Tetration

Tetration (also hyper4, power tower, super-exponentiation) is iterated exponentiation, the first hyper operator after exponentiation. It is used for the notation of very large numbers.

Tetration follows exponentiation in the sequence:

1. addition

   <math>a+b<math>

2. multiplication

   <math>{{\ } \atop a \times b = } {{b\mbox{ copies of }a} \atop {\overbrace{a + \cdots + a}}}<math>

3. exponentiation

   <math>{{\ } \atop a^b = } {b\mbox{ copies of }a \atop {\overbrace{a \times \cdots \times a}}}<math>

4. tetration

   <math>{a \uparrow\uparrow b = \atop {\ }} \!\!\!\!\!\!\!{{\underbrace{a^{a^{\cdot^{\cdot^{a}}}}}} \atop {b\mbox{ copies of }a}}<math>

where each operation is defined by iterating the previous one.

We can think of multiplication (<math>a \times b<math>) as B instances of A added together, and we can consequently think of exponentiation (<math>a^b<math>) as B instances of A multiplied together. So we can go a step further, and think of tetration (<math>a \uparrow\uparrow b<math>) as B instances of A exponentiated together.

Note that when solving multiple-level exponentiation, the exponentiation is done at the deepest level first (in the notation, at the highest level). In other words:

<math>\,\!2^{2^{2^2}} = 2^{(2^{(2^2)})} = 2^{(2^4)} = 2^{16} = 65,\!536<math>

<math>\,\!2^{2^{2^2}}<math> is not equal to <math>\,\! \left({(2^2)}^2\right)^2 = 256<math>

There is no standard notation for tetration. The notations in which it can be written (some of which allow further iteration) include:

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

For the Ackermann function we have <math>2 \uparrow\uparrow b<math> = A(4, b−3) + 3, i.e. A(4, n) = <math>2 \uparrow\uparrow (n-3)<math> − 3

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

Examples

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

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

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

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

Extension to low values of the second operand

Using the relation <math>n\uparrow\uparrow k = \log_n \left(n\uparrow\uparrow (k+1)\right)<math> (which follows from the definition of tetration), one can derive (or define) values for <math>n\uparrow\uparrow k<math> where <math>k \in {-1, 0, 1}<math>.

<math> \begin{matrix}

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

\\

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

\\

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

\end{matrix} <math>

This confirms the intuitive definition of <math>n\uparrow\uparrow 1<math> as simply being <math>n<math>. However, no further values can be derived by further iteration in this fashion, as <math>\log_n 0<math> is undefined.

Similarly, since <math>\log_{1} 1<math> is also undefined (<math>\log_{1} 1 = \ln 1{/}\ln 1 = 0/0<math>), the derivation above does not hold when <math>n = 1<math>. Therefore, <math>1\uparrow\uparrow{-1}<math> must remain an undefined quantity as well. (The figure <math>1\uparrow\uparrow{0}<math> can safely be defined as 1, however.)

Again, <math>0^0<math> is an undefined quantity, so values for <math>0\uparrow\uparrow{k}<math> cannot be defined directly. However, <math>\lim_{n\rightarrow0} n\uparrow\uparrow{k}<math> is well defined, and exists:

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

This limit holds for negative <math>n<math>, as well. <math>0\uparrow\uparrow{k}<math> could be defined in terms of this limit, but <math>0\uparrow\uparrow2 = 0<math> would conflict with the standard undefinedness of <math>0^0<math>.

Complex tetration

Since complex numbers can be raised to powers, tetration can be applied to numbers of the form <math>a + bi<math>, where i is the square root of −1. For example, <math>n\uparrow\uparrow k<math> where <math>n=i<math>, 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 <math>i\uparrow\uparrow (k+1) = a'+b'i<math> given any <math>i\uparrow\uparrow k = a+bi<math>:

a' = e^-bπ/2 cos(aπ/2) and b' = e^-bπ/2 sin(aπ/2)

The following approximate values can be derived, where <math>i\uparrow n<math> is ordinary exponentiation (ie. iⁿ).

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

Solving the relation yields the expected <math>i\uparrow\uparrow0<math> = 1 and <math>i\uparrow\uparrow-1<math> = 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.

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^^(b+1) = a^(a^^b)
• 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^^x=x+1 for −1<x<0, hence a^^x=a^x for 0<x<1, a^^x=a^a^(x−1) for 1<x<2, 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)=1e10, f(2.3)=googol, f(3)=10^1e10, f(3.3)=googolplex. It passes 10^10^x at x = 2.376, f(x) = 4.83e237.

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:

slog_aa^b = 1 + slog_ab

slog_ab = 1 + slog_a log_ab

slog_ab > −2

Examples:

• slog₁₀−3 = −1 + slog₁₀.001 = −1 + −.999 = −1.999
• slog₁₀3 = log₁₀3 = .477
• slog₁₀10^6e23 = 1 + slog₁₀6e23 = 2 + slog₁₀ 23.778 = 3 + slog₁₀ 1.376 = 3 + log₁₀ 1.376 = 3 + .139 = 3.139

Other attempts

When 10^^(1/2) is defined as the x with x^^2=10 then 10^^(1/2)=2.51. When 10^^(1/4) is defined as the x with x^^4=10 then 10^^(1/4)=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

<math>\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{..}}}}}}<math> converges to 2, and can therefore be said to be equal to 2. However, if e.g. we replace the 2s by 3s there is no convergence.