User:Vel!/How to count up to infinity/Annotated

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 *  How to count up to infinity


 * All we known,infinity is larger than all of any other numbers, so most of people think that infinity can’t be reached.But the fact is that?This article will tell us how to count up to infinity.

What? Infinity can't be reached by definition. ω is defined as the first ordinal unreachable through finite applications of 0 and the successor function.


 * At fisrst,all of matter in each universe can store informations,so each universe can be a super huge computer .Besides,there are about Chris Birds N(HH(3)(3) in Chirs Bird’s H(n)) universes in the sky,

Are you trying to say that we live in an infinite multiverse? Because the mathematical expression you dropped is very, very small in infinity's eyes. Also, said mathematical expression makes no sense.


 * and each universe is a super huge group-universe-computer,so it use quantum algorithm to compute.

I get the quantum algorithm bit, but what's a group-universe-computer?


 * In this way,the speed of this computer can reach {3*10^125N,3(1)2} per second,and the internal memory can reach {9.4670856*10^(10^10^10^10^1.1+125)N,3(1)2}bytes(1 byte=8 bits,and use 1 bit to express any symbol in this computer)Now, Let us to use DNA-computer mode,it can let the computer to do any operations.As follow,there is a way can count up to infinity

      Number                                               Notation Level

Lower Bound                 Upper Bound

0                           6                              Class 0

6                           1000000                        Class 1

1000000                     10^1000000                     Class 2

10^1000000                  10^10^1000000                  Class 3

10^10^1000000               10↑↑10                       Exponentiation

10↑↑10                    10↑↑↑11                     Tetration

10↑↑↑11                  hyper(10,10^100,100)             Up-arrow

hyper(10,10^100,100)          BOX_M~                        Chained arrow

BOX_M~                     {3,27(1)2}                       Linear array

{3,27(1)2}                    {10,10(1)10,10,10,10,10,100}       2 row array

{10,10(1)10,10,10,10,10,100}     3^2&10                        3 row array

3^2&10                      {10,10(gongulusplex)2}            Dimensional array

<p class="MsoNormal">{10,10(gongulusplex)2}          10↑↑(goppatoch)&10           Tetration array

<p class="MsoNormal">10↑↑(goppatoch)&10         {10,100}&10&10&10              Array of

<p class="MsoNormal">{10,100}&10&10&10            {100,100 100^100 100}            Legiattic array

<p class="MsoNormal">{100,100 100^100 100}          Meameamealokkapoowa oompa    Beyond legiattic array

<p class="MsoNormal">Meameamealokkapoowa oompa  HH(3)(3)                        Chris Birds H(n)

<p class="MsoNormal">There are recursive ordinal and the smallest non-recursive ordinal is ω1CK.Here we have completely exhausted BEAF.

<p class="MsoNormal">In order,the functions are Rado’s sigma function and its higher order cousins,Xi function and its higher order cousins,arxex function and its higher order cousins,Rayo’s function and going even further.

<p class="MsoNormal">fω1CK(n)≈Σ(n)

<p class="MsoNormal">f2ω1CK(n)≈Σ2(n)

<p class="MsoNormal">f3ω1CK(n)≈Σ3(n)

<p class="MsoNormal">fmω1CK(n)≈Σm(n)

<p class="MsoNormal">fαCK(n)≈Ξ(n)( α→ωα)

<p class="MsoNormal">Rayo(n)(at least ω1CK,unknown)

<p class="MsoNormal">Going even further

<p class="MsoNormal">fω^2CK(n)

<p class="MsoNormal">fω^3CK(n)

<p class="MsoNormal">fω^mCK(n)

<p class="MsoNormal">fω^^2CK(n)

<p class="MsoNormal">fω^^3CK(n)

<p class="MsoNormal">fω^^mCK(n)

<p class="MsoNormal">fε0CK(n)

<p class="MsoNormal">fε1CK(n)

<p class="MsoNormal">fεmCK(n)

<p class="MsoNormal">fεωCK(n)

<p class="MsoNormal">fζ0CK(n)

<p class="MsoNormal">fη0CK(n)

<p class="MsoNormal">fΨ(m,0)CK(n)

<p class="MsoNormal">fΨ(ω,0)CK(n)

<p class="MsoNormal">fΨ(ε0,0)CK(n)

<p class="MsoNormal">fΨ(ζ0,0)CK(n)

<p class="MsoNormal">fΨ(η0,0)CK(n)

<p class="MsoNormal">fΓ0CK(n)

<p class="MsoNormal">fΓ1CK(n)

<p class="MsoNormal">fΓωCK(n)

<p class="MsoNormal">fΓε0CK(n)

<p class="MsoNormal">fΓΓ0CK(n)

<p class="MsoNormal">fΓΓ1CK(n)

<p class="MsoNormal">fΓΓΓ0CK(n)

<p class="MsoNormal">fΨ(1,1,0)CK(n)

<p class="MsoNormal">fΨ(1,m,0)CK(n)

<p class="MsoNormal">fΨ(1, ω,0)CK(n)

<p class="MsoNormal">fΨ(1, ε0,0)CK(n)

<p class="MsoNormal">fΨ(1, Γ0,0)CK(n)

<p class="MsoNormal">fΨ(2,0,0)CK(n)

<p class="MsoNormal">fΨ(m1,m2,0)CK(n)

<p class="MsoNormal">fΨ(m1,m2,…,ma,0)CK(n)

<p class="MsoNormal">fθ(Ω^ω)CK(n)

<p class="MsoNormal">fθ(Γ0)CK(n)

<p class="MsoNormal">fθ(Ψ(2,0,0))CK(n)

<p class="MsoNormal">fθ(θ(Ω^ω))CK(n)

<p class="MsoNormal">…

<p class="MsoNormal">Go even further,we define that f(ω1CK)CK(n)=fω12CK(n),the same as ωm, εm, ζm, ηm, Γm, Ψ(m1,m2,…,ma), θ(εΩ+1), θ(Γ0) and so on,and we define that f(ω12CK)CK(n)=fω13CK(n),Suppose k=m,we can get f(ω1mCK)CK(n)=fω1(m+1)CK(n),and we define that fω1ω1CK(n),fω1ωmCK(n),fω1ω^mCK(n),fω1ε0CK(n),fω1Γ0CK(n),the same as ωm, εm, ζm, ηm, Γm, Ψ(m1,m2,…,ma), θ(Ω^ω), θ(θ(Ω^ω)), θ(θ(Ω^^ω)) etc.

<p align="left" class="MsoNormal">Let we define that fω1(ω1CK)CK(n)=(fω1CK(n))^2=2&fω1CK(n)={fω1CK(n),2(1)2},fω1(ω1<span style="mso-bidi-font-size:10.5pt;font-family:宋体;mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:宋体;mso-fareast-theme-font: minor-fareast;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin">（ ω1CK<span style="mso-bidi-font-size: 10.5pt;font-family:宋体;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin; mso-fareast-font-family:宋体;mso-fareast-theme-font:minor-fareast;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin">） CK)CK(n)=(fω1CK(n))^3=3&fω1CK(n)={fω1Ck(n),3(1)2},…, fω1(ω1<span style="mso-bidi-font-size:10.5pt;font-family:宋体;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-fareast-font-family:宋体;mso-fareast-theme-font: minor-fareast;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin">（ …(ω1CK<span style="mso-bidi-font-size:10.5pt;font-family:宋体;mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:宋体;mso-fareast-theme-font: minor-fareast;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin">） …)CK)CK(n)(There are m-1s)=(fω1CK(n))^m=m&fω1CK(n)={fω1CK(n),m(1)2}.Be similar as John Bowser’s array notation,we can get dimensional array of fω1CK(n),tetration array of fω1CK(n),pentation array of fω1CK(n),legiattic array of fω1Ck(n),beyond legiattic array of fω1CK(n),Chirs Bird’s H(n) array of fω1CK(n) and go even further.

<p align="left" class="MsoNormal">We define that fω1CK(n)&fω1CK(n)=2&&fω1CK(n),fω1CK(n)&fω1CK(n)&fω1CK(n)=3&&fω1CK(n),and we can get fω1CK(n)&fω1CK(n)&…&fω1CK(n)(there are m fω1CK(n)s)=m&&fω1CK(n).Be similar as John Bowser array notation and we can also get fω1CK(n)&&…&fω1CK(n)&&…&…&&…&fω1CK(n)(there are m fω1CK(n)s and p &s between each two fω1CK(n)s)=m&&…&(there p+1 &s)fω1CK(n),and we define it is equal to {m,p+1(1)2&fω1CK(n)}.Also, we define that is similar to John Bowser’s array notation,and we define {fω1CK(n),p+1(1)2&fω1CK(n)} is equal to p+12-&fω1CK(n),go along this way,we can call that each of one new function is defined is passing one step(The same as former part of this article ,”array of” in each step is similar to the John Bowser’s array naotation,the following part of this article too),the same asωm, εm, ζm, ηm, Γm, Ψ(m1,m2,…,ma) andθ.

<p align="left" class="MsoNormal">fθ(θ(…(θ(N&Ω))…))CK(fθ(θ(…(θ(N&Ω))…))(N)(There are N+1θs before the CK and N+1θs after the CK)=f(N+1) θ(N&Ω)CK(f(N+1) θ(N&Ω)(N)) is large enough or not?The answer is not.Let f(N+1) θ(N&Ω)CK(f(N+1) θ(N&Ω)(N))= α,and let f(α+1) θ(α&Ω)CK(f(α+1) θ(α&Ω)( α)) to pass f(α+1) θ(α&Ω)CK(f(α+1) θ(α&Ω)( α)) steps is A according to the above definition,let f(A+1) θ(A&Ω)CK(f(A+1) θ(A&Ω)(A)) pass f(A+1) θ(A&Ω)CK(f(A+1) θ(A&Ω)(A))+ f(α+1) θ(α&Ω)CK(f(α+1) θ(α&Ω)( α)) steps is B.Go along this way,we can get C,D,…,Z,a,b,…z,h0(0),h0(1),…,h0(n),…(For distinguish upper-case letters and lower-case letters,we define those letters were defined in this paragraph need to turn 30° to the right when we write them).

<p align="left" class="MsoNormal">Let h0h0(n)(n)=2&h0(n)={h0(n),2(1)2}(John Bowser’s array notation),go along with the &&,&&&,&&…&(There are n &s),…,2-&,3-&,…,n-&,…(it is similar to the John Bowser’s array notation,the following part of this article too),be similar to the former article,we can call that each of one new function is defined is passing one step,we are able to get h0(n) pass h0(n) steps is h1(n) according to the above definition,h1(n) pass h1(n) steps is h2(n) according to the above definition,…,hm(n) pass hm(n) steps is h(m+1)(n) according to the above definition(be similar to qm(n) to q(m+1)(n),(p)CqSm(n) to (p)CqS(m+1)(n),Dm(n) to D(m+1)(n),(p)SqDm(n) to (p)SqD(m+1)(n),Um(n) to U(m+1)(n),(p)SqUm(n) to (p)SqU(m+1)(n),Nm(n) to N(m+1)(n),(p)SqNm(n) to (p)SqN(m+1)(n), λm(n) toλ(m+1)(n)).And we define that hhn(n)(n)=1&h&hn(n),hhhn(n)(n)(n)=2&h&hn(n),be similar to the former article,we also go along the &&,&&&,&&…&,(There are n&s),…,2-&,3-&,…,n-&,…,and each of one new function is defined is passing one step,and we define that hn(n) pass hn(n) steps is q0(n) according to the above definition(be similar to the qn(n) to (0)C0S0(n),(p)CqSn(n) to (p)C(q+1)S0(n),(r)CnSn(n) to (r+1)C0S0(n),(n)CnSn(n) to D0(n),Dn(n) to (0)S0D0(n),(p)SqDn(n) to (p)S(q+1)D0(n),(r)SnDn(n) to (r+1)S0D0(n),(n)SnDn(n) to U0(n),Un(n) to (0)S0U0(n),(p)SqUn(n) to (p)S(q+1)U0(n),(r)SnUn(n) to (r+1)S0U0(n),(n)SnUn(n) to N0(n),Nn(n) to (0)S0N0(n),(p)SqNn(n) to (p)S(q+1)N0(n),(r)SnNn(n) to (r+1)S0N0(n),(n)SnNn(n) toλ0(n)).

<p align="left" class="MsoNormal">λQ(Q)(Q=qcghjlxy(20121220))is large enough or not?Either the answer is not.LetλλQ(Q)(Q)=1&λ&λQ(Q), λλλQ(Q)(Q)(Q)=2&λ&λQ(Q),be similar to the former article,we also go along the &&,&&&,&&…&(There are n &s),2-&,3-&,…,n-&,and each of one new function is defined is passing one step,and we define thatλλQ(Q) (Q)passλλQ(Q)(Q) steps is 1 time-unit(it means this computer need to do once-through operation ),and λ1 time-unit(1 time-unit) pass λ1 time-unit(1 time-unit) + λλQ(Q)(Q) steps is 2 time-units(Note(1 time-unit)*2=2(1 time-unit) ≠2 time-units,and 2 time-units means that this computer need to do twice-through operations,the same as second-unit,order-second-unit,orderplex-second-unit,super-orderplex-second-unit,superplex-ord-erplex-second-unit and those numbers are used any other symbol to express).Go along this way,we can get 3 time-units,4 time-units,…,n time-units,…,go on going along this way,we define that {3*10^125N,3(1)2} time-units=1 second-unit(It means that this computer need to spend 1 second on computing).Be based on that ,we can get n*{3*10^125N,3(1)2} time-units=n second-units.Go along this way,we define that if this number need to spend n second-units on computing this number,we call this number is n 2-order-second-units.We can note that is (n-second-units) second-units=n 2-order-second-units.Be based on above definition,we can get (n second-units) m-order-second-units=n (m+1)-order-second-units(Be similar to the orderplex-second-unit,super-orderplex-second-unit,superplex-orderplex-second-unit).Be similar to the former article,we can define that m (n second-units)-order-second-units=n 2-orderplex-second-units(Be similar to the orderplex-second-unit to super-orderplex-second-unit and super-orderplex-second-unit to superplex-orderplex-second-unit).Go on going even further,in n n-superplex-orderplex-second-units,if n>=1 second-unit,we can use one of any other symbols(except those were shown in former article) to express.Be similar to the second-unit,it obeys 1+1≠2(it means that if we suppose 1(1 second-unit)-superplex-orderplex-second-unit=1μ,2μ=1(2 second-unit) -superplex-orderplex-second-unit≠2*(1(1 second-unit)-superplex-orderplex-second-unit)),and 2*(1(1 second-unit)-superplex-orderplex-second-unit))=2(1 μ),be similar to the any other this kind of numbers) and it also goes along order,oderplex,super-orderplex and superplex-orderplex.Until all of symbols of this computer is used up,we need to use 2 symbols to express and it need 2 bits of internal memory inthis computers.Go on going along this way and until the internal memory of this computer is used up.As this way,we can count up to infinity.