HDTenEightyP/NetNews huggingface dataset:
From: kristoff@genbank.bio.net (David Kristofferson)
Newsgroups: sci.math,sci.chem,sci.engr.chem,sci.engr.mech,sci.physics,sci.bio,bionet.journals.contents
Subject: Re: Survey on the Future of Electronic Information Distribution
Date: 27 Jul 92 20:59:41 GMT
Distribution: usa
Organization: GenBank Online Service

bashford@scripps.edu (Don &) writes:

>>>>>> In article <1992Jul25.044321.18144@news.acns.nwu.edu>, jlemay@nwu.edu (John LeMay) writes:

>jlemay>            ELECTRONIC SURVEY ON THE FUTURE OF
>jlemay>             DIGITAL INFORMATION DISTRIBUTION 

>jlemay> The Boston Consulting Group is conducting a study on the 
>jlemay> future of electronic publishing and distribution, and would 
>jlemay> like to get your input....

>This looks to me like a blatant commercial use of the net.

>Boston Consulting Group is asking us to spend our time to
>give them information, yet there is nothing in their post to
>indicate that the results of the survey will be made available to
>the net community.  Evidently they intend to sell the information
>as part of their consulting business.  In other words, we are being
>asked to make a collective effort for the sake of their profit
>rather than our own collective benefit.


Just to elaborate on propriety a bit, NSF has ruled (as I posted in a
BIONEWS posting about a year ago) that marketing surveys **could** be
posted to the net provided that they were in support of products used
for research and educational purposes.  I think that at the very least
we would need to get a better idea from Boston Consulting as to whom
the consumers of this information will be before deciding whether or
not the posting is appropriate.  For those who may not be familiar
with business 8-), Boston Consulting Group is one of the premier
consulting firms in the business world, and I would assume that they
are being well-paid by not insignificant clients for this study.

				Sincerely,

				Dave Kristofferson
				GenBank Manager

				kristoff@genbank.bio.net
Xref: sparky sci.math:9368 soc.culture.british:10451
From: dj@ssd.kodak.com (Dave Jones)
Newsgroups: sci.math,soc.culture.british
Subject: Re: American mathematical hegemony
Date: 21 Jul 92 15:13:48 GMT
Article-I.D.: pixel.1992Jul21.151348.29570
Organization: :noitazinagrO

In article <2176@sjfc.UUCP> dlm@sjfc.UUCP (Don Muench) writes:
>
>Look me up in London and I'll give ya all an authentic upstate NY
>(in particular, RAH-chisster) accent, in exchange for your inviting me and
>my wife to your home for dinner.  
>
To potential takers of this offer:

Be warned that "dinner" should include the wings of about a dozen
chickens, properly detached from the breast portions, plucked and deep
fried (no batter please), served with a sauce whose pungency would bring
a smile to the face of Red Dwarf's lone human occupant.  The meal should
be accompanied by short sticks of celery and Blue Cheese salad dressing.
I've never been able to fathom the origins or purpose of those last
items.

Do this at your own risk.

-- 
||))) If you build it                          )))))))))))))))))))))))))))))|
||))) They will cancel it   - Field of Dweebs. )))))))))))))))))))))))))))))|
||))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))|
||Dave Jones (dj@ekcolor.ssd.kodak.com) | Eastman Kodak Co. Rochester, NY   |
Newsgroups: sci.math
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Subject: Re: run the hailstone backwards...
Organization: Rose-Hulman Institute of Technology
Date: Thu, 23 Jul 1992 19:09:15 GMT


In article <1992Jul22.134601.6735@nmt.edu> jefu@akbar.nmt.edu (Jeff  
Putnam) writes:
> 
> I was thinking the other day about the hailstone numbers
> (or whichever name you prefer to use) and thought about running
> them backwards as a programming exercise.  I wanted to
> give the numbers n and m and produce the m smallest precursors
> (a precursor to n is a number k, such that f(k), f(f(k)), and 
> (stuff deleted)
> jefu <=> Jeff Putnam - New Mexico Tech <=> jefu@nmt.edu 
> "Overturn the dominant paradigm"

David Klarner at the University of Nebraska-Lincoln Computer
Science Dept. did some interesting work along these lines:  
Take some (affine) functions (in the case of this sequence a:n -> 2n, 
and b:n->(n-1)/3 are the functions: they're the inverses of the
functions used in the statement of the problem) and apply them
repeatedly to the integer 1.  Think of a and b as generating a 
semigroup.  Each way of composing these functions gives a word
in the semigroup.  Relations appear since the sets of precursors
intersect.  Klarner was able to show that certain sets of affine
functions would "generate" the integers in this fashion.  

All I have are pre-preprints, so I can't give a good reference.
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: Choice and Measurability:  Con(ZF+M+CC) is true
Date: 27 Jul 92 22:31:13 GMT
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: carabalo@fine.Princeton.EDU (David G. Caraballo)

In article <1992Jul24.161921.9655@Princeton.EDU>, carabalo@fine (David G. Caraballo) writes:
>I have received numerous emphatic objections on sci.math regarding 
>my claim, yet I will make it once again, with renewed support. [...]
>[letter from Solovay omitted]

Solovay's statements are accurate.  Yours weren't.  And yes, I'm a
platonist too, but truth is not the same as accuracy!
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Newsgroups: sci.math
From: djvelleman@amherst.edu
Subject: Re: You know, the integers (was: Re: Stupid question about FLT)
Organization: Amherst College, Amherst Mass.
Date: Wed, 22 Jul 1992 14:01:21 GMT

In article <1992Jul21.150931.22637@sics.se>, torkel@sics.se (Torkel Franzen) writes:
> In article <14h7srINNnjc@function.mps.ohio-state.edu> edgar@function.
> mps.ohio-state.edu (Gerald Edgar) writes:
> 
>    >"You know, THE SETS."
> 
>   More precisely, the cumulative hierarchy of sets. We learn about the
> natural numbers as generated from 0 by repeated application of the successor
> function at such an early age, at such a basic level, that it's all but
> impossible to make it clearer once we're spouting mathematics or philosophy.
> Set theory is a different matter: even today we can point to Zermelo's
> "Ueber Grenzzahlen und Mengenbereiche" as a presentation unequaled in
> clarity of the cumulative hierarchy of sets as a model of ZFC.

  I find the cumulative hierarch of sets very *unclear*--especially in
comparison to the integers.  My reaction to the statement, "You know, THE
INTEGERS" would be "Oh yes, of course, I know those."  I'm afraid my reaction
to "You know, THE SETS" is "No, I don't know them.  I know some of them, and
I have some vague intuitions about others, but at some point the picture gets
too fuzzy for me to be sure what we're talking aobut."

  Dan Velleman
  Dept. of Mathematics and Computer Science
  Amherst College

Newsgroups: sci.math
From: ap@levangerhs.no (Andrei Prasolov)
Subject: Re: ALGEBRAIC NUMBER ARITHMETIC
Keywords: Algebraic numbers, Arithmetic operations
Organization: Hogskolen i Levanger
Date: Wed, 29 Jul 92 16:38:30 GMT

   This is the answer to article 22109.

Operating with algebraic numbers one does not need
to know the interval. For arithmetical operations
it does not matter WHICH root of the (irreducible!) 
equation is taken. It does matter ONLY if you wants
to calculate the root (approximately, of course).
First prove a very simple theorem:
TH. A complex number x is algebraic iff there exists
    a fin-dim Q-subspace V of C s.t. xV lies in V.
PF. => V is generated by 1, x, x^2, ..., x^(n-1) where n is 
       the degree of the polynomial.
    <= x is a root of the characteristic polynomial of the
       multiplication by x : V -> V.

Now if V serves for x, and W serves for y, then
VW serves for xy, x+y, x-y. For 1/x just take the mirror reflexion
of a polynomial.

              Andrei PRASOLOV
              Hoegskolen i Levanger
              Kirkegt 1, 7600 Levanger
              Norway

Tel. 47-76-89157 (office), 47-76-89688+2304 (home)
Telefax 47-76-89155, e-mail AP@LEVANGERHS.NO
From: fry@tara.harvard.edu (David Fry)
Newsgroups: sci.math
Subject: Re: A Divisibility Problem
Date: 27 Jul 92 04:51:28 GMT
Organization: Harvard Math Department

In article <9207270341.AA10422@.euclid.uoregon.edu.> s42234@DUCVAX.AUBURN.EDU (M Johnson) writes:
>
>HI, I encountered a problem relating to the divisibility of a sum of integers
>as follows, and I wonder whether the following is true:
>
>Given  N > 1  positive integers, it is always possible to find a subset of
>of the set of N positive integers such that its sum is divisible by N.
>
>
>Is this true? Thanks for your answer.

I hope this isn't a homework problem, but...

This is true.  Imagine adding the numbers up one by one, and writing
down, after adding each number, the remainder left over when you divide 
the running sum by N.

1) If you ever get a remainder of 0, then you've found a subset that
is divisible by N.  Suppose that this never happens.

2) If you ever write down a remainder that you've written down before,
then you must have in the meantime added a subset that is divisible
by N.  Suppose that this never happens.

Then, when you finish, you must have written down N distinct remainders
(numbers between 0 and N-1), none of which is 0.  This is impossible,
so one of 1) or 2) must have happened.

David Fry                                  fry@math.harvard.EDU
Division of Applied Sciences               fry@huma1.bitnet
Harvard University                      ...!harvard!huma1!fry
Cambridge, MA  02138
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Lattice points in a polytope? (How to find all of them?)
Date: 25 Jul 92 14:52:30 GMT
Organization: U of Cambridge, England

In article <1992Jul24.132125.155015@ns1.cc.lehigh.edu>, fc03@ns1.cc.lehigh.edu (Frederick W. Chapman) writes:

>                   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?

I'm sure you can prove it by induction on M, but I think there's a nice
combinatorial proof, something like this:

What we are asking for is sequences of n non-negative numbers whose sum is
most M. Equivalently, n *positive* numbers whose sum is at most M+n (write
N instead of M+n from now on).

So, draw N dots in a row:

  .   .   .   .   .   .   .   .

and consider the spaces after the dots:

  . | . | . | . | . | . | . | . |

and observe that choosing n positive numbers adding up to at most N is
the same thing as choosing n of those |s; then x_i is the number of dots
between the i'th | and the one before, and the sum of the x_i is the number
of dots left of the right-most |.

How many ways are there to choose n |s out of those N? Answer: "N choose n",
which is the answer given.

I think this proof is even prettier than the result itself.
Newsgroups: sci.math
From: mgh3@po.cwru.edu (mike hurley)
Subject: Re: Hailstone sequences
Followup-To: sci.math
Organization: case western reserve u
Distribution: usa
Date: Fri, 24 Jul 92 16:48:33 GMT

In article <1992Jul24.152238.12742@williams.edu>, ddailey@williams.edu
wrote:
> 
> <deletion>
> >
> >The result that on the positive integers, the function
> >  f(x)=3x+1 or x/2

<deleted> 

> I first heard of these sequences from Stan Ulam and assumed they originated
> with him. Did they?
> 
> David Dailey

J.C. Lagarias in article in the MAA Monthly (Jan. 1985, p3) discusses
the origin of the problem. It would appear that the problem (or at
least a somewhat similar problem) is attributable to Lothar Collatz 
in the 1930's.
From: cliff@WATSON.IBM.COM ("Cliff Pickover")
Newsgroups: sci.math
Subject: Binary Undulants
Date: 24 Jul 92 21:00:16 GMT

"Binary undulants" are numbers of the form 2**N that undulate
(alternate) the adjacent digits 1 and 0 in their decimal expansion.

For example, 2**49 is the "highest quality" binary undulant so far
discovered because it has the undulating sequence 101010 in it:

2**949 =
4758454107128905800953799994079681792420032645310062268978469949811010102913
Here N = 949 is called an undulation seed of order 6, since it
gives rise to a 6-digit undulation pattern of adjacent 1's and 0's.

Are undulation seeds of higher order easily found using either
mathematical prowess or computation power?

Thanks Cliff   cliff@watson.ibm.com
Organization: University of Illinois at Chicago
Date: Wednesday, 29 Jul 1992 21:28:53 CDT
From: <U53644@uicvm.uic.edu>
Newsgroups: sci.math
Subject: Re: Why are Some Journals so Expensive?
 <1992Jul23.004719.29335@src.umd.edu> <92209.164013U53644@uicvm.uic.edu>
 <1992Jul29.165858.24646@Princeton.EDU>

No, the mathematical community is not the only available collection of
consumers.

The problem here is that people forget that the same presses that can be used
to put out Math books and journals, can be used to put out publications on any
other subject as well. If the publisher doesn't like the Math market, he can
expand into others (which will probably be more lucrative anyway). The Math
market, as dear to all of our hearts as it may be, is a piddling little market.
No flame here, there's just no polite way to say it - some people here think
that they have more clout than they actually have. (Collectively, the Math
community has darn close to none, economically or politically - how much can a
collection of a few thousand middle class people have ?)

And in case one point wasn't clear - I am not suggesting that the quality of
the contents is a minor issue (far from it). Just that frills like having
nice, glossy paper, staples along the side rather than on the front, and other
cosmetic issues are. If one can produce a cheaper journal by sacrificing that,
I see no reason to not do so.

                                                Joseph B. Dunphy
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Computational aids.  Was:  Re: An interesting limit problem.
Date: 29 Jul 92 13:43:51 GMT
Organization: Purdue University Statistics Department

In article <1992Jul29.000223.27339@massey.ac.nz> news@massey.ac.nz (USENET News System) writes:
>In article <1992Jul28.191037.28756@gdr.bath.ac.uk>, mapsj@gdr.bath.ac.uk (Simon Juden) writes:

>> >mattm@lure.latrobe.edu.au writes:

>> [stuff deleted]

>> >Will someone please back me up, and say that no one learning
>> >Calculus should ever be allowed anywhere _NEAR_ Mathematica?
>> >(When I say 'learning Calculus' I mean the first two or three
>> >semesters)

>> I could hardly agree more. Mathematica should be viewed as a fancy calculator
>> (which, essentially, is what it is). Very good and very useful, BUT undergrads
>> HAVE to learn basic concepts and manipulative skills before they're let near
>> it. Otherwise they will never become mathematicians, but rather computer
>> operators. Also they'll never FULLY understand subtle concepts or follow harder
>> arguments used in difficult proofs if they haven't had to do things with their
>> bare hands first. Like any good similar system Mathematica can be very helpful
>> to the professional mathematician. But I believe such gadgets can also do
>> irreparable harm unless used appropriately. I also believe children should not
>> be given calculators until they can do sums without them. But doubtless these
>> views will get me flamed...

>Since flames were explicitly invited :-), I'll start the ball rolling.

>I believe that it is far more important to understand the meaning of
>the operations of arithmetic than to be able to work through the
>algorithms like trained monkeys without having a clue why they work.
>(I know some bright spark will ask how many monkeys I've trained to
>do long division - well, no actual monkeys, but a lot of great apes :-)).

I must concur.  There is no evidence whatever that learning to compute
gives any understanding of what the computation does, and why it is done.

A six-year-old can understand the Peano postulates with little difficulty;
they do not mention anything except counting.  Then the understanding of
what the arithmetic operations mean can be taught.  Those who have
manipulative ability will learn the time-saving mechanics, and those
who do not are far better served by using calculators.

The situation is worse at the calculus level.  As I reported earlier, 
80% of the undergraduates, having had the full calculus sequence, could
not use their calculus for relatively routine problems on a take-home
examination in a probability course I taught.  

Only researchers and teachers need to know the mechanics.  It is useless
to know how to differentiate if one does not know what a derivative means.
No amount of memorizing formulas and tricks helps in understanding.
-- 
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)
Xref: sparky sci.math:9701 rec.puzzles:5377 comp.theory:1717
From: kodiak@sportster.ksu.ksu.edu (Bryan D. Nehl)
Newsgroups: sci.math,rec.puzzles,comp.theory
Subject: Re: Ten equals 9.9999....
Date: 31 Jul 92 02:50:35 GMT
Organization: Kansas State University
NNTP-Posting-Host: sportster.ksu.ksu.edu

Proof that 10 = 9.9999...

  x 	=  9.999...
10x 	= 99.999...
10x -x 	= 99.999... - 9.999...
 9x 	= 90.0
  x 	= 10

  10.0 = x = 9.999...

Bryan.

/* === Bryan Nehl ========== kodiak@Kodiakpc.Manhattan.KS.US =========+
[            USDA-ARS-NPA-WERU          ][ bdn@chepil.weru.ksu.edu    ]
[     913.532.6233 or 913.532.6495      ][ kodiak@matt.ksu.ksu.edu    ]
+_______________ ...!rutgers!matt.ksu.ksu.edu!kodiak _______________ */
From: jankins@Aero.org (Mark Jankins)
Subject: A Question About Fourier Transforms
Organization: The Aerospace Corporation
Date: Thu, 30 Jul 1992 19:55:15 GMT

I would like to know if the following is true, and if it is, does anyone have
a reference to a proof.

Suppose f is in L^2(R) (square integrable over the reals) and tf^(t)
is also in L^2(R), where f^ is the Fourier transform of f.  Then f is
differentiable a.e.

I would appreciate replies by e-mail.

Thanks
Mark Jankins
(jankins@aero.org)
From: lrf@silverblue.fac.cs.cmu.edu (Lee Furnival)
Newsgroups: sci.math
Subject: Mathematica Help
Date: 21 Jul 92 22:23:51 GMT
Article-I.D.: cs.1992Jul21.222351.160911
Distribution: usa
Organization: School of Computer Science, Carnegie Mellon


I am having trouble with this small part of a mathematica function I am
writing, maybe someone here can help...

I want to define a vector, call it v*, such that the components of v*
are defined by the partial derivatives of a group of other functions.
Here is what the code looks like at this point:

	While [i < n,
		v*[[i]] = D[f[x1,x2,...,xn],x[[i]]];
		i++ ]

So, what I WANT this part do (but it doesn't) is give me this: 
	
	v* = {(dx1/dx1,dx2/dx1,...,dxn/dx1),...,(dx1/dxn,dx2,dxn,...,dxn,dxn)}

but when I try to execute the While loop, I get back something like
"v* not assigned an immediate value".  To solve this, I have tried to define
v* before the loop by using Array and Table, but I still get the same thing.
My problem is that I want some way of indexing v* in the While loop so
that its components can be defined as the loop iterates.

I don't know if this makes any sense, so if you have any questions or know
exactly what I am talking about, then please e-mail me!

Lee Furnival
lrf@silverblue.fac.cs.cmu.edu
From: ow0a+@andrew.cmu.edu (Oswald Wyler)
Newsgroups: sci.math
Subject: Prime quadruplet records?
Date: 28 Jul 92 06:06:20 GMT
Article-I.D.: andrew.AeRJFQW00WBNI1kVwn
Organization: Carnegie Mellon, Pittsburgh, PA

A quadruplet of primes occurs when four out of five consecutive odd numbers
are prime.  Except for the first two quadruplets, 3,5,7,11 and 5,7,11,13,
all quadruplets of primes are of the form 30n + 11, 30n + 13, 30n + 17,
30n + 19.  It is well known that there are only finitely many prime
quadruplets.
Question 1.  Is the largest prime quadruplet known?
Question 2.  If not, what is the largest known prime quadruplet?
Question 3.  Up to which n (see above) are all prime quadruplets known?
From: burt@aupair.cs.athabascau.ca (Burt Voorhees)
Newsgroups: sci.math
Subject: Re: American mathematical hegemony
Date: 22 Jul 92 16:37:51 GMT


  "Things fall apart, the centre cannot hold..."
Newsgroups: sci.math
From: martin@math.rwth-aachen.de (  Martin Schoenert)
Subject: Re: Quadratic Residue Question
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen
Date: 22 Jul 92 11:44:27 GMT

In his article of 18-Jul-92 bumby@no.rutgers.edu (Richard Bumby) writes:

    Euler's expression for the Legendre symbol, which only makes sense if
    p is prime, can be calculated in O(log p) steps by repeated squaring.
    It is fairly common for textbooks in elementary number theory to note
    that reciprocity leads to a fast computation of Legendre symbols, but
    it turns out not to be significantly faster than Euler's criterion.

In  GAP (a system for computational  group  theory, which  supports large
integers) computing the value  of $J(n/m)$  where  $n$  and $m$  are both
about 100 digits long takes roughly  600 ms  on a DECstation  5120.   But
computing $J(n/m)$ using quadratic reciprocity takes only 6 ms.  Thus for
numbers of this size it is already about 100 times faster.

This  is to be expected.   As you note Euler's expression can be computed
in $O(log p)$ steps.  However, it is important to note that each of those
steps involves  a  multiplication of two numbers  with $O(log p)$ digits,
and therefore about $O((log p)^2)$ bit operations.  So evaluating Euler's
expressions takes a total of $O((log p)^3)$ bit operations.  On the other
hand the algorithm that  uses quadratic reciprocity also takes $O(log p)$
steps, but the numbers are getting smaller with each step.  The behaviour
is very similar to the Euclidean algorithm, and in fact the total time is
$O((log p)^2)$ (just like the Euclidean algorithm).

You continue:

    Thus quadratic reciprocity must be important for some other reason :-).

    -- 
    R. T. Bumby **  Rutgers Math ||   Amer. Math. Monthly Problems Editor
    bumby@math.rutgers.edu       || P.O. Box 10971 New Brunswick, NJ08906-0971
    bumby@dimacs.rutgers.edu     || Phone: [USA] 908 932 0277 * FAX 908 932 5530

Indeed.

Martin.

--
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,  +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, D 51 Aachen, Germany
Newsgroups: sci.math