HDTenEightyP/NetNews huggingface dataset:
From: pkk36438@uxa.cso.uiuc.edu (Polygon)
Newsgroups: sci.math
Subject: Who are C.D. Pan and J.R. Chen ?
Keywords: Pan, Chen
Date: 17 Apr 92 20:40:47 GMT
Organization: University of Illinois at Urbana


	I am a beginner in number theory.  When I read the books, I notice
that the names of C.D. Pan and J.R. Chen are frequently mentioned and their
works on prime numbers are quoted and cited in almost every book.

	Can anyone please tell a little about themselves? (Which university
they are in?  What are their Chinese names?  Favorate food?  Favorate pop
singer?  Bulls fan?  Blah blah blah, blah blah blah... )

	Thank you in advance.

						Polygon 
						UIUC
From: erick@fraser.sfu.ca (Erick Bryce Wong)
Newsgroups: sci.math
Subject: Re: factorial function
Date: 14 Apr 92 02:31:53 GMT
Organization: Simon Fraser University, Burnaby, B.C., Canada

In article <1992Apr13.193906.2606@cs.ubc.ca> mic@cs.ubc.ca (Michelangelo Grigni) writes:
>Ok, what is the time to calculate n! mod p, where p is
>a little larger than n (say in the range n to n^2)?

Well, if p=n this is trivial. ;)
If p=n+1, and p is a prime, then Wilson's Theorem states (p-1)! = -1 (mod p).
Maybe this could also be used to derive a formula for p near 2n, 3n, etc...

-- 
 -- Erick (erick@sfu.ca, erickw@sfu.ca, or erick@cs.sfu.ca, take your pick)
From: gruntz@inf.ethz.ch (Dominik Gruntz)
Newsgroups: sci.math
Subject: limit of exp(exp(Psi(Psi(x))))/x
Keywords: limit
Date: 14 Apr 92 14:16:25 GMT
Organization: Dept. Informatik, Swiss Federal Institute of Technology (ETH)
Originator: gruntz@rutishauser.inf.ethz.ch

Can some one give me a tip on how to compute the limit of

               exp(exp(Psi(Psi(x))))/x

as x goes to infinity? (Psi is the first polylogarithm function)

I know, that the limit is finite, but I have not idea on how to
get that limit. Of course, I have a presumption what the limit
is, but how would YOU solve that problem?

Thanks in advance,
- Dominik
-- 
/*            Dominik Gruntz    Institut fuer Wissenschaftliches Rechnen  *
 *                              Department fuer Informatik                *
 *        gruntz@inf.ethz.ch    ETH-Zentrum                               *
 *      Tel: +41 1 254 74 77    CH-8092 Zuerich, Switzerland              */
Newsgroups: sci.math
From: gd1@cunixa.cc.columbia.edu (Gautam Dalmia)
Subject: List of Unsolved Prime Number Problems
Organization: Columbia University
Date: Tue, 7 Apr 1992 08:37:52 GMT

elkies@ramanujan.harvard.edu (Noam Elkies) writes:
(Original) Subject: Tangent^2 re: Mersenne
>Whoa --- one also needs that half the primes p *such that 2p+1 is prime*
>are congruent to 1 mod 4; it is still a big open problem to prove that
>there are infinitely many such p to begin with!  I certainly have no
>reasonable doubt that there are infinitely many, and that they are
>correctly distributed mod 4, but a "Theorem!" it isn't.

Does anyone know where I can find a list of "big open problems"
concerning prime numbers, like the problem given above? I have read
Apostol, Analytic Number Theory and looked over some other books, but
these restrict mention of unsolved problems to the few most famous
ones, and do not contain large enough lists.



===========================================================================
In high school I read number theory and realized the the beauty of
mathematics: the  simple unanswered questions and the elusive, but
simple, answers. Then, one day, I went to a library to look at
journals/ big books and alas ... mathematicians had invented abstruse,
ugly methods to attack these beautiful questions. At first I was
saddened but then I looked up, and I laughed and said "God, what fools
these mathematicians be".... and God and I have been laughing ever since!
From: ucak0cg@ucl.ac.uk (C Goutis)
Newsgroups: sci.math
Subject: Re: On expected convergence
Summary: improper probabilities
Date: 12 Apr 92 14:19:36 GMT
Organization: Bloomsbury Computing Consortium, London

In article <1992Apr10.095030.24447@leland.Stanford.EDU> algebra@leland.Stanford.EDU (Leong Weng Ng) writes:

>     [stuff deleted]
>                                                    My idea is that
>this identity can be obtained by probabilistic reasoning. Grant me that
>infinity can be treated as a random variable integer R. To evaluate the infinite
>series, take the Rth term as the last term (dangerous I know). Since R is
>random, it has equal probabilties of being odd or even. If R lands on a odd
>term, the sum is 1; if even, the sum is 0. So the expected sum of the series is 
>1/2(1)+1/2(0) = 1/2, thus agreeing with the binomial method.
>    [stuff deleted]
>	Comments and criticisms are much appreciated.

The problem is the danger that you mentioned. In some sense you can treat
R as random, having 'equal chance to be any number'. However it does not
follow that it has equal probabilities of being odd or even. If you try to
compute any probability as (number of favourable outcomes)/(total number of 
outcomes) you end up with either something finite/infinity or infinity/infinity
which are either 0 or indeterminate respectively. At some deeper level
the problem is equivalent: how to treat plus infinity minus infinity
or infinity/infinity.  The answer is unfortunately 'we do not know'.

People in statistics occasionally use pseudo-probabilities arising from
probabilities like the one you mention. For some things they work quite
well, but paradoxes or counterexamples are routinely constructed. All of
them involve showing some careless (or non-rigorous if you like) use of
infinity.

Costas
Newsgroups: sci.math
From: roth@gauss.enet.dec.com (Jim Roth)
Subject: Re: Calculating the # of primes < n
Organization: Digital Equipment Corporation
Date:  7 APR 92 20:20:40    


In article <1992Apr2.115308.8760@linus.mitre.org>, bs@faron.mitre.org (Robert D. Silverman) writes...
>In article <1992Apr2.014052.1514@cunixf.cc.columbia.edu> pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:
>:I was trying to compute what fractions of the positive integers are
>:prime. I was doing this by getting the fraction left if from the
>:positive integers we remove everything divisible by the primes
>:p_1,p_2,p_3,...  

I'd recommend looking thru Hans Riesel's book _Prime Numbers and Computer
Methods for Factorization_, Birkhauser 1985.

This book is easy to understand, and includes a discussion of a number
of interesting methods for counting the primes below a given number
that are, surprisingly, much faster than finding all the primes.
These techniques allow the accuracy of asymptotic prime counting
formulas (such Riemann's) to be experimentally compared to the exact counts.

By the way, there *is* an analytic "formula" for the exact number
of primes below a given number, but it's sort of cheating since
it uses the Riemann zeta function in an essential way - and that
function requires you to know the primes!  But truncating the resulting
series still gives a remarkable approximation to the stairstep like
jump discontinuities of the prime counting function.

- Jim
From: parson_r@cubldr.colorado.edu (Robert Parson)
Newsgroups: sci.math
Subject: Re: Zero, "Arabic" numerals, etc.
Date: 10 Apr 92 19:11:49 GMT
Organization: University of Colorado, Boulder

In article <1992Apr10.014241.14414@midway.uchicago.edu>, trivedi@rabi.uchicago.edu (Anil Trivedi) writes:
> In article <44588@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:
> 
>>What is being discussed is not the notion of zero, but a SYMBOL for 
>>zero.  
> 
> Well, you are half way there but not quite. This is one of the deepest
> misunderstandings. The key intellectual and practical leap was in
> USING ZERO TO OBTAIN A PLACE VALUE REPRESENTATION OF THE NUMBERS.
> 
> Think about it this way. Suppose that the Romans had the concept of
> zero as well as a symbol for it. And their number system looked like 
> this:
> 	0, I, II, III, IV, V, VI, VII, VIII, IX, X, XI, ...
> 
> Now that is not too much of an improvement, zero or not, is it?

 The Chaldeans (c. 300 BC - same place as Babylonians but later) had
 this concept. The Greeks and Romans seem not to have picked up on it.
A place to read about this is Neugebauer's "The Exact Sciences in Antiquity".
From: bolwidt@fwi.uva.nl (Erwin Bolwidt)
Newsgroups: sci.math,alt.irc
Subject: Re: electronic math seminar?
Date: 11 Apr 92 13:55:21 GMT
Distribution: sci
Organization: FWI, University of Amsterdam

hellmond@phoenix.Princeton.EDU (Peter H. Hellmonds) writes:

>jk87377@cc.tut.fi (Juhana Kouhia) writes:

[stuff deleted]

>->Also, IRC doesn't call you as does phone and talk programs, so we have
>->to announce the meeting time and maybe program in this group.

>IRC has also a /summon command which would call a user to the conference
>(provided the user is currently logged on to his/her computer).

At least some servers have disabled /summon because it can be irritating for
non-ircers on the servers. Also (correct me if i'm wrong) it isn't yet possible
to summon users on a host that is not an irc server.
But what about the following:

talk person@host.you.want.to.invite.to.irc
<Wait for connection to establish>
Hello! our irc meeting is about to start on irc channel #meeting.
Please come to irc as soon as you can :)
^C

[stuff deleted]

=====
Erwin Bolwidt
<bolwidt@fwi.uva.nl>
From: fcs$1224@altair.selu.edu
Newsgroups: sci.math
Subject: Problem with magic square  algorithm
Date: 9 Apr 92 00:02:29 -0600
Organization: Southeastern Louisiana University


I got this algorithm to generate odd magic squares off the net a
year or so ago.  There is only one thing wrong with it.  It doesn't
work!  Would someone mail to me an algorithm that will work?

This one may be almost correct.

Thanks,

Tom

higginbotham@selu.edu

Algorithm to generate magic squares which doesn't work follows.
-------------------------------------------------------------------

A magic square is an n by n array of integers having the same value
for all the row sums, column sums, and sums of the two main
diagonals.  For example;

     4    9    2
     3    5    7
     8    1    6

is a 3 by 3 magic square with row sums, columns sums, and diagonals
sums equal to 15.  The algorithm for generating a magic square when
n is odd is as follows:

a.   Generate the integers from 1 to n ** 2
b.   As this is done, place the first number in the middle of the
     bottom row in position k(i,j).
c.   Place the next number in position k(mod(i, n) + 1, mod(j, n)
     + 1), unless there is a number there, in which case, place it
     in position k(i - 1, j).
d.   Repeat step c until all integers have been place into the
     array.
From: tordm@vana (Tord Malmgren)
Newsgroups: sci.math
Subject: Re: Last Fermat Theorem
Date: 12 Apr 92 22:52:51 GMT
Reply-To: tordm@vana
Organization: Dept. of Physics, University of Stockholm -- Sweden

In article <1992Apr12.213416.174945@ipgaix.unipg.it>, peppe@ipgaix.unipg.it (G. Vitillaro) writes:

[text deleted]
>The conclusion is that Fermat stated some type of 
>geometrical/topological fact: from the class of curves
>x^n + y^n = 1 on the euclidean plane R^2 only the straight line
>and circle have rational positive points (p,q) in Q+ x Q+.

>The question is: have this any sense, do you already know the
>geometrical (if even exists) fact that cause this consequence?

 isn't this Falting's and Mordell's assumption?





---------------+--------------------------------
 Tord Malmgren | Internet: tordm@vand.physto.se | These opinions are my own,
               | BITnet  : TORDM@SESUF51.BITNET | and NOT of this department!
---------------+--------------------------------
 Department of Physics, University of Stockholm
From: tordm@vana (Tord Malmgren)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorum
Date: 11 Apr 92 10:25:51 GMT
Reply-To: tordm@vana
Organization: Dept. of Physics, University of Stockholm -- Sweden

In article <1992Apr11.042713.6065@athena.cs.uga.edu>, shepard@athena.cs.uga.edu (William Erik Shepard) writes:

>>One of the tantalizing things about FLT is the fabled note that Fermat
>>left before he died saying that he found an elegant proof.

>>There are two main possibilities:

>>(a) He really found the proof, and the historical record is lost

>>(b) He had a mental abberation, and this mistaken proof is lost too

>Actually, from what I understand, Fermat stated that he had discovered the
>elegant proof but as he was making notes in the margin of a book, he did
>not have enough room to write out the entire proof.  In either words, he
>either discovered the proof but ran out of room or THOUGHT that he discovered
>the proof but ran out of room.

 fermat was a frequent letter-sender/receiver and had a correspondance
about his problems and solutions; he mentioned in many letters about
proofs for n=3 and n=4, so if he really HAD a proof for all n, wouldn't
he have mentioned it in those letters to his friends? 350 years of
mathematicians as clever, not to say more clever than he, and with
more knowledge, more powerful methods haven't been able to solve it,
so if you ask me; he goofed or boosted or tried to get into history books...




---------------+--------------------------------
 Tord Malmgren | Internet: tordm@vand.physto.se | These opinions are my own,
               | BITnet  : TORDM@SESUF51.BITNET | and NOT of this department!
---------------+--------------------------------
 Department of Physics, University of Stockholm 
Newsgroups: sci.math
From: wdh@linus.mitre.org (Dale Hall)
Subject: Re: Question on Connectedness
Followup-To: sci.math
Summary: Partial solution is trivial
Keywords: Topology, quotient space, connected metric space
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Distribution: na
Date: Fri, 17 Apr 1992 16:54:59 GMT

In article <36937@gremlin.nrtc.northrop.com>
		jbarnett@nrtc.northrop.com (Jeff Barnett) writes: 
>Is the following true in the topology of Euclidean n space:
>
>P is a collection of non-empty (pairwise) disjoint closed sets.  P is
>a partition of S.  Claim -- if S is connected, then the cardinality
>of P is either 1 or uncountable.
>
	This is trivial if you also assube the members of P are
	compact.  What you do is consider the space S/P, in which
	the points are members of P, and the induced topology is
	the finest topology for which the canonical mapping
	S ---> S/P is continuous (it's also described by stating
	that a set X in S/P is open iff the union of the members
	of X is open in S).  If P consists of compact sets, then
	S/P inherits a metric (d(P1,P2) = min (d(x1,x2): xi in Pi)).
	Since S is connected, S/P is a connected metric space (you
	could probably show that it is a complete metric
	space if S is complete [I haven't]).  Then show that S/P is either a
	single point or has uncountably many points. This is not very hard:
	if a, b are distinct elements of S/P, let d(a,b) = D, and define 

		F: S/P --> [0,D] by F(x) = min(D, d(a,x)).

	(1) F is continuous :

		F^(-1) [0,r) = open disc of radius r, @ a
	and 	F^(-1) (r,s) = open disc of radius s
				- closed disc radius r

	(2) F is a surjection, since its image must be connected (F is
	    continuous and S/P is connected), and the image contains
	    both 0 and D.

	
	Since S/P has a point for each element of P, this finishes the
	job. 

	I don't know yet how to fix the situation for non-compact
	closed sets in P.  Presumably, the same proof applies if P
	contains finitely many non-compact elements (with some
	tweaking, I guess).  But if P is entirely composed of non-
	compact sets???  I'll think awhile longer.

							Dale.
From: allis@cs.rulimburg.nl (Victor Allis)
Newsgroups: sci.math
Subject: Re: Problem with magic square algorithm
Date: 14 Apr 92 11:36:18 GMT
Organization: University of Limburg

torbenm@diku.dk (Torben AEgidius Mogensen) writes:

>fcs$1224@altair.selu.edu writes:


>>I got this algorithm to generate odd magic squares off the net a
>>year or so ago.  There is only one thing wrong with it.  It doesn't
>>work!  Would someone mail to me an algorithm that will work?

>[Algorithm deleted]

>This is a well-known algorithm, but it only works if the side-length
>is odd. Even sided magic squares are more difficult to make, and in
>many cases you need back-tracking. There is no simple and efficient
>algorithm (that I know of) that will produce magic squares of any
>side-length above 2.

>	Torben Mogensen (torbenm@diku.dk)

This is not true. I have a book at home (which I read over 10 years ago)
in which an algorithm is described to generate a magic square for
each size of n. 

For n odd, it gives the standard algorithm, where you go down and right
diagonally each step (treating the square as a torus).
When you get stuck, you move up 1 square.
You start at the bottom middle square.

For n even, and a multiple of four, you just write down the numbers from
1 to n^2, from top-left to bottom-right. Each square with value x has
a complement n^2+1 - x (which is the mirror image with respect to the
center of the square). Now you swap of half the pairs in the square the
elements. Which half? Just look at the folloing pattern. The squares which
are marked are swapped:

----------
| XX  XX |
|X  XX  X|
|X  XX  X|
| XX  XX |
| XX  XX |
|X  XX  X|
|X  XX  X|
| XX  XX |
----------

It is easy to generalise this pattern.

For n even, but not a multiple of four there was also an algoritm, but
I have forgotten it, but could check it out.

BTW, a more interesting question is to find out HOW MANY different magic
squares there are of a given size.

n = 3 : 1 (or 8 counting symmetries)
n = 4 : 880 (or 7040 counting symmetries)
n = 5 : left as an exercise (:-)

Victor Allis.
allis@cs.rulimburg.nl
From: ks7@cunixb.cc.columbia.edu (Kamlendra Singh)
Newsgroups: sci.math
Subject: Looking for English tr. of Abel's Quintic Eqns paper
Date: 7 Apr 92 22:39:03 GMT
Organization: Columbia University

There has recently been some discussion of Abel's proof that there is
no general formula for roots of a quintic equation...

If anyone knows where one can find an English translation of this paper
please send me the reference. Thanks.
Xref: sparky sci.math:5823 sci.physics:6658
From: greil@guug.de (Anton Greil)
Newsgroups: sci.math,sci.physics
Subject: a special semigroup in SL(2,C)
Keywords: Picard's group, free product, word problem of semigroups
Date: 9 Apr 92 16:10:42 GMT
Article-I.D.: guug.greil.702835842
Followup-To: sci.math



I'm looking for the features of a special sub-semigroup of SL(2,C)

    where  SL(2,C) := "the special linear group ..."
           C       := "the set of the complex numbers" .

That semigroup is generated by the four matrices M1, M2, M3, M4:

        ( 1   1 )       ( 1   0 )       ( 1   i )       ( 1   0 )
        ( 0   1 )       ( 1   1 )       ( 0   1 )       ( i   1 )

            M1              M2              M3              M4

(where "i" is the imaginary unit: i*i = -1).

The matrices M1 and M2 alone generate the free semigroup F2.
All four matrices together don't generate a free semigroup, because
(M3*M4)**6 = the unit matrix, and so there are "relations".

The generated semigroup is contained in the "Picard Group", which is
generated by

        ( i   0 )       ( 1  -1 )       ( 0  -1 )       ( i  -1 )
        ( 0  -i )       ( 0   1 )       ( 1   0 )       ( 0  -i )

with some defining relations.
[e.g. Wilh. Magnus, "Noneuclidean Tessalations And Their Groups", 1974, p.153]

My main questions
-----------------

  (1)  can that semigroup be represented as a free product of semigroups ?

  (2)  is that semigroup known under a special name and in any special
       context (mathematical, physical, ...) ?

My optimism for a "good structure" of the semigroup is founded on the fact
that the word problem of the semigroup is very well solvable (in logspace !),

[ R.J. Lipton and Y. Zalcstein, "Word Problems Solvable in Logspace", 
                              Journal of the ACM, Vol. 24, 1977, pp. 522-526 ]

(as I was told earlier by the friendly "net").


Thank you for your interest!

Toni Greil, Muenchen (Germany)
From: kwhyte@daisy.uchicago.edu (Kevin Whyte)
Newsgroups: sci.math
Subject: Re: An elementary, "naturally occurring" problem
Date: 10 Apr 92 10:50:31 GMT
Distribution: sci
Organization: Dept. of Mathematics, Univ. of Chicago

In article <880@armltd.uucp> dseal@armltd.co.uk (David Seal) writes:
>In article <3767@wet.UUCP> sam@wet.UUCP (Sam Needham) writes:
>
>>In article <1992Apr1.201733.25777@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>>-In article <1992Apr1.183035.14916@Princeton.EDU> carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:
>>-:Prove or disprove that there exists a bijection f:[0,1] -> [0,1] that is
>>-:not continuous at any point. 
>>- 
>>-Doesn't the following function do it?
>>-f(x) = 1-x  if x is irrational.
>>-f(x) = x if x is rational.
>>-Bob Silverman
>>
>>Continuous at x = 0.5.
>>
>>The proof that there is no such function uses, I think, the Baire Category 
>>Theorem. Suppose f: [0,1] to [0,1] is discontinuous everywhere and f(a) =
>>0.5; then the family
>>
>>	[0,1] minus inverse-image-of-f( 0.5 - 1/n, 0.5 + 1/n) (n>=3)
>>
>>has union [0,1] but none of them contains an open neighbourhood of a. 
>
>What about the function:
>
>  f(x) = x        if x is rational
>         x + 1/2  if x is irrational and x < 1/2
>         x - 1/2  if x is irrational and x > 1/2
>
>This is a bijection [0,1] -> [0,1] (it is its own inverse) and is
>discontinuous everywhere. (Proof: any open interval contains an irrational
>x1. Having chosen x1, the interval also contains rationals x0 < x1 and x2 >
>x1. If x1 < 1/2, this gives us f(x1) - f(x0) = 1/2 + x1 - x0 > 1/2; if x1 >
>1/2, it gives us f(x2) - f(x1) = x2 - x1 + 1/2 > 1/2. So any open interval
>contains two points whose function values differ by more than 1/2.)
>
>As for how the above proof breaks down, it seems to me that
>inverse-image-of-f(0.5 - 1/n,0.5 + 1/n) never contains 1/2. So the union of
>the stated family is not [0,1] - it is in fact [0,1] - {0.5}.
>
>David Seal
>dseal@armltd.co.uk
>
>All opinions are mine only...

   The "proof" breaks down at two points.  The one place you point
out, namely if f is a bijection then f-inv(.5-1/n,.5+1/n) always
contains f-inv (.5) so its complement in [0,1] cannot.  Thus this
union is always [0,1]-{a}.  The Baire category theorem IS still
true for the space [0,1]-{pt}, but it no longer "proves" the "theorem".

  The other mistake is, in some sense, bigger.  The Baire category
theorem says (in one form) that if E(i) are closed sets, and union
E(i)= whole space  and the whole space has some nice properties
(complete metric spaces and compact Hausdorff spaces are both good
enough)  then for every x in X there is a nbhd of x in some E(i).
However, [0,1]-f inv(.5-1/n,.5 + 1/n) is not closed.  Since we
are not assuming f continuous  these sets need have no nice topological
properties. 

Kevin
kwhyte@math.uchicago.edu
Newsgroups: sci.math