An infinite exponentiation is something which is raised to a power which is something raised to a power ad infinitum> Suppose

G = a^aa…
which can be represented as
G = a^G

This equation might seem a puzzlement as to whether it has any solution other than the obvious one of a=1 and G=1. However a little manipulation turns it into a seemiongly trivial problem. The manipulation is to take the G-th root of both sides giving

G^1/G = a

Now if we want a value of a that gives G as a solution we need only take the G-th root of G and we have the answer. For example, for G=2, a=2^½=√2. Thus

√2^√2√2… = 2

Furthermore since (½)² = 1/4

¼^¼¼… = 1/2

To verify these relations consider an iterative scheme of the form

G_n = a^G_n-1

The results of the first 20 iterations are:

However everything is not as simple as the preceding. For one thing G^1/C does not have a single valued inverse.

Thus 4^1/4=2^1/2 so the infinite exponentiation of 4^1/4 does not converge to 4, instead it converges to 2.

Also the infinite exponentiation of the cube root of 3 ; i.e.,

³√^3√…

should give 3 as a result but the iteration scheme does not converge to 3, instead it converges to a value of 2.478.... which is a lower value of G such that G^1/G is also equal to the cube root of 3.

The function G^1/G reaches its maximum for G=e, the base of the natural logarithms 2.71828.. At that value of G, G^1/G is equal to 1.44466786100977…, call it ζ. This is the maximum value of a for which infinite exponentiation has a finite value and

ζ^ζζ… = e

For details see Maximum.

Thus the maximum value an infinite exponentiation can converge to is e and the maximum base for the infinite exponentiation is ζ=1.44466786100977 so this is why the procedure worked for G=2 and G=1/2 but not for G=3.

(To be continued.)

The Maximum Value of G^1/G

The maximum of the function is found by finding the value of G such that the derivative is equal to zero. The derivative is found for a function in which its argument variable appears in more than one place is to get the derivative for the variable in each place it appears treating it as a constant in the other places.

For a(G)=G^1/G suppose the function is represented as G₁^1/G₂, then

da/dG₁ = (1/G₂)G₁^1/G₁-1
and
da/G₂ = d(e^{ln(G₁)(1/G₂)}/dG₂ = ln(G₁)e^{ln(G₁)(1/G₂)}(−1/G₂²)
and therefore da/dG = G^{(1/G -2)} − ln(G)G^1/G/G²

Setting this equal to zero

G^{(1/G -2)} − ln(G)G^1/G/G² = 0
and dividing G^1/G yields
1/G² − ln(G)/G² = 0
multiplying by G²
further reduces it to
ln(G) = 1
which means
G = e

The maximum value of the function G^1/G is then e^1/e.