mm-1549
sum_{p <= n} 1/p = log log n + C + O(1/log n)
where
C = gamma + sum_p [log(1 - 1/p) + 1/p].
IIRC this is due to Mertens and is in Hardy and Wright.
You want to go up to p(n) which is ~ n log n by PNT, so
you'd replace log log n by log(log n + log log n)
which is still as-near-as-dammit log log n.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
  Elegance is an algorithm
    Iain M. Banks, _The Algebraist_
Hmm I guess thats |2t|<2  .Sorry,again Stuart M Newberger
http://www.cs.uu.nl/wais/html/na-dir/sci-math-faq/axiomchoice.html
Then the FAQ is mistaken?
 And these statements depend on AC (i.e., they cannot be proved in ZF
   without AC):
     * The union of countably many countable sets is countable.
     * Every infinite set has a denumerable subset.
     * The Loewenheim-Skolem Theorem: Any first-order theory which has a
       model has a denumerable model.
     * The Baire Category Theorem: The reals are not the union of
       countably many nowhere dense sets (i.e., the reals are not
       meager).
     * The Ultrafilter Theorem: Every Boolean algebra has an ultrafilter
       on it.
  No, the statements cited are not arithmetical.
  No, from G.9adel's work on the constructible hierarchy. The full story
is told by Platek in a paper in the Journal of Symbolic Logic,
vol 34 no 2 1969.
  In the particular case of AC or CH, there is no significant increase
in the length of proofs. But we know that extending a theory T by a
statement A undecidable in T will shorten the proofs of some
statements to any extent, in the sense that for any recursive function
f there is a statement B such that f(p)<p', where p is the length of
the shortest proof of B in T+A and p' the length of the shortest proof
of B in T. (G.9adel was the first to prove a result of this kind.)
Friedman has given dramatic actual examples of how non-finitary
principles make it possible to prove statements that otherwise have no
feasible proof. And of course we also know that there are statements
that cannot be proved without such principles.
What infinitary principles are you referring to?
Is what you say true for arithmetical statements?
  By the incompleteness theorem, statements asserting the existence of
large infinite sets have new arithmetical consequences.
His father was Lutheran and his mother Catholic with some
racial Jew in his ancestry. Saying that he was Catholic doesn't
indicate much. Quite a few Jews changed their names and
became Catholics because they were discriminated against
for university positions. Tarski for instance whose father's
name was Ignacy Teitelbaum, became a Catholic. Sol
Feferman isn't very fond of Cantor's Set Theory so I don't
think ancestry much to do with attitudes towards Set Theory. 
Maybe that's why I didn't say it, then.  Cantor was philo-Catholic,
not Catholic.
Just about the opposite of Cantor's case in both senses.  Rather
than becoming Catholic without genuine Catholic sympathies, he
*didn't* become Catholic *in*spite* of genuine Catholic sympathies.
And he *kept* a Jewish-sounding surname even though he wasn't
Jewish.
There isn't much difference between Lutheranism and Catholicism
especially regarding the role of Jesus Christ. So I don't think it
matters which one I picked for Cantor, it isn't relevant.
Technically, I think the word Hebrew has more to do with
racial origins/bloodlines/ethnic land of origin. Jewish used to be
reserved for religion. I decided to add this note because internet has
some picky users who might complain about jewish ancestry
meaning genetics rather than continued commitment to Judaism.
How do you explain black Jews? They are israelites in the biblical 
sense. If they keep the Commandments they are no different from Moses or 
Abraham.
Bob Kolker
The point of my post was about ancestralism:
Is that true?  It's a new one on me.  A little surprising, given
that Cantor wasn't Jewish.  Cantor was, I believe, Lutheran, ...
I meant that religion wasn't the key issue. Ancestry was the key issue.
I think about 1/3 of Germans had Jewish ancestry with the density
higher in cities. Cantor probably would have been included in the
pogrom because of his ancestry, not his current religious taste.
My post was actually about recognizing jewish mathematicians
from aryan due to their attitude toward Cantor's set theory. A
Nazi would primarily care about the bloodline, not the religion.
A Nazi might also expect Jewish mathematicains to react favorably
to Cantor's set theory, after all they ^^  or the Nazis might not know
what religion Cantor was or that he was say only 1/8 genetically Jewish.
I was pointing out that there might be some truth to this perception
of favoring a mathematical contribution because of ancestry or
perceived ancestry, it would hardly amount to accurate prediction;
but be boosted by Nazi perception of Jewish banking conspiracies.
I was at first disinclined to believe that Nazis would employ such an
illogical criteria. However, they also pursued religous relics as a
possible source of power, which is not logical ... Indiana Jones is
only an embellishment of this direction of thought, not totally fictional.
The practice of Jews changing their religion and last names in order
to secure university positions arose from the first world war and
the depression and financial penalties placed on Germany. Anti-
Semitism is hardly anything new, just starting with the Nazis.
Well, you can sort of see how this would be confusing, since
On what do you base this?
In The Light of Logic by Sol Feferman page 248:
Infinity in Mathematics: Is Cantor Necessary? (conclusion)
Independently of such detailed work which puts into question
the necessity of higher set theory for everyday mathematics*,
I am convinced that the platonism which underlies Cantorian
set theory is utterly unsatisfactory as a philosophy or our
subject, despite the apparent coherence of current set-theoretical
conceptions and methods. To echo Weyl, platonism is the
medieval metaphysics of mathematics; surely we can do better.
SH: It took me a few hours to find something compact and convincing.
Stephen 
  Yes, that certainly justifies describing Feferman as not very fond
of Cantor's set theory!
Much briefer than the longer post I sent before:
http://math.stanford.edu/~feferman/book98.html
In the Light of Logic. Author: Solomon Feferman. (Oxford University Press, 
1998, ISBN 0-19-508030-0, Logic and Computation in Philosophy series)
Description from the jacket flap:
In this collection of essays written over a period of twenty years, 
Solomon 
Feferman explains advanced results in modern logic and employs them to cast \

light on significant problems in the foundations of mathematics. Most 
troubling among these is the revolutionary way in which Georg Cantor 
elaborated the nature of the infinite, and in doing so helped transform the \

face of twentieth-century mathematics. Feferman details the development of 
Cantorian concepts and the foundational difficulties they engendered. He 
argues that the freedom provided by Cantorian set theory was purchased at a \

heavy philosophical price, namely adherence to a form of mathematical 
platonism that is difficult to support.
 Discussion, linux)
Feferman thinks that the mathematical theory ZF somehow requires
Platonism?
Bizarre.  I'll have to put this book on my list, because I sure can't
guess what his argument might be.
-- 
Jesse F. Hughes
Contrariwise, continued Tweedledee, if it was so, it might be, and
if it were so, it would be; but as it isn't, it ain't. That's logic!
                                                     -- Lewis Carroll
Feferman finds platonism philosophically unsatisfying...
ad(ii) The platonist position does not necessarily require the
existence of infinite sets ...(but of course that is essential to
Cantorian set theory.)
Is Cantor Necessary?, is answered with a resounding no.
SH: I've been reading a lot of Feferman lately and have a directory
for him as well as his  book In The Light Of Logic. I think Feferman
is not a platonist and he thinks Cantorian set theory requires platonism
when he quotes Cohen's reasons for rejecting the platonist realist
position and accepting a formalist position in the set-theoretical
foundations of mathematics.
Yes, it is in chapter 2 of In The Light of Logic (page 30)
Infinity In Mathematics: Is Cantor Necessary?
In order to explain the objections to Cantor's ideas in mathematics
that lead one to search for viable alternatives, one must first provide
some understanding of their nature and use. This will done here more
or less historically though necessarily in outline; we start with a rather 
innocent looking problem about the existence of certain special kinds
of numbers 2.
*the actual infinite is not required for the mathematics of the physical 
world*
Page 44:
ad(i). Views of mathematical objects as independently existing abstract
entitities are generally called a form of _platonism_. In its particular
set-theoretical manifestation, this reveals itself most obviously in such
principles as the Axiom of Extensionality and the Axiom of Choice. *
ad(ii). The platonist position _per se_ does not necessarily require the
existence of infinite sets (or, for that matter, of sets at all, but of 
course
that is essential to Cantorian set theory.
--------------------------------------
http://wwwpersonal.ksu.edu/~aarana/papers/FefermanReviewMathIntelligencer.pd\

f
Review of In The Light Of Logic
In Part I, Feferman raises as a problem the role of transfinite
set theory in mathematics. Since transfinite sets are supposed
to be infinite objects about which facts are true independently of
our abilities to verify them, it seems that these abstract entities
must exist independently of human thoughts or constructions. This
family of beliefs about sets is frequently called platonism.
Feferman finds platonism philosophically unsatisfying, and thus
presents three projects aimed at avoiding platonism:
L.E.J. Brouwer's 'intuitionism', David Hilbert's 'finitism', and
Hermann Weyl's 'predicativism'. [SH: Feferman prefers Weyl.]
In feferman.unlimited.pdf Some Formal Systems For The Unlimited
Theory of Structures and Categories
On page 1 he lists some informal statements in the naive theory
These sorts of statements cannot be accounted for directly in
currently (generally) accepted mathematics, i.e. as formulated
in systems such as ZF or its immediate extensions. But they do not
have an unreasonable or cooked-up look. Each of them arises
as a natural continuation of ordinary mathematical talk about
structures (in particular, categories).
without them.
SH: I've read that they are independent.
http://math.stanford.edu/~feferman/book98.html
In his concluding chapters, Feferman uses tools from the special
part of logic called proof theory to explain how the vast part, if
not all of scientifically applicable mathematics can be justified
on the basis of purely arithmetical principles. At least to that
extent, the question raised in two of the essays of the volume,
Is Cantor Necessary?, is answered with a resounding no.
SH: Also pages 68 and 69 of In the Light of Logic seem relevant.
Stephen
  True. Results about formal derivability doesn't tell us what
mathematics is actually useful or indeed indispensable.
   at 09:39 PM, zimmathematics@yahoo.com (Zim Olson) said:
ITYM delusions.
That depends on what is is.
For purely idiosyncratic meanings of makes sense and works.
-- 
Unsolicited bulk E-mail subject to legal action.  I reserve the
right to publicly post or ridicule any abusive E-mail.  Reply to
domain Patriot dot net user shmuel+news to contact me.  Do not
Daryl McCullough ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
JXStern ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Lester Zick ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
David C. Ullrich ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Larry Hammick ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Larry Hammick ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
JXStern ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Herman Jurjus ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Mike Oliver ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
robert j. kolker ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
David C. Ullrich ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
robert j. kolker ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
LordBeotian ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Torkel Franzen ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Mike Oliver ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
LordBeotian ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
JXStern ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Zim Olson ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Daryl McCullough ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
Jesse F. Hughes ,
oo coin flippers flip coins oo times each.  how many flips of the 
antidiagonal
have been flipped one after the other?
ANSWERS NOT EXCUSES
Herc
What's the reason why
(
et
)
implies
?
Note: x,y belongs to the normed space
      (V,|| ||), || denotes the absolute 
      value in R.
Hint: Apply the triangle inequality to || x + (y - x) || and....
--Lynn
For real numbers a, |a| is either a or -a.
So if you prove a < b and also -a < b, it follows
that |a| < b.
I would like to know from you if it is possible to find out the
solution of the below First oder differential equation.
dv(t)/dt + c1*v(t)^2 = c2*¤(t)
where,
c1 and c2 are constants
¤(t) can be any standard signal like Step or Ramp or 
Sinusoidal....
I need the solution of v(t) in terms of ¤(t)
Please mail me at your convinience
Ramesh Babu
This is called a Riccati equation.
For beta(t) = cos(2*t), using Maple I get
y(t) = (MathieuCPrime(0, 1/2*c1*c2, t)+_C1*MathieuSPrime(0, 1/2*c1*c2, t))
/(c1*(_C1*MathieuS(0, 1/2*c1*c2, t)+MathieuC(0, 1/2*c1*c2, t)))
where _C1 is an arbitrary constant.
For beta(t) = a*t + b, 
y(t) = -(-c1*c2*a)^(1/3)*_C1*AiryBi(1, -(-c1*c2*a)^(1/3)*(a*t+b)/a)/(c1*
(_C1*AiryBi(-(-c1*c2*a)^(1/3)*(a*t+b)/a)+AiryAi(-(-c1*c2*a)^(1/3)*(a*t+b
)/a)))-(-c1*c2*a)^(1/3)*AiryAi(1, -(-c1*c2*a)^(1/3)*(a*t+b)/a)/(c1*(_C1*
AiryBi(-(-c1*c2*a)^(1/3)*(a*t+b)/a)+AiryAi(-(-c1*c2*a)^(1/3)*(a*t+b)/a))
)
These can be pieced together to solve the equation with piecewise-linear
signals.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
on Thurs, Feb 10 2005 12:00 pm
Generally, I agree...
seq(residue(Pi^2*2*Pi*I*(u*coth(u) - 1)/u^2/(u^2+Pi^2),u=I*Pi*k),
k=1..20);
5/2, -1/3, -1/12, -1/30, -1/60, -1/105, -1/168, -1/252, -1/360,
-1/495, -1/660, -1/858, -1/1092, -1/1365, -1/1680, -1/2040,
-1/2448, -1/2907, -1/3420, -1/3990, -1/4620, -1/5313, -1/6072
Now you transformed the original task to the summation of an
infinite series
5/2 -1/3 -1/12 -1/30 -1/60 -1/105 -1/168 -1/252 -1/360 -1/495...
It is still not obvious for me (well... at least, I believe, for
your students and honest average playgoers :)  how to get 2 as
the result of summation of this series, so please tell what are
the next step(s)?
For example, how to force Maple/Mathematica/... display the
sacramental 2, and by this complete the proof?
Or how to transform it to the shape where 2 would be an obvious
answer?
With the deepest respect to your computational flair,
Vladimir Bondarenko
http://www.cybertester.com/
http://maple.bug-list.org/
http://www.CAS-testing.org/
QED!
gf := 1/2*(2*ln(1-x)*x^2-4*ln(1-x)*x-3*x^2+2*ln(1-x)+2*x)/x^3
2
Thus, the original wild integral
int(z*csc(z)*(Pi*coth(Pi*tan(z))*csc(z)+Pi^2*csch(Pi*tan(z))^2*
sec(z)-2*cot(z)*csc(z)), z=0..Pi/2);
is really equal to Pi.
In passing we calculated that
Int((z*coth(z)-1)/(z^2+Pi^2)/z^2, z=0..infinity)
=  1/Pi^2
Viva la team-work! ;)
Vladimir Bondarenko
http://www.cybertester.com/
http://maple.bug-list.org/
http://www.CAS-testing.org/
on Thurs, Feb 10 2005 12:00 pm
Generally, I agree...
seq(residue(Pi^2*2*Pi*I*(u*coth(u) - 1)/u^2/(u^2+Pi^2),u=I*Pi*k),
k=1..20);
5/2, -1/3, -1/12, -1/30, -1/60, -1/105, -1/168, -1/252, -1/360,
-1/495, -1/660, -1/858, -1/1092, -1/1365, -1/1680, -1/2040,
-1/2448, -1/2907, -1/3420, -1/3990, -1/4620, -1/5313, -1/6072
Now you transformed the original task to the summation of an
infinite series
5/2 -1/3 -1/12 -1/30 -1/60 -1/105 -1/168 -1/252 -1/360 -1/495...
It is still not obvious for me (well... at least, I believe, for
your students and honest average playgoers :)  how to get 2 as
the result of summation of this series, so please tell what are
the next step(s)?
For example, how to force Maple/Mathematica/... display the
sacramental 2, and by this complete the proof?
Or how to transform it to a shape where 2 would be the obvious
answer?
With the deepest respect to your computational flair,
Vladimir Bondarenko
http://www.cybertester.com/
http://maple.bug-list.org/
http://www.CAS-testing.org/
I'm working on a novel interpretation of quantum theory, and have come
across a stumbling block with the requirement for two asymmetric
functions that add up to zero.  I think I can make this work, but
require a new kind of number that is related to transfinite ordinals. 
Essentially, the numbers I have postulated are smaller than zero in an
analagous way that finite numbers are smaller than transfinites.  In
other words, these ('subfinite')numbers would occur within zero itself
(thus, for example, allowing us to discriminate 0/1 from say 0/7 as
different numbers).
Has anyone worked on such numbers before or are they entirely new? 
Can anyone give me more information on this matter or a definitive
explanation of why such numbers can't exist or be useful?
Maybe you need a system that has infinitesimals.
which, though vanishingly small, are certainly not smaller than zero.
I really am looking for a new kind of number here (or a convincing
argument as to why the whole concept is insane or impossible).  With
numbers that occur within zero in the same way that transfinites occur
beyond infinity we would have the tools to talk about a structure
within singularities.  This would open up entirely new fields of maths
and physics.  You can find some background information on the reasons
I'm seeking these 'subfinite' numbers at
http://www.soothsleuth.com/moebius/moebiushtml/the%20twist.htm.
I've recently worked on perhaps a better way of representing
transfinite ordinals and cardinals, I don't know weather it can help
you, but as a numeral system it differentiates between 0/1 and 0/7.
see messagerepresenting infinite numbers
I hope it would be of help.
Zuhair
helps me.  What I'm looking for is a way to do algebra with numbers
that are 'within' zero.  I don't see how your system helps me to do
that (may be I'm missing the point?).
Just about everything that Sarfatti posts makes sense. He's very fringe, 
but he does know his physics, unlike most of the regulars on sci.physics and \

sci.physics.relativity. His ideas are no more implausible than some of the 
crazier stuff in popular works by Roger Penrose and Stephen Hawking. People \

like Fred Alan Wolf are far crankier than Sarfatti.
-- Ben
says...
Of course. That's why NASA has been flying through stargates for years 
now, but keeps them secret from everyone.
   Both make perfect sense (from the POV of wanting to keep the serfs 
malleable).
   Fortunately, the Internet has no ruling Oligarchy.
   Yet...
   Ronaldus Magnus showed the Commies that the only true means of 
production is people who can see a reward coming for exerting 
themselves every day. Not being shipped to a Gulag does not constitute a 
reward.
   And look how both venues educate the candidates for said gangs.
   ROFLOLPIMP! Asimov's little-fingernal-clippings understood more about 
economics that Ilya does. Still, sometimes I wonder if Hari was Asimov's 
imagination speaking aloud, or a deliberate parody of Economists in 
general.
   Mark L. Fergerson
a correction note:
first: 5a/2.5=2a and not 5a/2.5a= 2a
second : the concept of closure that I've adopted for each 1/0,2/0
....,etc has no proof at all , so in an open system 2a -2a =2 , and
a/0=a^2,since 2a is 2/0 and ax1/0=a/0=axa. In reality I'm not sure of
both systems open or closed.but I'm almost sure that c/0 is an
0-D matrices. I want two of them, that are different.
Seems like a good challenge for you.
Eh,
Poly-poly man
The 0x0-matrices over R and over C are different.
Well, maybe that is not such a good example since R < C.
Make it the 0x0-matrices over R and over the field of order 2.
Derek Holt.
Last weekend I came across a simpler way to describe surrogate
factoring with a couple of quadratics and solutions for x and z, which
I promptly programmed--and it didn't seem to work.
I puzzled over that a bit, and posted the equations:
yx^2 + Ax - M^2 = 0
and
yz^2 + Az - j^2 = 0
where A, j, and M are integers greater than 0 chosen, where M is the
target to be factored, and you find that you can use T, where
T = M^2 - j^2
and substituting out y to solve for x and z gives
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and you *should* be able to always factor M, by iterating through
factors of Tj^2 to get the full set of of integer solutions for Ax.
But, as I said, I tried the idea last weekend, and I didn't see it
work, and also a poster named Rick Decker claimed to have a
counter-example.
However, I was missing something easy, as consider A (yes, I've screwed
this up before but here's the correct argument) as it's *chosen*
arbitrarily.
Now then, for any rational x and z, there must exist some A, such that
Ax and Az
are integers, right?  Like if x = 29343/3947387492 then
A = 3947387492, will force Ax to be an integer, equal to 29343.
Well, look at
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and notice then that an A can always be chosen such that Ax and Az are
integers.
So what, right?
Well, multiply both equations by A, and you get
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
so that now you have integers inside and out!!!
And looking at
sqrt((Az - 2M^2)^2 - 4TM^2)
it MUST be true that for some integer Az, M is factored.
In retrospect, with my tests last weekend, I, um, assumed that A=1, as
I just figured it didn't matter, and that's probably why they didn't
work, I guess, or also I might not have done plus and minus.
That is, the square roots give a positive and a negative solution, and
I don't remember if I checked both.
In any event, the proof that the method has to work is easy.  There
must exist integers Ax and Az, as shown above.
James Harris
No.. it did not work. You were certain that it was simple and proven, but 
you were wrong. And all who doubted you were, according to you, liars.
And if we do not believe you this time, we are again liars, right? 
[...]
[JSH]
[o[CapitalYAcute]in]
Of course -- that's always how it goes.
James is certainly right that for any rationals x and z, there exists an 
integer A such that Ax and Az are both integer.  Alas, there are an infinite \

number of such A, the set of all integer multiples of lcm(denom(x), 
denom(z)) (where lcm(a, b) is the least common multiple of a and b, and 
assuming x and z are both in normalized form (gcd(numerator, denominator) = \

1)).
Without a proof that an _appropriate_ A can be found efficiently, he may 
well have a correct method (don't know -- didn't check), but of no use.  For \

example, here's a much simpler method of that kind:
k*p = M.  Now switch to JSH-mode:
    So looping through *all* possible integer k as required by the product
    MUST factor M.
Quite true (although confusingly written), but quite useless too.  Since JSH \

didn't give any reason to suspect that an appropriate A can be found 
efficiently, there's no reason to presume that the latest twist will be more \

_effective_ than random trial division.  He does seem to have a talent for 
coming up with overly complicated ways to do just that. 
There
an
infinite
and
denominator) =
may
use.  For
Hmmm...an actual assertion in there against my proof?
Well, let me give just a bit of mathematics, and the theorem this time,
followed by proof, in line with the convention one poster keeps
bringing up.
Given a natural number M, a natural number j can be chosen such that
the factorization of M^2 - j^2 and j will factor M.
It can be shown that with non-zero integer A, and rationals x and z,
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where A is an integer such that Ax and Az are both integers.
Proof complete.
James Harris
Well, you left out a lot of steps at the end, but I'll stipulate
that your claim is correct. However, the claim is not what you
need to turn your proof into a useful factoring algorithm.
One major hole is that you haven't shown that the factors of M
you find are *nontrivial*.  Of course there are guaranteed to
be factors of -4Tj^2 that will lead to a factorization of M--
all one has to do is use (-2j)(M + j) and (2j)(M - j). The
problem with this choice is that the factor of M^2 you get
is nothing but M^2 itself. Tim has provided evidence that
suggests that as M gets larger, the number of j values that
lead to useful (i.e., nontrivial) factors of M gets smaller.
Might it be that for some values of M there are *no* j values
that lead to useful factors of M?
Frankly, I doubt it, which is to say that I think the conjecture,
    For each M, there is a j which, when used in Harris'
    construction, will give a nontrivial factorization
    of M.
Is true.  However, you haven't proved that, so it must be regarded
as an open question at this time.
There's another problem, though, and that comes from trying
to turn your construction into a useful factorization
algorithm.  Even if the conjecture above could be proven
to be true, it might make no practical difference if you
had to search through all (or most) factorizations of -4Tj^2
to find a useful factor of M.
I urge you to take a look at recent posts by Tim, Vedat, and
Rick
that
z,
Your objection is handled as follows.
Consider
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and let M = p_1 p_2, where p_1 and p_2 are prime.
Now let Az have p_1 as a factor but be coprime to p_2, so you have
Az = kp_1, then
sqrt((Az - 2M^2)^2 - 4TM^2)) = sqrt((kp_1 - 2M^2)^2 - 4TM^2))
so
p_1 sqrt((k - 2p_2^2)^2 - 4Tp_2^2))
and an integer k exists.
Your assertion is that k must have p_2 as a factor, but that is easily
proven false as there must exist a solution for every prime factor of T
where p_2 is not a factor of (k - 2p_2^2).
James Harris
which
but
liars.
Mathematical proof is what matters.
Here it's easy.  If you refuse to accept mathematical proof then there
is no room for discussion.
Here is the proof.
Given
yx^2 + Ax - M^2 = 0
yz^2 + Az - j^2 = 0
and
T = M^2 - j^2
it follows that
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and, you can multiply both equations by A, to
get
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where for any rational x and z there must be an integer A such that
both Ax and Az are integers, so looping through *all* possible integer
Ax and Ay as required by the square roots MUST factor M.
So, I can't be wrong, or, algebra is wrong.
If you disagree, show a false step.
The math is not hard, or do you disagree?
I'm asking you to do more than just blindly criticize, but to show you
are actually using rational thought.
Can you do math, or are you just someone disagreeing to be
disagreeable?
James Harris
James,
I'm going to compile Tim Peters' and Rick Decker's arguments in an
easy-to-follow message and include a lot of concrete examples. I hope you
don't disregard this reply, because it points the fatal flaw that you've
been ignoring up until now.
To start off, I'm not arguing the validity of your method. I'm not checking
it either, so you may have made a mistake or two in the signs (and then
again, you may not). For now let's just assume that it works. So basically,
you've reduced the problem of finding a factor of M to the problem of
finding a suitable A.
Now I'll introduce you to a related idea. Or maybe you know it already.
Suppose that I have a pair of integers x and y, such that x != y mod n and
x != -y mod n, but x^2 == y^2 mod n. Now gcd(x-y,n) is a nontrivial factor
of n. See how I've reduced the problem of finding a factor of M to finding
two suitable values of x and y?
However, there's a problem. Assuming that n = p*q is the product of two
prime factors, then for a given x, there are only 4 values of y such that
x^2 == y^2 mod n, and of those 4 values, two of these are x and -x, so
they're not valid. So, unless we have an efficient method to search for the
2 valid values, we'll have to search through a sizeable amount of the
interval [1,n] before finding an answer. Now you have to agree that I might
just as well generate primes from 1 to sqrt(n) and test whether these
primes divide n -- at most I'll have to look through sqrt(n)/(2 log n)
primes, which is much better than searching a range the size of n.
There's a very important difference between your idea and this idea that I
just described, which is known as congruence of squares. For the latter,
some very bright researchers have devised efficient ways of sieving through
values of x and y to find suitable values. You may have heard of these
methods -- they're called the Quadratic Sieve and the Number Field Sieve.
You think that you've solved the factoring problem just because one can, in
principle, search through all possible values of A until finding a suitable
one. But don't forget that's included in the cost of the algorithm! One
doesn't measure the cost of QS or NFS just as the time to compute
gcd(x-y,n), but rather as the whole cost of finding these suitable values
of x and y and then taking the gcd. So, if surrogate factoring ever becomes
a world-renowned and respected factoring algorithm, it won't be because of
the algebra you did above, but because of an efficient method that you
provide to find the suitable A that splits M.
QS and NFS aren't the only examples of this. The ECM method also gives an
easy way to factor n, as long as you find a suitable elliptic curve group
(which is to say, one of smooth order.) Now the big deal about ECM is that
some bright researchers have analyzed it and guaranteed us that by
generating elliptic curves at random, we'll somewhat quickly find a
suitable one -- not as quickly as we desired (i.e. in polynomial time), but
quicker than previous methods, and that's why ECM is so valuable.
Pollard rho works the same way. The basic idea is that if you find two
distinct values x and y (distinct modulo n, that is), which lie in the same
position in a pseudorandom cycle modulo one of the factors of n, then these
two values are congruent modulo p, and hopefully not congruent modulo n/p.
Thus gcd(x-y,n) should reveal a non-trivial factor of n. This by itself
isn't very useful, but Pollard suggested a somewhat efficient way to seek
values of x and y: use a pseudorandom function f(x) (usually f(x) = x^2 +
1), then iterate through the two sequences f(x) and f(f(x)). Eventually
those two sequences will collide, and using either Floyd's or Bren't
cycle-finding algorithms one can easily spot the collision between these
two sequences.
So, is it clear to you that your method is useless, unless you provide an
efficient way to find the required parameters? I have shown that one can
efficiently find j such that M + j and M - j are trivially factorable. Can
you efficiently find A? I suggest you concentrate your research on this,
because that's the big gap in your algorithm.
D.8ecio
How do you find them? I swear it's not clear from your algebraic
developments (which I refuse to refer to as proofs.) To me it just feels
like two equations dependent on each other -- i.e. x = f(z,A,M,T,j) and z =
g(x,A,M,T,j). I know M, T and j are all related, so we could trim it down
to x = f'(z,A) and z = g'(x,A), but that's still a circular dependence.
How do you break out of it? I think Rick Decker explained this in some 
other
post, but I'm too lazy to look it up now. And the most important question,
*can you find the required parameters in an efficient manner*? In the end
that's all that matters (at least if you're searching for an efficient
algorithm and not just an obfuscated curiosity.)
D.8ecio
Would you please start with stating the theorem you are trying to prove? As \

it stands, you have no proof. All you have is a bunch of stupid algebra 
expressions that come from nowhere and go nowhere, followed by the words 
So, I can't be wrong, or, algebra is wrong.
prove? As
algebra
words
Actually, a mathematical proof begins with a truth and proceeds by
logical steps to a conclusion which then must be true.
By *convention* yes, certain schools may teach their students to start
by stating the theorem being proven, but that's just that--a
convention--and it has no relevance to correctness.
As for my start, I begin with valid algebraic expressions:
Given
yx^2 + Ax - M^2 = 0
yz^2 + Az - j^2 = 0
and
T = M^2 - j^2
which are all valid algebraic expressions.
Or do you disagree?
Now then, trivial algebraic manipulations using those expressions gives
you
the equations
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and trivially, in algebra, you can multiply both by A, to get
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j ^2 - 2Az )
and
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
or do you disagree?
I have valid algebraic expressions and am using valid algebraic
operations, or do you disagree?
Now then, the rationality of Ax in the first equation is determined by
the rationality of the square root
sqrt((Az - 2M^2)^2 - 4TM^2)
or do you disagree?
That square root will give integers based on the prime factors of TM^2.
Or do you disagree?
The same basic argument applies to Az and the rationality of its square
root, where the difference is that rationality depends on the prime
factors of Tj^2.
Or do you disagree?
Now then, for any rational x and rational z, there must exist an
integer A such that Ax and Az are integers, or do you disagree?
Therefore, it follows that all integer Ax and Az are given by the
equations I've shown.
If that is a jump for you, then admit it, and I can go into further
detail.
Notice I'm pushing you to the mathematical argument, and away from
social tools.
Either you know algebra and can do math, or you can mindlessly chatter,
disagreeing to be disagreeable.
I am determining if you are willing and able to be reasonable.
James Harris
Derivation deleted.
That sounds like a theorem to me, though not one which is phrased in a
particularly conventional manner.
The question of whether the theorem is either profound or useful I
leave open.
Paul
-- 
Hanging on in quiet desperation is the English way.
The time is gone, the song is over.
Thought I'd something more to say.
Wrong.  Proofs begin with a theorem which is later proved in the
proof section [following potentially several lemmas].
start
It's just logical.  You can't prove something without first stating the
problem.    Just stating a proof doesn't make sense because the reader
won't know what you are proving.
Why?  What is y, x, A, M, z or j?  Where did you get them?
They are valid or not depending on whether or not solutions exist for
them.  You have yet to prove this true or not.  So far your results
are not complete.
gives
[not checking if it's true]... Why did you multiply by A?  Is a
solution in $GROUP provable to exist?
So far we don't know what you're doing so it's hard to say.  Is this
over Z, Q, R or some other field?
by
Have you proven that the difference is positive?
TM^2.
How do you conclude this?
Hard to say.  Recall it's up to YOU to prove that your statements are
true.
square
Same as above.
there must exist... this sounds like a theorem.  I don't see a proof
that follows though...
It certainly is a huge jump.  So far you haven't defined 6 variables, 1
field and are lacking at least 3 critical proofs.
j00 r0x0r.
Tom
the
LOL.  So you say proofs begin with a theorem, which is proved in a
proof section.
So, what is the proof?  The theorem, or the proof section?
If the proof section is the proof, then why do you have to state the
theorem first?  Isn't it extra then?
James Harris
Look in any text book: the usual arrangement is for the theorem to be
stated and then proved.  Sometimes there is a discussion followed by a
statement of the theorem that the discussion has proved.  But the
theorem is always _stated_ somewhere.
What social issue?  All we've been maintaining is put up or shut up.
There is a difference between equations which are true and proof of a
theorem.
if y = 2 and x = y^2 then x = 4, does this prove that there is a god?
That sounds stupid right?  ... now put yourself in our shoes.  You list
off a dozen equations [which may or may not be equivalent] then from
this you jump to the conclusion that factoring is solved.  Despite
the fact you haven't been able to factor [faster than just guessing the
factors] any trivial sized numbers.
If you want people to take your math serious you're going to have to
present it much more precisely [e.g. you have yet to actualy write an
algorithm with well defined start/terminate conditions that factors]
and some results.
chatter,
Can't it be both?
You honestly have to open your mind up a bit.  People can *both* know
math and enjoy reading your daily insanities.
Reasonable is not a word you get to use here.  Your reasonable
privileges have been revoked.
Tom
up.
god?
list
the
I'm in a weird mood today so I'm answering a lot of these posts.
First off, my post actually includes a theorem, which comes in at the
end, not stated at the beginning, which is just a convention.
I wonder about your statement claiming certain expression may or may
not be equivalent, as that's an actual mathematical statement, which
you embedded in a lot of chatter, and it's provably false.
I say, if you wish to make a challenge to my proof, then challenge the
proof.
The math is at the level of what in America is typically called high
school, or in Europe secondary school mathematics.
There does not need to be a lot of verbiage.  There is not room for
error to hide.
Chatter is a waste of time here.
If you have a challenge to the proof, make it directly.
James Harris
Great!
When it turns out that method still doesn't work, you won't forget to
tell us about it, right?!
Jose Carlos Santos
Huh?  Do you believe in algebra, or not?
Given
yx^2 + Ax - M^2 = 0
yz^2 + Az - j^2 = 0
and
T = M^2 - j^2
you must also have
x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
and, you can multiply both equations by A, to get
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where for any rational x and z there must be an integer A such that
both Ax and Az are integers, so looping through *all* possible integer
Ax and Ay as required by the square roots MUST factor M.
So, I can't be wrong, or, algebra is wrong.
Now that's a trivially easy proof.  And I want to emphasize that it is
a proof as I've seen weird stuff from sci.math'ers where proofs are
just discounted, as if they don't matter.
But they do.  A mathematical proof gives absolute certainty.
Surrogate factoring must work.  There is no room for doubt.
James Harris
 Discussion, linux)
[...]
James, the past several weeks, you've oscillated between the two
states:
(1) I've solved the factoring problem and my proof is absolute and if
you doubt me, then you hate Mother Mathematics.  Just wait for the
Hammer.
(2) It doesn't work.  Where did I go wrong?  
,----
| 
| Well, reality is I can't get a program to work, so I can't exactly hold
| on to that belief, now can I?
| 
| So, yes, I'm back-tracking as I can't get an implementation to work,
| and now find myself theorizing more, and often understanding less.
`----
Now aren't you a bit embarrassed saying, There is no room for doubt?
-- 
I'm starting to absorb information [...] as I find myself more and
more fascinated by my own prime counting function.  The rate of
information absorption is on an exponential scale, like it always has
been for me for things I'm interested in. -- James S. Harris
You have tried the argument that algebra is wrong but JSH is correct before, \

why not again today?
Usually a proof starts of with a theorem. I missed that somehow. What did 
you prove? That you have a polynomial time factoring algorithm? I missed 
that somehow too.
Not absolute.... every proof works relative to a set of axioms.
You said that many times before, and then you admitted that you were wrong. \

Why would today be a change?
before,
is
did
missed
Sounds like *you* missed algebra in school.
It's a trivially easy proof.  End of story.
If you wish to betray that you're either not bothering to read it, or
can't comprehend VERY basic algebra, then ok.
Reply again, and so betray.
James harris
Here we go again. The point has been reached - as many times before in 
history - where James Harris doesn't even understand what he is reading. 
Does anyone believe that James Harris has the slightest clue what 
polynomial time factoring algorithm even means? No wonder the only 
thing that poor child can do is answering with insults. 
It must be awful, coming up with all these brilliant mathematical ideas 
and the only responses you get are gibberish. That is the world that 
James Harris lives in, every day. 
If we wasn't such a nasty piece of , I would feel sorry for him.
State the therem that you think you proved.
State the therem that you think you proved.
State the therem that you think you proved.
What did
missed
And there you have it.
I've seen this mindless behavior go on for *hundreds* of posts.
Yup, hundreds of posts, where the fact that an actual proof has been
given and shown is lost in the shuffle.
What motivates these people?
I think their behavior is instinctive.
Easily this thread could in a few hours be over a hundred posts long,
despite my having given a very basic argument, using very basic
algebra, at the level that teenagers learn.
Oh, and note that the posters so far in this thread are obsessive
repliers.
The Last Danish Pastry is notable for starting a flame webpage
against me some years back.  Yes, years back.
Jose Carlos Santos is a current regular flamer.
The other two are as well.
And they will reply, reply, reply.
Notice that reasoning doesn't work with them.
James Harris
jstevh@msn.com says...
Move to strike your honor, nonresponsive.
-- 
Randy Howard (2reply remove FOOBAR)
Making it hard to do stupid things often makes it hard
 to do smart ones too. -- Andrew Koenig
Please state the theorem you were proving.
correct
it
are
What
That doesn't change the fact that you write a theorem first then a
proof....
The fact that you haven't formally performed EITHER action though
shouldn't stop you..
Betray who?  
Tom
that
or
It's a social convention to begin with the theorem being proven.
The proof is rather basic and short.
A lot of chatter around it just proves my point that many of you
dislike mathematics itself.
So you post on math newsgroups?  So?  If you can't handle a little
proof like mine then you can't actually like mathematics.
James Harris
Yes, it is a very standard convention, which serves a purpose. It is good to \

know up front where a proof is trying to get to. Would you please be kind 
and follow that convention here? If you cannot provide a theorem up front, I \

will make the conclusion that you have no idea what you are trying to prove. \

yes... adds to the drama in wiles' case... he must have had fun with the 
surprise punch line...
yes. that is really all i want from him .... but in JSH's case, I doubt that \

hidin it until the end of his proof woul add much exciting effect. 
Excellent.
So, what is the next project?
-- 
Clive Tooth
http://www.clivetooth.dk