Notations and Opinions

[bo198214]

\(
\begin{tabular}{c|l|l|l}
\text{inv} & \text{my} & \text{GFR's} & \text{name} \\
\hline
{z =}
& \mathtt{b[N]\^h(x)}
& \mathtt{b[N]<h>x}
& \text{iterated hyper-operations}
\\
b =
& \mathtt{z/[N]\^h(x)}
& \mathtt{z/[N]<h>x}
& \text{auxiliary hyper-roots}
\\
h =
& \mathtt{b[N]\^{\backslash}z(x)}
& \mathtt{b[N]< >x{\backslash}z}?
& \text{auxiliary hyper-logarithms}
\\
x =
& \mathtt{b[N]\^(-h)(z)}
& \mathtt{b[N]<-h>z}
& \text{negatively iterated hyper-operations}
\end{tabular}
\)

However, I object about line 3. In fact, among the Andydude's and my (GFR's) sequential codings, we have a strong correlation (and my formulas may also be slightly ... shorter), e.g.:
b[N]^h <---> b[N]<h>
b[N]^(-h) <--> b[N]<-h>

Therefore, if the third Andydude's formula is correct, we would also have:
h = b[N]^\z (x) = b[N]<>\z x

Gianfranco! There was a "?" behind your entry in the 3rd line! Which means here that Andydude is not sure how to express this with your notation (or means that it is not expressable with your notation).

Is ^ without operand?

Yes thats the point about about the ^ notation. That you can apply the slash/backslash modifications for example to the operator [N]^ in b[N]^h, while there is no such in between operator with your notation b[N]<h>.

However currently I am not that sure how Andrew uses this backslashed operator. If we remember the backslash rule then b [N]^\ (b [N]^ h) = h.
b[N]^h is a function, thatswhy you can write b[N]^h(x),
so [N]^\ takes as left operand a number, the base b, and as right operand a function and returns a number, but that does not match exactly how it is used in the third line.