Re: infinity


sci.math_031010:
An infinite binary sequence with a beginning is presented to you, and
you can modify it via this method: you can change any element from a
one to a zero, or a zero to a one. If you change a one to a zero, then
you must change that or some other zero to a one. If you change a zero
to a one instead, you must change that or some other one to a zero.
There may be allowed the interchange of identical elements as they
would not change the sequence.

sci.math_031010:
About the dual representation, consider a crazy model where there is
not dual representation and .1000... does not equal .0111.... If you
must have dual representation, then the infinite binary sequence is a
unique representation of a subset of the naturals but not a unique
representation of a real, it is a representation but not necessarily
unique, via dual representation.

sci.math_031010:
So anyways given a sequence (01)..., the subsequence 01 repeating
infinitely, it's possible to change it to (10)..., the subsequence 10
repeating infinitely, by the operation of changing the first, second,
third, etcetera elements of the sequence, as the first requires
changing the value of a symbol that occurs in the second place,
changing the third requires changing the value of a symbol that occurs
at the fourth place, etcetera, ad infinitum.

sci.math_031010:
Given an input sequence a and some output sequence b, there is a
function between the various elements they may represent, a bijection.
For example if the sequence a has finitely may ones so does b, the same
amount, and if sequence a has infinitely many ones so does b.

sci.math_031010_c:
I'm trying to think of some implications of that the sequence element
interchange algorithm yields an output infinite sequence from the input
infinite sequence with the same density of binary elements.



sci.math_031010_c:
One thing about it is that it allows me to define a sequence of a given
density and then to determine zero or infinitely many other sequences
with the same sequence. Then, I can relate the elements of the
sequence by their index to elements of the naturals and make claims
about the density of subsets of the integers within the integers, in a
way separate from but consistent with the asymptotic density of number
theory.

sci.math_031010_c:
Now I can say that "half of the non-negative integers are even" because
there the index of the sequence represents an element of N. To
represent Z I think that here I could consider an infinite sequence
with no beginning, ...(01)..., but also as I can biject N to Z. I can
map the index elements to Z. For example, this might have the first
element representing zero, then the second represents one and third
negative one: ( 0 1 -1 2 -2 3 -3 ...) When I construct a sequence
that has a one for each even integer and a zero for each odd integer it
is as so: 100110011001(1001).... The repeating sequence by sequence
element interchange can be modified to (1010), and thus reduced to (10)
as it contains duplicate subsequences, matching the canonical repeating
subsequence for one-density of one half! Half the integers are even.

sci.math_031011:
Yet, that might only slow the process and then still I must show if
that would not allow the conversion of any sequence (001)... through
(1)... to all other sequences with infinitely many ones and zeros.

sci.math_031011:
How about (011) to (110)...? The list of intermediate sequences
between them is infinite. A rule to convert them all at once for each
triplet is to convert (011) to (101) to (110). For the previous
example the elements are interchanged in a two element sliding window.

sci.math_031011_b:
Let's see here, I want an algorithm that allows to me convert any
(infinite, binary) sequence of a contrived density to any other
sequence of the same contrived density but no sequence of a different
contrived density.

sci.math_031011_b:
Where I am at right now is that a one can be exchanged with a zero
wihin q many places for contrived density p/q to the left or right and
then that or another one is interchanged with a zero within q places in
the opposite direction. What I'm trying to determine is that any
(infinite, binary) sequence of a contrived density can be thusly
converted to any other sequence of the same contrived density but no
sequence of a different contrived density.

sci.math_031011_b:
I can restrict the case to sequences of only the repeating subsequences
and then say it converts to one of the other subsequences of that
length but that doesn't handle the general case. That's about for any
finite sequence that you can interchange the ones and zeros and never
change either of their counts. That's not good for my purposes, I need
a method that works on infinite sequences. I could select an
arbitrarily large finite window the length of which is an integer
multiple of the length of the subsequence, but I don't know how that
would do me any good either. Yet, if there doesn't exist some
arbitrarily large finite window or finitely many arbitrarily large
finite windows, and only work on the infinite case for the subsequence,
that would help, but I think there would still be sequences that would
be intractable.

sci.math_031011_b:
For example that would stop (001)... from converting to (01)...,
because there is not finitely many finite contiguous subsequences of
(001)... that convert to (01).... It would allow (01)... to convert to
(10)..., because 01 and 10 are subsequences of the same length where
there are infinitely many of them.

sci.math_031011_b:
Basically this divides the canonical sequence into subsequences of
length q, or an integral multiple of q. Then, each can have
interchanged its elements. The resultant subsequences remain in the
same order in the output sequence. For example, the sequence could be
forty billion subsequences of 01, then 00001111, then the (infinitely)
repeating subsequence 10 , and that can convert to (01)... or (10)....
For the finite case a normal permutation is allowable.

sci.math_031011_b:
Let's see here, I want an algorithm that allows to me convert any
(infinite, binary) sequence of a given density to any other sequence of
the same density but no sequence of a different density.

sci.math_031011_b:
In the context of this thread that goes back to determining the
asymptotic probability of n choice n/2. Asymptotic: a word I use not
knowing the definition. I just think it means for a scalar infinite
variable.

sci.math_031012_seqconstants:
One well known example of this kind of summation is that of 1/2^x for
x=1 to infinity, which is equal to one.

sci.math_031012_seqconstants:
I got to thinking about this from the discussion about how an infinite
binary sequence represents a given subset of the natural numbers, or in
this case positive integers. I suppose the binary sequence could as
well represent any subset of any infinite subset of the naturals or
integers, or rationals, etcetera.

sci.math_031013:
About the compass and rule to trisect an angle, for example to form an
angle of pi/3 radians from lines intesecting a point on a straight
line, an angle of pi radians, it's easy to bisect an angle with the
compass, and 1/3 is the sum of 1 / 2^2x for each positive integer x, eg
..010101(01).... A theoretical rule and compass bijection that takes
infinitesimal time to bisect an angle may well trisect an angle in some
finite time. When you cross the room, you get all the way there.

sci.math_031013:
About y/x, say that all that is known is that y is greater than x. Is
y thus a dependent variable of x? All that is known is that y > x.
The value of y/x as x goes to infinity is indefinite. It might be a
finite value, it might diverge. Are x and y thus interdependent?

sci.math_031013:
About the itty bitty line segments, infinitesimal line segments might
be comprised of a pair of adjacent points. Yet, that's not acceptible.

sci.math_031014:
In this thread we've discussed an expression of n!/ (n/2)!^2 2^n as n
diverges to infinity. That would seem to imply that that many of the
sequences have equal densities of ones and zeros. Yet, in light of
that fact and possibly contradictorily to it while that might be the
case, any absolutely normal number has equal densities of zeros and
ones, and Borel says the Lebesgue measure of abnormal numbers is zero.
This is where I just thought that half the binary numbers had equal
densities of ones and zeros.

sci.math_031014:
Assuming that we have some mutual comprehension of what density is
within an infinite binary sequence, is it so that a number normal to
base two has zero-density and one-density of one half each?

sci.math_031015:
Half the integers are even: represent the non-negative integers as a
naturally or canonically ordered sequence by an infinite binary
sequence (10)..., that sequence has a density of one half. The other
half of the non-negative integers is odd, the odd integers are the
other half of the integers to the even integers. The multiples of
three are not the other half of the integers to non-multiples of three,
they're the other third to multiples of three plus one and multiples of
three plus two, there are three equal categories, thirds. To demand
otherwise in ignorance of that plain fact is ridiculous. To be sure
there is a sophisticated enough understanding of these as infinite sets
that a function is easy to define among the multiples of two and the
multiples of three, but a closer examination that a specification of
the function as being in the form x+C gives a much more intuitive match
to these well known sequences of odd and even numbers definitely
showing that being in line with expectation. The true sophisticate
demands more to reconcile the intuition with succinct provability.

.

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