Newsgroup sci.math.research 6091

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gerry@mpce.mq.edu.au (Gerry Myerson) writes:
[which integers are the sum of three cubes?]
>Nobody knows. It may be that every number not congruent to plus-or-minus 4 
>mod 9 is a sum of three cubes; on the other hand, no one has ever found 
>such an expression for 30. 
>See Richard K Guy, Unsolved Problems in Number Theory, D5. 
Don't have Guy at hand, what are the upper bounds people have searched?
A quick and dirty search indicates that if x^3 + y^3 + z^3 = 30 (x, y, z
integers), then at least one of x, y, z is larger than 3073 in absolute
value...
michael
brundage@ipac.caltech.edu

I have just updated the Cunningham Project tables held on ftp.ox.ac.uk
in the /pub/math/cunningham directory.  The tables are believed to be
correct and complete up to 15 September 1996.  I would appreciate
hearing of any errors that may be found, and even more appreciative if
corrections are included with your report.
For those not familiar with the Cunningham Project, I append a snippet
from /pub/math/cunningham/README
	   In 1925 Lt.-Col. Alan J.C. Cunningham and H.J. Woodall
   gathered together all that was known about the primality and
   factorization of such numbers and published a small book of tables.
   "These tables collected from scattered sources the known prime factors
   for the bases 2 and 10 and also presented the authors' results of
   thirty years' work with these and the other bases" (see [1])
	   Since 1925 many people have worked on filling in these tables.
   It is likely that this project is the longest, ongoing computational
   project in history. D.H. Lehmer, a well known mathematician who passed
   away in 1991 was for many years a leader of these efforts. Professor
   Lehmer was a mathematician who was at the forefront of computing as
   modern electronic computers became a reality. He was also known as the
   inventor of some ingenious pre-electronic computing devices
   specifically designed for factoring numbers. These devices are
   currently in storage at the Computer Museum in Boston.
	   For a history of this project I suggest you obtain a copy of:
   [1]:
   J. Brillhart, D.H. Lehmer, J. Selfridge, S.S. Wagstaff Jr., &
   B. Tuckerman Contemporary Mathematics vol 22, "Factorizations of
   b^n +/-1, b = 2,3,5,6,7,10,11,12 up to high powers", published
   by the American Math. Society 1983, 2nd Edition 1988
Paul
--
Paul Leyland          | Hanging on in quiet desperation is
Oxford University Computing Services     |     the English way.
13 Banbury Road, Oxford, OX2 6NN, UK     | The time is gone, the song is over.
Tel: +44-1865-273200  Fax: 273275        | Thought I'd something more to say.
PGP KeyID: 0xCE766B1F

In article <326C69C9.167E@boas5.bo.astro.it>,
luca ciotti   wrote:
>I'm not a mathematician, so my question probably is not formally
>well posed. 
>
>My question is: in R^3, the Poisson equation is Lapla(Psi)= F
>where Psi and F are functions of X=(x,y,z).
>
>I conjecture that Psi=Psi(F) if and only if Psi is spherically
>symmetric.
GK Batchelor Introduction to Fluid Dynamics (Cambridge 1967) p19
suggested that Psi might have to be one of (a) a function of one co-ord
of a rectangular system or (b) the radial coordinate of a cylindrical
polar system or (c) the radial coordinate of a spherical polar system.
Cases (a) and (b) are counter-examples to the conjecture. But I was
never happy with Batchelor's list: one can imagine an infinite line 
of equally spaced "stars" each of which is nearly but not exactly 
spherically symmetric, but I admit I never filled in the details.
John Harper Mathematics Dept. Victoria University Wellington New Zealand

Conway has noted that the simplest construction of the
24-dimensional lattice called the Leech lattice involves
the unique even unimodular Lorentzian lattice in 26 dimensions,
called II_{25,1}.  We can think of this as the vectors in
R^{26} whose components (x_1,...,x_{26}) sum to an even integer,
and are either all integers or all integers + 1/2.  
On the other hand, the smallest representation of the
exceptional Lie group F4 is 26-dimensional.  One may think
of it as acting on the traceless 3x3 hermitian octonionic
matrices, on which it preserves a trilinear form.  
So, has anyone tried to find a copy of II_{25,1} naturally
sitting in this representation of F4, and to do something
interesting with this?

Let f be a bounded, real-valued Lipschitz function on [0,1]^2. 
Of course, it is then differentiable a.e., perhaps due to Radamacher. 
(Well, _assume_ anyway that for a.e. x the partial derivative f_x(x,y) 
does exist for a.e. y)
Is the following incredibly reasonable assertion true? If so, why? 
For a.e. x, the set of y with f(x,y) = 0 and f_x(x,y) <> 0 has measure zero.
Thanks, Lones
  .-.     .-.     .-.     .-.     .-.     .-.    
 / L \ O / N \ E / S \   / S \ M / I \ T / H \   
/     `-'     `-'     `-'     `-'     `-'     ` 
 Lones Smith, Economics Department, M.I.T., E52-252C, Cambridge MA 02139
 (617)-253-0914 (work)  253-6915 (fax)   lones@lones.mit.edu 

In the book
Paul Erdoes & Ronald L. Graham: "Old and New Problems and Results in Combina-
torial Number Theory", L'Engseignement Mathematique Universite de Geneve,
1980
one can read on page 31:
> It is true that for any rational a/b, the closest strict underapproximation
> R_n (a/b) of a/b by a sum of n unit fractions is given by
>
> R_n (a/b) = R_{n-1} (a/b) + 1/m
>
> where m is the least denominator not yet used for which R_n (a/b) < a/b, 
> provided that n is sufficiently large.
The authors do not give any citation of the proof of this theorem. Does any-
body know in wich article it is proven or maybe know how it can be proven?
Stefan Bartels

luca ciotti  writes:
>I'm not a mathematician, so my question probably is not formally
>well posed. 
>My question is: in R^3, the Poisson equation is Lapla(Psi)= F
>where Psi and F are functions of X=(x,y,z).
>I conjecture that Psi=Psi(F) if and only if Psi is spherically
>symmetric.
>Any help/suggestions/counterexamples are REALLY appreciated.
This is not true.
If F=F(x) then there is a Psi=Psi(x) and so Psi=Psi(F)
Willi Mohring

In article <54f3e4$gh6@amanda.dorsai.org>, mantell@amanda.dorsai.org (Abe Mantell) writes:
> 
> Hello,
> 
> I have been quite a bit baffled by the following initial value problem:
> 
> y' = y^2 + x^2, y(0)=1
> 
> The slope field gives no hint of any singularities, only that the solution
> "grows" rapidly for increasing x & y.  However, the analytic solution 
> contains Bessel Functions...which yield a singularity at about 
> x=0.96981065995.  I cannot see why this occurs from examining the DE or
> the slope field!!!
By comparison, look at
   y' = y^2, y(0) = 1
which has the solution y = 1 / (1 - x).  It blows up at x = 1.
The solution to you ODE has larger slope, grows faster, and hence
blows up slightly sooner.
Roger

This is true. The set is contained in the boundary of 
either {x|f(x)>0} or {x|f(x)<0}. For a Lipschitz function f on R^2, such
boundaries are   1-dim Hausdorff measurable and the  measures are finite
locally.
Therefore both boundaries have the desired property. So does the set
you are interested in.
It is a well known fact in geometric measure theory. And it should be
somewhere in Federer's book.

G & S big book on tilings is full of intriguing research problems, marked with (**) I 
imagine several of the listed problems must have been solved by now. Is there any 
central clearinghouse/person/website where I can get status/references, etc?
Thanks
J. Currie

In article <54m4tr$k9@gap.cco.caltech.edu>,
Michael L. Brundage  wrote:
>Juergen Symanzik  writes:
>>Where can I find an algorithm that tests whether a given
>>directed graph is strongly connected? Actually, I do not need this
>>algorithm but just a reference that such an algorithm exists
>>and some ideas about its complexity. Then, how can I determine
>>how many steps it may take to return to any given node.
>
>Polynomial time.  Kuc'era, _Combinatorial Algorithms_ probably includes one,
>though I don't have it on hand to be certain.
Kucera's book gives a brief outline and reference to an O(e+v) algorithm for
  finding the strong components of a digraph.
The reference is:
	Tarjan, R. E. 1974 "Depth-first search and linear graph algorithms."
	SIAM J. Comput. 1. 146-60.
--Apollo
-- 
Apollo's .sig of the hour for Thu Oct 24 14:27:35 MDT 1996:
  Amusing Trivia Fact: In 1983, a Japanese artist made a copy of the Mona
      Lisa completely out of toast.
  Amusing Trivia Fact: The first known contraceptive was crocodile dung,

In article <54o0lh$q83@infosrv.rz.uni-kiel.de>,
stu30219@srv2.mail.uni-kiel.de (Bartels) wrote:
=> 
=> In the book
=> 
=> Paul Erdoes & Ronald L. Graham: "Old and New Problems and Results in 
=> Combinatorial Number Theory", L'Engseignement Mathematique Universite 
=> de Geneve, 1980
=> 
=> one can read on page 31:
=> 
=> > It is true that for any rational a/b, the closest strict 
=> > underapproximation R_n (a/b) of a/b by a sum of n unit fractions is 
=> > given by
=> >
=> > R_n (a/b) = R_{n-1} (a/b) + 1/m
=> >
=> > where m is the least denominator not yet used for which R_n (a/b) <
a/b, 
=> > provided that n is sufficiently large.
=> 
=> The authors do not give any citation of the proof of this theorem. Does 
=> anybody know in wich article it is proven or maybe know how it can be 
=> proven?
In the case where the rational you are trying to approximate is 1 (one), 
see D. R. Curtiss, On Kellogg's diophantine problem, Amer. Math. Monthly 
29 (1922) 380--387. However, Curtiss' proof, while entirely elementary, 
is not easy to follow and, according to a post to this group earlier this 
year, is incorrect. 
A simpler proof is given by O. Izhboldin and L. Kurliandchik, Unit
fractions, 
Proc. St. Petersburg Math. Soc. 3 (1995) 193--200. This appears in English 
translation in Amer. Math. Soc. Translations, Series 2, 166. 
They say nothing about the general rational, and I have no opinion as to 
whether their methods generalize. 
Gerry Myerson (gerry@mpce.mq.edu.au)

Dear Users,
thank you very much for your suggestions/answers/counterexamples.
Many of them was known also to me, and now I realise how 
mathematically approximate
was the formulation of my original post. I need to be more exact.
What I'm looking for is a potential-density pair usefull in galaxy
modelling. The Poisson eq. is 
                  1) Lap[Psi]= C * F
where C<0, Psi=Psi[x,y,z] is the potential , Psi>0, grad[psi]=0 in the 
origin, and Psi->0 at infinity, F=F[x,y,z] is the density, and 
F>0. In the astronomical literature many axisymmetric systems are 
explicitely known that verify 1) with the required conditions, but 
no one shows the property that the coordinates can be eliminated
giving
                  2) F=F[Psi] 
It is obvious that, for some "special" simmetry 2) is always true:
for plane parallel symmetry, cilindrical symmetry with no z-dependence,
spherical symmetry. 
On the contrary, I'm able to show (it is very trivial) that 
assuming Psi (or F) to be stratified on generalized homeoids
                  3)  Psi=Psi[m],
                  3') m = (x/a)^2p+(y/b)^2q+(z/c)^2r
(with p,q,r, integers), the only way to have 2) is p=q=r=1 and a=b=c,
i.e., spherical symmetry.
My strong feeling is that only spherically symmetric systems, once
excluded "peculiar" symmetries, can satisfy 2).
Any help is really appreciated.
Luca Ciotti

In 1990 the book `Open Problems in Topology' was published.
At regular intervals the editors announce solutions to problems
from that book in the journal `Topology and its Applications'.
My contribution to that book was co-authoring (with Jan van Mill)
the paper `Open problems in $\beta\omega$'.
I know of several solutions to the problems from our paper
but I may have missed some.
If anyone has any solutions to problems on $\beta\omega$
then please inform me of them either via e-mail or through
a posting in this newsgroup.
Thank you,
                K. P. Hart
                k.p.hart@twi.tudelft.nl
PS. Turn to
http://www.elsevier.com/catalogue/SAJ/220/08350/08354/504633/504633.html
For information on the book.

I don=B4t know if a got the problem, but I think you have a (probably=20
continuous) function $f$ defined in the vector space $gl(p,R)$ of $p\time=
s=20
p$ real matrixes. Then you consider the function $A$ defined in $SO(q)$=20
taking values in $gl(p,R)$ which assigns to each $q\times q$ orthogonal=20
matrix it=B4s upper left $p\times p$ corner. Then you wan=B4t to integrat=
e the=20
composition of $f$ and $A$, $f\circ A$ over the Lie group $SO(q)$ using=20
Haar measure.
I really don=B4t see why $p=3D1$ should be any easier than the general ca=
se,=20
since the problem arises when trying to calculate an integral over=20
$SO(q)$, no matter what the function is.
Usually when integrating over two-dimensional manifolds, it=B4s possible =
to=20
find a coordinate system which covers the manifold except for a zero=20
measure set. In higher dimensions, when this can=B4t be found, you have t=
o=20
write your manifold as the union of a countable number of disjoint=20
measurable sets, each of them contained in a coordinate system domain.
In the case of $SO(q)$ there is a quite simple coordinate system in a=20
neighborhood of the identity:
$$A\in SO(q)\mapsto A-A^t\in so(q)$$
I don=B4t really know in which domain this function is a coordinate syste=
m=20
but it=B4s probably no so hard to find one. Other coordinate systems can =
be=20
found by left translation. If $X$, $Y$ are matrixes tangent to $SO(q)$ in=
=20
some point then $tr.(X Y^t)$ gives you a bi-invariant Riemannian metric=20
which induces the Haar measure (except that probably the volume of the=20
group won=B4t be 1).
Another idea is to think of $SO(q)$ as a $SO(q-1)$-principal bundle over=20
the sphere $S^{q-1}$. Then by removing one point of the sphere (and=20
therefore one fiber of $SO(q)$, which has zero measure), you can calculat=
e=20
the integral over $SO(q-1)\times R^{q-1}$ (using stereographic projection=
=20
as a diffeomorphism from $S^{q-1}$ to $R^{q-1}$ - the fiber bundle over=20
$R^{q-1}$ must be trivial since the basis is contractible but you have to=
=20
find the actual trivialisation to perform the calculations, I don=B4t kno=
w=20
what is it now).
Then you will reduce an integration problem over $SO(q)$ to one over=20
$SO(q-1)$ and so one...
For $q=3D1$ the problem is really trivial since the group has only one=20
point. For $q=3D2$ it=B4s easy too, since the group is just the circle $S=
^1$.=20
For $q=3D3$ you can remember that $SO(3)$ admits a two-fold covering by=20
$S^3$ and then use degree theory to relate the integral over $SO(3)$ and=20
over $S^3$ (which is an easy integral, remove one point, use stereographi=
c=20
projection, etc.)
I=B4m not saying that this are simple calculations but I don=B4t think th=
ere=20
can be a simpler method (unless, perhaps, if you have more information=20
about the function $f$).
Daniel Victor Tausk - elibene@ibm.net or tausk@ime.usp.br

	Last week I wrote:
>         When, as in defining quotient groups, one defines an operation on
> classes by picking a member of each class to be operated on, and performing
> an operation on the members, (and then worrying about "problematic", (i.e.
> possibly ambiguous and possibly not) definitions), then one is defining an
> operation *******-wise.  I'm looking for this locution, filling in the
> row of asterisks with a one- or two-syllable word.  Does anyone know it?
	In a post that hasn't yet appeared at this site
	(I got it by gopher-ing to math.lfc.edu
	Andrew Stacey  replied:
> This is a complete stab in the dark, it's probably not recognised
> terminology but if you just want something that carries the idea try:
> 
> representative-wise
> 
> Since the 'x' in x+H is a representative element of the equivalence class
> defined by: x~x' iff x-x' is in H (using addition rather than mult as
> subtraction is easier to type than inverses!)
	Actually my question was something of a stab in the dark.  I don't
know that any such term is generally used, nor indeed that it has _ever_
been used.  But it seemed that it would sometimes by convenient.
"Representativewise", or "representative-wise" was also proposed by one
other person, via private e-mail not posted.  But he thought it was a bad
term and so do I.  It doesn't flow.  The rhythm is terrible.  The language
should flow smoothly and be suggestive.  Everyone who knows what is meant
should know what is meant (and I fear that might be the case with
"representativewise" -- some people who know what is meant might fail to
know what is meant!).  I suppose we need someone with both mathematical and
poetic abilities.  It now occurs to me that I know such a person -- I think
I'll ask him.
	Thanks are due to Andrew Stacey & the other respondent.
	Mike Hardy
Michael Hardy
hardy@stat.umn.edu

When you have a real algebraic manifold, I see what it is to xomplexify it,
you just tensorize by C, and get something of complex dimension equal to the
real dimension of the first thing. But Manin proposed a philosophical trick to
complexify things stranger than manifolds :
	He says that when you have Er and Ec its complexified, then, you should
consider Pc=Pi1(Ec) as the complexified of Pr=Pi0(Er). For example, Sn, the
symmetric group is Pi0(R^n\D) where D is the diagonal, and the pure braid group
Bn*=Pi1(C^n\D), so he claims that Bn* should be the complexified of Sn.
	Does someone knows what is behind this philosophical argument ?
	In particular, how could the true complexification be understood in 
this way ?
	Thank you, 		Christian Mercat

I would like to understand why the circle S1 should be the complexified
of Z2={-1,+1}. For that, I should find "something", a CW complex, a manifold,
(infinite dimensional), E such that S1=Pi1(E).
And after, if that E is by any chance the complexified of something Er, I would
love that Z2=Pi0(Er).

John Harper wrote:
> 
> In article <326C69C9.167E@boas5.bo.astro.it>,
> luca ciotti   wrote:
> >I'm not a mathematician, so my question probably is not formally
> >well posed.
> >
> >My question is: in R^3, the Poisson equation is Lapla(Psi)= F
> >where Psi and F are functions of X=(x,y,z).
> >
> >I conjecture that Psi=Psi(F) if and only if Psi is spherically
> >symmetric.
> 
> GK Batchelor Introduction to Fluid Dynamics (Cambridge 1967) p19
> suggested that Psi might have to be one of (a) a function of one co-ord
> of a rectangular system or (b) the radial coordinate of a cylindrical
> polar system or (c) the radial coordinate of a spherical polar system.
> Cases (a) and (b) are counter-examples to the conjecture. But I was
> never happy with Batchelor's list: one can imagine an infinite line
> of equally spaced "stars" each of which is nearly but not exactly
> spherically symmetric, but I admit I never filled in the details.
> 
> John Harper Mathematics Dept. Victoria University Wellington New Zealand
   No, of course the solution to the Laplace equation on R^n doesn't
have to be spherically symmetric. Whatever solution you have, if it
is spherically symmetric just add it e.g. x+2y+3z.
   I didn't quite understand what the converse says (any spherically
symmetric \psi solves \Lap u = F for some F? Of course it does, put 
F = \Lap \psi in the according Sobolev space, it has nothing to do with
symmetricity)
   What he probably meant is that the only second order partial
differential operator on R^n that is spherically symmetric and
translation invariant is the Laplace operator.
-- 
Jan Rosenzweig
e-mail: rosen@math.mcgill.ca
office:                                        home:
Department of Mathematics and Statistics       539 Rue Prince Arthur O. 
Burnside Hall, room 1132, mbox F-10            Montreal
805 Rue Sherbrooke O.                          Quebec H2X 1T6
Montreal, Quebec H3A 2K6