Calculating the fractional iterates of \(f(z) = e^z-1\) to reasonable accuracy
#2
Hi James -

 nice to see you're back!   

 I'm giving your procedure a try and to understand the elements of that procedure.  

 I've got 2 points at the moment:

 - the 'init_vartheta' routine, although it is called only once and thus its time-consumption is of marginal importance, can much be improved.
Using the construction of the coefficients in the BETA by Stirling numbers 2.nd kind, the time for the initialization is about
Code:
Limit      matrix  BETA (msecs)
  100         47    812
  200         62   3136
  300         78   7160
  400        109  12840
  500        124  20078
Moreover, they are integer/rational values now and thus exact to arbitrary precision.

- I'm not sure to have understood the procedure in the integral in the 'intnum()' in function 'F()'; there is also a double-substituion. Does the X substitute work as height-argument for the (interpolation of the) height of iteration of \( exp(z)-1 \) ? In the first view it looks for me similar to my ansatz for interpolation of the coefficients in the list of powerseries of increasing (integer) heights, as considered for instance here ? But it seems also different ... so I'm not sure about this.         
For instance, simply using \( X=-1 \) and evaluate I get the following paowerseries in \( z \) :
Code:
Ser(subst(BASE,'x,-1),'z)

%304 = 2.718281828*z + 2.718281828*z^2 + 2.944805314*z^3 + 3.794268386*z^4 + 5.689514883*z^5 + 9.688440945*z^6 + 18.41853923*z^7 + 38.56758979*z^8 + 88.00303333*z^9 + 216.9453279*z^10 + 573.7534217*z^11 + 1618.357771*z^12 + 4844.436029*z^13 + 15324.50278*z^14 + 51039.55211*z^15 + 178405.6437*z^16 + 652621.0682*z^17 + 2492135.375*z^18 + 9912046.967*z^19 + 40978660.41*z^20 + O(z^21)

or

exp(1)*( z + z^2 + 13/12*z^3 + 67/48*z^4 + 1507/720*z^5 + 61589/17280*z^6 + ...)

       and I've no idea what this is exactly.             

I'm confident that - using the replacement of the generating-procedure for the BASE by the S2-matrix-expressions - analoguously one can improve the time-consumption as well as in the above item.
Gottfried Helms, Kassel
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RE: Calculating the fractional iterates of \(f(z) = e^z-1\) to reasonable accuracy - by Gottfried - 11/15/2022, 12:53 PM

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