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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#29
Tommy1729, thank you for bringing this thread back to our/to my attention!

With my current knowledge I'm pretty sure, that my challenge for the alternating sum at the time of the original post was most likely false; a (still only "most likely") more meaningful version uses now the "polynomial method" for the tetration for bases outside the Shell-Thron-region, thus for instance of all real bases .

The results using this method gives the following table for some bases b and the argument x=I :
Code:
_
  base           log(base)           AS(base,I)                
  1.1000000  0.095310180   -0.45069485+0.91385251*I
  1.2000000   0.18232156   -0.40503707+0.85093158*I
  1.3000000   0.26236426   -0.36504196+0.80486309*I
  1.4000000   0.33647224   -0.33093412+0.77025573*I
  1.5000000   0.40546511   -0.30187962+0.74327224*I
  1.6000000   0.47000363   -0.27685754+0.72148441*I
  1.7000000   0.53062825   -0.25501392+0.70337461*I
  1.8000000   0.58778666   -0.23569667+0.68796239*I
  1.9000000   0.64185389   -0.21841589+0.67458888*I
  2.0000000   0.69314718   -0.20280135+0.66279493*I
  2.1000000   0.74193734   -0.18856952+0.65225041*I
  2.2000000   0.78845736   -0.17550012+0.64271115*I
  2.3000000   0.83290912   -0.16341960+0.63399285*I
  2.4000000   0.87546874   -0.15218883+0.62595235*I
  2.5000000   0.91629073   -0.14169584+0.61847841*I
  2.6000000   0.95551145   -0.13184894+0.61147999*I
  2.7000000   0.99325177   -0.12257089+0.60488549*I
  2.8000000    1.0296194   -0.11380069+0.59863552*I
  2.9000000    1.0647107   -0.10548489+0.59267760*I
  3.0000000    1.0986123  -0.097576064+0.58697104*I
  3.1000000    1.1314021  -0.090038424+0.58148334*I
  3.2000000    1.1631508  -0.082843993+0.57618212*I
  3.3000000    1.1939225  -0.075964723+0.57103646*I
  3.4000000    1.2237754  -0.069372706+0.56602343*I
from where my challenge should have been:
Code:
AS(2,I) =  -0.20280135+0.66279493*I

I haven't (in the year of the above posts) been aware of the problem of the attracting and repelling fixpoints and the sensitivity of the "Alternating Series with increasing heights" (which I call now "alternating iteration series") around which fixpoint the series is developed. Assuming that the "polynomial method" for tetration approximates (and generalizes) the Kneser-solution and also assuming that the Kneser-solution is the best for real tetration the AS()-function should be based on that "polynomial method".

The Pari/GP-procedures are even simpler than that of the original posting;
Code:
n=32  \\ constant gives dimension for matrices
default(realprecision,200)  \\ internal real-arithmetic computation uses 200 digits
default(format,"g0.12") \\ show 12 digits in user-interface
{ASinit(b)=local(a);   \\ define the matrix and coefficients for powerseries
  a=log(b);
  B = matrix(n,n,r,c, (a*(c-1))^(r-1)/(r-1)!);  \\ the carlemanmatrix
  C = matsolve(matid(n)+B,matid(n)[,2]);    \\ vector of coefficients
  return("ok");}
AS(x) = sum(0,n-1, x^k*C[1+k])  \\ the value for the (truncated) and approximate powerseries AS(base,x)



Note, that the coefficients in C form the (head of a) powerseries, which seems to have nonzero (and surprisingly interesting) range of convergence even if base b is outside of the Shell/Thron (resp Euler)-range!

After that, do the computation:
Code:
base=2
ASinit(base)
%486 = "ok"

AS( 1 )      \\ consider only x for which the truncated powerseries is convergent
%487 = 0.28740870

AS( I )   \\ reproduce the new result (if n is at least 32  )
%488 = -0.20280135 + 0.66279493*I

Note finally that for bases inside the Euler-interval one can use the builtin Pari/GP-function "sumalt" to simply do a summation for the (then conditionally converging) alternating iteration series and use this to crosscheck the AS(x)-evaluation taken by the "polynomial method" for tetration.

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 04/12/2014, 01:10 AM

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