Mizugadro, pentation, Book

[tommy1729]

Ok I understand now.

But I fear I have to tell you your construction is not new.

It's just a classical fixpoint method used on an nonstandard function.

Your claim that it does not depend on M is thereby true and easy to prove.

To give a big hint :

suppose an analytic function f has a fixpoint at 0 with f ' (0) = Y > 1.

THEN for any real k , as n goes to oo :

pent(z) = sexp^[z](x_0) = f^[n] ( Y^(z-k) slog^[n+k](x_0) ).

Or said differently for any distinct pair reals k_1,k_2 :

f^[n] ( Y^(z-k_1) slog^[n+k_1](x_0) ) = f^[n] ( Y^(z-k_2) slog^[n+k_2](x_0) ).

At least if both sides converge.

Once you can see that , you will understand.

SECOND HINT : plug in koenigs function.

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For those who can still follow , the real question is what if sexp(x-t) = x is a parabolic fixpoint ? Then how do we get GOOD convergeance for sexp(x-t)^[z].
Maybe thats not so hard either , but it seems a logical followup question.

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Again for those who still follow , My answer is illuminating but the convergeance speedup is not yet understood.

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I wonder about Eremenko's Conjecture regarding pentation.
I rediscovered Eremenko's Conjecture as a kid , guess that explains it.

regards

tommy1729