slog_b(sexp_b(z)) How does it look like ?
#15
Dear Sheldon

Its true that I have to read some old threads again.
But I think there are a few misunderstandings in our communication.
Forgive me for responding quicky without thinking deeply but time is Always against me.

I think its easiest to consider base "e" only for the moment being.

I have to think a bit more before I adress everything.
But lets let me start with this one :

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Do you agree that sexp(z) + 2pi i is another branch of sexp(z) ??
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I assume the logic of that idea is clear , even if its wrong.

Like if exp(sexp(u)+2 pi i) = exp(sexp(u)) = sexp(u+1) then it seems logical to conclude sexp(z) + 2pi i is another branch of sexp(z).

Also if sexp(z) + 2pi i is another branch of sexp(z) for some z near B, then by analytic continuation ( valid on a subset of a branch without singularities ) it must be so almost everywhere on that branch.

A related argument is : slog(exp(z)) = slog(exp(z+ 2pi i)) = slog(z)+1
although this equation is not satisfied everywhere.
[ more on this in another thread : ( http://math.eretrandre.org/tetrationforu...80#pid6880 ) although Im stuck there. Also the math below hints why its not possible that the equation holds everywhere ]

Another idea is that the values (range) of one branch relates to another in a logical way :

the fundamental branch that maps the positive reals to the positive reals and the solution L of exp(L) = L is at - oo , + oo i , - i oo. Or more accurately like you said ; approaches L as imag(z) increases, and as real(z) decreases.

(ok I ignored the complex conjugate , im well aware of it )

The idea is that the solution L(2) that is different from L and satisfies exp(exp(L(2)) = L(2) cannot be on the same branch as the fundamental one.

And thus values close to the reals or L are on the fundamental branch and values close to L(2) are on another.

By analogue one can construct L(n) and have n branches.

The reasons come from the slog rather than the sexp :

slog(exp(z)) = slog(z)+1
this cannot hold for z=L
slog(exp(exp(z)) = slog(z) + 2 ( by induction )
this cannot hold for z=L(2)
I hope it is clear why !

This suggests that L(2) is not on the same branch because we use solutions to tetration that have a minimal amount of singularities for both slog and sexp.

( I assumed the L(n) do not form a natural boundary otherwise the idea is in trouble )

Notice I did not yet claim for instance that L(n) represents the n th branch.
Or for instance that z + L(a) - L(b) is another branch.

But these ideas need to be adressed.

regards

tommy1729


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RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/28/2014, 09:01 PM



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