(10/24/2016, 11:50 PM)Gottfried Wrote: By some other idea I investigated the behave near above the real axis, and specifically I looked at the point and found, that the required height to iterate from to is nearly , so

That this is very near suggests, that here equality is intended....

Gottfried,

You're using a real base>exp(1/e). Either base(2) or base(e), so you don't really need the complex base support in tetcomplex. Kneser.gp would work just fine, and as you noted, tetcomplex won't work with the latest versions of pari-gp.

For any real valued sexp base as imag(z) increases, sexp(z) will go to the fixed point, since the Kneser tetration function is a 1-cyclic mapping of S(z) where S(z) is the Schröder solution superfunction from the complex fixed point. By definition,

. In the upper half of the complex plane, we use

and in the lower half of the complex plane we use

generated from the complex conjugate fixed point.

The solution in the upper half of the complex plane approaches

as

increases. This is because the periodic terms in the 1-cyclic

mapping quickly decay to zero as imag(z) increases leaving only the constant term of k. Try plugging in z=2*I into this equation for theta, to see how quickly the periodic terms in

decay.

The

functions are also periodic with a different period that decays more slowly. For base2 the Su(z) Schröder solution superfunction has a period of 5.584 + 1.0542*I, and Su(z) decays to the fixed point as

increases, but the periodic terms in Su(z) decay much more slowly than the

function.

At the real axis

has a really complicated singularity at integer values of z, since the Schröder superfunction solution never takes on the values of 0 anywhere in the complex plane, and yet sexp(-1)=0. So

is only used in the upper half of the complex plane; and its complex conjugate is used in the lower half of the complex plane.

So sexp_2(2*Pi*I) = 0.827783486030388 + 1.56642681578434*I; and the fixed point is 0.824678546142074 + 1.56743212384965*I.

abs(sexp_2(2*Pi*I)-L)=0.00326

abs(sexp_2(10*I)-L)=0.0000572

abs(sexp_2(20*I)-L)=1.101E-9

abs(sexp_2(30*I)-L)=2.105E-14

I hope this helps.