Forum:Why does Madore's Psi function get stuck at zeta 0 ?

I don't fully understand the reason why $$\psi(\alpha)=\zeta_0$$ when $$\zeta_0 \leq\alpha\leq\Omega$$, and though I asked, it still seems arbitrary to me. Consider that if $$\psi(0)=\epsilon_0$$, $$\psi(1)=\epsilon_1$$, $$\psi(2)=\epsilon_2$$, and so on, then in general, $$\psi(\alpha)=\epsilon_{\alpha}$$. So why should $$\psi(\zeta_0 +1)$$ equal to $$\zeta_0$$ instead of $$\epsilon_{\zeta_0 +1}$$ ?

Also, if $$\psi(\zeta_0)=\zeta_0$$, then logically $$\psi(\zeta_0 +1)$$ could be seen to be equal to $${\zeta_0}^{{\zeta_0}^{{\zeta_0}^{.^{.^{.}}}}}$$, according to the definition of the psi function. Therefore, would $$\epsilon_{\zeta_0 +1}={\zeta_0}^{{\zeta_0}^{{\zeta_0}^{.^{.^{.}}}}}$$ ?

Clearly this cannot be so, but I would like to know where I went wrong, and how, in complete and understandable terms, the Psi function gets stuck at $$\zeta_0$$, and why $$\psi(\zeta_0 +1)$$ should behave any differently than $$\psi(\Omega +1)$$. Edwin Shade (talk) 01:24, November 13, 2017 (UTC)