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Dave L. Renfro

Posts: 4,792
Registered: 12/3/04
Posted: Apr 8, 2002 6:49 PM
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I'm still working on an extensive revision of my sci.math

However, I thought I'd post a preliminary version of the first
three sections now. I'm hoping that some of you can double-check
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 Howard-Bachmann ordinal
level of the Grzegorczyk-Wainer 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






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 base-10 exponentiated tower form.
That is, we want an expression of the form 10^10^...^10^N, where
N is expressed in some ordinary base-10 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 order-of-magnitude of our number and the numerical
accuracy of our number is incorporated into the value of 'N'.
In the case of base-10 exponentiated tower form, the number of
10's appearing in 10^10^...^10^N defines a generalized
order-of-magnitude for our number and the numerical accuracy
of our number is incorporated into 'N'.



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 tetrated-scientific notation
as a single number written in tetrated-scientific notation. I'll
leave this for the interested reader to pursue.

There is at least one key difference between tetrated-scientific
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 many--in fact, most--integers 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,

Another path that an interested reader might wish to pursue is to
come up with a reasonable notion of what I'll call a "base-10
hyper-logarithm". In the same way that the base-10 logarithm of
N x 10^n is n + log(N), it might be worthwhile to investigate the
possibility of letting the base-10 hyper-log of N @ 10^^n be
n times the base-10 hyper-log of N. Of course, the real problem is
to define what the base-10 hyper-log 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 base-10, it will not be very
difficult to extend it to other bases. [As for notation, I suggest
using (in non-ASCII format) "log" with a left superscript '2' and
a right subscript '10' to denote base-10 hyper-logarithm. Note
this allows us to have a notation for hyper-hyper log's (use a left
superscript '3'), hyper-hyper-hyper logs (use a left superscript
'4'), etc., should these notions eventually arise.] For more about
this and some related topics, see Section E below ("Hyper-Roots").
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.



EXAMPLE 1: Evaluate 7^(35,348).

Take the base-10 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 base-10 numerical approximation of this is



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 "super-factorials". ]]

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.


3 ^ [1.539446 x 10^9407] = 10 ^ [7.345024 x 10^9406].

Now we need to evaluate the base-2 exponential of
10 ^ [7.345024 x 10^9406]. Its base-10 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 base-10 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 top-most 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.



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. ]]

Because we will be using large numbers, we'll need to take a
base-10 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 base-10 logarithm of (10^10)! is approximately

10^10 * [10 - .4342944819]

= 9.565705518 x 10^10.


(10^10)! = 10^(9.5657 x 10^10).

EXAMPLE 2: Evaluate (10^100)!.

The base-10 logarithm of (10^100)! is approximately

10^100 * [100 - .4342944819]

= 9.9565705518 x 10^101


(10^100)! = 10^(9.95657 x 10^101).

EXAMPLE 3: Evaluate (10^1000)!.

The base-10 logarithm of (10^1000)! is approximately

10^1000 * [1000 - .4342944819]

= 9.99565705518 x 10^1002.


(10^1000)! = 10^(9.995657 x 10^1002).

EXAMPLE 4: Evaluate googolplex!.

The base-10 logarithm of [10^(10^100)]! is approximately

10^(10^100) * [10^100 - .4342944819]

= 10^(10^100) * 10^100 = 10^(10^100 + 100).


[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.



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 hyper-cube 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 hyper-roots 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 [n-3 10's],

which is equivalent to solving

x^^3 = log(log(...(log b)...)) [n-3 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)...)) [n-4 log's],

which is equivalent to solving (take log's three times)

log [ (x^x)*log(x) + log(log x) ] = log(log(...(log b)...))

[n-1 log's].

I found that I didn't need this for the following except in the
case of the hyper-fourth 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 hyper-hyper 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 well-defined 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 well-defined 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 short-cut 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+n-1)

(10^^k) ^^ (1/n) = 10 ^^ (k-n+1) (k > n).

For more about defining x^^y for rational and real numbers y,
see Appleby [5], Bennett [12], Bromer [16] (pp. 172-173),
Knoebel [63] (pp. 247-248), and Wright [109].

Below are some web pages that deal with issues related to
this topic. The most relevant of these for hyper-roots is
the sci.math thread "Generalization of Radicals".

"1. Tetration (hyper-exponentiation)" by Constantin Rubtsov

sci.math -- "What is pi to the pi, pi times? (Was Q: `Half' a
logarithm?)" [Nov. 13, 1995]

sci.math -- "Re: Q: `Half' a logarithm?" [Nov. 14-16, 1995]

sci.math ''Hyper-Exponentiation" [April 23 to Oct. 31, 1999]

sci.math -- "Interpolation of iterated exponentiation?"
[Nov. 16, 1999]

sci.math -- "Generalization of Radicals" [Jan. 9-11, 2000]



[5] J. C. Appleby, "Hexponentiation", The Mathematical Gazette
79 #484 (March 1995), 84-87.

[12] Albert A. Bennett, "Note on an operation of the third grade",
Annals Math. (2) 17 (1915-16), 74-75.

[16] Nick Bromer, "Superexponentiation", Mathematics Magazine 60
(1987), 169-173.

[63] R. Arthur Knoebel, "Exponentials reiterated", Amer. Math.
Monthly 88 (1981), 235-252. [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), 163-171. [Reprinted on pp. 100-113 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.]

[109] E. M. Wright, "Iteration of the exponential function",
Quarterly J. Math. (Oxford) 18 (1947), 228-235.


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