Cyclic complex functions and uniqueness

[bo198214]

I dont know what you mean, Gottfried, the definition
\( \exp^0(t)=t \)
\( \exp^{n+1}(t)=\exp(\exp^n(t)) \)
uniquely defines \( \exp^n \) for any natural \( n \). This definition is equivalent to if you substitute in the second line
\( \exp^{n+1}(t)=\exp^n(\exp(t)) \).
So the finite iterations are equal and hence also the limit for \( n\to\infty \)!?

But lets continue this discussion of Dmitrii's article in the other thread. As this thread is about cyclic complex functions.

[Gottfried]

I dont know what you mean, Gottfried, the definition
\( \exp^0(t)=t \)
\( \exp^{n+1}(t)=\exp(\exp^n(t)) \)
uniquely defines \( \exp^n \) for any natural \( n \). This definition is equivalent to if you substitute in the second line
\( \exp^{n+1}(t)=\exp^n(\exp(t)) \).
So the finite iterations are equal and hence also the limit for \( n\to\infty \)!?

But lets continue this discussion of Dmitrii's article in the other thread. As this thread is about cyclic complex functions.

Hmm, Henryk -

I don't know how to say more than in the previous msg.

In

lim n->oo \( \exp^{n}(t)=\exp(\exp^{n-1}(t)) \)

we have

\( \exp^{\infty}(t)=\exp(\exp^{\infty}(t)) \)

and we have to evaluate the limit to evaluate the limit...
Perhaps another view makes my problem better visible. Look at D.F.Barrow's (*1) illustration

   

This images suggests a sequence of parameters, indexed from 0 to n, but which we had to evaluate beginning at n instead of 0 according to right associativity of the operation. However, the convention of partial evaluation to the approximate result would be a0, a0^a1, a0^a1^a2,...

Again hmm, I cannot say more. If it isn't sufficient to explain my concern, then I'm rather helpless, and am apparently facing another mystery of math from outside (of the game)...

Gottfried
(*1) D.F.Barrow; Infinite Exponentials;
The American Mathematical Monthly, Vol. 43, no. 3 (Mar., 1936), 150-160

(Sorry, I still appended my reply still in this thread - seemed a bit strange to change the place. You may reorder as you think it is appropriate)

[bo198214]

Gottfried, if the definitions agree on finite arguments then they also agree in the limit, this is a triviality. Everything else is a matter of taste.

If we define
\( \exp^0(x)=x \)
\( \exp^{n+1}(x)=\exp^n(\exp(x)) \)
then of course also
\( \exp^{n+1}(x)=\exp(\exp^n(x)) \)
and vice versa.
If you want, you can prove that by induction.