HDTenEightyP/NetNews huggingface dataset:
From: algebra@leland.Stanford.EDU (Leong Weng Ng)
Subject: Probability puzzle 
Organization: DSG, Stanford University, CA 94305, USA
Date: Fri, 10 Apr 92 17:12:52 GMT

Puzzle:

	Choose any straight line in a plane. The plane is then divided into two
half-planes by the line. Select any two points on the plane not lying on the
chosen line. What is the probability that both points lie on different half-
planes? Is it 1/2 or something else?

Cheers                                                                 Al
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: sci.math
Subject: Re: Problem with magic square algorithm
Date: 13 Apr 92 11:06:20 GMT
Organization: Department of Computer Science, U of Copenhagen

fcs$1224@altair.selu.edu writes:


>I got this algorithm to generate odd magic squares off the net a
>year or so ago.  There is only one thing wrong with it.  It doesn't
>work!  Would someone mail to me an algorithm that will work?

[Algorithm deleted]

This is a well-known algorithm, but it only works if the side-length
is odd. Even sided magic squares are more difficult to make, and in
many cases you need back-tracking. There is no simple and efficient
algorithm (that I know of) that will produce magic squares of any
side-length above 2.

	Torben Mogensen (torbenm@diku.dk)
From: bs@gauss.mitre.org (Robert D. Silverman)
Newsgroups: sci.math
Subject: Re: Series involving primes
Date: 1 Apr 92 15:22:02 GMT
Distribution: usa
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA

In article <1992Apr1.135207.23092@fid.morgan.com> sethb@fid.morgan.com (Seth Breidbart) writes:
:In article <1992Mar30.173245.16551@linus.mitre.org> bs@gauss.mitre.org
 
stuff deleted....


:|>Instead of summing d(i)/p(i), try summing d(i) copies of 1/p(i).  Now
:|>notice that it dominates the harmonic series, and therefore diverges
:|		^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
:|
:|How can you conclude that d(i) copies dominates the harmonic series??
:|I don't follow this. Please explain.
:
:The harmonic series:
:
:1  1  1  1  1  1  1  1  1   1   1   1   1   1   1   1   1   1   1   1
:-  -  -  -  -  -  -  -  -  --  --  --  --  --  --  --  --  --  --  -- ...
:1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20
:
:d(i) copies of 1/p(i)
:
:   1  1  1  1  1  1  1  1   1   1   1   1   1   1   1   1   1   1   1
:   -  -  -  -  -  -  -  -   -  --  --  --  --  --  --  --  --  --  -- ...
:   2  3  3  5  5  7  7  7   7  11  11  11  11  11  11  17  17  19  19
:
 
Yup! Sorry, I didn't see this. Some days you just feel stupid.

:This actually indicates that sum(d(i)/p(i)) > log p(n), since in the
:second series, p(n) terms correspond to n terms of the original series.
:(OK, I lost the original 1, but when we're talking about the order of
:divergence, who cares?)
:
:Note that since there is always a prime between n and 2n, this series
:is bounded above by twice the harmonic series.  That upper bound can
:undoubtedly be greatly tightened, but suffices to show divergence of
:order log p(n); since p(n) ~ n log n, log p(n) ~ log n + log log n, so
:the order of divergence is O(log n).
:
:|>as log p(i).  (That observation also shows the divergence whenever
:|>p(i) is replaced by any increasing unbounded sequence.)
:| What is  sum(i=1 to N  of d(i)/p(i))???
:|
:|Summing the reciprocals of the primes shows that this sum is at least 
:|O(log log N).  Seth Breidbart posts a rough heuristic suggesting that
:|the true order should be O(log N).  This is in fact the case, but
:|PROVING it takes 5-6 pages of some delicate estimates of some Stieltje's
:|integrals [or equivalently some Euler-Maclauren summations]. 
:
:Some of us are lazy and prefer to do things the easy way :-).
:
 
Your observation that sum(d(i)/p(i)) is bounded above by twice the harmonic
series is valid. I posted a sketch of the derivation that sum(1 to N) d(i)/p(i)
= C log N +  O(log N/exp(sqrt(log N))).  In fact, one can show that the constant
C is exp(gamma) ~= 1.78.  Doing so takes some delicate estimates involving
Brun's sieve to show that the gap between primes is bounded by any fixed
constant (indeed any fractional power of log N) only on a very thin set.
[thanks to Claudia for help with this.]

--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
From: lgardner@mac.cc.macalstr.edu
Newsgroups: sci.math
Subject: Re: World's Smallest Prime Discovered
Date: 9 Apr 92 03:10:53 GMT
Article-I.D.: mac.1992Apr8.211053.2402
Organization: Macalester College

In article <1992Apr1.182930.20444@mintaka.lcs.mit.edu>, rivest@theory.lcs.mit.edu (Ron Rivest) writes:
> 
> The New York Times of 4/31/92, in describing the discovery of the new
> Mersenne prime, felt it necessary to describe the series of primes,
> which, (according to them) goes:
> 
>         1, 2, 3, 5, 7, 11, 13, ...
> 
> You will notice that they have slipped in what must be another discovery
> (by one of their writers?) of the world's smallest prime: 1.  I'm sure
> the mathematicians of the world must be tearing their hair out for having
> missed this one.


Big chuckle from a math major who isn't all that impressed with a computer
findinga new prime.  I know it's important, but, there are other mathematical
pursuits that could have more benefit.


:)


Lisa Gardner
Newsgroups: sci.math
From: bs@gauss.mitre.org (Robert D. Silverman)
Subject: Re: factoid about new prime
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Thu, 9 Apr 1992 02:09:06 GMT

In article <SMITH.92Apr7102215@gramian.harvard.edu> smith@gramian.harvard.edu (Steven Smith) writes:
:In article <1992Apr1.201324.25572@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
:
:> In article <3987@crl.LABS.TEK.COM> scott@ferrari.LABS.TEK.COM (Scott Huddleston) writes:
:> :It was mentioned on CNN that the new prime number discovered
:> :recently is four times bigger then the previous record.
:> :
:>
:> This is a (slight) exaggeration. 756839/216091 = 3.5024 times as big.
:
:Pardon me, but
:
:                756839
:               2       - 1                                          162781
:               -----------  ~=  2.33397019895780128135136676956 x 10
:                216091
:               2       - 1
:
:which is larger than four.  It appears that CNN was conservative in
:their report.
 
Pardon me, but in Complexity theory and in number theory the SIZE of a number
is its number of digits or bits.
 
SIZE(N) = O(log N).

And even your arithmetic is wrong. The above ratio is around 2^540748

I'd like to know where 162781 came from.
--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
Newsgroups: sci.math
From: pfferrei@descartes.uwaterloo.ca (Paul Ferreira)
Subject: Putnum Mathematics Competition
Organization: University of Waterloo
Date: Mon, 6 Apr 1992 05:58:05 GMT


Does anyone know the official results of the Putnum cmpetition?  What were the top 10 schools..  I have heard different variations and would like to know what the official results were.


-- 
|     Paulo Ferreira                            |   University of Waterloo  |  
| INTERNET: pfferrei@descartes.waterloo.edu     |      Waterloo, Canada     |
|           pfferrei@descartes.uwaterloo.ca     |                           |
-----------------------------------------------------------------------------
From: dix@clinet.fi (Risto Kaivola)
Newsgroups: sci.math
Subject: the domain of a rational function
Keywords: discontinuity, L'Hospital
Date: 13 Apr 92 06:35:39 GMT
Organization: Clinet, Helsinki, Finland

The following question was on this year's matriculation examination here
in Finland:
               ax + b
Let f:  f(x) = ------
               cx + d

The coefficients of this function satisfy c <> 0,  d <> 0,  and  ad = bc.
Prove that f is a constant function for all x in Dom(f).


The given solution (solutions to all questions are published in at least
one newspaper afterwards) was:

The function f is defined whenever x <> -d/c.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [emphasis mine]
When x satisfies the above,

       ax + b   c(ax + b)    acx + bc    acx + ad
f(x) = ------ = --------- =  --------- = ---------  =
       cx + d   c(cx + d)    c(cx + d)   c(cx + d)

       a   cx + d       a
       - * ------    =  -
       c   cx + d       c   .     Thus, x is a constant function.

[end of the solution given]


Now, the expression manipulation part is, of course, absolutely correct.
What I disagree with is that R \ {-d/c} is given as the domain of the
function f.  Here is how I solved it:

Given the expression f(x) = (ax + b) / (cx + d) we first find out
whether the function is discontinous when the expression that forms the
denominator of f takes the value 0.
This is only possible when x equals -d/c.  When this happens, f takes
on the value
              a(-d/c) + b
              -----------
                   0       .

But, this value is not of the form k/0 where k <> 0.  That is, also
the nominator takes the value 0 at this point.  This can be proven as
follows:
                       ad = bc (assumption)  || : (-c)  (<>0 by definition)

                         <=>

              a(-d/c) + b = 0,  which was to be shown.

This is what I believe is called an indefinite form, so we have to use
L'Hospitals rule (I think it is known by this name) to find out the
real value of f in this point.  Dx means 'derivation with respect to
x'.
        Dx(ax + b)   a
f(x) =  ---------- = -
        Dx(cx + d)   c
                          [this formula applies if and only if x = -d/c].

To satisfy the requirement that f must have the same value irrespective
of the value of x, we have to show that f(x) equals a/c whenever x<> -d/c.
That is,

             ax + b   a
             ------ = -
             cx + d   c   .

Starting from the assumption  ad  =  bc,  we get
                        acx + bc  =  acx + ad

                        c(ax + b) =  a(cx + d)  || : (c*(cx + d))  (<>0)

                          ax + b     a
                          ------  =  -
                          cx + d     c  .

So, I see no reason to say that the domain of f isn't the set of all real
numbers.  Comments?



-- 
Risto Kaivola (dix@clinet.fi) or (Risto.Kaivola@f123.n220.z2.FIDONET.ORG)
Newsgroups: sci.math,sci.physics,sci.comp-aided
From: semmett@gmuvax2.gmu.edu (Steve Emmett)
Subject: Re: Complex Hermitian eigenvalues 
Organization: George Mason University, Fairfax, Va.
Date: Mon, 6 Apr 1992 21:04:05 GMT

I meant to say diagonal and not trace in my previous post.  the case is

|  1  i |
| -i  1 |.   As a further example, CH returns the incorrect eigenvalues
for 
       | 1  1+i |
       | 1-i  1 |

Steve Emmett
From: sichase@csa1.lbl.gov (SCOTT I CHASE)
Newsgroups: sci.math
Subject: Re: On expected convergence
Date: 10 Apr 92 18:36:03 GMT
Organization: Lawrence Berkeley Laboratory - Berkeley, CA, USA
Distribution: na
Reply-To: sichase@csa1.lbl.gov
NNTP-Posting-Host: 128.3.254.196
News-Software: VAX/VMS VNEWS 1.3-4   

In article <1992Apr10.170059.24735@pasteur.Berkeley.EDU>, luzeaux@campanile.berkeley.edu (Dominique Luzeaux) writes...

>definition of convergence. But this is a personal point of view; I do
>not know what is done exactly in renormalization in modern physics, may be

As with the formal series solutions of many differential equations, the 
perturbation expansions of covariant perturbation theory, which are interpreted
in term of Feynman diagrams, are formally divergent.  They are asymptotic
series, and as such, you assume that the error is of order of the last term
you have evaluated, and then you stop if the terms start getting larger. 
So far as I know, nobody has ever calculated that far, and all the finite
approximations done have agreed with experiment, so most people don't worry too
much about it.

But to be perfectly honest, I'm not sure what youre question has to do with the
topic at hand.

-Scott
From: kasdan@cunixa.cc.columbia.edu (John Kasdan)
Newsgroups: sci.math
Subject: Re: Art Gallery Problem
Date: 16 Apr 92 02:01:26 GMT
Reply-To: kasdan@cunixa.cc.columbia.edu (John Kasdan)
Organization: Columbia University School of Law


In article <1363@hydra.bucknell.edu> cwalker@client10.bucknell.edu (carleton walker `93) writes:
>
>	I'm looking for information on the art gallery problem. I more
>interested in books than papers on the subject, but any help is
>appreciated.
>

And what _is_ the art gallery problem?


/JK (.sig closed for repairs)
From: leranth@miller.cs.uwm.edu ( TOM )
Newsgroups: sci.math
Subject: Sparse systems over GF(2) [Don Coppersmith's reply]
Date: 10 Apr 92 20:14:44 GMT
Article-I.D.: uwm.1992Apr10.201444.19941
Distribution: na
Organization: Cyclists
Originator: leranth@miller.cs.uwm.edu

[This is being POSTED for  Don Coppersmith
		       (due to apparent lack of write access)]

Don Coppersmith [in reply to
                    message<1992Apr9.181436.29316@uwm.edu>] :

    ...............................
   A good starting point is
   
       D. H. Wiedemann,
       "Solving sparse linear equations over finite fields,"
       IEEE Trans. Information Theory, (1986) IT-32, pp. 54-62.
   
   For an NxN system with an average of K nonzeroes per row,
   he solves the equations in time roughly (NxNxK).
   
   I have done work subsequent to that,
   
       Don Coppersmith,
       "Solving Linear Equations over GF(2) II: Block Wiedemann Algorithm,"
       IBM Research Report RC 17293, October 17 1991.
   
   which improves Wiedemann's timings by about a factor of 32.
   The idea is to do 32 matrix-vector operations at a time, by packing
   32 field elements into a word.  There's a lot of juggling that goes
   on behind the scenes to make that all work out, though.
   I have tested this on a matrix of size 65,000 by 65,000 over GF(2),
   with an average of 20 nonzeroes per row, and solved it in 65 minutes
   on an IBM 390.
   
   Don Coppersmith, IBM Research, copper at watson.ibm.com
     ............................................
Newsgroups: sci.math
From: blanc@caroli.csis.dit.csiro.au (Jacques Blanc-Talon)
Subject: Re: On knowing who's who (was: Fermat's Last Theorem)
Reply-To: blanc@caroli.csis.dit.csiro.au
Organization: CSIRO Division of Information Technology
Date: Mon, 13 Apr 1992 23:48:43 GMT

In article <29E4E96C.13732@ics.uci.edu>, jeff@madeleine.ics.uci.edu (Jeff Erickson) writes:
> a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
> 
> | Erdos is probably the best known active mathematician, possibly the
				     ^^^^^^
> | best known living mathematician.
> Jeff Erickson 

  And what do you think of Elie Cartan, Jean Dieudonne, Schwartz
(yes, I'm french) ?

Jacques Blanc-Talon
*
From: gindrup@math.okstate.edu (Eric `'d'kidd' G..)
Newsgroups: sci.math
Subject: Re: closed from solution to d.e.
Date: 14 Apr 92 15:57:25 GMT
Organization: Oklahoma State University, Math Department

In article <1992Apr9.041250.1959@nuscc.nus.sg> fbagohkh@nuscc.nus.sg (Mark Goh (Dr)) writes:
>Is there a closed form solution for this d.e.

At first glance, I don't see one...
:-)
-Eric Gindrup ! gindrup@hardy.math.okstate.edu
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: sci.math
Subject: Re: Tiling problem
Date: 13 Apr 92 12:18:34 GMT
Organization: Department of Computer Science, U of Copenhagen

dwilson@cvbnet.prime.com (David Wilson x5694 4-1600) writes:



>    Will the pentomino

>        **
>         **
>         *

>    tile the plane minus a quadrant?

Yes, even without mirroring.

We start by making four sections, which combine to 3/4 of the plane.

First a series of vertical strips continuing down and left:

	    A    A
	 BAAA BAAA
    ... BBBAaBBBAa
	BbaaaBbaaa
	bbbaAbbbaA
	bBAAAbBAAA
    ... BBBAaBBBAa
	BbaaaBbaaa
	bbbaAbbbaA
	bBAAAbBAAA
    ... BBBAaBBBAa
	BbaaaBbaaa
	bbbaAbbbaA
	.
	.
	.

Then the same, rotated 90 degrees:

		.
		.
		.
	 AAaaAAaaAAaaAAaa
	  AAaaAAaaAAaaAAaa ...
	 BAbaBAbaBAbaBAbaB
	BBbbBBbbBBbbBBbbBB
	 BBbbBBbbBBbbBBbbBB
	 AAaaAAaaAAaaAAaa
	  AAaaAAaaAAaaAAaa ...
	 BAbaBAbaBAbaBAbaB
	BBbbBBbbBBbbBBbbBB
	 BBbbBBbbBBbbBBbbBB

Then a triangular section extending left and diagonally up-left:

	.
	 .
	  .
	   A
    . . . AA
	   AAa
	  BBaa
	  aBBaaA
	 aaBbbAA
    . . . aaAbbAAa
	 BBAAbBBaa
	  BBAAaBBaaA
	  BbbaaBbbAA
	    bbaa bbAA
	    b    b

And a similar one extending up and diagonally up-left:

	.	.
	 .	.
	  .	. AA
	      AA  BAA
	  AA  BAABBA
	   AABBAaaBBb
	   AaaBBbaabb
	     aabbaAAbb
	     aAAbbBAA
	       AABBA
	       AaaBBb
		 aabb
		 aAAbb
		   AA
		   A

We then combine these. I have numbered the sections in the order they
were introduced.

	     .
     .       .
    . .	     . 44
     . .   44  444	.
      .44  444444	.
       34444444444	.
 ...  3344444444442222222222222222
       3334444444442222222222222222 ...
      33334444444422222222222222222
      33333344444222222222222222222
     3333333444444222222222222222222
  ... 3333333344442222222222222222
     333333333444442222222222222222 ...
      33333333334422222222222222222
      33333333334222222222222222222
	3333133331222222222222222222
	3111131111
    ... 1111111111
	1111111111
	1111111111
	1111111111
    ... 1111111111
	1111111111
	1111111111
	1111111111
    ... 1111111111
	1111111111
	1111111111
	.
	.
	.

It is clear to see that the connection at the edges is periodic, so it
will fit until the limit.

	Torben Mogensen (torbenm@diku.dk)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
From: harper@kauri.vuw.ac.nz (John Harper)
Date: Tue, 7 Apr 1992 05:19:02 GMT
Organization: Victoria University of Wellington, New Zealand

 >Pardon my ignorance. Who is this Erdos dude?
Several people have been abusing one another instead of answering the
question. Paul Erdos is a very famous Hungarian mathematician (aged around
80 I'd guess) who occasionally offers cash prizes of a few hundred dollars
for solutions to problems that he hasn't solved himself and that interest him.
The problems I have seen are HARD and are on combinatorics and number theory.

John Harper Mathematics Dept. Victoria University Wellington New Zealand
Xref: sparky sci.math:6061 sci.physics:6774
From: davis@pacific.mps.ohio-state.edu ("John E. Davis")
Newsgroups: sci.math,sci.physics
Subject: shape functions
Date: 14 Apr 92 23:53:41 GMT
Reply-To: davis@pacific.mps.ohio-state.edu  (John E. Davis)
Followup-To: sci.math
Organization: "Dept. of Physics, The Ohio State University"

Hi,

   I am looking for a way of quantizing shapes.  For example, imagine a
probability distribution in one dimension p(x), 0 <= x <= 1.  Then, what are
some simple parameters that classify various p(x) functions? Obviously, the
moments

                   m_k :== integral from 0 to 1 of x^k p(x)

form such a set.  Are there other sets of parameters that are more suited to
numerical work?  I am sure that this question has been asked before.

   A related question:  Suppose there is a set of parameters that characterize
the shape, say {m_k}.  Then can one define a canonical form such that
f(x;{m_k}) forms a reasonable approximation to p(x)?  What is the canonical
form for the monments defined above?  (e.g. m_0, m_1, m_2 may be written as a
gaussian, throw in m_3 and you get a skewed one, etc...)

   Note that I am not asking for an existence proof, i.e, given p(x) of
compact support, etc... and its moments {m_k} (or other shape parameters),
does {m_k} ==> a unique p(x)?  Let's just assume it does.  I already have a
list of references where some Russian mathematicians have worked on this.

Thanks,

--
     _____________
#___/John E. Davis\_________________________________________________________#
#                                                                           #
#   /\/\/\   internet: davis@amy.tch.harvard.edu                            #
#  /\/\/\/\    bitnet: davis@ohstpy                                         #
#  \/\/\/\/    office: 617-735-6746                                         #
#   \/\/\/                                                                  #
#___________________________________________________________________________#
#                                                                           #
Newsgroups: sci.math
From: fbagohkh@nuscc.nus.sg (Mark Goh (Dr))
Subject: closed from solution to d.e.
Organization: National University of Singapore
Date: Thu, 9 Apr 1992 04:12:50 GMT

Is there a closed form solution for this d.e.
Newsgroups: sci.math
From: bs@gauss.mitre.org (Robert D. Silverman)
Subject: Re: Fermat's Last Theorem
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Thu, 9 Apr 1992 02:19:05 GMT

In article <3660@tuegate.tue.nl> wsadjw@urc.tue.nl writes:
:In article <STROMME.92Apr6130830@jeeves.math.utah.edu> stromme@math.utah.edu (Stein Stromme) writes:
:>Why not just answer the question? After all, the poster says "Pardon
:>my ignorance", and then he does not deserve to be run over like that.
:>
:>Stein 
:
:Well, the poster is the [expletive deleted] Derek Abbott who has 
:developed a habit of asking silly questions in a tone as if he is
:a kind of know-all, and that about a wide range of subjects, among
:which quantum mechanicss, free will, evolution, entropy, and 
:various instances in sci.math.  
:Let's say, some people are like that.
:The respondee, Bob Silverman is known for the somewhat caustic way
:of expressing himself (with a dash of jusitified superciliousness),
:but in this case his response showed remarkable restraint. Of course,
:ignoring D.A. altogether would have been even better.
:
:JWN


I am just surprised that someone with a presumed interest in math would
not know one of the most outstanding and prolific mathematicians of
the 20th century. If I were Erdos, I'd be insulted.

Paul Erdos has been one of the movers and shakers of the 20th century.
The fact that he is relatively unknown is just another example of the
piss-poor attitude the American public has towards math and mathematicians.

I'm sure the original poster knows who Michael Jordan is, whose only
claim to fame is his ability to jump high and stuff a ball through a hoop.
The contributions of Paul Erdos will have long lasting value to the
knowledge base of the human race and his work has provided stimulous for
many others. What long lasting value Michael Jordan's contributions have
would be a matter of serious debate.

And I did answer the question in a half joking manner. The way the question
was posed: "who is this Erdos dude" invites such because calling someone a
'dude' without knowing that person is in itself somewhat flip and insulting.
--
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"
Newsgroups: sci.math
From: dbeattie@cse.unl.edu (Derek Beattie)
Subject: interal
Organization: University of Nebraska - Lincoln
Distribution: usa
Date: Thu, 9 Apr 1992 06:30:02 GMT

I need some help.  Can someone integrate 1/(x^4+1)dx.  You have
to factor the bottom and use partial fractions but I can
get no further. You can change x^4+1 into (x^2+1)^2-2x^2 and then
that is a difference of squares and you can factor that then use
partial fractions and then I am lost....

Newsgroups: sci.math