# Convergence of a recursive sequence $z_{n+1} = \frac{1}{2} ( z_n - \frac{1}{z_n})$

Let $p$ be a complex number. Let $z_0 = p$ and, for $n \geq 1$, define $z_{n+1} = \frac{1}{2} ( z_n - \frac{1}{z_n})$ if $z_n \neq 0$. Prove the following:

i) If $\{ z_n \}$ converges to a limit $a$, then $a^2 + 1 = 0$

ii) If $p$ is real, then $\{ z_n \}$, if defined, does not converge

iii) If $p = iq$, where $q \in \mathbb{R} \backslash \{0\}$, then $\{ z_n \}$ converges.

I have been able to do the first two parts of this (the second is because the sequence would be real, but would have to have a complex limit). I am stuck on the third part, though. Any help would be greatly appreciated.

Thanks

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$a-\frac{a^2+1}{2a}=\frac{a}{2}-\frac1{2a}$. If you're familiar with Newton-Raphson... – J. M. Apr 16 '11 at 17:42
Unfortunately I'm not... – Anon445 Apr 16 '11 at 17:59
Then may I suggest looking it up? :) – J. M. Apr 16 '11 at 18:00

I'm not sure how much help this could provide but anyway. Look that what part i) tells you is that if the sequence $(z_n)$ converges then its limit must be $\pm i \in \mathbb{C}$. So in part ii) you show that if the first term $z_0 = p$ is real then the sequence does not converge because all terms are real so there's no way to get the imaginary $i$.

Now look at part iii). You can show that in that case all terms are purely imaginary and that the imaginary part of all the terms in the sequence has the same sign as that of $q$. For example by induction, if $z_n = i p_n$ for some $p_n \in \mathbb{R} \setminus \{ 0 \}$, then

$$z_{n+1} = \frac{1}{2} \left ( \frac{z^2_n - 1}{z_n} \right ) = \frac{i}{2} \left ( \frac{p^2_n + 1}{p_n} \right ) = i p_{n+1}$$

with $p_{n+1} \in \mathbb{R} \setminus \{ 0 \}$. Thus now you can concentrate on the sequence defined by $p_0 = p$ and

$$p_{n+1} = \frac{p^2_n + 1}{2p_n}$$ This sequence must have limit $\pm 1$, depending on the sign of $p$.

Then I guess is not that hard to just prove that this sequence $(p_n)$ converges by showing that it is monotonic and bounded. I think that you have to distinguish two cases depending if $p>1$ or not. So for example if $p>1$ then every term of the sequence $p_n > 1$ and

$$p_{n+1} = \frac{p^2_n + 1}{2p_n} = \frac{p_n}{2} + \frac{1}{2p_n} < \frac{p_n}{2} + \frac{p_n}{2} = p_n$$

so in this case the sequence is decreasing and bounded below by $1$. The other cases should be similar.

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Got it, thanks. – Anon445 Apr 16 '11 at 19:05
@Mathmo No problem. Glad it helped you =) – Adrián Barquero Apr 16 '11 at 19:10
When $p_0$ is positive, the sequence $(p_n)$ is nonincreasing and bounded below by $1$, after its first step. – Did Jun 30 '13 at 10:25

From the recursion relation we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{z_{n}+i}{z_{n}-i}\right)^2, \forall n\geq1$$ So iteratively, we get, $$\left(\frac{z_{n+1}+i}{z_{n+1}-i}\right)=\left(\frac{p+i}{p-i}\right)^{2^{n+1}}$$ Rearranging, we get, $$z_{n+1}=i\frac{1+w^{2^{n+1}}}{1-w^{2^{n+1}}}$$ where $$w=\frac{p-i}{p+i}$$ When $p=iq$ $$w=\frac{q-1}{q+1}$$ Clearly, $|w|<1$, since $q\ne 0$. Hence $$\lim_{n\rightarrow \infty}w^{2^{n+1}}=0.$$ Thus $$\lim_{n\rightarrow \infty}z_n=1 \hspace{0.6cm}\Box$$

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