Talk:Large numbers

From Wikipedia, the free encyclopedia

Mathematics Portal

This article is within the scope of WikiProject Mathematics, which collaborates on articles related to mathematics.

Mathematics rating:

B Class

Mid Priority

Field: General

Please update this rating as the article progresses, or if the rating is inaccurate. Please also add comments to suggest improvements to the article.

1 Old Comments
2 World Almanac
3 Examples
4 sorry
5 new comment
6 accumulated error in binary vs decimal exponents
- 6.1 About the universe computer analogy...
7 From article page
8 Citations?
9 Systematically creating ever faster increasing sequences
10 Improving this article
11 Graham's number?
12 This article

[edit] Old Comments

OK, this is a first stab at getting all the large number topics together, please feel free to kick this into shape. The Anome

Can I suggest that we include only pure numbers in this article, not distances and other measurements? Would anyone object if I deleted the astronomical distances, since they are only large numbers when expressed in small units? I suppose I should go further and say that Avogradro's number is also just an arbitrary unit, but I shan't, because I feel I'm on a slippery slope towards excluding everything! -- Heron

Then why do people call large numbers "astronomical", as the article informs us? Perhaps it's because astronomical distances are large when expressed in any human-sized scale. I think the concept of "largeness" needs to be explained. The whole article is subjective anyway -- I wouldn't call 10¹⁰ large, I deal with those sorts of numbers every day. $10^{10^{10}}$ is more like it. -- Tim Starling 09:26 18 Jun 2003 (UTC)

I agree with you about 10¹⁰. I wouldn't call the number of bits on a hard disk particularly large, either. It is certainly subjective. My point was that measurements of distance etc. are different from pure numbers. Measurements are, by definition, relative, whereas at least pure numbers are absolute. Largeness is another thing. Perhaps one definition would be "a number considered as large at a particular time by a particular culture". For example, I seem to remember that the Old Testament uses the number 40 as a generic large number in several places (e.g. "40 days and 40 nights"). -- Heron

Let me put this another way. I think the present article should be, as it mostly is, about the mathematics of large numbers. Other large quantities, such as astronomical distances, already have a place on the orders of magnitude pages (1e10 m etc.) Perhaps we should just link to them. -- Heron

Yes, you're quite right. Well, about most things. I could argue that physically distance is dimensionless but that would just be arrogant pedantry. The page title is "large number" not just "large", and the order of magnitude pages are pretty good for comparing distances. BTW did you see my reply for you on Wikipedia:Reference desk? -- Tim Starling 13:53 18 Jun 2003 (UTC)

[edit] World Almanac

What's special about the World Almanac year 2000?? Do any of you Wikipedians have an edition of the World Almanac year 2004?? Try it. 66.32.95.180 01:52, 27 May 2004 (UTC)

There's nothing at all special about it. It's a pretty lousy source, actually. But it's a source. But it was the only source I happened to have at hand for quattuordecillion, etc. If they're in the 2000 edition they're probably in the 2004 edition, too, but I didn't think it was appropriate to say "World Almanac" without identifying which edition, and I certainly didn't think it was appropriate to reference an edition I hadn't consulted.

I think these are in the Merriam-Webster Third and probably lots of other places. I may get around to making a trip to the library this weekend and finding out. Hopefully someone else will do it first. Dpbsmith 14:29, 27 May 2004 (UTC)

[edit] Examples

$10^{\,\!100^{10}} < 10^{10^{20}}$
$100^{\,\!10^{10}} < 10^{10^{10.3}}$

Also compare:

$1.1^{\,\!1.1^{1.1^{1000}}} > 10^{10^{1.02*10^{40}}}$

I'm reverting the changes that were made to these equations, the discussion surrounding the equations clearly delineates the purpose of each and each is construed to show a certain aspect of "power towers", the revision is completely misleading when reading the text (besides making the equations wrong).

$10^{\,\!100^{10}}=10^{10^{11}} << 10^{10^{20}}$

$100^{10^{\,\!10}} = 100^{10000000000}$

$10^{10^{\,\!10.3}} \approx\ 10^{19952623149} = 100^{19952623149}$ not even close

In the 1.1 problem, I simplified the last exponent, thus:

$1.1^{\!1.1^{\,\!2.4699*10^{41}}} ? 10^{10^{1.02*10^{40}}}$

$1.1^{24} \approx\ 10$ $1.1^{1.1^{22}} \approx\ 10$

thus:

$10^{\!10^{\,\!2.4699*10^{41}}} > 10^{10^{1.02*10^{40}}}$

note that

2.4699 * 10 41

is not larger than

2.4699 * 10 41 - 46

by enough to change the outcome of the main problem.

Dusty78 03:38, 12 May 2005 (UTC)

Dusty, I think you need to bone up on your high school algebra. (a^b)^c = a^(b*c) In particular,

$10^{\,\!100^{10}}=10^{10^{11}}$

is wrong. 100^10 is not equal to 10^11, it's equal to 10^20. Revolver 04:05, 12 May 2005 (UTC)

oops, right I've gone nuts spent too much time working out the rest to bother with checking all my work, still, the example doesn't really fit with the explination... oh well, I'm quitting while I'm behind Actually, I think I'll be reworking the probs untill I'm sure I'm right on the others..Dusty78 04:14, 12 May 2005 (UTC)

Ahh... Pride goeth before a fall... and screwy math before a bad post.... don't know what I was smoking ;) I'm just going to revert back to when it was actually right and correct the text 2nd example is wrong for same reason as first, 3rd is wrong but it took some actual number crunching to evaluate.Dusty78 04:25, 12 May 2005 (UTC)

[edit] sorry

Sorry, I wasn't thinking for a second about the relative errors of really large numbers. Hopefully the current explanation explains it well. Revolver 05:23, 13 May 2005 (UTC)

[edit] new comment

I don't see any reference to the enormous numbers you get when you calculate permutations and combinations? How many permutations are there in a googolplex? (I hope that's grammatically correct.) {roger} 29 June 2005

You're on the right track! Check out the article on Combinatorics referenced at the beginning of the "even larger numbers" section, or better yet the Permutations and combinations article. Then you'll know to ask "How many permutations are there of 1 googolplex objects without replacement?", and that the the answer is googolplex factorial (written "googolplex!"). Lunkwill 29 June 2005 06:33 (UTC)

[edit] accumulated error in binary vs decimal exponents

The examples in the 'rule of thumb for converting between scientific notation and powers of two' section are misleading I think. The small error inherent in $2^{10} \approx 10^3$ is magnified immensely at these scales, not to mention the horrible single step of $100 \approx 128$ . The former is covered in Binary Prefix but it bears noting here as well if this section is to remain. The examples indicates that $10^{47} \approx 2^{157}$ and $2^{101} \approx 2 \times 10^{30}$ , which have errors of 82.7% and 26.76% respectively. $\approx$ might be appropriate with a 27% error, but 83% is very far outside the acceptable bounds of 'approximately equal to' imho. rules of thumb like that are what can put a crater instead of a lander on mars

Good point. I've added a disclaimer/correction for larger exponents. Feel free to reword it. Owen× ☎ 17:00, 5 December 2005 (UTC)

Changed your example. I hope you'll be OK with it and find it more instructive. 75.4.107.1 06:23, 17 November 2006 (UTC)

what's wrong with logarithms?

[edit] About the universe computer analogy...

How many possible characters are there to be chosen from? one human language? all human languages? ascii? ansi? the entire argument is invalid if we don't know how many individual characters there are to choose from. not trying to be mean, just a friendly request for clarification :)

Shadowrunner340 02:02, 15 September 2006 (UTC)RobbyShadowrunner340 02:02, 15 September 2006 (UTC)

[edit] From article page

The heading "Uncomputably large numbers" is misleading, as no integer is uncomputably large (every value of a busy beaver function is an integer). It's a busy beaver function that's noncomputable, not the values in its range. Also, although Rado's sequences might have been the first ones, it's very easy to produce noncomputable functions that grow faster than every computable function. --r.e.s. 04:14, 14 October 2006 (UTC)

Tom Harrison ^Talk 13:32, 14 October 2006 (UTC)

Thanks for moving my comment (above) from the article page to here, where it belongs. Sorry for the mix-up.--r.e.s. 19:16, 14 October 2006 (UTC)

[edit] Citations?

Many sections of this article, e.g. Systematically creating ever faster increasing sequences; Standardized system of writing very large numbers, have no citations. Although they are very interesting, it appears they are original research which is not allowed by Wikipedia policy. True? Mytg8 20:44, 30 October 2007 (UTC)

[edit] Systematically creating ever faster increasing sequences

Moved from page (original research):

Let a strictly increasing integer sequence $f 0 (n)$ (n≥1) be given (written as a function for conveniently writing the functional powers), with $f 0 (1) > 1$ . Then a sequence of sequences can be found by $f_k(n)=f_{k-1}^n(1)$ , from which we can select the "cross-sequence" $f 0 (10), f 1 (10),..$ ^[1]

Together this is a process of creating a new sequence from a given one. This can be repeated (i.e., we can apply recursion), and again we can select from the matrix of numbers a single sequence, by taking the 10th element of each. We can apply the same whole process again and again.

Starting from $f_0(n)=10^n=(10 \to n)$ this process corresponds to adding an element 10 in the Conway chain before the variable n, which is at the end of the chain: we get $f_k(n)=f_{k-1}^n(1)=10 \uparrow^{k+1} n=(10 \to n \to (k+1))$ , and the new sequence selected from the matrix is that of which the kth element is $f_{k-1}(10)=f_k^{10}(1)=10 \uparrow^k 10=(10 \to 10 \to k)=(10 \to 10 \to k \to 1)$ . (See also Knuth's up-arrow notation and Conway chained arrow notation). ^[2]

Repeating this process we get $(10 \to 10 \to k \to n)$ for successive values of n, and selecting k=10 we get a single sequence $(10 \to 10 \to 10 \to n)$ .

Repeating this whole process we get ever longer chains. Selecting n=10 we get the sequence of (10→10), (10→10→10), (10→10→10→10),... This can be used as starting sequence to apply the process again, etc. Even the value $f 1 (2)$ for this sequence is already a Conway chain of length 10 billion plus one.

Each sequence in this whole process can be identified by its order type in the process:

(10→n→k) is the sequence with index n with order type k - 1
(10→10→n→k) is the sequence with index n with order type ω + k - 1
(10→10→10→n→k) is the sequence with index n with order type 2ω + k - 1
(10→10), (10→10→10), (10→10→10→10),... is the sequence with order type ω²

In a similar way this can be continued, and we get a set of sequences, well-ordered by the procedure of construction.

zero case: $f 0 (n) = 10 n$
successor ordinal: $f_{x+1}(n)=f_x^n(1)$
for limit ordinals: $f_{a}(n)=f_{a_{n-1}}(10)$ for a suitable sequence of ordinals $a n - 1$ tending to a.

Thus we have transfinite recursion as far as we define for each limit ordinal a suitable sequence of ordinals tending to it.

Consider the Cantor normal form $\omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k$ , where k is a natural number, $c_1, c_2, \ldots, c_k$ are positive integers, and $\beta_1 > \beta_2 > \ldots > \beta_k$ are ordinal numbers. The limit ordinals cover the case $β k > 0$ . For order types less than ε₀ (epsilon nought) $β k$ is less than the ordinal itself. The following rules provide for each limit ordinal a suitable sequence of ordinals:

If $a=(\omega\uparrow)^p (u+v)$ where p ≥ 0, u ≥ v, and v is a limit ordinal, we take $a_n=(\omega\uparrow)^p (u+v_n)$ for a suitable sequence of ordinals $v n$ tending to v.

For $(\omega\uparrow)^p v$ where p ≥ 0, and v is a limit ordinal, we take the sequence $(\omega\uparrow)^pv_n$ for a suitable sequence of ordinals $v n$ tending to v.

For $(\omega\uparrow)^{p+1} (a+1)$ where p ≥ 0 and a ≥ 0, we take the sequence $(\omega\uparrow)^p (\omega^a n)$ .

For example, this procedure defines for the order type $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot5+578$ a particular, very rapidly increasing, sequence of integers, and to specify a particular large integer, we can refer e.g. to the 37th element in this sequence. The sequence is defined in 578 steps from the sequence with order type $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot5$ , which in turn is defined from the sequences with order types $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot4+\omega^{\omega^n}$ . The twelfth element in this sequence is defined from the sequences with order types $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot4+\omega^{\omega^{11}n}$ , etc.

For ε₀ we can take the sequence $\omega\uparrow\uparrow n$ , and for ε₀(k+1) the sequence ε₀k + $\omega\uparrow\uparrow n$ , etc. We cannot reach omega-one (ω₁), the set of all countable ordinal numbers, and the smallest uncountable ordinal number: no sequence of ordinal numbers below ω₁ has that ordinal as limit. It is also clear from the fact that we define sequences of which each element is defined in a finite number of steps, so making use of only a finite number of auxiliary sequences. Therefore for any of our sequences the set of auxiliary sequences is countable.

We have $f a (n) < f a + 1 (n)$ , except that we have "=" for n = 1. Note however that for a < b we do not always have $f a (n) < f b (n)$ . For example:

$f 3 (n) < f ω (n)$ for n = 1, 3, 4, 5, 6, .., but we have "=" for n = 2.
$f 4 (n) < f ω (n)$ for n = 1, 4, 5, 6, .., but we have ">" for n = 2, 3.

[edit] Improving this article

I would like to see this page become an A-class mathematics article. Might I suggest that one of the changes that would improve this article is to split it up into Large numbers and Notation of large numbers? AJRobbins (talk) 02:50, 20 November 2007 (UTC)

You need a notation to talk about things. There is only a need for a separate page if there is enough to say about alternative notations which is not needed for talking about the numbers themselves.--Patrick (talk) 12:53, 20 November 2007 (UTC)

Ok, this is what I see. This article is huge. The section "Systematically creating ever faster increasing sequences" was taken out because it was original research, but someone put it back in for some reason; this section can be taken out. The following sections are all about representations, and not numbers themselves: "Using Scientific notation", "Standardized system of writing very large numbers", "Comparison of base values", "Accuracy", and "Notations". Either A) these should be moved to Notation of large numbers, or B) they should stay and all other sections be moved to List of large numbers (not just a redirect) because these are the only two kinds of sections in this article. Since all sections fit nicely into either the "Examples" topic or the "Representation" topic, I think that they should be in separate articles. I would actually prefer B, as this follows Wikipedia's "List of" convention, and have this article talk about representation only. If this is going to be a "Large number" portal, then it only makes sense to cover the notation here, and leave examples, history, and so on to other pages. AJRobbins (talk) 07:34, 22 November 2007 (UTC)

Only the second part of "Systematically creating ever faster increasing sequences" was taken out, and is has not been put back.

Examples clarify notation, so perhaps they can better be put together.

Patrick (talk) 23:40, 22 November 2007 (UTC)

[edit] Graham's number?

I noticed that the page said that Graham's number is between $10 \rightarrow 10 \rightarrow 64 \rightarrow 2$ and $10 \rightarrow 10 \rightarrow 65 \rightarrow 2$ . Shouldn't the 10's be 3's? And if that's true, where does it go? ZtObOr 22:50, 22 January 2008 (UTC)

That's covered in footnote 4, where it is explained "regarding the comparison with the previous value: $10\uparrow ^n 10 < 3 \uparrow ^{n+1} 3$ , so starting the 64 steps with 1 instead of 4 more than compensates for replacing the numbers 3 by 10".

Also related to Graham's Number, I'd like to keep it in "See Also" as was recently added -- even though it's linked by the article text -- either that or, perhaps we can diminish that long list of example numbers. If we keep 10 → 5 → 2 and 10 → 9 → 2, can't we remove the ones in between? (etc.) —Preceding unsigned comment added by Mrob27 (talk • contribs) 22:45, 13 February 2008 (UTC)

[edit] This article

I cannot deal with this article. —Preceding unsigned comment added by 68.46.238.3 (talk) 21:59, 31 May 2008 (UTC)