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s(n) map
Based on$$f^x(x)$$
Growth rate$$f_{\omega^\omega}(n)$$

s(n) map is a functional, which is a function which maps functions to functions. It was defined by Japanese googologist Fish in 2002[1] and used to define Fish number 3. The name of the map was taken from the Japanese word shazou, which means mapping.

## Definition

\begin{eqnarray*} s(1)f & = & f^x(x) \\ s(n)f & = & s(n-1)^xf(x) (\text{if } n>1) \end{eqnarray*}

## Analysis

Let $$f(x) = x+1$$, and the growth rate can be calculated as:

\begin{eqnarray*} f^2(x) & = & x+2 \\ f^3(x) & = & x+3 \\ s(1)f(x) & = & f^x(x) = x+x = 2x \\ s(1)^2f(x) & = & g^x(x) = 2^x x > 2^x\text{, where }g(x)=2x \\ s(1)^3f(x) & > & h^{x}(x) > 2 \uparrow^2x \text{, where } h(x)=2^x \\ s(2)f(x) & = & s(1)^{x}f(x) > 2\uparrow^xx>A(x,x)=A(1,0,x) \approx f_{\omega}(x) \end{eqnarray*}

Here, $$A$$ is Taro's multivariable Ackermann function, where the growth rate in FGH is: \begin{eqnarray*} A(..., a3, a2, a1, a0, n) \approx f_{... + \omega^3･a3 + \omega^2･a2 + \omega･a1 + a0}(n) \end{eqnarray*}

Let $$f(n) = A(X, b, n)$$ (X is a vector in any length), and: \begin{eqnarray*} A(X, b+1, n) & = & A(X, b, A(X, b+1, n-1)) \\ & = & f(A(X , b+1, n-1)) \\ & = & f^2(A(X, b+1, n-2)) \\ & = & … = f^n(A(X, b+1, 0)) \\ & \approx & f^n(n) \end{eqnarray*}

Therefore, comparing the 3 functions,

• $$s(1)f(x) = f^x(x)$$
• $$f_{\alpha+1}(n) = f^n_\alpha(n)$$
• $$A(X, b+1, n) = f^n(n)$$ where $$f(n) = A(X, b, n)$$

they all have similar growth rate. $$s(1)$$ map has the same effect of adding 1 to the ordinal in FGH and adding 1 to the second parameter from right in the Ackermann function. This results in:

\begin{eqnarray*} s(1)s(2)f(x) & \approx & A(1,1,x) \approx f_{\omega + 1}(x) \\ s(1)^2 s(2)f(x) & \approx & A(1,2,x) \approx f_{\omega + 2}(x) \\ s(1)^n s(2)f(x) & \approx & A(1,n,x) \approx f_{\omega + n}(x) \end{eqnarray*}

and by diagonizing $$s(1)$$ again,

\begin{eqnarray*} s(2)^2 f(x) = s(1)^x s(2)f(x) \approx A(1,x,x) = A(2,0,x) \approx f_{\omega \times 2}(x) \end{eqnarray*}

Here, the calculation of $$s(2)^2f(3)$$ goes as follows:

\begin{eqnarray*} s(2)^2f(3) &=& s(1)^3s(2) f(3) \\ &=& [s(1)^2 s(2)f]^3(3) \\ &=& [s(1)^2 s(2)f]^2[[s(1)s(2)f]^3(3)] \end{eqnarray*}

For this calculation, by changing $$s(2)^2 f$$ to $$f_{\omega \times 2}$$, $$s(1)^3 s(2)f$$ to $$f_{\omega+3}$$, $$s(1)^2s(2)f$$ to $$f_{\omega+2}$$, and $$s(1)s(2)f$$ to $$f_{\omega+1}$$, respectively, the following is obtained:

\begin{eqnarray*} f_{\omega \times 2}(3) &=& f_{\omega+3}(3) \\ &=& f_{\omega+2}^3(3) \\ &=& f_{\omega+2}^2(f_{\omega+1}^3(3)) \end{eqnarray*}

which shows exactly how FGH is calculated.

Calculation goes in the same way:

\begin{eqnarray*} s(2)^n f(x) & \approx & A(n,0,x) \approx f_{\omega \times n}(x) \\ s(3)f(x) & = & s(2)^{x}f(x) \approx A(x,0,x) = A(1,0,0,x) \approx f_{\omega^2}(x) \\ s(3)^2 f(x) & \approx & A(2,0,0,x) \approx f_{\omega^2 \times 2}(x) \\ s(3)^n f(x) & \approx & A(n,0,0,x) \approx f_{\omega^2 \times n}(x) \\ s(4)f(x) & = & s(3)^{x}f(x) \approx A(x,0,0,x) = A(1,0,0,0,x) \approx f_{\omega^3}(x) \\ s(1)^4 s(2)^3 s(3)^2s(4)f(x) & \approx & A(1,2,3,4,x) \approx f_{\omega^3+\omega^2 \times 2+\omega \times 3 + 4}(x) \\ s(5)f(x) & \approx & f_{\omega^4}(x) \\ s(6)f(x) & \approx & f_{\omega^5}(x) \\ s(n)f(x) & \approx & f_{\omega^{n-1}}(x) \\ s(x)f(x) & \approx & f_{\omega^\omega}(x) \end{eqnarray*}

Therefore, by applying $$s(x)$$ map, which diagonizes $$s(n)$$ map, to the function $$f(x)=x+1$$, the growth rate is $$f_{\omega^\omega}(x)$$, similar to array notation and Taro's multivariable Ackermann function.

## Sources

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Other: Taro's multivariable Ackermann function · TR function · Arai's $$\psi$$ · Sushi Kokuu Hen

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