} \left( z_n+\frac{a}{z_n} \right)$$

for $n \ge 0$. For which $z_0 \in \mathbb{C}$ is the above sequence well defined and if it is, what is it's limit?

Someone said that it suffices to reduce the above exercise to the exercise below:

Let $z_0 = x_0+iy_0 \ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as

$$z_{n+1} = \frac{1}{2} \left( z_n+\frac{1}{z_n} \right)$$

for $n \ge 0$. Show that if

  1. $x_{0} > 0$ then $\lim_{n \to \infty} \ z_n = 1$.

  2. $x_{0} < 0$ then $\lim_{n \to \infty} \ z_n = -1$.

  3. $x_{0} = 0$ and $y_{0} \ne 0$ then $(z_n)_n$ is not defined or divergent.

I do not understand how such a reduction should work. Could you please explain that to me?

Note: Here I posted major parts of my solution to the second exercise.

$\endgroup$
1

1 Answer 1

4
$\begingroup$

Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence

$$z_{n + 1} = \frac 1 2 \left(z_n + \frac{1}{z_n}\right)$$

and define a new sequence $w_n := \sqrt{a} z_n$. Notice that

$$\frac 1 2 \left(w_n + \frac{a}{w_n}\right) = \frac 1 2 \left(\sqrt a z_n + \frac{a}{\sqrt a z_n}\right) = \frac {\sqrt a} 2 \left(z_n + \frac{1}{z_n}\right) = w_{n + 1}.$$

From here, the reduction is immediate.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .