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?
Best Answer
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 $$R_0 \frac{\sin{t_0}}{t_0}=\frac{12}{\pi}$$