Subject: Re: Laws of Flight
From: "Jack W. Crenshaw"
Date: Thu, 14 Nov 1996 05:46:51 -0500
Gerry Quinn wrote:
>
> I wonder can any aeronautical experts answer these questions:
>
> (i) If the density of air increases, does flight become easier, and if so in
> what proportions? Or does it depend on the velocity?
Absolutely, it becomes easier. It also depends on the velocity, as
usual. Lift and drag both depend on dynamic pressure, which is
Q = (1/2) rho*v^2
where rho is the density. Double the density, and you can lower the
velocity. Pilots know this, and speak of density altitude -- the
equivalent altitude at standard temperature (density goes down as temp
goes up, of course). They know that takeoff speed and landing speed are
higher, at higher density altitudes.
Birds know it too, which is why they love clear, cold mornings, in the
middle of barometric highs.
>
> (ii) Suppose a large dome was held at high pressure so as to create the
> maximum safe atmospheric density ( 10-20 kg/m3 I suppose ). Would
> human-powered flight be possible inside such a dome?
Good question, and the answer is, of course, yes, since human-powered
flight is (just barely) possible without the dome. I think what you're
suggesting is, wouldn't it be much _EASIER_ inside. Answer: yes.
Especially if you also made it as cold as bearable.
Sounds like a new craze to replace bungee jumping!
Jack
Subject: Re: Autodynamics
From: tim@franck.Princeton.EDU.composers (Tim Hollebeek)
Date: 14 Nov 1996 10:29:51 GMT
In article <56drbo$4pj@bunyip.cc.uq.oz.au>, dean@psy.uq.oz.au writes:
> Erik Max Francis writes:
>
> >Dean Povey wrote:
>
> >> In AD gravitation, the perihelion advance for each planet is
> >> proportional to the square root of the division of the solar mass by
> >> the orbital radius power 3.
> >>
> >> Tp = sqrt(M / r^3) [ditto: DGP]
>
> >Care to derive this?
>
> >> If the Mercury value is taken as 43" . . . .
>
> >Do you _actually_ mean that Autodynamics can't predict Mercury's perhelion
> >precession without being given it? That's not very impressive. Right
> >there general relativity has a head start on you.
>
> From what I can gather from the web pages, the AD equation uses a constant
> which indicates the quantity of mass received from pico-gravitons
> per each gram of mass present, per second. This is a universal constant which
> is the same for all celestial bodies. Hence, the input of Mecury's perhelion
> advance is merely a method to calculate this constant. (You could predict
> Mercury's perhelion advance by using accurate observations of another body to
> calculate the constant.)
>
> I don't see much wrong with this, you find constants throughout physics,
> (eg. the GR equation uses G and pi).
Your forget the Fundamental Rule of Physics:
He who dies with the least unexplained constants wins.
BTW, pi is just a number, not a measured constant. It's a good distinction
to keep in mind.
---------------------------------------------------------------------------
Tim Hollebeek | Disclaimer :=> Everything above is a true statement,
Electron Psychologist | for sufficiently false values of true.
Princeton University | email: tim@wfn-shop.princeton.edu
----------------------| http://wfn-shop.princeton.edu/~tim (NEW! IMPROVED!)
Subject: Re: cuts and codes in models of arithmetic
From: ikastan@alumnae.caltech.edu (Ilias Kastanas)
Date: 14 Nov 1996 10:42:42 GMT
Organization: Caltech Alumni Association
Keywords:
Cc:
In article <2gohh1jzbd.fsf@pulsar.cs.wku.edu>,
Allen Adler wrote:
>
>I'm reading J.B.Paris' paper "A hierarchy of cuts in models
>of arithmetic". I don't understand some of the basic notions.
>
>First some notation.
>
>Let P denote Peano's 1st order axioms for arithmetic.
>Let I\Sigma_n be the same set of axioms but with induction
>confined to \Sigma_n formulas.
>Let N denote the standard model of arithmetic.
>Let M denote a countable nonstandard model of I\Sigma_1.
>Each element x of M is identified with the initial segment
>consisting of all of the predecessors of x in M.
>
>Definition: A subset I of M is called a cut if I contains 2,
>is closed under multiplication and is an initial segment of M
>other than M.
>
>Definition: Let I be a cut of M. Let A be a subset of I.
>We say that A is coded in M if there is some element b of M
>such that for all x in I, x belongs to A if and only if the
>x-th prime of M belongs divides b. (Presumably one speaks of
>b as being the code of A).
>
>Here is what I don't understand.
>(1) Paris writes: "Similarly we can talk of f:I->M being coded.
> Henceforth all subsets of I or functions on I
> etc. which are mentioned are assumed to be coded in M unless
> otherwise stated." Now, I can imagine how that might work,
> e.g. use a bijection between MxM and M to think of a function
> as a subset of M. But is that what is actually intended? Even
A pairing function might use unique factorization, or some other
method; in most cases the details do not really matter. Something ana-
logous happens when using nonstandard integers to encode infinite sets.
Consider: a subset A of w (omega) is called _standard on M_ (M a
model of PA) if for some formula P(x, y) and for some element c of M,
A = { n in w: M |= P(n, c) }. Let us take the canonical indexing of
finite sets, D_x (i.e. D_0 = empty, D_x = { x0, x1, ..., xi } where
x = 2^x0 + 2^x1 + ... + 2^xi, x0 < x1 < ... < xi) and apply it for
x = an infinite b; then D_b can be infinite, and can be thought as enco-
ding its standard part (D_b intersect w). So if M is nonstandard then
b in M codes a standard set A iff A = { n in w: M |= n in D_b } =
= D_b intersect w. It is one way of setting up codes.
It is easy to see that for any b in M what b encodes is standard on
M. Conversely, any A standard on M is encoded by some b... in fact, it
has arbitrarily small infinite codes.
D_x then is a reasonable way to code. Of course so is E_x = { n in w:
p_n divides x}, and its extension E_d for infinite d, following Paris's
approach. PA easily proves Ax Ey E_y = D_x; so the "choice of details"
is a matter of convenience. E.g. the E's are preferable when proving
Tennenbaum's theorem, that in any (countable) nonstandard M every A standard
on M is recursive in each of Add and Mul (of M) (and hence neither Add nor
Mul can be recursive).
> if it is, I still don't understand the following:
>(2) Paris writes: "Suppose f: I -> a is coded by e in M. Then by
> overspill there is a b > I [i.e. b > x for all x in I] such that
> e codes a map f': b->a with f' restricted to I = f Hence we
> have a continuation of f a little way above I."
>
>My vague intuition is that what is going on in (2) is that
>one can't actually define I in M, so whatever one can say about
>f using the code e will have to define something other than I,
>but necessarily containing I. According to that intuition,
>(2) seems to make sense. However, I would feel better about
>it if I could see a formal proof.
Your intuition is right; the simplest form of Overspill is that
in any nonstandard M, w is not = { a in M: M |= P(a, b) }, for any b in M.
That is, M |= P(n, b) for all n in w iff for some infinite c in M,
M |= Ax < c P(x, b).
You can parlay this into Strong Overspill: if f: M -> M is definable
by a formula with parameters from M, then: f(n) is infinite for all n in w
iff for some infinite c in M, f(a) is infinite for all a < c. Strong
Underspill also follows: f(a) is finite (in w) for all infinite a in M
iff there is an n in w so that f(c) is finite for all c > n. Our cup
runneth over.
The proof of Strong Overspill is easier than it seems: if for all n
f(n) is infinite, then for all n f(n) > n; apply Overspill.
One application of Strong Overspill, possibly relevant to what Paris
is doing, involves _indicators_ for a family of initial segments (an f so
that there is an initial segment including x but not y iff f(x, y) is
infinite). If the family contains w and has a definable indicator then it
contains an initial segment other than w; and f(x, y) = least z such that
y < (x + 2)^z is an indicator for initial segments closed under multi-
plication.
Ilias
Subject: Re: measure
From: JC
Date: Thu, 14 Nov 1996 10:53:17 +0000
David Ullrich wrote:
>
> JC wrote:
> >
> > JC wrote:
> > >
> > > Biblioteca matematica wrote:
> > > >
> > > > let C be Cantor set
> > > > can you find a subset D of R that is omeomorph with C and such that its
> > > > lebsgue measure isn't zero?
> > >
> > > No. Such a subset D would have to be compact (because C is) hence
> > > closed in R, and nowhere dense (because it must be 0-dimensional,
> > > hence interiorless). Such a set necessarily has 0 measure.
> > >
> > > JC
> >
> > Oops! Mea culpa. I was assuming that all copies of C in the reals
> > were homeomorphic via a homeomorphism which extends to the reals.
>
> Huh? Actually I believe this is true, so assuming it should
> not do any harm. How would this imply what you said about measure?
> (Hint: It doesn't.)
>
Okay, then I'm really confused. The jist of my argument was this.
Suppose you have copies of the Cantor set C1, C2 embedded in the
reals. We can assume they both lie within [0,1], and (you claim,
and I'm inclined to agree) there is a (monotone increasing)
homeomorphism h:[0,1]->[0,1] taking C1 to C2. But such a map is
uniformly continuous and therefore.... AH! The penny's dropped.
I was assuming such a map preserved 0-measure for the simple
reason that there is some K>0 s.t.
|x-y| < K\epsilon => |hx-hy| < \epsilon
which is of course bullsh*t.
Ho hum. Perhaps I should take up stamp collecting.
JC
Subject: Vietmath War: where the p-adics are essential in physics
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 14 Nov 1996 05:46:54 GMT
In article
David Kastrup writes:
> Ludwig, I like to call you Ludwig still, do you mind? I am
> divorced now because I would lecture my wife in bed instead
> of doing the physics she wanted. I have the scherr habit of
> lecturing even though I don't understand what I am lecturing
> about! I have been a pedantic lecturing fool all of my life,
> and it is an uncontrollable habit of mine.
In article <1993Dec4.013650.12700@Princeton.EDU>
wiles@rugola.Princeton.EDU (Andrew Wiles) writes:
>> Not quite. Mathematics does not care one hoot about reality or
>> applicability, it cares about consistency.
In article (Gerd
Faltings@Max-Planck-Institut.Bonn) Gerd Faltings writes:
>>> I can develop a number system in which 1+1=0 and work with it and
>>> derive theorems about it quite fine as long as I keep consistent. It
>>> does not matter for this that one sheep plus one sheep does not make
>>> no sheep. Sheep are not good for modulo 2 arithmetic. But if I look
>>> carefully, almost every mathematic system *can* be applied in some
>>> ways: calculation modulo 2 is quite well-suited to finding out whether
>>> the light is on depending on how many people happened to throw the
>>> switch.
In article <1993Dec4.013650.12700@Princeton.EDU>
wiles@rugola.Princeton.EDU (Andrew Wiles) writes:
>> That's the difference: in physics, different world models are sort of
>> "winner takes all" oriented (although no winner is up to now none,
>> only quite a lot of non-winners been thrown out of the race).
>> I am a winner.
In article (Gerd
Faltings@Max-Planck-Institut.Bonn) Gerd Faltings writes:
>>> In mathematics, Newtonian mechanics and relativistic mechanics could
>>> coexist quite nicely: different axiomatic systems do not need to obey
>>> the same laws as long as they obey their respective axioms.
>>>
>>> That one of them applies better to modern reality does not make it
>>> mathematically illegitimate, only physically. It just happens that
>>> *relative* speed counts in the universe, not absolute.
>>>
John.Coates@University.of.Cambridge
(John Coates) writes :
>>>> Depends on what you mean by "wrong". In mathematics you are allowed
>>>> to do crazy things (like allowing a fake proof to get published and
>>>> ignoring any opposition) and see where that would take you, as long as you
>>>> carefully watch that you are not mixing up your "real-world"
>>>> expectations with actual consequences of the changed systems.
In article (Gerd
Faltings@Max-Planck-Institut.Bonn) Gerd Faltings writes:
>>> Not at all, the link is one-way. It might, however, make more
>>> physicists interested in a branch of mathematics (p-adics) which they
>>> otherwise would rather choose to ignore.
>
> But all this is one-way: being able to apply real numbers or p-adic
> ones or whatever does not influence the validity of the use of natural
> numbers, but at most the interest taken in them.
>
> --
> David Kastrup Phone: +49-234-700-5570
> Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
> Institut fuer Neuroinformatik, Universitaetsstr. 150, 44780 Bochum, Germany
I changed my mind, Gerd, mind telling Witten tomorrow when you
telephone him that I now think the first case of where the p-adics are
found essential in physics and where the Finite Integers are inadequate
is ' harmonic oscillators ' such as springs and even the Coulomb force
law. I first thought that the Quantum Hall Effect of its bizarre math
numbers will be the first essential need for p-adics but now I think it
is harmonic oscillation. The p-adics in fact are numbers of harmonic
oscillation.
What does it feel like Gerd, to have the physicist show you the
correct mathematics of Naturals = p-adics = Infinite Integers and you
were playing with the silly fiction of Naturals = Finite Integers.
Please ask Witten for he knows physics. Need to find out where in
physics the p-adics are essential and simultaneously where the Finite
Integers are inadequate to do the job.
Subject: Re: 0! = 1 (was: Could life have evolved by chance?)
From: David Kastrup
Date: 14 Nov 1996 12:49:42 +0100
clesley@mesa7.MesaState.EDU (Chris C. Lesley) writes:
> Alan Silver (alan@consultancy-services.com) wrote:
> : David L Evens wrote ...
> : >Actually, it IS meaningful. How many ways can you list the elements o=
f
> : >the empty set? Exactly one. How many ways can you list the elements =
of
> : >a set with one member? Exactly one. The idea of an empty set is quit=
e
> : >well defined.
>
> My problem with this is that the statement "list the elements of the
> empty set" is not clearly defined. What does it mean to list a bunch of
> non-existent objects? How is that different from a list of ONE
> non-existent object?
>
> : This is *still* missing the point. From a mathematical viewpoint you ar=
e
> : correct, but I was not discussing this from a mathematical viewpoint as
> : the person (forgotten who it's been so long) did not express it that
> : way. I made a point of saying that I was approaching this from a point
> : of view of philosophy. I the "real" world (if such a thing exists), the
> : idea of arranging the members of an empty set is meaningless as the
> : empty set has no members.
>
> Exactly.
I disagree:
O =3D { }
and you cannot enumerate the members in any different way, as opposed
to, say
A =3D {1,2,3} =3D {1,3,2} =3D {2,1,3} =3D {2,3,1} =3D {3,1,2} =3D {3,2,1}
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: Fractional Differential
From: David Kastrup
Date: 14 Nov 1996 12:59:04 +0100
Simon Read writes:
> binklmj5@wfu.edu (Mathew J. Binkley) wrote:
> > n
> >d L
> >--- with n not an integer.
> > n
> >dx
>
>
> Take the Fourier transform, then multiply the resulting function
> by (iw)^n, then transform back.
>
> n can be fractional if you feel like it.
>
> That only differentiates a periodic function,
Misconception. This differentiates *any* function satisfying the
Dirichlet criteria (such as a finite number of discontinuities on any
given interval). A Fourier transform is defined on continuously
valued aperiodic functions or distributions. What you are mixing tis
up with is a Fourier series, of which the so-called "Discrete Fourier
transform" is a special case.
> The trouble with that
> method is, it might destroy the property that a derivative is a purely
> local operation.
Sort of. You can *still* regard it as a local operation and develop
it into a Taylor-like series of derivatives at just one point if the
function satisfies certain criteria, but considering all derivatives
of such a function at one point is sufficient for knowing the
function *everywhere*.
On the other hand, integration (that is, differentiation by a negative
degree) is non-local as well, so why bother?
--
David Kastrup Phone: +49-234-700-5570
Email: dak@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut f=FCr Neuroinformatik, Universit=E4tsstr. 150, 44780 Bochum, Germa=
ny
Subject: Re: errorcorrection formula
From: rgep@dpmms.cam.ac.uk (Richard Pinch)
Date: 14 Nov 1996 12:11:57 GMT
In article <01bbd1a1$0fa17540$3880099a@vianen1.involve.nl>,
"Rene Hagen" writes:
|> I need a pascal routine (or basic or c) that performs the following
|> calculations:
|> Each codeword has 21 information bits, which correspond to the coefficients
|> of a polynomial having terms from x30 down to x10. This polynomial is
|> divided, modulo-2, by the generating polynomial x10+x9+x8+x6+x5+x3+1. The
|> check bits correspond to the coefficients of the terms from x9 to x0 in the
|> remainder polynomial found at the completion of this division. The complete
|> block, consisting of the information bits followed by the check bits,
|> corresponds to the coefficients of a polynomial which is integrally
|> divisible in modulo-2 fashion by the generating polynomial. To the 31 bits
|> of the block is added one additional bit to provide an even bit parity
|> check of the whole codeword.
The following table may help; it shows the remainders of x^10,...,x^30 on division
by g(x) = x^10 + x^9 + x^8 + x^6 + x^5 + x^3 + 1 (mod 2 of course)
10 x^9 + x^8 + x^6 + x^5 + x^3 + 1
11 x^8 + x^7 + x^5 + x^4 + x^3 + x + 1
12 x^9 + x^8 + x^6 + x^5 + x^4 + x^2 + x
13 x^8 + x^7 + x^2 + 1
14 x^9 + x^8 + x^3 + x
15 x^8 + x^6 + x^5 + x^4 + x^3 + x^2 + 1
16 x^9 + x^7 + x^6 + x^5 + x^4 + x^3 + x
17 x^9 + x^7 + x^4 + x^3 + x^2 + 1
18 x^9 + x^6 + x^4 + x + 1
19 x^9 + x^8 + x^7 + x^6 + x^3 + x^2 + x + 1
20 x^7 + x^6 + x^5 + x^4 + x^2 + x + 1
21 x^8 + x^7 + x^6 + x^5 + x^3 + x^2 + x
22 x^9 + x^8 + x^7 + x^6 + x^4 + x^3 + x^2
23 x^7 + x^6 + x^4 + 1
24 x^8 + x^7 + x^5 + x
25 x^9 + x^8 + x^6 + x^2
26 x^8 + x^7 + x^6 + x^5 + 1
27 x^9 + x^8 + x^7 + x^6 + x
28 x^7 + x^6 + x^5 + x^3 + x^2 + 1
29 x^8 + x^7 + x^6 + x^4 + x^3 + x
30 x^9 + x^8 + x^7 + x^5 + x^4 + x^2
So given databits x_31,...,x_10, simply add together the remainders of the
corresponding powers of x.
--
Richard Pinch Queens' College, Cambridge
rgep@cam.ac.uk http://www.dpmms.cam.ac.uk/~rgep
Subject: Re: Interesting COUNTING PROBLEM: Unary central relations !!!
From: Paolo Ciaccia
Date: Thu, 14 Nov 1996 12:35:19 +0100
Alioune Ngom wrote:
>
> Here is the problem:
>
> Let K = {0, 1, ..., k - 1} (k > 1) be a set of k logic values.
> Let Union and Intersection be two operators defined on K. Union is
> defined as the bitwise OR operation between two elements represented in
> binary numbers (having each log(k) bits, the base of the log is 2).
> Intersection is defined as the bitwise AND operation between two
> elements represented in binary numbers.
> Let k be a power of 2 (i.e. k = 2^r, with r > 0). Unary central
> relations are the non-empty and proper subsets of K. A unary central
> relation R is closed under Union and Intersection if x Union y and
> x Intersection y are in R whenever x and y are in R. In other words:
> (x in R and y in R) implies (x Union y is in R and x Inter y is in R).
>
> Now the problem statement: For k = 2^r, how many unary central
> relations are closed under Union and Intersection ?
> ---
>
> I was not able to find a closed-form formula or even a recursive
> formula.
>
Nice problem, indeed. I've no solution yet, but a more general
formulation:
"How many "sub-lattices", closed under + and *,
can be obtained from a Boolean lattice over r elements?"
For instance, the same problem arises if you consider
a set of r elements and the lattice of all its susbsets,
with standard set Union and Intersection.
Just a partial hint for a recursive formula:
with r elements you can at least include all the
cases obtained for 1 <= s < r, thus retaining
only s out of r elements (=bit positions in your formulation), and
these elements can be chosen in comb(r s)=r!/((r-s)!s!) different ways.
The other (r-s) elements have fixed values.
E.g. In your example r=2. With s=1 (k=2), we have, in binary notation:
{0}{1}{0,1}, which can be extended to yield:
{0} -> {00}{01}{10} (these are trivial cases)
{1} -> {10}{11}{01}
{0,1} -> {00,10},{01,11},{00,01},{10,11}
(it seems that principle of inclusion-exclusion should be used here
to avoid counting twice or more a same configuration)
Let me know if you find a solution!
Paolo
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Paolo Ciaccia
DEIS - CSITE-CNR
University of Bologna - ITALY
mailto:ciaccia@cs.unibo.it
http://www.cs.unibo.it/~ciaccia
Subject: Re: Mission Impossible: Can probability=0 events occur?
From: ikastan@alumnae.caltech.edu (Ilias Kastanas)
Date: 14 Nov 1996 12:31:36 GMT
In article <56eave$601@cantuc.canterbury.ac.nz>,
Bill Taylor wrote:
>ikastan@sol.uucp (ilias kastanas 08-14-90) writes:
>
>|> As Bill's Dialogue Concerning Two New Sciences underscores,
>
>Hah! Very neat. But I rather follow in Lakatos' footsteps there...
>
>|> observing an infinite sequence of Bernoulli trials (i.e. a real number) is
>|> an abstraction. All physical experiments and observations yield rationals;
>|> ... ... Forget P = 0; such a thing is simply
>|> _impossible_... it cannot happen (!). Never has, and never will.
>
>Yep, that pretty well sums up the clincher. Nicely put.
Thanks; I said this once before and it got some flak. But even 'ra-
tional approximations to any desired accuracy' are not possible -- and
not necessarily by appeal to quantum theories.
>|> If we grant this abstraction, it is math, not physics.
>|> ... ...
>|> and x is in B^w, "x can not occur" is logically false.
>
>Hmmm... well if we're in math-not-physics, the word "occur" shouldn't really
>be there at all. You'd just say, "x is not in the sample space", which is
>false by definition. So we agree.
Surely the Cartesian product {H, T} x {H, T} x ... cannot be "miss-
ing" this or that element. Next thing you know, someone might assert that
such a product of nonempty sets is not necessarily nonempty!
>|> The remarkable part is focusing on _one_ countable set, the definable
>|> (recursive) x's, and deeming it "more justified".
>
>Yeah right. It's just laziness on our part. We can only be bothered
>talking about the things it's possible to talk about. ;-)
How unreasonable can one get?
There are very pretty uses of measure, or category, that reach things
it is not possible to talk about explicitly.
>|> A little-known fact:
>|> coins have tiny cellular telephones and talk to each other;...
>|> ... They will not be caught
>|> doing anything recursive, or recursively enumerable...
>
>Or even definable?
Eh, the ice is getting thin here... It could well be a definable
set is nonempty but has no definable elements. The wellorderings of R...
the non-constructible reals (R - L)... the non-measurable sets... I used
the most restrictive notion of 'definable'.
>|> Clever little devils.
>
>It is even LESS well known, (if that were possible), that all sets have
>a CantorNet connection with auto-antigeneric parallel many-world devices
>that prevent an infinite number of unspecified choices being made
>from their members.
>
>DAMN cunning...
Finitely many choices they don't mind... but move on to infinitely
many and suddenly the telephone lines flare up, an infinity of calls to
come up with strategy, resist the outrage, "hell no we won't go"... Some
sets must refuse to put forth a member, absolutely. But... which sets?
Curses! -- they _cannot choose_!
Ilias
Subject: Employment: South Africa, Remote Sensing Researchers
From: chris@bayes.agric.za (Christopher Gordon)
Date: 14 Nov 1996 12:41:28 GMT
AGRICULTURAL RESEARCH COUNCIL of SOUTH AFRICA
INSTITUTE FOR SOIL CLIMATE AND WATER
REMOTE SENSING DIVISION
The following positions are now on offer at this Pretoria, South
Africa based Institute with its well equipped digital image
processing facility.
The successful candidates will form part of a team of 12
researchers and support staff specializing in Remote Sensing.
Three persons are required to research the development and
application of Remote Sensing Techniques for obtaining
Environmental and Agricultural Resource Information and
Statistics.
In addition to the educational requirements set for each
position, a relevant post graduate qualification and/or
experience in Remote Sensing/Digital Image Processing and GIS
will serve as a strong recommendation in each instance.
The specific requirements for each position are as follows:
Post 1 Rangeland Applications: A university degree in Ecology,
Botany, Rangelands Science or related fields.
Post 2 RADAR Applications: A university degree in Physics,
Applied Mathematics, Statistics, Engineering or a related field.
Post 3: A university degree in Natural, Earth or Pure Science or
related field (Soil Science, Geography, Botany, Geology,
Environmental Studies)
Applicants for all posts may be required to undertake
psychometric tests.
The ARC offers challenging opportunities in a pleasant work
environment as well as competitive remuneration packages,
including standard fringe benefits, which will be negotiated in
accordance with qualifications and experience.
Please forward your application together with CV to:
The Director:ISCW, P.Bag X79, Pretoria, 0001. (Fax --27 12 323
1157)
Applications close on 22 November 1996
Enquiries:
Dr JF Eloff / Mr TS Newby ph (--27 12) 326 4205
E-Mail : TERRY@IGKW2.AGRIC.ZA
Subject: Re: ZEROS of the 8th degree polynomials
From: tleko@aol.com
Date: 14 Nov 1996 13:07:10 GMT
In article dik@cwi.nl (Dik T. Winter) wrote:
>
> Ah, within a year we will also see the zero's of a 365th degree
polynomial.
> Interesting.
In article <32337DBA.1080@dei.unipd.it> Enoch wrote:
>>
>> I am looking for an iterative algorithm to find the complex roots of
>> LARGE polynomials (degree >200) WHEN THE ROOTS (at least some of them)
>> ARE KNOWN TO BE VERY, VERY, VERY CLOSE TO EACH OTHER (i.e. I know >>
that
for several of them |xi - xj| << min |xi|,|xj|). Low complexity isn't
>> much of a requirement (as long as it doesn't soar up to exponential or
>> similarly absurd things); control of errors is. Could some kind soul
help
>> me and tell me where to find help? :-)
>> Thanks a lot in advance
In article <328A0519.269@cdf.toronto.edu> Peter Kanareitsev wrote:
:
: I am dying of curiosity. What is tleko? It's a computer program, right?
: It writes trivial MATLAB code and posts it to usenet.
: What else does it do? What is the goal of this project?
To reach 200th degree polynomials.
tleko@aol.com
Subject: Re: Cantor and the reals
From: "Spencer M. Simpson, Jr."
Date: Wed, 13 Nov 1996 08:24:50 -0500
David K. Davis wrote:
> I'm also not sure - now that I think about it - whether this proof
> doesn't involve the axiom of choice (which says one can form a
> new set by picking one element out of each set of a collection of
> sets). Help!
>
> BTW, Ilias K's amendment is right on - and the above may need more
> amending.
>
> -Dave D.
I looked it up in my handy dandy Bernays (1968, p. 117)
Apparently, the diagonalization proof doesn't require AC.
Bernays proves the equivalent for classes,
"Ft(F) & delta_1 (F) = a* & delta_2 (F) = c ->
(Ez)(z <=a & ~(z e C))"
(sets in lower case, classes in upper case, a* = {x | x e a} )
by considering the set a * {x | ~(x e F(x)) }
(here * is intersection -- let me know if there's a better Usenet
notation)
which (heedless of Ilias's advice :) translates well into
infinite sets of integers, digit expansions, and diagonalization.
This is all done _without_ using AC and even _without_ using
the potency axiom!
The addition of the potency axiom immediately makes
the set-only form true.
Hope I helped,
Spencer
Subject: Can maths be learned "linearly" ?
From: Peter Viktor Kohut
Date: 14 Nov 1996 13:43:42 GMT
The following question is on mind since a long time. Now I'm interested in
some opinions.
Can mathematics be learned in a linear way ?
What I mean by that is the following : I am very interested in various
fields of math (general interest). Now, when you dig deeper and
deeper into a certain area then sometimes you realize that there
is a piece of (sometimes basic) knowledge missing. You go back
and try to read about the missing piece. But then again there might
be something more basic missing.
A real life example :
I was reading a book about stochastic integration when I realized
that I didn't know enough about a topological property to understand
the full extent of the stated theorem. So I took a topology textbook and
tried to become familiar with the missing piece. But there, again, was
something which was based on set theory. So, I had to grab my set
theory textbook for that. And so on ....
Now I'm wondering if it would be possible to learn (serious) math in
a way so that no piece is missing when you cover more and more
advanced topics and so that the entire current mathematical knowlegde
is covered (or at least a given subset).
I guess I had to start somewhere between Goedel and Peano.
This question is also of interest because of my interest in artificial
intelligence. If such a line through mathematical knowledge can
be drawn then a smart computer program could learn math in a
fairly easy way. (I was very impressed by the Colossus computer
in the movie "The Forbin Project", when it extended the human
mathematical knowledge by several factors)
Any opinions ?
Regards
-Peter-
***********************************************************************
Peter Viktor Kohut
UBS Securities New York
Risk Measurement Global Fixed Income and Global Derivatives
e-mail : nykjz@ny.ubs.com
**********************************************************************
Subject: Re: Groove on a record?
From: mbc7051@hertz.njit.edu (matthew b charlap apmt stnt)
Date: Thu, 14 Nov 1996 14:12:30 GMT
In article <56b9lt$meq@niktow.canisius.edu>,
Don Girod wrote:
>Dru Morgan (drum@loop.com) wrote:
>: Is there a formula for finding the lenght of a groove that goes around a
>: spiral (such as a record album)? If you know the diameter of the record and
>: the distance (period?) between grooves, what would be the formula for the
>: length of the groove? You would also have to subtract the part in the
>done back when we still made records. So if you want to know how long
>the groove is, time the record. But if you want to predict how long
>the record will play, read the jacket.
Timing the record willnot give you the length of the record's groove.
The "easiest thing I could think of is (assuming grooves have constant spacing)
is to do the arc-length integral (probably in polar coordinates).
we know that r=R-T*s/(2*pi), where r is the current radius, R is the radius
of the largest groove, T is the angle from this starting point, and s is the
groove spacing.
now that you know r(T), set up the arc-length integral. It may be messy,
but that is what needs to be evaluated.
--Matthew
Subject: Re: finding teh limit value of 'x*ln(sinx)' at x=pi
From: mbc7051@hertz.njit.edu (matthew b charlap apmt stnt)
Date: Thu, 14 Nov 1996 13:58:09 GMT
In article ,
David Kastrup wrote:
>e8725229@stud1.tuwien.ac.at (godzilla) writes:
>> In article <569je8$gvr@helios.ccunix.ccu.edu.tw>,
>> g8544015@ccunix.ccu.edu.tw (Graduate) wrote:
>> >i want to find the limit value of 'x*ln(sinx)' at x=3Dpi
>> >can anyone tell me how to solve the problem
>> >please help me , thank you very much
>> since x approaches pi and ln(sin(x) approaches -infinity the limit of the=
>> whole expression is -infinity.
>> ithink that's it
>Then you think wrong. The limit does not exist, as there are
>differing limits when you approach from the left and from the right:
>from the right we get an additional term of pi*i*(pi+2k), k any
>integer.
Well, if this course allows for complex numbers (when I first came in
contact with limits like this, we were working exclusively with the reals),
then you are right- there is no two-sided limit (the left-side limit is
-inf+(2*k*pi)i, k any integer, the right-side limit is -inf+((2*k+1)*pi)*i, k
any integer.)
If you work with the reals, then the left-hand limit is -inf, and the right-
hand limit (since sin is negative here) does not exist.
In either case, the two-sided limit does not exist. I would answer with any of
the above as is appropriate for where the question arose.
--Matthew
Subject: Re: 7 November, PLutonium Day is the only future holiday
From: JC
Date: Thu, 14 Nov 1996 14:54:07 +0000
David Kastrup wrote:
>
> JC writes:
>
> > Zdislav V. Kovarik wrote:
> > > Right on! November 7 is the 79th anniversary of the Great October
> > > Socialist Revolution in Russia. It turned out to be a colossal failure,
> > > too.
>
> You think so? At least after the revolution the peasants owned their
> own bodies.
Apart from the dead ones of course.
> For most it had certainly been a relative improvement,
> although there has been a lot of stagnation (to put it very mildly)
> since. Free market has not turned out too hot in Russia as well up to
> now, BTW.
Oh dear, one of these apologists for Stalin et al who suffer a total
sense of humour failure when you point out that the revolution was one
of the worst things that ever happened to Russia.
> An interesting employer you must have... What's your job? Designing
> calendars for display in Star Trek episodes?
WHAT? You dare to laugh, but it will be I who laughs when the world
accepts the one totality of the single cheese atom. I had that Andrew
Wiles in the back of my cab once. Blah Blah Blah. Ha Ha Ha HA HA
HAHAHAHAHA...
Archipelago Plutedium
Subject: Re: World's second most beautiful syllogism
From: Mike Herauf
Date: Thu, 14 Nov 1996 06:57:26 -0800
> To quote Dr. Foakes-Jackson (1855-1941) of Cambridge ..."It's no use
> trying to be clever, we are *all* clever here. Just try to be kind."
>
> Darla
> ---who hereby thanks her brave champions, both public and private.
Hooray for you Darla. You know now, that you are damned to the river
Styx or something like that now, don't you? Archie Pootonium is quite a
(dishwasher) character. He hates anyone who disagrees with his
"theories", which are obviously idiotic, and have been proven so. They
carry no weight or importance here.
However, A.P. is quite intelligent. If you get him to coverse on a
worthwhile subject, he can be quite interesting and resourceful.
As far as his attacks upon you, forget them.
Mike
P.S.
Welcome to the club
Subject: One combinatorial problem
From: aburshte@mail1.sas.upenn.edu (Alexander Burshteyn)
Date: 14 Nov 1996 06:07:56 GMT
Hello,
I wonder if anyone can offer any solution (or even a suggestion) about the
following problem which came up in the course of my research:
For every positive integer $n>1$ and every permutation $\tau \in S(n)$
(i.e. of {1,...,n}, the following inequality holds:
$
\sum_{j=1}^{n} {
\sum_{k=1}^{n} {
\binomial{j+k-2,j-1} \times \binomial{2n-j-k,n-j} \times
\binomial{ \tau (j) + \tau (k) - 2, \tau (j) - 1} \times
\binomial{ 2n - \tau (j) - \tau (k), n - \tau (j)}
}
}
> \binomial{2n-1,n} ^ 2
$
Even a proof or a pointer for the case $\tau = id(n)$ would be great.
Thank you in advance.
Alex Burstein
aburshte@sas.upenn.edu
alexb@math.upenn.edu
--
AB
******************
145 = 1! + 4! + 5!