Zeration

[marraco]

the bracketing must be to the right. So
ao(ao(aoa))=a+4
ao(aoa)=a+3
aoa=a+2
a = a+1 ???

The answer to that question comes clear if zeration is defined this way:

Neutral element: \( ^{-\infty \,=\, 0 - \infty \,=\, -\infty} \)

\( \,\,\,-\infty \,=\, a \,+\, -\infty \)
\( \,\,\,-\infty \circ a \,=\, a \,+\, 0 \)
\( \,(-\infty \circ a) \circ a \,=\, a \,+\, 2 \)
\( ((-\infty \circ a) \circ a) \circ a \,=\, a \,+\, 3 \)

This is similar to addition:
Neutral element: \( ^^{0 \,=\, \frac{1}{\infty} \,=\ \div \infty} \)

\( \,\,\,\div\infty \,=\, a \,.\, 0 \)
\( \,\,\,\div\infty + a \,=\, a \,.\, 1 \)
\( \,\,(\div\infty + a) + a \,=\, a \,.\, 2 \)
\( ((\div\infty + a) + a) + a \,=\, a \,.\, 3 \)


Also is similar to product:
Neutral element: \( ^^{1 \,=\, n^{\frac{1}{\infty}} \,=\ \sqrt[\infty]{n}} \)

\( \,\,\,\,\,\sqrt[\infty]{n} \,=\, a^0 \)
\( \,\,\,\,\,\sqrt[\infty]{n} \,.\, a \,=\, a^1 \)
\( \,\,(\sqrt[\infty]{n} \,.\, a) \,.\, a \,=\, a^2 \)
\( ((\sqrt[\infty]{n} \,.\, a) \,.\, a) \,.\, a \,=\, a^3 \)

And is similar to exponentiation
Neutral element: \( ^^{1 \,=\, ?_{(\infty,n,b)} \) where I conjecture that "?" is the inverse function of tetration, and is not slog.

\( \,\,\,\,\,\,\,\,\sqrt[\infty]{n} \,=\, ^0a \)

\( \,\,\,\,\,a^{\sqrt[\infty]{n}} \,=\, ^1a \)

\( \,\,a^{a^{\sqrt[\infty]{n}}}\,=\, ^2a \)

\( a^{a^{a^{\sqrt[\infty]{n}}}} \,=\, ^3a \)

Note that the neutral elements are all related with the inverse of the higher ranked operation.
I think that it is a very important clue.

\( -\infty= 0-\infty \)
\( ^^{0 \,=\, \frac{1}{\infty} \,=\ \div \infty} \)
\( ^^{1 \,=\, n^{\frac{1}{\infty}} \,=\ \sqrt[\infty]{n}} \)


But why I choose \( ^{-\infty} \) as the neutral element of zeration?
-Because it is \( ^{-\infty= 0-\infty} \)
-Because \( ^{-\infty\,\circ\, -\infty \,=\, -\infty \,+\, 1 \,=\, -\infty} \)
-Because this sequence:
\( ln(a^1)=ln(a) . 1 \)
\( ln(a.1)=ln(a) + 0 \)
\( ln(a+0)=ln(a) \circ [-\infty=ln(0)] \)

I conjecture that zeration has a periodic component, so it can just add 1 no matter how high are his variables, and also

ln(a+b) could be equal to ln(a) ° ln(b)
ln(a+b)=ln(a)+ln(1+b/a)
if a>b => 0 < ln(1+b/a) < ln(2)
If a=b => ln(1+b/a) < ln(2)

this is suspiciously like zeration, adding a small number to the larger number, or adding 2 if a=b