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

[tommy1729]

The gaussian method idea variant method :

see : https://math.eretrandre.org/tetrationfor...p?tid=1339

R(s) = exp( tr(s) * R(s) )

and tr(s) strictly rises from 0 at s = -oo to 1 at s = + oo in a fast way.

The gaussian method had t(s) = ( 1 + erf(s) )/2.


R(s) uses tr(s) = ( 1 + erf(2sinh(c s)) )/2

For some real c that makes it analytic.

Now 2 sinh(c s) is periodic with period V.

So this V is isomorphic to a function fV(z) such that

fV(exp(x)) = exp(fV(x))

and fV(fV(x)) = x.


***

The interesting part is that this might relate to caleb ideas.

Maybe assuming the periodicity is continuation beyond a boundary, where boundary implies natural boundary ( not analytic ) , not converging or not satisfying the basis equations.

In that case one might get a " fake " solution.
( not to confuse with fake function theory )


regards

tommy1729

If c = 1 works that is ; 

not " fake "

and analytic.

then the 2sinh method might also give a valid solution if the 2sinh is analytic ( since we have the same period !)

the 2sinh method satisfies the semi-group iso and the other one might not ;

resulting in maybe 2 different valid solutions ...

But this would be equivalent to having 2 valid solutions for f(x).

Which bring the question :

apart from existance , do we have a proof of uniqueness for f(x) ??


Remember we only want a local solution for f(x) !!



regards

tommy1729