HDTenEightyP/NetNews huggingface dataset:
From: regeorge@magnus.acs.ohio-state.edu (Robert E George)
Subject: Re: Probably you know the probability ;-)
Organization: The Ohio State University
Date: Thu, 15 Oct 1992 00:13:53 GMT

In article <1992Oct14.193802.25129@noose.ecn.purdue.edu> kavuri@lips1.ecn.purdu
e.edu (Surya N Kavuri ) writes:
>  I am given a "rod" of fixed length l. I took a hammer and
>  broke it into three parts.
>
>   (a) what is the probability that they form a triangle ?

     Consider this as an order statistics problem, see H. David's
textbook "Order Statistics". Let Y1 = length of shortest piece, Y2 =
length of "middle" piece, Y3 = length of longest piece (these are the so-called
order statistics). Then integrate to find Prob( Y1 + Y2 > Y3).

     Of course, this will be complicated by the fact that the observations
themselves are *not* independent --

>   (b) what is the probability that they form an isoceles triangle ?
>
     Zero, since the lengths of the broken pieces is a continuous r.v. :-)
> 
>   Surya Kavuri
>
>

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: lusardi@sybil.cs.Buffalo.EDU (Christopher Lusardi)
Newsgroups: sci.math
Subject: Recursion and Rice's Theorems
Date: 7 Oct 92 20:30:28 GMT
Organization: State University of New York at Buffalo/Comp Sci
Originator: lusardi@sybil.cs.Buffalo.EDU

I'm not really good at proof techniques, so would someone send me examples
of showing a language is recursive or r.e. (recursively enumerable) with 
the Recursion theorem, with Rice's theorem, and with a diagonalization 
argument? These were used in a theory of computation course of mine about 
4 years ago, so I'm looking for something easy to read.

I'm really not looking for proofs, but what I'm looking for is a good 
explanation of these. I have 3 or 4 books on the subject, but I would like
an intuitive explanation of doing this.

							Thank you,
-- 
Christopher M. Lusardi     University At Buffalo  lusardi@sybil.cs.buffalo.edu
Newsgroups: sci.math,sci.physics
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: How do you draw a straight line?
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Wed, 7 Oct 1992 18:56:08 GMT

pcl@oxford.ac.uk (Paul Leyland) writes:

>I've known people who *define* "straight line" to be equivalent to "geodesic",
>on the grounds that "a straight line is the "shortest distance between two
>points".

How do these people define "distance" ?
I always wonder if mathematicians are the only people who might
understand that there's more than one definition of "distance".

>Paul Leyland <pcl@oxford.ac.uk>          | Hanging on in quiet desperation is
>Oxford University Computing Service      |     the English way.
>13 Banbury Road, Oxford, OX2 6NN, UK     | The time is come, the song is over.
>Tel: +44-865-273200  Fax: +44-865-273275 | Thought I'd something more to say.

I always wondered where O.W. got the term "Bunbury" from ...

Sven :)
Xref: sparky sci.math:13338 sci.physics:16667
From: Richard.Mathews@West.Sun.COM (Richard M. Mathews)
Newsgroups: sci.math,sci.physics
Subject: Re: "The Universe of MOTION" (crackpot index)
Date: 16 Oct 92 19:48:24 GMT
Organization: Sunsoft Inc., Los Angeles, CA.
NNTP-Posting-Host: astro
Originator: richard@astro.West.Sun.COM

hagerman@ece.cmu.edu (John Hagerman) writes:
>> 113.755 John Hagerman
>> 133.333 Scott Brigham, corp
>> 182.353 <RVESTERM@vma.cc.nd.edu>
>> 319.444 Brad Wallet
>>
>> The prize for best crackpot spoof goes to Brad Wallet.  Hagerman had a
>> chance at this prize, but blew it by submitting a couple serious articles.
>Hmm, as far as I can tell, Brad Wallet has posted only once in the
>last few weeks, and that post fully included my post, but the left
>Character was not '>'.  Computerization simplifies plagiarism! :-)

You're right.  Brad is disqualified.  As you keep up these serious posts,
I have to disqualify you too; so the prize now goes to RVESTERM@vma for
his posting illustrating the proper use of repeated exclamation points.

Now if you had properly claimed that PHYSICISTS have ignored your PROVOCATIVE PRINCIPLES and that my ARROW-neous award for Brad's COMMUNICATION was a CONSPIRACY by all SCIENTISTS to discredit your SUPER-GENIUS inspired by DR. PENROSE and maintain the STATUS QUO ACCORDING to DR. GALILEO, DR. NEWTON, DR. EINSTEIN, and DR. HAWKING, then I might have considered giving you the award!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!-)

Well, there goes my rating.  (exactly 1000 for this article per my program)

Richard M. Mathews	Richard.Mathews@West.Sun.COM		Laisve!
Newsgroups: sci.math
From: news@massey.ac.nz (USENET News System)
Subject: Re: Probably you know the probability ;-)
Organization: Massey University
Date: Fri, 16 Oct 92 01:30:13 GMT

In article <1992Oct14.220009.27453@noose.ecn.purdue.edu>, kavuri@lips.ecn.purdue.edu (Surya N Kavuri ) writes:
> 
> In article <1992Oct14.193802.25129@noose.ecn.purdue.edu>, kavuri@lips1.ecn.purdue.edu (Surya N Kavuri ) writes:
> >   I am given a "rod" of fixed length l. I took a hammer and
> >   broke it into three parts.
> > 
> >    (a) what is the probability that they form a triangle ?
> >    (b) what is the probability that they form an isoceles triangle ?

>      (c) what is the probability that they form an acute-angled traingle ?
> 
First, assume break points independent and uniform on (0, 1).

(a) 1/4
(b) 0
(c) ln(8) - 2

(Sorry, I can't fit the proof in the margin).

Terry Moore
Xref: sparky sci.math:13343 sci.physics:16669 sci.astro:10882 sci.bio:3611 sci.chem:3955 soc.men:17827 soc.women:17821
From: 890091b@ace.acadiau.ca (HEIDI JILL BURKHOLDER)
Newsgroups: sci.math,sci.physics,sci.astro,sci.bio,sci.chem,soc.men,soc.women
Subject: Re: Is Math Hard?
Date: 16 Oct 92 18:40:52 GMT
Organization: Acadia University


	Put it this way. In elementary school I looked forward to the end of 
summer vacation because I actually missed doing my math homework! Mind you, 
my 2000 level calculus course in university was someething Barbie could 
say "math class is tough" about, but hey, life goes on. We need math, so 
let's like it - expends a lot less energy than hating it! (i.e. It's 
thermodynamically favorable to like math!) 
Newsgroups: sci.math
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: FAREY
Organization: DSG, Stanford University, CA 94305, USA
Date: Fri, 16 Oct 92 21:19:40 GMT

In article <1992Oct16.170803.14345@nntp.uoregon.edu> scavo@cie.uoregon.edu (Tom Scavo) writes:
>In article <168827A63.DUENTSCH@dosuni1.rz.uni-osnabrueck.de> DUENTSCH@dosuni1.rz.uni-osnabrueck.de writes:
>>In article <98687@bu.edu> rehman@math.bu.edu (Naved Rehman) writes:

>Ian Stewart also mentions the Farey sequence in his _Does God Play Dice?_
>(Basil-Blackwell, 1989?).

On the other hand, I would not trust Ian Stewart about such things.
In one of his books he states that the Riemann hypothesis is
equivalent to

  pi(x) = x/log(x) + O(sqrt(x) log^2(x))

so he is clearly writing about something that he has no clue about.
In fact, de la Valle Poussin showed that 

  pi(x) = li(x) + O(x exp(-c sqrt(log(x))))

where 

  li(x) = int_2^x 1/log(x) dx 

and integration by parts gives

  li(x) = x/log(x) +  x/log^2(x) + ...  
           
          + (k-1)! x/log^k(x) + O(x/log^{k+1}(x))

so 

  pi(x) = x/log(x) + x/log^2(x) + O(x/log^3 (x)).

The point is that a popular account of something as well understood
and easy to look up as this should be extremely accurate.

Moreover, I find the way he writes his column in Scientific American
to be totally insulting. If the material were at all interesting then
there would be reason to dress it up as some fable.

-ilan
Newsgroups: sci.math
From: nissim@mary.fordham.edu (Leonard J. Nissim)
Subject: Halmos handshake problem (was: Re: The problem of Irish men)
Organization: Fordham University
News-Software: VAX/VMS VNEWS 1.41    
Distribution: usa
Date: Thu, 15 Oct 1992 17:45:00 GMT
X-Posted-From: mary.fordham.edu
NNTP-Posting-Host: sol.ctr.columbia.edu

In article <israel.719166972@unixg.ubc.ca>, israel@unixg.ubc.ca 
(Robert B. Israel) writes...
>In <1992Oct15.111727.11840@cs.tcd.ie> dherity@cs.tcd.ie (Herity D.) writes:
> 
>>In article <1992Oct15.063757.4956@noose.ecn.purdue.edu> 
kavuri@lips.ecn.purdue.edu (Surya N Kavuri ) writes:
>>>
>>>  Prove that there are at least two Irish men who shook 
>>>  hands the same number of times with other Irish men.
>>>
>>I can construct a case where that assertion is false. I'm an Irish man, so I'm
>>fortunate to have the opportunity for experimental verification :-).
> 
>>Let there be four Irish men, none of whom have shaken hands.
>>The first man ignores the second and shakes hands with the third. He then
>>shakes hands with the fourth twice. Then :
>>The first man has shaken hands three times.
>>The second has shaken hands zero times.
>>The third has shaken hands one time.
>>The fourth has shaken hands two times.
> 
>Perhaps what Surya meant was:
>  there are at least two Irish men who shook hands with the same
>  number of other Irish men.
>(i.e. multiple shakes with the same man don't count).  Actually you have
>to add another clarification:  "there are two Irish men" seems to imply
>that they are currently alive, while "shook" (in the past tense) would
>appear to include the possibility of shaking hands with someone who
>subsequently died.  To make it work, we have to either include both
>the living and the dead in both places, or restrict to the living in
>both places.
> 
>Once you do this, it's an easy application of the Pigeonhole Principle.
>-- 
>Robert Israel                            israel@math.ubc.ca

When you're done, try this one, known as the Halmos Handshake problem:

Professor Halmos and his spouse went to a party, at which there were
four other couples.  At the party:  Nobody shook hands with himself or
herself, and nobody shook hands with her or his own spouse.  All other
handshakes were mutual (A shook hands with B iff B shook hands with A).  
Halmos asked each other person at the party's end "How many different people  
did you shake hands with at this party?"  He received nine *different* an-
swers.  (For nitpickers, everyone had a perfect memory and told the truth.)  
With how many people did Halmos' spouse shake hands?
-------------------------------------------------------------------------------
Leonard J. Nissim  (nissim@mary.fordham.edu)   
Disclaimer:  "I speak only for myself."
-------------------------------------------------------------------------------
Newsgroups: sci.math
From: moeller@gwdgv1.gwdg.de
Subject: puzzling squares
Reply-To: moeller@gwdgv1.gwdg.de
Organization: GWDG Goettingen, F.R.Germany
Date: Thu, 8 Oct 1992 00:29:37 GMT

Have you ever seen a (rational) square number which,
when written in decimal notation, has the same sequence of digits
left and right of the decimal point?

To my surprise, I noticed that such numbers do exist, but are *large*.
The smallest one I know is the square of

	((10^136 + 1)/17) * 6 * 10^(-68)

To prove that this is indeed the smallest such number,
one had to show that  (10^(2*k) + 1)  is square-free, for (2*k) < 136.
Does the current state of the "art of factorization" allow for this check?

Wolfgang J. Moeller, GWDG, D-3400 Goettingen, F.R.Germany | Disclaimer ...
PSI%(0262)45050352008::MOELLER      Phone: +49 551 201516 | No claim intended!
Internet: moeller@gwdgv1.dnet.gwdg.de   | This space intentionally left blank.
From: cxm7@po.CWRU.Edu (Colin Mclarty)
Newsgroups: sci.math
Subject: Re: Mattel on women
Date: 7 Oct 1992 14:43:32 GMT
Organization: Case Western Reserve University, Cleveland, OH (USA)
Reply-To: cxm7@po.CWRU.Edu (Colin Mclarty)
NNTP-Posting-Host: slc5.ins.cwru.edu


In a previous article, dmi@peregrine.peregrine.com (Dean Inada) says:

>
>Censor?  Hmm, I think I'll have to agree...
>We shouldn't let Mattel 'censor' the counter-view that 
>hard subjects like math can be fulfilling and important
>by making Barbie deliver their 'Math Phobic' view exclusively.
>
	Now this is really odd.  Mattel says "math is tough",
and you see them as denying that "hard subjects like math can
be fulfilling and important"?
 
	You AGREE with what they said--you agree math is tough.  And
where did you hear them say "don't try tough things"?  I didn't hear
it at all.

	Another poster took the Mattell blurb as saying "math is
uncool".  Which I also find odd.  In fact, "tough" was slang for
"cool" a lot more recently than "cool" was.

	I did help a girl get through third through fifth grade
math.  She told me that addition facts are hard when you have to
carry.  Now, I could have said "no they aren't" and then she could
could have chosen among three possiblities: she was stupid, I was
lying, I didn't know what I was talking about.  What I did say was
"yeah, but you can do hard things" and by the time we got to 
division flash cards she thought this stuff was a blast--she still
didn't want to stop playing to do it, but when homework time came
she'd start with math.

	So I am honestly not sure.  Maybe this blurb is a bad
thing.  But you sure won't convince me by claiming that tough
things can't be cool.  Nor will I believe that girls won't try 
anything they hear is tough until you give me some specific
evidence.

Colin
From: kdq@quest.UUCP (Kevin D. Quitt)
Newsgroups: sci.math
Subject: Help needed
Date: Thu, 08 Oct 92 23:42:41 PDT
Reply-To: srhqla!quest!kdq
Organization: Job quest  (805) 251-8210,  So Cal: (800) 400-8210

Below is the standard equation for calculating monthly payments (m)
given the periodic interest rate (i), the principal (p), and the number
of payments (n).  I need to rearrange the equation to solve for the
periodic interest rate, given the other variables, but my math is sadly
no longer up to it.

If someone would be so kind as to solve for (i), or look up the
equation, or whatever, and respond by email, I would be eternally
grateful (well, a long time, anyway).  I ask for email since this site
does not receive sci.math.  Thanks in advance.

             i
m = p * ------------
                   -n
        1 - (i + 1)


 _
Kevin D. Quitt                                 Job Quest (805) 251-8210
srhqla!quest!kdq                               Learn How to Get a Job in SoCal

                96.37% of all statistics are made up.
Newsgroups: sci.math
From: gsmith@euterpe.uucp (Gene W. Smith)
Subject: Re: Area of five sided polygon needed
Organization: IWR, University of Heidelberg, Germany
Date: Tue, 13 Oct 92 13:42:57 GMT

In article <1992Oct13.040617.16321@husc3.harvard.edu>
kubo@birkhoff.harvard.edu (Tal Kubo) writes:
>In article <1992Oct12.200125.826@altair.selu.edu>
fcs$1224@altair.selu.edu writes:

>>Given a five sided polygon, is there a formula, in terms
>>of its sides, for the area?

>No, the sides don't uniquely determine the area, but...

>1. There is a formula (due to Moebius?) for the area, S, in terms of the
>   areas of the triangles A,B,C,D,E cut off by the diagonals:
>
>        S^2 - (A+B+C+D+E)S + (AB+BC+CD+DE+EA) = 0

A curious fact about the above is that it us a resolvent. If a,b,c,d,e
are the ordered roots of a polynomial over F with Galois group dihedral
or cyclic, then the above polynomial has coefficients in F.  If the
Galois group is cyclic, then the above polynomial factors over F. 
Hence an irreducible degree five polynomial can be tested for having
Galois group D5 or C5 by seeing if the  polynomial above has
coefficients in the field for some ordering of the roots.  

Supposing the roots are all real and positive and come from the areas
of of triangles, if the total area s is F-rational, but the areas of
the five triangles are of degree 5 over F, then the group is cyclic.

I presume there is a reason for this.  Maybe looking at what happens when
you try to find actual triangles which work, given the areas and the
fact that they fit together into a pentagon would help explain it.

>3. I've heard of Brahmagupta-type polynomial equations for the areas of
>   *cyclic* N-gons, in terms of their sides.

These work like Heron's formula.
-- 
     Gene Ward Smith/Brahms Gang/IWR/Ruprecht-Karls University 
               gsmith@kalliope.iwr.uni-heidelberg.de
Xref: sparky sci.math:13226 comp.lang.c:15001 comp.lang.pascal:6003 comp.os.os2.programmer:5672
From: kavuri@lips.ecn.purdue.edu (Surya N Kavuri )
Newsgroups: sci.math
Subject: Correction to the Irish men problem
Date: 15 Oct 92 18:15:59 GMT
Organization: Purdue University Engineering Computer Network


  
  Ok, here is the correct version


   Prove that there are atleast two Irish men who shook 
   hands with the same number of Irish men.


  Surya Kavuri
From: zhu@graphack.asd.sgi.com (Benjamin Zhu)
Newsgroups: sci.math
Subject: Simple convex polygon
Date: 16 Oct 1992 00:22:11 GMT
Organization: Silicon Graphics, Inc.
Distribution: world
NNTP-Posting-Host: graphack.asd.sgi.com


Hi!

I have a conjecture that I have not been able to either prove or
disprove so far. I want to utilize the net wisdom here.

Conjecture:
Given an arbitrary (might be simple, might be complex, i.e., self-
intersecting) polygon, p[0], p[1], p[2], ..., p[n-1], no two
vertices coincide, no three vertices are co-linear. Suppose the
first three vertices p[0], p[1], and p[2] go counter-clockwise,
i.e., we have a left turn at p[1]. Define the interior angle at
a vertex p[i] to be the sweep angle from vector p[i]p[(i-1)%n]
to vector p[i][(i+1)%n] clockwise. 

If every interior angle is less than \phi, and the summation of
all interior angles is (n-2)*\phi, the polygon in question is
convex and simple (non self-intersecting).

If this is indeed true, please either give me a pointer or a
reasonably-detailed sketch of proof. If not, can you give me a
counter-example? Please respond by emails.

Thanks a lot,

Ben

--
Benjamin Zhu             ;;; Priest of ucode and self-appointed bug terminator
Silicon Graphics, Inc.   ;;; ``Just call my name, I will be there.''
zhu@graphack.asd.sgi.com ;;; ``Projective geometry is all geometry.'' - Cayley
(415) 390-1187           ;;; ``Learn forever, young forever.'' - Anonymous
Xref: sparky sci.math:12915 nyc.seminars:78
From: bumby@no.rutgers.edu (Richard Bumby)
Newsgroups: sci.math,nyc.seminars
Subject: New York Number Theory Seminar
Date: 8 Oct 92 20:36:43 GMT
Followup-To: sci.math
Organization: Rutgers Univ., New Brunswick, N.J.

Here is the first announcement of the seminar.

                       New York Number Theory Seminar
                             33 West 42nd Street
                          New York, New York 10036

                     Thursdays at 3:00 p.m. in Room 732

                                Fall, 1992


October 15:  Gerald Myerson, Macquarie University, Australia
             "The opposite of uniform distribution"

October 29:  Ilan Vardi, Rutgers University, New Brunswick
             to be announced


Posted for the organizers by R. T. Bumby
-- 
R. T. Bumby **  Rutgers Math ||   Amer. Math. Monthly Problems Editor
bumby@math.rutgers.edu       || P.O. Box 10971 New Brunswick, NJ08906-0971
bumby@dimacs.rutgers.edu     || Phone: [USA] 908 932 0277 * FAX 908 932 5530
Newsgroups: sci.math
From: wjh@cbnewsl.cb.att.com (william.j.hery)
Subject: Re: 1, 2, 4, 8, 13...where have I seen it before?
Organization: AT&T
Distribution: na
Date: Thu, 8 Oct 1992 13:18:26 GMT

In an earlier posting, I wrote:
> In computing something in combinatorics, a sequence that started
> with 1, 2, 4, 8, 13 (not the expected 16) came up, and I'm sure I've
> seen it before, possibly in some kind of trick problem that is set
> up to make you think the next term is 16. Does anyone recognize something
> simple that starts like this? BTW, my sequence continues with 21, 31, 45(?)
> 
> Another (simpler) source might give some insight into the problem I'm looking
> at.

In response, chrisman@herb-ox.berkeley.edu (chrisman) writes:
>I don't know if this sequence has anything to do with
>yours, but here it is (an example of a sequence whose
>first few terms are misleading):
>
>Let A(n) be the maximum number of regions into which
>a circle can be divided by drawing segments between
>n points on the circle.                      
>
>The first few terms are (starting with n=1)
>
>      1, 2, 4, 8, 16, 31 (!).

That is the problem I had seen but couldn't think of that starts of with a 
similar series--thanks for reminding me of it. Thanks also to the netter's
who mailed the same problem to me.

Thanks also to the several netters who pointed me to N J A Sloanes book
on integer sequences, which included one that matched even more of my terms,
(but not all of them).

As for the answer to my problem, well there isn't one...I was actually trying
to construct a sequence of sets S(k) with k positive integers and some other
special properties (more involved is relevant at the moment), and the
sets I came up with were {1,2}, {1,2,4}, {1,2,4,8}, {1,2,4,8,13}... and
I was hoping that I could just add terms from a sequence to build the sets.
On closer examination, it turns out that for most k the sets are not even
unique, and I am not even sure if you can always an S(k) set to an S(k+1) set.
But I was able to construct S(10) (which was my immediate need) and have
found other connections to help me construct other sets if needed.

Bill Hery
Newsgroups: sci.math
From: ramos@engr.latech.edu (Alex Ramos)
Subject: Quick help needed: Matrix functions in C
Reply-To: ramos@engr.latech.edu
Organization: Louisiana Tech University
Date: Tue, 13 Oct 1992 16:22:50 GMT

Does anybody have, or know where I can find matrix libraries in C?  (I
need something that'll do Inverse and Multiplication of 100x100
matrices) -

Thanks in advance!
--
-------------------------------------------------------------------------------
                      |       Alex Ramos       | fala-se portugues, tche! 
ramos@engr.latech.edu | Electrical Engineering | se habla espanol         
                      |      LA Tech Univ.     | english spoken here
-------------------------------------------------------------------------------
Newsgroups: sci.math
Subject: Re: Question on real numbers
From: news@massey.ac.nz (USENET News System)
Date: Fri, 9 Oct 92 02:43:18 GMT
Organization: Massey University

In article <1992Oct8.233240.11257@wdl.loral.com>, mab@wdl39.wdl.loral.com (Mark A Biggar) writes:
> 
> In article <1992Oct8.211117.19295@rz.uni-karlsruhe.de> vhansen@ipfs.bau-verm.uni-karlsruhe.de (Wolfgang von Hansen) writes:
> >is it possible to express any real number x with the following term
> >
> >My idea is to write any real number as a pair of rational numbers
> 
> No you can't do this.  The set of all pairs of rationals (a,b) in only
> countably infinite, therefore the set {a+rb} is countable, But the set
> R in uncountably infinite, therefore R != {a+rb}.  As a counter example
> 
> 
This is true, but it is possible to find an analogy of sorts.
The set of ordered pairs of reals forms a 2 dimensional vector space
over the reals. The reals can be thought of as an infinite dimensional
vector space over the rationals.

See Angus E Taylor, "Introduction to Functional Analysis"
Wiley (1958), p44, where he defines the concept of a Hamel basis.

Unfortunately, you need the axiom of choice to prove the existence
of such a basis.

Terry Moore
Newsgroups: sci.math,sci.physics