HDTenEightyP/NetNews huggingface dataset:
From: kranzer@adx.adelphi.edu (Herbert Kranzer)
Subject: Re: More on Huygens' principle
Organization: Adelphi University
Date: Fri, 25 Dec 1992 04:38:00 GMT

columbus@strident.think.com (Michael Weiss) writes:


>John Baez has posted the outline of two computations that show why Huygens'
>principle holds for the wave equation with an odd number of spatial
>dimensions, except 1, but fails in even dimensions.

>I'm a little puzzled about the exception of 1.  Isn't u = f(x+t) + g(x-t) a
>general solution to the wave equation in 1 dimension?  This appears to
>propagate without leaving "echoes", i.e., if the supports of f and g for
>t=0 are contained in [-a,a], then the support at t>0 is contained in the
>union of [-a-t, a-t] and [-a+t, a+t].  This should work also for
>distributions.  What am I missing?

 
Close, but no cigar.  Imagine two functions f and g with support R such
that  f + g  has support in [-a, a]; for example, f(x) = sgn(x+a) and
g(x) = -sgn(x-a).  Then u(x,0) has support in [-a, a], while the support
of u(x,t) is the entire interval [-a-t, a+t].
 
Your argument does, in fact, tell half the story.  If you look at the
d'Alembert solution to the initial value problem  u(x,0) = v(x),
u_t(x,0) = w(x) for the one-dimensional wave equation  u_tt = u_xx,
you will find that v(x) propagates as a sharp front as in Huyghens' 
principle, but w(x) leaves echoes.


Herbert Kranzer                Dept. of Math. & Computer Sci.        
kranzer@adx.adelphi.edu        Adelphi University         
                               Garden City, NY 11530
From: otterw@prl.philips.nl (otter den w)
Newsgroups: sci.math
Subject: Re: a + bx + cx^2 + dx^3
Date: 22 Dec 92 15:14:05 GMT
Organization: Philips Research Laboratories Eindhoven, Netherlands

ghica@fig.citib.com (Renato Ghica) writes:

>does anyone know of a c/c++ algorithm to solve equations of 
>this type for a range of x's ? At some ftp site maybe?

>thanks.

Well, this problem can be solved very simple:

  a x^3 + b x^2 + c x + d = 0

Substitute  x = y - b/3a

This way we get rid of the quadratic term. We end up with
something like this:
  
  a y^3 + p y + q =0

Now substitute y = z - p/3az

After some rewriting, multiplying by z^3 and replacing z^3 by w
we end up with a quadratic equation. I guess you can solve that
one.

Performing all substitutions to return at x we end up with 6
solutions, of which a couple are identical. That's all.



I've heard there is a simular method for fourth-order
polynomials. Does anybody out there know this method?


__________________________________________________________________
Wouter den Otter,   otterw@prl.philips.nl
Xref: sparky sci.math:17367 sci.physics:21670
From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math,sci.physics
Subject: More on Huygens' principle
Date: 23 Dec 92 11:49:33
Organization: Thinking Machines Corporation, Cambridge MA, USA
NNTP-Posting-Host: strident.think.com


John Baez has posted the outline of two computations that show why Huygens'
principle holds for the wave equation with an odd number of spatial
dimensions, except 1, but fails in even dimensions.

I'm a little puzzled about the exception of 1.  Isn't u = f(x+t) + g(x-t) a
general solution to the wave equation in 1 dimension?  This appears to
propagate without leaving "echoes", i.e., if the supports of f and g for
t=0 are contained in [-a,a], then the support at t>0 is contained in the
union of [-a-t, a-t] and [-a+t, a+t].  This should work also for
distributions.  What am I missing?

If one assumes that Huygens' principle holds in 3 spatial dimensions, one
can deduce that it should fail in 2.  Hadamard called this argument the
"method of descent".  Briefly: consider a pulse disturbance at t=0 along
the entire z-axis.  It creates spreading waves that obviously will have no
dependence on z, and hence satisfy the 2-dimensional wave equation.  Now we
apply Huygens' principle in 3 dimensions--- that is, we add up spherical
waves spreading from each (0,0,z) (i.e., integrate over z).  An observer
positioned at (x,y,0) will of course "hear" the pulse at t = sqrt(x^2+y^2),
due to the spherical wave spreading out from (0,0,0) but will also hear
something at t = sqrt(x^2 + y^2 + z^2), thanks to the spherical wave from
(0,0,z). QED (I was hoping there was a similarly intuitive argument for why
Huygens' should work for 3 dimensions, but I guess not.)

Did the gods give us three spatial dimensions so music would sound nice?
Perhaps, but if so, they made Huygens' principle fail in two dimensions so
skipping stone on a still pond would produce pretty ripples!  Question for
the net: which is more important in determining the residual ripples, this
"anti-Huygens" effect, or the fact that water waves are more complex than
Laplacian(f) = f_tt?


Newsgroups: sci.math
From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
Subject: Re: need proof:  (1 + 1/n)^n ==> e
Organization: Computer Science Department,  Stanford University.
Date: Thu, 31 Dec 1992 07:30:54 GMT

In article <1992Dec31.045851.26887@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
>
>Problem.  Give comparably elementary proofs of the stronger result that
>X_n (not just X_N) has a limit, and the yet stronger result that X_n is
>monotone increasing [where X_n = (1+1/n)^n --- the point here is to do
>this without resorting to either the arithmetic-geometric mean inequality
>or the tedium of binomial expansion.].

Seems like a good dinner is almost as effective as sleeping on a
problem.

Lemma.  (1-1/n^2)^n >= 1-1/n, for integer n>0.

Proof.  Shaving one face of the unit n-cube to a depth of 1/n^2 removes
volume 1/n^2, so shaving n faces removes at most volume 1/n.

(Is there a comparably simple algebraic argument?  Disturbing if not.)


Proposition.  X_n increases monotonically.

Proof:  X_n / X_{n-1}  =  (1+1/n)^n * (1-1/n)^{n-1}
		       =  (1-1/n^2)^n / (1-1/n)
		       >=  1   by the Lemma.

This plus Lemma 2 (Y_N increases monotically---I don't see a comparably
simple proof that Y_n increases monotonically) from my previous message
then allows us to bound X_n via

	X_n  <  X_N  <  1/Y_N  <  1/Y_1.

where N is any power of 2 >= n (recall that 1/Y_N = (1+1/(N-1))^N).

I think this offers everything obtainable from the AGM argument.  Still
not as short as my elementary argument that X_N is monotone though.
-- 
Vaughan Pratt				There's safety in certain numbers.
From: eidlheit@bimacs.BITNET (Gadi Eidelheit)
Newsgroups: sci.math
Subject: Transferring Product to Sigma
Date: 30 Dec 92 10:59:55 GMT
Organization: Math & CS, BarIlan U, Ramat-Gan, Israel

Hello all!
        I need an algoritham or a method (or computer algoritham)
to convert a product of the type:
            n
         ------
          |  |   (X-Xi)
          |  |
           i=1

to A sigma in the from


       n
     -----
     \       CiX^i
     /
     -----
      i=0


what i look for is of course the constants Ci.
I find out by trying that each that Ci is the Sigma of all
combinations of products of (n-i) Xj
and that the sign of Cn is always positive (and equal 1)
and the sign of C(n-1) is begative and so on.

Its quite hard coding all of this into a program and i guess someone
has/know of a better way to do it

any suiggestions/ideas/corrections
        Please email me directly
                to eidlheit@bimacs.biu.ac.il

                Thanks in advance
                        Gadi Eidelheit
Newsgroups: sci.math
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Subject: Re: Probability and QM
Organization: Purdue University Statistics Department
Date: Thu, 24 Dec 1992 13:27:26 GMT

In article <1992Dec23.130300.6812@oracorp.com> daryl@oracorp.com (Daryl McCullough) writes:
>naturman@uctvax.uct.ac.za writes:

			..........................

>>The debate over whether QM uses "classical" probability would be aided if
>>someone actually bothered to define what they mean by "classical" or
>>"non-classical"

>Someone has actually bothered to define the difference. In short,
>classical probability theory assigns probabilities to sets of
>outcomes. (A set of outcomes is called an "event"). The sets that are
>assigned probabilities form a sigma-algebra. That is, they are closed
>under set difference and countable unions and intersections. 

>This allows us to think of events as statements, in a certain sense,
>and to think of the probability assigned as a generalized truth
>value. If A and B are events, then so are (A /\ B), and (A \/ B).

>Nonclassical probability theory is any departure from this scheme.
>There is more than one way to generalize things, of course, but in
>particular one may abandon the requirement that the measurable sets
>form a sigma-algebra. For example, there may be events A and B such
>that (A /\ B) is not an event (it is not assigned a probability).

In quantum mechanics, given a list of observations which can be 
made jointly, the distribution of the results satisfies the ordinary
rules of classical probability.  If an "alternate worlds" model could
explain the results, it would be possible to extend this even to the
case where the observations cannot be made jointly.  This cannot be
done; easily obtained contradictions arise.

Now we can restrict probability.  When we do this, do we have a 
model which is at least in principle one from which an initial state
will provide whatever can be stated for the future, as QM does?  If
probability is restricted too much, the predictions cannot be made.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@snap.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!snap.stat!hrubin(UUCP)
Xref: sparky sci.math:17411 alt.books.technical:387
From: ng4@husc10.harvard.edu (Ho Leung Ng)
Newsgroups: sci.math,alt.books.technical
Subject: Re: High Prices of Math Books. I am pissed.
Date: 25 Dec 92 07:52:55 GMT
Article-I.D.: husc.ng4.725269975


   Dover manages to put out lots of fairly specialized books in math
and physics and sell them in paperback editions for <$15.  How is it that
they can stay in business?

Ho Leung Ng
Newsgroups: sci.math
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Subject: Re: Measures, and Measurability
Organization: Purdue University Statistics Department
Date: Mon, 21 Dec 1992 13:53:05 GMT

In article <1992Dec21.053850.13489@news.media.mit.edu> minsky@media.mit.edu (Marvin Minsky) writes:
>In article <1992Dec18.191446.7806@panix.com> banana@panix.com (Walter Polkosnik) writes:
>>I am reading a few papers, and they mention the concept of measures
>>(particularily Lebesque measures) and the concept of measurability and the
>>non-measurability of a set. Can anyone provide me with references, or are the
>>concepts simple enough to explain in e-mail or a post?

>>Thanks. I'm just a Physicist, so be gentle

>This may be hard to believe, but one of the easiest introductions is
>in an old book by John von Neumann: "Mathematical Foundations of
>Quantum Mechanics".  I haven't seen it for ever so long, but I
>remember it as being very clear and intuitive.  Starts out with nice
>proof of metric density theorem, which says that if a set of an
>Euclidean space is measurable, then it contains rectangles whose
>measures are arbitrarily close to 1.

This certainly is not the case unless a very unusual definition of 
rectangle is given.  There exist quite a few sets of measure 0 which
intersect all rectangles.   But this is not all; it is not the case
that measurable sets must contain any rectangle except for a set of
measure 0.

There exist easily constructed measurable sets in the line which do not
contain any interval.  These can be constructed like the Cantor set, by
removing smaller parts.  As one can also add middle parts to the intervals
deleted, one can get measurable sets such that the intesection of the set
with any open interval, as well as the complement, have positive measure.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@snap.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!snap.stat!hrubin(UUCP)
Newsgroups: sci.math
From: ara@zurich.ai.mit.edu (Allan Adler)
Subject: Re: more math puzzles
Organization: M.I.T. Artificial Intelligence Lab.
Date: Wed, 23 Dec 1992 16:53:12 GMT



I'm not entirely convinced by Macolm Beattie's explanation that the tangent
bundle to L is nontrivial and there is another point he raises that I
would like to address.

Suppose the tangent bundle to L is nontrivial. Then as he points out
there is a nowhere vanishing vector field on L. I don't integrate
vector fields very often, but it seems to me that this implies that
there is an action of R on L, where R is the additive group of real
numbers, which I will refer to as translation on L. Translating by
a real number x will be a mapping from L to itself and any mapping
from L to itself fixes a closed cofinal subset. This is therefore
true for x=1, x= 1/2, x=1/4, and so on. The intersection of countably
many closed cofinal subsets of L is itself closed and cofinal,
so there is a closed cofinal (in particular nonempty) subset of L
every points of which is fixed by all translations. The vector field
vanishes at such a point.

It is not true that every continuous map from R to L is eventually constant.
R can be mapped homeomorphically onyo an open interval of R, hence of
one of the intervals used to make up L. It is, however, true that
any map from L to R is eventually constant. More generally, any
map from L to a metric space is eventually constant. To prove this,
note that L is sequentially compact and that any continuous map must
preserve sequential compactness. So if f: L--> M is continuous and
u is in L then, denoting by Lu the part of L above and including u,
f(L_u) is sequentially compact and hence compact, since M is a metric
space. Therefore, the intersection of f(L_u) is nonempty as u goes
to infinity in L. That  proves there is a u such that f(u) is
taken on cofinally as a value. Since any two cofinal closed subsets
of L meet, it follows that f is eventually constant and equal to f(u).

One application of this is that L is itself not metrizable. Here is another:
it is proved in Singer and Thorpe, e.g. that the product of continuum
many unit intervals is not metrizable. But using the long line, you can
prove that the product of aleph-1 unit intervals is not metrizable:
for each countable ordinal b, you can map L onto the b-th interval that
makes up the long half line by making the map the identity on the interval,
scrunching stuff before the interval into the left endpoint and scrunching
stuff after the interval into the right endpoint. That gives a family
of mappings into [0,1] indexed by the countable ordinals and these maps
separate points of the long half line(which is what I actually proved the 
property for, i.e. that maps into metric spaces are eventually constant).
The product of this family of maps is a continuous injective map from the
long half line into the product of aleph-1 copies of [0,1]. Therefore,
that product is not a metric space.

This proves that the long line is good for something. :)

Allan Adler
ara@altdorf.ai.mit.edu
From: israel@unixg.ubc.ca (Robert B. Israel)
Newsgroups: sci.math
Subject: Re: Two problems
Date: 30 Dec 92 22:07:45 GMT
Organization: The University of British Columbia
NNTP-Posting-Host: unixg.ubc.ca

In <1992Dec30.185116.29392@maths.tcd.ie> tim@maths.tcd.ie (Timothy Murphy) writes:

>israel@unixg.ubc.ca (Robert B. Israel) writes:

>>Suppose there are n couples.  
>>I'll assume that the men are seated in numerical order, so husband 
>>#i+1 is two places to the right of #i.

>I like your silent assumption of perfect etiquette!
>Thinks: must revise my idea of California ...

Actually, Canada.  

BTW, while I'm here I should correct a minor slip in my posting.  I
wrote that phi(s) -> exp(2 e^{i s}), which should of course be
exp(2 e^{i s} - 2).  And I forgot to answer the original question:
the probability is exp(-2).

Maybe I misread the original problem.  I thought that the assumption was
that men and women alternated around the table.  Looking back, I see that
that wasn't stated.  Actually, though, the original poster ( Samuel Gustaf 
Siren, SAMUEL@front.se, in <1992Dec28.220454.129@front.se>) appears to
have tacitly made that assumption as well, judging from his numerical
results.  E.g. for two couples, he has p(2,0) = 0, while if arbitrary
seating was allowed each man could have the other man on one side and the
other man's wife on the other. 

I'll have to check what the answer would be for arbitrary seating.  Well, I
can say what it _should_ be.  The probability of any particular man being 
on his wife's left would be 1/(2n-1), so the expected number of such men
would be n/(2n-1).  The expected number sitting next to their wives would
be 2n/(2n-1).  If the limiting distributions are still Poisson, the answers
would be exp(-1/2) for problem 1 and exp(-1) for problem 2.
 
-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             or israel@unixg.ubc.ca
University of British Columbia
Vancouver, BC, Canada V6T 1Y4
Newsgroups: sci.math
From: wielinga@newton.physics.uq.oz.au (Bruce Wielinga)
Subject: An Integral Problem
Organization: Physics Dept. The University of Queensland
Date: Mon, 4 Jan 1993 02:29:35 GMT


	Hello people,
 	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,
	
	Sorry about the formating. I have tried to find this in tables etc, but    
to no avail.

		Thanks,
		Bruce Wielinga. 
Newsgroups: sci.math
From: nissim@mary.fordham.edu (Leonard J. Nissim)
Subject: Re: Help wanted with equation !
Organization: Fordham University
Date: Mon, 21 Dec 1992 16:58:00 GMT
News-Software: VAX/VMS VNEWS 1.41    
Distribution: usa
X-Posted-From: mary.fordham.edu
NNTP-Posting-Host: sol.ctr.columbia.edu

In article <1992Dec21.155633.14071@fwi.uva.nl>, schavema@fwi.uva.nl (John Schavemaker) writes...
>	Hi,
> 
>	Does anyone know how to solve the following equation:
> 
>		7 * y + 3 = x * x
> 
>	Both the variables y and x are integers. Otherwise stated:
>	Can a square be the addition of a multiple of seven and
>	three ? Any help appreciated, e-mail prefered.
> 
> 
>				John Schavemaker
> 
>---
>John Schavemaker
>schavema@fwi.uva.nl
>-- 
>John Schavemaker
>schavema@fwi.uva.nl

There are no solutions to the equation. 
Brute force proof: 
Modulo 7, note that 0*0=0, 1*1=1, 2*2=4, 3*3=2, 4*4=2, 5*5=4, and 6*6=1.
So 3 is not a square in Z/7Z.

For an elegant proof (or the solution in general), look in a number 
theory book under "quadratic residues".

-------------------------------------------------------------------------------
Leonard J. Nissim  (nissim@mary.fordham.edu)   
Disclaimer:  "I speak only for myself."
-------------------------------------------------------------------------------
From: cooklev@ss.titech.ac.jp (COOKLEV)
Newsgroups: sci.math
Subject: Bernstein approximants
Date: 25 Dec 92 10:02:57 GMT
Distribution: sci
Organization: Tokyo Institute of Technology, JAPAN



What about the the following Bernstein approximant

 N
----                                  alpha(i)      beta(i)-alpha(i)
\       alpha(i)   Gamma[alpha(i)+1] x         (1-x)
/    f(---------) ---------------------------------------------
----    beta(i)   Gamma[alpha(i)+1] Gamma[beta(i)-alpha(i)+1]
i=0

Using the property of the Gamma function Gamma (x+1)=x Gamma (x), this
Bernstein approximant is readily identified as a generalization
of the Bernstein polynomial for alpha (i) = i and beta (i)=N. The
sampling pattern of the function is specified by the ratio
alpha(i) / beta(i). 

The above Bernstein approximant has been found useful in some applications
for simple functions alpha (i) and beta(i). 
However I am not a mathematician and for me it is difficult to evaluate it.
My questions are
1. Is the suggested generalization of the Bernstein polynomials important?
2. Is it new?
3. What properties does it have?

I would welcome any comments and suggestions.

Merry X'mas and Happy New Year to all,

Todor Cooklev, Dept. Physical Electronics    |
Tokyo Institute of Technolgy                 |
2-12-1 Ookayama, Meguro-ku, Tokyo 152, Japan |
E-mail:  cooklev@ss.titech.ac.jp             |
From: arodgers@dcs.qmw.ac.uk (Angus H Rodgers)
Newsgroups: sci.math
Subject: Re: Negative Zero
Keywords: -0
Date: 23 Dec 92 16:32:24 GMT
Organization: Computer Science Dept, QMW, University of London

In <1992Dec21.190158.4586@galois.mit.edu>
jbaez@riesz.mit.edu (John C. Baez) writes:

>[...] I would imagine that
>after a while he'd get bored, start thinking about the problem and come
>to some conclusion.  Maybe: "Infinity MUST exist.  If it didn't this
>program would have halted by now!"

Or: "1 exists. 2 exists. 3 exists. ... 10,677 exists. Hmm -- impressive
enough evidence. Conjecture: infinity exists. ... Proof? Hmm -- try
mathematical induction? Yes, that's great! 0 exists. And if n exists,
so does n+1. So n exists, for all n. But there are infinitely many n;
so infinity exists. QED."  :-)
--
Gus Rodgers,  Dept. of Computer Science, |  [Profound thought for 1993]
Queen Mary & Westfield College, Mile End |     "Philishave is about
Road, London, England.   +44 71 975 5241 |      personal choice" 
E-mail (JANET):   arodgers@dcs.qmw.ac.uk |             -- TV advert
From: gmmiller@mcs.drexel.edu (Michael R. Miller)
Newsgroups: sci.math
Subject: Applied Math - what do YOU do?
Summary: Survey of applied mathematicians
Keywords: applied math visualization modeling
Date: 30 Dec 92 06:40:59 GMT
Distribution: sci.math,sci.math.symbolic
Organization: Drexel University

I would like to gather information regarding some of the more 
interesting work done by applied mathematicians in industry and 
in academia.

I am especially interested in users of tools such as Mathematica, 
Maple, Matlab, et. al. in visualization, modeling, and data analysis work.

I welcome all responses, but am curious about the following areas:
     medical research
     image processing (esp. fractals)
     space, aeronautics
     weather forecasting
     physics (esp. lasers)
     signal processing
     information theory

Please e-mail your responses (short or long!) directly to me 
(gmmiller@mcs.drexel.edu).  The survey is for my own use (academic and 
career guidance).  If the responses warrant it, I will summarize and 
post the results.

If you love your job, here's a chance to brag about it!    :)

Thanks very much -- and happy 1993!

Mike Miller
(gmmiller@mcs.drexel.edu)
Xref: sparky sci.math:17452 alt.books.technical:397
From: hdev@dutiai.tudelft.nl (Hans de Vreught)
Newsgroups: sci.math,alt.books.technical
Subject: Re: High Prices of Math Books. I am pissed.
Date: 28 Dec 92 09:15:45 GMT
Organization: Delft University of Technology

gh@cs.toronto.edu (Graeme Hirst) writes:

>In article <1992Dec24.194233.10811@linus.mitre.org> bs@gauss.mitre.org (Robert D. Silverman) writes:
>>Why are people always surprised that technical books with a small audience
>>are expensive?

I'm not.

>This misses the point of the original posting, which was this:  Some
>publishers (e.g., McGraw-Hill, Prentice-Hall, Addison-Wesley, and other
>names with hyphens) will make the *same* book in different bindings at
>different prices in different parts of the world.   In particular, they
>will sell a cheap paperback edition only outside North America, while
>insisting that North American customers buy the hardcover edition at a
>much higher price.

I don't think that paperbacks aren't that cheap over here. Text books are
generally far more cheapier in the States than over here in Europe. Don't
forget that S&H, sales tax, and the "middle man" from the States to Europe
make books more expensive.

Well, since the States, Asia, and Europe are far apart, publishers can sell at
different prices in different parts of the world. Nice? No, but that's life.
--
Hans de Vreught                          | John von Neumann:
hdev@dutiba.twi.tudelft.nl               | Young man, in mathematics
Delft University of Technology (TWI-ThI) | you don't understand things,
The Netherlands                          | you just get used to them.