HDTenEightyP/NetNews huggingface dataset:
From: jerry@cs.mcgill.ca (Gerald (Jerry) KUCH)
Newsgroups: sci.math
Subject: Re: Movie mathematicians
Date: 26 Sep 92 14:38:17 GMT
Organization: McGill University, Montreal, Canada

In article <Bv5F08.LMH@unix.amherst.edu> mkrogers@unix.amherst.edu (MICHAEL K ROGERS) writes:
>: ags@seaman.cc.purdue.edu (Dave Seaman) writes:
>: >Another movie mathematician is Bonnie Bedelia in the movie she did with  
>: >Harrison Ford. I think the title was something like "Presumed Innocent."
>
>Bonnie Bedelia visited Columbia University and consulted a grad student
>there in preparing for her role.  Not having seen the movie, I can't
>say if this had any effect.

If memory serves, the sum of her math graduate student role consisted of
sitting around in the background with a bunch of books in front of her and
occasionally griping about how long it's taking to finish her degree.  I've
never seen anybody like that in real life... :-)


-- 
 Jerry Kuch (jerry@cs.mcgill.ca) ---- PUT YOUR HAND INSIDE THE PUPPET HEAD.
 "I was wrong to play God.  Life is precious, not a thing to be toyed with.
  Now take out that brain and flush it down the toilet."
                  --- M. Burns  "Treehouse of Horror II"
From: somos@ces.cwru.edu (Michael Somos)
Newsgroups: sci.math
Subject: Re: Z <-> Q formula
Date: 28 Sep 1992 05:26:55 GMT
Organization: Computer Engineering and Science, Case Western Reserve University
NNTP-Posting-Host: george.ces.cwru.edu

In article <1992Sep27.085952.7534@Csli.Stanford.EDU>
 hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:

>The set of intergers Z and the set of rationals Q have the same cardinality,
>i.e. there exist a bijection between them.  So we all know that.
>But then, is there a "simple" formula that expresses any such mapping?
>You may use any device you please, but note that the subjective little
>"simple" is the criterion.

We are trying to find a mapping  n -> p(n)/q(n) from the integers to
the rationals that is bijective.  Let p(-n)=-p(n), q(-n)=q(n) to take
care of negative numbers, and let p(0)=0, q(0)=1 to take care of zero.

There is a procedure to generate all positive rationals described in
detail in "Concrete Mathematics".  Essentially they are all generated
in a binary tree starting from 0/1 and 1/0 using mediants.  Thus:

 0/1                                                             1/0
  :                              1/1                              :
  :              1/2              :              2/1              :
  :      1/3      :      2/3      :      3/2      :      3/1      :
  :  1/4  :  2/5  :  3/5  :  3/4  :  4/3  :  5/3  :  5/2  :  4/1  :
  :1/5:2/7:3/8:3/7:4/7:5/8:5/7:4/5:5/4:7/5:8/5:7/4:7/3:8/3:7/2:5/1:
  : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :

Each fraction has its numerator and denominator the sum of the same
parts of two preceeding fractions (the mediant).  For example:

      4 = 1 + 3      5 = 3 + 2
      -   -   -      -   -   -     etc...
      7 = 2 + 5      8 = 5 + 3

Now number each layer in order starting from 1/1 to generate the list:

  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 ...

  1  1  2  1  2  3  3  1  2  3  3  4  5  5  4  1  2  3  3  4  5  5  4
  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  - ...
  1  2  1  3  3  2  1  4  5  5  4  3  3  2  1  5  7  8  7  7  8  7  5

The recursive formula for p(n)/q(n) is given by:

  p(1) = q(1) = 1, p(n) = f(n,0), q(n) = f(n,1)
  f(n,t) = f(floor(n/2),t) + g(ceiling(n/2),t,mod(n,2))
  g(1,t,e) = mod(t+e, 2),
  g(n,t,e) = f(floor(n/2),t) if n odd, g(n/2,t,e) if n even

How "simple" this is I leave to the reader.
--
Shalom, Michael Somos <somos@alpha.ces.cwru.edu> * My opinions only *
-- 
Michael Somos <somos@alpha.ces.cwru.edu> or <aj009@cleveland.freenet.edu>
From: bennett@math.ksu.edu (Andy Bennett)
Newsgroups: sci.math
Subject: Re: Mattel says "Math is tough."
Date: 5 Oct 1992 16:20:45 -0500
Organization: Dept. of Mathematics, Kansas State University
NNTP-Posting-Host: hilbert.math.ksu.edu

mroussel@alchemy.chem.utoronto.ca (Marc Roussel) writes:
>There is
>nothing intrinsically difficult about elementary school mathematics but
>the attitudes developed at this age will last a lifetime.  

No and Yes. Elementary school mathematics is clearly harder than calculus.
The difficulty is that we learned it so long ago that it seems so natural
now that we can't recognize the difficulty anymore. Place value and
regrouping are concepts as deep as anything in calculus and arithmetic
algorithms (especially with fractions) are at least as bad as the quotient 
rule and the chain rule. Algebra is harder than either subject, which
is why half of students drop math after algebra and why it is algebra
errors rather than calculus errors that predominate on calculus exams.
Of course you can argue that neither calculus nor elementary school
mathematics are intrinsically difficult and you might be right.

This is not meant to be a flame at Mr. Roussel whose posts I generally
enjoy. But mathematicians talking about helping young children need to
be aware that the material is decidely non-trivial for most people when
they first learn it. 

It is also definitely correct that student attitude is extremely important,
which is why my daughter's (extensive) Barbie collection won't include
Teen Talk Barbie. 

-- 
Andrew G. Bennett         bennett@math.ksu.edu         If you count too
Dept. of Mathematics      Voice: (913) 532-6750        much you turn
Kansas State University   Fax:   (913) 532-7004        purple.  -  SARAH
Manhattan, KS 66502       STRICTLY MY OWN OPINIONS                    
Xref: sparky sci.math:12439 sci.math.symbolic:2582
From: brnstnd@nyu.edu (D. J. Bernstein)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: Integral Tables
Date: 29 Sep 92 20:00:24 GMT
Organization: IR

In article <1992Sep28.173211.6153@alchemy.chem.utoronto.ca> mroussel@alchemy.chem.utoronto.ca (Marc Roussel) writes:
>      Macsyma is probably the most reliable of the symbolic packages.  In
> this case however, its behaviour is unsatisfactory.

It's been long enough since I used Macsyma that I don't recall whether
it would automatically handle range checks on the definite integrals you
mention; certainly it was never as ignorant as Mathematica. I do agree
that the available symbolic math packages just don't know enough math.

I'll know computer math is improving when I see a correct graph-drawing
program. Such a program would produce a _picture_ of any graph given by
an equation in two variables; by definition, a picture of a graph is a
set (of dots, of ink, whatever) which contains the graph. Of course, a
big black square is a picture of any graph, but a high-quality program
will produce pictures which are close to minimal.

---Dan
Xref: sparky sci.math:12255 sci.math.symbolic:2544
From: bumby@dimacs.rutgers.edu (Richard Bumby)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: Integral Tables
Date: 26 Sep 92 20:09:28 GMT
Followup-To: sci.math
Organization: Rutgers Univ., New Brunswick, N.J.

john@aquarius.unm.edu (John Prentice) writes:

>In article <1992Sep24.163552.12172@alchemy.chem.utoronto.ca> mroussel@alchemy.chem.utoronto.ca (Marc Roussel) writes:
>>     I have been thinking about this matter a little in the last year or
>>two and have come to the conclusion that symbolic algebra systems should
>>be more conservative than they now are. . . .

>I strongly endorse what Marc is saying here!  What is amazing is that it
>has to be said, it is simple good software design.

My own experience here is mostly limited to use of MapleV on a small
machine.  However, I have also quotes of integral tables or other
symbolic systems in connection with problems proposed in the American
Mathematical Monthly.  These sources are often not very careful about
selecting the base point (the point where the given integral will be
zero).  The most annoying effect is having a solution produced in a
portion of the complex plane which has been slit along the very
interval that you are considering.  In Maple, the base point seems to
be chosen to be infinity wherever possible, even when integrating
something involving sqrt(1-x^2), so this effect is fairly common.
There is some indication that the user will be given greater control
over this in versions currently being developed.  It may be that the
promise of having a general and accurate answer available on demand is
responsible for some thought being given to what kind of answer is
possible.  There is something that considers it unsatisfactory for an
"answer" to be "there are several possible answers, which do you
want?".

A good test question for any system would be Problem 6653 from the
Monthly (March 1991), whose solution is scheduled for publication in
December 1992.  The integral in question is a definite integral from 0
to Pi of a complicated expression involving an inverse cosine of a
complicated trigonometric expression.  The integral depends on a
parameter which is restricted to being a non-negative real number in
the statement of the problem.  However, some standard techniques break
down when the parameter is equal to 1.  When the parameter is greater
than 1, the result can be found in the table of Gradshteyn & Ryzhik.
Some of our more persistent solvers were able to evaluate the integral
for all nonzero real values of the parameter.
-- 
R. T. Bumby **  Rutgers Math ||   Amer. Math. Monthly Problems Editor
bumby@math.rutgers.edu       || P.O. Box 10971 New Brunswick, NJ08906-0971
bumby@dimacs.rutgers.edu     || Phone: [USA] 908 932 0277 * FAX 908 932 5530
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Logarithms - how did they start/
Date: 1 Oct 92 14:40:07 GMT
Organization: Purdue University Statistics Department

In article <1992Oct1.011035.59324@cc.usu.edu> sl870@uranus.ee.usu.edu (Ullas Gargi) writes:
>A sitter from a netter:
>
>How did the first guys (Napier et al) calculate the logs of numbers?
>If I wanted to sit down and manually calculate the log of a non-trivial no.
>(like 10), how would I do it? A Taylor expansion formula?

There is nothing obscure about this.  Napier originally used a base
1 + 10^(-7) and used multiplication to get the antilogarithms.  Later,
when common logarithms came into use, one used square roots to get 
logarithms, and then interpolation.  AC (after Calculus), Taylor 
expansions were used to get natural logarithms, and common logarithms
were obtained by division.

There are many ways to compute logarithms, and to compute the natural
logarithm of 10, one way to do this would be as 3*ln(2) + ln(1.25).
Another way would be to use (10*ln(2) - ln(1.024))/3.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
From: gore@pepper.rutgers.edu (Bittu)
Newsgroups: sci.math
Subject: uvw prime puzzle
Keywords: Solution wanted
Date: 30 Sep 92 14:57:35 GMT
Organization: Hill Center


A few weeks ago I saw a discussion on the uvw prime puzzle on this
newsgroup. The problem asked to show that there do not exist strings  
u,v,w of digits such that the infinite sequence obtained by pumping
the string v consists entirely of primes. If someone saved a copy of
the solution that was posted, could you please send it to me? My email
address is gore@cs.rutgers.edu.

Thanks,

--Vivek
Newsgroups: sci.math,sci.logic
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: sci.math
Subject: Re: Roman Numerals
Date: 5 Oct 92 21:42:48 GMT
Organization: CWI, Amsterdam

In article <1992Oct5.150029.29200@Princeton.EDU> tao@ginger.princeton.edu (Terry Tao) writes:
 > Some people hypothesize that brackets were put over numbers to symbolize
 > multiplication by 1000, hence (V) is 5000.  In fact, since (I) has a faint
 > resemblance to M, this could explain the symbol M (rather than the traditional
 > explanation that M is the first letter of mille, the latin for a thousand; this
 > doesn't hold water, as all the other letters are not initials of latin numbers,
 > except possibly C.)
 > 
Possibly.  (I)  (where the brackets prepresent C and reverse C respectively)
was used for 1000.  Symbols in use for 10,000 and 100,000 were ((I)) and
(((I))).
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: arodgers@dcs.qmw.ac.uk (Angus H Rodgers)
Newsgroups: sci.math
Subject: Re: Topos theory for mere mortals (?)
Date: 29 Sep 92 13:51:05 GMT
Organization: Computer Science Dept, QMW, University of London

In <19o64aINNb9f@usenet.INS.CWRU.Edu> cxm7@po.CWRU.Edu (Colin Mclarty) writes:

>	What an elementary topos IS, is a category something like
>SET.  Any two "sets" have a cartesian product, for any equation
>there is a "set" of solutions which may of course be empty, for any
>two "sets" thre is a "set of all functions" from the first to the
>second.  And there is a "set" of truth values. [...]

The following point, in essence, has just come up in our e-mail
correspondence (and I'm sorry that I haven't given you time to reply,
but in my original message this point may have been entangled with
other confusions of my own making); but it can be put so simply and
briefly that I think it may deserve some space on the Net.

The adjunction defining Cartesian closedness determines a (natural)
bijection between hom-sets: [(A x B) -> C] --> [A -> (C^B)].
^^^^^^^^^             ^^^^
If topos theory is a foundation for maths, independent of any (other)
form of set theory, does this mean that it is no longer Categorically
Correct to use such old-fashioned language as I have just underlined?
More to the point, perhaps, is this sort of language consistently
avoidable when doing elementary topos theory?

I can't remember all the possible definitions of the term
"adjoint", but I do know that at least one of them consists
in the assertion of the existence of such a bijection between
two hom-sets (which transforms "naturally", as the objects
involved vary).

And category theory gets nowhere fast without using adjunctions.
So it seems to need a supporting set theory from the start.

What mistake, if any, am I making now? (Presumably it results from
my ignorance of the definition of topos theory in elementary terms;
but even so, it might give other readers some more insight into the
theory to explain why I am wrong, so the bandwidth won't have been
entirely wasted.)
--
Angus H. Rodgers  (PhD student), | arodgers@dcs.qmw.ac.uk  
Dept. of Computer Science, Queen | [ +44 | 0 ] 71 975 5241   
Mary & Westfield College,  Mile  | This space has not been left blank,
End Rd., London E1 4NS, England  | intentionally or otherwise.
Newsgroups: sci.math,comp.theory,geometry.research
From: orourke@sophia.smith.edu (Joseph O'Rourke)
Subject: Convex hulls in higher dimensions: applications?
Organization: Smith College, Northampton, MA, US
Date: Wed, 30 Sep 1992 14:07:08 GMT

I am interested in learning of practical applications of algorithms
for constructing the convex hull of a set of points in dimensions
greater than three.  What I have in mind is the application
that Cohen & Hickey used a decade ago:  they analyzed the
branching probabilities in programs as ratios of volumes of
polytopes whose dimension depended on the complexity of the
code.  Any other such applications?  Thanks!
From: ad@tamdhu.st-and.ac.uk (Tony Davie)
Newsgroups: sci.math
Subject: Re: Opinions Sought (Why major in math?)
Date: 5 Oct 92 11:39:21 GMT
Organization: St.Andrews University

In article <stephen.717902097@mont>, stephen@mont.cs.missouri.edu (Stephen Montgomery-Smith) writes:
> 
> 
> A while ago, I phoned up alumni from our math dept, asking people
> what career they had persued.  While many said that their math
> degree helped them to think logically, none said that they actually
> used their math.

Surely this is the whole point of an education rather than a training.
After a good education (majoring in ANY subject) one should come out knowing
how to find things out for yourself. It's a second level activity. You may
not know just how to do a certain task, but you sure as hell know where to
look to find out or who to ask. The only purpose of a university education is
to learn how to use a library.

Tony Davie               Department of Mathematical and Computational Sciences
Tel: +44 334 76161 x8136 St.Andrews University
Fax: +44 334 77068       North Haugh
ad@dcs.st-andrews.ac.uk  St.Andrews
                         Scotland
                         KY16 9SS

Heard recently: Why do computer scientists confuse Halloween and Christmas?

Because oct 31 = dec 25.
Newsgroups: sci.math
From: hiraga@Csli.Stanford.EDU (Yuzuru Hiraga)
Subject: Re: Z <-> Q formula
Organization: Stanford University CSLI
Date: Sun, 27 Sep 1992 18:29:21 GMT

My problem was:
>The set of intergers Z and the set of rationals Q have the same cardinality,
>i.e. there exist a bijection between them.  So we all know that.
>But then, is there a "simple" formula that expresses any such mapping?

I should have noted there, but for those of you who have (or may have)
come up with a solution for an NxN -> N mapping, I must tell you that
that won't do.

Just as Michael Zeleny writes in <1992Sep27.120756.15885@husc3.harvard.edu>:
>Recall that the idea is to order the pairs first by their
>sum, and within the sum in a lexicographic manner.  So we get the
>enumeration of (x,y):
>
>J(x,y) = (\sum_{i<x+y}(i+1))+x = (x+y)^2+3x+y \over 2
>
>Note that this doesn't answer your question, as each rational number
>is counted infinitely many times.

And yes, that was exactly my point.
The rational numbers are the equivalence classes modulo reduction over
fractions, i.e. 1/2, 2/4, 3/6, ...(-1)/(-2), ... are all the one and
same rational number.

>In order to come up with a simple
>bijection between N and Q (which could be trivially modified to yield
>one between Z and Q), we might modify the above procedure to exclude
>the cases of pairs (x,y), in which x and y are not relatively prime.

And since this is possible as an effective procedure, it is possible
to encode that into a formula, using e.g., Goedelian techniques.
The problem is how simple this can be.

-Yuzuru Hiraga
Newsgroups: sci.math
From: rjk@world.std.com (robert j kolker)
Subject: A more virus resistant architecture
Summary: Babbage machine less susceptible to viruses
Keywords: computer,virus,Babbage,analytic engine
Organization: The World Public Access UNIX, Brookline, MA
Date: Mon, 5 Oct 1992 01:11:11 GMT

While reading a book on the Babbage Analytic Engine the other night, it
occurred to me that a Babbage machine might be less susceptable to virus
than a Von Neuman machine. In the Babbage machine, the program is stored
on an medium external to the data store (punched cards for example). The
Babbage Analytic Engine was patterned after the Jacquard Loom.

The question is this: Is a computer whose program space is totally
disjoint from the the data space less likely to be buggered by viruses
than a  Von Neuman Machine, where the data and the program are kept in the
same store.

Your opinions on this questions would be much appreciated.

(This note was also posted on comp.virus)

Conan the Libertarian  rjk@world.std.com
"If you can't love the Constitution, at least hate the Government"
From: epstein@rene.ma.utexas.edu (Paul Epstein)
Newsgroups: sci.math
Subject: primes
Date: 29 Sep 92 19:00:47 GMT
Distribution: sci
Organization: University of Texas at Austin Mathematics

Are there any primes greater than 5 for which (p-1)!+1 is a power of p?

(This is an exercise in Apostol's book on elementary number theory which
I am unable to solve.)

My email address is epstein@math.utexas.edu
From: frisinv@alleg.edu
Newsgroups: sci.math
Subject: Re: Mattel says "Math is tough."
Date: 6 Oct 92 17:47:10 GMT
Distribution: usa
Organization: Allegheny College

> > 
> >What would you have her say?: 

I'm no expert on toys, so I can't really answer this question. I also  
can't say for sure that what the doll says is/isn't woman-bashing.  
However, I do know that NO toy should include a message that may  
discourage a child from pursuing any discipline. I will not deny that math  
is a hard subject; I also will not deny that it is an enjoyable asnd  
rewarding topic for those who wish to pursue it. If, when he or she is  
older, the child decides that math is not the way to go in their case,  
fine. But negative messages such as this will not help anyone. If math is  
tough in their case, the child will find out. If it isn't, why bother them  
with Barbie's little message.

-------
Vincent
From: cn0cu8ah@triton.unm.edu (A. Bailey)
Newsgroups: sci.math
Subject: Math Book Suggestions
Summary: Looking for a good book on Boundary Value Problems (PDEs)
Keywords: Boundary Value Problems, PDEs
Date: 1 Oct 92 01:31:29 GMT
Distribution: usa
Organization: University of New Mexico, Albuquerque


	Hello,

	I'm looking for a book on boundary value problems (PDEs) with
	clear explanations and plenty of worked out examples. If you
	can suggest such a book, please e-mail me. The book I'm currently
	using does not suit my needs and I would like to know what is
	available on this subject. Thank you in advance.

	Alec Bailey (abailey@xenon.unm.edu)
Newsgroups: sci.math,news.answers
From: torkel@sics.se (Torkel Franzen)
Subject: Re: consequences of the Axiom of Choice
Organization: Swedish Institute of Computer Science, Kista
	<TORKEL.92Oct5044312@lludd.sics.se>
	<BvMu6n.1Hp@cantua.canterbury.ac.nz>
Date: Mon, 5 Oct 1992 07:03:52 GMT

In article <BvMu6n.1Hp@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz 
(Bill Taylor) writes:

   >To find one that definititely won't have a definable choice function, we
   >must be more cunning, I think.

 Actually, the request was only for a case where a choice function is not
obvious. Your comments re my suggestion are unclear: why does it seem
likely that the set I defined should have a definable choice function?
I don't see this at all.

  As for your suggestion, it makes no immediate sense as it stands.
You say

   >Now look at the set of: pairs of infinite sets of real numbers that are
   >z-constructible, but not m-constructible for any m<z.

where z is the first non-constructive ordinal. However, this set is empty,
by your own definition:

   >for limit ordinal l, call X  l-constructible if it is m-constructible for
   >some m<l.
Xref: sparky sci.math:12812 sci.math.symbolic:2639
From: gyu@rnd.GBA.NYU.EDU (George Gang Yu)
Newsgroups: sci.math,sci.math.symbolic
Subject: Help need to solve a system of ODE
Keywords: Urgent help needed for solve a simple ODE
Date: 6 Oct 92 19:06:44 GMT
Followup-To: poster
Organization: NYU Stern School of Business

I am seeking help from mathematicans or the like to solve a system of two 
ordingary differential equations. Please give me any suggestions on how to 
solve the problem, or give me the explicit solutions. Your help will be greatly 
appreciated. The following LaTeX file contains the problem.  


George
-------------------------


\documentstyle[twoside,11pt]{article}
\renewcommand{\baselinestretch}{1.4}
\addtolength{\topmargin}{-5pc}
\addtolength{\textheight}{8pc}
\addtolength{\oddsidemargin}{-3pc}
\addtolength{\textwidth}{6pc}
\setlength{\evensidemargin}{\oddsidemargin}
\newcommand{\rt}{\right}
\newcommand{\lt}{\left}

\begin{document}
{\bf Could you please help to solve the following system of ODE?}
\[ {d\over dS}\lt(\begin{array}{l} Y\\ Z \end{array}\rt)=\lt( \begin{array}{l} e^{Z-Y}\\ \rho e^{Y-Z}-{2\over\beta\sigma^2}(\beta r_1S+\beta\mu +Z-1) \end{array}\rt),  \]
where $Y,\,Z$ are functions of only $S$.
\end{document}
Newsgroups: sci.math