HDTenEightyP/NetNews huggingface dataset:
From: rags@triples.math.mcgill.ca (Robert A. G. Seely)
Newsgroups: sci.math,sci.logic
Subject: Re: The Continuum Hypothesis:  Must it be {True or False}
Date: 25 Dec 92 01:11:58 GMT
Organization: /etc/organization

In article <1992Dec24.161747.18827@husc3.harvard.edu> zeleny@husc10.harvard.edu (Michael Zeleny) writes:
>In article <1992Dec24.034938.11339@smsc.sony.com> 
>markc@smsc.sony.com (Mark Corscadden) writes:
>
>>In article <1992Dec14.200024.6435@nas.nasa.gov>
>>asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:

The practice of quoting in full is one of M Zelany's most irritating
characteristics - about 75% of the post referred to here was quote.  
Of course, he is not by any means alone in this - PLEASE SUMMARIZE or at the
very least leave the quote to the END of your reply.  (And mark it as such,
so we are not even tempted to scan it if we recall the points.)
I find too often I have to hit the "K" key because some folk prefer the "F"
to the "f" key (I am using trn - translate to your own news reader's
conventions - the sense I hope is clear:  responding with quote rather than
without.)  Bah humbug....

On the other hand, maybe that is an indication they are doing me a favour
;-}

Happy hol's to all,

- rags
Xref: sparky sci.math:17401 alt.books.technical:384
From: gh@cs.toronto.edu (Graeme Hirst)
Newsgroups: sci.math,alt.books.technical
Subject: Re: High Prices of Math Books. I am pissed.
Date: 24 Dec 92 22:58:43 GMT
Organization: Department of Computer Science, University of Toronto

In article <1992Dec24.194233.10811@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>(1) These books are difficult to typeset and thus are a lot more expensive
>    to print than your average novel. 
>(2) They have a small market, so the high setup cost of producing the plates
>    to print them must be amortized over a small number of customers.
>(3) It requires a lot more expertise to edit these books than your average
>    novel.
>
>Why are people always surprised that technical books with a small audience
>are expensive?

This misses the point of the original posting, which was this:  Some
publishers (e.g., McGraw-Hill, Prentice-Hall, Addison-Wesley, and other
names with hyphens) will make the *same* book in different bindings at
different prices in different parts of the world.   In particular, they
will sell a cheap paperback edition only outside North America, while
insisting that North American customers buy the hardcover edition at a
much higher price.

-- 
\\\\   Graeme Hirst    University of Toronto	Computer Science Department
////   gh@cs.utoronto.ca  /  gh@cs.toronto.edu  /  416-978-8747
Newsgroups: sci.math
From: sumner@math.scarolina.edu (David Sumner)
Subject: Re: Marilyn Vos Savant's error?
Keywords: savant
Organization: USC  Department of Computer Science
Date: 22 Dec 92 21:59:01 GMT


ac999239@umbc8.umbc.edu writes
 
>You have 3 pancakes. You choose 1 and see a side is brown. 
>You know you do not have the one with gold on both sides by 
>common sense! You know there is one which is brown/brown and 
>one that is brown/gold. Therefore the probability is
>1/2.
 
So much for common sense!
It's true that you have reduced the choice to two
possibilities. However, the two choices are not equally
likely. The brown-brown pancake is twice as likely to be
the one you chose as is the brown-gold pancake.
 
David Sumner
 
Newsgroups: sci.math
From: chisnall@cosc.canterbury.ac.nz (The Technicolour Throw-up)
Subject: Re: Centillion etc. (was Re: Negative Zero)
Organization: Computer Science,University of Canterbury,New Zealand
Date: Mon, 21 Dec 1992 13:39:13 GMT

From article <djoyce.724343714@black.clarku.edu>, by djoyce@black.clarku.edu (Dave Joyce):
> It's unfortunate, but it seems number names
> 
>      billion, trillion, quadrillion, ..., centillion, ...
> 
> do not have universal values.  Depending on the country, a billion can be
> either a thousand million or a million million, and so on for the rest.
> Only a million has a universal value.
> 
> I was thinking the other day about this very problem.  Perhaps a solution
> would be to "go metric".  That is, use the metric prefixes for large numbers.

Another solution, one that's probably occurred to many people, is to use a
recursive scheme for building larger -illion numbers.  You start with one
million having the normal value and define a billion to be a million million.
Next you define a trillion to be a billion billion, a quadrillion to be a
trillion trillion, and so on, defining each -illion number as the square of
the preceeding one.

Under british nomenclature the nth -illion number is 10^{6n} while under
american nomenclature it is 10^{3(n+1)}.  Under the recursive scheme outlined
above the nth -illion number is 10^{3*(2^n)}.  So a billion is now 10^12, a
trillion is 10^24, a quadrillion is 10^48 etc.

I don't know about other people but I've often felt that both the british and
american naming shemes rush through the numbers too quickly giving a new
-illion name to just about everything.  A billion trillion, for example, is a
sexillion.  Compare this with the recursive scheme which takes a much more
leisurely stroll through the numbers giving out new -illion names much more
sparingly.   Now we find that a billion trillion is a number in it own right,
lying between a trillion and a quadrillion.  The recursive notation is much
less redundant.

--
Just my two rubber ningis worth.
Name: Michael Chisnall  (chisnall@cosc.canterbury.ac.nz)
I'm not a .signature virus and nor do I play one on tv.
Newsgroups: sci.math
From: droth@silver.ucs.indiana.edu (David Roth)
Subject: Re: need help in calculus problem.
Organization: Indiana University
Date: Sun, 3 Jan 1993 09:04:08 GMT

Tangent line?  In 3-space, there are infinitely many tangent lines to a
surface at a point..

David
From: Barry.John.Good@brunel.ac.uk (Barry John Good)
Newsgroups: sci.math
Subject: OR software
Date: 23 Dec 92 09:50:55 GMT
Reply-To: C20249@PRIME-A.PLYMOUTH.AC.UK
Organization: Brunel University, West London, UK


The University of Plymouth have recently dropped SAS, since most of
its procedures are included in SPSS.  However, we have no equivalent software 
for SAS's OR (operational research) module.
 
So I'm looking for a package (preferably multiplatform,
Windows/DOS/Mac/mainframe)  that covers the area of mathematical programming
(linear/integer programming, travelling salesman, assignment problem, etc)

Any information will be most invaluable.

Please reply to C20249@PRIME-A.PLYMOUTH.AC.UK.
I am not a regular reader of this newsgroup.
If there is a more appropriate newsgroup,  please let me know.


Thanks.

Barry Good.
From: israel@unixg.ubc.ca (Robert B. Israel)
Newsgroups: sci.math
Subject: Re: Sums of reciprocals
Date: 22 Dec 92 21:32:35 GMT
Organization: The University of British Columbia
Distribution: sci
NNTP-Posting-Host: unixg.ubc.ca

In <ARA.92Dec21193839@camelot.ai.mit.edu> ara@zurich.ai.mit.edu (Allan Adler) writes:


>Let f(n)=1/n for every positive integer n. Let x be a positive
>real number. Let g(n,x) be defined as follows:

>(1) g(n,x) is a strictly increasing sequence of positive integers, possibly
>     a terminating sequence.
>(2) for all n, g(n,x) is the smallest integer consistent with (1)
>    such that the sum of the reciprocals of g(k,x) for k=1,...,n
>    does not exceed x.

>I have been told that if x is rational then this leads to a way of writing
>x as a finite sum of reciprocals of distinct positive integers. Let the number
>of terms in this sum be N=N(x). I have been told that N is also the number of
>terms in a minimal representation of x as a sum of reciprocals of positive
>integers allowing repetitions.

>Is this true and if so how or where is it proved?

>Allan Adler
>ara@altdorf.ai.mit.edu

It does lead to a finite representation.  To save typing, I'll write g(n,x)
as g(n) and 1/g(n) as h(n).  Let r(n) = x - sum_{i=1}^n h(n).  Thus
r(0) = x, and r(n+1) = r(n) - h(n+1).  Moreover, 
        g(n+1) = max(g(n)+1,ceil(1/r(n)))
(where ceil(y) = least integer >=y).  I claim that for sufficiently large
n we have g(n+1) > g(n)+1.  For suppose g(n) > 2 and g(n+1) = g(n)+1.  Then
r(n-1) = r(n+1) + h(n) + h(n+1) >= 1/g(n) + 1/(g(n)+1) > 1/(g(n)-1).
The only reason why g(n)-1 was not chosen instead of g(n) is that
g(n)-1 = g(n-1).  Thus if g(n+1) = g(n), we must have g(k) = g(k-1) + 1
whenever k <= n and g(k) > 2.  Thus x >= sum_{k=2}^n 1/k.  By the 
divergence of the harmonic series, this can't happen for large n.

Now suppose n is large enough that g(n+1) > g(n) + 1.  Let r(n) = p/q
with p and q positive integers.  For some integers k and r,
0 <= r < p, q = kp + r.  If r = 0, the representation ends with 
g(n+1) = k.  Otherwise, 1/(k+1) < r(n) < 1/k so g(n+1) = k+1.  Then
r(n+1) = p/(kp+r) - 1/(k+1) = (p-r)/((kp+r)(k+1)).  But this has
a smaller numerator than does r(n).  Therefore after at most p 
further terms, the representation must terminate.


It is not true, on the other hand, that N is minimal.  For example,
take x = 1/4 + 1/5, a sum of two reciprocals.  The representation
using g(n,x) is 1/3 + 1/9 + 1/180.

-- 
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,sci.philosophy.tech
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: Numbers and sets
Organization: Boise State University
Date: Sun, 3 Jan 1993 22:02:25 GMT

In article <1993Jan2.223636.18944@husc3.harvard.edu> zeleny@husc10.harvard.edu (Michael Zeleny) writes:
>In article <1992Dec28.165203.402@guinness.idbsu.edu>
>holmes@opal.idbsu.edu (Randall Holmes) writes:
>
>>I'm avoiding nested quotations here.

...again.

>Randall, you are conveniently neglecting the history of the matter.
>Cantor, Dedekind, and Zermelo were there long before Quine and Jensen;
>moreover, their use of the term is overwhelmingly prevalent among set
>theoreticians.  Your professional entitlement in no way absolves you
>from the responsibility for usurping a term possessed of a well-defined
>conventional meaning.  You have acknowledged that your reference to
>"sets" involves a radically unorthodox conception of the same; surely
>you do not fancy that there remains a practical chance that it would
>prevail over the Cantorian one.  Indeed, even with the Axiom of Choice,
>where one may cite genuine mathematical motivations for alternative
>assumptions (vs. the extramathematical considerations adduced by Aczel
>and Co in favor of rejecting the Axiom of Foundation), the consensus is
>that the descriptive set theorists who adopt the Axiom of Determinacy,
>are really investigating a *deviant* set theory.  Thus, inasmuch as I am
>conducting this discussion in accordance with the social conventions of
>the English language, I am perfectly well justified in assuming that
>sets are characterized as members of the cumulative hierarchy, and so
>likewise -- in claiming that Foundation and Choice are analytically true
>of sets.

On the definition "element of the cumulative hierarchy", Foundation
becomes analytic, but this is _not_ the case for Choice.  Choice has
always been regarded as somewhat dubious, and mathematical practice
reflects this; people generally take care to indicate when it is used,
which they would not do with an analytic principle.  (I subscribe to
the Axiom of Choice myself as true, but not analytic).

>
>>-- 
>>The opinions expressed		|     --Sincerely,
>>above are not the "official"	|     M. Randall Holmes
>>opinions of any person		|     Math. Dept., Boise State Univ.
>>or institution.			|     holmes@opal.idbsu.edu
>
>cordially,
>mikhail zeleny@husc.harvard.edu
>"Le cul des femmes est monotone comme l'esprit des hommes."


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
From: tim@maths.tcd.ie (Timothy Murphy)
Newsgroups: sci.math
Subject: Composition-commuting polynomials
Summary: Is there a general theory of polynomials commuting under composition?
Keywords: Polynomials commuting composition
Date: 28 Dec 92 17:51:28 GMT
Organization: Dept. of Maths, Trinity College, Dublin, Ireland.

Which polynomials f(x) satisfy

	f(x^2+1) = f(x)^2 + 1 ?

Evidently this is satisfied by f(x) = x^2 + 1,
and in general, if we define g^{[n]}(x) 
as the n-th compositional power of x^2+1, ie

	g^{[0]}(x) = x, g^{[n]}(x) = g^{[n-1]}(x^2+1),

then f(x) = g^{[n]}(x) will commute (in the above sense) with g(x).
But are these all the polynomials with this property?

I'm sure this subject must have been exhaustively discussed somewhere,
but where?
(It arose when I set the above problem in a Putnam-like competition here.
Fortunately, I worded it in the above form,
which you will notice does not require that the setter can solve the problem!
Long experience teaches that there are few questions
that cannot be stated in this agnostic manner.)

In general, which polynomials f(x) commute (compositionally)
with a given polynomial g(x)?


-- 
Timothy Murphy  
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Newsgroups: sci.math
From: victor@watson.ibm.com (Victor Miller)
Subject: Re: need proof:  (1 + 1/n)^n ==> e
Date: Mon, 28 Dec 1992 15:50:23 GMT
Reply-To: victor@watson.ibm.com
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM, T.J. Watson Research Center

Very nice.  I was going to recommend looking at the book
"Inequalities" by Korovkin.  It's filled with all sorts of beautiful
elementary inequalities, and their consequences.  One of them being
the proof of the limit (1+1/n)^n.
--
			Victor S. Miller
			Vnet and Bitnet:  VICTOR at WATSON
			Internet: victor@watson.ibm.com
			IBM, TJ Watson Research Center
Newsgroups: sci.math
From: ilan@seas.gwu.edu (Ilan Berkner)
Subject: Need: prpndcular dist from pt to line
Organization: George Washington University
Date: Tue, 29 Dec 1992 21:05:54 GMT


Given a line, or line segment (I can do both), I need to find the 
distance from a point (that may or may not be on the line) to that line.
If the point is on the line, then the distance is zero, obviously.

I know the endpoints of the line segment in cartesian space, and I also
know the slope and intercept, and the coordinates of the point.

This is for a computer application, so anything involving integer math
is preferable to floating-point.

Thanx very much in advance.  I know this is almost a trivial problem
for all you math geniuses out there, but I also want to make sure that
whatever solution I get, is also the best.
Xref: sparky sci.math:17586 sci.physics:22016
From: hougen@focus.csl.uiuc.edu (Darrell Roy Hougen)
Newsgroups: sci.math,sci.physics
Subject: Re: Bayes' theorem and QM
Date: 2 Jan 1993 03:15:38 GMT
Organization: Center for Reliable and High-Performance Computing, University of Illinois at Urbana-Champaign
NNTP-Posting-Host: focus.csl.uiuc.edu


% In article <1992Dec31.024039.2740@smsc.sony.com> markc@smsc.sony.com (Mark Corscadden) writes:
%% Blair, there's an excellent discussion of this in the sci.math faq:
%% 17Q.- Cutting a sphere into pieces of larger volume  

This is undoubtedly a FAQ, but where is the FAQ for this news group?
I've never seen it posted.


Darrell R. Hougen
Newsgroups: sci.math
From: mrob@world.std.com (Robert P Munafo)
Subject: Re: Two (I think) interesting Problems
Summary: Here's how to make the 1:sqrt(2) rectangle out of any rectangle
Keywords: paperfolding pentagon origami geometry long
Organization: The World Public Access UNIX, Brookline, MA
Date: Mon, 28 Dec 1992 06:02:31 GMT


> Fascinating.  We unhappy americans have no convenient source of
> root(2) paper.
 
Well, of course we could use a paperfolding method to generate a 1:sqrt(2)
from an arbitrary rectangle:
 - Fold a short edge of the rectangle so that it meets a long edge,
   forming a five-sided shape with a 45-degree angle at one end of the
   fold you have just created.  Call the point with the 45-degree corner
   "Point A".  Then, open it flat again.
 - One of the long edges of the rectangle has Point A as one of its
   endpoints.  Call the other endpoint of this edge "Point B".  Fold this
   edge so that it coincides with the crease you
   made in the first step, forming a 67.5-degree angle at Point A.  If
   the rectangle you started with is more square than 1:sqrt(2) (e.g.
   8.5x11 US Letter paper) you will now have a pentagonal shape; continue
   with the next step.  Otherwise (e.g. 8.5x11 US Legal paper) you have a
   non-convex polygon of some kind; skip down to step "*".
 - (Still in the pentagonal shape) fold again so that Point A coincides
   with Point B.
 - Part of the short edge of the original rectangle is now parallel to
   the unfolded long edge.  Fold this long edge forming a crease that
   coincides with the parallel short edge, extending the crease all the
   way past the folded-in corners.
 - Unfold the two previous folds, leaving only this last fold.  You now
   have a 1:sqrt(2) rectangle.
 * (For rectangles that were less square than 1:sqrt(2)): The long edges
   of the original rectangle cross; call this point "Point C".  Make a
   new fold so that one endpoint of the new crease is at Point C, and
   so that Point B lands on the edge AB.
 - Unfold both.  Part of the second crease you made will be parallel to
   the short edge of the rectangle.  Fold along this crease, extending
   the crease all the way to the other long edge.  The distance from
   this crease to the short edge containing Point A is sqrt(2) times the
   length of the short edge.  If your original was longer than 0.5:sqrt(2),
   you'll have to fold more times to get rid of the excess.
 
- Robert P. Munafo               (yup, the computer ate my .signature)
Xref: sparky sci.math:17527 sci.physics:21920
From: kbanerje@grad12.math.ncsu.edu (Kaushik Banerjee)
Subject: sci. comp., comp. math, & numerical analysis
Organization: NCSU Mathematics Dept.
Distribution: usa
Date: Wed, 23 Dec 1992 18:22:13 GMT

The following question is addressed to students, academics, and
people who work in industry (basically, everyone).

What is the difference between scientific computing, computational
mathematics, and numerical analysis?

If you have a few minutes to answer this question, please
e-mail me at kbanerje@math.ncsu.edu.

Thanks.

K. Banerjee
Newsgroups: sci.math.num-analysis
From: amal@mel.dit.csiro.au (Amal De Silva)
Subject: Re: Nonlinear systems of equations
Organization: CSIRO DIT (Melb.)
Date: Tue, 22 Dec 92 04:49:47 GMT


In article <92Dec17.174419.27750@acs.ucalgary.ca>, mgh@hobbes.phys.ucalgary.ca (Mike Henderson) writes:
|> I am trying to solve a set of 8 nonlinear equations in 8 unknowns.
|> I have tried the Newton-Raphson technique given in NR (I know
|> roughly what the solution should be) but havent had much luck. Are
|> there any more robust methods to choose from?
|> 
|> Mike Henderson
|> Dept. of Physics and Astronomy
|> University of Calgary
|> mgh@hobbes.phys.ucalgary.ca

The problem with the NR method is that it is not  globally convergent (It needs a
good starting point to get to the solution). One method of solving this problem is
to represent this as a minimization problem and use a globally convergent 
non linear programming (unconstrained) algorithm to solve it. I can
give  a reference to this. 

Numerical methods for unconstrained optimization and nonlinear equations by
J.R Dennis and Robert Schnabel, Prentice Hall (1983). page 147. 


Amal de Silva


Newsgroups: sci.math.num-analysis
From: phr@telebit.com (Paul Rubin)
Subject: Re: Numerical Recipes 2nd ed.?
Organization: Telebit Corporation; Sunnyvale, CA, USA
Date: 26 Dec 92 20:24:06


      I have seen Numerical Recipes for Fortan in the 2nd edition, with
    updated wavelet algorithms and such, but I was wondering if anyone
    knows the planned publication schedule for Numerical Recipes in C or
    Pascal, as there are already example books for the "2nd edition" in C
    and Pascal.

I've seen the 2nd edition of Numerical Recipes in C at my local bookstore.
Anyone want to buy my copy of the first edition?

Newsgroups: sci.math.num-analysis
From: edstrom@Elmer.hsc.ucalgary.ca (John Edstrom)
Subject: references on integration methods
Date: Mon, 28 Dec 92 17:45:44 GMT
Organization: Neuroscience Division, U of Calgary School of Medicine, Calgary, Alberta, Canada

A friend is doing some modeling that reiteratively integrates a set of
equations.  Different integration methods yield different answers
after a relatively few iterations, that is the model predictions diverge
rapidly depending on the integrator used.

Is there a general reference that discusses the strengths, weaknesses
and characteristic errors associated with different methods of
integrating differential equations?

JE | edstrom @ elmer.hsc.ucalgary.ca
--
 RM 2104, HSc Building,  Div. Neuroscience
 U. Calgary School of Medicine,  3330 Hospital Drive NW
 Calgary, Alberta       T2N 3Y4
 (403) 220 4493  (wk)
From: dhart@bronze.ucs.indiana.edu (dave hart)
Newsgroups: sci.math.num-analysis
Subject: Re: Chebyshev vs. Least Squares Polynomials
Date: 30 Dec 92 15:37:55 GMT
Organization: Indiana University

>	The idea of "least squares" is the same as orthogonal projection--
>"the shortest distance between a point and a line..."  This is tied up
>with geometry, ie a metric [inner product].  The Chebyshev polynomials
>form an orthonormal basis [q.v.] for the L^2 metric;  other functions
>form orthonormal bases for other metrics.
>
	_if_ you measure with dx/sqrt(1-x^2) [what else?]. [oops!:-]
From: hcobb@fly2.berkeley.edu (Henry J. Cobb)
Newsgroups: sci.math.num-analysis
Subject: Re: LU decomposition on very large matrix.
Date: 2 Jan 93 19:48:59
Organization: /rubin/hcobb/.organization
Distribution: inet
NNTP-Posting-Host: fly2.berkeley.edu
In-reply-to: hcobb@fly2.berkeley.edu's message of 23 Dec 92 14:38:56


	I'd first like to thank everybody for their input on this, there
didn't seem to be much of a consensus on this, so we're goning to handle it
like this:

	First a quick note: we're using a RS/6000 and we expect the matrix
to be around 500+ MB.

	Our game plan is to mmap a file, map the matrix to that address
space and fork off processes to handle chunks of it using MP style code.

	Our hope is that while some of the processes are blocked, waiting
for the parts of the matrix that they've touched to be paged in, other
processes will be able to continue crunching on pages that are curently in
core.

	We can (and will) take over the machine to run this, but are
wondering what the ratio should be between:

	Demand: The combined working set of the running processes. ( = Size
of each block * number of blocks active at a time for a process * total
number of processes)

	and Supply: Total system physical memory. (64 MB minus some amount
for the OS and per process overhead).

	It seems as though Demand should be about equal to Supply, but we'd
like to hear of any studies on this.

	It might prove to be useful to overcommit a little (for some sort of
read-ahead?) or undercommit by some factor (to provide "extra caching").
At the moment we're tempted to just provide args for block-size and number
of processes and test it out that way.
--
	Henry J. Cobb	hcobb@fly2.berkeley.edu	SFB Tyrant
	"A const pointer is a steady reference,
	while a pointer to const is a pledge not to tinker."
From: evans@ece.arizona.edu (Glen Evans)
Newsgroups: sci.math.num-analysis
Subject: Complex Error Function
Date: 22 Dec 92 21:35:51 GMT
Reply-To: evans@ece.arizona.edu (Glen Evans)
Distribution: world,local
Organization: University of Arizona Dept. of Electrical and Computer Engineering


Does anyone have any good Fortran routines to
compute the complex error function?
From: soubyran@ERE.UMontreal.CA (Soubyran Richard)
Newsgroups: sci.math.num-analysis
Subject: Real to fraction...
Date: 28 Dec 92 19:36:16 GMT
Organization: Universite de Montreal

Hello Mathers!

May be this one has already been posted... but I am taking the
chance...

Can anyone give me an algorithm to find the fractional
representation of a real number?

I thank you very much in advance,

Richard Soubyran

soubyran@alize.ere.umontreal.ca