It started as an exercise for my students. Calculate $i^i$, then $i^{i^{i}}$ and make a conjecture if we follow that pattern.
If we define $u_n=i$ and $$u_{n+1}=i^{z_n}=e^{i\frac{\pi}{2}z_n}$$
Then, naming $a_n$ and $b_n$ the real and imaginary part of $z_n$, we have: $$a_{n+1}+i b_{n+1}=e^{i\frac{\pi}{2}(a_n+i b_n)}=e^{-\frac{\pi}{2}b_n} \left( \cos \left(\frac{\pi}{2}a_n\right) +i \sin \left(\frac{\pi}{2}a_n\right) \right)$$ Thus, we have : $$a_{n+1}=e^{-\frac{\pi}{2}b_n} \cdot \cos \left(\frac{\pi}{2}a_n\right) \qquad b_{n+1}=e^{-\frac{\pi}{2}b_n} \cdot \sin\left(\frac{\pi}{2}a_n\right)$$
Suppose we manage to prove that those sequences converge then if $$\ell = e^{i\frac{\pi}{2}\ell}$$ Naming $W$ (Lambert function) the log product, we obtain : $$-\frac{i\pi}{2}\ell \cdot e^{-\frac{i\pi}{2}\ell}=-\frac{i \pi}{2} \Rightarrow w e^w =-\frac{i \pi}{2} \quad \text{where} \quad w=-\frac{i\pi}{2}\ell $$
$$\ell =\frac{2i}{\pi} W\left(-i\frac{\pi}{2}\right) \approx 0.43828 + 0.36059 i$$
This seems to be ok since the following Python program gives close results:
z = complex(0,1)
from math import exp, cos, sin, pi
l = complex(0.43828271856492007, 0.3605987950798971)
def cexp(z):
a, b = z.real, z.imag
return complex(exp(a) * cos(b), exp(a) * sin(b))
def f(z):
return cexp(complex(0,1) * pi/2 * z)
x, y = [[0], [0], [0]], [[1], [1], [1]]
for j in range(100):
z = f(z)
x[j % 3].append(z.real)
y[j % 3].append(z.imag)
for i in range(3):
plt.plot(x[i], y[i], "o" + "rgb"[i])
One of my brillant student (Esteban Sabatier) remarks this :
The sequence $w_n - \ell$ where $w_n=z_{3n}$ seems to be decreasing in module to $0$ forming a spiral, (it is also the case for $z_{3n+1}-\ell$, $z_{3n+2}-\ell$) but sadly $w_n-\ell$ is not geometric...
My question is how can I prove those sequences converge.
