- #36
ShayanJ
Gold Member
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So I'll post ## \infty ##!Ben Niehoff said:The product of the factorials of (the absolute values of) all the numbers that will be posted, then. :D
So I'll post ## \infty ##!Ben Niehoff said:The product of the factorials of (the absolute values of) all the numbers that will be posted, then. :D
Ben Niehoff said:A(G,G),
where A is the Ackermann function and G is Graham's number.
is only 16 characters. If you include definitions...A(A(G,G),A(G,G))
Now it's still only up to 51, so, by expanding on the same idea and being slightly more concise with the explanation you can get...A=Ackermann function
G=Gram's number
A(A(G,G),A(G,G))
Which by my count comes to 200 characters, including spaces.Ackermann function
Gram's number
G↑↑A(A(A(A(A(A(G,G),A(G,G)),A(A(G,G),A(G,G))),A(A(A(G,G),A(G,G)),A(A(G,G),A(G,G)))),A(A(A(A(G,G),A(G,G)),A(A(G,G),A(G,G))),A(A(A(G,G),A(G,G)),A(A(G,G),A(G,G))))),A(G,G))
Ackermann function
Gram's number
B(n,x)=A performed recursively n times with arguments x. I.E. B(2,3)=A(A(3,3),A(3,3))
C(n,x)=B performed recursively n times with arguments x.
C(C(G,G),C(G,G))
f(x)=10^x!
f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(9)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
ChrisVer said:Code:f(x)=10^x! f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(f(9)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
maybe... I haven't seen that number...micromass said:This is still dwarfed by Grahams number though...
Fair warning. It'll break your brain.ChrisVer said:maybe... I haven't seen that number...
Dembadon said:[itex]G^G[/itex]↑[itex]G^G[/itex][itex]G^G[/itex], where G is Graham's number.
Indeed. So I looked at mrspeedybob's post and cut a few characters out so I could fit in exponents for the arguments of C. Microsoft Word says it's 200 characters exactly including spaces:micromass said:I'm afraid that doesn't beat the hugeness of the Ackerman function. https://en.wikipedia.org/wiki/Ackermann_function
rootone said:∞-1
nolxiii said:11
nolxiii said:also, to clarify, this number is not written in base ten but in some much larger base size
Lets say binary for simplicity, although I do realize that that a computer memory containing the number would require an infinite number of bits.micromass said:In which number system are you working when you say ##\infty## ?
nolxiii said:no one else specified their base size
rootone said:Lets say binary for simplicity, although I do realize that that a computer memory containing the number would require an infinite number of bits.
lim_[n ->0] 1/n^2
ChrisVer said:what would happen in case I write:
??Code:lim_[n ->0] 1/n^2
micromass said:Not a real number.
ChrisVer said:really?
let G = graham's #
let ☺ mean G ↑'s in knuth notation
let ☻ mean G ☺'s
base G
13☻☻☻☻☻☻☻☻☻☻☻☻☻☻☻☻☻☻☻13
I really liked this part:micromass said:Time to come clean. I made this thread because I read a very interesting article about big numbers. It seems in this thread, many found their way to Graham's number and Ackermann function. But there is a function which increase even faster than those: the busy beaver function. Check it out:
http://www.scottaaronson.com/writings/bignumbers.html
So it seems a very large number can use the BB function with BB(G) recursions? I know there is a more elegant and rigorous way to write it, but I don't think I'm clever enough.Could early intervention mitigate our big number phobia? What if second-grade math teachers took an hour-long hiatus from stultifying busywork to ask their students, "How do you name really, really big numbers?" And then told them about exponentials and stacked exponentials, tetration and the Ackermann sequence, maybe even Busy Beavers: a cornucopia of numbers vaster than any they’d ever conceived, and ideas stretching the bounds of their imaginations.
Even a googolplex is very very very small compared to graham's number.TheQuietOne said:googol, period
yeah, suppose ##\lim_{x \rightarrow 0} 1/x^2=b##. A theorem says that ##\lim_{x \rightarrow a} f(x) = c## if and only if for every sequence ##x_n## which converges to ##a##, the sequence ##f(x_n)## converges to ##c##. So take the sequence ##\{1/n\}_{n \in \mathbb{N}}##, this sequence converges to 0, but ##f(1/n)=n^2## for ##f(x) = 1/x^2##. This sequence does not converge to any real number, so it won't converge to ##b##.ChrisVer said:really?
TheQuietOne said:googol, period