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.

-----

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