Zeration = inconsistant ?

[MphLee]

the solution of the equation

\( x \odot_e^{a}x=b \)

should be, as you say
\( \exp^{\circ a}( 2 ln^{\circ a} (x) ) = b \)
\( x={\exp}^{\circ a}(\frac{ln^{\circ a}(b)}{2}) \)
I guess is possible to do more here... I'll try to find a forumula only involving homomorphic operators defined via exponential...

I looked at it better and is easy to write.
The solution of the equation

\( x \odot_e^{a}x=b \)

is \( x=b\oslash_e^{1+a} {\exp}^{\circ a}(2) \)

where \( \oslash_e^{1+a} \) is the inverse operation of \( \odot_e^{a+1} \)

so we should have

\( x \odot_e^{a}x=x\odot_e^{a+1}{\exp}^{\circ a}(2) \)

Anyways i'm not 100% sure. I have to chek it with calm.

I already made the choice :
a [0] b = max(a,b) + 1 + kroneckerdelta(a,b)

regards

tommy1729

Ok i get it...
Well first of all \( (-1) \)-ation should have a non-empty intersection with RR-Zeration

RR-Zeration: \( a [0]_{RR} b = max(a,b) + 1 + \delta_{ab} \)

\( [0]_{RR} \cap [-1]\neq \emptyset \)

Their intersection has to contain a segment of the trivial zeration (the successor of the second argument) because in that segment \( (-1) \)-ation is the subfunction of RR-Zeration.

Anyways I think that if we chose RR Zeration \( (-1) \)-ation its gonna be a multivalued oepration (see Hyperstructures theory and multimaps and this Brief introduction by Viro, comment by Mphlee)

Why? Because its translations are not invertible functions. Anyways I guess that using the Litinov-Maslov's Limit Process we could find a formula that show to us the real shape/behaviour of the set \( [-1] \).