# Find limit of recursive complex sequence

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?

• Take the limit as $n \to +\infty$ of your definition $z_{n+1} = \frac{1}{2}(z_n + |z_n|)$. This gives an equation verified by $\lim z_n$. Commented Jul 21, 2023 at 13:08
• Hint: look at the form of $z_0$. Polar representation works better for this - both $\left|z_n\right|$ and $\arg{z_n}$ behave nicely; you end up with an infinite product to work out (but it's not too bad). Commented Jul 21, 2023 at 13:20
• @Kovomaka if I understood you correctly, solving $lim_{n \to \infty}{z_{n+1}} = lim_{n \to \infty}{z_n}$ doesn't really give anything except $lim_{n \to \infty}{z_{n}} = lim_{n \to \infty}{|z_n|}$, which just means that $z_n$ converges to $x_n$, but I still need to find that $x_n$ Commented Jul 21, 2023 at 13:23
• @ChrisLewis Ok, will try that, thought polar representation won't work here Commented Jul 21, 2023 at 13:24
• @Ercanayan yes, it was correct! somehow I messed up my numerical estimation in standard form, I later used python's built in cmath to properly work with it and got the same answer :) Commented Jul 23, 2023 at 15:13

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 $$R_0 \frac{\sin{t_0}}{t_0}=\frac{12}{\pi}$$

• Interesting, you have a different result to @abezhiko Commented Jul 21, 2023 at 15:53
• You can check: wolframalpha.com/… Commented Jul 21, 2023 at 15:58
• I love both yours and @Abezhiko solutions! Your seems more convinient in terms of final product convergance, which can be proven with simple trigonometric formula. Thank you very much, was able to learn a lot from both of you Commented Jul 21, 2023 at 17:01

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 $$r_\infty = r_0\,\mathrm{sinc}(\theta_0) = 4\cdot\frac{\sin(\pi/6)}{\pi/6} = \frac{12}{\pi}$$.

• Hm. We get a different result. I've checked numerically and am getting $\frac{12}{\pi}$. I don't think the factor of two is right in your sinc function. Here's a numerical verification: wolframalpha.com/… Commented Jul 21, 2023 at 16:00
• I mis-indexed the product actually, I’ll correct it soon. Commented Jul 21, 2023 at 16:26
• Thank you for your solution too! This problem was out of my league, but it is very interesting. It's a little bit unrelated, but do you know any collections of problems of similar kind, so I can practice? I have read couple of calculus books, but there are little to no tasks on complex sequences, especially recursive? :) Commented Jul 21, 2023 at 18:40
• @hr3nk I'm not aware of a book dedicated to complex sequences specifically $-$ even it probably exists. However, they are treated in the same as real sequences for the main part, that is why you can refer to books about discrete mathematics dealing with recurrence relations. Commented Jul 22, 2023 at 6:09