﻿ Infinitely Iterated Exponentiation of a Complex Number

San José State University

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Thayer Watkins
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 Infinitely Iterated Exponentiation of a Complex Number

## 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

However a little manipulation turns it into a seemingly 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

A previous study worked out the analysis when G is a real number It was found that there is convergence if and only if a<e1/e where e=2.721828….

When a and G are complex number the equation a=G1/G is actually two equations. Let G=R·e and a=r·. Then

#### r·iθ = (R·eiΦ1/R·eiΦor, expressed in terms of logarithms ln(r) + iθ = (e−iΦ/R)[ln(R) + iΦ) or, equivalently ln(r) + iθ = (cos(Φ) −i·sin(Φ)/R)[ln(R) + iΦ]

The RHS may be rearranged in terms real and imaginary components as

Thus

#### ln(r) = (cos(Φ)ln(R) + Φ·sin(Φ))/R and θ = (cos(Φ)ln(R) + Φ·sin(Φ))/R

The real component is maximized where d(ln(r))/dR = 0; i.e.,

#### (ln(R)/R² − 1/R² = 0 and that is where ln(R) − 1 = 0' and hence R = e

(To be continued.)

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.