Hyper operator

(Redirected from Hyper4)

The hyper operators forming the hypern family are as follows:

hypern (a, b) = $\begin{matrix} \mbox{hyper}(a,n,b)&&=&&a\uparrow^{n-2}b&&=&&a\to b\to (n-2)\\ \end{matrix}$

(See Knuth's up-arrow notation and Conway chained arrow notation.)

For n = 4 we have hyper4 or tetration, super-exponentiation or power towers in terms of an extension of standard operators:

$\operatorname{hyper4} (a, b) = \operatorname{hyper}(a, 4, b) = a ^ {(4)} b = {a \uparrow\uparrow b = \atop {\ }} \!\!\!\!\!\!\!{{\underbrace{a^{a^{\cdot^{\cdot^{a}}}}}} \atop {b\mbox{ copies of }a}}=a\to b\to 2$

Derivation of the notation

It can be seen as an answer to the question "what's next in this sequence: summation (+), multiplication (&times), exponentiation (^),…?" Noting that

$a + b = 1 + (a + (b - 1))$
$a \times b = a + (a \times (b - 1))$
$a ^ b = a \times (a ^ {(b - 1)})$

recursively define an infix triadic operator (making n=0 correspond to the successor function):

$a ^ {(n)} b= \left\{ \begin{matrix} a+1, & \mbox{if }n=0 \\ a, & \mbox{if }n=1,b=0 \\ 0, & \mbox{if }n=2,b=0 \\ 1, & \mbox{if }n>2,b=0 \\ a ^ {(n-1)} ( a ^ {n} (b - 1)) & \mbox{otherwise} \end{matrix} \right.$

then define $\operatorname{hyper\mathit{n}} (a, b) = a ^ {(n)} b$ and $\operatorname{hyper}(a, n, b) = a ^ {(n)} b$

This gives:

$\operatorname{hyper1} (a, b) = \operatorname{hyper}(a, 1, b) = a ^ {(1)} b = a+b$

$\operatorname{hyper2} (a, b) = \operatorname{hyper}(a, 2, b) = a ^ {(2)} b = ab$

$\operatorname{hyper3} (a, b) = \operatorname{hyper}(a, 3, b) = a ^ {(3)} b = a^b$

as further explained in the separate article tetration.

Known aliases for hyper4 include superpower, superdegree, and powerlog; other notation, $\operatorname{hyper4}(a,b)={}^{b}a$ .

The family has not been extended from natural numbers to real numbers in general for n>3, due to nonassociativity in the "obvious" ways of doing it.

Evaluation from left to right

An alternative for these operators is obtained by evaluation from left to right. Since

$a + b = (a + (b - 1)) + 1$
$a\times b = (a\times (b-1))+a$
$a^b = (a^{(b-1)})\times a$

define (with subscripts instead of superscripts) $a (n + 1) b = (a (n + 1) (b - 1)) (n) a$ with $a (1) b = a + b$ , $a (2) 0 = 0$ , and $a (n) 0 = 1$ for $n > 2$

But this suffers a kind of collapse, failing to form the "power tower" traditionally expected of hyper4: $a_{(4)}b = a^{(a^{(b-1)})}$

How can $a (n) b$ be so different from $a (n) b$ for n>3? This is because of a symmetry called associativity that's defined into + and × (see field) but which ^ lacks. It is more apt to say the two (n)s were decreed to be the same for n<4. (On the other hand, one can object that the field operations were defined to mimic what had been "observed in nature" and ask why "nature" suddenly objects to that symmetry…)

The other degrees do not collapse in this way, and so this family has some interest of its own as lower (perhaps lesser or inferior) hyper operators.

For example:

moser = (..(2^^2)^^..2)^^2 (258 numbers 2)

External links

The Dictionary of Large Numbers (Dictionary's author Robert Munafo claims the (n) notation as his own—it's all right to use it, but it's not a standard.)
Lynz and the Clarkkkkson
On extending hyper4 to nonintegers

Categories: Arithmetic

Last updated: 10-21-2005 09:44:19

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Encyclopedia

Dictionary

Quotes

Hyper operator

Derivation of the notation

Evaluation from left to right

See also

External links

The Online Encyclopedia and Dictionary

Encyclopedia

Dictionary

Quotes

Hyper operator

Derivation of the notation

Evaluation from left to right

See also

External links