tetration

Tetration (also exponential map, hyperpower, power tower, super-exponentiation, and hyper4) is iterated exponentiation, the first hyper operator after exponentiation. The word tetration was coined by Reuben Louis Goodstein. Tetration is used for the notation of very large numbers. Tetration follows exponentiation in the sequence:

addition
: $a+b$
multiplication
: $}$
exponentiation
: $}$
tetration
: ${\ ^ba = \atop {\ }} \!\!\!\!\!\!\!}}}} \atop {b\mbox{ copies of }a}}$

where each operation is defined by iterating the previous one. We can think of multiplication (

a \times b

) as B instances of A added together, and we can consequently think of exponentiation (

a^b

) as B instances of A multiplied together. So we can go a step further, and think of tetration (

a \uparrow\uparrow b

) as B instances of A exponentiated together. Note that when evaluating multiple-level exponentiation, the exponentiation is done at the deepest level first (in the notation, at the highest level). In other words:

\,\!2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{\left(2^4\right)} = 2^{16} = 65,\!536

\,\!2^{2^{2^2}}

is not equal to

\,\! \left({\left(2^2\right)}^2\right)^2 = 256

Though the latter is a form of iteration also, it is less interesting, since it can be written non-iteratively as

\,\! \left(\left(2^2\right)^2\right)^2 = 2^{2 \cdot 2 \cdot 2} = 2^{2^3}

. The notations in which it can be written (some of which allow further iteration) include:

Standard notation: ${}^ba$ — first used by Maurer; Rudy Rucker's book Infinity and the Mind popularized the notation.
Knuth's up-arrow notation: $a \uparrow\uparrow b$ — allows extension by putting more arrows, or equivalently, an indexed arrow
Conway chained arrow notation: $a \rightarrow b \rightarrow 2$ — allows extension by increasing the number 2 (equivalent with the extensions above), but also, even more powerfully, by extending the chain
hyper4 notation: $a^{(4)}b$ = hyper4 (a, b) = hyper (a, 4, b) — allows extension by increasing the number 4; this gives the family of hyper operators

For the Ackermann function we have

2 \uparrow\uparrow b

= A(4, b−3) + 3, i.e. A(4, n) =

2 \uparrow\uparrow (n+3)

− 3 The up-arrow is used identically to the caret (^), so that the tetration operator may be written as ^^ in ASCII: a^^b.

Examples

$1\uparrow\uparrow2$ = $1^1$ = 1
$2\uparrow\uparrow2$ = $2^2$ = 4
$3\uparrow\uparrow2$ = $3^3$ = 27
$4\uparrow\uparrow2$ = $4^4$ = 256
$5\uparrow\uparrow2$ = $5^5$ = 3,125
$6\uparrow\uparrow2$ = $6^6$ = 46,656
$7\uparrow\uparrow2$ = $7^7$ = 823,543
$8\uparrow\uparrow2$ = $8^8$ = 16,777,216
$9\uparrow\uparrow2$ = $9^9$ = 387,420,489
$10\uparrow\uparrow2$ = $10^{10}$ = 10,000,000,000

$1\uparrow\uparrow3$ = $\,\!1^{1^1}$ = 1
$2\uparrow\uparrow3$ = $\,\!2^{2^2}$ = 16
$3\uparrow\uparrow3$ = $\,\!3^{3^3}$ = 7,625,597,484,987
$4\uparrow\uparrow3$ = $\,\!4^{4^4}$ = $1.34078079\times10^{154}$
$5\uparrow\uparrow3$ = $\,\!5^{5^5}$ = $5^{3125}$ = $1.91\times10^{2184}$ (over 2,000 digits long)
$6\uparrow\uparrow3$ = $\,\!6^{6^6}$ = $6^{46656}$ = $2.66\times10^{36305}$ (over 35,000 digits long)

$1\uparrow\uparrow4$ = $\,\!1^{1^{1^1}}$ = 1
$2\uparrow\uparrow4$ = $\,\!2^{2^{2^2}}$ = 65,536
$3\uparrow\uparrow4$ = $\,\!3^{3^{3^3}}$ = $3^{7,625,597,484,987}$ (over three trillion digits long)

$1\uparrow\uparrow5$ = $\,\!1^{1^{1^{1^1}}}$ = 1
$2\uparrow\uparrow5$ = $\,\!2^{2^{2^{2^2}}}$ = $2^{65536}$ = $2.00\times10^{19728}$ (nearly 20,000 digits long)

Extension to low values of the second operand

Using the relation

n\uparrow\uparrow k = \log_n \left(n\uparrow\uparrow (k+1)\right)

(which follows from the definition of tetration), one can derive (or define) values for

n\uparrow\uparrow k

where

k \in {-1, 0, 1}

\begin{matrix}

   n\uparrow\uparrow 1     & = &   \log_n \left(n\uparrow\uparrow 2\right)     & = &   \log_{n} \left(n^n\right)     & = &    n \log_{n} n      & = &    n

   n\uparrow\uparrow 0     & = &   \log_{n} \left(n\uparrow\uparrow 1\right)     & = &    \log_{n} n     & & & = &   1

   n\uparrow\uparrow -1     & = &   \log_{n} \left(n\uparrow\uparrow 0\right)     & = &   \log_{n} 1     & & & = &    0

\end{matrix} This confirms the intuitive definition of

n\uparrow\uparrow 1

as simply being

n

. However, no further values can be derived by further iteration in this fashion, as

\log_n 0

is undefined. Similarly, since

\log_{1} 1

is also undefined (

\log_{1} 1 = \ln 1{/}\ln 1 = 0/0

), the derivation above does not hold when

n = 1

. Therefore,

1\uparrow\uparrow{-1}

must remain an undefined quantity as well. (The figure

1\uparrow\uparrow{0}

can safely be defined as 1, however.) Again,

0^0

is an undefined quantity, so values for

0\uparrow\uparrow{k}

cannot be defined directly. However,

\lim_{n\rightarrow0} n\uparrow\uparrow{k}

is well defined, and exists:

\lim_{n\rightarrow0} n\uparrow\uparrow k = \begin{cases} 1, & k \mbox{ even} \\ 0, & k \mbox{ odd} \end{cases}

This limit holds for negative

n

, as well.

0\uparrow\uparrow{k}

could be defined in terms of this limit, but

0\uparrow\uparrow2 = 0

would conflict with the standard undefinedness of

0^0

Complex tetration

Since complex numbers can be raised to powers, tetration can be applied to numbers of the form

a + bi

, where i is the square root of −1. For example,

n\uparrow\uparrow k

where

n=i

, tetration is achieved by using the principal branch of the natural logarithm, and noting the relation: i^(a+bi) = e^{iπ/2 (a+bi)} = e^-bπ/2 (cos(aπ/2) + i sin(aπ/2)) . This suggests a recursive definition for

i\uparrow\uparrow (k+1) = a'+b'i

given any

i\uparrow\uparrow k = a+bi

: a' = e^-bπ/2 cos(aπ/2) and b' = e^-bπ/2 sin(aπ/2) The following approximate values can be derived, where

i\uparrow n

is ordinary exponentiation (ie. iⁿ).

$i\uparrow\uparrow1$ = i
$i\uparrow\uparrow2$ = $i\uparrow\left(i\uparrow\uparrow1\right)$ = 0.2079
$i\uparrow\uparrow3$ = $i\uparrow\left(i\uparrow\uparrow2\right)$ = 0.9472+ 0.3208i
$i\uparrow\uparrow4$ = $i\uparrow\left(i\uparrow\uparrow3\right)$ = 0.0501+ 0.6021i
$i\uparrow\uparrow5$ = $i\uparrow\left(i\uparrow\uparrow4\right)$ = 0.3872+ 0.0305i
$i\uparrow\uparrow6$ = $i\uparrow\left(i\uparrow\uparrow5\right)$ = 0.7823+ 0.5446i
$i\uparrow\uparrow7$ = $i\uparrow\left(i\uparrow\uparrow6\right)$ = 0.1426+ 0.4005i
$i\uparrow\uparrow8$ = $i\uparrow\left(i\uparrow\uparrow7\right)$ = 0.5198+ 0.1184i
$i\uparrow\uparrow9$ = $i\uparrow\left(i\uparrow\uparrow8\right)$ = 0.5686+ 0.6051i

Solving the relation yields the expected

i\uparrow\uparrow0

= 1 and

i\uparrow\uparrow-1

= 0, with negative values of k giving infinite results on the imaginary axis. Plotted in the complex plane, the entire sequence spirals to the limit 0.4383+ 0.3606i, which could be interpreted as the value where k is infinite. Such tetration sequences have been studied since the time of Euler but are poorly understood due to their chaotic behavior. Most published research historically has focused on the convergence of the power tower function. Current research has greatly benefited by the advent of powerful computers with fractal and symbolic mathematics software. Much of what is known about tetration comes from general knowledge of complex dynamics and specific research of the exponential map.

Extension to real numbers

Extending x↑↑b to real numbers x>0 is straightforward and gives, for each natural number b, a super-power function f(x) = x↑↑b. (The term super is sometimes replaced by hyper: hyper-power function). As mentioned above, for positive integers b the function tends to 1 for x tending to 0 if b is even, and to 0 if b is odd, while for b=0 and b=−1 the function is constant, with values 1 and 0, respectively. Consider the problem of finding a super-exponential function or hyper-exponential function f(x )=a↑↑x which is an extension to real x>−2 to what was defined above, satisfying (for a>1):

$a\uparrow \uparrow(b+1) = a^{\left(a\uparrow \uparrow b\right)}$
it is monotonically increasing
it is continuous

When a↑↑x is defined for an interval of length one, the whole function easily follows for all x>−2 A simple solution is given by

a \uparrow\uparrow x = x+1

for

-1, hence a \uparrow\uparrow x = a^x for 0, a\uparrow \uparrow x=a^{a^{(x-1)}} for 1, etc.  However, it is only piecewise differentiable; at integer values of x the derivative is multiplied by ln a: 10↑↑.99 = 9.77, 10↑↑1 = 10, 10↑↑1.01 = 10.55.  Other, more complicated solutions may be smoother and/or satisfy additional properties.  A super-exponential function grows even faster than a double-exponential function; for example, if a=10: f(−1)=0, f(0)=1, f(1)=10, f(2)=10

¹⁰, f(2.3)=googol, f(3)=

10^{10^{10}}

, f(3.3)=googolplex. It passes

10^{10^x}

at x = 2.376, f(x) = 4.8310²³⁷. When defining a↑↑x for every a, another possible requirement could be that a↑↑x is monotonically increasing with a. The inverse functions are called super-root or hyper-root, and super-logarithm or hyper-logarithm slog_a defined for all real numbers, also negative numbers. The function slog_a satisfies:

\mathrm{slog}_a a^b = 1 + \mathrm{slog}_a b

\mathrm{slog}_a b = 1 + \mathrm{slog}_a \log_a b

\mathrm{slog}_a b > -2

Examples:

$\mathrm{slog}_{10} -3 = -1 + \mathrm{slog}_{10} .001 = -1 + -.999 = -1.999$
$\mathrm{slog}_{10} 3 = \log_{10} 3 = .477$
$\mathrm{slog}_{10} 10^{6\times 10^{23}} = 1 + \mathrm{slog}_{10} 6\times 10^{23} = 2 + \mathrm{slog}_{10} 23.778 = 3 + \mathrm{slog}_{10} 1.376 = 3 + \log_{10} 1.376 = 3.139$

Other attempts

When 10↑↑ is defined as the x with x↑↑2=10 then 10↑↑=2.51. When 10↑↑ is defined as the x with x↑↑4=10 then 10↑↑=1.73. However, there is no direct relation between the two. Thus this approach may not be suitable as a starting point to extend the definition of a↑↑b to real b. See http://home.earthlink.net/~mrob/pub/math/ln-notes1.html#real-hyper4 for attempts to extend tetration to real numbers. It arrives at e.g. 2↑↑1.2 = 2.22, and correspondingly, 2↑↑2.2 = 2^2.22 = 4.66, and 2↑↑3.2 = 2^4.66 = 25.3, approximately the same as with the definition above.

Infinitely high power towers

\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{..}}}}}}

converges to 2, and can therefore be said to be equal to 2. In general, let

r

be a positive real number. Let

x = r^{1/r}

. Then the infinite power tower

x^{x^{x^{..}}}

converges to

r

, provided that r is not more than Euler's number e, hence

x

is not more than

e^{1/e}

External links

Extension of the hyper4 function to reals Robert Munafo discusses extending tetration to the real numbers.
Mathematics brain teasers Ioannis Galidakis does extensive research on tetration. Galidakis’s web site contains the definitive list of references to tetration research. Lots of information on the Lambert W function, Riemann surfaces, and analytic continuation.
Power Tower From MathWorld--A Wolfram Web Resource. Nice website on background information relevant to tetration, with new content constantly being added.
Some Critical Points of the Hyperpower Function This site based on one of the few papers ever published specifically devoted to tetration.
Web pages for infinitely iterated exponentials Dave L. Renfro has compiled entries from questions about tetration on sci.math.

References

R. Knobel
Exponentials Reiterated, Amer. Math. Monthly 88, (1981), p. 235-252
Hans Maurer
ber die Funktion $y=x^{x^{[x(\cdots)}]}$ fr ganzzahliges Argument (Abundanzen), Mittheilungen der Mathematische Gesellschaft in Hamburg 4, (1901), p. 33-50; reference to usage of $\ ^ba$ from Knobel's paper.
Reuben Louis Goodstein
Transfinite ordinals in recursive number theory, Journal of Symbolic Logic 12, (1947)

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