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[AIS] (alternating) Iteration series: Half-iterate using the AIS?
#13
(12/13/2012, 02:49 AM)sheldonison Wrote:
(12/11/2012, 01:16 AM)Gottfried Wrote: Hi Sheldon -

... Then the empirical observation that (it) is sort of "hybrid", ... and the fractional iteration based on it should then as well be taken as "hybrid" of the developments at the two different fixpoints ....
I have no idea, how, for instance, could this process then be inverted and get a power series (even if only a formal power series)
...its an interesting summation, and probably leads to an analytic abel function via, .

Gottfried,

This new superfunction is intriguing to me, so I generated a Taylor series for it. This alternating sum function seems to be clearly a different solution than anything we've seen before, and it is analytic in the complex plane. I generated a Taylor series for the superfunction generated from the asum inverse abel function, such that . This is the inverse of the abel function generated via the formula I posted earlier for the abel function, . The taylor series is posted below, and you can cut and paste it into pari-gp. The code to generate the taylor series was involved and complicated, but the results below are accurate to about 32 decimal digits. My initial observations are that this function is 7x closer to the upper fixed point than it is to the lower fixed point. It is not at the midway point or the average of the two functions. The nearest singularity appears to be near z=-0.45075+2.5642i, though I don't understand exactly what causes the nearest singularity, but the graphs clearly show the singularity there, and I assume there are probably other singularities nearby.

I only worked with one case, which is also conjugate to tetration base so I was able to use kneser.gp code to generate the two schroder based super functions from the upper and lower fixed points. The algorithm I used to generate the alternating sum superfunction involves generating the inverse of the abel function such that for complex z. Emperically, the amplitude=0.012759644622895086738385654211903016. Start with the equation . Use an iterative search for the inverse asum to find the corresponding value of z, for each individual z in a unit circle in the complex plane. Then I used those results for a cauchy integral to generate the taylor series. My algorithm involved the intermediate step of generating a fourier series for such that . I could post the results separately if interested.
- Sheldon

Code:
{asumseries=
        4.4326681493968770505571112763721
+x^ 1* -2.3127880792234992209576805587190
+x^ 2* -0.14084258491474819923457781223460
+x^ 3*  0.20335747585717425959215788217730
+x^ 4*  0.039275901513416003532439382454487
+x^ 5* -0.017009465238482374843375971269215
+x^ 6* -0.0068287020794466075850802628591481
+x^ 7*  0.00076014644212133151176635681156923
+x^ 8*  0.00086248272210835707107513090941173
+x^ 9*  0.000071520117050006623094494822900916
+x^10* -0.000077629314222049548560922960618619
+x^11* -0.000022505555300052370351182879813873
+x^12*  0.0000035401828994520591869141075019558
+x^13*  0.0000031483563900756350868343285614899
+x^14*  0.00000032157140688878365331284588360017
+x^15* -0.00000028558473156302855917778930719633
+x^16* -0.00000010334120760488941005250412797733
+x^17*  0.000000012943612612189077177828480927098
+x^18*  0.000000015360572479134856021220865544942
+x^19*  0.0000000011036019290515980548941065490902
+x^20* -0.0000000016730425874414457995630327124141
+x^21* -0.00000000036721227769768456656696964685516
+x^22*  1.4162975891198095985445333625383 E-10
+x^23*  5.8592098019216975332064455548172 E-11
+x^24* -8.0401538929341986731328293884869 E-12
+x^25* -7.3416487850088416632084519252185 E-12
+x^26* -1.5722225445945495608665708043910 E-13
+x^27*  7.8120308439624867309714945272883 E-13
+x^28*  1.5236013728170180703553240190130 E-13
+x^29* -6.6440204264056669379101686704698 E-14
+x^30* -3.3256155209233008361117491546045 E-14
+x^31*  2.7883958953618392771683428904884 E-15
+x^32*  5.4048795650959664939329068427242 E-15
+x^33*  4.8862748181200020531353685294020 E-16
+x^34* -7.2850552648689176555739757750074 E-16
+x^35* -1.6903839033767469660939690985276 E-16
+x^36*  8.1615817130139369058635852175326 E-17
+x^37*  3.2441786112344646197997328754328 E-17
+x^38* -7.0117617473585778251979936641602 E-18
+x^39* -4.8770242396322354595201016762943 E-18
+x^40*  2.7736370929525121569426683591711 E-19
+x^41*  6.0867216158279521431841056375529 E-19
+x^42*  5.4221678758070323009283591239994 E-20
+x^43* -6.1144436818779516533573016247449 E-20
+x^44* -1.7898429546152626958145407293999 E-20
+x^45*  3.9506967114524661122627881771652 E-21
+x^46*  3.3861201509178531139520486671443 E-21
+x^47*  1.5559234721851701604010643412265 E-22
+x^48* -5.0631042609964385159471991321737 E-22
+x^49* -1.1394185585099365557198284259904 E-22
+x^50*  6.2904948098171521570692828155407 E-23
+x^51*  2.5827247624986099858687811008930 E-23
+x^52* -6.2471872957876823012867141466869 E-24
+x^53* -4.3587354373751214441358203074525 E-24
+x^54*  3.8884884436503892161650861946872 E-25
+x^55*  6.0943046646291791424395564521200 E-25
+x^56*  1.8439958337928800380369839903195 E-26
+x^57* -7.0859034973425997345978358446834 E-26
+x^58* -1.1632853102487036751983700912011 E-26
+x^59*  6.2601315833494088275748746969203 E-27
+x^60*  2.5199563986304743008057629022862 E-27
+x^61* -2.2323646504238188509264319710687 E-28
+x^62* -4.0557631158394751876949181749196 E-28
+x^63* -6.4627958522674708557831244169246 E-29
+x^64*  5.3241698056956264676968949735094 E-29
+x^65*  2.0886771988617750086971635596413 E-29
+x^66* -5.5811667984264629192614685069187 E-30
+x^67* -4.0897559936926817027829612691371 E-30
+x^68*  3.8119737006650043902694847671351 E-31
+x^69*  6.4047286261518541864231624031981 E-31
+x^70*  1.1022926017021090185661192397339 E-32
}
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Messages In This Thread
RE: Iteration series: Half-iterate using the infinite iteration-series? - by sheldonison - 12/15/2012, 04:47 AM

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