When arg 2sinh^[1/2] = pi / 4 we can expect singularities.
2 reasons ( arguments only )
1) ln( 2sinh^[1/2] (exp( pi i /4) A) ) ~ ln 2sinh^[1/2] (-A)
=> { symmetrie of 2sinh } ~ ln ( - 2sinh^[1/2] )
so we get near the branch cut of ln , so it can not be analytic.
2) since we have zero's at re 2sinh = 0 , 2sinh is not close to exp on the imag axis.
Likewise since the curve arg = pi/4 for the half-iterate maps to the imag axis after 2 iterations ( and we get zero's then ) , 2sinh^[1/2] Cannot both be analytic and close to exp^[1/2] at the same time at arg = pi/4 ; so a branch cut / singularity on that curve.
So basically this agrees with sheldon and his plot kinda.
But the idea was that for Some real x > 1 we would get a small radius where it is analytic.
Such as x = 15 , with a radius of 0,00001.
I see no objection to that by sheldon's post.
Assuming i got his idea correct , which i think so because of the plot and the feeling i explained his reasoning here a bit more.
---
It seems the nature of the 2sinh is such that if the functional equation fails around z (even after continuation - if possible ) then so does f(z,n) - f(z,n-1) ~ 0 , implying divergeance and hence :
For real(z) >1 : around z :
Functional equation satisfied IFF analytic.
That might help to understand the above.
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Perhaps not easy to see but this relates to fake function theory.
We had that the fake half-exp also failed the functional equation when arg = pi/4.
The resembleance is crystal clear !
Fake semi exp ~ fake semi 2 sinh near real x> 1.
So i conjecture that the ~ also holds for arg between - pi/4 and pi/4 OR between - pi/8 and pi/8.
the /8 comes from the analogue of argument 2 in the beginning of the post.
It depends on if the fake can distinguish between exp and 2sinh well.
( in particular away from the real axis ).
A few plots like Sheldon did in the fake exp thread would " visually settle this ".
-----
Regards
Tommy1729
Ps : edit : i made Some typo's and oversimplifications.
Basicly to keep things short i mixed Up pi/2 , pi/4 , pi/8 arguments.
Sorry.
But i guess you can still see basically what i meant.
In short for real(z) << 10 and abs arg(z) >= ~ pi/8 all bets are OFF.
Regards
Tommy1729
2 reasons ( arguments only )
1) ln( 2sinh^[1/2] (exp( pi i /4) A) ) ~ ln 2sinh^[1/2] (-A)
=> { symmetrie of 2sinh } ~ ln ( - 2sinh^[1/2] )
so we get near the branch cut of ln , so it can not be analytic.
2) since we have zero's at re 2sinh = 0 , 2sinh is not close to exp on the imag axis.
Likewise since the curve arg = pi/4 for the half-iterate maps to the imag axis after 2 iterations ( and we get zero's then ) , 2sinh^[1/2] Cannot both be analytic and close to exp^[1/2] at the same time at arg = pi/4 ; so a branch cut / singularity on that curve.
So basically this agrees with sheldon and his plot kinda.
But the idea was that for Some real x > 1 we would get a small radius where it is analytic.
Such as x = 15 , with a radius of 0,00001.
I see no objection to that by sheldon's post.
Assuming i got his idea correct , which i think so because of the plot and the feeling i explained his reasoning here a bit more.
---
It seems the nature of the 2sinh is such that if the functional equation fails around z (even after continuation - if possible ) then so does f(z,n) - f(z,n-1) ~ 0 , implying divergeance and hence :
For real(z) >1 : around z :
Functional equation satisfied IFF analytic.
That might help to understand the above.
-----
Perhaps not easy to see but this relates to fake function theory.
We had that the fake half-exp also failed the functional equation when arg = pi/4.
The resembleance is crystal clear !
Fake semi exp ~ fake semi 2 sinh near real x> 1.
So i conjecture that the ~ also holds for arg between - pi/4 and pi/4 OR between - pi/8 and pi/8.
the /8 comes from the analogue of argument 2 in the beginning of the post.
It depends on if the fake can distinguish between exp and 2sinh well.
( in particular away from the real axis ).
A few plots like Sheldon did in the fake exp thread would " visually settle this ".
-----
Regards
Tommy1729
Ps : edit : i made Some typo's and oversimplifications.
Basicly to keep things short i mixed Up pi/2 , pi/4 , pi/8 arguments.
Sorry.
But i guess you can still see basically what i meant.
In short for real(z) << 10 and abs arg(z) >= ~ pi/8 all bets are OFF.
Regards
Tommy1729