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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
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?


I can only say it diverges, and I understand You want me to find a method to sum it and a result. (...)

Well, diagonalization seems to give an answer here.

Remember, in my matrix-notation, 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 diagonal-matrix 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)

which can be described by

// "I" means here the identity-matrix

so the eigenvalues of

are not a vandermonde-vector, 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 (cesaro-and) Euler-summable bases and the Shanks-summable bases with values of the diagonalization-summable 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 "matrix-method" for "diagonalization")
The result looks pretty smooth...

Because of the nice eigenvalue-configuration, you may arrive at these results even without fixpoint-shifts and expression by the Ut-matrix; simply invert -using an empirical Bb-matrix- I+Bb of reasonable size, say 32x32 or 64x64, to get good approximations.

Gottfried Helms, Kassel

Messages In This Thread
RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/27/2008, 09:23 AM

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