Should tetration be a multivalued function?
#8
Let's call c₁ to the superior asymptotic limit, and c₂ to the inferior. There are interesting relationships between those numbers.

They can be deduced from the known relation that defines those numbers: The base of the function Y(x)=ˣa is \( \\[27pt]

{a=c_1\,^{\frac{1}{c_1}}} \), and also \( \\[27pt]

{a=c_2\,^{\frac{1}{c_2}}} \)

then:

\( \\[25pt]

{\frac{c_2}{c_1}=\frac{ln(c_2)}{ln(c_1)}}=a^{c_2-c_1} \)

\( \\[25pt]

{{\color{Red}c_1\,^{c_2}=c_2\,^{c_1}=a^{c_1.c_2}}} \)

The last one is analogous to the relationship between the asymptotes for bases between \( \\[15pt]

{ 0<a<e^{-e}} \)

(04/13/2015, 08:01 PM)marraco Wrote: for bases between \( \\[15pt]

{0<b<e^{-e}} \) it remains bounded between 0 and 1 (for x>0), and converges to c, were c is the solution of \( \\[15pt]

{a^{a^c}}\,=\,c \), which seems to have 2 roots, c₁ and c₂, with
\( \\[15pt]

{a^{c_1}\,=\,c_2} \)
\( \\[15pt]

{a^{c_2}\,=\,c_1} \)
\( \\[15pt]

{{ \color{Red} {c_1}^{c_1}\,=\,c_2^{c_2}\,=\,{a^{c_1.c_2}} }} \)
\( \\[15pt]

{a\,=\,{c_1}^{\frac{1}{c_2}}\,={c_2}^{\frac{1}{c_1}}} \)

I suspect that this relation is the key to solve tetration equations:

\( \\[15pt]

{a^{c_1}\,=\,a^{a^{c_2}}\,=\,a^{a^{a^{c_1}}}\,=\,... \)

Here is tetration base a=0.01:
c₁ = 0,941488369
c₂ = 0,013092521
[Image: svh6ykX.png?1]

c₁ is the limit to the left, and c₂ to the right.
I have the result, but I do not yet know how to get it.


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