Moderators: gmalivuk, Prelates, Moderators General
william wrote:Toeofdoom wrote:@will i raise you an aleph one... >.>
aleph xkcd number.
Toeofdoom wrote:@jesting rabbit... you arent allowed use previously defined numbers to get your record it explains that at the end of the clarkson number page.
Torn Apart By Dingos wrote: Instead, I will counter with:
The class of all sets.
The Mighty Thesaurus wrote:Why? It does nothing to address dance music's core problem: the fact that it sucks.
Torn Apart By Dingos wrote:The busy beaver function laughs in contempt at the Clarkkkkson function's utter puniness.
kakos wrote:Regardless of whichever number is biggest, I truly possess the largest number.
I define the kakos number to be as follows: sum(x, x \in K)+1, where K is defined as the set of all numbers conceived by human thought.
tgjensen wrote:kakos wrote:Regardless of whichever number is biggest, I truly possess the largest number.
I define the kakos number to be as follows: sum(x, x \in K)+1, where K is defined as the set of all numbers conceived by human thought.
then the tgjensen number is the equivelant, except replace K with T, T being defined as the set of all numbers conceived by human thought as well as all those that have not, are not and will not be conceived by human thought.
Hawknc wrote:Gotta love our political choices here - you can pick the unionised socially conservative party, or the free-market even more socially conservative party. Oh who to vote for…I don't know, I think I'll just flip a coin and hope it explodes and kills me.
Jack Saladin wrote:etc., lock'd
Mighty Jalapeno wrote:At least he has the decency to REMOVE THE GAP BETWEEN HIS QUOTES....
Sungura wrote:I don't really miss him. At all. He was pretty grouchy.
tgjensen wrote:then the tgjensen number is the equivelant, except replace K with T, T being defined as the set of all numbers conceived by human thought as well as all those that have not, are not and will not be conceived by human thought.
Macbi wrote:Does anyone know what the sum of the reciprocals of the BB sequence converges to? Bonus points if it's computable!
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
xkdvd wrote:Let Iota(x) = The largest number that can be defined mathematically in x symbols using arithmetic and being able to define functions.
(Note that this is not paradoxical because Iota(x) cannot be defined in this way.)
My number is Iota(Graham's number).
Alpha Omicron wrote:My contribution: Conway's Chained Arrow Notation sucks. Much prefer hyper function.
So I just had this awesome idea about large numbers and went to necro' this thread. However, on reading through the thread, i see that robheus has already had an idea similar to mine.robheus wrote:Let us first give a definition for a function F_{1} that "feeds-in" the result of another function f(n) a f(n) number of times, so that:
F_{1} f(n) = f(f(f( ... f(n) parenthesis .. f(n) .... f(n) parenthesis ... )))
and the recursive definition for higher subscripts:
F_{k} = f(f(f( ... F_{k-1} f(n) parenthesis ... f(n) ...F_{k-1} f(n) parenthesis ... ))).
Then
F_{BB(Clarkkkkson)} BB(Clarkkkson) would be quite large indeed!
And without problem, we can add to this scheme by defining a function F_{1,k} as follows:
F_{1,k} f(n) = F _{Fk f(n)} f(n)
The answer is no. The Clarkkkson is probably less than BB(1000)quintopia wrote:The question is whether Clarkkkkson will become larger than this number before the end of the universe. . .
Macbi wrote:If α is a limit ordinal of {β_{1},β_{2}...} then L_{α}(x) is L_{β_{L_0(x)}}(x)
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
The other problem with my idea is that it gives a function for every ordinal, and yet there are more ordinals then functions. However for most "small" limit ordinals there is an obvious sequence for which they are the limit. This should be a firm enough idea to shore up the definition of the number in my post. The sequence I think is "obvious" for ε_{0} would be {1,ω,ω^ω,ω^{ω^ω}...}, the sequence for ω^{ω^ω} would be {ω^{ω^0},ω^{ω^1},ω^{ω^2},...}. (This means my number is still the highest in the thread so far.) However there's no point going into this further because...antonfire wrote:Macbi wrote:If α is a limit ordinal of {β_{1},β_{2}...} then L_{α}(x) is L_{β_{L_0(x)}}(x)
This isn't well-defined. There is more than one sequence of which a particular ordinal is the limit ordinal.
Right you are, I'd quite forgotten about halting oracles.Besides, if you allow yourself scary uncomputable functions like BB, it's easy to beat anything like that in a simpler way.
Can we have a BB_{ω}? That is, the BB sequence for Turing machines that have an oracle for arbitrarily high BB_{n}. Would this idea in fact give us a BB function for each ordinal?Let BB_{1} be the busy beaver function for Turing machines which have access to an oracle that computes BB. Then anything you define in a systematic (computable) way in terms of the busy beaver function grows slower that BB_{1}(n). If I can write a program of reasonable length that will compute your number (allowing it access to the busy beaver function), then your number is smaller than BB_{1}(k) for a reasonably small k.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
quintopia wrote:I'm pretty sure Graham's number is bigger than that. For theoretical problems, larger numbers may be needed as upper bounds simply to prove something is finite.
Users browsing this forum: No registered users and 5 guests