Hi, OK, let us try to name the biggest natural (finite) number we can, and let us do that transfinite ascension on the growing functions from N to N.
We have already build some well defined sequence of description (code) of growing functions. Let us choose the Hall Finney sequence to begin with (but the one by Tom Caylor can be use instead). F1 F2 F3 F4 F5 ... With F1(n) = factorial(n), F2(n) = factorial(factorial n), etc. Note this: Hal gave us a trick for getting from a growing function f, a new function growing faster, actually the iteration of the function. That is, Hal gave us a notion of successor for the growing function. Now the diagonalization of the sequence F1 F2 F3 F4 ..., which is given by the new growing function defined by G(n) = Fn(n) + 1 gives us a growing function which grows faster than any Fi from Hal's initial sequence. Precisely, G will grow faster than any Fi on *almost all* number (it could be that some Fi will grow faster than G on some initial part of N, but for some finite value (which one?) G will keep growing faster. Technically we must remember to apply our growing function on "sufficiently big input' if we want to benefit of the growing phenomenon. We will make a rough evaluation on that input later, but let us not being distract by technical point like that. The diagonalization gives an effective way to take the "limit" of the sequence F1, F2, F3, ... G grows faster than any Fi. Mathematician will say that the order type of g, in our our new sequence F1 F2 F3 ... G, is omega (the greek letter). But G is just a well defined computable growing function and we can use Hall Finney "successor" again to get the still faster function, namely G(G(n)). The order type of G(G(n)) is, well, the successor of omega: omega+1 And, as Hall initially, we can build the new sequance of growing functions (all of which grows more than the preceding sequence): G(n) G(G(n)) G(G(G(n))) G(G(G(G(n)))) etc. which are of order type omega, omega+1, omega+2, omega+3, omega+4, etc. Now we have obtained a new well defined infinite sequence of growing function, and, writing it as: G1, G2, G3, G4, G5, G6, ... or better, as F_omega, F_omega+1, F_omega+2, F_omega+3 just showing such a sequence can be generated so that we can again diagonalise it, getting H(n) = Gn(n) + 1, or better H(n) = F_omega+n (n) + 1 Getting a function of order type omega+omega: we can write H = F_omega+omega And of course, we can apply Hall's successor again, getting F_omega+omega+1 which is just H(H(n), and so we get a new sequence: F_omega+omega+1, F_omega+omega+2, F_omega+omega+3, ... Which can be diagonalise again, so we get F_omega+omega+omega, and then by Hal again, and again ...: F_omega+omega+omega+1, F_omega+omega+omega+2, F_omega+omega+omega+3 ... Oh Oh! a new pattern emerges, a new type of sequence of well defined growing functions appears: F_omega, F_omega+omega, F_omega+omega+omega, F_omega+omega+omega+omega, And we can generated it computationnaly, so we can diagonalise again to get: F_omega * times omega, and of course we can apply Hal's successor (or caylor one of course) again, and again .... Oh Oh Oh Oh Oh .... A new pattern emerge (the Ackerman Caylor one, at a higher level). F_omega, F_omega + omega F_omega * omega F_omega ^ omega F_omega  omega (omega tetrated to omega, actually this ordinal got famous and is named Epsilon Zéro, will say some words on it later) F_omega  omega F_omega  omega F_omega  omega F_omega  omega F_omega  omega F_omega  omega F_omega  omega ... In this case they are all obtained by successive diagonalzations, but nothing prevent us to diagonalise on it again to get F_omega [omega] omega OK, I think the following finite number is big enough: F_omega [omega] omega (F_omega [omega] omega (9  9)) Next, we will meet a less constructivist fairy, and take some new kind of "big leap". Be sure to be convinced that, despite the transfinite character of the F_"alpha" sequence, we did really defined at all steps precise computable growing functions ... (if not: ask question please). Tricky Problem: is there a sequence in which all growing computable functions belong? Is it possible to dovetail on all computable growing functions, ... I let you think, Bruno http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to email@example.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list -~----------~----~----~----~------~----~------~--~---