Find limit of recursive complex sequence

Question

Sequence is defined as follows:

$$z_0 = 4\exp\left(i\frac{\pi}{6}\right)$$

$$z_{n + 1} = \frac{z_n + |z_{n}|}{2}$$

Task is to find $\lim_{n \to \infty}{z_n}$

I've decomposed the task of finding limit of $z_n$ to finding limits of $x_n$ and $y_n$ (real and imaginary parts of $z_n$). For $y_n$ it's quite obviously decreasing to $0$ as $n \to \infty$.

However for $x_n$ it gets complicated. I was able to proof that $lim_{n \to \infty}{x_n}$ exists due to sequence monotonically increasing and being a Cauchy sequence (which is equivalent to being bounded). But I can't find a way to evaluate the limit itself (only numerically with computer, but that doesn't count). Does anybody know how to explicitly find this limit?

Answer 1

To work out the limit, it's easiest to use a polar representation for $z_n$ (as hinted by the form of $z_0$). Let $z_n = R_n e^{it_n}$, where $R_n$ and $t_n$ are real.

Then $$z_{n+1}=\frac{R_n e^{it_n}+R_n}{2} = \frac{R_n}{2}\left(1+e^{it_n}\right)$$

Now, using $\cos{t_n}=2\cos^2{\frac{t_n}{2}}$ a couple of times, we can show $\left|1+e^{it_n}\right|=2\cos{\frac{t_n}{2}}$ and $\arg{\left(1+e^{it_n}\right)} =\frac{t_n}{2}$, so $$R_{n+1}=R_n \cos{\frac{t_n}{2}}$$ and $$t_{n+1}=\frac{t_n}{2}$$

The argument tends to zero as $n$ tends to $\infty$; so in the limit we're only interested in the value of $R_n$.

We have $R_{n+1}=R_0 \prod_{k=0}^n \cos{\frac{t_0}{2^k}}$. This product is equal to $$R_0 \frac{\sin{t_0}}{t_0}=\frac{12}{\pi}$$

Answer 2

If the sequence $z_{n+1} = f(z_n)$ converges, its limit will be a fixed point of $f(z)$, hence the equation $z = \frac{1}{2}(z+|z|)$ and $z = |z|$, so that we know straighforwardly that $z_\infty$ will be real, but we can't do more from there, that is why we will have to solve the sequence explicitly.

In polar coordinates, i.e. $z_n = r_ne^{i\theta_n}$, we get : $$ \begin{cases} r_{n+1}\cos\theta_{n+1} = \frac{1}{2}r_n(1+\cos\theta_n) \\ r_{n+1}\sin\theta_{n+1} = \frac{1}{2}r_n\sin\theta_n \end{cases} $$ Dividing the second recurrence relation by the first one leads to $$ \theta_{n+1} = \mathrm{tan}^{-1}\left(\frac{\sin\theta_n}{1+\cos\theta_n}\right) = \mathrm{tan}^{-1}\left(\mathrm{tan}\frac{\theta_n}{2}\right) = \frac{\theta_n}{2}, $$ and thus $\theta_n = 2^{-n}\theta_0$, with $\theta_0 = \frac{\pi}{6}$, hence $\theta_\infty = 0$ as found above. Combined with either the identity $\sin(2x) = 2\sin(x)\cos(x)$ or $\cos^2\left(\frac{x}{2}\right) = \frac{1+\cos(x)}{2}$, the initial recurrence relations become $r_{n+1} = r_n\cos\theta_{n+1}$, hence $$ \begin{array}{rcl} r_n &=& \displaystyle r_0\prod_{k=1}^n \cos\theta_k \\ &=& \displaystyle r_0\prod_{k=1}^n \cos(2^{-k}\theta_0) \\ &=& \displaystyle r_0\prod_{k=1}^n \frac{\sin(2^{k-1}\theta_0)}{2\sin(2^{-k}\theta_0)} \\ &=& \displaystyle \frac{r_0}{2^n} \frac{\sin(\theta_0)}{\sin(2^{-n}\theta_0)} \\ &=& \displaystyle r_0\frac{\mathrm{sinc}(\theta_0)}{\mathrm{sinc}(2^{-n}\theta_0)} \end{array} $$ where we used the fact that the last product is telescoping. Since $\displaystyle \lim_{n\to\infty} \mathrm{sinc}(2^{-n}\theta_0) = \lim_{x\to0} \mathrm{sinc}(x) = 1$, we end up with $r_\infty = r_0\,\mathrm{sinc}(\theta_0) = 4\cdot\frac{\sin(\pi/6)}{\pi/6} = \frac{12}{\pi}$.