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(09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...).
I'm also curious: What about exactly? You said it converges slowly, but how do you iterate it at all? What is the asymptotic as the tower ?
I used a complex fixpoint and could generate the matrices for regular iteration (in context of diagonalization). The fixpoint I used is
t0 = 0.1957457524880764  1.691199920910569*I
one of its logarithms is
u0 = 0.5320921219863799 + 4.597158013302573*I
where u0 = log(t0) + 2*Pi*I // log giving the principal branch
With this I create the triangular Bellmatrix and diagonalize.
The series has complex terms and is very difficult to evaluate  I accelerate slow converging series usually with Eulersummation, but the series has complex terms and it seems I need also complex order for Eulersummation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle.
However, the fractional iterates behave even worse, and two halfiterates reproduce the integer iterate just to two decimals...
The schröderterm s for schr(x') and x'=x/t0  1 at x=1 is, according to the last three partial sums of the series (128 terms):
Code: [126] 0.4119542792176348+1.439754774257274*I
[127] 0.4119542792176264+1.439754774257268*I
[128] 0.4119542792176181+1.439754774257268*I
...
where I assume s~ 0.411954279217... +1.439754774257...*I as correct decimals.
From here we can compute y' = schr°1(s * u0^h ) and with h=1 I reproduce exp_b°1(1) = b =exp(exp(1)) to 15 digits exact.
The last three partial sums of the series:
Code: ´
[126] 1.004456441437337+0.03850266639539727*I
[127] 1.004456441437337+0.03850266639539727*I
[128] 1.004456441437337+0.03850266639539727*I
...
From here y' = 1.004456441437337+0.03850266639539727*I, y=(y'+1)*t0 , yb = 7.405812 E17 + 2.57894 E15*I is an acceptable approximation. However  it is only the integeriteration. I've not yet found en Eulerorder or another convergence acceleration which stabilizes the partial sums for the halfiterate to an acceptable degree.
Gottfried
Gottfried Helms, Kassel
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Why do you need to use a complex fixed point for ? It converges to a real fixed point as the tower approaches infinity (namely, ).
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09/13/2009, 08:08 AM
(This post was last modified: 09/13/2009, 08:09 AM by bo198214.)
(09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...).
I dont think, one can say that the regular iteration didnt work.
Its rather that it doesnt match your requirement to have a singularity at 2.
Though this requirement is quite apparent for realvalued functions, because we necessarily use the real branch of the logarithm, the necessity is not so clear for complex valued functions, where there is free choice of the branch of the logarithm.
It sounds anyway strange to prefer a singular tetrational over an entire tetrational if there is not the demand of real values.
Btw.: Kouznetsov's Contour integral method is rather applicable between two conjugate nonreal fixed points. I doubt that you can use the idea for a real 2cycle.
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09/13/2009, 09:49 AM
(This post was last modified: 09/13/2009, 09:51 AM by mike3.)
Why not use the entire? Because the behavior is not consistent. It seems stranger to imagine a function that has singularities for only real bases , then suddenly it becomes free of them and entire for all . It's two wildly different analytical behaviors and that doesn't make much sense. Ideally it would be nice to be able to interpret the tetration at to be what you'd get if you did an analytical continuation in the base from through the complex plane. Finally, at the integer towers of this base, we are still dealing with real numbers: when we take the log of for to get as 0, we are still using a real logarithm of a real number to a real base. So why not continue using this principal real logarithm for the rest of the tet function at this base?
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09/13/2009, 09:57 AM
(This post was last modified: 09/13/2009, 09:59 AM by Gottfried.)
(09/13/2009, 07:24 AM)mike3 Wrote: Why do you need to use a complex fixed point for ? It converges to a real fixed point as the tower approaches infinity (namely, ).
The problem with the real fixpoint is, that the triangular Bellmatrices have (alternating signed) units on its diagonal. This prevents the computation of a matrixlogarithm as well of the diagonalization  at least in my implementations.
If I have no option for one of those, I can approximate tetration only via diagonalization of the square Bellmatrices (means: omitting the fixpointshifting). But here all coefficients depend on the size of the used matrix, they are in my view unpredictable and may only serve as rough approximations for a "first impression".
But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newtonbinomialformula and others...
Gottfried
Gottfried Helms, Kassel
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(09/13/2009, 09:57 AM)Gottfried Wrote: The problem with the real fixpoint is, that the triangular Bellmatrices have (alternating signed) units on its diagonal. This prevents the computation of a matrixlogarithm as well of the diagonalization  at least in my implementations.
That should not pose a problem. You need the first row of , where is the Bell matrix. You make a Jordan decomposition and then where for each Jordanblock for eigenvalue with multiplicity one sets . The sum is finite because is nilpotent: .
Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself.
Quote:But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newtonbinomialformula and others...
I dont think that the Newton formula helps. It is realvalued and hence can not return a suitable solution for a decreasing base function.
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(09/13/2009, 11:23 AM)bo198214 Wrote: Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself.
arrgh... I knew, one day I'd have to look at the jordan decomposition... why not now...<sigh>
Gottfried Helms, Kassel
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09/13/2009, 01:34 PM
(This post was last modified: 09/13/2009, 03:46 PM by Gottfried.)
(09/13/2009, 06:14 AM)Gottfried Wrote: The series has complex terms and is very difficult to evaluate  I accelerate slow converging series usually with Eulersummation, but the series has complex terms and it seems I need also complex order for Eulersummation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle.
However, the fractional iterates behave even worse, and two halfiterates reproduce the integer iterate just to two decimals...
The schröderterm s for schr(x') and x'=x/t0  1 at x=1 is, according to the last three partial sums of the series (128 terms):
Code: [126] 0.4119542792176348+1.439754774257274*I
[127] 0.4119542792176264+1.439754774257268*I
[128] 0.4119542792176181+1.439754774257268*I
...
where I assume s~ 0.411954279217... +1.439754774257...*I as correct decimals.
The general precision can drastically be improved if we insert a "stirlingtransform" of the schröder and the inverse schröderfunction.
Code: [126] 0.4119542792176179+1.439754774257279*I
[127] 0.4119542792176179+1.439754774257279*I
[128] 0.4119542792176179+1.439754774257279*I
...
ps[128]ps[127]=1.432629629141992 E55  1.024270322737871 E55*I
So the partial sums in that region differ only by values of order 1e55 and we get a reliable value for the schröderfunction up to at least 50 digits.
We compose the coefficients of the Schröderfunction using the factorially scaled Stirlingnumbers 2'nd kind (just premultiply the Bellmatrix of the schröderfunction by the Bellmatrix of exp(x)1 and postmultiply the Bellmatrix of the inverse schröderfunction by the Bellmatrix of log(1+x), in my notation fS2F*W and WI*fS1F )
If we call the new schröderfunction
eschr(log(1+x)) = schr(x)
then I got much better precision: denote the function f(h) the so constructed sexpfunction f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch ( =log(t0) + 2*Pi*I)
Code: bl = log(b) // = 2.718...
x' = log(x/t0)
y'=eschr°1 (u0^h*eschr(x'))
y = exp(y')*t0
then
f(1)  b = 2.244868624099733 E60 + 3.153343574801035 E60*I // order for Eulersum: 1.4  0.1*I
f(2)  b°2 = 1.922763821393618 E13 + 1.897277601645949 E13*I // order for Eulersum: 2.51.2*I
f(1)*bl  0 = 4.089597929065041 E56 + 6.283185307179586*I // order for Eulersum: 1 (direct sum, no acceleration needed)
However, as we see at the f(1)entry we needed a correction factor to get the correct real value  but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this...
Gottfried
Gottfried Helms, Kassel
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09/13/2009, 07:34 PM
(This post was last modified: 09/13/2009, 09:37 PM by Gottfried.)
(09/13/2009, 01:34 PM)Gottfried Wrote: then I got much better precision: denote the function f(h) the so constructed sexpfunction f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I)
Code: bl = log(b) // = 2.718...
x' = log(x/t0)
y'=eschr°1 (u0^h*eschr(x'))
y = exp(y')*t0
then
f(1)  b = 2.244868624099733 E60 + 3.153343574801035 E60*I // order for Eulersum: 1.4  0.1*I
f(2)  b°2 = 1.922763821393618 E13 + 1.897277601645949 E13*I // order for Eulersum: 2.51.2*I
f(1)*bl  0 = 4.089597929065041 E56 + 6.283185307179586*I // order for Eulersum: 1 (direct sum, no acceleration needed)
However, as we see at the f(1)entry we needed a correction factor to get the correct real value  but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this...
However, I just tried the (positive h=+0.5) halfiterate (hoping that it may not be required for positive heights to include this additional correctionfactor). I got an halfiterate, which when iterated reproduced the integeriterate well, I got about b°0.5(b°0.5(1))  b ~ 5.0e18
Code: \\ halfiterate, u=log(t0)+2*Pi*I
U_05=W*dV(u^0.5)*WI; \\ Bellmatrix for halfiterate; stirlingtransformation is already included in W and WI (=W^1)
x=1
ESum(1.70.7*I) *dV(log(x/t0)) * U_05 \\ Use complex Eulerorder for summation of 1.70.7*I, manually optimized
\\ y' = 3.903870288432298 + 1.032427916941400 *I \\ untransformed value as result of matrixsum
y = exp(y')*t0
\\ y = 0.02725344094115782  0.02087339685842267*I \\ transformed, final value
x = y \\ use found value and iterate again
ESum(2.40.6*I) *dV(log(x/t0)) * U_05 \\ another Eulerorder required
\\ y' = 3.250373950445425  4.597158013302573*I \\ untransformed matrixsum
y = exp(y') *t0
\\ y = 0.06598803584531253 + 5.687696437325240 E18*I \\ transformed, final value
y  b = 4.706607418746630 E18 + 5.687696437325240 E18*I
\\error <1e17
So the halfiterate from 1 by regular iteration using fixpoint t0 is y ~= 0.02725344094115782  0.02087339685842267*I .
Seems, one can find at least any solution using regular tetration for this base, however difficult.
What makes me still headscratch is the required correctionterm for the one negative height example
[update] I could improve this to err~1e25 when simply computed the schrödervalues separately as described in the mail before. Then I even did not need complexeulerorders; just for one summation (of four) I needed a small Euleracceleration of order 1.3 (which is nearly untransformed partial sums)
The above value for the halfiterate is confirmed and has even more stable digits in the partial sums using around 120 to 128 terms.
[/update]
[update 2]
The graph below is only a proposal. Only h=0.5 was used to reproduce the integer iteration so far. Other rational fractions should be verified, too.
[/update 2]
Gottfried
Gottfried Helms, Kassel
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09/14/2009, 02:02 PM
(This post was last modified: 09/14/2009, 02:06 PM by Gottfried.)
Hmm, I could verify the fractional heights for h=1/2 , h=1/3, ... h=1/6 by reinserting the values in the according powerseries (using the stirlingtransformation). Iterations with h=1/8 reproduce up to h=4/8 = 1/2 correctly, and fail at h=6/8 or h=7/8 .
(This does not neccessarily mean, that the results for the fractional iterates of the previous posting are wrong, but the series may be useless for multiple repeated applications)
With finer fractional heights there seem to be generally trouble which I didn't try to investigate yet. I'm nearly sure this is due to the modulus 2*Pi*I .
In another investigation I took especially care for the effect of the expfunction, which reduces the windings to (mod 2*Pi*I) and got an correct answer where the use of the expfunction led to a wrong result, so I'm beginning to consider whether we should build a library of functions/operators, where the operations keep track of the integer multiples of 2*Pi*I as well. Maybe this will give another improvement for the difficult bases. Don't know, whether I can proceed here...
Gottfried
Gottfried Helms, Kassel
