mm-207
===
Subject: Re: Group actions and SO(n)> If I define an
action SO(n+1) x S^n ----> S^n where G = SO(n+1) and X =
S^n,>Sorry, the action is (A,z) ----> Az> then what is the
stabilizer of an element x in S^n? Is it SO(n) and
anyone wishes to check my angle chase, heres one 
description
of it.Ive tried it three times now, so either 
its right or
Im hitting ablind spot. Start by d   arawing in a 
representative
diagram up to thatpoint. (Make it big -- this is important!)
Let the vertices inclockwise order be labelled L, M, N, O, P,
Q, so that LM is side A, MNis side B, etc. Let the photon
start at point R, and successive pointswhere it changes
direction be S, T,..., Z. (Z is the point on the lastD-F
segment, and I aim to show that Z does not actually
exist.)That is, R is on side A (LM), S is on side E (PQ), T is
on B (MN),U is on D (OP), V is on ST, W is on RS, X is on F
(QL), Y is on E (QS),and Z should be on WX.Let angle LRS =
theta. Then we can chase angles (all degree signsomitted) as
follows: QSR = 120 - theta PST = 120 - theta RST = 2*theta -
60 MTS = 120 - theta NTU = 120 - theta TUO = theta PUV = theta
UVS = 120 SVW = 120 VWS = 120 - 2*theta SWX = 120 - 2*theta RWX
= 60 + 2*theta LXW = 180 - 3*theta QXY = 180 - 3*theta WXY =
6*theta - 180 XYQ = 3*theta - 120 SYZ = 3*theta - 120 XYZ =
420 - 6*thetaBut then in triangle XYZ, angles XYZ + ZXY sum to
240 degrees, which isimpossible. We conclude that the ray is
actually heading at 60 degrees_away_ from the last segment
from D-F. So that ray must actually crosssegment WS rather
than WX. I believe it then crosses VS and hits Eagain,
possibly then going via UV, TV and maybe back to the
startingpoint. (I think, but am not sure, that there is enough
leeway for this.)So a modified version of the problem might 
have
Re: Reconsidering Halton Arp > You see, Dr. Arp is a
scientist, a world renowned scientist and he has> *data*,
real, hard astronomical data, which is more substantive in>
disproving the commonly taught Big Bang Theory, than the data
used to> support that theory.> Arps data is presented in
the _Atlas of Peculiar Galaxies_,> Astrophysical Journal
Supplement Number 123, Volume 14, November 1966.> In the
Atlas Arp presents galaxies that appeared abnormal. Follow-up>
observations showed that some, not all, of the galaxies were in
fact> two galaxies that are apparently interacting. What caused
the doubt> about the Big Bang was that some of these pairs have
very different> red-shifts. If the galaxies are close to each
other the different> red-shifts would sound the death knell
for expansion and the BB.> However, as observing technique
has improved weve determined that> most of these pairs are
simply close in the line of sight and are> at very different
distances. There are a few cases that have not> been
elucidated, the necessary obsrvations are, at best, 
difficult. These remaining cases do not constitute an overthrow of the
BB,> to do that would require high quality observations of
difficult> objects; big results require big data, obscure,
difficult cases> do not provide that.I went to Google, and
found a relevant link.<Quote In figure A below, there are four
objects. NGC 7603 is a spiral galaxy> with a redshift value of
0.029. Object 1 is a quasar with z = 0.057.> Objects 2 and 3
are quasar-like objects with z values of 0.243 and> 0.391
respectively. As L.97pez-Corredoira and Guti.8errez noted:>
Everything points to the four objects being connected among>
themselves, but how to explain the different redshifts? (p.
L17). How> to explain indeed? Gribbin lamented: That strikes
at the foundation> stone of received cosmological wisdom (p.
65). It certainly does! As> case where we once again are
experiencing a situation where data get> thrown out if they
dont fit the theory. Big Bang cosmology simply> 
cannot explain
Arps anomalies.> </Quote date November 1966, in what I see 
as
a not subtle attempt to skew> reader opinion.As Ive said, 
Dr.
Arp has DATA. Check for yourself.James HarrisDont be so
stooopis Harris--Arps anomalies are just that.Tests of Big
Bang Cosmology
http://map.gsfc.nasa.gov/m_uni/uni_101bbtest.htmlThe Big Bang
Model is supported by a number of importantobservations, each
of which are described in more detail onseparate pages:
Re: Selecting the correct graphIts a general question. Just
help on selecting the right graph for theright kind of data
Group actions and SO(n)>If I define an action>SO(n+1) x S^n
----> S^n(If you DO define an action, it would be easier to
answer yourquestion. Does the group act trivially, perhaps? Or
is this thenatural linear action of a matrix group, restricted
to a subsetof R^{n+1} ?)>then what is the stabilizer of an
element x in S^n? Is it SO(n) and why?Trivial action: Stab(x)
is all of SO(n+1).Natural action: Stab(x) is a conjugate of
SO(n). Specifically, when x is the vector x0=(1,0,0,...), the
stabilizer is the set of matrices in SO(n+1) which take x0 to
x0 (of course) and thus consists of matrices whose first 
column
is (1,0,0,...). The rest of the first row is then all zeros 
too,
but the remaining n-by-n matrix can be any special orthogonal
group. So this stabilizer is in a natural way isomorphic to
SO(n). Then if x is any other point of S^n, choose any matrix
M in your group such that x = M.x0 (this group action is
transitive) and then youll find that Stab(x) = 
M Stab(x0)
M^{-1}.Other action: It depends!>where G = SO(n+1) and X =
S^n,Youre assigning labels you dont use. I 
think it was Mary
Ellen Rudinwho related the story of a student whose proof
Car crash formula > Im unaware of dimensional analysis.> 
The
cars mass 1134 Kg> Objects Mass 68 Kg> length 22.861 M> 
Im
trying to find a formula for where I can measure the speed of
acar> when> it hits a still object based on cars weight,
objects weight, andtotal> distance the object was thrown. I
realize that there are many other> factors> such surface
friction, in this case road, but im just looking foran>
estimate, not to be as exact as posible.> Im unaware of
dimensional analysis.> The cars mass 1134 Kg> Objects Mass 
68
Kg> length 22.861 MSounds like a pedestrian being hit by a car.
Searles formula
isappropriate.http://www.aitsuk.com/download/Pedestrian%
20throw.pdfBy length, do you mean the distance the person was
thrown?If so, the simplest form Vmin = Sqrt(2u*g*S/1+u^2) will
give a minimumvelocity of 52km/hr.Read up on the text and
consider the information you know aboutthe accident to
(statistics)how to make date more like Laplacian
distribution should follow the shapeof Laplacian
distribution... the data obtained from measurement is of
coursea little off(not very symmtrical), how can I make the
measured data moreLaplacian distribution like(make it at least
a little more symmtrical)?Can anybody give me an example or
detailed explanation? I am kind of afraidof statistics...
Cosmology, De Sitter Group Lie AlgebraYou seem to be writing
to yourself a lot these days.Wonder what that means?MB> Let R
be a stringy Kaluza-Klein compactification scale of an extra
space> dimension Suppose (c/H(t))Lp*(t) = R^2(t) Use the
holographic Lp*(t)^2 = Lp^4/3(c/H(t))^2/3 with the world
hologram on the surface of a Planck sphere. Therefore,
(c/H(t))^4/3Lp*(t)^2/3 = R^2(t) This one puts the world
hologram on the past light cone wave front of> thickness
Lp*(t) back c/H(t) in time from t = now.> The first one is a
dual relation.> Do not reply to mindspring which is a dummy
address so that I can send> mail on my regular MacMail program
out from a WiFi Caffe. Use> sarfatti@well.com also I will be
moving soon so sarfatti@pacbell.net will be defunct.>
sarfatti@well.com is the one to use.> Jack, you ask: ï... 4
special conformal generators ...> What do they locally gauge
to?> My hunch is /zpf,u ... Yes, I think so too, and have
written some stuff about it> on my web page at>
http://www.innerx.net/personal/tsmith/coscongraviton.html The
basic reference for that work is a paper by Aldrovandi and
Pereira at> http://xxx.lanl.gov/abs/gr-qc/9809061> which
describes in some detail how the special conformal group>
gives rise to cosmological constant type terms. Yeah that
paper is interesting.> They seem to say that the deSitter
group limits to 15 parameter> conformal group when
cosmological constant limits to> infinity! Also there is an
interesting stringy duality between> infinite and zero
cosmological constant. It is the intermediate cases> that is
of interest. Also localizing - not just a space of constant>
curvature, i.e. locally gauge the Lie algebra of the De Sitter
group> which is more general than the conformal group? 
infinity
hence gravity as curvature is no longer possible. BTW the
world hologram idea Lp*^2 = Lp^4/3(c/Ho)^1/3 is alluring, but
has real problems of consistent interpretation such as>
cosmological time increase in the Regge slope i.e.> Wittens
alpha = Lp*^2 = 1/(string tension) means the string tension
decreases as the universe 3D space expands.> My guess is that
current astrophysics falsifies that? The electron rest mass
from Higgs field part of Vacuum Coherence is (h> = c = 1) m ~
e^2/zpf*^1/2 ~ e^2/(alpha)^1/2 What does this do to e/m? If
/zpf* = 1/Lp*^2 = 1/G* = (alpha)^-1 And if Blackett 
relation
for quantized trappedEM §ux in the Wheeler> micro wormhole> of
Mass without mass and Charge without charge: e = G*1/2 m =
(alpha)^1/2m m ~ (alpha)m^2(alpha)^-1/2 = 
(alpha)^1/2m^2
Ignoring m = 0 root. m ~ (alpha)^-1/2 e is then invariant,
but e/m is not. Ho = R(now),t/R(now) in the FRW metric. We
want G* on large scale > 10^-3 cm to be G(Newton) and it is
also> thought to be G(Newton) at the Planck scale> 10^-33 cm,
yet we want G* = 10^40G on the fermi scale of 10^-13
great trouble reading this post. It is always hard figure out
the statistical content when> the context is super-abstract. >
It becomes impossible to figure, when idiosyncratic 
terminology>
is combined with bad spelling -I have a question.> For a
randomly moving object in two-dimensional plane, > the object
has to move from point X to point Y. - *has* to move? Or, you
want to assume that does...> During the movement, there are
two random> processes posing on the object. - random forces,
pushing on the object?For example, one process is the
irregular geograph - geography?and the other process is the
varying weather. The two > processes may be correlated. Plz
give some suggestions on > where can I find the related
reference and which book > specify such problem.Okay.
Concretely, as I imagine it: You have a mountain goat > whose
wandering depends on the hilliness and weather (while>
exposure depends partly on hills, too). > Where will he go,
how fast? Look up animal husbandry.Concretely as you want:
What is the problem or question?Sorry for making you
confusing. But I think I should explain this.Each one of us
has his own research area, mine is the wirelesscommunication.
If I will simply copy my problem into this group, thenI have
to explain a lot of terms in my area. So I abstract
thepractical problem into an understandable math question,
which may notbe so suitable but should be easier for
process> I have a question.> For a randomly moving object in
two-dimensional plane, the object has to> move from point X to
point Y. During the movement, there are two random> processes
posing on the object. For example, one process is the
irregular> geograph and the other process is the varying
weather. The two processes> may> be correlated. Plz give some
suggestions on> where can I find the related reference and
> --> ZHANG Yan> http://www.ntu.edu.sg/home5/pg01308021Look
for the random walk topic in any advanced mathematical
statistics> book. You just have a random walk in 2 dimensions
here. Im not going to> suggest any of the old standard 
books
infinity ?[referring to improvements in version 5]> 
Theyve
improved FullSimplify too, then; in 4.2, Mathematica returns>
the sum of the sine series as -1/2 I (Log[1-E^-I] -
Log[1-E^I])> (approximately--this is from memory), and
FullSimplify doesnt help. Furthermore, FullSimplify 
wouldnt
touch ArcTan[Sin[t]/(1-Cos[t])]> either, although it has an
obvious simplification. Ill have to try> that 
in 5.0 (which I
have at work, but not at home).In version
5,FullSimplify[ArcTan[Sin[t]/(1-Cos[t])]] yields
ArcTan[Cot[t/2]].You say that it has an obvious 
simplification,
but Im not sure howobvious it is. Are you thinking about
something like (Pi - t)/2 + Pi*Floor[t/(2*Pi)]perhaps?
(However, that expression is defined for all real t,
whereasArcTan[Sin[t]/(1-Cos[t])] and ArcTan[Cot[t/2]] are not
polynomialI ran across a problem in a book to try and determine
if there was a> polynomial p(x) with at least 2 nonzero terms
such that p(x)^2 had> exactly the same number of nonzero terms
as p(x). I proved it> couldnt happen for linear, quadratic, 
or
cubic polynomials, and I> think I proved it couldnt happen 
for
quartic polynomials also. Then> I found a quintic where it is
true: Namely, p(x)=4x^5+4x^3-2x^2+2x+1,>
[p(x)]^2=16x^10+32x^9+28x^6+4x^3+x^2.Make that > 16 x^10 + 32
x^8 - 16 x^7 + 32 x^6 - 8 x^5 + 20 x^4 + 4x + 1.Oops, 
mustve
been a mistype. Tryp(x) = 4x^5 + 4x^4 - 2x^3 + 2x^2 +
EngineeringHow do you mean nonmetricity.?By nonmetricity I
understand the nonmetric part of an affine connection which is
generally composed of three terms: Levi Civita connection +
nonmetricity + torsion.Yes, OK.I use metricity in Hagen
Kleinerts senseguv^;v = 0i.e. Diff(4) divergence vanishes.I
use the term metricity in the same way as you: if the
covariant divergence of a connection vanishes, it is metric
or, when there is no torsion, a Levi Civita connection.In
general, for every Riemmanian manifold without torsion, Levi
Civita connecion {} and and an affine connecion A, there is 
one
and only one (1,2) tensor of nonmetricity S, that the the 
affine
connection is the sum of the Levi Civita and the nonmetricity
tensor (A = {} + S) and the curvature tensor of the affine
connection R(A) splits into a sum of a Riemmanian curvature
tensor on {} and a nonmetric curvature tensor of the same
structure wherein the affine connection A is replaced by the
nonmetricity tensor S. The Einstein tensor G likewise splits
into affine G(A) and nonmetric G(S) parts and can be rewritten
as G = G(A) - G(S). Thus the Einstein field equations take 
form
ofG(S) = -(8piG/c^4)T + G(A)In my view, the tensor of
nonmetricity describes gravity, while the affine connection
describes a chosen frame of reference. The second member on
the right side of the equation describes the contribution of
the inertial forces to the energy-momentum. In an inertial
frame it, of course, vanishes.These field equations are
completely covariant with a covariant energy-momentum (instead
of pseudotensor) tensor t.Alex.OK, I need to study what you say
above. :-)With regard to odd claims from Akimovs group in
Russia today about torsion fields based on Gennady 
Shipovs
theories:Again the physics issue here is whether or not:1. Do
torsion fields exist in Nature?2. Is the coupling of torsion
fields to electron and proton spins is 
sufficiently strong to
have a largebio-weapons effect on living matter as claimed by
the Akimov School in Russia today?3. Einsteins gravity 
field
guv(x) of curved spacetime is the local compensating gauge
force field from the 4-parameter translation group generated 
by
total energy-momentum. Obviously curved space-time breaks
space-time translational symmetry upon which the rigid Fourier
transforms depend (as distinct from wavelet transforms).
Curvature is the stringy topological disclination defect
density in the large scale  Lp^2 = hG(Newton)/c^3 ~ 10^-66
cm^2 (Hagen Kleinert).guv = Flat Minkowksi + huvThe degrees of
freedom of the local gauge force compensating huv restore local
conservation of stress-energy density current in the sense
thatTuv(matter/radiation)^;v = 0only when the two Bianchi
identities manage to hold so thatGuv(space-time geometry)^;v =
0;v is the Diff(4) covariant partial derivative using the
symmetric Levi-Civita connection for parallel transport of
tensor fields along vector fields in the curved 
base space-time
of the tangent bundle of Einsteins 
GR.Einsteins local
geometrodynamic field equation with zero torsion and no zero
point energy density exotic vacuum dark energy/matter term
isGuv(geometry) = -(string
tension)^-1Tuv(matter/radiation)trigger detectors to click.4.
Adding the exotic vacuum term givesGuv(geometry) + /zpfguv =
-(string tension)^-1Tuv(matter/radiation)Or, you can write
this in terms of the perfectly balanced local stress-energy
density tensorstuv(geometry) + tuv(exotic vacuum) +
Tuv(matter/radiation) = 0The Bianchi identities break down
which means direct cross-transfer of stress-energy density
currents between geometry, zero point vacuum energy and
ordinary matter-energy such that the total currents are
conserved.This is essential for the metric engineering of Star
Gates and Weightless Warp Drives as well as the Doomsday WMD of
Chapter 9 of Sir Martin Reess Our Final Hour ripping space 
in
a kind of Ice Nine effect.5. Gravity is emergent (Andrei
Sakharov) out of micro-quantum vacuum instability and is
non-perturbative like the BCS superconducting ground state
which cannot be derived in a finite analytic perturbation
series from the normal metal ground state.6. The torsion field
of skewed or twisted space-time is from locally gauging the
6-parameter Lorentz group O(1,3). This local gauging should
not be confused with the freedom to make local Lorentz
transformations of the tetrads in the tangent space. That
freedom is already in torsion-free Einstein 1915
geometrodynamics with a symmetric connection. The local
gauging in the sense here gives a compensating gauge torsion
force field that manifests as an additional Diff(4)
antisymmetric (3rd rank) tensor term added to the
non-tensorial symmetric Levi-Civita-Christoffel connection.
This torsion field corresponds to stringy topological
dislocation defect densities in the large scale limit (Hagen
Kleinert).7. The string is because gravity and torsion emerge
from a MACRO-QUANTUM Vacuum Coherence Field with O(2) symmetry
at least in the low-energy effective emergent More is different
(P.W. Anderson) ODLRO (Oliver Penrose) c-number super§uid local
field theory sense.8. One need not stop there. Look at the De
Sitter Group for space-times of constant curvature
corresponding to Einsteins cosmological constant There is 
it
appears from others work, a duality between zero and 
infinite
cosmological constant that brings in the 4 parameter special
conformal transformation of uniformly proper acclerating
hyperbolic motion of special relativity that is a subgroup of
the 15 parameter Penrose massless twistor conformal group as
the infinite cosmological constant limit of the De Sitter
Group.Locally gauge the De Sitter Group to get gravity +
torsion + ??? (new stuff like variable /zpf of exotic vacuum
unified dark energy/matter which has a Bondi-Terletskii vacuum
propeller weightless warp drive possibility built in.I rewrite
Einsteins zero torsion 1915 geometrodynamic classicallocal
field equatonGuv = -(8piG/c^4)TuvasGuv = 
-alphaTuvalpha =
(string tension)^-1 = Witten parameterInfinite string tension
means no gravity because space-time geometry istoo stiff to
bend.The local stress-energy density tensor of pure geometry
is thentriviallyTuv(Geometry) = 
(alpha)^-1GuvEinsteins field
equation is then simply the balanceTuv(Geometry) +
Tuv(Ordinary Mass-Energy) = 0Adding random micro-quantum zero
point energy density from all quantumfields of spin 1/2
lepto-quarks and spin 1 gauge force bosons givesadditional
termtuv(zpf) = (alpha)-1/zpfguv/zpf > 0 is exotic vacuum 
dark
energy with w = -1 negative pressure./zpf < 0 is exotic vacuum
dark matter with w = -1 positive pressureDark matter detectors
will never click except by false positives inmy
theory.Einsteins equation is thenTuv(Geometry) + 
Tuv(Ordinary
Mass-Energy) + tuv(zpf) = 0In the 1915 theory with /zpf =
0Tuv(Geometry)^;v = 0from the Bianchi identities.However these
identities FAIL IMHO when /zpf =/= 0 and is variableand if
there are torsion fields.In my theory (with 
Wittens h = c =1
convention)/zpf = 
(alpha)^-1[(alpha)^3/2[|MACRO-QUANTUM
VACUUM COHERENCE|^2 - 1]guv = Minkowski metric + Kleinert
World Crystal Lattice Strain TensorMake the Levi-Civita
connection from guv in the usual way.World Crystal Lattice
Distortion Field = du(x) = alpha(Goldstone Phaseof
MACRO-QUANTUM VACUUM COHERENCE),uStrain Tensor = du(x),v +
dv(x),uDiff(4) Landau-Ginzburg eq for VACUUM COHERENCE in a
two-way feedbackloop between IT World Crystal Lattice
Distortion Field andBIT VACUUM COHERENCE in sense of Bohms
interpretation of IT(hiddenvariable) + BIT(Pilot Wave of
Active Information)Torsion fields meandu(x),v - dv(x),u =/=
Arp has DATA. Check for yourself.> James HarrisBut JSH does
data distribution should follow the shape> of Laplacian
distribution... the data obtained from measurement is of
course> a little off(not very symmtrical), how can I make the
measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?Can anybody give me an example
or detailed explanation? I am kind of afraid> of statistics...
:=)Why do you think your data is Laplacian? If its not
symmetric,perhaps thats telling you that you 
dont understand
what the datareally should be?How was the data
generated?Ciao,Peter Apologies for answering a question with
questions! K.-- Peter J. KootsookosI will ignore all ideas for
new works [..], the invention of which has reached its limits
and for whose improvement I see no further hope.- Julius
question:> Suppose:> K2 = a + b + c> K1 = a b + b c + c a>
K0 = a b c> I want to acheive 2a - b - c, 2b - a - c, 2c - a
- b using any> combinations of K2, K1, K0 using any operations
(+, -, *, , roots, logs,> etc.) under real numbers.> 1) How
do I know if such combinations of K2, K1, K0 will get me to 2a
-> b -> c?> 2) If such a combination exists, how would I 
figure
it out?Your K0,K1 and K2 are symmetric functions of a,b,and c,
i.e., each of them is invariant under any parmutation of
(a,b,c).IICR, any formula generated from them must also be
symmetric in a,b and c in the same sense.Since2a - b - c, 2b -
a - c, and 2c - a - b are not symmetric functions of a,b and c,
I believe you are SOL.The best you can do is probably something
like 2a - b - c = 3a - K2, 2b - a - c = 3b - K2, 2c - a - b =
Reality> Equivalently, M*N is the same as M*N mod (M +
N - 1). Sorry, this should be multiplication of M digits
with N digits, base b,> is equivalent to multiplication modulo
b^(M + N - 1), i.e. M+N-1 digits.>Oh well, so FFT or not,
looks like multiplying M by N by any method meansMN
multiplications! Well, when these multiplications are
hardwired (as inhuman memory for single digits) the
computational issues (On*n) becomesreally irrelevant, for
they all are done in no time at. Like, the videoextraction
for radar data processing is done by NAND gates - its all done
inreal time!> You are in error. The number of
multiplications required for multiplying two numbers with
the FFT method is O(n*log(n) where n is the larger of the
two numbers; it is not m*n.>Fine, just multiply 12345 by
67809 using FFT with less than 25>multiplications. Do it
here.Seemingly you do not understand what the O() notation
signifies. When> one says that the FFT method is O(n*log(n))
one is saying that there> is some constant C such that for n
sufficiently large,(# of required multiplies) is less than
C*n*log(n)This does not mean that the cost of the FFT method
is less than n*n> for all n, just that it is for n 
sufficiently
large. Thus, your> proposed test is irrelevant to the point
under discussion. That said, the simple two point formula runs
as follows: 12*67 = 804 (4 one digit multiplies)> 345*809 =
279105 (9 one digit multiplies)> (12+345)*(67+809) => 357*876
= 312732 (9 one digit multiplies)Term 0 = 279105> Term 1 =
312732 - 279105 - 804 = 32823 > Term 2 = 80412345*67809 =
804000000 + 32823000 + 279105> = 8371021054 + 9 + 9 = 22
multiplies < 25Wrong! You have multiplied 804 with 1000000,
which is 10multiplications. Then again you have multiplied
32823 with 1000 whichis again 9 multiplications. So totally
you have done 22+19 or 41multiplications. Which is more than
25! I know that the last 18 areeasily done, but they are
multiplications all the same. Even if wecount the multiplies
with 10^x as 1 operation, then we still come downto 25
operations, just as O(n^2).And all these are pretty complex
operations, always needing the carryfor the internal
have shown that this is simpler than the Vedic method. In fact,
it is dreadful. I am sure that in due course people
willmultiply using the Vedic arithmetic. So much easier.
Properly done(with hard-wired single digit computation, that
will makemultiplication efforts meaningless) I have no doubt
that it will beatFFT hands down. In fact, FFT as you show it
is a clumsy approximationto the elegant Vedic multiplication.>
Be that as it may, the cross product method for multiplication
is> quite obvious and is regularly rediscovered. I discovered
it myself> as a child and even then was under no illusion that
I had done> anything remarkable.But no one demonstrated that
here before I did, with the example.People came up with all
sorts of ideas, but no one could do it. And Idid give them the
chance! I wanted others to show it, and none did. To say now
that you knew it, does not convince. Did you also knowabout
the one-line division method? Supposing someone were to
explainthat with an example (I may do that after I get that
book within anyear or so) would you then say that you knew
that method all along?Arindam Banerjee.> Richard Harter,
cri@tiac.net> http://home.tiac.net/~cri,
http://www.varinoma.com> We have people from every planet on
the earth in this State.> -- California Governor Gray
identity in a paper by theChudnovsky brothers: infinity -----
(n + 1) 2 pi + 2 = ) ---------------- / binomial(2 n, n) -----
n = 1My question is: is there any elementary wayto prove
(Q) How many possible combinationsHi StevenNo I am not
familiar with combinatorial and permutations? What does the
function C() do ?can you point to a site that perhaps has a
http://web.hamline.edu/~lcopes/SciMathMN/concepts/
cperm.htmlbut got lost on how it relates to a problem with
only two possible values all the examples calculated
permutations on multiple unique values such as 1,2,3,4 when
considering only two possible values the number of
combinations is substantially less do I have to iterate over
combinatorial and permutations? If so, your> answer is, if
z=zeros and o=ones, (z+o)C(o) or (z+o)C(z). They are> equal.
 I have a problem I would like to solve but the answer
son keeps coming home with these papers called > sum-part-part
product. The sum is filled in as is the product > and he must
fill in the part boxes. I cant, for the life of 
me, > figure
out a mathmatical way to calculate the parts....there must >
be a way. Can anyone help?This is the solution another poster
referred to. It requires high-school algebra.Let x and y be
the unknown parts, and s and p be the given sum and
product.Then x+y=s and xy=p. These two equations can be solved
for x and y as follows:y=p/x, so x+p/x=s, or x^2-sx+p=0. This
is a quadratic, which can be solved using the quadratic
formula, asx=(s+-sqrt{s^2-4p})/2 (+- means there are two
solutions, one with each sign)For example, you had s=14 and
p=49. s^2-4p=14*14-4*49=0, so x=s/2=7 (for both parts). A more
complicated example (not from your sheet), would be s=14 and
p=33. Then s^2-4p=196-132=64, and x=(14+8)/2=11 or
x=(14-8)/2=3, so the parts are 11 and 3.E-mail me if this is
cycles/transpositionsYou know, whenever one posts source
code in a NG that is notcomp.lang.thatlanguage or a
subhierarchy of it, it would be a VeryNice Thing(TM) to
specify in which language it is written!Its Python. You can
tell because it looks like pseudocode>but its not. I knew
its Python. I can tell because of the snippets of code
Iveseen before, and in particular because of the 
ïdefs
(being used toPerl and having been used to C/C++ a long time
ago, these somehowgrasped my attention).>You know, nobody ever
complains about not recognizing C.Hmmm... I admit its true!
But even in that case it would be a FineThing(TM) -a companion
product to the above mentioned one- to write atleast in C:n.
Its not expensive: not more than 6 bytes. And itwill make 
you
seem human!>Things are gonna be different after the
revolution...Uh-Oh...-- > Comments should say _why_ something
is being done.Oh? My comments always say what _really_ should
have happened. :)- Tore Aursand on
circles - divide by zero??> d is the distance between the
centerpoints (youll notice in my test> data, all the 
circles
are on the same y-axis, to simplfy things). If the circles do
not intercept, it doesnt perform this calculation> 
(theres a
conditional that looks at d and the radii of the two>
circles)What conditional are you using? Although, I managed to
figure out that its not a divide by zero> 
error...its a>
square root of a negative error.Good. That makes more sense to
me. To clarify: Center1 and Center2 contains (x, y) coordinates
of the centerpoints of> the respective circles. R1 and R2 are
the radii of the respective circles. d is the distance between
the centerpoints of the two circles> d = ABS(Center1.x -
Center2.x) ïexpression to find the XintersectA 
value> 
 
XintersectA = ((center2.X + Center1.X) / 2) + ((center2.X ->
Center1.X) * ((R1 ^ 2) - (R2 ^ 2))) / (2 * (d ^ 2)) +
(((center2.Y -> enter1.Y) / (2 * (d ^ 2))) * Sqrt(((((R1 + R2)
^ 2) - (d ^ 2)) * ((d> ^> 2) - ((R2 - R1) ^ 2))))) BREAKDOWN:>
a = (Center2.X + Center1.X) / 2> b = (Center2.X - Center1.X)>
c = ((R1 ^ 2) - (R2 ^ 2)) Numerator = a + (b * c) e = (2 * (d
^ 2))> f = (center2.Y - enter1.Y)> g = (2 * (d ^ 2))> h = ((R1
+ R2) ^ 2) - (d ^ 2))> j = (d ^ 2) - ((R2 - R1) ^ 2)) k = ((e +
f) / g)> l = Sqrt(h * j) ****This is where the problem
occursGood. Now the problem is that h*j is negative, right? So
start by figuringout, which of the two factors is the
negative.Looking once again at your test data, particularly
the line 50, (150, 150),10, (180, 150)** I notice that these
two circles do not intercept. Rather,the second is completely
contained within the first. This corresponds to anegative 
value
of j. The next test case 50, (150, 150), 20, (180,
150)**corresponds to two circles just touching each other at
the point (200,150).In this case, your value of j should be
zero. Perhaps you should considerdouble-checking your
expression for j. I mean, what values of j does yourprogram
Foundation (absymally stupid question)>Since were 
discussing
the axiom of foundation (in the textbooks Ive seen, 
>its
called the axiom of regularity), does anyone know what the
intuitive >justification for this axiom is?The intended
interpretation is known as the cumulative hierarchy.It has
ranks in an ordering, and the sets of a given rank are the
setsS whose elements all lie on lower ranks, except the S that
havealready appeared at some lower rank already. If the ranks
arewell-ordered (in particular, there is no infinite 
descending
sequenceof ranks) this gives a (transfinite) inductive 
definition
of the sets ofeach rank. The cumulative hierarchy consists of
all the sets definedthis way, with ranks allowed to be all
ordinals.So for instance, the only set of rank 0 is the empty
set, {}, sincethere are no sets of lower rank. The only set of
rank 1 is {{}}. Thesets of rank 2 are {{{}}} and {{},{{}}}.
Things get more interestingat rank omega, the first 
infinite
ordinal. There are finitely manysets at each 
finite rank, and
there are countably many of themaltogether, but at rank omega
all the sets of sets of finite rank(which 
havent already
appeared) appear-- a continuum of them.Well-foundedness
permits what is known as epsilon induction.If a property of
sets in the hierarchy holds for each set when itholds for all
the members of the set, then it holds for every setin the
hierarchy.If you do a web search for foundation axiom, a lot
of the hitsare to the anti-foundation axiom. The
antifoundation axiom saysroughly that every way that the
foundation axiom can fail it does.ZF with the antifoundation
axiom instead of the foundation axiomis in a sense the same
strength as ZF, so if for some reason onewants to deal in
non-well-founded sets, theres no real problemin doing
Rationals|How small can d(n) be? Clearly, d(n) < 1/n. But can
we make d(n) much|smaller than that?|| Q1: Can we find
arbitrarily large values of n such that d(n) < 1/n^2?Whenever
a number is rational, it can be approximated this well.Let 0,
x, 2x, 3x, ..., nx be the multiples of an irrational x,
andconsider their fractional parts 0, x-[x],
2x-[2x],...,nx-[nx]. Sincethere are n+1 of them, there are two
that are within 1/n ofeach other. If |(rx-[rx])-(sx-[sx])|<1/n,
then (r-s)x comes within1/n of an integer, and |r-s|<n.The
golden mean is one of the worst-approximable numbers.| Q2: Can
we find arbitrarily large values of n such that d(n) <
1/n^3?Probably not. What is expected to be true is that for
everyepsilon>0 and C>0, there are finitely many n such 
thatd(n)
< C/n^(2+epsilon). Apparently its simply very hard toprove
anything as strong as that. The reals that can beapproximated
to an exponent greater than 2 have measurezero, by the way, so
===
Re: (Q) How many possible combinations>Subject: Re: (Q) How
many possible combinations>Message-id:
<ec1a1c31fee14c2e20214e3bc8fe3f72@news.teranews.com>Hi
StevenNo I am not familiar with combinatorial and
permutations? What does the function C() do ?C(m,n): from m
items, choose n at a timeIf we have 4 binary bits and 1=chosen
0=not chosen, then the counts are0000 zero at a time (1)0001
one at a time (4)0010010010000011 two at a time
(6)010101101001101011000111 three at a time
(4)1011110111101111 four at a time (1)The counts add up to
2^m: 1+4+6+4+1 = 16You can get them from Pascals Triangle11
11 2 11 3 3 11 4 6 4 1where m is the row number (starting from
0) and n is the column number (alsofrom 0).Or you can compute
C(m,n) directly byC(m.n) = m!/((n!)*(m-n)!)How many 8-bit
binary numbers have exactly 4 ones?C(8,4) = 8!/(4!)*(4!) =
8*7*6*5*4*3*2 / 4*3*2 * 4*3*2 = 8*7*6*5 / 4*3*2 = 7*6*5 / 3 =
7*2*5 = 70 can you point to a site that perhaps has a good
descriptionNo, but heres a page with a couple
examples:http://members.aol.com/mensanator666/fun/playing.htmI
>http://web.hamline.edu/~lcopes/SciMathMN/concepts/
cperm.htmlbut got lost on how it relates to a problem with
only two >possible values all the examples calculated
permutations on multiple unique >values such as 1,2,3,4 when
considering only two possible values the number of
>combinations is substantially less do I have to iterate over
all posible values?Not if you only need the counts. To answer
my Cheese Puzzle, I just needed thecount, but to make the
animation that steps through all 70 solutions, I neededto
with combinatorial and permutations? If so, your answer is,
if z=zeros and o=ones, (z+o)C(o) or (z+o)C(z). They are
equal.  Hi I have a problem I would like to solve but
Topologist?E L E M E N T SBOOK 1Definition 1: a 
pointermorphism
is that which has no partmorphismDefinition 2: a lineamorphism
is breadthmorphismless lengthmorphismDefinition 3: the
endmorphisms of a lineamorphism arepointermorphismsDefinition
4: a straightmorphic lineamorphism is a lineamorphismwhich
lies evenly with the pointermorphisms on itself(SNIP more
definitions)Postulate 1: it is possible to draw a
straightmorphic lineamorphismfrom any pointermorphism to any
pointermorphismPostulate 2: it is possible to produce a
finitemorphicstraightmorphic lineamorphism
continuousmorphically in astraightmorphic
lineamorphismPostulate 3: it is possible to describe a
circlemorphism with anycentermorphism and
radiusmorphismPostulate 4: That all right anglemorphisms are
isomorphic toeachother(SNIP to Common Notion 2)Common Notion
2: If equalmorphisms are added to equalmorphisms, thenthe
wholemorphisms are isomorphic to eachother(SNIP to Proposition
1)Proposition 1: it is possible to construct an
equilateralmorphictrianglemorphism on a given finitemorphic
straightmorphiclineamorphismProof: Let AB be the given
finitemorphic straightmorphiclineamorphism.It is required to
construct an equilateralmorphic trianglemorphism onthe
straightmorphic lineamorphism AB.Describe the circlemorphism
BCD with centermorphism A andradiusmorphism AB. Again describe
the circlemorphism ACE withcentermorphism B and radiusmorphism
BA. Join the straightmorphiclineamorphisms CA and CB from the
pointermorphism C at which thecirclemorphisms cut one another
to the pointermorphisms A and B. Now,since the pointermorphism
A is the centermorphism of thecirclemorphism CDB, therefor AC
is isomorphic to AB. Again, since thepointermorphism B is the
centermorphism of the circlemorphism CAE,therefore BC is
isomorphic to BA. But AC was proved isomorphic to AB,therefore
each of the straightmorphic lineamorphisms AC and BC
isisomorphic to AB. And thingmorphisms which are isomorphic to
the samethingmorphism are also isomorphic to one another,
therefore AC also isisomorphic to BC. Therefore the three
straightmorphic lineamorphismsAC, AB, and BC are isomorphic to
one another. Therefore thetrianglemorphism ABC is
equilateralmorphic, and it has beenconstructed on the given
finitemorphic straightmorphic lineamorphismAB.
Q.E.F.-morphismYour friendmorphism,Nathaniel DeethAge
risk of causing more confusion, let me try to explain this
from adifferent angle. You might argue that Im trying to 
get
around what youretrying to prove, even though 
Im not. But
what the hell. Youre notlistening very much anyway. But 
just
try, ok? Ok, here goes.To start with, if you havent 
actually
read Isaac Newtons Principia, Istrongly recommend it. It
really is a terriffic book. You should understandthat it
describes itself as a work of philosophy, which at that time
waswhat physics was considered to be. Most importantly, it
starts off bydefining three axioms, which you should know 
well.
They are described asNewtons three laws. They are described
right at the start of the book, inthe first chapter of book I,
right after the definitions section, whichdefines 
some necessary
terms such as momentum, mass and so on.This is the important
part: all the rest of the book, which defines what wecall
Newtonian mechanics, is based entirely on these three axioms.
Thisintroduces two important but quite separate questions for
the reader:1. Is the model of mechanics so described
internally consistent? That is tosay, given the three posited
axioms, does everything else follow?2. Are the three axioms an
accurate re§ection of the real world?These are quite separate
questions. It would be quite possible to create,for the
purposes of philosophical study, an imaginary universe based
on adifferent set of axioms. The model of that imaginary
universe that wouldfollow could be entirely self consistent
and valid as a model of a form ofmechanics, and yet bear no
relation whatever to the universe we know.However, if you
accept that the model Newton derives from his posited axiomsis
self consistent, it only remains to determine if the three
axioms heposits actually re§ect our universe. If they do,
everything else follows.If they do not, Principia is just a
philosophical experiment.Here is the crux of your problem: you
are using Newtonian mechanics to tryto disprove one of the
axioms, namely the third law. Should you besuccessful, you
will prove only that Newtons model is wrong, because
thatcannot be done if his model is internally consistent.
However, you willemphatically not have proven that your device
works. If your device works,Newtonian mechanics is necessarily
bunk, in the sense that it would notdescribe the physical
universe. Therefore you cannot use Newtonian mechanicsto prove
that it works in the real world.Newtonian mechanics is used as
a model for the physical universe because twothings are
generally accepted to be true* for entirely different
reasons:1. The axioms are accepted to be an accurate re§ection
of the physicaluniverse by empirical means.2. Everything that
follows from those axioms is accepted to be correctlyinducted
mathematically.The conclusion is that the only way you can
prove that your device works isempirically, that is, build it
and show it works. If it does, all ofNewtonian mechanics is
necessarily wrong and must be rewritten from scratch.But this
is the only method you can use. It makes no sense to attempt
todisprove axioms by mathemetical (for which read theoretical)
means,because they are accepted as a good model empirically,
not mathematically.I hope this clears things up. But frankly,
Im not hopeful.Krill*ok, so Im leaving out 
very large scales
and very small scales, where thisis no longer true. But its 
no
longer true because the axioms dont 
fitobservation, and for no
other reason> Just for record. I went to Indian Institute of
Technology (IIT), Powai, Mumbai to> explain mechanism of my
Action Device and to seek technical help. I> met Dr. Amitay
Issac of Aerospace Engineering Department and I tried> to
explain very basic component/idea of this action device. I
have> given in my homepage what exactly I tried to convince
him. http://www.geocities.com/actiondevice But he insisted
that point B will shift its position along Y axis!.> I had to
return in few minutes. Now I tried to convince again to Dr. G
Arvind Rao of Aerospace> Engineering Department by email, but
he also said that point B will> shift its position along Y
axis !. Indian Institute of Technology is most prestigious
college in India.> This institute gives people for Aviation
Industry around the world.> And I just wonder, why so highly
educated people fail to understand> such simple problem. In
fact, this is not problem at all. But what a tragedy, I am
facing> such ridiculous problems. I can end my all problems
anytime, but I am following the rules of> this battle, waiting
game. I am just watching how the minds of highly educated
people around the> world are controlled by that Supreme Force
mathematics.> Isnt arithmetic a part of mathematics? Hardly, no (strange as 
that may sound).> It does sound
strange. And also unbelievable. Especially since> Arithmetic
is taught as a part of Mathematics in school.> The only
mathematics in arithmetic is when you ask yourself things>
like why does this method work, is there a quicker one, etc.
Fair enough, so you do agree that arithmetic is a part of
mathematics.The _calculus part_ of elementary
school-arithmetic can_hardly_ be called mathematics, thats
all im saying.> Applying the methods is then no longer 
maths,
it is arithmetic. One may think of arithmetic as the
computational arm of mathematics.> Without arithmetic,
mathematics becomes extremely wooly, and> restricted to a
closed few. Like, very few people can understand a>
mathematical theorem, just by itself. When there is a
working-out of> the theorem using arithmetic, people
understand the theorem a lot> better. To divorce arithmetic
from mathematics is to rarefy the scope> of mathematics, and
while it may put mathematicians on a high pedestal> (like
Einsteinian physicists) there is the risk that the public
will> be alienated. So such elevation may well be
temporary.Thats an important point, yes. But restricting to
numbers onlydoesnt make maths very exciting for laymen, 
im
afraid.Personally, i would never have gotten any interest in
mathematicsif not for geometry, reasoning from axioms, etc.
Historically,mathematical progress always ended fairly soon,
in all culturesthat restricted maths to ïdoing 
calculations.>
And of course, the anglosaxon world uses 
ïmathematics as a>
synonym of arithmetics, adding to the confusion. No, it is not
a synonym. It is a part.Yes. But i meant something different,
namely the everyday lifeuse of the word, not the order of
education.To add: the word mathematics is used with two
meanings:one denotes the academic discipline, the other a
synonym ofarithmetic. (Anyway thats the way it seems to 
me.) That may also explain> why the math teaching world is
not very fond of tricks like this.> It is not a trick,
it is a sound method for multiplication.> Yes. But ive
seen a site about Vedic maths showing some other things.> We
are talking about multiplication here, and nothing else.> But
it is used as pr for ïthe rest of Vedic 
maths... Is it? Have
you read the book in question? Is it marketed in Western>
world? Exactly who is doing the PR?Years ago i was present in
a lecture about Vedic maths, already.Someone visited
universities with it. Recently, ive seena website with the
same story. The impression they leave with meis that they are
plain PR, so i take the liberty to call it PR.(Your own posts
are also PR for vedic maths. BTW, I dontconsider PR to be a
dirty word.)> What is vital in math education is to give the
children a basis,> something to fallback on, when it gets less
directly intuitive. The Vedic multiplication method gives a
great deal of insight into the> relevance of place value.I
wasnt talking here about the multiplication method,but 
about
some of the other tricks, as far as ive seen them.> The
current method is certainly very clumsy> in contrast. It has
been agreed that my definition for that method is> the same as
for the advanced method of convolution, used for> multiplying
very large numbers. If primary schoolchildren use such>
advanced tricks at their very tender age, surely their
capabilities> will develop fast.The alternative multiplication
will not help mathematics a single bit. No, I do not agree.
Children will be freed from the torture of> learning
arithmetic the modern bad way, and they should have a more>
positive learning attitude towards mathematics as a result of>
incorporating Vedic arithmetic in primary schools.Ah... now we
suddenly are talking about more than thealternative
multiplication method?As said, the multiplication method is
welcome. But whative seen of the rest makes me hesitate a
closed map?> Hello group,I am trying to decide wether the
canonical projectionp_i: X -> X_i, X = X_1x...xX_n,which is an
open map, is also closed or not. I already found out that> the
closed subsets of the product space X are infinite
intersections> of sets like X-U_l, whereU_l = U^l_1 x ... x
U^l_n with all U^l_i open in X_i.But the problem ist the
intersection of all of these subsets:p_i( intersection(l
element L)(X-U_l)) subset intersection(l> element
L)p_i(X-U_l)Between the left and right side of the above
statement there not> always exists equality. The right side
seems to be a closed subset,> as p_i(X_U_l) = X_i, but this
doesnt help me.Can anyone give me a hint?Rene.> Think of a
simple case. RxR ---> R, say via projection onto the
firstfactor. It shouldnt be hard to come up 
with an example
that willconvince you of the answer to your question. BTW, the
answer is trivialif the factors are compact (theres your
SO(n+1) in any way?Yes. SO(n) is isomorphic to a subgroup of
SO(n+1).> That is, can I mod out SO(n+1) by a subgroup to get
functions)Hi all,Let I be an open interval of the reals that
contains a pointp and let f be a differentiable function from
I into the realssuch that f is bounded at some neighborhood
of p. Therefore,limsup f(x) at p and liminf f(x) at p are both
real numbers.Question: must f(p) be equal to their
Approximating Pi by Rationals >I feel that this is probably a
well-known subject for number theorists,but Ive never 
read
anything about it. The question is how closely canwe
approximate pi by rationals. More specifically:For integer
n>0, let f(n) be the largest integer m such m/n < pi.Let
d(n) = pi - f(n)/n.Then d(n) measures how accurately we can
approximate pi by a rationalwith denominator n.How small
can d(n) be? Clearly, d(n) < 1/n. But can we make d(n)
muchsmaller than that? Q1: Can we find arbitrarily large
values of n such that d(n) < 1/n^2? Q2: Can we find
arbitrarily large values of n such that d(n) < 1/n^3? Q3: In
general, for each p>1, can we find arbitrarily large values 
of
n such that d(n) < 1/n^p?>NO! Surprisingly, K. Mahler
showed that d(n) > 1/n^(42). That>exponent 42 has been
improved subsequently. I guess the best current>result is
8.0161, due to M. Hata.>For positive x, define g(x) = inf {c: 
n
x - §oor(n x) > n^-c for all positive integers n}. True or
More SO(n)> Is SO(n) related to SO(n+1) in any way?That is,
can I mod out SO(n+1) by a subgroup to get
observed asthose rotations of R^(n+1) that fix the n+1st
axis{(x1, ..., xn) | x1=x2=...=xn = 0}.SO(n) is not generally
a quotient of SO(n+1). While itis true that SO(4) ~ SO(3) x
S^3 (homeomorphic), Idont think the group structures 
(direct
productstructure on the right).The natural inclusion of SO(n)
into SO(n+1)is intimatelyrelated to the sphere S^n; this fact
is helpful inunderstanding the topology of the orthogonal
groups. Formore information, read the first few chapters of 
any
(differentiable functions)>Hi all,Let I be an open interval of
the reals that contains a point>p and let f be a differentiable
function from I into the reals>such that f is bounded at 
some
neighborhood of p. Therefore,>limsup f(x) at p and liminf f(x)
at p are both real numbers.>Question: must f(p) be equal to
their average?>Huh? Consider a constant function.--
Children will be freed from the torture of> learning
arithmetic the modern bad way, and they should have a more>
positive learning attitude towards mathematics as a result of>
incorporating Vedic arithmetic in primary schools.Lets not
forget that the Vedic principle crosswise and verticaldoes
also lie at the heart of this modern, bad, clumsy
way...Because the Vedic principle does not talk about the
Analysis question (correct question this time)Hi all,Heres 
my
question again; there was a stupid error in thequestion posted
a few minutes ago.Let I be an open interval of the reals that
contains a pointp and let f be a differentiable function from
I into the realssuch that f is bounded at some neighborhood
of p. Therefore,limsup f(x) at p and liminf 
f(x) at p are
both real numbers.Question: must f(p) be equal to their
Canonical projection a closed map?> Think of a simple case.
RxR ---> R, say via projection onto the first> factor. It
shouldnt be hard to come up with an example that will>
convince you of the answer to your question. BTW, the answer
is trivial> if the factors are compact (theres your
in (-Pi/2,Pi/2). Underprojection it becomes the open interval
(-Pi/2,Pi/2). Is that a validexample? I think so.Rene.--
Ren.8e MeyerStudent of Physics & MathematicsZhejiang
limits - Need Help!In sci.math, Jonathan
Miller<jonmillere1@comcast.net><krCdnZA88-uNzSeiRVn-uA@
comcast.com>: Note that both are a special case of
L^Hopitals rule (note spelling):What are we supposed to
notice? That you dont have a circum§ex?> That you 
dont know
that the (usual) substitution for a circum§ex> o is os? That
slepping falmes always contain mispleddings?Jon Miller> Search
engines are notorious for missing the point when a rulename is
misspelled. I did not intend insult. :-)Are you suggesting it
should be spelled LH.99pitals Rule? 
Idgo along with that,
from the limited amount of French I know(oui, non, je 
tadore,
je ne parle pas Fran.8dais, etc.)Its been too long. :-)In 
any
event,lim {x->0} f(x) / x = lim {x->0} f(x) / 1by a trivial
application thereof, assuming all expressions make sense.--
#191, ewill3@earthlink.netIts still legal to go
> What role does the Jacobson radical> ab = a+b - ab> That is not the 
Jacobson radical. The Jacobson radical
of a ring is the intersection> of it proper left (or right)
ideals. You need some conditions on the left ideals that
you> are taking the intersection of. Doh! I meant maximal
left ideals ...Just maximal left ideals is not enough. Take
the nilpotent algebra> A = R^2 with multiplication (x,y)(a,b)
= (0,xa). A has only three ideals:> A , {0}xR, and {(0,0)}.
But in this case the Jacobson radical is the> whole
algebra.Thats not a ring: they alway have 1 elements :-)--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to
say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14
doing some light (?) reading on topological groups for kicks.
Unfortunately, my brain is slowing down. A problem:Let G be a
topological group.If K,L subseteq G are closed in G, do we
necessarily have KL closed in G?Note that AB := {ab | ain A,
across a problem in a book to try and determine if there was
a> polynomial p(x) with at least 2 nonzero terms such that
p(x)^2 had> exactly the same number of nonzero terms as
p(x). I proved it> couldnt happen for linear, quadratic, or
cubic polynomials, and I> think I proved it couldnt happen
for quartic polynomials also. Then> I found a quintic where
it is true: Namely, p(x)=4x^5+4x^3-2x^2+2x+1,>
[p(x)]^2=16x^10+32x^9+28x^6+4x^3+x^2.Make that > 16 x^10 + 32
x^8 - 16 x^7 + 32 x^6 - 8 x^5 + 20 x^4 + 4x + 1.Oops, 
mustve
been a mistype. Tryp(x) = 4x^5 + 4x^4 - 2x^3 + 2x^2 + xNote
that since this latest p(x) is a multiple of x, p(x)/x = 4x^4
+ 4x^3 - 2x^2 + 2x + 1 will work,as will (x^n p(x)) for any
positive integer n. Also q(x) = x^4 + 2x^3 -2x^2 + 4x + 4,
reversing the order of coefficients of p(x)/x, will work, as
to SO(n+1) in any way?> That is, can I mod out SO(n+1) by
a canonical inclusion in SO(n+1), observed as> those rotations
of R^(n+1) that fix the n+1st axis> {(x1, ..., xn) |
x1=x2=...=xn = 0}.SO(n) is not generally a quotient of
SO(n+1). While it> is true that SO(4) ~ SO(3) x S^3
(homeomorphic), I> dont think the group structures (direct
product> structure on the right).Via quaternions, SO(3) is
isomorphic to SU(2)/<-I>and SO(4) is isomorphic to (SU(2) x
SU(2))/<-(I,I)>.Its clear then that SO(3) is isomorphic toa
quotient of SO(4). The quotiented subgroup isisomorphic to
SU(2).I dont think this extension 
splits.Whats going on here
is that SO(4) has Lie algebra with rootsystem D_2 (isomorphic
to A_1 x A_1) so locally SO(4)is a direct product.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
... mumble ...> SO(n) is not generally a quotient of SO(n+1).
While it> is true that SO(4) ~ SO(3) x S^3 (homeomorphic), I>
dont think the group structures (direct product> structure 
on
the right). ^^^ are compatible (i.e., the two groups are not
isomorphic). ... the end ...> Dale> I hate it when that
the number of subgroups of the symmetric group S_10 on ten
elements?Can someone provide a reference?Can GAP do
question this time)>Hi all,Heres my question again; there 
was
a stupid error in the>question posted a few minutes ago.Let I
be an open interval of the reals that contains a point>p and
let f be a differentiable function from I into the reals>such
that f is bounded at some neighborhood of p.
Therefore,>limsup f(x) at p and liminf f(x) 
at p are both
real numbers.>Question: must f(p) be equal to their 
average?No.Consider x^2sin(1/x).For this function f(0) is indeed 
the
average of the limsup (1) and liminf (-1) but you can tweakthe
function a bit so that the limsup becomes 2, say, and the
liminf becomes -1.KP-- E-MAIL: K.P.Hart@EWI<dot>TUDelft.NL
PAPER: Faculty EWIPHONE: +31-15-2784572 TU DelftFAX:
+31-15-2786178 Postbus 5031URL: http://aw.twi.tudelft.nl/~hart
Fiber Bundle Physics/Consciousness 2)]<snip B L U R
...http://www.astrocentral.co.uk/beagle.html
Constraining the Topology of the Universe
http://arxiv.org/abs/astro-ph/0310233I catch them on my
lightmach and I,ll throw them in the brig with the>
butting aliens who are always trying to sneak across the
underside> of the bridge.> You think the mexican border is
porous you should see the problem with> lightmach 24 right
above us.> ing humans and their god damm navel gazing
eccentricities.1998/12/11
Machinehead.1812.008C7B1E%40Fact-Factory.Org!i!i!i!i!i!i!i!i!i
!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i!i
!i!i!i!i!i!i!i!i!i!i <snip >Commentary 2[...]>One can
also imagine a fiber of strings of qubits.>1 qubit is a
parallel infinity of c-bits.>i.e.>|qubit> = |1
c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit Where there
is a continuous infinity of different c-bit bases>or
orthonormal frames each corresponding, for example,>the the
angular orientation of an inhomogeneous field>magnet in a
Stern-Gerlach filter for spin qubits>in the DARPA spintronics
project or like the billion billion>Single Electron
Transistors inside the human brain at the>sub-microtubular
protein dimer hydrophobic cage level forming>the hardware
interface with external world whose software is our stream>of
inner consciousness.>[[[[(({<http://www.thoughtpolice.com/images/norton/
nort50c.jpg>}}}]]]]>Each possible orientation is a primitive
parallel quantum universe.>The quantum computer computes in all
possible>orientations simultaneously like a 
continuous>infinity
of classical Turing machines in a>distributed network working
on the same problem> - or so the folklore goes.<snip
[...] http://www.stardrive.org/Sarmail11-6-00.shtml ~^~>By the year 2025, 
Earth could lose as manyas one fifth of
all species known to existtoday. In recent centuries,
hundreds ofspecies have disappeared, almost alwaysas a
result of human
activities.http://www.worldwildlife.org/news/pubs/
specieslist.html ~^~What our Dreams Tell UsDo our
dreams give us messages from our bodies about health
problemswe may not be aware of? The ancient Greeks thought
that dreamscontained information that could be used to
diagnose disease. Withsome diseases, specific dreams are 
more
likely to occur; however,people who have the most severe
cases of these diseases often saythey dont dream at all.>Dr. Trisha
Macnair<http://www.bbc.co.uk/health/profiles/trisha_
macnair.shtml that men with heart problems often dream about
death and dying, whilefemale heart patients dream of
separation. Migraine sufferers havedreams containing extreme
fears (perhaps because they fear the onsetof another
headache). People whose brain scans show signs of dementiaor
brain shrinkage often dream about losing something,
especiallymoney or food (ie. something essential).>Victims of stroke, 
epilepsy or Parkinsons disease have
noted changesin the amount of time they spend dreaming and
in the quality of theirdreams, which have fewer visual
images. Theyre also less able toremember their dreams.
People with high blood pressure have dreamsfilled with
hostility (one of the causes of their problem?)Patients
with narcolepsy, who find it hard to stay awake, dreamabout
strange and frightening events. People under the in§uenceof
alcohol and drugs (including sedatives and
antidepressants)have nightmares when the drugs are stopped.
Asthma patients havevery emotional dreams, perhaps because
not being able to breatheis such an emotional experience.>People with 
psychosomatic illnesses (who tend to think
theyresick when theyre not) have dreams 
filled with
aggression, fearand helplessness, which are probably the
underlying causes ofthis condition.Dreams occur during
REM (Rapid Eye Movement) sleep. This is whenthe brain is
most active and our sleep is the deepest. People whoare
deprived of REM sleep dont feel as if theyve 
slept
enough.It occurs roughly every hour to 1 hours, several
times a night.REM is tied to bodily changes in temperature,
pulse rate, andblood pressure, so the dreams that are
produced can actually setoff heart attacks, migraine and
asthma attacks. In South EastAsia there is a rare disorder
where men die mysteriously in theirsleep, called Pok-Kuri,
which may be caused by abnormal heartrhythms during REM
sleep. ~^~> Stuart Hameroffs Home Page:>
http://www.consciousness.arizona.edu/hameroff/index.html>
The Elegant Universe homepage>
http://www.pbs.org/wgbh/nova/elegant/> _________> The
Religious Experience of ip K. Dick by R. Crumb>
http://www.philipkdick.com/weirdo.htm> THE POWER of NOW *>
by Eckhart Tolle> http://www.eckharttolle.com/> ... It is
finding your true nature beyond> name and form. The inability
to feel this connectedness> gives rise to the illusion of
separation, from yourself> and from the world around you. You
then perceive yourself,> consciously or unconsciously, as an
isolated fragment.> Fear arises, and con§ict within and
without becomes> the norm. ...> -- Eckhart Tolle
http://www.eckharttolle.com/> http://adidam.org/> Yes! There
is no religion, no Way of God, no Way of Divine> Realization,
no Way of Enlightenment, and no Way of> Liberation that is
Higher or Greater than Truth Itself.> .... Therefore, Reality
(Itself) Is Truth,> and Reality (Itself) Is the Only Truth.>
-- Adi Da Samraj, a.k.a. Bubba Free John,> a.k.a. Da Free
John, a.k.a. Da Kalki,> a.k.a. the Ruchira Avatar, Adi Da
Love-Ananda Samraj,> a.k.a. Franklin Jones>
http://adidam.org/> http://www.daplastique.com/home.html>
Let me share with you this little model Ive worked out> 
about
who we are as human beings. I call it the> Three-Plane
Consciousness Model. If I were to take a> picture of who I see
you to be, the picture would show> three Is ---three 
different
levels of who you are,> planes on which you have an identity.>
Number One is what I call ego, thats the I we all> know 
very
well, the plane of the body, mind, and> personality; of all
those things we think we are.> Number Two I call the soul; the
soul measures time not> in days and years but in incarnations,
and its the I> that was around before we as egos were born
and that> will be around after we as egos die.> And Number
Three is ... just Number Three. We all have> different names
for it, and wars are fought over what> to call it, so I avoid
all that by just calling it> Number Three.> I see our task as
learning to live on more than one of> those planes
simultaneously, experiencing ourselves as> egos and souls at
the same time. And since you gotta> be one to see one, once we
are resting in our souls,> then we will see others as souls as
well. Then when we> look into another persons 
well say, Are
you in there?> Im in here. Far out!> When we are able to 
look
behind even that identity as> soul, well see that we have
still another identity> because we are also Number Three.>
Thats the mystic I, because in Number Three 
theres> actually
only one of us. Your Number Three isnt merely> like my 
Number
Three---Theyre the same thing.> -- Baba Ram Das, a.k.a.
Richard Alpert> http://ramdasstapes.org/index.htm> _________ Committee for 
Surrealist Investigation of Claims of the
Normal> [ CSICON ] http://www.rawilson.com/csicon.shtml> < C
O N T A C T > <http://neu-ark.net/contact.html > Uppaluri
Gopala Krishnamurti (Born 9 July 1918)>
http://www.well.com/user/jct/mystiq1.htm> THE MYSTIQUE OF
ENLIGHTENMENT> Part One [Excerpt]> U.G. Krishnamurti> People
call me an ïenlightened man -- I detest that 
term -- they>
cant find any other word to describe the way I 
am
functioning.> At the same time, I point out that there is no
such thing as> enlightenment at all. I say that because all my
life Ive searched> and wanted to be an enlightened man, and 
I
discovered that there> is no such thing as enlightenment at
all, and so the question> whether a particular person is
enlightened or not doesnt arise.> I dont 
give a hoot for a
sixth-century-BC Buddha, let alone all> the other claimants we
have in our midst. They are a bunch of> exploiters, thriving on
the gullibility of the people. There is> no power outside of
man. Man has created God out of fear. So the> problem is fear
and not God.> ______________> I discovered for myself and by
myself that there is no self to> realize -- thats the
realization I am talking about. It comes> as a shattering
blow. It hits you like a thunderbolt. You have> invested
everything in one basket, self-realization, and, in> the end,
suddenly you discover that there is no self to discover,> no
self to realize -- and you say to yourself What the hell have>
I been doing all my life?! That blasts you.> _______________>
All kinds of things happened to me -- I went through that, you
see.> The physical pain was unbearable -- that is why I say you
really> dont want this. I wish I could give you a glimpse 
of
it, a touch> of it -- then you wouldnt want to touch this 
at
all. What you are> pursuing doesnt exist; it is a myth. You
wouldnt want anything> to do with this.> UG: You see, I
maintain that -- I dont know, whatever you call> this; I
dont like to use the words 
ïenlightenment, 
ïfreedom,>
ïmoksha or 
ïliberation; all these words are loaded words,
they> have a connotation of their own -- this cannot be
brought about> through any effort of yours; it just happens.
And why it happens> to one individual and not another, I 
dont
know.> Questioner: So, it happened to you?> UG: It happened
to me.> Q: When, Sir?> UG: In my forty-ninth year.> But
whatever you do in the direction of whatever you are> after --
the pursuit or search for truth or reality -- takes> you away
from your own very natural state, in which you always> are.
Its not something you can acquire, attain or accomplish> as 
a
result of your effort -- that is why I use the word 
ïacausal.>
It has no cause, but somehow the search come to an end.> Q:
You think, Sir, that it is not the result of the search? I
ask> because I have heard that you studied philosophy, that
you were> associated with religious people ...> UG: You see,
the search takes you away from yourself -- it is in> the
opposite direction -- it has absolutely no relation.> Q: In
spite of it, it has happened, not because of it?> UG: In
spite of it -- yes, thats the word. All that you do makes> 
it
impossible for what already is there to express itself. That
is> why I call this ïyour natural state. 
Youre always in
that state.> What prevents what is there from expressing
itself in its own way> is the search. The search is always in
the wrong direction, so all> that you consider very profound,
all that you consider sacred, is> a contamination in that
consciousness. You may not (Laughs) like> the word
ïcontamination, but all that you consider 
sacred, holy> and
profound is a contamination.> So, theres nothing that you 
can
do. Its not in your hands.> I dont like to 
use the word
ïgrace, because if you use the> word 
ïgrace, the grace of
whom? You are not a specially chosen> individual; you deserve
this, I dont know why.> If it were possible for me, I would
be able to help somebody.> This is something which I cant
give, because you have it.> Why should I give it to you? It is
ridiculous to ask for a thing> which you already have.> Q:
But I dont feel it, and you do.> UG: No, it is not a
question of feeling it, it is not a question> of knowing it;
you will never know. You have no way of knowing> that at all
for yourself; it begins to express itself. There is> no
conscious.... You see, I dont know how to put it. Never 
does>
the thought that I am different from anybody come into my>
consciousness. [...]> ((({<snip>})))> Continued at:
<http://www.well.com/user/jct/Mystique.htm > The Archetype and
the Beast ï98> http://pw1.netcom.com/~mthorn/arcbeast.htm>
[~][^][~]> Disingenuous Demagogues Deteriorate Daily> All
Politicians are Demagogues, yet not all> Demagogues are
Politicians...> ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~>
Sagittarius assimilated by OUR Milky Way:> Resistence Is
Futile!> http://www.astro.virginia.edu/~mfs4n/sgr/>
~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ T h e Y e z
i d i s o f K u r d i s t a
nhttp://www.songsouponsea.com/Promenade/GnosisE.html>EARTHlights (by
NASA):http://antwrp.gsfc.nasa.gov/apod/image/0011/
earthlights_dmsp.jpg ~0~ Were like a few bacteria
waiting for the next §ush. -- Peter Prehn
www.contemposcribe.com/cbcwash/centwashgeology.htm ~0~>ICE MEMORY by 
ELIZABETH KOLBERT Does a glacier hold the
secret of how civilization began--and how it may end?>Excerpt:... Over the 
past decade or so, there has been
ashift--inevitably labelled a ïparadigm 
shift--inthe way
scientists regard the Earths climate. Thenew view goes
under the catchphrase ïabruptclimate change, 
although it
might moreevocatively be called neo-catastrophism, after
theold, Biblically inspired theories of §ood anddisaster.
Behind it lies no particular theoreticalinsight--scientists
have, in fact, beenhard-pressed to come up with a theory to
makesense of it--but it is supported by
overwhelmingempirical evidence, much of it gathered
inGreenland. The Greenland ice cores have shownthat it is
a mistake to regard our own, relativelybenign experience of
the climate as the norm. Bynow, the adherents of
neo-catastrophism includevirtually every climatologist of
any standing.Abrupt climate changes occurred long
beforethere was human technology, and therefore
havenothing directly to do with what we refer to asglobal
warming. Yet the discovery that for mostof the past hundred
thousand years the Earthsclimate has been in §ux, 
changing
not gradually, oreven incrementally, but violently and
withoutwarning, cant help but cast the
global-warmingdebate in new terms. It is still possible to
imaginethat the Earth will slowly heat up, and that
thelandscape and the weather will gradually evolve
inresponse. But it is also possible that the changewill
come, as it has in the past, in the form ofsomething much
worse. ~0~See also: North Greenland Ice core Project
(NGRIP)http://www.glaciology.gfy.ku.dk/ngrip/index_eng.htm>VOLUNTEERS NEEDED 
For 180 Light Year Round Trip
Excursion![Estimated time of departure & arrival still under
review]Scientists Discover Planetary System Similar to Our
Own: http://www.nsf.gov/od/lpa/news/03/pr0373.htm The
small so-called grays ... are simply mature human fetuses
grown and tailored within artificial wombs. I would assume
that the reptilian and insectoid humanoid entities
reportedly seen by some abductees are just other examples
of genetically engineered life forms culled from reptilian
and insect life forms on Earth. -- Raymond Fowler
http://www.nicap.dabsol.co.uk/bio-fowler.htm(Everybody
ïwonders about Raymond) 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~>B L U R 
...http://www.astrocentral.co.uk/beagle.html> An
Atlas of The Universe> http://www.anzwers.org/free/universe/>
SDSS: Sloan Digital Sky Survey> http://www.sdss.org/> WMAP:
Wilkinson Microwave Anisotropy Probe>
http://map.gsfc.nasa.gov/> http://lambda.gsfc.nasa.gov/>
The Origins of the Fear of Death and Dying>
http://primal-page.com/death.htm> Stuart Hameroffs Home
Page:>
http://www.consciousness.arizona.edu/hameroff/index.html>
http://www.mazepath.com/uncleal/sunshine.jpg> Ommmmmmmm....* E
c k h a r t T o l l e Stillness Speaks:When you lose touch with
inner stillness, youlose touch with yourself. When you lose
touchwith yourself, you lose yourself in the world.Your
innermost sense of self, of who you are,is inseparable from
stillness. This is the I Amthat is deeper than name and
form.Stillness is your essential nature. What is stillness?
The inner space or awareness in which the words on this page
are being perceived and become thoughts. Without that
awareness, there would be no perception, no thoughts, no
world.You are that awareness, disguised as a person.The
equivalent of external noise is the inner noise of thinking.
The equivalent of external silence is inner stillness.Whenever
there is some silence around you -- listen to it. Thatmeans
just notice it. Pay attention to it. Listening to
silenceawakens the dimension of stillness within yourself,
because it isonly through stillness that you can be aware of
silence.See that in the moment of noticing the silence around
you, you arenot thinking. You are aware, but not thinking.When
you become aware of silence, immediately there is thatstate of
inner still alertness. You are present. You have steppedout of
thousands of years of collective human conditioning. Look at a
tree, a §ower, a plant. Let your awareness rest upon it.How
still they are, how deeply rooted in Being. Allow nature
toteach you stillness.When you look at a tree and perceive its
stillness, you become stillyourself. You connect with it at a
very deep level. You feel aoneness with whatever you perceive
in and through stillness.Feeling the oneness of yourself with
all things is true love.Silence is helpful, but you dont need
it in order to find stillness.Even when there is noise, you 
can
be aware of the stillnessunderneath the noise, of the space in
which the noise arises. Thatis the inner space of pure
awareness, consciousness itself.You can become aware of
awareness as the background to allyour sense perceptions, all
your thinking. Becoming aware ofawareness is the arising of
inner stillness.Any disturbing noise can be as helpful as
silence. How? Bydropping your inner resistance to the noise,
by allowing it to be asit is, this acceptance also takes you
into that realm of inner peacethat is stillness.Whenever you
deeply accept this moment as it is -- no matterwhat form it
takes -- you are still, you are at peace.Pay attention to the
gap -- the gap between two thoughts, thebrief, silent space
between words in a conversation, between thenotes of a piano
or §ute, or the gap between the in-breath andout-breath.When
you pay attention to those gaps, awareness of somethingbecomes
-- just awareness. The formless dimension of pureconsciousness
arises from within you and replaces identificationwith
form.True intelligence operates silently. Stillness is where
creativity and solutions to problems are found.Is stillness
just the absence of noise and content? No, it isintelligence
itself -- the underlying consciousness out of whichevery form
is born. And how could that be separate from whoyou are? The
form that you think you are came out of that and isbeing
sustained by it.It is the essence of all galaxies and blades
of grass; of all §owers, trees, birds, and all other
forms.Stillness is the only thing in this world that has no
form. But then, it is not really a thing, and it is not of
this world.When you look at a tree or a human being in
stillness, who islooking? Something deeper than the person.
Consciousness islooking at its creation.In the Bible, it says
that God created the world and saw that itwas good. That is
what you see when you look from stillnesswithout thought.Do
you need more knowledge? Is more information going to savethe
world, or faster computers, more scientific or
intellectualanalysis? Is it not wisdom that humanity needs
most at this time?But what is wisdom and where is it to be
found? Wisdom comeswith the ability to be still. Just look and
just listen. No more isneeded. Being still, looking, and
listening activates thenon-conceptual intelligence within you.
Let stillness direct yourwords and actions.The human condition:
lost in thought.Most people spend their entire life imprisoned
within the confinesof their own thoughts. They never go beyond
a narrow,mind-made, personalized sense of self that is
conditioned by thepast.In you, as in each human being, there
is a dimension ofconsciousness far deeper than thought. It is
the very essence ofwho you are. We may call it presence,
awareness, theunconditioned consciousness. In the ancient
teachings, it is theChrist within, or your Buddha
nature.Finding that dimension frees you and the world from the
sufferingyou in§ict on yourself and others when the mind-made
little meis all you know and runs your life. Love, joy,
creative expansion,and lasting inner peace cannot come into
your life except throughthat unconditioned dimension of
consciousness.If you can recognize, even occasionally, the
thoughts that gothrough your mind as simply thoughts, if you
can witness yourown mental-emotional reactive patterns as they
happen, then thatdimension is already emerging in you as the
awareness in whichthoughts and emotions happen -- the timeless
inner space inwhich the content of your life unfolds.The stream
of thinking has enormous momentum that can easilydrag you along
with it. Every thought pretends that it matters somuch. It
wants to draw your attention in completely.Here is a new
spiritual practice for you: dont take your thoughtstoo
seriously.How easy it is for people to become trapped in their
conceptualprisons.The human mind, in its desire to know,
understand, and control,mistakes its opinions and viewpoints
for the truth. It says: this ishow it is. You have to be
larger than thought to realize thathowever you interpret your
life or someone elses life orbehavior, however you judge any
situation, it is no more than aviewpoint, one of many possible
perspectives. It is no more than abundle of thoughts. But
reality is one unified whole, in which allthings are
interwoven, where nothing exists in and by itself.Thinking
fragments reality -- it cuts it up into conceptual bits
andpieces.The thinking mind is a useful and powerful tool, but
it is also verylimiting when it takes over your life
completely, when you dontrealize that it is only a small
aspect of the consciousness that youare.Wisdom is not a
product of thought. The deep knowing that iswisdom arises
through the simple act of giving someone orsomething your full
attention. Attention is primordial intelligence,consciousness
itself. It dissolves the barriers created byconceptual
thought, and with this comes the recognition thatnothing
exists in and by itself. It joins the perceiver and
theperceived in a unifying field of awareness. It is the 
healer
ofseparation.Whenever you are immersed in compulsive thinking,
you areavoiding what is. You dont want to be where you are.
Here,Now.Dogmas -- religious, political, scientific -- arise
out of theerroneous belief that thought can encapsulate
reality or the truth.Dogmas are collective conceptual prisons.
And the strange thingis that people love their prison cells
because they give them asense of security and a false sense of
I know.Nothing has in§icted more suffering on humanity than its
dogmas.It is true that every dogma crumbles sooner or later,
becausereality will eventually disclose its falseness;
however, unless thebasic delusion of it is seen for what it
is, it will be replaced byothers.What is this basic delusion?
Identification with thought.Spiritual awakening is awakening
from the dream of thought.The realm of consciousness is much
vaster than thought cangrasp. When you no longer believe
everything you think, you stepout of thought and see clearly
that the thinker is not who you are.The mind exists in a state
of not enough and so is always greedyfor more. When you are
identified with mind, you get bored andrestless very easily.
Boredom means the mind is hungry for morestimulus, more food
for thought, and its hunger is not beingsatisfied.When you 
feel
bored, you can satisfy the minds hunger bypicking up a
magazine, making a phone call, switching on the TV,surfing the
web, going shopping, or -- and this is not uncommon--
transferring the mental sense of lack and its need for more
tothe body and satisfy it brie§y by ingesting more food.Or you
can stay bored and restless and observe what it feels liketo be
bored and restless. As you bring awareness to the feeling,there
is suddenly some space and stillness around it, as it were.
Alittle at first, but as the sense of inner space grows, the
feeling ofboredom will begin to diminish in intensity and
significance. Soeven boredom can teach you who you are and who
you are not.You discover that a bored person is not who you
are. Boredomis simply a conditioned energy movement within
you. Neither areyou an angry, sad, or fearful person. Boredom,
anger, sadness, orfear are not yours, not personal. They are
conditions of thehuman mind. They come and go.Nothing that
comes and goes is you.I am bored. Who knows this?I am angry,
sad, afraid. Who knows this?You are the knowing, not the
an identity with pi> I found the following identity in a paper
by the> Chudnovsky brothers: infinity> ----- (n + 1)> 2> pi + 
2
= ) ----------------> / binomial(2 n, n)> -----> n = 1My
question is: is there any elementary way> to prove
this?Consider the sumf(x) = sum_{n=1}^infinity (2x)^{2n-1}/[n
(2n choose n)].Prove that(1 - x^2)f(x) = 1 + x f(x) and
deduce thatf(x) = sin^{-1}(x)/sqrt(1-x^2).Your series is
essentially f(1/sqrt(2)).(This argument was stolen from
Borwein/Borweins_Pi and the AGM_).-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
which is 10>multiplications. Then again you have multiplied
32823 with 1000 which>is again 9 multiplications. So totally
you have done 22+19 or 41>multiplications. Which is more than
25! I know that the last 18 are>easily done, but they are
multiplications all the same. Even if we>count the multiplies
with 10^x as 1 operation, then we still come down>to 25
operations, just as O(n^2).Well, no, we arent multiplying 
804
by 1000000 etc; were just movingit over six places. I could
have just as well written the sum 279105 32823 804 ---------
837102105Be that as it may it is quite clear that you still
havent graspedwhat is meant by the O() notation or you
wouldnt have said anything> Seemingly you do not understand
what the O() notation signifies. When> one says that the FFT
method is O(n*log(n)) one is saying that there> is some
constant C such that for n sufficiently large,(# of required
multiplies) is less than C*n*log(n)This does not mean that the
cost of the FFT method is less than n*n> for all n, just that
it is for n sufficiently large. Thus, your> proposed test is
irrelevant to the point under discussion.Ponder that
statement. Think about it. Please dont follow up tothis
posting until you understand it. >And all these are pretty
complex operations, always needing the carry>for the internal
you have shown that this is simpler than the Vedic method.I
didnt say that it was simpler than the Vedic method. And,
ofcourse, it is not simpler. However it is much more efficient
if oneis multiplying large numbers, large being dozens or
hundreds of digitsin the multiplicands. >In fact, it is
dreadful. I am sure that in due course people will>multiply
using the Vedic arithmetic. So much easier. Properly
done>(with hard-wired single digit computation, that will
make>multiplication efforts meaningless) I have no doubt that
it will beat>FFT hands down. In fact, FFT as you show it is a
clumsy approximation>to the elegant Vedic multiplication.> Be
that as it may, the cross product method for multiplication
is quite obvious and is regularly rediscovered. I discovered
it myself as a child and even then was under no illusion that
I had done anything remarkable.But no one demonstrated that
here before I did, with the example.>People came up with all
sorts of ideas, but no one could do it. And I>did give them
the chance! I wanted others to show it, and none did. >To say
now that you knew it, does not convince. Did you also
know>about the one-line division method? Supposing someone
were to explain>that with an example (I may do that after I
get that book within an>year or so) would you then say that
you knew that method all along?very venue (rec.arts.books) on
found on my web site. Richard Harter,
cri@tiac.nethttp://home.tiac.net/~cri,
http://www.varinoma.comWe have people from every planet on the
earth in this State.-- California Governor Gray
of the cubic polynomialax^3 + bx^2 + cx + d = 0,in terms of
cubic roots? Assume real coefficients, looking for
need help with algebracan anyone explain how would I graph
absolute value of y plus absolutevalue of x equal to 2. what
kinda graph would it be and how would Iget there. plz help.
neighborhood -topology~~~hello.......i saw.......U is a
neighborhood(=nbd) of x<=> U is an open set containing x by
Munkres bookbut<=> U is an super set of open set G containing
x by Lipschutz book---------------------------------Define f:
Ty={0,Y,{b}}let f(1)=b ,f(2)=b , f(3)=aif nbd=open set(by
Munkres), f is discontinuous on 1because f(1)=bf^-1{b}={1,2} :
not in Tx....thus not openif nbd=super set of open set(by
Lipschutz), f is continuous on 1because f^-1{all nbd
containing b} is nbd of
1---------------------------------whats wrong??which of 
case
is right??um.......advice....please....sir.....thank
SoftwarePNN Ltd. announces the release of PNN DiscoveryClient
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neighborhood -topology~~~><=> U is an super set of open set G
containing x by Lipschutz bookWhat is a super set? I only know
the definition of a neighborhood of xas an open subset 
including
x.Rene.-- Ren.8e MeyerStudent of Physics & MathematicsZhejiang
-topology~~~super set is reverse of subsetif U is an super set
of open set G containing x,it is mean that x in G(open set) in
where I would need to solve for a system (specifically2) of
quadratic equations. My search on the net lead me to systems
ofpolynomial equations. Unfortunately my math knowledge does
not yet allow meCould anyone please point me to a text that
could be understood by ainterested technician?For completeness
I formulate my problem, and what I know so far:1)The
equations:transpose (x-a) * A * (x-a) = r0transpose (x-b) * B
* (x-b) = r1where x is the (column) matrix of unknowns, a and
b are column matrices andA,B are symmetric matrices. r0 and r1
are scalars.2) I know about how to transform the quadratic
forms into theireigensystems, but then still can see no
systematic way of how I couldeliminate one of the unknowns.3)
I guess the order of my (specific) problem is at most of 
degree
4.Any pointers appreciated.RolandBTW.: Of course I would be
interested how to solve for more than twoequations
countingX-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>[...]Basically,
summing the partial differential apparently gives you aclose
approach to the prime counting distribution, which is
closerthan li(x) itself!> And its been repeatedly
pointed out that youve given _no_ evidence that the
solution to the pde should have anything whatever to do with
counting primes.> Hint: The fact that you _say_ something
_appears_ to be so does not count as evidence, because you
have zero credibility left after making so many false
statements.Why should JSH have zero credibility, while--after
your blunders,>you, a Ph.D. in mathematics and professor of
same--retain yours?>Could it be that that what commands
respect on sci.math and sci.logic>is not mathematical prowess,
but a stop-at-nothing determination>to silence or neutralize
dissenters from the canon?Tee-hee. Yeah, thats exactly 
right.
Of course youre not going toget anyone to _admit_ that 
thats
what commands respect here, becauseafter all that would be
giving the game away.Giggle. I especially like the part about
stop-at-nothing determinationto silence dissenters, coming
from the only person Ive ever seen_explicitly_ threaten to
_kill_ people who disagreed with him
onusenet...*****************************John Correy
says:Christian (what an oxymoron!): Degrade, demean, goad and
bait meas Ullrich and the Boyz have done to JSH, and I wont
tri§e withwriting your employer: Ill come after you with 
an
more like Laplacian distribution?> If I already know a prior
that my data distribution should follow the shape> of
Laplacian distribution... the data obtained from measurement
is of course> a little off(not very symmtrical), how can I
make the measured data more> Laplacian distribution like(make
it at least a little more symmtrical)?Can anybody give me an
example or detailed explanation? I am kind of afraid> of
statistics... :=)Lets see if I understand. Your data does not
followyour prior beliefs. Therefore the data is wrong, andyou
want to know how to modify said data so that itdoes exactly
what you want. This makes sense - in avery distorted way.Why
did you bother with measuring that blasted datain the first
place? You have already decided theresult. Measurements just
get in the way.Please start by reading the book How to lie
withStatistics. Then return to your data, and learn fromit. Is
it just random noise that has given your datathis property you
did not expect? Or is this anindication of a problem in your
measurement? Perhapsit indicates something wrong with the
theory? Perhapsanother factor distorts the data? Maybe your
sampleis just too small!I can summarize the strong suggestions
I give to mystudents who deal with data in three words:Plot -
think - learn.Only after that do I tell them to do any
actualmodeling, or use their data in any way.HTH,-- There are
no questions ? about my real address.The best material model
of a cat is another, orpreferably the same, cat.A.
result akin to Brouwer, new?X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>Let S be the set of
(x,y,z) in R^3 such that>x + y + z = 1>and>x>=0 and y>=0 and
z>=0.>Let f be a continous map S to S. Define three closed
subsets of S by:>A = set of (x,y,z) such that f(x)>=x>B = set
of (x,y,z) such that f(y)>=y>C = set of (x,y,z) such that
f(z)>=z??? Those inequalities make no sense if f is a map from
S to S.What did you actually mean?>and a fourth by:>D = (A cap
B) cup (B cap C) cup (C cap A).>Show that D contains a
connected subset T which meets all three sides of S,>ie. T
contains the three points>(0,a,b)>(d,0,c)>(e,f,0)>for some
a,b,...,f, not necessarily nonzero.I think I can grind out a
proof, but maybe the result is already familiar
===
to>somebody?>TIA,>Larry> C. UllrichSubject: Re: Canonical
closed map? >I am trying to decide wether the canonical
projection >p_i: X -> X_i, X = X_1x...xX_n, >which is an open
map, is also closed or not. >Can anyone give me a hint?P = {
(x,1/x) | x > 0 } is closed subset R^2p2(P) = (0,oo) is open
Analysis question (correct question this
time)X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>Hi all,Heres my
question again; there was a stupid error in thequestion
posted a few minutes ago.Let I be an open interval of the
reals that contains a pointp and let f be a differentiable
function from I into the realssuch that f is bounded at
some neighborhood of p. Therefore,limsup f(x) at p and
liminf f(x) at p are both real numbers.Question: must 
f(p)
be equal to their average?No.>Consider x^2sin(1/x).>For
this function f(0) is indeed the average of the limsup (1)
and >liminf (-1) but you can tweak>the function a bit so that
the limsup becomes 2, say, and the liminf >becomes -1.To
expand a bit on that: Imagine an example where |f(x)| <= x^2,
sof(0) = 0. Take a piecewise-linear thing that bounces up 
and
downbetween points where it equals x^2 and points where it
equals -x^2;make it increase from -x^2 to x^2 with slope 2 and
decrease fromx^2 to -x^2 with slope -1. (Now smooth it out at
the corners.)>KP-- >E-MAIL: K.P.Hart@EWI<dot>TUDelft.NL PAPER:
Faculty EWI>PHONE: +31-15-2784572 TU Delft>FAX: +31-15-2786178
Postbus 5031>URL: http://aw.twi.tudelft.nl/~hart 2600 GA
===
Delft> the Netherlands C. UllrichSubject: Re: Cardinals,
Nodes? >Consider the Binary Node tree of naturals. > 0 > / > /
> 0 1 > / / > 0 1 0 1 > . > . >(I hope it is lined up.)If you
want it to line up for all readers, composeand advise reader,
to use monospace font. >Obviously if you consider the omegath
row, it will have beta-1 (C) >elements, but since omega is the
first limit ordinal, it is the first >row with an 
infinite number
of elements. >However, consider the previous row. What ordinal
does it represent? >omega? How many elements does it have?
beta-1? Let us move up the >tree until we find a row with
beta-0 (aleph-0) elements. What >ordinal corresponds to this
>row?There isnt an previous to omega, that is a
fantasy.Whats the next real just after 0? Similar
fantasy.Whats the rational just before 0? Another
fantasy.Whats the rational just after 0 in { 1/n | n in N
}?Now do map f(x) = 1/x and youve the questionwhat is the
integer just before 1/0 = omega? >Furthermore, if P=NP,
Re: Reconsidering Halton Arp > You see, Dr. Arp is a
scientist, a world renowned scientist and he has> *data*,
real, hard astronomical data, which is more substantive in>
disproving the commonly taught Big Bang Theory, than the data
used to> support that theory.> Arps data is presented in
the _Atlas of Peculiar Galaxies_,> Astrophysical Journal
Supplement Number 123, Volume 14, November 1966.> In the
Atlas Arp presents galaxies that appeared abnormal. Follow-up>
observations showed that some, not all, of the galaxies were in
fact> two galaxies that are apparently interacting. What caused
the doubt> about the Big Bang was that some of these pairs have
very different> red-shifts. If the galaxies are close to each
other the different> red-shifts would sound the death knell
for expansion and the BB.> However, as observing technique
has improved weve determined that> most of these pairs are
simply close in the line of sight and are> at very different
distances. There are a few cases that have not> been
elucidated, the necessary obsrvations are, at best, 
difficult. These remaining cases do not constitute an overthrow of the
BB,> to do that would require high quality observations of
difficult> objects; big results require big data, obscure,
difficult cases> do not provide that.I went to Google, and
found a relevant link.<Quote In figure A below, there are four
objects. NGC 7603 is a spiral galaxy> with a redshift value of
0.029. Object 1 is a quasar with z = 0.057.> Objects 2 and 3
are quasar-like objects with z values of 0.243 and> 0.391
respectively. As L.97pez-Corredoira and Guti.8errez noted:>
Everything points to the four objects being connected among>
themselves, but how to explain the different redshifts? (p.
L17). How> to explain indeed? Gribbin lamented: That strikes
at the foundation> stone of received cosmological wisdom (p.
65). It certainly does! As> case where we once again are
experiencing a situation where data get> thrown out if they
dont fit the theory. Big Bang cosmology simply> 
cannot explain
Arps anomalies.> </Quote date November 1966, in what I see 
as
a not subtle attempt to skew> reader opinion.As Ive said, 
Dr.
Arp has DATA. Check for yourself.examples have already been
disproved. What is left are some vague cases.As Tom Kirke
said: big results require big data, obscure, difficult cases 
do
<>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$-
GiZp>UN2a}!Fmw+%h}Y<?5]3mj~`n8?0wycf-nf(r8SAdWK`G=JC[<3fz48E[v
{Ns!r]MT;JPgLG7|pBA7=lP1oGgUt^>L`!h_XXr5Q>_nGsY2_> What is the
general solution of the cubic polynomialax^3 + bx^2 + cx + d =
0,in terms of cubic roots? Assume real coefficients, looking
<http://mathworld.wolfram.com/CubicEquation.html>It is
interesting that, even if the coefficients are real, and
allthree roots are real, the formula in terms of radicals may
involvenon-real complex numbers.-- G. A. Edgar
of quadtraticIve come across a real world problem which
involves the intersectionbetween a linear graph (y=ax+b) and
the graph of a square root of aquadratic (y =
sqrt(cx^2+dx+e)).Can anyone point me to any resources on this
- or in particular the latterfunction? I didnt come across 
it
at school or university and it has so farthrown up many
in journalswhereas sci.logic and sci.math provide forums (in
principle at least),for work in progress which §aunts these
norms, and in so doing§aunts?§aunt: to wave in the wind: 
to
move ostentatiously:to carry a gaudy or saucy appearence: to
display ostentatiously ....(Chambers 1983)-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
neighborhood(=nbd) of x <=> U is an open set containing x by
Munkres book but <=> U is an super set of open set G
containing x by Lipschutz book>HelloI dont have 
Munkres book
but he *certainly* didnt give that definition.X 
a topological
space, V a subset of X, and x a point in X. Then V is
aneighborhood of x <=> there exists an open set U, x is in U
===
and U isincluded in V.Subject: Re: is tan(n)/n
thread the sequence {tan(n)/n} arose (n=1,2,3...). >The
discussion showed that this sequence does not approach
zero.lim(n->oo) tan (pi/2)(n - 1/n) = lim(n->oo) cot pi/2n =
oolim(n->oo) (pi/2)(n - 1/n) = ooby lHospitallim(n->oo) 
(tan
n)/n = = lim(n->oo) (tan (pi/2)(n - 1/n)) / (pi/2)(n - 1/n) =
lim(n->oo) (sec (pi/2)(n - 1/n))^2 (pi/2)(1 + 1/n^2) /
(pi/2)(1 + 1/n^2) = lim(n->oo) (sec (pi/2)(n - 1/n))^2 =
Re: dumbass need help with algebra> can anyone explain how
would I graph absolute value of y plus absolute> value of x
equal to 2. what kinda graph would it be and how would I> get
there. plz help. got stuck. I know Im a dumbass, but help
plz.First, assume x and y are both positive and graph. Then
consider the effect of changing the sign of x or y or
neighborhood -topology~~~> hello.......i saw.......U is a
neighborhood(=nbd) of x<=> U is an open set containing x by
Munkres bookbut<=> U is an super set of open set G containing
x by Lipschutz book---------------------------------Define f:
Ty={0,Y,{b}}let f(1)=b ,f(2)=b , f(3)=aif nbd=open set(by
Munkres), f is discontinuous on 1because f(1)=bf^-1{b}={1,2} :
not in Tx....thus not open> if nbd=super set of open set(by
Lipschutz), f is continuous on 1because f^-1{all nbd
containing b} is nbd of
1---------------------------------whats wrong??which of 
case
is right??um.......advice....please....sir.....thank
you....The two definitions are NOT equivalent. If you use the
Lipschutz definition of neighborhood, you have to use a
corresponding definition of continuous, which mentions open
I NEED HELP BADLY (sorry, maths not psych) But what are YOU
saying?Not untill the electrode 1 km away feels the opposing
force?> IS THAT WHAT YOU ARE SAYING? NO PAUL. I am
saying that your claim infers that the effect of an injected
electron will be felt INSTANTLY at the far electrode.>I have
understood that you think my answer is wrong.>I am asking
you: WHAT IS YOUR ANSWER?>Since you claim that the force
cannot act instantly,>I am asking you, WHEN will the force
act?>You claim that the distance to the other electrode is
relevant.>WHY is it relevant?>How does this distance affect
the delay before the force>on the electron starts acting?This is a concrete 
scenario.>Please state what you think will
happen. You keep talking about the force on the electron. 
Im
more concered about teh> force at the far electrode.  Of
course, since the electron existed BEFORE it was injected,
the effect would have already been there even though the
near electrode was in the way. The only way this experiment
can even be hypothesized is by either 
ïannihilating a very
large number of electrons or by monitoring the force on the
far electrode with movement of the electron mass towards
it.>Say, what the hell are you babbling about?>Dont tell
fairytales about suddenly disappearing charges! What happens
after this: (e+) + (e-) = ? >ADDRESS THE SCENARIO GIVEN!>It
is a concrete scenario which in principle could be made.>(And
which EASILY could and IS made with shorter distance>between
the electrodes. You have it in your TV set!)>Let it be a hot
wire in the hole in the electrode.>Thermionic emission of
electrons.>When one of these electrons gets out of the
hole>and into the static field between the electrodes,>how 
long
time will it take before a force start acting on it? I could
probably write a whole book answring that. However I would
conficently say that, before it starts to move, the force is>
instant.Its settled then.We agree: as it enters a static
electric field.One can but wonder why it took you so long to
realizethe obvious.>My answer is:> as it enters a static
electric field.>WHAT IS YOUR ANSWER? I just gave it to you.;
Now please answer MY question. If a highly charged sphere is
moved, say, backwards and forwards between two> electrodes,
what happens to the force on those electrodes. Does it change>
instantly or is there a time lag?Why do you have to ask?There
is obviously a time lag.> Hint: neither you nor anyone else
knows the answer.Nonsense.Of course we know exactly what will
happen.There are no mysteries in electrodynamics.know the
Reconsidering Halton Arp> Rather naively I thought that if
you put out something simple enough> that most people could
understand it, then theyd be willing to at> least question
authority long enough to see when authority figures get> it
pointed out grave> procedural errors and §at out empirical
wrong results in your most> recent spews. You responded like a
diversity hire - turning your> back, screaming
ïDISCRIMINATION!, shuf§ing down the same incompetent> 
road
some more... all the while demanding a larger
paycheck.http://www.crank.net/harris.html> Its not every
braying jackass that gets a whole page at crank.netHey
stooopid, we enjoy cleaning up your messes like we enjoy
scraping> dog off our shoes on a curb, distasteful but
expedient. Hey> stooopid, the whole world is against you for
good and just objective> reasons. Take a pooper scooper to
yourself.> James,Tom Kirkes response above is an accurate
summary of the status of> Arps work. There are a few
instances where two galaxies with> different redshifts appear
to be dynamically interacting, which would> of course be
impossible under the accepted interpretation of redshift.>
This occurence is an artifact of looking through a three
dimensional> volume --- sometimes objects at different
distances will lie close to> each other (or even on top of
each other) on the sky plane. I dont> think that there is 
any
evidence that this happens more frequently> than expected.
There is no big conspiracy here ---- it has a very> simple and
expected geometrical explanation. By the way, most of Arps>
peculiar galaxies really are interacting systems where the
constituent> members have the same redshift.The problem I have
with that is that in looking at actual pictures youcan see
material curling from one galaxy to another that is
clearlypulling the matter over.Ill put in a quote and link
from a previous post so readers can gosee it for
themselves.<Quote>In figure A below, there are four objects.
NGC 7603 is a spiral galaxywith a redshift value of 0.029.
Object 1 is a quasar with z = 0.057.Objects 2 and 3 are
quasar-like objects with z values of 0.243 and0.391
respectively. As L.97pez-Corredoira and Guti.8errez
noted:Everything points to the four objects being connected
amongthemselves, but how to explain the different redshifts?
(p. L17). Howto explain indeed? Gribbin lamented: That strikes
at the foundationstone of received cosmological wisdom (p. 65).
It certainly does! Ascase where we once again are experiencing
a situation where data getthrown out if they dont 
fit the
theory. Big Bang cosmology simplycannot explain Arps
anomalies.</Quote>Here a picture is really worth thousands of
Ive come across a real world problem which involves the
intersection> between a linear graph (y=ax+b) and the graph of
a square root of a> quadratic (y = sqrt(cx^2+dx+e)).This graph
will be part of a conic section: ellipse, parabola
orhyperbola. > Can anyone point me to any resources on this -
or in particular the latter> function? I didnt come across 
it
at school or university and it has so> far thrown up many
suirpises to me.To solve this we need y = ax + b =
sqrt(cx^2+dx+e).Squaring gives (ax + b)^2 = cx^2 + dx +
e.Rearranging this gives a quadartic equation for x.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
Let S be the set of (x,y,z) in R^3 such thatx + y + z =
1andx>=0 and y>=0 and z>=0.Let f be a continous map S to
S. Define three closed subsets of S by:A = set of (x,y,z) 
such
that f(x)>=xB = set of (x,y,z) such that f(y)>=yC = set of
(x,y,z) such that f(z)>=z??? Those inequalities make no sense
if f is a map from S to S.> What did you actually mean?I
presume for A, the condition is that the first component of
f(x,y,z)is >= x etc.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
Arp is a scientist, a world renowned scientist and he has>
*data*, real, hard astronomical data, which is more
substantive in> disproving the commonly taught Big Bang
Theory, than the data used to> support that theory.Arps
data is presented in the _Atlas of Peculiar Galaxies_, >
Astrophysical Journal Supplement Number 123, Volume 14,
November 1966.In the Atlas Arp presents galaxies that appeared
abnormal. Follow-up> observations showed that some, not all, of
the galaxies were in fact> two galaxies that are apparently
interacting. What caused the doubt > about the Big Bang was
that some of these pairs have very different> red-shifts. If
the galaxies are close to each other the different> red-shifts
would sound the death knell for expansion and the BB.However,
as observing technique has improved weve determined that>
most of these pairs are simply close in the line of sight and
are> at very different distances. There are a few cases that
have not> been elucidated, the necessary obsrvations are, at
best, difficult.These remaining cases do not constitute an
overthrow of the BB, > to do that would require high quality
observations of difficult > objects; big results require big
data, obscure, difficult cases > do not provide that.I went to
Google, and found a relevant link.> <deleted Well I looked
around at the rest of the page and, um, it looks like ithas
some kind of religious affiliation.Oh well, I guess 
thats part
of the travails for Dr. Arp, as he getspromotion from some odd
people, which of course, includes me!!!Heres another
link.http://perso.wanadoo.fr/lempel/red_shift_NGC_7603_
maths not psych) >And you have not provided any theory
of E&M that allows any such>thing as a reverse field. Nor why
there should be any kind of>speed limit involved. Nor why it
should follow any such thing>as the kinetic energy formula
observed in accelerators. Nor have>you provided a relation
between energy and mass if you dont>accept
relativity.>Socks radiation from an acceleraed charge!
fields associated with a moving charge! The 
ïBack EMF
concept. I would be most amazed if a moving charge DID NOT
alter the field around> itself, wouldnt 
you?>Quite.are
accelerated.>You KNOW the following, Henry.In an
accelerator going at full efficiency, we KNOW thatbecause it
looses this energy as synchrotron radiation in the bendsof
the circuit.(Very obvious and easily measurable.)So we - and
YOU - know that the RF-cavities never ceasesis only few mm/s
below the speed of light.>So why do you keep pretending
that the E-field is notspeed approaches c, when you KNOW 
that
isnt true? I DID NOT SAY THAT.>Indeed you did. The
question is how much energy?>Forgotten that too? Come on
Paul, you are being silly now. You know what 
ïasymptote
means.Indeed, and that whats wrong.Here is what I have told
you many times,and what you once more have forgotten:time it
passes through the accelerating field, regardless of its
speed.The gained energy does NOT approach zero (or any other
value)asymptotically when the speed approached zero, because
it is constant!That is how much energy!The proof of that is
that when the accelerator is in steady state,the lost energy
in the bends is equal to the energy gained inthe RF-cavities.
The lost energy is radiated as synchrotronradiation, which is
easy to measure.This lost energy does NOT decrease when the
speed ofthe electrons increases, quite the contrary.Thus the
gained energy does NOT decrease when the speedof the electrons
approaches c.So whatever you think happens to the field in the
RF-cavitieswhen the speed approaches c, we KNOW for certain
thatWhich proves you WRONG.The radiation from an accelerated
charge!or the fields associated with a moving charge!or The
ïBack EMF concept.any less when the speed 
approaches c.You
know this, and have admitted to know this.But you keep
forgetting what you know,and restate the claims you know are
wrong over and over. You are making no attempt to answer
that one. In typical fashion, you pretend the relevant
question does not exist.>A blatant lie.... because you know
point B shifts its position along> Y axis, angle ABC or EBF
will change and as I am saying again and> again, angle ABC is
solid angle.I know but...That was very clever indeed, George.
I admit, the whole confusion wasgenerated because of use of my
word SOLID angle instead of FIXEDangle. I messed it up but you
knew what I was trying to say.> I am responding to your
diagram and there is> nothing in that to keep the angle the
same. What I am> saying first like everyone else is that, as
you have> drawn it, point B will move and the angle will
change.> Something extra will be needed to prevent it moving>
that is not shown in the diagram at the moment, and I> cannot
speculate on what you might add.Very good indeed. There are
many ways to prevent the angle changing sothat it remains
FIXED. We can insert V-shaped rigid rod inside thesesprings so
that angle V does not change. Or we may put these springsinside
hollow pipe. We can use shock-ups. That is the best solution.So
you also tried to trap me in my words. Like HIM who
iscontrolling things at this moment, you knew the meaning.OK,
above all, I made a mistake in explaining this damn thing. But
nowyou know what I am trying to say.Now, will the point B move
along Y axis if the angle ABC remains FIXEDthroughout
to make date more like Laplacian distribution?Hello Walala,How
big is your sample size? If you grab only100 samples, you
wouldntexpect it to exactly fit the generating 
distribution.
Can you grab 1,000,000samples? If still it doesnt 
fit, then
suspect your generating dist. isdifferent from a Laplacian.
You can look into bootstrapping methods to getan estimate of
data distribution should follow theshape> of Laplacian
distribution... the data obtained from measurement is
ofcourse> a little off(not very symmtrical), how can I make
the measured data more> Laplacian distribution like(make it at
least a little more symmtrical)? Can anybody give me an example
or detailed explanation? I am kind ofafraid> of statistics...
length and> stiffness and both springs are in relaxed state
initially. Angle ABC> is solid angle.Apparently this means you
have some rigid structure> forcing this angle to be
are. I admit, I should have used the word FIXED angle rather
than SOLIDangle. There are many ways to do this. Take two
shock-ups like thoseattached to front wheel of my bike and
connect one end of these twoshock-ups so that they form V
shaped structure. Now this angle Vremains FIXED even if we
pull free ends of shock-ups in direction ofarms of this V
shaped structure.The angle V does not change throughout
operation of this device. Thisis what I am trying to
of Archimedes has 536 solutions. You can maa.org.
http://www.maa.org/news/mathgames.htmlOver at MathWorld, you
can read a column about a solutionfor the order-5 perfect
magic cube.http://mathworld.wolfram.com/--Ed Pegg Jr,
READ WITH HALF A LIGHTBULB?> at 08:43 PM,
raydpratt@hotmail.com (raydpratt) said:>Let us remember that
for many centuries, the world was most assuredly §at.Let us
not remember things that never happened. That story is as
bogus>as the cherry-tree story.Are you sure it wasnt a case
of quantum geography? The earth was §atuntil we measured it,
BADLY (sorry, maths not psych) >HenriWilson <Henri@the.edge>
skrev i melding I still cannot see the philosophical reason
for using F=dp/dt rather than F=m.dv/dt where m=f(v)>Its
Newtons second law of motion in its original form.>The 
change
of motion is proportional to the force impressed;> and is made
in the direction of the straight line in which the force> is
impressed.>What did Newton mean by motion?>Those who have
studied his texts think he meant momentum.>PaulIt is
interesting because the term ïv.dm/dt is 
explains the energy
increase> that is normally associated with 
ïrelativistic
mass increase.Of course.The question is what is the
momentum?If the momentum is mv, then the mass _must_ increase
with the speed.In 1905 this was the accepted definition, and
was why Einsteinsaid that the mass increased with
speed.However, the modern approach is to say thatmomentum =
m*f(v) where m is the invariant massand f(v) =v/sqrt(1 -
v^2/c^2)> It actually supports my argument that this energy
really goes into the ïreverse> field 
bubble that forms around
a moving charge.Why not call your enigmatic bubble a
fairy?more invisible but massive fairies clings to it.When the
fairies loose their grip in the bends of the accelerator,their
mass is transformed to synchrotron radiation.Make perfect
prevent the angle changing so> that it remains FIXED. We can
insert V-shaped rigid rod inside these> springs so that angle
V does not change. Or we may put these springs> inside hollow
pipe. We can use shock-ups. That is the best solution.[...]>
Now, will the point B move along Y axis if the angle ABC
remains FIXED> throughout operation of this device?How do you
pull on A and C?If you attach strings to A and C, B will move
unless the force you exertis less than necessary to overcome
the friction between your triangle and
process>Each one of us has his own research area, mine is the
wireless>communication. If I will simply copy my problem into
this group, then>I have to explain a lot of terms in my area.
So I abstract the>practical problem into an understandable
math question, which may not>be so suitable but should be
easier for understanding.Except that your abstraction should
then be based on standardmathematical terms and notation to be
understandable. Its likesomeone posted the same question in
its original form in a forum forwireless communication and
someone replied with a theoreticalmathematical model full of
epsion-deltas and partial-differentialequations.If youre
unsure how to formulate the problem mathematically it
wouldhelp to see some general statements about the actual
problem sincechances are then someone in this group will
understand what you wantand will be able to formulate it
unbounded?X-DMCA-Notifications:
===
http://www.giganews.com/info/dmca.html>Subject: is tan(n)/n
unbounded? >In another thread the sequence {tan(n)/n} arose
(n=1,2,3...).>The discussion showed that this sequence does
not approach zero.lim(n->oo) tan (pi/2)(n - 1/n) = lim(n->oo)
cot pi/2n = oo>lim(n->oo) (pi/2)(n - 1/n) = ooby
lHospital>lim(n->oo) (tan n)/n =You cant 
apply LHopitals
rule here. Find a calculus book:there are _hypotheses_ that
need to be satisfied...Good example for the next 
LHopitals
rule: child of Satan?thread, though.> = lim(n->oo) (tan
(pi/2)(n - 1/n)) / (pi/2)(n - 1/n)> = lim(n->oo) (sec (pi/2)(n
- 1/n))^2 (pi/2)(1 + 1/n^2)> / (pi/2)(1 + 1/n^2)> = lim(n->oo)
(sec (pi/2)(n - 1/n))^2> = lim(n->oo) [1 + (tan (pi/2)(n -
a question:> Suppose:> K2 = a + b + c> K1 = a b + b c + c
a> K0 = a b c> I want to acheive 2a - b - c, 2b - a - c, 2c
- a - b using any> combinations of K2, K1, K0 using any
operations (+, -, *, , roots,logs,> etc.) under real
numbers.> 1) How do I know if such combinations of K2, K1,
K0 will get me to2a -> b -> c?> 2) If such a combination
exists, how would I figure it out?> Your K0,K1 and K2 are
symmetric functions of a,b,and c, i.e., each> of them is
invariant under any parmutation of (a,b,c). IICR, any formula
generated from them must also be symmetric in a,b> and c in
the same sense. Since2a - b - c, 2b - a - c, and 2c - a - b
are not symmetric> functions of a,b and c, I believe you are
SOL. The best you can do is probably something like> 2a - b -
c = 3a - K2, 2b - a - c = 3b - K2, 2c - a - b = 2c -
psych)HenriWilson <Henri@the.edge> skrev i melding
<paul.b.andersen@hia.no>HenriWilson <Henri@the.edge> skrev i
melding> I still cannot see the philosophical reason for using
F=dp/dt rather than> F=m.dv/dt where m=f(v)Its 
Newtons
second law of motion in its original form.The change of
motion is proportional to the force impressed; and is made
in the direction of the straight line in which the force is
impressed.What did Newton mean by motion?Those who have
studied his texts think he meant momentum.Paul It is
interesting because the term ïv.dm/dt is 
explains the energy
increase that is normally associated with 
ïrelativistic
mass increase.Of course.>The question is what is the
momentum?>If the momentum is mv, then the mass _must_ increase
with the speed.>In 1905 this was the accepted definition, and
was why Einstein>said that the mass increased with
speed.However, the modern approach is to say that>momentum =
m*f(v) where m is the invariant mass>and f(v) =v/sqrt(1 -
v^2/c^2)> It actually supports my argument that this energy
really goes into the ïreverse field 
bubble that forms around
a moving charge.Why not call your enigmatic bubble a fairy?more
invisible but massive fairies clings to it.>When the fairies
loose their grip in the bends of the accelerator,>their mass
is transformed to synchrotron radiation.Make perfect sense,
doesnt it?Neat! Hot fairies. /BAHSubtract a hundred and 
four
Hello all,I am doing some light (?) reading on topological
groups for kicks. > Unfortunately, my brain is slowing down. A
problem:Let G be a topological group.> If K,L subseteq G are
closed in G, do we necessarily have KL closed in G?No. Take G
= real numbers with the usual topology and the group
structuregiven by the addition. Take K = Z (the integers) and
L = a.Z, where a issome irrational number. Then K + L is a
dense subset of R, but it is notequal to R and therefore not a
neighborhood -topology~~~U is a neighborhood(=nbd) of x<=>
U is an open set containing x by Munkres bookbut<=> U is an
super set of open set G containing x by Lipschutz book> HelloI
dont have Munkres book but he *certainly* 
didnt give that
definition.> X a topological space, V a subset of X, and x a
point in X. Then V is a> neighborhood of x <=> there exists an
open set U, x is in U and U is> included in V.No, Im afraid 
he
actually does (have just double checked in my copy of his
book). In my opinion its a very bad definition 
to use, but for
some reason Im unable to fathom a lot of people *do* use 
it.
Doesnt really make a huge amount of difference in the 
maths,
but it means the terminology starts to get a fair bit messier
in places and one has to start talking about things like ïthe
closure of a neighbourhood rather than a closed
would need to solve for a system (specifically> 2) of 
quadratic
equations. My search on the net lead me to systems of>
polynomial equations. Unfortunately my math knowledge does not
yet allow me> Could anyone please point me to a text that could
be understood by a> interested technician?For completeness I
formulate my problem, and what I know so far:1)The equations:>
transpose (x-a) * A * (x-a) = r0> transpose (x-b) * B * (x-b) =
r1where x is the (column) matrix of unknowns, a and b are
column matrices and> A,B are symmetric matrices. r0 and r1 are
scalars.There is a method of Newton for finding the zeros
ofarbitrary functions. In one dimension, you draw the
tangentline at your current value of x, then go to where
thattangent line intersects the x-axis.This idea is easily
extended to functions of multiple variables and to
simultaneous nonlinear equationslike yours: make a linear
approximation near yourcurrent value of x, then solve that
linear systemfor your next value (or at least for the
directionto move).Heres one
discussion:http://il.water.usgs.gov/proj/feq/feqdoc/chap_9_2.
htmlOne critical thing is the behavior of your
secondderivatives, which is the matrixes A and B. If theyare
positive definite (all positive 
eigenvalues)Newtons method
works fantastically well. Otherwiseit may well diverge unless
you start close toa solution. More robust algorithms are out
there tohandle these cases.Im not clear if you want the
theory or just pre-cannedsoftware. Both are widely available
on the web. Doa search for nonlinear simultaneous equations.
Forsoftware, Netlib is a good place to look: www.netlib.org. -
explain how would I graph absolute value of y plus absolute>
value of x equal to 2. what kinda graph would it be and how
would I> get there. plz help. got stuck. I know Im a 
dumbass,
but help plz.|x| + |y| = 2With absolute value, its often
easiest to break itdown into specific cases.If x >=0, replace
|x| by x.If x <0, replace |x| by -x.If y >= 0, replace |y| by
y.If y < 0, replace |y| by -y.Thus, you have four separate
line segments to draw:1. x >= 0, y >=0, x + y = 2(this is the
part of the line x + y = 2 that lies in thefirst quadrant,
x>=0, y>=0).2. x >= 0, y < 0, x - y = 2(Similar to above.
Which quadrant is this in?)3. x < 0, y >= 0, -x + y = 24. x <
difficult for some people?I am a science teacher but I have
taught math for a few years. I tooagree tht it could be
possible that people have a math gene. Throughout my school
years I have always struggled with math, while somestudents
just seemed to get it right off. No matter how hard Istudied
it was a constant battle.While teaching math to my students I
see the same types of behaviors Ithink this makes me a better
math teacher cause I know what it feelslike not to understand,
which in turn makes me more patient with mystudents. I think
the book is worth reading and considering. Howeveryou have to
be careful when sharing this information with yourstudents
MATHEMATICIANS READ WITH HALF A LIGHTBULB? at 08:43 PM,
raydpratt@hotmail.com (raydpratt) said:>Let us remember that
for many centuries, the world was >most assuredly §at.Let us
not remember things that never happened. That story is as
bogusas the cherry-tree story.> Are you sure it wasnt a
case of quantum geography? The earth was §at> until we
measured it, at which point it became spherical.Sort of like
Calvins dads explanation of how the world 
used to be
blackand white.Also, the original claim didnt say anything
about *when* the world was§at. Surely, the prehistoric world
must have been §at.-- Dave SeamanJudge Yohns mistakes
revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
What you fail to understand that if point B shifts its
position along> Y axis, angle ABC or EBF will change and as
I am saying again and> again, angle ABC is solid angle.> I
know but... That was very clever indeed, George. I admit, the
whole confusion was> generated because of use of my word SOLID
angle instead of FIXED> angle. I messed it up but you knew what
I was trying to say. > I am responding to your diagram and
there is> nothing in that to keep the angle the same. What I
am> saying first like everyone else is that, as you have> 
drawn
it, point B will move and the angle will change.> Something
extra will be needed to prevent it moving> that is not shown
in the diagram at the moment, and I> cannot speculate on what
you might add. Very good indeed. There are many ways to
prevent the angle changing so> that it remains FIXED. We can
insert V-shaped rigid rod inside these> springs so that angle
V does not change. Or we may put these springs> inside hollow
pipe. We can use shock-ups. That is the best solution. So you
also tried to trap me in my words. Like HIM who is>
controlling things at this moment, you knew the meaning. OK,
above all, I made a mistake in explaining this damn thing. But
now> you know what I am trying to say. Now, will the point B
move along Y axis if the angle ABC remains FIXED> throughout
operation of this device?No.But this is a simple statics, it
easy to see that all forces will balance.I will give you a
hint:When you pull in the points A and C to bring them toE and
F, in which direction must you pull to make B not move?A second
===
hint:Moment.PaulSubject: Re: neighborhood
neighborhood(=nbd) of x ><=> U is an open set containing x by
Munkres bookI like this definition of nhood. To make clear 
that
I consider nhoods tobe open I often use the expression open
nhood. >but <=> U is an super set of open set G containing x
by Lipschutz >bookThis definition I dont like. 
It has little
use beyond the opennhood definition while adding some
complexity to usage of nhoods.Yes, theres two different
defitions for nhood. Theres also 
differentdefinitions of
normal, regular, T3, T4, depending upon if T2 is includedor
possiblecombinations have been used by one author or another,
so its important toknow how the author uses those terms.
Another convention to look out foris some authors consider all
spaces to be Hausdorff. Thus a theorem willget stated, even
quoted by student, without including Hausdorff, whenHausdorff
is intended.-- >Define f: X->Y >let X={1,2,3}
Tx={0(=empty),X,{1},{1,3}.{3}} >let Y={a,b,c} Ty={0,Y,{b}}
>let f(1)=b ,f(2)=b , f(3)=a >if nbd=open set(by Munkres), f
is discontinuous on 1 >because f(1)=b >f^-1{b}={1,2} : not in
Tx....thus not openNo, for f to be continuous at 1, its
necessaryfor all U nhood f(1), some V nhood 1 with f(V) subset
UThus 1 in {1} = V, f(V) = {b} subset U >if nbd=super set of
open set(by Lipschutz), f is continuous on 1 >because f^-1{all
nbd containing b} is nbd of 1Again f continuous at 1 by correct
definition given above.-- >whats 
wrong??Youve confused the
notion of continuouswith the notion of continuous at. >which
MATHEMATICIANS READ WITH HALF A LIGHTBULB?> raydpratt
<raydpratt@hotmail.com> grava .88 la saucisse et au marteau:
Please explain how ïconvergence refutes that 
logic.Because,
according to you, what is A = 1-1+1-1+1 .... ?Is it 0 because
A=(1-1)+(1-1)+...?But this is also 1-(1-1+1-1....) = 1-AA =
1-A, so A = 1/2But A = 1-(1-1)-(1-1)... = 1So, what do you
night tohelp me get to sleep, and I now understand several
arguments that havebeen raised in this thread about
convergence and the related conceptthat the associative
property of addition does not apply to anon-convergent 
infinite
sums series, which is what you are pointingout above with the
apparent paradoxes.My answer is that a bird is more than its
parts and that we cannotmangle a bird to prove that the
concept of a §ying bird is invalid. As to the present problem
[ 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) ],we cannot mangle
the left side without equally mangling the rightside, and
although algebra generally allows such mangling andcollapsing
of processes into ïequivalent expressions, we 
have to
becareful when dealing with infinity processes since the
processesgenerate a specific series of sums, not just any
series of sums.So, in essence, I CONCUR WITH YOU AND ALL
OTHERS, but our agreement ison different grounds. Algebra is
not violated by my qualificationsince we are dealing with an
extra logical element of an infinityprocess where the process
cannot be logically simplified withoutaccounting for the exact
same simplification on both sides of theequation, i.e., the
infinity process itself is a term on both sides ofthe 
equation,
an extra logical element that we have not studied wellenough to
understand and manipulate properly.Very
might sound like a dumb question, but im not a
mathematician...I am using Simpsons rule to numerically
integrate a function I wouldbe glad if you can help me finding
the best step size at which i haveto evaluate the function to
do the integration.my two integrals are: Vmax Vmax - V / / |
f(V) dV | 
f(V)dV / / V 0where Vmax is a constant; V is 
a
variable independent of V, and ischanging between 0 and 
Vmax
at a known step size(lets say Vmax/m -where m is known).In
fact the function studied can be a normal or
Re: Worlds oldest puzzle solved>The Loculus of Archimedes 
has
536 solutions. You can >maa.org.
http://www.maa.org/news/mathgames.htmlOver at MathWorld, you
can read a column about a solution>for the order-5 perfect
magic cube.http://mathworld.wolfram.com/--Ed Pegg Jr,
http://www.mathpuzzle.com Youre certain ther were no other
I NEED HELP BADLY (sorry, maths not psych)> time it passes
through the accelerating field, regardless of its speed.> The
gained energy does NOT approach zero (or any other value)>
asymptotically when the speed approached zero, because it is
constant!Should be: .. when the speed approaches c,
In sci.math, Jonathan Miller> <jonmillere1@comcast.net
<krCdnZA88-uNzSeiRVn-uA@comcast.com>: Note that both are a
special case of L^Hopitals rule (note spelling):> What 
are
we supposed to notice? That you dont have a circum§ex?> 
That
you dont know that the (usual) substitution for a 
circum§ex>
o is os? That slepping falmes always contain mispleddings?>
Jon MillerSearch engines are notorious for missing the point
when a rule> name is misspelled. I did not intend insult.
:-)Im sorry, I misinterpreted your remark.> Are you
suggesting it should be spelled LH.99pitals 
Rule? Id> go
along with that, from the limited amount of French I know>
(oui, non, je tadore, je ne parle pas Fran.8dais, 
etc.)Well,
either that or LHospital (he spelled it both ways himself, 
as
theAcademie Francais [note mispellings] hadnt yet 
standardized
the language).But sometimes correct spellings miss web pages,
because the page sponsor hasmisspelled things.Ive been
avoiding accents and cedillas because I was under the
impressionthat ASCII didnt support them, but I now realize
that that is incorrect.Theyre there, so 
theyll display
correctly on everyones machine. Today Iavoided them out of
sheer laziness. Theyre not on my keyboard.Jon
action>SO(n+1) x S^n ----> S^n (If you DO define an action, it
would be easier to answer your> question. Does the group act
trivially, perhaps? Or is this the> natural linear action of a
matrix group, restricted to a subset> of R^{n+1} ?) >then what
is the stabilizer of an element x in S^n? Is it SO(n) and why?
Trivial action: Stab(x) is all of SO(n+1).> Natural action:
Stab(x) is a conjugate of SO(n). Specifically, when x> is the
vector x0=(1,0,0,...), the stabilizer is the set of matrices
in> SO(n+1) which take x0 to x0 (of course) and thus consists
of matrices> whose first column is (1,0,0,...). The rest of 
the
first row is then> all zeros too, but the remaining n-by-n
matrix can be any special> orthogonal group. So this
stabilizer is in a natural way isomorphic> to SO(n).Why is
this true? Wouldnt the stabilizer in this situation be in a
naturalway isomorphic to a group larger than SO(n)? In
particular, you can fitan orthogonal matrix in the remaining n
by n matrix,,,so couldnt bythe same reasoning the 
stabilizer
be naturally isomorphic to O(n)?Then if x is any other point
of S^n, choose any matrix> M in your group such that x = M.x0
(this group action is transitive)> and then youll 
find that
Stab(x) = M Stab(x0) M^{-1}.> Other action: It depends!>where
G = SO(n+1) and X = S^n, Youre assigning labels you 
dont
use. I think it was Mary Ellen Rudin> who related the story of
a student whose proof began, Let X be a> topological space.
is tan(n)/n unbounded? >In another thread the sequence
{tan(n)/n} arose (n=1,2,3...).>The discussion showed that
this sequence does not approach zero.lim(n->oo) tan (pi/2)(n -
1/n) = lim(n->oo) cot pi/2n = oo> lim(n->oo) (pi/2)(n - 1/n) =
ooby lHospital> lim(n->oo) (tan n)/n => = lim(n->oo) (tan
(pi/2)(n - 1/n)) / (pi/2)(n - 1/n)> = lim(n->oo) (sec (pi/2)(n
- 1/n))^2 (pi/2)(1 + 1/n^2)> / (pi/2)(1 + 1/n^2)> = lim(n->oo)
(sec (pi/2)(n - 1/n))^2> = lim(n->oo) [1 + (tan (pi/2)(n -
1/n))^2] = ooHuh?!!!!LHopital? For a limit that does not 
even
exist??For a function that does not have a derivative? (sinceit
is not defined for real numbers, but integers).And besides,
since when LHopital can be applied atall for lim (x->oo)?? 
If
the derivative at infinityis something other than zero, then 
the
functionsvalue can not be zero (which is a condition to
Numerical Integration> Hi all,> it might sound like a dumb
question, but im not a mathematician...> I am using 
Simpsons
rule to numerically integrate a function I would> be glad if
you can help me finding the best step size at which i have> to
evaluate the function to do the integration.> my two integrals
are: Vmax Vmax - V > / /> | f(V) dV | 
f(V)dV> / /> V 0where
Vmax is a constant; V is a variable independent of V, and 
is>
changing between 0 and Vmax at a known step size(lets say
Vmax/m -> where m is known).> In fact the function studied can
be a normal or log-normal> distribution.I will appreciate any
help.Well, youre the only one that knows what is the amount
of errorthat you can tolerate. With that, just check the
formula forthe error upper bound:M * h^4 (b - a) / 180Where M
is an upper-bound for the fourth derivative of thefunction
being integrated, h is your step size, and (a,b) isthe
integration interval.Another thing you could do is to
iteratively reduce the sizeto half, until the difference in
the result from one iterationto the next one is below some
RationalsKRamsay says...|How small can d(n) be? Clearly, d(n)
< 1/n. But can we make d(n) much>|smaller than that?>|>| Q1:
Can we find arbitrarily large values of n such that d(n) <
1/n^2?Whenever a number is [ir]rational, it can be
approximated this well.>Let 0, x, 2x, 3x, ..., nx be the
multiples of an irrational x, and>consider their fractional
parts 0, x-[x], 2x-[2x],...,nx-[nx]. Since>there are n+1 of
them, there are two that are within 1/n of>each other. If
|(rx-[rx])-(sx-[sx])|<1/n, then (r-s)x comes within>1/n of an
integer, and |r-s|<n.The golden mean is one of the
===
argument.--Daryl McCulloughIthaca, NYSubject: Square root of a
About naturalsOk, Ignacio. La l.8ei. No parece generalizable
el patr.97n que siguen. Lossiguientes en la serie
fallan.SaludosIgnacio Larrosa Ca.96estro
<ilarrosaQUITARMAYUSCULAS@mundo-r.com> escribi.97> Peter L.
Montgomery <Peter-Lawrence.Montgomery@cwi.nl> escribi.97 en
el> You can get at least 24:> a = b = c = 5050505.> a+b =
a+c = b+c = 10101010> a+b+c = 15151515> S(a+b+c) = 24> An
upper bound is 60.> S(a + b + c) = S(10a + 10b + 10c) <=
S(5a + 5b) + S(5a + 5c) + S(5b +5c)> <= 5S(a + b) + 5S(a + c)
+ 5S(b + c) <= 15*4 = 60.> A solution is a = 4554554555> b =
5455455455> c = 5545545545 S(a+b) = S(a+c) = S(b+c) = 4;
S(a+b+c) = 51 (By Deschamps in es.ciencia.matematicas)> -->
Re: solving equationIm having the following equation :S = {
[(x%n) + 1] % p for (x/n) even> { [(x%n) + 1 + p/2] % p
for(x/n) odd> and p>1For non-C programmers: with a and b
integers,a % b means a mod b, and a/b means §oor(a/b).> Its
easily proven that S(a) = S(a+2n) because of the fact they are
both even> or both odd for starters and x%n = (x+2n)%n.Problem
is I wish to prove that S(a) != S(a+1) but i have no idea on
how to> handle that equation due to the multiple modulos in
it. I know you cant> just get rid of the modulo signs and
need to add an extra factor (say b*p)> but even then im 
still
stuck. Any hints?There are four cases (which may be reducible
totwo).I. a/n even, (a+1)/n oddII. a/n even, (a+1)/n evenIII.
a/n odd, (a+1)/n oddIV. a/n odd, (a+1)/n evenCases I and IV
can only happen if a mod n = (n-1)and so (a+1) mod n = 0.So
S(a+1) = 1 mod p = 1 or S(a+1) = (1 + p/2) mod pwhile S(a) =
(n-1) mod p or S(a) = (n-1 + 1 + p/2) mod p.But these could
easily be equal. For instance ifn = p+2, then clearly n-1 is
congruent to 1 mod p.Example: p = 3, n = 5, a = 9a/n = 1,
odd(a+1)/n = 2, evenS(9) = ((9 % 5) + 1) % 3 = (4 + 1)% 3 =
2S(10) = ((10 % 5) + 1 + 3/2) % 3 = (0 + 1 + 1) % 3 = 2 -
what I sent to Eric. It should answer your
question.-------------------------The program below is fairly
efficient. It lists successive maxima and minima> in the
sequence, preceded by the value of n giving that
extremum.$MaxExtraPrecision = 400; min = 0; max = 0; i = 0;>
While[i < 350, i = i + 1;> v = N[Tan[n]/n, 70];> If[v < min,
min = v; Print[{n, N[min, 6]}]];> If[v > max, max = v;
Print[{n, N[max, 6]}]]]...>
{
140278322695820130839421972690860545853324377324794798150266034
6275822,>
-74.7721}{
656663628873183465002720935077228312202585023816648634619965107
978725953831633144141594852184603577778946951537005011,>
18.0566}{
237245119117113587257974180846905994200670445824975057251225928
091938884235884204270870335868832984721112397578700716369398842
40354956727657904859772359124,> 556.306}, youre a dude!Is 
it
proven that the extrema have to be on CF convergents
numerators?I know thts the best place to look for them, but
there are a lot of numbers between 65666... and 23724... which
could be darned close to k*Pi/2, but arent on the CF.Anyway, 
I
ported the above to GP and ran it a bit further. Perhaps 
youd
like to independently verify these, and forward them to Eric?
(my GP code verified all your results, obviously.)Summary:
bounds -529.446126 to 10539.847388 from 5999
convergents.(numbers pushed below a .sig separator, so they
dont get included in any replies, as theyre 
big)-- (eek -
had to split the lines)-92.573200
164314003623502451783251403218809330051254587707144685188930087
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207734-113.121394
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119.470171
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877770529553959237844454123820638310437285744-- Unpatched IE
vulnerability: Click hijackingDescription: Pointing IE mouse
events at non-IE/system windowsReference:
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HijackClick-Content.HTMExploit:
http://safecenter.net/liudieyu/HijackClick/HijackClick2-
like Laplacian distribution?> If I already know a prior that
my data distribution should follow theshape> of Laplacian
distribution... the data obtained from measurement is
ofcourse> a little off(not very symmtrical), how can I make
the measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?Walala, others have already
responded that its unusual to modify data tofit 
a
distribution. However, it is common to fit a distribution to
data.That distribution may be a more symmetric model for the
process thatgenerated the data. Youd probably also want to
test your assumption thatthis is a good fit, either informally
(maybe with a graph) or formally (witha hypothesis test).
Could it be that you really want information about howto fit a
Laplace distribution?Heres a bit of MATLAB code that
generates some normal data and fits aLaplace distribution to
it, then compares a histogram of the data to there-scaled
Laplace density, which is symmetric.% Generate normal data,
just for exampleN = 50;x = 20 + 2*randn(N,1);% Fit the Laplace
parametersmu = median(x);sig = mean(abs(x-mu));% Compare a
histogram to the fitted
distributionhist(x,§oor(min(x)):ceil(max(x)))xlim =
get(gca,xlim);xx = 
linspace(xlim(1),xlim(2),200);yy = N *
exp(-abs( (xx-mu)/sig ))/(2*sig);hold on; 
plot(xx,yy,r-);
Does anyone know an approximation for Stirling Numbers of the
Second> kind, S(n,k), for very large values of n?S(n,1) =
S(n,n) = 1 are easy, as are S(n,2) and S(n,n-1). Its the>
intermediate values of k that are difficult. I 
cant use the>
natural numbers define Stirling numbers of second kind to
bethecoefficients S(n,k) from developmentx^n = SUM_{k=0 to
k=n}S(n,k)x(x-1)(x-2)...(x-k+1) .If e_0(t)=1 , e_j(t)=t^j for
j=1,2,..., then according toNewton interpolation polynomial it
may recognized that(1) S(n,k) = [0,1,...,k; e_n] where
[x_0,x_1,...,x_k; f] denotes the divided difference of a
certainfunctionf:I-->R at a sistem of knots (distinct)
x_0,x_1,...,x_k from I.Moreprecisely [x_0,x_1,...,x_k;
f]=SUM_{j=0 to j=k} f(x_j)/w(x_j) where
w(x)=(x-x_0)(x-x_1)...(x-x_k) .If we consider x_0<x_1<...<x_k
and f is in C^{k-1}[x_0,x_n] and thederivativef^{(k)} exists
on (x_0,x_k) , then there is at least a point c , c
in(x_0,x_k),such that(2) [x_0,x_1,...,x_k; f]= f^{(k)}(c)/k!
.(2) S(n,k) = (n!/(n-k)!)*(c^{n-k}) < n!*n^{n-k}/(n-k)!
.However this bound is not the
Touchard-polynomial t_n(x) by(3) t_n(x) = SUM_{k=0 to
k=n}S(n,k)*x^k = =e^{-x}SUM_{j=0 to j= infty}j^n *x^j/j!
(Dobinski formula).Note that t_0(x)=1 and t_n(0)=0 , n>= 1.The
numbers B_n :=t_n(1) , n=0,1,..., are called ,,Bell Numbers .In
other words(4) B_n = SUM_{k=0 to k=n}S(n,k).Because
t_n(x)=x*e^{-x}(t_{n-1}(x)*e^x) it may be shown that
allrootsof t_n(x) are real,distinct, distributed in (-inft,0]
.In other words , for n>=1 , Touchard polynomial t_n(x) has
only realroots.Take an arbitrary polynomial p(x)=SUM_{k=0 to
k=n}C_k *x^{n-k} whoseroots are all real. There are known some
inequalities satisfied bycoefficients c_k ,c_k in 
R, k=0,1,..,n.
For instance(5) c_0+c_1+...+c_n =< K(n)*max_{j=0,1,..,n}c_j
where K(n)=(n+1)^n/C(n,s)(n-s)^{n-s}(s+1)^s , with s=[n/2]and
C(n,s)=n!/(s!(n-s)!). Therefore(5) B_n =< K(n)*max{S(n,k)} 
.
L. Euler [Acta Petropolitana ,Vol. 13, p.105] has shown
that(6) c_k^2 >= (1+1/k)(1+1/s)c_{k-1}*c*{k+1} with s=n-k .As
a special case of (6) is the so-called Gua Theorem [Jean Paul
de Gua , Memoires de lAcademie des Sciences de Paris,1741],
namely(7) c_k^2 > C_{k-1}*c_{k+1} . In your situation when
p(x)=t_n(x) one finds
S(n-k,k)^2 > (1+1/k)*(1+1/(n-k))S(n-k-1,k)*S(n-k+1,k)
,k=1,2,...,n-1 . , n>=2 .
inequality. For x in [-1,1] we have|t_n(x)|=< SUM_{k=0 to
k=n}S(n,k)|x|^k =< t_n(1) = B_n . This imples that(8)
|t_n^{(k)}(x)|/k! =< n*B_n*C(n+k,2k)*2^k/(n+k) , x in [-1,1].
If we choose x=0 in (8), one finds
S(n,k)=|t_n^{(k)}(0)|/k! =< n*B_n*C(n+k,2k)*2^k/(n+k)
Bell numbers is known .For a >= 3 equation (9) x*e^x = a , x
in (0,infty)has exact one solution x=x(a) . If g(a)= ln(a)-
ln(ln(a)) , then x(a) =asymp.= g(a) , (for a-->infty) , i.e.
lim_{a-->infty}x(a)/g(a) = 1 .Some results concerning
asymtotic behavior of B_n are :i) c_1* ln(n)/n =< B_{n-1}/B_n
=< c_2*ln(n)/n , (n large),where c_1,c_2 does not depend on
n.ii) B_{n+1}/B_n - 1 =asymp.= n/(e*ln(n)) e being Napiers
constant.iii) Denote by N the root x(n) of equation (9).If
z(n)= (1-(2*N^2+7N +10)*N^2)/(24*n*(1+N)^3), then B_n =asymp=
exp(n*(N-1+1/N)*z(n))/sqrt(1+N) , (n--->infty) . If you want
to study the asymptotics of Stirlin numbers, please
see[1]-[4].Perhaps,help. References[1]L.H.Harper ,,Stirling
behavior is asymptotically normal , Ann.Math.Statist., 38
(1968)410-414.[2]L.Moser and M.Wyman ,,An asymptotic formula
for the Bell numbers, Trans.Roy.Soc.Canada 49 (1955)
49-53.[3]A.Reny ,,Probabilistic methods in Analysis I, II ,
Mat.Lapok 18 (1967) 5-35,175-194 .[4]E.Ja.Riecstinis
,,Asymptoticeskaia razlajenia integralov,(Russian), Riga, 1981
quadtratic> Ive come across a real world problem which
involves the intersection> between a linear graph (y=ax+b) and
the graph of a square root of a> quadratic (y =
sqrt(cx^2+dx+e)).Can anyone point me to any resources on this
- or in particular the latter> function? I didnt come 
across
it at school or university and it has so far> thrown up many
suirpises to me.y^2 = a^2.x^2 + 2.a.b.x + b^2y^2 = c.x^2 + d.x
+ e=> (a^2-c).x^2 + (2.a.b-d).x + (b^2-e) = 0-- Unpatched IE
vulnerability: mhtml wecerr CAB §ipDescription: Delivery and
installation of an executableReference:
===
Subject: Three-Dimensional Delta Functioni need help to deine
delta(g(x,y,z))=?in the form of
<>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$-
GiZp>UN2a}!Fmw+%h}Y<?5]3mj~`n8?0wycf-nf(r8SAdWK`G=JC[<3fz48E[v
{Ns!r]MT;JPgLG7|pBA7=lP1oGgUt^>L`!h_XXr5Q>_nGsY2_This (real or
complex) matrix has no square-root: 0 1 0 0In general, to do a
square-root for a complex matrix, use the Jordannormal form,
and then see whether the blocks have square-roots.-- G. A.
(statistics)how to make date more like Laplacian
distribution should follow the shape> of Laplacian
distribution... the data obtained from measurement is of
course> a little off(not very symmtrical), how can I make the
measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?Can anybody give me an example
or detailed explanation? I am kind of afraid> of statistics...
measured. If they dont suit you, do as corporations and
government agencies do: lie.Jerry-- Engineering is the art of
making what you want from things you can
definitness of quadratic formsIf q is a quadratic form on R^n
with a symmetric matrix Q =(q_{jk})_{j,k=0}^n let Q_m =
(q_{jk})_{j,k=0}^m then q is positivedefinite if and only if
define an action> SO(n+1) x S^n ----> S^n when x is the
vector x0=(1,0,0,...), the stabilizer is the set of matrices
in SO(n+1) which take x0 to x0 (of course) and thus consists
of matrices whose first column is (1,0,0,...). The rest of 
the
first row is then all zeros too, but the remaining n-by-n
matrix can be any special orthogonal group. So this
stabilizer is in a natural way isomorphic to SO(n).Why is
this true? Wouldnt the stabilizer in this situation be in a
natural>way isomorphic to a group larger than SO(n)? In
particular, you can fit>an orthogonal matrix in the remaining 
n
by n matrix,,,so couldnt by>the same reasoning the 
stabilizer
be naturally isomorphic to O(n)?You wanted the stabilizer _in
SO(n+1)_, didnt you? So sure, you cantake any block matrix 
of
the form (1) + M, but you need for this tobe (a) orthogonal
(requiring M to be orthogonal) and (b) of determinantequal to
1 (requiring det(M)=1). But yes, this trick also helps you
identify the stabilizer of a pointunder any other group
action, too; the question is just to determinethe possible
values of M which keep (1) + M in the original
action> SO(n+1) x S^n ----> S^n  when x is the vector
x0=(1,0,0,...), the stabilizer is the set of matricesin
SO(n+1) which take x0 to x0 (of course) and thus consists of
matrices whose first column is (1,0,0,...). The rest of the
first row is then all zeros too, but the remaining n-by-n
matrix can be any special orthogonal group. So this
stabilizer is in a natural way isomorphic to SO(n).>Why is
this true? Wouldnt the stabilizer in this situation be in
anatural>way isomorphic to a group larger than SO(n)? In
particular, you can fit>an orthogonal matrix in the remaining 
n
by n matrix,,,so couldnt by>the same reasoning the 
stabilizer
be naturally isomorphic to O(n)? You wanted the stabilizer _in
SO(n+1)_, didnt you? So sure, you can> take any block 
matrix
of the form (1) + M, but you need for this to> be (a)
orthogonal (requiring M to be orthogonal) and (b) of
determinant> equal to 1 (requiring det(M)=1). But yes, this
trick also helps you identify the stabilizer of a point> under
any other group action, too; the question is just to determine>
the possible values of M which keep (1) + M in the original
coefficient of variation can only be used ifthe random 
variable
is positive. I would like to know why cannot oneextend the use
of the same coefficient with not necessarily positiverandom
variables but with positive mean. Could somebody here
Number of Subgroups of S_10> What is the number of subgroups
of the symmetric group S_10 on ten> elements? Can someone
prime factorization of 10! = 2^8 * 3^4 * 5^2 * 7^1You can do
-topology~~~Originator: grubb@lola>U is a neighborhood(=nbd)
of x><=> U is an open set containing x by Munkres book>I like
this definition of nhood. To make clear that I consider nhoods
to>be open I often use the expression open nhood.I *dont*
like this definition. It adds no new information andadds new
terminology.>but <=> U is an super set of open set G
containing x by Lipschutz>book>This definition I 
dont like. It
has little use beyond the open>nhood definition while adding
some complexity to usage of nhoods.This is my prefered
terminology. I like to be able to talk aboutcompact
neighborhoods and connected neighborhoods, etc. It
adds§exibility and ease to exposition, which is a good
anyone explain how would I graph absolute value of y plus
absolute> value of x equal to 2. what kinda graph would it be
and how would I> get there. plz help. got stuck. I know Im 
a
dumbass, but help plz.If you know whether x or y is positive
or negative you can remove theabsolute value sign so ...There
are four cases: (1) x>=0, y>=0: The equation becomes x + y = 2
(2) x<=0, y>=0: -x + y = 2 (3) x<=0, y<=0: -x + -y = 2 (4)
x>=0, y<=0: x + -y = 2Now graph it in each
the number of subgroups of the symmetric group S_10 on ten
>elements?According to Magma, there are 1593 conjugacy classes
of subgroups, and29594446 subgroups altogether.>Can someone
provide a reference?I dont know - I 
cant.>Can GAP do it?I
expect so - I am just trying that as a check, but it is still
question (oops)  What role does the Jacobson radical>
ab = a+b - ab> That is not the Jacobson radical.> The
Jacobson radical of a ring is the intersection> of it proper
left (or right) ideals. You need some conditions on the
left ideals that you> are taking the intersection of. Doh!
I meant maximal left ideals ...> Just maximal left ideals is
not enough. Take the nilpotent algebra> A = R^2 with
multiplication (x,y)(a,b) = (0,xa). A has only threeideals:> A
, {0}xR, and {(0,0)}. But in this case the Jacobson radical is
the> whole algebra. Thats not a ring: they alway have 1
solve a system of quadratic equations?> I ran into a problem
where I would need to solve for a system(specifically> 2) of
quadratic equations. There is a method of Newton for finding
the zeros of> arbitrary functions. In one dimension, you draw
the tangent> line at your current value of x, then go to where
that> tangent line intersects the x-axis. This idea is easily
extended to functions of multiple> variables and to
simultaneous nonlinear equations> like yours: make a linear
approximation near your> current value of x, then solve that
linear system> for your next value (or at least for the
direction> to move).Hmm. As far as I know there is no
Convergence proved for the case ofarbitrary functions. Is this
different in my case?What about retrieving all roots? 
Doesnt
Newton converge (if it does) tojust one root? Heres one
discussion:>
you, I will try this one. Im not clear if you want the 
theory
or just pre-canned> software. Both are widely available on the
web.Both of course ;-)In the meantime I am more interested in
the theory, because I have solved myproblem at hand because I
made succesful use of additional properties of myequations.
But it might be the case, that I will need to solve for
othersimilar problems, where these specific structure is not
available. I justwant to be prepared.>Do> a search for
nonlinear simultaneous equations. For> software, Netlib is a
lot before posting to this group.What I was thinking of was
something like a theory about quadraticsimultaneous systems
extending of what I know about linear systems. Also Iwould
need some text that can be read by a technician who does not
There are many ways to prevent the angle changing so> that it
remains FIXED. We can insert V-shaped rigid rod inside these>
springs so that angle V does not change. Or we may put these
springs> inside hollow pipe. We can use shock-ups. That is the
best solution.Consider putting a rigid rod inside the springs.
Will the springs ever touch the rigid rods? Of course they
must, as otherwise you wouldnt need them. Do they press
against the rods? Of course they must, othewise you wouldnt
need them. Is this pressing a force? Of course it is. Have you
forgotten to take into account that in order to keep the angle
fixed you will be introducing additional forces that changethe
setup that you currently have. Of course you have.Dont be a
fool. Listen to what people are trying to teach you, and
learn. You seem utterly clue-resistant, to be honest. I reckon
teh best place for yous the killfile...-- 
Unpatched IE
vulnerability: Click hijackingDescription: Pointing IE mouse
events at non-IE/system windowsReference:
http://safecenter.net/liudieyu/HijackClick/
HijackClick-Content.HTMExploit:
http://safecenter.net/liudieyu/HijackClick/HijackClick2-
starting to play a little with MS-Excel for Math purposes.
Doesanyone know of a way to calculate several values for a
given function, i.e.Cos(1, pi/2, pi/3, ...n) ? And, can a cell
reference be given instead ofputting in the values? Such as,
Cos(A1:A10). (I have also cross-posted totwo Excel
hello.......i saw.......U is a neighborhood(=nbd) of x<=> U is
an open set containing x by Munkres bookbut<=> U is an super
set of open set G containing x by Lipschutz
book---------------------------------Define f: X->Ylet
X={1,2,3} Tx={0(=empty),X,{1},{1,3}.{3}}let Y={a,b,c}
Ty={0,Y,{b}}let f(1)=b ,f(2)=b , f(3)=aif nbd=open set(by
Munkres), f is discontinuous on 1because f(1)=bf^-1{b}={1,2} :
not in Tx....thus not open> if nbd=super set of open set(by
Lipschutz), f is continuous on 1because f^-1{all nbd
containing b} is nbd of
1---------------------------------whats wrong??which of 
case
is right??um.......advice....please....sir.....thank
you....The problem is with your definition of is continuous
at.It seems to be confused with the definition of is
continuous.A function can be continuous at a point without
being continuous.If f is not continuous then the f inverse
image of an (open) neightborhoodwont always be an (open)
neighborhood.Since continuous at does not require continuityit
needs to be defined in a way that does not rely on
continuity,i.e., without expecting the inverse image of an
open set to be open.continuity at x is just the generic
version of what is taught in calculus:for all e, there is d,
for all x (|x - x| < d -> |f(x) - 
f(x)| < e)this translates
to for every neighborhood V of y=f(x) ({y: 
|y-y| < e})there
is a neighborhood U of x ({x: |x-x| < 
d})such that V
contains the image of U (|f(x) - f(x)| < e)You may dictate
that U and/or V must be open neighborhoods in the
definition,but the definition cannot require the 
inverse image
of V to be a neighborhood of X. It can only requre X to
contain a neighborhhod that gets mapped into U.By the way,
your function is not discontinuous at 1. It is discontinuous
FunctionOriginator: grubb@lola>i need help to deine
delta(g(x,y,z))=?>in the form of
these types of things. Usually,g(x,y,z) is a real valued
function of three variables. Thenthe meaning of
delta(g(x,y,z)) is the surface area measureon the surface
g(x,y,z)=0. This cannot be written as a sumof the form you
want.If however, g(x,y,z) is a *vector* valued function of
threevariables (with three components), you can
interpretdelta(g(x,y,z)) as the sum of delta(xi) times the
inverse ofthe Jacobian of g at xi Jacobian where the xi are
thevector solutions of g(xi)=0. If your function does nothave
finitely many such points xi, the you are potentiallyin
trouble. If the Jacobian is 0 at those points, there is
alsodifficulty.The second version is obtained by considering
the change ofvariables formula, just like it is in one
is, can I mod out SO(n+1) by a subgroup to get SO(n)?> SO(n)
is not generally a quotient of SO(n+1). Since no one else
seems to be saying it:Youre not going to get any 
interesting
quotients at all from orthogonal groups G. There is a small
abelian quotient group G/G. There is a small normal abelian
subgroup Z(G). The remaining 
chunkG/Z(G) is simple in most
cases, including the orthogonalgroups of (real) Euclidean
spaces, as well as the orthogonalgroups of nondegenerate
quadratic forms over finite fields with 
theexception of some
cases of 3- and 4-dimensional inner-product spaces.Theres a
very nice geometrical discussion of this in JacobsonsBasic
Loculus of Archimedes has 536 solutions. You can >maa.org. 
http://www.maa.org/news/mathgames.html>Over at MathWorld,
you can read a column about a solution>for the order-5 perfect
magic cube.>http://mathworld.wolfram.com/>--Ed Pegg Jr,
http://www.mathpuzzle.com> Youre certain ther were no other
puzzlles before this one?> How do you know that?The Italian
police still havent solved the puzzle of who killed .85tzi
the Iceman, for example! But thats equivocating.-- 
Unpatched
IE vulnerability: Extended HTML Form AttackDescription: Cross
Site Scripting through non-HTTP ports, stealing cookies,
etc.Published: February 6th 2002Reference:
http://eyeonsecurity.org/advisories/
Gary - good stuff. Reformulating quantum field theory and
string theory in the wavelet transform generalization of the
Fourier transform is important. Note how complex spacetime
comes in. Possibly hypercomplex non-commutative spacetime
beyond that.http://www.arxiv.org/abs/math-ph/0303027Authors:
Gerald Kaiserof Physics A: Mathematical and General, this http
URLSubj-class: Mathematical Physics; Complex Variables;
Analysis of PDEsFor the first time, complete source
distributions for the emission andabsorption of acoustic and
electromagnetic wavelets are defined andcomputed, both in
spacetime and Fourier space. The biggest surprise is thegreat
simplicity of the Fourier sources as compared to the rather
convolutedspacetime expressions obtained from the original
wavelets. This suggeststhat the associated pulsed-beam
propagators may play a fundamental role inemission and
absorption processes including focus or directivity. It
alsoopens the way to constructing FFT-based algorithms for
pulsed-beam analysesof acoustic and electromagnetic
waves.Electromagnetic Wavelets as Hertzian Pulsed Beams in
Complex
Spacetimehttp://www.arxiv.org/abs/gr-qc/0209031Authors: Gerald
KaiserComments: 16 pages, 2 figures, Topics in Mathematical
Physics, GeneralRelativity and Cosmology conference (in honor
of Jerzy Plebanski) <thishttp URL>Subj-class: General
Relativity and Quantum Cosmology; Mathematical Physics;Complex
VariablesElectromagnetic wavelets are a family of 3x3 matrix
fields W_z(x)parameterized by complex spacetime 
points z=x+iy
with y timelike. They aretranslates of a sl basic rm wavelet
W(z) holomorphic in thefuture-oriented union T of the forward
and backward tubes. Applied to acomplex polarization vector p
(representing electric and magnetic dipolemoments), W(z) gives
an anti-selfdual solution W(z)p of Maxwells 
equationsderived
from a selfdual Hertz potential Z(z)=-iS(z)p, where S is the
slSynge function rm acting as a Whittaker-like scalar Hertz
potential.Resolutions of unity exist giving representations of
sourcelesselectromagnetic fields as superpositions of 
wavelets.
With the choice of abranch cut, S(z) splits into a difference
of retarded and advanced slpulsed beams rm whose limits as yto
0 give the propagators of the waveequation. This yields a
similar splitting of the wavelets and leads to theircomplete
physical interpretation as EM pulsed beams absorbed and
emitted bya sl disk source rm D(y) representing the branch
cut. The choice of ydetermines the beams orientation,
collimation and duration, giving beams assharp and pulses as
short as desired. The sources are computed as
spacetimedistributions of electric and magnetic dipoles
supported on D(y). Thewavelet representation of sourceless
electromagnetic fields now splits intorepresentations with
advanced and retarded sources. These representationsare the
electromagnetic counterpart of relativistic
coherent-staterepresentations previously derived for massive
Klein-Gordon and DiracNon-linear Vacuum Phenomena in
Non-commutative
QEDhttp://www.arxiv.org/abs/hep-th/0006209Authors: L.
Alvarez-Gaume, J.L.F. BarbonComments: LaTeX, 23 ppReport-no:
CERN-TH/2000-181Journal-ref: Int.J.Mod.Phys. A16 (2001)
1123-1146We show that the classic results of Schwinger on the
exact propagation ofbe readily extended to the case of
non-commutative QED. It is shown thatnon-perturbative effects
on constant backgrounds are the same as theircommutative
counterparts, provided the on-shell gauge invariant dynamics
isreferred to a non-perturbatively related space-time frame.
For the case ofthe plane wave background, we find evidence of
the effective extended naturescattering. Besides the known
`dipolar character of non-commutative neutralthey behave
-topology~~~> No, Im afraid he actually does (have just
double checked in my copy of> his book). In my opinion its 
a
very bad definition to use, but for some> reason 
Im unable to
fathom a lot of people *do* use it.Woa ... I leave this
trying to read Conant and Vogtmanns on a theorem of
Kontsevich.Now if you have two spiders S and T (as defined in
the paper) there isa mating operation o which requires the
specification of one leg,y, of S and one leg, x, of T giving a
new spider (S,y)o(T,x). Furtherdevelopment of the paper
indicates that this operation should becommutative i.e. we
should have that (T,x)o(y,S) = (S,y)o(x,T).However, for simple
cases involving the associative operad thisdoesnt seem to 
be
the case. Does anyone know if the mating operationas given in
Worlds oldest puzzle solved||> |>The Loculus of Archimedes 
has
536 solutions. You can |>maa.org.
|>|>http://www.maa.org/news/mathgames.html|>|>Over at
MathWorld, you can read a column about a solution|>for the
order-5 perfect magic
cube.|>|>http://mathworld.wolfram.com/|>|>--Ed Pegg Jr,
http://www.mathpuzzle.com|> |> |> Youre certain ther were 
no
other puzzlles before this one?|> How do you know that?||The
Italian police still havent solved the puzzle of who 
|killed
.85tzi the Iceman, for example! But thats equivocating.if i
know the italian police they were probably in on it.-- [e-mail
are some topology questions I was wondering about.Assume X is a
topological space and A is a subset of X.What is int(int(A))?
Is it the same as int(A)? My intuition tells me itis, but how
to prove it?This would then mean that int(ext(A)) is ext(A),
ext(int(A)) is ext(A),and ext(ext(A)) is int(A).bdry(int(A))
need not be the same as bdry(A). Example: considersingleton
sets in a Hausdorff space. But then, in some
cases,bdry(int((A)) is indeed the same as bdry(A).Likewise
bdry(ext(A)) might or might not be the same as bdry(A).But
what has got me stumped is what int(bdry(A)), ext(bdry(A))
andbdry(bdry(A)) are. My intuition tells me that the first is
empty,the second is the same as bdry(A), and the third is
Xbdry(A), but Icant figure out how to prove 
it.Please help me,
------------- Finland ----------
http://www.helsinki.fi/~palaste --------------------- rules!
--------/You could take his life and... - Mirja
===
Robertson versus ips screwdriver Re:Subject: Re: what spin 1/2
is in physical reality Re: Robertsonversus ips screwdriver Re:
Archimedes Plutonium <a_plutonium@dtgnet.com> NOdtgEMAIL whole
entire Universe is just one big atom where dots of the
electron-dot-cloud are galaxies sci.physics, sci.math,
sci.physics.electromagAnd if this ipsheadscrewdriver (btw, I
am beginning to like thehabit ofjoining words as if it was one
word instead of the hyphen and the screwdriver should be one
word).Anyway, I would need to have to explain spin up and spin
down and why2electrons fill an orbital. So that the 
geometrical
explanation answershowa spin up and spin down fill an
orbital.So if the interior of an electron has two diametrical
forces then whyshould4 diametrical forces of the interiors of
2 electrons accomodate one another in a filled orbital??The
best I can think of at this moment in time is that the Maxwell
Equations are 4 entities. And that EM is satisfied with 4
entities.Andso, if we can transfer each of the Maxwell
Equations as a force inand of itself and thus say that the
interior of 1 electron has only 2ofthe Maxwell Equations Force
and another placed in close proximity have the other 2 Maxwell
Equations Force would be inclusive ofall EM forces.I do not
know if anyone has made such a transfer where they take the 4
Maxwell Equations and replace them as Maxwell Forces. We are
familar with StrongNuclear, WeakNuclear, Gravity and Coulomb
forces but let us also add Antigravity and those make 5
conventional forces. But let us combine those forces so that
they comprise 4, each corresponding to one of the Maxwell
Equations. Whether any physicist or mathematician has done
such an enterprise of locking together the forces of physics
as Maxwell Equations. Obviously the Coulomb force is already a
Maxwell Eq in the Gauss law. But let us say the other 3 Maxwell
Equations bundled up the StrongNuclear, WeakNuclear, gravityand
antigravity.And since the Maxwell Equations offer a complete EM
description, or atleast hope to do such, then all I need to do
is say that the 2rectangularsolids inside each electron
ellipsoid which would be 4 rectangularsolidsinside 2 electron
ellipsoids fill an orbital. (Keep in mind eachrectangularsolid
is the analogy of one of the crosses in
aipsheadscrewdriver).Truthfully, I do not like the above, for
the picture is not comingclearerbut more messy and muddy.
Perhaps the screwdriverhead analogy is kaput at about here.
But I do like the Maxwell Equation thought path because it is
a limit in that 4 things seem to describe all of EM. Andwith a
Coulomb Unification of Forces, then a 4 limit would make sense
as to why spin-up and spin-down of 2 electrons fills an
orbital.Experiment: if the above is correct in part or whole,
then anexperimentshould reveal that no atomic orbital is
spherical but that all of themareellipsoid or distorted
spheres because the interiors ofelectrons that fill an orbital
are neverperfectly balanced by those 4 interior
forces.Archimedes Plutonium, a_plutonium@hotmail.comwhole
entire Universe is just one big atom where dotsof the
help with algebra> can anyone explain how would I graph
absolute value of y plus absolute> value of x equal to 2. what
kinda graph would it be and how would I> get there. plz help.
got stuck. I know Im a dumbass, but help plz.Since
replacement of x with -x or y with -y results in the
sameequation, the graph is symmetric with respect to the x-
and y- axes.So, it suffices to plot in the upper right
quadrant. There, theequation becomes x + y = 2. You should be
difficult for some people?> I am a science teacher but I have
taught math for a few years. I too> agree tht it could be
possible that people have a math gene. Through> out my school
years I have always struggled with math, while some> students
just seemed to get it right off. No matter how hard I> studied
it was a constant battle. While teaching math to my students I
see the same types of behaviors I> think this makes me a
better math teacher cause I know what it feels> like not to
understand, which in turn makes me more patient with my>
students. I think the book is worth reading and considering.
However> you have to be careful when sharing this information
with your> students cause it could turn into a self pity
issue.In fact, there is a book by that title. Listed on
Amazon.com (search Bookswith the words math gene.) I read it a
few years ago, and found itinteresting. The same part of the
brain involved in higher math (not basicarithmatic) does
(MultiVariable) class>Everyone;>Im trying to 
find sources for
reasonably priced, good math courses>online. Im 
specifically
looking for Calculus III and beyond. Im not>interested in 
any
of the pseudo schools that want your money to sell>you a
degree, I want the information.. Im limited from pursuing
this>in a formal school environment due to disabilty although
over the>years I did manage to struggle through the first of
the Physics>sequence and Calculus II. Is there an instructor
out there that might consider a self-study with>me? I am slow
so a self-paced program would work best. A directed>self-study
free be reasonably enough priced? I just retired in Mayfrom
ASU where I taught a lot of Calc III in recent years. I will
giveyou:1. A copy of my last syllabus.2. Copies of my lecture
notes in Adobe Acrobat pdf format. These arehand written
lectures that I gave.3. Problem assignments (from Stewart 4th
Edition for which you canpick up the 3rd semester break-out
version cheap because they are nowin edition 5, although there
is nothing wrong with Edition 4)4. When you are ready, copies
of the exams I gave plus solutions.If you are interested, post
compact Riemann SurfacesLet X be a compact Riemann
Surface.Suppose that x_1, ..., x_n in X, and let Y = X {x_1,
..., x_n}.Now let f: Y --> Y be biholomorphic, i.e. f is an
automorphism of Y.I believe f can then be extended to an
automorphism of X.But how do I prove this?Any suggestions are
topology questions I was wondering about.> Assume X is a
topological space and A is a subset of X.What is int(int(A))?
Is it the same as int(A)? My intuition tells me it> is, but
how to prove it?int = interior?If U is open, prove that int(U)
= U.Also, prove that int(A) is open.> This would then mean that
int(ext(A)) is ext(A), ext(int(A)) is ext(A),> and ext(ext(A))
is int(A).> bdry(int(A)) need not be the same as bdry(A).
Example: consider> singleton sets in a Hausdorff space. But
then, in some cases,> bdry(int((A)) is indeed the same as
bdry(A).> Likewise bdry(ext(A)) might or might not be the same
as bdry(A).But what has got me stumped is what int(bdry(A)),
ext(bdry(A)) and> bdry(bdry(A)) are. My intuition tells me
that the first is empty,> the second is the same as bdry(A),
and the third is Xbdry(A), but I> cant figure 
out how to prove
with the usual topology, A isthe rationals. Compute
interesting askthere is a fact , surely trivial. Not for
me.May you give me a reason for which two any fibers of a 
fiber
questionsA N Niel <qt4yn2oc02@sneakemail.com> scribbled the
following: But what has got me stumped is what int(bdry(A)),
ext(bdry(A)) and bdry(bdry(A)) are. My intuition tells me
that the first is empty, the second is the same as bdry(A),
and the third is Xbdry(A), but I cant figure 
out how to
X is the reals with the usual topology, A is> the rationals.
Compute int(bdry(A)), ext(bdry(A)), bdry(bdry(A)).Hmm... in
that case, A is both dense and codense, so it has an
emptyinterior and an empty exterior, so that makes bdry(A)=X.
This meansthat int(bdry(A))=X, and
ext(bdry(A))=bdry(bdry(A))=empty. Is thisright?-- /-- Joona
Palaste (palaste@cc.helsinki.fi) ------------- Finland
---------- http://www.helsinki.fi/~palaste
--------------------- rules! --------/All that §ower power is
Re: how to solve a system of quadratic equations?>I ran into a
problem where I would need to solve for a system
(specifically>2) of quadratic equations. My search on the net
lead me to systems of>polynomial equations. Yes indeed,
because the two problems are equivalent: ANY system
ofpolynomial equations can be transformed into a set of
quadratic equations(hint: x^3 can be written x y if we adjoin
another equation y=x^2).And the process of solving general
systems of polynomial equations iscomplicated. In principle
most such systems can be triangularizedso that the system is
rewritten as a system something like this: P1(x1)=0,
P2(x1,x2)=0, P3(x1,x2,x3)=0, ...for some polynomials P_i of i
variables. Then you solve first forx1, plug in and solve for
x2, etc. Unfortunately it is computationallyvery difficult to
transform the original to such a system, and thefinal target
could well be something much messier than this
hoped-fortriangularized system.But if youve only got _two_
_quadratic_ equations, life is not so complicated.Eliminate
one variable from the set and youll be reduced to one
low-degreeequation in n-1 unknowns. Now, what do you mean by
solve in this case?If n=2, the solution set is probably finite
and you can find the solutions easily. If n>2, the solution 
set
is probably infinite; whatkind of solution are you looking
exist any method to factorize a square matrix M such that
threads from sci.mathwhose subject contains square root of a
r4HZ74ZEwebVAs+Ecg-9Xq+VpGQEtEi6vfehl5iPXbJk4tjsWc0xkkThis
(real or complex) matrix has no square-root: 0 1> 0 0In
general, to do a square-root for a complex matrix, use the
Jordan> normal form, and then see whether the blocks have
square-roots.> Additionally, if you *know* (or can assume)
that a root exists,and your matrix has positive eigenvalues,
then you also can simply apply the newton-algorithm, which may
not be efficient, but straight-forward: Z -- your matrix
(positive eigenvalues) M0 -- unit-matrix of same dimension M1
= 0.5* ( Z * inv(M0) + M0) M2 = 0.5* ( Z * inv(M1) + M1) ...
M_x * M_x = Z (within numerically roundoff-errors)of this
Calculus III (MultiVariable) class> Everyone;> Im trying to
find sources for reasonably priced, good math courses> online.
Im specifically looking for Calculus III and 
beyond. Im not>
interested in any of the pseudo schools that want your money
to sell> you a degree, I want the information.. Im limited
from pursuing this> in a formal school environment due to
disabilty although over the> years I did manage to struggle
through the first of the Physics> sequence and Calculus II. Is
there an instructor out there that might consider a self-study
with> me? I am slow so a self-paced program would work best. A
you,> JoYou may find a portion of what you want by
visitinghttp://www.math.temple.edu and selecting Calculus on
grubb@lola>These are some topology questions I was wondering
about.>Assume X is a topological space and A is a subset of
X.>What is int(int(A))? Is it the same as int(A)? My intuition
tells me it>is, but how to prove it?Yes, it is. The proof
depends on the particular definitions youare using. Different
books/authors use different definitions, allof which are
equivalent.>This would then mean that int(ext(A)) is ext(A),
ext(int(A)) is ext(A),>and ext(ext(A)) is int(A).What
definition of ext(A) do you use? If you mean int(XA), then
thefirst is correct, but the other two are not.>bdry(int(A))
need not be the same as bdry(A). Example: consider>singleton
sets in a Hausdorff space. But then, in some
cases,>bdry(int((A)) is indeed the same as bdry(A).>Likewise
bdry(ext(A)) might or might not be the same as bdry(A).>But
what has got me stumped is what int(bdry(A)), ext(bdry(A))
and>bdry(bdry(A)) are. My intuition tells me that the first is
empty,>the second is the same as bdry(A), and the third is
Xbdry(A), but I>cant figure out how to prove 
it.Probably you
are wrong on all accounts. Consider X=reals, A=rationals.What
of same length and> stiffness and both springs are in
relaxed state initially. Angle ABC> is solid
angle.Apparently this means you have some rigid structure>
I was wondering where you are. About 39N latitude, 75 W
longitude. Why is that relevant?> I admit, I should have used
the word FIXED angle rather than SOLID> angle. There are many
ways to do this. Take two shock-ups like those> attached to
front wheel of my bike and connect one end of these two>
shock-ups so that they form V shaped structure. Now this angle
V> remains FIXED even if we pull free ends of shock-ups in
direction of> arms of this V shaped structure.The angle V does
not change throughout operation of this device. This> is what I
am trying to say.Therefore you have static equilibrium. Nothing
askOriginator: grubb@lola>there is a fact , surely trivial.
Not for me.>May you give me a reason for which two any fibers
of a fiber bundle are>disjoint?they are inverse images of
questionsOriginator: grubb@lola Try some examples. Say X is
the reals with the usual topology, A is the rationals.
Compute int(bdry(A)), ext(bdry(A)), bdry(bdry(A)).>Hmm... in
that case, A is both dense and codense, so it has an
empty>interior and an empty exterior, so that makes bdry(A)=X.
This means>that int(bdry(A))=X, and
ext(bdry(A))=bdry(bdry(A))=empty. Is this>right?Looks
one(s) ??... your simple explanation is well,that ïobjects in
teh background can seem to be overlappedby objects in hte
foreground -- quick thinking! > simple and expected
geometrical explanation. By the way, most of Arps> peculiar
galaxies really are interacting systems where the constituent>
members have the same redshift.The problem I have with that is
that in looking at actual pictures you> can see material
curling from one galaxy to another that is clearly> pulling
the dumb crap in journals§outs (sp.?) ?? > §aunt: to wave 
in
the wind: to move ostentatiously:--les ducs
digits> In the code for the Maple procedure quadres that I
previously posted there> was a procedure legendre_pow that is
not normally accessible. However,> Robert Israel and Alec
Mihailovs showed me two ways to get it. I have> attached the
method by Alec to the end of this message. --Edwin Heres 
the
Maple code for quadres:> showstat(numtheory:-quadres);>
numtheory:-quadres := proc(a, p)> local fac, ig, n, pf, s;> 1
if nargs <> 2 then> 2 error invalid arguments> end if;> 3 n :=
modp(a,p);> 4 if type(n,integer) and issqr(n) then> 5 return 1>
end if;> 6 if not (type(n,integer) and type(p,integer)) then> 7
return ïprocname(args)> end if;> 8 if p < 6 
then> 9 return -1>
end if;> 10 if isprime(p) then> 11 return
mods((ïpower)(n,1/2*p-1/2),p)> end if;> 12 ig 
:= igcd(n,p);>
13 if ig <> 1 and igcd(denom(n/ig^2),p/ig) = 1 then> 14 return
numtheory:-quadres(n/ig^2,p/ig)> end if;> 15 pf := ifactor(p);>
16 if type(pf,`^`) then> 17 return legendre_pow(n,pf)> end if;>
18 for fac in pf do> 19 if type(fac,`^`) then> 20 s :=
legendre_pow(n,fac)> else> 21 s :=
numtheory:-quadres(n,op(fac))> end if;> 22 if s = -1 then> 23
return -1> end if> end do;> 24 1> end proc> Heres the code
for legendre_pow following Alecs
method:kernelopts(opaquemodules=false):>
showstat(numtheory:-legendre_pow);numtheory:-legendre_pow :=
proc(nn, pow)> local b, n, r;> 1 r := op(2,pow);> 2 b :=
op(op(1,pow));> 3 n := modp(nn,b^r);> 4 if n <> 0 then> 5
while igcd(n,b) <> 1 do> 6 if not type(n/b^2,integer) then> 7
return -1> else> 8 n := n/b^2> end if> end do> end if;> 9 if
has({0, 1, 4, 9, 16},n) then> 10 return 1> end if;> 11 if b =
2 then> 12 if modp(n,2^min(3,r)) = 1 then> 13 return 1> else>
14 return -1> end if> end if;> 15 quadres(n,b)> end
procDoesnt H. Cohns book on 
algebraic numbers discuss the
quadres function?Also, wasnt there a Putnam problem posed
some years ago which askedcontestants to prove that 1444 is
the largest square ending in 3 identicalnonzero digits?In this
same vein, I would like to pose the following question, which
is aspecial case of a problem posed here some years ago:Does
there exist a square (in base 10) satisfying the following 2
conditions?1). Its last 4 digits are 1001.2). All its digits
Halton Arpthats a great explanation; first I 
read of it! maybe
I should say, scanned, sinceI didnt actually pick a book up
and move pages;ever see the movie, Scanners?its about a
select group of graduatesfrom the Evelyn Wood Reading Dynamics
course. >
http://perso.wanadoo.fr/lempel/red_shift_NGC_7603_uk.htm--les
and logic> B:> Pertti Lounesto> - invented quadratic
nature of trialityYou should compare the structure of his
octonions
http://redquimica.pquim.unam.mx/clifford_algebras/v11/aaca112/
loun112.pdfwith truth tables as switching functions (p
193).There are 16 truth-functional connectives comprised of 8
complementary pairs. The octonions are a numbersystem with 8
dimensions.Of the 8 complementary pairs of connectives, 7 are
pairs of threshold functions. Lounesto describes theoctonians
as a product of the form,O=RxR^7In *this room* (sci.logic in
Greene-ese) we are supposed to understand the identity
relation because wehave a paradigm that requires 3 identity
relations. In discussing the structure of the octonions,Thus O
is spanned by (1 e R) and the 7 imaginaryunits i, j, k, l, il,
jl, kl, each with square -1, so thatO=RxR^7. Among the subsets
of 3 imaginary unitsthere are 7 triplets, which associate and
span theimaginary part of the quaternionic subalgebra.
Theremaining 28 triplets anti-associate.The multiplication
table of the unit octonions canbe summarized by the Fano
plane, the smallestprojective plane, consisting of 7 points
and 7 lines,with orientations. The 7 oriented lines
correspondto the 7 quaternionic/associative triplets.I find it
curious that people who spent 100 years reducing everything to
completeness ended up with 3identity predicates in order to
have a competent paradigm. Well, then again, perhaps not. The
completeconnectives (NAND and NOR) are among the threshold
connectives that would be oriented with respect tothe triplets
of the projective plane.Of course, now that I have 
Lounestos
paper in front of me, I realize why no one understood
thecomparisons I have been making all of this time. *I* was
trying to explain things in terms of the affinegeometry dual 
to
the Fano plane. That is the plane whose representation in a
Euclidean space consists ofpoints labeled with (0,0), (0,1),
(1,0), (1,1).Of the 28 triples that anti-associate, you will
note that 28 is 7 times 4. The number 4 is of importbecause of
the fact,If one has available say 16 alternative messagesamong
which he is equally free to choose, thensince 16=2^4 so that
log_2(16)=4, one says thatthis situation is characterized by 4
bits of information.conveniently discussed by Warren Weaver in
his introduction to Shannons The Mathematical Theory
ofCommunication. One of the crucial issues is how you are
going to refer to those 4 bits. You cancertainly use the 
affine
labeling--(0,0), (0,1), (1,0), (1,1)--to get a geometry that is
directly relatedto truth tables. Or, you can use the quaternion
units--1, i, j, k--and apply Lounestos construction toget 
the
dual Fano plane.the sense of one of Weavers other
statements:One has the vague feeling that information
andmeaning may prove to be something like a pair ofcanonically
conjugate variables in quantum theory,they being subject to
some joint restriction thatcondemns a person to the sacrifice
of the oneas he insists on having much of the other.I have
been asking people on internet newsgroups to think about
understanding the foundations ofmathematics in terms of purely
dual geometric forms for a year now. But, that is kind of hard
when peoplehave investments in something that is neither
information nor meaning--namely knowledge.As for the one
dimension that Lounesto *does not* refer to as imaginary (that
is, the truth functionalconnectives not corresponding with
threshold functions), I will once again recommend that readers
take alook at the paper,
http://citeseer.nj.nec.com/feigelson97forbidden.htmlIf I
recall correctly, the statement of Theorem 4 has a footnote
about logical equivalence and
exclusivedisjunction.:-)mitchP.S. In fights over the
foundations of mathematics, it is usually wise to actually
unbounded?by lHospitallim(n->oo) (tan n)/n =You 
cant
apply LHopitals rule here. Find a calculus 
book:> there are
_hypotheses_ that need to be satisfied...Good example for the
next LHopitals rule: child of Satan?> 
thread,
though.Heh-heh-heh!-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
Group Lie Algebra> You [Jack Sarfatti] seem to be writing to
yourself a lot these days.> Wonder what that means?Business as
usual?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
solving linear systems...I have a system:A X = RI want to
solve for X with a least-square 
approach:X=inv(A*A)*A*RBut
how can I deal with inequality constraints (such as xi>0)I
know that for equality constraint it goes like thatfor C = XDX
=
inv(A.*A)*A.*Rinv(A.[Capita
lOTilde]*A)*C.*inv(C*inv(A.*A)*C.[CapitalOTi
lde])*(C*inv(A
.*A)*A.*R - D);but how about 
inequality ?EuhStart to fade
graph> Its a general question. Just help on selecting the
right graph for the> right kind of data involved. Thats as
simple as I can make it.Way too general a question to answer
here. Best I
canhttp://www.jwolsen.com/wtip071.htmhttp://
www.essentialitskills.com/
Decision_Support_Tools_web_document.PDFhttp://ontrack.ncsu.edu
/Materials/Cont_discrete_variables.dochttp://
www.sciencebuddies.org/mentoring/project_data_
analysis.shtmlhttp://www.math.yorku.ca/SCS/Gallery/Usually
there is either some point youre trying to make,or perhaps
you dont know what story the data is tellingyou and you are
looking for the data to tell you. Inthat case you may go
through several different formatsbefore you find the one that
by Rationals>For positive x, define g(x) = inf {c: n x - 
§oor(n
x) > n^-c for >all positive integers n}. True or false: g^(-1)
(0) has Lebesgue >measure 0?Looks to me like g(x) = infinity
(i.e. there are no such c). Namely,try n=1.But even if you
change all positive integers n to all integers n >= Nfor some
fixed N, youll have g(x) >= log_N(1/frac(Nx)) > 
0 sog^(-1)(0)
just starting to play a little with MS-Excel for Math
purposes. Does> anyone know of a way to calculate several
values for a given function, i.e.> Cos(1, pi/2, pi/3, ...n) ?
And, can a cell reference be given instead of> putting in the
values? Such as, Cos(A1:A10). (I have also cross-posted to>
two Excel ngs.)You can certainly say =cos(a1). I 
dont think
you can put the rangeaddress in, however. What you should do
Re: little interesting askthere is a fact , surely trivial.
Not for me.May you give me a reason for which two any fibers
of a fiber bundle aredisjoint?they are inverse images of
different points?...by an honest function, rather than
something like a rational function.I missed the original
question, but its conceivable that the originalposter is 
from
a world in which people would refer to something likea
Lefschetz pencil as a fiber bundle, or (in a similar way)
thinkthat the rational function z/w expresses C^2 as a fiber
Re: Using Excel for math> I am just starting to play a little
with MS-Excel for Math purposes. Does> anyone know of a way to
calculate several values for a given function, i.e.> Cos(1,
pi/2, pi/3, ...n) ? And, can a cell reference be given instead
of> putting in the values? Such as, Cos(A1:A10). (I have also
cross-posted to> two Excel ngs.)Set B1 = cos(A1), then use
the fill-down command to set B2 = cos(A2), ..., Bn =
Laplacian distribution? > If I already know a prior that my
data distribution should follow theshape> of Laplacian
distribution... the data obtained from measurement is
ofcourse> a little off(not very symmtrical), how can I make
the measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?> Can anybody give me an
example or detailed explanation? I am kind ofafraid> of
statistics... :=) Why do you think your data is Laplacian? If
its not symmetric,> perhaps thats telling 
you that you dont
understand what the data> really should be? How was the data
generated? Ciao, Peter Apologies for answering a question with
questions! K. -- > Peter J. Kootsookos I will ignore all ideas
for new works [..], the invention of which> has reached its
limits and for whose improvement I see no further> hope. -
to lie upon seeing my question? Oh, its my problemthat I 
did
not clearly present the background...Here is the story: in
deblocking of block DCT coded JPEG images, it wasknown that
the DCTed coefficients are Laplacian distributed... But now I
amlooking at low bit rate JPEG images, so there are someking
of artifacts...in order to reconstruct the original images...
many algorithms have beendevised... one possibility is to make
the image coefficients more Laplcianlike...So that came my
question: how to make data more Laplican like...Please giveme
some detailed explanation as I am not veteran in
not lieing government agency, issurance company,weapon dealer,
(statistics)how to make date more like Laplacian distribution?
> If I already know a prior that my data distribution should
follow theshape> of Laplacian distribution... the data
obtained from measurement is ofcourse> a little off(not very
symmtrical), how can I make the measured data more> Laplacian
distribution like(make it at least a little more symmtrical)? Can anybody 
give me an example or detailed explanation? I am
kind ofafraid> of statistics... :=)> Lets see if I understand.
Your data does not follow> your prior beliefs. Therefore the
data is wrong, and> you want to know how to modify said data
so that it> does exactly what you want. This makes sense - in
a> very distorted way. Why did you bother with measuring that
blasted data> in the first place? You have already decided 
the>
result. Measurements just get in the way. Please start by
reading the book How to lie with> Statistics. Then return to
your data, and learn from> it. Is it just random noise that
has given your data> this property you did not expect? Or is
this an> indication of a problem in your measurement? Perhaps>
it indicates something wrong with the theory? Perhaps> another
factor distorts the data? Maybe your sample> is just too
small! I can summarize the strong suggestions I give to my>
students who deal with data in three words: Plot - think -
learn. Only after that do I tell them to do any actual>
modeling, or use their data in any way. HTH,> -- > There are
no questions ? about my real address. The best material model
of a cat is another, or> preferably the same, cat.> A.
I want to lie upon seeing my question? Oh, its my 
problemthat
I did not clearly present the background...Here is the story:
in deblocking of block DCT coded JPEG images, it wasknown that
the DCTed coefficients are Laplacian distributed... But now I
amlooking at low bit rate JPEG images, so there are someking
of artifacts...in order to reconstruct the original images...
many algorithms have beendevised... one possibility is to make
the image coefficients more Laplcianlike...So that came my
question: how to make data more Laplican like...Please giveme
some detailed explanation as I am not veteran in
not lieing government agency, issurance company,weapon dealer,
(statistics)how to make date more like Laplacian
distribution?> Hello Walala, How big is your sample size? If
you grab only100 samples, you wouldnt> expect it to exactly
fit the generating distribution. Can you grab1,000,000>
samples? If still it doesnt fit, then suspect 
your generating
dist. is> different from a Laplacian. You can look into
bootstrapping methods to get> an estimate of the dist. Clay>
distribution should follow the> shape> of Laplacian
distribution... the data obtained from measurement is of>
course> a little off(not very symmtrical), how can I make the
measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?> Can anybody give me an
example or detailed explanation? I am kind of> afraid> of
people think I want to lie upon seeing my question? Oh, its
my problemthat I did not clearly present the background...Here
is the story: in deblocking of block DCT coded JPEG images, it
wasknown that the DCTed coefficients are Laplacian
distributed... But now I amlooking at low bit rate JPEG
images, so there are someking of artifacts...in order to
reconstruct the original images... many algorithms have
beendevised... one possibility is to make the image
coefficients more Laplcianlike...So that came my question: how
to make data more Laplican like...Please giveme some detailed
lot,-Walala(I am just a poor student, not lieing government
agency, issurance company,weapon dealer, lawyers, and
(statistics)how to make date more like Laplacian distribution?
distribution should follow theshape> of Laplacian
distribution... the data obtained from measurement is
ofcourse> a little off(not very symmtrical), how can I make
the measured data more> Laplacian distribution like(make it at
least a little more symmtrical)?> Can anybody give me an
example or detailed explanation? I am kind ofafraid> of
have are what you measured. If they dont suit you, do as>
corporations and government agencies do: lie. Why people think
I want to lie upon seeing my question? Oh, its my 
problemthat
I did not clearly present the background...Here is the story:
in deblocking of block DCT coded JPEG images, it wasknown that
the DCTed coefficients are Laplacian distributed... But now I
amlooking at low bit rate JPEG images, so there are someking
of artifacts...in order to reconstruct the original images...
many algorithms have beendevised... one possibility is to make
the image coefficients more Laplcianlike...So that came my
question: how to make data more Laplican like...Please giveme
some detailed explanation as I am not veteran in
not lieing government agency, issurance company,weapon dealer,
lawyers, and politicians... so please help me!)When will I
learn? The answers to lifes problems arent 
at the bottom of
a bottle. Theyre on TV! --Homer
DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?> raydpratt
<raydpratt@hotmail.com> grava .88 la saucisse et au marteau:>
Please explain how ïconvergence refutes that 
logic.Because,
according to you, what is A = 1-1+1-1+1 .... ?Is it 0 because
A=(1-1)+(1-1)+...?But this is also 1-(1-1+1-1....) = 1-AA =
1-A, so A = 1/2But A = 1-(1-1)-(1-1)... = 1So, what do you
night to> help me get to sleep, and I now understand several
arguments that have> been raised in this thread about
convergence and the related concept> that the associative
property of addition does not apply to a> non-convergent
infinite sums series, which is what you are pointing> out 
above
with the apparent paradoxes.My answer is that a bird is more
than its parts and that we cannot> mangle a bird to prove that
the concept of a §ying bird is invalid. > As to the present
problem [ 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) ],> we
cannot mangle the left side without equally mangling the
right> side, and although algebra generally allows such
mangling and> collapsing of processes into 
ïequivalent
expressions, we have to be> careful when dealing with infinity
processes since the processes> generate a specific series of
sums, not just any series of sums.What does exactly mangling
mean to you in this context? What isextra logical element for
that matter? Could you please providerigorous definitions for
with MS-Excel for Math purposes.Does> anyone know of a way to
calculate several values for a given function,i.e.> Cos(1,
pi/2, pi/3, ...n) ? And, can a cell reference be given
insteadof> putting in the values? Such as, Cos(A1:A10). (I
have also cross-postedto> two Excel ngs.) Set B1 = cos(A1),
then use the fill-down command to set B2 = cos(A2), ...,> Bn =
I have with that is that in looking at actual pictures you>
can see material curling from one galaxy to another that is
clearly> pulling the matter over.The problem is that you are
looking at one messy chaotic systemin the line of sight of
another messy chaotic system ( this is trueno matter how far
apart or close they are ). Having been burned a few times
astronomers no longer accept simple it looks like arguments.
These arguments are used to start orinform other studies of
the objects. For these next cases I dont have references in
front of me, an examination of the archives of _Astronomy &
Astrophysics_ will locate these and other studies. For example
Vera Rubin reports a case where a small spiral galaxy is close
to but behind an elliptical, there appears to be a bridge
betweenthe two. As the red shifts are different this could be
a counter caseto the BB. Dr Rubin measured the rotation
velocities in the spiraland through application of the
Tully-Fischer(sp?) relation ( maximum rotational velocity is
correlated with galaxy mass and size ) showedthat the spiral
was large and approximately at the red shift distance;ie: much
further than the elliptical.Radio astronomy was used in another
study, a bridge was shown tohave two velocity components, one
at low red shift, the other athigh red shift. There is nothing
between them. These are both difficult measurements, the 
objects
are small andfaint. The variation in signal is small compared
to the total.We should not be surprised that it has taken
decades to elucidatethese cases, when Dr Arp first published
the necessary measurementswere impossible.What JSH and others
seem to think is that there is a large quasi-religoushierarchy
of priest-astronomers that impose a rigid orthodoxy onthe 
field.
This ignores generations of intelligent, hard-working,ambitious
grad students ( and full professors for that matter ) who would
like nothing better than to upset the apple cart. That wouldbe
a professional meal-ticket; lots of big telescope time and
adepartment head-ship.> date November 1966, in what I see as a
not subtle attempt to skew> reader opinion.people who agree
with Arp ). 1966 is the date that started things. Subsequent
reports have found a few other cases, Arp was good enough to
find most, but the new cases have not altered the argument.To
date there have been *NO* cases where physicaly close ( not
line of sight close ) objects have been shown to have markedly
differentred shifts. Arp and others have provided a list of
where this mayhappen, in the cases we have examined it does
not.Dark skies,tom-- We have discovered a therapy ( NOT a cure
Numerical integration, prime counting>Basically, summing
the partial differential apparently gives you aclose
approach to the prime counting distribution, which is
closerthan li(x) itself!> And its been repeatedly
pointed out that youve given _no_ evidence that the
solution to the pde should have anything whatever to do with
professional astronomers would be truly excited to find that
thestandard interpretation of redshift is incorrect --- it
would be ahuge discovery. They arent dogmatically holding 
to
the Hubbleinterpretation. Rather, they know that this
interpretation has a lotof supporting data and describes many
observed phenomena. They alsoknow that it requires very clear
and compelling evidence to make sucha big change to our
physical description of the universe.Most astronomers are not
willing to overthrow the standardinterpretation of redshift
(and all of the other data that support it)because of a
handful of peculiar galaxies out of the billions ofobservable
galaxies in the sky.When we take deep images, there are many
mishapen galaxies --- somewith extended spiral arms, some with
long tails, etc. Sometimes thereis no readily apparent neighbor
to cause the asymmetry. Sometimesthere is a background galaxy
or quasar which appears to beinteracting. But at the moment,
there is no compelling data to provethat these associations
happen more frequently than we expect bychance. Of course, we
dont have complete statistics on thedistributions of 
quasars
and galaxies in redshift or space. As wegather more data, we
should be able to determine how significant 
(orinsignificant)
RationalsI am trying to generalize the result cited by GAE,
viz.,>For integer n>0, let f(n) be the largest integer m such
m/n < pi. Let d(n) = pi - f(n)/n.Surprisingly, K. Mahler
showed that d(n) > 1/n^(42). That>exponent 42 has been
improved subsequently. I guess the best current>result is
8.0161, due to M. Hata.As RI has pointed out, this must be
true not for all n, but for sufficiently large n. My 
first
thought that such a bound is surprising. But maybe something
like this is true for almost all reals. Let me try
again.Consider the set of positive numbers x with the
following property. For all positive c and m, there exists an
integer n > m such that n x - §oor(n x) < n^-c. Does this set
Topology questions Adjunct Assistant Professor at the
University of Montana.>These are some topology questions I was
wondering about.>Assume X is a topological space and A is a
subset of X.What is int(int(A))? Is it the same as int(A)? My
intuition tells me it>is, but how to prove it?By using the
definition of interior, of course. The interior is thelargest
open set that is contained in the given set. In particular:the
interior of a set is open.What is the interior of an open
set?>This would then mean that int(ext(A)) is ext(A),
ext(int(A)) is ext(A),>and ext(ext(A)) is int(A).Depends on
your definitions of interior and exterior. Is ext(A) thesame 
as
int(X-A)? If so, then int(ext(A)) is ext(A), but ext(int(A))
need not be thesame as ext(A). For you are
takingint(X-int(A)), and that could very well be all of X
(e.g., if A hasempty interior). So, for example, X=reals with
usual topology,A=rationals. Then int(A)=empty; and
ext(A)=int(R-A)=empty as well, soext(int(A))= X, which is
different from ext(A).Likewise, ext(ext(A)) would
beint(X-ext(A)) = int(X-int(X-A)), and there is no reason to
believethis will be int(A) always.>bdry(int(A)) need not be
the same as bdry(A). Example: consider>singleton sets in a
Hausdorff space. But then, in some cases,>bdry(int((A)) is
indeed the same as bdry(A).>Likewise bdry(ext(A)) might or
might not be the same as bdry(A).But what has got me stumped
is what int(bdry(A)), ext(bdry(A)) and>bdry(bdry(A))
are.Again, depends on your definitions. Is bdry(A) = A-int(A),
or is itext(A)-int(A)? Or what?> My intuition tells me that
the first is empty,If bdry(A) = A-int(A), then int(bdry(A)) is
indeed empty. For if Owere an open set contained in bdry(A),
then it would be contained in Aand disjoint from int(A), from
which you deduce that int(A) U O is anopen set, contained in
A, and which contains int(A), the largest openset contained in
A. Therefore, int(A)= int(A) U O, from which wededuce that O is
empty. So int(bdry(A)) is indeed empty.>the second is the same
as bdry(A),You mean the third, surely! bdry(bdry(A)). In that
case, yes, becauseif bdry(A) = A-int(A), and
int(bdry(A))=emtpy, thenbdry(bdry(A)) = bdry(A) - int(bdry(A))
= bdry(A).> and the third is Xbdry(A), but I>cant 
figure out
how to prove it.And here you mean the second, ext(bdry(A)). If
ext(A) = int(X-A), thenext(bdry(A)) = int(X - bdry(A)) = int(X
Apocalypse NOW!> Abhi replied to Jeff Root: > Have you done
elementary Geometry Laura? Take a look at my homepage.
http://www.geocities.com/actiondevice What is on that page
is silly. Point B is pulled downward by any downward force
applied to it unless support is provided to hold it up. You
provide no support, so point B is pulled down.> When I say
minds, thinking ability, intelligence of people around the>
world is being controlled, this is the reason.> I am not
applying any downward force. The force is applied to point A>
along the direction of line BE which makes 60 degree angle
with X axis> and the force is applied to point C along the
direction of line BF> which also makes 60 degree angle with X
axis. What makes you think that that is not a downward force?
Instead of saying downward, I will use your terms. You are
applying forces in the direction of BE and in the> direction
of BF. So point B is pulled toward point E and> toward point F
unless some opposing force holds it back. > Again I am talking
about V-shaped SPRING which can store elastic> potential
energy. Yes. But in order to stretch a spring, force must be>
applied to both ends. If you pull on only one end of a>
spring, the whole spring will move, and will not stretch. >
Think again.... I did. Now you act. Did you make and test your
device, yet?> If you havent, do it *now*! It is your idea.
*You* have to> make it work!>But that would make it impossible
for him to revel in his current positionas unrecognized genius
Is there a wayto solve it without using numerical methodsor a
computer? i.e by hand.2*pi*b*int[ sin(t)*Sqrt[a^2*sin^2(t) +
b^2*cos^2(t)] dt, 0, pi]where a and b are both positive
numbers.Id appreciate if someonecan give me the answeror an
of ALL headers otherwise we will be unable to process your
complaint properly.Can anyone help me in analysing the
recurrence relations:x[n+1]=a.x[n]-b.y[n]y[n+1]=c.y[n]+d.x[n]I
am interested in the long term behaviour of this model,
===
fordifferent values of a,b,c,d and x[0] and y[0].Subject: Re:
(sci.logic in Greene-ese) we are supposed to : understand the
identity relation because we : have a paradigm that requires 3
identity relations.No, were not.Were 
supposed to understand
that the Tarskian semanticsolution to the vicious circle
paradox was a hierarchyof languages, and that that means that
equality in theobject-language is going to be different from
equalityin the meta-language. And it further means, if termsin
the object-language are being interpreted as
elements-of-the-domain and relations-over-the-domain as
described in the meta-language, then the specification of an
interpretation as sayingthat some term in the object-language
is to be interpreted as (i.e.,is equal to) some element of the
domain of the model (asdescribed in the meta-langauge), then
using equality toexpress that interpretation is going to
entail treatingit as a sort of bridge predicate between the
two languages.But these are NOT REALLY different. The
underlyingidea is the same and IF you develop a theory rich
enoughto talk about both the terms and the model in the
samelanguage then you can unify them. The fact that
everythingis equal to itself and nothing else remains
analytic.Bothering to have 3 different type-signatures is
justhygiene. : In discussing the structure of the
octonions,Something complicated enough that for anybody to
suggest(as mitch is about to) that needing 3 identity
relationsis a good reason for condemning a paradigm as too
complicated,and that this deserves to be considered as an
alternative,is, well, comical. : Thus O is spanned by (1 e R)
and the 7 imaginary : units i, j, k, l, il, jl, kl, each with
square -1, so that : O=RxR^7.RxR^7 is inadequately
parenthesized. If you mean(RxR)^7, then just confirm that and
spare us thewhines about harassment. This seems like a
mis-quotein any case. A space spanned by (1 e R) and 7 other
thingsshould be 8-dimensional. Is RxR^7 just an
anyone help me in analysing the recurrence
relations:>x[n+1]=a.x[n]-b.y[n]>y[n+1]=c.y[n]+d.x[n]>I am
interested in the long term behaviour of this model,
for>different values of a,b,c,d and x[0] and y[0].>The
beheavior depends on the eigenvalues of the matrix [[a -b], [c
d]]. One good introduction is Luenberger, Introduction to
[referring to improvements in version 5]> Theyve improved
FullSimplify too, then; in 4.2, Mathematica returns> the sum
of the sine series as -1/2 I (Log[1-E^-I] - Log[1-E^I])>
(approximately--this is from memory), and FullSimplify 
doesnt
help.> Furthermore, FullSimplify wouldnt touch
ArcTan[Sin[t]/(1-Cos[t])]> either, although it has an obvious
simplification. Ill have to try> that in 5.0 
(which I have at
work, but not at home).In version
5,FullSimplify[ArcTan[Sin[t]/(1-Cos[t])]] yields
ArcTan[Cot[t/2]].You say that it has an obvious 
simplification,
but Im not sure how> obvious it is. Are you thinking about
something like (Pi - t)/2 + Pi*Floor[t/(2*Pi)]perhaps?
(However, that expression is defined for all real t, whereas>
ArcTan[Sin[t]/(1-Cos[t])] and ArcTan[Cot[t/2]] are not defined
at> t = 2*N*Pi for integer N.)I was thinking of (Pi - t)/2,
since I was imagining 0 <= t < 2 Pi. Irealize this is hard for
CASs to do, since they deal with genericvariables (and
Mathematica doesnt even assume t is real), but
thiscalculation is too important to be totally abandoned by a
CAS.Mathematica CAN do such things. There is a package,
TrigSolve, whichsuccessfully solves trigonometric equations,
taking into account theperiods. In this case, its merely a
question of computingArcTan[Tan[...]], which should be
integral> for several days. Is there a way> to solve it
without using numerical methods> or a computer? i.e by hand.
2*pi*b*int[ sin(t)*Sqrt[a^2*sin^2(t) + b^2*cos^2(t)] dt, 0,
pi] where a and b are both positive numbers.Hint: sin^2(t) = 1
- cos^2(t), d(cos(t)) = -sin(t) dtThis will at least reduce the
problem to integratingsqrt(a^2 + (b^2-a^2)u^2), which can
probably be done by furthersubstitution.-- P.A.C. SmithThe
vast majority of Iraqis want to live in a peaceful, free
world.And we will find these people and we will bring them to
EASY way of finding out the orders of elements of anysymmetric
group, and the number of elements of each order?E.g., Can one
tell what the orders of elements in S_5 are (AND howmany of
each order) without doing a lot of mindless calculations?I
should very much appreciate a short step-by-step
Flawed beyond belief.> Also, the original claim didnt say
anything about *when* the world was> §at. Surely, the
prehistoric world must have been §at.Im not sure that
prehistoric folks had a coherent notion of theworld such that
the question makes sense. But lets assume they did,and that
they thought the world was §at.It still does not follow that
the prehistoric world was §at. Why Iwas just watching Nova
about paleomagnetism, and they were quite surethat the world
Re: Using Excel for math> Hi all, I am just starting to play a
little with MS-Excel for Math purposes. Does> anyone know of a
way to calculate several values for a given function,i.e.>
Cos(1, pi/2, pi/3, ...n) ? And, can a cell reference be given
instead of> putting in the values? Such as, Cos(A1:A10). (I
also getting something strange when I try to calculate
cos(pi/2). Ikeep getting the result 6.12574E-17. Is that
suppossed to be essentially 0?Cant Excel give you 0? I know
that from programming, but I thought MScould have just
different than my subjective views.> I disagree. Your approach
is limited to functions which are one-to-one> (what you term
non-reversing) in the region in question, which makes> it
annoying to apply to a general function: you would first need
to> figure out the intervals on which it is one-to-one before
your> approach can be used. That makes the approach certainly
less useful> than the standard approach. Why should it be in a
textbook, absent> ->actual evidence<- that it is useful for a
significant proportion of> students?> - Students 
dont already
know where the calculus is heading, butyou do. Thats why 
You
find it annoying. I know my approach islimited but for a
student two small steps are better than one big one. > (No, I
did not check it very carefully, though in general it seems>
about right; - cmon, thats not a good lead 
to the
following...>... your function M(x) is closely related to
Newtons Method, - close doesnt count>... and 
a formula for
it could be developed directly by using the> derivative. - and
thats what I did, Im after the integral> 
Without the
derivative, you end up having to do> approximations: your
claim that you can find it graphically really> amounts to
claiming that you can draw accurate tangents by> hand. 
Thats
really hardly true in practice, so your claims towards> the
end really end up being that you can find M(x) and R(x) by
using> the derivative, which means that you are really just
running around in> circles.... - as above, my point is that
the integral is a function [f(x)]multiplied by a function
[M(x)], no more, no less. Try that on yourstudents.Also, I may
be wrong but I think you believe that limits andinfinitesimals
are the same thing. They are not. A limit is a fixedor 
variable
quantity that can be used to produce the derivativeamongst
other things. An infinitesimal is a creation like my
littlestrait(?) part of the curve. Like anything that does not
challenge>Im not a mathematician. I dont 
know any
mathematicians. >Mathematicians always delete my e-mails.Maybe
its because youre condescending to them. 
DougI hear you Doug
- no matter what I do - I cant Make you read it!:) TR
question--HELP Adjunct Assistant Professor at the University
of Montana.>Is there an EASY way of finding out the orders of
elements of any>symmetric group, and the number of elements of
each order?Every element of the symmetric group S_n can be
written uniquely as aproduct of disjoint transpositions. Each
product of disjointtranspositions corresponds to a partition
of n: a way to write n as asum of positive integers.For each
partition of n, n = a_1+...+a_r, a_1>=...>=a_r, the order ofan
element with disjoint cycle decomposition of type a_1,
a_2,...,a_rhas order lcm(a_1,....,a_r).For example, take n=4.
The partitions of 4 are:4 = 44 = 3+14 = 2+24 = 2+1+14 =
1+1+1+1So there are five kinds of elements in S_4:(a) The
4-cycles; they have order 4.(b) The 3-cycles; they have order
lcm(3,1) = 3.(c) The product of two disjoint 2-cycles; they
have order lcm(2,2) = 2.(d) The 2-cycles; order
lcm(2,1,1)=2.(e) The identity, of order lcm(1,1,1,1)=1.How
many of each? Youll need to do a counting argument forthem.
Writen = k_1*b_1 + ... + k_r*b_r,where b_i,k_i>0; interpret
this as k_1 cycles of length b_1, k_2cycles of length k_2,
etc. So above, we would have4 = 1*44 = 1*3 + 1*14 = 2*24 = 1*2
+ 2*15 = 4*1.Then you want to distribute the n elements in the
different cycles. Inprinciple there are n! ways to distribute
them in order. But eachb_i-cycle can be written in any of b_i!
ways (depending on whatelement you start with); and you can
order the different b_i cycles ink_i! factorial ways. So you
haven!/prod_{i=1}^r([b_i!]^{k_i}*[k_i!])So, for example, how
many elements of type 4*1 are there in S_4? Well,there are
4!/([1!]^1*4!) = 1. How many 2-cycles? Since the
2-cyclescorrespond to 4 = 1*2 + 2*1, you have4!/(
2!^1*1!*1!^2*2!) = 4!/2!2! = 6. Etc.>E.g., Can one tell what
the orders of elements in S_5 are (AND how>many of each order)
without doing a lot of mindless calculations?Depends on your
infinitesimals exist> The barbarians are at the gate 
again.just
===
ignore me maybe ill go awaySubject: Re: octonions, 
triality,
paper in front of me, I : realize why no one understood the
comparisons I have been making : all of this time. *I* was
trying to explain things in terms of : the affine geometry 
dual
to the Fano plane.Speaking of any affine geometry as dual to a
projective oneis inviting confusion; the whole defining 
feature
of the projectiveones inTRA is that points and lines are dual
TO EACH OTHER.Using dual inTER the realms requires noting that
you can turnan affine plane of appropriate order into the
projective plane ofthe same order by adding 1 new line at
infinity, each point of whichis a new-shared-meeting point for
each parallel-class of the affine plane.Inverting that by
removing the right line gets you duality in the
otherdirection, usually, I guess, but WHY? I mean, thats
WORK! : That is the plane : whose representation in a
Euclidean space consists of points : labeled with (0,0),
(0,1), (1,0), (1,1).Euclidean space does not need numbers.The
affine plane of order 2 has 4 points and 6 lines.Saying you
have this in Euclidean space is terriblymisleading because
this plane is NOT planar, in Euclidean space.If you make the 4
points co-planar anddraw all 6 lines as straight in Euclidean
space, then 2 of themwill appear to intersect in a 5th
point.It looks like this, REGARDLESS OF ANY LABELS on the
points:o_____o| /|| X | The 4 points are the os; there is 
no
point at the X.|/ | The 6 lines are the 4 edges of the outer
box and the 2 inner diagonals,o_____o which do NOT intersect
in a point at the X.If this is in Euclidean space then either
the plane is not planar orthe lines are not straight. If
claiming that the diagonals dont intersectdoes too much
violence to your planar intuition, you can (if curved
lineswould do less) EQUALLY well represent it this way, as a
baseball infieldwith home plate at the left): ___ o / / 
o-----o
) / / / / o___/ : Of the 28 triples that anti-associate,The
affine plane of order 2 does NOT obviously give rise TO
ANYtriples that anti-associate, let alone 28 of them. you will
note that 28 is 7 times 4. The number 4 is of import : because
of the fact, : : If one has available say 16 alternative
messages : among which he is equally free to choose, then :
since 16=2^4 so that log_2(16)=4, one says that : this
situation is characterized by 4 bits of information.Excuse me,
but since we are doing all this on computers, we DID KNOWthat 4
bits of information yield 16 alternatives. Dropping nameslike
Weaver and Shannon IS NOT relevant to this. : conveniently
discussed by Warren Weaver in his introduction to Shannons
The Mathematical Theory of : Communication. One of the crucial
issues is how you are going to refer to those 4 bits.This is
NOT a crucial issue.There are 24 different orderings of them,
no matter HOW you choose,and it does NOT matter which one you
choose, as long as you are consistent.We could call them the
red bit, the blue bit, the green bit, and the yellowbit. We
could call the mitchs 4 curled left toes. The most usual
thingwe do is index them, 0,1,2, and 3. If you think that
indexing them asordered truth-value-pairs would illuminate
something about logic, thenplease feel free to keep
clarifying. : You can certainly use the affine 
labeling--(0,0),
(0,1), (1,0), (1,1)Idiot: There is NOTHING WHATSOEVER ABOUT
THESE 4 bit-vectors-of-length-2that is in ANY way connected to
the affine anything, beyondthe fact that there are 4 of them 
and
there are 4 points in anaffine plane of order 2. But the 
affine
plane of order 2 SHARESTHAT property with all OTHER 4-element
sets. And this isnt evena unique labeling: given any 
4-point
affine plane and these 4 labels,there are 24 DIFFERENT
labelings of that plane by these labels.More to the point, why
did you feel no need to pause to ponder justwhat sort of
LABELING you were going to use for the TWO
differentbit-positions in your two-bit labels? I mean, if it
was an importantissue for 4, its got to be that much more 
so
for 2, right?There are 16 different truth tables.The affine
plane of order 2 has 4 points, 6 lines, and no obvious
relationship tothose truth tables. The projective plane, by
contrast, COULD yield an obviousrelationship, because its 7
points and 7 lines, totaling 14, are dual toeach other,
meaning you could make 7 of the truth functions points,their
denials lines, and leave out the other 2 completely BECAUSE
THEYARE UNIFORM/trivial/not-threshold. : Or, you can use the
quaternion units--1, i, j, k--and apply Lounestos
construction to : get the dual Fano plane.If you want
something from the Fano plane that is related to truth tables
thensurely the easiest thing to do is just pick 14 of the 16
binary boolean functionsto be the 14 things in the
variation can only be used if> the random variable is
positive. I would like to know why cannot one> extend the use
of the same coefficient with not necessarily positive> random
variables but with positive mean. Could somebody here please>
help me?> To my way of thinking, the CoV is mainly of value
whenthe underlying data are log-normal. So, yes, the
valueswill be positive. - I do not recall reading about it
that way, butwhat you describe sounds like admirable
advice.That is, the SD needs to *stay* proportional to the
mean, or there is hardly any value in talking about its value,
right?You could compute a ratio and call it a CoV. But will you
haveanything worthwhile, if you have values that are negative,
andnot natural amounts of something?So, I guess the reason
that you cant, is if your journal editors and grant 
reviewers
are apt to see it that way, too.If someone ever has a good use
for a CoV on number-linedata from some source, I figure they
could argue for the exception for that source. This is a
practical matter.-- Rich Ulrich,
wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are
Reconsidering Halton Arp> They say its settled. The data 
says
that its not.I want to emphasize that point! The issue here 
is
that certain people> are working to dismiss alternative
theories, other explanations in> favor of what is now the
current Big Bang Theory.That is a natural defensive mechanism
for established theories.If the fundamental assumptions had to
be evaluated all of thetime, science would be reduced to
discussions about science,without any time to do real
science.You may find some answers to your questions about how
scienceworks in a good book on the theory of science. I
recommendWhat is this thing called Science? by Chalmers. You
may inparticular find the chapters on Kuhn and Lakatos
HELP BADLY (sorry, maths not psych)Expires: 28 daysHow long
after the electron enters the field will>it be
affected?Randy, havent you noticed that whenever Paul is 
in
trouble, he always attemptsto hijack the converstaion by
diverting the subject down a side track.I have noticed this in
conversations between you and Paul. Between you>and me, too.
However, it isnt Paul who does not. Nor is it me.When 
does
the velocity start changing from 100 m/sec? How>far does the
electron get before this happens?That is not the question I
have raised.I know. Its the question he was asking you, 
that
you didnt answer.>At least you admit that rather than 
answer
the question, you changed>the subject. You raised this
question in lieu of giving an answer. - Randy>BULL!See the
Stupidity of
A challenge Adjunct Assistant Professor at the University of
than my subjective views.Well, of course. However, you
presented your claims as OBJECTIVE. Youclaimed that limits
->are<- confusing, not that they confuse->you<-. You claim
that your approach belongs on textbooks, not thatyou ->think<-
it does.  I disagree. Your approach is limited to functions
which are one-to-one (what you term non-reversing) in the
region in question, which makes it annoying to apply to a
general function: you would first need to figure 
out the
intervals on which it is one-to-one before your approach can
be used. That makes the approach certainly less useful than
the standard approach. Why should it be in a textbook,
absent ->actual evidence<- that it is useful for a
significant proportion of students?> - Students 
dont already
know where the calculus is heading, but>you do. And I dont
know where YOUR approach was heading either. Your
approachseems completely ad hoc and unmotivated. Why would an
ad hocunmotivated approach make more sense to me? Or to anyone
else? Itwould not.YOU find your approach better because YOU 
know
where it isheading. But the student will not know where you are
heading->either<-. > Thats why You find it 
annoying. I did not
say I found your approach annoying. I said your approachwould
be ANNOYING TO APPLY TO A GENERAL FUNCTION. Please do ->try<-
toread what I write, not what you think I am writing. >I know
my approach is>limited but for a student two small steps are
better than one big one.I do not see two small steps. I see
unmotivated jumps. Why would Iever consider the quantities
M(x) and R(x)? There is absolutely NOmotivation for the
construction. They are geometric artifacts thathappen to give
(apparently), the right answers. They are no moreintuitive
than the complicated diagrams from Euclid, which
students->certainly<- do not find intuitive.And in your reply,
you did not address the most important point: YOUHAVE NO
EVIDENCE WHATSOEVER THAT YOUR METHOD IS USEFUL FOR
ASIGNIFICANT NUMBER OF STUDENTS. Thats just ->your<- guess.
As such,it is not sufficient to warrant the sort of claims you
make. (No, I did not check it very carefully, though in
general it seems about right; - cmon, 
thats not a good
lead to the following...Take what you get.... your function
M(x) is closely related to Newtons Method, - close 
doesnt
countDoes not count for what?The intersection of the line with
the x-axis is the next approximationfor Newtons Method. One
can derive M(x) directly from the Newtonsmethod formula
approximation.... and a formula for it could be developed
directly by using the derivative. - and thats what I did,
Im after the integral> Without the derivative, you end up
having to do approximations: your claim that you can find it
graphically really amounts to claiming that you can draw
accurate tangents by hand. Thats really hardly true in
practice, so your claims towards the end really end up being
that you can find M(x) and R(x) by using the derivative, 
which
means that you are really just running around in circles....
- as above, my point is that the integral is a function
[f(x)]>multiplied by a function [M(x)], no more, no less. Try
that on your>students.Sorry, but I am not about to experiment
on ->MY<- students, andcertainly not with an ad-hoc method
that has no motivationwhatsoever. Bring back solid data from
experiments, and I may considerit. But the say so from someone
who admits having had trouble with theconcept, and who admits
to not being well-versed on the subject, andwho has NOTHING
but his own subjective impression to back up that sayso, is
simply not sufficient for me to risk confusing the students.
>Also, I may be wrong but I think you believe that limits
and>infinitesimals are the same thing. You are indeed
wrong.>They are not. A limit is a fixed>or variable quantity
that can be used to produce the derivative>amongst other
things. No. A limit is NOT a fixed or variable quantity that
can be used toproduce the derivative amongst other things.> An
infinitesimal is a creation like my little>strait(?) part of 
the
curve.No, an infinitesimal is not a creation like [your] 
little
straiGhtline.Since you are obviously ignorant about these
basic concepts, why isyour say so of what is useful and what
Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28
daysBut what are YOU saying?>Not untill the electrode 1 km
away feels the opposing force? IS THAT WHAT YOU ARE
SAYING? NO PAUL. I am saying that your claim infers that the
effect of an injected> electron will be felt INSTANTLY at the
far electrode.I have understood that you think my answer
is wrong.I am asking you: WHAT IS YOUR ANSWER?Since you
claim that the force cannot act instantly,I am asking you,
WHEN will the force act?You claim that the distance to the
other electrode is relevant.WHY is it relevant?How does
this distance affect the delay before the forceon the
electron starts acting?This is a concrete
scenario.Please state what you think will happen. You keep
talking about the force on the electron. Im more concered
about teh force at the far electrode.> Of course, since
the> electron existed BEFORE it was injected, the effect would
have already been> there even though the near electrode was in
the way. The only way this> experiment can even be
hypothesized is by either ïannihilating a very 
large> number
of electrons or by monitoring the force on the far electrode
with> movement of the electron mass towards it.Say, what
the hell are you babbling about?Dont tell fairytales 
about
suddenly disappearing charges! What happens after this: (e+)
+ (e-) = ?ADDRESS THE SCENARIO GIVEN!It is a concrete
scenario which in principle could be made.(And which EASILY
could and IS made with shorter distancebetween the
electrodes. You have it in your TV set!)Let it be a hot
wire in the hole in the electrode.Thermionic emission of
electrons.When one of these electrons gets out of the
holeand into the static field between the electrodes,how
long time will it take before a force start acting on it? I
could probably write a whole book answring that. However I
would conficently say that, before it starts to move, the 
force
is instant.Its settled then.>Paul, it is by no means
settled.>We agree:> as it enters a static electric field.One
can but wonder why it took you so long to realize>the
obvious.Read what I said! Can you see the words before it
starts to move?Do you know what they mean?Do you also know
what the electron is doing for the rest of the time? IT
ISMOVING! >My answer is: as it enters a static electric
field.WHAT IS YOUR ANSWER? I just gave it to you.; Now
please answer MY question. If a highly charged sphere is
moved, say, backwards and forwards between two electrodes,
what happens to the force on those electrodes. Does it
change instantly or is there a time lag?Why do you have to
ask?>There is obviously a time lag.Why didnt you say so
earlier. One can but wonder why it took you so long torealize
the obvious.But are you quite sure of your answer? After all,
you DID say that electrostatic forces acted instantaneously.
Haveyou changed your mind?Can you refer to an experiment that
shows the time lag? > Hint: neither you nor anyone else knows
the answer.Nonsense.>Of course we know exactly what will
happen.>There are no mysteries in electrodynamics.know the
answer to questions like this?Well you obviously dont. You
have already changed your mind once.Maybe you will do it
again.Paul>See the Stupidity of
Cardinals, Ordinals, Binary Nodes? > Consider the Binary Node
tree of naturals.> 0> / > 0 1> / / > 0 1 0 1> .> .> .> (I hope
it is lined up.)> Obviously if you consider the omegath row, I
dont know what that means. I can certainly write down the
n-th row> for any natural number n. But I have no way of
knowing what you mean by> the omegath row. Its not
well-defined.Row #omega, and omega is Cantors 
smallest infinite
ordinal.> it will have beta-1 (C) elements, beta than none at
all I guess.I am sorry, Bet-sub-1. It is the second infinite
cardinal in the series ofcardinals where 2^Bet-n=Bet-(n+1).
Since 2^bet0=bet1, it will have bet-1elements.> but since
omega is the first limit ordinal, it is the first 
row with an>
infinite number of elements.You havent 
defined what you mean by
the omegath row. For example if I> ask you what is the 47th
node of row 2000, you could tell me. But you> cant tell me
whats the 47th node of row omega.Sure I can. It is one. >
However, consider the previous row. Well clearly if you intend
the omegath row to be the result of some type> of limiting
process, there is no immediately previous row. You can> define
omega to be the first ordinal that follows all the 
finite>
ordinals, but you cannot then speak of the ordinal immediately
preceding> omega.If it is written down on a list, then some
omnipresent being could walk toomega and take one step back.
What I am saying is: let us say that theheight of this being
is omega (when he lies down against the rows, his feetare at
omega if his head is at one). Imagine pegs that were at the
ends ofeach row. If he laid in the middle of row omega, and
started rolling downtowards one, where would he get caught?
I.e., where is the length, ororder, of the row omega?) It has
to be somewhere because he will, in acountable time, come to a
row of finite length, but the row above was 
alsofinite, so he
couldnt have gotten there. On the other hand, where 
doesrows
of length omega start? The row above him must also have
Re: (statistics)how to make date more like Laplacian
upon seeing my question? Oh, its my problem> that I did not
clearly present the background... ...> (I am just a poor
student, not lieing government agency, issurance company,>
weapon dealer, lawyers, and politicians... so please help
me!)Walala,I believe that I was the only one who mentioned
lying. I hope you understood that I was being facetious. The
point I and others hammered on was that measured data are what
you measure. Altering them after the fact is distortion.I 
dont
know the answer to your problem -- where is JJ when we need
him! I hope you can figure out how to distort a blocky image 
so
it seems pleasanter to the eye.Jerry-- Engineering is the art
of making what you want from things you can
HELP BADLY (sorry, maths not psych)Expires: 28 daysHenriWilson 
<Henri@the.edge> skrev i melding> I still cannot
see the philosophical reason for using F=dp/dt rather than>
F=m.dv/dt where m=f(v)Its Newtons second 
law of motion
in its original form.The change of motion is proportional to
the force impressed; and is made in the direction of the
straight line in which the force is impressed.What did
Newton mean by motion?Those who have studied his texts think
he meant momentum.Paul It is interesting because the term
ïv.dm/dt is explains the energy increase that 
is normally
associated with ïrelativistic 
mass increase.Of course.>The
question is what is the momentum?>If the momentum is mv, then
the mass _must_ increase with the speed.>In 1905 this was the
accepted definition, and was why Einstein>said that the mass
increased with speed.However, the modern approach is to say
that>momentum = m*f(v) where m is the invariant mass>and f(v)
=v/sqrt(1 - v^2/c^2)> It actually supports my argument that
this energy really goes into the ïreverse field 
bubble that
forms around a moving charge.Why not call your enigmatic
bubble a fairy?more invisible but massive fairies clings to
it.>When the fairies loose their grip in the bends of the
accelerator,>their mass is transformed to synchrotron
radiation.Make perfect sense, doesnt it?It makes just as 
much
sense as saying ïmass increases for no apparent 
reason.Where
is your supporting PHYSICAL evidence? How can mass simply
appear toincrease? What does ïmass increase 
mean Paul?You
people are really funny.Paul>See the Stupidity of
(statistics)how to make date more like Laplacian
distribution?Hello Walala,I dont think anyone has accused 
you
of lying. But I see the problem as youhave stated it is that
theory predicts your data should be Laplacian, andfor some
reason you think your sample data is inconsistent with this
notion.It probably is consistent but at first blush it appears
otherwise.Now for a simple example. Lets say you have a 
fair
die. Each time you tossit, the probabily of each outcome is
1/6. So theory says we have a uniformdistribution. Now toss
the die 100 times and count how many times each valueshows up.
Now theory says on average each value shows up 1/6 of the
time.However in any given sample (set of measurements), you
cant expect thisperfect a distribution every time. Why you
ask? That is because each trialis independent of the prior
ones. And in this situation I only tossed it 100times and 100
is not a multiple of six, so not all bins can have the
samenumber of counts. Since the counts must be integers and
100/6 = 16.6666666,you can see the one problem.Now to
illustrate another problem, we will use simple coin tosses as
thishas a smaller sample space than the dice problem. In a
simple coin tossexperiment where on average heads or tails
shows up 50% of time, the resultsof the next toss cant 
depend
on the result of the last toss. Otherwise wecould accurately
predict the coin toss result for every throw and we knowthat
is not true.But lets look at the results of a coin toss 
trial
and try to fit adistribution to it. If you toss a single coin 
4
times or just simply toss 4coins once, we know there are
2^4=16 possible outcomes. And if you enumerateall 16 sets you
will notice that only 6 out of the 16 will have heads halfof
the time and tails half of the time. Wow more than half of the
time, thesample distribution doesnt fit the 
theoretical
distribution. So if I go andtoss 4 coins and get 3 heads and 1
tail, can I proclaim my coins are unfair?Clearly I cant,
because probability says that will happen 25% of the time.What
all I done so far is show how a sample statistic (extracted
from oneset of data) is not very good at predicting a
population statistic(extracted from all possible sets of
data). And when you are trying todetermine a distribution you
are in effect trying to find all of thepopulation
statistics.Now for your problem. The 1st thing is the theory
correct in saying thedistribution is Laplacian? That can be
answered by going through the detailsof the derivation. I
dont have your papers, so I cant help you 
here. Nowyou can
test this hypothesis statistically by one of several goodness
of fittests. A common one is a chi square test. But these test
work well withlarge amounts of data. But since you are
reducing pictures and the world hasa lot pictures, you
shouldnt be lacking for data. Another way is to
developconfidence intervals for the Laplacian distribution, 
for
example 90%. Now ifyou repeat the experiment many times, will
your measurements fall inside oroutside with too high a
frequency. I.e., If my 3 heads and 1 tailcombination showed up
50% of the time, instead of the expected 25%, in alarge number
of trials, Id be suspicious that my distribution
isntuniform.Now for the question you posed about how to 
make
your data more Laplacianlike. First you will have to decide
what distribution it does have. Then youcan find a
transformation that turns it into a Laplacian and you will
haveto account for the Jacobian here. But if you know the
distribution thereisnt much point is transforming it into
another distribution.To find the distribution when you 
dont
have a theoretical way to get thereis by the process of
resampling. Ie., you extract the distribution byprocessing
subsets of a large chunk of sample data. Some good
resamplingmethods that are used to find the distributions are
known collectively asbootstrapping methods. These really hit
the stats scene during the 1970s. Asimple example of
resampling can be found in the jackknife method.I hope this
helps clarify some of the issues. I think it is time to
studysome stats.Clay > Hello Walala,> How big is your sample
size? If you grab only100 samples, you wouldnt> expect it 
to
exactly fit the generating distribution. Can you grab>
1,000,000> samples? If still it doesnt fit, 
then suspect your
generating dist. is> different from a Laplacian. You can look
into bootstrapping methods toget> an estimate of the dist.>
distribution should follow the> shape> of Laplacian
distribution... the data obtained from measurement is of>
course> a little off(not very symmtrical), how can I make
the measured datamore> Laplacian distribution like(make it
at least a little moresymmtrical)?> Can anybody give me an
example or detailed explanation? I am kind of> afraid> of
Why people think I want to lie upon seeing my question? Oh,
its myproblem> that I did not clearly present the
background... Here is the story: in deblocking of block DCT
coded JPEG images, it was> known that the DCTed coefficients
are Laplacian distributed... But now Iam> looking at low bit
rate JPEG images, so there are someking of artifacts...> in
order to reconstruct the original images... many algorithms
have been> devised... one possibility is to make the image
coefficients more Laplcian> like... So that came my question:
how to make data more Laplican like...Pleasegive> me some
detailed explanation as I am not veteran in statistics...
government agency, issurancecompany,> weapon dealer, lawyers,
Symmetric group question--HELPNNTP-Posting-User:
,;upNQ$t/)!L]^R9Po%swv4<z?~Yuu~AaN37_ZF?7GZA5A_j|];J9NBVl@Yc+
sha1:uHYW6/VuLROZUDAIVHnZRs8k4QI=>Is there an EASY way of
finding out the orders of elements of anysymmetric group, 
and
the number of elements of each order? Every element of the
symmetric group S_n can be written uniquely as a> product of
disjoint transpositions. Each product of disjoint>
transpositions corresponds to a partition of n: a way to write
n as a> sum of positive integers.I think you should say cycle
instead of transposition in theprevious paragraph.> For each
partition of n, n = a_1+...+a_r, a_1>=...>=a_r, the order of>
an element with disjoint cycle decomposition of type a_1,
a_2,...,a_r> has order lcm(a_1,....,a_r). For example, take
n=4. The partitions of 4 are: 4 = 4> 4 = 3+1> 4 = 2+2> 4 =
2+1+1> 4 = 1+1+1+1 So there are five kinds of elements in S_4:
(a) The 4-cycles; they have order 4.> (b) The 3-cycles; they
have order lcm(3,1) = 3.> (c) The product of two disjoint
2-cycles; they have order lcm(2,2) => 2.> (d) The 2-cycles;
order lcm(2,1,1)=2.> (e) The identity, of order
lcm(1,1,1,1)=1. How many of each? Youll need to do a 
counting
argument for> them. Write n = k_1*b_1 + ... + k_r*b_r, where
b_i,k_i>0; interpret this as k_1 cycles of length b_1, k_2>
cycles of length k_2, etc. So above, we would have 4 = 1*4> 4
= 1*3 + 1*1> 4 = 2*2> 4 = 1*2 + 2*1> 5 = 4*1. Then you want to
distribute the n elements in the different cycles. In>
principle there are n! ways to distribute them in order. But
each> b_i-cycle can be written in any of b_i! ways (depending
on what> element you start with); and you can order the
different b_i cycles in> k_i! factorial ways. So you have
n!/prod_{i=1}^r([b_i!]^{k_i}*[k_i!])> So, for example, how
many elements of type 4*1 are there in S_4? Well,> there are
4!/([1!]^1*4!) = 1. How many 2-cycles? Since the 2-cycles>
correspond to 4 = 1*2 + 2*1, you have 4!/( 2!^1*1!*1!^2*2!) =
4!/2!2! = 6. Etc.>E.g., Can one tell what the orders of
elements in S_5 are (AND howmany of each order) without
doing a lot of mindless calculations? Depends on your
definition of mindless calculation!
I accept as reality.> --- Calvin (Calvin and Hobbes)>
PalmieriDept of Mathematics, Box 354350
mailto:palmieri@math.washington.eduUniversity of Washington
http://www.math.washington.edu/~palmieri/Seattle, WA
to understand that if point B shifts its position along> Y
axis, angle ABC or EBF will change and as I am saying again
and> again, angle ABC is solid angle.> I know but... That
was very clever indeed, George. I admit, the whole confusion
was> generated because of use of my word SOLID angle instead
of FIXED> angle. I messed it up but you knew what I was trying
to say.It happens often in the group, you are not alone. I
alwaystry to respond to what the poster meant. After all, I
makemistakes too.> I am responding to your diagram and there
is> nothing in that to keep the angle the same. What I am>
saying first like everyone else is that, as you have> drawn 
it,
point B will move and the angle will change.> Something extra
will be needed to prevent it moving> that is not shown in the
diagram at the moment, and I> cannot speculate on what you
might add. Very good indeed. There are many ways to prevent
the angle changing so> that it remains FIXED. We can insert
V-shaped rigid rod inside these> springs so that angle V does
not change. Or we may put these springs> inside hollow pipe.
We can use shock-ups. That is the best solution. So you also
tried to trap me in my words.No, I responded as if you had
said ïfixed but pointedout the 
confusion so you could avoid
the problem in future.> Like HIM who is> controlling things at
this moment, you knew the meaning. OK, above all, I made a
mistake in explaining this damn thing. But now> you know what
I am trying to say.Sure. As you say there are several ways you
could do itand they will all produce similar results. The 
ïV
shapedrod inside the springs is the easiest for me to
describebecause that way you can still get at the springs to
pullthen from A and C to E and F. If you use a pipe, I
cantsee how you wou get at them.> Now, will the point B 
move
along Y axis if the angle ABC remains FIXED> throughout
operation of this device?If you put a rod inside the springs,
fixed to the springsat B but so they are free to move over the
rest of theirlength, then you can stretch the springs without
point Bmoving for this reason: You have pulled the springs
downfrom A and C to E and F so you are holding the last loopof
each spring fixed at those points. If point B and therod were 
to
move down the Y axis, the points E and F wouldneed to get
closer together, but they cannot because youare holding the
spring loops apart. The rod will then presson the held loops
of the spring by just enough to balancethe force shown by the
blue arrow and point B will not move.The upwards force on
point B from the rod that stops itmoving is being transmitted
as a downwards force on the lastloop of the spring. If that
was all, the ends of the springsat E and F would move along
the X axis towards O. You haveto hold the ends apart to
prevent this and that force,together with the downward force
of the rod on the bottomloops will exactly balance the pull of
the spring towards
B.http://www.dishman.me.uk/George/Abhi/abhi_rod.gifB will not
move and every force will be balanced by an equaland opposite
square, given radius, find maximum number of
circle packingproblem will be greatly appreciated: Given a unit
square, and a radius R, how does one find the maximum number 
of
circles, each of radius R, that can be packed into the
Re: An apparently simple problem>Does anyone has an ideia how
can this be solved?>I have two differential
equations:>X(t)=a-b*X(t)>Y(t)=c-d*Y(t)>I 
want to know when
the solutions intersect. By doing it straight forward>(just
X(t)=Y(t) ) I end up with the transcendental equation:You
could just solve the two equations separately, then see when
the two solutions are equal. Of course the answers will depend
on the initial conditions, which you didnt specify.Hmmm...
does when here mean for what values of t, or for what values
of the initial conditions? The second interpretation may be
HALF A LIGHTBULB? Also, the original claim didnt say
anything about *when* the world was §at. Surely, the
prehistoric world must have been §at.> Im not sure that
prehistoric folks had a coherent notion of the> world such
that the question makes sense. But lets assume they did,> 
and
that they thought the world was §at.> It still does not follow
that the prehistoric world was §at. Why I> was just watching
Nova about paleomagnetism, and they were quite sure> that the
world has been a sphere for a very very long time.You mean the
world did not literally change shape?Gee!-- Dave SeamanJudge
Yohns mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
Professor at the University of Montana. [.snip.] (No, I did
not check it very carefully, though in general it seems
about right; - cmon, thats not a good lead 
to the
following...If you want a detailed criticism, youll have to
pay for my time. Butheres a sample of what you will
get:Looking at http://www.precalculus.netfirms.com/on Wed. 
Nov.
19, at 2pm MST(1) Proportions and Change, line 1: Mathematics
is the study of proportion and change. No, thats not what
mathematics is. This line does not belong there. Line 2:
Anything we can measure is a dimension. Again, this is an
incorrect use of dimension, and the meaning of measure is
unclear. This line is either content-free or wrong. Line 2:
The measurements we take are simply relative differences in
proportion. No, they are not. Line 2: Proportion is the
quality that forms a continuum from zero to infinity with one
being the base to compare the proportion. Unclear at best,
incorrect at worst. Is that what you want? It will take
several hours of my time, which Iam not at liberty to spend
for free. If you are willing to pay, Illbe happy to provide
you with a complete list of errors and suggestedcorrections.
But I doubt that you are. A more serious problem is that you
are claiming that your method isa more intuitive of
understanding the Fundamental Theorem ofCalculus. I disagree.
There is no intuition behind M(x) or R(x). Theyjust simply
happen to work. But, if I do not know the fundamental theorem
of calculus, why onEarth would I ever draw the line you
suggest? Why on earth would Iever consider the function M(x),
or the function R(x)? What is theirrelation to the problem I
am ->supposed<- to be working on, namely,that of finding the
area under the curve of f(x)?The answer is simple: you would
not. There is absolutely no intuitivereason for me to draw the
tangent to F(x) at the point a if I aminterested in finding 
the
area under the curve of f(x). In fact, thereis no reason for
me to go looking at F(x) in the first place, UNLESS Iapproach
can hardly count as an intuitive way of getting to
thefundamental theorem of calculus, if the only justification
you canfind for taking your first step (looking at 
an
antiderivative of f(x))is that you ->already<- know that the
Fundamental Theorem of calculuswill connect integration with
derivatives, and in what way it willconnect them. Whats
worse, you convert the problem of finding thearea under a 
curve
into a problem of finding a volume of a3-dimensional solid.
Thats a step back, not a step forward. Worse: you do not
PROVE the fundamental theorem of calculus. YouASSUME it in
order to state how your M(x) and R(x) relate to theintegral of
f(x). So, why is this a better way to teach students the
fundamentaltheorem of calculus? You dont ->prove<- it, you
->use<- it. Toexplain it, you introduce two completely ad-hoc
functions, and tryto convert the simple problem of figuring 
out
an area into adiscussion of figuring out a volume of a
3-dimensional solid. Your argument seems to be that you were
confused, and this explanationhelped you. Therefore, if
someone is confused, this explanation willhelp them. Alas,
that is not a valid logical inference.Finally, your final
paragraph: In essence, calculus is the method of using rules
for simplifying and differentiating a function and then
applying them to your advantage to solve problems involving
changing quantities. If that is what you think calculus is in
essence, then I have newsfor you: it is not that you ->were<-
confused, you ->still are<-. Youhave no clue what the subject
matter of calculus is, and you areconfusing the method with
the subject. So, why does your stuff belong in textbooks? It
does not proveanything, it is not intuitive, it is not
illuminating, it is ad hoc,it works only in a very limited
situation, and it assumes as true thething you claim it
forwarding the After my talk I hope I answered (and also
understood) your question--the charges in my electron and
photon models are moving faster than light. The mass of the
electron corresponds to the circulating photon-like object
having the Compton wavelength. is moving superluminally? Do
you have any general or specific comments on my superluminal
electron/photon models?I really do not understand what you are
saying, but I have not spent enough time trying to understand
your model.Note that Burinskii and Kaiser ,in independent
work, have disks of charge but the rim of the disk is limited
to speed o§ight and is not superluminal.Charges moving faster
than light should Cerenkov radiate?If lepto-quarks confined
inside hadrons are off mass shell they could be space-like
with imaginary virtual rest mass.But I think charges are
confined to speed of light in rotating open string models of
hadrons?How does my electron model in your opinion compare
with Milo Wolffs space resonance electron model? (he spoke
after you.)I cannot understand Milo Wolf. To me his ideas are
not even wrong (W. Pauli) gibberish. As soon as one says, like
he does, that he does not know the math but - my eyes glaze
over. :-) However, you are seeing something interesting I
sense, it may not be right in the end, but it is interesting
to see what it is you are seeing even if I cant see it at
this time. Do you think my electron model could be held
together by a mini-black hole at its center as you 
suggested
in your talk about what might hold an electron together?I do
not understand what your model is enough. It did look
interesting, but I could not understand it in the time
available and I missed the beginning of your talk.I am
basically finishing what Wheeler started in the 
1950s in his
book Geometrodynamics.To get it to work, i.e. to derive
lepto-quarks from vacuum geometry as multiply-connected geons
with quantized trapped EM §ux (color and weak §ux perhaps
disguised EM §ux) you need at least1. Variable G, i.e. G* =
10^40G(Newton) at fermi 1Gev scale.2. Instability of
micro-quantum globally §at lepto-quark/electroweak-strong
vacuum so that both Einsteins gravity and exotic vacuum
residual zero point energy-density phases emerge as
MACRO-QUANTUM vacuum structures.3. O(2) symmetry of the
MACRO-QUANTUM vacuum coherence demands string defects like
quantized vortices in Type II superconductor ground state
where coherence -> 0 at the phase singularities inside the
core, e.g. Kerr-Newman ring that rotates at c. Do you favor
the Compton vortex electron model or any other particular
electron model? I understand from your mini black hole comment
in your talk that you dont consider an electron to be
point-like as is now commonly accepted (though I think not by
Dirac himself).It is not pointlike.At low energy it is ring
e^2/mc^2 ~ 1 fermi surrounded by plasma of §uctuating virtual
electron-positron pairs and virtual photons reaching out to
h/mc ~ 10^-11 cm.The huge micro space-warp from G* ~ 10^40G
means that this extended structure looks more and more like a
point as the momentum transfer p in the scattering Heisenberg
microscope increases i.e. ratio of circumference C to radius R
shrinks from the Euclidean value with radius fixed at 
Heisenberg
uncertainty h/mc, with effective size defined as C/2pi.For
example, in a toy model micro-geon without charge and
spinC/2pi = (1 - 2G*mp/c^2h)^1/2This micro-geon looks like a
point at the horizon2G*mp/c^2h = 1This heuristic picture is
complementary to a string picture sincec^4/8piG* = string
tension = Wittens (alpha)^-1mc^2 ~
e^2/zpf(core)^1/2Therefore2G*e^2/zpf^1/2p/c^4h =
===
best wishes, Richard----- Original Message -----Subject: Re:
===
superluminal talkSubject: Re: Physical wavelets and their
missing from Proofs from THE BOOK>obviously has a bias towards
content related to Erdos. And of course >beauty is relative
(some of the proofs I find ...er... inaccessible).>But I 
cannot
fault the authors at all; deep, readable, meaty, true to
>purpose (they put in beautiful -proofs-, not necessarily
beautiful >-truths-, though of course there is overlap).>Now
to my issue: of the sections in the book (number theory,
geometry, >analysis, combinatorics, graph theory), the authors
seem to cover the >range of modern mathematics fairly well, a
bit lopsided, but still a >good cover (probability is in a few
of the proofs, even algebra (a >finite division ring is a 
field),
topology is really just a fancy word >for geometry,
right?).Topology is NOT a fancy word for geometry.>-But- there
is one field which is entirely missing: Logic.>Are there no
beautiful proofs in the field of logic?>Goedels 
first
incompleteness theorem has a large part to it that is >totally
ugly algebra/number theory/encoding, but the gist of it is
simple.The algebra/number theory/encoding is totally
unnecessary.All that is needed is to have every
formula/proposition/theorem/proof have a positive integer
assigned to it; oneway to do this is to use a direct ASCII (or
other such)representation.>What about the related halting
problem or (almost identical) >nondenumerability of
reals?Nondenumerability of the reals is set theory, not
logic.There are lots of such proofs in set theory; many of
theseelegant and somewhat paradoxical proofs are quite
accessibleto the non-expert.>Sure there are lots of important
theorems of logic, Im just trying to >think of those that
have proofs worthy of THE BOOK.>Mitch-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
counting> Make no mistake, Ullrich is a dedicated and
persistent liar.He also is a math professor at Oklahoma State
University.> James HarrisWhereas James S Harris is a dedicated
and persistent liar about a lot of things in genteral and aboud
in particular.He, JSH, also is not a professor of anything at
any unversity, and, according to his last known statement on
the subject, is unemployed, so has lot of time to get into
Reconsidering Halton Arp [JSH crank alert]> James Harris>
Rather naively I thought that if you put out something simple
enough> that most people could understand it,>
www.crank.net/harris.html> or (more fun) do your own googling
with shrewd combinations of keywords. I> tried> moron
factorization bull> and Google found James Harris in 0.24
seconds.This is just great. Now JSH can claim to have Google
against him inthis massive international conspiracy to silence
him. Eventually,hell be the lone hero battling an entire
universe controlled by Satanor something else of similar
result akin to Brouwer, new?Let S be the set of (x,y,z) in
R^3 such thatx + y + z = 1andx>=0 and y>=0 and z>=0.Let
f be a continous map S to S. Define three closed subsets of S
by:A = set of (x,y,z) such that f(x)>=xB = set of (x,y,z)
such that f(y)>=yC = set of (x,y,z) such that f(z)>=z> ???
Those inequalities make no sense if f is a map from S to S.>
What did you actually mean? I presume for A, the condition is
that the first component of f(x,y,z)> is >= x etc.Yes, right,
sorry. Shouldve saidA = set of (x,y,z) such that
f_1(x,y,z)>=xB = set of (x,y,z) such that f_2(x,y,z)>=yC = set
of (x,y,z) such that f_3(x,y,z)>=zwhere
sha1:JCIb+kK6OZuSwjc3gdVa2y6TZwU=Consider the following
proposition:For every natural numbers n,k there is a bijection
f between the setsA={1,2,3,...,k} and B={n+1,n+2,...,n+k} such
that x divides f(x) forall 1<=x<=k.Proof: Induction on k.
Clear for k=1, since the only map {1}->{n+1}satisfies the
condition since 1 divides n+1.Assume valid for k-1. Now in any
set of k consecutive numbers there isone that is divisible by
k. Put it away from B, remove k from A andcomplete the
bijection using the induction hypotheses. QEDI find this
example instructive since the statement is easy to grasp,and
not only works for k=1, but also for k=2. NOTE: Im posting
this only to share, I know where the induction goeswrong, but
Symmetric group question--HELP Adjunct Assistant Professor at
the University of Montana.Is there an EASY way of finding 
out
the orders of elements of any>symmetric group, and the number
of elements of each order? Every element of the symmetric
group S_n can be written uniquely as a product of disjoint
transpositions. Each product of disjoint transpositions
corresponds to a partition of n: a way to write n as a sum
of positive integers.I think you should say cycle instead of
pointing that out. Errare stupidum est[1]. The nextparagraph I
used the correct disjoint cycle decomposition. For each
partition of n, n = a_1+...+a_r, a_1>=...>=a_r, the order of
an element with disjoint cycle decomposition of type a_1,
Square root of a square matrix> Is there exist any method to
factorize a square matrix M > such that given any M, M=A*A ?If
M is diagonalizable, i.e. there exists S such thatinv(S)*M*S =
D is diagonal, then A = S*sqrt(D)*inv(S)where sqrt(D) is a
diagonal matrix whose entriesare square roots of D.Proof:A*A =
S*sqrt(D)*inv(S)*S*sqrt(D)*inv(S) = S*sqrt(D)*sqrt(D)*inv(S) =
S*D*inv(S) = S*(inv(S)*M*S)*inv(S) = MM is diagonalizable if
and only if you can form a setof n linearly independent
eigenvectors of M, in whichcase S is a matrix whose columns
are those eigenvectors.I dont know how youd 
search for a
square root ofa non-diagonalizable matrix, though certainly
atleast some have one. (Trivially, the zero matrixhas itself
algebra> can anyone explain how would I graph absolute value of
y plus absolute> value of x equal to 2. what kinda graph would
it be and how would I> get there. plz help. got stuck. I know
Im a dumbass, but help plz.In any region, such as a 
quadrant,
in which x and y do not change signs, the graph must be a
straight line. So consider the intercepts of such
something strange when I try to calculate cos(pi/2). I> keep
getting the result 6.12574E-17. Is that suppossed to be
essentially 0?> Cant Excel give you 0? I know that from
programming, but I thought MS> could have just programmed that
to represent 0. Whats up?> You dont have 
exactly pi/2. You
have a §oating-pointnumber which approximates pi/2 to 16
places, and thusthe cosine of that number approximates
polynomial> I ran across a problem in a book to try and
determine if there was a> polynomial p(x) with at least 2
nonzero terms such that p(x)^2 had> exactly the same number
of nonzero terms as p(x). I proved it> couldnt happen for
linear, quadratic, or cubic polynomials, and I> think I
proved it couldnt happen for quartic polynomials also. 
Then I found a quintic where it is true: Namely,
p(x)=4x^5+4x^3-2x^2+2x+1,>
[p(x)]^2=16x^10+32x^9+28x^6+4x^3+x^2.> Make that > 16 x^10
+ 32 x^8 - 16 x^7 + 32 x^6 - 8 x^5 + 20 x^4 + 4x + 1.Oops,
mustve been a mistype. Tryp(x) = 4x^5 + 4x^4 - 2x^3 + 2x^2 
+
xNote that since this latest p(x) is a multiple of x, > p(x)/x
= 4x^4 + 4x^3 - 2x^2 + 2x + 1 will work,> as will (x^n p(x))
for any positive integer n. Also q(x) = x^4 + 2x^3 -2x^2 + 4x
+ 4, reversing the order of > coefficients of p(x)/x, will
work, as will (x^n q(x)).I do not know whether it may be
relevant in searching for other examples, but q(x) = x^4 +
2x^3 -2x^2 + 4x + 4 is more easily solved than a general
quatric. As a difference of two squares:q(x) = x^4 + 2x^3
-2x^2 + 4x + 4 = (x^2 + x + 2)^2 - 7x^2Thus it has general
solution x = (-1 - s1 sqrt(7) + s2 sqrt( s1 2
sqrt(7)))/2,where s1 and s2 are sign options, each being +1 or
-1,and all 4 roots generated by the 4 joint options for s1 and
all,> I am doing some light (?) reading on topological
groups for kicks.  Unfortunately, my brain is slowing down.
A problem:> Let G be a topological group. If K,L
subseteq G are closed in G, do we necessarily have KL closed
in G? No. Take G = real numbers with the usual topology and the
group> structure given by the addition. Take K = Z (the
integers) and L => a.Z, where a is some irrational number.
Then K + L is a dense subset> of R, but it is not equal to R
SantosI see... I figured as much, but had trouble coming up
recurrence relations> Can anyone help me in analysing the
recurrence relations: x[n+1]=a.x[n]-b.y[n]
y[n+1]=c.y[n]+d.x[n] I am interested in the long term
behaviour of this model, for different values of a,b,c,d and
x[0] and y[0].> The beheavior depends on the eigenvalues of the
matrix [[a -b], [c > d]]. One good introduction is Luenberger,
Introduction to Dynamic > Systems.By the way, if the maximal
modulus of the eigenvalues is less than one, the system tends
to the origin in the long term. Greater than one, it generally
goes off to infinity.If your four coefficients are 
positve, the
matrix has only complex eigenvalues. This implies that the
that, but I am not used tousing Excel. I usually use
Mathematica, or something. I am just trying tolearn Excel so
that I can add it to my resume.Lurch> I am also getting
something strange when I try to calculate cos(pi/2).I> keep
getting the result 6.12574E-17. Is that suppossed to
beessentially 0?> Cant Excel give you 0? I know that from
programming, but I thought MS> could have just programmed that
to represent 0. Whats up?You dont have 
exactly pi/2. You have
a §oating-point> number which approximates pi/2 to 16 places,
and thus> the cosine of that number approximates cos(pi/2) up>
Rationals>I am trying to generalize the result cited by GAE,
viz.,For integer n>0, let f(n) be the largest integer m such
m/n < pi. Let>d(n) = pi - f(n)/n.Surprisingly, K. Mahler
showed that d(n) > 1/n^(42). Thatexponent 42 has been
improved subsequently. I guess the best currentresult is
8.0161, due to M. Hata.>As RI has pointed out, this must be
true not for all n, but for >sufficiently large n. My 
first
thought that such a bound is >surprising. But maybe something
like this is true for almost all >reals. Let me try again.I
believe that Mahler proved his result for all integers n >= 2,
whileHata proved his for n sufficiently large. 
Hatas bound is
probablytrue for all n >= 2. I dont know if anyone has
calculated how largeis sufficiently large in 
Hatas
proof.>Consider the set of positive numbers x with the
following property. >For all positive c and m, there exists an
integer n > m such >that n x - §oor(n x) < n^-c. Does this set
have measure 0?Yes. In fact, consider the inequality (*) |n x
- p| < f(n)where f is positive and continuous on (0,infinity)
with x f(x) non-increasing. If int_1^infinity f(x) dx
converges, then foralmost all x the inequality (*) has only
finitely many solutions inintegers (n,p). If the integral
diverges, then for almost all xthe inequality has infinitely
many solutions in integers (n,p). See e.g. Khinchin, Continued
Fractions, theorem 32. In particular, youre looking at f(q) 
=
q^(-c) with p = §oor(n x). If c > 1, the integral converges so
for almost all x, your inequalityhas finitely many integer
anyone know what the difference between a ring and a fieldis?
As far as I can tell they are subject to the same axioms, yet
Ihave been told that the set of integers are members of the
set ofrings, but not members of the set of fields. Also, if
there is adifference, is it true that rings are a subset of
fields> Hey, does anyone know what the difference between a
ring and a field> is?A field is a commutative ring 
with unit,
different from {0}, in which everynon-zero element is
invertible.If you throw away the commutativity, one speaks of
a division ring.A commutaitve ring can be considered to be a
subring of a field (i.e. isisomorphic to a subring of a 
field)
if, and only if, it is integral (thatis, different from {0}
and the product of two non-zero elements is non-zero,which is
difference rings and fields>Hey, does anyone know what the
difference between a ring and a field>is? As far as I can tell
they are subject to the same axioms, yet I>have been told that
the set of integers are members of the set of>rings, but not
members of the set of fields. Also, if there is a>difference,
is it true that rings are a subset of fields?>A 
field is a
commutative ring with 1 each of whose nonzero elements has a
multiplicative inverse. Put another way, the nonzero elements
under multiplication form an abelian group.Thus, all fields 
are
and fields Adjunct Assistant Professor at the University of
Montana.>Hey, does anyone know what the difference between a
ring and a field>is? A field must be commutative, 
AND every
nonzero element of the ringmust have a multiplicative
inverse.A ring need not be commutative, nor must every nonzero
elementhave a multiplicative inverse.>As far as I can tell they
are subject to the same axioms,There are two axioms that a 
field
satisfies but which a ring need notsatisfy: a 
field F is a ring,
which, in addition to all the ringaxioms, satisfies: x*y = y*x
for all x, y in Fand For all x in F, if x is not 0, then there
exists y in F such that xy=1.So the axioms of a ring are a
proper subset of the axioms of afield. Every field 
is a ring,
but not every ring is a field.> yet I>have been told that the
set of integers are members of the set of>rings, but not
members of the set of fields.This is nonsense as written. What
you mean to say, presumably, is thatthe integers, with the
usual addition and multiplication, are a ringbut not a
field.(There is no set of rings and there is no set of
fields)Yes, that is correct. Because it is not true that for
every nonzerointeger x there exists an integer y such that x*y
= 1. For instance,there is no integer y such that 2*1 = 1, but
2 is a nonzero integer.> Also, if there is a>difference, is it
true that rings are a subset of fields?Every field 
is a ring, but
not every ring if a field. So you would betrying to say that 
the
collection of all fields is contained in thecollection of all
multiplication>First, consider positive integers k and j with
m and n digits>respectively. Assume moreover that neither k
nor j is equal to 0. >Can we determine exactly the number of
digits in the product kj? Its>either m+n-1 or m+n, but 
which
one it is changes depending on the>nature of k and j. I wanted
to use this to determine the number of>digits in, say 2^64.
Now, this one might not be so difficult because>many people
know 2^32 off the top of their head so they know it has
10>digits. So they would know immediately that 2^64 either has
20 or 19,>but still its not clear (to me anyway) if 
its 20 or
19 without>multiplying it out. But still, what about 3^64? And
what about the>number of digits of a^b where a and b are any
positive integers?a^b has n decimal digits iff 10^(n-1) <= a^b
< 10^n. So n = 1+§oor(log_10(a^b)) = 1+§oor(b 
log_10(a)).For
top-of-the-head computations its useful to know somebounds 
on
these logs. If L < log_10(a) < U, then this willbe enough to
determine n unless ceil(bU) - §oor(bL) > 1.For powers of 2,
3/10 < log_10(2) < 28/93 which is sufficient to determine the
number of digits in 2^b if b < 103. For example,§oor(64*3/10)
= 19 = ceil(64*28/93)-1 so there are 20 digits in 2^64.For
powers of 3, 31/65 < log_10(3) < 21/44 which is sufficient to
determine the number of digits in 3^b if b < 109.