
2
According to Knuth Arrows Theorem by Sbiis Saibian,
for all \(a≥2, b≥1, c≥1, k≥2: (a\uparrow^kb)\uparrow^kcn+3\)
\(a\uparrow^kM(n)\)
\(=a\uparrow^k(M(n1)+a\uparrow^{k+1}(n+1))\)
\(>(a\uparrow^ka\uparrow^{k+1}(n+1))\uparrow^kM(n1)\)
\(=(a\uparrow^{k+1}(n+2))\uparrow^kM(n1)\)
\(>a\uparrow^{k+1}(n+2)\times M(n1)\)
\(>a\uparrow^{k+1}(n+2)\times (n+2)\)
\(>M(n+1)\)
Therefore,
\((a\uparrow^k)^{b+c3}M(1)\)
\(>(a\uparrow^k)^{c1}M(b1)\)
\(=(a\uparrow^k)^{c1}(M(b2)+a\uparrow^{k+1}b)\)
lemma:
for all \(n∈N:a\uparrow^k(M(b2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)>M(b2)+(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)\)
proof:
\(a\uparrow^k(M(b2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)\)
\(=a\uparrow^k(M(b3)+a\uparrow^{k+1}(b1)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)\)
\(=(a\uparr…
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Proofs not included. Some of the following have formal proof, some don't.
1. \((a\uparrow^nb)\uparrow^nc=a\uparrow^n(b\times c)\)
The equation holds when n=0 or 1 (n=0 for multiplication)
It's a simple property but it doesn't have a name.
This is useful for extention to rational numbers.
\(a\uparrow^n(q/p)=(a\uparrow^nq)\uparrow^n{1/p}\)
Also, \((a\uparrow^n{1/p})\uparrow^np=a\), so \(a\uparrow^n{1/p}=a\downarrow_R^np\)
Therefore, \(a\uparrow^n(q/p)=(a\uparrow^nq)\downarrow_R^np\)
(Notation explanation in 10.)
2.\((a\uparrow\uparrow b)\uparrow\uparrow c1\)
6.for all \(n \geq 1\), \(y=a\uparrow^n x\) passes (0,1), (1,0), ..., (1n,2n), n points in total. All on the magic line of y=x+1
Define that \(K_n\) is the biggest solution smaller than zero t…
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Originally I wanted to put the proof that uparrow notation is strictly increasing, but it turned into an extremely long proof. I cut it off and will probably put it in another post.
For all a,b,c > 2
When m = 1, \(a\uparrow(b\uparrow c) > (a\uparrow b)\uparrow c = a\uparrow(b\times c) \) holds.
Assume that : \(a\uparrow^m(b\uparrow^mc)> (a\uparrow^mb)\uparrow^mc\) for \(m\geq1\)
\((a\uparrow^{m+1} b)\uparrow^{m+1} c\)
\(=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}\)
\(
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The basic definitions:
1. \(f_0(n)=n+1\)
2. \(f_{\alpha+1}(n)=f_\alpha^n(n)\)
FGH here is smaller than \(f_\omega(n)\), so the situation of when \(\alpha\) is a ordinal is ignored.
FGH and intuitional upper & lower bounds
lower FGH upper
1 \(2n\) \(2n\) \(2\uparrow n\)
2 \(2\uparrow n\) \(n2^n\) \(2\uparrow\uparrow n\)
3 \(2\uparrow\uparrow n\) \(...\) \(2\uparrow\uparrow\uparrow n\)
The most obvious lower bound of \(f_m(n)\) is \(2\uparrow^{m1}n\), so let's prove it.
1. Take \(f_2(n)\) as our starting point, for \(n\geq1\) , \(f_2(n)=n2^n\geq 2\uparrow n\) , the inequality holds.
2. Assume that \(f_m(n)\geq2\uparrow^{m1}n\) for \(n\geq1\),
\(f_{m+1}(n)=\underbrace{f_m(f_m(...(f_m}_{n}(n))...))\)
\(2\uparrow^mn=\underbrace{2\uparrow^{m1}2\uparrow^…
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There may be some informal or invalid step in my proof. Just point it out. There's still a lot for me to learn.
Feeding FGH into itself was considered before. I just want to find the limit of it and probably some usage of it.
It looks like this:
1. . It's theoretically the limit of FGH itself. But with Ordinal Collapsing Function that is more powerful and inaccessible ordinals, it is possible for FGH to go beyond its limit.
D57799 (talk) 06:24, October 5, 2014 (UTC)
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