Solving tricky Knuth Up Arrow Notations

Question

How would I solve something like
$2\uparrow\uparrow n$? when n ≤1?
Or
$2\uparrow^{-2}2$?

Thanks!

Answer 1

Here is a possible answer:

We know that for every positive integer n, m↑↑n = log_m[m↑↑(n+1)]. If we assume the same equation to hold for nonpositive n, we get:

m↑↑0 = log_m(m↑↑1) = log_mm = 1

m↑↑-1 = log_m(m↑↑0) = log_m1 = 0

m↑↑-2 = log_m(m↑↑-1) = log_m0 = -infinity (in other words, it is undefined)

This seems to be the only logical extension of the double-arrow operator to nonpositive integer "exponents".

As for the extension to a "nonpositive number of arrows", it is even simpler:

m↑⁰n = m*n (ordinary multiplication)

m↑^-1n = m+n (ordinary addition)

But the next one is somewhat surprising:

m↑^-2n = (m+n+4)/2 (you might want to try and verify on your own that this is correct, by proving that m+n is indeed equal to this function iterated n times on m)

So the expression in your original question, 2↑^-22, evaluates to (2+2+4)/2 = 4. Of-course, this is hardly surprising, since 2[any-number-of-arrows]2 always evaluates to 4.

Answer 2

I'm not aware of an interpretation with a negative number of arrows, so I'll just address the case where $n \geq 1$.

With up arrow notation, you can "strip away" an arrow and then write $n$ copies of 2, each separated by one fewer arrows than you had previously. In your case, we'd have $$ 2 \uparrow \uparrow n = \underbrace{2 \uparrow 2 \uparrow \cdots \uparrow 2}_{n \text{ copies of } 2}. $$

A single uparrow is just regular exponentiation, so we'll get an exponent tower of 2's having height $n$.

For example, when $n = 3$, this all works out to $$ \begin{align*} 2 \uparrow \uparrow 3 &= 2 \uparrow 2 \uparrow 2\\ &= (2 \uparrow 2) \uparrow 2\\ &= (2^2) \uparrow 2\\ &= 2^{2^2}. \end{align*} $$