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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.

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    $\begingroup$ Given that $i$ has a magnitude of $1$, we would have to conclude that $|x| = 1$. Any other value causes decay to zero or a blowup. But multiplication by any such $x$ represents a fixed rotation by some angle $\theta$. Then $x^n$ represents a rotation by $n\theta$. Sending $n$ to infinity, this clearly has no limit (unless $\theta = 0$ ,$x=1$), so no such $x$ can exist which satisfies your desired property. $\endgroup$
    – Hyperion
    Jun 26, 2020 at 5:10
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    $\begingroup$ If $x=e^{\frac{\pi}{2}}$ then $x^{x^{x^{...}}}$ is definitely NOT $i$. There are few mistakes in that clip. $\endgroup$
    – N. S.
    Jun 26, 2020 at 5:15
  • $\begingroup$ Thanks @Hyperion that is a very intuitive answer! $\endgroup$
    – PMaynard
    Jun 26, 2020 at 5:16
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    $\begingroup$ Using the method shown in that YouTube video, you could show $$1 \cdot 1 \cdot 1 \cdots = a$$ for any complex number $a$ (since $1\cdot a = a$), because all it proposes (with some errors) is that $x^i=i$ is true for $x=e^{\pi/2}$. $\endgroup$
    – Brian Moehring
    Jun 26, 2020 at 5:16
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    $\begingroup$ Next, an interesting video that proves that $\sum _{n=1}^{\infty} = -\dfrac{1}{12}$ :-) $\endgroup$
    – Déjà vu
    Jun 26, 2020 at 5:32

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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.

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    $\begingroup$ Wow I did not expect that... I see now someone has called him out in the comments. That is such a popular video too I wonder why he wouldn't take it down or at least leave a note :( Thanks good catch $\endgroup$
    – PMaynard
    Jun 26, 2020 at 5:17
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    $\begingroup$ To make the mistake in the clip a bit more obvious, maybe emphasize that the clip is argumenting that IF $x^{x^{x^{x^{...}}}}=i$ has a real solution then the solution must be $x=e^{\frac{\pi}{2}}$. But this doesn't necesarily mean this is a solution, one still needs to check that, and it is easy to argue that the LHS is actually $\infty$ in that case.... Ironically, if fixed, the argument in the clip actually shows that there is no real solution to that equation. $\endgroup$
    – N. S.
    Jun 26, 2020 at 5:20
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    $\begingroup$ @PMaynard I can understand that. It after all looks natural to do the replacement. However, within the framework of mathematics, some intuitive things don't work, how much ever we'd like it to. One example is as you've seen above. Of course, the video would not be taken down : it is entertaining even if wrong! $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jun 26, 2020 at 5:21
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    $\begingroup$ @PMaynard I just "called him out" in the comments. :) He is an enthusiastic math poster, but lacks rigour in my opinion. $\endgroup$
    – Deepak
    Jun 26, 2020 at 5:24
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    $\begingroup$ @PMaynard Popularity does not imply quality. Best witness is the famous "equation" $1+2+3+\cdots=-1/12$ which has caused immense damage confusing countless innocient youtube-watchers. Studying the wikipedia-articles or pdf-articles is much more helpful to learn mathematics than watching youtube-videos. $\endgroup$
    – Peter
    Jan 20, 2021 at 15:05

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