Jabotinsky's iterative logarithm

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Jabotinsky's iterative logarithm

andydude Offline

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#16

05/24/2008, 06:37 AM

bo198214 Wrote:For me it looks as if there always \( t=1 \) must be set ...

oops
\( \mathcal{J}[f](f^{\circ t}(x)) = (f^{\circ t})'(x) \mathcal{J}[f](x) \)

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RE: Jabotinsky's iterative logarithm - by andydude - 05/24/2008, 06:37 AM

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