User:Kyodaisuu/Sandbox

This is a personal sandbox to prepare for publishing at this wiki. ja:利用者:Kyodaisuu/砂場

Fish number 5

m(n) map

m(n) map
m(n) map is a, which is a function which maps maps to maps. It was defined by Japanese googologist Fish in 2003 and used to define Fish number 5. It has a growth rate of \(f_{\varepsilon_0}(n)\), similar to Hydra function and tetrational array of BEAF. The name of the map was taken from a word mapping.

m(1) is a normal function which maps from numbers to numbers, and m(2) is a function which maps from functions to functions, defined as

\begin{eqnarray*} m(1)(x) & = & x^x \\ m(2)f(x) & = & f^x(x) \end{eqnarray*}

where \(m(2)\) matches \(s(1)\) of s(n) map; \(m(2)=s(1)\).

m(3) map is a function which maps "function which maps from functions to function" to "function which maps from functions to function", defined and calculated as (parenthesis is written verbosely here to indicate the order of calculation)

\begin{eqnarray*} ((m(3)m(2))f)(x) & = & (m(2)^xf)(x) \\ & = & (s(1)^xf)(x) \\ & = & (s(2)f)(x) \end{eqnarray*} Therefore, \(m(3) m(2) = s(2)\) in \(s(x)\) map \begin{eqnarray*} ((m(3)^2m(2))f)(x) & = & ((m(3)(m(3)m(2)))f)(x) \\ & = & ((m(3)m(2))^xf)(x) \\ & = & (s(2)^xf)(x) \\ & = & (s(3)f)(x) \end{eqnarray*}

Calculation goes on similarly, with FGH approximation for \(f(x)=x+1\),

\begin{eqnarray*} m(3)^3 m(2) f(x) & = & s(4)f(x) \approx f_{\omega^3}(x) \\ m(3)^4 m(2) f(x) & = & s(5)f(x) \approx f_{\omega^4}(x) \\ m(3)^n m(2) f(x) & = & s(n+1)f(x) \approx f_{\omega^n}(x) \\ m(3)^x m(2) f(x) & = & s(x)f(x) \approx f_{\omega^\omega}(x) \\ \end{eqnarray*}

In general


 * \(M_0\) = set of nonegative integers
 * \(M_{n+1}\) = set of functions which map \(M_n\) to \(M_n\)

\(m(n)\) map \((n \ge 1)\) belongs to \(M_n\) and is defined as follows.
 * For \(f_n \in M(n)\), \(m(n+1)(f_n) = g_n\) is defined as follows.
 * For \(f_{n-1} \in M(n-1)\), \(g_n(f_{n-1}) = g_{n-1}\) is defined as follows.
 * For \(f_{n-2} \in M(n-2)\), \(g_{n-1}(f_{n-2}) = g_{n-2}\) is defined as follows.
 * For \(f_0 \in M(0)\), \(g_1(f_0) = g_0\) is defined as follows.
 * \(g_0 = (..((f_n^{f_0}f_{n-1})f_{n-2})...f_1)f_0\)
 * \(g_0 = (..((f_n^{f_0}f_{n-1})f_{n-2})...f_1)f_0\)

Therefore, by denoting \(f_1)\) as f and \(f_0\) as x, \begin{eqnarray*} m(1)(x) & = & {x}^{x} \\ (m(2)f)(x) & = & ({f}^{x})(x) \\ (..((m(n+1)f_n)f_{n-1})...f_1)(x) & := & ((..({f_n}^{f_0}{f_{n-1}})...f_2)f)(x) \end{eqnarray*}

And it works out like this.

\begin{eqnarray*} m(4) m(3) m(2) f(x) & \approx & f_{\omega^\omega}(x) \\ m(3) [m(4) m(3)] m(2) f(x) & \approx & f_{\omega^{\omega+1}}(x) \\ m(3)^2 [m(4) m(3)] m(2) f(x) & \approx & f_{\omega^{\omega+2}}(x) \\ m(3)^a [m(4) m(3)] m(2) f(x) & \approx & f_{\omega^{\omega+a}}(x) \\ \bigl[ m(4) m(3) \bigr]^2 m(2) f(x) & = & f_{\omega^{\omega×2}}(x) \end{eqnarray*} ここまでが、$F_3$の定義で記述できるところです. さらに、$F_3$の拡張である多変数化したs(x)を使うことで、 \begin{eqnarray*} \bigl[ m(4) m(3) \bigr] ^2 m(2) & = & ss(n) = s(1,3,1) = F \bigl[ +\omega^{\omega×2} \bigr] \\ \bigl[ m(4) m(3) \bigr] ^3 m(2) & = & s(1,4,1) = F \bigl[ +\omega^{\omega×3} \bigr] \\ \bigl[ m(4) m(3) \bigr] ^a m(2) & = & s(1,a+1,1)= F \bigl[ +\omega^{\omega×a} \bigr] \\ \bigl[ m(4)^2 m(3) \bigr] m(2) & = & s(2,1,1) = F \bigl[ +\omega^{\omega^2} \bigr]  \\ m(3) \bigl[ m(4)^2 m(3) \bigr]  m(2) & = & s(2,1,2) = F \bigl[ +\omega^{\omega^2+1} \bigr]  \\ \bigl[ m(4) m(3) \bigr]  \bigl[ m(4)^2 m(3) \bigr]  m(2) & = & s(2,2,1) = F \bigl[ +\omega^{\omega^2+\omega} \bigr]  \\ \bigl[ m(4) m(3) \bigr] ^2 \bigl[ m(4)^2 m(3) \bigr]  m(2) & = & s(2,3,1) = F \bigl[ +\omega^{\omega^2+\omega \times 2} \bigr]  \\ \bigl[ m(4) m(3) \bigr] ^3 \bigl[ m(4)^2 m(3) \bigr]  m(2) & = & s(2,4,1) = F \bigl[ +\omega^{\omega^2+\omega \times 3} \bigr]  \\ \bigl[ m(4)^2 m(3) \bigr] ^2 m(2) & = & s(3,1,1) = F \bigl[ +\omega^{\omega^2 \times 2} \bigr] (x) \\ \bigl[ m(4)^2 m(3) \bigr] ^3 m(2) & = & s(4,1,1) = F \bigl[ +\omega^{\omega^2 \times 3} \bigr] (x) \\ m(4)^3 m(3) m(2) & = & s(x,1,1) = F \bigl[ +\omega^{\omega^3} \bigr] \\ m(4)^4 m(3) m(2) & = & s(x,1,1,1) = F \bigl[ +\omega^{\omega^4} \bigr] \\ m(5) m(4) m(3) m(2) & = & s(1,1,...(1がx個),1,1) = F \bigl[ +\omega^{\omega^\omega} \bigr] \end{eqnarray*} このように計算されます.

これまでの計算から、 \begin{eqnarray*} m(3) m(2) & = & F[+\omega] \\ m(4) m(3) m(2) & = & F[+\omega^\omega] \\ m(5) m(4) m(3) m(2) & = & F[+\omega^{\omega^\omega}] \end{eqnarray*} となり、同様に \begin{eqnarray*} m(6) m(5) m(4) m(3) m(2) & = & F[+\omega^4] \\ m(7) m(6) m(5) m(4) m(3) m(2) & = & F[+\omega^5] \\ & … & \end{eqnarray*} となることが期待されます. この時、