mm-3169
====
 
If n has the form: n = (p^x)k + 1, and n = n1*n2
What form must n1 and n2 each have?
 
what do you mean by for?  you could equate the two expressions since n
equals n1*n2 and it also equals kp^x+1 but what does that solve fo you, are
looking for dinmensions?
Jay
Subject: Re: Quick Math Guide to core error issues
A literate person, I see. The Dower House, Lakey Hill actually.
http://www.south-borsetshire-hunt.org.uk/officials.html
Subject: Collatz Conjecture Community Web Site
I have the site back online after the move. I went ahead and snagged a
proper URL for the site.
endevor.
The new URL -
www.collatzconjecture.com
-or-
collatzconjecture.com
------------------------------
I have been interested in this problem for many years and have noticed
a need, or at least a desire on my part, for a community type website
dedicated to this problem. So, I created one.
This site serves these features;
1. A Links page for you to post pages about this problem with the
ability to rate the pages for their usefulness.
2. A discussion board for a focused discussion on this problem.
3. A Books review section to suggest published material to assist with
the topics related to this problem.
4. An events announcement page, where events can be put forward to a
targeted community of people interested in this problem.
5. A simple image gallery for images of interest about this problem.
6. A downloads section where users can get a central listing of PDF,
Mathematica Notebooks, Maple Files, etc.. all related to this problem.
This site is brand new, but should be fairly stable, though slow for a
while until I find a good site to host it. I would be interested in
anyone who would like to be a moderator to keep the site clean and
targeted and assist with the sites overall design and content.
Please check it out. You can comment directly to me at the link on the
site in the header.
I think a site like this would serve as a needed gathering place for
any and all individuals interested in the Collatz Conjecture.
The url is a bit odd as I am borrowing one I currently have and am not
using. If you have a suggestion for the url, please let me know.
The URL is:
(it was for my mother, who does not really care to do anything with it
:)
Subject: Re: Antipodal Points and S^1
is continuous then there are x, y such that f(x)=f(y) and d(x,y) is
maximum?
Subject: Re: Factorial/Exponential Identity, Infinity
No, it's not.
Considering the unit interval, if some subset of those points has a
measure of one half, then, half of the points on the unit interval are
elements of the subset.  Consider for example half the unit interval.
A problem with measure is that it says that many sets that are proper
subsets and infinitely proper subsets of the unit interval's points
have measure equal to one, in the above sense implying that all the
elements of the set are elements of the proper subset, which would be
a contradiction in terms, which is why it is necessarily not defined
that way.
Borel said that almost all of the elements of reals are absolutely
normal, it's been shown that uncountably many elements of the reals
are abnormal, not normal.  I believe that this also applies to finite
intervals in the reals.  So we consider the unit interval, and a
characteristic of it is that almost all of the elements are absolutely
normal.  To be absolutely normal the number has equal zero and one
density in its binary representation for it to be normal to base two,
four, eight, etcetera. There are further conditions upon it but that
is a necessary condition.  We evaluated the combinatoric expression of
how many sequences have equal numbers of zeros and ones and it
evaluates to an asymptotic expression of 1/sqrt(n) ~ 0.  I wouldn't
call that almost all real numbers, but then again as I've made clear
my thought was that about half of the sequences would be of that form.
So we have two in a way opposing statements:  
1. Almost all of the real numbers are normal
2. Almost none of the real numbers are normal
Simlarly:
1.  Almost all of the real numbers have equal zero- and one-densities.
2.  Almost none of the real numbers have equal zero- and
one-densities.
There are those and my unfounded assumption:
3.  Half of the real numbers have equal zero- and one-densities.
A key element of that logical parallel is that a normal number has
equal zero- and one-densities.
Grams and meters are SI units of mass and length.  A gram is based
upon the mass of a standard bar in France, or as well a mole of H^1. 
A meter is based upon the distance traveled by an unladen photon
cesium atom, ie, the unit of length is a unit derived from standardly
recognized velocity and time.
Then here we are talking about how many elements of an infinite binary
sequence, in terms of an asymptotic value that in terms of ever
increasing finite lengths that are not infinite, the variable n, how
many of those sequences exhibit certain characteristics.  Borel says
one thing about that, combinatorics says another about that.  I
suggest another.
About irrationals, an irrational is plainly a number that can not be
represented as the ratio of two integers.  That's its definition.
 
About the convergence, it reminds me of the discussion of the method
to preserve the density of a canonical sequence:  preserve the density
in any finite sequence.
There are a lot of rational numbers.
Ross
Subject: Re: Factorial/Exponential Identity, Infinity
But no rational Ross's?
Subject: Re: Factorial/Exponential Identity, Infinity
Is there any parsing of this paragraph that makes any kind of sense?
Subject: Re: Factorial/Exponential Identity, Infinity
Yes, there is, that sentence is readable, although I admit it was
somewhat of a task to finish after I started writing it.
Virgil, y'all, what I'm trying to understand here is whether there is
a discrepancy between almost all reals being normal and binary
sequences with densities of one half being very few.
Generally this is considering something like this:
very few          half          almost all
Combinatorics has there being very few sequences with equal densities.
very few          
Borel in his statement about normal numbers says almost all sequences
have equal densities.
                                almost all
  ____________________________________
                                       
What's the deal with that?  That last bit there is to deter Ullrich. 
How does one result say one thing and another another?  Can't they all
just get along?
Well anyways, let's suppose we resolve this.  Then I want to know
this: why can I replace k with n/2 in the binomial coefficient in the
large, but not z with n/2 in the Gamma function?
Then I want you to tell me if there are rational functions g(n, x)
s.t. (sum n)^x- sum (n^x) / g(n, x) = s(n+1, n-x+1), besides for x=0,
sum(n^n)-(sum n)^n ) / n!^2 = 1, and that I found another asymptotic
expression for n!, if you would.
Let's suppose we don't resolve this.  How is it resolved?  It's OK for
you to say something like that's kind of strange, what do you think?
 Preferred is here is an avenue to approach this cognitive impasse,
or even there is no reason to cause confusion and dilemna, please
sweep this under the rug and stop talking about it, as that would be
humorous.
I think that some large cardinals would be in different order based on
their discovery dates.
I read about A-D-E on one of Sarfatti's sci.math posts, it is talk
about lattices, I read Conway and Sloane's sphere packing book, about
Euclidean n-spheres in Euclidean n-d, many of the highest densities of
packings and converings occur on those lattices, and the laminated
lattices, including the Leech lattice, Lambda_24.
This is about something else, it's about the set of all sets that do
not contain themselves.  It doesn't exist.
What are two sets codense in the reals with equal density in the
reals?
Ross
Subject: Re: Factorial/Exponential Identity, Infinity
I can read it, but I can't parse it in any way that makes any kind 
of sense.
Subject: Re: Sign of F(t) , 1=<1<infty
Why not study the graphics of this function?
If it's growing or not, (local and absolute) maxima and minima, and limit
as t tends to infinity.
I beleive that this helps you. It's what I should do if I have your
problem.
What did you try?
Xan.
Subject: Re: Primes as number of solutions of a diphantine equation.
I only want to appoint an unusual point of view of primes as propierties
of the cardinal of the integer points on which a specific curve passes.
Actually, a prime p is only a number p such that the curve y = f(x,n) =
n/x passes only for two integer points (or two if we count the negative
numbers), that is,
p is prime iff S(p) := |{(x,y) in Z^2 | y = n/x}| = 4
And I want to see the characteristics of S(p) with another functions
f(x,n) different of n/x.
Xan.
PS: I propose f(x,n) the next simple function: parabola.
Subject: Re: Primes as number of solutions of a diphantine equation.
I sent you a private email yesterday and have not heard from you. did you
get it as I did not get a delivery failure.
If you are interested in discussing the matter further, please post your
private email
tiny lizard
Subject: Re: Primes as number of solutions of a diphantine equation.
I only want to appoint an unusual point of view of primes as propierties
of the cardinal of the integer points on which a specific curve passes.
Actually, a prime p is only a number p such that the curve y = f(x,n) =
n/x passes only for two integer points (or two if we count the negative
numbers), that is,
p is prime iff S(p) := |{(x,y) in Z^2 | y = n/x}| = 4
And I want to see the characteristics of S(p) with another functions
f(x,n) different of n/x.
Xan.
PS: I propose f(x,n) the next simple function: parabola.
Subject: Re: Primes as number of solutions of a diphantine equation.
Conjecture:
is allways finite.
can be infinite.
Is it true?
Xan.
Subject: Re: Sum of unique prime squares?
squares of primes, see 54:7373 below. Also below are reviews of some
papers on this and related topics (by no means complete).
-Bill Dubuque
----------------------------------------------------------------------------

--
10:283d  10.0X
Sprague, R.  Uber Zerlegungen in ungleiche Quadratzahlen.
Math. Z. 51 (1948). 289--290.
----------------------------------------------------------------------------

--
The author shows in a quite elementary way that every positive integer than
128 is the sum of distinct squares. The proof depends on showing with a 
small
list of such representations that the integers between 129 and 256 are sums 

of
distinct squares not exceeding 100. There are exactly 32 integers
2,3,6,7,8,11,...,128  which are not the sum of distinct squares.
                         Reviewed by D. H. Lehmer
----------------------------------------------------------------------------

--
10:514h 10.0X
Sprague, R.  
Uber Zerlegungen in  n-te Potenzen mit lauter verschiedenen Grundzahlen.
Math. Z. 51 (1948). 466--468.
----------------------------------------------------------------------------

--
Let  P_n(k)  denote the number of partitions of  k  into distinct parts 
taken
from  1, 2^n, 3^n, ... . The author proves that there is a positive integer
words, for each  n  there are only a finite number of integers which are 
not
the sums of distinct  n th powers. This result for  n = 2  has already been
obtained by the author [same vol., 289-290 (1948); these Rev. 10:283]. 
The method of proof is an application of the Tarry-Escott problem.
                         Reviewed by D. H. Lehmer
----------------------------------------------------------------------------

--
11:646a  10.0X
Richert, Hans-Egon
Uber Zerlegungen in paarweise verschiedene Zahlen.
Norsk Mat. Tidsskr. 31, (1949). 120--122.
----------------------------------------------------------------------------

--
The author generalizes some theorems by Sprague [Math. Z. 51, 289-290, 
466-468
(1948); these Rev. 10:283,514] and himself [Math. Z. 52, 342-343 (1949); 
these
Rev. 11:502] about the partition of numbers into distinct parts. The 
general
theorem may be stated as follows. Let M be an infinite set of positive
integers  m_1, m_2, ... , with the following properties: (A) there exist
may be partitioned into distinct parts taken from the first  k  members of  

M;
than  N  may be partitioned into distinct parts taken from  M. The proof is
quite straightforward. As an example, if  M  is the set of triangular 
numbers
then (A), (B) and (C) are satisfied with  k=8, N=33, S=45. Hence every 
integer
except  2,5,8,12,23,33  is the sum of distinct triangular numbers.
                Reviewed by D. H. Lehmer
----------------------------------------------------------------------------

--
22:26  10.00
Birch, B. J.
Note on a problem of Erdos. 
Proc. Cambridge Philos. Soc.  55  1959 370--373.
----------------------------------------------------------------------------

--
The following result is proved: given any positive coprime integers  p, q,
a sum of the form  n = p^a1 q^b1 + p^a2 q^b2 +...,    where the (ai,bi) are
distinct pairs of positive integers. The proof is self-contained and
elementary, but far from trivial.
                Reviewed by L. Mirsky
----------------------------------------------------------------------------

--
24:A103  10.46 (10.48)
Cassels, J. W. S.
On the representation of integers as the sums of distinct summands taken
from a fixed set. 
Acta Sci. Math. Szeged  21  1960 111--124.
----------------------------------------------------------------------------

--
Let A  be a sequence of distinct positive integers a_1 < a_2 < a_3 <... ;
A(n) denotes the number of members of A less than or equal to  n. For a 
real
number x, write |x| for the difference between x and the nearest integer. 
The
main theorem proved is that if (1) lim[A(2n)-A(n)]/log(log n) = oo, and if
(2) sum|a_n t|^2 = oo for every real  t  in  0<t<1, then every sufficiently
large integer is the sum of distinct elements of A. The author proves his
theorem by an original adaptation of the Hardy-Littlewood circle method.
An outstanding feature of this result is the weakness of the growth 
condition
(1); this gives the theorem great generality. Earlier work by Roth and 
Szekeres
[Quart. J. Math. Oxford Ser. (2) 5 (1954), 241-259; MR 16:797] gave 
estimates
for the number of partitions of a large integer  n  as the sum of distinct
elements of A, but only at the cost of much stronger conditions 
corresponding
to (1) and (2).
Erdos and others had conjectured that the conclusion of the theorem would
remain true if condition (2) were replaced by the condition that the 
sequence
 A  contains infinitely many numbers in every arithmetic progression. As 
his
a sequence  C  such that (i) c_n = O(n^{2+e}), (ii) C  contains infinitely
many elements in every arithmetic progression, but (iii) the set of 
integers
which are sums of distinct elements of  C  has density less than  e. In 
fact,
a suitable sequence  C  may be constructed from the Fibonacci numbers.
                Reviewed by B. J. Birch
----------------------------------------------------------------------------

--
26:2387  10.45
Erdos, P.
On the representation of large integers as sums of distinct summands taken
from a fixed set. 
Acta Arith.  7  1961/1962 345--354.
----------------------------------------------------------------------------

--
Let  {a_i}  be an infinite sequence of positive integers such that every
arithmetic progression contains at least one integer which is the sum of
distinct  a's;  A(n)  is the number of  a's not exceeding  n . The author
proves by elementary means that if  A(n)  increases faster than  
n^{(sqrt(5)-1)/2}, then every large enough integer is a sum of distinct  
a's.
This appears much weaker than a result recently obtained by the circle 
method
by Cassels [Acta Sci. Math. (Szeged) 21 (1960), 111-124; MR 24:A103]; but 
the
author dispenses with a regularity condition essential to Cassels' 
argument,
and in fact the theorem as stated becomes false if the exponent 
1/2(sqrt(5)-1)
is reduced to  1/2-eps; it may well be possible to reduce the exponent to
1/2+eps  by improving the method of this paper. The crux is the 
construction
of a fairly long representable arithmetic progression with not too large
difference; this is achieved by means of Lemma 1, the rest is mopping-up.
                Reviewed by B. J. Birch
----------------------------------------------------------------------------

--
29:63  10.60 (12.30)
Graham, R. L.  Complete sequences of polynomial values.
Duke Math. J. 31 1964 275--285.
----------------------------------------------------------------------------

--
Let  S = (s_1,s_2,...)  be a sequence of real numbers and  P(S)  the set of
sums of the form  sum_{k=1}^oo e_k s_k , where  e_k  is 0 or 1 and all but 
a
finite number of e_k are 0. If all sufficiently large integers belong to 
P(S),
then S is said to be complete. It is proved in the paper in a simple and
elementary manner that if f(x) is a polynomial with real coefficients,
f(x) = a0 + a1(x 1) +...+ an(x n), an != 0, (x r) = (x(x-1)...(x-r+1))/r!
then the sequence S(f)=(f(1),f(2),...)  is complete if and only if 
(1)  ak = p_k/q_k  for some integers  p_k  and  q_k  with  (p_k,q_k)=1  and  

     q_k !=  0  for  0 <= k <= n
(3)  gcd(p_0,p_1,...,p_n) = 1 .
                Reviewed by M. J. Wonenburger
----------------------------------------------------------------------------

--
33:7318  10.45
Folkman, Jon
On the representation of integers as sums of distinct terms from a fixed
sequence. 
Canad. J. Math.  18  1966 643--655.
----------------------------------------------------------------------------

--
Let  A  be a sequence of positive integers  a_1 <= a_2 <= a_3 <=...  and
denote by  P(A)  the set of integers which are the sums of any number of
distinct terms of  A , that is, P(A) = {sum_{n=1}^oo e_n a_n : e_n = 0 or 
1,
almost all e_n = 0}. It is proved that  P(A)  contains all sufficiently 
large
integers provided that (i) it contains an element of every arithmetic
progression, and (ii) either  a_n = O(n^a)  or  a_n  is strictly 
increasing
and  a_n = O(n^{1+a}) , where a is a fixed number in the range  0 <= a 
< 1.
conjecture of Erdos, who was able only to cope with  a <= (sqrt(5)-1)/2
[Acta Arith. 7 (1961/62), 345-354; MR 26:2387]. The proofs are elementary 
in
the technical sense.
                Reviewed by J. W. S. Cassels
----------------------------------------------------------------------------

--
48:5994  10A45 (10-04)
Dressler, Robert E.; Parker, Thomas
12,758.
Math. Comp. 28 (1974), 313--314.
----------------------------------------------------------------------------

--
Authors' introduction: R. Sprague [Math. Z.  51 (1948), 289-290; MR 
10:283]
proved that every positive integer greater than 128 can be expressed as a 
sum
of distinct perfect squares. He also proved [ibid. 51 (1948), 466-468; 
integer  r_n  that is not expressible as a sum of distinct  n th powers (of
positive integers). In the light of the first reference above, we have  
r_2 = 128. Unfortunately, the technique used in the second reference above 
is
purpose here is twofold: (1) we use a computer to find explicitly the value
of  r_3  (it turns out to be 12,758), and (2) in the process we give a new
quantitative proof of Sprague's theorem for  n=3.
----------------------------------------------------------------------------

--
50:9759  10A40
Dressler, Robert E.; Parker, Thomas
An addendum to: 12,758 (Math. Comp.  28 (1974), 313--314). 
Math. Comp.  29 (1975), 675.
----------------------------------------------------------------------------

--
announced by R. L. Graham [Duke Math. J. 31 (1964), 275-285; MR 29:63].
----------------------------------------------------------------------------

--
54:7373  10B35
Dressler, Robert E.; Pigno, Louis; Young, Robert
Sums of squares of primes.
Nordisk Mat. Tidskr. 24 (1976), no. 1, 39--40.
----------------------------------------------------------------------------

--
It is known [R. Sprague, Math. Z. 51 (1948), 289-290; MR 10, 283] that the
largest integer not representable as a sum of distinct squares is 128. In 
this
paper, the authors announce that the largest integer not representable as a
sum of distinct squares of primes is 17163. The authors use a lemma due to
H.-E. Richert [Nordisk Mat. Tidskr. 31 (1949), 120-122; MR 11:646] with the
computational verification for the numbers  n in  17163 < n <= 17163 + 
503^2.
                Reviewed by K. Thanigasalam
----------------------------------------------------------------------------

--
90h:11011  11B13 (11D85 11P99)
Zannier, U.(I-SLRN)
On a theorem of Birch concerning sums of distinct integers taken from
certain sequences. 
Math. Proc. Cambridge Philos. Soc.  106 (1989), no. 2, 199--206.
----------------------------------------------------------------------------

--
B.J. Birch [same journal 55 (1959), 370-373; MR 22:26] has proved that, for
as a sum of distinct terms  p^r q^s. J.W. Cassels [Acta Sci. Math. (Szeged)
21 (1960), 111-124; MR 24:#A103] established a more general result by 
making
use of the circle method of Hardy and Littlewood. Here a different
infinite set of positive integers such that the quotient of distinct 
elements
of A is never a power of p. If every arithmetic progression contains a sum 
of
A' of A such that every sufficiently large integer is a sum of distinct 
terms
of) Birch's result since, as is easy to see, every arithmetic progression
contains a sum of distinct powers of  p ,  q  and  pq . The proof is quite
elementary but by no means trivial. {In (2) the central term must be 
replaced
by the fractional part of  log K/log p+h/m .}
                     Reviewed by G. Turnwald
----------------------------------------------------------------------------

--
92j:11115  11P05 (11B13)
Mimura, Yoshio(J-KOBEE)
Partial sums of sequences. 
Math. Japon.  36 (1991), no. 4, 677--679.
----------------------------------------------------------------------------

--
The author gives a generalization of a method used by Sprague to show that 
all
integers larger than 128 are sums of distinct squares. H.-E. Richert [Norsk
Mat. Tidsskr. 31 (1949), 120-122; MR 11:646], J. L. Brown, Jr. [Amer. Math.
Monthly  83 (1976), no. 8, 631-634; MR 55:2741] and R. L. Graham [Duke 
Math.
J. 31 (1964), 275-285; MR 29:63] have all dealt with similar questions and 
the
author's theorem is very close to a generalization given by Brown [op. 
cit.].
A number of special cases, some of which will be found in the 
above-mentioned
papers, are worked out. For example [...]
                Reviewed by Joe Roberts
----------------------------------------------------------------------------

--
98a:11016  11B25 (11B05)
Hegyvari, Norbert(H-EOTVO)
Subset sums in   N^2 . (English. English summary)
Combin. Probab. Comput.  5 (1996), no. 4, 393--402.
----------------------------------------------------------------------------

--
A well-known result of J. Folkman [Canad. J. Math. 18 (1966), 643-655; 
MR 33:7318] is that given an increasing sequence of positive integers 
A = {a_i}_{i=1}^oo <= N  whose counting function A(n) = sum_{a_i <= n} 1
positive  d, the subset sums set  P(A) = {sum e_i a_i : e_i in {0,1}}
always contains an infinite arithmetic progression.
In his earlier paper [European J. Combin. 17 (1996), no. 8, 741-749; 
MR 97g:11006], the author showed that no similar result holds for sequences 

on
the plane: there exists a sequence  A < N^2  such that the number  A(n)  of
for all sufficiently large  n  and a positive constant a, but  P(A)  
contains
no infinite arithmetic progression.
In the paper under review it is shown, however, that a conclusion of this 
type
(and in fact, a much stronger conclusion) holds provided that the density
restriction imposed on  A  is expressed in the correct way. The main 
result
is essentially the following: Theorem. Let  A <= N^2 , and let d  be a 
fixed
positive number. Assume that for any  n in N^2  there is an  a in A  for 
which
|a-n| <= |n|^{1/8-d}. Then there exist x_0 in N^2  and linearly 
independent
d1, d2 in N^2  such that {x0 + x1 d1 + x2 d2 : x1, x2 in N} <= P(A) .
It is also shown that  d1  and  d2  can always be chosen in such a way that
the angle between these two vectors is large and that the exponent  1/8 
in the conditions of the theorem cannot be replaced by anything greater 
than
(sqrt{13}+1)/6  ~= 0.7675.
        Reviewed by Vsevolod F. Lev
----------------------------------------------------------------------------

--
2001c:11014  11B25 (11B75)
Hegyvari, Norbert(H-EOTVO)
On the representation of integers as sums of distinct terms from a fixed 
set. 
Acta Arith. 92 (2000), no. 2, 99--104.
----------------------------------------------------------------------------

--
An infinite set of distinct positive integers is said to be subcomplete if 
its
set of subset sums contains an infinite arithmetic progression. P. Erdos 
asked
for the maximal density of a non-subcomplete set. He proved [Acta Arith. 7
(1961/1962), 345-354; MR 26:2387] that, if the counting function  A  of a 
set
result was improved by J. Folkman [Canad. J. Math. 18 (1966), 643-655; 
MR 33:7318], who relaxed the condition on the counting function to  
in view of Cassels' construction [J. W. S. Cassels, Acta Sci. Math. Szeged
21 (1960), 111-124; MR 24:A103].
the subcompleteness. The key tool is a result obtained by A. Sarkozy [J. 
Number
Theory 48 (1994), no. 2, 197-218; MR 95k:11010] and G. A. Freiman [Discrete
Math. 114 (1993), no. 1-3, 205-217; MR 94b:11013] on the structure of the 
set
of subset sums for sufficiently dense sets (the result is that under such a
kind of hypothesis it contains a long arithmetic progression).
Let us finish this review with two references. Firstly, there is an 
addendum
to Freiman's quoted paper, namely [Discrete Math. 126 (1994), no. 1-3, 447
under review) by T. Luczak and T. Schoen [On the maximal density of 
sum-free
sets, Acta Arith., to appear] contains essentially the same result as the
paper under review.
                Reviewed by Alain Plagne
----------------------------------------------------------------------------

--
Chen, Yong-Gao(PRC-NJN)
On subset sums of a fixed set. 
----------------------------------------------------------------------------

--
Let  A  be an infinite set of positive integers. The set of subset sums is
defined by  P(A) = {sum_{a in B} a : B <= A; |B|<oo}. The sequence is 
said
to be subcomplete if  P(A)  contains an infinite arithmetic progression.
of  A , where  a = sqrt{5}-1/2  and  A(n)  is the counting function of  A 
.
In 1966 Folkman proved that a = 1-eps, and recently the reviewer [Acta 
Arith.
92 (2000), no. 2, 99-104; MR 2001c:11014] and independently T. Luczak and 
T. Schoen [Acta Arith. 95 (2000), no. 3, 225-229; MR 2001k:11018] improved 
this
In the present paper the author improves our result. He proves that if
The reviewer should mention that Szemeredi has also proved a similar 
result,
and his proof will appear in the near future.
                   Reviewed by Norbert Hegyvari
Subject: Re: Sum of unique prime squares?
This intrigued me so I ran a quick program to test this and found that 
there
seem to be many possible answers - I found 21 variations for 4^2 = 16 just
using the first 10 primes (2 3 5 7 11 13 17 19 23 29).
Eg:
16 = 529 - 361 - 289  + 121 + 25 - 9   (23, 19, 17, 11, 5, 3)
16 = 841 - 361 - 289 - 121 - 49 - 9 + 4 (29, 19, 17, 11, 7, 3, 2)
However there is nothing special about squares, I generated numerous
variations for every integer, positive and negative. I ran the first
2,000,000 results (out of the 3^20 possible) using the first 20 primes and
saw no special pattern - every integer appears to be getting represented
repeatedly by different permutations.
Manning
Sydney
----
----
than
small
sums of
----
----
taken
integer
other
not
been
----
----
466-468
these
general
N+S,
M;
greater
is
numbers
integer
----
----
as
are
----
----
real
The
sufficiently
condition
Szekeres
estimates
corresponding
sequence
his
is
integers
fact,
----
----
a's.
method
the
argument,
1/2(sqrt(5)-1)
construction
----
----
of
but a
P(S),
and
----
----
1,
large
increasing
< 1.
confirms a
in
----
----
10:283]
sum
positive
(of
is
value
----
----
been
----
----
this
a
the
and
503^2.
----
----
for
written
(Szeged)
making
elements
of
subset
terms
form
replaced
----
----
all
[Norsk
Math.
Math.
the
cit.].
above-mentioned
----
----
sequences on
of
contains
type
result
fixed
which
independent
that
than
----
set.
----
its
asked
set
This
expected
ensure
Number
[Discrete
set
a
addendum
sum-free
----
----
said
subcompleteness
.
Arith.
this
subcomplete.
result,
Subject: Re: Sum of unique prime squares?
A related question:
n by the minimum number of distinct squares of primes.
The program only searched for a maximum of 4 terms. The results for
n<=20:
  n Terms Primes in sum
  0 4   7 -11 -17  19 (means 0 = 7^2 -11^2 -17^2 +19^2)
  1 3   7  11 -13
  2 4   5   7  17 -19
  3 4   2  -7 -11  13
  4 1   2
  5 2   3  -2
  6 ?
  7 ?
  8 4   3  -7 -11  13
  9 1   3
 10 4   3   7  11 -13
 11 4  -2  -3  -5   7
 12 3  -2  -3   5
 13 2   3   2
 14 4  -3  -5 -11  13
 15 3  -3  -5   7
 16 2   5  -3
 17 ?
 18 ?
 19 4   2  -3  -5   7
 20 3   2  -3   5
List for n<=200 at
http://www.randomwalk.de/scimath/sumpsq.txt
Can we always find a solution with 5 terms for those n marked with ?.
The list of n where I didn't find a solution with <=4 tems continues:
6,7,17,18,31,41,42,55,60,66,84,89,90,102,103,113,114,127,132,138,162,
Hugo Pfoertner
Subject: Re: Sum of unique prime squares?
In the meantime Edwin Clark confirmed that 5 terms are suficient. He
<<
I find that just using the first 12 primes all numbers from 1 to 1123 
can be written in the form sum(s[i]*p[i]^2, i=1..k) where s[i] is 1 or
-1, 
the primes p[i] are distinct and k <=5
Hugo
Subject: Re: Sum of unique prime squares?
My guess is that the primality is a red herring. What matters is that
the sequence has a growth rate that is not too fast and is moderately
regular.
Martin Cohen
Subject: Re: Sum of unique prime squares?
This surely look plausible. A possibly related result due to
Sprague states that any integer larger than 128 can be expressed
as a sum of distinct squares. Using only squares of primes will
make things a little bit more difficult, but there are plenty of
primes around, so one might be able to prove the same result
using squares of primes only, and for a limit larger than 128.
Then you could simply run a computer check for the (finite set of)
excluded values.
Look up e.g. Joe Roberts, Elementary Number Theory - a Problem
Oriented Approach, ch. XI for ideas leading to Sprague's result.
I don't know, if such a simple approach will work in this case.
Jyrki Lahtonen, Turku, Finland
Subject: Re: Sum of unique prime squares?
Was this result derived from another result or brute-forced?
Another related question is, is it possible to express the square of
any prime as differences of sums of other squares of distinct primes?
It's certainly possible if one includes 1, but what about without it?
Subject: Re: Sum of unique prime squares?
 
Computed, not quite brutally, but now here's a proof (well, almost).
Let p_n be the n'th prime.  
Note (by explicit computation) that all integers up to 642 can be expressed
using the first 8 primes, e.g. 642 = 19^2 + 13^2 + 11^2 - 3^2.
The bottleneck, btw, is 2, which can't be expressed using the first 7 
primes,
but is 19^2 - 13^2 - 11^2 - 7^2 - 5^2 + 3^2 - 2^2.
Let X_k be the largest positive integer x such that all integers from 1 to 
x
(since for any integer x from X_k + 1 to X_k + p_{k+1}^2, we can subtract
p_{k+1}^2 and obtain an integer that can be expressed using the first k 
primes).
p_{k+2} <= sqrt(2) p_{k+1}.  Now I'm quite sure that this is true
Number Theorem, but I don't know of explicit estimates.  Assuming it is
large.  Moreover, in the expression any negative terms will come from the 
first 8 primes.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
Subject: Re: Sum of unique prime squares?
Clap - clap - clap - bravo!!
Well, they say there are similarities between math and music.
This is pretty true: You can enjoy the performance even if
you do not really understand :-)
(Am I right, when I see a distant connection between your
conjecture (...well, almost...) and Andrica's conjecture?)
Rainer Rosenthal
r.rosenthal@web.de
Subject: Re: Need help for this integral
Hint: 
Let f(a,s,k) = integral (x=0..oo) exp(ikx - ax) x^s dx
Subject: Re: Need help for this integral
Try the integral with a=1,k=1,x'=0 ...
 f = exp(I*x - abs(x))*log(1/abs(x));
With Maple, I get
                             1         1   
                     gamma + - ln(2) + - Pi = 1.709187419
                             2         4   
where gamma is Euler's constant.  For a=1/10 I get
20          20          20          10            200           
--- gamma - --- ln(2) - --- ln(5) + --- ln(101) + --- arctan(10)
101         101         101         101           101           
 = 3.028409420
                                 /1               
                         2 arctan|-|     /     2  
             2 gamma a           a/   ln1 + a / a
             --------- + ----------- + ------------
                   2            2              2   
              1 + a        1 + a          1 + a    
Now put in xprime.  For the integral I get:
    1    /                  /     2                       /1
  ------ |I a sin(xprime) ln1 + a / + 2 cos(xprime) arctan|-|
       2                                                  a/
  1 + a                                                       
                                                     /1
     + 2 cos(xprime) gamma a + 2 I sin(xprime) arctan|-|
                                                     a/
                                                 /     2
     + 2 I sin(xprime) gamma a + a cos(xprime) ln1 + a /|
                                                         /
  Pi*exp(I*xprime)
This is still k=1.  I'll stop now.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
Subject: Re: Probability Question
Well at least he didn't deceive you (I mean with a name like Yanx, and
all)
Subject: Re: Probability Question
I think that your friend is your probability teacher.
Seriously, though, what sort of friend would come up to you and say
Hey, Bobby - check out this interesting problem.  Suppose Dave and 
Mike are going from destination...?  
Or to bring this back on-topic, what's the probability that someone
would actually do this in public?
Doug
Subject: Re: Probability Question
I suspect that this is an open question - have you tried
sci.math.research?
************************
David C. Ullrich
Subject: Re: What is 0/0 ?
0/0=NaN=Not a Number, true for constants, for numbers.
For variables:
(x square - a square)/(x - a)=x+a, it's also valid, when x=a,
and then it's 2*a.That's the derivate of x square at the point a.
Most simple example:The distance from the finish of two
sprinters may  be a and b. They run side by side and both
have the same speed, so a / b will be 1, and that's true in 
the finish as well. For the rest: calculus, as the derivate is
always 0 / 0 and can be any number.
der Natur and Karl Marx in his Mathematische Manuskripte 
(Kronberg Taunus 1974)about this too.
Have fun
Hero
Subject: Re: What is 0/0 ?
considered
I sure as hell hope that every CPU will honor that undefined.
I'd hate to have to check the contents of every variable in
every program for zero because my hardware may not follow
standards.
NO!!!!!  The whole point of having a higher level programming
language is so the code doesn't have to know how the hardware
(it's running on) acts.
Oh, good grief no.  In bank accounts, data that can't be nonzero
should be checked before it gets record and not at the
instance of divide execution.  If all of this data checking has
been done and a job run gets a divide exception, I got bad
hardware.
make
These are exceptions and can be programmed to do execptions.  Why should
the rest of the computing world change techniques just because
one program in the whole existence wants to break a standard?
Nope.  You want everybody else to acquire your eccentricity.
I declare your name to be Mud.  Using your logic, you now have to
change all documents to reflect that.
/BAH
Subtract a hundred and four for e-mail.
Subject: Re: What is 0/0 ?
0/0 is an indeterminate form, if you want to do something with that you 
need
to use L'Hopital's Rule (differentiate the top and bottom separately) and
take the limit of the function as x goes to zero...
Jay
*always*
Subject: Re: What is 0/0 ?
need
I'm not arguing about how to do it.  I am stating that hardware
and programs shipped by general distribution should not do
what they feel like.  The poster, with whom I yakked, proposed
that a CPU shouldn't flag any division by zero.  That implies
that a CPU will give a consistent result that implies no errors
detected.  This is not a Good Thing.
/BAH
/BAH
Subtract a hundred and four for e-mail.
Subject: Re: Game of pieces on 2 linear-boards
I forgot some possibilities in one of the choices for each player at a 
move.
(see below within original message.)
Also (in option A):
If both colors match but are of player's opponent's color,
or either or both stack at a position has no pieces but
the other is of the opponent's color (if both stacks
are not empty), then leave the stacks at this position alone.
And, if one stack has player's own color, and other stack is empty,
a player (if choosing option A) must move a piece from the
occupied stack to the empty stack.
Subject: mathematician.org
Back online: www.mathematician.org
Subject: Transcendental numbers
I know that rational and irrational numbers are everywhere dense in
the real number continuum, but what about transcendental numbers? Do
intervals exist in the real number continuum that contain no
transcendental numbers? Has this problem even been solved and is there
a proof? (Maybe this belongs in another math group?)
Subject: Re: Transcendental numbers
There are more transcententals in any real interval than rational 
or even algebraic numbers. Transcendentals are , if anything, more 
dense, since there are uncontably many in any real interval whereas 
there are only countably many algebraics (including the rationals).
Subject: Re: Transcendental numbers
Try this out. Between any two transcendental numbers there exists a 
transcendental number. Suppose this were not the case. Then there would 
exist transcendental numbers a, b such that for all x in (a,b) x is 
algebraic. But this would imply there are uncountably many algebraic 
numbers. Not so.
So there is always a trascendental number between any two transcendtal 
numbers which is sufficient to prove the set of transcendental numbers 
is dense.
Bob Kolker
Subject: Re: Transcendental numbers
So transcendental numbers are dense in the real numbers? As all rational
numbers are algebraic, this means that the transcendentals must also be
codense in the real numbers, as any set with a dense subset is dense.
But are the transcendentals codense in the irrational numbers?
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
I am not very happy acting pleased whenever prominent scientists 
overmagnify
intellectual enlightenment.
   - Anon
Subject: Re: Transcendental numbers
Yes.  In fact, the set { sqrt(2)+q : q in Q } is a subset of the
algebraic irrationals that is dense in R.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
Subject: Re: Transcendental numbers
If you take just a single transcendental number and multiply that
with all rational numbers, then those numbers are all transcendental.
Just as the rational numbers, they form a dense subset of the real
numbers.
-- 
State University, Las Cruces, NM, USA. |        f(x + y) = f(x) + f(y)
Subject: Re: Transcendental numbers
Every nondegenerate interval contains an uncountable number of points, of
which only countably many are algebraic.  Almost all of the real numbers
are transcendental.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
Subject: f(x) = exp(alteration of f's own Taylor-series)
Let f(x) be an analytic function (about x=0).
Let f_k = k_th derivative of f(x) at x=0.
(So, f(x) = sum{k=0 to oo} (x^k/k!) f_k .)
What is f(x), if f(x)  = 
exp(sum{k=0 to oo} (x^(n^k)/(n^k)!) f_k ) ?    
 
Now,
f_0 = 1.
f_{j+1} =
sum{k=0 to floor(ln(j+1)/ln(n))} binomial(j,n^k -1) f_{j+1-n^k) f_k,
I think.
If n=2, say, I get (by-hand and carelessly, so may be wrong):
Leroy Quet
Subject: Conditional Expectaion Question
The following problem was on a midterm I had in Random Processes recently.
I have referenced several texts (as well as my professor) and can't seem
to arrive at the answer without handwaving. Any rigorous answers or 
pointers
would be appreciated.
Let X and Y be two random variables. Assume that E[Y|X] = c, where c is
a constant. Show that X and Y are uncorrelated.
-- 
%  Randy Yates                  % ...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall.
http://home.earthlink.net/~yatescr
Subject: Re: Conditional Expectaion Question
The conditional expectation formula, viz.,
EX = EE[X|Y]
implies the conditional covariance formula,
Cov(X,Y) = E Cov(X,Y | Z) + Cov(E[X|Z], E[Y|Z]).
Letting  Z = X,
Cov(X,Y) = Cov(X, E[Y|X]).
(You do know that  Cov(X,c) = 0,  right?)
-- 
Stephen J. Herschkorn                        herschko@rutcor.rutgers.edu
Subject: Re: Conditional Expectaion Question
Hello Stephen,
some more comments below.
I don't see how you got there. I try and get stuck:
   E[cov(X, Y|X)] + cov(E[X|X], E[Y|X]) (since E[Y|X] = c and cov(anything, 

c) = 0)
     = E[cov(X, Y|X)]
     = E[E[(X)(Y|X)]] - E[X]E[Y|X]]
     = E[E[(X)(Y|X)]] - (m_x)(m_y)
This is as far as I get. Can you please show in detail step-by-step how
to get there?
I also don't know how the conditional expectation formula implies the
conditional covariance formula.
Yes.
-- 
%  Randy Yates                  % ...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall.
http://home.earthlink.net/~yatescr
Subject: Re: Conditional Expectaion Question
I think it is easier for you to go the way that your teacher suggests - 
you don't need the generality of the conditional covariance formula.  As 
RI has pointed out, you should know that
EX = EE[X|Y]
and
E[f(X)Y|X] = f(X) E[Y|X]
(Check these with discrete and/or continuous disitrbutions if you 
haven't already, but you should have.)
Thus,
EY = EE[Y|X] = Ec = c
and
E[XY] = EE[XY | X] = E[X E[Y|X]] = E[X c] = c EX = EX (EY)
i.e.,  X  and  Y  are uncorrelated.
-- 
Stephen J. Herschkorn                        herschko@rutcor.rutgers.edu
Subject: Re: Conditional Expectaion Question
Let me give a little more info. I know that E[Y] = E[E[Y|X]] = E[c] = c.
   E[XY] = E[X*E[Y|X]],
but I don't buy this step.
-- 
%  Randy Yates                  % ...the answer lies within your soul
%% Fuquay-Varina, NC            %       'cause no one knows which side
%%% 919-577-9882                %                   the coin will fall.
http://home.earthlink.net/~yatescr
Subject: Re: Conditional Expectaion Question
It's not a therefore.
You know E[XY] = E[E[XY|X]].  The point now is that E[XY|X] = X E[Y|X].
This should follow pretty directly from whichever version of the 
definition of conditional expectation you're using.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
Subject: equicontinuity
I want to show that a sequence of functions {f_n} on a connected domain G 
is
equicontinuous on compact subsets of G. I'm given that the partial
derivatives of f_n are equibounded on compact subsets of G, and so by the
mean value theorem from vector calculus I can write
f_n(x)-f_n(y) = grad f_n(x_0) (dot) (x-y) for some x_0 lying on a line
segment connecting x and y. Since I can use the equiboundedness of the
partial derivatives to get a bound on sup |grad f_n|, I can write
|f_n(x)-f_n(y)| <= M |x-y|, when x and y lie on the same line, where M is
the bound, and then I have the equicontinuity I want. But I don't know what
to do in the general case, where x and y are connected by finitely many 
line
segments. The natural thing for me to do was to write
|f_n(x_1)-f_n(x_N)|  <= |f_n(x_1)-f_n(x_2)| + ... + |f_n(x_{N-1})-f_n(x_N)|
<= M(|x_1-x_2| + ... + |x_{N-1}-x_N|), where x_i, x_{i+1} are connected by 
a
line segment. But I have the right-hand side in terms of all the individual
distances, rather than |x_1-x_N|, so I can't get equicontinuity...how to 
get
around this?
Subject: Re: equicontinuity
K_r = {x in R^n: dist(x,K) <= r} is in G.  Note that K_r
contains the line segment joining two points in K if ...
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
Subject: Re: equicontinuity
But what I'm saying is what if there _is_ no line segment in G joining two
points x and y. I'm only given that sup grad f_n is bounded on compact
subsets of G, so if the line connecting x and y goes outside of G, I can't
say anything about sup grad f_n there....
is
what
line
Subject: Re: equicontinuity
But you don't need to.  Look again at the definition of equicontinuity. 
You only need to worry about the line connecting x and y if the distance
between them is ...
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
Subject: Re: equicontinuity
Let K be compact in G. Let d be the distance from K to the boundary of G. 
Then K* = {x in G : d(x,K) <= d/2} is a compact subset of K. If x and y are 

in K and d(x,y) < d/2, then x and y lie in a ball contained in K*. That 
ball is convex and you can apply your straight line argument there.
Subject: Re: equicontinuity
G is
the
is
what
are
Subject: Re: equicontinuity
Time to get a good night's sleep.
Subject: Re: equicontinuity
Also, as you have it, K* is not a subset of K. It includes K and then some
(the points outside of K but still less than d/2 from it).
Subject: Re: equicontinuity
K* is a compact subset of G, is it not? Therefore the gradients are 
uniformly bounded on K*, are they not? And you drank one too many beers 
last night while doing this homework, did you not?
 electron-dot-cloud are galaxies
Subject: Re: should Gauss's Law of Magnetism be: Integral B dot dA = -uq/p + 

 q/p_t
I had a look up of Graham Lee's remark of the London Equations. I was able 
to find this site
on the www.
quoting: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/chrlen.html#c4
Meissner Effect from London Equation
 The London equation relates the curl of the current density J to the
 magnetic field:
 By relating the London equation to Maxwell's equations, it can be shown
 that the Meissner effect
 arises from the London equation. One of Maxwell's equations is
 Using the vector calculus identity
 along with the curl of the Maxwell equation above
 and by substitution
 Since  by Maxwell's equations, the value for B inside the superconductor
 must be identically zero unless the penetration depth is infinite (i.e., 
not
 a superconductor). This is one of the
 theoretical approaches to explaining the Meissner effect.
                                                                             

          Index
Superconductivity
concepts
Reference
Kittel,
                                                                             

          Ch 12
*** end quoting ***
I was not able to copy paste the math equations as is typical in physics 
that the keyboard
makes translation too difficult. We should in physics adopt a new means of 
representing math
concepts such that the keyboard of a computers is easily able to transfer. 
Get away from
Greek symbols and have the symbols of physics equations begot from the 
regular keys of a
computer keyboard.
I do not know what persons are behind the London Equations, perhaps Mr. 
Kittel is one of
those London Equationers involved.
And it says above that the London Equations is  ... one of the theoretical 
approaches to
explaining the Meissner effect end quoting.
What I am saying is that the Meissner Effect is a Additional Term that needs 

insertion into
the Maxwell Equations themselves. Specifically the Gauss Law of Magnetism 
needs revision and
to insert a term. The end result will be a symmetrical Maxwell Equations.
So what the London School of Equations has started, I need to finish with a 

symmetrical
Maxwell Equations and a modernized and revised Gauss Law of
Magnetism. I could use some help from mathematicians to add an extra term 
into Gauss's Law
of Magnetism to cover the Meissner Effect.
Archimedes Plutonium, a_plutonium@hotmail.com
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
Subject: Re: Quotient Space help
Subject: Re: Finishing argument, core error proven
can you explain the @MSN thing?...  did he just initially start it
to create a strawman, but forgot?
 
 
--les ducs d'Enron!
Subject: Re: Finishing argument, core error proven
nice transfer of the additive inverse to the situation
of the exponents, I guess.  so,
do you use symbolic software?
 
--les ducs d'Enron!
Subject: Re: Finishing argument, core error proven
            Adjunct Assistant Professor at the University of Montana.
   [.snip.]
www.berkeleyuniversity.net
I am not a graduate of Berkeley University (which does not appear to
have a Ph.D. program in mathematics; or an undergraduate one for that
matter; which would make true your statement that Berkeley
University [I assume it was a typo, of course] is indeed NOT the
place to get a math degree, but it has nothing to do with what I have
shown or not)...
Why do you take so much trouble to expose such a reasoner as
 Mr. Smith? I answer as a deceased friend of mine used to answer
 on like occasions - A man's capacity is no measure of his power
 to do mischief. Mr. Smith has untiring energy, which does 
 something; self-evident honesty of conviction, which does more;
 and a long purse, which does most of all. He has made at least
 ten publications, full of figures few readers can criticize. A great
 many people are staggered to this extent, that they imagine there
 must be the indefinite something in the mysterious all this.
 They are brought to the point of suspicion that the mathematicians
 ought not to treat all this with such undisguised contempt,
 at least.
   -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
Subject: Re: Finishing argument, core error proven
I'm pleased to see that they put scare quotes around their title at the
top of that page.
-- Richard
-- 
Spam filter: to mail me from a .com/.net site, put my surname in the 
headers.
FreeBSD rules!
Subject: Int. Journal of Mathematics - TOC alert
International Journal of Mathematics
Articles are available in electronic format at 
http://www.worldscinet.com/ijm.html
Table-of-contents:
Orbifold Cohomology as Periodic Cyclic Homology
by Vladimir Baranovsky
Complexification and Hypercomplexification of Manifolds with a Linear 
Connection
by Roger Bielawski
Local Simple Connectedness of Resolutions of Log-Terminal Singularities
by Shigeharu Takayama
Singularities on the 2-Dimensional Moduli Spaces of Stable Sheaves on K3 
Surfaces
by Nobuaki Onishi and Kota Yoshioka
Amenability and Co-Amenability for Locally Compact Quantum Groups
by E. Bedos and L. Tuset
On q-Basic Multiple Gamma Functions
by Nobushige Kurokawa and Masato Wakayama
********************
For more information go to http://www.worldscinet.com/ijm.html
********************
Subject: arc (was: New Maple bug in series)
]Follow-up to sci.math]
     ARC = A Reversed Computation.  
-- 
ARNOLD = Anagram of RONALD         ENEGGER = Backwards mis-pronounced 
REAGAN
This is a black -- I mean SCHWARZ -- period in California.
        Peter-Lawrence.Montgomery@cwi.nl    Home: San Rafael, California
        Microsoft Research and CWI
Subject: Re: JSH: So much fun
my guess is that either the real Jimmy Dean Harris
-- you don't want to see how mathematics is made
over the course of ten years, either --
is either dead, or has found another avocation.
    inless he's being paid to do this, of course.
that's a good idea, though, as it seems that
we merry few are stuck in a tiny cell
of the googolplex.  (lucky for you,
Georgia's the fourth (?) state
to agree not to agree to go into the Matrix database,
perhaps; eh?...  that was in today's paper.
 
 
--les ducs d'Enron!
Subject: Re: JSH: So much fun
It's later than you think, James.
Subject: Re: JSH: So much fun
It doesn't look like it to me.  What are you muttering about Jim
Ferry?
My problem is that people seem to be confused on constant terms not
varying based on a variable, even when they're looking at actual
numbers!!!
It's amazing.
Readers, you need to look at numbers as numbers, and think of them as
little rocks floating around in the factorization, immune to the value
of m.  Changes to the constant terms have to happen *independent* of
m.
So you have
P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078
Note: The constant term is 1078, and 1078 = 7(7)(22)
Then I do some manipulations to factor it.
P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)
which is
P(m)= 7^2(2401 m^3 - 147 m^2 + 3m) (5^3) - 3(-1 + 49 m )(5)(7^2) + 7^3
an now I can factor to get
P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7).
Now suddenly people act like algebra just got VERY iffy and
wish-washy, but algebra didn't change, your minds tell you that maybe
the same rules no longer apply.
Yup, somehow looking at that factorization makes many of you wiggy and
you start doubting the rules of algebra.
The constant factors are STILL independent of m, but things are
somewhat obscured by the a's.
You need to BELIEVE in algebra.
The rules don't change just because you find yourself feeling lost.
James Harris
Subject: Re: JSH: So much fun
so, if we are to believe in algebra,
then what is it that you'll accept as a rostrum
of textbook definitions therefrom?...
to bolster our faith, of course.
    for instance,
I'm not sure that rocks floating around is going to pass
as a viable definition of independent term,
as nice of a metaphor as it is.
 
 
--les ducs d'Enron!
Subject: Re: Boolean Algebra - Arithmetic Relationship
What's the minimum number of axioms a deductive system can be reduced
to and still produce an infinite number of proofs?
long before Euclid explicity defined the axioms.  Did they just use
derived axioms/rules that made reasonable sense to them? Did Euclid
later keep infering back until he was left with only five?
Are most deductive systems discovered initially in this way? Only to
be formalized by a later generation?
-Steve
Subject: Re: Boolean Algebra - Arithmetic Relationship
How about one rule:
  For any formula A, A is an axiom.
If the definition of formula allows infinitely many formulae then
there are infinitely many proofs--none of them interesting.
-- 
G.C.
Subject: Re: Boolean Algebra - Arithmetic Relationship
I don't know.  Why are you asking?  I'll hand-wave: for propositional
calculus, one axiom and two rules (I know that for a fact); for first
order logic, rules only, no extra axioms (I'm guessing); for set theory,
one axiom (I think: just conjoin all the usual axioms).  
Yes, and not just the Greeks.
I don't know.
I dont know
-- 
G.C.
Subject: Re: Boolean Algebra - Arithmetic Relationship
Do I know that for a fact?  One can get by with no axioms.
-- 
G.C.
Subject: Re: Boolean Algebra - Arithmetic Relationship
What operations?
A rule is something which allows a valid statement (the conclusion) to
be inferred from valid statements (the premises).  For example
              A, if A then B                A
              --------------      or      ------
                    B                     A or B
An axiom is a statement assumed to be valid (usually with good reason). 
For example
              if p (if q then p)
but a rule with no premises has the force of an axiom, or perhaps an
axiom schema.  For example
              ----------
              A or not-A
Possibly, but it would have no connective not, because if  A  where
true  not-A  would have to be meaningless.  So neither
              p or not-p     nor      not-(p and not-p)
would be true or even meaningful.  There _are_ negation-free calculi but
we cannot even have the connective if-then at the same time as having
two or more statements, else we'll have
              if p then q
which isn't valid.  Weird.
-- 
G.C.
Subject: Re: tangential quadrilateral
a circle can be constructed through any three points;
try it with compasses, because it's not hard.
--les ducs d'Enron!
e hauled away by campus security on more than one occasion. My
friends and I refer to him as Crazy Number Man.
I've been posting to (and about) Shawn for over two years with big
 
 if it comes from anyone not already in my addressbook.
 I'll never even see it)
Subject: Re: recurrence relation
Here's how I would do this:
x_n/(n+1) = x_{n-1}+[(n-1)/3]n!
so, dividing by n!,
x_n/(n+1)! = x_{n-1}/n!+[(n-1)/3]
Let y_n = x_n/(n+1)!.
Then
y_n = y_(n-1) + (n-1)/3.
Find y_n, and then get x_n.
Martin Cohen
Subject: Re: product of arithmetic sequence terms
This can be made a little simpler:
           = (2n)!/(2^n*n!).
Then apply Stirling.
Martin Cohen
Subject: Re: cross-section
Oooops, you are absolutely correct.
The phrase I intended was Geometric Tomography.
Sorry.  I have now located my remaining brain cell and will
attempt to use it.
-Mike
Subject: Arcs as beziers?
Hi all,
my head hurts. I'm trying to draw an arc with a beziercurve given a
centerpoint (x,y), radius r and angles alphaStart and alphaEnd
I can draw a quarter circle with
  Const kappa = 0.5522847498307936
 
  k = kappa * radius
  a.x = center_x  
  a.y = center_y - radius
  a.x2 = center_x + radius
  a.y2 = center_y  
  a.order = 2
  a.controlx(0) = center_x + k
  a.controly(0) = center_y - radius
  a.controlx(1) = center_x + radius
  a.controly(1) = center_y - k
and I think I've figured out that for angles alphaStart and alphaEnd I need
  a.x = center_x + (radius * cos(alphaStart))
  a.y = center_y - (radius * sin(alphaStart))
  a.x2 = center_x + (radius * cos(alphaEnd))
  a.y2 = center_y - (radius * sin(alphaEnd))
  a.order = 2
  a.controlx(0) = center_x + ???
  a.controly(0) = center_y - ???
  a.controlx(1) = center_x + ???
  a.controly(1) = center_y - ???
but that's where my 'school maths leave me hanging.
How can I calculate the x and y coordinates of the control points for this?
Please reply directly to me.
Markus
-- 
Dr. Markus Winter
Tel: 0064 (0)9 373 7599 (wait for message then type extension) 83960
 
 
               |//
               (o o)
-. .-.   .-oOOo~(_)~oOOo-.   .-. .-.   .-. .-.   .-. .-.   .-. .-.   .-.
||X||| /|||X||| /|||X||| /|||X||| /|||X||| /|||X||| 
/|||X|||/|||X
|/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ 
|||X|||/
'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'
  
Subject: Re: Arcs as beziers?
To be more precise:
The problem as I see it is that all the calculations I've seen go one way -
e.g. you have the defining points of a Bezier curve and that draws the
points of the figure. So it should be possible to go the other way - use a
few points to determine the defining points.
The problem is that a circle cannot be accurately represented by a Bezier
curve - so I wonder what is going to happen to the reverse calculation 
which
uses points on the circle.
Given an arc is it actually possible to calculate the control points???
Markus
-- 
Dr. Markus Winter
Tel: 0064 (0)9 373 7599 (wait for message then type extension) 83960
 
 
               |//
               (o o)
-. .-.   .-oOOo~(_)~oOOo-.   .-. .-.   .-. .-.   .-. .-.   .-. .-.   .-.
||X||| /|||X||| /|||X||| /|||X||| /|||X||| /|||X||| 
/|||X|||/|||X
|/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ 
|||X|||/
'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'   `-' `-'
Subject: Topics in 21st Century ADEX Theoretical Physics
Part II The Physical and Mathematical Foundations
Reply to Angelides's useful clarification on conformal group is below:
Jack,
A brief summary of a big topic, Reality, Mind and the Physical world:
Yes, that was my essential question.
The Question is: What is The Question? (JA Wheeler)
[A much improved version of the talk I gave in Paris at Vigier IV is now at
http://qedcorp.com/APS/StarGate.mov
Quick Time movie made in Keynote with all the animation that enhances 
comprehension of the ideas
http//qedcorp.com/APS/WheelerAustin.ppt
interactive version Power Point
Version on-line now is close to 99% of what I will present this Friday 
in Austin, Texas to
Society for Literature and Science
Tim Ferris is giving a talk on same subject BTW at that same meeting.
Remember this multi-dimensional presentation in conceptual hyperspace is 
to a literate audience of academics from the Two Cultures hence I have 
jazzed it up with reference to great Art, Drama, and lots of humor. It's 
a show, it's edutainment, but the physics is real not watered down to 
New Age psychobabble as is too common today.]
My approach to a theory of reality, which includes consciousness, is to
postulate that reality is the vast entity (cf. Phillip K. Dick's VALIS)
that underlies the entire set of A-D-E Coxeter graphs (and even goes 
beyond these graphs as V.I. Arnold has led the way).
I think this is close to the Level IV Super Platonism in Max Tegmark's 
is implemented physically is his idea.  You agree?
You further think that V.I. Arnold's math may be the Mother of All Math 
- The Mathematical Theory of Everything.
Then we have Wigner's The Unreasonable Effectiveness of Mathematics in 
Physics.
On the VALIS theme, its historical roots are well-presented in Erik 
Davis's Tech Gnosis.
This means that each of the many types of mathematical object that is 
A-D-E classified is a separate
window into this vast reality.  [Note: Mathematicians are wondering what
mathematical entity underlies the A-D-E graphs. I say it is not a
mathematical entity, but reality itself--physical, mental, and perhaps 
more than these (much debated) categories entail. The mathematics is the 
set of mathematical categories classified by the A-D-E graphs, each 
providing a different map of the territory of reality. The overall 
name for my approach is ADEX-theory -- the study and application of all 
the A-D-E graphs, with the X indicating the (mostly unknown) reality 
underlying the set of A-D-E classifications; and X also indicates the 
aspect of going beyond the A-D-E graphs via the three E type graphs as 
doorways into the enormous extension of the graphs (which has been only 
partly explored by V.I. Arnold and his students).
OK. Good. This is the clearest you have been on this. I would like you 
to include this in the third book of the Space-Time and Beyond Series
Star Gate Universe.
On eigenvalues: I believe that the deepest principle of quantum 
mechanics is that only eigenvalues (of measurement operators) can be 
observed.
Yes, that is true for orthodox micro-quantum theory like one sees in all 
the text books.
No question that Von Neumann measurement theory, that you summarize here 
,works in its proper domain of validity, which is essentially scattering 
experiments using probe beams that are statistical ensembles of single 
Green's functions like in Bohm-Pines-Nozieres et-al work in the 50's on.
However, P.W. Anderson's More is different principle came in 1967 and 
was never absorbed into the thinking of the quantum measurement Pundits. 
For example, I see no evidence that Stapp and Penrose have any inkling 
of it? I could be wrong.
Note my paper Goldstone Theorem and the Jahn-Teller Effect 1966 with 
Marshall Stoneham at Theoretical Physics Division of UKARE, Didcot, 
Berks, Oxfordshire published in 1967 Proceedings of the Physical Society 
of London and cited in AIP Resource Letter on Symmetry in Physics, 
anticipates PW Anderson's More is different. I was at Anderson's first 
talk on the subject at UCSD in 1967 where Wheeler would often also come 
to give long seminars that I attended.  This was partly because I was in 
Keith Bruckner's group and they often had meetings of the DOD ARPA 
JASONs there. I also met Frohlich at UCSD. Frohlich told me What is 
Vigier on that one. Below I show how a spatially extended electron 
micro-geon in strong short range gravity induced by partially cohered 
zero point fluctuations (mainly vacuum polarization not electromagnetic) 
Heisenberg micro-quantum microscope is increased by increasing the 
energy in the center of mass of the probe-target frame. The basic 
explanation here is in Kip Thorne's picture P3 p. 31 in the Prologue of 
the circumference of the thick black circle is far less than 2pi times 
its radius
This space warp increases as the momentum transfer p in the scattering 
increases since the Schwarzschild radial coordinate outside the event 
horizon
r ~h/p  (Heisenberg microscope equation)
where the effective micro-gravity coupling on the scale of 1 fermi and 
less is G* ~ 10^40 G(Newton)
I say:
1. Orthodox eigenvalue quantum measurement theory, i.e. results of 
measurement are real eigenvalues of Hermitian generators of Unitary 
groups breaks down completely for emergent complex systems with 
generalized phase rigidity.
2. Of course, orthodox eigenvalue quantum measurement theory, 
equivalent to Stapp's the statistical predictions of orthodox quantum 
dressed electrons in metals, and collective excitations like plasma 
oscillations) above the ground state (or vacuum in which case the 
elementary excitations are I suppose fundamentally the transverse string 
vibrations of the 2D world sheet you mentioned, but at lower energy they 
are the lepto-quarks and gauge force bosons etc. as first pointed out by 
Nambu in the 60's using the relativistic ODLRO super conducting vacuum 
non-perturbative BCS gap of form we^-1/z. But the eigenvalue 
measurement folklore breaks down completely for the LOCAL MACROQUANTUM 
complex number at low energy):
PSI(x) = (HIGGS FIELD(x))e^i(GOLDSTONE PHASE(x))
It's the rigidity in the GOLDSTONE PHASE local field (x is an event in 
4D space-time that can be generalized to hyperspace with fermionic 
dimensions and even PSI itself may become a hypercomplex number when x 
is also a hypercomplex number as in non-commutative spacetime geometry) 
that prevents the collapse into eigenvalues of this huge 
non-classical local wave with the substantiality of a classical 
electromagnetic field as PW Anderson emphasizes in his series of More is 
different writings in A Career in Theoretical Physics (World 
Scientific).
For example, suppose I Fourier transform PSI(x) to get PSI~(p), in no 
sense can you PHYSICALLY do a kind of Stern-Gerlach filter measurement 
to get a power spectrum |PSI(p)|^2 as a measurement for p. There is no 
diffraction grating big enough to do the job! See Bohm's classic text 
Quantum Theory on the use of diffraction gratings to measure p etc.
The Born probability interpretation needed for the eigenvalue 
measurement folklore simply breaks down completely for MACRO-QUANTUM 
things. That is one of the implications for the More is different 
principle that is for emergent order of complex systems what the 
equivalence principle is for Einstein's gravity and what Heisenberg's 
uncertainty principle is for micro-quantum theory.
signal nonlocality e.g. the presponse of the MACRO-QUANTUM conscious 
mind is observed by Dick Bierman et-al
4. P.W. Anderson's phase rigidity is, for the unstable vacuum of the 
micro-quantum Dirac electron field in globally flat space-time is 
precisely Andrei Sakharov's metric elasticity for the background 
independent emergence of c-number Einstein gravity + unified exotic 
vacuum dark energy/matter out of the zero point vacuum fluctuations in 
a way consistent with t'Hooft-Susskind world hologram + Hagen Kleinert 
world crystal lattice + Bohm's realist interpretation.
5. All of the key phenomena of the new precision cosmology of dark 
energy and dark matter as well as of high energy physics i.e.
1. stability of the electron (Abraham-Poincare stress problem for 
self-energy that Feynman failed to solve)
2. zero point fluctuation induced rest mass of electron ~ 0.5 Mev ~ 
(e^2/c)/*zpf^1/2
2. point-like appearance of lepto-quarks in deep-inelastic SLAC 
scattering i.e. space-warp micro-geon effect
(close relation of quasi micro-black holes to your vibrating strings of 
M theory)
3. Universe Regge slope (string tension) of hadronic resonances - basic 
data of string theory
J ~ G*M^2/hc
(consistent with my f-gravity rotating black hole work with Abdus 
Salam at ICTP 1973-74) published in Collective Phenomena edited by 
Frohlich & Cummings)
G* = c^3Lp*^2/h
/*zpf = 1/Lp*^2
Lp* = Lp^2/3(c/H)^1/3  (world hologram) H is the Hubble cosmology parameter
Lp^2 - hG(Newton)/c^3
4. Blackett effect (actually astrophysics)
    e ~  G*^1/2m
With respect to the A-D-E graph (read as classifications of Lie algebras)
each graph node corresponds to a set of basic mirrors (which generate a
full set of mirrors) in what I call reflection space, but could also be
called eigenvalue space. This is because, attached to each mirror are two
vectors called positive and negative roots. These roots carry a set of
eigenvalues (one eigenvalue for each dimension of the reflection space,
which is the number of nodes in the graph.
There is no doubt this applies to micro-quantum measurement theory in 
its proper domain of validity, which is part of The Question here. It 
may also have a more universal significance in MACRO-QUANTUM theory as 
the unfolding (topology change) of Rene Thom catastrophes on the complex 
hypersurfaces of PSI(x,L) as a wavelet transform at scale L. When 
PSI(x,L) is the human conscious mind field the topology changes of the 
PSI(x,L) landscape (as in complexity theory, e.g. Murray Gell-Mann's 
The Quark and the Jaguar) signal emergence and fading away of 
conscious qualia.
When PSI(x,L) is the physical vacuum MACRO-QUANTUM PARTIAL COHERENCE of 
ALL the zero point fluctuations of the lepto-quark/gauge boson quantum 
fields, or ALL the transverse vibrations on the 2D string world sheet at 
the deeper level, then the topology changes are in the structure of the 
physical vacuum itself.
This immediately takes us to the ICE 9 Doomsday WMD problem that Sir 
Martin Rees describes in Chapter 9 of his Our Final Hour book. Our 
book Destiny Matrix, Saul-Paul, has reference to this ICE 9 WMD 
issue and our book came out in November 2002 several months before 
about this issue in 1973-74 from Abdus Salam's ICTP in Trieste. I 
lectured on it at the Nuclear Institute in Lubljana, Yugoslavia. Fred 
Alan Wolf was with me. One of the people who heard me talk in Lubljana 
there. I think it was Domitan Popescu?
http://www.mindspring.com/~cerebroscopic/Program.html
Lev Okun published and also told me that Andrei Sakharov was very 
worried about this problem in the same period the 1970's that Rees makes 
public. Okun's writing is in my Austin APS talk at http://qedcorp.com/APS/
Though the conventional nuclear, bio-chem WMD in Iraq may be Neo-Conned 
Deception rather than Destruction, the WMD in the sky, whether 
asteroids colliding with earth or dark energy/matter bottled for 
small-scale technology use, is real in the not too distant future not 
imagined IMHO.
     These roots are lattice points in the root lattice, and root
Dual to the root lattice is the weight lattice which carries fermionic
eigenvalues.
OK at micro-quantum level of application of ADEX - nice insight.
Each root(as a set of eigenvalues) corresponds to a single
eigenvector, which is a basis element of the Lie algebra corresponding to
the Coxeter (Dynkin) graph. This is according to the Lie algebra
eigenvector equation:
     AV = [A,V] = AV - VA = eV
where A runs through a set of basis elements of the Cartan subalgebra of
dimension equal to the rank r of the Lie algebra, i.e. the number nodes in
the graph. Thus there will be r eigenvalues corresponding to a particular
eigenstate V.
     The adjoint representation of the Lie algebra provides the root
eigenvalues. Whereas other representations provide various weight
eigenvalues. For example in E7, there is a 7-d Cartan subalgebra (and
reflection space), a 133-d adjoint representation (corresponding to the
dimensionality of the Lie algebra itself). So there are 126 non-zero 
roots, and 7 zero roots (corresponding to the 7 elements of the Cartan 
subalgebra that commute with each other). These 126 non-zero roots form 
the vertices of a 7- polytope (which defines the sphere packing of 126 
6-d spheres around a central sphere in the 7-d reflection space. This is 
a lattice packing, and the lattice also generates the Hamming-7 
error-correcting code. [Note: the IT (eigenvector) from BIT (binary 
error-correcting code), or BIT from IT, is evident in this mathematical 
structure.]
     The 126 roots carry bosonic eignevalues, so the IT in this case is
bosonic.  The fermionic eigenvalues are carried in a 56 vertex polytope
(corresponding to the 56-d representation of the E7 Lie algebra.
     The ultimate IT from BIT must be in the E8 lattice, which is 
self-dual and thus does not distinguish between bosonic and fermionic 
eigenvalues. This lattice generates the Hamming-8 code (which is merely 
the Hamming-7 code with an extra parity check digit). And the E7 lattice 
structure is a sub-structure within the E8 lattice.
Yes, I agree this smells like something very fundamental as in
The Word made Flesh (Genesis Old Testament)
IT FROM BIT (Wheeler)
We are such dreams that stuff is made from.  (Variation on 
Shakespeare's Tempest Prospero)
Explicating the Super-Implicate Order (Bohm-Hiley)
The finite group generated by the mirrors is the Coxeter reflection 
group, and is (in Lie theory) called the Weyl group. This reflection 
group sends vertices of root polytopes into each other, and weight 
are obeyed.
On collapse of the wave function (or rather projection of the state 
vector): I speculate that there is an infinitesimal blip of awareness 
associated with every reflection--i.e., with every change of eigenstate. 
  A nervous system is a means of concatenating an enormous quantity of 
such blips of awareness into the rich experience we call individual 
consciousness.
Here I differ with you. What you suggest I think is too vague and cannot 
be mathematized.
This is where More is differentcomes in essentially.
Micro-quantum theory is
IT FROM BIT without BIT FROM IT is nonlocal micro-quantum theory with 
signal locality as shown in Bohm & Hiley p. 30 corresponding to
Antony Valentini's sub-quantal heat death with Von Neumann measurement 
theory of projection to micro-quantum eigenstates - that all works well
where it should for simple enough systems below the consciousness 
threshold IMHO.
No action without reaction.
BIT = MIND
IT = MATTER
Generation of inner conscious qualia is back-action BIT FROM IT
Just as mind moves matter in IT FROM BIT
So matter induces consciousness in mind in BIT FROM IT.
Just as geometry moves matter along the timelike geodesic (neglect 
hyperspace Kaluza-Klein for now to keep it simple),
So matter warps geometry to induce gravity.
No action without reaction.
Now, because of the other A-D-E classifications, such as catastrophe
structures (via V.I. Arnold's singularity theory approch), other aspects 
of the mind-body relationship come into the picture.  The action of the
reflection group on the reflection space generates the orbit space which 
is the base space of a catastrophe bundle. The mirrors of the reflection 
space get mapped onto the separatrix in the base space. The identity 
fiber is C^2/g (where g is a finite subgroup of SU(2). This is a 
singularity containing 4-d space, which when lifted up into the Lie 
algebra (in just the right way) becomes desingularized.  This is a 
hyper-Kahler manifold, and is a particular type of gravitational 
instanton whose infinity is the orbit space SU(2)/g. (This discription 
is due to Peter Kronheimer,
1989,1990).
OK this is getting closer to what I am talking about above in a more 
informal Wheeler way.
     The identity fiber C^2/g can also be unfolded by choosing a 
series of points on some path in the base space. Moving the along this 
path entails picking a new fiber at each point of the path -- and moving 
along the path corresponds to a time advancing. Whenever the path 
crossed the separatrix a topological change occurs in the fiber.
Yes, cool. Does this explain Brian Greene's topology change in current 
Sci Am 11 Dimensions?
     There is much more to say on this topic of catastrophe 
structure--base on the work of Rene Thom, V.I. Arnold and others. I will 
mathematical model of the interaction of soul and body in a single 
A-D-E classifications as a 'manification of the mysterious unity of all 
things.'
Yes, that is essentially the correct way to go IMHO.
     In addition to the mathematical categories described above,there 
are at least a dozen other mathematical categories classified by the 
A-D-E graphs that can usefully be brought into the picture. I will 
mention only a half dozen.
     Conformal field theories (which live on the 2-d string world sheet)
     Heisenberg subalgebras of the A-D-E Lie algebras.
     Korweg de Vries hierarchies of non-liner equations
     Quantizing lattices (for analog-to-gigital transforms)
     Generalized braid groups (related to knots and links)
     Wave fronts.
Nuff said 
Saul-Paul
Correct me if I am wrong Saul-Paul but what is your picture of the
connection of this group structure to consciousness. Do you have in
mind orthodox quantum measurement theory with the results of
measurement being eigenvalues along lines of Wigner & Stapp i.e.
collapse. I have argued against that view for consciousness. However,
the A-D-E structure also applies to catastrophies and in my picture
there is a MACRO Q-BIT landscape in ordinary space. Think of a giant
Bohmian real quantum potential Q in ordinary space made from PSI(x) the
coherent order parameter that is literally the physical mind field.
This PSI may have a fractal structure. Actually what we have is not
PSI(x) but a wavelet transform PSI(x,L) at scale L and so we have a
family of complex functions PSI at different scales L and there are
topology changes of this what sheaf of landscapes with coupling at
different scales - gets very complex and that's I think how V.I.
Arnold's theory of singularities come in. Indeed the dynamics of the
mental landscape, say in the real representation of Q(x,L) ~Q(x,L)^-1
Grad^2Q(x,L) (or maybe the wave propagation generalization) can
actually be pictured in computer simulations allowing us to literally
read thoughts!  There is a great simplification in that the
MACRO-QUANTUM mental field of the living mind in my theory is a local
complex field in ordinary 3D space and a lot of its features can be
pictured in terms of the dynamics of the real MACRO-QUANTUM potential
Q. Of course both the amplitude and phase of PSI will also have
physical importance. Since MACRO-QUANTUM PSI is literally a fusion of
IT and qubit in the condensate one has to ask if there is any analog to
a localized ball rolling on it as there is in micro-quantum theory. The
answer I think is no. But there is a distributed say electric dipole
moment IT distribution D(x,L) in two-way relation to the BIT PSI(x,L).
D(x,L) at some scale L will be directly related to nerve impulse firing
patterns for example. On smaller scales to MRI and so on.
Subject: Re: Conformal Group & Exotic Vacuum Dark Energy/Matter Link?
..............
There is, of course, one force missing in the above unification -- gravity.
There you need Diff(4) from breaking the 4-parameter translation 
subgroup of the 10 parameter Poincare group. Gravity as curved spacetime 
is a broken symmetry of global translational invariance. The Poincare 
group, in turn, is from breaking I suppose a 5-parameter subgroup of the 
15 parameter conformal group of the light cone used in Penrose's 
twistors. One parameter is dilation. I forget - what are the other 4? 
  What is the physical interpretation of the 5 additional generators of 
the Conformal group Lie algebra? 4-momentum from translation sub-group 
and space-time rotations from 6-parameter Lorentz sub-group. What are 
the others? Shipov in his torsion theory seems to want to break the 
symmetry of the 6 Lorentz group parameters to get the torsion twists 
generalization of space-time rotations! That makes sense actually. What 
about those 5 conformal generators?
Starting from 15 parameter conformal group of the light cone,
C(15) = ?(5) T(4)O(1,3)
Is ?(5) = D(1)?(4)?
D(1) a one parameter dilation whose breaking of symmetry introduces the 
rest mass m into E^2 = (mc2)^2 + (pc)^2 mass shell
What are the 4 missing generator charges to add to 4-momentum and 
space-time rotations?
What about the dilatation and the 4 other things? What are they 
physically? They should already be there in global special relativity 
without gravity hidden as rest masses of quantum fields in some way?
The 4 other charge generators of the conformal group, must, therefore, 
be the 4-gradient of the Higgs Field! That sort of makes physical sense. 
Will there be a compensating gauge field?
to be continued
Jack and Saul-Paul, the 4 missing elements of the special conformal 
group are known as inversions (actually, they are products of inversions 
and translations).
Lie algebra of the missing stuff is not too different from that of the 
Pu Lie algebra. I was worried about that.  I am glad to see that 
essentially we have another copy of the translation group since I need 
two such copies to modulate both the Goldstone phase and Higgs amplitude 
of my QED vacuum polarization partial coherence ODLRO parameter 
PSI(x,L).  I get Einstein's gravity from locally gauging one of those 
translation sub-groups in the usual way Kibble did it at Imperial 
College in the 60's and I think we should get the unified w = -1 exotic 
vacuum dark energy/matter field by locally gauging this second copy of 
the translation group with the extra inversion. Is this related to 
Brian Greene's mirror symmetry with two geometries and one physics 
in the current Sci Am?
Geometrically, this is done by the successive application of an 
inversion f^1 through a unit circle centered at the origin O, an 
inversion f^2 through a unit circle
at some point, say, p, and a translation f^3 by the vector Op. That is, 
f^3 o f^2 o f^1 (o stands for composition).
The conformal inversion maps infinity, which is not a point of Minkowski 
spacetime, into a finite point and conversely. The action of the 
conformal group is
defined on the so-called conformally compactified Minkowski spacetime.
Right which is important in Penrose's Twistor theory that I vaguely 
remember from attending his seminars at Birkbeck in 1971.
BTW Thomas I will have access to a 2Bdr apartment in Kensington section 
of London soon, so I will be seeing you and Simon. Will take Saul-Paul.
What are they physically ? In the physics literature, these elements 
have been referred to as acceleration transformations. Whether these 
can be linked to
gravity, I am not so sure...
Any URLs on this. This is something I must understand much better since 
it is the modulation of the Higgs in my model that gives the lepto-quark 
rest masses like
m(electron) ~ (e^2/c^2)/zpf*^1'2
where /zpf* = 1/Lp*2
in the quantized circulation vortex core of the micro-geon ring 
One last point of logic: Since all these small groups are subgroups 
of Diff(4), then whenever Diff(4) holds, so does each of its subgroups a 
fortiori.
I hope this helps.
What I mean here is that in Einstein's curved space-time the translation 
subgroup of the global Poincare group no longer holds. Of course, you 
can say that once the translation sub-group is locally gauged then the 
local conservation of Pu is restored including the curvature as a 
dynamical degree of freedom as in the Bianchi identities
Now all that comes from Kibble's locally gauging only the translation 
sub-group of the 10 parameter Poincare group.
I think Utiyama in 60's locally gauged the full 10 parameter Poincare 
group to get the torsion fields from the 6 parameter Lorentz group years 
before G. Shipov got there.
My question now is what is the physics of locally gauging the dilaton + 
the 4 translations with these inversions?
If my theory is right what we get physically are dark energy/matter 
exotic vacuum fields.
That is gravity + torsion + exotic vacuum dark energy/matter + dilation 
field
We know that Universe is 96% exotic vacuum at large FRW scale.
We did not really know that last year for sure.
All good wishes
Thomas.
PS Jack, your references to the so-called quantum nonlocality and signal 
locality (a rather silly idea) etc need to be corrected (but I am not 
going to start
quoting chapter and verse;  you know what I mean...). 
On that we really need to Pow Wow at length with The Pundits including 
Hiley, Valentini, Stapp, Bierman (the experimentalist) and Duke it all 
out. :-)
A good place to do this is Simon's Private London Club on Portman Square 
which has wireless for our laptops.  I will get to London ASAP probably 
early next year for an extended stay of a week or two or three depending 
on how long I can get the Kensington apartment at that time.
I suppose you mean quantum nonlocality is the silly idea since you 
argue for signal locality. The big issue is signal nonlocality for 
which there is evidence IMHO. How good that evidence is, is also part of 
The Question.
Jack & Thomas,
On the role of the conformal group SU(4) or SU(2,2):
Yes, for Penrose twistor spinor of conformal space-time.
The extended E7 Coxeter graph is
     1-2-3-4-3-2-1
           |
           2
(Kac-Moody)Lie algebra and OD (the octahedral double group as a subgroup 
of SU(2)) thatthe complex group algebra C[OD] is equivalent to the 
direct sum of total
matrix algebras:
     C[OD] = M1 + M2 + M3 + M4 + M3 + M2 + M1 + M2
Since the squares of the numbers on the Coxeter graph (which I call 
balance numbers) sum to 48 (in accord with the 48 elements of OD), we 
see that the matrix algebras span the full 48-d vector space of C[OD].
In my Consciousness: a Hyperspace View paper (published in 1993 as a 
40 page appendix in Jeffrey Mishlove's book *Roots of Consciousness*) I 
     In the scheme of this paper E7 is the high-energy symmetry group, 
but because of the McKay correspondence between E7 and OD, the unitary 
group P in C[OD] is a 48-d principal bundle projecting down to a 10-d 
base space S, each point of which is a copy of the 38-d fiber G. We 
write these spaces out as the following Lie groups:
         S = U(2) x T^6;
         G = U(1) x SU(2) x SU(3) x SU(3) x SU(4) x SU(2)
where U(2) is a 4-d spacetime called *conformally compactified Minkowski
space*; T^6 is a 6-d torus = U(1) x U(1) x U(1) x U(1) x U(1) x U(1).
     We consider S as the 10-d spacetime of superstring theory and G as 
the symmetry group of the following six forces, which we identify in 
sequence as: electromagnetism, weak, strong (color), hyperweak, gravity 
and perhaps the feeble force.
     We can write: P = S x G. Since U(2) = U(1) x SU(2), we can 
rearrange S as S = SU(2) x T7. We regard SU(2), i.e., spherical 3space 
S^3, as the space of cosmology -- the space in which we as macroscopic 
bodies appear to live. Every point of a macroscopic body is a point of 
S^3. Thus, if we view S as a fiber bundle, every point of a macroscopic 
body is actually a 7-d spce which is a copy of T^7. Note that the 
7=torus T^7 incorporates the factor U(1) from the U(2) spacetime, and 
thus includes time. Now T^7 corresponds (via McKay's theorem) to the 7-d 
reflections space of E& as follows:
         T^7 = R^7/Lr
where R^7 is the *real* part of the E7 complex reflection space C^7, and 
Lr is the E7 root lattice. This means that all the points of the lattice 
are identified as a single point, the identity element of T^7, and every 
other point of T^7 is a copy of Lr.
     [end of quote from my 1993 paper]
     Note that SU(4) is the 15-d conformal group, and if you need a 
fancier version of this group it is sitting in C[OD] as M4 -- which is 
the set of all 4 by 4 complex matrices. The set of all invertible 
elements in M4 is GL(4,C), and the set of all unitary elements is U(4) = 
U(1) x SU(4).
     This is the paper that John McKay referred to in his short paper A
rapid introduction to ADE theory, which can be found at:
http://math.ucr.edu/home/baez/ADE.html
[Note:
     It may be that the cosmological space of this paper S^3 = SU(2) 
should be replaced by SU(2)/g where g is a finite subgroup of SU(2). 
McKay's correspondence consists of a profound A-D-E classification of 
these subgroups, which I have called McKay groups.
     Cf. October 9 *Nature* paper by Luminet et alia.
     Morover, in 1989 Peter Kronheimer (a graduate student of Michael 
Atiyah) proved the A-D-E classification of C^2/g as gravitational 
instantons (closely related to twistor structures). The infinity of 
any such instanton is SU(2)/g.
     In addition, it is very interesting that C^2/g is the identity 
fiber of an A-D-E classified (via V.I. Arnold) Thom type catastrophe 
bundle.
     BTW: There is a different conformal invariance involved in string
theory. The 2-d world sheet is conformally invariant (and 2-d Lorentz
invariant) so that a 2-d conformal field theory lives on this space -- 
and is a basic entity in superstring theory. And it just so happens that 
2-d conformal field theories are A-D-E classified.
     For this and many other reasons, I believe that the study of all
classifications of the A-D-E graphs (which I call ADEX theory) is the 
key to the deep study of reality.
All for now.
Saul-Paul
I have not yet absorbed your last message Saul-Paul. Leaving for Austin, 
Texas soon.
Part II
....
     I met Charles Muses in October 1973 at the Institute for the Study 
of Consciousness in Berkeley, CA. This was when the Institute was first 
being set up by Arthur Young, who had asked me to be his research 
associate. Charles was very keen on the hyperspace idea. Arthur was not 
so happy with going beyond the 4 dimensions of spacetime. Arthur 
encouraged me to read Eddington's works and finite group theory. In 
particular, Arthur wanted me to work out the symmetry group of the 
tetrahedron. I did this in early 1974, and found that this group T has 
12 elements of rotations, but when reflections are allowed this is 
enlarged to the full 24 element group consisting of all permutations of 
4 things (also called the symmetric-4 group S4). This is equivalent to 
the Octahedral group O of rotations of the octahedron (and its dual the 
cube).
     In process of generating the group table for O (= S4), I found that 
6 cosets of K4 (the Klein-4 group) within S4, seemed to map perfectly
elements in S4 (the number of elements in the classes being 1, 3, 8, 6, 
and 6). The first 3 classes correspond to the T (=A4) subgroup of 12 
weakons, 8 gluons. Moreover the odd permutations consisting of the two 
remaining classes map onto 3 families of fermions, with two quarks and 
two leptons in each
family).
     This was all finite group modeling, yet I was very intrigued by 
Muses's work on hyperspace, especially his paper, The Amazing 24th 
Dimension, published in the *Journal for the Study of Consciousness* 
(which he edited for Arthur Young), Vol. 5, No. 1 (1972), pp. 82-24. 
There Muses describes the packing of spheres in 24 dimensional space, 
via the Leech Lattice. There are 196,560 spheres packed around a central 
sphere in this maximally dense packing.
     Since S4 is a 24 element group, I thought there might be a 
connection between this group and the magical 24-d space. So my next 
move was to construct the 24-d group algebra of S4. This is done by 
regarding the 24 elements of S4 as vectors in a vector space. Moreover, 
if we make this a complex vector space, then via various theorems (such 
as Wedderburn's) we can see that
         C[S4] = M1 + M2 + M3 + M3 + M1
where C[S4] is the complex group algebra of S4, and Mn is the total 
matrix algebra of n x n complex matrices. Then the unitary elements 
within C[S4] is of necessity the 24-d unitary Lie group:
           U = U(1) x U(2) x U(3) x U(3) x U(1)
     which we can rewrite as T4 x U(1) x SU(2) x SU(3) x SU(3), and thus 
(plus and extra SU(3) which may function as a hyperweak gauge group 
acting on the three family labels -- thus breaking CP symmetry -- but 
this is speculative.
     We are still a long way from the Leech lattice in 24-d vector space 
that
Charles Muses described as important for both physics and consciousness.
     There is, of course, one force missing in the above unification --
gravity. In order to make contact with gravity gauge groups (assuming 
such an approach makes sense), I went to the double group version of S4, 
which could be written as DS4, but I prefer to write it as OD, the 
octahedral double group (of the crystallographers). Just as O is a 
finite subgroup of SO(3), OD is a finite subgroup of SU(2). In other 
words, just as SU(2) double covers SO(3), and we write SO(3) = 
SU(2)/{1,-1}, so we can write
  O = OD/{+1,-1}.
Then if we construct the group algebra of the 48 element OD group, we 
find that:
     C[OD] = C[O] + M2 + M2 + M4
     But this can be written out and reordered as:
      M1 + M2 + M3 + M4 + M3 + M2 + M1 + M2
The reason for writing the direct sums of these matrix algebras in this
order is that there is a 1-1 correspondence between these matrix 
dimensions and the numbers on the nodes of the extended Coxeter graph 
for the E7 Lie algebra (and Coxeter-Weyl group).
     1--2--3--4--3--2--1
              |
              2
That there is indeed a deep and beautiful connection between these two 
very different mathematical structures, a finite group algebra and a Lie 
algebra is the great discovery of John McKay in 1979. The connection is 
that the 8-d center of C[OD] is equivalent to the 8-d Cartan subalgebra 
of the (infinite dimensional) extended E7 Lie algebra.
     I call these numbers on the Coxeter graph balance numbers because 
the positive mirror vectors in the E7 reflection space (which Coxeter 
calls the kalleidoscope), when lengthened by these balance number 
factors will all be in balance -i.e., as vectors they sum to zero. These 
number then also define the setting up a a lattice in the 7-d Cartan 
subalgebra space of the 133 dimensional E7 Lie group. Moreover this 
lattice generates a binary error-correcting code -- the Hamming-7 code.
     So how do we get to the 24-d Leech lattice?  Well, the E7 lattice 
is a sub-lattice of the E8 lattice in the Cartan subalgebra of the 248-d 
E8 Lie algebra. And this E8 lattice generates the Hamming-8 code.
     This is important for the E8 x E8 heterotic superstring theory. The
lattice structure in this Lie algebra provides for the anomaly
cancellations for the heterotic string theory. And it was this 
even-self dual lattice structure that provides this anomaly 
cancellation. What physicists call anomaly cancellation is, in effect, 
what the mathematicians call error-correcting -- every even-self dual 
lattice corresponds to an even self-dual error-correciting code.
Physical reality as software? At least the BIT part?
     Here's where the 24-d Leech lattice comes in: the Leech lattice is 
also an error-correcting code lattice; in fact it is the lattice for the 
even self-dual 24-bit Golay code.  This code can be generated by three 
copies of the E8 lattice: E8 + E8 + E8.  This is important because in 
the heterotic string theory, the bosons see the full 26 dimensions of 
the original 26-d string theory of 1970,  in which the fermions see only 
the 10 dimensions of the superstring theory of 1971.  Since the strings 
sweep out a 2-d world sheet, it is the transverse vibrations of this 
transverse vibrations are in a 24-d space, while the fermionic 
transverse vibrations are in an 8-d space. The 24-d space has the Leech 
lattice corresponding to the Golay code (or E8 + E8 + E8 lattice), while 
the 8-d fermionic vibratory space corresponds to the Hamming-8 code of 
the a single E8 lattice.  The E8 + E8 lattice (16-d) interpolates 
between the Leech lattice and the single E8 lattice in the
lattice direct sum E8 + E8 + E8, from which one can derive the 24-d 
Leech lattice.
     So although, my 24-d group algebra C[O] seemed to contain much of 
direct contact with the 24-d Leech lattice. However, there is a kind of 
apotheosis for the E7 Lie algebra (which is closely associated with 
C[OD] via McKay's correspondence (mentioned above)).
     The E8 x E8 heterotic string theory is just one of five competing 
string theories. But since 1995, due to the work of Edward Witten (and 
others) there is an overarching theory called M-theory which unifies the 
5 competing string theories as subtheories. Moreover, the 11-d 
11-d M-theory. As a consequence E7 is brought in as the basic symmetry 
group of the ll-d theory.
     So I feel that I was barking up the right tree after all. My 
starting point in all this was the work I did at the ISC under the 
influence (in somewhat different direction) of Arthur Young and Charles 
Muses.
     In any case, it is now abundantly clear that Muses's fascination 
with hyperspace in general and the 24-d Leech lattice in particular was 
a move in the right directing for advance in physics.
     I would even say that Muses's speculation that consciousness should 
be modelled via hyperdimensional mathematics is on the right track. My 
approach to this is deeply involved in E7 and indeed in the complete set 
of A-D-E Coxeter graphs.
     This is written up in great detail in my appendix paper 
Consciousness: a Hyperspace View in Jeffrey Mishlove's book, *Roots of 
Consciousness* (2nd Edition), Marlowe, 1993.
     A summary (and update) is A Mathematical Strategy for a Theory of
Consciousness pp, 579-588 of *Toward a Science of Consciousness: the 
First Tucson Discussions and Debates*, edited by S.R. Hameroff, A.W. 
Kaszniak, & A.C. Scott (MIT, 1996).
     I should mention that Jeffery Mishlove also worked with Arthur 
Young at the ISC, and interacted with Charles Muses there and elsewhere. 
And there is material on both Charles Muses and Arthur Young his book 
*Roots of Consciousness*
     You can find out more the McKay correspondence between finite 
subgroups
of SU(2) and Lie groups at
     http://math.ucr.edu/home/baez/ADE.html
where John McKay has the short paper A Rapid Introduction to ADE 
Theory.
There is a reference to *Roots of Consciousness* at the end of McKay's
paper.
     My mathematician friend Tony Smith had sent McKay of copy of my long
appendix paper of 1993. So that is apparently why McKay refers to *Roots 
of Consciousness*
BTW: Tony Smith has also become interested in the work of Charles Muses, 
and you can find some of Tony's thinking on this in the long paper Del 
Pezzo Surfaces and the A-D-E series at:
     http://www.innerx.net/personal/tsmith/NDxNDalg.html
On Superstring theory, see the November issue of *Scientific American* 
is hosting a Nova series on string theory. Parts I & II on Oct. 28, Part 
III on Nov. 4.
All for now.
Saul-Paul
Correct me if I am wrong Saul-Paul but what is your picture of the 
connection of this group structure to consciousness. Do you have in mind 
orthodox quantum measurement theory with the results of measurement 
being eigenvalues along lines of Wigner & Stapp i.e. collapse. I 
have argued against that view for consciousness. However, the A-D-E 
structure also applies to catastrophies and in my picture there is a 
MACRO Q-BIT landscape in ordinary space. Think of a giant Bohmian real 
quantum potential Q in ordinary space made from PSI(x) the coherent 
order parameter that is literally the physical mind field. This PSI may 
have a fractal structure. Actually what we have is not PSI(x) but a 
wavelet transform PSI(x,L) at scale L and so we have a family of complex 
functions PSI at different scales L and there are topology changes of 
this what sheaf of landscapes with coupling at different scales - gets 
very complex and that's I think how V.I. Arnold's theory of 
singularities come in. Indeed the dynamics of the mental landscape, say 
in the real representation of Q(x,L) ~Q(x,L)^-1 Grad^2Q(x,L) (or maybe 
the wave propagation generalization) can actually be pictured in 
computer simulations allowing us to literally read thoughts!  There is a 
great simplification in that the MACRO-QUANTUM mental field of the 
living mind in my theory is a local complex field in ordinary 3D space 
and a lot of its features can be pictured in terms of the dynamics of 
the real MACRO-QUANTUM potential Q. Of course both the amplitude and 
phase of PSI will also have physical importance. Since MACRO-QUANTUM PSI 
is literally a fusion of IT and qubit in the condensate one has to ask 
if there is any analog to a localized ball rolling on it as there is in 
micro-quantum theory. The answer I think is no. But there is a 
distributed say electric dipole moment IT distribution D(x,L) in two-way 
relation to the BIT PSI(x,L). D(x,L) at some scale L will be directly 
related to nerve impulse firing patterns for example. On smaller scales 
to MRI and so on.
On Charles A. Muses (1919-2000) and Hyperspace:
Muses gave at the Parapsychology Research Group in San Francisco, 
December 1, 1986.
     The Parapsychology of Time
     In 1967 quantum physicist and Nobel laureate Eugene P. Wigner 
predicted a convergence of psychology and physics, in his book 
*Symmetries and
Reflections*.
Jack commented:
Wigner accepted Von Neumann's and London's notion of collapse of the 
micro-quantum information wave. Bohm and Hiley in The Undivided 
Universe show why collapse is an illusion. It is a muddled idea.
Any attempt to connect consciousness with collapse is a badly 
mistaken idea IMHO even though many of my heroes like John Wheeler have 
advocated it. Great men make great mistakes. The modern theory of 
decoherence by W. Zurek also shows why collapse is really only a 
superficial description. You can say as if collapse in a crude sense. 
There is no collapse in the many worlds approach either.
The basic simple insight missed by Wigner, Stapp, Wheeler, Penrose et-al 
is that the living conscious mind is a BIG THING. It's not like the wave 
of a single electron or even a small number of electrons. As P.W. 
Anderson says
More is different.
For Penrose to think that a single graviton has anything to do with 
the creation of a conscious moment is completely absurd IMHO even though 
Penrose is a great mathematical genius no doubt about it.
While the conscious mind is a BIG THING it is not a classical physics 
thing.
Stapp is correct on that score. Also Stapp is correct on making the 
distinction between rocklike things and thoughtlike things that 
correspond to Wheeler's IT and BIT, to Bohm's hidden variable 
and 
pilot wave, to Flesh and Word, to material and mental.
The conscious mind is a MACRO-QUANTUM thing not a micro-quantum thing.
The laws of MACRO-QUANTUM things are not the same as the laws of 
micro-quantum things.
More is different.
The laws of micro-quantum things are
1. A linear nonlocal unitary dynamical equation of motion in 
configuration (or phase) space for the wave (or the Wigner density) when 
there is entanglement.
  I mean this in the sense of second-quantized theory in Fock occupation 
number space where the Hamiltonian is a linear operator in Fock space 
even though configuration space representations look nonlinear as in 
Hartree-Fock theory for example.
2. Signal locality,  i.e. EPR entanglement cannot be used as a 
stand-alone communication channel.  It can be used with classical 
signals for information storage and processing like in quantum 
computing, cryptography and teleportation. This is called passion at a 
distance by Abner Shimony.
Note A: The projection operator formalism, associated with collapse, 
Stern-Gerlach experiment.
Note B: Antony Valentini has clarified passion at a distance as an 
artifact of sub-quantal equilibrium in which uncontrollable quantum 
randomness is viewed, as in Vigier's models, as a sub-quantal Brownian 
motion. In this case signal-locality is only found in the Brownian case, 
not in colored noise situations of sub-quantal non-equilibrium open 
pumped complex systems.  No living system with a conscious mind can be 
closed in maximal randomness at any level classical, quantum or 
sub-quantum. Therefore, signal-nonlocality is a necessary condition 
for life as Brian Josephson first glimpsed in his Biological Use of 
Nonlocality with Fontini Pallikari.
The sub-quantal Brownian noise case is the Born probability distribution 
with signal locality, i.e. all reduced quantum density matrices for the 
parts of entangled wholes are completely random in the sense that they 
do not depend on control parameters of distant parts to which the part 
in question is entangled by the EPR effect.
This is what is entailed by what Stapp calls the statistical 
predictions of orthodox quantum theory.
An instability in a micro-quantum ground state for real on-mass-shell 
quanta, or in the vacuum for virtual off-mass-shell quanta makes a 
relative non-classical entropy lowering phase transition to an emergent 
more complex order via the formation of a boson mode (if initially a 
fermion system) that condenses into a much smaller volume of the 
original phase space. This is a non-classical phase transition from 
micro-quantum stuff to MACRO-QUANTUM STUFF (either real or virtual).
MACRO-QUANTUM THINGS do not obey  the statistical predictions of 
orthodox quantum theory because of what P.W. Anderson calls 
generalized phase rigidity.
The latter in the special case of unstable quantum field theory of 
electrons (at least if not lepto-quarks in general) in globally flat 
space-time leads directly to Andrei Sakharov's metric elasticity for 
the emergence of Einstein's c-number curved space-time 
geometrodynamics along with the observed unified dark energy/matter 
exotic vacuum fields out of the MACRO-QUANTUM PARTIAL COHERING of the 
formerly completely random micro-quantum zero point fluctuations of all 
the quantum fields of the standard model or of its string theory 
generalizations to hyperspace and M-theory.
This is highly scale-dependent. The non-gravitating non-exotic 
MACRO-QUANTUM vacuum is completely coherent with zero residual random 
micro-quantum zero-point fluctuations at the FRW scale. However the new 
precision cosmology shows that only at most 4% of the universe at that 
scale in non-exotic. 96% is exotic - approximately 70% anti-gravitates 
as dark energy and approximately 26% percent gravitates as 
non-hadronic non-radiation non-lepton dark matter.
In general the dynamics of MACRO-QUANTUM THINGS is Landau-Ginzburg not 
Schrodinger-Dirac i.e.
NONLINEAR LOCAL NONUNITARY for a ROBUST ground state (or vacuum) wave 
with generalized Goldstone phase rigidity that completely disobeys the 
projection operator formalism of orthodox micro-quantum measurement theory!
The living conscious mind is such a MACRO-QUANTUM THING with the 
signal-nonlocality that Dick Bierman sees in his presponse data.
The local GIANT MIND WAVE is coupled in two-way relation (Bohm & 
Hiley's term) to the 10^18 hydrophobically caged electron qubits at the 
sub-microtubular level made popular by Stu Hameroff. Max Tegmark has 
completely shot down the Hameroff-Penrose single-graviton collapse 
model not of a single electron qubit inside the cage, but of the 
superposition of the heavy dimer 2-state configurations. Every 
nano-technologist knows that classical mechanics works for the molecular 
configurations - this is the Born-Oppenheimer approximation from me/mp ~ 
10^-3. This boundary has been pushed with atom interference, but I doubt 
you can have that for the dimers?
bcc
...
We are still a long way from the Leech lattice in 24-d vector space 
that Charles Muses described as important for both physics and 
consciousness.
You were working for the consciousness side from Piaget's idea that K4 
described some of the functional aspects of learning in kids. Is that 
correct? In that case, the group structure was pure phenomenology from 
the empirical study of certain psychological tests. Correct? I mean no 
physics model here as yet - except by some kind of formal analogy.
There is, of course, one force missing in the above unification --
gravity.
There you need Diff(4) from breaking the 4-parameter translation 
subgroup of the 10 parameter Poincare group. Gravity as curved spacetime 
is a broken symmetry of global translational invariance. The Poincare 
group, in turn, is from breaking I suppose a 5-parameter subgroup of the 
15 parameter conformal group of the light cone used in Penrose's 
twistors. One parameter is dilation. I forget - what are the other 4? 
  What is the physical interpretation of the 5 additional generators of 
the Conformal group Lie algebra? 4-momentum from translation sub-group 
and space-time rotations from 6-parameter Lorentz sub-group. What are 
the others? Shipov in his torsion theory seems to want to break the 
symmetry of the 6 Lorentz group parameters to get the torsion twists 
generalization of space-time rotations! That makes sense actually. What 
about those 5 conformal generators?
Starting from 15 parameter conformal group of the light cone,
C(15) = ?(5) T(4)O(1,3)
Is ?(5) = D(1)?(4)?
D(1) a one parameter dilation whose breaking of symmetry introduces the 
rest mass m into E^2 = (mc2)^2 + (pc)^2 mass shell
What are the 4 missing generator charges to add to 4-momentum and 
space-time rotations?
Breaking symmetry in this context means that the vacuum does not have 
the symmetry of the dynamical action. I mean More is different 
spontaneous symmetry breaking not dynamical symmetry breaking by adding 
weird terms to the action of the local field theories. When you locally 
gauge you restore the symmetry of the action, though not the vacuum.
field that also spontaneously breaks global flatness of the vacuum, i.e. 
tidal tensor curvature not locally eliminated the way the non-tensor 
g-force connection for parallel transport can be.
the dislocation twist to space-time.
What about the dilatation and the 4 other things? What are they 
physically? They should already be there in global special relativity 
without gravity hidden as rest masses of quantum fields in some way?
OK in my theory, gravity comes from ripples in Goldstone phase, and 
exotic vacuum dark energy/matter come from ripples in Higgs field where
Higgs Field(x) e^i(Goldstone Phase(x))
Higgs Field = |PSI|
This is physical here since |PSI|^2 is the density of virtual 
electron-positron bound states in the vacuum condensate. Dilation 
symmetry is broken here.
The 4 other charge generators of the conformal group, must, therefore, 
be the 4-gradient of the Higgs Field! That sort of makes physical sense. 
Will there be a compensating gauge field?
to be continued
We are pleased to send you the following information about this
newly published book:
Eight Preposterous Propositions
Warming
Robert Ehrlich
To read the introduction, please visit:
http://www.pupress.princeton.edu/titles/7588.html
  Placebo cures. Global warming. Extraterrestrial life.
Psychokinesis. In a time when scientific claims can sound as
strange as science fiction--and can have a profound effect on
individual life or public policy--assessing the merits of a
far-out, supposedly scientific idea can be as difficult as
it is urgent. Into the breach between helpless gullibility
and unyielding skepticism steps physicist Robert Ehrlich,
with an indispensable guide to making sense of scientific
claims. A series of case studies of some of the most
controversial (and for the judging public, deeply vexing)
topics in the natural and social sciences, Ehrlich's book
serves as a primer for evaluating the evidence for the sort
of strange-sounding ideas that can shape our lives.
  A much-anticipated follow-up to his popular Nine Crazy
Ideas in Science, this book takes up issues close to
readers' everyday reality--issues such as global warming,
the dangers of cholesterol, and the effectiveness of
placebos--as well as questions that resonate through (and
beyond) civic life: Is intelligent design a scientific
alternative to evolution? Is homosexuality primarily innate?
Are people getting smarter or dumber? In each case, Ehrlich
shows readers how to use the tools of science to judge the
accuracy of strange ideas and the trustworthiness of
ubiquitous experts. As entertaining as it is instructive,
his book will make the work of living wisely a bit easier
and more reliable for scientists and nonscientists alike.
  Robert Ehrlich is Professor of Physics at George Mason
University and a Fellow of the American Physical Society. He
is the author of twenty books, including Nine Crazy Ideas in
Science: A Few Might Even Be True, Why Toast Lands
Jelly-Side Down, and Turning the World Inside Out & 174 Other
Simple Physics Demonstrations (all Princeton).
0-691-09999-5 Cloth  $27.95 US and L18.95
352 pages. 14 halftones. 43 line illus. 65 tables. 5 x 8.
This message is for Members of Princeton University Press's E-mail List for
Biological Sciences, Physics, Mathematics.
receiving more announcements of this kind as new books are released in
the subject areas you have selected. You may un-subscribe from this list
at any time by sending a message to webmaster@pupress.princeton.edu
We're very interested in your comments and suggestions on this service.
Feel free to e-mail us at webmaster@pupress.princeton.edu
Nicht danke - ich habe kein nottink.
I thank you for the compliment of proclaiming me to be the
highly-intelligent hacker Indrid Cold. Actually, I'm not that clever.
Yes, I corrected that error. It was my unintentional slip because I was 
having problems sending with pacbell.
It happens a few times a year because I multi-task very fast. You know I 
got turbo-charged upstairs in 1953
when I was Taken! :-) JOKE GRIN - well who really knows?
 I think that Indrid Cold went back to Lanulos, or wherever he flew in 
from. Yes, I did send that message out about Wheeler's anti-PSI stance. 
I wanted to make sure that your other readers would see it. Yes, it's 
true that some of your emails use the blind cc's, while others have sent 
the addresses out in the clear. So I did nothing really clever here - 
just a cut and paste. I tried to send it using my mail server at 
debunker.com (hosted by Lanset), but it complained 'too many 
recipients. So I tried pacbell, and voila.
As for this trixcleverspacealien, I know nothing about him, her, or 
it. Perhaps you've been hacked by the Illuminati? :)
I am all that is left of them. Saul-Paul too. Also Stephen Schwartz! :-)
re: http://qedcorp.com/APS/StarGate1.mov
http://qedcorp.com/APS/WheelerAustin.ppt
I tried to look at your advertised entertainment below, but even though 
I have DSL and it is working OK, somehow nothing comes up for me. - Hold 
the presses - as I am typing this, the first page of your PPT came up, 
after about 10 mins. Are you looking to set a world record in internet 
bandwidth use?
Something is wrong with your system. The Quick-Time file is 10 megs, the 
Power Point is 8 Megs. But with streaming that should not matter much.
It loads instantly on Safari on my MAC with DSL. One can see the 
streaming ahead of the presentation on the meter provided.  My brother 
got it on his windows machine without waiting for 10 minutes. At least 2 
are a large number of people simultaneously downloading it at the time 
you tried? Yours is the only such complaint so far.
You are correct, I am not a physicist and do not claim to be. However, 
I do know that the hypotheses most likely to be true are those agreed 
upon by a majority consensus of professional physicists.
Red Herring. Irrelevant. True, but irrelevant.
It's not an infallible guide
to truth - nothing is - but it's the best we have. And I'm perfectly 
willing to concede that there's a possibility that good old Sarfatti may 
indeed hit upon some important discovery in physics that everyone else 
has missed. These things happen occasionally. That 'dark energy' and 
'vacuum energy' stuff is bizarre beyond belief, but it seems to be good 
physics.
It's only bizarre to you because you do not understand enough physics. 
It's quite ordinary to a physicist who understands both the equivalence 
Physics Today.
  The mainstream of physics has embraced it, in all its strangeness. 
The Casimir effect is undeniably real. Perhaps if you play around with 
that wild stuff
long enough, you may hit upon pay dirt.
There is no perhaps if you are able to understand 
http://qedcorp.com/APS/StarGate1.mov IN ITS FULLNESS. I could be wrong 
of course. If not, it would not be science.
(Or you may hit nothing, no matter how long you try.) Kepler was a 
half-crazy astrologer, 
Newton was an Alchemist and very cruel to his competitors like Hooke. He 
was almost as bad as the Trotksyite in Neo Con Cloak, Stephen Schwartz,, 
author of The Two Faces of Islam. :-)
but his good work was unbelievably good,
The comparison of Newton with Schwartz stops there. :-)
and Tesla for all his brilliance was more than a little wacky. Does 
that mean that every scientist who goes off the deep end finds something 
valuable? Hell, no! But occasionally somebody who goes out on a limb 
sees something that others didn't. Only time will tell if what you're
doing will have any lasting value to science. However, I'd have a lot 
more confidence in it if UFOs and ESP didn't keep popping up. That 
suggests to me that your 'nonsense filter' needs to be seriously 
tightened.
You can battle that out with Jacques Vallee, Hal Puthoff, Eric Davis and 
Bruce Maccabee on UFOs and with Nobel Prize Laureate Brian Josephson on 
ESP.  What you do not get is that thinking about UFOs and ESP is a great 
mental exercise, a gedankenexperiment, for getting out of the box to 
make the breakthrough - see Brian Greene in latest Sci Am on 11 
Dimensions. Your mind is much too closed and that is the problem with 
all of you pro-Skeptics. On the other side the New Age starry-eyed 
believers are too open. Both camps do not understand enough physics and 
science in general. I consider both sides extremists with me in the middle.
Still, I doubt very much if Wheeler would agree with you when you say:
'Our mind is a GIANT MACRO-quantum bit local wave in two-way relation 
with electrical dipole configurations that couple to nerve impulses, 
chemical messengers etc. Direct back-reaction from the electric dipoles 
to this mind wave generates inner experience.'
Irrelevant. For one thing Wheeler is now 93 and is not the man he used 
to be - though he is still fantastic. However, if I had a few hours with 
him I am sure he would understand what I am proposing, which is not very 
far out BTW but it is quite mainstream in condensed matter physics.
You should ask him about that.
Do you think the mind is a little bitsy teensy weensy micro-quantum 
thing? Speak for yourself. :-)
           best,
             Roberto, il tenore
Hey if we did the Meistersinger thing, I would win. :-)
You should be devoting your Psychic Bad Vibrations column in Skeptical 
Inquirer to trying to shoot down my stuff on UFOs and ESP with a 
responsible intellectually honest debate between me and any Pundit who 
thinks he can stand up to me. I mean if you had real courage like in The 
Wizard of Oz you would do that. We are all bored with the same old same 
old beating of dead horses on the obviously goofy New Age mindless stuff 
on ESP and UFOs in the pop media which only serve to coverup the Right 
Stuff.
Detailed exegesis not polemical vague smears is what is called for if 
you really want to serve the public and weed out the mis and 
disinformation in ESP and UFOs which most of the world's thinking people 
are fascinated with. You pro-debunkers are too simplistic it's time to 
      Robert Sheaffer - roberto@debunker.com - Skeptical to the Max!
Visit the Debunker's Domain - http://www.debunker.com
Resources Debunking All Manner of Bogus Claims
Also: Skepticism / Astronomy / Opera / more
On Charles A. Muses (1919-2000) and Hyperspace:
Muses gave at the Parapsychology Research Group in San Francisco, 
December 1,
1986.
     The Parapsychology of Time
     In 1967 quantum physicist and Nobel laureate Eugene P. Wigner 
predicted a convergence of psychology and physics, in his book 
*Symmetries and Reflections*.
Jack commented:
Wigner accepted Von Neumann's and London's notion of collapse of the 
micro-quantum information wave.Bohm and Hiley in The Undivided 
Universe show why collapse is an illusion. It is a muddled idea.
Any attempt to connect consciousness with collapse is a badly 
mistaken idea IMHO even though many of myheroes like John Wheeler have 
advocated it. Great men make great mistakes. The modern theory of 
decoherenceby W. Zurek also shows why collapse is really only a 
superficial description. You can say as if collapse in a
crude sense. There is no collapse in the many worlds approach 
either.
The basic simple insight missed by Wigner, Stapp, Wheeler, Penrose et-al 
is that the living conscious mind is a BIG THING. It's not like the wave 
of a single electron or even a small number of electrons. As P.W. 
Anderson says
More is different.
For Penrose to think that a single graviton has anything to do with 
the creation of a conscious moment is completely absurd IMHO even though 
Penrose is a great mathematical genius no doubt about it.
While the conscious mind is a BIG THING it is not a classical physics 
thing.
Stapp is correct on that score. Also Stapp is correct on making the 
distinction between rocklike things and thoughtlike things that 
correspond to Wheeler's IT and BIT, to Bohm's hidden variable 
and 
pilot wave, to Flesh and Word, to material and mental.
The conscious mind is a MACRO-QUANTUM thing not a micro-quantum thing.
The laws of MACRO-QUANTUM things are not the same as the laws of 
micro-quantum things.
More is different.
The laws of micro-quantum things are
1. A linear nonlocal unitary dynamical equation of motion in 
configuration (or phase) space for the wave (or the Wigner density) when 
there is entanglement.
  I mean this in the sense of second-quantized theory in Fock occupation 
number space where the Hamiltonian is a linear operator in Fock space 
even though configuration space representations look nonlinear as in 
Hartree-Fock theory for example.
2. Signal locality,  i.e. EPR entanglement cannot be used as a 
stand-alone communication channel.  It can be used with classical 
signals for information storage and processing like in quantum 
computing, cryptography and teleportation. This is called passion at a 
distance by Abner Shimony.
Note A: The projection operator formalism, associated with collapse, 
Stern-Gerlach experiment.
Note B: Antony Valentini has clarified passion at a distance as an 
artifact of sub-quantal equilibrium in which uncontrollable quantum 
randomness is viewed, as in Vigier's models, as a sub-quantal Brownian 
motion. In this case signal-locality is only found in the Brownian case, 
not in colored noise situations of sub-quantal non-equilibrium open 
pumped complex systems.  No living system with a conscious mind can be 
closed in maximal randomness at any level classical, quantum or 
sub-quantum. Therefore, signal-nonlocality is a necessary condition 
for life as Brian Josephson first glimpsed in his Biological Use of 
Nonlocality with Fontini Pallikari.
The sub-quantal Brownian noise case is the Born probability distribution 
with signal locality, i.e. all reduced quantum density matrices for the 
parts of entangled wholes are completely random in the sense that they 
do not depend on control parameters of distant parts to which the part 
in question is entangled by the EPR effect.
This is what is entailed by what Stapp calls the statistical 
predictions of orthodox quantum theory.
An instability in a micro-quantum ground state for real on-mass-shell 
quanta, or in the vacuum for virtual off-mass-shell quanta makes a 
relative non-classical entropy lowering phase transition to an emergent 
more complex order via the formation of a boson mode (if initially a 
fermion system) that condenses into a much smaller volume of the 
original phase space. This is a non-classical phase transition from 
micro-quantum stuff to MACRO-QUANTUM STUFF (either real or virtual).
MACRO-QUANTUM THINGS do not obey  the statistical predictions of 
orthodox quantum theory because of what P.W. Anderson calls 
generalized phase rigidity.
The latter in the special case of unstable quantum field theory of 
electrons (at least if not lepto-quarks in general) in globally flat 
space-time leads directly to Andrei Sakharov's metric elasticity for 
the emergence of Einstein's c-number curved space-time 
geometrodynamics along with the observed unified dark energy/matter 
exotic vacuum fields out of the MACRO-QUANTUM PARTIAL COHERING of the 
formerly completely random micro-quantum zero point fluctuations of all 
the quantum fields of the standard model or of its string theory 
generalizations to hyperspace and M-theory.
This is highly scale-dependent. The non-gravitating non-exotic 
MACRO-QUANTUM vacuum is completely coherent with zero residual random 
micro-quantum zero-point fluctuations at the FRW scale. However the new 
precision cosmology shows that only at most 4% of the universe at that 
scale in non-exotic. 96% is exotic - approximately 70% anti-gravitates 
as dark energy and approximately 26% percent gravitates as 
non-hadronic non-radiation non-lepton dark matter.
In general the dynamics of MACRO-QUANTUM THINGS is Landau-Ginzburg not 
Schrodinger-Dirac i.e.
NONLINEAR LOCAL NONUNITARY for a ROBUST ground state (or vacuum) wave 
with generalized Goldstone phase rigidity that completely disobeys the 
projection operator formalism of orthodox micro-quantum measurement theory!
The living conscious mind is such a MACRO-QUANTUM THING with the 
signal-nonlocality that Dick Bierman sees in his presponse data.
The local GIANT MIND WAVE is coupled in two-way relation (Bohm & 
Hiley's term) to the 10^18 hydrophobically caged electron qubits at the 
sub-microtubular level made popular by Stu Hameroff. Max Tegmark has 
completely shot down the Hameroff-Penrose single-graviton collapse 
model not of a single electron qubit inside the cage, but of the 
superposition of the heavy dimer 2-state configurations. Every 
nano-technologist knows that classical mechanics works for the molecular 
configurations - this is the Born-Oppenheimer approximation from me/mp ~ 
10^-3. This boundary has been pushed with atom interference, but I doubt 
you can have that for the dimers?
Subject: prime order subgroup
if H and K are two subgroups of a group G
and the order of H and K are the same: a prime number p
now I wanna prove the intersection of H and K is the identity
can't figure out how to do this
can anyone help me
Subject: Re: prime order subgroup
            Adjunct Assistant Professor at the University of Montana.
Is the intersection of two subgroups a subgroup?
If A and B are subgroups of G, and A is contained in B, is A a
subgroups of B?
What does Lagrange's Theorem say about the relation between the orders
of a subgroup and the group?
What are the divisors of a prime number p?
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
Subject: Re: prime order subgroup
You can't, you may have intersection H/K = H
H and K are cyclic as their order is prime.
What if H / K /= {e} ?
Some j,k in Z with a^j = b^k /= e.
j,p coprime.  Some x,y in Z with xj + yp = 1
a = a^xj a^yp = b^jk a^yp = b^jk
a in K, thus H subgroup K;  likewise K subgroup H
H = K
Thus H/K = {e} or H/K = H
Subject: Re: prime order subgroup
How can you derive this 
a^xj = b^kx 
Subject: Re: prime order subgroup
Suppose that H and K have at least one element in common which is not 
the identity element; say
we call the element  f.  Then, the order of f in H divides the order of 
H, and similarly the order
of f in K divides the order of K.
Now, what are the divisors of the prime p?  Clearly, 1 and p and nothing 
else.
Could the order of f in H be 1 ?  No, because we assumed that f =/= 
identity.  So, the
order of f in H is p.  The same reasoning shows that the order of f in K 
is p.
So, H = K.
assumed.....
So, we get a contradiction.
This shows that our assumption :
`` Suppose that H and K have at least one element in common which is not 
the identity element;  must
be false.    If it is false, we conclude that H and K have precisely one 
element in common, namely
the identity element.  So the intersection of H and K is {e}, where e is 
the identity element of G.
David Bernier
Subject: Re: prime order subgroup
Is it because that every group has prime order must be a cyclic group.
so that H=K
Subject: Re: prime order subgroup
It is true that every group of prime order is cyclic.  The group of 
integers modulo 10 with addition is
cyclic, since the congruence class [1] generates Z/(10*Z); but the 
element [5] in Z/(10*Z) generates
a subgroup of Z/(10*Z) with only two elements:  [0] and [5].  So how do 
we know that
In more detail:
order of a subgroup of K divides the
K, then we must have
= K.
(The only divisors of a prime p are 1 and p. If a divisor is not 1, then 
it must be p...)
David Bernier
Subject: Re: prime order subgroup
[...]
I think there's a proof which is simpler,   using:
(a) the intersection of two subgroups H, K is a subgroup of H, and of K 
also
(b) the order of a subgroup of H (resp. K) divides the order of H (resp. K)
(c) any prime p has as divisors 1 and p and nothing else....
You could use as a maxim:  what theorem(s) could I use today?
David Bernier
Subject: Re: Magidin is too many
you've been presenting some argument about logic,
which I'm not really into, but, seeing as predicate logic or
boolean algebra or just higher-order syllogisms map so well
to ordinary arithmetic, why don't you just buckle-up and
do it -- prove that JSH has found some core error?
http://members.tripod.com/~american_almanac/inverse.htm
thus quoth:
Had any second leading Democratic Presidential candidate, or former
President Bill Clinton, associated himself with me in my fight to
bring about the defeat of recall in California, the Democratic Party
would have carried the state. In effect, when faced with a brutally
nasty adversary, George Shultz's muscle-bound Hitler, Arnie the
Beast-man Schwarzenegger, the Democratic National Committee Chairman
reacted like a scared rabbit, and, figuratively, froze and died a
political death, on that spot.
 
--les ducs d'Enron!
http://members.tripod.com/~american_almanac/inverse.htm
Subject: Re: Mind, Brain, Consciousness, Math & Physics
yo.  whoah, on the first step excerpted, below;
what is the justification for this?
 
 
 
 
--les ducs d'Enron!
http://members.tripod.com/~american_almanac/inverse.htm
Subject: any good curve fitting model/data analysis software that can extend 

data point?
I have big trouble on this curve fitting for a few days.
This is related to signal processing. I have a bunch of data, y, x, and w,
where w is a 5-element vector. I want to find the relationship between y, 
x,
and w. Here is how I obtained these data.
First for image A:
I measured w to be:
1.75E+04
26.395
2.5283
0.75389
0.47193
Under this w, I sample 7 data points for x and y:
x=
6.3795
1.8693
0.9923
0.6455
0.4612
0.3494
0.2753
y=
0.068
-0.0131
-0.0301
-0.0449
-0.0504
-0.0506
-0.0578
For image B,
w=
1.61E+04
54.845
41.778
9.7303
7.8343
I sampled x and y to be the following:
x=
7.5179
2.1687
1.0814
0.6556
0.4505
0.3321
0.2586
y=
0.0521
-0.0119
-0.0387
-0.05
-0.0484
-0.0459
-0.0508
For image C:
W=
1.46E+04
250.53
128.03
50.119
23.137
x=
7.6963
2.0372
1.0318
0.6505
0.4576
0.3373
0.2606
y=
-0.1174
-0.1379
-0.1191
-0.11
-0.0992
-0.0889
-0.0836
For image D:
W=
6.98E+03
40.405
28.292
18.178
10.712
x=
8.9289
3.8078
2.2907
1.5374
1.0899
0.8161
0.6354
y=
0.0715
0.0176
-0.0078
-0.0225
-0.035
-0.043
-0.0496
For image E:
W=
1.84E+04
145.86
96.09
33.04
26.242
x=
8.5197
2.9649
1.6112
1.0238
0.709
0.5226
0.3975
y=
0.0611
0.0025
-0.0262
-0.0365
-0.0479
-0.0521
-0.0557
----------------------------------------------------
I did a order-7 multinomial fit, with 6 variables(x, w1, w2, w3, w4, w5),
i.e., fit by multinomial... then use the curve to predict the y for the
following test image:
W=
1.15E+04
117.95
108.05
85.947
70.185
x=
10.8565
4.9287
2.9304
1.9591
1.4218
1.0899
0.862
y should be:
0.0193
-0.025
-0.0336
-0.0426
-0.0441
-0.042
-0.0428
--------------------------------------------------------
But the result is way-off with over 100% error. I guess the problem is that
poly-nomial fit using the least square method(the  in matlab) is good 
at
interpolate data, but not good at predict extended data. The test 
image
has a x=10.8565, which is out of the range of the training data. Hence the
predictor cannot correctly predict it.
I am quite dispointed at this. There are two things I should do I guess,
1.)obtain more training set and do more training; 2) use a better fit 
model.
A model that can extend... or a model can deal with little training
information. Maybe neural network?
Anybody give more ideas/suggestions/comments?
-Walala
Subject: Re: any good curve fitting model/data analysis software that can 
extend data point?
I'm still having trouble even understanding how many variables,
dependent and independent, that you have.
Does this mean that you fixed the w-vector, then varied x
and measured one value of y(x,w1,w2,w3,w4,w5) for each x?
You have either 5 data points (your 5 images) or something
more. I'm not sure.
Again, is this ONE test point or a bunch of test points?
ALL models have that property, but especially polynomial fits.
It does? Then what does this mean:
This seems to say there are 7 x-values associated with the
image, or that x is arbitrary and you choose 7 test values.
No, any model given limited input data is going to be
statistically invalid when applied to data outside the training
range.
First let's see if we can clarify what your model is, how many
variables you have, and how many data points you have. I'm really
confused.
             - Randy
Subject: Re: any good curve fitting model/data analysis software that can 
extend data point?
y=f(x, w1, w2, w3, w4, w5), w is a parameter vector with 5 elements.
Yes, fix w1, w2, w3, w4, w5 first and change x and get 7 x-y pairs.
Currently 5 images. But I am going to generate more. Because some people
told me my effort is useless and the data is not predictable, so I hesitate
to generate more. It took me one night to get 5 images done... I am slow...
The above x are 7 samples sampled at a fixed w=(w1, w2, w3, w4, w5)
I agree.
The above is the 7 sample points of a new image for testing, i.e., I have a
w=(w1, w2, w3, w4, w5) for this test image. And then I use the same method
to sample 7 data points, these are x's. Then I try to use the above model 
to
predict y(7 data points).
Then the method is dead. Because nobody can guarantee that real data is
always within the test range... what can I do? just throw this away?
-Walala
Subject: Re: Evil Western people stole our numbers
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9MEYKp17714;
I would bet 10 to 1 that OBLj has no affiliation with 
Muslims whatsoever- but just thinks saying stupid things
in their name makes them somehow look stupid...which to me 
is obviously the authors primary goal (and whose real first
name is probably Jack or Bob...).
  
Subject: Re: Simple principle in core error proof
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9MHUPg30018;
  You didn't provide an answer to my implied question.  
Do you conclude here that a_1 or a_2, or both, are divisible
by 5?
  Andrzej
Subject: Re: Simple principle in core error proof
JSH does not answer questions.
Gib
Subject: Re: Simple principle in core error proof
Well, sometimes he does; it just comes with an insult as a bonus.
David Moran
Subject: Re: Simple principle in core error proof
No.
Subject: Re: JSH: So much fun
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9N3wAY08507;
  Hey - I get it - you're writing *** poetry ! ***
  Nora B.
Subject: A monotone function.
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9N6Dpd17274;
Consider polynomial P_n(x)=a[1]*x+a[2]*x^2+...+a[n]*x^n with absolute
term a[0]=0, i.e. P_n(0)=0.
Suppose P_n(x) admits the following properties:
3. P_n(1)=1, i.e. a(1)+...+a(n)=1.
4. Coefficients a[1],...,a[n] takes values in R^n cube:
[-1,1]x...x[-1,1].
Suppose now that coefficients a[1],...,a[n] are uniformly distributed
on set, which is deteminated from properties (2)-(4).
Question:
1. What is mathematical expectation of random vector a[1],...,a[n]?
of some known function?
Subject: Re: pow(x,y) approximation
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) id h9N5OAw14271;
antilog2(Y*log2(x))
The best strategy is to use a base of two for the log and antilog.
Also, such a function should test for a variety of special cases to
calculate with other methods that are faster and/or apply over a domain
that other methods don't cover adequately.  For example, test whether
a power of two is being raised to a positive integer power.  Only the
exponent needs to be adjusted for that case.  There are a number of
other special cases.
In the general case, a scaling step is needed for the log and antilog
to bring it within a standard range, such as 1 to 2 for log or 0 to 1
for antilog.  The reason for using base two is so that this scaling
step can be done without loss of acuracy.
To make the general case computation faster, chop up the interval from
1 to 2 into small pieces and use formulas optimized for each of the
small pieces.  An initial table lookup can send the program to the
right formula and/or parameters.
The natural log and antilog functions can also make use of these same
base-two subroutines by using a scaling factor.
Subject: Re: elliptic integral of first kind
Indeed,  
This follows from
The above formula also implies, that your expression for e(r) will
not give [2] (btw, do you really call the potential differentiated
wrt r the electric field? For me, the electric field is the gradient
of the potential, so the electric field in this case should be
the potential differentiated wrt r times the gradient of r.)
What I get when [1] is differentiated wrt r is
1/(pi*r^2) ((a^2+2*r^2)*E[z^2]-(2*a^2+r^2)*K[z^2]).
HTH,
Michael.
-- 
&&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
Dr. Michael Ulm
FB Mathematik, Universitaet Rostock
michael.ulm@mathematik.uni-rostock.de
Subject: Gears and sprockets
I've been trying to obtain some info on the geometry of gears and gear
teeth, mainly because I need to do some CAD plotting but also because they
are extremely nifty objects. I'm struggling because I don't know the 
correct
terminology so my Googling comes up empty.
It there a proper mathematical term for these objects? Does anyone know
where I might go to find the formulae that underpin the sprockets in their
various ratios?
Gratefully, Manning
Subject: Re: Gears and sprockets
You need to do a search on the word involute - this is the name
given to the curve that is used for the gear tooth shape. A good
reference on gears is Fundamentals of Machine Elements by B.J.
Hamrock, B. Jacobson & S.R. Schmin, published by McGraw-Hill
Ian Taylor
Subject: Re: Gears and sprockets
Egads - you can achieve so much when you know the correct term.
little bit of maths-engineering.
http://easyweb.easynet.co.uk/~chrish/geardata.htm
Manning
they
correct
their
Subject: Conjecture related to simple arithmetic
I've stumbled on a little fact or (rather it seems to be factual for
small k).  Although it seems clumsy to state (I've probably overlooked
a more direct statement) it seems fundamental and has probably been
observed (and proven) before.
First I'll try to state it in math terms, then give a more intuitive
statement.
Let N_k = {1,2,3,4,...,k}
Let P(N_k) = {A | A subset N_k}
Let P(P(N_k)) = {D | D subset P(N_k)}
Let c(A) = {x | x in N_k and x not in A}
Define a partial ordering on P(N_k) by
Say D in P(P(N_k)) is happy  iff
        For each A in P(N_k), either A in D, or c(A) in D, but not both;
        and For each A in D, A <= B implies B in D
(Check our work: I find 1, 1, 2, 3, 7, 21, 135, 2470 distinct
happy sets when k = 1, 2, 3, 4, 5, 6, 7, 8 respectively.)
Now, consider a sequence of weights
        0 <= w(1) <= w(2) <= ... <= w(k)
and define w(A) = sum {w(a) | a in A}
Define w() as cheerful iff no subset of the weights
sum to exactly w(N_k)/2, i.e. w(A) = w(c(A)) never occurs. 
Remark: If w() is cheerful then F(w) is happy.
Prove (or find counterexample):
For every happy set D, there is cheerful w() with D = F(w)
My computer program has shown this exhaustively for k <= 7, I think.
 - - - - - - - - - - - - - - - - - - - -
Problem rephrased in English:
In a certain game, each of k tokens will be captured by exactly
one of two players.  Each token has a weight; the player whose
tokens' weights sum to the larger total is the winner.  The
details of the game do not concern us; the puzzle is simply
How many distinct ways are there to assign weights to tokens?
(First fix k.  k < 5 is easy; k = 5 is good for hand solution;
k = 6 is for computer programmers.)
For example, when k=3, the bundle of weights {2,3,4} is equivalent
to {1,1,1} -- in either case a player simply needs to capture
a majority of the tokens.  The bundle {2,3,5} will be outlawed --
see Restriction 2 below.  The bundle {2,3,6} is equivalent to {0,0,1},
and these two cases, {1,1,1} and {0,0,1}, seem to be
the only distinct ways to assign weights to 3 tokens.
Let's impose the following restrictions:
Restriction 1:  Token weights must be positive numbers.
Restriction 2:  Ties must be impossible; i.e. no subset of the weights
can sum to precisely half the total.
There is an obvious approach using a partial ordering among the
subsets of N_k to find possible classes of such weights.
Does that approach yield any possible solutions which cannot be
realized by any assignment of weights?
James D. Allen
Subject: biparametric representation of odds (collatz significance)
In studying the Collatz conjecture I came across an interesting
representation, possibly already documented but I haven't seen it.
The odds are the base from which the 3x+1 step is taken. The result
of this operation is obviously even and here's where the
representation has significance. To what power of 2 must these even
numbers be divided in order to become odd again and the process
continue? This area of study arose when looking at the form of x after
n iterations (an iteration is either a 3x+1 or division by greatest
power of 2 to keep whole and odd) of the collatz process. It can be
shown that after n iterations:
x = 3^(n-1) + 3^(n-2)*2^(r1) +...+ 2^(r1+r2+...+r(n-1))
    ---------------------------------------------------
                    2^(r1+r2+...+rn) - 3^n
where x positive odd, n positive integer, r1,r2,...,rn are the
greatest powers 2 to divide after each 3x+1 step.
Now for small n its easy to show no such x exists, but ultimately
we're looking to prove for all n. Induction seems obvious but leads
nowhere on the surface. So I looked into the behaviour of the sequence
This [mess] represents every odd, each represented by a unique ordered
pair (x,k):
               ( 2^k * 5^((k+1) mod 2) + 1)
x * 2^(k+1) -  ----------------------------
                            3
The uniqueness and inclusivity of all odds is easily proven, but for
convenience and to keep the message size low I will not include it
(unless I hear disbelief in the crowd :).
So the above representation generates an infinite series of series for
all k.
I.e. for a given k, a series is generated.
E.g. k=1 generates the series 4x-1. So for all x an odd number is
produced.
The significance is that for a given series (i.e. a given k), all odds
produced will be divided by 2^k after the 3x+1 step.
So for k=1, the odds generated by the series 4x-1 will all be divided
by 2. This is easily shown:
the odd becomes: 3( 4x-1 ) + 1 = 12x - 2
Therefore for all x, this is only divided by 2, becoming 6x-1 which is
odd for all x.
So any given odd falls into only one series (i.e. has only one x and
one k that could represent it), and therefore it is known how many
times it is to be divided by 2 after the 3x+1 step. This may bring
some order to the chaos of the sum r1+r2+...+rn, I haven't had time to
do much more.
Let me know what you think,
Davin Lunz.
Subject: Business&Games Calculator - ideal for profitable work and  pleasant 

rest
                     New release from QUDATA.COM
B&G Calculator combines ease of use and tiny size with a lot of powerful 
features, memory, history, statistical functions, conversion tools, 
numerous skins. You can even play games on this calculator! 
The B&G Calculator has a flexible interface that can be easily adjusted 
to your taste. It operates on various types of calculations.
* easily adjusted form
* you can work through your PC keyboard
* convenient work with memory
* can be used as the reference book on different constants
* various skins
* play an enthralling game if you are tired
Read more about these features:
- Form that can be easily adjusted. You can work either only with the narrow 

strip, or you can activate different calculator panels. 
- You can enter numbers and functions not only with the graphic 'keyboard' 
pad, but also from the PC keyboard using the calculator line of input.
 
- Working with the memory in QuData calculator is very convenient. One 
can use a set of cells (variables) for storage of intermediate results. 
You can see the current values of all memory cells on the panel Memory.
 
- The list of calculation history allows you to return to the earlier 
calculated expressions. The history can be printed or saved in the HTML 
file.
 
- High-capacity set of mathematical, statistical, special and financial 
functions allows you to carry out diversified calculations. QuData 
calculator also functions as the reference book on different mathematical 
and physical constants. You can always convert pounds into kilos, insert 
the distance between the Earth and the Sun into your calculations and 
enter your own constants easily and without any problem. 
- Skins allow you to change calculator's appearance. For each skin you 
can choose font, size and type. It is also possible to set the mode of 
random skin, and then calculator will please you every day with new design.
 
- In case you are tired and need a rest, you can play the most enthralling 
games Bubble Shooter and Gem Slider that are strongly recommended by 
psychologists as the best way to relax. Bubble Shooter is placed in one 
of the calculator panels.
                         
In addition to the aforesaid you are provided with:
- timely technical support by e-mail
- friendly interface
- all new versions are free for you if you are registered
Evaluation Copy Available for free download: 
http://qudata.com/calculator/bgcalculator.exe
Limitations: expires after 10 uses
Registration cost is US$ 25
A full version for reviewers is available upon request
More information: info@qudata.com
Subject: Re: finite groups
I'm really impressed with your esoteric knowledge!
But as a matter of interest,
is there any general method of determining whether a subgroup
of one of these groups is maximal?
-- 
Timothy Murphy  
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Subject: Re: finite groups
subgroups of finite groups has been one of my primary research projects
for the past three years or so. This problem can be reduced to finding
the maximals of the almost simple groups, which are those groups G
with S <= G <= Aut(S) for a nonabelian simple group S, so I have
managed to acquire a reasonable working knowledge of the area!
The code that we are developing works by family - currently we can handle
families like PSp(4,p), PSU(3,q), PSU(4,q), PSL(n,q) for n<=5, plus all
groups up to some order.
For the general families of groups there are lengthy papers and books 
around,
for example by Kleidman and Liebeck, which classify maximal subgroups.
After the classification of finite simple groups, finding their maximal
subgroups was one of the most natural problems.
Derek Holt.
Subject: Re: finite groups
Is this problem not yet completely solved?
Anvita
Subject: Re: finite groups
[snip much useful and interesting info]
buckle down and learn GAP.  One common thread in your responses
was lists of the *maximal* primitive subgroups of A_n, whereas
all I can easily get from Dixon & Mortimer are *all* of the
primitive (not necessarily maximal) subgroups of S_n (not A_n).
I assume there must be a way in GAP to pull out this more
detailed info.
-- 
Jim Heckman
Subject: Re: Andrica's Conjecture
The proof is very difficult to read, so I had a hard time trying to
figure out what you were saying.  However, I'm saying that you need
ALL to make the next step of the argument valid, whereas you're only
showing ANY.  Just because you can assert that you can't use an
ascending list of prime factors doesn't really mean anything, and the
argument (at least, what I think is your argument; you really left a
whole lot out) doesn't follow from the any case.
       If you really think you have a proof here, why not write it up
formally?  After all, anyone who actually came up with a valid proof
of this would become famous in mathematical circles.  More
importantly, it'd be *much* easier to discuss.
Subject: Re: How much longer must physics put up with F=ma?
What's happened to Donnie?
Double-A
Subject: Re: How much longer must physics put up with F=ma?
He discovered that his garden was infested with skugs.
--
Mensanator
Ace of Clubs
Subject: Re: Formula
product(k=0..n) 2k-1
product(k=0..n) 2k-1 = -product(k=1..n) 2k-1
        = -(2n-1)! / (2^(n-1))(n-1)!
product(k=0..n) 2k+1 = (2n+1)! / (2^n)n!
Subject: Polygon barycenters
Given is a polygon as ordered list of coordinates
(x1,y1),(x2,y2),... (no convexity required).
You can compute the barycenter of
a) the vertices
b) the sides
c) the area.
(b doesn't concide with a or c already for a triangle)
Since I'm too lazy to evaluate the integral definitions
and surely someone has done this already - can anyone
give a formula as F[sum(xi**2 or whatever)]?
-- 
Hauke Reddmann <:-EX8    
Private email:fc3a501@math.uni-hamburg.de
For our chemistry workgroup,remove math from the address
For spamming, remove anything else
Subject: Re: Polygon barycenters
xm = Sum(x_i, i, 1, n)/n
ym = Sum(y_i, i, 1, n)/n
xm = Sum(Sqrt((x_(i+1) - x_i)^2 + (y_(i+1) - y_i)^2)(x_(i+1) + x_i)/2, i, 
1,
n)/
         Sum(Sqrt((x_(i+1) - x_i)^2 + (y_(i+1) - y_i)^2), i, 1, n)
Where  x_(n+1) = x_1  and y_(n+1) = y_1
Similarly for y_m
I think that it is a more difficult matter ... The easy method to do it
manually, is to descompose the polygon in disjoint triangles and do as for
the sides. But I don't know how program it.
-- 
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
Subject: An interesting statistics problem
Hi folks,
We came across an interesting statistics problem at work and I was
wondering if anyone knew how to solve it:
You have an unfair coin to flip.  This coin comes up heads 51% of
the time.  Assuming that you always bet on heads, how many times would
you have to flip this coin in order to be guaranteed with 99%
confidence that you will make money?
I appreciate any responses.
Ellis Horowitz
Subject: Re: An interesting statistics problem
Set it up as a normal approximation to the binomial distribution,
and find the smallest n such that the 99% confidence interval lies entirely
on the far side of even money.
Doug
Subject: Re: An interesting statistics problem
This is not particularly difficult. Assuming that you play even odds,
you will make money whenever the sample proportion p of wins is
greater than 0.5. Now p has mean 0.51 and variance 0.51*0.49/n,
to take you a lot of time before you can take advantage of that
unbalanced penny!
Subject: Easy middle school problem that I can't get?
A company has 350 employees; women get a $10 bonus, men get $8.15 bonus;
some of the women refuse the bonus; the total amount of bonus is 
independent
of the number of women in the group; what's the total bonus amount and how
much did the men get?
This is how the problem was stated.  The way I see it, there is not enough
information to solve the problem.  Because if the total amount of bonus is
independent of the # of women, you might as well set the # of women at 0,
and so the total bonus amount of men is 350 *10.  But if you set the women
at 50, then the total bonus of men is 300*10.  Am I missing something?
John
Subject: Re: Easy middle school problem that I can't get?
Can't you give an answer in terms of the number of women?
Jeff Kish
Subject: Re: Easy middle school problem that I can't get?
I think the answers are $2852.50 and $1222.50, respectively.
I'm imagining m=men on the vertical axis and t=women takers of the bonus
on the horizontal.  Then we're concerned with the lattice points in the
triangle in the first quadrant bounded by the line from (0,350) to 
(350,0).
Every point represents a possible number of m+t<=350.  
Now there's a diophantine equation 815m+1000t=B, where B is the unknown
bonus.  Dividing by 5 gives 163m+200t=B/5.  (Hmmmm....163.....fancy 
that.)
This is a line that hits darned few lattice points inside the triangle
regardless of which B is chosen.  If we sketch a few representative
lines, we see that it has negative slope, steeper than the hypoteneuse
of the triangle.  One line passes through the top vertex of the triangle
and intersects the horizontal axis.  Call this one L.
If zero women work at the factory, then all the men must take the bonus.
The only one of our lines that allows for this possibility is line L,
so this must be the right value of B which allows for independence from
the number of women.  Since the m-intercept is both B/815 and 350, we
compute B to be 285250 cents.
Well, the problem indicates that there are some women. There are only
two nonnegative lattice points on L, (0,350) and (163,150).  So if
we take the latter, we compute that the men get 150*8.15=$1222.50.
(And certainly we've made a couple of guesses here.)
Bart
Subject: Re: Easy middle school problem that I can't get?
There does not appear to be sufficient information.
x = #men
y = # women accepting bonus
z = #women rejecting bonus
B$ = amount of bonus
Hence
350 = x + y + z
B$ = 8.15x + 10y  (... + 0z)
As given, you have four variables and only 2 equations. The bonus amount 
can
be anywhere from 0 (where z = 350) to $3500 (where y = 350).
independent
Subject: Re: Easy middle school problem that I can't get?
can
how
enough
is
0,
women
Could the answer to the problem be that the bonus is 0?  Therefore, there
are 0 men and all of the women refuse.  Rememberr that the total bonus is
said to be independent of the # of women.
Subject: Re: Easy middle school problem that I can't get?
Well as the total number of women = y + z, the only possible answer is
That is certainly one of a multitude of possible answers. $2852.50 is
another (350 men @ $8.15 each), etc.
My suspicion, given that this is a middle school question, is that the
teacher meant to include the bonus amount. If (as an example) the question
said the bonus amount was $1815.00, then the answer involves basic 
algebraic
manipulation to come up with x = 100, y = 100 and z = 50. That would seem a
reasonable question to ask middle school students.
Without a specified bonus amount, there are any number of answers possible.
The statement that the total bonus being paid is 'independent' of the total
number of women is so vague as to not really mean anything at all as far as
I can see.
Manning
Subject: this is a simple complex question....
my question is....
A complex fuction is differentiable at one point(p)
and not differentiable all other point except point(p).
---------------------------------
it is true??
i need your hot-advice.
thank you very much......
Subject: Re: this is a simple complex question....
In precise, f has derivatives of all order if f is analytic on a region.
(It is given by the Analyticity of Derivatives)
Also, f is differentiable at one point only does not
mean f is analytic at one point.
Furthermore, if f is differentiable on a region, then
we can say that f is analytic on the region. Thus
we can say that f is analytic at every point in the region.
Michael Leung
:bn8jju$ejm$1@news.hananet.net...
Subject: Re: this is a simple complex question....
No. Of course not - by _definition_, if f has a second derivative
at p it must have first derivative in a neighborhood of p.
************************
David C. Ullrich
Subject: Re: Formulae for Latin squares of size 2^n
I just realized that both formulae given in the original
post can be subsumed under:
    u(i,j) = 2*m*i*j + i + j   mod 2^n
with an arbitrary m.
M. K. Shen
Subject: How to prove?
Subject: Re: How to prove?
     You've had various correct answers, but to me it seems simpler just
to transfer everything to one side and factorize.
     (x^2 + x) - (y^2 + y) = 0
so   (x^2 - y^2) + (x - y) = 0
so   (x - y)(x + y + 1) = 0.
     Since x and y are non-negative the second factor can't be 0, so the
first one must be.
          Kebn Pledger.
Subject: Re: How to prove?
First suggestion:
Rearrange; complete the square, put into standard form for the relation of 
this
conic section, interpret values with standard knowledge of this conic 
section
relation.  
I suspect this is intermediate algebra in level.  
G   C
Subject: Re: How to prove?
Kosken vaikka en saisi.
-- 
Subject: Re: How to prove?
I don't know if you meaningfully can *prove* this - you are simply
rephrasing one of Euclid's axioms, 'that which are equal to the same are
equal to each other'. It's true for all values of x and y, including 
complex
values.
Bertrand Russell probably spent 322 pages proving this in the Principia
Mathematica, which would be worth researching if you have no friends or
hobbies.
Manning
Subject: Re: How to prove?
I hesitate to paraphrase Euclid, but I think his axioms means:
For all x, y, z,  if  x = z  and  y = z  then  x = y.
Which seems not to help in this case.
Otoh, he probably didn't.
-- 
G.C.
Subject: Re: How to prove?
Say what?
I can't tell if this is a case of a) Me dramatically misunderstanding 
something (I'm pretty sure it's not), b) You misreading the OP, or c) a 
joke...
Anyway. Onto the original question.
It is not, as Manning said, true for all numbers; it's not even true for 
all real numbers.
There are two easy ways to answer the question. First of all if you 
always positive. This means that the function is strictly increasing, so 
tiny bit more work)
Alternatively you can rewrite it as f(x) = (x + 1/2)^2 - 1/4 by 
completing the square. Then see what happens if you have f(x)=f(y).
David
(E-mail address spamblocked in the obvious way)
Subject: Re: How to prove?
You need to differentiate it to see that it's increasing ?!
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
 Alan Partridge, _Bouncing Back_ (14 times)
Subject: subsequences
I don't know if I'm asking a question that is out of my league, but here 
goes.  I'm studying subsequences and it seems to me that if one is able 
to cover the entire sequence with a number of converging 
subsequences(that is, every element of the sequence is contained in one 
(or more than one?) of the subsequences), then finding the limits of 
these converging subsequences is the same as finding all the 
subsequential limits of the sequence.
I think if the number of converging subsequences is finite, I should be 
able to prove that myself.  If the converging subsequences are infinite 
(presumably countably infinite) I might be able to work this out if the 
limit points of the subsequences do not group themselves into one or 
more cluster points.  But some sequences will have subsequential limits 
that group around cluster points.  For example 1, 1, 1/2, 1, 1/2, 1/3, 
1, 1/2, 1/3, 1/4 ... has subsequential limits grouping densely around 0 
and zero itself should be a subsequential limit point.  I don't know how 
to think about this last situation.
Tom Adams
Subject: Re: subsequences
Yes.
Actually there can be uncountably many convergent subsequences,
all with different limits. (See one of the examples in Ross: If (r_n)
is a sequence containing every rational then _every_ real number
is a subsequential limit!)
Yes.
Not sure what the question is...
It's a fact that there's no simple procedure someone can give
you which will allow you to identify all the subsequential limits
of any sequence whatever. Sounds like maybe you're asking
for someone to explain that method - it doesn't exist.
************************
David C. Ullrich
Subject: Re: subsequences
Right, I did not ask a question.
I am assuming that somehow I have succeeded in finding a number of 
converging subsequences that cover every element of the sequence.  I 
believe you are saying that I can now claim that I have thereby 
succeeded in finding every subsequential limit of the sequence.  I have 
trouble getting started on a proof for this proposition because some 
sequences will have subsequential limits that group around cluster 
points.  For example 1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4 ...
Please remember that it is difficult for beginners such as myself to 
comprehend a world with an infinite number of things in play.
I would be happy with a hint or a reference to a helpful text.  Is Ross 
such a text?
I appreciate that you have even helped me this much.
Tom Adams
Subject: Re: subsequences
If I understand what you mean by this then yes.
That sequence has infinitely many subsequential limits. But
it's also not the union of finitely many convergent subsequences.
For exactly this question, no. I assumed for some reason that
Ross was the text you were using.
Say (s_n) is a sequence. Say we have finitely many convergent
subsequences that cover the sequence. So we have sets of
integers A_1, ... A_N, such that the union of the A_j is all the
natural numbers, and such that if we restrict (s_n) to the
values of n in A_n the resulting subsequence converges.
be an element of A_j, is L_j.
holds: |s_n - L_1| < epsilon, ... or |s_n - L_N| < epsilon.
Then once you show that, try to figure out why it implies
that the only possible subsequential limits are L_1, ... L_N.
************************
David C. Ullrich
Subject: Re: subsequences
This was how I approached this problem at the outset, but I think it 
only works for finite number of subsequences.  I believe this shows that 
we can infer that we have gotten all limit points of a sequence if we 
identify converging subsequences that use every element of the sequence. 
The limit points of the sequence are then the same as the limits of the 
covering subsequences. I do not assume these converging subsequences are 
disjoint
A. Assume every element of a sequence a_n is contained in at least one 
of a finite (infinite?) number of sequences b_1_n, b_2_n, . . . .
B. Assume further that every subsequence b_k_n converges to a limit L_k. 
Call the set of all these limits S.
Now set up the straw man.
C. Assume another subsequence c_n exists and it converges to a limit L 
different than any limit in S.
Define the set T = {|x - L| | x element of S} and select 4 eps < inf T. 
(inf T exists because T has a lower bound, 0)
true (apparently not true if the subsequences are infinite in number 
because inf T can be 0, and therefore 4 eps can be zero):
| c_n - L | < eps
| b_k_n - L_k | < eps
| c_n - L | + | -b_k + L_k | < 2 eps
| c_n - L - b_k_n + L_k | < 2 eps
But c_n1 = b_k_n2 for some n1 and n2 greater than n*
in which case
| L_k - L | < 2 eps
and we have a contradiction of the form
or its logical equivalent
Hopefully this is correct for finite numbers of subsequences.  But what 
I still don't see is how this be extended to infinite subsequences.
Subject: Re: subsequences
            Adjunct Assistant Professor at the University of Montana.
b_k_i converge ever more slowly?
So you need either some sort of uniformity condition on the b_i_n (so
rate), or only a finite number of values of k, so you can find a
common bound for all.
You can extend if you assume that  L is not a limit point of S (so
the subsequences. But the latter may be false.
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
Subject: Re: subsequences
Don't take my last post into account, I misread the question.
--
JS
Subject: Re: subsequences
That's wrong, and you can find a counterexample easily. (taking u_n =
(-1)^n)
Subject: Re: An Ownership Puzzle
rules.
will
to
twelve
half
can
the
Subject: Re: An Ownership Puzzle
This seemst to be fairly similar to Zeno's Paradox. Zeno's Paradox
states that if you travel from point A to point B by travelling
_exactly_ half of the remaining distance, you will never reach point
B, unless you have an infinite amount of time. It seems to me that,
assuming perfect mathematical time, that at _precisely_ 1:00, the gold
brick would already be travelling at an infinite speed. Also, if you
halve the amount of time remaining, you are doubling the speed of the
gold brick every time it is passed from one person to another. If the
brick is travelling at an infinite speed, then that seems to me at
least that both Mary, Peter and Paul would all have the gold brick at
_exactly_ 1:00.
I hope that this helps!
Anthony
Subject: Re: An Ownership Puzzle
Some guy named Zeno walks in at the last minute and steals the gold bar.
-- 
A.
Subject: Re: An Ownership Puzzle
coming from Artful Dodger!
and Tralfaz (traffic or traffic~false) talking about broken maps......
why does noone believe my theory on numerology, people around *me*
tend to reply insync with their name, ateast 10,000 words trying to
explain this simple simple empirical concept.  meanwhile everyone laughs
because you won't believe the government spy on me because of it
so thinking you're on the TRUEman show is a complex now.
100,000 witnesses you can't all hide forever. sit a 2 minute match the name
test and prove numerology, 200 requests for over a year.
Herc
Subject: Re: An Ownership Puzzle
Nobody.  They will be passing the bar back and forth so fast that when in
the end they are passing the bar in an infinitesmial time, they'll all be
passing the bar at, within an infinitesmial, the same time.
Subject: Re: An Ownership Puzzle
line),
has to
agreed to
be
converge.
Great try William but there will never be a time when the gold bar is being
passed around infinitely frequently.  The frequency with which the gold bar
is being passed around increases without bound as one approaches one 
o'clock
but at one o'clock the frequency with which it is being passed around is
zero.  So the frequency is always finite, kind of like the function f such
that f(x)=0 when x=0 and f(x)=1/x for all x different from zero.  The
function f(x) never takes on an infinite value but, for arbitrarily small
The non-existence of the limit of p(t) as t tends to one o'clock just means
that taking generalized limits is not the way to find out where the gold 
bar
is at one o'clock.  A function can have a value at a point without that
value being a limit point.  For example, consider the function f such that
f(x)=1 if x is a negative rational, f(x)=2 if x is a negative irrational,
and f(0)=3.  Thus, the function p(t) has no left-handed limit as t
approaches one o'clock but the right-handed limit of p(t), as t approaches
one o'clock, must equal the final owner of the gold bar.  So the question
remains: who will be the final owner of the gold bar?
Randy
http://www.rlgerl.com
Subject: Re: An Ownership Puzzle
As you've already been informed, Zeno took the bar at t = 1 and
he has had it every since.  Thus the right hand limit of p(t) = Zeno
Further more it makes no difference if he had it at t = 1, that he
Subject: Re: An Ownership Puzzle
bar?
in
be
approaches
question
Zeno is not the final owner of the gold bar nor did he take it at t=1.
Randy
http://www.rlgerl.com
Subject: Re: treatment of Lebesgue integral
Here's a proof for skeptics that the functions that are integrable by
this definition are the same as the Lebesgue integrable functions.
Let's agree that a step function is a finite sum of characteristic 
functions of bounded intervals, where the intervals can be open,
closed, or half-open - we've seen elsewhere that this really 
doesn't matter much, we could do the same thing with a more
restricted definition but it would be more complicated.
Say a function f is an M-function if there is a sequence of step
functions (f_n) such that (a) and (b) holds; in this case say
(f_n) gives f. [Note that if (f_n) and (g_n) both give f then
sum int f_n and sum int g_n both equal the Lebesgue integral;
hence the M-integral is well-defined.]
a sequence (f_n) which gives f, such that the sum of int|f_n|
we can get sum int|f_n| = int f. Let chi(S) be the characteristic
function of S.
and sum int f_n < infinity then sum f_n is a good M-function.
Pf: Let (f_n,k)_k give f_n, with 
     sum_k int |f_n,k| < eps/2^n + int f_n.
Note that if sum_n,k |f_n,k(x)| < infinity then sum_k |f_n,k(x)|
is finite for each n... QED.
Lemma 2: If O is an open set of finite measure then chi(O)
is a perfect M-function.
Pf: Well, duh. QED.
Lemma 3: If A is a measurable set of finite measure then
chi(A) is a good M-function.
Pf: Let K_n be an increasing sequence of compact sets,
each contained in A, and O_n a decreasing sequence of
open sets, each containing A, such that 
           sum m(O_nK_n) < eps.
Consider the sum
(*)    chi(O_1) - chi(O_1K_1)
     + chi(O_2K_1) - chi(O_2K_2)
    + chi(O_3K_2) - chi(O_3K_3)
   + ... .
Write each of those open sets (O_1, O_jK_k) as
a disjoint union of open intervals, and let (h_j) be
an enumeration of plus-or-minus the characteristic
function of each of the intervals you get. Then (h_j)
gives chi(A), with waste less than eps, or 2*eps
or something:
First, if x is in the union of the K_n then the sum in
(*) equals 1 = chi(A)(x). For example, if x is in K_1 then 
the sum is (1 - 0) + (0 - 0) + (0 - 0) + ... = 1, while
if x is in K_2K_1 then the sum is 
(1 - 1) + (1 - 0) + (0 - 0) + ... = 1.
Similarly, if x is not in the intersection of the O_n then
the sum is 0 = ch(A)(x). For example, if x is not in O_1
then all the terms vanish, while if x is in O_1 but not
in O_2 then (noting that x cannot lie in any K_n) the
sum is (1 - 1) + (0 - 0) + ... = 0, etc.
And if x is in intersection(O_n)union(K_n) then
x lies in every O_nK_n, hence the sum of |h_j(x)|
is infinite, so we're allowed to iignore this x.
So (h_j) gives chi(A). And the sum of |h_j| is
      m(O_1) + m(O_1K_1)
     + m(O_2K_1) + m(O_2K_2)
    + m(O_3K_2) + m(O_3K_3)
   + ...;
m(O_1) is no larger than m(A) + eps and the
sum of the other terms is less than eps or
2*eps or whatever. QED.
It follows that any non-negative simple function is
a good M-function. And if f is a non-negative function
in L^1 then f is the (pointwise!) sum of a sequence of
non-negative simple functions, so Lemma 1 shows f
is an M-function.
Since M is a vector space (cf the proof of Lemma 1)
M contains all of L^1.
************************
David C. Ullrich
Subject: Re: treatment of Lebesgue integral
[*]
Well, as I pointed out in my first reply to you, the fact that you're 
not familiar with something does not prove that it doesn't exist.
One reason for defining step function as in that book, and as
I was above, is to make the characterization of integrable functions
above correct.
Yes, they do indeed give your definition of step function and then
state that a function is Lebesgue integrable if and only if (a) and
(b) for some sequence of step functions. I have not been claiming
that the definition on that web site is correct.
But I have no reason to think that it's wrong. I haven't been thinking
about whether the reasoning in your counterexample is correct,
because I've been distracted by the question of how you know
that step functions are _always_ defined to be linear combinations
of characteristic functions of half-open intervals. Having established
that that's not so (in Asplund & Bungart the definition includes
the things I was calling step functions above) I decided to think
about whether your example was correct given your definition
of step function. It simply isn't - you've asserted some things
without proof which are simply wrong. Honest:
Let's say a sstep function is a linear combination of characteristic
functions of half-open intervals. Suppose that f is the characteristic
function of a countable set (r_n). Then there _is_ a sequence
of sstep functions such that (a) and (b) hold:
First, let (I_n) be a countable collection of half-open intervals
with the following properties:
(ii) For every j there exist infinitely many n such that I_n is equal
(iii) The sum of the lengths of the I_n is finite.
Now for each n, let (J_n,k) be a pairwise disjoint collection
of half-open intervals such that the union of the J_n,k, for
a fixed n, is precisely the interior of I_n (ie, (r_n, s).)
Let (S_n) be an enumeration of the I_n's and J_n,k's,
in any order you like, doesn't matter. Let f_n be the
characteristic function of S_n if S_n = I_j for some j,
and let f_n be the negative of the characteristic
function of S_n if S_n = J_j,k for some j,k.
Then this sequence (f_n) satisfies condition (b):
Suppose that x is such that the sum of |f_n(x)|
is finite. The x is not one of the r_j, so f(x) = 0,
and we need to show that the sum of f_n(x) is 0.
But x is in only finitely many S_n, and for every
I_j such that x is in I_j there is exactly one k
such that x is in J_j,k (and conversely, if x is 
in J_j,k for some j,k then x is in I_j). So the
non-zero terms in sum(f_n(x)) consist of
finitely many 1's and -1's, with the same number
of each; hence sum(f_n(x)) = 0, qed.
(Note that I'm _not_ claiming that the sum of f_n(x)
equals f(x) for all x! That's not required in (b).)
(a) and (b) do not imply that f is a sum of step functions(!)
For that f it's incredibly easy to get a sum of what I and
Asplund-Bungart call step functions satisfying (b), and
it's not quite so easy, but not all that hard, to get a
sequence of what you've been calling step functions
satisfying (b) (see above).
No. Your cardinality argument is invalid. The reason is this: Given a
sequence of step functions satisfying the conditions above there
can be more than one f, in fact uncountably many f, which satisfy
(a) and (b), with exactly that one sequence of step functions.
Read the characterization again, more carefully.
Ok, don't read it, I'll explain: Let (f_n) be the step functions 
I defined above, at the spot marked (*). Suppose that f is 
_any_ function with the property that f(x) = 0 for all irrrational 
x. Then condition (b) holds. So there you have uncountably 
many f's which satisfy (b), with the same sequence of step functions.
That example only gives c f's for one sequence of step functions.
One can easily concoct examples where 2^c f's work for
a fixed sequence of step functions.
You don't seem to have noticed this paragraph in my post,
btw. Your explanations of why this is impossible don't
hold water - you should point out the error in the proof
of Lemma 2.1.10 in the book.
************************
David C. Ullrich
Subject: Re: treatment of Lebesgue integral
as
infty
Lebesgue
are
your
I had it so that f(x)=1 when x was irrational, so you probably want to 
alter
f_n to be the negative of the value you currently have and add the
characteristic function of [0,1]. I think I've spotted an easier proof, if
f(x) is the characteristic function of any set obtainable via this method
then this procedure shows that the complement if also obtainable. If the
c.f. of sets A and B are obtainable as sum f_n and sum g_n then A U B 
is
obtainable as sum (f_n+g_n - f_n*g_n) and A-B as sum(f_n - f_n*g_n) so 
we
have a ring. It looks a bit messy but we probably have a sigma ring too. 
The
c.f of an interval is obtainable, therefore the Borel sets are obtainable.
The set of all rationals is a Borel set, as is its complement. QED.
Use the integral of the c.f of a set A as its measure etc.
than
work,
can
is
mean
I think the argument is valid, it's proving what it says, namely that most
Lebesgue measurable functions cannot be written as a countable sum of step
functions, but as you point out I misread condition (b) so the proof is
irrelevant. Nevertheless I'm pleased with the simplicity of it.
I don't have access to the book and I can see now how it could very well be
true. It's obviously a mapping from countable collections of step functions
to the equivalence class of Lebesgue integrable functions.
Subject: Re: treatment of Lebesgue integral
See a post I inserted at the top of the thread. (By all means go ahead
and adapt the argument to a _proper_ notion of step function...)
************************
David C. Ullrich
Subject: Re: treatment of Lebesgue integral
Not quite, for two reasons. Let's say that the sequence (f_n) gives 
f if (a) and (b) hold. Say f is the c.f. of A, g is the c.f. of B, 
(f_n) gives f and (g_n) gives g:
(i) You want to take sum(h_n), where (h_n) is some enumeration of
the functions f_n, g_n and -f_n*g_n. Why: Because then if x is such
that sum|h_n(x)| is finite then it follows that sum|f_n(x)| and 
sum|g_n(x)| are finite, so f(x) = sum f_n(x) and g(x) = sum g_n(x).
If you take h_n = f_n + g_n - f_n*g_n you have no guarantee that
f = sum f_n at points where sum|h_n| is finite.
(ii) You got the algebra wrong - the sum of a product is not the
product of the sums. You actually want to let (h_n) be an
enumeration of the functions f_n, g_n, and -f_n*g_m
(not f_n*g_n).
Same two comments.
Not that it matters whether this is a simpler proof or not, but it's
easy to give something that _looks_ like a simple proof if you
allow statements like it looks messy, but probably... as part
of a proof... if you insert an actual proof of this assertion is
it still a simpler proof?
Actually it's clear that the obtainable sets do not form a
sigma-ring, since any obtainable set must have finite measure.
The whole point to this approach, at least according to
Asplund & Bungart, is to concentrate on integrals of functions
instead of measures of sets.
Anyway: Whether it's simpler or not, it _is_ clear that arguing
something like what's above is better in that it's the sort
of argument that might be used in showing that what we
get here is exactly the Lebesgue integrable functions. But
there are details that it's not entirely clear to me how to
work out: Say A_n is obtainable and the A_n are disjoint.
Suppose that the sum of the measures of A_n is finite, so
there's some hope. Say the sequence (f_n,k) gives A_n. 
_If_ we knew that the sum over n and k of int|f_n,k|
was finite then I can imagine showing that the union
of the A_n was obtainable, but it's not so clear to me
how to show that that sum is finite.
Yes, it's simply amazing.
I made a point of not looking at any of the proofs in the book the
other day - now that I'm a grownup I should be able to figure this
out for myself. Say M is the class of functions integrable according
to (a) and (b), and L^1 is defined as usual (except that we're going 
to regard it as literally a space of functions, not equivalence 
classes modulo almost every equality). It's clear that M is a subset
of L^1, and that the M-integral of any function in M equals the
Lebesgue integral. To show M = L^1 I imagine doing something 
like this:
(i) Show that M is a vector space (argument above does this).
(ii) Show that f in M implies |f| in M (???)
(iii) Define the norm of f to be the M-integral of |f|; show that this
is a seminorm on M.
(iv) Show that M is complete(???)
QED.
I don't see how to do (ii), and the problem with (iv) is
it leads to a question why can we get the double sum
int|f_n,k| finite? as above.
Hmm, another possibility:
(ii) Define p(f) for f in M to be the inf of sum int|f_n| over
all sequences (f_n) which give f.
(iii) Show that p is a seminorm on M, and let M~ be
M modulo the subspace with p(f) = 0. Define ||f|| on
M~ by p.
(iv) Show that M~ is complete (this may be messy but
unless I'm missing something it's actually quite
straighforward.)
(v) Note that the Lebesgue integral of |f| is <= p(f),
so we actually have a (bounded) inclusion of M~ in L^1.
(vi) Finish the proof somehow...
************************
David C. Ullrich
Subject: Re: treatment of Lebesgue integral
Sorry I ment to post this to the NG.
----- Original Message -----
Subject: Re: treatment of Lebesgue integral
alter
if
is
Yes it's amazing what I can 'prove' at 3 O'clock in the morning! In my
defence what I think I had in mind was: let F_k= sum f_n for n=1..k and
look at H_k=F_k+G_k - F_k*G_k. Doesn't the sequence of  functions
h_n=H_{n+1}-H_n work - in fact isn't that exactly what you got?
OK, but if the space is [0,1] or of finite measure?
obtainable.
set
Lebesgue
Something
they
most
step
be
What about looking at  F_k(x)=sum f_n  n=1..k?
let h_n(x)=|F_{n+1}(x)| - |F_n(x)|
then |h_n(x)|=|f_n(x)|
therefore sum h_n(x)=|f(x)| where sum|h_x(x)|<infty
To show L^1 is a subset of M what's wrong with bludgeoning the problem to
death?
i) Show c.f of any Lebesgue measurable A set is in M (provided
mu(A)<infty). [Looks tricky but it has to be true.]
ii)Any simple function is in M (provided its integral is finite).
iii)Show that if f is the limit of a sequence of simple functions, then f 
is
in M [have to have some restrictions here - maybe look at L^1(E) where E is
of finite measure].
iv)Show every Lebesgue measurable function is the limit of a sequence of
simple functions.
v)Finish off somehow.
Subject: Euclidean-Cartesian geometry
With the advent of Riemannian geometry: Which is a non-Euclidean geometry 
in
which straight lines are replaced by geodesics and in which the parallel
postulate is replaced by the postulate that every pair of straight 
(geodesic)
lines intersects; physics began to loose it's semblance to reality, and
became like a ship adrift at sea.
Without Euclidian geometry and Cartesian coordinates, there is no way to
signify, or establish positions and directions.
Patooey to non-Euclidian; non Cartesian geometry: Physics is nonsensical
without them.
Makes me think all physicists are nimbuses; are you?
  http://newsone.net/ -- Free reading and anonymous posting to 60,000+ 
groups
   NewsOne.Net prohibits users from posting spam.  If this or other posts
made through NewsOne.Net violate posting guidelines, email 
abuse@newsone.net