Posts: 9
Threads: 3
Joined: Nov 2012
Halfiterate can only be defined if we define someting in third hyper operation called "power level".
because we always assume these levels as integers
for example 2^x , 2^2^x , etc..
in the example, number 2 is powered to the first power level and also to the second power level.
but we shall study it also to nonintegers level.
this study must be related to third hyper operation (the exponential function) before we jump to fourth hyper opertation (tetration).
if we can find solutions for that, it means halfitrated could be defined.
for me it is clear that:
(n^^x)^(*)(n^^x) = x , while * means "to the half power level"
Posts: 758
Threads: 117
Joined: Aug 2007
12/14/2012, 07:16 PM
(This post was last modified: 03/26/2015, 03:12 PM by Gottfried.)
Here is some explanation in terms of Pari/GPcode, how the serial alternating iteration sums asum(x) can be expressed/computed with the help of power series.
Preliminaries: Globale variable for base, log of the base and dim for dimension. Then the stdcall for the exponential/logarithm with a variable (but integer) heightparameter:
Code: [tb = 1.3, tl=log(tb)]
dim=64 \\ for power series expansion to "dim" terms, and matrixsize
\ps 64
\\ for direct access to the iteration height
exph(x,h=1) = for(k=1,h, x= exp(x*tl)1);for(k=1,h,x=log(1+x)/tl);return(x);
glx = 1.0 \\ for sequential access to the iteration by one step
{nextx(x,h=0)= if (h==0, glx=x;return(glx));
glx=if(h>0,exph(glx,1)
,exph(glx,1));
return(glx); }
First, the formula for the alternating iteration series; this is "asum(x)". This is simply the summation of the iterated exponentials/logarithms.
Because we want to compare this with a true power series solution (via the Carlemanmatrixconcept) where we have to separate this in two parts, we do this here too; the alternating iteration series towards fp0 is asump(x), and that towards the fp1 it is asumn(x). After adding both two segments we have to reduce the joint sum by x because it was doubly accounted for:
Code: asump(x)= sumalt(k=0,(1)^k*nextx(x, k))
asumn(x)= sumalt(k=0,(1)^k*nextx(x,k))
asum(x) = asump(x)+asumn(x)  x
\\ 
Now we create Carlemanmatrices for that sums, that means we shall get power series for that sums. First we need the Carlemanmatrices for tb^x1 developed around fp0 and that developed around fp1. These are the "Carl0" and the "Carl1"matrices.
Then the matrices, which provide the coefficients for the powerseries for the alternating sums, are created using the Neumannseriesexpression for that Carlemanmatrices (CarlAsp,CarlAsn)
Note the small difference in the formula for the CarlAsn compared to that of CarlAsp.
\\ Series and Carlemanmatrix related to fixpoint fp0 (= 0)
Code: coeffs0 = polcoeffs(exp(tl*(x+fp0))1  fp0, dim);coeffs0[1]=0;
print( coeffs0) \\ that statement to display coeffs0 in the gpdialogue
Carl0 = matfromser(coeffs0); \\ Carlemanmatrix for tb^x1 by powerseries around fp0 (=0)
CarlAsp = (matid(dim) + Carl0)^1 \\ Newtonseriesmatrix for Carl0 giving seriescoefficients for "asump" (= the altern. series towards fp0)
coeffs_asp =CarlAsp[,2] \\ that coefficients in a vector, for computation of "asp_mat(xfp0)+fp0" = "asp_mat(x)" because fp0=0
\\ Series and Carlemanmatrix related to fixpoint fp1
coeffs1 = polcoeffs(exp(tl*(x+fp1))1  fp1, dim);coeffs1[1]=0;
print( coeffs1 ) \\ that statement to display coeffs1 in the gpdialogue
Carl1= matfromser(coeffs1); \\ Carlemanmatrix for tb^(x+fp1)1fp1 by powerseries around fp1 (=???)
CarlAsn = (matid(dim) + Carl1^1)^1 \\ Newtonseriesmatrix for Carl1^1 giving coefficients for "asumn" (= the altern. series towards fp1)
coeffs_asn = CarlAsn[,2] \\ that coefficients in a vector, for computation of "asn_mat (xfp1)+fp1"
{asump_mat(x)=local(h,su);
while(abs(xfp0)>0.5, \\ move the argument x for the powerseries towards fp0
su=su+x; x=exph(x,1);
su=sux; x=exph(x,1);
);
x = xfp0;
su= su + sum(k=0,dim1,x^k*coeffs_asp[1+k]) ; \\ evaluate asp_mat at x_h
su= su + fp0/2;
return(su); }
{asumn_mat(x)=local(h,su);
while(abs(xfp1)>0.5, \\ move the argument x for the powerseries towards fp1
su=su+x; x=exph(x,1);
su=sux; x=exph(x,1);
);
x = xfp1;
su= su + sum(k=0,dim1,x^k*coeffs_asn[1+k]) ; \\ evaluate asn_mat at x_(h)
su= su + fp1/2;
return(su); }
asum_mat(x)=asump_mat(x)+asumn_mat(x)x
\\ ==============================================
Let us ignore the integerheight iterations to shift an argument x sufficiently towards the fixpoints, then the formulae for the powerseriessolutions are
where the coefficients p are taken from the coeffs_asparray.
where the coefficients n are taken from the coeffs_asnarray and t denotes the upper fixpoint fp1.
Now we test these functions:
Code: \\ test that at x0=1.0
x0=1.0
[aser=asump(x0),amat=asump_mat(x0),err=aseramat]
[aser=asumn(x0),amat=asumn_mat(x0),err=aseramat]
[aser=asum (x0),amat=asum_mat (x0),err=aseramat]
\\ serial matrixbased difference
%1269 = [0.764698042872, 0.764698042872, 7.30924528016 E132]
%1270 = [0.245204215222, 0.245204215222, 1.43545677556 E118]
%1271 = [0.00990225809450, 0.00990225809450, 1.43545677556 E118]
The errors with power series truncated to 64 terms are far smaller than 1e100. So we should accept this as very good approximation and I think it is a reasonable hypothese to assume it a true analytical solution.
However, for me it is somehow unusual to have two power series to include in the computation, and also the initial "shift" of the xargument towards the resp. fixpoints by integer iterations, such that the power series converge. 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) assigned to...
Here are the first few coefficients for the power series for reference for you, if you want to check this. Column 1 and 2 for the b^x1 resp log(1+x)/log(b)expressions, then column 3 for asump_mat and column 4 for asumn_mat .
Gottfried Helms, Kassel
Posts: 633
Threads: 22
Joined: Oct 2008
12/15/2012, 04:47 AM
(This post was last modified: 12/23/2012, 05:53 PM by sheldonison.)
(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 parigp. 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 E10
+x^23* 5.8592098019216975332064455548172 E11
+x^24* 8.0401538929341986731328293884869 E12
+x^25* 7.3416487850088416632084519252185 E12
+x^26* 1.5722225445945495608665708043910 E13
+x^27* 7.8120308439624867309714945272883 E13
+x^28* 1.5236013728170180703553240190130 E13
+x^29* 6.6440204264056669379101686704698 E14
+x^30* 3.3256155209233008361117491546045 E14
+x^31* 2.7883958953618392771683428904884 E15
+x^32* 5.4048795650959664939329068427242 E15
+x^33* 4.8862748181200020531353685294020 E16
+x^34* 7.2850552648689176555739757750074 E16
+x^35* 1.6903839033767469660939690985276 E16
+x^36* 8.1615817130139369058635852175326 E17
+x^37* 3.2441786112344646197997328754328 E17
+x^38* 7.0117617473585778251979936641602 E18
+x^39* 4.8770242396322354595201016762943 E18
+x^40* 2.7736370929525121569426683591711 E19
+x^41* 6.0867216158279521431841056375529 E19
+x^42* 5.4221678758070323009283591239994 E20
+x^43* 6.1144436818779516533573016247449 E20
+x^44* 1.7898429546152626958145407293999 E20
+x^45* 3.9506967114524661122627881771652 E21
+x^46* 3.3861201509178531139520486671443 E21
+x^47* 1.5559234721851701604010643412265 E22
+x^48* 5.0631042609964385159471991321737 E22
+x^49* 1.1394185585099365557198284259904 E22
+x^50* 6.2904948098171521570692828155407 E23
+x^51* 2.5827247624986099858687811008930 E23
+x^52* 6.2471872957876823012867141466869 E24
+x^53* 4.3587354373751214441358203074525 E24
+x^54* 3.8884884436503892161650861946872 E25
+x^55* 6.0943046646291791424395564521200 E25
+x^56* 1.8439958337928800380369839903195 E26
+x^57* 7.0859034973425997345978358446834 E26
+x^58* 1.1632853102487036751983700912011 E26
+x^59* 6.2601315833494088275748746969203 E27
+x^60* 2.5199563986304743008057629022862 E27
+x^61* 2.2323646504238188509264319710687 E28
+x^62* 4.0557631158394751876949181749196 E28
+x^63* 6.4627958522674708557831244169246 E29
+x^64* 5.3241698056956264676968949735094 E29
+x^65* 2.0886771988617750086971635596413 E29
+x^66* 5.5811667984264629192614685069187 E30
+x^67* 4.0897559936926817027829612691371 E30
+x^68* 3.8119737006650043902694847671351 E31
+x^69* 6.4047286261518541864231624031981 E31
+x^70* 1.1022926017021090185661192397339 E32
}
Posts: 758
Threads: 117
Joined: Aug 2007
12/15/2012, 06:37 AM
(This post was last modified: 12/15/2012, 07:11 AM by Gottfried.)
(12/14/2012, 07:16 PM)Gottfried Wrote: Here is some explanation in terms of Pari/GPcode, how the serial alternating iteration sums asum(x) can be expressed/computed with the help of power series. Just a short, but useful addendum: if we assume identity of the serial comnputation of asum(x) via the Pari/GPsumaltprocedure and that via the Neumannseriesmatrices asum_mat(x) and its power series then one should consider to use that second method as its standard basis. I toyed a bit around with differentating and integrating using the asum(x) and found, that the asum_mat(x) needs only about 1/20 of the computation time, so my example integral needed 80 000 msec with the serial implementation of the asum(x) but only 4 000 msec using the powerseries implementation.
This should also be useful for the computation of the inverse of asum(x) as long as we need to interpolate it by binary search/Newtonmethod, where many function calls are needed.
I think moreover, that this shall prove useful, once we shall step further to analytically continue the range for the x and for the base b outside the "safe intervals" and enter the realms of truly divergent series for the asum(x).
Gottfried
Additional readings:
An early(2008 ) discussion of this method and some of the problems, which we seemingly can resolve now, but also a (very natural) view into regions of bases outside the Eulersummable range for the serial computation of the asum(x) is here http://go.helmsnet.de/math/tetdocs/Tetr...roblem.pdf
An involved discussion (2007) about the ability of the Neumanntype matrix for the asum(x) to represent an analytical continuation for the divergent cases  the matrixansatz was crosschecked against a shankssummation in the range, where the shankssummation was computable: http://go.helmsnet.de/math/tetdocs/Iter...tion_1.htm
Gottfried Helms, Kassel
Posts: 758
Threads: 117
Joined: Aug 2007
(12/15/2012, 04:47 AM)sheldonison Wrote: 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.
Hi Sheldon 
this sounds really great; at the moment (saturday morning, 7 o'clock) it's a bit over my head, but I'll come back to it later. Cool, that you already did the view into the complex plane, that is all much curious...
What about the cosine instead of the sine? You introduced the cosine in the formula: is there some optimization argument for it? I think it shifts the heightzerodefinition from to the position and must be compensated elsewhere in the formulae?
Gottfried
Gottfried Helms, Kassel
Posts: 633
Threads: 22
Joined: Oct 2008
12/15/2012, 01:43 PM
(This post was last modified: 12/15/2012, 02:25 PM by sheldonison.)
(12/15/2012, 06:49 AM)Gottfried Wrote: Hi Sheldon 
this sounds really great ...
What about the cosine instead of the sine? You introduced the cosine in the formula: is there some optimization argument for it? I think it shifts the heightzerodefinition from to the position and must be compensated elsewhere in the formulae?
Gottfried Hey Gottfried,
I think using cosine vs sine is completely arbitrary. has a period of 2, so to switch from cosine to sine requires a shift of 0.5. To generate the function, I needed to empirically calculate a very accurate maximum amplitude for your asum function, so I think that's when I started centering on the cosine(z) maximum amplitude of asum, as opposed to the sin(z) zero crossing of asum. I also noticed the maximum occurs approximately at the midway point between the two fixed points, so I kept working with cosine after that. Here is the equivalent series, shifted by 0.5, for the . If you want the series centered anywhere else, +/2.35i in the complex plane, I can provide that too, with equivalent accuracy.
Code: {asumsine=
5.5313086758882852195782696632367
+x^ 1* 2.0430674012650642389618054781319
+x^ 2* 0.37225775792009989386677189320438
+x^ 3* 0.099658659136533363777013664824123
+x^ 4* 0.056164249815831421690267039207707
+x^ 5* 0.0024341739796340503006223611990660
+x^ 6* 0.0048821136236917176881522235140457
+x^ 7* 0.0013625033138659058581259860969190
+x^ 8* 0.00013860229935952324746169580217982
+x^ 9* 0.00016031197013218517260959597928241
+x^10* 0.000026233716766403365591594754596210
+x^11* 0.0000078397741621291476048490513062752
+x^12* 0.0000041654570665036359076289439137305
+x^13* 0.00000043429845750234876856499448438055
+x^14* 0.00000020220255943526951948556412163547
+x^15* 0.00000010469795440929394604162385515811
+x^16* 0.000000022104640668047825697063550269138
+x^17* 0.0000000050752778367022374662152498065127
+x^18* 0.0000000054268147119071794970623507258578
+x^19* 0.00000000088377028240205109643729838693901
+x^20* 0.00000000053724808383268656860483599781710
+x^21* 2.2695303649482206760147509826254 E10
+x^22* 2.0755300318667016920518969613493 E11
+x^23* 2.8606106192156205211842427520580 E11
+x^24* 2.4115669661617778343439526060037 E12
+x^25* 2.4097061226690732664291635948851 E12
+x^26* 5.3024239886580580464040016659410 E13
+x^27* 1.3352860389778966943126997119541 E13
+x^28* 4.4676066586863015132548102097061 E14
+x^29* 5.6825914165455658434390405670152 E15
+x^30* 4.8592335192576129248775913955152 E16
+x^31* 1.2100990011140146543285064592901 E15
+x^32* 7.6111952883565188094575760424171 E16
+x^33* 4.0350523000744874273361378890307 E16
+x^34* 1.2914109236182899564695839116964 E16
+x^35* 8.9756371000000830488253273137664 E17
+x^36* 1.1989065889716564053858711596068 E17
+x^37* 1.4933359276677119637947130427580 E17
+x^38* 2.6806360741610229880740798420225 E19
+x^39* 2.0268972881933589191981413633055 E18
+x^40* 1.2667505083540508761525831833105 E19
+x^41* 2.3992792073090946823641233544269 E19
+x^42* 2.7280436906032629343367213777381 E20
+x^43* 2.7155583649715904622682398258307 E20
+x^44* 3.3406372025307868881770171768099 E21
+x^45* 3.3583042746061851597680641775725 E21
+x^46* 2.8315393198749754587922156142229 E22
+x^47* 4.9201511971234663439574586786040 E22
+x^48* 1.9937960921207882784670984182728 E23
+x^49* 7.9867782170632207111309500434223 E23
+x^50* 2.8277790598852084735133034329599 E24
+x^51* 1.2866120668308479467449444353412 E23
+x^52* 8.0009133087978621182694530691203 E25
+x^53* 1.9452214442357667538533694067711 E24
+x^54* 1.9749461280904030199079122668892 E25
+x^55* 2.7421549570850430614725716263261 E25
+x^56* 3.8853147981092745620598414279056 E26
+x^57* 3.6832634896487287902937856325878 E26
+x^58* 6.4643562945915383055244186623733 E27
+x^59* 4.8954244803017602327236947884728 E27
+x^60* 9.6822083644359538571851444600130 E28
+x^61* 6.6905004127327106773176137800281 E28
+x^62* 1.3903608890787339354290348605861 E28
+x^63* 9.5633563001034297889480127198725 E29
+x^64* 2.0299634580180042177140372334119 E29
+x^65* 1.4105236281490162329783786025153 E29
+x^66* 3.1114882040622484025316768678442 E30
+x^67* 2.0873558246659861647640896462484 E30
+x^68* 4.9639057430754083874748766760233 E31
+x^69* 3.0323328554893664432512738086894 E31
+x^70* 7.9943649080493773011295509130754 E32
+x^71* 4.2928725325635863410096873706367 E32
+x^72* 1.2676013386130031721900449722132 E32
}
Posts: 1,358
Threads: 330
Joined: Feb 2009
I agree with sheldon.
regards
tommy1729
Posts: 758
Threads: 117
Joined: Aug 2007
(12/15/2012, 01:43 PM)sheldonison Wrote: Hey Gottfried,
I think using cosine vs sine is completely arbitrary.
Hi Sheldon 
Ok, it might have been the case that there was some advantage that I had not seen ... For me the best option seemed to be the sine, because I feel that norming x0 where asum(x0)=0 has some flair of naturality, and then to search left and right from there to the extrema to find the halfiterates alludes much to a sine wave.
One day I must learn to understand the cauchyintegral and riemannmapping subjects  perhaps I can ask you another day for some tutoring? Most often I need only the first keystep into a matter and can then proceed on my own. (However, at the moment it could be I don't get my head free enough, the two classes need some attention and special preparation before the break around christmas and new year, but let's see)
So far only for a short response 
Gottfried
Gottfried Helms, Kassel
Posts: 758
Threads: 117
Joined: Aug 2007
(12/15/2012, 01:43 PM)sheldonison Wrote: Code: {asumsine(x)= 5.5313086758882852195782696632367
+x^ 1* 2.0430674012650642389618054781319
+x^ 2* 0.37225775792009989386677189320438
+x^ 3* 0.099658659136533363777013664824123
(...)
Hi Sheldon 
I just crosschecked your function with my asum_hgh(x)function:
Code: h1 = 0.5
x1 = asumsine(h1)
h_chk = asumhgh(x1)+1 \\ I'd centered h0 to be one period distant from asumsine(0)
h1  h_chk
%1290 = 0E125
perfect match! Extremely cool :)
It's sunday, I'll come back to this perhaps in the evening. Thanks so far!
Gottfried
Gottfried Helms, Kassel
Posts: 633
Threads: 22
Joined: Oct 2008
12/17/2012, 10:19 PM
(This post was last modified: 12/17/2012, 11:01 PM by sheldonison.)
I decided to compare the asum function for tetration base sqrt(2), to the bummer thread on this forum. has an upper fixed point of 4 and a lower fixed point of 2. You can compare this image to the last image in the post I linked to. The base(sqrt(2)) has been discussed a great deal on this forum, so I thought it would be a good comparison study. The differences between these functions are extremely tiny, so the differences are scaled by 10^25 so as to fit on the same plot which shows a little bit of the superfunction going from the upper fixed point of 4 down towards the lower fixed point of 2, with the graph centered at 3.
Once again, the asum super function function is not an average of the superfunctions from the two fixed points. As you can see from the graph, the asum superfunction is much closer to the upper fixed point superfunction, which is entire, than to the lower fixed point superfunction, which is also imaginary periodic in the complex plane, but has logarithmic singularities.
At the real axis near the neighborhood of 3, the asum itself has a very small amplitude, of approximately 7.66E12, as compared to an amplitude of 0.012 for Gottfried's asum(1.3^z1) example. Also, the asum needs a lot more iterations for the sqrt(2) base than for Gottfried's example because of slower convergence towards the fixed points. If you take the asum of the upper fixed point superfunction, than the asum has a period of 2, and the third harmonic harmonic has a magnitude of 1.186E36. The even harmonics of the asum of the superfunction from the fixed point are zero since by definition asum(z)=asum(z+1), and this is only true for odd harmonics, not odd harmonics.
As imag(z) increases towards >8.5*I, the differences between these three functions becomes larger and large in amplitude, becoming macroscopic instead of microscopic. The difference increase with a scaling factor of . At some point, the higher harmonics become visible too, and one expects the three functions to diverge once the upper fixed point superfunction is no longer converging towards 2 as real of z increases. Eventually the asum superfunction is probably dominated by singulariies; but I conjecture the asum superfunction is not imaginary periodic like the two fixed point superfunctions.
 Sheldon
