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fractional powers of function inversion (was: changing terminology)
#5
(08/11/2009, 12:02 PM)Gottfried Wrote: and fractional "iterates of inversion" is then multivalued with complex heights according to the complex roots of -1

inv°[s](f(x)) = f°[(-1)^s](x)

Ya true, for iteration operators we dont need a new technique for general operator iteration, it can be reduced to powers of the exponent .
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RE: fractional powers of function inversion (was: changing terminology) - by bo198214 - 08/11/2009, 01:07 PM

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