Does real*real*real... = imaginary? $x\cdot x\cdot x\cdot x\cdot x\ ...\ =\ i, x \in \mathbb{R}$

Question

Please be advised as is pointed out below, the video was incorrect and this: $$ x=e^{\frac{\pi}{2}} \Rightarrow x^{x^{x^{x^{...}}}} = i$$Is completely false!

I recently watched the video by real^real^real^... = imaginary? by blackpenredpen and he shows that this is possible:

$$ x=e^{\frac{\pi}{2}} \Rightarrow x^{x^{x^{x^{...}}}} = i$$

This made me wonder if it is possible to find a similar real number for repeated multiplication rather than exponentiation? $$x\cdot x\cdot x\cdot x\cdot x\ ...\ =\ i, x \in \mathbb{R}$$

My initial thoughts were that repeated multiplication is just exponentiation so maybe we could look at the problem like this:

$$\lim_{n\rightarrow \infty}x^n = i, x \in \mathbb{R} $$

So is this possible? If not it would be nice to see a proof.

Answer 1

The youtube video is wrong.

$x^{x^{x^{x^{...}}}}$ is a so-called power-tower, or hyperpower function, also known as infinite tetration.

It should be noted that the power tower only converges for $x \in (\frac 1{e^e}, e^{\frac 1e})$.

To solve $x^{x^{x^{x^{...}}}} = k$, a non-rigorous "trick" is to write $x^k = k \implies x = k^{\frac 1k}$.

But for $x^{x^{x^{x^{...}}}} = k$ to give the valid solution $x = k^{\frac 1k}$, you must have $k < e$ (based on the radius of convergence I gave above).

So, the solution ($x = \sqrt 2$) is valid for $k=2$, but the solution $x = 3^{\frac 13}$ is not valid for $k = 3$ (as $3 > e$).

Similarly, saying $x^{x^{x^{x^{...}}}} = i \implies x = e^{\frac{\pi}2}$ is simply nonsense. The video is wrong.

An "infinite product" of reals (assuming convergence) has to be real. But it makes little sense to speak of $x\cdot x\cdot x \dots$ because that value is either $0$ for $|x|<1$, $1$ for $x = 1$ and undefined otherwise.