HDTenEightyP/NetNews huggingface dataset:
From: chepan@gsusgi1.gsu.edu (Parthasarathy Nambi)
Subject: Re: A puzzle!
Keywords: Math, puzzle
Organization: Georgia State University
Date: Fri, 24 Jul 1992 21:17:03 GMT

chepan@gsusgi1.gsu.edu (Parthasarathy Nambi) writes:



> Here is something for puzzle lovers:

> Below is a sequence of numbers, can you find the next two numbers and 
> the general formula for generating numbers in this sequence:

> The numbers are:

> 4, 72, 21600, 190512000, 58090919040000,   etc 


> If you know the answer please post a response in this group.


> P.Nambi

 Oops! I made a mistake. A zero is missing from the fifth number
 above. The correct sequence of numbers is:

 4, 72, 21600, 190512000, 580909190400000, etc

 I apologize for the error.

 Good hunting!

 P.Nambi

Newsgroups: sci.math,soc.culture.british
From: sbrown@symcom.math.uiuc.edu (Scott Brown)
Subject: Re: American mathematical hegemony
Organization: University of Illinois at Urbana
Date: Thu, 23 Jul 1992 02:19:23 GMT

dj@ssd.kodak.com (Dave Jones) writes:

>To potential takers of this offer:

>Be warned that "dinner" should include the wings of about a dozen
>chickens, properly detached from the breast portions, plucked and deep
>fried (no batter please), served with a sauce whose pungency would bring
>a smile to the face of Red Dwarf's lone human occupant.  The meal should
>be accompanied by short sticks of celery and Blue Cheese salad dressing.
>I've never been able to fathom the origins or purpose of those last
>items.
  
   Mainly, to cool off the burn from the sauce, which raises blisters
if it is properly prepared.  Clears the sinuses, too. 

>Do this at your own risk.

   Oh...Of course...food served with sauces that have _flavor_.
  No wonder you brits don't get it.


Newsgroups: sci.math
From: benzvi@riemann.geom.umn.edu (David Ben-zvi)
Subject: Re: Fibre Bundle Theory
Organization: Geometry Center, University of Minnesota
Date: Mon, 27 Jul 1992 17:05:55 GMT

In article <s#T-+B+@engin.umich.edu> pradeep@engin.umich.edu (Pradeep Jain) writes:
>Hi,
> 
>I understand that Fibre Bundle Theory is about formulating geometric 
>problems as Fibre Bundle Theoritic problems (and then as algebric 
>problems?). Can someone please give me some introductory refrences to
>Fibre Bundle Theory? I would greatly appreciate the help. Also, what 
>are the pre-requisites for gaining an understanding of Fibre Bundle
>Theory?
>
>Thanks in advance.
>
>Pradeep
>

Here are some (rather standard) texts on Fibre Bundles. 
It is recommended that the reader be somewhat familiar with the
rudiments of Algebraic Topology (i.e. has seen homology before, and
most certainly knows about the fundamental group),
though most of these books cover the essentials:

Steenrod, The Topology of Fibre Bundles. This is the original text
       on the subject, from its founder. Somewhat old-fashioned, but classic.
	Princeton University Press.
Milnor, Characteristic Classes. Develops the theory of vector bundles (probably
	the most important fibre bundles) through the use of (the title),
	which are algebraic measures of the twisting of a bundle.
	Beautifully written, with an excellent introduction to cohomology
	as an appendix, and a summary of Thom's cobordism theory,
	Chern classes, the Hirzebruch signature theorem and the Chern-Weil
	thoery relating characteristic classes to curvature.
	Annals of Math Studies.
Bott&Tu, Differential Forms in Algebraic Topology. Another great book
	which introduces deRham cohomology (no real knowledge of homology 
	necessary), and uses it to explore various topics in topology,
	with a lot of fibre bundles (mainly vector and sphere bundles)
	and reaching all the way to spectral sequences in homotopy
	theory, the Grothendieck approach to characteristic classes,
	and other wonders of the field. Springer-Verlag, Graduate
	Texts in Mathematics (everyone's favorite yellow books).
Husemoller, Fibre Bundles (2nd Edition). I don't know much about this
	book, but it has been recommended to me.. Graduate Texts
	in Math #20.
Spanier, Algebraic Topology. The hard-core algebraic topology text. Covers
	a lot of Fibre Bundle theory. Extremely condense, thorough 
	and abstract (unless you've seen Grothendieck's EGA!!), and covers
	some rather beautiful ideas (e.g. the Brown representability Theorem). 
	Recommended for the head-banging hard hitting mathematician in
	all of us. (?!)

Good Fibrations,  (sorry!)
David Ben-Zvi

	an overwhelming urge for abstraction and generality
From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
Newsgroups: sci.math
Subject: Choice and Measurability
Date: 21 Jul 92 18:32:09 GMT
Organization: Princeton University Mathematics Department
Originator: news@ernie.Princeton.EDU

In article  <1992Jul20.171849.17294@Princeton.EDU> I wrote

> Let M be the axiom  "All subsets of R^n are measurable".  Then the 
> following is true:
> M is consistent with ZF and Countable Choice, but not with ZF and  
> AC.   Does someone have a reference?

I have been given a reference that supports this claim.  I will include 
it later in this post.  

In article  <ARA.92Jul20200638@camelot.ai.mit.edu>  Allan Adler wrote

> David Carabalo believes that Solovay proved that the nonexistence of
> nonmeasurable sets is consistent with ZF+DC.

For the record, David Caraballo believes no such thing.  The assumption 
that there exists an inaccessible cardinal is essential here, as Shelah 
proved.  I wrote "Countable Choice", not DC ("Dependent Choice").  I believe 
it was Bernays (Is this the correct attribution?) who proved that 
AC => DC => Countable Choice.  Consistency with ZF+DC is, of course, much 
more useful (DC allows us to prove all sorts of useful results, such as 
"The union of a countable number of countable sets is countable.").  

The relevant result that actually appears in Solovay's paper mentions 
DC, not Countable Choice.  The complete reference, by the way, is
Annals of Math, 92; 1970, pp. 1-56.   However, consider the following 
reference which someone sent to me in response to my post:  

"Solovay in 1964 proved that the assertion 'all sets are Lebesgue
measurable' is consistent with ZF and a restricted version of choice.
This restricted version of choice, called the countable axiom of choice,
asserts that every countable set of non-empty sets has a choice function."
                         Malitz, _Introduction to Mathematical Logic_, p.49

I had something similar in my own notes (which is why I posted something to 
this effect -- included above).  I am not prepared to disregard my notes 
(and now, the above reference) without seeing an actual proof that my claim 
"M is consistent with ZF and Countable Choice" is false.  If someone has a 
proof, I would love to see it.  Thank you.  

David G. Caraballo
From: fritz@gems.vcu.edu (Fritz)
Newsgroups: sci.math
Subject: Do we know the integers? (was: You know, the integers)
Date: 25 Jul 92 17:04:47 -0400
Organization: Medical College of Virginia

Just to muddy the waters a bit, I will now proceed to demonstrate that
the actual, genuine, accept-no-substitutes integers are non-standard :-).

We need to show that there is an integer   Z   such that all of the
following are true:   0 < Z, 1 < Z, 2 < Z, ...   .

Let us say that an integer   n   is _effectively mentioned_ when someone
has written it down precisely enough that a sufficiently motivated
reader could, in principle, rewrite the integer in unary notation--as  
0   followed by   n   primes, for instance.  So the number   10^10^10^10
is effectively mentioned (even though in practice nobody could make that
many marks on a piece of paper).  Even the number   m   defined by

        m = 1   , if the continuum hypothesis is true
        m = 0   , if the continuum hypothesis is false

is effectively mentioned, since both   0   and   1   are effectively
mentioned--we don't know which of   0   or   1   is an effective mention
of   m   , but we do know that one of them is.

However, given the limited supply of matter in the universe, there
is a limit to the size of the formulae that we can write down, and
consequently there are only finitely many integers that we could
effectively mention.   Let us call an integer _huge_ if it is larger
than every integer that could be effectively mentioned.  Let   Z   be
the smallest huge integer.  Then   Z   has the desired property:   0 < Z  
since   0   is effectively mentioned,   1 < Z   since   1   is
effectively mentioned, and so forth.  To write a number as a numeral
numeral is to effectively mention it:  thus every number which can be
written as a numeral is less than   Z   .                             :-)


Of course the argument above cheats by switching between possibility
``in principle'' and possibility in the real world (whatever that may
be).

                 - Fritz    (fritz@ruby.vcu.edu,fritz@vcuvax.bitnet)
                             Disclaimer: nothing I say represents my employer
From: brnstnd@nyu.edu (Dan Bernstein)
Newsgroups: sci.math
Subject: Re: ALGEBRAIC NUMBER ARITHMETIC
Date: 31 Jul 92 02:29:09 GMT
Organization: IR

In article <aet.712408532@munagin> aet@mullian.ee.mu.OZ.AU (bert thompson) writes:
> i want to perform arithmetic on algebraic numbers.

See Lovasz's Algorithmic Theory of Numbers, Graphs and Convexity.

---Dan
Newsgroups: sci.math
From: hdev@ph.tn.tudelft.nl (Hans de Vreught)
Subject: Re: Some musings on expensive journals
Organization: Delft University of Technology
Date: Thu, 23 Jul 1992 07:58:55 GMT

sbrown@symcom.math.uiuc.edu (Scott Brown) writes:

>      In response to the suggestion that putting journals
>on-line instead of hardcopy would solve the problem, I
>hope that I am not alone in voicing a resounding *NONONO!*
>Pleez O Pleez no! Why? For starters, I cannot imagine that
>it would be nearly as conducive to perusing journals as the
>hardcopy is.  There is a joy in wandering through a crowded
>library stacks with half an hour to kill, selecting a journal
>at random from a field of interest, and skimming the contents;
[rest deleted]

That is only a smart thing to do when you are checking the latest issues.
But you must know that al titles + abstracts of almost every journal is
collected in for instance: Zentralblat fur Mathematiek (problem some typo in
there). Don't be afraid of the German title: it is almost entirely in English
(since that's the language that is normally used in science).

And of course if you have to much money, you can acces several databases as
well. But then I have to agree with you, just paging through some journals can
be rewarding: you find that article that you wouldn't have found otherwise.
-- 
Hans de Vreught                          | John von Neumann:
hdev@dutiaa.tudelft.nl                   | Young man, in mathematics
Delft University of Technology (TWI-ThI) | you don't understand things,
The Netherlands                          | you just get used to them.
Newsgroups: sci.math
From: torkel@sics.se (Torkel Franzen)
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Organization: Swedish Institute of Computer Science, Kista
	<14h7srINNnjc@function.mps.ohio-state.edu>
Date: Tue, 21 Jul 1992 15:09:31 GMT

In article <14h7srINNnjc@function.mps.ohio-state.edu> edgar@function.
mps.ohio-state.edu (Gerald Edgar) writes:

   >"You know, THE SETS."

  More precisely, the cumulative hierarchy of sets. We learn about the
natural numbers as generated from 0 by repeated application of the successor
function at such an early age, at such a basic level, that it's all but
impossible to make it clearer once we're spouting mathematics or philosophy.
Set theory is a different matter: even today we can point to Zermelo's
"Ueber Grenzzahlen und Mengenbereiche" as a presentation unequaled in
clarity of the cumulative hierarchy of sets as a model of ZFC.
Xref: sparky sci.math:9356 soc.culture.british:10437
From: mmh@cs.qmw.ac.uk (Matthew Huntbach)
Newsgroups: sci.math,soc.culture.british
Subject: Re: American mathematical hegemony
Date: 21 Jul 92 13:52:06 GMT
Organization: Computer Science Dept, QMW, University of London, UK.

From: sooz@iastate.edu (Susan F Bradley)
> I lived in Leicestershire for 1 1/2 years, with the same people day in,
> day out. After all that time, I still couldn't say the words "water" or
> "tomato" without sending them all into paroxysms of laughter. One day, an
> elderly guest at the place I worked pitched a fit at me when she heard me
> say that I had "gotten" something. She did this in front of maybe 20-25
> people and serves as a good argument against another point of discussion --
> this idiotic notion that the British are more polite than Americans.
> This woman was decidedly middle class, supposedly well-bred. Where I come
> from, what she did was not only rude, but mean-spirited and completely
> uncalled for.

I doubt the intention was to be deliberately rude. Probably
they just found your accent quaint and amusing. Remember that
in the UK we are familiar with American accents mainly through
cartoons and TV shows, particularly in areas like Leicestershire,
outside the tourist circuit. Hearing your accent would be rather like
having a character come alive and step out of the TV. I think
you are also running into the common phenomenon that it is
considered acceptable to poke fun at groups of people who are
in an economically and politically superior position to
yourself, but not vice versa. Anti-Americanism is acceptable
because Americans tend to be wealthy, and the USA bosses the
world around. It is not so acceptable to laugh at disadvantaged
groups.

Matthew Huntbach
From: ralf@ark.abg.sub.org (Ralf Stephan)
Newsgroups: sci.math
Subject: Re: High Order Undulation Seeds and Binary Undulants
Date: 25 Jul 92 14:11
Organization: Private Coherent 3.2.1 site waiting for 4.0

cliff@watson.ibm.com ("Cliff Pickover") writes:
> 
> 2**949 =
> 4758454107128905800953799994079681792420032645310062268978469949811010102913
> Here N = 949 is called an undulation seed of order 6, since it
> gives rise to a 6-digit undulation pattern of adjacent 1's and 0's.

Isn't this just a probability question?  Just curious. I'm no expert.

--ralf
------------------------------------------------- please ignore next sig!
From: bennett@math.ksu.edu (Andy Bennett)
Newsgroups: sci.math
Subject: Re: Math or Maths
Date: 24 Jul 1992 11:53:05 -0500
Organization: Dept. of Mathematics, Kansas State University
NNTP-Posting-Host: hilbert.math.ksu.edu

mcfarlan@corona.math.washington.edu (Thomas J. McFarlane) writes:

>mckay@alcor.concordia.ca ( John Mckay ) writes:
>>
>> The Greek is MATHEMATIKOS - when did the abbreviations first occur?
>>
>My Greek Lexicon gives
>    ho mathematicos     a mathematician
>    ta mathematika      mathematics
>It is derived from the verb manthano, which comes from the root math.

I always like to note to my classes that in ancient greek, mathematics means
knowledge or learning in general as well as the specific learning of 
the subject of mathematics. Thus (from Liddel & Scott Abrdiged)

MANTHANO            to learn by inquiry, to ascertain

to mathema          that which is learnt, a lesson
ta mathemata        mathematics
on mathematicos     I.  fond of learning
                    II. belonging to the sciences, esp. mathematics
ho mathematicos     a mathematician
he mathematice      mathematics

By contrast, sophia means cleverness or skill, in addition to (if not before)
wisdom. In particular, sophists debated to show their cleverness and skill
and their wisdom was subject to dispute. Mathematics was not debated but was
learned. It was not subject to disputation.
Anyone familiar with greek philosophers can see 
that for something to no longer be debated required it meet some fairly
stringent tests. It is not surprising then that mathematics as we know
it today is the study of things which can be rigorously proven, or that
it was the Greeks who introduced the concept of proof. 

-- 
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
From: trm+@CS.CMU.EDU (Thomas Mathies)
Newsgroups: sci.math
Subject: Re: Percolation on the Plane
Date: 26 Jul 92 04:39:42 GMT
Article-I.D.: cs.1992Jul26.043942.272572
Organization: Carnegie Mellon University

In article <CHALCRAFT.92Jul10104635@laurel.uk.tele.nokia.fi>, chalcraft@uk.tele.nokia.fi (Adam Chalcraft) writes:
...
>For tilings where four or more pieces can meet, it would be nice to know
>which results still hold.
...

For a tiling with squares this is a site percolation problem. Each square is colored
white with probability p and black otherwise. For what values of p does there exist
an infinite contiguous set of white squares (with prob. 1)?

Consider two types of adjacency when the plane is tiled with squares:
  1) Nearest neighbor rule (each square has four neighbors); and
  2) Next-nearest neighbor rule (each square has eight neighbors).

Using the nearest neighbor rule, the critical probability is about 0.59275
according to Stauffer [1] (p. 17). There are three regions of interest:
    0       <= p <  0.40725  --> infinite contiguous set of black squares
    0.40725 <= p <= 0.59275  --> no infinite contiguous set of either color
    0.59275 <  p <= 1        --> infinite contiguous set of white squares

I just got a paper [2] on the next-nearest case which gives the critical probability
as 0.391. Again there are three regions of interest (with the middle one differing
from above):
    0     <= p <= 0.391  --> infinite contiguous set of black squares
    0.391 <  p <  0.609  --> infinite contiguous sets of each color
    0.609 <= p <= 1      --> infinite contiguous set of white squares

Notes:
  1) The values for the critical probabilities were obtained by computer simulation
     and extrapolation.
  2) I've arranged the inequalities based on a statement by Grimmet [3] (p. 164)
     that no infinite contiguous set exists at the critical probability.

References:

1. D. Stauffer, INTRODUCTION TO PERCOLATION THEORY, Taylor and Francis, 1985.
2. Shaohua Qu, K.L. Yao, and Boming Yu, "Study of the two-dimensional next-nearest-
   neighbor percolation model," CHINESE PHYSICS, vol. 11 #4, Oct-Dec 1991,
   pp 806-811.
3. G. Grimmett, PERCOLATION, Springer-Verlag, 1989.

------------
Tom Mathies <mathies@cs.cmu.edu>
"If tofu adds years to your life, they probably wouldn't be the best years."
				-- Garrison Keillor
------------------------
Newsgroups: sci.math
From: algebra@leland.Stanford.EDU (Leong Weng Ng)
Subject: Re: A puzzle!
Keywords: Math, puzzle
Organization: DSG, Stanford University, CA 94305, USA
Date: Sat, 25 Jul 92 00:36:19 GMT

In article <chepan.712011867@gsusgi1.gsu.edu> chepan@gsusgi1.gsu.edu (Parthasarathy Nambi) writes:
>
>
> Here is something for puzzle lovers:
>
> Below is a sequence of numbers, can you find the next two numbers and 
> the general formula for generating numbers in this sequence:
>
> The numbers are:
>
> 4, 72, 21600, 190512000, 58090919040000,   etc 

nth term =	(P)^(P) * (P)^(P) * ... * (P)^(P) 
         	  1   n     2   n-1         n   1
where P is the nth prime.
       n

> If you know the answer please post a response in this group.
>
>
> P.Nambi


Xref: sparky sci.math:9455 soc.culture.british:10567
From: kellyc@ul.ie
Newsgroups: sci.math,soc.culture.british
Subject: Re: American mathematical hegemony
Date: 23 Jul 92 14:58:50 GMT
Organization: University of Limerick, Ireland

In article <6738@charon.cwi.nl>, dik@cwi.nl (Dik T. Winter) writes:
> In article <1992Jul21.212224.18859@cl.cam.ac.uk> cet1@cl.cam.ac.uk (C.E. Thompson) writes:
>  > 
>  > Well, the sources seem to be in some doubt here. I seem to recall that when
>  > I was taught such things (c.1960), the two systems were distinguished as 
>  > "American" and "Continental" (which I took to mean "French"). However, the
>  > OED seems to be sure that French custom matches the American in making
>  > "billion" = 10^9, "trillion" = 10^12. On the other hand, they also believe
>  > that French "milliard" = 10^9. Is this really a case of two words for the
>  > same number?
> Darn.  I have now two French dictionaries, one dated 1953 and one dated 1988.
> They disagree!  The 1953 one translates "billion" to "miljard" while the
> 1988 one translates "billion" to "biljoen".  But there is a rescue.  The
> 1988 dictionay tells us for "trillion": "biljoen", "triljoen" (after 1948).
> So apparently there was a change in 1948 (amazing that the 1953 dictionary
> did not pick it up).

The Oxford Dictionary for Writers and Editors says that since 1948 the 
French, British and German definition of a billion has been one million 
millions. A 1960's printing of the OED says that the french billion is a
thousand billions.

Does anybody know hwat happened in 1948? who decided on wahtever change 
occurred?
  
Cathal

---------------------------------------------
|     Cathal Kelly                             
|     Ollscoil Luimnigh (Uni. Limerick),    For the benefit of 
|     National Technological Park,          the non-Irish the
| IRL Limerick                              't' in 'Cathal' is       |
|                                           silent and the name      |
      (Ireland)                             is attached to the       |
                                            male of the species.     |
      kellyc@ul.ie                                                   | 
                                --------------------------------------
From: cliff@WATSON.IBM.COM ("Cliff Pickover")
Newsgroups: sci.math
Subject: The Pi Sequencing Initiative
Date: 24 Jul 92 21:01:00 GMT

I'm curious about mathematicians' answers to the following 3 questions.

1.  Would you be in favor of a project, the goal of which was to
determine one trillion digits of pi?

2.  Would you like the NSF to fund such a project for 1 million dollars?

3.  If, as a result of this study, the distribution of digits in pi were
found to be nonrandom, would you consider this to be a profound and
important finding?

Please send comments to cliff@watson.ibm.com