Finding ${i^i}^{i\cdots}$

Question

We know that surprisingly enough, $i^i=\frac1{e^{\frac\pi2}}$.

But what about finding the value of ${i^i}^{i\cdots}$? Is it possible?

My attempt: Let $${i^i}^{i\cdots}=x$$ $$i^x=x$$ Or $$\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^x=x$$ $$e^{\frac{\pi}{2}ix}=x$$ But can we take the $\log$ of both sides? Even if we take, $$\frac{\pi}{2}ix=\log x$$ Now how to solve this equation?

Edit: OKay, as pointed by @JackD'Aurizio, that we have to first check whether the given sequence converges or not. But I have no idea for how to check that? (Please answer in elementary terms)

Answer 1

Here is a treatment of the problem that does not make use of the $W$-function.

For $z\in{\mathbb C}$ we choose the following definition of $i^z\>$: $$i^z:=\exp\bigl(z\>{\rm Log}(i)\bigr)=e^{i\pi z/2}\ .$$ We then have to analyze the sequence $(z_n)_{n\geq0}$ defined by $$z_0:=1,\qquad z_{n+1}:=e^{i\pi z_n/2}\quad(n\geq0)\ .$$ The first terms of the sequence are $1$, $i$, $e^{-\pi/2}\doteq 0.207$, all in the set $A:=\overline\Omega$, where $$\Omega:=\{z=x+iy\in{\mathbb C}\>|\>|z|<1, \ x>0,\ y>0\}\ .$$ The well known properties of the exponential map show that the function $$f(z):=e^{i\pi z/2}=e^{-\pi y/2}\cdot e^{i\pi x/2}$$ maps $A$ bijectively onto the domain $B$ shaded red in the following figure, and it maps $\Omega$ onto the interior of $B$, which is a proper subset of $\Omega$. As $B\subset A$, by Brouwer's fixed point theorem $f\restriction A$ has at least one fixed point $z_*\in A$. Since no point of $\partial A$ stays fixed it follows that $z_*\in \Omega$.

The Riemann mapping theorem allows to conjugate $f\restriction\Omega$ with a map $g:\>D\to D$, the unit disk of the $w$-plane, that keeps $0$ fixed. Since $g(D)\subsetneq D$ Schwarz' Lemma then guarantees $|g(w)|<|w|$ for all $w\in \dot D$. This implies that there can be no other fixed point, and more: The sequence $(w_n)_{n\geq0}$ will converge to $0$ (the assumption $\inf_{n\geq0} |w_n|=:\rho>0$ would lead to a contradiction). We therefore can conclude that $\lim_{n\to\infty} z_n=z_*$. Note that $$\bigl|f'(z_*)\bigr|=\bigl|g'(0)\bigr|<1$$ shows that $z_*$ is indeed attractive. I computed the first $200$ points $z_n$ and obtained the estimate $z_*\doteq 0.438283 + 0.360592\> i$. At the fixed point one has $\bigl|f'(z_*)\bigr|\doteq0.89$. This reveals that the convergence is not particularly fast.

Answer 2

You can't, in general, solve equations like $i^x=x$ in closed form using elementary functions.

You can solve this with the Lambert W-function.

Let $z=-\frac{i\pi x}{2}$. Then $ze^z = \frac{i\pi}{2}$. So $z=W(i\pi/2)$ and $$x=-\frac{2}{\pi i} W(i\pi/2) = \frac{2i}{\pi}W(i\pi/2)$$

That solves the equation $i^x=x$, but it doesn't prove that this is the limit.

And the $W$-function is actually multi-valued on the complex plane, so it doesn't give a single value.

Answer 3

Let $f(z) = \exp(\frac {i\pi}2 z)$. We have $|f'(z)| = \frac \pi 2 |f(z)|$.

We would like to find out when a sequence (defined by $z_{n+1} = f(z_n)$) gets stuck in a small disk where $f$ is contractible.

Suppose we want it to get stuck in $B(z,\rho)$.

Let $K = \sup_{|y-z| \le \rho} |f'(y)| = \frac {\pi} 2|f(z)|\sup_{|y|\le \rho} |f(y)| = \frac \pi 2 |f(z)| \exp(\frac \pi 2 \rho)$. First, we need $K \lt 1$.

Then we need $f(B(z,\rho)) \subset B(z,\rho)$. Since we have $f(B(z,\rho)) \subset B(f(z),K\rho)$ it is enough to have $|f(z)-z|+K\rho \le \rho$ hence $K \le 1 - \frac 1 \rho |f(z)-z|$.

so we need to find $\rho$ such that $\frac \pi 2 |f(z)| \exp(\frac \pi 2 \rho) + \frac 1 \rho |f(z)-z| \le 1$

If the sequence does converge then we can expect the possible $\rho$ to range from a $O(|f(x)-x|)$ lower bound to a $O(1)$ upper bound.

So if we pick $\rho = - \frac 1 \pi \log (\frac \pi 2 |f(z)|)$ (half what is needed to make the left term $1$) we should eventually succeed.

It turns out that at the $55$th iteration, this choice of $\rho$ succeeds in giving a proof that the sequence stay bounded in $B(z_{55} = 0.439273\ldots+ 0.361095\ldots i,\rho = 0.036804\ldots)$, which is stable by $f$ and where $f$ is contractible. So it converges to the unique fixpoint of $f$ in this ball.

Answer 4

OKay, as pointed by @JackD'Aurizio, that we have to first check whether the given sequence converges or not. But I have no idea for how to check that? (Please answer in elementary terms)

One way to prove convergence is through fixed point iteration (There are other ways using Shell's and Baker-Rippon's Theorems as well).

We iterate the function:

$$g_c(z)=c^z$$

Using the principal branch of complex exponentiation for $c=i$, it becomes:

$$g_i(z)=i^z=\exp(\ln(i)\cdot z)$$

Solving the equation $g_i(z)=z$, there are infinitely many fixed points of $g_i(z)$ given by:

$$z_k=\frac{W(k,-\ln(i))}{-\ln(i)},\,\,k\in\mathbb{Z}$$

If we check the derivative at the fixed points, we get:

$$|g_i'(z_k)|=|-W(k,-\ln(i))|,\,\,k\in\mathbb{Z}$$

For $k=0$ we get:

$$|g_i'(z_0)|=|-W(-\ln(i))|=\left|-W\left(-\frac{i\pi}{2}\right)\right|\doteq 0.8915\lt 1$$

Therefore the iterates of $g_i(z)$ as defined converge to the limit:

$$h(i)=\frac{W(-\ln(i))}{-\ln(i)}=\frac{2iW(-\pi i/2)}{\pi}\doteq 0.43828+0.36059 i$$

Small addendum (after @mercio's question)

For the iterated exponential sequence $\{c,c^c,c^{c^c},\ldots\}$ we use the following notation, using the principal branch of the complex $\log$ map:

$${^n}c = \begin{cases} c, & \text{if $n=1$} \\ c^{^{n-1}c}=\exp(\ln(c)\cdot ^{n-1}c), & \text{if $n>1$} \end{cases} $$

When examining convergence of sequences of the form: $\{{^n}c\}$, $n\in \mathbb{N}$, it is more convenient to examine convergence of the iterates of $g_c(z)=c^z=\exp(\ln(c)\cdot z)$ instead, i.e. the iterates $g_c^{(n)}(z)$, $n\in \mathbb{N}$: $\{c^z,c^{c^z},c^{c^{c^z}},\ldots\}$, with:

$$g_c^{(n)}(z) = \begin{cases} c^z, & \text{if $n=1$} \\ g_c(g_c^{(n-1)}(z)), & \text{if $n>1$} \end{cases} $$

in order to be able to apply fixed point iteration.

Using the above definitions it follows that the sequence $\{{^n}c\}$ converges if and only if the sequence $\{g_c^{(n)}(1)\}$ does and the limits (if they exist) are going to be the same (by definition), i.e.:

$${^n}c=g_c^{(n)}(1)\Rightarrow$$ $$\lim\limits_{n\to\infty}{^n}c=\lim\limits_{n\to\infty}g_c^{(n)}(1)$$.

Summarizing:

From fixed point iteration we now know that:

$$\lim\limits_{n\to\infty}g_c^{(n)}(z)=h(c),\,\,\,\text{(for all $c,z$ in the basin of attraction)}$$

We also know that $z=1$ is a super-attractor:

$$|-W(-\ln(1))|=0<1,\,\,\text{(1 belongs to the basin of attraction)}$$

Consequently, afortiori:

$$\lim\limits_{n\to\infty}{^n}i=\lim\limits_{n\to\infty}g_i^{(n)}(1)=h(i)$$

Answer 5

Maybe I'm wrong but I'd like to know where. I'm starting from your step $i^x = x$

Consider $i^x = x$ 1. Raise 4th power on both sides. It gives us

$i^{4x} = x^4$

2. It's equivalent to:

$i^{{(4)}x} = x^4$ equals $1^x = x^4$

Therefore, $x = 1^{1/4}$ That means x can be any 4th root of unity.