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In article <5681p9$4rk@vixen.cso.uiuc.edu>,
 sanders@titan.iwu.edu (Robin Sanders) writes:
[...]
|> Let G be a bipartite graph, and let V(G) = A \cup B be the
|> partition of the vertices.  Let \pi be an automorphism of G.
|> We'll call \pi a switching automorphism if \pi takes the 
|> vertices in A to vertices in B and vice versa.  Clearly any 
|> switching automorphism is of even order.
|> 
|> Our question:  Can a bipartite graph have switching automorphisms,
|> but have no switching automorphism of order 2?  If so, can you
|> give us an example of such a graph?
Just arrange for a cyclic group of order 4, say, of automorphisms,
such that the aut's of order 4 are switching.
Enjoy, Gerhard
SPOILER below:
(Ever looked at a chess board?  Take the fields as vertices, with pairs
of adjacent fields as edges, and the bipartition obvious from the coloring.
Now this won't do as it stands, due to the mirror symmetries, but it's
easy to remove mirror symmetry whilst preserving cyclic symmetry by removing
four fields.  Don't remove too many fields or you'll introduce automorphisms
not coming from geometric motions.  A 6x6 board is enough, removing one of
the middle fields from each edge.)
-- 
* Gerhard Niklasch  *** Some or all of the con-
* http://hasse.mathematik.tu-muenchen.de/~nikl/ ******* tents of the above mes-
* sage may, in certain countries, be legally considered unsuitable for consump-
* tion by children under the age of 18.  Me transmitte sursum, Caledoni...  :^/

I've got this problem about Ornstein-Uhlenbeck processes:
compute the distribution of the first passage time to a given level b.
The OU process is the solution of the SDE dX=K(a-X)dt+sdW
Is it well known?
Easy with standard machinery?
Hard?
Any idea will be wellcome.
Thank you very much.

In article <5681p9$4rk@vixen.cso.uiuc.edu>,
Robin Sanders  wrote:
>Let G be a bipartite graph, and let V(G) = A \cup B be the
>partition of the vertices.  Let \pi be an automorphism of G.
>We'll call \pi a switching automorphism if \pi takes the 
>vertices in A to vertices in B and vice versa.  Clearly any 
>switching automorphism is of even order.
>
>Our question:  Can a bipartite graph have switching automorphisms,
>but have no switching automorphism of order 2?  If so, can you
>give us an example of such a graph?
Take a cycle of twenty vertices, numbered 1,2,...,20, and add edges
connecting the following pairs of vertices:
     (1,6)    (6,11)     (11,16)     (16,1)
     (1,4)    (6,9)      (11,14)     (16,19)
The resulting bipartite graph has only two switching automorphisms
(inverses of each other), both of order 4.
Randall Dougherty                       rld@math.ohio-state.edu
Department of Mathematics,  Ohio State University,  Columbus, OH 43210  USA
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated."  Poul Anderson, "Call Me Joe"

In article <2gloc82vqp.fsf@pulsar.cs.wku.edu>, adler@pulsar.cs.WKU.EDU
(Allen Adler) wrote:
> Let f : X -> Y be a surjective morphism between complete algebraic
> surfaces, where X is nonsingular and Y is normal. I allow f to have
> degree greater than 1. Let p be a singular point of Y (necessarily
> isolated) and let C_1,...,C_n be the irreducible components of the
> preimage of p in X. Let M be the intersection matrix of the curves
> C_1,..,C_n, i.e. the i,j entry of M is the intersection number of
> C_i and C_j.
> 
> True or false: M is nonsingular.
True. This is well known when f has degree 1; in fact M is to definite
negative. See for example D. Mumford: Topology of normal singularities and
a criterion for simplicity, Publ. Math. IHES 9 (1961), 5-22.
In general, let Y be the normalization of Y in the function field of X;
then f factors as X->Y'->Y, where Y' is normal, X->Y' has degree 1, and
Y'->Y is finite (in technical terms, we are doing a Stein factorization of
f.) Then if {p'_1,...,p'_r} is the inverse images of p in Y', the matrix M
is the direct sum of the intersection matrices of the inverse images in
each of the p'_i, and so it is also negative definite.
Angelo Vistoli

In my recent research, I'm trying to estimate the shape of distribution of
the average of n i.i.d U(0,1) random variables.  Clearly the central limit
theorem applies, and so I expect the distribution of the average to be 
~Normal(0.5,\sqrt{n/12}) for large n, since the uniform(0,1) distribution
has mean 0.5 and variance 1/12.
This is fine, but I want to look at the 
lim_{n -> infinity} [2^n * Prob(average < x)]
So now I can't use the central limit theorem, since I'm taking a limit over
a product of 2^n (very large) * Distribution.
More concretely, for what 'x' is the above limit equal to precisely 1?
Is there a general literature addressing these sorts of questions on tails
of distributions?
thanks for any help,
-- 
    -********* *-
   -*  Vince  *-
  -* *********-

Let f be a supermodular and Lipschitz function (f_{xy}>0) on [0,1]X[0,1].
Let h be a Lipschitz function on [0,1], and put g(x,y) = f(x,y) + h(x) + h(y).
Define the indicator function I(x,y) = 1 if f(x,y)>=0 and g(x,y)<0 or
f(x,y)<0 and g(x,y)>=0, and zero otherwise.
CLAIM       For all epsilon>0, there exists delta>0 such that
sup(|h(z)|)  ess-sup_x \int I(x,y) dy < epsilon
The motivating idea is that for x10. But I find it hard to deduce the above
strong form of continuity. (If it helps, observe that g_{xy}=f_{xy}.)
Thanks for any help with the claim! -- 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 article <5681p9$4rk@vixen.cso.uiuc.edu>,
Robin Sanders  wrote:
>I and a colleague are looking at bipartite graphs.  We need
>to know some information about certain kinds of automorphisms.
>
>Let G be a bipartite graph, and let V(G) = A \cup B be the
>partition of the vertices.  Let \pi be an automorphism of G.
>We'll call \pi a switching automorphism if \pi takes the 
>vertices in A to vertices in B and vice versa.  Clearly any 
>switching automorphism is of even order.
>
>Our question:  Can a bipartite graph have switching automorphisms,
>but have no switching automorphism of order 2?  If so, can you
>give us an example of such a graph?
Yes, it can.  Take a cycle of length 12 with vertices labelled
0 through 11 around the cycle.  To each vertex  i  add  i mod 3
new vertices adjacent to  i.  The automorphisms of this graph on
24 vertices (you can get fewer by being a bit more clever) are
exactly gotten by adding 3 mod 12 to the vertices in the cycle,
i.e. by rotations through three vertices.
The two automorphisms of order 4 are switching automorphisms, but
the unique automorphism of order 2 is not one.
David Moulton
University of Wisconsin - Madison
moulton@math.wisc.edu

Particulars relating to the position of
RESEARCH OFFICER, Level A,  
in the Department of Mathematics within THE UNIVERSITY OF QUEENSLAND
Duties: The  successful  applicant will  be required to  work with  Drs Phil
        Diamond and Darryn Bryant, of The University of Queensland, and Professor 
	Nikolai Kuznetsov of the Russian Academy of Sciences, on the Australian 
	Research Council funded project "Statistical Laws for Computational
	Collapse of Chaotic Systems". The applicant must be prepared to make 
	a commitment to work FULL-TIME on the project. 
Salary: $37,170 -- $38,587 -- $40,004 (Annual increments, in Australian 
	Dollars). It is expected that there will be a 5%--8% increase in 
	these figures, granted by mid 1997.
Qualifications: Applicants should  have research  interests in Dynamical 
                Systems Theory or a closely related field of Mathematics, 
		with strengths in measure theory, probability, asymptotics 
		and functional analysis.  They should either hold a  Ph.D. 
		or be nearing completion of their Ph.D. They should be able  
		to work  independently,  albeit under supervision.
Date of 
commencement: This will  be as  soon  as  possible  after  January 1st  1997.
              Applicants who would  wish to commence  late in 1996 SHOULD NOT
              BE DETERRED FROM APPLYING. If the successful applicant does not
              hold a Ph.D., he/she must have submitted their thesis for exam-
              ination prior to taking up the post.
Period of
appointment: The post  will be  offered for  a maximum of  three years, on a
             one-year  renewable  basis.  The  post will  terminate on  31st 
             December 1999. For applicants without a Ph.D., appointment to a
             second year will be dependent on their successfully  completing
             the requirements of the degree.
Method of
application: Applications  should  be  forwarded  as soon as  possible to  Dr
             Phil Diamond at the  address below,  preferably by e-mail (a
             speedy  acknowledgement  will follow) or  FAX, and  by no  later
             than  9th December 1996.  Applications   should  include a  full
             curriculum  vitae (resume),  together with the names,  addresses
             and telephone numbers of THREE referees.  It would be helpful if
             e-mail addresses and/or  FAX numbers  could be provided for each
             referee.
Relocation
expenses:    A single one-way economy airfare to Brisbane will be provided.
Interested parties are urged to make contact as soon as possible.
+------------------------------+--------------------------------------------+
| Phil Diamond                 | Telephone        (+61 7) 3365 3253         |
| Department of Mathematics    |                                            |
| The University of Queensland | e- mail          pmd@axiom.maths.uq.oz.au  |
| Queensland 4072              |                                            |
| AUSTRALIA                    | Fax              (+61 7) 3365 1477         |
+------------------------------+--------------------------------------------+

Robert Israel (israel@math.ubc.ca) wrote:
: In article <562s4a$827@senator-bedfellow.MIT.EDU>,
: Lones A Smith  wrote:
: >Let U(x,y) be a bounded function on [0,1]X[0,1], with U(x,0)=0 for all x,
: >and U(x,y) monotone nondecreasing in all y>=0. Suppose that the sum of 
: >all discontinuities of U in x, the sum taken over all (x,y) in [0,1]X[0,1],
: >is at most 1. I presume this means that there are countably many such 
: >discontinuities (x_1,y_1),(x_2,y_2),... with the left or right jump in x 
: >at (x_i,y_i) equal to a_i, with |a_1|+|a_2|+... <= 1. 
: >
: >CLAIM: For all epsilon>0 there exists delta>0 s.t. for almost all x, 
: >U(x,y) 0, U(0,0) = 0 and U(0,y) = 1
: for y > 0.  
: Robert Israel                            israel@math.ubc.ca
Stupid omission on my part! In my problem I forgot to say that I have 
lim_{y->0} U(x,y)=0 for all x. This is my reason for considerable hope 
in its truth. It kills this counterexample, and all that a friend and 
I have brainstormed.
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 article <5604em$dti@wiscnews.wiscnet.net>, Alex Smith  
wrote:
>If f is an isometry between two Banach
>spaces, the must f necessarily be linear?
>
I think this is not true in complex spaces, since the complex
conjugation in the complex number field C is an isometry, but
not linear. Otherwise one would get
-i = conjugate(i*1) = i * conjugate(1) = i
using linearity at the second equal sign. With this it should
be possible to construct nonlinear isometries in all complex
Banach spaces or at least in all complex Hilbert spaces (using
orthonormal bases).
Bye, Silvio
--
Silvio Martin     This message represents
FB Mathematik, Gerhard Mercator Universitaet   neither my employer's
47048 Duisburg, Germany                        opinion nor my own.

Robin Sanders  wrote:
> I and a colleague are looking at bipartite graphs.  We need
> to know some information about certain kinds of automorphisms.
> Let G be a bipartite graph, and let V(G) = A \cup B be the
> partition of the vertices.  Let \pi be an automorphism of G.
> We'll call \pi a switching automorphism if \pi takes the 
> vertices in A to vertices in B and vice versa.  Clearly any 
> switching automorphism is of even order.
> Our question:  Can a bipartite graph have switching automorphisms,
> but have no switching automorphism of order 2?  If so, can you
> give us an example of such a graph?
Yes, it can happen.  I don't have the paper with me at the moment,
but you can find an example in:
D. McCarthy and B. D. McKay, Transposable and symmetrizable matrices,
 Journal of the Australian Mathematical Society (Series A),
 29 (1980) 469-474.
A related complexity result appears in:
C. Colbourn and B. D. McKay, A correction to Colbourn's paper on 
 the complexity of matrix symmetrizability, Information Processing
 Letters, 11 (1980) 96-97.
Namely: to determine if there is a switching automorphism is
polynomially equivalent to the graph isomorphism problem, but
to determine if there is one of order 2 is NP-complete.  The
latter result rested heavily on work of Lubiw.
Brendan McKay.

In finite dimensional vector spaces
diagonable operators are dense in space of all operators.
Is this true in infinite-dimensional Hilbert,Banach ,.... spaces ?
>From Russia with Love :)
Alexander Chervov.

In article <5681p9$4rk@vixen.cso.uiuc.edu> sanders@titan.iwu.edu (Robin Sanders) writes:
>Our question:  Can a bipartite graph have switching automorphisms,
>but have no switching automorphism of order 2?
                       o
                       |
                    o  o
                    |  |
                 o--o--o--o
                 |        |
           o--o--o        o--o
                 |        |
              o--o        o--o--o
                 |        |
                 o--o--o--o
                    |  |
                    o  o
                    |
                    o
-- 
   /\   Greg Kuperberg        greg@math.ucdavis.edu
  /  \ 
  \  /  Recruiting or seeking a job in math?  Check out my Generic Electronic
   \/   Job Application form, http://www.math.ucdavis.edu/~greg/geja/

In  Hilbert space Trace of operators is independet from choice of
ORTONORMAL base.
What is situation if base is NOT ORTONORMAL ?
If space is not Hilbert then :
    Can we give definition of trace for some class of operarors ?
If base exists in this space (in Schauder or other sense )
    Can we define base sum of diogonal elements and will this sum be independant
from choice of base?
Really i'm not interested  about proofs or details of this results.
I proved  formula for trace of some operators and i'm sure that in my
case there is no dependence of choice of base and it will be useful
for me if there exists some general results which i can use.
Best regards
Alexander Chervov.

Hello all, I'm a third year grad student at the university of wash,
seattle, and I have a rather vague question I was hoping someone
might be able to answer for me.
I've been playing around with algebraic geometry for a little while
now and I've recently began playing with the Jacobian Conjecture
(f: C^n \to C^n is such that df is everywhere invertible, then the
conjecture says f with invertible).  Along the way I've run into
a very particular class of coordinate rings I'd like to consider.
Namely those coordinate rings who can be embedded into the coordinate
ring for affine space.  I was wondering if anyone out there had ever
heard of anyone studying the objects?
Thanks Much,
Bill Adams
P.S.  If you could send your responses to adams@math.washington.edu, 
I'd very much appreciate it.

In article ,
Vince Darley  wrote:
>In my recent research, I'm trying to estimate the shape of distribution of
>the average of n i.i.d U(0,1) random variables.  Clearly the central limit
>theorem applies, and so I expect the distribution of the average to be 
>~Normal(0.5,\sqrt{n/12}) for large n, since the uniform(0,1) distribution
>has mean 0.5 and variance 1/12.
>This is fine, but I want to look at the 
>lim_{n -> infinity} [2^n * Prob(average < x)]
>So now I can't use the central limit theorem, since I'm taking a limit over
>a product of 2^n (very large) * Distribution.
>More concretely, for what 'x' is the above limit equal to precisely 1?
>Is there a general literature addressing these sorts of questions on tails
>of distributions?
There is an extensive literature on large deviation theory, starting
more than 60 years ago.  In general, for continuous distributions,
P(average < x) ~ [c(x)]^n/[h(x)sqrt(2n\pi)]
when x is less than the mean.  The literature usually discusses
P(average > x) for x greater than the mean.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu	 Phone: (317)494-6054	FAX: (317)494-0558

Dear all,
we're facing a problem with a simple Gaussian function.
Does anybody know the closed form of:
(LATEK) \frac{d^m}{dx^m}(e^{-x^2})
(Visual)
.             2
.        m  -x
.       d  e
.      --------
.           m
.        d x
for any m in N?
Any help will be greatly appreciated.
Kind regards,
Marco magix@dibe.unige.it
Giorgio nemo@dibe.unige.it

adler@pulsar.cs.WKU.EDU (Allen Adler) writes:
> Let f : X -> Y be a surjective morphism between complete algebraic
> surfaces, where X is nonsingular and Y is normal. I allow f to have
> degree greater than 1. Let p be a singular point of Y (necessarily
> isolated) and let C_1,...,C_n be the irreducible components of the
> preimage of p in X. Let M be the intersection matrix of the curves
> C_1,..,C_n, i.e. the i,j entry of M is the intersection number of
> C_i and C_j.
> 
> True or false: M is nonsingular.
True, the earliest reference I know of is 
@Article{Mu61,
  author =       "David Mumford",
  title =        "The topology of normal singularities of an algebraic
                  surface and a criterion for simplicity",
  journal =      "Publ. Math. IHES",
  year =         1961,
  number =       9,
  pages =        "5--22"
},
the statement that the intersection matrix is in fact negative
definite is to be found at page 6.

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!

magix  wrote:
>Dear all,
>we're facing a problem with a simple Gaussian function.
>Does anybody know the closed form of:
>(LATEK) \frac{d^m}{dx^m}(e^{-x^2})
>(Visual)
>.             2
>.        m  -x
>.       d  e
>.      --------
>.           m
>.        d x
>
>for any m in N?
Doesn't  the Rodriguez formula for Hermite polynomials give at once that 
the derivative is
\exp(-x^2) H_m(x) (-1)^m
?
(and there are explicit formulae for thesepolys)
-- 
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* Dr W B Stewart            phone +44 1865 279628      *
* Exeter College            fax   +44 1865 279630      *
* Oxford                                               *
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********************************************************

magix (magix@dibe.unige.it) wrote:
: Dear all,
: we're facing a problem with a simple Gaussian function.
: Does anybody know the closed form of:
: (LATEK) \frac{d^m}{dx^m}(e^{-x^2})
: (Visual)
: .             2
: .        m  -x
: .       d  e
: .      --------
: .           m
: .        d x
: for any m in N?
: Any help will be greatly appreciated.
: Kind regards,
: Marco magix@dibe.unige.it
: Giorgio nemo@dibe.unige.it
Faa di Bruno's formula strikes again!
This is the second posting on the same topic belonging to
Calculus (!) that skips past the moderators.
Felipe

In article <56bp2a$ai0@senator-bedfellow.MIT.EDU>, lones@lones.mit.edu (Lones A Smith) writes:
|> : In article <562s4a$827@senator-bedfellow.MIT.EDU>,
|> : Lones A Smith  wrote:
|> : >Let U(x,y) be a bounded function on [0,1]X[0,1], with U(x,0)=0 for all x,
|> : >and U(x,y) monotone nondecreasing in all y>=0. Suppose that the sum of 
|> : >all discontinuities of U in x, the sum taken over all (x,y) in [0,1]X[0,1],
|> : >is at most 1. I presume this means that there are countably many such 
|> : >discontinuities (x_1,y_1),(x_2,y_2),... with the left or right jump in x 
|> : >at (x_i,y_i) equal to a_i, with |a_1|+|a_2|+... <= 1. 
|> : >CLAIM: For all epsilon>0 there exists delta>0 s.t. for almost all x, 
|> : >U(x,y) Stupid omission on my part! In my problem I forgot to say that I have 
|> lim_{y->0} U(x,y)=0 for all x.
Ah... Then you don't need to say anything about the "sum of the discontinuities",
only that there is a sequence y_n decreasing to 0 such that none of the 
discontinuities are at points with y component y_n (in particular this will
be true if there are countably many discontinuities), and the conclusion
will be true for all x, not just almost all.
Proof: u_n(x) = U(x,y_n) are a sequence of continuous functions on [0,1] 
decreasing monotonically to 0.  By Dini's Theorem, they converge uniformly 
to 0.  So given epsilon > 0, there is N so that u_N(x) < epsilon for all x.  
This means you can take delta = y_N, and you will have 
U(x,y) <= U(x, y_N) < epsilon for all x and all y < delta.
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4