mm-282
Subject: Re: Minimal perimeter of flattened skin of unit
dodecahedron ?>>Suppose I have a paper model of a common
regular dodecahedron of >>fixed dimensions.>>Suppose I cut it,
either at edges or even through the middle of >>faces, until I
have a compact surface that I can make flat.>>What's the
minimal value of the perimeter of the flattened surface ? >
The cuts you make must form a tree on the dodecahedron's
surface, that > connects all 20 dodecahedron vertices, and the
perimeter of the > flattened surface is just twice the total
length of the tree edges.> For instance cutting only along
edges gives perimeter 38 times the edge > length.Conversely
any Steiner tree lets you unfold the dodecahedron into a flat
> compact surface that may however overlap itself when
unfolded onto the > plane. So, ignoring overlap, the problem
is just one of finding an > optimal Steiner tree of the
dodecahedron's vertices, with the length of > the tree edges
measured along surface geodesics. Ignoring overlap might > not
be too unreasonable, I wouldn't be surprised if the optimal
Steiner > tree gives you a solution that doesn't overlap when
you unfold it.For instance, one could form a Steiner tree by
combining three-, four-, > and five-vertex subtrees within
some of the faces of the dodecahedron. > Each of these has
less length than the corresponding non-Steiner tree, > so one
could get a smaller perimeter this way than what you would get
by > cutting only along edges, but I don't know what
combination of face > subtrees would work best (a small matter
of calculation), whether that > combination can be arranged so
that it unfolds to a non-overlapping > surface (seems pretty
likely), or whether the optimal solution only > involves
subtrees that stay within one face.Thanks for putting this
clearly.The surface of the outer shell of the great
dodecahedron and that of the first stellation of the
icosahedron (two self-intersecting polyhedra with icosahedral
symmetry covering the sphere thrice and twice respectively),
can be refold to fit on the surface of the common dodecahedron
through a cut that is along edges or pseudo-edges of these
polyhedra while corresponding to a cut on the surface of the
dodecahedron belonging to the family you describe, as a connec
sum of 5-vertices-connecting single-pentagon-face Steiner
sub-trees.>>How large is the space of corresponding shapes ?>
Infinite.> :) I can summarize my interest to the value a
particular constant : the optimal -shape- factor for a
rectangular box around the compactly flattened dodecahedron
skin - optimal in the sense that the rectangle has minimal
surface in initial units under the sta constraint. To prune
the search, I want to consider only shapes of minimal
perimeter. Lemmas setting an upper bound to the error entailed
by this choice, welcome :)This leaves me with a review of
observations and questions.(1) Consider the full
self-intersecting shape of [(a)] the great Kepler-Poinsot
dodecahedron [, (b) the first stellation of the icosahedron],
what sorts of further optimal Steiner trees does this [or
that] shape allow ? Do/should all these Steiner trees have a
straightforward interpretation as cutting the polyhedron
-flat- ? How does the matter of these polyhedra
self-intersecting, harmonize to the matter of their net,
self-intersecting, if/when flattened under the action of an
optimal Steiner tree ?(2) Consider the 4D 120-cell polytope.
Do the questions, up to here, asked of the 3D regular
12-hedron admit clear analogues, about the 4D 120-cell ?(3)
Consider the S(5,6,12) Steiner system. I believe there are
ways to distribute the 12-set to the faces of the regular
dodecahedron, and elements of the Steiner system to
corresponding 2-colorings, s.t. the distribution commutes with
the rotation group of the dodecahedron. If we could somehow
find common substructure to otoh such distributions, and otoh
the system of optimal Steiner trees to slice the dodecahedron
surface flat, we would strike something fa to the name of
Steiner Dodecahedron ! Boris Borcic--What's F(Syracuse) if
F(Eureka) is the = in E=mc^2 ?
===
Subject: math pinballAssume A
and B are two obstacles in the plane (e.g., compact
disjointconvex subsets with smooth boundary), and there is a
ball (of radius0) bouncing between them. Of course no
friction, constant speed, andthe bouncing follows the rule of
reflection of light (i.e., theincoming angle with the normal
is equal to the outcoming angle). Thisis the setting.Clearly
there is a periodic trajectory: the ball bounces back andforth
between the closest points of A and B. Question: is this
theonly trapped trajectory? or else, do there exist other
trajectories inwhich the ball bounces between A and B
forever?It seems easy to show that you can construct two
convex sets A and Bfor which such a trajectory exists (it
approaches asymptotically theperiodic one). But if A and B are
given, what can be said?E.g., take two disks of radius 1. Can
you find a trajectory with aninfinite numer of bounces
different from the periodic one?Clearly the curvature is
important, and probably the dividing line isbetween degenerate
or non degenerate points...Piero
===
Subject: Re: math
pinball>Assume A and B are two obstacles in the plane (e.g.,
compact disjoint>convex subsets with smooth boundary), and
there is a ball (of radius>0) bouncing between them. Of course
no friction, constant speed, and>the bouncing follows the rule
of reflection of light (i.e., the>incoming angle with the
normal is equal to the outcoming angle). This>is the
setting.>Clearly there is a periodic trajectory: the ball
bounces back and>forth between the closest points of A and B.
Question: is this the>only trapped trajectory? or else, do
there exist other trajectories in>which the ball bounces
between A and B forever?>It seems easy to show that you can
construct two convex sets A and B>for which such a trajectory
exists (it approaches asymptotically the>periodic one). But if
A and B are given, what can be said?>E.g., take two disks of
radius 1. Can you find a trajectory with an>infinite numer of
bounces different from the periodic one?Yes, in principle.
Suppose the two disks of radius 1 are centered at (0,0) and
(0,2h) with h > 1.For convenience, put a straight mirror along
y=h, so we only have to dealwith one disk. Consider the mapping
g defined on a neighbourhood of (-pi/2, pi/2) asfollows: if a
ray of light in direction given by the unit vector
[cos(phi),sin(phi)] hitting the point [cos(theta),sin(theta)]
on the unit circle is reflec from the circle and the straight
mirrorto direction [cos(phi'),sin(phi')] and hits the circle
again at point [cos(theta'),sin(theta')], then g(phi,theta) =
(phi',theta'). The preciseformula is a bit complica, but
clearly g should be analytic with (-pi/2, pi/2) as an unstable
fixed point. If I'm not mistaken that should be a hyperbolic
fixed point, and there should be stable andunstable manifolds
consisting of points whose orbits are infinite.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2 
===
Subject: Reply to Re: A
number theory problem (equation)Epigone-thread: shirfroykyThe
equation should be x^2 - d^2 xy + y^2 = 1, which in general
hasinfinitely many non-trivial solutions. Solutions can be
genera fromM^n x where M =[{0,1},{-1,d^2}] (2x2 matrix) and x
= (0,-1) (2 x 1vector) and n > 1 integer.For n = 5, this gives
solutions like {{1, d^2}, {d^2, -1 + d^4}, {-1 + d^4, d^2*(-2 +
d^4)}, {d^2*(-2 +d^4), 1 - 3*d^4 + d^8}, {1 - 3*d^4 + d^8,
d^2*(3 - 4*d^4 + d^8)}}.We can also generate solutions using
M=[{d^2,-1},{1,0}] and x=(0,1).I used the algorithm in --->
http://www.alpertron.com.ar/QUAD.HTM togenerate the solutions
above.The equation is a Thue equation, see
--->http://mathworld.wolfram.com/ThueEquation.html . Seee also
--
>http://mathworld.wolfram.com/
DiophantineEquation2ndPowers.html .The challenge is to solve
the equation above when x^2 + y^2 is asquare. I have performed
extensive computer search for such solutionswithout success.So,
I suspect that there are no such (non-trivial) solutions.Any
comments are welcome!Of course, the solutions of the equations
x^2 + d^2 xy + y^2 = 1 and x^2 - d^2 xy + y^2 = 1 rare rela. A
solution (x,y) of the formergives the solutions(-x,y) and
(x,-y) of the latter etc..
===
Subject: Re: A number theory
problem (equation)> Find integer solutions, if any, of the
equation x^ + d^2 xy + y^2 = 1> for given integer d and when
x^2 + y^2 is a square.I think I've proved that a necessary
condition for x^2 + D.xy + y^2 = 1with x^2 + y^2 = z^2 to have
non-trivial (xy != 0) integer solutionsfor an integer D != 0 or
1 is that there exists an integer E such thatD^2 - 3.E^2 =
4.I'll check this again tomorrow, and post it if it seems OK.
But meanwhileif anyone has done computer searches for various
D > 0 (not necessarilysquare) it would be interesting to see
any
solutions.Cheers----------------------------------------------
-----------------------------John R Ramsden
(jr@adslate.com)----------------------------------------------
-----------------------------Eternity is a long time,
especially towards the end. Woody Allen
===
Subject: Re: A
number theory problem (equation)Epigone-thread: shirfroykyKen
Holing asks:> Find integer solutions, if any, of the equation
x^ - d^2 xy + y^2 =1 for given integer d and when x^2 + y^2 is
a square.Since you require that x^2 + y^2 = z^2, the solution
(Diophantinetriple) is x = (a^2 - b^2) y = 2ab z = (a^2 +
b^2)Obviously there are no solutions to the original equation
unless (a,b)= 1. Substituting for x and y in your equation, we
have (a^2 + b^2)^2 - d^2*(a^2 - b^2)*2ab = 1We can also take
(a^2 - b^2)*2ab as square-free, by incorporating anysquare
factors into d if necessary, and this becomes the
well-knownPell equation of the form A*X^2 + 1 = Y^2.For the
Pell equation to have any solutions, (X,Y) = (A,Y) = 1.i.e.1)
d must be relatively prime to (a^2 + b^2) and2) (a^2 -
b^2)*2ab relatively prime to (a^2 + b^2). I am unable to
proceed further to determine whether there are
anysolutions.More on Pells equation can be found
on:http://www.ieeta.pt/~tos/pell.html#r1Joseph
Sinyor
===
Subject: Kernel TheoremI have a proof of the Kernel
theorem at http://www.geocities.com/masegandand am looking for
ways to improve the method.
===
Subject: Great Lakes K-theory
Conference, X Great Lakes K-theory Conference, Xat the
University of Illinois at Urbana-Champaign. The conference is
beingorganized locally by Dan Grayson and Randy McCarthy with
the help andscientific advice of Eric Friedlander, Rick
Jardine, and Manfred Kolster.If you might come, please let Dan
Grayson <dan@math.uiuc.edu> know.This time we are pleased to
announce that the conference is generously fundedby the
National Science Foundation and the University of Illinois. At
leasthalf of the funds made available by the NSF grant must go
to US-based graduatestudents, junior faculty, underrepresen
groups and/or otherwise unsupporindividuals, so we are eager
to receive applications for financial supportfrom such
individuals.To apply for support, please send email to Dan
Grayson <dan@math.uiuc.edu>,estimating your expenses and
stating what other monetary support ispotentially available to
you. Please include references to publications and/orsolicit a
brief email letter of reference from an advisor or mentor.
US-basedGraduate students are especially encouraged to apply.
For full considerationdon't hesitate to apply after that date,
for there may be cancellations or notenough
applications.Schedule:We will have six talks, with four on
Saturday, May 8, beginning at 11am, andtwo on Sunday, May 9,
in the morning, ending by 11:30am. There will be ampletime
between talks for interaction with the speakeand we'll go out
todinner together Saturday evening.Speakers:The following
mathematicians have agreed to speak. o Gunnar Carlsson,
Stanford University, on Derived representation theory and the
K-theory of fields. o Christian Haesemeyer, University of
Illinois at Urbana-Champaign, on Homotopy K-theory of
blow-ups. Abstract: We give a proof that Weibel's homotopy
invariant K-theory satisfies the expec descent property for
arbitrary blow-ups, at least over a base field of
characteristic zero. We give applications regarding the
negative K-groups of singularities, obtaining partial results
on the relevant conjecture of C. Weibel. o Marc Levine,
Northeastern University, on The Postnikov tower in motivic
stable homotopy theory. Abstract: Voevodsky has defined a
version of the Postnikov tower in the motivic stable homotopy
category and has shown that the slices in this tower have the
natural structure of motives. We will describe how the
homotopy coniveau tower used by Friedlander-Suslin in their
interpretation and generalization of the Bloch-Lichtenbaum
motivic cohomology to K-theory spectral sequence generalizes
further to give another construction of Voevodsky's Postnikov
tower. This gives a direct relation of the motivic Postnikov
tower with the Friedlander-Suslin tower, showing that the
slices of the motivic Postnikov tower for K-theory are motivic
cohomology. o Steve Mitchell, University of Washington. o
Holger Reich, University of Muenster. o Jonathan M. Rosenberg,
University of Maryland.Housing: o We have reserved (until April
7) a block of 40 rooms at the Illini Union, the building just
East of Altgeld Hall. Call 217-333-1241 to make a reservation.
Rates are $72-$83 for 1 person, $83-88 for 2 persons, $93 for 3
persons, $97 for 4 persons, and $165 for a suite. Hotel tax is
11% extra. Parking is available nearby. Refer to our
conference when making a reservation, and send me email, too,
so I can confirm your identity to the hotel, if necessary. o
We have also reserved (until April 23) a block of 15 rooms at
Hampton Inn at University of Illinois, about 0.2 miles east
and then 0.4 miles north of Altgeld Hall, 217-337-1100. Rates
are $60 for 1 person and $65 for 2 persons. Refer to our
conference when making a reservation, and send me email, too,
so I can confirm your identity to the hotel, if necessary. o
See http://www.math.uiuc.edu/~dan/travel.html for a listing of
other hotels.Getting there:See
http://www.math.uiuc.edu/~dan/travel.html for instructions on
traveling tothe University of Illinois.Web page:See
http://www.math.uiuc.edu/K-theory/Calendar/GL10/ for up to
dateinformation.
===
Subject: irrational numbersEpigone-thread:
yodimclimIrrational numbers are divided into two classes,
algebraic (roots ofpolynomial equations with integer
coefficients) and transcendental. I'm interes in the following
questions.1. Who made this distinction and why? Also when?2. If
we made the coefficients algebraic numbers instead of
integers(in the polynomials), would we get a bigger class?
It's clears thatup to fourth degree we won't.
===
Subject: Re:
irrational numbers> Irrational numbers are divided into two
classes, algebraic (roots of> polynomial equations with
integer coefficients) and transcendental.> I'm interes in the
following questions.> 1. Who made this distinction and why?
Also when?As for the who and when:Mathematicsat
http://members.aol.com/jeff570/mathword.html I copy the
followingrelevant statements. See the website for a few more
details: Leibniz coined the word transcendental in
mathematics, using transcendensin the fall of 1673 in
Progressio figurae segmentorum circuli aut
eisygnotae.According to Paulo Ribenboim in My NumbeMy Friends,
LEIBNIZ seems to bethe first mathematician who employed the
expression 'transcendental number'(1704).the first writer to
use the terms 'rational' and 'irrational' in the sensenow
current in arithmetic and algebra.ALGEBRAIC NUMBER. According
to an Internet web page, this term was used byAbel.Algebraic
quantity appears in 1673 in Elements of Algebra (1725) by
JohnKersey: Two or more Algebraic quantities (OED2).Algebraic
(in the sense of an algebraic number) is found in 1840
inMathematical Dissertations (1841) by J. R. Young [James A.
Landau]. Clark
===
Subject: Re: irrational numbers> 1. Who made
this distinction and why? Also when? I'm not sure who defined
transcendental numbebut Cantor's theorem was the first proof
of their existence, and this happened sometime in the mid
1800s. Why? They are a natural generalization to rational
numbers since any rational is the solution to a linear
equation ax+b=0. Any intersection of circles and line (conics
in general, but moreimportantly circle and lines since those
are the locii of points generaby compass and straighge) are
represen as a solution of quadratics and so it can be shown
that all constructible quantities form an algebraic closure of
solutions to quadratics. Take these a step further and you get
algebraic numbers.> 2. If we made the coefficients algebraic
numbers instead of integers> (in the polynomials), would we
get a bigger class? It's clears that> up to fourth degree we
won't. Take the following polynomial [sqrt(2)-sqrt(5)] x +
cuberoot(3) = 0 (**) Can you form a new polynomial P(x) with
integer coefficients such that the solutions to (**) are also
solutions to P(x) ? If there is a process, can you generalize
it?J
===
Subject: Re: irrational numbers>> 1. Who made this
distinction and why? Also when?> I'm not sure who defined
transcendental numbebut Cantor's theorem >was the first proof
of their existence, and this happened sometime in the >mid
1800s.No. Liouville was the first to prove their existence, in
his paper, Sur des classes tr`es-'etendues de quantit'es dont
la valeur n'est ni alg'ebrique, ni m^eme reductible `a des
irrationnelles alg'ebriques, C.R. 18 (1844) 883-5. Since he
used that circumlocutionrather than the word transcendentales,
I would infer that this terminology had not been used, or at
least was not commonly used, beforethat time. Hermite proved
the transcendence of e in 1873; Cantor didn't introduce the
concept of countability until 1874.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: irrational
numbers> I'm not sure who defined transcendental numbebut
Cantor's theorem > was the first proof of their
existenceLiouville construc an infinite class of
transcendental numbers in1844
(mathworld.wolfram.com/LiouvillesConstant.html). Cantor
wasborn in 1845. Strangely, Mathworld repeats the incorrect
statementon mathworld.wolfram.com/TranscendentalNumber.html.--
Chris Jeris cjeris@oinvzer.net Apply (1 6 2 4)(3 7) to domain
to reply.
===
Subject: Re: irrational numbers>> 1. Who made this
distinction and why? Also when?> I'm not sure who defined
transcendental numbebut Cantor's theorem >was the first proof
of their existence, and this happened sometime in the >mid
1800s. Why? They are a natural generalization to rational
numbers >since any rational is the solution to a linear
equation ax+b=0.> Any intersection of circles and line (conics
in general, but more>importantly circle and lines since those
are the locii of points genera>by compass and straighge) are
represen as a solution of quadratics >and so it can be shown
that all constructible quantities form an algebraic >closure
of solutions to quadratics.> Take these a step further and you
get algebraic numbers.>> 2. If we made the coefficients
algebraic numbers instead of integers>> (in the polynomials),
would we get a bigger class? It's clears that>> up to fourth
degree we won't.> Take the following polynomial>
[sqrt(2)-sqrt(5)] x + cuberoot(3) = 0 (**)> Can you form a new
polynomial P(x) with integer coefficients such that >the
solutions to (**) are also solutions to P(x) ? If there is a
process, >can you generalize it?>J It needs to be a monic
polynomial. sqrt(2) - sqrt(5) is not 1.If we are given x^2 +
(sqrt(2) - sqrt(5))*x + cuberoot(3) = 0then we can eliminate
the cube root by [x^2 + (sqrt(2) - sqrt(5))*x]^3 + 3 = 0Next
eliminate sqrt(5) by replacing sqrt(5) by -sqrt(5)(and
sqrt(10) by -sqrt(10)). Then eliminate sqrt(2).The result is
monic, degree 24. We first replaced X + cuberoot(3) by (X +
cuberoot(3))*(X + w*cuberoot(3))*(X + w^2*cuberoot(3))where w
is a primitive cube root of unity. Then we replaced sqrt(5)by
its conjugate and multiplied the results. Lastly we did
sqrt(2).-- John Adams served two terms as Vice President and
one as President, but lostreelection. Later his son became
President despite losing the popular vote.That son lost his
reelection attempt badly. Now history is repeating
itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San
Rafael, California
===
Subject: Re: irrational numbers
<Hs6s68.F59@cwi.nl>> 1. Who made this distinction and why?
Also when?>[...]>> 2. If we made the coefficients algebraic
numbers instead of integers>> (in the polynomials), would we
get a bigger class? It's clears that>> up to fourth degree we
won't.[followed by an example with quardatic and cubic
irrationalities]That was fine, but what if one of the
coefficients is, for example, thepositive root of the
polynomial x^5-x-1 (I checked that it is irreducibleover
integers)? No radical formula exists for roots of
irreduciblequintics.Here is one general idea (constructive,
but try it only if you have plentyof time and a rich gullible
sponsor):Every algebraic number of degree p (the smallest
degree for which anon-zero polynomial over integers exists
that makes that number its root)has (p-1) conjugates. Every
symmetric rational function in the roots overintegers will
simplify as a rational function over integers of
thecoefficients. (Recall the elementary symmetric polynomials
in the roots ofx^2+a*x+b: x_1 + x_2 = -a and x_1 * x_2 = b,
and this extends to higher degrees.)Along with the polynomial
with algebraic coefficients, consider allpolynomials where you
replace every coefficient with its conjugates, oneat a time.
That will make a long list of polynomials with
algebraiccoefficients. Now: Consider the product of all those
polynomials, including the originalpolynomial; it is a
symmetric polynomial in the conjugates of allcoefficients, so
it simplifies to a polynomial over integewhich hasthe original
roots, among others. ZVK(Slavek).
===
Subject: Re: irrational
numbers<Pine.LNX.4.44.0401272153150.4760-100000@eva117.
cs.ualberta.ca>,1. Who made this distinction and why? Also
when? I'm not sure who defined transcendental numbebut
Cantor's theorem > was the first proof of their existence, and
this happened sometime in the > mid 1800s. Liouville gave the
first transcendentality proof, around 1850, about 30 years
before Cantor's existence proof.-- 
===
Subject: Re: irrational
numbers> Liouville gave the first transcendentality proof,
around 1850, > about 30 years before Cantor's existence proof.
Interesting... I was using MathWorld as a reference, and I will
quote the relevant section: Georg Cantor was the first to prove
the existence of transcendental numbers. Liouville subsequently
showed how to construct special cases (such as Liouville's
constant) using Liouville's approximation theorem. But
according to this biograph of Cantor:
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/
Cantor.html it says: Liouville established in 1851 that
transcendental numbers exist. Twenty years later, in this 1874
work, Cantor showed that in a certain sense 'almost all'
numbers are transcendental by proving that the real numbers
were not countable while he had proved that the algebraic
numbers were countable. As usual, we have discrepancies in
math history.J
===
Subject: Re: irrational numbers> As usual, we
have discrepancies in math history.The math history experts
(not to say there aren't some reading thisnewsgroup) may be
found at the unusually civil and scholarly mailing
listHistoria-Mathematica:
http://mathforum.com/epigone/historia_matematicaIt might be a
good idea to bring up this question there.Edwin
Clark
===
Subject: Genus of Bivariate Polynomial Curves - How
Many of Each Genus?I'm wondering about how many bivariate
polynomials of a given genusthere are.Rela is: Are the number
of Genus 0 curves more than Genus 1 etc? I realise that the
number of curves is infinite and hence the numberof curves of,
say, Genus 0 is infinite too but is there somethingknown about
the density of the various Genus's, or scaling with thesize of
the polynomial coefficients (rational of course)?Perhaps these
are naive questions but I've read nothing on this? Can anyone
enlighten me? Pete
===
Subject: when are polynomials equal mod
q?Let f(z) and g(z) be polynomials in z with coefficients in
Z_qand that both factor over Z_q. I'm looking for necessary
andsufficient conditions in terms of the multiplicities ofthe
roots so that f(z) = g(z). For example, mod
4,(z-1)^2*(z-2)^2*(z-3)^6 = (z-0)^2*(z-1)^8. Are such
conditionsknown? If not in general what about for prime
powers?-Frank R.-- ----------------------Frank Ruskey e-mail:
fruskey (AT) cs.uvic.caDept. of Computer Science fax:
250-721-7292University of Victoria office:
250-721-7232Victoria, B.C. V8W 3P6 CANADA WWW:
http://www.cs.uvic.ca/~fruskey