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Topic:
BIG NUMBERS #2
Replies:
8
Last Post:
Feb 3, 2003 1:58 PM




BIG NUMBERS #2
Posted:
Apr 8, 2002 6:49 PM


I'm still working on an extensive revision of my sci.math post "GRAHAM'S NUMBER AND RAPIDLY GROWING FUNCTIONS" at
http://groups.google.com/groups?selm=28ae5e5e.0203041003.2b10abad%40posting.google.com
However, I thought I'd post a preliminary version of the first three sections now. I'm hoping that some of you can doublecheck my computations and/or give me suggestions for topics that I could discuss in these sections. I would especially like to know of any interesting examples in which large numbers arise that I haven't already dealt with.
My plan is to carefully work up to the HowardBachmann ordinal level of the GrzegorczykWainer Hierarchy, and then touch on some of the constructible ordinal notations that go beyond this. At present I have 10 sections (in various stages of completion), but there may be more than this by the time I'm done.
Dave L. Renfro
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2. HOW TO CALCULATE WITH LARGE NUMBERS
A. BASE10 EXPONENTIAL TOWERS B. TETRATEDSCIENTIFIC NOTATION C. EXPONENTIAL EVALUATIONS OF LARGE NUMBERS D. FACTORIALS OF LARGE NUMBERS E. HYPERROOTS F. REFERENCES FOR SECTION 2
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A. BASE10 EXPONENTIAL TOWERS
In this section I will show how some of the computations in Section 1A can be made. In many places below I will use the "=" sign for approximate equality. [To do otherwise would tend to obscure some of the main ideas and, in the present context, this usage of "=" should not cause confusion.]
When I say "evaluate", "find the value of", etc. of a numerical expression involving large numbers, what I really mean is to rewrite the expression in base10 exponentiated tower form. That is, we want an expression of the form 10^10^...^10^N, where N is expressed in some ordinary base10 manner, such as N = 3.14159, N = 1024, N = 6.023 x 10^23, etc. This is analogous to putting numbers into scientific notation. In the case of scientific notation, we write numbers as N x 10^n, where 'n' specifies the orderofmagnitude of our number and the numerical accuracy of our number is incorporated into the value of 'N'. In the case of base10 exponentiated tower form, the number of 10's appearing in 10^10^...^10^N defines a generalized orderofmagnitude for our number and the numerical accuracy of our number is incorporated into 'N'.
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B. TETRATEDSCIENTIFIC NOTATION
To carry the above analogy further, we can write 10^10^...^10^N as N @ 10^^n, where 'n' is the number of 10's used in the former expression. Note that exponentiation is neither commutative nor associative. Therefore, the obvious way to generalize scientific notation, namely by writing numbers in the form N ^ 10^^n, doesn't give what we want. This is why I'm using the symbol '@'. We want to use a symbol whose presence in this context would not cause any notational ambiguities.
It might be interesting to see if there are any identities that would allow us to rewrite expressions involving products and/or exponentiations of numbers written in tetratedscientific notation as a single number written in tetratedscientific notation. I'll leave this for the interested reader to pursue.
There is at least one key difference between tetratedscientific notation and ordinary scientific notion. Every integer can be expressed exactly in the form N x 10^n for an appropriate pair of integers N and n, but manyin fact, mostintegers cannot be expressed exactly in the form N @ 10^^n for any integers N and n. This will not be very important for our use of this notation, however.
Another path that an interested reader might wish to pursue is to come up with a reasonable notion of what I'll call a "base10 hyperlogarithm". In the same way that the base10 logarithm of N x 10^n is n + log(N), it might be worthwhile to investigate the possibility of letting the base10 hyperlog of N @ 10^^n be n times the base10 hyperlog of N. Of course, the real problem is to define what the base10 hyperlog of a number is. I'm simply suggesting that the definition be compatible with tetrated scientific notation in this way. I suspect that once the notion has been satisfactorly defined for base10, it will not be very difficult to extend it to other bases. [As for notation, I suggest using (in nonASCII format) "log" with a left superscript '2' and a right subscript '10' to denote base10 hyperlogarithm. Note this allows us to have a notation for hyperhyper log's (use a left superscript '3'), hyperhyperhyper logs (use a left superscript '4'), etc., should these notions eventually arise.] For more about this and some related topics, see Section E below ("HyperRoots"). At the end of Section E there is a list of web pages that anyone interested in these things would probably want to look at.
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C. EXPONENTIAL EVALUATIONS OF LARGE NUMBERS
EXAMPLE 1: Evaluate 7^(35,348).
Take the base10 logarithm of this number. Using properties of logarithms, we can rewrite the result so that what we have is numerically manageable:
log [7^(35,348)] = (35,348)*log(7)
A base10 numerical approximation of this is
29872.5255184239504520154203087.
Therefore,
7^(35,348) = 10 ^ (29872.5255184239504520154203087)
= 10 ^ (29872 + .5255184239504520154203087)
= [10^(.5255184239504520154203087)] * [10 ^ (29,872)]
= 3.3536553 x 10^(29,872).
EXAMPLE 2: Evaluate 2^{3^[4^(5^6)]} and 6^{5^[4^(3^2)]}.
[[ Expressions such as these are sometimes called "superfactorials". ]]
2^{3^[4^(5^6)]} = 2 ^ {3 ^ [4^15625] }
... now use the method in Example 1 ...
= 2 ^ {3 ^ [1.539446 x 10^9407] }
Let's focus on 3 ^ [1.539446 x 10^9407]. The logarithm of this number can be written as
(1.539446)*log(3) x 10^9407
= .7345024 x 10^9407
= 7.345024 x 10^9406.
Therefore,
3 ^ [1.539446 x 10^9407] = 10 ^ [7.345024 x 10^9406].
Now we need to evaluate the base2 exponential of 10 ^ [7.345024 x 10^9406]. Its base10 logarithm is
{ 10 ^ [7.345024 x 10^9406] } * log(2)
= { 10 ^ [7.345024 x 10^9406] } * {10 ^ [log(log 2)]}
= { 10 ^ [ log(log 2) + 7.345024 x 10^9406] }
= { 10 ^ [7.345024 x 10^9406] }.
Note that we can completely ignore the additive term log(log 2) in the presence of 7.345024 x 10^9406. More generally, 2^(10^n) is, for all intents and purposes, the same as 10^(10^n) when n is large enough that we can ignore log(log 2) in the presence of n. Still more generally, since two applications of a base10 logarithm to a^(b^n) gives n*log(b) + log(log a), we have
a^(b^n) = 10 ^ { 10 ^ [n*log(b) + log(log a)] }.
If n is reasonably large compared to 'a', then we can replace a^(b^n) with 10^(b^n). Moreover, if n is also fairly large compared to b (e.g. n = 10^k, where k is sufficiently large that we can additively ignore log(log b) in the presence of k), then we can replace a^(b^n) with 10^(10^n).
What this means is that if we have an exponential tower such as a^b^c^d^... (a finite tower), then we can typically replace all but the top three powers with 10's, or with any other numbers of our choosing, as long as the replacement numbers are not "too large". [We can probably do this whenever all of the replacement numbers can be additively ignored in the presence of the value of the topmost evaluation of ^, but I haven't looked into this matter very closely.]
Littlewood [71] (p. 103) puts it this way >>>
"We may sum up these considerations as the 'principle of crudity': the practical upshot is that in estimating a number a^b^c^... it is worth taking trouble to whittle down the top index, but we can be as crude as we like about things that bear only on the lowest terms."
Getting back to our example, we found that
2^{3^[4^(5^6)]} = 10^(10^(7.345024 x 10^9406)).
Evaluating 6^{5^[4^(3^2)]} in the same way (details omitted so that you can use this as a practice example), we get
6^{5^[4^(3^2)]} = 10^(10^(1.83231376 x 10^5))
= (1.83231376 x 10^5) @ 10^^2.
This is a much smaller number than what we got before, and it illustrates the principle that the upper exponents in a tower are overwhelmingly more important than the lower exponents in a tower.
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D. FACTORIALS OF LARGE NUMBERS
Stirling's formula says that for large values of n, n! is approximately equal to
sqrt(2*Pi) * n^(n + 1/2) * exp(n).
[[ This formula gives a result that is about 25/(3n) percent less than the correct value. ]]
http://mathworld.wolfram.com/StirlingsApproximation.html http://www.sosmath.com/calculus/sequence/stirling/stirling.html
Because we will be using large numbers, we'll need to take a base10 logarithm at some point (re: the method explained above). Doing so now will allow us to find a simpler expression for n! when n is large.
log(n!) = (1/2)*log(2*Pi) + (n + 1/2)*log(n)  n*log(e)
For reasonably large values of n, we can ignore the first term and the "+ 1/2" term. This gives
log(n!) = n*log(n)  n*log(e)
= n * [log(n)  log(e)].
Since we will be working with powers of 10, let's incorporate this as well. If n = 10^k, then we have (approximately)
log(n!) = n * [k  log(e)].
EXAMPLE 1: Evaluate (10^10)!.
The base10 logarithm of (10^10)! is approximately
10^10 * [10  .4342944819]
= 9.565705518 x 10^10.
Therefore,
(10^10)! = 10^(9.5657 x 10^10).
EXAMPLE 2: Evaluate (10^100)!.
The base10 logarithm of (10^100)! is approximately
10^100 * [100  .4342944819]
= 9.9565705518 x 10^101
Therefore,
(10^100)! = 10^(9.95657 x 10^101).
EXAMPLE 3: Evaluate (10^1000)!.
The base10 logarithm of (10^1000)! is approximately
10^1000 * [1000  .4342944819]
= 9.99565705518 x 10^1002.
Therefore,
(10^1000)! = 10^(9.995657 x 10^1002).
EXAMPLE 4: Evaluate googolplex!.
The base10 logarithm of [10^(10^100)]! is approximately
10^(10^100) * [10^100  .4342944819]
= 10^(10^100) * 10^100 = 10^(10^100 + 100).
Therefore,
[10^(10^100)]! = 10^(10^(10^100)))
= 100 @ 10^^3.
Looking at this last example we see that if n = 10^(10^m) with m > 100, then n! is indistinguishable from 10^n. Algebraically, if n = 10^(10^m), then we have
log(n!) = n * [10^m  log(e)] = n * 10^m
= 10^(10^m) * 10^m = 10^(10^m + m)
= 10^(10^m)
= n,
where I've made two clearly appropriate approximations.
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E. HYPERROOTS
Consider the following:
b*(1/3) is the solution to the equation x*3 = b
b^(1/3) is the solution to the equation x^3 = b
Let's see what happens if we continue in the obvious way and define:
b^^(1/3) is the solution to the equation x^^3 = b
[[ The operation ^^ (tetration) is discussed in Section 3. For now it will be enough to know that x^^n represents an exponentiated tower of x's of height n. For example, x^^4 = x^[x^(x^x)]. ]]
Assume x > 1, b > 1, and n is a positive integer. Then x^^n = b has a unique solution for x because x^^n is a strictly increasing continuous function from the interval (1, infinity) onto the interval (1, infinity).
With these restrictions in mind, we will define b^^(1/n) to be the unique solution to x^^n = b.
When using a MATHEMATICA, MAPLE, etc. to solve x^(x^x) = b for b > 1, it is sometimes more convenient to solve the equivalent equation (obtained by taking log's twice)
x*log(x) + log(log x) = log(log b).
This form is especially recommended for larger values of b. [The actual size of b where this form becomes more useful will vary probably vary with the software.] Here are some hypercube roots:
10 ^^ (1/3) = 1.9235840
1000 ^^ (1/3) = 2.3849098
(10^10) ^^ (1/3) = 2.8942439
(10^100) ^^ (1/3) = 3.8304829
(10^1000) ^^ (1/3) = 4.7117508
googolplex ^^ (1/3) = 56.849709
Skewes number ^^ (1/3) = 3.0780337 x 10^32
Note this shows rather dramatically how much larger Skewes number is compared to a googolplex.
The evaluation of higher hyperroots is easy when we recall an observation made in Example 2 of Section C above:
What this means is that if we have an exponential tower such as a^b^c^d^... (a finite tower), then we can typically replace all but the top three powers with 10's, or with any other numbers of our choosing, as long as the replacement numbers are not "too large".
What this means is that if n > 3 and we want to solve x^^n = b (for b sufficiently large), then we will not be very far off if we solve 10^(10^(...(10^(x^(x^x)))...)) = b [n3 10's],
which is equivalent to solving
x^^3 = log(log(...(log b)...)) [n3 log's].
Even better, and probably to an accuracy well beyond anything you'd need when b is large, would be to solve
x^^4 = log(log(...(log b)...)) [n4 log's],
which is equivalent to solving (take log's three times)
log [ (x^x)*log(x) + log(log x) ] = log(log(...(log b)...))
[n1 log's].
I found that I didn't need this for the following except in the case of the hyperfourth root of Skewes number. For Skewes number I had to resort to solving the equation
log [ (x^x)*log(x) + log(log x) ] = 34.
10 ^^ (1/4) = 1.7343125
1000 ^^ (1/4) = 1.9319788
(10^10) ^^ (1/4) = 2.0900767
(10^100) ^^ (1/4) = 2.2974808
(10^1000) ^^ (1/4) = 2.4396281
googolplex ^^ (1/4) = 3.8314042
Skewes number ^^ (1/4) = 24.403004
It would be useful to have a "well behaved" definition of x^^y for y > 0, at least when x > 1 if not more generally, because that would allow us to define, for instance, the hyperhyper cube root of b, e.g. b^^^(1/3), as the solution to the equation
x^^^3 = x^^(x^^x) = b.
However, it is not obvious how to do this. We would, of course, begin by defining x^^y for positive rational numbers y. If the resulting function (x fixed, y rational) is continuous (that is, whenever y_m > y for positive rationals y_m and y, then we have x^^(y_m) > x^^y), then we can get a welldefined and continuous extension of x^^y to positive real numbers y. Unfortunately, we encounter some serious problems in the first step of this process.
If we want x^^y to behave well under limits for rational y's, then at the very least we need to have x^^y welldefined when y is a rational number. The following evaluations show that the two obvious ways that one might use to define x^^y for rational y's fail even on this account. [Warning: The shortcut method I mentioned earlier for evaluating b^(1/n) when n > 3, where we replace all but the top three x's in x^x^...^x = b (n x's) with 10's (or even the more accurate approximation where all but the top four x's are replaced with 10's), doesn't work for the last two evaluations below. The values of b in these two calculations wind up not being large enough for this approximation to be valid.]
10 ^^ (1/2) = 2.5061841 googol ^^ (1/2) = 56.961248
[10 ^^ 2] ^^ (1/4) = 2.0900767 [googol ^^ 2] ^^ (1/4) = 2.2989170
[10 ^^ (1/4)] ^^ 2 = 2.5984930 [googol ^^ (1/4)] ^^ 2 = 6.7603517
[10 ^^ 3] ^^ (1/6) = 1.8273653 [googol ^^ 3] ^^ (1/6) = 2.2992298
[10 ^^ (1/6)] ^^ 3 = 2.6537973 [googol ^^ (1/6)] ^^ 3 = 9.1140260
Even more striking is the following >>>
googolplex ^^ (1/2) = 1.0203172 x 10^98
[googolplex ^^ 2] ^^ (1/4) = 56.849709
[googolplex ^^ (1/4)] ^^ 2 = 171.82257
Note that [googolplex ^^ 2] ^^ (1/4) agrees with the value we found earlier for googolplex ^^ (1/3). Moreover, these two values are not very different from googol ^^ (1/2). This is related to the fact that for k > 3 the following "identities" hold:
(10^^k) ^^ n = 10 ^^ (k+n1)
(10^^k) ^^ (1/n) = 10 ^^ (kn+1) (k > n).
For more about defining x^^y for rational and real numbers y, see Appleby [5], Bennett [12], Bromer [16] (pp. 172173), Knoebel [63] (pp. 247248), and Wright [109].
Below are some web pages that deal with issues related to this topic. The most relevant of these for hyperroots is the sci.math thread "Generalization of Radicals".
"1. Tetration (hyperexponentiation)" by Constantin Rubtsov http://numbers.newmail.ru/english/01/1e.html
sci.math  "What is pi to the pi, pi times? (Was Q: `Half' a logarithm?)" [Nov. 13, 1995] http://groups.google.com/groups?selm=487vle%24jeq%40news.cis.okstate.edu
sci.math  "Re: Q: `Half' a logarithm?" [Nov. 1416, 1995] http://groups.google.com/groups?hl=en&th=834d1208768cfaec&rnum=4
sci.math ''HyperExponentiation" [April 23 to Oct. 31, 1999] http://mathforum.org/epigone/sci.math/wermzoujerm
sci.math  "Interpolation of iterated exponentiation?" [Nov. 16, 1999] http://mathforum.org/epigone/sci.math/sworkanddwee
sci.math  "Generalization of Radicals" [Jan. 911, 2000] http://mathforum.org/epigone/sci.math/brelsmozhing
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F. REFERENCES FOR SECTION 2
[5] J. C. Appleby, "Hexponentiation", The Mathematical Gazette 79 #484 (March 1995), 8487.
[12] Albert A. Bennett, "Note on an operation of the third grade", Annals Math. (2) 17 (191516), 7475.
[16] Nick Bromer, "Superexponentiation", Mathematics Magazine 60 (1987), 169173.
[63] R. Arthur Knoebel, "Exponentials reiterated", Amer. Math. Monthly 88 (1981), 235252. [Has an extremely complete bibliography of 125 references. Bennett's paper does not appear, however.]
[71] John E. Littlewood, "Large Numbers", Mathematical Gazette 32 #300 (July 1948), 163171. [Reprinted on pp. 100113 of Bela Bollobas, LITTLEWOOD'S MISCELLANY, Cambridge Univ. Press, 1986. The largest number in Archimedes' "The Sand Reckoner" is given incorrectly as 10^(8 x 10^15) on p. 163 of the article and on p. 100 of the book reprint.] http://uk.cambridge.org/mathematics/catalogue/052133702X/default.htm
[109] E. M. Wright, "Iteration of the exponential function", Quarterly J. Math. (Oxford) 18 (1947), 228235.
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