Talk:Slow-growing hierarchy

Origin
Does anyone know who originally proposed the slow-growing hierarchy? I learned of it from (Gallier 1991) and it doesn't cite any source. Cheers, — sligocki (talk) 23:02, 7 December 2009 (UTC)


 * Schwichtenberg's "Classifying Recursive Functions" (1997) says, in reference to the ε0-recursive functions ...


 * "Girard[1981] has shown that one might as well associate a far bigger ordinal with the functions provably recursive in arithmetic, the Bachmann-Howard ordinal.
 * To this end, Girard has introduced the so-called slow growing hierarchy Gα, α < the Bachmann-Howard ordinal ..."


 * I suppose, however, that that doesn't eliminate the possibility of some earlier-defined version of the hierarchy up to an ordinal smaller than the B-H ord. (BTW, I'm accessing this via Google preview of pp. 541-542 of "Handbook of computability theory", 1999, E. R. Griffor, ed.)
 * — r.e.s. 16:48, 16 December 2009 (UTC)

something is wrong?
If g0(n)=0, then g1(n)=g0(n)+1=0+1=1; g2=g1(n)+1=1+1=2 and so on.

I think something is wrong written in article, because it has no sense to insert counting sequence in this article —Preceding unsigned comment added by 83.21.60.126 (talk) 17:59, 27 April 2011 (UTC)


 * No, that's the correct result — so you can see why this hierarchy is called "slow growing". In fact, if α is represented in Cantor normal form, then gα(n) is the result of replacing ω by n everywhere in the Cantor normal form. Thus the functions with integer index i are just constants (gi(n) = i for all n), and the first non-constant function in this hierarchy occurs when the index is ω (gω(n) = n).  Similarly, gω·i + j(n) = n·i + j, and so on. — r.e.s. (talk)  02:54, 28 April 2011 (UTC)
 * The actual entries for the counting sequence are f0(-1) = 0, f0(0) = 1, f0(1) = 2 and so on, as f0(n) = n + 1.

Catching ordinal
I'm pretty sure that the slow-growing hierarchy catches up the fast-growing one at large Veblen ordinal. 31.42.233.14 (talk) 11:39, 21 April 2013 (UTC)


 * The slow-growing hierarchy catches up the fast-growing one at $$\theta(\Omega_\omega)$$, as it can be seen here.
 * $$ f_0(n) = g_{\omega+1}(n)$$
 * $$ f_1(n) = g_{\omega2}(n)$$
 * $$ f_2(n) = g_{2^{\omega}\omega}(n)$$
 * $$ f_{\omega+1}(n) = g_{\Gamma_0}(n)$$
 * $$ f_{\omega2+1}(n) = g_{\theta(\Omega2)}(n)$$
 * $$ f_{\omega^2+1}(n) = g_{\theta(\Omega^2)}(n)$$
 * $$ f_{\omega^\omega+1}(n) = g_{\theta(\Omega^\Omega)}(n)$$
 * $$ f_{\omega^{\omega^\omega}+1}(n) = g_{\theta(\Omega^{\Omega^\Omega})}(n)$$
 * $$ f_{\omega^{\omega^{\omega^\omega}}+1}(n) = g_{\theta(\Omega^{\Omega^{\Omega^\Omega}})}(n)$$
 * $$ f_{\epsilon_0+1}(n) = g_{\theta(\Omega\uparrow\uparrow\omega)}(n)$$
 * $$ f_{\theta(\Omega)}(n) = g_{\theta(\Omega_2)}(n)$$
 * $$ f_{\theta(\Omega_2)}(n) = g_{\theta(\Omega_3)}(n)$$
 * $$ f_{\theta(\Omega_3)}(n) = g_{\theta(\Omega_4)}(n)$$
 * When the fast-growing hierarchy reached Ω, the slow-growing hierarchy reached Ω2 and therefore, the slow-growing hierarchy index adds only 1 to the first Ω-subscript of the fast-growing hierarchy index.
 * $$ f_{\theta(\Omega_m)}(n) = g_{\theta(\Omega_{m+1})}(n)$$
 * However, its significance becomes less and less as this subscript grows bigger.
 * $$ f_{\theta(\Omega_{10})}(n) = g_{\theta(\Omega_{11})}(n)$$
 * $$ f_{\theta(\Omega_{100})}(n) = g_{\theta(\Omega_{101})}(n)$$
 * $$ f_{\theta(\Omega_{1000})}(n) = g_{\theta(\Omega_{1001})}(n)$$
 * $$ f_{\theta(\Omega_{10000})}(n) = g_{\theta(\Omega_{10001})}(n)$$
 * So, it is safe to say, that $$ f_{\theta(\Omega_\omega)}(n) = g_{\theta(\Omega_\omega)}(n)$$, as ω ≫ 1 and therefore, ω + 1 is basically the same as ω.
 * So, the slow-growing hierarchy and the fast-growing hierarchy will meet together at Ωω. Therefore, when the fast-growing hierarchy reaches ΩΩ, the slow-growing hierarchy will also have ΩΩ, as Ω ≫ ω ≫ 1.
 * $$ f_{\theta(\Omega_\Omega)}(n) = g_{\theta(\Omega_\Omega)}(n)$$ — Preceding unsigned comment added by 84.151.248.103 (talk) 13:03, 13 September 2020 (UTC)

exception
The slow-growing hierarchy and the fast-growing hierarchy will meet together at Ωω. But, there is an exception.

The exception is additions.

If there are additions, the slow-growing hierarchy and the fast-growing hierarchy will meet together, if the first ordinal is Ω and has more than 1 subscript.

$$ f_{\theta(\Omega_{\omega+1})}(n) = g_{\theta(\Omega_{\omega+2})}(n)$$

$$ f_{\theta(\Omega_{\epsilon_{m+1}})}(n) = g_{\theta(\Omega_{\epsilon_{m+1}})}(n)$$

$$ f_{\theta(\Omega_{\Omega+1})}(n) = g_{\theta(\Omega_{\Omega+2})}(n)$$

$$ f_{\theta(\Omega_{\Omega_{\Omega+1}})}(n) = g_{\theta(\Omega_{\Omega_{\Omega+1}})}(n)$$

Inconsistent?
Maybe I'm reading it wrong, but the first paragraph and the end of the second paragraph of the section "Relation to the fast-growing hierarchy" seem to contradict each other. The first paragraph seems to assert (regardless of what choice of fundamental sequences is made) that fε 0 is only matched by the growth of g at the Bachmann-Howard ordinal. The end of the second paragraph seems to assert that for a certain choice of fundamental sequences, they match (at the same ordinal) as early as $$\omega^2$$. Does the first statement implicitly use one of the standard definitions of fundamental sequences? And what is the (counterintuitive?) choice of fundamental sequences that would make them match up at $$\omega^2$$? Or am I missing something? I don't have easy access to the references, or I'd look it up. Spireguy (talk) 04:24, 3 April 2017 (UTC)

Slow-growing hierarchy, but the greek letters are replaced by well-known large numbers.
gωω2(4) = 65536

How large are gGraham's number(4), gTREE(3)(4), gSSCG(3)(4), gSCG(13)(4), gLoader's number(4), gRayo's number(4) and gFish number 7(4)? — Preceding unsigned comment added by 84.154.66.177 (talk) 07:55, 9 June 2020 (UTC)

gGraham's number(4) ≈ 2262144 ≈ 1.61×1078913