User blog:B1mb0w/Zeta Nought of Two

What is \(f_{\zeta_0}(2)\) equal to ?
This blog will provide a detailed calculation of this function. Refer to my Fundamental Sequences blog for more information.

Diagonalising \(\zeta_0\)
\(\zeta_0[2] = \varphi(2,0)[2] = \varphi_{1+1}(0)[2]\)

\(= \varphi^{\omega}(1,0_*)[2] = \varphi^2(1,0_*) = \varphi(1,\varphi(1,0)) = \epsilon_{\epsilon_0}\)

Detailed Calculation of \(f_{\zeta_0}(2)\)
\(f_{\zeta_0}(2) = f_{\varphi(2,0)}(2) = f_{\varphi^2(1,0_*)}(2) = f_{\varphi(1,\varphi(1,0))}(2)\)

\(= f_{\varphi(1,\varphi^2(1))}(2) = f_{\varphi(1,\varphi(\varphi(1)))}(2)\)

\(= f_{\varphi(1,\varphi(\omega))}(2) = f_{\varphi(1,\varphi(2))}(2)\)

\(= f_{\varphi(1,\omega^2)}(2) = f_{\varphi(1,\omega.2)}(2) = f_{\varphi(1,\omega + 1)}(2)\)

The result \(\varphi(1,0) = \omega.2 = \omega + 1\) will be used again throughout this calculation.

\(= f_{\varphi(1,\omega)\uparrow\uparrow 2}(2) = f_{\varphi(1,\omega)^{\varphi(1,\omega)}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,2)}}(2) = f_{\varphi(1,\omega)^{\varphi(1,1)\uparrow\uparrow 2}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,1)}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)\uparrow\uparrow 2}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\varphi(1,0)}}}}(2)\)

Using the previous result for \(\varphi(1,0)\) we get:

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega + 1}}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\varphi(1,0)}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.(\omega + 1)}}}(2) = f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0)^{\omega}}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0)^2}}}(2) = f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0).\varphi(1,0)}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0).(\omega+1)}}}(2) = f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0).\omega + \varphi(1,0)}}}(2)\)

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0).\omega + \omega + 1}}}(2)\)

The result \(\varphi(1,1) = \varphi(1,0)^{\omega}.\omega + \varphi(1,0).\omega + \omega + 1\) will be used again throughout this calculation.

\(= f_{\varphi(1,\omega)^{\varphi(1,1)^{\varphi(1,0)^{\omega}.\omega + \varphi(1,0).\omega + \omega}.varphi(1,1)}}(2)\)

Work In Progress