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How do I approach the problem?

Q: Let $ \displaystyle z_{n+1} = \frac{1}{2} \left( z_n + \frac{1}{z_n} \right)$ where $ n = 0, 1, 2, \ldots $ and $\frac{-\pi}{2} < \arg (z_0) < \frac{\pi}{2} $. Prove that $\lim_{n\to \infty} z_n = 1$.

vonbrand
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hasExams
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3 Answers3

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One common trick to deal with recurrence equation involving reciprocal is to rewrite $z_n$ as ratio of two sequence $p_n, q_n$ to be determined. Notice,

$$\frac{p_{n+1}}{q_{n+1}} = z_{n+1} = \frac12 (z_n + \frac{1}{z_n}) = \frac12(\frac{p_n}{q_n} + \frac{q_n}{p_n}) = \frac{p_n^2 + q_n^2}{2p_nq_n}$$

We can take $p_{n+1} = p_n^2 + q_n^2$ and $q_{n+1} = 2 p_{n} q_{n}$. From this, we get

$$p_{n+1} \pm q_{n+1} = ( p_{n} \pm q_{n} )^2 \implies (p_{n} \pm q_{n}) = (p_{0} \pm q_{0})^{2^n} $$

Choose $p_0 = z_0$ and $q_0 = 1$, we get $p_n \pm q_n = (z_0 \pm 1)^{2^n}$ and hence:

$$z_n = \frac{(z_0 + 1)^{2^n} + (z_0 - 1)^{2^n}}{(z_0 + 1)^{2^n} - (z_0 - 1)^{2^n}}$$

For $z_0 \in \mathbb{C}\setminus\{0\}$ such that $-\frac{\pi}{2} < \arg( z_0 ) < \frac{\pi}{2}$, we have:

$$|z_0 + 1| > |z_0 - 1| \implies |\frac{z_0-1}{z_0+1}| < 1$$

Let $\alpha = \frac{z_0-1}{z_0+1}$, we get:

$$\lim_{n\to\infty} z_n = \lim_{n\to\infty}\frac{1 + \alpha^{2^n}}{1 - \alpha^{2^n}} = 1$$

A similar argument shows that if you start with a $z_0$ such that $-\frac{\pi}{2} < \arg( -z_0 ) < \frac{\pi}{2}$, $|\alpha|$ will be greater than $1$ and $z_n$ converges to $-1$ instead.

achille hui
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  • I had completely forgotten this trick, if I ever knew it. Very nice, +1. – Julien Mar 17 '13 at 19:04
  • +1 It's especially nice that this handles existence of the limit and evaluation of the limit at the same time. – 2'5 9'2 Mar 17 '13 at 19:07
  • I remember that you have applied this trick before. Still nice to encounter it again. +1 – user1551 Mar 17 '13 at 19:09
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    The most useful place of this trick is not for this. Sometimes, you can use this trick to turn a non-linear recurrence equation on a scalar sequence $z$ into a linear recurrence equation on a vector sequence $(p,q)$. This allow one to use matrix techniques to investigate the behavior of the original sequence. – achille hui Mar 17 '13 at 19:18
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If the limit exists, then $$\lim_{n\to\infty}z_{n+1}=\lim_{n\to\infty}z_n$$ Substitute it in, you can get $$\lim_{n\to\infty}z_n=\pm1$$ Since there is one more constrain on arg$(z)$, you can conclude that the limit must be $1$.

It is not hard to prove the existence of limit.

Assume $|z_n|>1$, then $|\frac{1}{z_n}|<1$. By the triangle inequality, $|z_n+\frac{1}{z_n}|\leq|z_n|+|\frac{1}{z_n}|<2|z_n|$. Therefore, $|z_{n+1}|<|z_n|$.

NECing
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Let $f(z) = \frac12\left(z+\frac1z\right)$. Clearly $f(1)=1$ and $f'(1)=0$. So suppose $z=1+\Delta z$. Then $$ f(z) = 1 +f'(1)\,\Delta z + \text{higher-degree terms in $\Delta z$}, $$ so $f(z)$ is closer to $1$ than $z$ is. You have an attractive fixed point.

Later edit:

Or put it this way: $f'(1)=0$ and $f'(z)$ is close to $0$ when $z$ is close to $1$. In particular, certainly $f'(z)$ is between $\pm1/2$ when $z$ is close enough to $1$. Therefore $f(z)$ is changing less than half as fast as $z$ is changing when $z$ is close to $1$. That means $f(z)$ is less than half as far from $1$ as $z$ is. If you keep cutting the distance in half, the distance approaches $0$ as a limit.

Michael Hardy
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