04/03/2008, 11:28 PM

Hey, Henryk!

Concerning the proposals for [n]\ (hyper-log) and /[n] (hyper-root) "reducing" operators:

I mean, taking, for instance:

y = b[s]n, with b: base, n: (hyper)exponent and s: rank, we get:

b = nth-[s]hyper-root (y) = y /[s]n = b[s]n /[s]n .... /[s]n simplifies from the right;

n = [s]hlog(base b) (y) = b[s]\ y = b[s]\ b[s]n ........ b[s]\ simplifies from the left.

Divine surprise?

GFR

Concerning the proposals for [n]\ (hyper-log) and /[n] (hyper-root) "reducing" operators:

bo198214 Wrote:However your idea to put [n]\ instead of \[n] is a better one as you can better memorize the ruleDo you realize, Henryk, that our proposal to use these right (hyper-root) and left (hyper-log) reducing operators practically .... simplify ??

b[n]k /[n] k = b and b [n]\ ( b[n] k ) = k

as "The thing to be reduced is on the (reducing) operation side (i.e. the side with the / or \ attached)"

............

I mean, taking, for instance:

y = b[s]n, with b: base, n: (hyper)exponent and s: rank, we get:

b = nth-[s]hyper-root (y) = y /[s]n = b[s]n /[s]n .... /[s]n simplifies from the right;

n = [s]hlog(base b) (y) = b[s]\ y = b[s]\ b[s]n ........ b[s]\ simplifies from the left.

Divine surprise?

GFR