HDTenEightyP/NetNews huggingface dataset:
From: geversti@oasys.dt.navy.mil (Gordon Everstine)
Newsgroups: sci.math,sci.physics
Subject: Re: tensors: How about 3rd, 4th rank?
Date: 22 Aug 92 12:26:35 GMT
Reply-To: geversti@oasys.dt.navy.mil (Gordon Everstine)
Followup-To: sci.math
Organization: David Taylor Model Basin, Bethesda, Md.

Another example of a tensor of rank 4 is the material property tensor
in elasticity.  Generalized Hooke's law says that each of the nine
components of stress is a linear combination of the nine components
of strain, yielding 81 possible material constants (before accounting
for symmetries).  Thus,
    s(i,j) = c(i,j,k,l) e(k,l)
with summation (from 1 to 3 in 3-D) over the repeated indices k,l,
where s is stress, e is strain, and c is a tensor of rank 4.
It turns out that 21 of the 81 constants are independent in the worst
case.
Newsgroups: sci.math
From: ara@zurich.ai.mit.edu (Allan Adler)
Subject: address of Boltzmann archive
Organization: M.I.T. Artificial Intelligence Lab.
Distribution: sci
Date: Wed, 19 Aug 1992 08:25:53 GMT



Can someone tell me the address of the Boltzmann archive in Vienna and
the name of the person to write to there?

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math
From: me@csri.toronto.edu (Daniel R. Simon)
Subject: Re: mathematician's marrying ages.
Organization: CSRI, University of Toronto
Date: 19 Aug 92 15:35:22 GMT

In a previous article, guckes@math.fu-berlin.de (Sven Guckes) says:
>Galois and Abel never married. So there.

Abel very nearly did marry at one point, but his intended, who lived
in another city, would have been expected to move in with him, and he 
decided that he would rather commute.


"There *is* confusion worse than death"		Daniel R. Simon
			     -Tennyson		(me@theory.toronto.edu)
From: toon@moene.indiv.nluug.nl (Toon Moene)
Newsgroups: sci.math
Subject: Re: Calculus and Mathematica
Date: 18 Aug 92 20:32:40 GMT
Organization: Moene Computational Physics, Amsterdam, The Netherlands

In article <1992Aug3.204132.2552@cs.rose-hulman.edu>  
goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:

[ ... How to teach Calculus to DE students (what are DE students,
      for heavens' sake ?) ... ]

I won't go into that, I simply assume that you're right and that  
Mathematica is useful for that purpose, especially on a NeXT (Read My  
Signature :)
Now my younger brother comes to me (a M.Sc. in Tropical Agriculture and  
Meteorology): Say, I need

     z0
    /       m 
    |    log  z dz
    /
   0

Now you have that marvelous symbolic algebra package - so just give me the  
result - NOT.

I spent two weeks forcing Mathematica to do these tricks - in general to  
do the sort of integrals that give a 1 or 2 level recurrence relation on  
partial integration.

My advise: Unless you're absolutely sure MMA (or the other symbolic  
algebra package you're looking at) does what you want it to do, or you  
have enormous amounts of time doing it the hard way (i.e. programming it  
in MMA's language), don't buy into symbolic algebra - it is no Artificial  
Intelligence.

> Bart Goddard
> Asst. Professor of Mathematics
> Rose-Hulman Inst of Tech.    goddard@nextwork.rose-hulman.edu

--
Toon Moene (toon@moene.indiv.nluug.nl)
Kantershof 269, 1104 GN  Amsterdam, The Netherlands
Tel.: + 31 20 6982029; Fax: + 31 20 6003411
No Disclaimers; a NeXT at home protects against this occupational hazard.
From: boshuck@triples.math.mcgill.ca (William Boshuck)
Subject: Re: The Converse of Kaplansky's Therorem
Organization: Dept Of Mathematics and Statistics
Date: Wed, 12 Aug 92 16:52:34 GMT

In article <1349@newsserver.cs.uwindsor.ca> tarokh@server.uwindsor.ca (TAROKH VAHID             ) writes:
>Would somebody please tell me if the converse of Kaplansky's Theorem
>is true or not? That is if R is a ring with identity such that all the
>projective R modules are free, then is R necessarily Local? If we assume
>that R does not have identity what could be said?
>
>Thank You,
>Vahid Tarokh
>

The answer to the first question is NO. Look at the ring of integers.
I haven't thought about the second question since I have absolutely
no experience with rings without identity.
From: sarwate@uicsl.csl.uiuc.edu (Dilip V. Sarwate)
Newsgroups: sci.math
Subject: Re: Littlewood's Three Principles
Date: 12 Aug 1992 18:49:01 GMT
Organization: Center for Reliable and High-Performance Computing, University of Illinois at Urbana-Champaign
NNTP-Posting-Host: uicsl.csl.uiuc.edu

edgar@function.mps.ohio-state.edu (Gerald Edgar) writes:

>Where do I find the original (published) formulation of Littlewood's
>three principles, or at least a verbatum statement of them?

In Section 3.6 of his book, Real Analysis (2nd ed., MacMillan, 1968),
H. L. Royden gives a verbatim quotation from J. E. Littlewood's Lectures on
the Theory of Functions (Oxford, 1944, p. 26)

"The extent of knowledge required is nothing like so great as is sometimes
supposed.  There are three principles, roughly expressible in the following
terms: Every (measurable) set is nearly a finite union of intervals; every
[measurable] function is nearly continuous; every convergent sequence of
[measurable] functions is nearly uniformly convergent.  Most of the results
of [the theory] are fairly intuitive applications of these ideas, and the
student armed with them should be equal to most occasions when real variable
theory is called for.  If one of the principles would be the obvious means to
settle the problem if it were `quite' true, it is natural to ask if the `nearly'
is near enough, and for a problem that is actually solvable it generally is."

I do not know how the passage actually reads in Littlewood's book, and have
always assumed that the words in brackets and parentheses were inserted by
Royden.
Newsgroups: sci.math
From: mgh3@po.cwru.edu (mike hurley)
Subject: Re: Still another problem.
Followup-To: sci.math
Organization: case western reserve u
Date: Mon, 17 Aug 92 17:08:24 GMT

In article <1992Aug14.142149.16686@mcs.drexel.edu>, dmagagno@mcs.drexel.edu
(David Magagnosc) wrote:
> 
> In article <1992Aug12.075304.28486@newssrv.edvz.univie.ac.at> pm@katz.cc.univie.ac.at (Peter Marksteiner) writes:
> >Bill Taylor writes:
> >
> >
> >>Prove that
> >>
> >>               n     n     n    n
> >>     1  /     2     3     4    5        \  
> >>     - |  1 + -- +  --  + -- + -- +....  | 
> >>     e  \     1!    2!    3!   4!       /   
> >>
> >>  is an integer for all positive integer n.
> >
    << text deleted >>
> 
> 
> While we're at it, prove that the following sum is always an
> integer:
> 
> 	 n    n    n    n
> 	1    2    3    4
> 	-- + -- + -- + -- + ...
> 	 1    2    3    4
> 	2    2    2    2
> 
> and find reasonable generalizations.
> 
> D. Magagnosc
> -- 


The new problem can be solved by the same method as the original 
one, namely by expanding using binomial coefficients, changing 
the order of summation, and using induction to obtain a 
recursive formula for the values of the sums:

Let a_n = sum_{j=1}^{infty}(j^n)/(2^j).

Clearly a_0 = 1 is an integer.
Let B[n,m] denote the binomial coefficient n over m. 

We can write 
a_n = (1/2)*\sum_{j=1}^{infty}(j^n)/(2^(j-1))
    = (1/2)*\sum_{j=0}^{infty}((j+1)^n)/(2^j)
    = (1/2)*\sum_{j=0}^{infty}\sum_{m=0}^n B[n,m](j^m)/(2^j)
    = (1/2)*\sum_{m=0}^n B[n,m]\sum_{j=0}^{infty}(j^m)/(2^j).
For simplicity break off the m=0 term:
    = (1/2)\sum_{j=0}^{infty}(j^0)/(2^j) 
        + (1/2)*\sum_{m=1}^n B[n,m]\sum_{j=1}^{infty}(j^m)/(2^j)
    = 1 + (1/2)*\sum_{m=1}^n B[n,m]*a_m.
Combine both the a_n terms on the left:
(1/2)a_n = 1 + (1/2)*\sum_{m=1}^(n-1) B[n,m]*a_m
a_n  =  2 +  \sum_{m=1}^(n-1) B[n,m]*a_m
     =  1 +  \sum_{m=0}^(n-1) B[n,m]*a_m
(since B[n,0]*a_0 is always 1). 

Thus by induction every a_n is an integer; the
sequence begins 1,2,6,26,150,... .

mike hurley   mgh3@po.cwru.edu
dept of mathematics
case western reserve univ. 44106-7058
Newsgroups: sci.math
From: israel@unixg.ubc.ca (Robert B. Israel)
Subject: Re: Mapping points in space to a plane
Organization: University of British Columbia, Vancouver, B.C., Canada
Distribution: usa
Date: Thu, 20 Aug 1992 20:47:39 GMT

In <BtAIpG.5pL@wpi.WPI.EDU> jac@cs.WPI.EDU (Jeffrey A Choate) writes:


>Hello,

>I am having a problem mapping a point to a plane, 
>which I hope someone can help with.

>The problem is as follows:

>I have a point which lies on a plane with the
>equation: x + y + z = 1 (point represents 
>probabilities).  After I perform some non-linear 
>transformation, the point no longer necessarily lies 
>on the "probabilistic" plane.

>I have to map this resulting point, which lies 
>somewhere in 3-space back to the probabilistic 
>plane.

>My question is how to best map this point back to
>the plane?

>One suggestion is to find the orthogonal projection 
>of the point on the plane.  Another suggestion is to 
>use the normal to the plane (1,1,1), and starting at 
>the point follow the normal vector back until it 
>intersects the plane, and use the intersection point.

>Are any of these ideas good?  Is there an algorithmic 
>way of doing this? (I have to perform this mapping for 
>every point in a very large data set) Any help would be 
>greatly appreciated.

>Thank you,

>Jeff Choate

>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>~ Jeff Choate         | (508) 831-5006    | Computer Science Department     ~
>~ jchoate@cs.wpi.edu  | AI Research Group | Worcester Polytechnic Institute ~
>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

You're in a Research Group at a Computer Science department and you don't
know first-year Linear Algebra?  What's this world coming to?

Both suggestions amount to the same thing.  The formula for the
projection is P v + b where v is your original point (as a column vector),
P is the matrix 
   [ 2/3 -1/3 -1/3 ]
   [ -1/3 2/3 -1/3 ]
   [ -1/3 -1/3 2/3 ]
and b is the vector (1/3, 1/3, 1/3) (transposed).

-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
Newsgroups: sci.math
From: orourke@sophia.smith.edu (Joseph O'Rourke)
Subject: Re: What is the EATCS?
Organization: Smith College, Northampton, MA, US
Date: Tue, 18 Aug 1992 15:35:12 GMT

In article <4752@balrog.ctron.com> wilson@web.ctron.com (David Wilson) writes:
>
>    Here I have a paper in the EATCS and I still do not know what it
>    stands for.

European Association for Theoretical Computer Science.  They publish
a Bulletin edited by G. Rozenberg, Leiden Univ., Dept. of Computer
Sci., PO Box 9512, NL-2300 RA Leiden, The Netherlands.
Newsgroups: sci.math
From: Deane Yang <yang@fields.waterloo.edu>
Subject: Re: Need Titles (for Differential Geometry textbooks)
Organization: The Fields Institute for Research in Mathematical Sciences
X-Useragent: Nuntius v1.1
Date: Tue, 18 Aug 1992 13:45:11 GMT

In article <Bt4vAo.73y@helios.physics.utoronto.ca> Adrian Butscher,
butscher@helios.physics.utoronto.ca writes:
>Can anyone give me some titles of modern, undergraduate level
differential
>Geometry textbooks (I know there are lots!)?  I'd like a book which
covers
>Manifolds, Differential Forms, Topology.  In fact, I'm looking for a book
>which would follow 'Calculus on Manifolds' by Michael Spivak (my 2nd year
>textbook).  Something with Group Theory applied to DG (if it's not too 
>advanced) would be nice, too.

Guillemin and Pollack, Differential Topology

Boothby, An introduction to differentiable manifolds and Riemannian
                   geometry

Gallot, Hulin, Lafontaine, Riemannian Geometry

DoCarmo, Riemannian Geometry

Munkres, Analysis on Manifolds

Of these, I'm most familiar with Guillemin and Pollack, which
is one of the best ungraduate math textbooks I've ever seen,
and Boothby, which I used as a text
when teaching an introductory graduate course. It covers a lot of
what I consider essential basic material. The new books by Munkres
and DoCarmo look very nice.

The book by Gallot, et al has a wonderful table of contents. It covers
a lot of ideas and techniques that are of current interest in Riemannian
geometry. Unfortunately, the book itself is a bit sketchy; you use the
book to get a rough idea of what's interesting and then you go look up
papers and other books for more details. Or you work it all out yourself.
I have to admit that I learned a lot of differential geometry by getting
only a rough idea from a paper, a book, a lecture, or a conversation
and then working out the details myself.

One problem with differential geometry books is that they contain mostly
definitions and technical lemmas, rather than big theorems. If you want 
some interesting theorems to go along with the geometry, look at the 
following books on the topology of manifolds:
Guillemin and Pollack, Differential Topology
Milnor, Morse theory
Milnor, Topology from a Differentiable Viewpoint
(Guillemin and Pollack's book is essentially a rewrite of this
short monograph)
Warner, Foundations of differentiable manifolds and Lie groups

One last book on Riemannian geometry that is out of print but is a
modern classic:
Cheeger and Ebin, Comparison theorems in Riemannian geometry
Deane Yang
Polytechnic University
Newsgroups: sci.math
Subject: Re: Erthstns' Sv
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Date: 13 Aug 92 17:37:24 EDT
Distribution: world
Organization: Harvard Math Department

In article <1992Aug12.224159.14681@husc3.harvard.edu> O wrote:
>Some years ago I read of a variation of Erathostenes' Sieve that
>manages to compute the primes up to N in time o(N) [perhaps order
>N/log(log(N))?] by compressing the sieve using the first few
>primes (e.g. pre-sifting the multiples of 2,3,5,7 so one is
>dealing with only 1*2*4*6=48 potential primes out of every
>2*3*5*7=210 integers).  Unfortunately I have lost the reference.
>Can any member of the computational number theory contingent on
>sci.math provide a pointer to this paper?

Thanks to all who replied.  I have six references now, which should
be plenty.  I list them below, each one with the respondent who first
suggested it.

--Noam D. Elkies (elkies@zariski.harvard.edu)
 Dept. of Mathematics, Harvard University

SvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSvSv

[From Todd Fine (fine@sctc.com):]

Some New Upper Bounds on the Generation of Prime Numbers, Harry G. Mairson,
Communications of the ACM, September 1977, Volume 20, Number 9.

A Linear Sieve Algorithm for Finding Prime Numbers, David Gries and
Jayadev Misra, Communications of the ACM, December 1978.

A Sublinear Additive Sieve for Finding Prime Numbers, Paul Pritchard,
Communications of the ACM, January 1981, Volume 24, Number 1.


[From Jeff Shallit (shallit@graceland.uwaterloo.ca):]

Paul Pritchard:

Explaining the wheel sieve, Acta. Infor. 17 (1982), 477-485.

A sublinear sieve for finding prime numbers, Comm. ACM 24 (1981), 18-23.
Corrigenda, 24 (1981) 772.

Linear prime number sieves:  a family tree. 
Sci. Computer Prog. 9 (1987), 17-35.


[From Andrew Odlyzko (amo@research.att.com):]

Fast compact prime number sieves (among others),
{\em J. Algorithms 4} (1983), 332--344.
From: bill@waikato.ac.nz
Newsgroups: sci.math
Subject: Primes
Date: 17 Aug 92 16:02:52 +1200
Organization: University of Waikato, Hamilton, New Zealand

How many primes are there less than 2 to the power of p?

Is there a way of calculating this, or does anyone know of where I can get my
hands on a list of primes so I can find out how many primes there are
less than 2^16 or 2^32 for example?

Is there some way of determining (or estimating) the expected magnitude of the
nth prime? Or of the growth of the difference between the nth and (n+1)th
prime?

Bill Teahan,
Systems Programmer,
University of Waikato,
Hamilton, New Zealand.
"I have never been lost but I will admit to being confused for several weeks."
-- Daniel Boone
Newsgroups: sci.math
From: daf@herky.cs.uiowa.edu (David Forsyth)
Subject: Re: Collineations
Date: Wed, 12 Aug 1992 20:58:43 GMT
Organization: University of Iowa, Iowa City, IA, USA

From article <1992Aug12.194204.24356@uwm.edu>, by radcliff@csd4.csd.uwm.edu (David G Radcliffe):
> Suppose f is a one-to-one function from the plane to itself
> which maps lines into lines, and suppose the image of f is 
> not contained in a line.  Must f be affine?
> 
> I have found some partial results.  The restriction of f to
> Q^2 is equal to the restriction of a projective transformation.
> If one also assumes that f is surjective, or continuous, or that
> it preserves betweenness, then f must be affine. 
> 
> --
> David Radcliffe                                
> radcliff@csd4.csd.uwm.edu 
If memory serves me correctly, for the projective plane, the requirement
that f be a collineation and one-to-one guarantees that f is a projective
transformation.  This would mean that for the affine case, f would be a projective
transformation that kept the line at infinity at infinity, hence an affine
transformation.  I think the first fact (result?) is in Semple and Kneebone, 
Algebraic Projective Geometry, Oxford University press, but they probably didn't
prove it all that rigorously.

Regards

David Forsyth
.

:wq
From: wilson@web.ctron.com
Newsgroups: sci.math
Subject: Re: u(v^n)w prime puzzle (SPOILER)
Date: 21 Aug 92 15:11:07 GMT
Reply-To: wilson@web.ctron.com ()
Organization: Cabletron Systems INc.
Originator: wilson@web



    Let B >= 2.  For string s over the base-B digits, let val(s) be the
    integer value of s as a base-B numeral (if s is the empty string,
    val(s) = 0).

    Choose strings u, v, and w of base-B digits, and for integer n >= 0,
    let

                     n
	f(n) = val(uv w)

    Since val(s) >= 0 for all s, f(n) >= 0 for all n.  We wish to show
    the either f(n) is a constant function of n, or that some f(n) is
    composite.

    It is not hard to show that there exist nonnegative integers a, b, c
    such that a >= 0, b >= 0, and

	f(n) =	a		if n = 0
		b*f(n-1)+c	if n > 0.

    (The values of a, b, and c in terms of u, v, and w are left to the
    interested reader).

    If b = 0, or if b = 1 and c = 0, f(n) is constant.

    Otherwise, f(n) is increasing in n, and we wish to show that
    some f(n) is not prime.  Suppose, then, that all f(n) are prime.

    Since f(n) increases in n, choose n with f(n) > b.  Since f(n) is
    prime, we have (b, f(n)) = 1.  Let p = f(n).

    Now observe the residues of the infinite sequence f(n), f(n+1), ...
    mod p.  Because of the recurrence f(n) = b*f(n-1)+c, and the number
    of residues mod p is finite, the sequence of residues eventually
    becomes cyclic.

    Suppose the residue cycle starts at f(n+j), j >= 1.  Clearly, the
    cycle has some length k <= p.  Since f(n+j) is in the cycle, we have

	f(n+j) == f(n+j+k) (mod p)

    However, f(n+j-1) is not in the cycle, so

	f(n+j-1) ~== f(n+j+k-1) (mod p)

    Because (b, p) == 1

	b*f(n+j-1)+c ~== b*f(n+j+k-1)+c (mod p)

    and by the recursion for f(n),

	f(n+j) ~== f(n+j+k) (mod p)

    a contradiction.  We must therefore conclude that j = 0, and the
    cycle starts at f(n), so that

	f(n) == f(n+k) (mod p).

    But f(n) = p, so

	f(n+k) == 0 (mod p).

    also, f(n) is increasing, so f(n+k) > f(n) = p.  This means that
    f(n+k) is a composite multiple of p, as required.

		-------<>------

    Let us look at an example.  Let B = 10, u = "1", v = "36", w = "1".
    We then have

	f(0) = 11
	f(1) = 1361
	f(2) = 136361
	f(3) = 13636361
	f(4) = 1363636361
	...

    all primes so far.  How do we locate the composite?

    First, we find the recursion

	f(0) = 11
	f(n) = 100 f(n-1) + 261

	(a = 11, b = 100, c = 261)

    Now, we choose n so that (f(n), b) = 1.  In this case, n = 0, and
    f(n) = 11 works fine.  Let p = f(n) = 11.

    Next, look at the residues of f(n), f(n+1), ... mod 11.  They are

    n			0  1  2  3  4  5  6  7  8  9 10 11
    f(n) (mod 11)	0  8  5  2 10  7  4  1  9  6  3  0

    At n = 11, we see that f(n) == 0 (mod 11), as guaranteed by the
    theorem, and so 11 | f(11).  Since f(n) increases, f(11) > 11,
    so f(11) is composite.

-- 
David W. Wilson (wilson@ctron.com)

Disclaimer: "Truth is just truth...You can't have opinions about truth."
- Peter Schikele, introduction to P.D.Q. Bach's oratorio "The Seasonings."
Newsgroups: sci.math
From: algebra@leland.Stanford.EDU (Leong Weng Ng)
Subject: Re: Still another problem.
Organization: DSG, Stanford University, CA 94305, USA
Date: Wed, 12 Aug 92 23:09:56 GMT

In article <1992Aug11.170858.275@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>Prove that
>
>               n     n     n    n
>     1  /     2     3     4    5        \  
>     - |  1 + -- +  --  + -- + -- +....  | 
>     e  \     1!    2!    3!   4!       /   
>
>  is an integer for all positive integer n.
>
>---------------------------------------------------------------------
>      Bill Taylor              wft@math.canterbury.ac.nz 
>---------------------------------------------------------------------
>              Stamp out silly signatures!
>---------------------------------------------------------------------
>
The expression above is just the nth derivative of 
e^(e^x+x-1) at x = 0. The nth derivative is 
(n*e^x+1)*e^(e^x+x-1), as can be verified by induction.
Substituting x = 0 shows that the expression reduces 
to n+1.

Weng Leong
Stanford U.
Newsgroups: sci.math
Subject: Re: u(v^n)w prime puzzle - HINTS
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 20 Aug 92 17:33:17 GMT

In <1992Aug20.161823.20996@wri.com> roach@bikini.wri.com (Kelly Roach) writes:

>In article <1992Aug20.043232.5087@nuscc.nus.sg> bhonsle@bhonsle.iss.nus.sg  
>(Shailendra K Bhonsle) writes:
>> |>      (2) Let a=|v|=length of v, b=|w|=length of w.
>> ...
>> As I mentioned in my last article let us take prime p="uw".
>> ...
>> Now we show that p cannot divide 10^a -1 to complete the proof.
>>      case 2: b <= a
>>              SIMPLE, Can you do it ?
>> ...


>     Consider:

>     u="1",v="36",w="1"
>     11, 1361, 136361, 13636361, 1363636361, 136363636361

>Here, a=2, p=11, 10^a-1=99, and p | 10^a-1.  So, yes, it
>is possible for p to divide 10^a-1.  Some more work on this
>proof needs to be done before it is complete I think.

uvw can't divide 10^a-1; it's larger!

--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; if you want to be sure I see a post, mail it.