A Limit-Question
WM
A Limit-Question
Consider the sets of positive even numbers E, of prime numbers P, of
Ulam's lucky numbers L, and of tetration numbers T
E = { 2, 4, 6, ... }
P = { 2, 3, 5, ... }
L = { 1, 3, 7, ... }
T = { 2, 2^2, 2^2^2, ... }
each of which may be denoted in the following by
{ x_1, x_2, x_3, ... }.
In every case* the initial segment
S(n) = { x_1, x_2, x_3, ... , x_n}
of the first n numbers contains such numbers which are greater than
the cardinal number |S(n)| of the segment. Now consider the cardinal
number of the set of those greater numbers and compare it to |S(n)|:
|{ x_k | x_k in S(n) & x_k > |S(n)| }| / |S(n)| = 1 - eps(n)
In the case of positive even numbers we have
eps(n) = 1/2 for even n, and
eps(n) = (n-1)/2n for odd n.
Both sequences converge to the limit
lim{n-->oo} eps(n) = 1/2.
Because of the asymptotic density n/logn for both P and L, we have in
these cases
lim{n-->oo} eps(n)*log(n) = 1 yielding lim{n-->oo} eps(n) = 0.
In case of tetration numbers T, we have of course also the lim{n-->oo}
eps(n) = 0. But is there a more informative short expression available
- in the sense of eps(n) --> 1/logn (or Li(n)/n or even (Li(n) +
k_1*n*exp(-k_2*sqrt(logn)))/n) being more informative than simply
writing eps(n) --> 0?
Regards, WM
*with exception of the first segment {1} of L.
[Moderator's Note: instead of eps(n) --> 1/logn it is
better to say eps(n) is asymptotic to 1/logn
or eps(n) ~ 1/logn .]
A. Bell
Perhaps the Lambert W-function could help in case T?
http://mathworld.wolfram.com/LambertW-Function.html
- A. Bell
WM
> > eps(n)
>
> > In the case of positive even numbers we have
> > eps(n) = 1/2 for even n, and
> > eps(n) = (n-1)/2n for odd n.
> > Both sequences converge to the limit
> > lim{n-->oo} eps(n) = 1/2.
>
> > Because of the asymptotic density n/logn for both P
> > and L, we have in
> > these cases
> > lim{n-->oo} eps(n)*log(n) = 1 yielding lim{n-->oo}
> > eps(n) = 0.
>
> > In case of tetration numbers T, we have of course
> > also the lim{n-->oo}
> > eps(n) = 0. But is there a more informative short
> > expression available
> > - in the sense of eps(n) --> 1/logn (or Li(n)/n or
> > even (Li(n) +
> > k_1*n*exp(-k_2*sqrt(logn)))/n) being more informative
> > than simply
> > writing eps(n) --> 0?
> Perhaps the Lambert W-function could help in case T?http://mathworld.wolfram.com/LambertW-Function.html
Thanks for your help. Alas, Lambert's W-function gives only the
limiting value for the complete power tower - not for the asymptotic
behaviour of the sequence of tetration numbers. But there is a simple
method to calculate eps(n) for the initial segments
{2}, {2, 2^2}, {2, 2^2, 2^2^2}, ...
I use the abbreviation T(n) for a power tower 2^2^...^2 consisting of
n two's such that the tetration numbers are now denoted by T(1), T(2),
T(3), ... . The cardinal number of an initial segment
S(n) = {T(1), T(2), T(3), ... , T(n)}
is just
|S(n)| = n.
In the special case
n = T(m)
being a tetration number, we have m numbers, T(1), T(2), ..., T(m),
less than or equal to |S(n)|. So we obtain
|{ T(i) | T(i) in S(n) & T(i) > |S(n)| }| / |S(n)| = (T(m) - m) / T(m)
such that in these cases
eps(n) = m / T(m).
This result is sufficient to determine the asymptotic behaviour of
eps(n).
Regards, WM
A. Bell
> But there is a simple
> method to calculate eps(n) for the initial segments
>
> {2}, {2, 2^2}, {2, 2^2, 2^2^2}, ...
>
> I use the abbreviation T(n) for a power tower 2^2^...^2 consisting of
> n two's such that the tetration numbers are now denoted by T(1), T(2),
> T(3), ... . The cardinal number of an initial segment
>
> S(n) = {T(1), T(2), T(3), ... , T(n)}
>
> is just
>
> |S(n)| = n.
>
> In the special case
>
> n = T(m)
>
> being a tetration number, we have m numbers, T(1), T(2), ..., T(m),
> less than or equal to |S(n)|. So we obtain
>
> |{ T(i) | T(i) in S(n) & T(i) > |S(n)| }| / |S(n)| = (T(m) - m) / T(m) [*]
>
> such that in these cases
>
> eps(n) = m / T(m).
>
> This result is sufficient to determine the asymptotic behaviour of
> eps(n).
By setting n = T(m) you get one of very few simple results for tetration numbers. It can be generalized: By adding j further elements to the segment S(n), such that n=T(m)+j but n<T(m+1) we get
|{ T(i) | T(i) in S(n) & T(i) > |S(n)| }| / |S(n)| = 1 - m/(T(m)+j)
and for T(m)+j=T(m+1) we obtain [*] with T(m+1) instead of T(m).
Therefore, the general result is
|{ T(i) | T(i) in S(n) & T(i) > |S(n)| }| / |S(n)| =< 1 - m/[T(m)]
where [T(m)] is the greatest tetration number less than or equal to |S(n)|=n.
Is there any special reason for calculating the number of numbers which are larger than the cardinal number?
- A. Bell
WM
> where [T(m)] is the greatest tetration number less than or equal to |S(n)|=n.
So it is. (I only corrected a typo in your final result [#].)
>
> Is there any special reason for calculating the number of numbers which are larger
> than the cardinal number?
For large n, nearly all elements T(i) are larger than the cardinal
number of the set.
If we obtain a theorem, as this one, which, with increasing strength,
holds for all finite cases, we use to say that it also holds in the
limit, i.e., in the infinite case.*) But in the present example this
would lead to the result that most tetration numbers must be greater
than the cardinal number of the set of all tetration numbers.**)
On the other hand, it is simple to show that the number of all
tetration numbers cannot be equal to or less than any natural (or
tetration) number.
This dilemma can only be solved by recognizing that the set of all
tetration numbers has no cardinal number which could be compared with
the natural numbers: aleph_0, like some complex number or like a
tensor or like a triangle or like a flavour, is not in trichotomy with
the natural numbers. I wanted to shows this.
*) Sometimes we say so even in case of decreasing strength. Consider
for example Cantor's well known diagonal argument.
By observing that the k-th digit b_k of the diagonal number differs
from the k-th digit a_k of the k-th list entry we know that the
diagonal number differs from the k-th list entry by at least (a_k -
b_k)*10^-k. We generalize this being-different from the first k
entries by saying that there is a difference between the diagonal
number and every list entry (although, because of lim {k --> oo} (a_k
- b_k)*10^-k = 0, the difference decreases with increasing k).
**) Without concluding from all finite cases to the infinite case we
could not infer that the infinite harmonic series Sum(1/n) is
infinite, not even that an infinite geometric series like Sum(1/2^n)
has a real value. We merely could make statements about finite sums
(with arbitrarily many terms though).
Regards, WM
victor_me...@yahoo.co.uk
(The notation in force is that T(1) = 2 and T(i+1) - 2^T(i).
The OP calls any number of the form T(n) a "tetration number".)
> For large n, nearly all elements T(i) are larger than the cardinal
> number of the set.
This sentence is unclear. It uses vague terms ("large",
"nearly") and the quantification is obscure
("for all n" then a statement without a free "n" within). Also
it is unclear what the final "set" is.
> If we obtain a theorem, as this one, which, with increasing strength,
> holds for all finite cases, we use to say that it also holds in the
> limit, i.e., in the infinite case.*)
Again "increasing strength" is a vague term. And "we" do
not always assert this. There are many theorems which
hold for all finite sets but not for infinite sets.
A trivial instance is from "all finite sets are finite" one cannot
infer that "all infinite sets are finite".
A less trivial example is that although every injection from a finite
set to itself is a surjection, it is false that
every injection from N to itself is a surjection.
> But in the present example this
> would lead to the result that most tetration numbers must be greater
> than the cardinal number of the set of all tetration numbers.**)
Which shows the absurdity of your vague assertion labelled *)
The cardinality of the set of tetration numbers is aleph_0
which is greater than each tetration number
(not less than "most" of them).
There is a bijection between N and the set {T(n):n in N}
of "tetration numbers", namely the map T. That is
what "the cardinality of the set of tetration numbers is aleph_0".
> On the other hand, it is simple to show that the number of all
> tetration numbers cannot be equal to or less than any natural (or
> tetration) number.
The number (actually the cardinality) of all tetration numbers
is aleph_0, which exceeds each natural number (finite cardinal)
or tetration number (since each is a natural number).
> This dilemma
There is no dilemma here.
> can only be solved by recognizing that the set of all
> tetration numbers has no cardinal number which could be compared with
> the natural numbers: aleph_0, like some complex number or like a
> tensor or like a triangle or like a flavour, is not in trichotomy with
> the natural numbers.
This alleged dilemma can be resolved by careful mathematical
reasoning, recognizing that the set of "tetration" numbers
is countably infinite, that is it has cardinality aleph_0
(contradicting your assertion **))
> I wanted to shows this.
Alas you didn't :-)
> **) Without concluding from all finite cases to the infinite case we
> could not infer that the infinite harmonic series Sum(1/n) is
> infinite, not even that an infinite geometric series like Sum(1/2^n)
> has a real value.
We can do this without your notion of generalizing
"from all finite cases to the infinite case"
by carefully defining the notion of the limit of an infinite.
series. This was done rigorously in the nineteenth century,
and can be found in all standard introductory texts on
analysis.
Victor Meldrew
"I don't believe it"
WM
> The OP calls any number of the form T(n) a "tetration number".)
>
> > For large n, nearly all elements T(i) are larger than the cardinal
> > number of the set.
>
> This sentence is unclear. It uses vague terms ("large",
> "nearly") and the quantification is obscure
> ("for all n" then a statement without a free "n" within). Also
> it is unclear what the final "set" is.
This sentence has been stated clearly by my equation [#] which you
deleted:
|{ T(i) | T(i) in S(n) & T(i) > |S(n)| }| / |S(n)| >= 1 - m/[T(m)]
[#]
where S(n) is the initial segment of n tetration numbers, and m is the
number of tetration numbers which are less than |S(n)| = n.
> > If we obtain a theorem, as this one, which, with increasing strength,
> > holds for all finite cases, we use to say that it also holds in the
> > limit, i.e., in the infinite case.*)
>
> Again "increasing strength" is a vague term. And "we" do
> not always assert this. There are many theorems which
> hold for all finite sets but not for infinite sets.
> A trivial instance is from "all finite sets are finite" one cannot
> infer that "all infinite sets are finite".
If you prefer a careful definition of limits, then apply it to [#]
please. You will see that the limit of m/T(m) = 0.
>
> > But in the present example this
> > would lead to the result that most tetration numbers must be greater
> > than the cardinal number of the set of all tetration numbers.**)
>
> Which shows the absurdity of your vague assertion labelled *)
> The cardinality of the set of tetration numbers is aleph_0
We may call it so.
> which is greater than each tetration number
> (not less than "most" of them).
What does support your assertion? Is it the fact that aleph_0 cannot
be smaller than any tetration number? That support is not sufficient
as you can see from the fact that there are numbers like i which are
not in trichotomy with the natural numbers. With regard to [#] this
assertion is obviously as unjustified as the statement that [งง] is
infinite.
>
> There is a bijection between N and the set {T(n):n in N}
> of "tetration numbers", namely the map T. That is
> what "the cardinality of the set of tetration numbers is aleph_0".
Of course. That is the definition. But the interpretation of aleph_0
as being a number in trichotomy with natural numbers is wrong.
>
> > On the other hand, it is simple to show that the number of all
> > tetration numbers cannot be equal to or less than any natural (or
> > tetration) number.
>
> The number (actually the cardinality) of all tetration numbers
> is aleph_0,
As I said already, I do not oppose.
> which exceeds each natural number (finite cardinal)
> or tetration number (since each is a natural number).
No set of tetration numbers can reach a cardinal number which is
larger than all numbers in the set. For finite sets this is obvious by
[#]. For the infinite set this follows from the limit n --> oo of
[#].
>
> > This dilemma
>
> There is no dilemma here.
Is it so hard to recognize what lim_(n --> oo) m/T(m) = 0 proves?
Regards, WM