HDTenEightyP/NetNews huggingface dataset:
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Subject: Re: SL(2,Z) fundamental domain
Organization: U of Cambridge, England
Date: Sun, 6 Sep 1992 23:06:03 GMT

It's asked: What semicircles (radius and centre) are there in the
usual tesselation of the upper half-plane by copies of the standard
fundamental region for SL(2,Z)?
Well, every such semicircle is the image under an element of SL(2,Z)
of Re(z)=0; in other words, of z+z*=0 (where z* is the conjugate of z).
So we get everything of the form (az+b)/(cz+d) + (az*+b)/(cz*+d) = 0,
which is to say, errm, (az+b)(cz*+d) + (cz+d)(az*+b) = 0,
i.e. ac.zz* + (ad+bc)Re(z) + bd = 0,
i.e. x^2+y^2 + (ad+bc)/ac.x + bd/ac = 0,
i.e. (x-(ad+bc)/2ac)^2 + y^2 = (ad+bc)^2/4(ac)^2 - bd/ac
except that I've probably made a mistake by now. (Um, when ac=0 we get
a vertical line. This is a boring case.)
So, the possible radii are those of the form
  root[ ((ad+bc)/2ac)^2 - bd/ac ]
and the centres are those of the form
  (ad+bc)/2ac;
and <centre,radius> can occur if there are a,b,c,d with ad-bc=1 giving
that pair.
Determining what radii can occur for given centre is the same as
determining what bd/ac can be, given (ad+bc)/2ac.
... Oh, I've just noticed that the radius is equal to
  (ad-bc)/2ac, give or take a change of sign. In other words, 1/(2ac).
So all radii are of the form 1/q, and then the centre is p/q for some
integer q. And q must be even.
Can all these occur? In other words, given ac, can ad+bc be anything
we like? Well, since ad-bc=1 this is the same as: given ac, can ad be
anything we like? Can we solve ac=x,ad=y?
Well, let's think. We have (x,y)=(ac,ad)=a(c,d)=a. So we know what
a has to be. Then we know also what c,d have to be. Now there will
be an appropriate b iff ad == 1 mod c; in other words, if y==1
mod x/(x,y).
What are x,y in the present context? 2x=q; 2y-1=p. So we need
(p+1)/2 == 1 mod (q/2)/(q/2,(p+1)/2)
i.e. mod q/(p+1,q). Weird. I probably really have made a mistake
by this point, so I'll now leave it up to those of you with pencil
--
From: mazzarel@wish-bone.berkeley.edu (Ariel Mazzarelli)
Newsgroups: sci.math
Subject: Re: Couple of questions
Date: 9 Sep 1992 20:45:38 GMT
Organization: U.C. Berkeley Math. Department.
NNTP-Posting-Host: wish-bone.berkeley.edu

In article <1992Sep9.102457.15049@news.columbia.edu> pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:
>Could someone please explain to me a few basic things:
>
>What is the logic used to answer questions of the below type?
>
>1) If the finite group G contains a subgroup of order 7 but no element
>(other than the identity) is its own inverse then the order of group G
>could be (a)27   (b)28   (c)35   (d)37   (e)42.
>

If a prime p divides the order of a group,  then the group must contain
an element of order p (Cauchy's theorem).  Since G has no element of
order 2,  |G| cannot be even,  so b & e are eliminated as options.  Since
7 must divide |G|,  a & d are eliminated.  That leaves c.

Ariel
Newsgroups: sci.math
From: takriti@engin.umich.edu (samer Takriti)
Subject: Re: Need Help on Matrix Differentiation
Date: Tue, 08 Sep 92 09:31:35 EDT
Organization: University of Michigan Engineering, Ann Arbor

In article <18h8pnINNdp2@matt.ksu.ksu.edu> bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:
>Hi,
>
>I was just wondering if somebody could help out regarding matrix differentiationThe problem is to minimize:   
>
>y= x'Ax + 2x1 + 3x2 - 10
>
>where A is a 2x2 matrix :   25  7
>                             7  13
>
>
>and x is the (x1 x2) column vector and x' denotes the transpose.
>
>It's easy if one expands the x'Ax term but I was wondering if there was some way doing it using matrix differentiation.
>
>Thanks for any help in this regard.
>

Write the problem as:
y = x'.A.x + c'.x
(The constant is not important), 
c = (2, 3)'.
You need the derivative to be zero, i.e., 
2.A.x + c = 0
x = -0.5 A^(-1) c
x is the solution for your problem. Make sure that A is
positive definite, this will guarantee that x is a minimum
otherwise it may be a maximum.
-Samer
Newsgroups: sci.math
From: marty@amaterasu.physics.uiuc.edu (Marty Gelfand)
Subject: Re: Jokes in papers (was Re: Einstein's Erdos number)
Organization: Dept. of Physics, Univ. of Illinois at Urbana-Champaign
Date: Wed, 9 Sep 1992 15:49:14 GMT

In article <neuhaus.716039445@vier> neuhaus@vier.informatik.uni-kl.de (Stephan Neuhaus (HiWi Mattern)) writes:
>I vaguely recall from a math course that there was a paper published
>by three soviet mathematicians, the third of which having signed the
>paper only to make the three mathematicians' names sound like ``alpha,
>beta, gamma''.  The third mathematician's name was Gamow.  Does
>anybody know the names of the other two?
>
I also vaguely recall such a story, but it was about three physicists...
Alfven, Bethe, and Gamow.  Sorry, I don't have a citation.
--marty gelfand   marty@amaterasu.physics.uiuc.edu
From: sassoj@lib102.its.rpi.edu (John J. Sasso Jr.)
Newsgroups: sci.math
Subject: NEED TEXT ON ORTHONORMAL BASES
Date: 11 Sep 92 14:47:00 GMT
Organization: Rensselaer Polytechnic Institute, Troy NY


   Does anyone know of any good a math books which pertain mostly to Hilbert
space and discuss quite a bit about orthogonality & orthonormality, basis
functions, and anything related in this nature.  I have looked through books
on functional analysis and real analysis but they seem to only briefly discuss
the subject.  Also, any references to articles in journals such as American
Journal of Math and such (journals which are fairly easy to read) which discuss
related areas would also be appreciated.

                    John
From: flash@unx.sas.com (Gordon Keener)
Subject: Re: Prime Factor Numbers
Date: Thu, 10 Sep 1992 21:53:59 GMT
Organization: SAS Institute Inc.

In <1992Sep8.040307.6288@cronkite.ocis.temple.edu> sklar@picasso.ocis.temple.edu (Dave Sklar) writes:


>	Here's something I was thinking about.  Take a number, n, and write
>down its prime factors [Excluding n and 1] from left to right, forming a new number.  Repeat.
>	When you get to a prime number, you end. Let's call the function
>d(n).  So, d(1001)=71113, d(5)=d(2)=d(11)=[Define something here, say]0.
>	Does every number eventually go to a prime, and then 0?

Well, I did a (slow) little shell script using the poor UNIX "factor" program,
and discovered that every n < 150000 either goes to a prime or exceeds 10^15,
at which point factor can't deal with it. Quite a few of them did overflow,
though, (including 1001; the smallest number that did was 91), so I would
guess that some of them will keep growing to infinity. I did notice quite a
few 12- and 13-digit primes, though, so not all sequences terminate in small
primes.

I did determine that there are no infinitely looping sequences of numbers
<10^15 that contain a number <150000. Big deal. Maybe someone with a decent
MP package can do better. :-)

>	What happens if you include the 1 and n?  What happens if you write
>prime factors that occur more than once once for each time they occur
>instead of once for any number of times they appear? [d(20)=225 instead of
>d(20)=25]

I didn't check these out. You'd get big numbers a lot quicker, though. :-)
--
Gordon Keener         SAS Institute, Inc.
flash@unx.sas.com     SAS Campus Dr, Cary,
+1 919 677 8000       NC 27513-2414 USA
"There is a heppy lend, fur, fur a-wa-ay." - _Krazy_Kat_, George Herriman
Organization: Central Michigan University
Date: Friday, 4 Sep 1992 16:51:06 EDT
From: <32WENPW@CMUVM.CSV.CMICH.EDU>
Newsgroups: sci.math
Subject: 2n = difference of primes?

Question:  Is it true that every positive integer 2n is the difference of some
pair of primes?   (2 = 5-3, 4 = 7-3, 6 = 11-5, 8 = 13-5, 10 = 13-3, ...?)

           ken

(ken w. smith, math dept., central mich univ., mt pleasant mi 48859)
Newsgroups: sci.math
From: martin@lyra.cis.umassd.edu (Gary Martin)
Subject: Re: Set-Theory Doubts: Got some conceptual glitches
Organization: University of Massachusetts Dartmouth
Date: Sat, 12 Sep 1992 14:29:15 GMT

In article <18s2mtINN2gf@matt.ksu.ksu.edu> bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:

   Hello there,

   I was just wondering if somebody on the net could help clarify a few conceptual glitches I have regarding very elementary concepts of set-theory.

   a) According to my textbook, the ORDERED PAIR of two objects x and y is the set <x,y> = {{x}, {x,y}}. It also goes on to state that by this definition, the ordered pair is determined by x and y; and the order, x first and y second, is important unless x=y.

   What throws me off is the definition; what is the motivation behind it?                     
Well, you want the definition of <x,y> to be a set from which you can 
recover x and y.  With this standard definition, you can recover x as the
intersection of the elements in <x,y>.  Recovering y seems trickier.  If
<x,y> has two elements, then y is their symmetric difference.  If it has
one element, then y = x, so you can recover y.  (Of course, when I said
above that x is the intersection, I meant that it's the only element of
the intersection.)
 
   b) Let f: X --> Y be an injective mapping. By definition, this means:
     for all x, x' in X: f(x) = f(x') ==> x=x'.

   Equivalently, x not equal to x' ==> f(x) not equal to f(x').

   My question here is: for the firstmentioned implication why is the 
   CONVERSE not true? (I know it's not but am having trouble finding an
   intuitive/logical answerfor it).

But the converse is true.  That's part of the definition of 'function'.
It simply says that if you feed the same input into the function twice,
you'll get the same output each time.

--
Gary A. Martin, Assistant Professor of Mathematics, UMass Dartmouth
Martin@cis.umassd.edu
From: kubo@zariski.harvard.edu (Tal Kubo)
Newsgroups: sci.math
Subject: Jokes in papers (was Re: Einstein's Erdos number)
Date: 8 Sep 92 09:12:36 GMT


Niall Graham mentioned:

>Source: From Figure 1 of "On properties of a well-known graph or
>what is your ramsey number?"  Annals of the New York Academy of
>Sciences, Vol. 328. The author calls himself Tom Odda of Xanadu University.
                                             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

This is reminiscent of one of Andre Weil's early papers
at the end of which he lists his affiliation as 
"Univ. of Erewhon, Erewhon, Ut."
From: dlb@fanny.wash.inmet.com (David Barton)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 12 Sep 92 18:54:13 GMT
Organization: Intermetrics Inc., Washington Division, USA

In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org
(Carter Bennett) writes:

      So there are 4426165368 possible ways to place the 8 rooks on the 
   board.  My understanding is that no two rooks can attack each other 
   only when they are arranged such that they line up on one of the two 
   diagonals of the board.  Meaning there are 2 ways out of 4426165368 
   possible setups that meet the criteria.  Putting that in normalized odds:

No.  Each rook has to be in a different row AND a different column
from every other rook.  There are far more than two ways to do this.

Let's see.  The first rook can go anywhere; it eliminates fifteen
squares from possible next placement, leaving 49.  The second rook has
49 out of 64 squares to choose from and eliminates thirteen squares
from the next possible placement set, leaving 36.  Continuing:

  p = 49/64 * 36/64 * 25/64 * 16/64 * 9/64 * 4/64 * 1/64 

or, whipping out my handy dandy emacs calculator:

  p = 5.77565515416e-6

or

  p = .00000577565515416

If I can do this, with my distance from my prob/stat course, it can't
be THAT hard.

					Dave Barton
					dlb@hudson.wash.inmet.com
Date: Monday, 14 Sep 1992 14:43:06 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: sci.math
Subject: Re: Chess Problem

someone said, "B2 as given is only correct if the bishop is placed on a
corner."

this, too, is untrue.  consider the bishop being placed ANYWHERE on a side.
it is only when the bishop is placed in the interior of the board that
B2 is not correct as given.
From: cet1@cus.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Chess Problem (8 bishops)
Date: 14 Sep 92 16:07:43 GMT
Organization: U of Cambridge, England

In article <1992Sep12.222402.14408@nmt.edu>, jmarlan@titan.nmt.edu
(Jon Marlan) writes:
|>
|> In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org
|> (Carter Bennett) writes:

It was, indeed, the saving grace in an otherwise utterly useless posting.

|> >  FOLLOW-UP QUESTION FOR EXTRA CREDIT:
|> >
|> >  Determine the odds for doing the same with eight bishops!  ;-)
|> 
|> I'll give this one a try.
  
[Details omitted: as others have already pointed out your errors.]

There are 22522960 ways of arranging 8 bishops on a chessboard so that
none attacks any of the others. The probability in question is therefore
22522960/4426165368 = 1/196.5179...

From the fact that there are (1,32,356,1704,3532,2816,632,16) ways
of arranging 0..7 non-attacking bishops on the 32 black squares
(or, of course, the white ones) it is easy to deduce that there are 
(1,64,1736,26192,242856,1444928,5599888,14082528,22522960,22057472,
12448832,3672448,489536,20224,256) ways of arranging 0..14 non-attacking
bishops on the full board. 

I admit to using a program to get the first set of numbers, but it is
perfectly possible, if a little tedious, to do by hand. I find it
helpful to think of arranging non-attacking rooks on a board shaped

            X
          X X X
        X X X X X
      X X X X X X X
      X X X X X X X
        X X X X X
          X X X  
            X

Chris Thompson
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 13 Sep 92 00:38:56 GMT
Organization: U of Cambridge, England

In article <1992Sep12.174545.23214@ima.isc.com>, karl@ima.isc.com (Karl Heuer) writes:

> Here's a more advanced chessboard question.
> 
> A white king, a black king, and a white pawn are randomly placed on an NxN
> chessboard.  (Independent, uniform distributions.)  What is the limit, as
> N -> infinity, of the probability that the position is a win for white?
> 
> (Note that it doesn't matter what you do about collisions or illegal
> positions, since they occur with probability zero in the limit.)

Eek. Let's think... (This is much better than the last one, to which the
answer is of course 8!/(64 choose 8) = 560/61474519, or about 9.109e-6)

If the pawn is a "rook's pawn", funny things happen; but that's only
probability 2/N, so we can forget that.
If the black king is behind the white pawn [but not next to it, but
that's only 8/N^2] then White wins, always.
If the black king is in front of the white pawn, but "too far from" it
(i.e. can't reach the promoting square before the pawn does) then White
wins by rushing the pawn forwards.
If the black king is in front of the pawn, and in range, then it all depends
on how the two kings are placed. If the white king is further away from the
pawn than the black one, then Black clearly draws. If the white king is closer
to the pawn, then White wins (except in O(1/N) of the cases, where it depends
on parity considerations) by getting in front of the pawn and keeping the
opposition.

[thinks: if I haven't made a mistake, all the interesting chess is in O(1/N)
of the cases...]

Right. So we want the probability that the black king is (i) within the pawn's
"light cone" and (ii) closer to the pawn than the white king. A moment ago I
thought this was really easy, but it's not quite as nice as I thought...
I think it might be preferable to place the kings first, so to speak,
and then we want the probability that the pawn is within the "reverse light
cone" of the black king and closer to it than to the white king. Ugh.

I don't think I have the patience for this right now, at least not without
pencil and paper to hand. Still, I've reduced the problem to the point at
which it's just a matter of summing some fairly easy (ha!) series. 

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]
From: bach@jalapeno.cs.wisc.edu (Eric Bach)
Newsgroups: sci.math
Subject: Re: Couple of questions
Date: 9 Sep 92 22:16:26 GMT
Organization: University of Wisconsin, Madison -- Computer Sciences Dept.

In article <18lnpiINN1qk@agate.berkeley.edu> mazzarel@wish-bone.berkeley.edu (Ariel Mazzarelli) writes:
>In article <1992Sep9.102457.15049@news.columbia.edu> pvl2@cunixb.cc.columbia.edu (Priscilla V Loanzon) writes:
>>Could someone please explain to me a few basic things:
>>
>>What is the logic used to answer questions of the below type?
>>
>>1) If the finite group G contains a subgroup of order 7 but no element
>>(other than the identity) is its own inverse then the order of group G
>>could be (a)27   (b)28   (c)35   (d)37   (e)42.

Since this is from an ETS test, we can eliminate (a) using the following 
principle: if you can get one of the answers by naively combining some of the 
numbers in the problem, that answer has to be wrong.  The question tells
us implicitly there is no element of order 2 (x^{-1} = x ==> x^2 = 1), and we 
know there is one of order 7, so 27 is out.

37 is out because it is a prime having nothing to do with the numbers
in the question (remember this is group theory, divisibility is important).

This leaves 35 and 42 as the only plausible contenders; at this point it
is worthwhile to bring in some actual algebra.  Suppose x^2 = 1 in a group
of size 35.  Raise both to the 18th power, to get x^{2*18} = x^36 = 1.
So there are no elements of order 2 in a group of size 35, hence we
know in our heart that 35 is the answer.

To *prove* that 35 is the answer, use Lagrange's theorem (the order
of a subgroup divides the order of the group) and Cauchy's theorem
(if a prime divides the order of the group, there has to be an element
of that order).  ETS probably wants you to do it that way.

Everyone who is studying for the GRE should read "None of the Above,"
by David Owen.

--Eric Bach
  bach@cs.wisc.edu
From: RVESTERM@vma.cc.nd.edu
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 12 Sep 92 19:32:00 GMT

i am not sure how to include your text in my followup...can someone tell me
how?

anyway, you said something like, "my understanding is that the only way to put
eight rooks on a chessboard without any rook attacking another is to line them
up along one of the diagonals."

your understanding is wrong.   there are eight factorial ways, not two, to
satisfy that condition.  consider: put a rook on the first column, any of eight
ranks.  you can then put a rook in the second column on any of seven ranks,
such that it is not attacking the first.  then a rook in the third column on
any of six ranks, such that it is not attacking either of the first two.  and
so on, until you get to the last column; then there is one rank to put a rook
in such that it does not attack any of the first seven.


                                           bob vesterman
From: kosowsky@hall.harvard.edu (Jeffrey J. Kosowsky)
Newsgroups: sci.math
Subject: Re: curves of degree 2
Date: 10 Sep 92 22:01:23 GMT
Organization: Harvard Robotics Lab, Harvard University


>Given any 5 points in the plane, there exists a curve of degree 2
>passing thru all 5 points.

>ie: There exist constants a,b,c,d,e,f such that the curve
>	ax^2 + by^2 + cxy + dx +ey +f = 0 
>passes thru all 5 points.

Thanks for all who responded. Of course, as pointed out by several
posters, this is equivalent to a system of 5 linear equations in 6
unknowns and hence has at least a 1-dimensional subspace of solutions.

In retrospect, this solutions is "trivial" and I feel quite humbled
for having wasted much time looking for "deeper", "more sophisticated"
solutions. Linear algebra triumphs again!

Jeff Kosowsky
Newsgroups: sci.math
From: hporopudas@tnclus.tele.nokia.fi
Subject: Re: Proof of God's Existence
Organization: Nokia Telecommunications.
Date: Sat, 5 Sep 1992 18:09:02 GMT


In article <8815@amsaa-cleo.brl.mil>, matt@amsaa-cleo.brl.mil
(Matt Rosenblatt) writes:
> In article <1992Sep1.090011.1@tnclus.tele.nokia.fi>
> hporopudas@tnclus.tele.nokia.fi writes:
>
>>   I think that God's existence is an axiom. He says in the Holy Bible
>> (Second book of Moses 3:14-15 in Old Testament and also in Johannes
>> 8:28-29 in New Testament) that He is "Existence Himself".  [Hannu]
>
> Er, not quite.  What He says in 2 Moses 3:14 is, "I AM WHAT I AM."

  You did not read the point 2 Moses 3:14 where God continues His speech
  to the prophet by saying say to them that "I AM sent me ...". I don't
  have english Bible but in finnish Bible it is like this. "I AM" means
  "ehjee" in hebrean, which is in finnish "Min{ Olen" (=Existence).

>
>>   If someone is arguing that "Existence Himself" does not exist, I think
>> that at the same time he/she is arguing that he/she himself/herself does
>> not exist.  [Hannu]
>
> If "I AM WHAT I AM" amounts to an existence proof, then we have just
> proved that Popeye The Sailor Man exists, too.
>
> 					-- Matt Rosenblatt
> 					(matt@amsaa-cleo.brl.mil)

  Contents of your sentence is empty.


              Hannu.
From: kim_sh@jhunix.hcf.jhu.edu (Sung Hyun Kim)
Newsgroups: sci.math
Subject: Re: Jokes in papers (was Re: Einstein's Erdos number)
Date: 9 Sep 92 15:47:05 GMT
Organization: HAC - Johns Hopkins University, Baltimore

In article <18l5lsINN176@hilbert.math.ksu.edu> bennett@math.ksu.edu (Andy Bennett) writes:
>neuhaus@vier.informatik.uni-kl.de (Stephan Neuhaus (HiWi Mattern)) writes:
>
>>I vaguely recall from a math course that there was a paper published
>>by three soviet mathematicians, the third of which having signed the
>>paper only to make the three mathematicians' names sound like ``alpha,
>>beta, gamma''.  The third mathematician's name was Gamow.  Does
>>anybody know the names of the other two?
>
>As I heard the story, it was Ahlfors, Bethe, and Gamow and Bethe was the
>one who signed just to make it come out right. Did this really happen? Does
>anyone have a reference to the article (which should be easy to find if
>you have access to the cd rom version of math reviews)?
>
>
>-- 
>Andrew G. Bennett         bennett@math.ksu.edu         If you count too

I remember reading this story from somewhere...it was either a biography of
Landau(Russian Physicist) or the automathography of Halmos.. Though I think
it must have been from Landau book, I should check again.. :-)
BTW, I thought they were  physicists..did they do some math also? or did
they publish that specific article in math journal?

-- 
 Sung Hyun Kim                                                         
         kim_sh@jhunix.hcf.jhu.edu  // kim_sh@jhuvms.hcf.jhu.edu 
Dept. of Economics, Johns Hopkins University, Baltimore, MD 21218               
..... graduate of Korea Institute of Technology.........
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Warm and Fuzzy?  Koalas vs. Math
Date: 14 Sep 92 13:34:13 GMT
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: voros@vaxc.cc.monash.edu.au (Joe Voros)

In article <1992Sep14.112820.90086@vaxc.cc.monash.edu.au>, voros@vaxc (Joe Voros) writes:
>>>>If you want something warm and fuzzy, raise koala bears.

>Since this is sci.math, which is noted for its pedantic tone, I must
>state that koalas are *not* bears.  

I think it was Spivak who footnoted his definition of "measure zero"
with the "red herring principle": in mathematics, you can define what
a "red herring" is without defining "red" or "herring".
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)