HDTenEightyP/NetNews huggingface dataset:
Newsgroups: sci.math
Subject: Re: A Non-Cantorian Set Theory question
From: kubo@widder.harvard.edu (Tal Kubo)
Date: 14 Aug 92 13:52:55 EDT
Organization: Dept. of Math, Harvard Univ.

In article <1992Aug12.113415.1648@gacvx2.gac.edu> kiran@gacvx2.gac.edu writes:
>Since we know that non-Cantorian set theories are possible, is there a
>one-dimensional shape which can be written some aleph times where that aleph is
>between aleph-nought and _c_? If so, what would the shape be?

The proof that only countably many homeomorphic copies of a letter T
can be placed on the plane can be done using only the intermediate
value theorem, and the fact that the circle and the plane both have
countable dense subsets.  Except for the axiom of (countable?) choice
for the intermediate value theorem, the proof could probably be carried
out in ZF.

This doesn't rule out the possibility that a "long line" of unusual
order-type could be written at most X times, where X is somewhere
between countably and continuum many.

-tk
Newsgroups: sci.math
Subject: Re: Is Card(R)=Card(R^2)?
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: 14 Aug 92 10:16:22 +1200
 <1992Aug13.000928.12631@unidus.rz.uni-duesseldorf.de> 
 <1992Aug13.011522.11161@informix.com> <16dihrINNpet@function.mps.ohio-state.edu>
Distribution: world
Organization: Department of Mathematics, University of Canterbury

|> >I believe that there is even an everywhere-continuous 
|> >bijective mapping from the unit line to the unit square.
|> 
|> No, there is not.  The interval and the square are not homeomorphic.
|> Such a continuous bijective map would be a homeomorphism,
|> since the two spaces are compact Hausdorff spaces.

Interestingly though, there *is* a homeomorphism from the unit line to
"most of" the unit square.      
i.e. so that the image in the unit square has measure arbitrarily close to 1 .
Newsgroups: sci.math
From: bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle)
Subject: Can you prove it?
Keywords: Number theory
Organization: Institute of Systems Science, NUS, Singapore
Date: Thu, 20 Aug 1992 05:56:33 GMT

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}




I have some sort of solution but I am not very sure.


Shailendra

(bhonsle @ iss.nus.sg)


--
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Is Card(R)=Card(R^2)? (yes, and a question)
Date: 13 Aug 92 14:50:21 GMT

In <1992Aug13.044351.7123@cs.brown.edu> dzk@cs.brown.edu (Danny Keren) writes:

>If I recall correctly, the fact that for *every* infinite cardinality
>alpha it is true that alpha = alpha X alpha is equivalent to the
>axiom of choice (help, anyone? I can prove that the axiom of choice
>implies alpha = alpha X alpha, what about the other direction?).

See Rubin & Rubin, _Equivalents of the Axiom of Choice_ for many and proofs
equivalents. (The Rubin's are my parents).  My recollection of the proof:

Let x be a set.  (Without loss of generality, x contains no ordinals).  Let
y be H(x) = {al: al an ordinal and al <= x} (Hartog's function).  (It
follows from elementary considerations that ~(y <= x).)  Consider x U y. 
If (x U y) X (x U y) = (x U y), then we can select f:x X y <= x U y.  If 
(Ez in x)(A(al)<y) (f(z,al) in x), then f(z,.):y <= x, a contradiction. 
Otherwise, define g on x by g(z) = f(z,first al<y such that f(z,al) in y). 
Clearly g: x <= y, so x can be well-ordered.  (It seems clear to me that
the well-ordering principle implies the axiom of choice.  Of course, I've
been working in the field on and off for 20 years.)

I don't have TeX here; with a little work I could convert that proof into
something that would appear reasonable in TeX, but then _I_ couldn't read
it.


--
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.
Our news system is unstable; if you want to be sure I see a post, mail it.
Organization: Middle East Technical University - Computer Center
Date: Wednesday, 19 Aug 1992 15:02:08 TUR
From: <E48453@TRMETU.BITNET>
Newsgroups: sci.math
Subject: anbody interested in Fermat's Last Theorem?

hello folks.let me directly go through the point,i have a friend who is
working on fermat's last theorem for more than 10 yaers,but as he doesnot
have any possibility to let other peolpe raed his work and judge,i decided to
send a msg to sci.math and ask for help;is there anybody STILL intersted in
the matter?if there is,send e-mail to:
c44590 at trmetu.bitnet.

we thank you all.bye
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: mathematician's marrying ages.
Date: 12 Aug 92 20:13:34 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <hubert.713379330@spica> hubert@spica.ucsb.edu (Chang;Hung-Hsien;H.;;;COMS;Graduate;930630;;;8-402525-19906-7;500) writes:
>Any counterexample? ( Marry late and have quite an achievement )

Erdos.

Later,
Andrew Mullhaupt
From: snorman@vulcan.den.mmc.com ( Stephen P. Norman)
Newsgroups: sci.math,sci.math-num-analysis
Subject: FFT routines in C/C++?
Keywords: FFT, DFT, C, C++
Date: 14 Aug 92 14:56:07 GMT
Distribution: usa
Organization: Martin Marietta Astronautics

Am posting this for a co-worker who is looking for
some FFT/DFT algorithms written in either C or C++. Apparently,
the ones contained in FFTPACK and VFFTPK are inadequate for
their purposes. Any suggestions are appreciated - replies
preferred over e-mail.

****************************************************************
Stephen P. Norman
Martin Marietta Astronautics Group
Computer Technology Center, Mailstop L0330
PO Box 179
Denver CO 80201
303-971-6066
****************************************************************
Newsgroups: sci.math
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Subject: Re: Can you prove it?
Organization: Rose-Hulman Institute of Technology
Date: Thu, 20 Aug 1992 14:46:18 GMT


In article <1992Aug20.130921.3437@linus.mitre.org> bs@gauss.mitre.org  
(Robert D. Silverman) writes:
> :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.

Well, 2 has order d which divides p-1.  Case 1 is OK with p-1
replaced by d.  Case 2 shows d=q-1, so by definition of d
we have 2^(q-1) == 1 (mod q^2)

>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.
This is only true if d=p.
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Help - non-integral power of a matrix?
Keywords: matrix
Date: 15 Aug 92 15:30:45 GMT
Organization: Purdue University Statistics Department

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.

>I imagine if such a thing is defined, then there are conditions
>on A...

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)
Newsgroups: sci.math
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: mathematician's marrying ages.
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Thu, 13 Aug 1992 00:03:29 GMT

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.
Newsgroups: sci.math
From: fast@spot.Colorado.EDU (FAST BRUCE ALAN)
Subject: Complex symbols -- zConj[z] =? x^2+y^2
Summary: be sure this is what you want
Organization: University of Colorado, Boulder
Distribution: usa
Date: Wed, 12 Aug 1992 16:04:52 GMT

bhattas@athena.mit.edu asks about using complex symbols:
  > I have been trying to manipulate complex symbols...
  >                                          ^^^^^^^
  > ...if we do z=x + I y  and then compute z*Conjugate[z], mathematica,
  > instead of giving the desired answer x^2 + y^2, outputs something like
  > (I x + y)*Conjugate(I x + y)....

It isn't clear to me that we all understand that (x^2+y^2)
is NOT the desired answer (and Conjugate[z] does NOT equal (x - I y))
unless  x  and  y  both happen to be real, not imaginary or complex.
Those who are thus aware, please ignore!  :-)
Newsgroups: sci.math.symbolic
From: jbuske@mountain.rmnug.org (Jeff Buske)
Subject: Info on method of Collocation with partial Diff Equations. 
Keywords: Collocation
Organization: Rocky Mountain NeXT Users' Group
Distribution: usa
Date: Thu, 13 Aug 1992 06:32:24 GMT

Looking for MMA program or good description of the method of  
Collocation with respect to partial Diff Equ.  I have not heard  
of this method before.  Apparently will enable one to convert  
partial diff. equations to a system of ordinary diff equations?
Thanks in advance.
Jeff Buske
From: Randy.Baer@f310.n3603.z1.FIDONET.ORG (Randy Baer)
Newsgroups: sci.math.symbolic
Subject: derive
Date: Wed, 12 Aug 92 18:02:55 EDT
Organization: FidoNet node 1:3603/310 - Tetragrammatton, St Petersburg FL

 
  In Derive 2.07 aftering loading Ode1.mth
 
  To solve the linear DE of the form:
  y'+p(x)y=q(x)    when y(1)=0  or whatever init condits
 
  I type:
 
  Linear1(p,q,x,y,x0,y0)
 
   I want to  be able to solve a DE in just the general form, without
the initial conditions of x0 and y0
 What should I put in for x0 and y0 to get the answer.

--  
Internet: Randy.Baer@f310.n3603.z1.FIDONET.ORG
UUCP: ...!myrddin!tct!psycho!310!Randy.Baer
Note:psycho is a free gateway between Usenet & Fidonet. For info write to
     root@psycho.fidonet.org.
From: greg@brooks.async.vt.edu (Greg Foliente)
Newsgroups: sci.math.symbolic
Subject: Random Vibration Analysis using Mathematica
Date: 17 Aug 92 19:25:41 GMT
Reply-To: greg@brooks.async.vt.edu OR pol2@vtvm1.cc.vt.edu

Has anyone written Mathematica packages and/or functions for random  
vibration analysis? I'm interested in packages that can be used for both  
teaching and research; thus, I'm interested in all areas of random  
vibrations.

--
.................................................................
Greg Foliente                   NeXTmail: greg@brooks.async.vt.edu 
Virginia Tech, Blacksburg, VA                 pol2@vtvm1.cc.vt.edu
.................................................................
From: tequila@ccwf.cc.utexas.edu (Mario Garcia)
Newsgroups: sci.math.symbolic
Subject: structs with MathLink ?
Keywords: MathLink
Date: 21 Aug 92 13:22:38 GMT
Reply-To: tequila@ccwf.cc.utexas.edu (Mario Garcia)
Organization: The University of Texas at Austin, Austin TX
Originator: tequila@dopey.cc.utexas.edu

Greetings,

I am using Mathematica 2.0 with MathLink 1.3 on a NeXTstation.  I have been
able to exchange some information between my C programs and Mathematica, but
I have been having trouble with structs.  I know the return type should be
declared as { Manual }, but how do I get MathLink to put the struct elements
in a List ?  Thanks in advance.

----------------------------------------------------------------------------
Mario L. Garcia
The University of Texas at Austin        tequila@ccwf.cc.utexas.edu
Applied Reaserch Laboratiories           tequila@truth.asg.arlut.utexas.edu
Advanced Sonar Division
---------------------------------------------------------------------------



Newsgroups: sci.math.symbolic
From: whitmore@iastate.edu (Kurt D Whitmore)
Subject: Re: MMA help on the Mac.
Organization: Iowa State University, Ames IA
Date: Mon, 17 Aug 1992 18:38:14 GMT

In <Bt4rJo.3yu@helios.physics.utoronto.ca> butscher@helios.physics.utoronto.ca (Adrian Butscher) writes:

>We have MMA on both a workstation (text interface) and on the Mac.  To get
>help on a command on the workstation, you just type ?Command or ??Command,
>and up comes the usage message and the notes.  On the Mac, however, the input
>?Command produces Attributes[Command] = something, even though we have the 
>MMA help file installed.  What is the problem here?  

>Adrian



I'm guessing that your msg.m package is not loading if you are using Mma 1.2 .

If you are 2.0 or higher go into the Edit - Preferences - File locations menu and select the Find button on the Help notebook.


If that doesnt work get back to me through email.



Kurt D. Whitmore
whitmore@iastate.edu
Newsgroups: sci.math.symbolic
From: victor@watson.ibm.com (Victor Miller)
Subject: PrimePi in Mathematica
Date: Thu, 20 Aug 1992 18:27:20 GMT
Reply-To: victor@watson.ibm.com
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM, T.J. Watson Research Center

I've been told that there is a function in some of the latest version
of Mathematica called PrimePi, which calculates the exact values of
pi(x) (number of primes <= x).  Can someone tell me what algorithm it
uses (I was wondering if it was mine: "Calculating pi(x): the
Meissel-Lehmer Method", by Lagarias, Miller, and Odlyzko, Math. Comp.
1985).

--
		Victor S. Miller
		Bitnet: VICTOR at WATSON
		Internet: victor@watson.ibm.com
		IBM, TJ Watson Research Center
		"Great artists steal; lesser artists borrow" Igor Stravinsky
Newsgroups: sci.math.symbolic
From: mrs@.waterloo.edu (M. R. Sridhar)
Subject: Mathematica - Output-Filename
Reply-To: mrs@.waterloo.edu (M. R. Sridhar)
Organization: University of Waterloo, Waterloo, Ontario
Date: Thu, 20 Aug 1992 02:09:46 GMT





stream = OpenWrite["filename", FormatType->OutputForm];

Write[stream, XX, "  ", YY];

Close[stream];


In the above set of statements how do i make the output filename independent
of the program? i.e. i would like to specify the output filename before
these statements are executed. 

Thanks in advance.

Sridhar.
Organization: University of Illinois at Chicago
Date: Friday, 21 Aug 1992 17:23:11 CDT
From: <U09207@uicvm.uic.edu>
Newsgroups: sci.math.symbolic,uiuc.mathematica
Subject: Help in evaluating an integral

********* I apologize if this is a repost *******************************
I am a relatively new Mathematica user. I will appreciate any comments and
suggestions on the following integral I am trying to evaluate:

           / H
           |
           | ( j beta/ r1^2 + 1/r1^3) exp( -j beta r1) dz
           |
          / -H
where r1=sqrt(z^2 + r^2 - 2 r z cos(theta)) and j=sqrt(-1).
This is a typical integral found in electromagnetics and antennas.
I tried to use Integrate[f,{z,-H,H}], but Mma responded with
Integrate[...] where ... was the integrand with the substitutions for r1 and j.
When I used NIntegrate[f,{z,-H,H}] and specified some typical values for
the constants, I got a value for the integral. But I would like to have a
closed form solution (or an approximate version). Does anyone have any
suggestions on this? Is there some kind of specialized Package available
for these type of problems.
I am using Mma on a Sparc2.

Thanks in advance.
Vahe Caliskan (vahe@uicbert.eecs.uic.edu)
Newsgroups: sci.math.symbolic,ba.seminars,ucb.seminars,ba.general,ucb.general
From: jake@skatter.usask.ca
Subject:  Anyone have an SMP that can solve this?
Organization: University of Saskatchewan
Date: Tue, 18 Aug 1992 02:36:06 GMT

Here it is (in Mma syntax)

Integrate[t^(-1/2) Sqrt[t - 4 m^2] Exp[ -t tau], {t,4 m^2, s0}]

I should point out that yes, I know that it is in Gradshteyn and Ryzhik
(on page 319) for s0 -> Infinity, and I also know that Mma can solve it
for that case.

If indeed it is known that the integral cannot be done for a finite upper
limit, I would appreciate someone enlightening me as to the precise reason.

Thank you in advance for your effort.

jake@skatter.usask.ca


Newsgroups: sci.math.symbolic
From: magnus@Fisher.Stanford.EDU (Magnus Nordborg)
Subject: Best way to factor out roots
Keywords: factor, polynomials
Organization: DSO, Stanford University
Date: Mon, 17 Aug 92 22:10:02 GMT

Given an enormous multivariate polynomial and a number of *known* factors,  
which is the best way to factor them out of the polynomial? I have access  
to MACSYMA, Mathematica, and Maple, but have had very little succes with  
all. Just asking the programs to "factor" does not work due to the size of  
the expression, but there must be some way of telling these programs  
*which* factors to look for.

I welcome all suggestions, as well as long lectures on algorithms if they  
are intended to teach me something ;-)

Thanks, Magnus
--


		Magnus Nordborg
		Department of Biological Sciences
		Stanford University
		magnus@fisher.stanford.edu (NeXT mail preferred)
From: cc3z+@andrew.cmu.edu (Christopher N. Carlson)
Newsgroups: sci.math.symbolic
Subject: Help! Quirky MMa Solver
Date: 13 Aug 92 07:38:12 GMT
Organization: Doctoral student, Architecture, Carnegie Mellon, Pittsburgh, PA

Can anyone explain why variable order makes a difference to the
Mathematica solver -- why it finds solutions for some orders of
variables and not for others?  Better yet, can anyone suggest a fix or
workaround?

For example,

    Reduce[
        {
        var[1] == 48.02 + var[2],	
        var[4] == (-2. + var[3])^2 ,
        (81.3016 + 1.84776 var[2])^2  - (-2. + var[3])^2  == var[2]^2 ,
        -2500. + (50. - var[2])^2  == (-2. + var[3])^2 
        },
        variabes
    ] 

does not yield solutions for any permutation of variables = {var[1],
var[2], var[3], var[4]} in which var[3] is the first element.

I'm desperate!  Any and all suggestions gratefully received!

--Chris Carlson
From: hearn@randvax.rand.org (Tony Hearn)
Newsgroups: sci.math.symbolic
Subject: New Versions of REDUCE Now Available
Date: 16 Aug 92 16:40:16 GMT
Article-I.D.: randvax.3721
Organization: RAND Corp., Santa Monica, Ca.


On July 15 this year, a new version of REDUCE, 3.4.1, was released.  Most
of the changes in this release are bug fixes.  However, updates of
packages in the network library are included, and there have been some
major improvements in the SOLVE code.

In response to requests from users who want to run REDUCE on their home
computers, Codemist Limited and the Konrad Zuse Center have released
personal versions of REDUCE for PC-class machines.  These implementations
support all REDUCE algebraic capabilities, but omit source files, among
other things, since these are readily available in the professional
versions running on office workstations. U.S. users can obtain these
versions from CALCODE Systems at a price of $295 per copy.  The addresses
of these and other distributors and a general description of REDUCE are
given in the REDUCE information package, which can be obtained by sending
the message "send info-package" to one of the network library servers,
reduce-netlib@rand.org, reduce-netlib@can.nl, redlib@elib.zib-berlin.de or
reduce-netlib@pi.cc.u-tokyo.ac.jp.  Further information on these servers
can be obtained by including the word "help" on a separate line in your
message.

Finally, a major hole in the machine coverage of REDUCE has been filled
with the release by Codemist of REDUCE 3.4.1 for the Apple Macintosh
series.  Both personal and professional versions are available.

Tony Hearn
Organization: University of Illinois at Chicago
Date: Friday, 21 Aug 1992 16:46:32 CDT
From: <U09207@uicvm.uic.edu>
Newsgroups: sci.math.symbolic,uiuc.mathematica,
Subject: Help on an integral

I am a relatively new Mathematica user. I will appreciate any comments and
suggestions on the following integral I am trying to evaluate:

           / H
           |
           | ( j beta/ r1^2 + 1/r1^3) exp( -j beta r1) dz
           |
          / -H
where r1=sqrt(z^2 + r^2 - 2 r z cos(theta)) and j=sqrt(-1).
This is a typical integral found in electromagnetics and antennas.
I tried to use Integrate[f,{z,-H,H}], but Mma responded with
Integrate[...] where ... was the integrand with the substitutions for r1 and j.
When I used NIntegrate[f,{z,-H,H}] and specified some typical values for
the constants, I got a value for the integral. But I would like to have a
closed form solution (or an approximate version). Does anyone have any
suggestions on this? Is there some kind of specialized Package available
for these type of problems.
I am using Mma on a Sparc2.

Thanks in advance.
Vahe Caliskan (vahe@uicbert.eecs.uic.edu)
Newsgroups: sci.math.symbolic,ba.seminars,ucb.seminars,ba.general,ucb.general