San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

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·e1/R·e
or, 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

(cos(Φ)ln(R) + Φ·sin(Φ))/R


ln(r) = (cos(Φ)ln(R) + Φ·sin(Φ))/R
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.

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