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Solving tetration for base 0 < b < e^-e
(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 Euler-summation, but the series has complex terms and it seems I need also complex order for Euler-summation. 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 half-iterates reproduce the integer iterate just to two decimals... Sad

The schröder-term 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):
[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 "stirling-transform" of the schröder- and the inverse schröder-function.
[126]  -0.4119542792176179+1.439754774257279*I
[127]  -0.4119542792176179+1.439754774257279*I
[128]  -0.4119542792176179+1.439754774257279*I
ps[128]-ps[127]=-1.432629629141992 E-55 - 1.024270322737871 E-55*I
So the partial sums in that region differ only by values of order 1e-55 and we get a reliable value for the schröder-function up to at least 50 digits.

We compose the coefficients of the Schröder-function using the factorially scaled Stirling-numbers 2'nd kind (just pre-multiply the Bell-matrix of the schröder-function by the Bell-matrix of exp(x)-1 and postmultiply the Bell-matrix of the inverse schröder-function by the Bell-matrix of log(1+x), in my notation fS2F*W and WI*fS1F )

If we call the new schröder-function

eschr(log(1+x)) = schr(x)

then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I)
bl = log(b) // = -2.718...
x' = log(x/t0)
y'=eschr°-1 (u0^h*eschr(x'))
y = exp(y')*t0
f(1)     - b    = -2.244868624099733 E-60 + 3.153343574801035 E-60*I  // order for Eulersum: 1.4 - 0.1*I
f(2)     - b°2  =  1.922763821393618 E-13 + 1.897277601645949 E-13*I  // order for Eulersum: 2.5-1.2*I
f(-1)*bl - 0    =  4.089597929065041 E-56 + 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 Helms, Kassel

Messages In This Thread
RE: Solving tetration for base 0 < b < e^-e - by Gottfried - 09/13/2009, 01:34 PM

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