[AIS] (alternating) Iteration series: Half-iterate using the AIS?
#30
Hmm, perhaps I have made some basic error now, or something is around which I did not understand correctly from the beginning.
It is clear, that the \( \operatorname{asum}(x) \) is 2-periodic, that means \( \operatorname{asum}(x_{-1})=\operatorname{asum}(x)=\operatorname{asum}(x_1)=... \)

The same seems obvious to me for the asum-derivatives in that periods. But I get different values for the first and second derivatives when I simply shift the center by 2 iterations. Here is a numerical protocol, where I use x1=3.2 (just a random value), and then compute the zero'th, first and second derivative:
Code:
.     [       x2=exph(x1,0),      asum(x2),     asum_deriv(x2,1), asum_deriv(x2,2)]      
      %697 = [3.20000000000, -0.00119822450167, 0.0175377529574, 0.000817628416425]        
      
      [      x2=exph(x1,2),      asum(x2),      asum_deriv(x2,1), asum_deriv(x2,2)]        
     %698 = [0.412136407584, -0.00119822450167, 0.0779236358328,   -0.129878114856]

Can someone crosscheck and possibly explain that? Or do I have only a knot in my head?


Well I asked that question also in MSE, and it might be a bit more instructive. Possibly I'm beginning to understand - but still not getting it in all of its consequences. I'll continue this observation/thoughts here later again... See the question in MSE
http://math.stackexchange.com/questions/...d-infinite

Hmm, after some hours thinking about it becomes a bit similar to the situation, when a little kitten finds its own tail first time and begins to run after it in circles... :-) Obviously I must have the answer for this already in my own analytic formulae for the evaluation of the asum and its derivatives

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
RE: Iteration series: Half-iterate using the infinite iteration-series? - by Gottfried - 01/03/2013, 07:07 AM

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