andydude Wrote:A comparison between the notation I used and the notation that GFR used:You surely mean "around the h". 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.:
\( \mathtt{b[N]\^h(x)} = \mathtt{b[N]<h>x} \)however, since GFR's notation requires angle-brackets around the 'y', it prevents it from being used with slash-notation, especially for some inverse hyperops. To illustrate the difficulties, I will use GFR's instead.
\(
\begin{tabular}{c|l|l|l}
\text{inv} & \text{my} & \text{GFR's} & \text{name} \\
\hline
{z =}
& \mathtt{b[N]\^h(x)}
& \mathtt{b[N]<h>x}
& \text{iterated hyper-operations}
\\
b =
& \mathtt{z/[N]\^h(x)}
& \mathtt{z/[N]<h>x}
& \text{auxiliary hyper-roots}
\\
h =
& \mathtt{b[N]\^{\backslash}z(x)}
& \mathtt{b[N]< >x{\backslash}z}?
& \text{auxiliary hyper-logarithms}
\\
x =
& \mathtt{b[N]\^(-h)(z)}
& \mathtt{b[N]<-h>z}
& \text{negatively iterated hyper-operations}
\end{tabular}
\)
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
which doesn'y prevent the (... moderate
) use of slashes. By the way, why <> is void? Is ^ without operand?Nevertheless, I don't understand the formulas of this line, which seem to me incomplete. Actually, if we want to "extract" b or h, the problem is difficult, unless I miss something. In fact, and just for my understanding, let us suppose that we mean the following (priority to the right):
z = b[N]^h (x) = b[N]<h> x = b[N] b[N] b[N]...b[N]x , h times, then:
x = b[N]^(-h) (z) = b[N]<-h> z = b[N] b[N] b[N]...b[N]\ z , h times = b[N]<h>\ z.
Right! Then, let us take the following hyperop: z = b[N]x.
The x-th N-root of z is: b = z /[N]x.
If we suppose, now, to have:
z = (b[N]x) [N]x , the x-th N-root of the x-th N-root of z can be put as:
b = z / [N]x [N]x = z / ([N]x)^2.
Then, in case of h iterations of the [N]x operation acting on its left on a base b, we should have (priority to the ... left!):
z = b ([N]x)^h = b [N]x<h> = b [N]x [N]x ...[N]x , h times, then:
b = z /([N]x)^h = z /[N]x<h> = z /[N]x [N]x ...[N]x , h times.
I don't know if this corresponds to the hyperroot auxiliary operations needed by Gottfried. Moreover, the "extraction" of h could indeed (I agree) be given by:
h = b[N]\z - b[N]\x (To be, anyway, also discussed).
I don't even know if what I just wrote is right or wrong. Let's hope for the best.
GFR