Talk:Iota function

This function may not be well-defined but it is an interesting thought experiment and wholly relevant to googology. FB100Z &bull; talk &bull; contribs 18:12, June 5, 2013 (UTC)

This function indeed is not well-defined. According to definition from article iota function set contains all functions \(f_a(n)=n+a\). Then \(I(n)>F_a(n)\) for all a, and we define \(F_a(n)=f_a(n)\). So Iota is undefined even for 1. But if we alter the definition so that we can only use functions definable with at most n English symbols and we can compose at most n functions, it all becomes finite. LittlePeng9 (talk) 18:56, June 5, 2013 (UTC)

I believe that I(n) is an attempt to create a function with order type \(\omega_1\). Ikosarakt1 (talk ^ contribs) 19:00, June 5, 2013 (UTC)