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Should tetration be a multivalued function?
#12
(01/09/2016, 06:20 AM)sheldonison Wrote:
(01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically:

ok, first lets define y and z as follows:



Then substitute these values of y and z into the Op's equation above, noting that


This is the Op's equations with the substitutions
this equation holds for all values of a,y,z
^^ Good. That's far more elegant that my reasoning, which used the commutative x^ln(y) and goofy induction.

Maybe is the key to calculate the derivative at the origin? (taking limit of r → 0)

Here is a graphic illustrating the equality. The exponentiation of the blue arrow is equal to the one in the red arrow.
The expression inside the rectangles are equal.

[Image: pqj5CIo.jpg?1]

By induction, if we invert the red arrow, it is equal to a larger arrow; I mean . I wrote it thinking that n was entire, but since r is real, and °a is arbitrary, it is obvious that is for any real n.

But is even more general, because this is valid on all branches, and for any definition of °a.
Graphically, you can move the arrows to the left or right, and the equality remains valid.

For example,

[Image: YrmCqFc.jpg?1]

and also , , etc.
I have the result, but I do not yet know how to get it.
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