HDTenEightyP/NetNews huggingface dataset:
Newsgroups: sci.math
Subject: HELP!
From: broggi@prsun0in.cineca.it (Alberto Broggi)
Date: 6 Oct 92 19:28:24 +0100
Distribution: world

        I'm desperately seeking an article about mathematics
        for a friend of mine.
        I definitely do not know what it is about, but I do know the
        complete reference:

        The actual name of the magazine is:

                'Acta Scientiarum Szegedensis'

        It was:

        T.RADO`,   'Uber der Begriff den Riemanschan Flache'
                   Acta Litt. Scie. Szeged, 1925, 2, pag.101-121


        IF ANYBODY HAS ACCESS TO THIS MAGAZINE, PLEASE LET ME KNOW.

        Thanks in advance,

                                        A.Broggi



-- 
  Alberto Broggi    
  Dipartimento di Ingegneria dell'Informazione   phone: +39 (521) 580707
  Universita` degli Studi di Parma                 fax: +39 (521) 580723
  43100 Parma - Italy                    E-Mail broggi@prsun02.cineca.it
From: thf@cbnewsm.cb.att.com (thomas.h.foregger)
Newsgroups: sci.math
Subject: Re: Z <-> Q formula
Summary: simple bijection from Q to Z
Date: 30 Sep 92 03:36:58 GMT
Organization: AT&T

In article <1992Sep27.085952.7534@Csli.Stanford.EDU>, hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
> 
> # This may be a FAQ, but...
> 
> The set of intergers Z and the set of rationals Q have the same cardinality,
> i.e. there exist a bijection between them.  So we all know that.
> But then, is there a "simple" formula that expresses any such mapping?
> You may use any device you please, but note that the subjective little
> "simple" is the criterion.
> 
> -Yuzuru Hiraga


Let a and b be relatively prime and not zero.
if a and b are positive send a/b to a/b, where
a/b on the right is an integer in base 11 notation (/ is one
of the digits)
Send 0 to 0.
If one of a or b is negative send a/b to -a/b,
where again we use base 11 notation on the right.

I think this solution or something like it appeared
in the American Math. Monthly not too long ago.

It also has the nice feature that the map is easily invertible.
tom foregger
From: dak@kaa.informatik.rwth-aachen.de (David Kastrup)
Newsgroups: sci.math
Subject: Re: Fourier bignum multiplication
Date: 26 Sep 92 09:56:04 GMT
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen

torbenm@diku.dk (Torben AEgidius Mogensen) writes:


>I was told that you would have to have about a million bits in your
>number before the FFT method is better than the traditional O(x^2)
>algorithm.
Rubbish.
FFT has O(n lg n) half butterfly operations that may be say 10 times
more complicated. So the magic size would be 10  n lg n = n^2 approx.
16. But FFT-like and similar algorithms (polynomial interpolation)
can be used to speed up ANY multiplication, the small multiplications being
of special interest. And there is a very plain, simple method to
reduce almost trivially to O(x^1.5) the traditional Algorithm.
See Knuth, Seminumerical Algorithms from the Art of Computer Progr.
What else.
Newsgroups: sci.math,rec.puzzles
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: EASY NUMBER SEQUENCE PUZZLE...
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Fri, 2 Oct 92 03:57:39 GMT

In article <1992Oct1.175503.10957@kth.se> TordM@VanA.PhySto.SE writes:
>In article <aet.717945133@munagin>, aet@mullian.ee.mu.OZ.AU (bert thompson) writes:
>
>>25, 1729, 635318657, ???
>>what's ??? ?
>
> could it be 0 ? is there anybody but me who thinks these kind of
>"puzzles" are totaly non-interesting?? the "???" could be just ANY number,

I find them uninteresting too, probably just 'cause I'm not too good
at them.  But there *are* better and worse answers to these things and
you know it - the very fact that you wrote 0, an "unlikely" answer, 
shows you know that there is some pattern to this thing.  

I also don't like football, but I don't complain, "What's the point
of running around with that prolate spheroid?  What difference does
it make which end of the field it's near?"
From: davidb@triangle.Berkeley.EDU (David T. Blackston)
Newsgroups: sci.math
Subject: Easy (?) probability question
Date: 1 Oct 1992 20:10:17 GMT
Organization: University of California, Berkeley
Distribution: world
NNTP-Posting-Host: triangle.berkeley.edu

Hi all,

Here's a problem that came up recently in an analysis of a sorting
algorithm.  It seems really obvious, but I can't come up with a nice
proof and I was wondering if anyone out there had one.

Let X be a random variable, and define the random variable Z by
Z=min{X, k} where k is some constant.  Show that var(Z) <= var(X).

I have shown this in the case where X is a bounded, integer valued r.v.,
but the proof is messy and I feel that there has to be a better way.
Is there?


					David Blackston
					davidb@cs.berkeley.edu

Newsgroups: sci.math
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: "Solvability" in Chess
Date: Tue, 29 Sep 1992 15:28:49 GMT
Distribution: sci
Organization: Security APL, Inc.

In article <7468@armltd.uucp> dseal@armltd.co.uk (David Seal) writes:
>In article <1992Sep28.155640.152096@Cookie.secapl.com>
>frank@Cookie.secapl.com (Frank Adams) writes:
>>In article <1992Sep25.014035.1269@cbnewsl.cb.att.com> ed@cbnewsl.cb.att.com (edward.m.hummel) writes:
>>>I think this argument rest on the assumption that chess is a zero sum game.
>>>How do you prove that?  Suppose the value of an infinite game is less than
>>>a draw for each player (but more than a loss), then neither player
>>>would choose a draw *with best play*.
>>
>>As noted, this was intended to read "more than a draw for each player (but
>>less than a win)".  However, it is actually closer to correct as written:
>>the value of an infinite game *is* less than a draw; not as specified by the
>>rules, but in fact: no one wants to spend the rest of their life playing a
>>single game of chess.  This isn't *in* the rules for chess; it is *assumed*
>>by them.
>
>In normal circumstances, I would agree with you. But it isn't provable, and
>isn't even true in abnormal circumstances. Imagine a game of chess with the
>following playing conditions:
>
>[...]

This is chess; this is chess plus strange external conditions.

My key point is the last sentence: that infinite games are draws (and that
chess is a zero-sum game) are not *in* the rules for chess, but are
*assumed* by those rules.
Newsgroups: sci.math
From: berling@infko.uni-koblenz.de
Subject: references on algebraic geometry
Keywords: algebraic geometry, geometry
Organization: University of Koblenz, Germany
Date: Fri, 02 Oct 1992 09:36:41 GMT

Hello!

I'd like to get the full reference (title, publisher, ISBN)
of G. Salmon's books on algebraic geometry.

Thank you for your help!

Roland


+----------------------------+-----------------------------------------+
|                            |                                         |
|  Roland Berling            |  Tel.:   +49 261 9119 436               |
|  Universit"at Koblenz      |  Fax.:   +49 261 9119 499               |
|  Institut f"ur Informatik  |  e-mail: berling@infko.uni-koblenz.de   |
|  Rheinau 3-4               |                                         |
|  D-5400 Koblenz            |                                         |
|  Germany                   |                                         |
|                            |                                         |
+----------------------------+-----------------------------------------+



Newsgroups: sci.math,sci.logic
From: torkel@sics.se (Torkel Franzen)
Subject: Re: ZFC etc. (was Re: Report on Philosophies of Physicists)
Organization: Swedish Institute of Computer Science, Kista
	<TORKEL.92Sep26103351@bast.sics.se>
	<1992Sep28.154237.148977@Cookie.secapl.com>
Date: Mon, 28 Sep 1992 18:55:55 GMT

In article <1992Sep28.154237.148977@Cookie.secapl.com> frank@Cookie.secapl.
com (Frank Adams) writes:

   >I'll go you one better: this is a *better* thing to say.  It's still a very
   >different thing from a simple "that happens to be false", which disproof of
   >the other conjecture would have elicited.

  Saying that a disproof of "all formulas have a disjunctive normal
form" would only have elicited a reaction on the lines of "that
happens to be false" is not really by itself an intelligible
description of a possible scenario. Are you stipulating that
"classical propositional logic" means the same in your hypothetical
world as it does here and now? And if so, are you not assuming that it
could still be proved in your hypothetical world that every formula
does have a disjunctive normal form? Then the only difference
between "propositional logic is inconsistent" and "there is a formula
in propositional logic that has no disjunctive normal form" would be
that people tend to react automatically to "propositional logic is
inconsistent" as "catastrophic". The actual inconsistency in our
reasoning would be exactly the same in the two cases.  Or are you
assuming that "classical propositional logic" means something different in
your hypothetical world? Then there is not necessarily anything even
mildly disturbing, in that world, about propositional logic being
inconsistent.
From: aaronm@cybernet.cse.fau.edu (aaron meyerowitz)
Newsgroups: sci.math
Subject: Re: Z <-> Q formula
Date: 1 Oct 92 01:28:19 GMT
Organization: Cybernet BBS, Boca Raton, Florida

hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
> The set of intergers Z and the set of rationals Q have the same cardinality,
> i.e. there exist a bijection between them.  So we all know that.
> But then, is there a "simple" formula that expresses any such mapping?
> -Yuzuru Hiraga

My favorite is this: Use the fact that any rational has a unique
prime factorization. Map p^(2n) to p^n and p^(2n-1) to p^(-n).
Extend multiplicitively. Hence 5400=2^3 3^3 5^2 maps to
2^(-2) 3^(-2) 5^1 = 5/36. Preserve sign to deal with negatives.
Newsgroups: sci.math
From: sassoj@rs6412.ecs.rpi.edu (John J. Sasso Jr.)
Subject: Lebesgue measure in Fractal Dimension
Organization: Rensselaer Polytechnic Institute, Troy NY
Distribution: usa
Date: Mon, 5 Oct 1992 22:22:56 GMT

Newsgroups: sci.math,
Subject: sci.math.research, sci.fractals
Summary: 
Followup-To: 
Distribution: usa
Organization: Rensselaer Polytechnic Institute, Troy NY
Keywords: Lebesgue measure, Cantor set, Fractals, Hausdorff dimension


   In Real Analysis, I learned that the set {x : 0 < x < 1} has Lebesgue 
measure 1 for the Lebesgue measure defined in R^1, but is 0 for the Lebesgue
measure defined in R^2 (thus, the measure of a set is relative to what 
space, i.e., R^1, R^2, etc..., the Lebesgue measure is defined in).

   Now, the Cantor set has Lebesgue measure 0 for the Lebesgue measure defined
in spaces R^1, R^2, etc...  However, the Cantor does have a fractal dimension
(or Hausdorff dimension) between 0 and 1 (I cannnot recall the value), and
I'll call this dimension Df.  Is it possible to define a Lebesgue measure in
a space of dimension Df (R^Df) such that the measure of the Cantor set is a 
value other than 0?

                        John



Newsgroups: sci.math,sci.math.symbolic
From: edp@math.zk3.dec.com (Eric Postpischil)
Subject: Re: Integral Tables
Reply-To: edp@math.zk3.dec.com (Eric Postpischil)
Organization: Digital Equipment Corporation
Date: Wed, 30 Sep 1992 14:02:36 GMT

In article <4528.Sep2920.00.2492@virtualnews.nyu.edu>, brnstnd@nyu.edu
(D. J. Bernstein) writes:

>It's been long enough since I used Macsyma that I don't recall whether
>it would automatically handle range checks on the definite integrals you
>mention; certainly it was never as ignorant as Mathematica.

Here's a plug for Derive; I've been checking the examples given in this
discussion in Derive on the HP-95 palmtop, and Derive seems to do very
well in applying only valid reductions.  It does not check integrals for
singularities inside the bounds of a definite integral, and it warns of
this in the manual, but it otherwise appears quite robust.

Interestingly, I believe it accomplishes this _without_ range checks on
the definite integrals.  For example, the integral of sqrt(x^2) dx
simplifies to x|x|/2.  This expression is valid regardless of whether x
is positive or negative.  If it is given as a definite integral from a
to b, the result is b|b|/2-a|a|/2 unless Derive can figure out that b or
a is positive or negative, in which case the result may be (when both
are non-negative) b^2/2-a^2/2.  But this simplification comes after the
integral has been resolved; it is simply a general transformation that
is known to be valid, not a range check on the integral.


                                -- edp (Eric Postpischil)
                                "Always mount a scratch monkey."
                                edp@alien.enet.dec.com
From: wsadjw@rw7.urc.tue.nl (Jan Willem Nienhuys)
Newsgroups: sci.math
Subject: Re: Mattel says "Math is tough."
Date: 5 Oct 92 11:28:37 GMT
Reply-To: wsadjw@urc.tue.nl
Organization: Eindhoven University of Technology, The Netherlands

In article <92278.183626RVESTERM@vma.cc.nd.edu> RVESTERM@vma.cc.nd.edu writes:
>uh, as a math grad student, i must agree with barbie.

Would people get upset if barbie said "Running marathons is tough"?
Or "relativity is tough"?  Or "Homer is tough"?
Or is it that nowadays in North American culture anything more advanced
than 1+1 is considered (a) math (b) tough ?

JWN
Newsgroups: sci.math
From: tao@potato.princeton.edu (Terry Tao)
Subject: Re: Irrational?
Originator: news@nimaster
Organization: Princeton University
Date: Tue, 6 Oct 1992 16:20:43 GMT

I think the english version of irrational, meaning nonsensical, comes from the
Pythagorean point of view that ratios of integers were divine, for example
music tones should be perfect ratios.  Something that is rational (i.e. sensical
can be divided up into manageable bits (rationed), whereas something irrational
cannot be cut up and so is perverse.

Terry
From: watson@clark.edu (Dennis Watson)
Newsgroups: sci.math
Subject: Re: Another Cute Math Joke (explanation of)
Date: 6 Oct 92 22:26:09 GMT
Article-I.D.: clark.1992Oct6.222609.24994
Organization: Clark College, Vancouver, Wa.  USA

In article <a_rubin.718387119@dn66> a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
>In <1992Oct4.205137.29695@clark.edu> watson@clark.edu (Dennis Watson) writes:
>
>>Sometime back someone posted the following  joke?   
>
>>\epsilon < 0   
>
>There's a \delta for every \epsilon,
>That's a fact the can be relied upon,
>There's a \delta for every \epsilon,
>...and now and again, there's also an \n.
>
>But one condition I musta give,
>The \epsilon must be positive,
>A lonely life all the others live,
>.
>.
>.
>
>-- Tom Lehrer
>
Ok, now that I have had the "joke" explained to me, here is my reason
for asking:  I don't have a typesetter program that uses \anything
in that I have a mac and don't need weird precharacters so thought
the "joke" was related to "slash-epsilon-less-than-zero" and tried
tipping it on the side to see a face :-) or something!!  I think
I like the vector X scalor joke better!! I know, it is all my fault!!
I am used to jokes that are funny!  I will try better next time!

:-) watson@clark.edu
Newsgroups: sci.math,rec.puzzles
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: EASY NUMBER SEQUENCE PUZZLE...
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Fri, 2 Oct 92 20:16:33 GMT

In article <1992Oct2.113505.22990@kth.se> TordM@VanA.PhySto.SE writes:
>In article <1992Oct2.035739.14663@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:

>>>>25, 1729, 635318657, ???
>>>>what's ??? ?

>>> could it be 0 ? is there anybody but me who thinks these kind of
>>>"puzzles" are totaly non-interesting?? the "???" could be just ANY number,

>>I find them uninteresting too, probably just 'cause I'm not too good
>>at them.  But there *are* better and worse answers to these things and
>>you know it - the very fact that you wrote 0, an "unlikely" answer, 
>>shows you know that there is some pattern to this thing.  

> I see the pattern
> x_n=(-1905950809n^3+13341650526n^2-26683285691n)/6+2541264354
> which would yield the above series AND x_4=0

I too can fit polynomials to points, but I know (as do you) that that's
not what's intended.  It is a GAME.  But there is a serious point here.
You may spend most of your time on math, but in other sciences one is
often trying to guess what the world will do next.  This is a problem
rather analogous to guessing the next digit - one seeks a maximally
parsimonious and elegant fit to the data, not just "any old" fit.  If
you try to fit everything with polynomials, nature will snicker at you
because you don't get the point.  

I hope everyone knows the "grue" paradox.  "Grue" means "green until the
year 2000 and blue afterwards."  Emeralds are grue... they've always
been grue... so by induction (empirical, not mathematical) they should
always remain grue.  Right?  Wrong!  But why?  Well, our only answer is
that "grue" is an "unnatural concept," a crummy way of slicing the
world.  But what does THAT mean?  Tord here could insist on his right to
use "grueness" while we use "greenness" and argue that it is just as
correct to guess that emeralds will all be grue after the year 2000 as
it is to guess that they'll be green.  If he wishes to be stubborn we
would have a lot of trouble convincing him he's wrong - just as it would
be a lot of trouble convincing him his continuation of the series above
is wrong.  And yet, it's wrong.  Philosophers are still busy trying to
explain why.  Reality has a "grain" to it and you have to learn how to
cut along the grain, that's all I can say.

> these series are like having a cross-word that would allow ten different
>letters in each tiny square... "fill it in like you wish! if you fill it
>in like the author wants it you win the car!"

>>I also don't like football, but I don't complain, "What's the point
>>of running around with that prolate spheroid?  What difference does
>>it make which end of the field it's near?"

> why don't you? it feels good to bitch about nothing every now and then!
>you know what they say "supressed feelings lead to frustration and anger"
>:-)

Thanks, I just took the opportunity to complain about your complaints.  :-)
From: dak@tabaqui.informatik.rwth-aachen.de (David Kastrup)
Newsgroups: sci.math
Subject: Re: Easy (?) probability question
Date: 3 Oct 92 16:18:20 GMT
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen

davidb@triangle.Berkeley.EDU (David T. Blackston) writes:

>Hi all,

>Here's a problem that came up recently in an analysis of a sorting
>algorithm.  It seems really obvious, but I can't come up with a nice
>proof and I was wondering if anyone out there had one.

>Let X be a random variable, and define the random variable Z by
>Z=min{X, k} where k is some constant.  Show that var(Z) <= var(X).

>I have shown this in the case where X is a bounded, integer valued r.v.,
>but the proof is messy and I feel that there has to be a better way.
>Is there?
Well suppose X is realvalued. Then let's arbitrarily set k to 0
(variance does not change under shifting, so we could always shift
our original distribution so that this is the case, and shift back
afterwards). Let's call the new variable X'. Now we introduce
the variable X^+, which shall be X under the constraint X>0,
and X^-, which shall be X-X^+.

Some manipulation suffices to show that
var(X') = var(X) - var(X^+) + 2 E(X^+) E(X^-)
Now var(X^+) is positive, E(X^+) positive, E(X^-) negative, so
var(X') must be smaller than var(X).

Since I am not familiar with complex variances, I cannot make any
guesses.
Newsgroups: sci.math
From: dak@kaa.informatik.rwth-aachen.de (David Kastrup)
Subject: Elegant proof to Easy (?) probability question
Organization: Rechnerbetrieb Informatik  /  RWTH Aachen
Date:  5 Oct 92 01:28:14 GMT

demanded was proof that var(min(X,k)) < var(X).
Define a boolean valued variable with Y = X<k.
Now for real X we have
var(X) = E(var(X|X<k)) + var(E(X|X<k))
And that is
prob(X<k) var(X|X<k) + prob(X>=k) var(X|X>=k) +
prob(X<k) prob(X>=k) (E(X|X<k) - E(X|X>=k))^2
Now if we replace X|X>=k by k, var(X|X>=k) vanishes,
and E(X|X>=k) rises to k, so that the squared value decreases.

The above formula is still valid if you replace X<k by X in any set, and
X>=k by X not in that set.

Don't ask about complex variances. I do not know them. It might be that the
above formula holds, with squares replaced by squares of absolute
values, but I am only guessing. Looks not improbable, though.
Newsgroups: sci.math
From: cyeomans@ms.uky.edu (Charles Yeomans)
Subject: Re: Mattel says "Math is tough."
    <1992Oct5.191034.27195@Csli.Stanford.EDU>
Date: Tue, 6 Oct 1992 15:02:56 GMT
Organization: University Of Kentucky, Dept. of Math Sciences

In article <1992Oct5.191034.27195@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>>:         Mattel has just released a new product. A Barbie doll that says
>>: "Math is tough".
>>
>>What would you have her say?: 
>
>Not what I'd have her say, but if all of us filed a protest to Mattel,
>it'll certainly be changed to "Mathematicians are tough".	:-)
>
>-YH

I would have Barbie say "Getting a job in math is tough"

Charles Yeomans
cyeomans@ms.uky.edu
yeomans@austin.onu.edu

Newsgroups: sci.math,rec.puzzles
From: ferenbaugh-charles@CS.YALE.EDU (Charles Ferenbaugh)
Subject: Re: Conway's Recursive Sequence
Organization: Yale University Math Dept.
Date: Fri, 2 Oct 1992 18:56:24 GMT


In article <1992Sep30.220851.15985@husc3.harvard.edu> kubo@zariski.harvard.edu (Tal Kubo) writes:

>I am looking for information about a recursive sequence
>defined by John Conway in 1988:

>   a(n)= a(a(n-1)) + a(n-a(n-1))        for n>2
   
>   a(1) = 1, a(2)=1

>         .... Pointers to relevant literature, or
>any information whatsoever, would be most welcome.


I spent several hours playing with this sequence yesterday, and came up
with the following pretty way of presenting it:

Let b(n) = a(n) - a(n-1).  Then b(n) = 0 or 1 always (easy result).
Write + for 1.  Then b(n) obeys the following pattern:

  n=3,4                         +    0

  n=5..8                     +    +0    0

  n=9..16               +    ++0      +00    0

  n=17..32          +   +++0    ++0+00    +000    0

                     4     3  2   2   2   2  3     4
  n=33..64      +   + 0   + 0+ 0+0   + 0+0 +0    +0    0

                   5                                 5
  n=65..128  +    + 0          ...etc...           +0      0

where each string is the concatenation of the two above it,
a la Pascal's triangle.

I checked with Conway by email, and he suspects that this is probably
equivalent to one of the combinatorial presentations already known.
But at any rate, it's prettier than the presentation in the Mallows
article.

- Charles Ferenbaugh
  crf@math.yale.edu
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
Newsgroups: sci.math
Subject: Re: Roman Numerals
Date: 6 Oct 1992 18:24:43 GMT
Organization: Computer Science Department at Rose-Hulman
Reply-To: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
NNTP-Posting-Host: l130a4.nextwork.rose-hulman.edu

I was once informed by an Indian Mathematician (educated in the old
British system) that as long as one followed the rules of "construction
of roman numerals", any configuration was correct, so 
4000 = IIIIIIIIIIII...III = VVVVVVV...V are perfectly correct, if  
inane.
                   __       _
So, I suppose both VI  and MV are correct.  After all, neither is
ambiguous.

bart
From: g5000005@nickel.laurentian.ca
Newsgroups: sci.math
Subject: Wanted: Statistica 3.1 manual
Date: 30 Sep 92 19:05:07 GMT
Organization: Laurentian University

Hello netters!

	I am looking for a manual/book for the statistical analysis program 
called Statistica 3.1 (for DOS). I am currently using VAX/VMS SPSSx but I will
be graduating soon (what time is it now?) and have to decide between SPSS-PC+ 
and Statistica for my PC.

	If anyone has used this package I would be interested to hear your 
comments/criticisms regarding this package.

	Thanks in advance!

Norm
Newsgroups: sci.math