Posts: 1,386
Threads: 90
Joined: Aug 2007
04/08/2008, 07:47 AM
(This post was last modified: 04/08/2008, 07:51 AM by bo198214.)
GFR Wrote:(priority to the right ... whenever necessary) :
b[3]<h>x = b^...^b^b^x
b[2]<h>x = b*...b*b*b*x ,
b[1]<h>x = b+...b+b+b+x,
... (and I stop here, for ... avoiding problems ...)
Thats exactly the proposal I was waiting for as we still needed an ASCII notation for . This perfectly fits with this proposal:
=f<n>(x)
In your above example we can directly interpret the function as f=b[k] which also is used in Haskell (I think) where you write (a*) for the function x>a*x and (*a) for the function x>x*a, where * is an arbitrary binary operation.
So when writing b[k]<n> we just let out the parens in (b[k])<n> for simplification.
Posts: 174
Threads: 4
Joined: Aug 2007
04/08/2008, 10:52 AM
(This post was last modified: 04/08/2008, 10:57 AM by GFR.)
RightYiHow!! (^^). I am happy that you liked it!
Posts: 174
Threads: 4
Joined: Aug 2007
04/08/2008, 03:31 PM
(This post was last modified: 04/08/2008, 03:35 PM by GFR.)
Once again, sorry!
Thinking of the negative <1> iteration of the exponential, which (if I am not mistaken) gives the log, I just realized that we might have:
y = b[3]x = exp_b (x), which gives the following rightinverse:
x = b[3]\ y = log_b (y), which is the <1> iteration of exp_b on y:
x = b[3]<1> y = log_b (y).
Therefore, we might have the following relation between these operations:
b[3]<1> x = b[3]\<1> x = b[3]\ x = log_b (x)
And (perhaps) in general, showing only the operators, we indeed have:
b[s]<n> = b[s]\<n> = shexp_b <n> = shlog_b <n> , with back and forward iterations.
Right? ... or ... wrong !!!
GFR
Posts: 1,386
Threads: 90
Joined: Aug 2007
04/08/2008, 04:22 PM
(This post was last modified: 04/08/2008, 04:25 PM by bo198214.)
GFR Wrote:b[s]<n> = b[s]\<n> = shexp_b <n> = shlog_b <n> , with back and forward iterations.
Right? ... or ... wrong !!!
Right. We have
f<r>(f<s>(x)) = f<r+s>(x)
f<0>(x) = x
As a derivation:
b [r]\ x = b[r]<1> x and
x /[r] n = ([r]n)<1>(x)
Though I think the mind has somewhat to cope with the last line in daily calculation.
And this does also not solve Gottfried's problem of expressing the height h such that b[r]<h> x = y.
Posts: 509
Threads: 44
Joined: Aug 2007
OK, I think we need to look at all the possibilities before we go too far into determining notation.
We seem to be stuck on two things that are very closely related: auxiliary hyperoperations and iterated hyperoperations, which are practically the same things, but from two different viewpoints (an example difference is that iterated hyper3 is auxiliary hyper4). Since the term "auxiliary" is new and "iterated" is old and venerable, it is more appropriate to call them "iterated" hyperoperations, although either term could suffice. A comparison between the notation I used and the notation that GFR used:
however, since GFR's notation requires anglebrackets around the 'y', it prevents it from being used with slashnotation, especially for some inverse hyperops. To illustrate the difficulties, I will use GFR's instead.
Fortunately, however, we do not need a notation for auxiliary hyperlogarithms, because:
So if neccessary, this can be written which means we really don't need either my notation, nor GFR's notation for auxiliary hyperlogarithms. Also, as you can see, we also don't need a notation for the auxiliary inverse, because these can be represented by negatively iterated hyperoperations. What this means is that we only need a notation for auxiliary hyperroots.
Andrew Robbins
Posts: 1,386
Threads: 90
Joined: Aug 2007
04/09/2008, 11:30 AM
(This post was last modified: 04/09/2008, 11:33 AM by bo198214.)
We have a bit of a dilemma here. Though the ^n has advantages with respect to applying the \ and / notation, basicly its ambiguous with taking the nth power.
f^n(x) = f(x)f(x)....f(x) or f^n(x)=f(f(...f(x)...)).
Of course you can say we can write the first one as f(x)^n, however then there is no operation to denote the nth power of a function.
So we need a different symbol instead of ^. For example in maple there is this notation with $n when taking the nth derivative with respect to x, you write diff(f,x$n). So I propose  if we want to give up the <n> notation for the advantage of taking (\ and) / 
f$n= .
correspondingly
b[k]$n(x), b/[k]$n(x), etc as you already listed.
As now the $ is available because we have the notation [4] and [5] and dont need dedicated symbols like #, $ or § for them anymore.
Quote:Fortunately, however, we do not need a notation for auxiliary hyperlogarithms, because:
So if neccessary, this can be written which means we really don't need either my notation, nor GFR's notation for auxiliary hyperlogarithms.
Thats really clever and again reminds me on Szekeres consideration of the Abel function as an integral in "Scales of infinity and Abel's functional equation", 1984.
Posts: 174
Threads: 4
Joined: Aug 2007
04/09/2008, 04:54 PM
(This post was last modified: 04/09/2008, 05:15 PM by GFR.)
andydude Wrote:Also, GFR uses x$y*a or something like that, which I find confusing.
Hey Andrew! I think that we have a misunderstanding here. Your citation, without ... warnings is also misleading. As a matter of fact, we adopted a special notation, at rank 4, for the representation of large numbers, based on an expression of a number z, such as:
z = b^(b^(b^ ...(b^x)...)), where b^ was supposed to be iterated n times (h?). This gives, for instance (n=3, x<b):
z = b^(b^(b^x)) = x @ (b#3) = b # (3+q), with q = slog_b x.
[in fact: 3+q = slog_b z, then: q = slog_b z 3 = slog_b x].
@ stands here for the "last" exponent appended to the tower height, corresponding to "mantissa" q (q<1). It is only a notation of numbers similar to the floating point scientific notation (always x<b):
z = b*b*b*x = x* (b^3) = b ^ (3+q), with q = log_b x.
So, indeed, I could write:
z = x @ (b#n)
for a "tower" with base b, height n and "tower extension" x. It's not confusing, in this particular application. In case of "integer" h iterations (..!) of a variable x, we could perhaps also write:
z = x @ (b#h) [x "appended" at the "top" of b#h]
But this was not a general notation proposal. It was an adhoc tetrational notation for the special purpose that I just mentioned.
And, of course, I agree that h = slog_b z  slog_b x, i.e.:
h = b[4]\ z  b[4]\ x
GFR
Posts: 174
Threads: 4
Joined: Aug 2007
04/09/2008, 05:10 PM
(This post was last modified: 04/09/2008, 05:12 PM by GFR.)
bo198214 Wrote:As a derivation:
b[r]\ x = b[r]<1> x and
x /[r]n = ([r]n)<1>(x)
Though I think the mind has somewhat to cope with the last line in daily calculation.
I would better put the last line as:
x /[r]n = x [r]n<1>
Keeping in mind the fact that the operators act from the left and from the right, respectively and that their operaional environment is not commutative. In particular, I am supposing the following equality among operators:
/[r]n = [r]n<1> = ([r]n)^(1).
GFR
Posts: 174
Threads: 4
Joined: Aug 2007
04/09/2008, 11:55 PM
(This post was last modified: 04/09/2008, 11:58 PM by GFR.)
andydude Wrote:A comparison between the notation I used and the notation that GFR used:
however, since GFR's notation requires anglebrackets around the 'y', it prevents it from being used with slashnotation, especially for some inverse hyperops. To illustrate the difficulties, I will use GFR's instead.
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.:
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 xth Nroot of z is: b = z /[N]x.
If we suppose, now, to have:
z = (b[N]x) [N]x , the xth Nroot of the xth Nroot 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
Posts: 1,386
Threads: 90
Joined: Aug 2007
GFR Wrote:Andydude Wrote: 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).
Quote: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.
