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Let $\{u_n\}$ be a sequence defined by:

  • $\alpha \in \mathbb{C}$ (complex number)
  • $u_0 = \alpha$
  • $u_{n+1} = \frac{1}{2} \left( u_n + \lvert u_n \rvert \right)$

Find the limit of $\{u_n\}$.

Note:

  • no issue if $\alpha \in \mathbb{R}$
  • no issue showing the imaginary part goes to $0$
  • I cannot even solve it for $\alpha = i$. Numerically, I can check it converges to $\frac{2}{\pi}$, but even with the result, I still have no idea how to approach this specific example.
  • I tried to look this up on the web for a solution, but searching for sequences is notoriously hard... no luck.

Backstory: I was skimming through by old maths lessons from when I was in university. While I can still solve most exercises, I got stuck on the aforementioned one.

Question: How to solve this?

Riemann
  • 11,801
BadSequence
  • 23

1 Answers1

11

The key idea is to use polar coordinates. Let $f(z)=\frac{z+|z|}2$. If $r\geq0$ and $\theta\in(-\pi,\pi]$, we may compute $$f(re^{i\theta})=\frac{r+re^{i\theta}}2=r\cos\left(\frac{\theta}2\right)e^{i\theta/2};$$ our choice of $\theta$ guarantees that $r\cos(\theta/2)\geq 0$. So, if $(r,\theta)\in[0,\infty)\times(-\pi,\pi]$, the function $f$ maps the point $(r,\theta)$ to $(r\cos(\theta/2),\theta/2)$. As a result, $$f^n(re^{i\theta})=r\cos\left(\frac\theta2\right)\cos\left(\frac\theta4\right)\cdots\cos\left(\frac\theta{2^n}\right)e^{i\frac{\theta}{2^n}}.$$ So, if $u=re^{i\theta}$, the sequence $\{u_n\}$ converges to $$r\prod_{k=1}^\infty\cos\left(\frac\theta{2^k}\right).$$ This is a fairly common infinite product, which has been evaluated a number of times (see here, for example), and the result is $$\lim_{n\to\infty}u_n=r\operatorname{sinc}\theta=\frac{r\sin\theta}\theta=\frac{\operatorname{Im}z}{\operatorname{arg}z},$$ where $\operatorname{arg}$ gives outputs in $(-\pi,\pi]$.

Carl Schildkraut
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