HDTenEightyP/NetNews huggingface dataset:
From: rodney@garfield.cs.mun.ca (Rod Hutchings)
Subject: Re: Need: prpndcular dist from pt to line
Organization: CS Dept., Memorial University of Newfoundland
Date: Wed, 30 Dec 1992 19:23:43 GMT

In article <1992Dec29.210554.5350@seas.gwu.edu> ilan@seas.gwu.edu (Ilan Berkner) writes:
>
>Given a line, or line segment (I can do both), I need to find the 
>distance from a point (that may or may not be on the line) to that line.
>If the point is on the line, then the distance is zero, obviously.
>
>I know the endpoints of the line segment in cartesian space, and I also
>know the slope and intercept, and the coordinates of the point.
>
>This is for a computer application, so anything involving integer math
>is preferable to floating-point.
>
>Thanx very much in advance.  I know this is almost a trivial problem
>for all you math geniuses out there, but I also want to make sure that
>whatever solution I get, is also the best.

If you have the slope and intercept of the line then you can write it
as y = mx + b where m is the slope and b is the intercept.  This can be 
converted to the form 
	ax + by + c = 0
by writing the equation y = mx + b as mx - y + b = 0.  

Now there is a general formula for the distance from the point (x0,y0) to 
the line Ax + By + C = 0.  It goes as follows:

		d = abs(A*x0 + B*y0 + C) / sqrt(A^2 + B^2)      

where (1) the coordinates of the point are put into the equation of the line
for x and y, (2) the coefficients of x and y are squared and summed, then
the results of (1) and the square root of (2) are divided out.

In this case we substitute A = m, B = -1, and C = b to give the form as
follows:

		d = abs(m*x0 - y0 + b) / sqrt(m^2 + 1)      

This will give the required distance, and can be easily implemented on a
computer.

--
regards,
the Rodster
rodney@garfield.cs.mun.ca

"Give the Clown credit, he beats up cowboys AND Indians."
				- Christopher Robin Zimmerman 
				(A fellow poster to rec.sport.pro-wrestling)

-- 
--
regards,
the Rodster
rodney@garfield.cs.mun.ca
Xref: sparky sci.math:17314 sci.physics:21600
From: jbaez@riesz.mit.edu (John C. Baez)
Newsgroups: sci.math,sci.physics
Subject: Re: Bayes' theorem and QM
Date: 20 Dec 92 05:05:44 GMT
Organization: MIT Department of Mathematics, Cambridge, MA

In article <1992Dec18.134107.24536@oracorp.com> daryl@oracorp.com (Daryl McCullough) writes:
>jbaez@riesz.mit.edu (John C. Baez) writes:
>
>>Classical mechanics and quantum mechanics are fundamentally different in
>>that the latter is fundamentally probabilistic. The former is only
>>probabilistic in the sense that computational limitations may prevent us
>>from computing the precise future state that is in principle determined by a
>>given present state.
>
>When you say that quantum mechanics is fundamentally probabilistic, do
>you mean (A) QM is a probabilistic theory with no known deterministic
>completion, or (B) QM is a probabilistic theory that is known *not* to
>have a deterministic completion?

Neither, since I'm not interested in so-called "completions" of
quantum mechanics, which is already complete enough for me.

What I mean is this.  In classical mechanics, all pure states are
dispersion-free.  In quantum mechanics this is not so.

Less tersely: in classical mechanics, in a pure state one can calculate
a numerical value for every observable; an ideal measurement of this
observable should give this number as an answer.  Probability
distributions for the value of an observable are only different from
delta functions in the case of mixed states (i.e., states in which one
doesn't know a maximal amount of information about what's going on, as
are used in statistical mechanics.)  We can say that probability theory
is only needed classically if you have some ignorance about what the
system is up to.  In quantum mechanics, even in a pure state one can
only calculate a probability distribution for the value of an
observable, and generically this is not a delta function, but has
nonzero "dispersion" or standard deviation.

(I'm sure Daryl knows all this but I want to be painfully clear.)

More mathematically: a state of a C*-algebra is "pure" if it is not a
mixture of two different states.  A state mu is "dispersion-free" if
mu(a^2) = mu(a)^2 for every self-adjoint a.  (The difference is the
variance of a in the state mu.)  It is a theorem that for commutative
C*-algebras, pure implies dispersion-free.  But not for noncommutative
C*-algebras.  In the language of C*-algebras, "classical" means
commutative, while "quantum" is noncommutative.
From: naturman@uctvax.uct.ac.za
Newsgroups: sci.math
Subject: Re: Bayes' theorem and QM
Date: 28 Dec 92 15:02:15 +0200
Keywords: qm
Organization: University of Cape Town

In article <TED.92Dec27145845@lole.nmsu.edu>, ted@nmsu.edu (Ted Dunning) writes:
> 
> In article <1992Dec26.080126.12238@monu6.cc.monash.edu.au> darice@yoyo.cc.monash.edu.au (Fred Rice) writes:
> 
>    >It seems likely
>    >to me that some form of deterministic chaos also underlys the seemingly
>    >fundamental nature of probability in quantum mechanics.
> 
>    Interesting... I have had the same thought of this possibility too.  Does 
>    anyone know if any work on how chaos and fundamental QM might relate has 
>    been done?  If so, can anyone give any helpful references or names?
> 
> 
> this is called the hidden variable interpretation of quantum
> mechanics.
> 
> 

Wrong. Without going into too much details, hidden variable interpretations are
interpretations which require QM probabilities to be recoverable by integration
over a space of "hidden variables".  The hidden variables model possible
unknown physical quantities whose specified values uniquely determine all
the values of all physical quantities. Hidden variables interpretations were
born out of an attempt to show that QM worked like the kinetic theory of gases.
However, one cannot have a hidden variables interpretation without violating
locality.

The attempts to explain QM in terms of chaos theory have nothing to do with
hidden variable theories.
From: kullmann@urz.unibas.ch (Peter Kullmann)
Newsgroups: sci.math
Subject: Harmonical, homogeneous polynomials
Date: 29 Dec 92 13:42:05 MET
Organization: University of Basel, Switzerland


I'm looking for references about the construction of harmonical, homogeneous
polynomials in n unknowns. I know the dimension of the space but it would be
nice to have some sort of "standard basis".

Any help greatly appreciated.
-- 
Peter

Newsgroups: sci.math
From: thf@cbnewsm.cb.att.com (thomas.h.foregger)
Subject: Re: Combinatorial? problem
Organization: AT&T
Date: Tue, 29 Dec 1992 04:19:21 GMT
Summary: soln of the problem
Keywords: binomial coefficients

In article <Dec.23.18.15.49.1992.22657@pepper.rutgers.edu>, gore@pepper.rutgers.edu (Bittu) writes:
> 
> Someone please help me with the following problem. I have pretty much
> given up on it after a lot of thinking. We want to show the following:
> 
> given m,n nonnegative integers, the quantity
> 
>      (2m)! (2n)!
>     -------------      is an integer.
>      m! n! (m+n)!
> 
> 
> Note that this is very easy to show by the standard argument where for
> every prime p, you find the highest power of p (say p^k) that divides
> the denominator and then show that p^k divides the numerator as well.
> 
> I want a combinatorial proof of this. I have tried rewriting the above
> as C(2m,m)*C(2n,n)/C(m+n,m) where C(a,b) is "a choose b" and also in
> other ways, but I still haven't come up with a combinatorial proof.
> 
> --Bittu


Without loss of generality assume m<=n so m+n <= 2m.
Write the ratio as

(2m)! . n! . (2n)!
----   ---   -----
(m+n)!  m!   n! n!

Each of the 3 factors is clearly an integer, so the ratio is an integer.

tom foregger
From: edp@bsdbld.zk3.dec.com (Eric Postpischil)
Newsgroups: sci.math
Subject: Mathematica price?
Date: 30 Dec 92 18:19:22 GMT
Reply-To: edp@bsdbld.zk3.dec.com (Eric Postpischil)
Organization: Digital Equipment Corporation

Wolfram wants $475 for Mathematica for IBM PC compatibles, even at the
student rate.  I thought I had seen an offer, perhaps a year or so ago,
for Mathematica at a better price.  It was on an advertising card that
came inside a math textbook, but I did not keep the card.  Does anybody
remember this?  I'd appreciate any information about this or other
legitimate ways to obtain Mathematica for a better price.


                                -- edp (Eric Postpischil)
                                "Always mount a scratch monkey."
                                edp@rusure.enet.dec.com
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math
Subject: Re: more math puzzles
Date: 24 Dec 92 13:50:23 GMT
Reply-To: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Organization: The Wistar Institute of Anatomy and Biology
In-reply-to: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)

In article <103098@netnews.upenn.edu>, weemba@sagi (Matthew P Wiener) writes:
>L is of course not sequentially compact.

Whoops.  Herman Rubin corrected me.  Given a countable sequence, it is
bounded wrt to w_1, so its supremum exists.
-- 
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
Xref: sparky sci.math:17430 sci.logic:2482 sci.philosophy.tech:4619
From: zeleny@husc10.harvard.edu (Michael Zeleny)
Newsgroups: sci.math,sci.logic,sci.philosophy.tech
Subject: References and Books Wanted
Date: 26 Dec 92 23:33:23 GMT
Expires: December 31, 1999
Organization: The Phallogocentric Cabal

Seeking references on the relationship between the Cohen and
Scott-Solovay truth (or meaning) definitions and modal logic,
especially possible world semantics.

Also looking for copies of Levy's _Basic Set Theory_, and other old
\Omega-series titles, logic colloquia and symposia, and other old
titles in the _Studies in Logic_ series, the set theory and recursion
theory AMS colloquia, volumes 2-4 of Tarski's _Collected Papers_, the
1967 edition of Birkhoff's _Lattice Theory_, Kunen's _Handbook of
Point Set Topology_, old volumes of the _Journal of Symbolic Logic_,
and what have you.  Your choice of greenbacks or a pound of flesh
(preferably adipose tissue) on delivery.
 
cordially,
mikhail zeleny@husc.harvard.edu
"Nothing can be said truly of what does not exist."
Xref: sparky sci.math:17431 sci.philosophy.tech:4620
Newsgroups: sci.math
Subject: Re: Sums of reciprocals
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: Wed, 23 Dec 1992 00:46:00 GMT
Distribution: sci
Organization: Department of Mathematics, University of Canterbury

In article <ARA.92Dec21193839@camelot.ai.mit.edu>, ara@zurich.ai.mit.edu (Allan Adler) writes:
|> 
|> Let x be a positive real number.
|> Let g(n,x) be defined as follows:
|> 
|> (1) g(n,x) is a strictly increasing sequence of positive integers, possibly
|>      a terminating sequence.
|> (2) for all n, g(n,x) is the smallest integer consistent with (1)
|>     such that the sum of the reciprocals of g(k,x) for k=1,...,n
|>     does not exceed x.
|> 
|> I have been told that if x is rational then this leads to a way of writing
|> x as a finite sum of reciprocals of distinct positive integers. Let the number
|> of terms in this sum be N=N(x). I have been told that N is also the number of
|> terms in a minimal representation of x as a sum of reciprocals of positive
|> integers allowing repetitions.
|> 
|> Is this true and if so how or where is it proved?

The second statement is false.

Let x = 9/20.
Then the sequence of g(n,x) is  3, 9, 180.

So N=3, but this is not minimal as  x = 1/4 + 1/5 . 


The first statement is true. That is, for x rational, the process of constructing
the g(n,x) is certain to terminate. This is easily seen by noting that the
successive remainders after subtracting the 1/g(n,x) terms are fractions whose
numerators strictly decrease, so must eventually reach 1.

If  x (or its current remainder) is p/q, and is approximated below by 1/n,

i.e.  1/n < p/q < 1/(n-1)  ,  ....(A) 

then next remainder = p/q - 1/n
                    = (pn-q)/qn    and (pn-q) < p  from (A).       Q.E.D.
-------------------------------------------------------------------------------
              Bill Taylor              wft@math.canterbury.ac.nz
-------------------------------------------------------------------------------
                   All finite math is essentially trivial.
                   All continuum math is essentially false.
-------------------------------------------------------------------------------



Newsgroups: sci.math
From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: more math puzzles
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Mon, 21 Dec 92 18:54:16 GMT

In article <1992Dec21.103233.723@black.ox.ac.uk> mbeattie@black.ox.ac.uk (Malcolm Beattie) writes:
>In article <1992Dec18.101409.4666@black.ox.ac.uk> mbeattie@black.ox.ac.uk (Malcolm Beattie) writes:
>>In article <24341@galaxy.ucr.edu> baez@ucrmath.ucr.edu (john baez) writes:
>>>3)  Prove that the tangent bundle of the long line is nontrivial.

>>First, the answer. Let L be the long line (more of which, later
>>in this post.) Assume for a contradiction that TL is trivial.
>>Then TL has a nowhere-vanishing section. Integrate this to get
>>a monotonic map from the reals to L. Any such map from the
>>reals to L must eventually become constant at some point of L
>>and the section therefore vanishes at that point. Contradiction.

>Rubbish :-) ...
>Surely there must be a nicer proof than this?

I'll relent and give Hirsch's one-sentence proof.  If the tangent bundle
of the long line were trivial, the long line would admit a Riemannian
metric, hence be metrizable, but it's not.

Perhaps some clarification might help those who aren't differential
geometry junkies.  The long line is not metrizable, i.e. it does not
admit a metric inducing its standard topology, essentially because it's
way too long.  It's also not paracompact.  Usually one only considers
paracompact smooth manifolds.  Paracompact smooth manifolds always admit
a Riemannian metric and hence, if they are connected, are metrizable.
But this fails without paracompactness.  If the tangent bundle of the
long line were trivial, all the tensor bundles would be, so one could
find a constant section of the bundle of (0,2) tensors (or is it (2,0)
tensors - I refuse to learn this convention!) giving the long line a
Riemannian metric - a contradiction.

If one has never shown the following, one should: every sum of
uncountably many positive real numbers diverges.  In a heuristic sort of
way this is why one might expect the long line not to be metrizable:
it's built out of uncountably many intervals, and if all of these had
positive length one could find two points that were infinitely far away
because there were uncountably many intervals between them.  That's no
proof, mind you.  

I don't feel I understand this all as well as I might.  Someone with
good German might try Koch and Puppe, Differenzbar Strukturen auf
Mannifaltigkeiten ohne abzahlbare Basis, Arch. D. Math. 9 (1960) p. 104.
Xref: sparky sci.math:17397 sci.logic:2477
From: wrightd@math.okstate.edu (David J. Wright)
Subject: Re: Question on Putnam B6
Keywords: Putnam Exam, Problems
Organization: Oklahoma State University, Math Department
Date: Thu, 24 Dec 1992 03:55:33 GMT

wrightd@math.okstate.edu (David J. Wright) writes:



>Well, I've been thinking about B6 now, and can't figure out
>the truth of the matter. So I might as well ask. Of course,
...

Oops, I made a hash of my post because in vi I use 4-character
tabstops.

>	 M' = { ( +- A   0         (  0   +- A
>				0    A )  and     A     0  ) ,  A \in M }
This should be:
M' = { ( +- A   0         (  0   +- A
           0    A )  and     A     0  ) ,  A \in M }

>M must contain    ( 0  -I
>					I   0 )
Should be:
 ( 0  -I
   I   0 )

>contains ( -I  0
>			0  I )
Should be:
 ( -I  0
    0  I )

Sorry about the typography.
From: naturman@uctvax.uct.ac.za
Newsgroups: sci.math
Subject: Re: Bayes' theorem and QM
Keywords: qm
Date: 30 Dec 92 22:06:47 GMT
Article-I.D.: uctvax.1992Dec30.200647.203436
Organization: University of Cape Town

> 
> Does this mean that work on how chaos might relate to fundamental QM has
> been done?  
> 
> If so, does anyone have any references or names?
> 
> (None have been given so far.)

True, there seems to be a lot of talk but no one seems to know where the
results are. I remember some remarks about chaos and QM in Scientific American
articles several years ago. When I studied Advanced Classical Analysis (Fourier 
stuff etc etc) I remember the lecturer mentioning work on explaining QM using
the notions of chaos and "non-differentiable motion". 

I'll be attending a conference on dynamical systems and chaos next month,
hopefully the experts in the field will be able to answer a few questions on
QM and chaos.

Colin
Newsgroups: sci.math
From: wayman@lamar.ColoState.EDU (Jeffrey Wayman)
Subject: Re: Math ( and other ) education.
Date: Sat, 02 Jan 1993 05:22:19 GMT
Distribution: na
Organization: Colorado State University, Fort Collins, CO  80523

In article <4932@execu.execu.com> dont@execu.execu.com (Don Thompson) writes:
>
>	Hello,
>
>	Regarding the recent thread discussing the state of mathematical
>	education in the U.S.
>
>	I decided about a month ago to try to salvage the wreckage of my
>	math education after realizing that the things I wanted to do
>	require more than + - /  and *. Some of the math I need to
>	to learn is the higher levels of Trig, Calc, and Geometry.
>
>	Now I live in Austin, Texas. We have at least four public or private
>	universities within driving distance and an extensive Community college
>	system. I work full time and my company is willing to foot the bill of 
>	any higher education including Masters and PhDs. 
>
>	"Great, I have a B.S. I'll just enroll in a few night classes, or
>	maybe even get into a Masters program". Well, It would work for the
>	community college, except that their math classes stop at "intro
>	to trig". "Try the University", I thought. Crash and Burn. My   
>	SAT-GREP-XYZ-ZZTOP-YOUR-TEST-GOES-HERE scores are not in the top
>	.00025% of the nation, I don't have a 4.2 grade average, and besides

Are the schools that selective?  You might do some good taking some test
a few times and get an outstanding score.  Tests like the ACT & GRE get a lot
easier after a try or two; many schools will take your highest score.

>	there are no night classes for the "serious" subjects. You want to
>	learn that stuff? Quit your job, move into the student slums, and 
>	repeat all your general ed. THEN you can learn the math. 
>
>	So, here I am. A productive member of society. I pay taxes, I vote,
>	I'm willing to give up my evenings with my family, to study hard,
>	to PAY, and what do I get? "Sorry, you not good enough. Return to
>	Go and pay $200". All because those pin heads did'nt have a slot that 
>	I fit into.
>
>	I always thought that education was a continuing process. It seems  
>	that the educational establishment disagrees. The only thing that
>	people want to learn after they get their degree is dBase or MSWord
>	and other "fun" classes. Unless of course you completely trash your
>	life and start over again.
>
>	Any suggestions?
> 
Not really a suggestion, but a point you might be able to turn into some
action:  Having taught a few "continuing education" type classes, I get
the idea that they will offer just about anything if enough people want
to take it (translated: a money maker).  Profs aren't hard to find.  If
there are more folks like you that want to take these classes, get them
to call the school and let the school know.  If there aren't more folks
like you, then you may want to look into commuting because the school
can't afford to pay a teacher for an unpopulated class.  

Good luck!!

Jeff          wayman@lamar.colostate.edu
From: hougen@focus.csl.uiuc.edu (Darrell Roy Hougen)
Newsgroups: sci.math
Subject: Re: An Integral Problem
Date: 4 Jan 1993 04:03:47 GMT
Organization: Center for Reliable and High-Performance Computing, University of Illinois at Urbana-Champaign
NNTP-Posting-Host: focus.csl.uiuc.edu

wielinga@newton.physics.uq.oz.au (Bruce Wielinga) writes:

%  	I would like help with this Intergral please 
%       if anyone is willing to spare the time.
% 	Basically it is:
% 		
% 	Integral of { [ A + B x^2 ]^(1/2) - [ C + x^2 ] }^(1/2) w.r.t x,

I think you just need to make several substitutions.  I made the
following substititutions to simplify the integral:

(1)  Let y = sqrt(A/B)x
(2)  Let tan(theta) = y
(3)  Let z = sec(theta)
(4)  Let t = z - A*sqrt(A)/(2B)
(5)  Let s = t/k where k = sqrt(AC/B - A^3/(4B^2) - 1)
(6)  Let tan(alpha) = s

Its too tedious to type in all the intermediate steps, but you should
get the point.  After all that, the integral I get is:

sqrt(-1) Integral sec(alpha)(k*tan(alpha) + m)^2 sec^2(alpha) d_alpha

where m = A*sqrt(A)/(2B).  This is just an ordinary trigonometric
integral.  Sorry about the sqrt(-1); I hope its supposed to be there.

Darrell R. Hougen

Newsgroups: sci.math
From: rivin@SAIL.Stanford.EDU (Igor Rivin)
Subject: Re: How's your geometric intuition?
Organization: Computer Science Department,  Stanford University.
Date: Sat, 2 Jan 1993 20:33:39 GMT

In article <1992Dec31.222328.22744@samba.oit.unc.edu> Jim.Buddenhagen@launchpad.unc.edu (Jim Buddenhagen) writes:
>Start with a regular tetrahedron circumscribed by a sphere.  Subdivide
>each triangle into 4 triangles by connecting edge midpoints, and project
>to the sphere (radially), to obtain a new polyhedron with all vertices
>on the sphere and with 16 triangular faces.
>
>Repeat this procedure with the new polyhedron to get a third polyhedron.
>
>True or false:  this third polyhedron is convex.
>
>(I found the answer surprising, what about you?)
>--

A simpler example of this phenomenon can be gotten by taking a regular
octahedron inscribed in the sphere and stellating the faces (that is,
adding a vertex in the middle of each face, connected to the vertices
of each face), and then pushing the new vertices out to the sphere.
This will not be convex, no matter where you put the new vertices.
What's more, there is no convex polyhedron inscribed in the sphere
combinatorially equivalent to the stellated icosahedron (which only
has 24 faces, but is not the smallest example). 

These kinds of questions go back a long way (to Descartes), and seem
to be finally well understood, through some work of mine on hyperbolic
geometry; see the research announcement in the most recent (Oct 92)
Bulletin of the AMS. I can send you the actual preprints/reprints, if
you are interested. 


	Igor Rivin


Newsgroups: sci.math,rec.puzzles
From: freier@mail.physics.utah.edu (rodney james freier)
Subject: Re: Integral Puzzle (Cute, not Evil)
Organization: University of Utah - Physics Department
Date: Sat, 26 Dec 92 08:00:20 GMT

In article <1992Dec19.011244.2780@Csli.Stanford.EDU>, hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
|> 
|> Let f be any function such that:
|> 	* for 0<=x<=1; 0<=f(x)<= a
|> 
|> 	  /1
|> 	* |  f(x)dx = A
|> 	  /0
|> where a and A are given constants (obviously, A<=a).
|> 
|> What are the maximum and minimum values that
|> 
|> 	/1       2
|> 	|  {f(x)}  dx
|> 	/0
|> 
|> can take?
|> The values can be easily induced, but come up with
|> a simple (cute) proof.
|> 
|> -Yuzuru Hiraga


Eeep.  Doesn't the second expression become undefined if we choose f(x) to
be A*delta(x-.5) or the like?  I guess my point is that there may be functions
for which the first expression is just peachy, but the second becomes pathological.
From: Van@cup.portal.com (Sylvan Van Jacques)
Newsgroups: sci.math
Subject: Re: help with vectors
Date: Mon, 21 Dec 92 19:07:03 PST
Organization: The Portal System (TM)

If i,j, and k are the unit vectors along the 3 axes, and t is some
parameter, call the first vector r1 and the 2nd r2.

Then find r3 = r1 - r2 by subtracting the components.

Let r3(i), i = 1,2,3 be the 3 components of r3. The distance between the
vectors r1 and r2 is
 
 d = sqrt[r3(1)^2 + r3(2)^2 + r3(3)^2]

 van@cup.portal.com