(09/13/2009, 01:34 PM)Gottfried Wrote: then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I)

However, as we see at the f(-1)-entry we needed a correction factor to get the correct real value - but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this...Code:`bl = log(b) // = -2.718...`

x' = log(x/t0)

y'=eschr°-1 (u0^h*eschr(x'))

y = exp(y')*t0

then

f(1) - b = -2.244868624099733 E-60 + 3.153343574801035 E-60*I // order for Eulersum: 1.4 - 0.1*I

f(2) - b°2 = 1.922763821393618 E-13 + 1.897277601645949 E-13*I // order for Eulersum: 2.5-1.2*I

f(-1)*bl - 0 = 4.089597929065041 E-56 + 6.283185307179586*I // order for Eulersum: 1 (direct sum, no acceleration needed)

However, I just tried the (positive h=+0.5) half-iterate (hoping that it may not be required for positive heights to include this additional correction-factor). I got an half-iterate, which -when iterated- reproduced the integer-iterate well, I got about b°0.5(b°0.5(1)) - b ~ 5.0e-18

Code:

`\\ half-iterate, u=log(t0)+2*Pi*I`

U_05=W*dV(u^0.5)*WI; \\ Bell-matrix for halfiterate; stirling-transformation is already included in W and WI (=W^-1)

x=1

ESum(1.7-0.7*I) *dV(log(x/t0)) * U_05 \\ Use complex Euler-order for summation of 1.7-0.7*I, manually optimized

\\ y' = -3.903870288432298 + 1.032427916941400 *I \\ untransformed value as result of matrix-sum

y = exp(y')*t0

\\ y = 0.02725344094115782 - 0.02087339685842267*I \\ transformed, final value

x = y \\ use found value and iterate again

ESum(2.4-0.6*I) *dV(log(x/t0)) * U_05 \\ another Euler-order required

\\ y' = -3.250373950445425 - 4.597158013302573*I \\ untransformed matrix-sum

y = exp(y') *t0

\\ y = 0.06598803584531253 + 5.687696437325240 E-18*I \\ transformed, final value

y - b = -4.706607418746630 E-18 + 5.687696437325240 E-18*I

\\error <1e-17

Seems, one can find at least any solution using regular tetration for this base, however difficult.

What makes me still headscratch is the required correction-term for the one negative height example

[update] I could improve this to err~1e-25 when simply computed the schröder-values separately as described in the mail before. Then I even did not need complex-euler-orders; just for one summation (of four) I needed a small Euler-acceleration of order 1.3 (which is nearly untransformed partial sums)

The above value for the half-iterate is confirmed and has even more stable digits in the partial sums using around 120 to 128 terms.

[/update]

[update 2]

The graph below is only a proposal. Only h=0.5 was used to reproduce the integer iteration so far. Other rational fractions should be verified, too.

[/update 2]

Gottfried

Gottfried Helms, Kassel