HDTenEightyP/NetNews huggingface dataset:
From: mingos@vax.oxford.ac.uk
Newsgroups: sci.math
Subject: multiples of 9 - THANKS!!
Date: 19 Dec 92 17:52:56 GMT
Organization: Oxford University VAX 6620


Thanks to all of you who responded to my query about multiples of 9!!
 I started to thank people individually, but in the end it became a toss up
between thanking everyone, and writing my christmas cards...!

 I've now got more proofs than I expected, many different, but all on the same
basic approach.
  The solution was along the lines that I'd gone, and ended up as easier to
prove than I expected.  However, I didn't manage to get a proof myself, which
is why I'm a chemist, and you guys are mathematicians.  

  Thanks to everyone who responded, I really appreciate your time and effort

				Cheers,

					Gavin
Newsgroups: sci.math
From: dennis@cadkey.com (Dennis Paul Himes)
Subject: Re: The golden number
Organization: cadkey
Date: Sat, 19 Dec 1992 19:45:28 GMT

In article <BzF76u.B7x@unccsun.uncc.edu> sbardhan@uncc.edu writes:
>Will someone start discussing about it. I'm not sure  about its properties.
>

     If you mean the golden ratio, it is .5 + .5 * (5^.5).  It has the
property that if the ratio between the length and width of a rectangle is
the golden ratio and a square whose side is equal to the length of that
rectangle is joined to it, then the new rectangles' length and width will
also have the golden ratio.

==============================================================================

               Dennis Paul Himes    <>    dennis@cadkey.com
 
"In watermelon sugar the deeds were done and done again as my life is done
in watermelon sugar."           - Richard Brautigan
Newsgroups: sci.math
From: lallu@cc.utah.edu (Upmanu Lall, Civil Eng)
Subject: shortest chain on a sphere / shortest around the world trip
News-Software: VAX/VMS VNEWS 1.4-b1  
Organization: University of Utah Computer Center
Date: 13 Dec 1992 16:42 MST  




Does anyone know if this is a solved problem ?
 It seems like it ought to be.
There are n arbitrary locations on the surface of the earth, each of which must
be visited
once and only once, starting from any one of them.
At the end of the trip one is back at the starting point.
It seems to me that the choice of the starting point does not matter,
i.e. there is a unique shortest chain, and you should be able to find it
irrespective of where you start on it.
Is this assertion correct ?
Is there an algorithm for determining the shortest path efficiently,
and the second shortest, etc ?
I recall from a network optimization class taken 2 decades ago, that
a chain network is the dual problem to a flow network
Can this be used to setup the problem such that one could
establich whether the primal or the dual is easier to solve, and to then solve
the problem ?
Distances are Euclidean on the sphere (or the Earth). LAts and Longs of the
points are known , and the number of locations is also known a priori (i.e.
this is not a stochastic process).

Any references, or suggestions to bodies of literature shall be appreciated.
Deterministic  or Monte Carlo techniques would be fine.


Thanks very much for your help
Please respond to
ulall@pub.uwrl.usu.edu
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: sci.math
Subject: Re: Tiling problem
Date: 14 Dec 92 12:19:09 GMT
Organization: Department of Computer Science, U of Copenhagen

choy@skorpio.usask.ca (I am a terminator.) writes:

>In article <israel.723837962@unixg.ubc.ca>, israel@unixg.ubc.ca (Robert B. Israel) writes:
>|> In <rcbaaw.723228012@rwb.urc.tue.nl>, rwb.urc.tue.nl (Angelo
>|> Wentzler) writes:
>|> 
>|> >Given a grid of squares, tile it with black and white tiles and make
>|> >the largest possible square so that no square contained in it has four
>|> >equally colored corners.


>Would the following be a recursive solution?

>0

>1

>Kinda small--let's double the size.

>00
>00

>Um ... no internal squares in these little squares so no 4 colored corners.

Note that the posting did not say INTERNAL squares. The whole square
also counts.

>Doubling again:

>0011
>0011
>1100
>1100

>Now your starting to worry. The internal squares have 4 (colored) corners.
>In general, the internal 0 (or the internal 1) has two 0 corners and two
>1 corners. So that's OK.

Note also that ALL subsquares must be considered, not just those that
are centered. The top-left 2x2 subsquare (and many more) has for
identical corners.

	Torben Mogensen (torbenm@diku.dk)
From: t4d192@rick.cs.ubc.ca (Bruce Paul Dow)
Newsgroups: sci.math
Subject: Re: Philosophy of Pi
Date: 12 Dec 1992 08:35:55 -0800
Organization: Computer Science, University of B.C., Vancouver, B.C., Canada
NNTP-Posting-Host: gambier.rick.cs.ubc.ca



Thanks for all of your relies. I found them all interesting and
helpful. I guess it was kind of worded like a 'stupid' or
'obvious' question but I meant it in all sincerity.

So thanks for not flaming me (yet)!

Actually there's a related point I haven't been able to grasp
intuitively. How about that -other- number: e

Okay, I won't ask why it's between 2 and 3, irrational, etc...

It's just that Euler identity... I can't seem to wave my hands
around in the right way to imagine what it means to multiply
e (or any number) by itself an imaginary number of times!

Okay, I know we can justify raising a base to an irrational
exponent using logarithms, but it doesn't seem to me to justify
the same thing for imaginary exponents.

Of course, I have seen the Taylor series proof of Euler's
identity but I can't seem to pull any intuitive feeling out 
of it. It seems to me as well that having proved Euler's
identity, you then have a strong link between e and Pi.
Is this not so? I was wondering whether anyone knew how strong
this link might be and whether we can express e and Pi in terms
of each other in any useful way?

In engineering we seem to get enough math to be able to use it
to solve problems but it's not deep enough to get comfortable
with it. By the way I just found out today I passed the
Partial Differential Equations exam after writing it for the
third time.

Congrats, flames, comments?

Bruce


Newsgroups: sci.math
From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Re: TO ALL !!!
Organization: Computer Science Department,  Stanford University.
Date: Sat, 19 Dec 1992 20:03:48 GMT

In article <BzIFH7.9AL@watdragon.uwaterloo.ca> deghare@daisy.uwaterloo.ca (Dave Hare) writes:
>I think I speak for a sizable minority of this newsgroup when I state that
>I would like anybody who wants to wish me a nice day to just go to hell.
>Not all of us have "nice days" and it is simply damnably insensitive
>of all you jerks to suggest that we should have nice days.

This would be the sizable minority living north of Lake Erie by choice,
I take it.  Wishing you lot a nice day is about as meaningful as
wishing the rest of us should win the lottery.

To the ESFJ's cheery "Good morning," INTP's have been responding on
their bad days with "Go to hell" since before sailors were cheerfully
greeting Prometheus each morning as they sailed past his rock.

The script continues, "Can I get you some coffee?"---"Bah, humbug."
These popular rhetorical devices for establishing mood can be enlivened
by an occasional interchange in the order of "Go to hell" and "Bah,
humbug".

Adding insult to your injury is the dreadful prospect that the ESFJ
greeter might really mean it and be depressed at the thought that a
fellow creature might not be having a simply wonderful morning.  INTP
greeters are much more tolerable in this regard: you know that they
derive a certain satisfaction from your disclosure that you are having
an even harder time getting your morning jump-started than they.

Season's greetings/Pi is transcendental/Drop dead (circle one)
-- 
Vaughan Pratt						Fun is not having it
Organization: Penn State University
Date: Tue, 15 Dec 1992 14:55:01 EST
From: <B7D@psuvm.psu.edu>
Newsgroups: sci.math
Subject:       '-' operation

Recently my daughter brought back Problem Set II of Wisconsin Mathematics
Science and Engineering Talent Search, and Question 4 in it is as following:
         Operation @ satisfies the conditions that
                X @ (Y @ Z) = X @ Y + Z and X @ X = 0 for any real numbers
          X, Y, Z. Show that @ must be subtraction.

   Since my daughter does not care much whether she has talent so I tried to
 solve it myself:-). It took me more than an hour but still I am not sure I get
 a right answer. The problem I have is: How is operation '-' defined? How about
 other operations like '+', '*', and '/'? Although I do not have Ph.D
 in math but I consider myself one of the best in learning math in those
 math classes. It is sort of ashameful that I never learned or remember
 how the basic operation is defined. For the above question, is only
 thing I have to prove is X @ 0 = X?, which is not very difficult because
 X @ 0 = X @ (X @ X) = X @ X + X = 0 + X = X.

 I hope no one will blame me for bringing 'homework' to the net or trying
 to be 'talent' since the deadline was passed and it is already too late
 to send the answer in for that purpose:-). I just like to know what you
 have to do to say some operation MUST be one of the basic operations.

 Duane
Newsgroups: sci.math,sci.physics,sci.chem,sci.bio
From: snyder@henry.ece.cmu.edu (John Snyder)
Subject: Re: TIME HAS INERTIA--INDEPENDENT VALIDATION OF ABIAN
Organization: Physics Department, Carnegie Mellon University
Date: Tue, 15 Dec 1992 06:41:03 GMT

In article <234323@tamsun.edu> echinter@flash.tamsun.edu (Eric Chinter) writes:
>I was extremely SCEPTICAL that any of ABIAN's claims were TRUE, but
>after I E-MAILED him for a more thorough DESCRIPTION of his THEORY, I
>was able to determine that it is CONSISTENT of itself and EXPLAINS
>adaquately many WELL-KNOWN and ACKNOWLEDGED PHENOMINONS.  ALL of his
>EQUATIONS follow neatly and cleverly from his several stated AXIOMS
>and POSTULATES.  FURTHERMORE, his THEORY makes several PROFOUND
>STATEMENTS that apparently have never before come to the attention of
>OTHER SCIENTISTS!!!  They are in fact so PROFOUND and EARTH-SHATTERNIG
>in their IMPLICATIONS that I will not reveal them except in private
>e-mail, but SEVERAL LABS have been notified and are setting up
>EXPERIMENTS to test WHETHER the UNFORSEEN IMPLICATIONS are indeed
>TRUE!!!  Preliminary results from one of the labs support the brilliant
>THEORY of DOCTOR ABIAN, but of course only time will VINDICATE him
>COMPLETELY.
>
>Anyone who desires to see the document of INDEPENDENT VERIFICATION whihc
>I will have prepared within ONE (1) week should send me PRIVATE E-MAIL.
>
>Thank you, and GOD-WILLING we'll set this mess straight!
>
>Eric CHINTER, PhD Berekeley


Please send me a copy.  My email eoes not seem to be able to reach you.
Also, please forgive this atrocious waste of bandwidth.

John
snyder@henry.ece.cmu.edu
From: membrilo@vax.oxford.ac.uk
Newsgroups: sci.math
Subject: Re: A word problem
Date: 15 Dec 92 10:01:05 GMT
Organization: Oxford University VAX 6620

In article <1992Dec12.162349.29729@dcs.qmw.ac.uk>, arodgers@dcs.qmw.ac.uk 
(Angus H Rodgers) writes:
> Are there any finitely generated infinite semigroups in which
> the idempotent law holds?

  You mean the square of each element is the identity?

  If yes, then it is enough to look at the 3 generators semigroup (in which the
idempotent law holds), since it is infinite. This is a direct consecuence of
the fact that you can construct square-free infinite words using only three 
letters
I give you a reference for a proof: "Combinatorics in Words" by Lottarie.

Hope this helps.



--------------------------------------------------------------------------------

F.H. Membrillo-Hernandez
Mathematical Institute
Oxford University
From: hougen@focus.csl.uiuc.edu (Darrell Roy Hougen)
Newsgroups: sci.math
Subject: Re: Factor 69 digit integer
Date: 11 Dec 1992 22:39:48 GMT
Organization: Center for Reliable and High-Performance Computing, University of Illinois at Urbana-Champaign
NNTP-Posting-Host: focus.csl.uiuc.edu

>In article <1992Dec10.132040.1186@walter.cray.com>, slow@ferrari (David Slowinski) writes:
>>> Please help me.  I need the prime factors of 
>>>  132686104398972053177608575506090561429353935989033525802891469459697

Write your own infinite precision arithmetic program to solve the
problem.  Here's a start:
(1) Store each number in an array, e.g., A[1..69] where each element
of the array is a digit, ie., 0,1,...,9.
(2) Implement a divide and carry algorithm.  For each remainder,
you'll have to try all dividends from 0 through 9 up the first that
doesn't work.

I'll let you fill in the details.

Darrell

Newsgroups: sci.math
From: barnes@bert.eecs.uic.edu (Craig Barnes)
Subject: Projecting a vector onto a plane
Organization: University of Illinois, Chicago
Date: Thu, 17 Dec 1992 15:04:09 GMT


This is a trivial question I'm sure, but I don't have 
any good references handy. How do I project an arbitrary
vector on to a plane defined by two other vectors?



Thanks


Craig Barnes
From: choy@skorpio.usask.ca (I am a terminator.)
Newsgroups: sci.math
Subject: Re: Lines in plane problem
Date: 16 Dec 1992 02:24:54 GMT
Organization: University of Saskatchewan, Saskatoon, Canada
Distribution: world
NNTP-Posting-Host: skorpio.usask.ca

In article <1992Dec9.110737.507@eve.comlab.ox.ac.uk>, mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
|> 
|>    Is it possible to place a countable number of unit length
|>    straight line intervals on the plane such that :

Hmm. I can put zero intervals or 1 interval on a plane and stop. That's
countable. I guess the problem is to place more than a countable number?

Can I have the intervals placed on an uncountable number of parallel lines,
such as the lines sweeping up the y-axis?

Would it be better to restrict the intervals to a line?

|>    i) No two lines touch (i.e. they're pairwise disjoint), and
|>   ii) No more such lines can be placed on the plane subject to cond (i)? 
|> 
|>   There are three cases to consider :
|> 
|>   a) All the lines are open intervals (i.e. exclude the endpoints)
|>   b) All the lines are closed intervals (keep the endpoints)
|>   c) All the lines are half-open (keep one endpoint)
|> 
|> I know how to do case (a), but as far as I'm aware, the other two
|> cases are open problems.

What's the difference between cases a and c?

How do you do case (a)?

Henry Choy
choy@cs.usask.ca
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Vertex Operator Algebras
Date: 20 Dec 92 13:18:38 GMT
Distribution: sci
Organization: M.I.T. Artificial Intelligence Lab.


From time to time, I pick up the book on Vertex Operator Algebras and
the Monster, by Frenkel, Lepowski and Meurman, and while the results
are admirable, I am nevertheless repelled each time by the notation
and the exposition. Apart from such discourtesies as changing their
notation and conventions in the middle of a book whose  technical details
would be hard enough to follow with consistent notation (something like
Grothendieck changing the meaning of the word scheme in the middle of
writing EGA), the authors often remind us that they are not giving
us the general picture (suggesting that the carefully learned technical details
will have to be unlearned later, either in the same book or in other sources).
The overview is essentially lacking while instead
the authors torture one kind of quadratic form after another, assuming that
the reader will get the general idea of how to proceed without a general
definition of the torturous procedure. Maybe it is the fate of books
on such technical material to be like that but I am not convinced.

There are two hints of a more general setting at the end of the book and
I would like to know where this generalization is carried out, in the
hope that it is done more conceptually, more simply and in a way that
is easier for me to read.

p.469, Remark A.2.3 "In this case, analytic continuation yields a family
of operators parametrized by a configuration of points on a Riemann surface."
(No reference is given)

p.481 Remark A.3.4 "Using the rule (A.3.10) it is not difficult to show
generally that for any homologically trivial sum of configurations of 
contours one obtains a formal variable identity."


The first hint suggests that we need a notion of vertex operator living on a
compact Riemann surface or a complex manifold. The second one suggests that we 
get relations among vertex operators from 1-cycles on the manifold.


The first hint reminds me of a lecture that Beilinson gave at Harvard,
in which he took compact Riemann surfaces and had a Virasoro algebra
at each point of the Riemann surface and associated an algebra to finite
subsets of the Riemann surface by taking products of the algebras at the
points of the finite set. He then considered functors between the categories
of modules of the algebras associated to different finite sets. But beyond
the vague similarity, I don't know if what he was talking about is really 
pertinent to the hints given above.

Allan Adler
ara@altdorf.ai.mit.edu
Xref: sparky sci.math:17143 comp.theory:2730
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Newsgroups: sci.math
Subject: Re: Integral Puzzle (SPOILER)
Date: 20 Dec 92 17:25:00 GMT
Organization: University of Michigan EECS Dept., Ann Arbor

In article <Dec.20.11.00.28.1992.16004@romulus.rutgers.edu> clong@romulus.rutgers.edu (Chris Long) writes:
>In article <1992Dec19.011244.2780@Csli.Stanford.EDU>, Yuzuru Hiraga writes:
>
>> Let f be any function such that:
>> 	* for 0<=x<=1; 0<=f(x)<= a
>
>> 	  /1
>> 	* |  f(x)dx = A
>> 	  /0
>> where a and A are given constants (obviously, A<=a).
>
>> What are the maximum and minimum values that
>
>> 	/1       2
>> 	|  {f(x)}  dx
>> 	/0
>
>> can take?
>
>Let I = \int_0^1 f(x)^2 dx; the fact that I >= A^2 follows immediately
>from the Cauchy-Schwarz inequality.  For the other (more interesting)
>direction, consider that if f(x) is non-negative, then so is a-f(x),
>and so (f(x)+a-f(x))^2 >= f(x)^2+(a-f(x))^2, hence \int_0^1 a^2 dx >=
>\int_0^1 f(x)^2 dx + \int_0^1 (a-f(x))^2 dx ==> a^2 >= I + a^2 - 2aA
>+ I ==> I <= aA.  To finish, such functions exist.  For the first case,
>consider f(x)=A.  For the second, consider f(x)=a for 0 <= x <= A/a,
>f(x)=0 elsewhere.
>-- 
>Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
>
>"S.B., with an I.Q. of 161, failed to complete his course of study,
>running away instead with his professor's wife."
>H. J. Eysenck, _Know Your Own I.Q._

The fact that I must be <= aA can be deduced more easily as follows :

f(x)^2 = f(x)*f(x) <= a * f(x), hence :

 	/1       2       /1
 	|  {f(x)}  dx <= | a * f(x) = aA
 	/0               /0

Kanad
Newsgroups: sci.math
From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
Subject: Re: Calculus Puzzle
Reply-To: thompson@atlas.socsci.umn.edu
Organization: Economics Department, University of Minnesota
Date: Fri, 18 Dec 1992 15:04:10 GMT

sbardhan@uncc.edu (Soumendu Bardhan) writes:

>X + X + X + X +X.... X times = X^2
>Differtiating both sides w.r.t X we get

This formula is only valid for positive integer values of X, so
differentiating with respect to X is not well-defined.

>1 + 1 + 1 + .....X times = 2X

Wrong!  You must also differentiate with respect to the "X" in the
".... X times" of the original expression.  This operation is not
defined.

>or , X = 2X
> as X may not be Zero

Why not?  I don't see anything in the original problem that precludes X = 0.

>1 = 2.

How do you get "1 = 2" from "X = 2X" and "X may not be Zero"?  Perhaps
you have a new theorem: "X .ne. 0" => "X = 1".  If so, please show us
the proof!

>--Soumendu, UNCC

--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209
Newsgroups: sci.math,comp.edu,misc.education
From: regeorge@magnus.acs.ohio-state.edu (Robert E George)
Subject: BreakUp Freshman Calculus ?WAS Re: Student attitudes
Organization: The Ohio State University
Date: Fri, 11 Dec 1992 23:38:45 GMT

In article <11DEC199208420044@cycvax.nscl.msu.edu> burtt@cycvax.nscl.msu.edu (B
RIAN BURTT) writes:
[deletions]
>
>An idea for a college/university large enough to offer several calculus
>sections:
>    Split the calculus sequence into three parts:
>    a) a sequence oriented to physical scientists and engineers that
>targets the math useful to them-- this is essentially freshman calculus
>as we know it
>    b) a sequence oriented to economists, business students, statisticians,
>etc., which give them calculus without relying on "physics" problems
>that are what calc textbooks normally have
>        c) a sequence oriented to mathematicians, that covers the theoretical
>aspects of calculus and offers a very early introduction to proofs and
>developing maturity of mathematical thought.
>    Before I get flamed:  I not advocating either insulating scientists
>totally from mathematical theory, or mathematicians from the applications
>of their work.  I just think the classes should be a little more relevant
>to those who take them.
>    I am a student, not an educator, so I'm no expert and welcome
>comments and criticism.
>
>--Brian Burtt
>Undergraduate major in electrical engineering, Michigan State University
>burtt@lynch.nscl.msu.edu

What about the engineering students who decide near the end of their sophomore
year that they just can't abide any more lab work and want to switch to
mathematics? What about the mathematics students who decide near the end of
their sophomore year that they just can't abide any more abstract work and
want to switch to engineering or physics? Can (a) and (c) be close enough
to substitute for each other?

One "advantage" of the current system is that it doesn't really force
students to make a firm decision as to what they're studying until fairly
well into their undergraduate career. As I understand, most European
and Asian students know what they want to study when they enter university?
Not so for Americans, Americans seem to like it this way.

If "we" decide we want to force students to make the decision earlier in their
career, I acknowledge that there are good arguments for doing so (and good
arguments against -- mainly that the entire secondary curriculum isn't
designed for people to decide that early). But it's necessary to at least
think about this issue before separating up freshman calculus.
Robert George
(speaking only for myself)

"The very essence of individual freedom is equal justice under a rule of
law, a law to which every man shall be subject and which no executive
can modify."
    Senator Robert Taft , March 11, 1944
From: feuerman@parc.xerox.com (Ken Feuerman)
Newsgroups: sci.math
Subject: Re: Interesting Discrete Math Question
Date: 17 Dec 92 20:52:17 GMT
Organization: Xerox PARC

[munch...]
> Here is the proof. (Pigeonhole principle).
> ~~~~~~~~~~~~~~~~~
> Partition the integers into 51 subsets:
> 
> {1,4}   {2,5}   {3,6}
> {7,10}  {8,11}  {9,12}
> .....
> {91,94} {92,95} {93,96}
> {97,100}
> {98}
> {99}
> 
> Now when 52 numbers are drawn, at least 2 must come from one subset, and thus
> be 3 apart.
[more munch...]

This is all well and fine, except, how do you know you've found the best partition
of those integers?  That is, how do you know there isn't some other partition Q = 
{q_1, q_2, ... q_n} of the numbers 1-100 such that

(i) for all i, j, and for all a in q_i, b in q_j, |a - b| != 3; and
(ii) n > 51?

I bring this up since someone else answered 51, having found a way to choose 50
elements before guaranteeing that two would be 3 apart.  At first blush that
seemed plausible, until others got a little more clever.  Is

1 2 3, 7 8 9, ....., 97 98 99

as clever as we need to be?

For the record, I believe so (QED :-)).

--Ken.