mm-34
===
Subject:
Yes, a long time ago.The business cycles has a varying
period and amplitude. Also cycles are notalways that clear when
you get 'double dips' and the such like.>> Has anybody ever
studied business cycle data with FFT-based spectral>
techniques?
===
Subject:
I think that the idea would then be to simply
examine the spectral function and see if it obeys a power law.
Has that ever been observed in the data?Power law dependence
would mean that it is aperiodic, that basis function
expansions are a waste of time (not counting wavelets) and
that active control wouldn't work. Passive control might still
work, (ie. setting interest rates to neutral and forgetting
about them) but I wouldn't bet on that either. I'm not up on
the math, but a few years ago there was a breakthrough in
aerodynamics, where engineers realized when a fighter jet went
into a tumble (chaotic trajectory) the pilot was causing it by
fighting the stick. If he let go of the stick, the plane would
spontaneously come out of the tumble.A sloppy person might be
inclined to simply try the same solution for stabilizing the
business cycle, but that would be a bad idea. It might be that
there is no stable limit cycle in the available range of
parameters except the one where all activity stops dead.This
would be the same as if the time constant for recovery were so
long that everybody starved before the economy came back. In
the absence of actual state equations for the process model, I
think you'd have to assume that this is the way things are and
that the business cycle is the best solution allowed by the
system.> Yes, a long time ago.The business cycles has a
varying period and amplitude. Also cycles> are not always that
clear when you get 'double dips' and the such> like.
===
Subject:
power
law ... would you mind to give a little bitmore than you
already said?I think that the idea would then be to simply
examine the spectral function> and see if it obeys a power
law. Has that ever been observed in the data?Power law
dependence would mean that it is aperiodic, that basis
function> expansions are a waste of time (not counting
wavelets) and that active> control wouldn't work.Passive
control might still work, (ie. setting interest rates to
neutral> and forgetting about them) but I wouldn't bet on that
either. I'm not up> on the math, but a few years ago there was
a breakthrough in aerodynamics,> where engineers realized when
a fighter jet went into a tumble (chaotic> trajectory) the
pilot was causing it by fighting the stick. If he let go> of
the stick, the plane would spontaneously come out of the
tumble.A sloppy person might be inclined to simply try the
same solution for> stabilizing the business cycle, but that
would be a bad idea. It might be> that there is no stable
limit cycle in the available range of parameters> except the
one where all activity stops dead.This would be the same as if
the time constant for recovery were so long> that everybody
starved before the economy came back. In the absence of>
actual state equations for the process model, I think you'd
have to assume> that this is the way things are and that the
business cycle is the best> solution allowed by the system.>
Yes, a long time ago.>> The business cycles has a varying
period and amplitude. Also cycles> are not always that clear
when you get 'double dips' and the such> like.
===
Subject:
If you do
not insist in FFT you may find a lot on timeseries analysis +
econometrics (i never had the patientsto work through J.
Hamilton) and even in TA it is used.The problem - as Rod says
- is data are disturbed andnot clearly follow one rule (you
need 'windowing').A concurrent approach is to set up a process
model andestimate parameters from market data. Your FFT willbe
translated to characteristic functions (take careto get not
lost in mathematical physics). For the timevarying aspect you
may dig for wavelets + econometrics.A third may be adaptive
low pass filter (being slow iprefer that). Anyway you need
lots of data. And maythink about ARMA / GARCH according to
your intention.Just my 2 Euro cents ...Yes, a long time
ago.The business cycles has a varying period and amplitude.
Also cycles are not> always that clear when you get 'double
dips' and the such like.>> Has anybody ever studied business
cycle data with FFT-based spectral> techniques?
===
Subject:
understand
what I am missing. The problem is as follows: The
Feige-Fiat-Shamir identification scheme is modified s.t.: -
Peggy sends each of y_1,...,y_k individually, where every y_i
=r*(s_i)^(b_i) - The verification in Step 4 is replaced with
checking (y_i)^2 *(v_i)^(b_i) = r^2 mod n. As before, there
are t rounds of the above protocol, so we can usethe notation
y_(i,j) for the y_i of round j, and similarily, b_(i,j)is the
b_i round of j. Of course, Victor can find out k-1 of the
s_i's if t=1, and canfind out all of the s_i's if t > 1 by a
judicious choice of theb_(i,j)'s, but we assume Victor is not
interested in doing so, so thateach b_(i,j) is random with
equal probability of being 1 or 0, andthen the b_(i,j)'s are
independent of each other. In this case, what is the
probability that Eve, who is recordingall the t rounds between
Peggy and Victor, can obtain s_1? This is my line of thinking:
The probablity we can obtain s_1 is based on two things: 1.
b_1 = 1, i.e. y_1 = r*s_1 2. b_k = 0, i.e, k != 1 i.e some y_i
= r P(1) = 1/2 P(2) = (2^(k-1)-1)/(2^(k-1)) P(pass in a single
round) = P(1) * P(2) = (1/2) - (1/2^k) P(pass in t rounds) =
P(1-fail in t rounds) = 1-(1-((1/2) -(1/2^k))) = 1-((1/2) +
1/2^k)^t This logic is somehow flawed. I have the posted
solution of 1 -2^(-t) - 2^(t-kt) + 2^(-kt) without a good
explination, which variesfrom my answer by one term. Can
anyone help me find the error inhere? -Chu
===
Subject:
>> But how do we
know (from theoretical considerations) that the batteries >
aren't using up their chemical energy fighting each other to
keep a current> from flowing???They aren't. The space charge
in the diode is doing that. There is a static field caused by
the accumlation of charge that creates the space charge.
Basically you form a capacitor in the diode.
===
Subject:
> There are
instances in physics where we encounter transcendental>
equations of the form:exp(-constant*x) = x-r1where r1 is a
constantHas anyone encountered real-life cases
of:exp(-constant*x) = (x-r1)*(x-r2)where r1 and r2 are real
numbers?> Well, I don't think that's exactly the kind of
example you would liketo get, but once I had an experince
almost like the one you mentioned.I work on the planning of
the expansion of electric systems and I wastrying to adjust,
by the least square method, a curve that representedthe
marginal cost C of a thermal plant as a function of it's
factor ofutilization, f. Through simulation programs I got a
table relating thevalues of C to several values of f. The
curve I found was divided into2 branches, the fist
comprehending the interval [0.2 , k] and theother the interval
[k, 0.96]. In the first branch the curve was givenby a 3rd
degree polynomial and in the second by a negativeexponencial.
Since the curve was supposed to be differentiable, at thepoint
k we had an equation almost like yours.I used an Excel
spreadsheet to find the the value of k, the values ofthe
coefficients of the polynomial the constants of the
exponentialcurve.Artur
===
Subject:
> I bet you already knew that cos pi
= -1 and such. But I was shocked to find,> just now, that there
appears to be a _vegetable_ in the domain of cos x.> (Yes, you
read that correctly. Obviously this is a silly
pseudo-mathematical> post, but then you could tell that from
the thread's title.)What is that vegetable? What is the value
of cos x there?(I guess I should wait for other people to try
to answer first. But this is> far too silly to be drawn out
that way, and so I've given the answers below.)DavidThe
vegetable is lettuce.According to my dictionary:cos lettuce =
romaineYou're wrong, lettuce is NOT in the domain of cos. It
would be if itwas cos(lettuce) = romaine, which is not the
case.Amanda
===
Subject:
>>I have never reviewed material for exams
when I took the>course.>> I like to say that a test
which can be studied for is not worth> giving (or
taking).Been to a medical doctor lately? I'm glad mine studied
for their exams.What is the graduate with the lowest GPA in a
Medical School class> called?Doctor.What do you call the
person with the least achievement to graduatefrom Stanford
Chem with a Ph.D.?Doctor Schwartz> A license to practice
medicine isn't meaningful until the doctor does> years of
clinical work supervised by nurses (re green lieutenants and>
seargents) and senior doctors. An MD is not a license to
practice medicine. What you are trying todiscuss above is
called a Residency. After MD, before license. Without it, you
don't see patients unsupervised.
===
Subject:
> An MD is not a license
to practice medicine. What you are trying to> discuss above is
called a Residency. After MD, before license. > Without it, you
don't see patients unsupervised.Here's something not to do:When
lying on the emergency room table and the ER Dr. is stitching
up your hand (where there are a LOT of nerve endings)and he
tries to distract you from the fact that the pain-killing shot
hasn't really taken effect, by making small talkwith you, by
saying:So, what do you do for a living?DO NOT SAY I'm a math
prof.Because odds are that Calculus almost kept the poor
bastardout of med school and he may inform you of that fact
whilehe continues to ram that needle in and out of your
hand.Indeed, sewing may become throwing stitches. Case in
point, my ER Dr. put stitches through my thumbnail.In one side
and up through the other. I'd have paid extranot to have that
done. What the heck was wrong with adhesivetape. Or duct tape.
I didn't give him his stupid C-, but Isure took the hit for it.
The ONLY saving grace of the entire experience was that it
wasn't a groin injury.
===
Subject:
What are your credentials Niles? Are
you only capable of whining andbitching?
===
Subject:
> Anyone know why
Grad Schools will not accept your application unless> you
have a Bachelor's? Is it not possible to have the required>
knowledge and no degree? Is it so very hard to verify that
one's claim> of knowledge.>> some reasons:>> 1. it
invalidates the whole college education schtick.> 2. lack of
institutional admissions exams.> 3. the general population
lacks the discipline to pull off such> thing as
self-education.>> now to address your latter question.>> it
wouln't be very hard to verify knowledge, but it is
politically> incorrect and somewhat costly and
time-consuming.Which is why grad schools can't be bothered
unless you indicate that you are> special. You don't get
special consideration unless you're a special> person.well, it
seems that our advise seeker is reporting that the
typicalgraduate programme does NOT offer such special
consideration, even tospecial persons.> Jon Miller
===
Subject:
Suppose
we take the chess rules and add a new one:A1: You have the
option of moving two pieces simultaneously, as long as each
move considered separately was legal in 'single move' chess.
The two pieces can't land on the same square, obviously.So you
can't move a pawn, then the bishop behind itwhich was
previously blocked by the moved pawn. Or,if your king is
checked, you move your king,and move another piece to kill the
check.I'm just wondering how drastic would a change be?Would a
typical chess game have half the number ofturns for each
player? Would the games be faster?Would there be some opening
trick that white can playthat will always guarantee a win in a
few turns?Now, if we were to look at the entire decision
treefor double-move chess, it would be bigger thansingle-move
chess, because double-move chess wouldhave all the moves of
single-move chess too.But now, let's look at the decision tree
for thosedouble-move chess games where two pieces were
alwaysmoved if legal to do so. Would the tree be roughlytwice
as wide but half as deep as the single-movechess decision
tree?I know silly questions... just curious to know
theanswer.
===
Subject:
> Suppose we take the chess rules and add a new
one:A1: You have the option of moving two pieces
simultaneously,> as long as each move considered separately
was legal in> 'single move' chess. The two pieces can't land
on the> same square,
obviously.http://homepages.stayfree.co.uk/gpj/gvc.htmand
scroll down to Two-Move Chess.Marco--
===
Subject:
> Suppose we take
the chess rules and add a new one:A1: You have the option of
moving two pieces simultaneously,> as long as each move
considered separately was legal in> 'single move' chess. The
two pieces can't land on the> same square, obviously.> So you
can't move a pawn, then the bishop behind it> which was
previously blocked by the moved pawn. Or,> if your king is
checked, you move your king,> and move another piece to kill
the check.> I'm just wondering how drastic would a change be?>
Would a typical chess game have half the number of> turns for
each player? Would the games be faster?Would there be some
opening trick that white can play> that will always guarantee
a win in a few turns?> It would probably go much faster. One
thing that might help is to give white one move, then black
gets 2 and they alternate 2 moves from there on.Two moves in a
row is such a huge advantage that in Monster Chess (white gets
Ke1, Pc2d2e2f2 and two moves per turn, black gets the normal
setup and one move per turn) is pretty much a forced win for
white.--Harold BuckI used to rock and roll all night, and
party every day. Then it was every other day. . . . -Homer J.
Simpson
===
Subject:
InHarold Buck <no_one_knows@attbi.com> typed:> Two
moves in a row is such a huge advantage that in MonsterChess>
(white gets Ke1, Pc2d2e2f2 and two moves per turn, black
getsthe> normal setup and one move per turn) is pretty much a
forced winfor> white.On the contrary, with careful play, Black
should win. It isn'teasy, and if you never seen it before,
Black is very hard toplay, but nevertheless it's better for
Black.-- Ken BlakePlease reply to the newsgroup
===
Subject:
> In>
Harold Buck <no_one_knows@attbi.com> typed:Two moves in a row
is such a huge advantage that in Monster> Chess> (white gets
Ke1, Pc2d2e2f2 and two moves per turn, black gets> the> normal
setup and one move per turn) is pretty much a forced win> for>
white.> On the contrary, with careful play, Black should win.
It isn't> easy, and if you never seen it before, Black is very
hard to> play, but nevertheless it's better for Black.Really? I
thought I'd read that it was hugely in white's favor. In any
case, the fact that it's even playable shows how important
those extra moves are.--Harold BuckI used to rock and roll all
night, and party every day. Then it was every other day. . . .
-Homer J. Simpson
===
Subject:
InHarold Buck <no_one_knows@attbi.com>
typed:> In>> Harold Buck <no_one_knows@attbi.com>
typed:>> Two moves in a row is such a huge advantage that in
Monster>> Chess>> (white gets Ke1, Pc2d2e2f2 and two moves per
turn, blackgets>> the>> normal setup and one move per turn) is
pretty much a forcedwin for>> white.>> On the contrary,
with careful play, Black should win. It isn't>> easy, and if
you never seen it before, Black is very hard to>> play, but
nevertheless it's better for Black.> Really? I thought I'd
read that it was hugely in white's favor.In any> case, the
fact that it's even playable shows how importantthose extra>
moves are.Yes, I agree.-- Ken BlakePlease reply to the
newsgroup
===
Subject:
>> Find the natural numbers k, as many as
possible, so that all the> solutions of the scalar equation
x'' = x^k can be extended> infinitely far?Do you mean you want
there to be a solution x(t) with a domain ofthe form (a, infty)
?> d(x^(k+1))/dt = (k+1)(x^k)x'> = (k+1)(x'')(x')> =
((k+1)/2)(d((x')^2)/dt)This tells you that x^(k+1) - (k+1)/2
(x')^2 is a constant, C. Thus dx/dt = sqrt( 2/(k+1) x^(k+1) -
C ). You can solve this byseparation of variables: t will be
an antiderivative of1 / sqrt( 2/(k+1) x^(k+1) - C ). More
accurately, if (a0, b0) and (a1, b1) are two points on
thegraph of x(t), then a1 - a0 = int_b0^b1 dx / sqrt( 2/(k+1)
x^(k+1) - C ). Since you want the domain to extend to
infinity, you need theintegral to grow unbounded as b1
increases. But your integrand isbig-O of x^{-(k+1)/2}, and so
the integral stays bounded if(k+1)/2 > 1, i.e. if k > 1.For
k=1, of course, all solutions can be defined across the whole
line.dave
===
Subject:
>> Find the natural numbers k, as many as
possible, so that all the> solutions of the scalar equation
x'' = x^k can be extended> infinitely far?>> d(x^(k+1))/dt =
(k+1)(x^k)x'> = (k+1)(x'')(x')> = ((k+1)/2)(d((x')^2)/dt)>>
How could I do this?>> --What do you mean by extended
infinitely far?Lurch
===
Subject:
> Also, can we have things such as
e^(quat) or ln(quat) where quat is a>> quaternion? Has anybody
ever figured out how to do this? Also, what>> about using lie
algebras which work on quaternions?>Any quaternion q can
be written a+bu, where a,b are real and u^2 = -1.>So the set
of {x+yu:x,y real} for this fixed u is isomorphic>to the
complex numbers. Compute e^q and ln(q) in it just as in
the>complex numbers.>>Generalizing functions of two variables
is where problems come, if you>use two quaternions that don't
commute.>>-- >G. A. Edgar
http://www.math.ohio-state.edu/~edgar/This proves that
representing quaternions as 3x3 matrices is notexact, since
they can at most have 2 eigenvalues, whereas 3x3
matricesusually obey a cubic.
===
Subject:
>Problem (Calculus of
variations)> b> />minimise s= | (1+u^2)^0.5 dx for u(x)> />
a>>where s>0 and b>a, subject toDid you mean to write u > 0?>>
b> /> |u dx - c = 0 c>0> /> a>>and b> / du> | --- dx - m = 0
m>0> / dx> a>when m = 0, the solution is trivial, namely
u=constant. >>I am interested in the case m>0.>You can use the
standard variation formula to show that the value ofthe
functional for any non-constant u(x) can be decreased by
changingu near two points where the value of u differs, in
such a way as toleave the integral of u equal to c (and
leaving the end points fixed).This shows that, in general,
there is no continuous minimizingfunction. A generalized
solution is a function u(x) that equalsc/(b-a) everywhere
except at x = b, where the value is chosen tosatisfy the last
condition.John Mitchell
===
Subject:
>I have a partial difference
equation whose sum over>> a certain range can be used to count
prime numbers.>>2) Presumably you are going to justify why
that's true at some point>Who cares if Mr. Harris can justify
it? Are you really interested in hisexplanation? The
interesting questions about his formula are: (1) Over
whatrange does the method work? and (2) Is it faster than
other methods on thisrange? Simple experiments should answer
both questions. rich
===
Subject:
>I have a partial difference
equation whose sum over> a certain range can be used to
count prime numbers.>>2) Presumably you are going to justify
why that's true at some point>>Who cares if Mr. Harris can
justify it? Are you really interested in his> explanation? The
interesting questions about his formula are: (1) Over what>
range does the method work? and (2) Is it faster than other
methods on this> range? Simple experiments should answer both
questions. richI care. Sufficiently many people whose opinions
I respect havedecided it will count primes, in a manner similar
to Legendre's method. Iwould like to see James, who has
consistently denied it has anything to dowith Legendre (in
fact for a while he even refused to admit Legendre'sformula
could be used to count primes, so how he knew it was different
fromhis is a mystery), prove he understands what's going on and
can justifyhis claims that it is an all new and powerful method
that no one else 'inthe history of mankind' has discovered. And
just proving a range in which it works is a start. Unlike all
the rest of his maths that he's shown recently, I think he may
well have something here. I just don't see that it is new,
exciting, or useful. He has also in the currently running
debate on his crankhood repeatedlysaid he wants a point by
point critique by a mathematician, withoutinsults, so that he
can rebut them all. So there were some observations by someone
who's not a number theorist. Ifanyone who does number theory
seriously wants to point out where he or Iam misunderstanding
something (which possibly I am too) please tell me.As it is
the thread currently gives the lie to James assertion that
hewould respond to any fair criticisms. I'm not remotely
interested in insulting him, or debating his mentalhealth; it
seems counterproductive. I am interested in people learning
more mathematics, and that includes me.
===
Subject:
>Dear Group
Members,>>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.
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) = r1>>where
x is the (column) matrix of unknowns, a and b are column
matrices and>A,B are symmetric matrices. r0 and r1 are
scalars.>>2) I know about how to transform the quadratic forms
into their>eigensystems, but then still can see no systematic
way of how I could>eliminate one of the unknowns.>>3) I guess
the order of my (specific) problem is at most of degree
4.>>Any pointers appreciated.>>Roland>>BTW.: Of course I would
be interested how to solve for more than two>equations
too.>>First expand both equations and rewrite them as M1 v =
v1, M2 v = v2, where in x has dimension n, v has dimension
equal to all quadratic, plus all linear, plus one constant,
terms.Now if M1 is invertable, M2 (M1^-1) v1 = v2. So it is
possible you have no solution at all.
===
Subject:
> (x,y)-->
sqrt{x^2+y^2} cos (arctg y/x) if x=/=0> comes from thinking
about (r,t)-->r cos(t)> it isn't lineartime to review polar
coordinates
===
Subject:
Mystery remains as journal withdraws paperJOHN
WHITFIELD[LONDON]A mathematics journal has withdrawn a paper
that claimed to crack one of thediscipline's great mysteries
after reviewing and accepting the work andpublishing it
online.On 18 November, Nonlinear Analysis published a paper by
Elin Oxenhielm - apostgraduate student in mathematics at the
University of Stockholm, Sweden -sixteenth problem, one of a
set of challenges laid out by GermanIf a solution were
validated, mathematicians agree, it would be asignificant step
towards a complete solution to the problem. Oxenhielmpredicts
just that: We could find one in a year or so, if we're lucky,
shesays.The work was described in a 24 November press release
from Oxenhielm andcovered in several media outlets including
the BBC. But the paperimmediately came under fire from
mathematicians. It's completelyinadequate - I can't imagine
who would have thought it was a proof, saysJohn Mather of
Princeton University, New Jersey.Critics include Oxenhielm's
supervisor, Yishao Zhou, who put a statement onher website
saying: The paper is incomplete and contains serious
mistakes.research. Solving any one of them is almost
guaranteed to make amathematician's name, and by 2000 all but
three had been solved.The sixteenth, the problem of the
topology of algebraic curves and surfaces,deals with the
territory where geometry meets algebra. Its second
partinvolves showing that the number of periodic solutions to
a differentialequation is finite.Such periodic solutions are
also known as limit cycles - stable, oscillatingtrajectories
to which a system will return if perturbed. Limit cycles
arecommon in nature, and a proof of the second part could lead
to a betterunderstanding of heartbeats, animal movements and
the kind of runawayvibrations that can shake a structure to
bits.Oxenhielm formulated her proof using 'describing
functions' - which canpredict roughly the presence of limit
cycles in nonlinear equations.A few minutes' scrutiny is
enough to show that her reasoning is false, saysmathematician
Grigori Rozenblioum of Chalmers University of Technology
inGothenburg, Sweden. The approximate solutions studied by
Oxenhielm cannotprovide the exact answers demanded in a proof,
he says, and some of herequations contain exact terms where
approximate ones should be used.The work should never have
been published, Rozenblioum says: It'simpossible to understand
the behaviour of the journal, which is one of theleaders in its
field.Nonlinear Analysis pulled the paper on 4 December.
Publication has beenhalted until a thorough investigation into
the matter has been handled,says editor-in-chief V.
Lakshmikantham, a mathematician at the FloridaInstitute of
Technology in Melbourne.Originally approved by one reviewer,
the paper has now been sent to two moremathematicians for
further round of review, along with a defence byOxenhielm, who
says that the critics do not understand her methods.She refuses
to comment further. Nonlinear Analysis' editors have
evaluatedthe paper, they accepted it for publication and they
have the copyright ofits contents - and thus they are
responsible for its correctness, she toldNorwegian newspaper
Aftenposten.--------------------------------------------------
------------------------------
===
Subject:
> [The author] refuses to
comment further. Nonlinear Analysis'> editors have evaluated
the paper, they accepted it for publication> and they have the
copyright of its contents - and thus they are> responsible for
its correctness, she told Norwegian newspaper>
Aftenposten.Remarkable. Watch out, James. Another eminently
.siggable quotefactory.But can she maintain this standard?--
Run mathematicians, RUN!!! I'm coming for you. It may take a
fewmonths, but I'll get [computer verification of my proof]
and then yourlives will be ended as you previously knew it. --
JSH meets PVS
===
Subject:
>That is the theorem...but the converse
is:>>If a triangle is inscribed in a semicircle with the angle
opposite to the>hypoteneuse as 90 then the hypoteneuse is the
diameter...>>How would you prove that ? I tried it many times
and I get stuck>Well that is why I put a question mark at my
statement of theconverse. I wondered if you asked what you
meant to ask. Here is whatyou asked in the original post:>How
do you prove the converse of this theorem?>If a right triangle
is inscribed in a semicircle then the hypoteneuse is
the>diameter of the circleSo you meant to ask how to prove
this theorem, not its converse.--Lynn
===
Subject:
>> Dear
Nutjob,>> Stay out of alt atheism with your insane claims.>>
There I said something else to you. What's my full legal
name?>> 100,000 witnesses is a well founded claim, just
examine the evidence>> someone to guess the names, its all in
google.>You are the one making the claim you can tell a
persons name by what they>said to you.>Your answer is
meaningless.> 6 witnesses admit I'm the Truman, all in
google. I've got evidence and>> witnesses, you're just a
heathen.>Sadly people like you need help but probably don't
qualify under 5150H&S.>However I bet that if you live in
California you are familiar with the>section.> He lives in
Townsville, Queensland. A population of about 140,000> people,
businesses, and major tourism. Strangely though, he seems>
unable to find employment, or a woman.>> That should tell you
something. Under 5150 if you are a danger to yourself or
others or can't take care ofyourself you can be committed for
72 hours. A doctor can over ride this. Itis a good and
powerful law that will probably be overturned one day
becauseof it's abuse. I once had some cops that wanted me to
commit a woman becauseshe didn't agree with her boy friend.
The cops (I used to be one) were notsmart enough to know they
could do it.This guy is sick and I should not have responded
to him.He is plonked.
===
Subject:
> is a good and powerful law that
will probably be overturned one day because> of it's abuse. I
once had some cops that wanted me to commit a woman because>
she didn't agree with her boy friend. The cops (I used to be
one) were not> smart enough to know they could do it.at least
you admit you're not smart and corrupt>> This guy is sick and
I should not have responded to him.> He is plonked.You can't
silence the truth, but you can PLONK it!after he actually
contacted his famiily in townsville, i assume because
theyconfirmed my storythis was his last post in aus.tv If what
you say is true, wouldn't I be able to confirm what you say by
'hearing' your thoughts or watching a broadcast of your life
on television? I have satellite cable, why can't I get the
Herc show? My brother in law and his wife life in Townsville,
and, amazingly enough, they can't seem to find this show on
their television either. Nor have they ever heard of you or
your demented claims. My partner's sister who lives in
Rockhampton seems surprisingly unfamiliar with The 24 / 7 Herc
Show!, starring everyone's favouriteWhy not use the imprison
without charge law on all applicants to skepticcompanies?
They're all insane right, cut out the middle man of
formalapplicaitons, test procedure, just put up a big million
dollar prize andsend all the names and addresses to the mental
lock up! Pigs are soentrenched with the skeptic companies
already wave a INC at a pigand you can lock up any unemployed
person you want.Noone examining my claim of 3 extraordinary
posts on the 1 day, just groupabuse from 5 newsgroups I've
spent numerous months at each not one person to giveme any
benefit of doubt for a few minutes investigation into my
claim, doesthe 3 question test have an answer, are replies to
herc's authors guessable?Very well, if YOU wont cooperate with
fellow man then you can't be ruled with religion,you will be
ruled with force like you are now under a real nutcase Bush
who floodsmy life with noise from a spy satellite, believe me
you would have preferred the former.HercEVIDENCEB Rust 3
attacks metal http://tinyurl.com/nd56E Rich Shewmaker 1 rich
showmaker James Randi http://tinyurl.com/nd52H See You In Hell
My Friend. 2 It all depends http://tinyurl.com/nd53predictions
that paranormal will be demonstrated on 02
02.http://tinyurl.com/yo1b one year
beforehttp://tinyurl.com/nw1r one day
before-----------------------------Then I have 100,000
witnesses that I'm paranormal, here they admitthat I am the
TRUEmanhttp://tinyurl.com/iky5 http://tinyurl.com/iky8
http://tinyurl.com/iky9http://tinyurl.com/iky4
http://tinyurl.com/rv5f http://tinyurl.com/v1yfHey
Trueman...love the show. YOU ARE the Truman I heard him.Very
spooky! We're sick of you You rule Truman.>Is the truman is
living in Townsville? I've been hearing stuff, yeah.and watch
here live in Google as I predict back in 2000 Jim Carrey's
next romance costarLaurie Holden with http://tinyurl.com/fuf8
she looks exactly like Laurie Holden of the x-fileswith bright
green eyes.Just like any Truman could have predicted, Majestic
hey? Adam and Eve
===
Subject:
Let X be compact Hausdorff, let {A_i} be
a collection of closed connectedsubsets of X that is simply
ordered by proper inclusion. Prove Y ={intersection of A_i
over all i} is non-empty and connected.How can I prove
this?The way I want to start is to suppose U,V are disjoint
non-empty opensubsets of X. Then I want to analyze A_i - (U
union V) over all i....canI get anywhere with this? How can I
complete the proof?Steven
===
Subject:
= some K1,..Kn with K1 /../ Kn
subset UNow assume some open disjoint U,V with /C subset U/Vand
use the exercise. I don't think Hausdorff is needed./ = cap =
intersection----
===
Subject:
>Let X be compact Hausdorff, let {A_i} be
a collection of closed connected>subsets of X that is simply
ordered by proper inclusion. Prove Y {intersection of A_i
over all i} is non-empty and connected.>How can I prove
this?>The way I want to start is to suppose U,V are disjoint
non-empty open>subsets of X. Then I want to analyze A_i - (U
union V) over all i....can>I get anywhere with this? How can I
complete the proof?Suppose that U union V contain the
intersection of the A_i. Show thatit must contain some A_i.
Compactness is helpful.--Dan Grubb
===
Subject:
> Let X be compact
Hausdorff, let {A_i} be a collection of closed connected>
subsets of X that is simply ordered by proper inclusion. Prove
Y  {intersection of A_i over all i} is non-empty and
connected.How can I prove this?The way I want to start is to
suppose U,V are disjoint non-empty open> subsets of X. Then I
want to analyze A_i - (U union V) over all i....can> I get
anywhere with this? How can I complete the proof?Consider
this: let U be an open subset of X such that cap A_i subsetU.
Prove that A_i subset U for some i (hence, for all
sufficientlylarge i--in the sense of the order). Why? Because
if not, A_i capU^c is nonempty, compact and has the finite
intersection property,hence has empty intersection.--Ron
Bruck
===
Subject:
Suppose that the random variables X_i are independent
and P(X_i = 2^k) =2^(-k) for all k >= 1.Show that (X_1 + X_2 +
... + X_n)/(n*log(n)) converges in probability tolog(2).Prove
or disprove that limsup (X_1 + ... + X_n)/(n*log(n)) =
infinitywith probability 1.Can anyone prove this
problem?Sincerely,Rob
===
Subject:
>> Suppose that the random variables
X_i are independent and P(X_i = 2^k)  2^(-k) for all k >=
1.>> Show that (X_1 + X_2 + ... + X_n)/(n*log(n)) converges in
probability to> log(2).> Prove or disprove that limsup (X_1 +
... + X_n)/(n*log(n)) = infinity> with probability 1.>> Can
anyone prove this problem?Well, Feller can.For the first part,
see Feller's An Introduction to ProbabilityTheory and Its
Applications: Volume 1. In my 3rd edition (1968),it's in
section X.4 starting on p. 251.Feller also gave a very
complete answer to the second problem in a1946 paper (A limit
theorem for random variables with infinitemoments. Amer J.
Math. 68, 257-262), but that result is much moregeneral than
you need.You should be able to prove/disprove the second part
fairly directlyusing Borel-Cantelli. Think about the events
{X_n > n log n}. Dothey occur infinitely or finitely often? If
the former, you should beable to easily show that the limsup is
infinite. If the latter, youcan't quite prove that the limsup
is finite, but you can get close.If you replace n log n with
something smaller such that the events*still* happen only
finitely often, you ought to be able to do it.Does that
help?-- Kevin <buhr@telus.net>
===
Subject:
>> Suppose that the random
variables X_i are independent and P(X_i = 2^k)  2^(-k) for
all k >= 1.>> Show that (X_1 + X_2 + ... + X_n)/(n*log(n))
converges in probabilityto> log(2).> Prove or disprove that
limsup (X_1 + ... + X_n)/(n*log(n)) = infinity> with
probability 1.>> Can anyone prove this problem?>> Well, Feller
can.>> For the first part, see Feller's An Introduction to
Probability> Theory and Its Applications: Volume 1. In my 3rd
edition (1968),> it's in section X.4 starting on p. 251.>Where
can I see your 3rd edition? What is your textbook called?(more
text below)> Feller also gave a very complete answer to the
second problem in a> 1946 paper (A limit theorem for random
variables with infinite> moments. Amer J. Math. 68, 257-262),
but that result is much more> general than you need.>> You
should be able to prove/disprove the second part fairly
directly> using Borel-Cantelli. Think about the events {X_n >
n log n}. Do> they occur infinitely or finitely often? If the
former, you should be> able to easily show that the limsup is
infinite. If the latter, you> can't quite prove that the
limsup is finite, but you can get close.> If you replace n log
n with something smaller such that the events> *still* happen
only finitely often, you ought to be able to do it.>I don't
see how you would do this. How about this approach :What we
want to show is P(limsup (X_1 + X_2 + .... + X_n)/nlog(n)
=infinity) = 1where log is base e. (correct?)Now, is the
following proposition true?For any sequence of independent
random variables {Z_k}, ifSummation (from k = 1 to infinity)
of P( |Z_k| > epsilon) = infinity forall epsilon > 0,then Z_k
converges almost surely to infinity. (Maybe I can prove this
withborel-cantelli lemma of independent events?)If true, then
I'd like to use Markov's inequality, by saying thatP((X_1 +
X_2 + .... + X_n)/nlog(n) > epsilon) <= (1/epsilon) * E((X_1
+X_2 + .... + X_n)/nlog(n))) or some variationof markov's
inequality, and then sum up these values to possibly
getinfinity and our proof would be done.If this doesn't work,
how can I view the events {X_n > nlog(n)}? I agreethis would
be helpful. I don't see how I can tellif they occur infinitely
often. Furthermore, if this were true, how wouldthis imply that
the limsup is infinite? And which Borel-Cantelli lemmawould you
use?(a bit lost)> Does that help?>> -- > Kevin
<buhr@telus.net>
===
Subject:
>> For the first part, see Feller's An
Introduction to Probability>> Theory and Its Applications:
Volume 1. In my 3rd edition (1968),>> it's in section X.4
starting on p. 251.>> Where can I see your 3rd edition? What
is your textbook called?> (more text below)Sorry, I meant my
3rd edition copy of Feller's An Introduction toProbability
Theory and Its Applications: Volume 1.> Now, is the following
proposition true?>> For any sequence of independent random
variables {Z_k}, if> Summation (from k = 1 to infinity) of P(
|Z_k| > epsilon) = infinity for> all epsilon > 0,> then Z_k
converges almost surely to infinity. (Maybe I can prove this
with> borel-cantelli lemma of independent events?)No, it's not
true. Constant random variables (i.e., randomvariables that are
always equal to a specific constant) areindependent of
anything, so any deterministic sequence is a sequenceof
independent random variables, too, but your proposition fails
forZ_k=-k (where the limit is -infinity instead of infinity)
or Z_k=k fork even and =0 for k odd (where there is no
limit).It *is* true that a sequence of nonnegative,
independent randomvariables {Z_k} satisfying
sum_{k=1}^infinity P(Z_k > epsilon) = infinityfor all epsilon
> 0 will have limsup Z_k = infinity, and this can beproved by
Borel-Cantelli. (Note that we have added the hypothesis
ofnonnegativity and changed the conclusion to specify the
limsup ratherthan the limit.)> If true, then I'd like to use
Markov's inequality, by saying that> P((X_1 + X_2 + .... +
X_n)/nlog(n) > epsilon) <= (1/epsilon) * E((X_1 +> X_2 + ....
+ X_n)/nlog(n))) or some variation> of markov's inequality,
and then sum up these values to possibly get> infinity and our
proof would be done.The problem is that the inequality goes the
wrong way. To apply theabove result, you need to show that the
sum of the probabilities isinfinite, but the inequality will
only let you conclude that this sumis less than or equal to
something, and that won't help.> If this doesn't work, how can
I view the events {X_n > nlog(n)}? I agree> this would be
helpful.Well, if n=6, say, then nlog(n) is about 10.75. So,
how would you goabout calculating P(X_n > nlog(n)) = P(X_6 >
10.75)? You should beable to generalize this to an arbitrary
n.> I don't see how I can tell if they occur infinitely
often.> Furthermore, if this were true, how would this imply
that the limsup> is infinite?Well, okay, I spoke too glibly.
It still wouldn't be *quite* enough,but it would tell you
that, infinitely often (i.e., whenever X_n >nlog(n)), you'd
have to have: (X_1 + ... + X_n)/nlog(n) > X_1/nlog(n) > 1Now,
if you could do the same thing with {X_n > a n log(n)} for
everya > 0 (instead of just for a=1), *then* that would be
enough. Do yousee why?> And which Borel-Cantelli lemma would
you use?Which do you have? The two I know about are the one
that applies whenthe sum of the probabilities is infinite and
the other that applieswhen the sum of the probabilities is
finite. So, once you'vedetermined whether the sum of the
probabilities is infinite or finite,you'll know.-- Kevin
<buhr@telus.net> <br8sum$bg57$1@netnews.upenn.edu>
===
Subject:
> If
this doesn't work, how can I view the events {X_n > nlog(n)}?
I agree> this would be helpful. I don't see how I can tell> if
they occur infinitely often. Furthermore, if this were true,
how would> this imply that the limsup is infinite? And which
Borel-Cantelli lemma> would you use?> Consider the events A_n
= {X_n > a n log n} where a is a constant.These are
independent events. Show that sum P(A_n) = infty(for any
choice of a).
===
Subject:
> If this doesn't work, how can I view the
events {X_n > nlog(n)}? Iagree> this would be helpful. I don't
see how I can tell> if they occur infinitely often.
Furthermore, if this were true, howwould> this imply that the
limsup is infinite? And which Borel-Cantelli lemma> would you
use?>> Consider the events A_n = {X_n > a n log n} where a
is a constant.>My apologies. These are discrete random
variables. Now it appears that sum P(A_n) = infty, as I plug
numbers into my calculator, but I need tocome> These are
independent events. Show that sum P(A_n) = infty> (for any
choice of a).>>Cancel-Lock:
sha1:fo+GBypctlzmhtte3qiDdDA9jI4
===
Subject:
> My apologies. These
are discrete random variables. Now it appears> that sum P(A_n)
= infty, as I plug numbers into my calculator, but> I need to
come up with a rigorous proof of this.Yes, and calculators can
be deceiving. Rumour has it that log(log(n))goes to infinity as
n does, but just try proving that with acalculator.-- Kevin
<buhr@telus.net>
===
Subject:
> If this doesn't work, how can I view the
events {X_n > nlog(n)}? Iagree> this would be helpful. I don't
see how I can tell> if they occur infinitely often.
Furthermore, if this were true, howwould> this imply that the
limsup is infinite? And which Borel-Cantelli lemma> would you
use?>> Consider the events A_n = {X_n > a n log n} where a
is a constant.>This is where my problem is....how do I
calculate P(A_n)? I understandthey are independent events and
then by Borel-Cantelli I am golden.But all I have is that
P(X_i = 2^k) = 2^(-k)...how can I calculateP(X_n >
an*log(n))?I really think this must be a stupid question I am
asking but I am justforgetting somethingfrom my elementary
probability...isn't P(X_i = 2^k) = 2^(-k) my
densityfunction?Then do I just integrate from -infinity to
an*log(n)? but I must change thedensityfunction to P(x) =
2^{-log(x)/log(2)} since I need a density function forP(X_i =
x), not P(X_i = 2^k)...I think I am just getting more and more
confused...> These are independent events. Show that sum P(A_n)
= infty> (for any choice of a).>>
===
Subject:
: : > : Anyway, my point
had nothing to do with path categories specificallyThis almost
looks like a mis-attribution; despite the fact that my nameis
the last one occurring before it, the above line was written
by the otherarguer, as was the following one: : > : it was
entirely about your refusal to accept the general phenomenon
that : > : different ways of looking at the same mathematical
objects can lead to : > : better insights about those objects.
: > : > All I can say to that, frankly, is that you are
starting to deserve : > insults at the same level at which you
are dishing them out. : : What insult?This insult: your refusal
to accept the general phenomenon that different ways of looking
at the same mathematical objects can lead to better insights
about those objects.I have NOT so refused and I am not so
refusing now.Or, as I said it better the first time, : > I DO
NOT refuse to accept the general phenomenon that : > different
ways of looking at the same mathematical objects can : > lead
to better insights about those objects. ... : > In projective
geometry, points : > and lines are REALLY dual. : Perhaps your
understanding of projective duality is different from
mine.Duality is one thing. The fact that it actually occurs in
projectivegeometry is another. But it does NOT occur (not
between points and linesanyway) in affine geometry. It is a
DIFFERENCE between affine and projective geometries that
points&lines ARE dual in projective geometry, but ARE NOT
dualin affine geometry.Unfortunately (terminology-wise), it is
also the case that for any givenaffine plane, there are some
fairly simple things you can do to it toturn it into a
projective one. This transition is SO small and simple
thatsometimes people are tempted to say that the affine plane
that you startedwith, and the projective one that you get to
(from applying this simple transformationto it), are
THEMselves dual to each other. But THAT is a BAD use of
dual.Or at least a distracting one, in the context of the more
real duality,which is between the projective plane's points and
the projective plane's lines. : In my understanding, the dual
of a configuration of points and lines : consists of the same
objects, but the objects that were points now get : called
lines and the objects that were lines now get called
points.You can always perform this rename/exchange but you
canNOT always CALLit the dual. In projective geometry, points
and lines ARE dual, whichmeans that that re-calling works in
the relevant sense. But where itdoesn't work, there simply IS
NO duality. : A projective geometry is a system of objects
with incidence relations : satisfying certain axioms and when
you rename the objects they still : satisfy the appropriate
axioms.Right, i.e., when you exchange points with lines, the
same statements all keepthe same truth-values. AFTER that
happens, THEN you can say that points andlies are dual, withIN
that axiomatic framework. But affine geometry is governedby
DIFFERENT axioms and under THEM, you canNOT perform this
renaming withoutchanging truth-values. : The only difference
is in the human : terminology for the objects, not in their
mathematical behavior or : identity.There is no difference in
projective geometry, yes; thatby definition is why you CALL it
projective. But we were talkingabout the affine case. : To get
back to an earlier point, one possible way of forming objects
: satisfying the set of projective geometry axioms is to have
points that : are just points (atomic from the point of view
of the geometry) while : the lines are sets of points and the
point-line incidence relation is : set membership. Projective
duality tells us that we could instead view : the lines as
being atoms, and the points as being sets of lines through :
them, with incidence again being equivalent to set
membership.Of course. : As someone who has personally gained
insight into geometry problems by : applying projective
duality,Oh, SHUT UP!The AFFINE plane IS NOT PROJECTIVE and
DOES NOT EXHIBIT projective duality!In order to apply
projective duality, you would FIRST need to APPLY
SOMETRANSFORMATION TO the affine plane so that the result
WOULD IN FACT BEprojective! ONLY AFTER that do you get to
exploit duality! : despite duality's inability to change the :
underlying mathematical objects involved, I find your claim You
will : not gain any insight into the nature of a point by
thinking about it as : an infinite collection of lines to be
counterfactual and strange.No, you don't.I was talking about
the affine case.In the projective case, obviously, points and
lines are dual.AND I SAID THAT. : Why are you trying to tell
me not to apply a technique that I know from : personal
experience has worked?SHUT the   up, bitch.THAT is, AGAIN,
the sort of insult up with which I need not put.-- --- It's
difficult ... you need to be united to have any strength, but
internal issues have to be addressed. --- E. Ray Lewis, on
liberalism in America
===
Subject:
> : Why are you trying to tell me not
to apply a technique that I know from > : personal experience
has worked?SHUT the   up, bitch.> THAT is, AGAIN, the sort
of insult up with which I need not put.You keep using that
word insult. I do not think it means what you think it means.I
also do not think you are making a positive contribution to
this newsgroup.*plonk*-- David Eppstein
http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine,
School of Information & Computer Science
===
Subject:
David Eppstein
insultingly asks, : > : Why do you want to blinker us?I
politely replied, : > I don't. : Give it a break George. Go
back to the hell that spawned you, mitch. : You know what you
are doing.Of course I do, but you don't, so whyare you
presuming to pontificate about it? : > The general thrust of
this thread is about : > representation theorems, not
blinkering. : : Now who is lying?You are, fool. I FOUNDED this
thread.I POSTED several messages in it about representation
theorems. : You bring ridiculous concepts like burden of proof
concerning metaphysical : positions like foundationalismWHERE?
QUOTE ME OR SHUT THE * * UP, BITCH. : to mathematical
discussionsI STARTED this discussion, IDIOT.IF I WANT to
discuss foundations, THEN I CAN DO THAT,both on sci.logic and
sci.math,AND THERE ISN'T * * that *YOU* can do about it. :
and aren't honest about : it to the people with whom you are
in discussion.About what, specifically, am I being
dishonest?Will you please QUOTE SOMETHING I SAID and then
EXPLAINwhy it is false? THEN you will become entitled to say
whatyou just said. UNTIL you can do that, you are perfectly
welcometo eat   and die.
===
Subject:
> David Eppstein insultingly
asks,>> : > : Why do you want to blinker us?>> I politely
replied,>> : > I don't.>> : Give it a break George.>> Go back
to the hell that spawned you, mitch.>> : You know what you are
doing.>> Of course I do, but you don't, so why> are you
presuming to pontificate about it?>I'm not. If after having
changed the topic in this thread you had redirected itunder
sci.logic alone (as you have repeatedly stated that you do
because you do notwant to be on math threads) I would have
left my simple lol response withoutfurther comment, if even
that.But you are in the business of a separate debate having
nothing to do withmathematics proper. My presumption comes
from a year of experience with your postson sci.logic. It is
inappropriate to engage people with a specific disinterest
inthe background motivation for your questions (forcing
ontological commitment),blinker them, and then claim that is
not what you are doing.As for the hell that spawned me--it was
a course on intermediate logic in aphilosophy department. I
received an 'A' and a recommendation for graduate school.As
for why I have been on sci.logic... that is because it is an
appropriateunmoderated newsgroup for posting ideas about the
foundations of mathematics (giventhe general disinterest in
these matters among mathematicians as you yourself
havenoted).Look George. I am sorry for coming down on you like
I did. But, there is adifference between opinions concerning
foundations and formal philosophicproposals. Likewise, there
is a difference between a discussion and
aninterrogation.:-)mitchHell hath no bounds,Nor is
circumscribed in any one self place.For where we are is
hell,And where hell is,There must we ever be.And to
conclude,When all the world dissolvesAnd every creature shall
be purified,All places shall be hell that is not heaven.
--Mephistopheles The Tragedy of Doctor Faustus by Christopher
MarloweIt is, perhaps, true. Ignorance is bliss. One thing I
find admirable about peopleof faith is that they begin with
the principle that they cannot know the essentialtruths of the
universe and then use that as a foundation for ethical
guidance.
===
Subject:
: > : > : You know what you are doing.I replied,
> Of course I do, but you don't, so why > are you presuming to
pontificate about it? : I'm not.Yes, you are. I will QUOTE you
doing so, in a second. : If after having changed the topic in
this thread you had redirected it : under sci.logic alone (as
you have repeatedly stated that you do because you do not :
want to be on math threads) I would have left my simple lol
response without : further comment, if even that.This is
idiotic. I already had an exchange with Tim Chow where I
EXPLAINED WHYthis thread is still on sci.math. And you OF ALL
PEOPLE are NOT qualified tojudge what does or doesn't belong
on sci. math. : But you are in the business of a separate
debate having nothing to do with : mathematics proper.That is
an idiotic lie and MORE OF the same
pontification-from-ignoranceabout what I am doing that you
just falsely alleged about yourself that youdon't engage in.
This debate is not separate from math. It does MATTERwhat you
think about concreteness and it does MATTER what
pedagogicallyuseful prototypical categories might be. : My
presumptionWhat EXACTLY are you presuming?Do you mean
presumption as in presuming a logical premise,or presumption
as in daring to DO something you are notentitled to do? :
comes from a year of experience with your posts on
sci.logic.Which, tragically, has taught you nothing. : It is
inappropriate to engage people with a specific disinterest in
: the background motivation for your questionsA motivation
which you are personally both intellectually
andpsychologically completely UNABLE to compute : (forcing
ontological commitment),and that was a really  ty guess at
it, since I am NOT trying toforce any kind of ontological
commitment. I already posteda reply to James Dolan in which I
completely agreed with himthat this was about noticing that
tons of viewpoints are ontologicallyrelevant, whether anyone
wants them to be or not, and that far fromcommitting to some
of them, we were rather about recognizingisomorphisms and
representation-theorems that would allow us tothink in terms
of 1 representative viewpoint. In this thread,I am about
electing representatives, NOT ontological commitment.But more
to the point,They can ALWAYS JUST IGNORE IT. I ADVERTISED that
I was gear-shiftingthe discussion. I did not lure anybody into
this under false pretenses,or engage anyone against their
will. Insinuations and pontificationsto the effect that I
*have* done that are just bull  and shouldbe flushed ALONG
with the fool who was fool enough (or hateful enough)to assert
them. : blinker them,I have not blinkered anybody.I can't help
it if Dave Eppstein thinks you can have a vertical linein an
affine geometry (I suppose next he will be telling us thatyou
can have an origin). I can't help it if the fact that there
areeasy transformations from affine to projective geometry has
himwrongly speaking of a point/line duality in the affine case,
when infact such only obtains in the projective case. I am NOT
blinkeringanybody. If you are going to publicly claim that I
am then you are justgoing to make yourself part of the
problem. : and then claim that is not what you are doing.Well,
I RE-claim, IT'S NOT. : As for the hell that spawned me--it was
a course on intermediate logic in a : philosophy
department.Just because you got it THEN doesn't mean you get
it NOW.In your own opinion, you have grown up a lot since
then, andyou now feel free to embrace perspectives that were
NOT taught inTHAT course. Not that that course would even be
EXPECTED to givea   about philosophy of math in ANY case. :
I received an 'A' and a recommendation for graduate
school.Well, that makes two of us. : Look George. I am sorry
for coming down on you like I did.You would do well to just
never apologize to me again, period.It won't matter. : But,
there is a difference between opinions concerning foundations
: and formal philosophic proposals.Yeah, but since I'm the one
with the philosophy degree here, maybe youshould ASK me instead
of TELLING me what that difference is. : Likewise, there is a
difference between a discussion and an interrogation.No, not
likewise, not at all.And interrogation is going to remain
appropriate as long as youkeep trying to evade basic
questions. I mean, you don'tsee *me* whining about getting
interrogated.
===
Subject:
Download a free beta version of my new proof
checking software, DC Proof 1.0athttp://www.dcproof.com/(for
MS Windows 95 and later)It is a learning aid to teach the
fundamentals of logic and proof. Afterentering a premise or
assumption, the user can select rules of inference forLogic,
Set Theory and Number Theory from easy-to-use, drop-down
menus. Itshould be impossible to write an invalid proof in
this system! Feedback isimmediate with each new line.A
tutorial with examples and exercises demonstrating the main
features of DCProof is included along with the user manual.
(Click the Help** Button.)DC Proof is interesting from both
pedagogical and theoretical perspectives.The axioms for set
theory, for example, are a simplified version of thestandard
ZF model. Yet they are robust enough to avoid the
knowncontradictions of naive set theory, and, I believe, to
develop much ofmodern-day mathematical theory.Please let me
know what you think of my new system.Dan ChristensenToronto,
Canada
===
Subject:
> Download a free beta version of my new proof
checking software, DC Proof1.0> at>> http://www.dcproof.com/>>
(for MS Windows 95 and later)>> It is a learning aid to teach
the fundamentals of logic and proof. After> entering a premise
or assumption, the user can select rules of inferencefor>
Logic, Set Theory and Number Theory from easy-to-use,
drop-down menus. It> should be impossible to write an invalid
proof in this system! Feedback is> immediate with each new
line.>> A tutorial with examples and exercises demonstrating
the main features ofDC> Proof is included along with the user
manual. (Click the Help** Button.)>> DC Proof is interesting
from both pedagogical and theoreticalperspectives.> The axioms
for set theory, for example, are a simplified version of the>
standard ZF model. Yet they are robust enough to avoid the
known> contradictions of naive set theory, and, I believe, to
develop much of> modern-day mathematical theory.>> Please let
me know what you think of my new system.>> Dan Christensen>>
Toronto, CanadaNeither the Help menu nor the Help** button
produced any help. Here is whatgot installed...C:Program
FilesDC Proofasycfilt.dllC:Program FilesDC
Proofcomcat.dllC:Program FilesDC ProofCOMDLG32.OCXC:Program
FilesDC ProofDC Proof.chmC:Program FilesDC ProofDC
Proof.exeC:Program FilesDC ProofDefaultDir.binC:Program
FilesDC Proofe.gifC:Program FilesDC Proofhhctrl.ocxC:Program
FilesDC Proofmsvbvm60.dllC:Program FilesDC
Proofoleaut32.dllC:Program FilesDC Proofolepro32.dllC:Program
FilesDC ProofRecentFiles.binC:Program FilesDC
ProofRICHTX32.OCXC:Program FilesDC Proofstdole2.tlbC:Program
FilesDC Proofunins000.datC:Program FilesDC Proofunins000.exe--
Clive Toothhttp://www.clivetooth.dk
===
Subject:
>> Download a free beta
version of my new proof checking software, DC Proof> 1.0> at>>
http://www.dcproof.com/>> (for MS Windows 95 and later)>> It is
a learning aid to teach the fundamentals of logic and
proof.After> entering a premise or assumption, the user can
select rules of inference> for> Logic, Set Theory and Number
Theory from easy-to-use, drop-down menus.It> should be
impossible to write an invalid proof in this system!
Feedbackis> immediate with each new line.>> A tutorial with
examples and exercises demonstrating the main featuresof> DC>
Proof is included along with the user manual. (Click the
Help** Button.)>> DC Proof is interesting from both
pedagogical and theoretical> perspectives.> The axioms for set
theory, for example, are a simplified version of the> standard
ZF model. Yet they are robust enough to avoid the known>
contradictions of naive set theory, and, I believe, to develop
much of> modern-day mathematical theory.>> Please let me know
what you think of my new system.>> Dan Christensen>> Toronto,
Canada>> Neither the Help menu nor the Help** button produced
any help. Here iswhat> got installed...>> C:Program FilesDC
Proofasycfilt.dll> C:Program FilesDC Proofcomcat.dll>
C:Program FilesDC ProofCOMDLG32.OCX> C:Program FilesDC ProofDC
Proof.chm> C:Program FilesDC ProofDC Proof.exe> C:Program
FilesDC ProofDefaultDir.bin> C:Program FilesDC Proofe.gif>
C:Program FilesDC Proofhhctrl.ocx> C:Program FilesDC
Proofmsvbvm60.dll> C:Program FilesDC Proofoleaut32.dll>
C:Program FilesDC Proofolepro32.dll> C:Program FilesDC
ProofRecentFiles.bin> C:Program FilesDC ProofRICHTX32.OCX>
C:Program FilesDC Proofstdole2.tlb> C:Program FilesDC
Proofunins000.dat> C:Program FilesDC Proofunins000.exe>> -- >
Clive Tooth> http://www.clivetooth.dk>>Somehow the HTML Help
viewer isn't being made available to my program as itis
downloaded. I will investigate and report back.In the mean
time, the download is temporarily not available. Sorry for
theinconvenience.Dan
===
Subject:
>> Download a free beta version of my
new proof checking software, DC Proof> 1.0> at>>
http://www.dcproof.com/>> (for MS Windows 95 and later)>> It
is a learning aid to teach the fundamentals of logic and
proof.After> entering a premise or assumption, the user can
select rules of inference> for> Logic, Set Theory and Number
Theory from easy-to-use, drop-down menus.It> should be
impossible to write an invalid proof in this system!
Feedbackis> immediate with each new line.>> A tutorial with
examples and exercises demonstrating the main featuresof> DC>
Proof is included along with the user manual. (Click the
Help** Button.)>> DC Proof is interesting from both
pedagogical and theoretical> perspectives.> The axioms for set
theory, for example, are a simplified version of the> standard
ZF model. Yet they are robust enough to avoid the known>
contradictions of naive set theory, and, I believe, to develop
much of> modern-day mathematical theory.>> Please let me know
what you think of my new system.>> Dan Christensen>> Toronto,
Canada>> Neither the Help menu nor the Help** button produced
any help. Here iswhat> got installed...>> C:Program FilesDC
Proofasycfilt.dll> C:Program FilesDC Proofcomcat.dll>
C:Program FilesDC ProofCOMDLG32.OCX> C:Program FilesDC ProofDC
Proof.chm> C:Program FilesDC ProofDC Proof.exe> C:Program
FilesDC ProofDefaultDir.bin> C:Program FilesDC Proofe.gif>
C:Program FilesDC Proofhhctrl.ocx> C:Program FilesDC
Proofmsvbvm60.dll> C:Program FilesDC Proofoleaut32.dll>
C:Program FilesDC Proofolepro32.dll> C:Program FilesDC
ProofRecentFiles.bin> C:Program FilesDC ProofRICHTX32.OCX>
C:Program FilesDC Proofstdole2.tlb> C:Program FilesDC
Proofunins000.dat> C:Program FilesDC Proofunins000.exe>> -- >
Clive Tooth> http://www.clivetooth.dk>Clive,Perhaps you do not
have the Windows HTML Help viewer. For the latest versionof it,
go
tohttp://msdn.microsoft.com/library/default.asp?url=/library/
en-us/htmlhelp/html/
hwHTMLHelpFrequentlyAskedQuestions.aspClick on Downloads. Read
the notes carefully. Different versions ofWindows may require
different solutions.If this does not seem applicaple, see the
FAQ's at this link. Please let meknow if you continue to have
problems.Dan
===
Subject:
> Clive,Perhaps you do not have the Windows
HTML Help viewer. For the latest version> of it, go
tohttp://msdn.microsoft.com/library/default.asp?url=/library/
en-us/htmlhelp/html/
hwHTMLHelpFrequentlyAskedQuestions.aspClick on Downloads. Read
the notes carefully. Different versions of> Windows may require
different solutions.If this does not seem applicaple, see the
FAQ's at this link. Please let me> know if you continue to
have problems.DanHaven't had a look at the software yet, but
you've usually got to writethe path to the required help file
in the registry.... Does this registryentry
exist?Bruce.
===
Subject:
within my diploma thesis I read chapter 2 in
the bookNumberTheory in Function Fields from Michael Rosen. I
followedthe advice to translate some results for $Z$ into the
context of thepolynomial ring $A=F_{p}[x]$. I did this also
for the following function:Let $m=p_{1}^{e_{1}} cdot
p_{2}^{e_{2}} cdot ldots cdotp_{t}^{e_{t}}$ where $p_{i}$
prime. Then define $g(m)=1$ if $m=1$and
$g(m)=max(e_{1},e_{2},ldots,e_{t})$ if $m>1$. If one
calculatesthe average value of $g$, one gets 1.705221, the
constant of Niven.I translated the function to the ring $A$ in
the following way: Let$f=P_{1}^{e_{1}} cdot P_{2}^{e_{2}} cdot
ldots cdotP_{t}^{e_{t}}$ where $P_{i}$ irreducible polynomials
in A. Then I defined$g(f)=1$ if$deg(f)=0$ or $deg(f)=1$ and
$g(f)=max(e_{1},e_{2},ldots,e_{t})$ if$deg(f)>1$. I calculated
$frac{1}{p^{n}} cdot sumlimits g(f)$(summing over all monic $f$
of degree $n$) for different $p$ andincreasing $n$. I got a lot
of numerical values, which led me to thefollowing statement:
$lim_{n to infty} frac{1}{p^{n}} cdot sumlimitsg(f) = 1
+frac{1}{p-1}$. Now I`d like to know if one of you out there
knowsthis result and if so, could tell me, where I can find
it. I`dlike to compare my prove to the known one.Yours
sincerlyStefan Wagenbrenner
===
Subject:
>I've never posted before,
but as it is, it's 4:48am and for some>>stupid reason, I've
been trying to solve a double integral for almost>>3 hours
now.>>It's just personal now (not sure of notation here so
i'm going to>>display it twice):>>/ 1 / arccos y>>| | exp
(sin(x)) dx dy>>/ 0 / 0>>or>>integrate relative to y
{0..1}>>integrate relative to x {0..arccos(y)}>>the equation:
exp(sin(x)) dx dy>>My trusty calculator can give me an
answer, but for some reason>>(fatigue for example) I can't
figure out how to come up with the>>answer.>>any help?>Change the order of integration: x from 0 to pi/2 and y
from 0 to cos x.>>KPThat's the wrong area, y goes from cos x
to 1, which still leaves aproblem.--Lynn
===
Subject:
No, y goes from 0
to cos x. If 0 < =x <= arccos y, then upon takingcosines (a
decreasing function when 0 <= x <= pi/2) 1 >= cos x >= y or
y<= cos x <= 1. This is NOT the same as cos x <= y <= 1!cos x
<= 1 is assured in the real domain. leaving y <= cos x. The
outerlimits in the original integral imply y >= 0, so we end
up with 0 <= y<= cos x.--OL
===
Subject:
>No, y goes from 0 to cos x. If
0 < =x <= arccos y, then upon taking>cosines (a decreasing
function when 0 <= x <= pi/2) 1 >= cos x >= y or y><= cos x <=
1. This is NOT the same as cos x <= y <= 1!>>cos x <= 1 is
assured in the real domain. leaving y <= cos x. The
outer>limits in the original integral imply y >= 0, so we end
up with 0 <= y><= cos x.>>--OLOops, you're correct of course.
Guess I was a little hasty on thekeyboard.--Lynn
===
Subject:
: (What I
presented was an incidence matrix for the trivial affine plane
because : it was easy to cut-and-paste from recent material I
had written. Like a good : mathematician, George visualized
the affine geometry from the presentation. My : words
explaining otherwise made no difference.)That is a lie about
what actually happened, not that any of you should care.If in
fact the incidence matrix posted was the incidence matrix for
the smallestaffine plane, then no words EVER COULD POSSIBLY
explain otherwise, for thesole and simple reason that the
TRUTH is NOT otherwise. If mitch wanted tosay something about
this matrix OVER AND ABOVE, IN ADDITION TO (as OPPOSED
tootherwise than) its being the one for this little plane,
well, I read it,and it is NOT the case that it made no
difference to me. It is, however,the case that what he was
saying about it was ridiculous.-- --- It's difficult ... you
need to be united to have any strength, but internal issues
have to be addressed. --- E. Ray Lewis, on liberalism in
America
===
Subject:
> .... > the axis are the eigenvector of B, and the
'diameters' are> d^2= 1/ eigenvalue> ( is it correct? I think
it is)> .... Not quite. You've missed a factor of 2 or 4. The
semi-axis (_half_ the diameter you're looking at) is
1/sqrt(eigenvalue). There's a handy little reference written
by a statistician, giving lots of facts about such things:
M.G. Kendall, A course in the what you want. Ken
Pledger.
===
Subject:
>and sorry for the silly q., but i'm not a
mathematician >>preamble( I hope this is correct)>let >
A[a11 a12 0;>a21 a22 0;>0 0 -1]>>A is an ellipse>a11 x^2 +
a22 y^2 +2 a12 x y =1 >if, let > B[a11 a12 ;>a21 a22]>>- B
is symmetric>- B is positive definite>>the axis are the
eigenvector of B, and the 'diameters' are>d^2= 1/ eigenvalue>(
is it correct? I think it is)>>I wish to know if I can
generalize in Rn ;>>A= [a11 .... a1n 0;> ..... >an1 an2 ...
ann 0;>0 0 0 .... 0 -1]>>B= [a11 .... a1n ;>....>an1 an2 ...
ann]>What about the eigenvalues and eigenvector of B?>>Are
they related to the axes of the ' hyperellipsoid', like in R2
??>Yes. You can see this by using an eigenbasis of B as a
coordinatesystem. In that (orthonormal) coordinate system, the
equation ise1*x1^2 + e2*x2^2 + ... + en*xn^2, where e1, e2,
etc. are theeigenvalues and x1, x2, etc. are the coordinates
(with respect to theeigenbasis). This describes an ellipsoid
with the coordinatedirections as axes and with diameters as
you indicated.John Mitchell
===
Subject:
> and sorry for the silly q.,
but i'm not a mathematicianpreamble( I hope this is correct)>
let> A [a11 a12 0;> a21 a22 0;> 0 0 -1]A is an ellipse> a11
x^2 + a22 y^2 +2 a12 x y =1> if, let> B [a11 a12 ;> a21
a22]- B is symmetric> - B is positive definitethe axis are the
eigenvector of B, and the 'diameters' are> d^2= 1/ eigenvalue>
( is it correct? I think it is)Yes.I wish to know if I can
generalize in Rn ;A= [a11 .... a1n 0;> .....> an1 an2 ... ann
0;> 0 0 0 .... 0 -1]B= [a11 .... a1n ;> ....> an1 an2 ...
ann]> What about the eigenvalues and eigenvector of B?Are they
related to the axes of the ' hyperellipsoid', like in R2 ??You
can use exactly the same formulas. Perhaps you may want to
look up theterm quadratic form over R. This has to do with the
so-called spectraltheorems.-- Phyics is much too hard for
physicists.
===
Subject:
> One thing I don't get, is a big picture of
how well I'm> doing. My investment dollars are spread across
multiple> IRA/401K/SEP/Roth accounts, as well as six
different> non-retirement accounts at 2 brokers. (don't ask --
this> can't currently be changed). Anyway, aggregating the>
data from these various sources is easy, analysing the> data,
in aggregate, is *not*.>The software 'KBH Investor Accounting'
outputs a year- to-date gain ofrealized and un-realized
positions combined together...for each date ofportfolio
activity. It calls this a mark-to-market accounting.
Theapplication also shows a year-to-date portfolio
percentagegain relative the net of deposits and
withdrawals...and it calls this aprofit margin. Finally, the
application shows a year-to-date portfoliopercentage gain
relative to a time-weighted net of deposits and withdrawalsand
it calls this a time-weighted profit margin. Note that this
last featureis an annualization or an IRR within the yearly
period.Obviously, a consolidated accounting should be input. I
would recommend aconsolidated accounting for the taxable
portfolio and a consolidatedaccounting for the tax-deferred
portfolio.Here is a user link to the
application:http://pages.prodigy.net/halsteadinvest/kbh.htm
===
Subject:
 This may be better placed in one of the investing
newsgroups,> but I find that most of the posters there can't
handle basic> arithmetic, let alone what I'm trying to do....
You could find it worth while to follow some of the links from
Ken
Pledger.
===
Subject:
http://online.itp.ucsb.edu/online/cmb02/caldwell/
http://qedcorp.com/APS/EmergentGravity.pdfhttp://qedcorp.com/
APS/StarGate1.movCaldwell does not seem to realize that the
attractive dark matter can also bea zero point energy /zpf
effect in addition to the repulsive dark energy ondifferent
scales.Latest data is dark energy/dark matter split is 0.7 to
0.3 in spatially flat sumOmega = 1 of FRW cosmology metric
with uncertainty 0.04. Ordinarymatter (stars, planets,
radiation, dust clouds etc are inside that 0.04
uncertainty).My key new idea here is that dark matter
detectors will not click with thethat is on mass shell in
terms of quantum field theory Feynman propagators,
whizzingaround in space like neutrinos passing through our
bodies. That is wrong IMHO.Dark matter is, like dark energy a
virtual, i.e. off mass shell exotic vacuum effectwith w =
(pressure)/(energy density) = -1 because the total local zero
point stress-energy densityvacuum Diff(4) symmetry group
tensor istuv(Vacuum) = (String Tension)/zpf(x)guv(Curved
Space-Time)where/zpf (x) = Lp*^-2[Lp*^3/2|Vacuum
Coherence(x)|^2 - 1]Lp*^2 = hG*/c^3hc/Lp*^2 = String TensionG*
is the variable scale-dependent effective gravity coupling
coefficient.Something likeG*/G(Newton) = e^<dilaton> ?The
NON-EXOTIC i.e. non-gravitating vacuum region of space-time is
when/zpf = 0When Vacuum Coherence = 0, like in the inside of a
vibrating string of pure energy (Brian Greene NOVA Elegant
Universe)we have a strongly attractive dark energy core
/zpf(core) = -1/Lp*^2 that, for example, explains the
self-cohesiveness of thespatially extended electron, which is
a micro-geon Kerr-Newman ring singularity of size ~ e^2/mc^2 ~
1 fermi surrounded bya virtual plasma of electron-positron
pairs and virtual photon extending out to ~ h/mc ~ 10^-11 cm.
This whole spatially extendedcomplex looks more and more like
a Bohmian hidden variable point enormous space-warping since
G*m^2/hc ~ 1 where m ~ 10^-27 gms.realism is consistent with
J.P. Vigier's idea for a non-nuclear release of atomic energy
from tight atomic states like the experiments in Beograd,
Serbia were looking
for.
===
Subject:
http://online.itp.ucsb.edu/online/cmb02/caldwell/>
http://qedcorp.com/APS/EmergentGravity.pdfhttp://qedcorp.com/
APS/StarGate1.movCaldwell does not seem to realize that the
attractive dark matter can > also be> a zero point energy /zpf
effect in addition to the repulsive dark energy on> different
scales.Latest data is dark energy/dark matter split is 0.7 to
0.3 in spatially > flat sum> Omega = 1 of FRW cosmology metric
with uncertainty 0.04. Ordinary> matter (stars, planets,
radiation, dust clouds etc are inside that 0.04 >
uncertainty).> My key new idea here is that dark matter
detectors will not click > with the> that is on mass shell in
terms of quantum field theory Feynman > propagators, whizzing>
around in space like neutrinos passing through our bodies. That
is wrong > IMHO.> Dark matter is, like dark energy a virtual,
i.e. off mass shell > exotic vacuum effect> with w =
(pressure)/(energy density) = -1 because the total local zero
> point stress-energy density> vacuum Diff(4) symmetry group
tensor istuv(Vacuum) = (String Tension)/zpf(x)guv(Curved
Space-Time)where/zpf (x) = Lp*^-2[Lp*^3/2|Vacuum
Coherence(x)|^2 - 1]Lp*^2 = hG*/c^3hc/Lp*^2 = String TensionG*
is the variable scale-dependent effective gravity coupling
coefficient.Something likeG*/G(Newton) = e^<dilaton> ?The
NON-EXOTIC i.e. non-gravitating vacuum region of space-time is
when/zpf = 0When Vacuum Coherence = 0, like in the inside of a
vibrating string of > pure energy (Brian Greene NOVA Elegant
Universe)> we have a strongly attractive dark energy core
/zpf(core) = -1/Lp*^2 > that, for example, explains the
self-cohesiveness of the> spatially extended electron, which
is a micro-geon Kerr-Newman ring > singularity of size ~
e^2/mc^2 ~ 1 fermi surrounded by> a virtual plasma of
electron-positron pairs and virtual photon extending > out to
~ h/mc ~ 10^-11 cm. This whole spatially extended> complex
looks more and more like a Bohmian hidden variable point >
enormous space-warping since G*m^2/hc ~ 1 where m ~ 10^-27
gms.realism is consistent with J.P. Vigier's idea for a
non-nuclear release > of atomic energy from tight atomic
states like the experiments in > Beograd, Serbia were looking
for.I love it when Sarfatti uses IMHO. It means just the
opposite.
===
Subject:
Here is the question:> Find the distance between
the lines L(1) = (1, 1, 1) + s (1, 2, 3) and L(2) = (0, 1, 1)
+ t (1, -1, -1). I would really appreciate a full solution
because many people have> given me tips on another message
board and I have been unable to solve> this. I know that is
not the style of this newsgroup but I would really> appreciate
it as I have an exam tomorrow morning and I am sure a> question
almost identical to this will be on it.ZackUsing vector
notation , with P*Q as dot product of vectors P and Q > and
P^Q as cross product of P and Q (see below), let A = (1,1,1),
> U = (1,2,3), B = (0,1,1), V = (1,-1,1-) then U is parallel
to line > L(1) and V is parallel to line L(2) so that C =
(A^B) is > perpendicular to both lines.Then (A -
B)*C/sqrt(C*C) is the projection of A-B parallel to C ( > and
therefore perpendicular to both lines) and is its absolute
value > is desired perpendicular distance between the lines.>
Note, dot product is given by > (a,b,c)*(c,d,e) = ac + bd + cf
> and cross product by> (a,b,c)^(d,e,f) = (bf - ce, cd - af, ae
- bd)Urq.It has always seemed to me that this kind of problem
can be solved witha lot less machinery and a lot more clarity
by simply considering thedistance (squared) between a point on
line L(1) and a point on L(2): f(s,t) = || (1+s,1+2s,1+3s) -
(t,1-t,1-t)||^2 = (1+s-t)^2 + (2s+t)^2 + (3s+t)^2.This
function is a minimum when its partial derivatives are both
zero. (And it does take a minimum, because f is convex and -->
+infty as s^2+ t^2 --> +infty.)That leads to simultaneous
equations 2(1+s-t) + 4(2s+t) + 2(3s+t) = 0 i.e. 28s + 8t = -2,
-2(1+s-t) + 2(2s+t) + 2(3s+t) = 0, i.e. 8s + 6t = 2.These are
easily solved to give unique solution s = -7/26, t =
9/13(hence, there is a unique pair of closest points). We have
f(-7/26, 9/13) = 1/26,and therefore the distance between the
lines is 1/sqrt(26).I would MUCH prefer that students be able
to use a seat-of-the-pants,common-sense method like this
instead of the cookbook procedure taughtin section 8.2 of
chapter 4, on page 167.Besides, your solution fails as soon as
you go to R^4.--Ron Bruck
===
Subject:
> Here is the question:>> Find
the distance between the lines > L(1) = (1, 1, 1) + s (1,
2, 3) and L(2) = (0, 1, 1) + t (1, -1, -1). > I would
really appreciate a full solution because many people have>>
given me tips on another message board and I have been unable
to solve>> this. I know that is not the style of this
newsgroup but I would really>> appreciate it as I have an exam
tomorrow morning and I am sure a>> question almost identical to
this will be on it.> Zack>>Using vector notation , with P*Q
as dot product of vectors P and Q >and P^Q as cross product of
P and Q (see below), let A = (1,1,1), >U = (1,2,3), B =
(0,1,1), V = (1,-1,1-) then U is parallel to line >L(1) and V
is parallel to line L(2) so that C = (A^B) is >perpendicular
to both lines.>Of course, you mean C = U^V, not
A^B.--Lynn
===
Subject:
> Here is the question:>> Find the distance
between the lines > L(1) = (1, 1, 1) + s (1, 2, 3) and
L(2) = (0, 1, 1) + t (1, -1, -1). > I would really
appreciate a full solution because many people have>> given me
tips on another message board and I have been unable to solve>>
this. I know that is not the style of this newsgroup but I
would really>> appreciate it as I have an exam tomorrow
morning and I am sure a>> question almost identical to this
will be on it.> Zack>>Using vector notation , with P*Q as
dot product of vectors P and Q >and P^Q as cross product of P
and Q (see below), let A = (1,1,1), >U = (1,2,3), B = (0,1,1),
V = (1,-1,1-) then U is parallel to line >L(1) and V is
parallel to line L(2) so that C = (A^B) is >perpendicular to
both lines.>Of course, you mean C = U^V, not A^B.--LynnRight!
Proof reading is such a bore!
===
Subject:
> Huh? How in the world does
it suggest that? None of the replies> have agreed with you...
I felt there might be a bit of defensive feeling behind some
of theposts. Is the geometrical approach the best one for
anyone who justwants to study functions? Then as a technique
analytical continuationis only practical with the aid of a
computer program. If cuts areallowed I can treat some
algebraic functions (I do not claim all)without using
it.
===
Subject:
>The fact that my original post has to date attracted
19 replies>suggests that there are others who do not see
Riemann surfaces as an>entirely finished subject. >> Huh?
How in the world does it suggest that? None of the replies>>
have agreed with you...>> I felt there might be a bit of
defensive feeling behind some of the>posts. Huh?>Is the
geometrical approach the best one for anyone who just>wants to
study functions? Huh? What does this have to do with anything
that's beendiscussed here? The question was whether there
arenecessarily arbitrary choices involved in constructing
theRiemann surface for a function...>Then as a technique
analytical continuation>is only practical with the aid of a
computer program. Huh????????>If cuts are>allowed I can treat
some algebraic functions (I do not claim all)>without using
it.Huh? Nobody's said anything about cuts not being
allowed.The question was [see
above].************************David C. Ullrich
===
Subject:
large n
(say n > 100 digits). Since n is that large, it's not
veryefficient to store the lookup table.I know it's not easy
to find a polynomial that works. I am lookingfor a
deterministic algorithm (likely to be recursive) probably
withsome finite lookup table storage. Would studying the
distributionof primes help? As I know given a particular
prime, it is possible tofind the next m primes -- is that
true?>> in which I need to map an integer identity (ID) to a>
prime number.>> I want h( ): Z_n -> Z-m (m > n) s.t. for every
integers in> Z_n, I can easily find a prime in Z_m that is
distinct from> the image of another integer.>> i.e.
h(ID_i)=h(ID_j)  ID_i = ID_j.>> Please note that I don't
want a lookup table for the mapping ---> that's what I meant
efficient.>> What's not efficient about table lookup?>> Do you
want to do this for one particular n? or do you want something>
that works for all n?>> If the latter, you're looking for a
simple (efficient) formula> f(a) that returns a different
prime for each positive integer a.> You're unlikely to find
anything to your liking. Certainly there> is no polynomial
that works.>> If n < 40 then you can use Euler's polynomial,
n^2 + n + 41 or> something like that.>> --
===
Subject:
> large n (say
n > 100 digits).Whoa! You want a function that gives you more
than 10^100 primes? and only primes? You're dreaming. > As I
know given a particular prime, it is possible to> find the
next m primes -- is that true?Yes, but only by testing each
number after your prime until you find m more. Suppose you
know some 100-digit prime, p, and you want to find the next 10
primes. Well, roughly one number in each 230 is prime up there,
so, to be on the safe side, maybe you want to look at p + k for
k from 1 to 5000. Now you can sieve out a lot of those k by
deleting multiples of 2, of 3, of 5, etc, and then subject the
remaining numbers to primality tests. Exactly how much sieving
to do, I'll leave to the experts (who may know a better
approach than what I'm outlining, but I don't see how one can
do fantastically better).--
===
Subject:
> If I have 2 - sqrt(12)i
how do I find the polar form?>> So far I've worked out
that r = 4 and the argument is 5pi/6.>> The next step I'm
not sure about. Do I put it in the form r(cos(theta) +>
isin(theta))? Making it 4(cos(5pi/6) + isin(5pi/6))? Then what
do I do?>> BTW - what is the correct notation on here for
angle theta?>> TIA> You've done it perfectlySorry I
should have elaborated. How do I get it into polar form with
cis> (whatever that's called)?cis(theta) is merely an
abreviation for cos(theta)+i*sin(theta), and is equivalent to
exp(i*theta) and e^(i*theta)Thus given x = r*cos(theta) and y
= r*sin(theta), one can write x + i*y = r*cis(theta) =
r*exp(i*theta) = r*e^(i*theta)
===
Subject:
In sci.math,
Michael<mick@microsoft.com><br5j50$28q8oq$1@ID-83837.
news.uni-berlin.de>:> If I have 2 - sqrt(12)i how do I find
the polar form?>> So far I've worked out that r = 4 and the
argument is 5pi/6.(Note: the mathematical symbol for theta
looks a bit like an Owith a horizontal hyphen through it.)If z
= a + bi, then r = sqrt(a^2 + b^2) and theta = atan2(b,a),where
atan2() is a fairly widespread function on computer
systems.More conventionally, theta = atan(b/a) if a > 0,
atan(b/a) + pi if a < 0,pi/2 if a = 0 and b > 0, -pi/2 if a =
0 and b < 0. (For a = b = 0,theta is conventionally taken to
be 0. Since r is 0 it doesn'tmatter all that much in that
case.)So r = sqrt(4 + 12) = 4 and theta = atan(-sqrt(12)/2)=
atan(-sqrt(3)) = -pi/3.>> The next step I'm not sure about. Do
I put it in the form r(cos(theta) +> isin(theta))? Making it
4(cos(5pi/6) + isin(5pi/6))? Then what do I do?Some use the
notation (r,theta) but that's obviouslyslightly confusing if
accidentally taken out of context.I think most simply state r
= value, theta = value.Converting back to Cartesian isn't
difficult:z = r*cos(theta) + i*r*sin(theta). On occasionone
can also use the identity z = r*exp(i*theta)or your form z =
r*(cos(theta) + i*sin(theta)).>> BTW - what is the correct
notation on here for angle theta?By convention, just theta.>>
TIA>-- #191, ewill3@earthlink.netIt's still legal to go
.sigless.
===
Subject:
> ... stuff deleted ...> ... given a map f : X
--> Y between> sufficiently nice spaces (homotopy type of a CW
complex should be nice> enough), the path-space construction:
P_f = {(x,p) in X x map(I,X) | p(0) = x }yields a space
homotopy-equivalent to X ... stuff deleted ...> ... and the
map j> P_f -------> Y (x,p) |--> p(1)is a fibration. Further,
the diagram f> X --------------->_Y> /|> /> /> i /j> /> /> />
_| /> P_fis commutative.In other words, every map is a
fibration, up to> homotopy. Well, just as long as the spaces
are> sufficiently well-behaved to permit the above>
construction, and it's been long enough for me> since I worked
with that construction that I> don't recall just what's
needed.Dale> Just to tie up this loose end, I consulted
Spanier (AlgebraicTopology, now published by Springer-Verlag).
He claims it's truefor any spaces X,Y, and any (continuous) map
f. I would be surprisedif some further technical condition
(e.g., normality) weren't necessary,but Spanier doesn't
mention it. The introduction to the text does saythat, absent
any explicit restriction, the term space means anytopological
space, so it could be that this construction holds
forarbitrary topological spaces X,Y.I would, of course,
welcome any correction on this matter.Dale.
===
Subject:
I need help on
the following problem. thatKerv={h in H| v(h)=0}={0} and
v(H)is dense in H but does not coincidewith it. Is it true
that v(H)=H or there can exist such an operator??I am looking
forward to receiving your answers .Greetings Mladen
===
Subject:
>I need
help on the following problem. >that>Kerv={h in H| v(h)=0}={0}
and v(H)is dense in H but does not coincide>with it. Is it
true that v(H)=H or there can exist such an operator??>I am
looking forward to receiving your answers .>Greetings
Mladenbasis |n>, n = 1, 2, 3, ..., and define the linear
operator v by v |n> = 1/n |n> for n = 1, 2, 3, .... Then v is
continuous, Ker(v) = 0, and the range of v is dense, but the
element sum_{n=1}^infty 1/n |n> Hope this helps.David
McAnally--------------
===
Subject:
>>I need help on the following
problem. >that>Kerv={h in H| v(h)=0}={0} and v(H)is dense in H
but does not coincide>with it. Is it true that v(H)=H or there
can exist such an operator??Certainly there can exist such an
operator. For example say H isL^2(R) and v f(x) =
f(x)/(1+x^2).>I am looking forward to receiving your answers
.>>Greetings Mladen************************David C.
Ullrich
===
Subject:
> I need help on the following problem. > that>
Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not
coincide> with it. Is it true that v(H)=H or there can exist
such an operator??square summable sequences.
===
Subject:
>[...]>>Well
this morning I was pissed I didn't have a solution to B6 -
so>you can imagine how I felt when you recorded my
pissed-off>non-solution there following an actual solution. So
I thought>about it. It's easy if int f = 0, which seems like it
should be>the hard case, for example it's trivial if f >= 0.
This morning>I was going to say just as a joke that you could
get a solution>in general by interpolating between these two
cases. But>it turns out that you can do exactly that. The part
you want>to record for posterity starts here:Aargh. What's
below is wrong. Wrong, wrong wrong.I still think it's curious
that the case int f = 0 is easy,> and it still seems that the
general case should follow,> since if f has a lot of
cancellation that should make it> harder. But what's below is
nonsense.>It's clear if f >= 0. Suppose that int f = 0.
Choose a function>g so that |g| = 1 and gf >= 0. Then>> int
int |f(x) + f(y)| >= int (f(x) + f(y)) g(x)>> = int f(x) g(x)
= int |f(x)|.>>You can do the general case by interpolating
between>these special cases: Suppose that int f > 0. Then>f =
g + h, where int g = 0, h >= 0, and g and h have>disjoint
support. Say g is supported on A and h is>supported on B.
Then>> int int |f(x) + f(y)| >>= int_A int_A |f(x) + f(y)| +
int_B int_B |f(x) + f(y)|> Let me try this again.int int |f(x)
+ f(y)| = int_A int_A |g(x)+g(y)| +int_B int_B (h(x)+h(y)) + 2
int_A int_B |g(x)+h(y)| >= int_A int_A |g(x)+g(y)| + int_B
int_B (h(x)+h(y)) + 2 int_{x in A} int_{y in B} (g(x)+h(y))Now
David's argument shows that the first integral on the extreme
right is >= int_A |g(x)|. The second is equal to 2m(B) int_B
h, and the third is equal to 2m(A) int_B h. This yieldsint int
|f(x) + f(y)| >= int_A |g| + 2m(B) int_B h + 2m(A) int_B h =
int_A |g| + 2int_B h >= int_A |g| + int_B h = int |f|.--
A.
===
Subject:
>>[...]>>Well this morning I was pissed I didn't
have a solution to B6 - so>>you can imagine how I felt when
you recorded my pissed-off>>non-solution there following an
actual solution. So I thought>>about it. It's easy if int f =
0, which seems like it should be>>the hard case, for example
it's trivial if f >= 0. This morning>>I was going to say just
as a joke that you could get a solution>>in general by
interpolating between these two cases. But>>it turns out that
you can do exactly that. The part you want>>to record for
posterity starts here:> Aargh. What's below is wrong.
Wrong, wrong wrong.> I still think it's curious that the
case int f = 0 is easy,>> and it still seems that the general
case should follow,>> since if f has a lot of cancellation
that should make it>> harder. But what's below is nonsense.>>It's clear if f >= 0. Suppose that int f = 0. Choose a
function>>g so that |g| = 1 and gf >= 0. Then>> int int
|f(x) + f(y)| >= int (f(x) + f(y)) g(x)>> = int f(x) g(x) =
int |f(x)|.>>You can do the general case by interpolating
between>>these special cases: Suppose that int f > 0. Then>>f
= g + h, where int g = 0, h >= 0, and g and h have>>disjoint
support. Say g is supported on A and h is>>supported on B.
Then>> int int |f(x) + f(y)| >>= int_A int_A |f(x) +
f(y)| + int_B int_B |f(x) + f(y)|>>Let me try this again.>>int
int |f(x) + f(y)| >> = int_A int_A |g(x)+g(y)| +int_B int_B
(h(x)+h(y))> + 2 int_A int_B |g(x)+h(y)|>>= int_A int_A
|g(x)+g(y)| + int_B int_B (h(x)+h(y))> + 2 int_{x in A} int_{y
in B} (g(x)+h(y))>>Now David's argument shows that the first
integral on the extreme right >is >= int_A |g(x)|. Does it? It
looks more like >= m(A) int_A |g(x)| to me (althoughconsidering
my record here what it looks like to me is of
littleinterest...)>The second is equal to 2m(B) int_B h, and
the third >is equal to 2m(A) int_B h. This yields>>int int
|f(x) + f(y)| >>= int_A |g| + 2m(B) int_B h + 2m(A) int_B
h>> = int_A |g| + 2int_B h>>= int_A |g| + int_B h>> = int
|f|.************************David C. Ullrich
===
Subject:
>[...]>>Well this morning I was pissed I didn't have a
solution to B6 - so>>you can imagine how I felt when you
recorded my pissed-off>>non-solution there following an actual
solution. So I thought>>about it. It's easy if int f = 0, which
seems like it should be>>the hard case, for example it's
trivial if f >= 0. This morning>>I was going to say just as a
joke that you could get a solution>>in general by
interpolating between these two cases. But>>it turns out that
you can do exactly that. The part you want>>to record for
posterity starts here:> Aargh. What's below is wrong.
Wrong, wrong wrong.> I still think it's curious that the
case int f = 0 is easy,>> and it still seems that the general
case should follow,>> since if f has a lot of cancellation
that should make it>> harder. But what's below is nonsense.>>It's clear if f >= 0. Suppose that int f = 0. Choose a
function>>g so that |g| = 1 and gf >= 0. Then>> int int
|f(x) + f(y)| >= int (f(x) + f(y)) g(x)>> = int f(x) g(x) =
int |f(x)|.>>You can do the general case by interpolating
between>>these special cases: Suppose that int f > 0. Then>>f
= g + h, where int g = 0, h >= 0, and g and h have>>disjoint
support. Say g is supported on A and h is>>supported on B.
Then>> int int |f(x) + f(y)| >>= int_A int_A |f(x) +
f(y)| + int_B int_B |f(x) + f(y)|>>Let me try this again.>>int
int |f(x) + f(y)| >> = int_A int_A |g(x)+g(y)| +int_B int_B
(h(x)+h(y))> + 2 int_A int_B |g(x)+h(y)|>>= int_A int_A
|g(x)+g(y)| + int_B int_B (h(x)+h(y))> + 2 int_{x in A} int_{y
in B} (g(x)+h(y))>>Now David's argument shows that the first
integral on the extreme right >is >= int_A |g(x)|. Does it? It
looks more like >= m(A) int_A |g(x)| to me (although>
considering my record here what it looks like to me is of
little> interest...)You and me both.Uncle.--
A.
===
Subject:
>[...]>>Now David's argument shows that the first
integral on the extreme right >>is >= int_A |g(x)|. > Does
it? It looks more like >= m(A) int_A |g(x)| to me (although>>
considering my record here what it looks like to me is of
little>> interest...)>>You and me both.>>Uncle.Yeah, but there
_must_ be a one or two line proof along these lines.I mean,
it's very simple if int f = 0. And int f = 0 says f has a
lotof cancellation, which would make int int |f(x) + f(y)|
small, sothe case int f > 0 must follow, by a similar
argument.(Shoulda just said that the case int f > 0 followed
and seemwhether anyone wanted to admit they didn't see
how...)************************David C. Ullrich
===
Subject:
>Rather,
continue as:> = int_A int_A |g(x) + g(y)| + int_B int_B (h(x)
+ h(y))>= int_A int_A |g(x) + g(y)| + int_B h(x)Nuh-uh, this
step is still wrong. You pick up a factor of 2 meas(B) onthe
right-hand side. I suspect that no solution which ignores the
termscoming from the cross-regions A x B and B x A can be made
to work forfunctions f with int f <> 0: if f is large when it's
positive and smallwhen it's negative, without the contributions
|f(x) + f(y)| when f(x) ispositive and f(y) is negative, you
won't get the inequality that you want.Dave
===
Subject:
>Rather,
continue as: = int_A int_A |g(x) + g(y)| + int_B int_B (h(x) +
h(y))>= int_A int_A |g(x) + g(y)| + int_B h(x)Nuh-uh, this
step is still wrong. You pick up a factor of 2 meas(B) on> the
right-hand side. Ack!> I suspect that no solution which ignores
the terms> coming from the cross-regions A x B and B x A can be
made to work for> functions f with int f <> 0: if f is large
when it's positive and small> when it's negative, without the
contributions |f(x) + f(y)| when f(x) is> positive and f(y) is
negative, you won't get the inequality that you want.--
A.
===
Subject:
>> One of their favorite labels is crank, where
the idea is that you> have some person who is unreasonable,
who has ideas that don't work,> who refuses to acknowledge
the truth.> Let's see now...> ...some person who
is unreasonable, who has ideas that don't work,> who refuses
to acknowledge the truth.> Sounds just like you,
James!>>Your opinion is noted Alec McKenzie.>>I do admit
that if enough people think that I'm just some nut, then
I>>have to wonder.> _If_ enough people thought that? What
reason do you have to>> think that _anyone_, with only one
exception that I can think of>> right now, doesn't think of
you as just some kind of nut?>Note that I made a
*specific* challenge in my previous post, which has>so far
been ignored. I'll talk about it more a little later.And note
that you're not answering the specific question I asked,in
regard to a specific thing you said.>Now then, I'm reasonable,
as I recognize that hey maybe I am nutty,>but then again, I
also know enough math to realize that mathematicians>haven't
made a logical case for total dismissal of my prime
counting>discovery, and I know enough about human nature to be
very wary when>people rely on insults, like here David Ullrich
is imputing that>everyone with one exception thinks I'm a
nut!!!Again: What reason do you have to think that anyone
thinks otherwise?************************David C. Ullrich
===
Subject:
> One of their favorite labels is crank, where the
idea is that you> have some person who is unreasonable, who
has ideas that don't work,> who refuses to acknowledge the
truth.> Let's see now...> ...some person who is
unreasonable, who has ideas that don't work,> who refuses to
acknowledge the truth.> Sounds just like you,
James!>>Your opinion is noted Alec McKenzie.>>I do admit
that if enough people think that I'm just some nut, then
I>>have to wonder.> _If_ enough people thought that? What
reason do you have to>> think that _anyone_, with only one
exception that I can think of>> right now, doesn't think of
you as just some kind of nut?>Note that I made a
*specific* challenge in my previous post, which has>so far
been ignored. I'll talk about it more a little later.And note
that you're not answering the specific question I asked,> in
regard to a specific thing you said.That's a rude question.My
family doesn't think I'm insane you sick turd. So far you've
shown just how despicable of a human being you can be,without
any regard for common decency or caring, or ethics.If you have
an issue with my prime counting discovery then why can'tyou
keep to the mathematics?I have a *specific* challenge to you
David Ullrich, I dare you to givean outlined dispute of the
value of my prime counting discoveryproviding *all* the
information necessary for readers to evaluate yourclaim,
without trying to rely on links or insults.What kind of math
professor are you if you can't handle a *math*challenge but
need to insult instead? If you are a math expert, proveit.Give
a point-by-point exposition to support your case without
insults,innuendo, or any other attempts at sly or back-handed
insults, butwith *mathematics* and logic.I don't think you
have the capacity to engage in an honest and opendebate
without trying to use tricks or insults David Ullrich. I
thinkyou lack the capacity to have a *reasoned* discussion,
but insteadrely on childish behavior.The challenge is to you
now, show that you can reason and discussmathematics
reasonably, if you can.James Harris
===
Subject:
[snip]>Note that I
made a *specific* challenge in my previous post, which hasso far been ignored. I'll talk about it more a little
later.The specific challenge you mention above has been
ignored because it represents a non-issue. In your previous
post youinvite readers to attack a particular point when there
is *already* a specific attack on a particular point on the
table.You have so far *refused* to either acknowledge or
defend the challenge to your claim that the so-called
'partialdifferential equation' solves the prime counting
problem -- a claim you made repeatedly.C'mon, James Harris. It
is *you* who claimed that the difference equation you posted
does things which no other methoddoes do or can do. Name them!
What things? Where is the proof that the so-called 'partial
differential equation' solves theprime counting problem? Where
is your evidence? Where is your data?Note that no one is fooled
or tricked by your childish attempt to divert attention from
the *real* issues to some side showby inviting certain
challenges. You ARE ALREADY CHALLENGED ON YOUR OWN CLAIMS. Put
up or SHUT UP.--There are two things you must never attempt to
prove: the unprovable -- and the obvious.--Democracy: The
triumph of popularity over
principle.--http://www.crbond.com
===
Subject:
>> One of
their favorite labels is crank, where the idea is that you>>
have some person who is unreasonable, who has ideas that don't
work,>> who refuses to acknowledge the truth.> Let's
see now...> ...some person who is unreasonable, who has
ideas that don't work,>> who refuses to acknowledge the
truth.> Sounds just like you, James!>>Your opinion
is noted Alec McKenzie.>>I do admit that if enough people
think that I'm just some nut, then I>have to wonder.>
_If_ enough people thought that? What reason do you have to>
think that _anyone_, with only one exception that I can think
of> right now, doesn't think of you as just some kind of
nut?>>Note that I made a *specific* challenge in my
previous post, which has>>so far been ignored. I'll talk about
it more a little later.> And note that you're not answering
the specific question I asked,>> in regard to a specific thing
you said.>>That's a rude question.I was rude to _you_?
Golly.>My family doesn't think I'm insaneRight. Actually I was
talking about your correspondents on usenetand elsewhere in the
mathematical community. And I askedwhether you had any reason
to think that any of those peopledid not think of you as some
kind of nut - there's a significantdifference in strength
between some kind of nut and insane.Look. _You_ said if enough
people think that I'm just some nut, thenI have to wonder. This
makes the question of how many peoplewould be enough, and why
you think there are any exceptions,a perfectly natural
question.> you sick turd. Uh, the expression you're looking
for is  ing dog .I like it when you complain about
people being rude to you...>So far you've shown just how
despicable of a human being you can be,>without any regard for
common decency or caring, or ethics.>>If you have an issue with
my prime counting discovery then why can't>you keep to the
mathematics?>>I have a *specific* challenge to you David
Ullrich, I dare you to give>an outlined dispute of the value
of my prime counting discovery>providing *all* the information
necessary for readers to evaluate your>claim,I and many other
people have done this, many times.> without trying to rely on
links or insults.>>What kind of math professor are you if you
can't handle a *math*>challenge but need to insult instead? If
you are a math expert, prove>it.>>Give a point-by-point
exposition to support your case without insults,>innuendo, or
any other attempts at sly or back-handed insults, but>with
*mathematics* and logic.>>I don't think you have the capacity
to engage in an honest and open>debate without trying to use
tricks or insults David Ullrich. I think>you lack the capacity
to have a *reasoned* discussion, but instead>rely on childish
behavior.>>The challenge is to you now, show that you can
reason and discuss>mathematics reasonably, if you can.>James
Harris************************David C. Ullrich
===
Subject:
First, I'll
directly address the title Common sense: If crank labelwere
believed. Some already believe the label of crank applies,
sodialogue present, it appears that the majority believe you
are a crankto some degree. If you wish to reverse that image,
your best optionis to conduct yourself in a professional
manner, not insult anyone forany reason, and fully respond to
all questions and challenges aboutyour claims.> Some of you
may have noticed these free-wheeling mathematical> discussions
I've been having with all these people, and wondered.Maybe
mostly you were wondering what they were doing on your>
newsgroup.Well I've been someone who has dabbled in
mathematical research in> full knowledge that I'm NOT a
mathematician, but partly because it's> fun, and partly
because I have confidence that people can miss things.> Well,
I'm not a mathematician either.> Mathematicians don't believe
they could miss anything simple enough> that a
non-mathematician could find that's important. And I'm not>
just saying that as I've read it more than once from
mathematicians.> Thats just wrong. I have corrected
mathematics professors on severaloccasions and they thanked
me. I have yet to find any mistake in apublished paper by a
mathematician, though when I find one, I do notexpect to be
belittled.> Now as I've talked about my various ideas and
research, math people> have talked a bit about the math, but
mostly they've retaliated by> insulting me and saying nasty
things about me on webpages.> The post archive on google tells
a different story. I looked at theearly posts, and there were
no insults.> Yes, it's odd but true.One of their favorite
labels is crank, where the idea is that you> have some person
who is unreasonable, who has ideas that don't work,> who
refuses to acknowledge the truth.Only problem is that I enjoy
tracing out the steps of my mathematical> arguments, and I can
do things like give you a Java program that> works.> I have
seen your mathematical arguments. Your tracing out the stepsis
actually rather poor. In your Advanced Factorization/Core
Errorthread, you only put in seven steps when I could see as
many asthirty.You should try programming in C. Its much better
for heavymathematical computation.> The truth is that I've
dabbled in a lot of mathematical areas, poking> around looking
for something dramatic, and pissed off some math> people.Does
that make me a crank?> What you need to address is what pissed
off the other sci.math users. > When you contact mathematicians
about specifics, what are theresponses?> There's some guy who
was going around quoting from me all the time in> his
posts!Now then, use your common sense.Is it simpler that I'm
really some crank, or that I've just upset a> lot of
oversensitive people who think that negative labeling and>
insults are the way to go?Now the problem is bad lessons
learned on Usenet. I'm sure lots of> dabblers who had some
ideas here or there were rapidly chased off of> the sci.math
newsgroup when they were ganged up on and insulted. So>
posters learned that they could control people by insults!
They were> happy!!! They could insult people and get rid of
them! But I came> along and decided that I didn't want to let
people control my speech> by insults.So because I refused to
be insulted into silence you have a weird> religion that's
developed in the math community which is based on> obsessively
insulting me!> You CAN stop the exchange of insults by not
responding to them.> And I wanted others on other newsgroups
to see that oddity.It's rather freaky, eh? But then again, who
really thought math> people were normal? And don't worry, this
post is very unlikely to> change anything.Yup, believe it or
not, it will all just keep going. I'll talk about> math, there
will be some who will argue with the math, while most will>
just lob insults.> Stick to just the math and ignore the
insults. When the insultersdont get a response, they will
eventually stop.> Fascinating.> James Harris
===
Subject:
> Granted,
some might think I do the same, but he comes to me. That is,>
David Ullrich makes sure to come to my threads and reply to my
posts,> and then whines when I tell him to go away.>> He's a
tag-along that just won't go away when he's told that he's
not> wanted.This is usenet, a public service. You don't get to
choose who participatesin your threads.
===
Subject:
>> James Harris,
barely-literate yard-ape.>> Yard ape, curtain climber, carpet
shark, ankle biter, nose miner,> sprog, linoleum lizard, loin
monkey, fartling, crotch fruit,> shriekling, flesh loaf, womb
dropping, little sticky person, crotch> trophy, howling
 -machine.>> But give James Harris credit: he fought off
the rusty coat hanger to> evenutally emerge as a twat lugie.
Hey stooopid loud troll James> Harris, put up or shut up,>LOL!
Git 'im!
===
Subject:
<snip>I *dare* any posters who disagree to give
just a point-by-point attack> on my prime counting function,
numbering out each point, and providing> all the information
in their post rather than just posting links.> That's just
silly. Your prime counting function is correct, as hasbeen
demonstrated several times.As I pointed out in an earlier post
(quoting)# Thus we see that the proposed prime counting method#
is indeed correct, though that wasn't in doubt.## Is this
Legendre's method? That depends on one's point of view.# It's
certainly not an exact copy, but it is not difficult to# see
that Legendre's prime counting formula may be derived from#
this method and conversely, so in that sense the two
techniques# are equivalent. If one were inclined to be
unchartiable, one# could call this obfuscated Legendre;
otherwise, it could be# designated as streamlined Legendre.##
Is it better than Legendre's method? Well, it does not
require# knowing the primes, since they're computed on the
fly.# Unfortunately, the resulting method is several orders
of# magnitude slower than some existing methods, at least
when# implemented as written.## Does it have any applicability
other than to produce a slow# method of counting primes? I
don't see any, though it is# tidy and compact. It might have
some use as a pedagogical# device, perhaps as an exercise in a
text, but I have to admit# that I don't see any use beyond
that.## Is it publishable? Perhaps as a classroom note, but
the journal# would have to be selected with care. It would
face the same# difficulty as an improvement on an algorithm
like bubble sort--# the response by a great number of
reviewers would be, I suspect,# This is mildly interesting,
but it hardly advances the state# of the art.Rick
===
Subject:
A
geometry can be expressed by a form w: T(M^n) -> C^n. We can
defineinvariant vector fields by,w_i ( v_j) = d_ijA * operator
by,*(w) e = sum(i1 .. in = 0 .. n - 1) sign_of_permutation(i1
... in)e(v_i1, ..., v_ip) w_i(p+1) ^ ... ^ w_in / (p!
(n-p)!)Some special types of geometries obey, ( l1 *(w) d *(w)
d + l2 d *(w) d *(w) ) w = 0.For example, smooth groups have l1
= 0.A value of the parameters might approximate gravitation. l1
= l2 seemslikely.Other values of the parameters might be
expressible as an algebraicrelation. The rules of physics
might.A generalization of the equation is,( l0 + sum(i = 1,
infinity) ( l_(2i - 1) ( *(w) d *(w) d )^i + l_(2i)( d *(w) d
*(w) )^i ) ) w = 0.mihai
===
Subject:
let us assume, we have an compact
topological space X, where thetopology comes from an absolut
value |.| on X. let us assume furtherwe have an surjective
continous function f : X^n -> X^m and let usmeasure X^i by the
ith product of the haar measure comming from X. myquestion is:
if N has measure 0 in X^m, does f^(-1)(N) then also
havemeasure 0? is there any literature dealing with these
problems? thankyou for any answer...tanja
===
Subject:
>let us assume,
we have an compact topological space X, where the>topology
comes from an absolut value |.| on X. let us assume further>we
have an surjective continous function f : X^n -> X^m and let
us>measure X^i by the ith product of the haar measure comming
from X. my>question is: if N has measure 0 in X^m, does
f^(-1)(N) then also have>measure 0? is there any literature
dealing with these problems? thank>you for any answer...There
are many things wrong with your set-up. First of all, a
generaltopological space doesn't have an absolute value. Do
you want X to beasubset of the complex numbers?Second, Haar
measure is something on a topological *group*, in fact,a
locally compact one.I am going to make a wild guess about what
you mean. Please correctme if I am wrong. I am assuming that X
is a subset of the complexesand that Haar measure is just
Lebesgue measure restricted to X. Even then,the answer to your
question is trivially no. Just consider anyfunction f:X->X
which is constant on a set of positive measure and is
stillsurjective. Then the inverse image of a point will be a
set of positivemeasure.I doubt there is much literature on a
problem so broad as this.--Dan Grubb
===
Subject:
>>let us assume, we
have an compact topological space X, where the>topology comes
from an absolut value |.| on X. let us assume further>we have
an surjective continous function f : X^n -> X^m and let
us>measure X^i by the ith product of the haar measure comming
from X. my>question is: if N has measure 0 in X^m, does
f^(-1)(N) then also have>measure 0? Well, I have no idea what
an absolute value on a compacttopological space is, nor what
haar measure on a compacttopological space is. But whatever
you mean, the answerto your question is almost surely _no_ -
inverses of null setsunder continuous functions are not going
to be null sets,except under very special conditions.>is there
any literature dealing with these problems? thank>you for any
answer...>>tanja************************David C. Ullrich
===
Subject:
>
I need a suggestion. I took linear algebra this semester up
here in Montana> (MSU-Billings). I thought that I could handle
it, but I'm really> struggling. In fact, I'm having to take an
incomplete in the course because> it's just not making any
sense to me. I've always been strong in math, but> matrices
and vectors spaces have got me entirely stumped.Are there any
good resources (either in print or online) out there that>
anyone knows of that would help me out. Something like Linear
Algebra for> Complete Morons would be nice . . .:) Okay, this
is pretty simplistic, but you asked for
it:http://ceee.rice.edu/Books/LA/contents.htmlIt's intended
for highschool students as a precalculus course. Doesn'tassume
any background. (I used it to teach myself.)-- - Laurel * * *
http://amberdine.com
===
Subject:
no, trust me, very basic is very good.
i don't know why i'm so lost withthis, but i feel like a
bumbling idiot.and to everyone: thank you very much for all
the responses, both here in-- matt> I need a suggestion. I
took linear algebra this semester up here inMontana>
(MSU-Billings). I thought that I could handle it, but I'm
really> struggling. In fact, I'm having to take an incomplete
in the coursebecause> it's just not making any sense to me.
I've always been strong in math,but> matrices and vectors
spaces have got me entirely stumped.>> Are there any good
resources (either in print or online) out there that> anyone
knows of that would help me out. Something like Linear
Algebrafor> Complete Morons would be nice . . .>> :) Okay,
this is pretty simplistic, but you asked for it:>>
http://ceee.rice.edu/Books/LA/contents.html>> It's intended
for highschool students as a precalculus course. Doesn't>
assume any background. (I used it to teach myself.)> -- > -
Laurel * * * http://amberdine.com>
===
Subject:
>no, trust me, very
basic is very good. i don't know why i'm so lost with>this,
but i feel like a bumbling idiot.>>and to everyone: thank you
very much for all the responses, both here in<pins>I'm
curious. Are you having problems with matrix arithmetic?I also
hit a very large brick wall when my book went from talking
about arithmetic to vectors and I still don't seemto be able
to get past the section where the unit vectoris defined. I
haven't spent much time on figuring outwhy yet but it has
something to do with the notation |v|.It is possible that my
brain is hardwired to map thetwo vertical glyphs to absolute
value, which also causesme enormous problems. /BAH
===
Subject:
My reply
is interwoven within original post below.> Here is an
unoriginal game (inspired by Dots and Boxes{name?}, Go,> and
by Scrabble somewhat).> But this post is not just a game-idea,
but a question about strategy. > The game can be played with
any number(m) of players, m>=2,> where each player has a black
pencil/pen and a colored pencil/pen of a > color different than
the player's opponents' colors.> The game is played on an
n-by-n grid, where n^2 is divisible by m.> I would suggest an
n of at least 6 or 8 for 2 players. The players take turns
each writing (in black) 1, 2, 3, ...(n^2/m), each > integer
being written into any still-unoccupied grid-square.> (with
his/her colored pencil) all grid-squares within any rectangle
in > the grid, IF, and only if,each grid-square in the
rectangle has an integer written in it,the SUM of every
integer in the rectangle is any PRIME number,and all the
squares in the rectangle are not yet colored in.> After each
player has moved (ie written an integer and perhaps colored in
> a rectangle)> (n^2/m) times, the winner is the player with
the most grid-squares of > that player's pencil-color.>
Variations:Players can only fill in rectangles which contain
the square they just >I thought this would be the best at
first. But now I believe thatbeing able to color in rectangles
not necessarily including the lastinteger written might be
better.For then the players can, for one thing, leave
prime-rectangles forlater, hoping their opponent(s) do not
fill them in.Anyway, either way would affect strategy
differently. > Players can write their integers into the grid
in any order they wish.> (I like this variation, but it
requires remembering which integers have > already been
played.)Players draw at-random their integers from a deck of
cards. > (Either from 4 decks of 1 through (n^2/m), or from
one master deck of 1 > through n^2.)> In this game, with or
without any of the variations (or with any > variations of
your own), what would be a good strategy???> One strategy
might involve putting down consecutive integers inadjacent
squares, so as to ensure that the line of integers iscomposite
(when length of line is >= 3).Players can ensure that certain
rectangles are composite until a givenpoint in the game, where
the rectangle is finished with an integerwhich makes it
prime.But the stategy depends upon what is known about an
opponent'sunplayed integers (if integers are not necessarily
playedsequencially).(If you do not know what I ma talking
about above, then okay. I do notknow exactly myself...) > By
the way, ideally, the game board will look aesthetically
interesting > when the game is done, especially with the right
m and n.> (Any suggestions for m and n from this standpoint?)>
Actually what really matters, probably, regarding the beauty
of thefinal board are the colors used, rather than the number
of coloursused.And I bet most players would use x's and o's or
something else as suchin-practice, instead of filling the
squares with colored pencil, sincethen they only need one
pencil.> Leroy Quet
===
Subject:
> Here is Eucleides's proof that there
are infinitely many Pythagorean> triplets.> If n is a natural,
then n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. This> means that
every odd natural (>1) is the difference between two>
squares.> Some of these odd naturals are squares themselves.
There are (at least)> as many Pythagorean triplets as there
are odd squares. And because there> are infinitely many odd
numbers, and an odd number's square is odd, then> there are
infinitely many Pythagorean triplets.> Do I remember this
proof correctly?Just for grins, here is a way to get every odd
number as the smallestterm of a Pythagorean triplet. I noticed
this when I hadn't yet hadany algebra in school, so I found
the proof very difficult. Now ofcourse it is extremely easy.We
were always given (3, 4, 5) and (5, 12, 13) as Pythagorean
triplesin homework problems. What I noticed was that 4 + 5 = 9
= 3^2 andthat 12 + 13 = 25 = 5^2, and of course that 4 and 5 as
well as 12 and13 differed by 1. After trying to extend this
pattern to (7, 24, 25),(9, 40, 41) and perhaps a few more, I
saw that it always worked. Iasked my big brother to prove it
for me and the rat made me do it formyself. I had a hard time
figuring out that the pattern was (2n+1,2n^2+2n, 2n^2+2n+1),
but once that was done, it was easy enough tocalculate that
(2n^2+2n+1)^2 = (2n^2+2n)^2 + 2(2n^2+2n) + 1 =(2n^2+2n)^2 +
(4n^2+4n+1) = (2n^2+2n)^2 + (2n+1)^2 as desired.Now it is
actually easy to see that this triple is primitive, forclearly
the last two terms differ by 1 and thus cannot have a
commonfactor, anything dividing the first two terms would also
divide2n^2+n, hence n, hence 1, so they are relatively prime,
and anythingdividing the first and third terms would have to
divide 2n^2+1, hencen+1, hence n, hence 1, and so they are
also relatively prime and thetriplet is thus primitive. I just
figured that out now, not way backwhen.It would not surprise me
if this result dates back to the ancientGreeks. Does anyone
know?Achava
===
Subject:
> Here's Euclid's proof that there are
infinitely many Pythagorean triples.>> If n is a natural, then
n^2 - (n-1)^2 = (n+n-1)(n-n+1) = 2n-1. This>> means that every
odd natural (>1) is the difference between two squares.>> Some
of these odd naturals are squares themselves. There are (at
least)>> as many Pythagorean triplets as there are odd
squares. And because there>> are infinitely many odd numbers,
and an odd number's square is odd, then>> there are infinitely
many Pythagorean triplets.>> Do I remember this proof
correctly?Just for grins, here is a way to get every odd
number as the smallest> term of a Pythagorean triplet. I
noticed this when I hadn't yet had> any algebra in school, so
I found the proof very difficult. Now of> course it is
extremely easy.We were always given (3, 4, 5) and (5, 12, 13)
as Pythagorean triples> in homework problems. What I noticed
was that 4 + 5 = 9 = 3^2 and> that 12 + 13 = 25 = 5^2, and of
course that 4 and 5 as well as 12 and> 13 differed by 1. After
trying to extend this pattern to (7, 24, 25),> (9, 40, 41) and
perhaps a few more, I saw that it always worked. I> asked my
big brother to prove it for me and the rat made me do it for>
myself. I had a hard time figuring out that the pattern was
(2n+1,> 2n^2+2n, 2n^2+2n+1), but once that was done, it was
easy enough to> calculate that (2n^2+2n+1)^2 = (2n^2+2n)^2 +
2(2n^2+2n) + 1  (2n^2+2n)^2 + (4n^2+4n+1) = (2n^2+2n)^2 +
(2n+1)^2 as desired.It's very easy to derive directly,
requiring only trivial algebraand some parity: n^2 odd  n
odd (< even^2 = even ); to wit:If (a,b,c) is a Pythagorean
triple with c = b+1 (so primitive)then (b+1)^2 = b^2 + a^2 
2b+1 = a^2 odd  a odd, say a = 2n+1  2b+1 = (2n+1)^2 =
4n(n+1)+1  b = 2n(n+1), a = 2n+1, c = b+1Fibonacci did it by
summing 2n+1 = (n+1)^2 - n^2, as is described in his entry at
the MacTutor math history archive: Liber Quadratorum, written
in 1225, is Fibonacci's most impressive piece of work,
although not the work for which he is most famous. The book's
name means the book of squares and it is a number theory book
which, among other things, examines methods to find
Pythagorean triples. Fibonacci first notes that square numbers
can be constructed as sums of odd numbers, essentially
describing an inductive I thought about the origin of all
square numbers and discovered that they arose from the regular
ascent of odd numbers. For unity is a square and from it is
produced the first square, namely 1; adding 3 to this makes
the second square, namely 4, whose root is 2; if to this sum
is added a third odd number, namely 5, the third square will
be produced, namely 9, whose root is 3; and so the sequence
and series of square numbers always rise through the regular
addition of odd numbers. Thus when I wish to find two square
numbers whose addition produces a square number, I take any
odd square number as one of the two square numbers and I find
the other square number by the addition of all the odd numbers
from unity up to but excluding the odd square number. For
example, I take 9 as one of the two squares mentioned; the
remaining square will be obtained by the addition of all the
odd numbers below 9, namely 1, 3, 5, 7, whose sum is 16, a
square number, which when added to 9 gives 25, a square
number.-Bill Dubuque
===
Subject:
Bruce Harvey
<newsforbruce@bearsoft..co.uk> scribbled the following:>> Here
is Eucleides's proof that there are infinitely many
Pythagorean>> triplets.>> If n is a natural, then n^2 -
(n-1)^2 = (n+n-1)(n-n+1) = 2n-1.> Apart from the algbra you
have it. Although the three expressions are indeed> equal the
middle one should be> n^2 -(n^2-2n+1) 12 year old's
algebra!!!!!!Why should it be that? I was using the rule a^2 -
b^2 =(a+b)(a-b), with b being equal to (n-1).-- /-- Joona
Palaste (palaste@cc.helsinki.fi) ------------- Finland
---------- http://www.helsinki.fi/~palaste
--------------------- rules! --------/That's no raisin - it's
an ALIEN! - Tourist in MTV's Oddities
===
Subject:
> Bruce Harvey
<newsforbruce@bearsoft..co.uk> scribbled the following:>> Here
is Eucleides's proof that there are infinitely many
Pythagorean>> triplets.>> If n is a natural, then n^2 -
(n-1)^2 = (n+n-1)(n-n+1) = 2n-1.>> Apart from the algbra you
have it. Although the three expressions areindeed> equal the
middle one should be>> n^2 -(n^2-2n+1) 12 year old's
algebra!!!!!!>> Why should it be that? I was using the rule
a^2 - b^2  (a+b)(a-b), with b being equal to (n-1).>but you
did say proof and to my simple mind as a maths teacher, a
proof isset out line by line stating any other theorems
invoked.> -- > /-- Joona Palaste (palaste@cc.helsinki.fi)
------------- Finland --------> --
http://www.helsinki.fi/~palaste --------------------- rules!
--------/> That's no raisin - it's an ALIEN!> - Tourist in
MTV's Oddities
===
Subject:
> Bruce Harvey
<newsforbruce@bearsoft..co.uk> scribbled the following:>> Here
is Eucleides's proof that there are infinitely many
Pythagorean>> triplets.>> If n is a natural, then n^2 -
(n-1)^2 = (n+n-1)(n-n+1) = 2n-1.>> Apart from the algbra you
have it. Although the three expressions areindeed> equal the
middle one should be>> n^2 -(n^2-2n+1) 12 year old's
algebra!!!!!!>> Why should it be that? I was using the rule
a^2 - b^2  (a+b)(a-b), with b being equal to (n-1).example
google?I asked the first original reference. Nobody replied.
I'm interested in mathhistory.Here is a partial copy of my
post (equations) dated above:For example:1) For ODD a , a
general solution exists:p^2 - (p-1)^2 = q^2 
p^2=(p-1)^2+q^2, becausep^2-(p^2-2p+1) 2p-1
=q^2Examples:5^2-4^2= 3^2 , where 3 is odd in Eq. in the
title.13^2-12^2= 5^225^2-24^2= 7^241^2-40^2= 9^2....etc.
infinitely for (odd)^2 on rhs.2) For EVEN b , a general
parametric solution exists:s^2-(s-2)^2=t^2  s^2-(s^2-4s+4)
=4(s-1) =t^2Examples:5^2 -3^2=4^2, where 4 is even in Eq.in
the title.10^2-8^2=6^217^2-15^2=8^2.......etc. infinitely for
(even)^2 on rhsMy point was: Because any square, even or odd,
is the difference of twosquares, then there are infinite many
Pythagorean triplets so that any evenor odd integer is a
member in some Pythagorean triplet. As there areinfinite many
primes, then there are infinite many different
non-trivialPythagorean triplets.But who invented that first.
Was it really Euclides? Somebody else?Tapio>> --> /-- Joona
Palaste (palaste@cc.helsinki.fi) ------------- Finland
--------> -- http://www.helsinki.fi/~palaste
--------------------- rules! --------/> That's no raisin -
it's an ALIEN!> - Tourist in MTV's Oddities
===
Subject:
>> Bruce Harvey
<newsforbruce@bearsoft..co.uk> scribbled the following:>>
Here is Eucleides's proof that there are infinitely many
Pythagorean>> triplets.>> If n is a natural, then n^2 -
(n-1)^2 = (n+n-1)(n-n+1) = 2n-1.>> Apart from the algbra you
have it. Although the three expressions are> indeed> equal
the middle one should be>> n^2 -(n^2-2n+1) 12 year old's
algebra!!!!!!>> Why should it be that? I was using the rule
a^2 - b^2  (a+b)(a-b), with b being equal to (n-1).>>
example google?> I asked the first original reference. Nobody
replied. I'm interested inmath> history.>> Here is a partial
copy of my post (equations) dated above:>> For example:> 1)
For ODD a , a general solution exists:> p^2 - (p-1)^2 = q^2 
p^2=(p-1)^2+q^2, because>> p^2-(p^2-2p+1) 2p-1 =q^2>>
Examples:>> 5^2-4^2= 3^2 , where 3 is odd in Eq. in the
title.> 13^2-12^2= 5^2> 25^2-24^2= 7^2> 41^2-40^2= 9^2> ....>
etc. infinitely for (odd)^2 on rhs.>> 2) For EVEN b , a
general parametric solution exists:> s^2-(s-2)^2=t^2 
s^2-(s^2-4s+4) =4(s-1) =t^2>> Examples:>> 5^2 -3^2=4^2, where
4 is even in Eq.in the title.> 10^2-8^2=6^2> 17^2-15^2=8^2>
.......> etc. infinitely for (even)^2 on rhs>> My point was:
Because any square, even or odd, is the difference of two>
squares, then there are infinite many Pythagorean triplets so
that anyeven> or odd integer is a member in some Pythagorean
triplet. As there are> infinite many primes, then there are
infinite many different non-trivial> Pythagorean triplets.>
But who invented that first. Was it really Euclides? Somebody
else?>> Tapio>I am not happy about the even numbers because
infinity is a difficultconcept: withx^2 as x aproaches
infinity being of a different order than x as xaproaches
infinity.Or even (x^x)! as x aproaches infinity.They are
different concepts of infinity.So in your series>13^2-12^2=
5^2> 25^2-24^2= 7^2> 41^2-40^2= 9^212, 24, 40 ...... is
increasing faster than 5, 7, 9 .......so its a different order
of infinityAs a concept, infinity needs pinning to a definite
conceptual entity such asa number line or a grid of unit cubes
formed by the grid lines of acartesian co-ordinate system
sapnning infinite space.-- Bruce Harveybruce@bearsoft.co.ukThe
Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft>>
--> /-- Joona Palaste (palaste@cc.helsinki.fi) -------------
Finland --------> -- http://www.helsinki.fi/~palaste
--------------------- rules! --------/> That's no raisin -
it's an ALIEN!> - Tourist in MTV's Oddities>>
===
Subject:
Michigan
State University grad student Michael Shafer has succeeded
inidentifying the largest known prime number to date, using a
distributedcomputer network of more than 200,000 computers
located around the world.The new number is 6,320,430 digits
long and is only the 40th Mersenne primeto have ever been
discovered (Mersenne primes are an especially rare breedthat
take the form of 2-to-the-power-of-P, where P is also a prime
number).
===
Subject:
> Michigan State University grad student Michael
Shafer has succeeded in> identifying the largest known prime
number to date, using a distributed> computer network of more
than 200,000 computers located around the world.> The new
number is 6,320,430 digits long and is only the 40th
Mersenneprime> to have ever been discovered (Mersenne primes
are an especially rare breed> that take the form of
2-to-the-power-of-P, where P is also a primenumber).Sounds
like a waste of processor power (and electricity) to me. There
is noreasonin discovering these large numbers (even if you want
to use them forencryption.I believe the encryption process
would need veeeery much time ;-}} ).Karl
===
Subject:
> Michigan State
University grad student Michael Shafer has succeeded in>
identifying the largest known prime number to date, using a
distributed> computer network of more than 200,000 computers
located around the world.> The new number is 6,320,430 digits
long and is only the 40th Mersenne> prime> to have ever been
discovered (Mersenne primes are an especially rare breed> that
take the form of 2-to-the-power-of-P, where P is also a prime>
number).Sounds like a waste of processor power (and
electricity) to me. There is no> reason> in discovering these
large numbers (even if you want to use them for> encryption.>
I believe the encryption process would need veeeery much time
;-}} ).I don't think Mersenne primes would be very useful for
encryption. For example in the RSA method, what you need is
the product of two large primes with the idea that nobody can
factor that product. Give me any product of two primes of
which one is a Mersenne prime, and I can factor it
easily.
===
Subject:
Mike Parker> Michigan State University grad student
Michael Shafer has succeeded in> identifying the largest known
prime number to date, using a distributed> computer network of
more than 200,000 computers located around the world.> The new
number is 6,320,430 digits long and is only the 40th
Mersenneprime> to have ever been discovered (Mersenne primes
are an especially rare breed> that take the form of
2-to-the-power-of-P, where P is also a primenumber).My mother
phoned me today to tell me that she had seen something on
CNNabout a breakthrough in mathematics. As soon as I got home
I scoured CNN'swebsite, finding nothing. Turns out it's just a
6-million digit prime. Whata letdown.LH
===
Subject:
Larry Hammick
<larryhammick@telus.net> scribbled the following:> Mike
Parker>> Michigan State University grad student Michael Shafer
has succeeded in>> identifying the largest known prime number
to date, using a distributed>> computer network of more than
200,000 computers located around the world.>> The new number
is 6,320,430 digits long and is only the 40th Mersenne>
prime>> to have ever been discovered (Mersenne primes are an
especially rare breed>> that take the form of
2-to-the-power-of-P, where P is also a prime> number).> My
mother phoned me today to tell me that she had seen something
on CNN> about a breakthrough in mathematics. As soon as I got
home I scoured CNN's> website, finding nothing. Turns out it's
just a 6-million digit prime. What> a letdown.Finding new
Mersenne primes is just a question on spending more and
moretime and processor power on distributed computer
networks.Now a real mathematical breakthrough would be for
example solving afamous conjecture, like Goldbach's
conjecture, either proving ordisproving it.-- /-- Joona
Palaste (palaste@cc.helsinki.fi) ------------- Finland
---------- http://www.helsinki.fi/~palaste
--------------------- rules! --------/'It can be easily shown
that' means 'I saw a proof of this once (which I
didn'tunderstand) which I can no longer remember'. - A maths
teacher
===
Subject:
Isn't there a power of 2 that is in fact
prime?--OL
===
Subject:
>Isn't there a power of 2 that is in fact
prime?>>--OL>2^1
===
Subject:
Chergarj schrieb:>>Isn't there a power of
2 that is in fact prime?>>2^1 Fascinating! Two of the most
fundamental mathematicalconstants in a simple formula...
giving a
prime...<G>
===
Subject:
<28085-3FD7D447-106@storefull-2315.
public.lawson.webtv.net>,> Isn't there a power of 2 that is in
fact prime?--OL> Yes, but it is not a large power, and was
discovered to be prime long
ago.
===
Subject:
><28085-3FD7D447-106@storefull-2315.
public.lawson.webtv.net>,> Isn't there a power of 2 that is
in fact prime?>>Yes, but it is not a large power, and was
discovered to be prime >long ago.You mean we won't be seeing
this in the papers then?Newsgroup participant discovers
smallest prime.
Ever.
===
Subject:
<28085-3FD7D447-106@storefull-2315.
public.lawson.webtv.net>,Isn't there a power of 2 that is in
fact prime?Yes, but it is not a large power, and was
discovered to be prime> long ago.I wonder if we could
generalize the result. There may beothers of these, hmmm,
not-large powers that turn out tobe prime.Jim
Burns
===
Subject:
<28085-3FD7D447-106@storefull-2315.
public.lawson.webtv.net>,> Isn't there a power of 2 that is in
fact prime?Yes, but it is not a large power, and was discovered
to be prime> long ago.I wonder if we could generalize the
result. There may be> others of these, hmmm, not-large powers
that turn out to> be prime.I would be interested in a list of
all known primes that are powers of Mersenne primes. I bet
there are not many of them, probably not more than fourty at
the moment.
===
Subject:
>
<28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>,> Isn't there a power of 2 that is in fact prime?>>
Yes, but it is not a large power, and was discovered to be
prime> long ago.>> I wonder if we could generalize the
result. There may be> others of these, hmmm, not-large powers
that turn out to> be prime.>> I would be interested in a list
of all known primes that are powers of> Mersenne primes. I bet
there are not many of them, probably not more> than fourty at
the moment.How about zero? As in There are no prime numbers
which arepowers of Mersenne primes. None at all. Ain't gonna
happen.End of story. Next.
===
Subject:
>
<28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>,> Isn't there a power of 2 that is in fact prime?>>
Yes, but it is not a large power, and was discovered to be
prime> long ago.>> I wonder if we could generalize the
result. There may be> others of these, hmmm, not-large
powers that turn out to> be prime.>> I would be interested
in a list of all known primes that are powers of> Mersenne
primes. I bet there are not many of them, probably not more>
than fourty at the moment.How about zero? As in There are no
prime numbers which are> powers of Mersenne primes. None at
all. Ain't gonna happen.> End of story. Next.p^n for p =
1.
===
Subject:
Mike Parker <None@myisp.not> scribbled the following:>
<28085-3FD7D447-106@storefull-2315.public.lawson.webtv.net>,> Isn't there a power of 2 that is in fact prime?>>
Yes, but it is not a large power, and was discovered to be
prime>> long ago.>> I wonder if we could generalize the
result. There may be>> others of these, hmmm, not-large powers
that turn out to>> be prime.>> I would be interested in a
list of all known primes that are powers of>> Mersenne primes.
I bet there are not many of them, probably not more>> than
fourty at the moment.> How about zero? As in There are no
prime numbers which are> powers of Mersenne primes. None at
all. Ain't gonna happen.> End of story. Next.Depends on your
definition of power. If p is a Mersenne prime, thenp^1 sure as
heck is a prime.-- /-- Joona Palaste (palaste@cc.helsinki.fi)
------------- Finland ----------
http://www.helsinki.fi/~palaste --------------------- rules!
--------/It's time, it's time, it's time to dump the slime! -
Dr. Dante
===
Subject:
> Michigan State University grad student Michael
Shafer has succeeded in > identifying the largest known prime
number to date, using a distributed > computer network of more
than 200,000 computers located around the world. > The new
number is 6,320,430 digits long and is only the 40th Mersenne
prime > to have ever been discovered (Mersenne primes are an
especially rare breed > that take the form of
2-to-the-power-of-P, where P is also a prime number).When I
read it the first time I found the most surprising not that
hehad discovered a large power of 2 but that he had discovered
one thatis prime. There must be something wrong with the core
of mathematics.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject:
>Message-id:
<HppK8p.8t3@cwi.nl> Michigan State University grad student
Michael Shafer has succeeded in> identifying the largest known
prime number to date, using a distributed> computer network of
more than 200,000 computers located around the world.> The new
number is 6,320,430 digits long and is only the 40th
Mersenne>prime> to have ever been discovered (Mersenne primes
are an especially rare breed> that take the form of
2-to-the-power-of-P, where P is also a prime>number).>>When I
read it the first time I found the most surprising not that
he>had discovered a large power of 2 but that he had
discovered one that>is prime. There must be something wrong
with the core of mathematics.Yup.>-- >dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam,
nederland,>+31205924131>home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/--MensanatorAce
of Clubs
===
Subject:
> Dear all,It is easy to prove that if the first
number in a row of Pascal's> triangle is a prime then all
other numbers (except 1) in that row are> multiples of that
prime.However, several sources say that this is also a
sufficient condition> for this phenomenon, ie if the first
number n is composite, then there> exists a binomial
coefficient in that row not a multiple of n.I've tried looking
for a proof online, but with no success. I would> greatly
appreciate it if you could help me out.Apologies if this is
rather trivial.RedIf we have n!/(k!(n-k)!) =
binomial(n,k),then binomial(n,k) =(n/k) binomial(n-1,k-1), for
k and n = positive integers.So, if GCD(n,k) = j, andn = n' j, k
= k' j,then binomial(n,k) = (n' j)/(k' j) binomial(n-1,k-1) =
multiple of n' / k'.And so, for n and k = positive integers,
binomial(n,k) is divisible by n'/k'.But k' does not divide n',
so k' divides binomial(n-1,k-1),since binomial(n,k) is an
integer.So,n' = n/GCD(n,k)always divides binomial(n,k),for n
and k = positive integers.Specifically, if n and k are
relatively prime, then n always dividesbinomial(n,k).Yet n MAY
still sometimes divide the binomial if it is not coprime with
k.(What is the first example??? Is there an example
??)thanks,Leroy Quet
===
Subject:
> Yet n MAY still sometimes divide the
binomial if it is not coprime with k.> (What is the first
example??? Is there an example ??)binomial(10,4) = 210--
Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject:
> Dear all,It is easy to prove
that if the first number in a row of Pascal's> triangle is a
prime then all other numbers (except 1) in that row are>
multiples of that prime.However, several sources say that this
is also a sufficient condition> for this phenomenon, ie if the
first number n is composite, then there> exists a binomial
coefficient in that row not a multiple of n.I've tried looking
for a proof online, but with no success. I would> greatly
appreciate it if you could help me out.Apologies if this is
rather trivial.Red Here's something I posted several years
ago:
===
Subject:
===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:==
===
Subject:
===
Subject:===
Subject:===
Subject:===
Subject:=|> |> Is it true that GCD of
{mC1,mC2,...,mC[m-1]} is p if m=p^n with p prime and|> 1
otherwise?|> Yes. |> Here mCn is m combination n, or the nth
entry in the mth row of Pascal's|> triangle.|> |> If so, is
there a really obvious proof? More important: can anybody name
a|> reference (textbook, paper etc..)?|> Using a result due to
Kummer which is nicely presented in [1], it's not _too_ hard
to prove. Let me change your notation a bit: it'll be easier
to follow the argument if m combination n is denoted by
C(m,n). That said, Kummer's result is as follows: given n, k >
0, let c_p(n,k) = the number of carries generated when adding n
& k using base p. Then Theorem(Kummer). If p is a prime, then
p^{c_p(n,k)} is the highest power of p which divides C(n+k,k).
If you grant me this, then I can argue as follows. Since C(m,1)
= m, the gcd is always a divisor of m. So, if m=p^n, the gcd of
the set is p^r for some r >= 0. But, given n and k (0 < n,k <
p^n) such that n+k = p^n, there _has_ to be at least 1 carry
generated when we add the base p versions of n and k -- that's
the only way to generate the '1' that's the coefficient of p^n
in the result. Hence, every element of our set is divisible by
(at least) p. On the other hand, if 0 < k < p, then Kummer's
result shows that C(p^n,kp^{n-1}) is _not_ divisible by p^2.
Hence, the gcd is p in this case. For the other case, suppose
that m is not a power of a prime, and let p be any prime which
divides m. In the base p representation of m, there is either a
term rp^u (with 1 < r < p) or a pair of terms rp^u +
sp^v.(There are probaly a *bunch* of other terms, as well -- I
don't care about those). In the first case, C(m,(r-1)p^u) is
_not_ divisible by p; in the second case, C(m,rp^u) is not
divisible by p. Since, for every prime p dividing m, there's
some C(m,n) in the set that is not divisible by p, it follows
that the gcd of the set is 1. 1. Peitgen,H.-O. _et_al._, Chaos
and Fractals/New Frontiers of Science, Springer-Verlag, 1992.
ISBN 0-387-97903-4 |> Flo.|> |> |> -- |>
----------------------------------------------------------|>
Florian Breuer at University of Stellenbosch, South Africa|>
reply to: s9520155@itu3.sun.ac.za|>
----------------------------------------------------------
===
Subject:
===
Subject:
===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:===
Subject:==
===
Subject:
===
Subject:===
Subject: The result that you're wanting to prove follows
easily -- unless n is a prime, some C(n.k) with 1 < k < n is
not a multiple of n.
===
Subject:
Dear all,Trying to figure out the
1-dimensional representations of D_n, the dihedralgroup of
order 2n.Should be easy...Let r be rotation, s be
reflection.Then r^n = 1, s^2 = 1, rs = sr^-1...Now let F be
the representation from D_n to k* , F : D_n --> k* wherek =
complex numbers.Then F(r)F(s) = F(s)F(r^-1) <= F(r) =
F(r^-1) since these numberscommute , so F(r^2) = 1 <
===
Subject:
> F(r)
= 1 or -1....likewise since s^2 = 1,F(s) = 1 or -1...So there
are 4 1-dimensional representations of D_n for all n?Rob
===
Subject:
>
Dear all,Trying to figure out the 1-dimensional
representations of D_n, the> dihedral group of order 2n.>
Should be easy...Let r be rotation, s be reflection.Then r^n =
1, s^2 = 1, rs = sr^-1...Now let F be the representation from
D_n to k* , F : D_n --> k* > where k = complex numbers.Then
F(r)F(s) = F(s)F(r^-1) <= F(r) = F(r^-1) since these
numbers> commute , so F(r^2) = 1 <
===
Subject:
> F(r) = 1 or
-1....likewise since s^2 = 1,> F(s) = 1 or -1...So there are 4
1-dimensional representations of D_n for all n?Are there four
1-dimensional reps of D_3?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
Subject:
>>
Dear all,>> Trying to figure out the 1-dimensional
representations of D_n, the> dihedral group of order 2n.>
Should be easy...>> Let r be rotation, s be reflection.>> Then
r^n = 1, s^2 = 1, rs = sr^-1...>> Now let F be the
representation from D_n to k* , F : D_n --> k*> where k =
complex numbers.>> Then F(r)F(s) = F(s)F(r^-1) <= F(r) =
F(r^-1) since these numbers> commute , so F(r^2) = 1 <
===
Subject:
>
F(r) = 1 or -1....likewise since s^2 = 1,> F(s) = 1 or -1...>>
So there are 4 1-dimensional representations of D_n for all
n?>> Are there four 1-dimensional reps of D_3?Ouch, you are
right. There are only 2 1-dimensional reps of D_3. Then howcan
I generalize to D_n?>> -- > Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html> Needless to say, I had the
last laugh.> Alan Partridge, _Bouncing Back_ (14 times)
===
Subject:
> Dear all,>> Trying to figure out the 1-dimensional
representations of D_n, the>> dihedral group of order 2n.>>
Should be easy...>> Let r be rotation, s be reflection.>>
Then r^n = 1, s^2 = 1, rs = sr^-1...>> Now let F be the
representation from D_n to k* , F : D_n --> k*>> where k =
complex numbers.>> Then F(r)F(s) = F(s)F(r^-1) <= F(r) =
F(r^-1) since these numbers>> commute , so F(r^2) = 1 <
===
Subject:
>
F(r) = 1 or -1....likewise since s^2 > 1, F(s) = 1 or
-1...>> So there are 4 1-dimensional representations of D_n
for all n?>> Are there four 1-dimensional reps of D_3?Ouch,
you are right. There are only 2 1-dimensional reps of D_3.
Then> how can I generalize to D_n?So far you have F(r) = +- 1
and F(s) = +-1. Are there any morerestrictions you can find on
these? Have you used all the relationsin the group?-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
Subject:
>> Dear all,>> Trying to figure out the
1-dimensional representations of D_n, the>> dihedral group of
order 2n.>> Should be easy...>> Let r be rotation, s be
reflection.>> Then r^n = 1, s^2 = 1, rs = sr^-1...>> Now
let F be the representation from D_n to k* , F : D_n --> k*>>
where k = complex numbers.>> Then F(r)F(s) = F(s)F(r^-1)
<= F(r) = F(r^-1) since thesenumbers>> commute , so F(r^2) =
1 <
===
Subject:
> F(r) = 1 or -1....likewise since s^2 > 1, F(s) = 1 or
-1...>> So there are 4 1-dimensional representations of D_n
for all n?>> Are there four 1-dimensional reps of D_3?>>
Ouch, you are right. There are only 2 1-dimensional reps of
D_3. Then> how can I generalize to D_n?>> So far you have F(r)
= +- 1 and F(s) = +-1. Are there any more> restrictions you can
find on these? Have you used all the relations> in the group?I
guess the last thing I would have to do is analyze the
conjugacy classesof D_n and that is all. (I can do this on my
own)>> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html>
Needless to say, I had the last laugh.> Alan Partridge,
_Bouncing Back_ (14 times)
===
Subject:
>> Dear all,>> Trying to
figure out the 1-dimensional representations of D_n, the>>
dihedral group of order 2n.>> Should be easy...>> Let r be
rotation, s be reflection.>> Then r^n = 1, s^2 = 1, rs =
sr^-1...>> Now let F be the representation from D_n to k* ,
F : D_n --> k*>> where k = complex numbers.>> Then F(r)F(s)
= F(s)F(r^-1) <= F(r) = F(r^-1) since these numbers>>
commute , so F(r^2) = 1 <
===
Subject:
> F(r) = 1 or -1....likewise since
s^2 = 1,>> F(s) = 1 or -1...>> So there are 4 1-dimensional
representations of D_n for all n?>> Are there four
1-dimensional reps of D_3?Ouch, you are right. There are only
2 1-dimensional reps of D_3. Then how> can I generalize to
D_n?>> -- >> Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html>> Needless to say, I had the
last laugh.>> Alan Partridge, _Bouncing Back_ (14 times)1,
time consuming: by learning some more representation theory or
2. lazy option: reading chapter 18 in james and liebeck:there
is an obvious 2-d rep. show, that there are in fact many more
such.you need to consider n odd and even separately. that odd
and evenexplanantion will tell you why there are 2 1-d reps n
odd, 4 n even.you're on the right track. if you want i could
write out the proof, buti'd rather not have to. oh, the n odd
and even thing means there aredifferent 'types' of conjugacy
classes. if you can read the book cos it'smuch more clearly
laid out than i'll manage to make it on here.
===
Subject:
> I know
guys like Abel and Galois did their best work early and then
died,> but can people give me some examples of mathematicians
who did their best> work in their early 20s (say), then
continued to work for a long time but I would like to see an
*objective* study on this.First, a group of unknown
mathematicians' work would be evaluated forgreatness (whatever
'greatness' means...),without those doing the evaluating
knowing any biographical info aboutthe mathematicians at
first.And *then* the mathematicians' ages (and sexes and race
and educationlevel and economic status and...) would be
revealed.Perhaps even the evaluators' own biases can be
studied as well, wherethey each fill out a survey (after
evaluating the math work, butbefore learning anything personal
about the mathematicians) as to whatthey GUESS what the
mathematicians' biographical info might be.... So...by looking
over MY past work posted here to sci.math over thelast few
years, how old do you all *think* I might be???.....;)(And I
know I have posted my age here on occasion. So no fair
peekingat that info!)thanks,Leroy Quet
===
Subject:
>> I know guys like
Abel and Galois did their best work early and then died,>> but
can people give me some examples of mathematicians who did
their best>> work in their early 20s (say), then continued to
work for a long time but> I would like to see an
*objective* study on this.>[...]> So...by looking over MY past
work posted here to sci.math over the> last few years, how old
do you all *think* I might be???.....> ;)>> (And I know I have
posted my age here on occasion. So no fair peeking> at that
info!)We don't have to peek.You react negatively to the
suggestion that mathematical genius peaksat an early age.
Conclusion: you're old.Me too. At least I'm old enough to be
past that peak. Of course, Ifirst realized I wasn't going to
hit the greatness for which I wasdestined when, at
twenty-four, I learned that Robert Johnson died atage
twenty-four[1]. Meanwhile, I had never had a hit blues
recordlike I expected. Of course, I didn't play an instrument
or singeither, but evidently it was too late to learn.Or to
quote Tom Lehrer, When Mozart was my age, he had been dead
fortwo years.Footnotes: [1] Evidently, I was misinformed. He
died at twenty-seven, but I'mpast that, too.-- So, at this
time, I'd like to assure you that I am not interested inI'll
have prosecutors knocking on your doors. I have no problem
with
===
Subject:
>message>> What with the war on terror, and Canada's
puny military and corrupt>> police force, the only way to
insure peace and safety for that>> country, is for it to be
annexed by America.>> It is the only way foward for Canada
aka Canuckistan.>> --------------------------------->>
Well, another solution is the liberation of U.$.A by Canadians
and the>> return of democracy to the Americans peoples.>> I am
sure Canadians will be welcome with (french) kisses and
flowers>when>> they will topple the Little Dictator and his
gang of corporate crooks.>> Patriot act will be scraped; free
press will be restaured, Health Care>> will be available to
all citizens,>>Yeah, like it is in Canada. If you wait. And
wait. And wait. And... oops,>died already.Or you can join an
American HMO where they deny you care because theydon't want
to pay out the money. ooopps one dead wife...mine!Capitalist
medicine is an abomonation and a human rights
violation.THOM>>But at least the bigtime politicians and
bureaucrats get their healthcare!>(By paying cash in Yankee
hospitals...)>>-- >zimriel sbc dot> at global
net>.>http://pages.sbcglobal.net/zimriel/blog/zimblog.html>
because everyone else is doing it>>
===
Subject:
> What with the war
on terror, and Canada's puny military and corrupt>> police
force, the only way to insure peace and safety for that>>
country, is for it to be annexed by America.>> It is the
only way foward for Canada aka Canuckistan.>>You seem to
forget that the Canadians kicked our ass in two
conflicts....>Really? Which two?
===
Subject:
>> What with the war on
terror, and Canada's puny military and corrupt>> police force,
the only way to insure peace and safety for that>> country, is
for it to be annexed by America.>> It is the only way foward
for Canada aka Canuckistan.>>You seem to forget that the
Canadians kicked our ass in two conflicts....>> Really? Which
two?>1775http://www.civilization.ca/cwm/chrono/1000invasion_
e.html1812http://www.civilization.ca/cwm/chrono/1774invasion_
repelled_e.html
===
Subject:
>> What with the war on terror, and
Canada's puny military and corrupt> police force, the only
way to insure peace and safety for that> country, is for it
to be annexed by America.>> It is the only way foward for
Canada aka Canuckistan.>>You seem to forget that the
Canadians kicked our ass in two conflicts....>Really? Which
two?he'se probably talking about ice hockeyTHOM>
===
Subject:
>
What with the war on terror, and Canada's puny military and
corrupt> police force, the only way to insure peace and
safety for that> country, is for it to be annexed by
America.>> It is the only way foward for Canada aka
Canuckistan.>>You seem to forget that the Canadians kicked
our ass in twoconflicts....>Really? Which two?>> he'se
probably talking about ice hockeyThose ignorant of history are
bound to repeat it.
===
Subject:
Let c(1,k) = 1, for all positive
integers k.For all m >= 2,Let c(m,k) = ---- ln(k) ------- >
c(j,k) ln(m) / --- j|m j<m (which, in linear-mode,
is...)c(m,k) =-(ln(k)/ln(m)) sum{j|m, j<m} c(j,m)(the sum is
over the positive PROPER divisors, j, of m.)So, let, for every
real x > 1, f(x) = oo--- --- > ( > c(k,m/k) ) /m^x / /---
---m=1 k|m(which in linear mode is...)f(x) = sum{m=1 to oo}
(sum{k|m} c(k,m/k) ) /m^x.(Inner-sum is over all positive
divisors, k, of m) Puzzle:What is a closed-form (in terms of
known functions) for f(x)??thanks,Leroy Quet
===
Subject:
> If you
 ing morons think that I will let you get away with not>
giving me credit for my  ing math discoveries then you have
another>  ing thing coming.As far as I know, you don't get
paid for mathematical discoveries.Certainly you don't get paid
by announcing them on a newsgroup. Consider -who's going to pay
you, and for what? There exist a few prizes inmathematics that
I've heard of, but you have to submit your work to thesejudges
and so forth. To my knowledge, these prizes aren't
altogetherspectacular. And you can forget about a Nobel Prize
in Mathematics. Thereisn't one, as I'm sure you know.I have
never heard of any corporation paying for a mathematical
discovery,with the exception perhaps of those made in the R&D
departments. Same withthe government, which probably wouldn't
be able to even judge a discoveryanyway (they'd need a
committee, and it probably would be composed ofpoliticians,
not mathematicians).I can understand you being upset if you
think you'd be denied money due toyou, or if you think someone
else would be able to steal it from you byplagiarizing your
work. But as I said - who's going to pay *anyone* for
amathematical discovery?Be well.Baruch
===
Subject:
synaptic energy
redistribution (audio) filter
http://apollonicon.com/music/gold.wavhttp://apollonicon.com/
music/black.wavhttp://apollonicon.com/music/gold.mp3http://
apollonicon.com/music/black.mp3google video
filter?http://apollonicon.com/series/pierre_21.jpg
===
Subject:
a good
score. I know the word good is fairly arbitrary, but if
weconsider the three cases:1) Want to get into a top 10
school2) Want to get into a top 20 school3) Want to get into
an upper Tier 2 schoolThen for each of the above, what would
be the subject testscore/percentile that one would
recommend?
===
Subject:
>Then for each of the above, what would be the
subject test>score/percentile that one would recommend?I can
tell you that a lot of mathematics graduate programs won't
evenlook at your score; just that you took it. Doug
===
Subject:
> a
good score. I know the word good is fairly arbitrary, but if
we> consider the three cases:>> 1) Want to get into a top 10
school> 2) Want to get into a top 20 school> 3) Want to get
into an upper Tier 2 school> Then for each of the above,
what would be the subject test> score/percentile that one
would recommend?Get a 900, and I think that you can call that
good. Otherwise, I would sayat least a 700. Most universities
want the top 1/3 of test takers. That'swhat I have found
anyway.Lurch
===
Subject:
> a good score. I know the word good is fairly
arbitrary, but if we> consider the three cases:1) Want to get
into a top 10 school> 2) Want to get into a top 20 school> 3)
Want to get into an upper Tier 2 school> Then for each of the
above, what would be the subject test> score/percentile that
one would recommend?Get good recommendations from your profs.
If you did especially well in a course taught by a famous
prof, that's the best.
===
Subject:
a good score. I know the word good
is fairly arbitrary, but if we> consider the three cases:1)
Want to get into a top 10 school> 2) Want to get into a top 20
school> 3) Want to get into an upper Tier 2 school> Then for
each of the above, what would be the subject test>
score/percentile that one would recommend?Get good
recommendations from your profs. If you did especially well in
> a course taught by a famous prof, that's the best.I realize
that, but for the purposes of this message I am concernedonly
about GRE scores.Why are people so resistant to answering
questions about GRE scores? Of all the times I've asked this
question, I've never once gotten anumber back as an answer.
Even my advisor was like Don't worry somuch about GRE scores.
Just worry about doing your best. Well ifGRE scores weren't
important, then why does practically everyuniversity in the
country require them to get into their PhD program? Clearly at
least one person at each graduate school must know what
GREscores are good and which are not, otherwise nobody would
ever takeGRE scores into consideration. That's all I'm asking
for here. Iknow GRE scores aren't the final word on
everything, but I hope thatfor once someone can finally give
me a straight answer, and hopefullyrespond with 3
numbers/intervals corresponding to the above 3 rangesof
schools.
===
Subject:
The Prophet Nobody known to the wise as
toolshed37@yahoo.com, opened the Book of Words, and read unto
the people:>Why are people so resistant to answering questions
about GRE scores? >Of all the times I've asked this question,
I've never once gotten a>number back as an answer. Even my
advisor was like Don't worry so>much about GRE scores. Just
worry about doing your best. Well ifPeople don't answer it
because, frankly, it's an ill-formedquestion. GRE scores
_don't_ get you into good universities.
Goodrecommendations/research history gets you into good
universities. Youneed at least a good GRE to get into a
mediocre university, but even agreat GRE won't get you into a
great university. The subject GREcovers competence, not
excellence; a GRE score above a certain levelguarantees you
know a good survey of undergrad math -- knowing moreabove that
simply means you have a good memory. The general GRE
mayhonestly be of more value to math programs, since the main
thing isnot so much what you know as how you think, and the
general ishonestly a better predictor of that.But, y'know,
don't let GRE scores bother you. If you think
they'reimportant, study hard. Apply everywhere you'd apply
anyways. I don'tsee how knowing the relationship between GREs
and acceptance (onewhich is actually tenuous, as I say above)
helps you in any way.>know GRE scores aren't the final word on
everything, but I hope that>for once someone can finally give
me a straight answer, and hopefully>respond with 3
numbers/intervals corresponding to the above 3 ranges>of
schools.The numbers/intervals you're asking for don't exist.
If it would makeyou happy, I could say '600-1000' for all
three (I'm guessing at thelower bound at which 'competence' is
assessed).+----------------------------------------------------
---------+| D. Jacob Wildstrom -- Math monkey and freelance
thinker || Graduate Student, University of California at San
Diego || A mathematician is a device for turning coffee into
|| theorems. -Alfred Renyi
|+------------------------------------------------------------
-+The opinions expressed herein are not necessarily endorsed
by theUniversity of California or math department
thereof.
===
Subject:
> The Prophet Nobody known to the wise as
toolshed37@yahoo.com, opened the Book of Words, and read unto
the people:>Why are people so resistant to answering questions
about GRE scores? >Of all the times I've asked this question,
I've never once gotten a>number back as an answer. Even my
advisor was like Don't worry so>much about GRE scores. Just
worry about doing your best. Well ifPeople don't answer it
because, frankly, it's an ill-formed> question. GRE scores
_don't_ get you into good universities. Good>
recommendations/research history gets you into good
universities. You> need at least a good GRE to get into a
mediocre university, but even a> great GRE won't get you into
a great university. The subject GRE> covers competence, not
excellence; a GRE score above a certain level> guarantees you
know a good survey of undergrad math -- knowing more> above
that simply means you have a good memory. The general GRE may>
honestly be of more value to math programs, since the main
thing is> not so much what you know as how you think, and the
general is> honestly a better predictor of that.But, y'know,
don't let GRE scores bother you. If you think they're>
important, study hard. Apply everywhere you'd apply anyways. I
don't> see how knowing the relationship between GREs and
acceptance (one> which is actually tenuous, as I say above)
helps you in any way.>know GRE scores aren't the final word
on everything, but I hope that>for once someone can finally
give me a straight answer, and hopefully>respond with 3
numbers/intervals corresponding to the above 3 ranges>of
schools.The numbers/intervals you're asking for don't exist.
If it would make> you happy, I could say '600-1000' for all
three (I'm guessing at the> lower bound at which 'competence'
is assessed).So with a GRE score of 600 you'd get into MIT?
Wrong. What aboutwith a GRE score of 700? Wrong.People seem to
confuse my question with what scores will get you intoa good
university? However, nowhere did I ask that question. I
saidwhat scores are good if you want to go to such and such
university? Hence a score which would cause the reviewer to
throw away yourapplication would not be considered good. So
the correct responseshould be an interval which cause the
reviewer to probably not throwaway your application. These
intervals do exist, as a few people havepointed out that. For
example, if you want to go to a top 10university your GRE
score should be in the top 95 percentile or so. Yes, I
understand that doesn't mean anything other than getting in
thetop 95 percentile will cause them to not eliminate you. But
afterall, that's what I was asking about.
===
Subject:
Do negative
numebrs exist in the real world or are they just a
purelysymbolical and mathematical construct?--nethlek
===
Subject:
> Do
negative numebrs exist in the real world or are they just a
purely> symbolical and mathematical construct?Do even positive
numbers exist? To answer that, you need to defineexist I think.
Something for philosophy rather than mathematics. Here is a
start:Realism in Mathematicsby Penelope Maddy Paperback: 224
pagesOxford Univ PrISBN: 019824035X-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject:
>Do negative numebrs
exist in the real world Negative numebrs are the opposite of
positive numebrs, so if you can provethe existence of positive
numebrs, then you immediately see negative numebrs.Doug
===
Subject:
>Do
negative numebrs exist in the real world or are they just a
purely>symbolical and mathematical construct?What are the
respective charges of the electron and proton?
===
Subject:
> Do
negative numebrs exist in the real world or are they just a
purely> symbolical and mathematical construct?Do positive
numebrs exist in the real world or are they just a
purelysymbolical and mathematical construct?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
Subject:
> Do
negative numebrs exist in the real world or are they just a
purely> symbolical and mathematical construct? --> nethleki
think it just abstract. consider 3 apples. does it really
three? is thesize of the apples is the same?
===
Subject:
> Do negative
numebrs exist in the real world or are they just a purely>
symbolical and mathematical construct?> When you run a credit
balance on a credit card or your phone bill, in other words
when you pay more than you owe, they send you a bill for a
negative amount. If the banks and credit card companies
believe in negative numbers, who are the rest of us to
disagree?
===
Subject:
> Do negative numebrs exist in the real world or
are they just a purely> symbolical and mathematical
construct?Numbers (even couting numbers) don't exist in a
physical sense. You can'tgo out to the shop and buy yourself a
box of 7s for example...But if you pay for a box containing 8
apples and you notice that there isonly 7 in there you've
applied this abstract notion of a counting numberto a real
world phenomena.The same thing goes for negative
numbers...They are useful for example in determining
direction. e.g The stockmarket is going down today!Numbers
provide a conceptual framework that has proven to be useful.
Bruce.
===
Subject:
>Numbers (even couting numbers) don't exist in a
physical sense. You can't>go out to the shop and buy yourself
a box of 7s for example...But I live at 7777 Seventy-Seventh
street. How am I supposed to
===
Subject:
>>Numbers (even couting
numbers) don't exist in a physical sense. You can't>>go out to
the shop and buy yourself a box of 7s for example...But I live
at 7777 Seventy-Seventh street. How am I supposed to
===
Subject:
>> Do
negative numebrs exist in the real world or are they just a
purely> symbolical and mathematical construct?>> Numbers (even
couting numbers) don't exist in a physical sense. You can't> go
out to the shop and buy yourself a box of 7s for example...As a
teenager I used to buy boxes of Number 6s!
===
Subject:
> Do negative
numebrs exist in the real world or are >>they just a purely>>
symbolical and mathematical construct?>> Numbers (even
couting numbers) don't exist in a physical >>sense. You
can't>> go out to the shop and buy yourself a box of 7s for
>>example...>As a teenager I used to buy boxes of Number
6s!>At the beginning of every school year, I buy my kids a box
of No. 2s. haha hehe. OK, I'll stop now.
===
Subject:
Do negative numebrs
exist in the real world or are they just a purely> symbolical
and mathematical construct?Numbers (even couting numbers)
don't exist in a physical sense. You can't> go out to the shop
and buy yourself a box of 7s for example...But if you pay for a
box containing 8 apples and you notice that there is> only 7 in
there you've applied this abstract notion of a counting number>
to a real world phenomena.The same thing goes for negative
numbers...> They are useful for example in determining
direction. e.g The stock> market is going down today!> Yes but
you can have a box with five apples in it. You can't have a box
with negative five apples in it. I think that's what the
original poster was getting at.
===
Subject:
> Do negative numebrs
exist in the real world or are they just a purely>
symbolical and mathematical construct?Numbers (even couting
numbers) don't exist in a physical sense. You can't> go out to
the shop and buy yourself a box of 7s for example...But if you
pay for a box containing 8 apples and you notice that there
is> only 7 in there you've applied this abstract notion of a
counting number> to a real world phenomena.The same thing goes
for negative numbers...> They are useful for example in
determining direction. e.g The stock> market is going down
today!> Yes but you can have a box with five apples in it. You
can't have a box > with negative five apples in it. I think
that's what the original > poster was getting at.Yes, but if I
own a warehouse I can!This is very easy to apply to real life
situations.Hypothetically ---- I own a warehouse that stocks
among other things x, y and z widgets.Due to shortages
experienced by my supplier my supply of z widgets isz = -230
units.In other words my orders to date for z widgets outstrips
my z widgetstock by 230 units.I am sure stock status reports'
for many manufacturers or distributorsdeal with negative
numbers all the time. Also in many areas of financenegative
numbers are the norm.Gee that reminds me, I have to get on my
suppliers' ass because I amloosing $ on these z widget sales.
;-)Dan
===
Subject:
> Yes but you can have a box with five apples in it.
You can't have a box > with negative five apples in it. I think
that's what the original > poster was getting at.I can't recall
who it was first pointed it out, but there is nothinginherently
natural about integer arithmetic. Yeah, integerarithmetic works
for apples.But for clouds? If you add one cloud to one cloud,
you don't get twoclouds. The individuation and the physics of
clouds are differentthan for apples. It is a matter of
experimentally determined factthat integer arithmetic works
for apples, and not some automaticthing.This means that
positive integers are not somehow the simple
countingnumbers--they are the simple *apple* counting numbers.
Or forcounting other discrete objects that work like apples
do.Other things that don't work like apples are crumbs. If you
cut anapple in half, you get two half-apples. But if you break
a crumb intwo, you don't get two half-crumbs, you get two
crumbs. So crumbs addlike apples do, but they don't divide
like apples do.So nature uses all different kinds of numerical
entities, and onlysome kinds of counting use positive numbers.
If we are counting, forexample, generations, then it is
perfectly natural to countdescendent generations with positive
numbers and ancestor generationswith negative numbers (or vice
versa).Or, if you are counting electrons, then you can have -5
electrons:when you have 5 positrons.Indeed, you can do that
with apples too, which makes the quotedobservation above
wrong. A little physics, and it turns out thatapples do *not*
behave such that natural numbers are the rightcounting tools,
but rather it turns out, integers! Because (howeverrare),
there is such a beastie as an anti-apple.Thomas
===
Subject: