mm-391
===
Subject:
: Re: Maximal number of ones in a nilpotent
matrix>What is the maximal number of ones in a nilpotent nXn
matrix >over GF(2) and in a nilpotent nXn (0,1) matrix over
the reals ?If n is even, the square of the matrix of all 1's
is 0 over GF(2).If n is odd, the cube of the matrix of all 1's
except for one diagonal entry is 0 over GF(2). So the answer to
the first question is n^2 ifn is even, n^2-1 if n is odd.Over
the reals, consider an nxn (0,1) matrix as the incidence
matrixof a directed graph (with arc i->j if A_{ij} = 1; arcs
i->i are allowed).The matrix is nilpotent iff the directed
graph has no cycles. In such a graph, there is a partial order
on the vertices such that if i->j isan edge, i>j in the partial
order. If the rows and columns are sortedusing ts partial
order, the matrix is strictly lower triangular.Conversely, a
strictly lower triangular matrix is nilpotent. So theanswer
over the reals is the number of entries below the diagonal,
namely (n^2-n)/2.=== 
===
Subject:
: lbert's 16thcan someone tell me
the status of the proof of the 2nd part of lbert's16th
problem? According to news reports a 22-year old student from
Sweden(Elin Oxenelm) has managed to prove it. However, Grigori
Rozenblum(Chalmers Univ. of Techn, Goeteborg, Sweden) has
apparently found an error.George Szpiro----------George
SzpiroNeue Z.9frcher Zeitung (Switzerland)POB 6278Jerusalem
91060Israel===
===
Subject:
: Re: lbert's 16th > can someone tell me
the status of the proof of the 2nd part of lbert's> 16th
problem? According to news reports a 22-year old student from
Sweden> (Elin Oxenelm) has managed to prove it. However,
Grigori Rozenblum> (Chalmers Univ. of Techn, Goeteborg,
Sweden) has apparently found an error.There's a copy of
Rozenblum's comments at
http://www.unstruct.org/arcves/000186.htmlincluding s
statement thatTs lies far below generally accepted
mathematical dards of proofOliver son-- otj1000@cam.ac.uk
http://www.statslab.cam.ac.uk/~son 01223 337946 Christ's
College and Statistical Laboratory, University of
Cambridge===
===
Subject:
: Re: lbert's 16th+ George Szpiro
<george@netvision.net.il>:| can someone tell me the status of
the proof of the 2nd part of| lbert's 16th problem? According
to news reports a 22-year old| student from Sweden (Elin
Oxenelm) has managed to prove| it. However, Grigori Rozenblum
(Chalmers Univ. of Techn, Goeteborg,| Sweden) has apparently
found an error.I cannot comment on the latter part, but the
proof - whether corrector not - is not quite for the full
problem. Here is the abstract: Let k be an integer such that k
is larger than or equal to zero, and let H be the lbert number.
In ts paper, we use the method of describing functions to prove
that in the Li.8enard equation, the upper bound for H(2k +1) is
k. By applying ts method to any planar polynomial vector field,
it is possible to completely solve the second part of lbert's
16th problem.The paper will appear in Nonlinear Analysis.
Those who have onlineaccess to the journal can find it under
Articles in press.-- * Harald Hanche-Olsen
<URL:http://www.math.ntnu.no/~hanche/>- Debating gives most of
us much more psychological satisfaction than tnking does: but
it deprives us of whatever chance there is of getting closer
to the truth. -- C.P. Snow===
===
Subject:
: An approximation of
first zeta zeroEpigone-thread: frayyeoiIn case ts means
sometng: 2*pi*e^sqrt(2) / (2*sqrt(2) - 1) is approximately
14.13472486539888320737873054362765411254519... The first
non-trivial zeta zero is approximately
14.13472514173469379045725198356247027078425... The difference
is approximately
-.000000276335810583078521439934816158239053... or around
-1/3618785.4114...Here are some other calculations,where z1 =
first zeta zero, g = gamma: z1/pi =
4.4992227511047348484865414231801574... sqrt(z1/2*pi) =
1.4998704529233074482191259501034442... g - g^pi =
3.9929698161176810954018841193514217... z1/pi - (g-g^pi) =
4.0999257694929667389463530112450152... ===
===
Subject:
: Monte
CarloEpigone-thread: proyyanwenTs may not answer your
question. However there is a simple methodused in Monte Carlo
for calculating what you want. It goes by thename rejection
method. Here one simply chooses a point uniformly inthe
interval [0,1], then keeps it if it is witng the measurable
setof interest, otherwise ignore it. The points selected will
have auniform distribution in the set of interest.===
===
Subject:
:
commutators!Suppose you have two sets of projectors in a
finite lbert space :