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05/26/2008, 07:43 AM
(This post was last modified: 05/26/2008, 08:47 AM by Ivars.)
Gottfried Wrote:[quote=Ivars]
[quote=Gottfried]
Hmm, for 1+I1I+1+I1I+.... I get 1/(1I) = 1/2 + 1/2I by the formula just invoking Pari/GP on that, and also 1/2 + 1/2I by conventional Eulersummation.
0 + Ix +0x^2  Ix^3 + 0 x^4 + I x^5 .... = II+II+..= Ix *(1/(1+x^2)) > 1/2 I
or
I  Ix + Ix^2  Ix^3 .... = I*(1/(1+x)) > 1/2 I
Gottfried
That seems logical. Thanks, I forgot the trick with generating series where x>1. Instead, and BTW, why can't it go to I as well (x>I) ? Certainly putting x=I will also lead to some sort of summation, albeit slightly different generating series. E.g.
0 + x 0x^2 + x^3  0 x^4 + x^5 .... = II+II+.. = x* (1/(1x^2)) >x=I> I*(1/(1+1))= 1/2* I
But also I/2= sin(I*ln(GoldenMean)) = sin(I*ln(1+0.618033..)= sin (I*( 1+0.618+0.618^2/2+0.618^3/3+0.618^4/4+...+0.618^n/n) = sin (I+I*0.618+I/2*0.618^2+I/3*0.618^3+..+I/n*0.618^n+..) > plus extend sinus in series.
See here Formula 26.
while I try to understand increasing height series, what would be Your take on harmonic imaginary series Sum : I/n? Or easier case, I/n^2?
I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?
so b[4]n= b^b^b^b... =(a1*b)^1*(a2*b)^2*(a3* b)^3*(a4* b)^4*...*(an*b)^n = ?= c1*b^1+c^2*b^b+c^3*b^b^b+..?
You probably have a more clear view on that.
Ivars
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05/26/2008, 09:01 AM
(This post was last modified: 05/26/2008, 09:03 AM by Gottfried.)
Ivars Wrote:while I try to understand increasing height series...
I think, that is not too difficult. Just take a base, where the trajectory converges (for instance b=sqrt(2)) and alternate sum the points of the trajectory. (In pari/gp there is the sumalt()function).
The problem is then only, if b>e^(1/e). Then the trajectory begins to strongly diverge  but for a certain range of b you may still be able to apply one common summationmethod, for instance the Shanksmethod.
But while even this method allows bases a little bit greater, the diagonalization method seems to be able to sum that to arbitrary bases...
:cool:
Best is, you try with convergent example  just add the points of your trajectory in the listmatrix with alternating signs (even Cesarosum should be sufficient for this)
Quote:I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?
Yes, sometimes I was speculating about this. But I doubt it is of any use... Infinite productseries can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...
Gottfried
Gottfried Helms, Kassel
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05/26/2008, 10:30 AM
(This post was last modified: 05/27/2008, 11:42 AM by Ivars.)
Gottfried Wrote:Infinite productseries can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...
Gottfried
Hmm. But if we have tetration of , then we can continue taking logarithms forever, reducing all heights? No.
Then my beloved just ends where everything else:
(both branches), where is Lambert W function.
So would than divergent products (and even some sums) similarly to divergent tetration end up in complex plane?
Ivars
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05/26/2008, 01:11 PM
(This post was last modified: 05/27/2008, 11:41 AM by Ivars.)
If we continiue like this, then we can check:
tetration of
Here we have : Interestingly, the same logarithms appeared here where I changed the issue by always using module of ln(x):
Iterations of ln(mod(x))
Anyway, if we take logarithms of negative numbers via complex numbers, we again end at :
Everything ends at that point. But also infinitely iterated logarithm of any number AT ALL if allowed to go into complex number when it first gets negative seems to end at this point(s). Is it true, or just oscillations die out so fast that all numbers give the same result... Anyway, why exactly close to W(1) which reminds me of Eulers Lambert function W(z)? His work about it is probably only in Latin.
Jaydfox reffered to these 2 fixed points here: Imaginary Iterates of exponentiation
This number and are kind of symmetric since they are solutions of equations:
and
respectively.
Also,
Ivars
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05/27/2008, 08:10 AM
(This post was last modified: 05/27/2008, 08:11 AM by Ivars.)
Gottfried Wrote:Refering to the header of this thread and your obvious love to iteration and trajectories, I challenge you to compute the alternating sum of all the intermediate steps of the trajectory
I, 2^I, 2^(2^I), 2^(2^(2^I)...
AS(2,I) = I  2^I + 2^2^I  2^2^2^I + ... ... = ??
My proposal is
AS(2,I) = 0.440033096027+0.928380628227*I
What do you think?
Gottfried
I can only say it diverges, and I understand You want me to find a method to sum it and a result. Unfortunately, I can not do it yet, but I am interested in 2^I in general and divergent series as well as they might be doing the same as tetrationsomehow sending real numbers to complex domain , though the example I made with infinite logarithms ended in a kind of disaster as every base b ( except h(0) and h(1)) atracted to single value W(1). So let us keep Your result in mind while learn the basics in my strange waysbackwards
Ivars
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05/27/2008, 09:23 AM
(This post was last modified: 05/27/2008, 10:12 AM by Gottfried.)
Ivars Wrote:Gottfried Wrote:AS(2,I) = I  2^I + 2^2^I  2^2^2^I + ... ... = ??
My proposal is
AS(2,I) = 0.440033096027+0.928380628227*I
What do you think?
Gottfried
I can only say it diverges, and I understand You want me to find a method to sum it and a result. (...)
Ivars
Well, diagonalization seems to give an answer here.
Remember, in my matrixnotation, we have
or more general
and using digonalization
Now the alternating sum is (using the small letter "i" for the imaginary unit to prevent confusion with "I" as the identity matrix)
where we have to determine the entries of the diagonalmatrix X. But they can all be determined by the rule of geometric series:
...
(Note, that we need no fractional powers, so the multiplications in the exponents are commutative and can be reordered)
Then
which can be described by
// "I" means here the identitymatrix
so the eigenvalues of
are not a vandermondevector, btw.
In most short form we may simply write, again invoking the identities of diagonalization
// [,1] means: use column 1 (the second) only for coefficients
The nice aspect now is, that even for b>e^(1/e), or abs(u)>1, which leads to divergent trajectories, the eigenvalues of the matrix (I+Bb)^1 form a convergent sequence (well, I've to check for the precise range) and AS() gives a reasonable result.
I've given a plot for some bases, where x=1 (instead of I as in the case here), which compares the (cesaroand) Eulersummable bases and the Shankssummable bases with values of the diagonalizationsummable bases, where the diagonalization extends the summablity to arbitrary high bases. (Note: this is an old picture, for instance I used "s" for base and "matrixmethod" for "diagonalization")
[attachment=77]
The result looks pretty smooth...
Because of the nice eigenvalueconfiguration, you may arrive at these results even without fixpointshifts and expression by the Utmatrix; simply invert using an empirical Bbmatrix I+Bb of reasonable size, say 32x32 or 64x64, to get good approximations.
Gottfried
Gottfried Helms, Kassel
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05/27/2008, 11:45 AM
(This post was last modified: 05/27/2008, 11:52 AM by Ivars.)
I corrected the mistakes of taking logarithms from h(z) in previous posts.
But one could still ask if there exists transformation H such that :
Then
would always be a divergent sum, related to h(z). Instead of all z being similar, one can have e.g z=2^n or other general term, then these sums will coincide with usually known divergent sums like 1+2+4+8+16+32........ or 1+1+1+1+1+1....
Similarly 1/h(z) could be transformed in z/z/z/z/z/z/z.......... and sums would be zzzzzz...?
Ivars
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Im fascinated by this thread...
regards
tommy1729
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04/12/2014, 01:10 AM
(This post was last modified: 04/12/2014, 11:32 AM by Gottfried.)
Tommy1729, thank you for bringing this thread back to our/to my attention!
With my current knowledge I'm pretty sure, that my challenge for the alternating sum at the time of the original post was most likely false; a (still only "most likely") more meaningful version uses now the "polynomial method" for the tetration for bases outside the ShellThronregion, thus for instance of all real bases .
The results using this method gives the following table for some bases b and the argument x=I :
Code: _
base log(base) AS(base,I)
1.1000000 0.095310180 0.45069485+0.91385251*I
1.2000000 0.18232156 0.40503707+0.85093158*I
1.3000000 0.26236426 0.36504196+0.80486309*I
1.4000000 0.33647224 0.33093412+0.77025573*I
1.5000000 0.40546511 0.30187962+0.74327224*I
1.6000000 0.47000363 0.27685754+0.72148441*I
1.7000000 0.53062825 0.25501392+0.70337461*I
1.8000000 0.58778666 0.23569667+0.68796239*I
1.9000000 0.64185389 0.21841589+0.67458888*I
2.0000000 0.69314718 0.20280135+0.66279493*I
2.1000000 0.74193734 0.18856952+0.65225041*I
2.2000000 0.78845736 0.17550012+0.64271115*I
2.3000000 0.83290912 0.16341960+0.63399285*I
2.4000000 0.87546874 0.15218883+0.62595235*I
2.5000000 0.91629073 0.14169584+0.61847841*I
2.6000000 0.95551145 0.13184894+0.61147999*I
2.7000000 0.99325177 0.12257089+0.60488549*I
2.8000000 1.0296194 0.11380069+0.59863552*I
2.9000000 1.0647107 0.10548489+0.59267760*I
3.0000000 1.0986123 0.097576064+0.58697104*I
3.1000000 1.1314021 0.090038424+0.58148334*I
3.2000000 1.1631508 0.082843993+0.57618212*I
3.3000000 1.1939225 0.075964723+0.57103646*I
3.4000000 1.2237754 0.069372706+0.56602343*I
from where my challenge should have been: Code: AS(2,I) = 0.20280135+0.66279493*I
I haven't (in the year of the above posts) been aware of the problem of the attracting and repelling fixpoints and the sensitivity of the "Alternating Series with increasing heights" (which I call now "alternating iteration series") around which fixpoint the series is developed. Assuming that the "polynomial method" for tetration approximates (and generalizes) the Knesersolution and also assuming that the Knesersolution is the best for real tetration the AS()function should be based on that "polynomial method".
The Pari/GPprocedures are even simpler than that of the original posting;
Code: n=32 \\ constant gives dimension for matrices
default(realprecision,200) \\ internal realarithmetic computation uses 200 digits
default(format,"g0.12") \\ show 12 digits in userinterface
{ASinit(b)=local(a); \\ define the matrix and coefficients for powerseries
a=log(b);
B = matrix(n,n,r,c, (a*(c1))^(r1)/(r1)!); \\ the carlemanmatrix
C = matsolve(matid(n)+B,matid(n)[,2]); \\ vector of coefficients
return("ok");}
AS(x) = sum(0,n1, x^k*C[1+k]) \\ the value for the (truncated) and approximate powerseries AS(base,x)
Note, that the coefficients in C form the (head of a) powerseries, which seems to have nonzero (and surprisingly interesting) range of convergence even if base b is outside of the Shell/Thron (resp Euler)range!
After that, do the computation:
Code: base=2
ASinit(base)
%486 = "ok"
AS( 1 ) \\ consider only x for which the truncated powerseries is convergent
%487 = 0.28740870
AS( I ) \\ reproduce the new result (if n is at least 32 )
%488 = 0.20280135 + 0.66279493*I
Note finally that for bases inside the Eulerinterval one can use the builtin Pari/GPfunction "sumalt" to simply do a summation for the (then conditionally converging) alternating iteration series and use this to crosscheck the AS(x)evaluation taken by the "polynomial method" for tetration.
Gottfried
Gottfried Helms, Kassel
