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 Generalized recursive operators
11/23/2007, 06:42 AM
Post: #1
 Whiteknox Junior Fellow Posts: 3 Joined: Nov 2007
Generalized recursive operators
If the operators +, *, ^, ^^, can be considered consecutive values of a sequence (with + as 1, I guess), is it possible to construct a generalized recursive function where R(1) is +, R(2) is *, and so on? In this case, can you find R(x) where x is fractional or real? I.e., is there an operator in between + and *?

[I am assuming that R(x) can be defined for positive integer x as follows: (where a and b are integers)
1. a R(x+1) (b+1) = a R(x) (a R(x+1) b)
2. a R(x+1) 1 = a
]

Additionally, has anyone considered investigating the properties as x->inf of R(x)? I would imagine that 2 R(inf) 2 is still 4 as well as x R(inf) 1 = x.
11/23/2007, 08:41 AM (This post was last modified: 11/26/2007 10:35 AM by bo198214.)
Post: #2
 bo198214 Administrator Posts: 1,365 Joined: Aug 2007
RE: Generalized recursive operators
Whiteknox Wrote:If the operators +, *, ^, ^^, can be considered consecutive values of a sequence (with + as 1, I guess), is it possible to construct a generalized recursive function where R(1) is +, R(2) is *, and so on? In this case, can you find R(x) where x is fractional or real? I.e., is there an operator in between + and *?
Thats an interesting question.

Quote:[I am assuming that R(x) can be defined for positive integer x as follows: (where a and b are integers)
1. a R(x+1) (b+1) = a R(x) (a R(x+1) b)
2. a R(x+1) 1 = a
]

Here is however a difficulty: a R(x) 1 != a for x=1. So if we want f(x):= a R(x) 1 being a continuous function then f(1)=a+1, f(2)=a, f(3)=a, f(n)=a for $n\ge 2$. Which is somehow strange to have it first a decreasing function afterwards being constant.

Quote:Additionally, has anyone considered investigating the properties as x->inf of R(x)? I would imagine that 2 R(inf) 2 is still 4 as well as x R(inf) 1 = x.

Yeah, you can show that 2 R(n) 2 = 4 by induction. So the limit should be the same.

And a warm welcome to the forum (though I am nearly absent and just found some time to read and write).
11/23/2007, 03:57 PM
Post: #3
 Whiteknox Junior Fellow Posts: 3 Joined: Nov 2007
RE: Generalized recursive operators
I carefully sidestepped the problem by using a R(x+1) 1 = a instead of a R(x) 1 in my definition. Since only positive integer values for this are straightforwards to define, the smallest x is 1 and the smallest (x + 1) is 2. This does become problematic for generalization to fractional x.
11/25/2007, 01:02 AM
Post: #4
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators
If I remember correctly, Daniel Geisler had a discussion about that. In fact he showed that
$b{\uparrow}^{\infty}\infty = b$
which can be found in his powerpoint presentation about it at
http://tetration.org/Resources/Files/Ackermann.ppt

Andrew Robbins
11/29/2007, 04:45 AM (This post was last modified: 11/29/2007 04:47 AM by andydude.)
Post: #5
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators
I wonder if hyper-operations after multiplication can be approximated by:
$a R(n) b \approx a^b$ for $0\le b \le 1$ and $n \ge 3$

If so, then maybe this could be used to extend the right hyper-operations to at least a function defined over all real N, but I don't think its continuous.

Here is some Mathematica code that I used to investigate this idea:

Code:
Hy[1] := Plus; Hy[2] := Times; Hy[3] := Power; Hy[n_][x_, 0] /; (n > 3) := 1; Hy[n_][x_, 1] /; (n > 2) := x; Hy[n_][x_, y_] /; (1 < n < 3) :=   Simplify[Evaluate[InterpolatingPolynomial[{   {1, x + y},   {2, x*y},   {3, x^y}}, n]]]; Hy[n_][x_, y_] /; (n > 3) := Piecewise[{   {HyLog[n-1][x, Hy[n][x, y + 1]],  y < 0},   {x^y,                        0 <= y <= 1},   {Hy[n-1][x, Hy[n][x, y - 1]],     y > 1}   }];    HyLog[1] := Subtract; HyLog[2] := Divide; HyLog[3] := Log; HyLog[n_][x_, 1] /; (n > 3) := 0; HyLog[n_][x_, x_] /; (n > 2) := 1; HyLog[n_][x_, z_] /; (1 < n < 3) :=   Simplify[Evaluate[InterpolatingPolynomial[{   {1, z - x},   {2, z/x},   {3, Log[x, z]}}, n]]]; HyLog[n_][x_, z_] /; (n > 3) := Piecewise[{   {HyLog[n][x, Hy[n-1][x, z]] - 1,    z < 1},   {Log[x, z],                    1 <= z <= x},   {HyLog[n][x, HyLog[n-1][x, z]] + 1, z > x}   }];

It makes some pretty amazing graphs, but I'll have to do some more investigating before I can come up with any conclusions.

For those of you who are confused by Mathematica code, a little primer: "_" means "any", i.e. a variable, "/;" means "where", and in order to use the function you would write "Hy[3.5][E, Pi]" for example would calculate $e {\uparrow}^{1.5} \pi$ in Knuth notation. I have no idea what this would be, but this function gives somewhere around 256 thousand.

Andrew Robbins

11/29/2007, 05:55 AM (This post was last modified: 05/04/2009 05:24 PM by andydude.)
Post: #6
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators
Wow! I can't believe I gave that away for free! I would pay hundreds to get my hands on this code! Good thing I didn't have to... I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:

\begin{align}
\lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\
\lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\
\lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6
\end{blign}

where the box is GFR's box notation for hyperops, and a is sufficiently close to e, because thats the number I used.

I suppose you could see this from the integer versions of these operators, but I think the continuous (or if not continuous, mostly real-valued) versions make it easier to see.

[update]These limits are false, the true value for the first limit is closer to -1.85, not -2... sorry about that.[/update]
11/29/2007, 06:20 AM (This post was last modified: 11/29/2007 06:23 AM by andydude.)
Post: #7
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators

This makes it appear that:

$\lim_{N\rightarrow\infty} (a \begin{tabular}{|c|}\hline N \\\hline\end{tabular} b) = b+1$ for all $a>1, b<0$

meaning, in the limit, all hyper-operators return to the successor operation, like the circle of life...

Andrew Robbins.
11/29/2007, 08:14 AM (This post was last modified: 11/29/2007 08:18 AM by Gottfried.)
Post: #8
 Gottfried Long Time Fellow Posts: 721 Joined: Aug 2007
RE: Generalized recursive operators
andydude Wrote:This makes it appear that:

$\lim_{N\rightarrow\infty} (a \begin{tabular}{|c|}\hline N \\\hline\end{tabular} b) = b+1$ for all $a>1, b<0$

meaning, in the limit, all hyper-operators return to the successor operation, like the circle of life...

Andrew Robbins.
Wow!
This is one of the crazy great things, that we seem to find here. Pretty productive diggers, we are. I hope, it will come out to be true gold, what we fiddle out of the mines of Klondyke in the hyper-west of number-theory.

:-)

Gottfried

Gottfried Helms, Kassel
11/30/2007, 06:12 PM
Post: #9
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators
I actually made a few errors in that code. HyLog is wrong for two values.

Code:
HyLog[2][x, z] == z/x but Divide[x, z] == x/z and HyLog[1][x, z] == z - x but Subtract[x, z] == x - z

Andrew Robbins
11/30/2007, 09:18 PM
Post: #10
 andydude Long Time Fellow Posts: 504 Joined: Aug 2007
RE: Generalized recursive operators
@Whiteknox

Sorry for getting ahead of myself. Welcome to the forum, and feel free to ask more questions. Concerning your specific question, it has definitely been considered before, but there are no known "nice" extensions. You can always use Lagrange interpolation to connect the integer values of any function. The problem arises with continuity, differentiability, and whether or not it still satisfies a formula.

For example, if you were to use Lagrange interpolation between all known hyper-operations then you would end up with a function $a\ R(n)\ b$ as you describe for real a, b, n, BUT doing this would prevent it from satisfying:
$a\ R(n)\ b = a\ R(n-1)\ (a\ R(n)\ (b - 1))$
for all non-integer n, since Lagrange interpolation does not take this into account. A method that works very well for tetration, also works well for hyper-operations in general: choosing an interval of n, for example 2 < n < 3 so that everything between multiplication and exponentiation are approximated by some kind of interpolation (maybe not Lagrange interpolation), and use the formula of hyper-operators to extend the domain of n from this interval to all real n. The problem with this method is that doing this prevents the function from being differentiable (and in some cases continuous).

So in order to find a "nice" extension of hyper-operators to real n, one would need to start with an "unknown" interpolation, apply the formula, then "solve" for the interpolation with respect to the formula in such a way that additional requirements are met, for example: continuity, differentiability, and so on.

Andrew Robbins
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