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
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 :
-------------------------------------------------------------------------
Do you agree that sexp(z) + 2pi i is another branch of sexp(z) ??
-------------------------------------------------------------------------
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