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m(m,n) map
Based on$$f^x(x)$$
Growth rate$$f_{\zeta_0}(n)$$

m(m,n) map is a function which maps maps to maps. It was defined by Japanese googologist Fish in 2007[1] and used to define Fish number 6. It is a 2 variable extension of m(n) map and has a growth rate of $$f_{\zeta_0}(n)$$.

## Definition Edit

$$M[m,n]$$ map $$(m=0,1,...; n=1,2,...)$$ is defined as follows.

• $$M[0,1]$$ is a set of function of natural numbers to natural numbers.
• $$M[m+1,1]$$ is a set that has one element from each $$M[m,n] (n=1,2,...)$$; $$M[m+1,1]$$ is a direct product of $$M[m,1], M[m,2], M[m,3], ...$$.
• Element of $$M[m,1]$$ also works as a function in $$M[0,1]$$ that is element of element of ... element of $$M[m,1]$$.
• $$M[m,n+1] (n=1,2,...)$$ is a set of maps from $$M[m,n]$$ to $$M[m,n]$$.

Element of $$M[m,n]$$, $$m(m,n)$$ is defined as follows. Here, $$a_i, b_i,f_i$$ are elements of $$m(m,i)$$ and strict structure of definition is same as m(n) map.

\begin{eqnarray*} m(0,1) (x) & := & x+1 \\ m(m,n+1) f_n f_{n-1} ...f_1 (x) & := & f_n^xf_{n-1}... f_1 (x) \\ & & (m=0; n=1,2,... \text{ or } m=1,2,...; n=2,3,…) \\ m(m+1,1) & := & [m(m,1),m(m,2),m(m,3),…] \\ m(m+1,2)[a_1,a_2,...] & := & [b_1,b_2,…] \text{ where } b_n \text{ is defined as:} \\ b_n f_{n-1}...f_1(x) & := & a_y a_{y-1}...a_n f_{n-1}…f_1(x) (y=max(x,n)) \end{eqnarray*}

## Calculation Edit

\begin{eqnarray*} m(1,1)(x) & = & [m(0,1),m(0,2),m(0,3),…](x) \\ & = & m(0,1)(x) = x+1 \\ \end{eqnarray*} Let $$m(1,2) m(1,1) = [a_1,a_2,a_3,…]$$ and \begin{eqnarray*} a_1(x) & = & m(0,x) m(0,x-1) … m(0,1) (x) \\ & \approx & f_{\varepsilon_0}(x) \\ \therefore m(1,2) m(1,1)(x) & \approx & f_{\varepsilon_0}(x) \\ m(0,2) a_1(x) & \approx & f_{\varepsilon_0 + 1}(x) \\ m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega}(x) \\ m(0,4) m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega^{\omega}}(x) \\ m(0,5) m(0,4) m(0,3) m(0,2) a_1 (x) & \approx & f_{\varepsilon_0 + \omega^{\omega^{\omega}}}(x) \\ a_2 a_1(x) & = & m(0,y) m(0,y-1) …m(0,2) a_1(x) (y=max(x,2)) \\ & \approx & f_{\varepsilon_0 \times 2}(x) \\ \end{eqnarray*}

Then,

\begin{eqnarray*} m(0,3) a_2 a_1 (x) & \approx & f_{\varepsilon_0 \times \omega}(x) \\ m(0,4) m(0,3) a_2 a_1 (x) & \approx & f_{\varepsilon_0 \times \omega^{\omega}}(x) \\ m(0,5) m(0,4) m(0,3) a_2 a_1 (x) &\approx & f_{\varepsilon_0 \times \omega^{\omega^{\omega}}}(x) \\ a_3 a_2 a_1(x) & = & m(0,y) m(0,y-1) ... m(0,3) a_2 a_1 (x) (y=max(x,3)) \\ & \approx & f_{\varepsilon_0 ^2}(x) \end{eqnarray*} As for $$a_4$$, \begin{eqnarray*} m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^\omega}(x) \\ m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega \times 2}}(x) \\ m(0,6) m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega ^ 2}}(x) \\ m(0,7) m(0,6) m(0,5) m(0,4) a_3 a_2 a_1 (x) & \approx & f_{\varepsilon_0 ^{\omega^{\omega}}}(x) \\ a_4 a_3 a_2 a_1(x) & = & m(0,y) m(0,y-1)...m(0,4) a_3 a_2 a_1(x) (y=max(x,4)) \\ & \approx & f_{\varepsilon_0 ^{\varepsilon_0}}(x) \end{eqnarray*}

And similarly, $$a_5$$, $$a_6$$ can be calculated as \begin{eqnarray*} a_5 a_4 a_3 a_2 a_1(x) & \approx & f_{\varepsilon_0 ^{\varepsilon_0^{\varepsilon_0}}}(x) \\ a_6 a_5 a_4 a_3 a_2 a_1(x) & \approx & f_{\varepsilon_0 ^{\varepsilon_0^{\varepsilon_0^{\varepsilon_0}}}}(x) \\ \end{eqnarray*}

Therefore, \begin{eqnarray*} m(1,2)^2 m(1,1) (x) &=& m(1,2)[a_1,a_2,...](x) \\ & \approx & f_{\varepsilon_1}(x) \end{eqnarray*}

Now let \begin{eqnarray*} m(1,2)^3 m(1,1)(x) = [b_1,b_2,b_3,...](x) \end{eqnarray*} and for calculating $$b_i$$, $$\varepsilon_0$$ in the calculation of $$a_i$$ is changed to $$\varepsilon_1$$. Therefore,

\begin{eqnarray*} m(1,2)^3 m(1,1)(x) & \approx & f_{\varepsilon_2}(x) \\ m(1,2)^4 m(1,1)(x) & \approx & f_{\varepsilon_3}(x) \\ m(1,2)^n m(1,1)(x) & \approx & f_{\varepsilon_{n-1}}(x) \\ \end{eqnarray*}

Then calculation goes on similar to $$m(n)$$ map. \begin{eqnarray*} m(1,3) m(1,2) m(1,1) (x) & \approx & f_{\varepsilon_\omega} \\ m(1,4) m(1,3) m(1,2) m(1,1) (x) & \approx & f_{\varepsilon_{\omega^\omega}} \\ m(1,5) m(1,4) m(1,3) m(1,2) m(1,1) (x) & \approx & f_{\varepsilon_{\omega^{\omega^\omega}}} \\ m(2,2) m(2,1) (x) &=& m(1,x) m(1,x-1) ... m(1,2) m(1,1) (x) \\ & \approx & f_{\varepsilon_{\varepsilon_0}} \\ \end{eqnarray*}

And then \begin{eqnarray*} m(2,2)^2 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_1}}(x) \\ m(2,2)^3 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_2}}(x) \\ m(2,2)^4 m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_3}}(x) \\ m(2,3) m(2,2) m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_\omega}}(x) \\ m(3,2) m(3,1) (x) &=& m(2,x) m(2,x-1) ... m(2,2) m(2,1) (x) \\ & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_0}}}(x) \end{eqnarray*}

As we have calculated to $$m(3,2) m(3,1) (x)$$, it is easy to see that \begin{eqnarray*} m(1,2) m(1,1) (x) & \approx & f_{\varepsilon_0} (x) \\ m(2,2) m(2,1) (x) & \approx & f_{\varepsilon_{\varepsilon_0}} (x) \\ m(3,2) m(3,1) (x) & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_0}}} (x) \\ m(4,2) m(4,1) (x) & \approx & f_{\varepsilon_{\varepsilon_{\varepsilon_{\varepsilon_0}}}} (x) \\ \end{eqnarray*} and therefore

$m(x,2)m(x,1)(x) \approx f_{\zeta_0}(x)$

## Sources Edit

1. Fish, Googology in Japan - exploring large numbers (2013)