Tetration

Tetration (also hyper4, power tower, super-exponentiation) is iterated exponentiation, the first hyper operator after exponentiation.

Tetration follows exponentiation in the sequence:

1. addition

   ${\displaystyle a+b}$

2. multiplication

   ${\displaystyle {{\ } \atop a\times b=}{{b{\mbox{ copies of }}a} \atop {\overbrace {a+\cdots +a} }}}$

3. exponentiation

   ${\displaystyle {{\ } \atop a^{b}=}{b{\mbox{ copies of }}a \atop {\overbrace {a\times \cdots \times a} }}}$

4. tetration

   ${\displaystyle {a\uparrow \uparrow b= \atop {\ }}\!\!\!\!\!\!\!{{\underbrace {a^{a^{\cdot ^{\cdot ^{a}}}}} } \atop {b{\mbox{ copies of }}a}}}$

where each operation is defined by iterating the previous one.

We can think of multiplication (${\displaystyle a\times b}$) as B instances of A added together, and we can consequently think of exponentiation (${\displaystyle a^{b}}$) as B instances of A multiplied together. So we can go a step further, and think of tetration (${\displaystyle a\uparrow \uparrow b}$) 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:

${\displaystyle \,\!2^{2^{2^{2}}}=2^{(2^{(2^{2})})}=2^{(2^{4})}=2^{16}=65,\!536}$

${\displaystyle 2^{2^{2^{2}}}}$ is not equal to ${\displaystyle \,\!\left({(2^{2})}^{2}\right)^{2}=256}$

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: ${\displaystyle \ ^{b}a}$.
• hyper4 notation: ${\displaystyle a^{(4)}b}$ = 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: ${\displaystyle a\uparrow \uparrow b}$ - allows extension by putting more arrows, or equivalently, an indexed arrow
• Conway chained arrow notation: ${\displaystyle a\rightarrow b\rightarrow 2}$ - allows extension by increasing the number 2 (equivalent with the extensions above), but also, even more powerfully, by extending the chain

For the Ackermann function we have ${\displaystyle 2\uparrow \uparrow b}$ = A(4, b-3) + 3, i.e. A(4, n) = ${\displaystyle 2\uparrow \uparrow (n-3)}$ - 3

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

Examples

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

• ${\displaystyle 1\uparrow \uparrow 3}$ = ${\displaystyle \,\!1^{1^{1}}}$ = 1
• ${\displaystyle 2\uparrow \uparrow 3}$ = ${\displaystyle \,\!2^{2^{2}}}$ = 16
• ${\displaystyle 3\uparrow \uparrow 3}$ = ${\displaystyle \,\!3^{3^{3}}}$ = 7,625,597,484,987
• ${\displaystyle 4\uparrow \uparrow 3}$ = ${\displaystyle \,\!4^{4^{4}}}$ = ${\displaystyle 1.34078079\times 10^{154}}$
• ${\displaystyle 5\uparrow \uparrow 3}$ = ${\displaystyle \,\!5^{5^{5}}}$ = ${\displaystyle 5^{3125}}$ = ${\displaystyle 1.91\times 10^{2184}}$ (over 2,000 digits long)
• ${\displaystyle 6\uparrow \uparrow 3}$ = ${\displaystyle \,\!6^{6^{6}}}$ = ${\displaystyle 6^{46656}}$ = ${\displaystyle 2.66\times 10^{36305}}$ (over 35,000 digits long)

• ${\displaystyle 1\uparrow \uparrow 4}$ = ${\displaystyle \,\!1^{1^{1^{1}}}}$ = 1
• ${\displaystyle 2\uparrow \uparrow 4}$ = ${\displaystyle \,\!2^{2^{2^{2}}}}$ = 65,536
• ${\displaystyle 3\uparrow \uparrow 4}$ = ${\displaystyle \,\!3^{3^{3^{3}}}}$ = ${\displaystyle 3^{7,625,597,484,987}}$ (over three trillion digits long)

• ${\displaystyle 1\uparrow \uparrow 5}$ = ${\displaystyle \,\!1^{1^{1^{1^{1}}}}}$ = 1
• ${\displaystyle 2\uparrow \uparrow 5}$ = ${\displaystyle \,\!2^{2^{2^{2^{2}}}}}$ = ${\displaystyle 2^{65536}}$ = ${\displaystyle 2.00\times 10^{19728}}$ (nearly 20,000 digits long)

Extension to low values of the second operand

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

${\displaystyle {\begin{matrix}n\uparrow \uparrow 1&=&\log _{n}(n\uparrow \uparrow 2)&=&\log _{n}(n^{n})&=&n\log _{n}n&=&n\\n\uparrow \uparrow 0&=&\log _{n}(n\uparrow \uparrow 1)&=&\log _{n}n&&&=&1\\n\uparrow \uparrow -1&=&\log _{n}(n\uparrow \uparrow 0)&=&\log _{n}1&&&=&0\end{matrix}}}$

This confirms the intuitive definition of ${\displaystyle n\uparrow \uparrow 1}$ as simply being ${\displaystyle n}$. However, no further values can be derived by further iteration in this fashion, as ${\displaystyle \log _{n}0}$ is undefined.

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

Again, ${\displaystyle 0^{0}}$ is an undefined quantity, so values for ${\displaystyle 0\uparrow \uparrow {k}}$ cannot be defined directly. However, ${\displaystyle \lim _{n\rightarrow 0}n\uparrow \uparrow {k}}$ is well defined, and exists:

${\displaystyle \lim _{n\rightarrow 0}n\uparrow \uparrow k={\begin{cases}1,&k{\mbox{ even}}\\0,&k{\mbox{ odd}}\end{cases}}}$

This limit holds for negative ${\displaystyle n}$, as well. ${\displaystyle 0\uparrow \uparrow {k}}$ could be defined in terms of this limit, but ${\displaystyle 0\uparrow \uparrow 2=0}$ would conflict with the standard undefinedness of ${\displaystyle 0^{0}}$.

Complex tetration

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

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

The following approximate values can be derived, where ${\displaystyle i\uparrow n}$  is ordinary exponentiation (ie. iⁿ).

• ${\displaystyle i\uparrow \uparrow 1}$  = i
• ${\displaystyle i\uparrow \uparrow 2}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 1)}$  = 0.2079
• ${\displaystyle i\uparrow \uparrow 3}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 2)}$  = 0.9472+ 0.3208i
• ${\displaystyle i\uparrow \uparrow 4}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 3)}$  = 0.0501+ 0.6021i
• ${\displaystyle i\uparrow \uparrow 5}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 4)}$  = 0.3872+ 0.0305i
• ${\displaystyle i\uparrow \uparrow 6}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 5)}$  = 0.7823+ 0.5446i
• ${\displaystyle i\uparrow \uparrow 7}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 6)}$  = 0.1426+ 0.4005i
• ${\displaystyle i\uparrow \uparrow 8}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 7)}$  = 0.5198+ 0.1184i
• ${\displaystyle i\uparrow \uparrow 9}$  = ${\displaystyle i\uparrow (i\uparrow \uparrow 8)}$  = 0.5686+ 0.6051i

Solving the relation yields the expected ${\displaystyle i\uparrow \uparrow 0}$  = 1 and ${\displaystyle i\uparrow \uparrow -1}$  = 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.

Extension to real numbers

Extending x^^b to real numbers b is straightforward and gives, for each natural number b, a super-power function f(x) = x^^b.

For a super-exponential function f(x)=a^^x one may require:

• it is correct for natural numbers x
• a^^(b+1) = a^(a^^b)
• it is a smooth function, monotonically increasing

When this is defined for 0<x<1 the whole function easily follows for all x>-2

If, for example 10^^0.5 = log 2, then 10^^1.5 = 2, 10^^ 2.5 - 100, etc.

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.

The inverse functions can be called super-root and super-logarithm.