HDTenEightyP/NetNews huggingface dataset:
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: Why don't we split the group?
Date: 8 Sep 92 14:56:29 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Sep4.135255.19682@infodev.cam.ac.uk> cet1@cus.cam.ac.uk (C.E. Thompson) writes:
>You have forgotten

>sci.math.existence-of-god

Actually, several of the posters in that thread have invited moving the
discussion to other groups - (sci.philosophy.meta, for example). However
the thread originated in sci.math as a question about what can be proved
mathematically. I think the concensus is that a correct mathematical proof
of God is unlikely, certainly bound to be controversial, and not terribly
convincing to theologians, although perhaps of some interest.

Maybe there should be an entry in the FAQ, since I suspect this one will
come up again, and being the FAQ seems to be a more effective way of
controlling the subjects in the group.

Later,
Andrew Mullhaupt
From: pa@verano.sba.ca.us (Pierre Asselin)
Newsgroups: sci.math
Subject: Reference for de Rham's theorem?
Summary: Need introductory reference to de Rham's theorem.
Keywords: closed exact differential forms
Date: 13 Sep 92 21:07:29 GMT
Organization: None.  Santa Barbara, CA

I need a reference for the following facts:

  -In a differentiable manifold where every closed curve is
   continuously deformable to a point, a 1-form is a gradient
   iff its exterior derivative is zero.

  -In a differentiable manifold where every closed surface is
   continuously deformable to a point, a 2-form is the exterior
   derivative of a 1-form iff its exterior derivative is zero.

In other words, I need sufficient conditions for closed k-forms to be
exact, for k= 1,2.  This is for a manuscript where algebraic topology is
the *least* of my concerns.  I just need to get it out of the way.  The
Encyclopaedia Britannica tells me that I'm looking for de Rham's
theorem.

What would be the standard, vanilla, everybody-knows-that, available-
everywhere, reference for de Rham's theorem?  The more elementary the
better.

My site doesn't receive sci.math;  please reply by email.  Thanks.
-- 

--Pierre Asselin, Santa Barbara, California
  pa@verano.sba.ca.us
Xref: sparky sci.math:11273 comp.lang.misc:3025 comp.lang.fortran:3519 comp.std.c:2607 comp.lang.c:13533
Newsgroups: sci.math,comp.lang.misc,comp.lang.fortran,comp.std.c,comp.lang.c
Subject: Re: Scientists as Programmers (was Re: Small Language Wanted)
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 11 Sep 92 15:30:08 GMT
 <1992Sep4.162938.29237@dscomsf.desy.de> <Bu2r42.5xr@mentor.cc.purdue.edu> 
 <id.I12T.GP4@ferranti.com> <a_rubin.716152347@dn66> <1992Sep11.012559.17662@nntpd.lkg.dec.com>

In <1992Sep11.012559.17662@nntpd.lkg.dec.com> diamond@jit345.bad.jit.dec.com (Norman Diamond) writes:

>In article <a_rubin.716152347@dn66> a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
>>In <id.I12T.GP4@ferranti.com> peter@ferranti.com (peter da silva) writes:
>>>In article <Bu2r42.5xr@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:
>>>>[nxn -> 2n multiplication]
>>>	int a, b;   long c;
>>>	c = a * b;
>>(peter commented further that this works on a PDP-11).
>>Aside from there being no guarantee that "long" is longer than "int",

>>I don't think it is LEGAL for ANSI C to do this, although, I, also have seen
>>it work in some C implementations.

>ANSI C leaves the result of overflow undefined.  But I think the standard
>could pretty well define whatever it wanted without breaking higher laws.

I don't have a copy of the C standard, but,...

My interpretation (from reading K&R 2) would be that c = a * b should be
equivalent to c = (int)(a * b).
--
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; please mail anything important.
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Re: 2n = difference of primes?
Date: 5 Sep 92 03:35:19 GMT
Article-I.D.: husc3.1992Sep4.233520.15379
Organization: Harvard Math Department

In article <92248.16510732WENPW@CMUVM.CSV.CMICH.EDU>
32WENPW@CMUVM.CSV.CMICH.EDU writes:
>Question:  Is it true that every [even] positive integer 2n is the
>difference of some pair of primes?
>(2 = 5-3, 4 = 7-3, 6 = 11-5, 8 = 13-5, 10 = 13-3, ...?)

Yes.  Or rather, it is next-to-certainly true.  But nobody at present
knows how to prove it for even one value of n.  For instance (n=1) the
twin-prime conjecture is still open.

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Dept. of Mathematics, Harvard University
From: bradtke@greed.cs.umass.edu
Newsgroups: sci.math
Subject: Terminology question
Date: 10 Sep 92 13:23:47 GMT
Reply-To: bradtke@greed.cs.umass.edu ()
Distribution: usa
Organization: University of Massachusetts, Amherst


	Is there a standard name and symbol for the operation of
multiplying two matrices element by element?  That is, if 
A op B = C, then C_ij = A_ij * B_ij.

Thanks,

Steve

bradtke@cs.umass.edu


Newsgroups: sci.math
From: syrjanen@tnclus.tele.nokia.fi
Subject: Re: Chess Problem
Organization: Nokia Telecommunications.
Date: Mon, 14 Sep 1992 10:22:12 GMT

In article <92256.155713RVESTERM@vma.cc.nd.edu>, RVESTERM@vma.cc.nd.edu writes:
> by "a win for white," do you mean immediately, or a guaranteed win for white
> if the game continues?
.. A good point!  BTW, if you promote the pawn into a knight or bishop,
    it's most likely a draw  ;-)
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Help with Calculus
Date: 4 Sep 92 21:48:16 GMT
Organization: U of Cambridge, England

In article <18531.2aa39479@levels.unisa.edu.au>, 8321207d@levels.unisa.edu.au writes:

> 1)  Let Phi(a)=Int from 0 to Inf of [exp(-ax)*Sinx/x]dx. Explain briefly
>     why it is plausible to write Phi(a)=Pi/2 - ArcTan a. Could this 
>     result be used to evaluate Int from 0 to Inf of [Sinx/x]dx ?

Well, how about differentiating under the integral sign? You get (d/da)Phi
equal to integral of -x.exp(-ax).sin(x)/x dx
       = integral of -exp(-ax).sin(x) dx
and you can do this, for instance, with complex numbers. (sin x is the imaginary
part of exp(ix), so you want the imaginary part of the integral of -exp((a+i)x);
this is dead easy.)
Since you only ask for "Why it is plausible" I don't need to prove that it's
*valid* to differentiate under the integral sign; but there are tests for when
you can do it, and I'm sure one of them applies here...

Um, you still have to work out what the constant is; try letting a tend to
infinity; then the integral tends to 0 because it is dominated by the integral
of exp(-ax), which tends to 0.

If needed I can provide all the gory details.

> 2)  Let Psi(r)=Int from 0 to Inf of [Sin rx/x(1+x^2)]dx for r >= 0.
>     Explain briefly why it is plausible to suggest that
>     Psi''(r) - Psi(r) = Pi/2 with Psi(0)=0 and Psi'(0)=Pi/2. Hence
>     deduce that Psi(r) = Pi/2*(exp(r) - 1).

Again, let's try differentiating under the integral sign.
Psi'(r)  = integral of x.cos(rx)/x(1+x^2) dx
         = integral of cos(rx)/(1+x^2) dx
Psi''(r) = integral of x.-sin(rx)/(1+x^2) dx
         = integral of -sin(rx).[x/(1+x^2)] dx.
So Psi''-Psi = integral of sin(rx)/(1+x^2).[x+1/x] dx
             = integral of sin(rx)/x dx
and we know what this is by the previous part.
The boundary conditions can't be hard. When r=0 the integrand for Psi is
0 everywhere, and that for Psi' is 1/(1+x^2); we know how to integrate that,
or ought to.

Hope all this helps... Good luck with whatever further studies you're
intending.
From:  (m19701)
Newsgroups: sci.math
Subject: looking for old calculus book
Date: 11 Sep 92 13:28:13 GMT
Followup-To: sci.math
Organization: mitre.org

Greeting y'all.  I need help.

My officemate, Rosemary (m22698), is in search of her old Calculus book. 
She has looked around the DC area with no luck.  My ribbing her about her
search has forced me to put up or shut up.  She bet me a freshly baked cake
(or pie) that I couldn't find a copy while I bet her lunch out that I
could.  Now I'll probably take her out to lunch anyway but I really want a
freshly baked blueberry pie and the season is almost over.

Name of book:		Calculus: With Analytic Geometry
															by Roland LARSON and Robert HOSTETLER
															copyright 1980  by D.C. Heath and Company

It comes with a 3 volume Solution Guide which is also desired.  Solution
Guide has one additional auther  David HEYD.

Apparently there may also be a Student Solutions Manual by HEYD.

Rosemary is willing to buy these books.

Thanks in advance.


ed casey

(insert standard disclaimer here)
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: sci.math
Subject: Re: That Collatz Flu
Date: 14 Sep 92 15:19:16 GMT
Organization: Department of Computer Science, U of Copenhagen

levine@symcom.math.uiuc.edu (Lenore Levine) writes:

>(Collatz function: f(n) = n/2 if n is even
>                        = (3n + 1)/2 if n is odd.

>It is conjectured that for every n > 0, there is a k such that f^k(n)
>(the kth iterate of f) = 1.)

>What I'm wondering is: Has it been shown that there are no n > 2 such 
>that f^k (n) = n?

Obviously not, as this would disprove the conjecture: if f^k(n) = n,
then either there is no j s.t. f^j(n) = 1, or else n=1 or n=2.

In general, a sequence of integers will either be non-repeating (and
thus grow arbitrarily large) or else eventually cycle with period p
(where p may be 1). The only known cycle in Collatz sequences is
1->2->1.

	Torben Mogensen (torbenm@diku.dk)
Newsgroups: sci.math
From: rabin@CS.Yale.Edu (Dan Rabin)
Subject: Re: computable reals
Reply-To: rabin-dan@cs.yale.edu
Organization: Computer Science, Yale University, New Haven, CT 06520-2158
Date: Sat, 12 Sep 1992 15:52:27 GMT

Bruce Lucas asks about the set of computable reals.  

Unfortunately, if you're hoping to use them in programs, numeric
comparisons are undecidable: Asking whether any arbitrary computable
real is zero is exactly the same as asking whether its defining
digit-producing program ever cranks out a non-zero digit.  That's not
decidable.  Oops.

Of course, this is not a deficiency with respect to the entire set of
real numbers, which has this problem plus the problem that some
(almost all, actually ;-))of its
elements aren't finitely representable in the first place.  It's only
a deficiency if you thought that moving from floating point to
computable-reals in a computer program would be a smooth transition.

There is a literature on this subject within computer science.  
Here are a couple of initial pointers:

@incollection{boehm-cartwright:exact-reals,
  author = 	{Hans Boehm and Robert Cartwright},
  title = 	{Exact Real Arithmetic: Formulating Real Numbers as
                Functions},
  editor =	{David A. Turner},
  chapter = 	{3},
  booktitle = 	{Research Topics in Functional Programming},
  year = 	{1990},
  series = 	{The UT Year of Programming Series},
  publisher = 	{Addison-Wesley Publishing Company, Inc.},
}

@InProceedings{vuillemin:exact-real,
  author = 	"Jean Vuillemin",
  title = 	"Exact Real Computer Arithmetic with Continued
		 Fractions",
  pages = 	{14--27},
  booktitle = 	"Proceedings of the 1988 {ACM} Conference on {L}isp
		 and Functional Programming",
  year = 	"1988",
  month = 	"July",
}

  -- Dan Rabin
     rabin-dan@cs.yale.edu
Newsgroups: sci.math
From: frank@csri.toronto.edu (Frank Kschischang)
Subject: D^k f^n ?
Keywords: differentiation, combinatorics
Organization: Electrical Engineering, University of Toronto
Date: 10 Sep 92 21:26:04 GMT


Let f(x) denote some function of x.  Let D=d/dx denote the
operation of differentiation with respect to x.

Question:
Has anybody ever worked out a formula for D^k f^n, the
k'th derivative of the n'th power of f(x) with respect to x?

Here `n' is a positive integer, and D^k f is assumed to
exist for all positive k.

-- Frank Kschischang
   frank@comm.utoronta.ca
From: masfeb@gdr.bath.ac.uk (F E Burstall)
Newsgroups: sci.math
Subject: Re: Couple of questions
Date: 9 Sep 92 18:23:28 GMT
Organization: School of Mathematics, University of Bath, UK

In article <1992Sep9.102457.15049@news.columbia.edu> pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:
>Could someone please explain to me a few basic things:
>
>What is the logic used to answer questions of the below type?
>
>1) If the finite group G contains a subgroup of order 7 but no element
>(other than the identity) is its own inverse then the order of group G
>could be (a)27   (b)28   (c)35   (d)37   (e)42.
>
>It says that the correct anwer is (c). I know why we can eliminate (a)
>and (d) but don't know how to proceed further.

Two things are happening here:

1) Good old Lagrange's theorem says the order of the subgrp divides the order
of the grp so the order is a multiple of 7--this, as you know, eliminates (a)
and (d).

2) Since no element is its own inverse except the identity, we can pair off
elements with their inverses and conclude that there is an EVEN number of
non-identity elements and thus an odd number of elements overall.  This
eliminates (b) and (e). Voila!


>Also could someone explain what is elementary Riemann and Lebesgue
>integration, and why fuctions can be integrable but now Riemann
>itegrable etc? I looked over a couple of books but did not quite get
>it.

This is a long story and comes down to the definitions of the different types
of integral.  Given you understand these definitions, here is a quick example
of a non-Riemann integrable function on [0,1]: define f to be 0 on rational
numbers and 1 on irrational numbers.  This fn is not Riemann integrable
(whatever the partition, the upper sum is 1 and the lower sum is 0) but has
Lebesgue integral 1 (since f=1 off a set of measure zero)

>Thanks

Da nada.

Fran
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: exotic R^4
Date: 12 Sep 92 17:52:20 GMT
Distribution: sci
Organization: M.I.T. Artificial Intelligence Lab.

At the moment I am in touch with this Donaldson-Freedman stuff only by rumor.
I am under the impression that the following assertions are true:
(1) There is a manifold X homeomorphic to R^4 but not diffeomorphic to R^4.
(2) All manifolds homeomorphic to R^5 are diffeomorphic to R^5.

Assuming that is correct (which I do not guarantee), then it seems to me that
XxR is necessarily diffeomorphic to R^5, i.e. X is diffeomorphic to a
hypersurface in R^5. So has anyone written down an explicit mapping
f:R^5->R such that X is the preimage of 0 under f? How good or bad
could such a mapping f be? For example, could it be a polynomial?

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math
From: platt@watson.ibm.com (Daniel E. Platt)
Subject: Re: Lebesgue integral (was: Couple of questions
Date: Tue, 15 Sep 1992 15:38:11 GMT
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM T.J. Watson Research Center

In article <1992Sep15.020445.26398@unixg.ubc.ca>, ramsay@unixg.ubc.ca (Keith Ramsay) writes:
|> In article <1992Sep11.130033.26063@watson.ibm.com> 
|> platt@watson.ibm.com (Daniel E. Platt) writes:
|> [...taking out some layers of quotes...]
|> |> In other words, how badly would you miss them if you threw out the
|> |> non-Riemann-square-integrable functions from L^2(R^3)?
|> ...
|> |I think you would end up flat on your back.  The problem is that many
|> |(most) of the techniques revolving around Fourier series and
|> |integrals, completeness of a basis, etc, ultimately involve being able
|> |to evaluate 'improper' integrals as a limit of an integral of a
|> |sequence of functions.
|> 
|> I've heard that much (nearly all, perhaps) of what is done with L^2,
|> regarded as Lebesgue square-integrable functions, modulo functions
|> supported on sets of measure zero, is just as nicely done (if not
|> better) by regarding L^2 as the formal completion (relative to the L^2
|> norm) of a convenient dense subset (where the Lebesgue measure theory
|> is not needed). I'm not enough of an analyst to confirm this.

That's true if you are only talking about proving convergences; if you
want to get more formal about how/why Dirac delta's work/what they mean,
and how the broad diversity of the limits under the integral work, then
some comment should be made about the added convergence that Lebesgue
offers that Reimann doesn't.  Most of the time, physics majors play
with these delta 'functions' and don't even realize they aren't functions
per se.  When something acts pathological, they don't know why.

Dan
Newsgroups: sci.math
From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
Subject: Conway Joke?
Organization: cpd
Distribution: sci.math
Date: Thu, 10 Sep 1992 08:09:52 GMT

Can anyone explain the following joke?

In Winning Ways, by Birkamp, Conway and Guy, in Volume 2, where it talks
about topological puzzles, magic mirrors, magic movies and things, there
is a double page full of puzzles, lettered (a)-(j). They all seem fairly
easy [with the possible exception of (g), which I can't do in my head],
except for (j), which is as follows (somewhat distorted):
        ___________
       |           |
       | Ring will |
       |  not fit  | These square things are actually
       | over this | shiny round spheres
       |_____  ____|
           __||__
          |  __  | Eye
        __| |__| |__
       / _| |__| |_ \
      ( (_|_|__|_|_) ) Ring
       \____________/
          | |  | |
        ,-+-+--| |-.
        | | |__| | | String (long)
        | |______| |
      __|___    ___|__
     |      |  |      |
     |  or  |  |  or  | More shiny round spheres
     | this |  | this |
     |______|  |______|
       _||_      _||_
      | __ |    | __ |
      ||__||    ||__|| These are clearly useless
      |____|    |____|

The ring will fit through the eye.
Everything except the string is rigid.
The puzzle is to remove the ring.

The book, for those who haven't come across it, is riddled with riddles
and choked with jokes, so this could be another, but if so then I'm
suffering from a sense of humour failure. Maybe it's a (bad) picture of
one of the authors by another??

Any ideas?
--
  ____________________________________________________________________________
 /  _Name: Adam|Names are linguistic constructs expressed in some language._  \
|\_|/|Chalcraft|They correspond to objects in some universe of discourse. |\|_/|
|`   |         |The corresponence between names (in the language) and     |`   |
|`   |Opinions:|objects (in the universe of discourse) is the relation of |`   |
|`   |     Mine|identifying. A name identifies the object to which it is  |`   |
|`   |_________|bound.__________________________ISO_7498-3_:_1989_(E)_5.1_|`   |
 \__/                                                                      \__/

Newsgroups: sci.math
From: kmc@netcom.com (Kevin McCarty)
Subject: Re: Lebesgue integral (was: Couple of questions
Date: Thu, 10 Sep 92 21:22:00 GMT
Organization: Self

In article <1992Sep10.173619.24343@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>In article <18neu6INN32k@function.mps.ohio-state.edu> edgar@function.mps.ohio-state.edu (Gerald Edgar) writes:
>
>>I have been told that the Lebesgue integral is not needed in physics--
>>presumably Riemann integral is enough.  Do you agree?
>
>Last I heard, the Hilbert space of a free particle was L^2(R^3), the
>space of all functions whose absolute value squared is LEBSEGUE integrable.

But is this a question of convenience or necessity?
In other words, how badly would you miss them if you threw out the
non-Riemann-square-integrable functions from L^2(R^3)?
Could still you do physics with but a slight limp, or would you be
flat on your back?
-- 
Kevin McCarty (kmc@netcom.com)
Newsgroups: sci.math
From: rusin@mp.cs.niu.edu (David Rusin)
Subject: Re: exotic R^4
Organization: Northern Illinois University
Distribution: sci
Date: Mon, 14 Sep 1992 17:47:01 GMT

In article <ARA.92Sep12125220@camelot.ai.mit.edu> ara@zurich.ai.mit.edu (Allan Adler) asks 
if it is possible to find exotic R^4's as hypersurfaces in R^5. I am
not a person  who would know but I can point to a related example. The
first exotic manifolds were the 27 non-standard structures on S^7
deduced by Milnor. (1957 I think). It turns out that they can be described
easily in the way proposed. Indeed, let  V  be the hypersurface in C^5
described by the equation

           4k+1    3     2     2     2
         z      + z   + z   + z   + z   = 0
          1        2     3     4     5

Then V  has one singular point at the origin, so that V - {0} is a
complex 4-manifold (real 8-manifold). You can show it meets the unit
sphere S^9 in  R^10 transversely, so that the intersection is a
real 7-manifold  M.  That  M  is a topological 7-sphere is not hard
(see Milnor's Singular Points of Complex Hypersurfaces). However,
the manifolds  M  are only diffeomorphic if the  k's are comgruent
mod 28. (I think this is due to Brieskorn. maybe it was
Hirzebruch.)
	I love to show this example to beginning topology grad students.
If a similar example is available for R^4 I would like to see it.

dave rusin@math.niu.edu

Newsgroups: sci.math,comp.lang.misc,comp.lang.fortran,comp.std.c,comp.lang.c
From: meissner@osf.org (Michael Meissner)
Subject: Re: Scientists as Programmers (was Re: Small Language Wanted)
Organization: Open Software Foundation
	<1992Sep4.162938.29237@dscomsf.desy.de>
	<Bu2r42.5xr@mentor.cc.purdue.edu> <id.I12T.GP4@ferranti.com>
	<a_rubin.716152347@dn66>
Date: 11 Sep 92 16:44:41

In article <a_rubin.716152347@dn66> a_rubin@dsg4.dse.beckman.com
(Arthur Rubin) writes:

| Using, instead, 
| 
|     c = (long)a * b;
| 
| it becomes a "quality of implementation" issue; and using the 4/2->2,2
| divide instruction would incorrect on 80x86 machines, as the divide
| instruction would overflow in cases where the C operations would be legal.

With most optimizing compilers, this is pretty simple to put in, since
most optimizations are essentially pattern matching (you match the two
converts, and long multiply).  Some time ago, I put long long multiply
code into the MIPS port of GCC for instance in about 2 minutes of
time.  The code:

	long long mul (long a, long b) { return (long long)a * b; }

produces the following code:

	mul:
		mult    $4,$5
		mfhi    $3
		mflo    $2
		j       $31
--
Michael Meissner	email: meissner@osf.org		phone: 617-621-8861
Open Software Foundation, 11 Cambridge Center, Cambridge, MA, 02142

You are in a twisty little passage of standards, all conflicting.
Xref: sparky sci.math:11077 sci.philosophy.tech:3133