(03/23/2015, 09:00 AM)MphLee Wrote: He means where is the xth iteration of the function f and [q] is the q-rank hyperoperation.

Oh. Thanks.

If the minus symbol do ever means subtraction, and not the inverse operation of [q] (which one?, some [q] have many inverses and even multivalued inverses), then it does not make sense, as tommy1729 says:

(03/21/2015, 11:11 PM)tommy1729 Wrote: a*b = b*aThat would mean

"add+b"^[a-1](1*b) = "add+a"^[b-1](1*a)

b*(a-1)=a*(b-1)

but if "-" is an inverse operator of [q], such that x-y is defined as x-y=x[q]-y, and -y is defined as y[q]-y=N(q,y) then

for [q]=product, "-" would be division, so

(03/21/2015, 11:11 PM)tommy1729 Wrote: "add+b"^[a-1](1*b) = "add+a"^[b-1](1*a)

would be

"add+b"^[a/1](1*b) = "add+a"^[b/1](1*a)

That would mean

b*(a/1)=a*(b/1)

(03/21/2015, 11:11 PM)tommy1729 Wrote: However notice that an expression likethen it would make sense if "-" is the inverse operator of zeration.

f^[a - oo](...) does not make sense as a nonconstant function.

-∞ is used on all the basic operations:

the neutral of addition is ÷∞=1/∞=0

the neutral of product is ∞√=∞√n=n^÷∞=1

the neutral of exponentiation is

from product viewpoint, all numbers smaller than 0 are transfinite.

from exponentiation of n viewpoint, all numbers smaller than 1 are transfinite.

from tetration of n viewpoint, all numbers smaller than are transfinite.