Jabotinsky's iterative logarithm
#14
bo198214 Wrote:Gosh, I should learn proper reading first. Jabotinsky states:
If \( F(G(z))=G(F(z)) \) it can be shown [5] that \( G(z) \) is some iterate of \( F(z) \) and hence that [translated into our style]:
\( \text{ilog}(F\circ G)=\text{ilog}(F)+\text{ilog}(G) \)

[5] J. Hadamard, Two works on iteration, Bull. Amer. Math. Soc. 50 (1944), 67-75
Reading the above with the "matrix-filter", and for F and G are resp. matrix-operators FM and GM thought. then the above equality (F°G = G°F) holds only, if the matrices commute - and this is also required, if the matrix-logarithms are to be added: log(FM) + log(GM) = log(FM*GM) = log(GM*FM) <=> FM and GM commute.
For finite matrices it is also known, that matrices are commuting, if their eigenvectors are identical. But if the eigenvectors are identical, then FM and GM differ only by their sets of eigenvalues FV and GV.

Hmm. What is surprising now is, that if FM and GM provide different iterates, then, for tetration, the eigenvalues in FV and GV must be appropriate powers of the same base - this seems much more restrictive than the statement of Jabotinski: there I can't find such a restriction (but maybe I'm just missing this at the moment)

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
Jabotinsky's iterative logarithm - by bo198214 - 05/21/2008, 06:00 PM
RE: Jabotinsky's iterative logarithm - by Ivars - 05/22/2008, 09:02 AM
RE: Jabotinsky's iterative logarithm - by Ivars - 05/22/2008, 01:13 PM
RE: Jabotinsky's iterative logarithm - by Ivars - 05/23/2008, 07:05 AM
RE: Jabotinsky's iterative logarithm - by Gottfried - 05/23/2008, 11:06 PM

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