HDTenEightyP/NetNews huggingface dataset:
From: andrew@rentec.com (Andrew Mullhaupt)
Newsgroups: sci.math
Subject: Re: Retarded log question
Date: 14 Sep 92 00:16:29 GMT
Organization: Renaissance Technologies Corp., Setauket, NY.

In article <1992Sep11.145135.20098@infodev.cam.ac.uk> gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:
>where "log" is to any base you want. This is your answer.
>
>  Proof of the base-conversion thing above:
>  it's the same as log_b(x).log_c(b) = log_c(x);

There is one true logarithm. Thou shalt not bow down before false logarithms...

The "natural" logarithm is all you ever need to solve any problem. For
example 2^1,000,000 = 10^x can be done

	log(2^1,000,000) = 1,000,000 log(2) = x log(10) = log(10^x)

and then solve for x:

	x = 1,000,000 * (log 2)/(log 10).

And of course there is no need to sully ourselves with the ln foolishness,
either.

Later,
Andrew 'one logarithm' Mullhaupt
Newsgroups: sci.math
From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Re: Report on Philosophies of Physicists
Organization: Computer Science Department,  Stanford University.
Date: Tue, 15 Sep 1992 05:01:18 GMT

In article <1992Sep15.035832.7576@cbnewsm.cb.att.com> mls@cbnewsm.cb.att.com (mike.siemon) writes:
>	- it is VERY important to "break" students from reliance on the
>	  geometry and algebra of their earlier studies as *the* only
>	  matter of a generalized "geometrical" reasoning.  There is such
>	  a wealth of counter-example to expectations in the basic stuff
>	  of a point-set topology course as to open previously closed minds.

This familiar assertion about topology should be contrasted with the
following conclusion to the prefatory "Note to the Reader" from
Munkres' "Topology", Prentice-Hall 1975.

	Now it is all too easy in studying topology to spend too much
	time dealing with "weird counterexamples."  Constructing them
	requires ingenuity, and is often great fun.  But they are not
	really what topology is all about.  Fortunately, one does not
	need too many such counterexamples for a first course; there is
	a fairly short list which will suffice for most purposes.  Let
	me give it here:

	[product, uniform, box topologies on R^n
	 R with basis all [a,b)
	 the order type S_Omega (better known to logicians as omega_1)
	 the lexicographic square]

I would add to this perspective that the "weird counterexamples" (at
least the ones I vaguely recall being put through) are for the most
part the *fault* of a set-theoretic perspective.  Topological spaces
and their continuous maps don't form a closed category (one such that
the set of continuous functions from S to T forms a topological space
in the "obvious" way), forcing one to jump around between spaces and
sets when thinking about maps f:SxT->U and fixing a point in S or T.
For ordinary consumers of topological spaces it's healthiest to go
right to compact Hausdorff spaces without messing around with
non-Hausdorff or non-compact counterexamples---not only is CGHaus
closed but very remarkably it is algebraic, i.e. essentially a variety
or equational class, just like groups or lattices, only with operations
at arities of every cardinality.  For those needing the whole of Top
but cleaner, locales are the clean way to go.  (Cleanliness is next to
godliness.  God I hate it when I get religious.)

Mathematicians who don't see the connection of their stuff with
foundations are muscle-bound programmers who write large grungy codes
without even being aware there's a problem.

-- 
======================================================| God found the positive
Vaughan Pratt   pratt@cs.Stanford.EDU   415-494-2545  | integers, zero was
======================================================| there when He arrived.
Newsgroups: sci.math
From: frank@Cookie.secapl.com (Frank Adams)
Subject: Re: measures of the `size' of infinite sets
Date: Fri, 11 Sep 1992 17:58:29 GMT
Organization: Security APL, Inc.

In article <1992Sep9.062701.8487@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>In article <1992Sep9.042345.7472@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>
>>  But for your problem the best answer involves not measure
>>but "density".  Given a set of integers, we calculate its density as
>>follows.  Figure out how many integers in your set lie between -n and n
>>- say that m do.  Take the ratio n/2m ... this just tells us what
>>fraction of the integers between -n and n lie in your set.  Now take the
>>limit as n goes to infinity!  If the limit exists and equals d (some
>>number between 0 and 1), we say your set has density d.
>
>Okay, I got that one messed up all right.  Take the ratio m/(2n+1),
>since there are 2n+1 integers between -n and n, counting the endpoints.

Since you're taking the limit as n goes to infinity, it doesn't really
matter.  The limit is always exactly the same either way.
Newsgroups: sci.math
From: brock@NeXTwork.Rose-Hulman.Edu (Bradley W. Brock)
Subject: Re: n doesnt divide 2^n-1
Organization: Rose-Hulman Institute of Technology
Date: Mon, 7 Sep 1992 20:03:35 GMT


In article <Sep.7.13.02.21.1992.5036@pepper.rutgers.edu>  
gore@pepper.rutgers.edu (Bittu) writes:
> Conj: For n >= 1, let 2^n = a.n + b where 0 <= b < n so that a is the
> quotient and b the remainder. Then the following hold:
> 
> 1. b is always even.
> 2. (even stronger) b is always a power of 2.
> 3. a is always even.
> 
> Note that 1 implies that n doesn't divide 2^n - 1 for n > 1. 2 gives
> us a slightly stronger theorem though I don't know how to say it. I
> don't know what 3 implies. My question is: Is any of the three
> statements true?
If n=25, a= 1342177 and b=7.  Hence, all three statements are false.
--
Bradley W. Brock, Department of Mathematics
Rose-Hulman Institute of Technology  | "Honor one another above yourselves."
brock@nextwork.rose-hulman.edu       | -Paul of Tarsus to the Romans
From: ara@zurich.ai.mit.edu (Allan Adler)
Newsgroups: sci.math
Subject: Re: Couple of questions
Date: 10 Sep 92 01:25:30 GMT
Article-I.D.: camelot.ARA.92Sep9202530
	<1992Sep9.174910.12677@galois.mit.edu>
Organization: M.I.T. Artificial Intelligence Lab.

There is now a generalization of the Riemann integeral, called the Generalized
Riemann Integral, which is powerful enough to integrate all Lebesgue integrable
functions. A positive function is integrable by this function if and only
if it is Lebesgue integrable. However, there are functions which are not
Lebesgue intergrable, by virtue of the fact that their absolute values are
not, but which are integrable by the Generalized Riemann Integral. There is
a Carus monograph on this. I think McShane has a book that handles the
Generalized Riemann Integral in R^n and for for the infinite dimensional
integrals that arise in stochastic processes. There is a very recent book
by Henstock which seems to achieve the greatest generality, axiomatizing
the structures necessary to carry out constructions of this type. 

There are some people who work in integration theory who believe that the
Lebesgue integral is now obsolete and that the generalized Riemann integral,
including its general formulation, should be the foundation that is taught.

I think the generalized Riemann integral does one other thing: every 
differentiable function is the generalized Riemann integral of its 
derivative.

Allan
Newsgroups: sci.math
From: levine@symcom.math.uiuc.edu (Lenore Levine)
Subject: That Collatz Flu
Organization: University of Illinois at Urbana
Date: Fri, 11 Sep 1992 14:51:04 GMT

I've gotten interested in the Collatz function. (I guess it's something
every math person has to go through for a week, like the flu.)

(Collatz function: f(n) = n/2 if n is even
                        = (3n + 1)/2 if n is odd.

It is conjectured that for every n > 0, there is a k such that f^k(n)
(the kth iterate of f) = 1.)

What I'm wondering is: Has it been shown that there are no n > 2 such 
that f^k (n) = n?

Also, any major results later than the 1985 survey article?

Lenore Levine 
Newsgroups: sci.math
From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
Subject: Re: Conway Joke?
Organization: cpd
Distribution: sci.math
Date: Mon, 14 Sep 1992 12:35:44 GMT

In article <CHALCRAFT.92Sep10100952@laurel.uk.tele.nokia.fi> chalcraft@uk.tele.nokia.fi (Adam Chalcraft) writes:

>	   ___________
>	  |           |
>	  | Ring will |
>	  |  not fit  | These square things are actually
>	  | over this | shiny round spheres
>	  |_____  ____|
>	      __||__
>	     |  __  | Eye
>	   __| |__| |__
>	  / _| |__| |_ \
>	 ( (_|_|__|_|_) ) Ring (fits through the eye)
>	  \____________/
>	     | |  | |
>	   ,-+-+--| |-.
>	   | | |__| | | String (long)
>	   | |______| |
>	 __|___    ___|__
>	|      |  |      |
>	|  or  |  |  or  | More shiny round spheres
>	| this |  | this |
>	|______|  |______|
>	  _||_      _||_
>	 | __ |    | __ |
>	 ||__||    ||__|| These are clearly useless
>	 |____|    |____|

It turns out that Winning Ways is wrong (gasp!), and the correct puzzle is:

The ring fits over the lower spheres,
The ring does not fit over the "useless" rings,
The "useless" rings fit through the eye,
The lower spheres do not fit through the eye.

A couple of people told me how to solve the (corrected) puzzle.
A couple of people told me they had solved it, and offered to email me the
 solution if required.
One person read the puzzle sufficiently closely to notice that it is not
 possible, and told me what the correction is.
One person checked in Winning Ways, compared with the real thing, and
 verified the mistake.

Thankyou to everyone who responded.

--
Adam Chalcraft -- Opinions are mine only, and probably wrong anyway.
Email: chalcraft@uk.tele.nokia.fi
       If that confuses your (uk) mailer, try chalcraft@cpdapo.tele.nokia.fi
       If that doesn't work, try ntl02.decnet.nokia.fi (zero-two)
       Alternatively, try dac11@phx.cam.ac.uk
Newsgroups: sci.math
From: holmes@opal.idbsu.edu (Randall Holmes)
Subject: Re: ZFC etc. (was Re: Report on Philosophies of Physicists)
Organization: Boise State University Math Dept.
Date: Tue, 15 Sep 1992 23:10:19 GMT

In article <COLUMBUS.92Sep15171406@rachmaninoff.think.com> columbus@rachmaninoff.think.com (Michael Weiss) writes:
>In article <1992Sep15.184930.10080@guinness.idbsu.edu>
>holmes@opal.idbsu.edu (Randall Holmes) answers the question:
>   >
>   >Are the proofs by Cohen and Godel are formal proofs? (I would think
>   >so.) Can we identify the formal system in which the proofs were
>   >performed?
>
>thus:
>
>   They are rigorous arguments.  No one ever does proofs in formal
>   systems, in practice.  The Godel proof is a construction in ZF (normal
>   set theory without choice).  The Cohen proof is a construction in ZFC
>   (Choice is used, I believe).
>
>The relative consistency results of Godel and Cohen can be regarded as
>purely combinatorial statements about pushing symbols around,
>and as such can be expressed in the language of Peano arithmetic.
>
>In principle, the proofs of these relative consistency statements could be
>carried out in Peano arithmetic.  Cohen discusses this briefly in the last
>chapter of his book "Set Theory and the Continuum Hypothesis".
>I believe the treatment in Shoenfield's book "Mathematical Logic" makes the
>same point.

I know that, and I thought of pointing it out, but I think it would be
deceptive to say that Godel and Cohen carried out their arguments in PA.

>
>In practice, the proofs become far more intuitive if we adopt a less
>puritan attitude, and talk about models of ZF.  Cohen makes free use of
>axiom SM (="there exists a model of ZF whose universe is a set and whose
>element-of relation is the standard element-of relation".)  Cohen proves
>with ZF+V=L+SM that there is (for example) a model of ZFC + not CH.
>Naturally this implies the relative consistency result, Con(ZF) -->
>Con(ZFC+not CH).  But if all you want is the relative consistency result,
>then a much weaker set of axioms (such as Peano arithmetic) will do.


-- 
The opinions expressed		|     --Sincerely,
above are not the "official"	|     M. Randall Holmes
opinions of any person		|     Math. Dept., Boise State Univ.
or institution.			|     holmes@opal.idbsu.edu
From: loader@vax.oxford.ac.uk
Newsgroups: sci.math
Subject: Re: Partitioning of uncountable sets
Date: 8 Sep 92 15:29:39 GMT
Organization: Oxford University VAX 6620

In article <1992Sep8.182706.90039@vaxc.cc.monash.edu.au>, kevin@vaxc.cc.monash.edu.au writes:
> A proof that every uncountable set can be partioned into two uncountable
> sets.
> Let X be an uncountable set. Consider the set of ordered pairs, (x,0),(x,1)
> where x is in X. Call this set Y. Then Y is also uncountable, moreover,
> {(x,0) with x in X}=Y(0) and {(x,1) with x in X}=Y(1) are both uncountable.
> But the cardinaltiy of X is the cardinality of Y. Thus, there is a bijection
   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
How exactly does one prove this without AC? If I remember correctly,
"for all infinite cardinals a; a = 2a" is independent of ZF (without AC), and
ZF+"2a=a, a infinite" is strictly weaker than ZFC. Of course, the difference
between uncountable and infinite is irrelavant here as ZF proves
    2 . omega = omega

> f from Y to X. f(Y(0)) and f(Y(1)) form the desired partition of X.
> The proof is of course easier if one assumes A.C, which I have avoided.
> 
> Love,
> Kevin Davey.
Newsgroups: sci.math
From: mcphee@cda.mrs.umn.edu (Nic McPhee)
Subject: Searching for a quote
Keywords: quote analytic geometry
Organization: University of Minnesota - Morris
Date: Sat, 12 Sep 1992 21:43:27 GMT

I once heard a quote that I really enjoyed, but have since lost it.  It  
was (if I remember correctly, which I may not) was from a review of a  
writing on analytic geometry from several hundred years ago, which  
referred to the subject as something like "a scabrous collection of  
symbols".  If anyone out there can either

	0.  Give me a reference for this, or
	1.  Tell me that I'm making it all up :-)

I'd be most appreciative.

Thanks in advance,

	Nic McPhee
	University of Minnesota, Morris
	mcphee@cda.mrs.umn.edu
Newsgroups: sci.math
From: dzk@cs.brown.edu (Danny Keren)
Subject: Re: help me, please
Organization: Brown University Department of Computer Science
Date: Thu, 10 Sep 1992 19:53:59 GMT

kyee@bnlux1.bnl.gov (kenton yee) writes:
#I need to figure out what the following integral is:
#
#  int dx_1 dx_2 ... dx_N  1    =   ??
#
#where  0 < x_i < 1    subject to the condition  
#   
#  x_1 + x_2 + ... + x_N  is less than or equal to 1.  

I guess you could compute the integral, but there is a cute 
geometric solution. Define two sets:

S1 = {(x1,x2....xn)/xi>=0, x1+x2+...xn<=1}

S2 = {(x1,x2....xn)/0<=x1<=x2...<=xn<=1}

What you are asking is the volume of S1. Now, look at the following
transformation from S1 to S2:

(x1,x2,....xn) ----> (x1,x1+x2,x1+x2+x3,...x1+x2+...+xn)

it is trivial that it is 1-1 and onto; also, its determinant is 1,
so it preserves volume. Its also trivial that the volume of
S2 is 1/n! (symmetry considerations) so your integral is equal
to 1/n! .

-Danny Keren.
Newsgroups: sci.math
Subject: Re: Question on divergence of series given condition
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: 8 Sep 92 15:21:28 +1200
Distribution: world
Organization: Department of Mathematics, University of Canterbury

In article <1992Sep4.174051.3223@eagle.lerc.nasa.gov>, sshws@convx1.lerc.nasa.gov (Herb Schilling) writes:
|> 
|>   If the terms, a  , of a series have the following property:
|>                  n
|> 
|>          a                               A
|>           n+1                1            n
|>         ------ =    1   -  -----   +   -----    as n -> oo
|>          a                   n           n
|>           n
|> 
|>
|>     where A    ->   0 as n -> oo
|>            n
|> 
|> 
|>   then does sum of the a    diverge  ?
|>                         n 
---------------------------------------------------------
Usually, it seems, but not necessarily, alas.

For example, if   a  =  1
                   n   ------   2
                        n(log n)

then the series converges, but the above criterion still holds,

with   A          2
        n  ~  -  ------
                 log n       .
------------------------------------------------------------------------
           Bill Taylor              wft@math.canterbury.ac.nz
------------------------------------------------------------------------
  The "faith" meme discourages the excercise of the sort of critical
judgment that might decide that the idea of faith was a dangerous one.
------------------------------------------------------------------------


Newsgroups: sci.math
From: bs@fatima.mitre.org (Robert D. Silverman)
Subject: Re: measures of the `size' of infinite sets
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Date: Thu, 10 Sep 1992 13:11:41 GMT

In article <18nem0INN2vg@function.mps.ohio-state.edu> edgar@function.mps.ohio-state.edu (Gerald Edgar) writes:
:
:Here is an old gem along these lines.  A positive integer, in its
:decimal representation, begins with one of nine possible
:digits.  What proportion of them begin with the digit '1'?
 
The 'distribution of first digit' problem is well-known and is discussed 
in Knuth vol. 2 [isn't everything? :-)]  I won't post a spoiler.
--
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"
From: jbaez@riesz.mit.edu (John C. Baez)
Newsgroups: sci.math
Subject: Re: measures of the `size' of infinite sets
Date: 11 Sep 92 21:59:06 GMT
Organization: MIT Department of Mathematics, Cambridge, MA

In article <1992Sep11.175829.79520@Cookie.secapl.com> frank@Cookie.secapl.com (Frank Adams) writes:
>In article <1992Sep9.062701.8487@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>>In article <1992Sep9.042345.7472@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
>>
>>>  But for your problem the best answer involves not measure
>>>but "density".  Given a set of integers, we calculate its density as
>>>follows.  Figure out how many integers in your set lie between -n and n
>>>- say that m do.  Take the ratio n/2m ... this just tells us what
>>>fraction of the integers between -n and n lie in your set.  Now take the
>>>limit as n goes to infinity!  If the limit exists and equals d (some
>>>number between 0 and 1), we say your set has density d.
>>
>>Okay, I got that one messed up all right.  Take the ratio m/(2n+1),
>>since there are 2n+1 integers between -n and n, counting the endpoints.
>
>Since you're taking the limit as n goes to infinity, it doesn't really
>matter.  The limit is always exactly the same either way.

Heh.  Take a close look at what I wrote the first time, and you will
see that even in the limit, it would not give the right answer.

(Someone else, who will remain nameless here, emailed me with the same
remark.  It's a good example of "reading what was meant, not what was
written.")  
Xref: sparky sci.math:11223 comp.lang.misc:3007 comp.unix.programmer:4617
From: siva@bally.com (Siva Chelliah)
Newsgroups: sci.math,comp.lang.misc,comp.unix.programmer
Subject: geometric theorem proving program
Keywords: theorem, geometry,program
Date: 10 Sep 92 17:51:53 GMT
Organization: Bally Systems


Hi,
  Is there any geometric theorem proving program available for IBM PC or Unix?
The program should prove theorems such as :
     In a triangle ABC, if AB = AC then        
          angle ABC = angle ACB

etc. etc.

Thanks
siva
From: myhre@plains.NoDak.edu (Robert Myhre)
Newsgroups: sci.math
Subject: Looking for Computer Aid
Date: 11 Sep 92 21:04:16 GMT
Organization: North Dakota Higher Education Computing Network


  HI all.  I was wondering if anyone knew of a program for tutoring in
Calculus, mainly integrationg?  I could use some refresher/aid!

Thanks

Robert Myhre
Computer Engineer in Training... and Training... and Training
North Dakota State University
myhre@plains.nodak.edu   (Internet)     
myhre@plains             (Bitnet)      
uunet!plains!myhre       (UUCP)         
 
Newsgroups: sci.math
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Subject: Re: Concordia's prof killer's message
Organization: U of Cambridge, England
Date: Fri, 4 Sep 1992 23:22:39 GMT

In article <Bu2IEH.Mss@math.uwaterloo.ca>, hstitan@neumann.uwaterloo.ca (Hari S. Titan) writes:

> Hi everyone,
> 
> I heard that the prof. killer at Concordia had sent an e-mail message to 
> many mathematicians complaining about problems he had with his job a week
> before he began the killings.  My understanding is that he meant this to 
> be public letter and wouldn't mind more people reading it.  If someone 
> still has his message could they e-mail me a copy.  
> 
> Thanks very much.
> 
> H. Titan 
> (hstitan@neumann.uwaterloo.ca)

I have done this. Actually there was more than just one letter -- the thing
I've sent hstitan is over 200k long and consists of 6 chunks. I haven't
actually read it all...
Newsgroups: sci.math
From: prakash@cs.mcgill.ca (Prakash Panangaden)
Subject: Re: Einstein's Erdos number
Organization: SOCS, McGill University, Montreal, Canada
Date: Mon, 7 Sep 1992 18:16:16 GMT

In article <2aa7a8a2wnr070@ark.abg.sub.org> ralf@ark.abg.sub.org (Ralf Stephan) writes:
>niall@nmsu.edu (Niall Graham) writes:
>> 
>> Correct.   Einstein's co-authors were 
>> Ehrenfest, Lorentz, Minkowski, Somerfeld, Straus, Weyl.
>
>and Rosen, Podolsky ?
>
>--ralf

Yes what about Podolsky and Rosen?  What about Leopold Infeld?  I have read
both these papers so I am certain that they were co-authors of Einstein.
Also I am very surprised to see that Lorentz and Minkowski are listed as
co-authors of Einstein.  Is it possible that this is an error stemming from
the fact that there is a collection of early papers on relativity that 
includes papers by Einstein, Lorentz and Minkowski (all solo authored).
I have this book and noticed that the authors are listed as Einstein,
Lorentz, Minkowski,...  I am also curious about papers that he may have
written with Weyl since I have another collection of papers that includes
solo authored papers by Einstein and by Weyl.  Of course I am not saying that
Einstein did not write papers with all these people only that it is 
surprising especially since he said once how much he disliked Minkowski's
work!

Newsgroups: sci.math