[MSE] f(f(z)) = z , f(exp(z)) = exp(f(z))

[tommy1729]

Tetrational Geometry

The equation \[f(exp(z)) = exp(f(z))\] tells us we want to be looking at tetration. Now add the constraint \[f(f(z))=z\] in the context of tetration. This means we are looking at period 2 tetration. The orange disk in the following image and the red disk in the image after is the area where tetration displays period 2 behavior. See Pseudocircle The number 0 lies in the disk and is period 2 under tetration,
\[0^0=1, 0^1=0\]





\[f(exp(z)) = exp(f(z)) \implies f(z) = \, ^{z+n}e\] which is inconsistent with the complex base being within the pseudocircle.
The question could be generalized to \[f(a^z) = a^{f(z)} \textrm{ and } f(f(z))=z \textrm{ where } a \textrm{ is in the pseudocircle.}\]. Since \[a \ne e\] the question has no solution beyond the trivial solution.

Im sorry, I do not get what you are doing ?

f(exp(z)) = exp(f(z)) does suggest that f being an iteration of tetration base e is a solution.

However not being base e and having f(f(z)) suggest not being an iteration of exp(x).
But it does not show that f(x) does not exist ?

By analogue :

g(h(h(x))) = g(h(x)) = g(x)

does not imply h(x) is an iteration of g.

for instance

h(x) = 1/x

g(x) = (x^4 + 3 x^2 + 1)/(x^3 + x)

Although It is a fact that g(h(x)) =/= h(g(x)).
So the analogue is not perfect.
 
***

I see you could say that 

e F(x) = F(exp(x))

Implies that

- F(x) = F(f(x))

where f(f(x)) = x
exp(f(x)) = f(exp(x))

And therefore 

sexp(x + pi i) = f( sexp(x) )

for a 2pi i periodic sexp at least in a strip containing Real +/- 2pi i.

Im not talking about global solutions , valid everywhere and analytic everywhere.
But local solutions valid somewhere and locally analytic.

***

f(exp(x+2pi i)) = f(exp(x)) = exp(f(x))

so f(x) = f(x + 2pi i) + 2 pi i K.
or not valid or analytic in such a range.

***

f(exp(x)) = exp(f(x))

implies LOCALLY 

ln( f(exp(x)) ) + 2 pi i L = f(x)

All of these observations are based on the invariant of exp(x).

***

Another idea I had is this one :

We know exp(x) = lim (1+ x/n)^n

And every solution f_n(x) to

f_n( (1+x/(n+3))^(n+3) ) = (1 + f_n(x)/(n+2))^(n+2)

exists for every finite positive integer n.

If f_n(x) converges for increasing n, and has a non-zero radius and validity for some x ,  the limit is a solution.

Another idea is truncated taylors

f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_m x^m 

f(f(x)) = x  mod x^n
f(exp(x)) = exp(f(x)) mod x^n
 
for  m/2 < n < m and increasing m.

Assuming the error terms can be neglected and the equation converges.

And perhaps this MOD idea is relevant too ?

https://math.eretrandre.org/tetrationfor...p?tid=1735

Now you may be right and I think sheldon once proved a similar statement.
Maybe I forgot.

Sorry for the dumb question maybe

regards

tommy1729