HDTenEightyP/NetNews huggingface dataset:
From: prechelt@i41s14.ira.uka.de (Lutz Prechelt)
Newsgroups: sci.math.stat,comp.sources.wanted
Subject: where to get BLSS package ?
Date: 28 Aug 92 15:00:39 GMT
Organization: University of Karlsruhe, FRG
NNTP-Posting-Host: i41s14.ira.uka.de

I have found on a Berkeley FTP server that the BLSS statistics package
(made in Berkeley) is "available off-campus".

Does this mean that it is available for free ?
If yes, where and how can I get it ?

Please reply by email, I'll summarize.

  Lutz

-- 
Lutz Prechelt   (email: prechelt@ira.uka.de)            | Whenever you 
Institut fuer Programmstrukturen und Datenorganisation  | complicate things,
Universitaet Karlsruhe;  D-7500 Karlsruhe 1;  Germany   | they get
(Voice: ++49/721/608-4317, FAX: ++49/721/694092)        | less simple.
From: jgk@osc.COM (Joe Keane)
Newsgroups: sci.math.stat
Subject: Re: Degrees of Freedom  Was: Re: Standard Deviation.
Summary: It's OK.
Keywords: information
Date: 28 Aug 92 05:03:07 GMT
Reply-To: Joe Keane <jgk@osc.com>
Organization: Versant Object Technology, Menlo Park, CA
Weather: sunny, high 83, low 56
Moon-Phase: new

In article <thompson.714338397@kiyotaki.econ.umn.edu>
thompson@atlas.socsci.umn.edu (T. Scott Thompson) writes:
>To put out a concrete example that illustrates the nature of my
>qualms, consider the following.  Suppose my data consist of a single
>number, say X.  Suppose that I am given two pieces of information
>about X (but not X itself):
>
>           (1)  I can observe Y := X**2
>           (2)  I can observe Z := {  1 if X > 0
>                                   { -1    otherwise

It seems to me that the concept of degrees of freedom is closely related to
that of dimension.  Given that, i would say that X and Y have one degree of
freedom, and Z has *no* degrees of freedom, because it's discrete.

On the other hand, it's often more useful to talk about bits of information.
If X is symmetrically distributed, Z gives you one bit of information.  If we
observe Y to some accuracy, we can divide that accuracy by dY/dX to get the
accuracy of X.  Then we compare that with our previous accuracy of X to see
how much information we have gained.  Believe it or not, this actually works
out in the end.

--
Joe Keane, amateur mathematician
jgk@osc.com (uunet!amdcad!osc!jgk)
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Another induction problem
Date: 30 Aug 92 15:35:47 GMT
Distribution: sci
Organization: M.I.T. Artificial Intelligence Lab.



Hironaka's proof of resolution of singularities depends on a complicated
induction scheme. I heard a rumor once that the scheme is so complicated
that someone else published a paper proving that such a scheme could work.

Is this rumor true and if so, what is a reference?

Allan Adler
ara@altdorf.ai.mit.edu
From: matt@amsaa-cleo.brl.mil (Matt Rosenblatt)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Summary: . . . and that's all what I am.
Date: 1 Sep 92 18:55:33 GMT
Organization: U.S. ARMY MATERIEL SYSTEMS ANALYSIS ACTIVITY (AMSAA), APG, MD.

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."

>   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)
From: rcollins@cs.umass.edu (Bob Collins)
Newsgroups: sci.math
Subject: Re: HELP!! Inscribed Circle Problem
Date: 27 Aug 92 16:18:00 GMT
Reply-To: RCollins@cs.umass.edu
Organization: CS Dept, UMass at Amherst
News-Software: VAX/VMS VNEWS 1.41

The easiest way to compare distances on the sphere is first
to convert from spherical coordinates to unit vectors.  You
should double check the formulas since this is off the top
of my head.

   (x,y,z) = (Sin[A] Cos[B], Sin[A] Sin[B], Cos[A])

where  A = Pi/2 - Latitude  (colatitude) and  B = Longitude,
measured in radians. The cosine of the arcdistance d between 
any two points is then 

            cosd(p1,p2) = x1 x2 + y1 y2 + z1 z2

The KMart with the LARGEST cosine of arcdistance to the
customer is the CLOSEST.  That is cosd(p,p) = 1, monotonically
decreasing to cosd(p,-p) = -1 (the point on the other side of
the world).

To convert arcdistance to miles, multiply by the radius of the 
earth (you have to look that one up).

         miles = arcdistance * radius-of-earth

Rather than check all 6 million points every time, it would
be best to set up a 2D histogram by latitude and longitude,
then only check those KMarts that fall in the same bucket as
the customer.  

To find all KMarts within R miles, convert it to cos of arcdistance

         cosr = Cos[R / radius-of-earth]

then find all points p such that cosd[p,customer] > cosr.

Hope that helps.  I can emphasize with the dog and pony show
scramble.  As for the term "math whiz" ... kind of derogatory.

  --Bob Collins
  --RCollins@cs.umass.edu
Newsgroups: sci.math
From: stieger@cco.caltech.edu (Ronald David Stieger)
Subject: Re: n doesnt divide 2^n-1
Organization: California Institute of Technology, Pasadena
Date: Fri, 28 Aug 1992 22:37:23 GMT

goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:

>If someone could prove the following for me, I would be forever in 
>his debt.

>If n > 1, then n doesnt divide 2^n-1.

>This _is_ a problem in a book, but I don't _DO_ homework anymore, I'm  
>just curious.

OK. I had number theory spring term so let's see what I remember.

(1) a^p = a^q (mod m) iff p = q (mod t)
      where t is the order (?) of m, that is the smallest positive 
      number k (< m) s.t. a^k = 1 (mod m)

The proof of this fact is fairly straightforward.  p = q+rt so
 a^p = a^(q+rt) = (a^q)*(a^(rt)).  a^(rt) = 1 (mod m) so a^p = a^q (mod m)

So... n divides 2^n - 1 implies 2^n = 1 = 2^0 (mod n) which implies by (1)
n = 0 (mod t) so n divides t, the order of n.

Now, the order of n always divides n-1.  I don't remember how to prove this.
Maybe you know, or someone else can post it, or maybe I'll think of it
later.  You can test it if you like, a^(n-1) = 1 (mod n) for all a and n.
So t divides n-1, but we also know that n divides t.  This 
implies that n divides n-1, or n = 1 (mod n), which is only true if
n = 1.

So, if n divides 2^n - 1, n = 1, and thus, as you asked,

if n > 1, n does not divide 2^n - 1.  QED.

-- 
--Ron Stieger
  stieger@cco.caltech.edu
"Everybody wants a rock to wind a piece of string around."-They Might Be Giants
Newsgroups: sci.math
Subject: Re: Contents of n-dimensional sphere.
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 31 Aug 92 15:26:29 GMT
Keywords: n-dimensional sphere.

In <lee.715264502@dutiag> lee@dutiag.tudelft.nl (Marcel van der Lee) writes:


> Hello,

>There exists a formula to compute the contents of an n-dimensional sphere.

>        pi^(n/2)
>C(n) = ------------- r^n
>        gamma(n/2+1)

>Where C denotes the contents, n denotes the dimension and r the radius of 
>the sphere. I am wondering why the gamma function occurs in this formula.
>If you have a proof or a good reference, please post it on Newsnet or send
>it by e-mail to me.
>Thanks.


It is clear that C(n) == A(n) r^n, where A(n) is a function of n only.

Hence:

C(n+1) = Integral(s=-r to r, A(n) (r^2-s^2)^(n/2) ds) =
         2 Integral(s=0 to r, A(n) (r^2-s^2)^(n/2) ds) = (substitute s = r t^(1/2))
         2 Integral(t=0 to 1, A(n) r^(n+1) (1-t)^(n/2) 1/2 t^(-1/2) dt) =
         A(n) r^(n+1) Integral(t=0 to 1, t^(-1/2) (1-t)^(n/2) dt) = 
         A(n) r^(n+1) beta(1/2,n/2+1) =
         A(n) r^(n+1) gamma(1/2) gamma(n/2+1) / gamma((n+1)/2+1)

             gamma(1/2)^n
so C(n) = K ------------- r^n
             gamma(n/2+1) 

C(0) = 1, so K = gamma(1) = 1; and gamma(1/2) = pi^(1/2) (many references,
including the duplication formula), giving the desired result.
--
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.
My interaction with our news system is unstable; if you want to be sure I see a post, mail it.
Xref: sparky sci.math:10865 sci.math.num-analysis:2617 sci.math.symbolic:2302
From: mcroberts@titan.ksc.nasa.gov
Newsgroups: sci.math,sci.math.num-analysis,sci.math.symbolic
Subject: Anyone doing robot manipulator analysis with Mathmatica?
Date: 2 Sep 92 09:57:04 EST
Organization: NASA, Kennedy Space Center

From:  Mike Sklar at MCDONNELL DOUGLAS SPACE SYSTEMS CO.

Has anyone tried to do any robot manipulator analysis using Mathmatica.
We are currently evaulating Mma for Windows (Version 2.0) for its use as 
a general analytic tool.  We are interested in setting up user friendly,
complete packages for doing manipulator forward and reverse positioning, 
complete calibration procedures, dynamic analysis, evaluation of path 
planning algorithms etc.

If anyone has any experience with this please post the information here
or e-mail me directly at sklar%tardis.decnet@meglos.ksc.nasa.gov

Thanks.
From: jxh@math.ksu.edu (James C. Hu)
Newsgroups: sci.math
Subject: Re: p prime, p divides ab => pdivides a or b
Date: 3 Sep 1992 00:36:07 -0500
Organization: Dept. of Mathematics, Kansas State University
NNTP-Posting-Host: hilbert.math.ksu.edu

Charles Blair (ceblair@ux1.cso.uiuc.edu) wrote:
>   My recollection is that a number theory course I took presented
>this as a difficult result, only proved after doing some stuff
>with the Euclidean algorithm.  Is there a proof which avoids that?

I don't recall using the Euclidean algorithm, but the proof I had
learned used a theorem we called Diophantine's Result:

	Let a, b be non-zero integers, where gcd(a,b)=d.
	Then, there exists integers x,y such that
	ax + by = d.

The proof to this is not hard, but it depends upon the Division
Algorithm, which may be what you meant instead.  Using this, and the
definition of a prime, it is quite easy to show that:

	p prime, then p | ab => p | a or p | b.

The proof basically assumes p does not divide either.  Then,
gcd(p,a) = 1 and gcd(p,b) = 1.  We write down the Diophantine
equations which are assured to exist, multiply them together, and
arrive at an contradiction.

There are doubtless many other ways to prove this, but this result is
required to prove UFT, and so I don't think any proofs of this using
UFT would be wise.  (Of couse, prime decomposition may be enough alone
to show the result)
-- 
James C. Hu (jxh@math.ksu.edu), 1804 Denholm Dr., Manhattan, KS 66502
"If it was so, it might be; and if it were so, it would be; but as it
isn't, it ain't.  Thats logic." - Lewis Carroll
Opinions expressed are mine.  Let me know if they are someone else's.
Xref: sparky sci.math:10888 sci.math.num-analysis:2620
From: takriti@engin.umich.edu (samer Takriti)
Newsgroups: sci.math,sci.math.num-analysis
Subject: SparcePack
Date: 2 Sep 92 14:38:22 GMT
Organization: University of Michigan Engineering, Ann Arbor

I mentioned SparcePack in a previous message. Some people are asking
about this package. Here is the reference:
J. A. George and J. W. H. Liu "Computer Solution of Large
Sparce Positive Definite Systems", Prentice Hall.
-Samer
Xref: sparky sci.math:10530 sci.physics:13373
From: matmcinn@nuscc.nus.sg (Mcinnes B T (Dr))
Newsgroups: sci.math,sci.physics
Subject: Re: tensors: How about 3rd, 4th rank?
Date: 24 Aug 92 03:21:47 GMT
Organization: National University of Singapore

Tensor is what a differential geometer becomes when physicists talk
about general covariance.
From: bennett@math.ksu.edu (Andy Bennett)
Newsgroups: sci.math
Subject: Re: Godel and the US constitution
Date: 3 Sep 1992 16:16:15 -0500
Organization: Dept. of Mathematics, Kansas State University
NNTP-Posting-Host: hilbert.math.ksu.edu

cyeomans@ms.uky.edu (Charles Yeomans) writes:
>The Constitution explicitly gives the president the power to suspend
>habeas corpus when martial law has been declared.  But the army has no
>such power unless the president himself calls for it.

Actually, the constitution explicitly gives Congress the power to suspend
habeas corpus (It is subsection 3 of Section 9 of Article I dealing with
Legislative authority. Presidential authority is dealt with in Article II)
On this basis the supreme court ruled Lincoln's suspension of habeas corpus
unconstitutional during the Civil war. Lincoln supposedly remarked that since
it was the supreme court's ruling, they could enforce it and continued to
deny habeas corpus. The army mainly used the suspension of habeas corpus
to lock up Marylanders who advocated seccession. Early in the Civil war it
appeared Maryland would secede. Since that would have placed Washington DC
between two states of the confederacy, it would have been a catastrophe for
Lincoln and the north. Therefore he overstepped his authority to prevent it.
Maryland's state song refers to "the tyrant" meaning Lincoln for his actions.

Now that I've had a turn too, can we get back to mathematics?


-- 
Andrew G. Bennett         bennett@math.ksu.edu         If you count too
Dept. of Mathematics      Voice: (913) 532-6750        much you turn
Kansas State University   Fax:   (913) 532-7004        purple.  -  SARAH
Manhattan, KS 66502       STRICTLY MY OWN OPINIONS
From: mauax@warwick.ac.uk (Brazil)
Newsgroups: sci.math
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Date: 27 Aug 92 16:50:01 GMT
Organization: Computing Services, Warwick University, UK

>>In article <1992Aug11.190119.23518@ariel.ec.usf.edu> mccolm@darwin.math.usf.
>>edu. (Gregory McColm) writes:
>>
>>   >From there, we get more problems.  Suppose that you 
>>   >believe that the model of the integers exist.  Is PA 
>>   >true on this model?
>>
>>  What is PA? My explanation of the integers is simple: the integers
>>are 0, s(0), s(s(0)) and so on. Any obscurity or ambiguity that you
>>find in this explanation carries over to any explanation of what is
>>meant by "PA".
>

Don't forget their additive inverses.



     /-------------------------------------------------------------------\
   /           Believe me I would love to let you know, but.....           \
 /---------------------------------------------------------------------------\
|                 I'm not at liberty to tell you anything.                    |
 \---------------------------------------------------------------------------/
   \             Brazil : mauax@lily.csv.warwick.ac.uk :)))                /
     \-------------------------------------------------------------------/
Newsgroups: sci.math
From: esh@curly.tekbspa.com (Edward S. Hirgelt)
Subject: Re: p prime, p divides ab => pdivides a or b
Organization: Direct Interfaces Corporation @ Teknekron
Date: Wed, 2 Sep 1992 21:37:47 GMT


On Wed, 2 Sep 1992 17:21:47 GMT,
ceblair@ux1.cso.uiuc.edu (Charles Blair) said:

Charles>    My recollection is that a number theory course I took presented
Charles> this as a difficult result, only proved after doing some stuff
Charles> with the Euclidean algorithm.  Is there a proof which avoids that?

Well, seems to me this follows pretty directly from the decomposition of
a composite into powers of primes. Something like:

1. ab is a product of powers of primes.
2. p | ab implies that p appears in the set of prime powers of ab at
   least raised to first power.
3. Since a divides ab and b divides ab, we expect that one or both of a,
   b have p in their set of prime factors. All the p[i] that are in ab
   have to go to one or the other of a and b.
4. Thus p divides either a or b.

OK, its been a long time since my number theory...

Ed
 
--
Ed Hirgelt			esh@tss.com
Direct Interfaces Corporation
San Jose, Ca.
Newsgroups: sci.math
From: rwc124@rsphy1.anu.edu.au (Roderick Vance)
Subject: Unitary Matrix Goups
Reply-To: rwc124@rsphy1.anu.edu.au (Roderick Vance)
Organization: Optical Sciences Centre, Australian National University
Date: Wed, 2 Sep 92 03:47:01 GMT

Keywor

My question is about the group, (call it U) of all NxN unitary matrices (N a
constant for a given discussion)  and on the structure of subgroups
generated by a restricted set of matrices.

To illustrate with a concrete example, consider the set T of all tridiagonal,
symmetric matrices. Form from them the set of unitary matrices:

T1 = {exp(i H), H belongs to T}

and then find the smallest subgroup (call it U1)  of U containing T1. I
should like to find
out the following things about U1:

Is U1 all of U?

If not, can one come up with a simple test of whether or not a matrix 
belongs to U1 ?

What bits of group/matrix theory should i be reading about to try to
solve these problems?

I realise that i may face an horrendously complex task. Even if  you
consider an
innocent looking animal such as the smallest subgroup of U containing
two unitary matrices
A and B, where A and B do not commute, then it seems to me that, for
pathological choices
of A and B, you might wind up  with something as complex as the free
group generated  by
{A,B}. In general, there would seem to be no  way of  reducing arbitrary
products A^r1*B^r2*...
A^rn*B^rn, {r1..rn are integers}. I am probably wrong about this last
conjecture since i know
next to nichts about this subject.


Many thanks  in advance

Rod Vance
Optical Sciences Centre
Australian National University
From: wilson@web.ctron.com (David Wilson)
Newsgroups: sci.math
Subject: Re: Solitaire games
Date: 31 Aug 92 13:25:52 GMT
Reply-To: wilson@web.ctron.com (David Wilson)
Organization: Cabletron Systems INc.



    I used to put three hours a day on the road to and from work.  My
    game was to factor license plate numbers in my head (preferably
    before I lost sight of the car).  A simpler game was to try to
    break the numbers on the license plate into two groups whose sum is
    the same. 



-- 
David W. Wilson (wilson@ctron.com)

Disclaimer: "Truth is just truth...You can't have opinions about truth."
- Peter Schikele, introduction to P.D.Q. Bach's oratorio "The Seasonings."
Newsgroups: sci.math
From: cyeomans@ms.uky.edu (Charles Yeomans)
Subject: Re: Godel and the US constitution
    <Btx5vK.33x@usenet.ucs.indiana.edu> 
    <92Sep1.205802edt.637@neuron.ai.toronto.edu>
Date: Wed, 2 Sep 1992 21:05:31 GMT
Organization: University Of Kentucky, Dept. of Math Sciences

In article <92Sep1.205802edt.637@neuron.ai.toronto.edu> radford@cs.toronto.edu (Radford Neal) writes:
>In article <1992Sep1.175844.16825@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:
>
>>Out of curiosity, does anyone know how Godel envisioned a
>>consitutional dictatorship possibly arising in the U.S. ???
>
>
>Well... My recollection is fuzzy, but can't the US congress create new
>states just by passing a bill? It should be no problem for them to
>create several hundred of them, each containing only a single
>resident, namely, one of themselves. They can then run in these new
>states come next election, elect themselves, and out-vote any
>uncooperative representatives from the old states...
>   
>    Radford Neal

Not quite.  To form a new state out of pieces of one or more old states
requires both the consent of congress and the consent of the legislatures
of the states involved - this is explicitly stated in the Constitution.

Charles Yeomans
yeomans@austin.onu.edu
cyeomans@ms.uky.edu

Newsgroups: sci.math
From: nickh@CS.CMU.EDU (Nick Haines)
Subject: Re: Solitaire games
Originator: nickh@VOILA.VENARI.CS.CMU.EDU
Organization: School of Computer Science, Carnegie Mellon University
	<1992Sep3.125317.8741@cs.rulimburg.nl>
Date: Thu, 3 Sep 1992 19:37:16 GMT

In article <4906@balrog.ctron.com> wilson@web.ctron.com (David Wilson) writes:
>    I used to put three hours a day on the road to and from work.  My
>    game was to factor license plate numbers in my head (preferably
>    before I lost sight of the car).  A simpler game was to try to
>    break the numbers on the license plate into two groups whose sum is
>    the same. 

I play a game with triplets of digits (such as one finds on 99% of
British numberplates), similar to the `four fours' game: combine them
using some set of operators to generate the positive integers. You
don't have to use all the digits, and the set of operators includes
+,-,*,/ (exact division),! and can includes sqrt(), rem(), div(), or
any others that strike my fancy.

E.g. late-night driving Cambridge -> Newcastle, behind a truck with
the plate DEG 652 B (say):

1  = 6-5
2  = 2
3  = 6/2
4  = 6-2
5  = 5
6  = 6
7  = 5+2
8  = 6+2
9  = 6+5-2 
10 = 5*2
11 = 6+5
12 = 6*2
13 = 6+5+2
14 = hmm, nope

So I overtake the truck and go on to the next vehicle.

I've never done a search for the `best' triple, or even kept a record
of scores: that would spoil the game.

Nick
From: gossett@bethel.edu (Gossett)
Newsgroups: sci.math
Subject: source of quote
Date: 25 Aug 92 16:12:04 GMT
Article-I.D.: bethel.1992Aug25.161204.5076
Organization: Bethel College (Minnesota)

The BSD 'fortune' program contains the following quote:

  "The most exciting phrase to hear in science, the one
   that heralds new discoveries, is not 'Eureka!' (I found it)
   but 'That's funny...'"  - Isaac Asimov

Does anyoone know the original source document for this?

Eric Gossett    gossett@bethel.edu

-- 

--------
Dr. Eric Gossett             INTERNET   gossett@bethel.edu
Bethel College               UUCP  
Newsgroups: sci.math
From: Dave Dodson <dodson@convex.COM>
Subject: Re: HELP!! Inscribed Circle Problem
Originator: dodson@bach.convex.com
Date: Thu, 27 Aug 1992 16:26:10 GMT
Reply-To: dodson@convex.COM (Dave Dodson)
Organization: Engineering, CONVEX Computer Corp., Richardson, Tx., USA
X-Disclaimer: This message was written by a user at CONVEX Computer
              Corp. The opinions expressed are those of the user and
              not necessarily those of CONVEX.

In article <27AUG199211184436@cs.umass.edu> RCollins@cs.umass.edu writes:
>[...]
>Rather than check all 6 million points every time, it would
>be best to set up a 2D histogram by latitude and longitude,
>then only check those KMarts that fall in the same bucket as
>the customer.  

Good posting, and good idea, but you would have to check the customer's
bucket and the buckets that are adjacent to the customer's bucket, since
the customer and nearest KMart might be near a bucket boundary but on
opposite sides of it.

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

Dave Dodson		                             dodson@convex.COM
Convex Computer Corporation      Richardson, Texas      (214) 497-4234
From: ric@hpspdla.spd.HP.COM (Ric Peregrino)
Date: Thu, 3 Sep 1992 19:26:29 GMT
Subject: asymmetric metric?
Organization: HP Stanford Park - Palo Alto, CA