What does $i\uparrow\uparrow i$ equal to?(or, it has no result?)

Question

What does $i\uparrow\uparrow i$ (or, i↑↑i) equal to?

• $i+i=2i$
• $ii=-1$
• $i\uparrow i=i^i≈0.20787957635076193$

Answer 1

Maybe some explanation as to what should be done before making sense of $i\uparrow\uparrow i$. As noted in the comments, there isn't a conventional way to define $n \uparrow\uparrow x$ where $n$ is a positive integer and $x$ is a positive rational. How might one try to get around to doing that?

Knuth's uparrow notation is defined for positive integers. The first arrow is simple exponentiation. We can take some inspiration from the positive integers: at some point, we are told that $4^5$, for example, if $4$ multiplied to itself five times. From this point-of-view, though, maybe $4^{1.6344}$ doesn't feel very sensical. How do I multiply $4$ to itself $1.6344$ times?

One way is to develop the notion of exponentiation using roots. Once I determine that I can multiply two real numbers together, I can fast-forward to see that $x^n$ on the interval $(0,\infty)$ is continuous and injective. So it has an inverse function, $x^{1/n}$, which also happens to be continuous. With some more fast-forwarding, I arrive an extension of $x^{n/m}$ and can eventually define $x^y$ for any real number $y$.

A better way might be to develop it using the exponential and logarithm functions. $e^x$ is continuous and injective on $\mathbb R$ so it has a nice continuous inverse function $\ln(x)$. Then I can represent, for a positive real $x$ and any real $y$, $x^y = (e^{\ln(x)})^y = e^{y\ln(x)}$. Now this is really the useful representation to extend to complex numbers.

Let's take a closer look at $i^i$. As pointed out in the comments, there is not one unique value. To see why, consider the fact that $i = e^{i\pi/2} = e^{i5\pi/2}$, for example. Indeed, $i = e^{i*(4k+1)\pi/2}$ for all integers $k$. So then $i^i$ could be $(e^{i \pi /2})^i = e^{-\pi/2} \approx 0.20787957635$. But $i^i$ could also be $(e^{i 5\pi /2})^i = e^{-\pi/2} \approx 0.0003882032$. However, we can pick principal values and generally define a complex principal logarithm, $\mathrm{Log}(z)$. (For more on this, see https://en.wikipedia.org/wiki/Complex_logarithm) At any rate, we can make sense of complex exponentiation this way.

Returning to the double arrow notation, is there a similar development for non-integral values? I know that $4\uparrow\uparrow 3 = 4^{4^4}$, so $4$ raised to itself three times. We first have the question of how to make sense of $n \uparrow\uparrow \dfrac{p}{q}$. When dealing with simple exponentiation, we were able to use roots to make sense of rational exponents in a natural way. Is there an obvious choice here? A good first step might be to establish some (naturally inspired) function $f : (0,\infty) \to \mathbb R$ so that $f(x) = x \uparrow\uparrow n$ is continuous and injective and make use of its continuous inverse, etc. (For example, in the case when $n = 3$ and we have a rational input $p/q$, the function $f$ should output $(p/q)^{(p/q)^{(p/q)}}$.)

Edit: As I should have mentioned before, $x\uparrow\uparrow 2 = x^x$ is not injective on $(0,\infty)$, so this adds the subtlety of the problem of defining $\uparrow\uparrow$ for non-integral values.

If we can manage this for the real numbers, perhaps there could then be a way to find some function akin to $e^x$ to make sense of $z\uparrow\uparrow w$ for complex $z$ and $w$.