Hi Tommy!
I'm not going to discuss the details of James approach, but I can say something about some lateral matters.
That identity you found, is long known as the recursion law for right-associative hyperoperations. The oldest reference of this I'm aware of dates back to 1953 by Giuseppe Arcidiacono, "On the extension of arithmetic operations".
It is a variant of Goodstein's original definition and they both agree for ranks 1, 2 and 3.
Said that, idk if it is a nicer choice. Ofc, if James's strategy of pasting together the Bennets spectrum of operation interpolating +, x and ^ works for extending Goodstein's (left-associative) will works too for (Arcidiacono's) because the ranks beyond are extended by piecewise method...
Third note: maybe I'm mistaken, but James is using his beta method algorithm for computing fractional iterations of exp, this post was born exactly from that.
Fourth note: sorry but I missed that part... where we he is talking about non commutativity?
I'm not going to discuss the details of James approach, but I can say something about some lateral matters.
That identity you found, is long known as the recursion law for right-associative hyperoperations. The oldest reference of this I'm aware of dates back to 1953 by Giuseppe Arcidiacono, "On the extension of arithmetic operations".
It is a variant of Goodstein's original definition and they both agree for ranks 1, 2 and 3.
Said that, idk if it is a nicer choice. Ofc, if James's strategy of pasting together the Bennets spectrum of operation interpolating +, x and ^ works for extending Goodstein's (left-associative) will works too for (Arcidiacono's) because the ranks beyond are extended by piecewise method...
Third note: maybe I'm mistaken, but James is using his beta method algorithm for computing fractional iterations of exp, this post was born exactly from that.
Fourth note: sorry but I missed that part... where we he is talking about non commutativity?
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)