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Andrew Robbins' Tetration Extension
#22
so , if at the fixed point A ( = exp(A) ) slog is defined and the " main identity " still holds

slog(z) = slog(exp(z)) - 1

then abs(slog(A)) needs to be oo.

but there is another logical value for slog(A).

of course , only when slog(A) exists and the main identity does not hold.

A = exp(A) = sexp(slog(A) + 1)

we know that a half - iterate should have the same fixpoints thus

A = hexp(A) = sexp(slog(A) + 1/2)

repeating that we get

lim n -> oo

A = sexp(slog(A) + 1/(2^n) ) = sexp(slog(A))

in fact

A = sexp(slog(A) + 1/(2^n) ) = A = sexp(slog(A) + 1/(2^m) )

holds for all positive integer n and m.

in fact , after some work ; all positive real n and m actually.

( assuming Coo )

this suggests that sexp is not an entire function either.

( hint : sexp(slog(A) + any real x ) = A => straith line on complex plane with constant value !! )

thus sexp(z) is not entire UNLESS abs ( slog(A) ) = oo !

( UNLESS = neccessary condition not neccessarily sufficient )

...


the subject gets more complicated , i wont continue to elaborate , but i have reasons to assume slog(A) = sexp(A) = A = exp(A)

i know slog has a fixed point that is not a fixed point of exp.


to save time and errors ,
i want andrew robbins to compute slog(A) ( exp(A) = A )

anybody else who has an slog function is also welcome to give his result for slog(A).

how many slog's are there anyway ??

regards

tommy1729
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Messages In This Thread
Andrew Robbins' Tetration Extension - by bo198214 - 08/07/2007, 04:38 PM
RE: Andrew Robbins' Tetration Extension - by tommy1729 - 06/29/2009, 08:20 PM

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