HDTenEightyP/NetNews huggingface dataset:
From: horn@schaefer.math.wisc.edu (Jeffrey Horn)
Newsgroups: sci.math
Subject: Cardinality of sigma-rings
Date: 21 Jul 92 17:50:55 GMT
Organization: Univ. of Wisconsin Dept. of Mathematics

Is there a way to prove that no sigma ring can have the cardinality of the
natural numbers without referring to the Hausdorf maximality theorem?
Newsgroups: sci.math
From: hhassler@iaik.tu-graz.ac.at (Hannes Hassler)
Subject: Re: A Divisibility Problem
Organization: Technical University of Graz, Austria
Date: Tue, 28 Jul 92 11:57:47 GMT

In article <9207270341.AA10422@.euclid.uoregon.edu.> M.Johnson writes:
>
>I wonder whether the following is true:
>     Given  N > 1  positive integers, it is always possible to find a subset
      of the set of N positive integers such that its sum is divisible by N.
>Is this true? Thanks for your answer.

(1) The claim (trivially) holds if you allow the empty set as subset.
(2) Otherwise, you may argue as follows:  Let a_1,a_2,...,a_N be an enumeration
    of the set, and define S_i := a_1+a_2+...+a_i, for 1<=i<=N.  If all sums S_i
    are different values modulo N, one of them is zero and we are done.  If two
    of the sums are equal, say S_j=S_k modulo N, then their difference is zero
    modulo N and also corresponds to some nonempty subset.  (qed)
(3) A more interesting result is the following:  Given a set of 2n integers, one
    can always choose a subset S of cardinality n and with sum divisible by n.
    The proof of this is left to the reader as an exercise.

*******************************************************************************
Hannes Hassler
*******************************************************************************

Newsgroups: sci.math
From: prasanth@gn.ecn.purdue.edu (Ravi K. Prasanth)
Subject: Re: An interesting limit problem.
Organization: Purdue University Engineering Computer Network
Date: Sun, 26 Jul 92 13:16:37 GMT

In article <1992Jul25.212844.1@lure.latrobe.edu.au>, mattm@lure.latrobe.edu.au writes:
> A challenge to all mathematicians. A 100 years ago, this would probably have 
> been solved fairly simply in a natural way, but can you? I think that this
> problem was first posed by the Russian mathematician Arnold. Hope you find this
> problem as interesting as I did when I first solved it.
> 
>             sin(tan x)    -    tan(sin x)
> 	lim  ----------------------------------  = ???
> 	x->0 arcsin(arctan x) - arctan(arcsin x)
> 
> Tim    email: mattm@lure.latrobe.edu.au

Suppose y = arcsin(arctan(x)). Then x = tan(sin(y)). Note that
y(x) is continuous at x = 0 and x(y) is conitnuous at x. Also,
if x -> 0 from above (below), then so does y -> 0.

Substituting for x in the expression whose limit is to be 
found we get 

	    arcsin(arctan y) - arctan(arcsin y)
 	     ----------------------------------  
             sin(tan y)    -    tan(sin y)

But this is exactly the inverse of the expression whose limit
is to be found. Therefore, the limit exists and is equal to 1.

             sin(tan x)    -    tan(sin x)
 	lim  ----------------------------------  = 
 	x->0 arcsin(arctan x) - arctan(arcsin x)
 
	    arcsin(arctan x) - arctan(arcsin x)
 	lim ----------------------------------  = 1 
      x->0   sin(tan x)    -    tan(sin x)

Ravi K. Prasanth
From: dmassey@sdcc3.ucsd.edu (Daniel Massey)
Newsgroups: sci.math
Subject: Group Theory Question
Summary: group question from Lang's Algebra
Keywords: group, coset
Date: 23 Jul 92 21:47:55 GMT
Article-I.D.: sdcc12.36049
Expires: 1 Sep 92 07:00:00 GMT
Distribution: usa
Organization: University of California, San Diego

Hi,
I'm reviewing algebra and have gotten stuck on the following
seemingly simple question:

G is an abelian group and H a subgroup  of G. Prove their exists
a subgroup of G which is iso. to G/H.

Anyway, I'm not making any progress on this one so any suggestions
would be greatly appreciated.
Thanks,
Dan Massey
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Help needed for a Proof
Keywords: Help
Date: 26 Jul 92 00:00:35 GMT
Organization: U of Cambridge, England

In article <1992Jul25.062401.29682@uniwa.uwa.edu.au>, ycchin@tartarus.uwa.edu.au (Chin Yih Chong) writes:
> Recently, I came across a very useful formula for finding determinant,
> and It goes something like this:
> 
> Given an n*n square matrix    |  A   B  |
>                               |  C   D  |
> 
> where A is of dimension r*r and D is of dimension s*s, such that 
> r>0,s>0 and r + s = n, then
> 
>       det | A  B |  =  det(D)*det(A - B*R*C) , where R = inverse of D
>           | C  D |
> 
> Can anyone enlighten me on the proof of this formula.....
> Any help is greatly appreciated.

Sure. Remember that det PQ = det P det Q, and that determinants aren't
changed by adding a multiple of one row or column of a matrix onto another.

Then det | A B |  =  det D det | A   B |
         | C D |               | D'C 1 |

where ' means inverse. (Multiply by | 1  0 |
                                    | 0 D' |. )

Now use lots of "column operations" to subtract the right-hand bit times D'C
from the left-hand bit (exercise: check that this doesn't change the determinant,
using the second fact recalled above), to get

det D det | A-BD'C  0 |
          |   0     1 |

which is what you wanted.

(I've used without comment the fact that det | P 0 | = det |P|. It's easy.)
                                             | 0 1 |
From: cet1@cl.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Hailstone sequences
Date: 26 Jul 92 00:09:25 GMT
Reply-To: cet1@cl.cam.ac.uk (C.E. Thompson)
Organization: U of Cambridge Computer Lab, UK

In article <CHALCRAFT.92Jul24095243@zebedee.uk.tele.nokia.fi>,
chalcraft@uk.tele.nokia.fi (Adam Chalcraft) writes:
|> 
|> The result that on the positive integers, the function
|>   f(x)=3x+1 or x/2
|> seems to eventually send every positive integer to 1, whereas the function
|>   f(x)=3x-1 or x/2
|> does not is rather strange. Clearly the first function is "larger" than
|> the second. All that this shows, of course, is that the word "larger" has
|> no useful meaning here, and so any proof that the first function always
|> goes to 1 must use something a bit more subtle.

What one would hope to prove (if one were insanely optimistic) would be 
that all sufficiently large integers have eventual iterates smaller than
themselves. How many cycles there are that they might then converge to
would be a matter of "chance".

One way of looking at this is to observe that a cycle with a rises and
b falls multiplies the original number by 3^a/2^b, with a little 
pertubation due to the 1s. So one looks for cases where 3^a ~= 2^b:

    3 > 2    3x-1 needed: cycle does exist (1,2)
    3 < 4    3x+1 needed: cycle does exist (1,4,2)
    9 > 8    3x-1 needed: cycle does exist (5,14,7,20,10)
   27 < 32   3x+1 needed: cycle doesn't exist ... ah well
  243 < 256  3x+1 needed: cycle doesn't exist ... life is unfair
 2187 > 2048 3x-1 needed: cycle does exist (17,50,25,74,37,110,55,164,82,41,
                                            122,61,182,91,272,136,68,34)

It isn't too difficult to see how one can bound the perturbation caused
by the +1 or -1, and thus put an upper bound on the smallest member of
any posssible cycle that has particular values of a and b.

The same sort of argument applies to the Conway permutation sequence 
generated by 2n->3n, 4n+1->3n+1, 4n-1->3n-1 that I mentioned in another
article, except that here the perturbations are smaller and are both 
positive and negative. A cycle of length r contaniing s odd numbers should
make 3^r ~= 2^(r+s). I showed the cycles of lengths 1, 2 and 5 and 
invited readers to find the other known one, but I did reveal its length.
I should have kept that concealed as well and suggested that anyone wanting
to guess it should look at the nearest piano...

Chris Thompson
Cambridge University Computing Service
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk
Newsgroups: sci.math,sci.physics
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Consistency of ZFC, or anything else useful
Date: 22 Jul 92 01:11:01 GMT
Organization: Purdue University Statistics Department

In article <1992Jul21.132554.152734@ns1.cc.lehigh.edu> fc03@ns1.cc.lehigh.edu (Frederick W. Chapman) writes:
>In article <1992Jul20.173716.6310@galois.mit.edu>,
>tycchow@riesz.mit.edu (Timothy Y. Chow) writes:

|>Perhaps you might try to use the fact that in ZFC one can formulate a
|>proof that any two Peano structures are isomorphic.  In that case,
|>consider nonstandard models of ZFC...

>If I am not mistaken, there are no *KNOWN* models for ZFC, standard or
>otherwise!  The consistency of ZFC set theory is not known; if a model
>for ZFC were to exist, then ZFC would be consistent.

>I can't speak for anyone else, but I find the notion that the
>consistency of ZFC has not yet been established to be the most
>singularly disturbing mathematical news to ever reach my ears, given
>that ZFC is intended to serve as a foundation for the rest of
>mathematics.

I suggest then that you take a good look at what Godel showed.

He showed that any system of axioms adequate for Peano arithmetic
which is consistent has propositions which can be neither proved
nor disproved.  One of these propositions is always the consistency.

So you must accept this disturbing news; any remotely useful foundation
will never have its consistency demonstrated.  Only relative consistency
can be shown. 	
-- 
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)
Newsgroups: sci.math
From: das50852@uxa.cso.uiuc.edu (dan)
Subject: Re: Computational aids. Was: Re: An interesting limit problem.
Organization: University of Illinois at Urbana
Date: Thu, 30 Jul 1992 05:45:18 GMT

mhosea@seas.smu.edu (Mike Hosea) writes:

>This question pops up from time to time, and the strongest reactions
>occur when someone suggests a total abandonment of hand calculation
>or a total abandonment of modern computational tools (in the context
>of education).  Either extreme is a poor choice IMO.  Why must we 
>go to an extreme?

Yes, this is true.  I never meant to imply that there were only two
ways to teach Calculus.  In my own personal opinion as a professional
student (aren't I though?) concepts are learned best, when they're
read, heard, and written.  This in no way implies that Mathematica
(or whatever) is a poor tool, since my rational for it's use would be
it's ability to illustrate more complex concepts than a textbook ever 
could, and the ability for students to experiment with more complex 
ideas.  However, since these are really quite powerful there are problems 
which require ingenuity/patience to solve by "hand" (as if the hand were 
doing the thinking), that Mathematica reduces to a burp and a "Out[15] =".
It's the power that makes it both so attractive as an education tool, and 
yet so dangerous to the student.

					dan simms

Note: I in no way am trying to flame the "Calculus&Mathematica" series;
	...it just looks that way.
-----
artie@uiuc.edu
From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
Newsgroups: sci.math
Subject: Re: Cardinality of Borel field
Date: 29 Jul 92 17:53:29 GMT
Organization: Princeton University
Originator: news@ernie.Princeton.EDU

A very early, quite flawed version of a post with a similar title might reach 
some sites.  I was unable to interrupt the posting, and although I sent a 
cancellation request immediately it is possible someone might see it.  The 
article listed under "References" above is the corrected article.  Please do
not bother responding to the other one.  Thank you.

DGC
Newsgroups: sci.math
From: delliott@src.umd.edu (David L. Elliott)
Subject: Re: Why are Some Journals so Expensive?
Organization: Systems Research Center, U. of Maryland, College Park
Date: Thu, 23 Jul 1992 00:47:19 GMT

As any librarian can tell you, the [expensive] journals are in the University
library by author and editor demand, not reader demand. When people stop
serving on editorial boards and sending papers to the highpriced science
journals, we may see more books in our libraries.  
	Talk to your science librarian about the problem!

David
From: fc03@ns1.cc.lehigh.edu (Frederick W. Chapman)
Newsgroups: sci.math
Subject: Re: Lattice points in a polytope? (How to find all of them?)
Date: 24 Jul 92 13:21:25 GMT
Organization: Lehigh University

In article <BruKLH.DBz@watdragon.waterloo.edu>, xjzhu@jeeves.waterloo.edu 
(Xiao Jun Zhu) writes:

>Hi, there:
>
>   I am in need of an efficient program or algorithm(prefer in C) which can 
>find all the integer points in a polytope defined by sets of inequalities.
>(Suppose that I know that the number of lattice points is not so huge.)
>If you know such a thing exists, please send a message to me. Thanks very
>much for your help.
>
>Regards, Xiaojun.

When you find your program, you might want to try it on the following
simple test case:  Let S be the simplex (the simplest example of a convex
polytope) defined by the inequalities

                       x_i >= 0 for i = 1, 2,..., n
                        x_1 + x_2 + ... + x_n <= M

where n is a positive integer, and M is a non-negative integer.  The number
of integer lattice points inside S is given by the binomial coefficient

                                ( M+n )
                                (     )
                                (  n  )

which is an n-th degree polynomial in M.

...........................................................................

Pardon my boyish exuberence, but I think this a nifty fact.  How does one
go about proving this -- by induction on M?

-- 

o ------------------------------------------------------------------------- o
|  Frederick W. Chapman, User Services, Computing Center, Lehigh University |
|    Campus Phone:  8-3218     Preferred E-mail Address:  fc03@Lehigh.Edu   | 
|  "I do comedy and magic; what you don't find funny -- that's the magic."  |
o ------------------------------------------------------------------------- o
Newsgroups: sci.math
From: luzeaux@bellini.berkeley.edu (Dominique Luzeaux)
Subject: Re: Integer Sequence Problem
Reply-To: luzeaux@bellini.berkeley.edu (Dominique Luzeaux)
Organization: University of California, Berkeley
Date: Tue, 28 Jul 1992 22:56:05 GMT

In article <1992Jul28.173452.24364@news.tu-graz.ac.at>,
hhassler@iaik.tu-graz.ac.at (Hannes Hassler) writes:
|> For S an infinite, increasing sequence of positive integers (1=a_1 < a_2 <
|> a_3 < ... < a_i < a_{i+1} < ...), we define the VALUE val(S) of S 
|> 
|>         val(S)  :=    sup_{n>1}  \sum_{i=1}^n a_i   / a_{n-1}
|> 
|> (in words: For every n, sum up the first n terms of S and divide this sum
|> by the (n-1)st term.  Then val(S) is the supremum of all these values).
|> 
|> PROBLEM: Find the minimum (infimum) of val(S) over all possible sequences S.

Let us call : s_n=\frac{\sum_{i=1}^n}a_i}{a_{n-1}}
and : q_n=\frac{a_n}{a_{n-1}}

Then : s_n - s_{n-1} = q_n - (1-1/q_{n-1}) s_{n-1}
Let us suppose val(S) is finite for some sequence S; in other words val(S)
is an adherence value of (s_n); we can then consider that s_n converges to
val(S) (it is always possible to find a subsequence converging to val(S) and
we consider then this particular subsequence).

If s_n converges to some limit L, this difference converges to 0; this
implies that:
        q_n
     -------------  --> L
     1 - 1/q_{n-1}
which can be rewritten:  q_n = (L + e_n)(1 - 1/q_{n-1})  where e_n is a small
perturbation.
Draw the function x \mapsto (L+e)(1-1/q_{n-1}) where e is small; after some 
iterations, q_n will be negative which is impossible by definition of S; this
holds even if e is allowed to change for each n, but remains in a bounded small
interval. The only way to avoid that situation is to have q_n = q_{n-1}
starting
from some n; let q be this constant value. As q has to be integer and
greater or
equal than 2, let us try q=2. This yields L=4.

In fact, the sequence (1,2,...,2^n,...) satisfies all the conditions and yields
val(S)=4

Furthermore, we can say that val(S) takes as only finite values q^2/(q-1) 
where q is an integer >=2, hence 4 is the minimum of val(S) over all
sequences S.


	Dominique Luzeaux



Newsgroups: sci.math
From: bula@snoopy (W. Bula)
Subject: Re: Homeomorphism
Keywords: glitch, bicontinuous, hmmm?
Organization: University of Saskatchewan, Saskatoon, Canada
Date: Thu, 23 Jul 1992 21:34:52 GMT

In article <_1gmcfj@lynx.unm.edu> weishaup@vesta.unm.edu () writes:
>I was looking at Gelbaum's book of problems in Analysis (Springer, ~1990),
>and i found a problem that I don't understand:
>
>1.)Show that the set [0,1) is homeomorphic to the Real Line...
>At first I thought that this was impossible (The real line is open, the inverse
>image of it under a continuous mapping must also be open), but I found that
>was not insurmountable (after all, the Real line is also closed), but I don't
>understand how Gelbaum's Homeomorphism works... He lets
>
>f:[0,1) -> R
>
>be defined by 
>
>f(x) = (1/(1-x)) * Sin[1/(1-x)]
>
>I don't get this at all... f(x) does not appear to be invertible, so I don't
>see how it can be a Homeomorphism.
>
>An explanation would be welcome for this non-homework problem's answer either
>thru e-mail or (if the author thinks it merits the attention) through this
>newsgroup; Alternate Homeomorphisms would also be handy.
>
>Thanks,,,,
>Ben Jones
>(weishaup@carina.unm.edu
>)
> (...deleted...)

I do not know this book, and so maybe someone could tell what the author
means by a homeomorphism. If we stick to the usual definition
(i.e. an invertible, onto, continuous, inverse function continuous), then
obviously, there is NO homeomorphism between the two spaces. The 
simplest reason I can see is that in [0,1), the point 0 does not separate
the space (i.e. [0,1)\{0} is connected), while there is no point in R
with the same property.

Regards,

Witold D. Bula
Dept. of Math. and Stats.
University of Saskatoon
Saskatoon, SK
S7N 0W0
bula@snoopy.USask.Ca




Newsgroups: sci.math
From: palais@binah.cc.brandeis.edu
Subject: Re: Homeomorphism
Reply-To: palais@binah.cc.brandeis.edu
Organization: Brandeis University
Date: Thu, 23 Jul 1992 18:30:03 GMT

If you remove any point from the Real Line, the result is a space
with two connected components. If you remove 0 from [0,1) the 
result is (0,1), a connected space. Hence they cannot be homeomorphic.
From: torkel@sics.se (Torkel Franzen)
Newsgroups: sci.math
Subject: Re: Stupid question about FLT
Date: 21 Jul 92 15:02:20 GMT
	<1992Jul20.080932.24570@sics.se> <1992Jul20.234036.9686@mixcom.com>
Organization: Swedish Institute of Computer Science, Kista

In article <1992Jul20.234036.9686@mixcom.com> ttyytt@mixcom.com (Adam Costello)
 writes:

   >Okay, I see that you could imagine "non-standard" models of Th N, but since
   >these models make exactly the same sentences true that N does, this brand
   >of non-standardness isn't very upsetting.  It's the possibility of a
   >non-standard model that actually disagrees with N about the truth of some
   >sentences that provokes discussions like the one we're in the middle of.

  Why the quotes? Th N has nonstandard models in the ordinary sense of this
term. As for the questions surrounding incompleteness and the limitations of
first order logic, it is my impression that it is often unclear what people
have in mind. Hence my opinion that it is a good idea to make clear that
"and so on" is not a first order concept in the characterization of the
integers as "0, s(0), s(s(0)), and so on". Those who believe that "we cannot
say" what the natural numbers are, or that there is some problem regarding
what are "the real" natural numbers should try to clarify, for themselves and
for others, why they believe that only first order notions make unambiguous
sense.
Newsgroups: sci.math
From: cxxh@cco.caltech.edu (Chris H. Hall)
Subject: Game theory/Optimization question
Keywords: game
Organization: California Institute of Technology, Pasadena
Date: Wed, 29 Jul 1992 15:43:37 GMT



   The game tree...

                       E4
                       V4
                      /W4
                  P24/
                    /
                E2 /
                V2<
               /W2 \
              /     \
          P12/    P25\
            /         \E5
           /           V5
       E1 /            W5
       V1<
       W1 \            E6
           \           V6
            \         /W6
          P13\    P36/
              \     /
               \E3 /
                V3<
                W3 \
                    \
                  P37\
                      \E7
                       V7
                       W7

   En = Expectation (in % of wager) at branch point n.    n = 1, ... ,7
   Vn = Variance at branch point n.
   Wn = Wager placed at branch point n.
  Pnm = Probability of going from n to m.
           P12 + P13 = 1
                .
                .
           P36 + P37 = 1

   Note: Expectations can be negative.


  The game is the sequential game tree given above and is
  repeated over and over.  The problem is to determine
  W1...W7 with the following constraints.

     1. Minimize risk of ruin.

     2. All wagers must be > 0.

     3. Wagers cannot differ by more than a factor
        of 2 between branches. Thus,
      
        .5*W2 <= W1 <= 2*W2
        .5*W3 <= W1 <= 2*W3

        .5*W1 <= W2 <= 2*W1
        .5*W4 <= W2 <= 2*W4
        .5*W5 <= W2 <= 2*W5
                 .
                 .
        .5*W3 <= W7 <= 2*W3


   Risk of Ruin:

                      B + Et*x
              RoR = ------------
                    (Vt*x)^1/2

     RoR = Risk of ruin measured in standard of deviations
           from the mean.

       B = Bankroll (constant)

      Et = Expectation (in % of bankroll) of the entire game tree.

      Vt = Variance of the entire game tree.

       x = The number of wagers when RoR is minimized.


   With a little work, it can be shown that...

                    B
              x = ----
                   Et

   Thus,
                   B + Et*(B/Et)
            RoR = ---------------
                  (Vt*(B/Et))^1/2

                     (B*Et)^1/2
                = 2*(------)
                     ( Vt )

   So, my real question is how does one maximize the ratio (Et/Vt)
   given the constraints above?

   
   The method is more important than the answer to the above example; it
   was only used to hopefully give a clearer picture.

   If the answer is too long or complicated to post, or is so trivial
   that it is taught is Math 101, a reference would be appreciated.

   Also, if I have left something vague or unclear, please email me
   and I will try and clear it up.

   Thanks,
   Chris Hall

Newsgroups: sci.math,sci.math.num-analysis,sci.math.symbolic
From: bs@fatima.mitre.org (Robert D. Silverman)
Subject: Re: Lattice points in a polytope? (How to find all of them?)
Keywords: polytope, integer points
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Thu, 23 Jul 1992 16:05:52 GMT

In article <BruKLH.DBz@watdragon.waterloo.edu> xjzhu@jeeves.waterloo.edu (Xiao Jun Zhu) writes:
:Hi, there:
:
:   I am in need of an efficient program or algorithm(prefer in C) which can 
:find all the integer points in a polytope defined by sets of inequalities.
:(Suppose that I know that the number of lattice points is not so huge.)
:If you know such a thing exists, please send a message to me. Thanks very
:much for your help.
 
There are probabalistic methods that work in polynomial time (in RP).
However, they only give approximate answers.

If you need an exact answer I'm afraid that the problem is NP-Complete.
There is no 'efficient' way of doing it.

See, for example, 

Theory of Linear and Integer Programming
Alexander Schrijver
John Wiley & Sons. ISBN 0 471 90854 1

I might also suggest Schrijver's book Polyhedral Combinatorics

--
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"