HDTenEightyP/NetNews huggingface dataset:
From: kosowsky@minerva.harvard.edu (Jeffrey J. Kosowsky)
Newsgroups: sci.math
Subject: Re: Help - non-integral power of a matrix?
Date: 17 Aug 92 21:51:35 GMT
Article-I.D.: minerva.KOSOWSKY.92Aug17165135
	<56923@mentor.cc.purdue.edu>
Organization: Harvard Robotics Lab, Harvard University

In article <56923@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:

>   In article <Aug.10.15.45.34.1992.26563@clam.rutgers.edu> gonzalez@clam.rutgers.edu (Ralph Gonzalez) writes:
>
>   >Hi.  Does anyone know of an algorithm to find a non-integral
>   >power of a matrix, e.g. A^.5 or A^1.3? Thus, A^2 is the same
>   >as AxA and A^0 is the identity.
>
>   There have been N postings in reply to this, which give partial
>   results, but none which I would consider sufficiently complete.
>
>   There are conditions, and it even depends on exactly what is 
>   wanted.  The simplest way to look at this is by considering a
>   reduction to one of the canonical forms.  A singular component
>   can be considered as having 0 down the diagonal and possibly
>   something above the diagonal; if this block is nxn, all powers
>   greater than n can be considered 0, but below n, it has to be
>   looked at carefully.
>
>   If one takes a block with a positive constant diagonal element h,
>   the Taylor series expansion around hI gives h^k*I plus a FINITE
>   series in (A-hI).
>
>   If a characteristic root is negative or complex, the same can be
>   done, but if a real solution is wanted, this may or may not be 
>   possible.  For example, the matrix 
>
>	   cos x	-sin x
>	   sin x	 cos x
>
>   is essentially exp(ix).  So powers of orthogonal matrices with
>   determinant 1 can be defined non-uniquely as real matrices.   
>   However, if the determinant is negative, there clearly are problems.
>   -- 
>   Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
>   Phone: (317)494-6054
>   hrubin@pop.stat.purdue.edu (Internet, bitnet)  
>   {purdue,pur-ee}!pop.stat!hrubin(UUCP)
>

Is this result really the definitive "sufficiently complete" answer
that we have all been waiting for?
Newsgroups: sci.math
From: roh@spot.Colorado.EDU (ROH JOHN J)
Subject: Any good method to find complex zeroes?
Organization: University of Colorado, Boulder
Date: Thu, 20 Aug 1992 07:44:44 GMT


 Does anyone know of any good method of finding complex zeroes
of an arbitrary transcental equation?  Specifically, I'm looking
for a subroutine in "C", but any help will be appreciated.

Thanks,
John
From: rmurphy@aludra.usc.edu (Bob Murphy)
Newsgroups: sci.math
Subject: Re: u(v^n)w prime puzzle
Date: 19 Aug 1992 13:56:16 -0700
Organization: University of Southern California, Los Angeles, CA
NNTP-Posting-Host: aludra.usc.edu

In article <1992Aug19.153755.14825@wri.com> roach@bikini.wri.com (Kelly Roach) writes:
>In article <l93702INNaq9@aludra.usc.edu> rmurphy@aludra.usc.edu (Bob  
>Murphy) writes:
BM>> If there were such values of u, v, and w, then one could construct
BM>> arbitrarily large prime numbers.  Since I have often heard
BM>> reference to "the largest known prime number", I would guess that
                               ^^^^^ 
BM>> there are no such values.  Now if only I had a proof.

KR>     Not quite.  The largest known prime number is 2^756839-1.
KR>But emphasize the word "known" here. 

If you look carefully at my posting I said 'known'.  I am well aware
that there is NO largest prime number.  I simply took the fact that a
Mersenne number is the largest KNOWN prime and used it to GUESS that
no such u,v,and w existed.  I specifically said that I don't have a
proof.

KR>     There ARE infinitely many prime numbers and they do
KR>get arbitrarily large.  A classic proof, due to Euclid is to
KR>take
KR>
KR>     P = p1*p2*p3*...*pN+1
KR>
KR>where p1,p2,p3,...,pN are the primes you already know.
KR>The number P is not divisible by any of p1,p2,p3,...,pN.
KR>All the primes dividing P are new primes. 

Actually, Euclid's proof is slightly different than above.

Euclid's proof:

Suppose that p1 < p2 < ... < pn are all the primes.
Let P = p1*p2*...*pn + 1 and let p be a prime dividing P;
then p cannot be any of p1,p2,...,pn, otherwise p would divide
the difference P - p1*p2*...*pn = 1, which is impossible.
So this prime p is still another prime.
This is a contradiction.
Hence there are an infinite number of primes.


                   Bob Murphy (rmurphy@aludra.usc.edu)


p.s. For those interested, Euclid's proof can be found as 
     Proposition 20 in Book IX of The Elements.  However, for a
     more readable proof, I would suggest The Book Of Prime Number Records
     by Paulo Ribenboim.  This book gives, as the author puts it,
     9 and 1/2 proofs on the infinitude of primes.
From: crippa@inf.ethz.ch (Davide Crippa)
Newsgroups: sci.math
Subject: Re: What is the EATCS?
Date: 18 Aug 92 12:01:12 GMT
Organization: Dept. Informatik, Swiss Federal Institute of Technology (ETH), Zurich, CH

In article <4752@balrog.ctron.com> wilson@web.ctron.com (David Wilson) writes:
>
>    Here I have a paper in the EATCS and I still do not know what it
>    stands for.
>
>
EATCS : european association of theoretical computer science

DC
From: cxm7@po.CWRU.Edu (Colin Mclarty)
Newsgroups: sci.math
Subject: Re: mathematician's marrying ages.
Date: 13 Aug 92 20:25:10 GMT
Reply-To: cxm7@po.CWRU.Edu (Colin Mclarty)
Organization: Case Western Reserve University, Cleveland, OH (USA)


In a previous article, guckes@math.fu-berlin.de (Sven Guckes) says:

>hubert@spica.ucsb.edu
>(Chang;Hung-Hsien;H.;;;COMS;Graduate;930630;;;8-402525-19906-7;500)
>[what's in a name ?]
>writes:
>
>>My friend (yu@math.harvard.edu) keep advocating that marrying
>>early is good for mathematician.  He gave several examples
>>( most of them are number theorist, according to his statemnet)
>>Examples he gave are Delinge, Gauss, Milnor, Serre, Zariski.
>>He told me that he got these information from bibliography
>>(maybe from introduction to authors...etc)
>
>If he uses four examples to prove his statement - tell him to quit math!
>
>>Any counterexample? ( Marry late and have quite an achievement )
>
>Galois and Abel never married. So there.
>

	Of course Abel and Galois did nothing late.
Newsgroups: sci.math
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: Puzzles
Date: Mon, 17 Aug 1992 22:24:12 GMT
Organization: Security APL, Inc.

In article <1992Aug13.130820.313@csc.canterbury.ac.nz> wft@math.canterbury.ac.nz (Bill Taylor) writes:
>John presents the Almost-Paradox
>
>[A-P] (1) it is provable that "for almost all n, P(n)".
>      (2) for any particular N, it is not provable that "P(N)".
>
>In this case, P(n) is the statement that n cannot be produced by a program
>of length <= E.  E is unspecified, but a value can be calculated (in principle).
>
>As a final bit of nonsensical musing on [A-P] as given; I got into the
>following grand wild goose chase.                     I was trying to
>imagine finding a number G, such that we could be reasonably sure (short of
>a proof), that *all* numbers greater than G would need programs of length > E.
>
>Clearly it would have to be at least ...
>
If the formal system involved is at all powerful, it would have to be bigger
than anything you can come up with.  We're talking about the busy beaver
problem here, into the range where even the size of the program is (by
conventional standards) rather large.  You can be sure that the smallest
such G is absolutely, unimaginably, large.
From: bs@gauss.mitre.org (Robert D. Silverman)
Newsgroups: sci.math
Subject: Re: Can you prove it?
Keywords: Number theory
Date: 20 Aug 92 13:09:21 GMT
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA

In article <1992Aug20.055633.19046@nuscc.nus.sg> bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle) writes:
:Hi
:I have a simple problem, can you prove or disprove it? I will appreciate all replies:
:
:Problem:
:
:Prove that for primes p & q, q<p
:
:if 2^(p-1) == 1 (mod q^2)
:
:then either 2^(q-1) == 1 (mod q^2) or q | p-1  { q divides (p-1}
 
We have that 2 has order p-1 in the group of units of Z/q^2Z.
Hence, p-1 | phi(q^2) by Lagrange's theorem.

Consequently, p-1 | q(q-1),  or  q(q-1) = k(p-1) for some integer k.
Case 1.
	k != q

We must have k | q-1,  hence  p-1 | q(q-1)/k  or  p-1 | q k'  for some
integer k' and hence q | p-1 as required.

Case 2.
	k = q.

Therefore p-q = q-1  and we have p=q. The first known case where 2^p-1 = 1 mod p^2  is p=1093. 

Since p =q in this case and we were given 2^(p-1) = 1 mod q^2, then we
also have 2^(q-1) = 1 mod q^2.

QED

--
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: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Subject: Collineations
Organization: UW-Milwaukee Math Department
Date: Wed, 12 Aug 1992 19:42:04 GMT

Suppose f is a one-to-one function from the plane to itself
which maps lines into lines, and suppose the image of f is 
not contained in a line.  Must f be affine?

I have found some partial results.  The restriction of f to
Q^2 is equal to the restriction of a projective transformation.
If one also assumes that f is surjective, or continuous, or that
it preserves betweenness, then f must be affine. 

--
David Radcliffe                                
radcliff@csd4.csd.uwm.edu            
Newsgroups: sci.math,sci.physics
From: frisinv@alleg.edu(Vincent Frisina)
Subject: math and physics degrees
Organization: Allegheny College
Date: Mon, 17 Aug 1992 12:59:11 GMT

   I'm an undergrad in math and physics and eventually want to get a Ph.D.  
Since my intrest shifts from one to the other periodically, I wonder if I  
can get two Ph.Ds simulataniously without skimping on either. If this is  
possible, which grad schools are best for it and how long will it take? If  
not, what's the closest I can get?
----
Vince
From: rmurphy@aludra.usc.edu (Bob Murphy)
Newsgroups: sci.math
Subject: Re: u(v^n)w prime puzzle
Date: 18 Aug 1992 17:55:30 -0700
Organization: University of Southern California, Los Angeles, CA
NNTP-Posting-Host: aludra.usc.edu

In article <1992Aug18.171532.14274@wri.com> roach@bikini.wri.com (Kelly Roach) writes:
>     Prove or disprove:  There are three non-empty
>     strings of digits u,v,w such that all the
>     numbers in
>          L = {u(v^n)w | n is a natural number}
>            = {uw, uvw, uvvw, uvvvw, uvvvvw, ...}
>     are prime numbers.
>
>  I am definitely not looking for any solutions involving u=v="0".  



If there were such values of u, v, and w, then one could construct
arbitrarily large prime numbers.  Since I have often heard
reference to "the largest known prime number", I would guess that
there are no such values.  Now if only I had a proof.


So far I have only been able to show that if a solution exists 
then v must be a multiple of 3.


Bob Murphy (rmurphy@aludra.usc.edu)




Xref: sparky sci.math:10399 sci.psychology:3513
From: zorrilla@cattell.psych.upenn.edu (Eric Zorrilla)
Newsgroups: sci.math,sci.psychology
Subject: Re: mathematician's marrying ages.
Date: 19 Aug 92 15:37:13 GMT
Followup-To: sci.math
Organization: University of Pennsylvania

>It is researched that those who are not married does have more
>mental, say, problems.

In almost every study I know of, this is simply a correlation.  How many people
do you know of that want to get married to someone with a "mental" problem?
(co-dependents and masochists excluded  :)   )

>Lonely people usually die more early than
>non-lonely people.

Same issue as above (though there is slightly better research in psychoneuro-
immunology suggesting an effect of social support in humans and controlled
studies of changes in "social environment" [e.g., housing status, gender-housed
status, etc.] in animals (though interpreting any of this as being "lonely" is
beyond my view)

>You see, human is not meant to be lonely.

If you mean lonely, probably true.  If you mean alone, that's a whole different
story.


[--------------------------------------+--------------------------------------]
| Eric Zorrilla - Dept. of Psychology  | zorrilla@cattell.psych.upenn.edu     |
| University of Pennsylvania           | 3815 Walnut, Philadelphia, PA 19104  |
[--------------------------------------+--------------------------------------]
Newsgroups: sci.math,sci.physics,sci.astro
From: kosowsky@schottky.harvard.edu (Jeffrey J. Kosowsky)
Newsgroups: sci.math
Subject: Re: Help - non-integral power of a matrix?
Date: 14 Aug 92 17:24:03 GMT
Article-I.D.: schottky.KOSOWSKY.92Aug14122403
	<a_rubin.713653963@dn66> <1992Aug12.231708.3644@galois.mit.edu>
Organization: Harvard Robotics Lab, Harvard University

In article <1992Aug12.231708.3644@galois.mit.edu> jbaez@nevanlinna.mit.edu (John C. Baez) writes:

   In article <a_rubin.713653963@dn66> a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
   >In <carle.713599691@vex> carle@vex.ugcs.caltech.edu (Matthew Thomas Carle) writes:
   >
   >>gonzalez@clam.rutgers.edu (Ralph Gonzalez) writes:
   >
   >
   >>>Hi.  Does anyone know of an algorithm to find a non-integral
   >>>power of a matrix, e.g. A^.5 or A^1.3? Thus, A^2 is the same
   >>>as AxA and A^0 is the identity.
   >
   >>>I imagine if such a thing is defined, then there are conditions
   >>>on A...

   >log "obviously" converges if all eigenvalues of A are strictly within 1 of 1,

   To clarify, perhaps, let me add that this is not only "obvious", it's
   true, at least if A is diagonalizable.  If A is a
   not-necessarily-diagonalizable matrix log A is defined if ||A - 1|| < 1;
>   I don't think the eigenvalue condition above is sufficient.


Aside:

The theory of taking analytic functions of a matrix (or more generally
any bounded linear operator, T, on a Banach space, X) is called the
Riesz (or functional) analytic calculus. Given a holomorphice
(ie: analytic) funtion, f, defined on an open neighborhood of the
spectrum, \sigma(T) of a bounded linear operator, T, we can define
f(T) by an extension of the Cauchy integral formula. If the Taylor
expansion of the function about some point contains the spectrum in
its region of convergence, then f(T) can equivalently be written as
the sum of the Taylor series terms where powers of T replace the
independent variable in the Taylor series. Much more could of course
be said about the properties of the functional analytic calculus. 
On normal elements of C* algebras (eg:  L(H) is a C* algebra where H
is a Hilbert space), there is a more general extension called the
Borel functional calculus. Suppose T is a normal element of a C*
algebra, A, and f is a continuous function on the spectrum of T, then
we can define f(T) via the inverse of the Gelfand transform.

end of aside:

In any case, the original poster was presumably talking about finite
dimensional matrices so that the spectrum \sigma(T) is equal to the
set of eigenvalues. Since the region of convergence of the logarithm
about 1 is 1, Arthur Rubin's characterization is obviously correct.
Assuming that John Baez is using the standard (2-norm) on linear
operators, then his criterion is equivalent since ||A -1|| equals the
largest magnitude eigenvalue of (A-1), so ||A-1|| < 1 iff all
eigenvalues lie within the unit circle around 1.

Jeff Kosowsky
Newsgroups: sci.math
From: rmurphy@aludra.usc.edu (Bob Murphy)
Newsgroups: sci.math
Subject: Re: Continued Fractions
Date: 20 Aug 1992 22:41:09 -0700
Organization: University of Southern California, Los Angeles, CA
Distribution: na
NNTP-Posting-Host: aludra.usc.edu
Keywords: Help, Continued Fractions

In article <1992Aug20.165623.11309@eliot.uucp> andyc@eliot.uucp (Andy Collins) writes:
>I am looking for information on continued fractions. I would
>hope that there are books on the subject out there, but have
>not had the time to do a full-out search. I would appreciate
>any information.
>
>Thanks
>
>andy

You may wish to look at "Continued Fractions" by C. D. Olds.
It was written in 1963 and is part of The New Mathematical Library
series put out by the MAA.  This series is geared toward beginners
so does not get very deep into the subject.  But it is also 
extremely well written!  I would highly recommend it.

I don't know much about the specifics of Ramanujan's work but I do
know that he did some work on continued fractions.  Bruce Berndt
supposedly covers this in one of his books but I don't recall
the exact title.  It may just be called "Ramanujan's Notebooks."
I think the continued fraction stuff is in book two of the series.

                            Bob Murphy (rmurphy@aludra.usc.edu)



Newsgroups: sci.math
From: butscher@helios.physics.utoronto.ca (Adrian Butscher)
Subject: Need Titles
Organization: University of Toronto Physics/Astronomy/CITA
Date: Mon, 17 Aug 1992 15:06:24 GMT

Can anyone give me some titles of modern, undergraduate level differential
Geometry textbooks (I know there are lots!)?  I'd like a book which covers
Manifolds, Differential Forms, Topology.  In fact, I'm looking for a book
which would follow 'Calculus on Manifolds' by Michael Spivak (my 2nd year
textbook).  Something with Group Theory applied to DG (if it's not too 
advanced) would be nice, too.

Thanks

A.
From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math
Subject: space-eating curves
Date: 14 Aug 92 16:51:16
Organization: Thinking Machines Corporation, Cambridge MA, USA
Distribution: sci
NNTP-Posting-Host: strident.think.com

As recent posts have pointed out, there is a continuous map from the unit
interval, I, onto the unit square, IxI.  The first one was discovered by
Peano; the best known is due to Hilbert.  Pictures of the first few
approximations to the Hilbert curve can be found in many math
popularizations, e.g., Kasner and Newman's "Mathematics and the
Imagination".  It's a cute and fairly easy exercise to write a
program that inputs n and outputs the order n approximation to the Hilbert
curve.  (Just a little more interesting than the cliched Towers of Hanoi.)

The Hahn-Mazurkiewicz theorem asserts that a Hausdorff space T is a
continuous image of I if and only if T is connected, locally connected,
compact, and second countable.  I^2 and more generally I^n (n any integer)
qualify.

Special case of interest: the Hilbert cube I^{aleph_null} also qualifies
(call it H).  Topologically, H is the product of countably many copies of I
with the product topology.  Equivalently, for x in R^{aleph_null}, define
||x|| to be sqrt(sum (x_i)^2) (setting ||x|| = infinity of course if the
series diverges; otherwise this is the usual l^2 norm); then sets of the
form N_e(x) = {y : ||y-x|| < e} form a neighborhood base (with e>0, of
course.)

A few years ago I noticed that this special case of the H-M theorem enjoys
a short and (IMHO) elegant proof-- once you have the continuous map
from I onto IxI.  Here goes:

    Let  x --> (f(x), g(x))   be the continuous map from I onto IxI.
    Construct the map from I to I^{aleph_null} as follows:

       x-->  (f(x), f(g(x)), f(g(g(x))), ...)

    Motivation: think of (f,g) as a tool for splitting a element x of I
    into two elements.  Starting with the list (x), we obtain longer and
    longer lists by repeatedly splitting the last element.  The following
    diagram illustrates the process:

	x
        |\
        | \
        |  \
     f(x)  g(x)
           |   \
           |    \
           |     \
        f(g(x))   g(g(x))
                  |      \
                  |       \
                  |        \
             f(g(g(x))     g(g(g(x)))
                           |        \
                           .         .
                           .          .
                           .           .

    It follows immediately from the definition of the product topology that
    this map is continuous and its image is dense in H.  However, its image
    is the continuous image of a compact set and is hence compact, hence
    closed, hence all of H!

Returning the general H-M theorem: if T satisfies 1)-4), then by the
Urysohn imbedding theorem, T can be imbedded as a subspace of H.  One can
in fact turn this observation into a proof the H-M theorem.
Newsgroups: sci.math
From: jbaez@nevanlinna.mit.edu (John C. Baez)
Subject: Re: Category theory funded by DOD?
Organization: MIT Department of Mathematics, Cambridge, MA
Distribution: na
Date: Wed, 12 Aug 92 22:59:29 GMT

In article <25401@dog.ee.lbl.gov> sichase@csa2.lbl.gov writes:
>In article <1992Aug11.160727.27151@galois.mit.edu>, jbaez@nevanlinna.mit.edu (John C. Baez) writes...
>> 
>>The problem with physicists is that they got so good at blowing things up
>>that we no longer need bigger bombs.  So while a "functor bomb" would be
>>amusing it would not bring in grants.
>
>That's a little too harsh a stab at physicists, considering that one of the big
>players at Los Alamos was Stan Ulam - a mathematician who contributed as much to
>the bomb effort as any man did. 

It wasn't meant to be a stab at physicists; certainly the fact that many
mathematicians are bad at inventing bombs doesn't give them any moral
superiority over physicists, and as you point out there were some
mathematicians that WERE good at inventing bombs - didn't von Neumann
also play a key role.  (Both von Neumann and Ulam were good at too many
things to really fit as a mathematician or a physicist, actually.)   My
point was simply that even if one could build a big bomb using category
theory, it wouldn't be clear that the government should fund it.






Newsgroups: sci.math,sci.math.num-analysis
From: ccrick@brolga.cc.uq.oz.au (Rick Ernst)
Subject: Modelling loudspeaker arrays??
Keywords: loudspeaker, array
Organization: Prentice Centre, University of Queensland
Date: Mon, 17 Aug 1992 06:57:10 GMT

Hi Netters,

I am involved in a project that requires loudspeaker arrays
to be modelled. It involves using the model to set up a series
of speakers in large halls and rooms. I am currently looking
at using Bessel coefficients but haven't had much success.
Any help or pointers to relevant info would be greatly appreciated.

If enough info is shown I will summarise for the net.

Please reply via e-mail.

Thanks in advance,

Rick.



--
------------------------------------------------------------
Rick Ernst      	email: ccrick@brolga.cc.uq.oz.au
Prentice Computer Centre
University of Queensland, St Lucia 4072, Austalia
Newsgroups: sci.math
From: regeorge@magnus.acs.ohio-state.edu (Robert E George)
Subject: RE: Equation Editor in WordforWindows -- Help!!!
Organization: The Ohio State University
Date: Sat, 22 Aug 1992 21:32:56 GMT

                                                                               
Has *anybody* been able to create mathematical formulae with Equation          
Editor in WordforWindows 2.0/Windows 3.1 and then have them successfully       
printed out on paper by a printer? Note successfully -- this does              
*not* mean some strange, squiggly symbols that one might eventually            
guess were integral signs, but rather integral signs that look                 
much like those we see in textbooks?                                           
                                                                               
If so, please email me and let me know what printer you are using.             
I seem to have very, very great difficulty finding a printer that can          
actually handle the mathematical symbols (such as integral sign).              
                                                                               
(I apologize for posting to sci.math, since I realize that this message        
is not really appropriate for that newsgroup, but I have exhausted so          
many avenues and wasted so much time that I feel a need to go immediately      
to people who do this kind of wordprocessing regularly.)                       
                                                                               
Thanks,                                                                        
Robert George                                                                  
regeorge@magnus.acs.ohio-state.edu                                             
(speaking only for myself)                                                     
                                                                               
Xref: sparky sci.math:10497 sci.physics:13291