HDTenEightyP/NetNews huggingface dataset:
From: ec120wee@sdcc4.ucsd.edu (Anthony Minkoff)
Newsgroups: sci.math
Subject: nonexistance of counterexample -> existance of proof?
Date: 9 Apr 92 03:50:11 GMT
Organization: University of California, San Diego

Forgive a little question from a poor, ignorant undergraduate who's
not even a math major:

An acquaintance and I were recently discussing Goldbach's conjecture
(that every even number greater than 2 is the sum of two primes).
He made the statement "If no counterexample exists, then a proof
exists."  Of course, I can't prove him wrong (I accused him of
making a "safe" statement, since clearly he could not be proven
wrong in the case of the Goldback conjecture), but I want to know,
in general, is his statement correct?  (I was hesistant to apply
Godel's theorem, since neither of us really understands it).

Anyway, the story continues, but it gets kinda silly from here.  I'm
mostly interested in an answer to this first part, but I'll share
the rest with you, and you may have some interesting insights...

Anyway, the story continues...  I then tried to present an example
of a theorem with no proof, and no counterexamples.  First, I made
the claim that there must be real numbers with no finite expression.
(A finite expression does not necessarily imply a finite expansion.
For example, sqrt(2) is a finite expression.)  I offered the
following proof:

Any finite expression can be encoded as a finite sequence of bits,
according to the following scheme:  Begin the sequence with an ASCII
text description of how to decode the part of the sequence following
the description.  (Of course, the entire description may be the
ASCII text.)

Since a one-to-one correspondence can be established between finite
sequences of bits and integers, but not between reals and integers,
there are reals with no finite expressions.

Then I proposed the theorem, "All reals *do* have some finite
expression."  Clearly, no proof exists, since the statement is not
true.  Then we got into an argument over whether counterexamples
exist.  Certainly he could not give me one, and nobody ever could.
So it became a question of what is really meant by the statement, "a
counterexample exists."

Well, I hope you haven't found this too boring...  What do you
think?

Should this be posted to a philosophy newsgroup?

Tony Minkoff
From: dseal@armltd.co.uk (David Seal)
Newsgroups: sci.math
Subject: Re: An elementary, "naturally occurring" problem
Date: 10 Apr 92 03:01:35 GMT
Distribution: sci
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <3767@wet.UUCP> sam@wet.UUCP (Sam Needham) writes:

>In article <1992Apr1.201733.25777@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>-In article <1992Apr1.183035.14916@Princeton.EDU> carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:
>-:Prove or disprove that there exists a bijection f:[0,1] -> [0,1] that is
>-:not continuous at any point. 
>- 
>-Doesn't the following function do it?
>-f(x) = 1-x  if x is irrational.
>-f(x) = x if x is rational.
>-Bob Silverman
>
>Continuous at x = 0.5.
>
>The proof that there is no such function uses, I think, the Baire Category 
>Theorem. Suppose f: [0,1] to [0,1] is discontinuous everywhere and f(a) =
>0.5; then the family
>
>	[0,1] minus inverse-image-of-f( 0.5 - 1/n, 0.5 + 1/n) (n>=3)
>
>has union [0,1] but none of them contains an open neighbourhood of a. 

What about the function:

  f(x) = x        if x is rational
         x + 1/2  if x is irrational and x < 1/2
         x - 1/2  if x is irrational and x > 1/2

This is a bijection [0,1] -> [0,1] (it is its own inverse) and is
discontinuous everywhere. (Proof: any open interval contains an irrational
x1. Having chosen x1, the interval also contains rationals x0 < x1 and x2 >
x1. If x1 < 1/2, this gives us f(x1) - f(x0) = 1/2 + x1 - x0 > 1/2; if x1 >
1/2, it gives us f(x2) - f(x1) = x2 - x1 + 1/2 > 1/2. So any open interval
contains two points whose function values differ by more than 1/2.)

As for how the above proof breaks down, it seems to me that
inverse-image-of-f(0.5 - 1/n,0.5 + 1/n) never contains 1/2. So the union of
the stated family is not [0,1] - it is in fact [0,1] - {0.5}.

David Seal
dseal@armltd.co.uk

All opinions are mine only...
Newsgroups: sci.math
From: price@linus.serum.kodak.com (Kevin Price (x37128))
Subject: Factoring an algebraic expression ---  HELP?
Organization: Clinical Products Division, Eastman Kodak Company
Distribution: usa
Date: Thu, 9 Apr 1992 19:02:34 GMT


Hello,
	I've been trying factor down an expression and have gotten stuck.
The symmetry of the remaining terms in the numerator leads me to suspect that 
there just may be more factoring possible, but I can't seem to get it.  If anyone
would like to help me out, please send e-mail to the address at the bottom
of the post.

NOTATION: '^' is the exponent operator
	  'a', 'b', 'c', 'd', are the variables


	ab^2(a+b)[b(c-d)+(c^2+d^2)] + c^2d(c+d)[c(a-b)+(a^2+b^2)]
	---------------------------------------------------------
			abcd(a+b)(b+c)^2(c+d)




-- 
-Kevin Price		price@serum.kodak.com             
                                                          
 (please use the address above ^   The mailer may call me 
  nobody@kodak.com, but IMHO, I am not.)                  
Newsgroups: sci.math,comp.theory
From: bs@faron.mitre.org (Robert D. Silverman)
Subject: Re: factorial function
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Distribution: usa
Date: Sat, 11 Apr 1992 00:58:05 GMT

In article <1992Apr10.201515.14401@ringer.cs.utsa.edu> djimenez@ringer.cs.utsa.edu (Daniel Jimenez) writes:
:Are there any good (e.g., O(log n)) algorithms for finding n! ?
:I am aware of Stirling's approximation:
:
:n! ~ [sqrt (2*pi*n) * (n/e)^n] * [1 + T(i/n)],
:
:where T represents "big theta."  I haven't been able to find
:a suitable function for T to get an accurate (enough) 
:approximation, except for T(x) = C*x where C is some
:maddening number around 11.95 that doesn't seem to have anything
:to do with pi, e, sqrt(something), etc.
:
:What I'd really like would be a simple O(log n) algorithm
:for n!, but I suspect there isn't one out there.
 
It's easy to show that an O(log n) algorithm is impossible.
n! is about (n/e)^n so log(n!) is about n log n. It takes just
this long to merely write down the answer.

By writing n! in a canonical form (2^a 3^b...  i.e. its factorization)
and combining pairs of factors in a clever way I know I can do at
least as well as O(n log^2 n loglog n) using FFT multiply techniques.

I believe Peter Borwein found an even better result.
You must remember that for even moderate n, n! has a LOT of digits
and that multi-precise multiplication is expensive.
--
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: psb@remarque.berkeley.edu (psb)
Newsgroups: sci.math
Subject: Re: Fermat's Last Theorem
Date: 7 Apr 1992 10:14:24 GMT
Organization: University of California, Berkeley
NNTP-Posting-Host: remarque.berkeley.edu

In article <1992Apr6.210950.26305@midway.uchicago.edu> kwhyte@zarquon.uchicago.edu (Kevin Whyte) writes:
>  Am I the only one who thought Bob's response was tounge-in-cheek?
>Kevin

So who exactly is this guy Bob? :-)

							--psb
-- 
/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\
|* Partha S. Banerjee				   Sic volo, Sic jubeo;	     *|
|* <psb@Berkeley.EDU> || <{spine}!ucbvax!psb>	   Stat pro ratione voluntas *|
\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/*\*/
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Does Goedel's work apply to number theory?
Date: 14 Apr 92 20:49:24 GMT
Organization: Beckman Instruments, Inc.

In <sei7hINNd77@agate.berkeley.edu> reid@durban.berkeley.edu (michael reid) writes:

>In article <1992Apr9.224106.4243@wdl.loral.com> mab@wdl39.wdl.loral.com
>(Mark A Biggar) writes:
>>I believe that the following problem has been shown to be undecidable in first
>>order logic + peano axioms (which should be sufficient for number threory):

>>Call a finite set in positive integers LARGE if the smallest member
>>of the set is greater then the number of elements in the set.

>>Take the set of numbers 1..N and partition them as follows: randomly
>>select a pair out of the set and color (label) one of the pair BLUE
>>and the other RED.  Continue selecting pairs without replacement and coloring
>>until the set is empty (if N is odd randomly color the left over element).

>>Now for any given k, show that there exists a sufficiently large N such that
>>it is guarenteed that the will exist a LARGE subset of the RED numbers
>>containing k elements.

>>It is possible (using higher order logic) to determine the function
>>N=f(k) and determines the smallest N for a given k, but this function grows
>>so fast (it's faster the ack(N,N)) that it is impossible to prove that it
>>the the solution to the problem when limited to the above axiom system.

>what am i missing here?  what's wrong with taking  N = 4k + 2 ?
>that way, we have  2k + 1  red numbers, the  k  largest of which
>are all greater than  k.

Actually: I think it was: for any coloring of the set of pairs (or maybe
triples) of numbers 1..N, there is a LARGE subset X of 1..N such that each
pair (triple) of elements of X is RED or each pair (triple) is BLUE.

I'm not sure whether pair was sufficient.
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
Ich bin ein Virus.  Mach' mit und kopiere mich in Deine .signature.
Xref: sparky sci.math:6025 alt.irc:1365
From: jk87377@cc.tut.fi (Juhana Kouhia)
Newsgroups: sci.math,alt.irc
Subject: Re: electronic math seminar?
Date: 14 Apr 92 10:50:44 GMT
Distribution: sci
Organization: Tampere University of Technology


In article <JOSHUA.92Apr11072658@wiretap.wiretap.Spies.COM>
joshua@wiretap.wiretap.Spies.COM (Joshua Geller) writes:
>
>emacs irc clients are available on anonymous ftp at various places;
>nic.funet.edu, 

A small correction: nic.funet.fi, this is in Europe.

Juhana Kouhia
Newsgroups: sci.math
From: victor@tuamotu.wri.com (Victor Adamchik)
Subject: Re: fractional derivatives
Organization: Wolfram Research, Inc.
Date: Wed, 8 Apr 1992 16:39:05 GMT

In sci.math article <1992Apr2.225037.9980@mailer.cc.fsu.edu> you wrote:
> 
> 
> Hi. I saw a refrence here to fractional derivatives.  A long time ago, I 
> tried to define fractional derivatives, and did not come up with consistent 
> results.  Could someone please e-mail me a short definition of fractional 
> derivative?
> Kermit@FSU or rose@fsu1.cc.fsu.edu.bitnet
> If you want to be sure I see your response, please e-mail me.


There are about 20 definitions of the fractional derivative (or the
fractional integration).
Xref: sparky sci.math:5696 sci.crypt:1822
From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math,sci.crypt
Subject: Re: New Mersenne Prime
Keywords: Mersenne, prime, primitive root, quadratic residue, reciprocity
Date: 7 Apr 92 03:04:21 GMT
Organization: Harvard Math Department

In article <1992Apr1.182446.21988@linus.mitre.org>
bs@gauss.mitre.org (Robert D. Silverman) writes:

>The theorem [any prime factor of 2^p-1 is congruent 1 or -1 mod 8]
>works because 2 is a primitive root. Any prime that has 2 as a
>primitive root must be congruent to 1 or -1 mod 8. Here, we have 2p+1 = 8n+7.

Um, for "primitive root" read "quadratic residue"; 2 is not a
primitve residue of any factor of 2^p-1 for any odd prime p,
nor can 2 be a primitive root of any prime congruent to 1 or -1
mod 8 (quadratic reciprocity).

--Noam D. Elkies (elkies@zariski.harvard.edu)
  Department of Mathematics, Harvard University
From: sachs@tanner.berkeley.edu (Rainer Sachs)
Newsgroups: sci.math.stat
Subject: Re: Ergodic theory for general Markov chains
Date: 10 Apr 1992 14:30:22 GMT
Organization: U.C. Berkeley Math. Department.
NNTP-Posting-Host: tanner.berkeley.edu

In article <1992Apr9.141813.21760@lth.se> tobias@maths.lth.se (Tobias Ryden) writes:
>Let Y_n be a Markov chain on a general state space E for which there
exists a stationary distribution v, but the initial distribution of
Y, i.e. the distribution of Y_1, need not necessarily equal v.
Under which conditions does

       N
  1   ---
  -   \
  N   /    f(Y_n) -> v(f)
      ---
      n=1

hold, where f is in L_1(v) and the convergence is w.p.1, in L_1 or in
>probability?

email bounced.  See, among many other places, RW Wolff, Stochastic modeling
and the theory of queues, section 3-6.  The theory of such averages
is considerably simpler than the theory of functions; for example
in a cyclic irreducible positive recurrent discrete time
Markov chain f may fail to converge but (1/n)sigma f (y sub i)
still converges.
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math.stat
Subject: Re: C Matrix Inversion from SPlus
Date: 6 Apr 92 03:50:54 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Apr3.231735.27376@unixg.ubc.ca> turkel@unixg.ubc.ca (Bill Turkel) writes:
>I have the C routines for matrix inversion from _Numerical Recipes in C_
>and I would like to dynamically link them into SPlus and call them from
>S. ...

This sounds like a very do-able but bad idea. Why not use the Splus qr or
svd functions?

>   The matrices are stored dynamically as (float **) in C.  The S manual 
	....
>... If I use S_alloc, then
>I get error messages to the effect that S is unable to allocate 15 million
>bytes for a 2x2 matrix.  Has anyone done this?  What am I doing wrong?

Sounds like you are passing things by value which have to be passed by
reference. You could use malloc(), but then the storage will not be
available to you after the call. Allocation which should persist across
calls (either Splus or C) should be memory-mapped to files. This is
the best way to avoid Splus's horrifying memory inefficiency (which it
inherited from S and is entirely AT&T's fault not to mention any names...)

>
>We are running SunOS 4.1.1 and Splus (the manual has a copyright of 1990,
>but I could find no version number).

You should go through an example from the blue book. This will show you how
dynamic loading is supposed to work. If it doesn't, then maybe your copy
of Splus is broken. You might consider their excellent technical support,
which is not too expensive.


Later,
Andrew Mullhaupt
From: jpuranen@cc.helsinki.fi
Newsgroups: sci.math.stat
Subject: NordSSS Second announcement
Date: 6 Apr 92 12:22:58 GMT
Organization: University of Helsinki

NordSSS
Second Nordic Symposium on Statistical Software
in Education, Research and Practice

June 1-3, 1992

University of Helsinki, Finland

SECOND ANNOUNCEMENT

SUBMISSION OF CONTRIBUTED PAPERS, POSTERS, DEMONSTRATIONS
AND EXHIBITIONS IS POSSIBLE BEFORE MAY 10, 1992

FOR FURTHER INFORMATION CONTACT

NordSSS
Department of Statistics
University of Helsinki
Aleksanterinkatu 7
SF-00100 Helsinki
FINLAND

Telephone: 358-0-1912518
Telefax:   358-0-1912124
E-mail:    nordsss@noppa.helsinki.fi

PROGRAM COMMITTEE

Hannu Niemi        University of Helsinki (Chairman)
Markku Korhonen    University of Helsinki
Ilkka Mellin       Finnish Statistical Society
Simo Puntanen      University of Tampere
Juha Puranen       University of Helsinki
Hans Stenlund      University of Ume}

OFFICIAL TRAVEL AGENT OF THE SYMPOSIUM

CONGRESS TEAM/Area Travel Agency Ltd.
P.O.Box 227
SF-00131 Helsinki
FINLAND

Telephone: 358-0-176866
Telex:     122327 arkau sf
Telefax:   358-0-654987

OBJECTIVES OF THE SYMPOSIUM

To present new developments in statistical software to teachers,
research-workers, practitioners and scientists of statistics as well
as to those who apply statistics at universities, research institutes,
government offices or industry.

SYMPOSIUM LANGUAGE

English


PROGRAM (at the end of March):

PRINCIPAL LECTURE
Peter Naeve, University of Bielefeld
     "Recent advances in statistical computing"

INVITED LECTURES
Seppo Mustonen, University of Helsinki
     "Editorial interface in statistical computing and related areas"
Dag S|rbom, University of Uppsala
     "Recent developments of the LISREL model and
      the LISREL computer program"
Sanford Weisberg, University of Minnesota
     "Graphical methods for model comparison"

WORKSHOPS
Seppo Mustonen, University of Helsinki
     "SURVO 84C"
Sanford Weisberg, University of Minnesota
     "Graphical regression using LISP-STAT"

CONTRIBUTED PAPERS
Robert Jonsson      Gothenburg University
John Josefson       SPSS Scandinavia
Jon H. Knudsen      University of Bergen
Gunnar Kulldorff    University of Ume}
Simo Puntanen       University of Tampere
Lennart R}de        Chalmers University of Technology
Dmitrii Silvestrov  University of Ume}
Ene-Margit Tiit     University of Tartu

DEMONSTRATIONS
Ky|sti Huhtala      University of Jyv{skyl{
Erik A. Johansson   Ume}
Peter S. Mortensen  Aarhus Business School

POSTERS
Stefan Cenner       Stockholm University
Aslak Herva         Joensuu
Erik A. Johansson   Ume}
Jon H. Knudsen      University of Bergen

EXHIBITIONS
The following statistical software will be exhibited during the
symposium:
     BMDP
     SPSS
     SURVO 84C
Newsgroups: sci.math.stat
From: Andrew.Robinson@anu.edu.au
Subject: Re: SLLN and CLT for dble. index RV ?
Reply-To: Andrew.Robinson@anu.edu.au
Organization: Australian National University
Date: Wed, 15 Apr 92 05:34:51 GMT

In article <1992Apr15.081401.1@cc.newcastle.edu.au> eemn@cc.newcastle.edu.au
(Brett Ninness) writes:
[...]
>The only twist is that my sequence {X_k} is double indexed: {X_{k,N}}.
>That is, the variances of the X_j in
>
>X_1,X_2, ....... X_k
>
>are different to the variances of the X_j in 
>
>X_1,X_2,.........X_k,X_{k+1}.
>
>To put it another way, the variance of a particular X_k depends on how
>may terms I have in the sequence, hence the double indexing.  All the terms
>are zero mean and independent no matter how long the sequence.
>
I'm intrigued.  How can you get a sequence of random variables whose values are
independent, but whose variances depend on the number of values in the
sequence?

regardless,
&rue

Andrew.Robinson@anu.edu.au   <= forestry statistician 
(if it moves, count it; if it doesn't move then CHOP IT DOWN and count it.)
My employers don't care what I say, in fact I don't even think they know I
exist.

Newsgroups: sci.math.stat
From: dow@nova2.stanford.edu (Keith Dow)
Subject: Re: SPSS export files
Organization: Stanford University
Date: Fri, 17 Apr 1992 04:47:00 GMT

Copy the file to a new test file. Delete the last line, i.e.
the one with garbage on it.  You should be okay now.
From: sumargoh@serss0.fiu.edu (H. Sumargo)
Newsgroups: sci.math.stat
Subject: Hotelling Transform Algorithm
Keywords: transform
Date: 16 Apr 92 03:10:47 GMT
Organization: Florida International Univ.


Does any of you out there know a complete information/book tht explain the above topic?  How about a FAST algorithm for the above topic?  Please email your response to sumargoh@fiu.edu

thank you.
From: pm@isds.duke.edu (Peter Mueller)
Newsgroups: sci.math.stat
Subject: abstracts
Date: 15 Apr 92 14:02:59 GMT


We are considering initiating a newsgroup for paper and seminar 
abstracts in statistics.

The purpose would be to make recent advances promptly available to
scholars and practicioners in the field, and favor informal 
circulation of results.

We would welcome any comments/suggestions/experiences.

Giovanni Parmigiani 
Peter Mueller

------------------------------------------------------
Peter Mueller,	ISDS, Duke University, Durham NC 27706
e-mail: pm@isds.duke.edu          Fax:  (919) 684 8594
Office: (919) 684 3437 	          Home: (919) 419 0572
------------------------------------------------------
From: rdnelson@phoenix.Princeton.EDU (Roger D Nelson)
Newsgroups: sci.math.stat
Subject: Re: IQRs and hypothesis testing
Date: 16 Apr 92 15:36:28 GMT
Organization: Princeton University
Originator: news@ernie.Princeton.EDU

In article <1992Apr15.233135.2693@newshost.anu.edu.au> Andrew.Robinson@anu.edu.au writes:
>In article <OLIVER.92Apr15054328@ballard1.mit.edu> oliver@athena.mit.edu (James
>D. Oliver III) writes:
>>I was using a packaged statistical program to do some simple means testing,
>>and notice that in several cases the program didn't do a straightforward
>>t-test, but instead "computed the standard error of the difference of means
>>assuming IQR(pop.1)/IQR(pop.2)= X".
>>
>>I found out what IQR is (interquartine(sp?) range) and have a general
>                          ^^^^^^^^^^^^^  [interquartile]?
>>understanding of what it is, but can find no references on the specifics of
>>this particular adjustment and/or the justification for using it.
>>
>Sounds to me like the package is trying to determine the relative spreads of
>the populations because the t-test fails when testing on populations with
>differing variances.  The ANOVA does as well, but I don't know of a comparable
>procedure.  In fact, that's the first time I've heard of a package which tries
>to do this.  Nice.
>
>But I have _no_ idea where to go from here.
>
If indeed the variances are so different the package sees the t-test as
inappropriate, you might consider looking at the distributions
themselves, then looking toward some applicable non-parametric test.
From: desj@idacrd.UUCP (David desJardins)
Newsgroups: sci.math.stat
Subject: Re: help needed with Markov flipflop
Date: 16 Apr 92 01:24:14 GMT
Organization: IDA Center for Communications Research

In article <JIL.92Apr2170245@mozart.linus.mitre.org> jil@linus.mitre.org (Jon Leivent) writes:
>I need some help finding an analytic solution for a seemingly simple
>Markov chain problem.  The chain (in discrete time) consists of two
>states, A and B, arranged as a flip-flop.  The one-step probabilities
>are: A to A is q, A to B is 1-q, B to B is r, B to A is 1-r.

>The initial condition is that we spend the first time step in state A.
>I'm trying to find the probability distribution function P(C,S), which
>gives the probability that after S steps, the count of the number of
>steps spent in state A is C (C is thus the "total sojourn time" in
>state A).

Let K be the number of times that we take the transition A->B.
Clearly we take B->A either K or K+1 times.  Separate the problem into
these two cases.  Let L = C-K be the number of times that we take the
A->A transition.  L must be less than or equal to S-2K in the first
case and S-2K-1 in the second case.

Given the number of times each of the four transitions is taken, we
can count the number of ways of dividing the L A->A transitions into K
runs as a binomial coefficient.  Similarly for the number of ways of
dividing the B->B transitions.  Thus we compute the number of state
sequences and hence the probability of K and L.  We sum this over K.
The result:

                 K      C-K  K      S-C-K
P(C,S)  =  Sum  q  (1-q)    r  (1-r)       B(C,K) B(S-C-1,K-1)
            K

                 K+1      C-K  K      S-C-K-1
        +  Sum  q    (1-q)    r  (1-r)        B(C,K) B(S-C-1,K)
            K

where B(N,M) is the binomial coefficient N!/M!(N-M)! and the sums are
over all terms for which the binomial coefficients are defined.

It is standard to express hypergeometric sums such as these in terms
of the hypergeometric function.  If I haven't made any mistakes:

                    C-1      S-C-1                         q r
P(C,S) = C q r (1-q)    (1-r)       F  ( 1-C, 1-S+C, 2; ---------- )
                                   2 1                  (1-q)(1-r)

                      C      S-C-1                         q r
             + q (1-q)  (1-r)       F  (  -C, 1-S+C, 1; ---------- )
                                   2 1                  (1-q)(1-r)

For fixed S and C this expression reduces to a polynomial in q and r,
as it surely must from the statement of the problem.  If we fix either
C or S-C, the expression reduces to a relatively simple function of q,
r, and S.  For general S and C no further simplification is possible.

                                        David desJardins
Newsgroups: sci.math.stat