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 [4] omega (omega tetrated to omega, actually this ordinal got
famous and is named Epsilon Zéro, will say some words on it later)
F_omega [5] omega
F_omega [6] omega
F_omega [7] omega
F_omega [8] omega
F_omega [9] omega
F_omega [10] omega
F_omega [11] 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] 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/
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Received on Thu May 25 2006 - 10:41:23 PDT