HDTenEightyP/NetNews huggingface dataset:
From: jac@ds8.scri.fsu.edu (Jim Carr)
Newsgroups: sci.math
Subject: Re: Complex Hermitian eigenvalues
Date: 6 Apr 92 18:50:15 GMT
Reply-To: jac@ds8.scri.fsu.edu (Jim Carr)
Organization: SCRI, Florida State University

In article <8062@sun13.scri.fsu.edu> jac@ds8.scri.fsu.edu (Jim Carr) writes:
>In article <1992Apr6.165255.17487@gmuvax2.gmu.edu> semmett@gmuvax2.gmu.edu (Steve Emmett) writes:
>>
>>CH correctly gives the eigenvalues as +/- 1.  Now, put 1, say, in the
>>trace.  You would expect the eigenvalues to be 2 and 0.  However, CH
>>gives you -.618 and 1.618 (the golden mean).  There is something going
>>on that I don't understand.  
>
>I get that 
>                   | 1-E  i | 
>               det |        | = (1-E)(-E) - (i)(-i) = E^2 - E - 1
>                   | -i  -E |
>
>for the characteristic equation.  Now b^2-4ac = 1+4 = 5; E= (1 +/- sqrt(5))/2, 
>which is the result CH and the computer gave you.  

And then I sensibly added (but did not contemplate long enough) 

>                                                   Granted it is monday and 
>I havent done this by hand in a decade, but I do not see where you got 2 and 0.

Argh.  As soon as I hit return I realized what you meant and what must be 
your problem.  I interpreted your sentence as meaning "a" 1 on the diagonal, 
but you wanted it to have a trace of 1.  The former gives the result I got 
above so that must be what the computer thought you wanted from your input. 

It is the case that 
                         | 1-E  i |
                     det |        | = E^2 - 2E  giving 2 and zero. 
                         | -i 1-E |

All I can imagine is that you inadvertantly ran what I did by accident above 
instead of what you intended to run as a check. 

--
J. A. Carr                                    |  "The New Frontier of which I  
jac@gw.scri.fsu.edu                           |  speak is not a set of promises
Florida State University  B-186               |  -- it is a set of challenges."
Supercomputer Computations Research Institute |   John F. Kennedy (15 July 60)
From: einarr@ifi.uio.no (Einar Andreas R|dland)
Newsgroups: sci.math
Subject: Re: Series involving primes
Date: 7 Apr 92 16:28:49 GMT
Organization: Dept. of Informatics, University of Oslo, Norway
Originator: einarr@skinneggi.ifi.uio.no


In article <1992Mar31.200531.9836@linus.mitre.org>, bs@gauss.mitre.org (Robert D. Silverman) writes:
> In article <1992Mar30.173346.1602@ifi.uio.no> einarr@ifi.uio.no (Einar Andreas R|dland) writes:
> :However, I'd like to know if
> :
> :    oo
> :   ---      1 
> :   >    ---------
> :   ---  p(i) d(i)
> :   i=1
> :
> :converges.
> 
> Yes it does. We have the above sum equal to the following Stieltjes
> integral:
> 
> integral from 2 to infty of  f(x) d[pi(x)]  where f(x) is:
> 
> 	1
>  -------------------------
> x log^2 x (1 + O(1/log^k x))
>  
> f(x) is obtained by using the PNT. p(i) ~ i log i   and
>  d(i) ~ log i (1+O(1/log^k i))  for some constant k.
>  
> and pi(x) is the prime counting function.
> 
> The above integral is finite, so the sum is finite.
> 
> --
> 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"

Well, if d(i) ~ log i (1+O(1/log^k i)) there would, for instance, have to
be a finite number of prime couples (which, of course, there might ... I
don't know...). The difference d(i)=p(i+1)-p(i) does not increase regularly;
it bounces up and down in a seemingly irregular fashion. In short,
expressions including O(f) (for some function f) should not be
differensiated; at least not so recklessly.

This still leaves the question unanswered.


EinAR
-- 
.__________
|  Einar Andreas R|dland (Roedland)  ---  E-mail: einarr@ifi.uio.no
|  Mathematician (student)
|  University of Oslo, Norway (Northern Europe)
Newsgroups: sci.math
Subject: Re: the best known living mathematician (was: On knowing who's who)
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 9 Apr 92 21:36:53 GMT
Distribution: na
Organization: Beckman Instruments, Inc.

In <COLUMBUS.92Apr9163155@strident.think.com> columbus@strident.think.com (Michael Weiss) writes:

>	<COLUMBUS.92Apr9091341@strident.think.com> <a_rubin.702838383@dn66>
>	<22604@dog.ee.lbl.gov> <a_rubin.702846754@dn66>
>NNTP-Posting-Host: strident.think.com
>In-reply-to: a_rubin@dsg4.dse.beckman.com's message of 9 Apr 92 19:12:34 GMT

>In article <a_rubin.702846754@dn66> a_rubin@dsg4.dse.beckman.com (Arthur
>Rubin) writes: 

>   Erdos is probably the best known active mathematician, possibly the best
>   known living mathematician....

>Hmm... probably depends a lot on one's background.  Charlie Fefferman, Paul
>Cohen, Gerd Faltings, John Milnor, Persi Diaconis.... take your pick.  Add
>to the list.  (This could on for as long as "name your favorite
>textbooks").

I suppose the statement is ambiguous, but I meant best known within the
mathematics community....and I've met most of the ones on your list.  In
fact, there are "serious" discussion of a mathematician's "Erdos number". 
What is the Erdos number, you ask.

Well, consider the graph with points being mathematicians; there is an edge
between two points if the two mathematicians have a join paper.  (There is
some argument about whether the paper must have ONLY those two
mathematicans as authors, or not.)  Then X's Erdos number is the distance
from X to Erdos in that graph.
--
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.
Ich bin ein Virus.  Mach' mit und kopiere mich in Deine .signature.
From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math
Subject: Differentiable everywhere, f' continuous nowhere (was: An elementary, "naturally occurring" problem
Date: 7 Apr 92 09:16:24
Organization: Thinking Machines Corporation, Cambridge MA, USA
	<1992Apr1.201733.25777@linus.mitre.org> <3767@wet.UUCP>
NNTP-Posting-Host: strident.think.com
In-reply-to: sam@wet.UUCP's message of 7 Apr 92 07:45:32 GMT


Along the same lines, prove that there does not exist a function
f: [0,1] --> [0,1] which is differentiable everywhere, but whose derivative f' is
continuous nowhere.
Newsgroups: sci.math
From: mab@wdl39.wdl.loral.com (Mark A Biggar)
Subject: Re: An elementary, "naturally occurring" problem
Organization: Loral Western Development Labs
Date: Thu, 9 Apr 1992 20:12:15 GMT

In article <1992Apr9.133716.24702@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>In article <michaelg.702602167@Xenon.Stanford.EDU> michaelg@Xenon.Stanford.EDU (Michael Greenwald) writes:
>:bs@gauss.mitre.org (Robert D. Silverman) writes:
>:
>:>In article <1992Apr1.183035.14916@Princeton.EDU> carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:
>:>:Prove or disprove that there exists a bijection f:[0,1] -> [0,1] that is
>:>:not continuous at any point. 
>:>Doesn't the following function do it?
>:>f(x) = 1-x  if x is irrational.
>:>f(x) = x if x is rational.
>:>Bob Silverman
>:I'll admit I'm rusty, but I'm having trouble seeing why this isn't
>:continuous at x = .5.  I'm probably missing something obvious...
>Yup. This has been pointed out by numerous people. However, a trivial
>change fixes the objection. Just let f(1/4) = 1/2, f(1/2) = 3/4 & f(3/4) = 1/4

While working on the harder version of this problem (the version with the line
intersection condition) an interesting additional problem came up.  Notice
that all the solutions proposed so far involve partitioning the interval [0,1]
in to at least two dense sets, one of which in uncountable and the others all
countable (the above solution, has three sets, one uncountable and dense,
one countablely infinite and dense, and one finite {1/4, 1/2, 3/4}).

Question: is it possible to partition the interval [0,1] into a finite
number of uncountable sets that are all dense on the interval [0,1]?
If so, can you give a rule to determine partition membership?

--
Mark Biggar
mab@wdl1.wdl.loral.com
From: palais@binah.cc.brandeis.edu
Newsgroups: sci.math
Subject: Re: Math songs?
Date: 17 Apr 92 19:18:48 GMT
Article-I.D.: news.1992Apr17.191848.27113
Reply-To: palais@binah.cc.brandeis.edu
Organization: Brandeis University

This is an old Tom Lehrer ditty (sung to the tune, most 
appropriately, of "There'll be some changes made): 

You take a function of x and you call it y
Pick any x-nought that you care to try
make a little change and call it delta x
the cooresponding change in y is what we find next
Now you take the quotient, and most carefully
send delta x to zero, and I think you'll see
that what the limit gives us (if our work all checks)
is what we call dy dx, its just deee why  dee eks.
From: cet1@cl.cam.ac.uk (C.E. Thompson)
Newsgroups: sci.math
Subject: Re: Did Archimedes prove radial-projection measure-preserving?
Date: 14 Apr 92 15:05:26 GMT
Reply-To: cet1@cl.cam.ac.uk (C.E. Thompson)
Organization: U of Cambridge Computer Lab, UK

In article <SQUASH.92Apr13100014@cantor.msri.org>,
 squash@cantor.msri.org (Jonathan King) writes:
|> 
|> One of the standard area-faithful projections that cartographers use
|> is reminiscent of Mercator projection of the sphere:
|> 
|>   Take a cylinder whose HEIGHT and DIAMETER each equal the diameter of
|> the sphere.  Place the cylinder with axis vertical so that it is
|> tangent to the sphere, and so that the sphere's equator and the
|> cylinder's "equator" coincide.
|>   Project the sphere onto the cylinder radially outward from the
|> line-segment joining the North and South poles.
|> 
...
|> 
|> QUESTION:  This mapping, f, preserves area.  Did ARCHIMEDES prove
|> this? 

Effectively, but not, as far as I can see, explicitly.

|> I am aware that Archimedes wrote a book on the sphere and cylinder and
|> I have indeed found a *different* area-preserving map proven there
|> -the mapping to the disk that cartographers sometimes use for polar
|> regions.

I take it that here you are refering to Propositions 42 and 43 of Book 1
of `On the Sphere and the Cylinder', in which it is shown that the area
of a spherical segment (cap) is equal to that of a circle with radius
equal to the distance between the vertex and the edge of the segment.

|> There are three standard treatments of the known writings of
|> Archimedes and I have consulted two of them without success.

(I am relying on Heath.) What you want is *almost* covered in
Proposition 3 of Book 2 of OtSatC. Here is it is shown that dividing
a spherical surface (with a plane) in a given ratio is achieved by
dividing the axis in the same ratio. It is a very small step from that
to showing that your f preserves area *ratios*.

The missing ingredient, that a cylinder whose diameter and height are
both equal to that of a sphere has a surface area (excluding the bases)
equal to that of the sphere, is implicit in the proof of the corollary
to Proposition 34 of Book 1. However what that Corollary actually states
is that the surface area of the cylinder *including* the bases is 3/2
times that of the sphere, and that the volume of the cylinder is 3/2
times that of the sphere.

It would seem that this double appearance of 3/2 seemed so `neat' to
Archimedes that what appears to us to be the more natural equality
is somewhat overlooked. The inscription on his tomb is supposed to
have commemorated this result, and Cicero claimed to have rediscovered
and identified it thereby.

Chris Thompson
JANET:    cet1@uk.ac.cam.phx
Internet: cet1@phx.cam.ac.uk
From: sichase@csa2.lbl.gov (SCOTT I CHASE)
Newsgroups: sci.math
Subject: Re: shape functions
Followup-To: sci.math
Date: 16 Apr 92 19:29:13 GMT
Organization: Lawrence Berkeley Laboratory - Berkeley, CA, USA
Distribution: na
Reply-To: sichase@csa2.lbl.gov
NNTP-Posting-Host: 128.3.254.197
News-Software: VAX/VMS VNEWS 1.3-4   

In article <DAVIS.92Apr14195341@pacific.mps.ohio-state.edu>, davis@pacific.mps.ohio-state.edu ("John E. Davis") writes...
>Hi,
> 
>   I am looking for a way of quantizing shapes.  For example, imagine a
>probability distribution in one dimension p(x), 0 <= x <= 1.  Then, what are
>some simple parameters that classify various p(x) functions? Obviously, the
>moments
> 
>                   m_k :== integral from 0 to 1 of x^k p(x)
> 

People who study fluctuations sometimes define "factorial moments" defined as

Fq = (1/M) SUM(m=1 to M) <Nm*(Nm - 1)...(Nm - q + 1)>/<N>^q

where M is the number of bins into which you divide your distribution,
Nm is the number of counts in the mth bin, and <> denotes the average over
your "events".  In a typical case in which a particle physicist wants
to study multiplicity fluctuations, the events are invidual interactions,
and the rapidity interval is divided into the M bins.

The higher factorial moments are sensitive to multiparticle correlations,
which is to say, "spikes" in the distribution.

In addition, I would say that any appropriate set of Fourier parameters, using
whatever basis set is most useful, are a good set of shape parameters. 
I would choose a basis set whose lowest frequency member most resembles your
distribution, so that higher order terms are "corrections".  

-Scott

Newsgroups: sci.math
From: ramsay@unixg.ubc.ca (Keith Ramsay)
Subject: Re: Fermat's Last Theorem
Organization: University of British Columbia, Vancouver, B.C., Canada
Date: Thu, 9 Apr 1992 23:45:35 GMT

In article <1992Apr9.141933.26609@linus.mitre.org> 
bs@gauss.mitre.org (Robert D. Silverman) writes:

>:I understand that Erdos offered $3000 for a solution to the following
>:problem:  let {n_i} be a sequence of positive integers such that
>:SUM(1/n_i) diverges.  Then the sequence contains arithmetic progressions
>:of arbitrary length.  Try this one in your spare time.

>A mathematician far more competent than myself told me that he had looked
>at this problem and that there were much easier ways to earn $3K.
>Working at McDonald's, for example :-)

I would say so! Erdos is not wealthy (carrying his belongings with him
as he travels about, letting Graham keep some things) and always gets
his money's worth on these. He also seems to believe this one is true.

Some of the special cases of the problem give an indication of its
difficulty. For example, if the conjecture is true, then there are
arbitrarily long arithmetic sequences of primes, which we haven't
proven either.

Szemeredi's theorem is the case is when the n_i have positive density.
Szemeredi's proof can be found in Acta Arithmetica vol. 27; the
existence of 3-term progressions (still non-trivial) is proven in
_Ramsey_Theory_ by Graham, Rothschild, and Spencer. Just imagine
trying to show the same for a sequence which is thinning out....
--
Keith Ramsay            Even if a proof from given axioms can, in
ramsay@unixg.ubc.ca     principle, be found mechanically, is it not
                        often found in ways not all that different
                        from finding a new axiom? -Georg Kreisel
Newsgroups: sci.math
From: jwlee@comsci.yonsei.ac.kr (Jaewoong Lee)
Subject: recursive functions
Keywords: recursive
Organization: CS Dept., Yonsei University, Seoul
Date: Thu, 16 Apr 92 16:16:25 GMT


Dear netters,

I have a question about recursive functions.
We know that recursive sets are ubsets of non-recursive sets.
And alos that recursive enumerable sets are subsets of non-recursive
enumerable sets..

How about recursive functions? Are they a subset of nen-recursive functions?



--
Jaewoong Lee, AI Lab, Computer Science Dept., Yonsei University,
              Seoul, 120-749, Korea
Phone ) +82-2-361-2713 (lab) +82-2-401-4569 (home)
E-mail) jwlee@comsci.yonsei.ac.kr (Internet) JWLEE@KRYSUCC1 (BITNET)
Newsgroups: sci.math
From: luzeaux@roma.berkeley.edu (Dominique Luzeaux)
Subject: Re: integral
Reply-To: luzeaux@roma.berkeley.edu (Dominique Luzeaux)
Organization: University of California
Distribution: usa
Date: Thu, 9 Apr 1992 20:18:19 GMT

In article <1992Apr9.063002.14913@unlinfo.unl.edu>, dbeattie@cse.unl.edu
(Derek Beattie) writes:
|> I need some help.  Can someone integrate 1/(x^4+1)dx.  You have
|> to factor the bottom and use partial fractions but I can
|> get no further. You can change x^4+1 into (x^2+1)^2-2x^2 and then
|> that is a difference of squares and you can factor that then use
|> partial fractions and then I am lost....

1/(x^4+1) = \frac{(1-\sqrt{2})/2}{x^2(1-\sqrt{2}+1}+
            \frac{(1+\sqrt{2})/2}{x^2(1+\sqrt{2}+1}
this looks like
1/(x^2 + a^2)  and 1/(x^2 - a^2) where a is some constant
In the second case, you factor once again and get things like 1/(x-a_1)
and 1/(x-a_2)
which yield logarithms after integration
In the first case, you take y=x/a and have something like 1/(1+x^2)
which yields
Arctan x after integration

	Dominique Luzeaux
From: mayne@sun16.scri.fsu.edu (Bill Mayne)
Newsgroups: sci.math
Subject: Combinatorial choice function (?)
Date: 10 Apr 92 19:29:49 GMT
Reply-To: mayne@sun16.scri.fsu.edu (Bill Mayne)
Organization: SCRI, Florida State University

What is the proper name and symbol for the function which gives
the number of distinct ways to choose i objects from a set of N
objects, where order is not important? C(i,N)=N!/(i!*(N-i)!),
e.g. there are (52*52*50*49*48)/(1*2*3*4*5) different 5 card
poker hands (five cards selected from 52, order not important).

I vaguely recall that in printed texts it is written something
like ${C_N}^i$ (my attempt to render C subscript N superscript i.)

Is there a name or standard symbol for the product all
C(i,N), where i=1 to N? How do you characterize the growth of this
function? (It seems to be faster than O(N!).)

Be gentle. This isn't my field, but is relavent to a problem
I'm working on.

Bill Mayne
Newsgroups: sci.math
Subject: Re: Paul Erdos
From: harper@kauri.vuw.ac.nz (John Harper)
Date: Mon, 13 Apr 1992 21:46:04 GMT
Organization: Victoria University of Wellington, New Zealand

In article <ARNDT.92Apr13153433@napoleon.c3consult.comm.se> arndt@c3consult.comm.se (Arndt Jonasson) writes:
>
>Since no one else did: let me just add that the correct spelling of
>his name isn't "Erdos", but since the proper glyph doesn't have an
>ASCII representation (nor an ISO Latin-1 one), "Erdos" will have to do
>in many circumstances. The proper spelling has two "grave accents" on
>top of the 'o'. I think (but am not certain) it would be "Erd\Hos" in
>TeX. 

Acute accents. The TeXbook p 52 says "long Hungarian umlaut"; p 309 gives 
P\'al Erd\H os
unless you wish to use typewriter type in which case it's 
{\tt P\'al Erd{\bf\H{\tt o}}s}

>Something not very close to "Airdersh" (second syllable long).

But the "r" in "ersh" is not to be pronounced (i.e. treat the "r" in 
"Air" as in American or Scottish English, the other "r" as in RP or
what used to be called the Queen's English).

John Harper Mathematics Dept. Victoria University Wellington New Zealand
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: electronic math seminar?
Date: 8 Apr 92 18:03:44 GMT
Article-I.D.: camelot.ARA.92Apr8130344
Distribution: sci
Organization: M.I.T. Artificial Intelligence Lab.


I am currently living in a place where I can conveniently use the facilities
of a nearby university, and I am grateful for that. However, the interests
of the department are so different from mine that it appears unlikely that
I can find anyone to attend some lectures I was thinking of giving. Indeed,
almost all of the mathematical conversations I have had for almost two years
have been over the networks (I haven't been here for two years, but the
situation was essentially the same at the last place I was).

On the other hand, I know that on many computer systems at universities,
they have a party line program, which might go by many names, e.g. PHONE
or TALK. Let's call it PHONE.

Anyone on such a system who types PHONE is connected to everyone
else who has typed PHONE and what one types appears on the screen of
all the other people, and reversely. Actually, PHONE usually has channels
so that people can separate into disjoint discussion groups and it is also
possible to make them private. 

There is no reason why such a discussion group could not discuss mathematics,
using TeX notation if necessary. So electronic seminars are entirely 
feasible. There are a few problems with this idea, though.
(1) You want to make it open to the "public", i.e. anyone wanting to attend
    the seminar should be able to log in to do so.
(2) You don't want an arbitrary time limit on the meeting.
(3) It would be nice to be able to save a transcript of the seminar.
(4) It would be nice to be able to feed previously prepared files into
    the discussion.
(5) One should be able to control the scrolling of stuff as it goes by,
    so it doesn't go by too quickly.

Cleveland Freenet answers question (1), but there is a 1 hour time limit on
logins. I don't think it meets the others.

I don't think this would be hard to implement. All that is required is
for some sites to offer their hospitality to host such events and to install
the appropriate software. 

If anyone wants to pursue this idea further, I would be glad to hear from
them.

Allan Adler
ara@altdorf.ai.mit.edu
Newsgroups: sci.math
From: peppe@ipgaix.unipg.it (G. Vitillaro)
Subject: Last Fermat Theorem
Summary: Fermat Theorem and Circle
Keywords: Fermat
Organization: Universita' di Perugia
Date: Sun, 12 Apr 1992 21:34:16 GMT

 Recently I saw many news related to last Fermat Theorem.
I have a degree in math and some time I spent my time thinking
(from a philosophycal point of view I believe) at math questions.

Then let me to make my excuses in advance about any stupid and
ingenuos thing I will say.

Well if I remember correctly the Fermat assumption is that the
equation
		x^n + y^n = z^n             (1)
have not positive integer (x,y,z in N) solution for any n>2.
Well if you rewrite this equation as
               (x/z)^n + (y/z)^n = 1        (2)
dividing all the equation by z the assumption become that 
(2) have no solution p,q belonging to Q+ (rational positive
numbers) such that
		p^n + q^n = 1               (3)
for any n>2.

Well this is equivalent to state that the equations
		x + y = 1
		x^2 + y^2 = 1
are the only of the class x^n + y^n = 1, n>=1 with positive
rational (p,q) solution. In some sense both of them represent
a circle (a straight line is only a particular case of the circle,
is that not true?).

The conclusion is that Fermat stated some type of 
geometrical/topological fact: from the class of curves
x^n + y^n = 1 on the euclidean plane R^2 only the straight line
and circle have rational positive points (p,q) in Q+ x Q+.

The question is: have this any sense, do you already know the
geometrical (if even exists) fact that cause this consequence?

Thank in advance,
      G.Vitillaro
-- 
Giuseppe Vitillaro - IBM SEMEA      |  E-Mail : peppe@ipgaix.unipg.it 
University of Perugia Italy         |  06100 Perugia  Phone:+39.75.585-2200
---------------------------------------------------------------------------
All comments/opinions are mine and don't represent those of IBM
Newsgroups: sci.math