Jabotinsky's iterative logarithm
#6
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

Sorry, for the inconvenience, this answers also Ivars question.
(And btw. it should be very rare to find so obvious mistakes in peer reviewed journals ...)
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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 bo198214 - 05/22/2008, 06:48 PM
RE: Jabotinsky's iterative logarithm - by Ivars - 05/23/2008, 07:05 AM

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