HDTenEightyP/NetNews huggingface dataset:
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: What a weird group!!!
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Thu, 15 Oct 92 04:53:12 GMT

No, I'm not referring to sci.math.  I'm referring to the Bohr
compactification of the real numbers, something I just learned about.
This is, of course, due not to Niels Bohr but to his brother Harald.

From an exercise in a book:  The Bohr compactification of the real
numbers is a compact topological group which contains the real numbers
as a dense set.  The embedding of the reals into the Bohr group is not a
homeomorphism.

Well, actually that in itself is not so weird now that I think of it.
One can embed the real numbers in the torus as a subgroup, namely a line
of irrational slope, and then it's dense but the image of the reals in
the torus with the induced topology is not homeomorphic to the reals
with their usual topology.  (I think this is what was meant by the
exercise above, by the way.)  

But I bet the Bohr group is a lot weirder than the torus.  

Here's how you get it.  Recall that there is a 1-1 correspondence
between compact Hausdorff spaces and a certain class of algebras called
C*-algebras.  The map one way is easy.  You just take your space X and
form C(X), the algebra of continuous complex-valued functions on X.
The algebras you get have certain properties and have been described
axiomatically, which is the point of the definition of (commutative
unital) C*-algebras.  Then the Gelfand-Naimark theorem says that any
algebra satisfying this definition really is C(X) for some compact
Hausdorff X, and tells you how to "construct" X.

Okay, so Bohr was interested in almost periodic functions on R.
These may be defined as follows.  Consider the space of functions on R
that are finite linear combinations of functions exp(ikx).  Then take
the closure of this space in L^infty(R).  These are the almost periodic
functions.  There's another definition too, that makes it clearer in
exactly what sense they are "almost" periodic.  

The point here is that they clearly form a subalgebra of L^infty(R) and
because this algebra is a closed subspace of L^infty(R) one easily
checks that it is a (commutative unital) C*-algebra.  Thus by
Gelfand-Naimark there is some compact Hausdorff space X out there such
that this algebra is really just C(X).  This X is the Bohr group.  Of
course, one has to show it's a group.  I imagine one simply extends the
group structure from R to this group in a unique continuous manner,
using the fact that R is dense in X.  

But what does this group *look like*??  That's what I'd like to know!!
From: wsadjw@rw7.urc.tue.nl (Jan Willem Nienhuys)
Newsgroups: sci.math
Subject: Re: Zeno
Date: 8 Oct 92 16:17:12 GMT
Reply-To: wsadjw@urc.tue.nl
Organization: Eindhoven University of Technology, The Netherlands

In article <1992Oct8.000340.1@opie.bgsu.edu> bc205cs@opie.bgsu.edu writes:
>
>Now here is my question:  The rule of the game is that you can only step half
>of the remaining distance on each step.  Well, if you are standing at Point B,

No.  There is no rule that says that Achilles must do a "step" whenever
he has covered a distance computed by Zeno.  Of course IF you make such rule,
you run into all kinds of paradoxes, precisely because that doesn't 
resemble the way "running races with tortoises" work.  JWN
From: dcj@dcs.ed.ac.uk (Mr Damian C Jackson)
Newsgroups: sci.math
Subject: Re: Do you know ?
Date: 16 Oct 92 12:34:00 GMT
Reply-To: dcj@dcs.ed.ac.uk (Mr Damian C Jackson)
Organization: Laboratory for the Foundations of Computer Science, Edinburgh U


Surya writes:  
>  Can you give me a function on an interval [a b] which is
>  not differentiable anywhere on that interval ?

Are you sure you don't want to qualify this? - it seems
a little straightforward. 

Damian "or am I missing something" Jackson
dcj@dcs.ed.ac.uk
From: cet1@cus.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Looking for fast methods of computing PI
Date: 16 Oct 92 14:50:24 GMT
Organization: U of Cambridge, England

In article <1992Oct16.111222.4303@CSD-NewsHost.Stanford.EDU>,
pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes, as part of a 
very interesting posting:
|> 
|> Before Lagrange-Gauss-Salamin, all improvements over Archimedes were
|> mere constant factors, as follows.
|> 
|>      Inventor          Speed in bits/op     Method
|> 
|>      Archimedes, c.250 BC    1       half-angle method
|>      Vieta, 1593             1       product(r=sqrt((r+1)/2))
|>      Descartes, < 1650       1       constant perimeter method
|>      Newton, 1666            2       series related to arcsin x
|>      Sharp, 1705             1.585   Gregory(sqrt(3))
|>      Machin, 1706            3.589   Gregory(5, 239)
|>      Stoermer, 1896          3.168   Gregory(8, 57, 239)
|> 
|> Here Gregory(a1,a2,...,an) is the formula pi/4 = a linear integer
|> combination of the arctan(1/ai)'s, e.g.  pi/4 = 6 arctan(1/8) + 2
|> arctan(1/57) + 4 arctan(1/239).  Note that both Sharp's and Stoermer's
|> "improvements" weren't, showing what happens when you don't do an
|> explicit complexity analysis.  Both Sharp's and Machin's methods were
|> used in early computer evaluations of pi.  For some reason Stoermer's
|> method was later chosen by Shanks and Wrench over Machin's, perhaps due
|> to inappropriate focus on the dominant arctan: arctan(1/8) must have
|> seemed more promising than arctan(1/5).

Do not the detailed mechanics of dividing by small integers with particular
hardware provide some reasons why theoretical bit complexity might be
ignored? Certainly Machin's formula was considered particularly suitable
for hand computation (in decimal, of course) because the divisions by 5 were
easy. I can imagine a similar argument being used in favour of Stoermer's
formula w.r.t. division by 8 on binary computers.

One would also like to know: are there Gregory(a1,...,an) methods with
arbitrarily high bits/op? If not, are there any better than Machin's?

Chris Thompson
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk
From: vitale@athena.cas.vanderbilt.edu (Michael A. Vitale)
Newsgroups: sci.math
Subject: Re: Is Math Hard?
Date: 16 Oct 92 21:53:29 GMT
Article-I.D.: news.1992Oct16.215329.20038
Organization: Vanderbilt University Mathematics Department


Do people write these things just to start flame wars?
--
--
Mike Vitale  vitale@athena.cas.vanderbilt.edu   vitalema@ctrvax.vanderbilt.edu
Box  120092                1926-C  18th Ave South     7853 Behm Road
Nashville  TN  37212-0092  Nashville  TN  37212-3804  West Falls  NY  14170-9743
(615) 322-6653             (615) 297-4150             (716) 662-4878
Xref: sparky sci.math:13026 sci.physics:16311
From: eder@hsvaic.boeing.com (Dani Eder)
Newsgroups: sci.math,sci.physics
Subject: Re: How do you draw a straight line?
Date: 9 Oct 92 13:48:29 GMT
Followup-To: sci.math
Distribution: na
Organization: Boeing AI Center, Huntsville, AL

Take an edge that is reasonably straight.  Draw a line tracing the
edge.  Flip the edge over keeping the ends at the same location.
Draw a second line.  Now bisect the gap between the two lines, and
the result will be straight.

Dani Eder

-- 
Dani Eder/Boeing/Advanced Civil Space/(205)464-2697(w)/232-7467(h)/
Rt.1, Box 188-2, Athens AL 35611/Member: Space Studies Institute
Physical Location: 34deg 37' N 86deg 43' W +100m alt.
***THE ABOVE IS NOT THE OPINION OF THE BOEING COMPANY OR ITS MANAGEMENT.***
Xref: sparky sci.math:12849 sci.math.symbolic:2647
From: huang@deakin.OZ.AU (Weiguang Huang)
Newsgroups: sci.math,sci.math.symbolic
Subject: RE: SymbMath: a symbolic math system learning from user
Date: 8 Oct 1992 01:16:28 +1000
Organization: Deakin University, Victoria, Australia
Distribution: world
NNTP-Posting-Host: sol.deakin.oz.au

Subject: Re: SymbMath: a system learning from user
From: Ice <skynet!ice@uunet.uu.net>

 
I've used it a little bit. I think it is a fascinating and powerful
demonstration of the  program's adaptive abilities. My questions
would concern things like the speed of Maple versus SymbMath, the
language it is written in and whether or not commercial symbolic
computation systems can really benefit from learning ability.


Ice()

Newsgroups: sci.math
From: borbor@divsun.unige.ch (BORIS Borcic)
Subject: Matrices as Group Sub-Algebras ?
Organization: University of Geneva, Switzerland
Date: Mon, 12 Oct 1992 13:51:27 GMT


Is it possible, for any (or all) N>1, to specify
f : {1..N}x{1..N} -> E, E subset of G, G a group,
f one-to-one,

and

c : G-E -> F, F some field,

so as to define

p : NxN matrices over F -> G algebra elements over F
           M            -> a : G -> F      

by a(g) = Mij  when g in E and f(i,j) = g
   a(g) = c(g) when g in G-E 

and so that

p(M*M') = p(M)*p(M') for M,M' any two NxN matrices over F

??

---

c would preferably range over {0,1}


 NOTE  (a*a')(g) =   Sum a(h)a'(h')  for a,a' any two G-algebra elements
                    hh'=g            and h,h',g  in G

---


Thanks in advance for any info,


- Boris Borcic
From: litow@csd4.csd.uwm.edu (bruce e litow)
Newsgroups: sci.math
Subject: some more exponential sums
Date: 15 Oct 1992 18:36:52 GMT
Organization: Computing Services Division, University of Wisconsin - Milwaukee
Distribution: world
NNTP-Posting-Host: 129.89.7.41
Originator: litow@proclus.csd.uwm.edu

Thanks to those who pointed out what I missed. Here is another
question which seems to be more algebraic in nature. Let alpha be a
primitive n-th root of unity and let P(x) be a polynomial over the
integers of degree < PHI(n) the magnitude of whose coefficients is
bounded above by 2^k. Clearly P(alpha) != 0 but is there a lower
bound on |P(alpha)| in terms of n and k? Please mail responses to
me. Thanks,


Bruce Litow
Computing Services Division
P.O. Box 413
Univ. Wisconsin-Milwaukee, Milwaukee, WI, 53201
litow@proclus.csd.uwm.edu
Newsgroups: sci.math
From: martin@lyra.cis.umassd.edu (Gary Martin)
Subject: Re: Order -> Algebraic Structure
Organization: University of Massachusetts Dartmouth
Date: Sat, 10 Oct 1992 14:46:47 GMT

In article <BvvuD8.I26@world.std.com> rjk@world.std.com (robert j kolker) writes:

   Let H be a denumerable completely ordered set, where the ordering is
   dense, and there are no maximum or minumum elements. It is well known that
   any two sets having these properties are order isomorphic. 

   The set of rationals Q is a fortiori this set (up to an isomorphism).
   Clearly the seemingly innocent densely ordered set inherits its algebraic
   properties via this isomorphism or does it? 

   Is there someway of showing independent of this coincidental isomorphism ,
   that denumerable,densely ordered -> the algebraic structure of Q, i.e. Q
   is a denumerable field or characteristic 0.

I don't believe so - the field structure on the underlying ordered set is
_extremely_ non-canonical.  The automorphism group of the ordered set is
enormous (it embeds the full direct product of countably many copies of
itself!), while the rational field is rigid.  In other words, the field
structure is not invariant under the action of the group of order
automorphisms, so it isn't definable from the order.  I don't know if
that's what you mean, though.





--
Gary A. Martin, Assistant Professor of Mathematics, UMass Dartmouth
Martin@cis.umassd.edu
From: scavo@cie.uoregon.edu (Tom Scavo)
Newsgroups: sci.math
Subject: Re: FAREY
Date: 16 Oct 92 17:08:03 GMT
Article-I.D.: nntp.1992Oct16.170803.14345
Organization: University of Oregon Campus Information Exchange

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:
> 
>>are there any applications of FAREY SEQUENCE?
> 
>Look into: Schroeder's book, approximate title: Number Theory in Science
>and Communications, Springer Verlag. There may be what you want.

That's Manfred Schroeder and the full title is _Number Theory in Science
and Communication, with Applications in Cryptography, Physics, Digital
Information, Computing, and Self-Similarity_ (2nd edition, Springer-
Verlag, 1990) some of which appears in his popular book _Fractals,
Chaos, Power Laws:  Minutes from an Infinite Paradise_ (Freeman, 1991).

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

-- 
Tom Scavo
scavo@cie.uoregon.edu
Newsgroups: sci.math
From: vhansen@ipfs.bau-verm.uni-karlsruhe.de (Wolfgang von Hansen)
Subject: Question on real numbers
Organization: IPF, University of Karlsruhe
Date: Thu, 8 Oct 1992 21:11:17 GMT

Hi everybody,

is it possible to express any real number x with the following term

x = a + rb;  a, b \in Q; r \in R, r const.

Some more words to explain what I mean:
I was wondering if there is an analogon between the real numbers and
the complex numbers.

It is well known that any complex number c can be
written as an ordered pair of real numbers a, b: c = (a, b).
Operations can be done by using i := (0, 1) to write c = a + ib. 
Knowing that i * i = -1 one can perform complex arithmetics by using
only the rules for real numbers.

My idea is to write any real number as a pair of rational numbers
one of them multiplied with a constant real number r (see above).
a is not necessarily (spelling? ;-) different from null. A useful value
for r may be \sqrt(n), n \in N, because r * r = n is easy to handle.
This representation of the real numbers might improve the speed
and/or accuracy of algorithms on computers since all calculations are
done with rational numbers.

There are some things left to do:
1. Proove if it is (not) possible.
2. Find a good value for r. (How many different values for r are
   existing? none, one, finite, infinite?)
3. Find algorithm(s) to convert real numbers.

I'd be very happy if someone could give me some hints how to deal with
these tasks. I'm afraid that there are some non-trivial problems
to solve.

Thanks for paying attention, Wolfgang
Newsgroups: sci.math
From: dzk@cs.brown.edu (Danny Keren)
Subject: Re: Question on real numbers
Organization: Brown University Department of Computer Science
Date: Fri, 9 Oct 1992 05:03:02 GMT

vhansen@ipfs.bau-verm.uni-karlsruhe.de (Wolfgang von Hansen) writes:
#Hi everybody,
#
#is it possible to express any real number x with the following term
#
#x = a + rb;  a, b \in Q; r \in R, r const.

That is impossible, as such a set is countable, hence cannot
be equal to R. It will simply span a 2D sub-space of R (if r
is irrational). Now, R is a vector space over Q, and hence
(if one allows the axiom of choice) have a basis, but that basis
has much more than two elements. The existence of such a basis
allows to construct interesting counter examples, such as a
function f:R--->R satisfying f(a+b)=f(a)+f(b) where f is not linear.

-Danny Keren.

Newsgroups: sci.math
From: plu@math.psu.edu (Todd Andrew Simpson)
Subject: Re: Catalan NUmbers
Organization: Department of Mathematics, Pennsylvania State University
Date: Wed, 7 Oct 1992 22:23:42 GMT

In article <BvrFrn.53A@cs.columbia.edu> thanasis@cs.columbia.edu (Thanasis Tsantilas) writes:

   Is there a combinatorial ("bijective") way to prove that the 
   Catalan numbers are equal to C(2n,n)/(n+1)? 
   The answer looks simple, so I thought there must be something.
   I'm only aware of the generating function solution of the recurrence.
   Any pointers to books/articles are appreciated.

There is a very elegant proof due to D. Andr\'e sometime in the 19th
century: First, C(2n,n)/(n+1) = C(2n,n)-C(2n,n-1).  Then:

    The number of paths from (0,0) to (n,n) in the plane, with unit steps
(1,0) and (0,1), is C(2n,n).

    The Catalan number C_n is the number of these paths that stay in the
region {(x,y): y<=x}.

    The number of paths from (0,0) to (n+1,n-1) with unit steps (1,0) and
(0,1) is C(2n,n-1).

    There is a bijection between paths (with unit steps (1,0) and (0,1))
from (0,0) to (n+1,n-1) and those from (0,0) to (n,n) that do not stay in
the region {(x,y): y<=x}: given the latter type of path, with the k-th
step being the first to go into the region y>x, for i from 1 to k if the
i-th step is (1,0) make it (0,1); if it is (0,1) make it (1,0).

This is described in Knuth's book, The Art of Computer Programming,
Volume I (which is where I found it).  Another approach might be to
associate each of a set of C_n objects (there are lots of sets
enumerated by the C_n) with paths from (0,0) to (n,n) in such a way
that there are n+1 paths for each of the objects.  I haven't seen such
a proof, though it'd be interesting since one would not need to know
_a priori_ that C(2n,n)/(n+1) was equal to C(2n,n)-C(2n,n-1).

   Thanasis Tsantilas
   Columbia University
   thanasis@cs.columbia.edu

- Todd Andrew Simpson
From: holmes@garnet.idbsu.edu (Randall Holmes)
Newsgroups: sci.math
Subject: Re: Question on real numbers
Date: 9 Oct 92 16:30:13 GMT
Organization: Boise State University

In article <1992Oct8.211117.19295@rz.uni-karlsruhe.de> vhansen@ipfs.bau-verm.uni-karlsruhe.de (Wolfgang von Hansen) writes:
>Hi everybody,
>
>is it possible to express any real number x with the following term
>
>x = a + rb;  a, b \in Q; r \in R, r const.
>
>Some more words to explain what I mean:
>I was wondering if there is an analogon between the real numbers and
>the complex numbers.
>
>It is well known that any complex number c can be
>written as an ordered pair of real numbers a, b: c = (a, b).
>Operations can be done by using i := (0, 1) to write c = a + ib. 
>Knowing that i * i = -1 one can perform complex arithmetics by using
>only the rules for real numbers.
>
>My idea is to write any real number as a pair of rational numbers
>one of them multiplied with a constant real number r (see above).
>a is not necessarily (spelling? ;-) different from null. A useful value
>for r may be \sqrt(n), n \in N, because r * r = n is easy to handle.
>This representation of the real numbers might improve the speed
>and/or accuracy of algorithms on computers since all calculations are
>done with rational numbers.
>
>There are some things left to do:
>1. Proove if it is (not) possible.
>2. Find a good value for r. (How many different values for r are
>   existing? none, one, finite, infinite?)
>3. Find algorithm(s) to convert real numbers.
>
>I'd be very happy if someone could give me some hints how to deal with
>these tasks. I'm afraid that there are some non-trivial problems
>to solve.
>
>Thanks for paying attention, Wolfgang

This is impossible.  The easiest argument is from cardinality; there
are too few numbers a + br (for any r).  They can all be listed and
then Cantor diagonalization can be carried out on them.  Even a list
of r_{i}'s for all i E n does not span the reals.  A spanning set for
the reals has uncountable cardinality.


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
From: fc03@ns1.cc.lehigh.edu (Frederick W. Chapman)
Newsgroups: sci.math
Subject: Re: Matrices as Group Sub-Algebras ?
Date: 13 Oct 92 02:16:16 GMT
Organization: Lehigh University

In article <1992Oct12.142642.9575@news.unige.ch>,
borbor@divsun.unige.ch (BORIS Borcic) writes:

>There are probably other ways to look into. My question
>boils down to : are there any (standard, simple) ways
>to construct endomorphisms from matrix algebra into
>group algebra ?

As stated, your question seems to ask: does an arbitrary group algebra
contain a subalgebra isomorphic to a matrix algebra; or put another
way, can we embed a matrix algebra in a given group algebra.  I really
have no idea.  If you ask the question the other way, however, the
answer is easy: can we embed a group algebra in a matrix algebra?
Yes, always.

Let A be ANY n-dimensional algebra with identity 1 over a field F, and
let M(n,F) denote the algebra of n x n matrices over F.  The regular
matrix representations of A give embeddings of A in M(n,F) as
follows.  Let x and y belong to A, and define lambda_x : A --> A
by labmda_x (y) := xy; similarly, define rho_x : A --> A by 
rho_x (y) := yx.  lambda_x is left-multiplication by x, and rho_x is
right-multiplication by x; both mappings are linear transformations
from A to itself (where A is viewed as an F-vectorspace).  Fix a basis
g_1, ..., g_n of A (the choice of a basis is obvious when A is a group
algebra -- the elements of the group, by definition, form a basis of
the group algebra).  Let L_x denote the matrix of the linear
transformation lambda_x with respect to the fixed basis; likewise, let
R_x denote the matrix of the linear transformation rho_x.  The mapping
A --> M(n,F) given by x |--> L_x is called the left regular matrix
representation of the algebra A; the mapping given by x |--> R_x is
called the right regular matrix representation.  Both left and right
regular representations are algebra homomorphisms.  Furthermore, the
presence of an identity element 1 in A makes both the left and right
regular representations "faithful", which means one-to-one/injective.
Thus, the left and right regular matrix representations give
embeddings of the algebra A into the matrix algebra M(n,F).

I know this is not exactly what you asked, but I hope the information
is of some use.  For more information, look for books on linear
representations of finite groups and linear representations of
finite-dimensional algebras over a field.
-- 

o ------------------------------------------------------------------------- o
|  Frederick W. Chapman, User Services, Computing Center, Lehigh University |
|    Campus Phone:  8-3218     Preferred E-mail Address:  fc03@Lehigh.Edu   | 
|         "The day after yesterday is the second-to-last day before         |
|               the rest of your life the day after tomorrow."              |
o ------------------------------------------------------------------------- o
Xref: sparky sci.math:12981 sci.classics:774
From: israel@unixg.ubc.ca (Robert B. Israel)
Newsgroups: sci.math,sci.classics
Subject: Re: Irrational?
Date: 9 Oct 92 08:12:45 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca


In <1aqoe7INN5r7@agate.berkeley.edu> chrisman@wheatena.berkeley.edu (chrisman) w
rites:

>Does anybody know if the common-use word "irrational" (meaning
>nonsensical) is actually derived from the mathematical term?


In <9210071709.hoey@aic.nrl.navy.mil> hoey@AIC.NRL.Navy.Mil (Dan Hoey) writes:

>The dictionary (I peeked) says that the Latin word _ratio_ means
>reason, computation, or reasoning, from which we get the words ratio
>and reason.  I believe that long division was a canonical example of
>computation, from which come the words having to do with division, and
>the rest of the words deal with thought process

>So irrational numbers are numbers you can't get by dividing integers,
>while irrational people are the ones who can't do the division.

>Dan Hoey
>Hoey@AIC.NRL.Navy.Mil

Well, not quite.  As you say, "ratio" is a Latin word.  The literal
meaning is "calculation", and using it figuratively you get
"reason" and "reasoning".  BUT this "calculation" would
originally refer to something more on the level of 2+2=4, rather
than more modern meanings of "ratio" - neither the Greek ratio of
geometric quantities, nor long division (which is a completely different
subject, and was considered extremely difficult before the development
of hindu-arabic notation).

BTW, according to Cajori's History of Mathematics, the first use
of the words "rational" and "irrational" in their modern mathematical
sense was by Cassiodorius (6th century).  This would be much
later than the non-mathematical meanings.

I'm cross-posting this to sci.classics, where we may hear from people
who actually know something about word origins.

-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
Newsgroups: sci.math