San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

Infinitely Iterated Exponentiation

Infinite Exponentiation

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

G = aaa
which can be represented as
G = aG

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

G1/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=2½=√2. Thus

√2√2√2 = 2

Furthermore since (½)² = 1/4

¼¼¼ = 1/2

To verify these relations consider an iterative scheme of the form

Gn = aGn-1

The results of the first 20 iterations are:

1 1
0.25 1.4142135623731
0.707106781186548 1.63252691943815
0.375214227246482 1.76083955588003
0.59442699715242 1.84091086929101
0.438651165125217 1.89271269682851
0.544384414863917 1.9269997018471
0.470162431706881 1.95003477380582
0.52111552337359 1.96566488651732
0.485575976841481 1.9763417544097
0.510098600013408 1.98366839930382
0.49304895344828 1.98871177341395
0.504841387135777 1.99219088294706
0.49665544235752 1.9945944507121
0.502323653394958 1.99625666626586
0.498391957558689 1.99740700114134
0.501115853363732 1.9982034775087
0.499227147304516 1.99875513308459
0.500535987744584 1.99913731011939
0.499628619597765 1.999402118325
0.500257487555813 1.99958562293568
0.499821555076828 1.99971279632964
0.500123703895513 1.99980093549297
0.499914262345387 1.99986202375778
0.500059432345487 1.99990436444334
0.499958806334303 1.9999337115821
0.50002855408854 1.99995405289782
0.499980208205761 1.99996815214924
0.500013718814578 1.99997792487387
0.499990490932778 1.9999846987471

However everything is not as simple as the preceding. For one thing G1/C does not have a single valued inverse.

Thus 41/4=21/2 so the infinite exponentiation of 41/4 does not converge to 4, instead it converges to 2.

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


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 G1/G is also equal to the cube root of 3.

The function G1/G reaches its maximum for G=e, the base of the natural logarithms 2.71828.. At that value of G, G1/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

ζζζ = 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 G1/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)=G1/G suppose the function is represented as G11/G2, then

da/dG1 = (1/G2)G11/G1-1
da/G2 = d(eln(G1)(1/G2)/dG2 = ln(G1)eln(G1)(1/G2)(−1/G2²)
and therefore da/dG = G(1/G -2) − ln(G)G1/G/G²

Setting this equal to zero

G(1/G -2) − ln(G)G1/G/G² = 0
and dividing G1/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 G1/G is then e1/e.

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