We’re doing this again!

I’m skipping Fish Number 2 because I’m really confused by it. I’ll do it next time.

Remember Fish Number 1? With the S and SS map? We have to define a new map. The s(n) MAP!

This is mainly just copied from the Googology Wikia.

Define s(n) map as:

- s(1)f := g; g(x) = f
^{x}(x) - s(n)f := g; g(x) = [s(n-1)
^{x}]f(x) (if n > 1)

where s(n) is a functional, and the rate of growth in the fast-growing hierarchy is s(x)f(x) ≈ f_{ω^ω}(x).

Define ss(n) map to be:

- ss(1)f := g; g(x) = s(x)f(x)
- ss(n)f := g; g(x) = [ss(n-1)
^{x}]f(x) (if n > 1)

and its growth rate is:

- ss(1)f(x) = s(x)f(x) ≈ f
_{ω^ω}(x) - ss(n)f(x) ≈ f
_{ω^(ω+n+1)}(x)

Huge.

Definition and the growth rate of Fish function 3:

- F
_{3}(x) := ss(2)^{63}f; f(x) = x + 1 - F
_{3}(x) ≈ f_{(ω^(ω+1)) * 63}

Fish Number 3 is:

F_{3} := F_{3}^{63}(3) ≈ f_{ωω+1 * 63 + 1}(63)

Wat

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