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I need somebody to help me clarifiy the elementary knowledge for tetration
#8
(07/13/2019, 06:49 PM)Ember Edison Wrote: Can you introduce the base for 3 Singularity, 0, 1, eta?
Is all problems come from log?
Why andydude say he can evaluate the base-infinity? He didn't reply to my email.

iterating
  
is congruent to iterating the cleaner problem with fixed point z=0


Now it turns out that there is a formal half iterate solution around 0 for exp(x)-1; but the formal equation doesn't converge!  See Will Jagy's thread on math overflow on the formal power series for f(f(x))=sin(x).  btw, I noticed a couple of our tetration forum folks on this thread, Daniel and Gottfried.

Here are the first 20 terms of the formal half iterate series for exp(x)-1.  half(half(x))~=exp(x)-1
Code:
{half=  x
+x^ 2*  1/4
+x^ 3*  1/48
+x^ 4*  0
+x^ 5*  1/3840
+x^ 6* -7/92160
+x^ 7*  1/645120
+x^ 8*  53/3440640
+x^ 9* -281/30965760
+x^10* -1231/14863564800
+x^11*  87379/24222105600
+x^12* -13303471/7847962214400
+x^13* -54313201/40809403514880
+x^14*  10142361989/5713316492083200
+x^15*  2821265977/7617755322777600
+x^16* -10502027401553/5484783832399872000
+x^17*  1836446156249/5328075722902732800
+x^18*  2952828271088741/1220613711064989696000
+x^19* -1004826382596003137/680288708300220923904000
+x^20* -7006246797736924249/1943682023714916925440000 }
Even though the asymptotic series is eventually divergent, the half iterate itself is analytic (except at zero), and the series can be used to calculate the half iterate arbitrarily accurately!  If z is negative, iterate  to get close enough to zero before evaluating the half iterate series h, and then iterate  the same number of times, and you get the half iterate.

For the other petal of the Leau-fatou flower, do the iteration in reverse order. Iterate  and then calculate the half iterate h, and iterate  the same number of times.   So even though the half iterate series is not analytic at zero, it is useful.  Cool stuff, the parabolic case.

As far as Kneser and base eta, if the limit of Kneser half iterate were the base eta sexp as the base arbitrarily approaches close to eta, the problem would seem to be which petal?  The two petals are different so maybe not.  It is a singularity, where the fixed points switch from real valued to complex valued for bases>eta.  So, as Henryk pointed out, the formal half iterate family of functions is different for bases<eta then it is for bases>eta.  The Koenig's half iterate for bases<eta probably have the lower fixed point goes to the lower petal, and the upper fixed point goes to the upper petal.  Kneser can be analytically extended to bases<eta but it is no longer real valued, so it has a different half iterate.  Henryk's graph shows the change on either side of the parabolic fixed point.
- Sheldon
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RE: I need somebody to help me clarifiy the elementary knowledge for tetration - by sheldonison - 07/13/2019, 09:06 PM

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