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Super-exponentiation

Tetration (also power tower, super-exponentiation) is iterated exponentiation.

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.

Just as exponentiation ("a raised to the power of b") can be thought of as a multiplied by itself b-1 times, tetration can be thought of as a raised to the power of itself b-1 times.

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

• Knuth's up-arrow notation, in which tetration is written <math>a \uparrow\uparrow b<math>.
• Conway chained arrow notation, in which tetration is written <math>a \rightarrow b \rightarrow 2<math>.
• Rudy Rucker's notation, in which tetration is written <math>\ ^b

a<math>.

• Hyper operators

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)

Properties

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}

\\

\\

\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>.