FANDOM

D57799

aka huehuehue

  • I live in somewhere that is omega kilometers away from you
  • I was born on April 14
  • D57799

    According to Knuth Arrows Theorem by Sbiis Saibian,

    for all \(a≥2, b≥1, c≥1, k≥2: (a\uparrow^kb)\uparrow^kcn+3\)

    \(a\uparrow^kM(n)\)

    \(=a\uparrow^k(M(n-1)+a\uparrow^{k+1}(n+1))\)

    \(>(a\uparrow^ka\uparrow^{k+1}(n+1))\uparrow^kM(n-1)\)

    \(=(a\uparrow^{k+1}(n+2))\uparrow^kM(n-1)\)

    \(>a\uparrow^{k+1}(n+2)\times M(n-1)\)

    \(>a\uparrow^{k+1}(n+2)\times (n+2)\)

    \(>M(n+1)\)

    Therefore,

    \((a\uparrow^k)^{b+c-3}M(1)\)

    \(>(a\uparrow^k)^{c-1}M(b-1)\)

    \(=(a\uparrow^k)^{c-1}(M(b-2)+a\uparrow^{k+1}b)\)

    lemma:

    for all \(n∈N:a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)>M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)\)

    proof:

    \(a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)\)

    \(=a\uparrow^k(M(b-3)+a\uparrow^{k+1}(b-1)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)\)

    \(=(a\uparr…

    Read more >
  • D57799

    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), ..., (1-n,2-n), 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…



    Read more >
  • D57799

    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 \(m\geq1\)

    \((a\uparrow^{m+1} b)\uparrow^{m+1} c\)

    \(=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}\)

    \(

    Read more >
  • D57799

    Ranging FGH before omega

    October 7, 2014 by D57799

    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 \(2\uparrow^{m-1}n\), 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 \(f_m(n)\geq2\uparrow^{m-1}n\) for \(n\geq1\),

    \(f_{m+1}(n)=\underbrace{f_m(f_m(...(f_m}_{n}(n))...))\)

    \(2\uparrow^mn=\underbrace{2\uparrow^{m-1}2\uparrow^…








    Read more >
  • D57799

    feeding FGH into FGH

    October 5, 2014 by D57799

    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)

    Read more >

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.