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Solving tetration for base 0 < b < e^-e
#19
(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)
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, 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...

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
So the half-iterate from 1 by regular iteration using fixpoint t0 is y ~= 0.02725344094115782 - 0.02087339685842267*I .

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
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Messages In This Thread
RE: Solving tetration for base 0 < b < e^-e - by Gottfried - 09/13/2009, 07:34 PM

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