mm-136
 Please look at my thread in:> >
http://www.physicsforums.com/ 
Dont bother. It merely links to a passowrd protected 
pdf
le.-- Robin Chapman, 
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless
to say, I had the last laugh. Alan Partridge, _Bouncing Back_
(14 times)
 >suppose (i) that f is C^infty and (ii) that
the Taylor series has> >positive (or in'nite) radius of
convergence at any point. Can E be> >that large in this
case?> > No. Suppose f is C^in'nity and its Taylor series has
positive (or > in'nite) radius of convergence everywhere. For
any positive integer > N let E_N = {x: for all n, |f^(n)(x)|
<= N^(n+1) n!}. Then E_N, being> an intersection of closed
sets, is closed. The condition on the > radius of convergence
implies that union_N E_N = R. By Baire Category, > some E_N
has nonempty interior. But f is analytic on any open interval
> contained in E_N.IOm sorry to keep adding new posts to 
this
thread, but I have one morequestion: suppose (ii) above is
replaced by the relaxed condition thatthe Taylor series has a
null radius of convergence at most at isolatedpoints. If x is
an accumulation point of E, then the radius ofconvergence in
x is 0, right?PS: for those who didnOt read the rest of the
thread, E is the set ofpoints where f (smooth) fails to be
analytic.Michele-- DOOh, Google doesnOt a 
provide a
.sig!
 IOm sorry to keep adding new posts to this thread,
but I have one more> question: suppose (ii) above is replaced
by the relaxed condition that> the Taylor series has a null
radius of convergence at most at isolated> points. If x is an
accumulation point of E, then the radius of> convergence in x
is 0, right?No. If a < b, you can de'ne f(x) =
exp(-1/[(x-a)(b-x)]) on (a,b), f = 0 on R (a,b). Then f is
C^oo(R) with all derivatives 0 on R (a,b). For each n, choose
f_n as above with respect to (1/(n+1), 1/n), n = 1,2, ...
There exist constants c_n > 0 such that sum_(n=1,oo) c_n*f_n
converges in C^oo(R) to some f. For this f, E = {1/n : n =
1,2,...} U {0}, 0 is an accumulation point of E, and the
radius of convergence of the Taylor series is oo at each
point of E, including 0.
IOm sorry to keep adding new
posts to this thread, but I have one more>question: suppose
(ii) above is replaced by the relaxed condition that>the
Taylor series has a null radius of convergence at most at
isolated>points. If x is an accumulation point of E, then the
radius of>convergence in x is 0, right?>PS: for those who
didnOt read the rest of the thread, E is the set of>points
where f (smooth) fails to be analytic.No. For example, let g
be a function thatOs C^in'nity on R andanalytic 
nowhere, and
let f(x) = exp(-1/x^2) g(x) for x <> 0, 0 forx = 0. Then f(x)
= O(x^n) as x -> 0 for every n, so the Taylor seriesof f at
x=0 is 0, which has in'nite radius of convergence. But 
's
nowhere analytic.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
 Luigi
that I read, the addition and subtraction formulas (sine> >
and cosine) come all you beginning from the cos(x+y).> >> >
then the cos(x+y) formula has proved through some geometric considerations.> >> > The problem that I have is this:> > this proof is valid only if x and y are inclusive between
zero and> > 45 (so that x+y is inclusive between 0 and 90).> > I know that every angle can be reduced to the 'rst
octant, but to> > prove that the cos(x+y) formula is true for
every x and every y, I> > would need to analyze various
cases...> >> > they are some about 'fty!> >> > 0 <= X <=45 0
<= Y <=45> > 0 <= x <=45 45 <y <=90 x+y <=90> > 0 <= x <=45
45 <y <=90 x+y>90> > 0 <= x <=45 90 <y <=180> > ...> >> > Is
there a quicker way?> >this is probably not what you want,
but if you have some complexanalysis under your belt and this
equation:e^(ix) = cos x + i*sin x has been proved to your
satisfaction, then you have a method forrigorously deriving
those trig identities for all angles and withvirtually no
work,just takee^i(x+y) = e^(ix) * e^(iy), and as a bonus you
get both the sin andthe cos formulas at the same time.this is
how i derive them on tests, i donOt bother trying to
memorizethem.sorry if this is totally
unhelpful.justin
>Luigi <eliliano@libero.it> escribi.97
and subtraction formulas (sine>and cosine) come all you
beginning from the cos(x+y).>>then the cos(x+y) formula
has proved through some
geometric>considerations.>>[...]> >this is
probably not what you want, but if you have some
complex>analysis under your belt and this equation:>e^(ix) =
cos x + i*sin x >has been proved to your satisfaction, then
you have a method for> [...]While weOre on the subject, the
addition formula can be proved directly from the fact that
sine and cosine are the solutions to the differential
equation y + y = 0 satisfying the appropriate initial
conditions. Of course, Luigi was looking for a geometric
proof.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
 This geomatric proof is
valid for any angle:>
http://www.geocities.com/scroussette/addtrig.htmlthank you
very much...your help has been fundamental to meLuigi
 > this proof is valid only if x and y are inclusive between
zero and> > 45 (so that x+y is inclusive between 0 and 90).>
If the sum of two non-negative numbers is no more than 90 it
DOES NOT follow> that the two numbers are less than 90!!
Consider 70 + 20 or 89.999 + .001> >Sorry I intended
(0<=x<=45) ^ (0<=y<=45)
 [0<=(x+y)<=90]not (0<=x<=45) ^
(0<=y<=45) <
 [0<=(x+y)<=90]Luigi>a sub 0 mod c = a^0
mod c and>a sub 1 mod c = a^1 mod c and>a sub 2 mod c = = (
a^1 * a ) mod c and>a sub 0 = a^0 and>a sub 1 = a^1 but, a<c
implies>a sub 2 mod c <= a^2 and 2<n implies>a sub n mod c <=
a^n, that is, itOs possible that>a sub n mod c # a^n so for
either a or b, 1<=a<b<c+1<=c<a+b, n>1 implies>either>a sub 3
mod c # a^3 or>b sub 3 mod c # b^3 so>0<=n<=2>ThatOs about
right. Let me add a bita sub n = ( (a sub n-1* a ) mod ca sub
0 = a^0 mod c so a.0 + b.0 = c implies a solution
(suf'ces),while a<=b<c means no such solution exists, anda
sub 1 = a^1 if a<c so a.1 + b.1 = c suf'ces, while such
solutions are pretty trivial, buta sub 2 <= a^2 but ifa,b <
sqrt (c), then a.2+b.2=c suf'ces, while all such solutions
are nontrivial, but3,4 > sqrt(5), but1,1 < sqrt(2) but n<=c
so this is the only onea sub 3 <= a^n but ifa,b <
cuberoot(c), then a.3 + b.3 = c suf'ces, and any such
solution would be a prize, but while1,1 < cuberoot(3) BUT1,2
<> cuberoot (3) and for n <= c this is seems to bemore than a
coincidence, that isn <= c implies n <= 2 because2,2 <
fourthroot4) and1,3 <> fourthroot (4) and sofor n >2, for c =
cmin = n anda <= b < c = nE a,b,c | b > nthroot(c) in other
wordsn>2 implies E b < c <= n | b>nthroot(c) and a.n + b.n =
c does not suf'ce 3>n implies E b < c <= n | b<nthroot(c) and
a.n + b.n = c suf'cesa clear distinction between the relevant
cases,so if you believe me when I say n <= c, this will be
worth reading, and I doprovide some argument that we need not
consider n>c.Yours,Doug Goncz (at aol dot com)Replikon
Research Read the RIAA Clean Slate Program Af'davit and
Description at http://www.riaa.org/I will be signing an
amended Af'davit soon.
=Aside from knowing and
understanding that P^n is a vector space, which mighthelp, I
canOt help.P^1 is the space of all 'rst order 
polynomialsP^n
is the space of all nth order polynomials.Yours,Doug Goncz
(at aol dot com)Replikon Research Read the RIAA Clean Slate
Program Af'davit and Description at http://www.riaa.org/I
will be signing an amended Af'davit soon.
 > >> They
couldnOt quite get the platinum-irridium artifact to weigh
exactly> >> 9.866 65 newtons - nohow - so they gave up and
adopted it by proclamation:> >> Now theyOre looking to
rede'ne that adopted kilogram:> >> >I believe 
itOs 9.80665 N.
(I may be wrong). m. ;-) > > You are wrong, but in a different
way than you think you might be.> > That value has nothing
whatsoever to do with the de'nition of a> kilogram as a unit
of mass.> > Furthermore, if you stick to SI units, any
standard acceleration of> free fall is worthless as tits on a
boar.I thought that free fall meant that acceleration was
zero?For taking F = ma, if F = 0, m > 0, then a has to be
They couldnOt quite get the platinum-irridium
artifact to weigh exactly> > >> 9.866 65 newtons - nohow - so
they gave up and adopted it by proclamation:> > >> Now 
theyOre
looking to rede'ne that adopted kilogram:> > >> > >I believe
itOs 9.80665 N. (I may be wrong). m. ;-) > > > > You are
wrong, but in a different way than you think you might be.>  > That value has nothing whatsoever to do with the 
de'nition
of a> > kilogram as a unit of mass.> > > > Furthermore, if you
stick to SI units, any standard acceleration of> > free fall
is worthless as tits on a boar.> > I thought that free fall
meant that acceleration was zero?No. An object falling
ballistically in gravity is infree fall, as is a satellite in
orbit. One onlinedictionary gives this: The ideal falling
motion of something subject only to a gravitational 'eld. -
Randy
 > > > > > > > >> They couldnOt quite get the
platinum-irridium artifact to weigh exactly> > > >> 9.866 65
newtons - nohow - so they gave up and adopted it by
proclamation:> > > >> Now theyOre looking to 
rede'ne that
adopted kilogram:> > > >> > > >I believe itOs 9.80665 N. (I
may be wrong). m. ;-) > > > > > > You are wrong, but in a
different way than you think you might be.> > > > > > That
value has nothing whatsoever to do with the de'nition of a>  kilogram as a unit of mass.> > > > > > Furthermore, if you
stick to SI units, any standard acceleration of> > > free
fall is worthless as tits on a boar.> > > > I thought that
free fall meant that acceleration was zero?> > No. An object
falling ballistically in gravity is in> free fall, as is a
satellite in orbit. One online> dictionary gives this: The
ideal falling motion of > something subject only to a
gravitational 'eld.The above does not contradict acceleration
<kullstj-nn@comcast.net> sysengr
=Suppose I have a real 3x3
rotation matrix, R. So, R is orthogonal andROR=I, where I is 
a
3x3 identity matrix.I need to compute the singular value
decomposition of (R - I), and I wouldlike to do it
symbolically (as opposed to numerically).Does the fact that R
is orthogonal help with this problem, or will theanswer be
ugly (symbolically)?Anybody already know of an answer
(published or otherwise)?John
Suppose I have a real 3x3
rotation matrix, R. So, R is orthogonal and>ROR=I, where I 
is
a 3x3 identity matrix.>I need to compute the singular value
decomposition of (R - I), and I would>like to do it
symbolically (as opposed to numerically).>Does the fact that
R is orthogonal help with this problem, or will the>answer be
ugly (symbolically)?rotation, and it will work out pretty
nicely.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
=Robert,
YouOre killing me. IOm old and tired. Do you 
really expect me
to work atthis if you already know the answer? ;-) Any more
(detailed) hints?John> >Suppose I have a real 3x3 rotation
matrix, R. So, R is orthogonal and> >ROR=I, where I is a 3x3
identity matrix.>> >I need to compute the singular value
decomposition of (R - I), and Iwould> >like to do it
symbolically (as opposed to numerically).>> >Does the fact
that R is orthogonal help with this problem, or will theanswer be ugly (symbolically)?>> rotation, and it will work
out pretty nicely.>> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T
1Z2
= - tired of spending all your free time trying to 'nd
the right job? - have you met your goals of sending a certain
number of resumes each week? - are you ready fr success in
your career search? We have answers to these questions for
you, come and visit us atwww.KareerSearch.com!
=S^c is not
open < Ae>0,Ey in X:y in (U(x,e) n S)you say that itOs not
true but then it means that the following isalso not trueS^c
is open < Ee>0,Ay in X:y in U(x,e)  y in S^cBut this is a
de'nition of open set or is it?No it isnOt, but 
if you insist
on writing it like that it would be:S^c is open < Ax in
S^c, Ee>0: U(x,e) contained in S^cDo you mean that the
following is correct?S^c is open < Ax in S^c, Ee>0, Ay in
similarity between the set of facts and the set of parts
intechnological order.For the set of parts, in technological
order, from Finite Mathematics withBusiness Applications,
there exists a vector, a, and a square lower lefttriangular
quantity matrix, Q with zero diagonal q.nn, such that:The
elements of a are in technological order, in other words, an
item a.i doesnot appear on the list until all of the parts
and subassemblies that must gointo it have already appeard in
elements a.1 through a.(i-1), andq.ij is the number of units
of item a.j directly needed to assemble one itema.i., that is
if we had on hand q.i1 units of item a1, q.i2 units of iterm
a2,etc., and q.in units of a.n then we could immediately
assemble one unit of a.iand have no parts left over. a.i is
listed once in terms of any or no parts a.1through a.i on row
i with q.j1 through q.j(i-1) the parts needed, if any,
orotherwise zero.Stage requirements exists on Q. Q^k = 0 for
some k<=n. The entries of Q^1 givethe one-stage requirments
for a.imax and its subassemblies. The entries of Q^2are the
two stage requirements for a.imax and etc. The entries of
Q^(n<-imax)are the n-stage requirements for a.imax, that is
the 'nal product, and anyneed subassemblies. Therefore since
the zero diagonal, Q^2 has a zero diagonaljust below the zero
diagonal, and kmin is the order of Q, not the rank orsize.I
believe but have no example that this matrix is as useful for
baking a cakeor sheet of n cookies per sheet as it is for
proving a theorem with numberedsingle stage axioms and other
postulates, corrolaries, and sub-proofs. The bookgives a
clear example of the assembly of A tripod mast constructed
from half-legs assembled from rods and bolts, legsassembled
from half-legs and bolts, and a plate and some bolts.Now for
a factory taking orders for individual items of manufacture
such as oneVCR with a free tape and manual, and three VCR
tapes, each with a free label,it is easy to extend Q to this
theorem:x = x1 + x2 + ... x.imax is the unknown production
vector, how many of eachitem to pull before assembly of the
order begins. The elements can be whollyseparated, such as a
VCR and a CD player, or inherently linked as above,including
orders for spare parts if offered. The production vector
times theprice vector can be the price of the order if
nothing is double billed, andthatOs just a restriction on 
the
price vector for a particular catalog.But E d = d1, d2, ...,
dn, the outside demand vector, or the order.How are x, Q, and
d, related?x = xQ +dd = x*(I-Q) and if I-Q has an inversex =
(d*(I-q)^-1)They show the inverse exists.(I-Q)^-1 = I + Q +
Q^2 + ... Q^(k-1), k<=imax the 'rst k for which Q^k = 0The
lower left triangular structure of Q makes this particularly
easy.By precomputing (I-Q)^(-1) per catalog, any order d from
that catalog producesx = d * ((I - Q)^(-1))by one sole matrix
multiplication, very fast in particular in Mathcad.Perhaps
this will produce by analogy a more subtle proof of God. I
think youmay have over-linearized the problem At least you
didnOt attempt to prove thereis a Minkowski space inside the
psyche, resulting from EinsteinOs GeneralRelativity, as a
crackpot did in talk.origins a few years ago! I 'nd
youranalysis intruiging.However, your analysis seems to imply
there is a 'rst reason for everything,or did I misunderstand
it?While thatOs not necessarily true, you may not have
implied it. I just seemsthat if E x<y for all reasons, the
list is in'nite and assumes there is a'rst 
reason for all
facts in the list F. If that is true, you have merelyassumed
the existence of a 'rst reason, and a 'rst 
cause, which is
either theBig Bang, something else, or God, right? Oh, wait,
the 'rst thing you say isyou can have heirarchical causality
_without_ a 'rst cause. IOd like to hearmore 
because this is
really well written, and I must read it fully
beforeattempting to understand the reasoning and the
valididty of the conclusion.Doug (sig
below)-
-
--A) INTRODUCTION I have decided
to do a rewrite forclari'cation purposes. The original post
was not a proof of god.Instead i tried to explain that we
canhave hiearchical causality without a singleminimal cause,
and yet with convergence,bringing all causes toward each
other.I attempted to demonstrate that math mightbe used to
solve this philosophical problem, usinga limit idea similar
to that used in calculus,and probably other branches of
math.Notice that i do *not* de'ne god. I onlypoint out how
the symbol god can be used in a meaningfulway, much as the
in'nity symbol is used in calculuswithout 
de'ning any set
with the name in'nity.(I understand that set theory has
in'nities but thatOsbeside the point.). In this 
revision I
haveattempted to make the math a lot clearer.Most of it
should be obvious, which all statementsare taken as fact
without proof. I can attemptto supply any proof of any
mathematical statementI made, should any one of them seem
inobvious to anyone.Note that if you see previous versions of
thisrevision, I did attempt to delete them, whetheror not all
servers deleted them i canOt say.I donOt mean 
wanted to make a fewchanges to things.B) INTRODUCTORY
DEFINITIONSDe'nition 1) F := set
of factsObviously this set doesnOt really exist in the
universeof ZFC, but since this is an informal discussion, we
canjust ignore that fact and move on.For the time being we
may as well consider F to beany set in ZFC satisfying
whatever conditions i specify.What is a fact? Well I am not
entirely sure.I think for the moment we should look at F as
somethingakin to a set of logical assertions, as ingrass is
green and grass exists.De'nition 2) CH := the causal
hierarchy.This is a relation on F^2.if (x,y) is in CH we
write x < y.CH is 1) anti-re§exive 2) anti-cyclic 3)
transitive 4) anti-symmetric 5) Also for all y in F, there is
a x in F such that x < y, in other words every fact has a
cause. 6) For y and z both in F, there is some x in F such
that x < y and y < z, which means that every two facts must
have a common cause. C) DISCRETE MODEL--In
this model F is assumed to be *discrete* (when orderedby
CH).De'nition of discrete:Supposing that a set S is discrete
(when ordered by some relation)then this would imply that for
every z in S,should there be some q in S such that q < z, then
there issome x in S such there is no y in S such that x < y <
z.Since we have supposed F to be discrete,then it is trivial
to prove thatfor every z in F there is a *unique* x in F such
that thereis no y in F such that x < y < z, and we
de'neReason(z) to be this x.In other words Reason(z) is the
immediate reason for or theimmediate cause of x. Reason^2(x)
could be said to be the reason for the reason,and so on. It
is clear that for all n in N (the naturals)Reason^n(x) is
de'ned.D) CONTINUOUS MODEL-First we begin
by dropping our conditionthat F be discrete. In its placewe
impose a new condition on F.The condition is that the
function d,be well-de'ned.d must satisfy these conditions: 1)
Domain(D)=CH 2) if both x and y are in F and x<y, then d(x,y)
is in R+ (the positive reals). 3) if x is in F d(x,x)=0 4) if
x,y, and z are all in F and x <= y <= z, then
d(x,z)=d(x,y)+d(y,z). 5) if x and z are in F and x<=z and
then for every m in R satisfying 0 <= m <= d(x,z), there is
some y in F such that x <= y <= z and d(y,z)=m Now we
re-de'ne de'ne the function Reason.We 
de'ne it as having
domain Domain(Reason)={(n,y) | x <= y and d(x,y)=n for some x
and y both in F and n in RR-}Then for all pairs (n,y) in
Domain(Reason)we de'ne Reason^n(y) to be the uniquex in F
such that x < y and d(x,y)=n. Note that in the continuous
model, it is possiblethat for some y in F and for some m in
RR-, forevery x in F such that x < y, dist(x,y) <=mIn this
case we might say that x is distance-bounded.(IOll cut this
down to d-bounded, to save space).If one element of F is
d-bounded, it is possible to prove thatevery element in F
must be d-bounded.Therefore if one element of F is d-bounded
it isreasonable to say that F itself is d-bounded.So suppose
that for some x, itOs god-limit is m,then we can write
lim[n->m]Reason^n(x)=god,in the same sense that one can
sometimes writelim[x->c]f(x)=inf in calculus.E) OBVIOUS
CORALLARIES- Every x in F has a
god-limit which can be thoughtof a distance from god.Suppose
that x has god-limit m,then for each y,there will be an n
satisfying 0 <= n < m such that x<=Reason^n(x) and
y<=Reason^n(x),Very roughly speaking,a common reason for x
and any fact y can always be foundbetween x and its
god-limit.Yours,Doug Goncz (at aol dot com)Replikon Research
Read the RIAA Clean Slate Program Af'davit and Description at
http://www.riaa.org/I will be signing an amended Af'davit
soon.
=I am suprised that anyone reads that stuff from so
far back.I had psychological problems at the time and was a
crank evenif I was unaware of it at the time. Its over with
andhopefully IOll never return to crankdom. But nowthat were
talking about all that stuff I posted long agoIOll mention
that, for the time being, IOve tried tostay away from topics
like this, although I donOtwant to renounce my old views
completely. I thinkthat reality needs a prime uni'er
(conciousness),and I guess it could also be construed as the
reasonfor everything. I donOt even know if math 
couldapproach
a topic like this because math talks aboutthings that donOt
really exist. Everything inthe phenomenal world never really
*is* because itundergoes change, and math speaks of
absolutely existingentities. Indeed, thatOs probably not a
failing of math.Its a failing of intellect itself when we
divide the worldinto beings. Even talking about the contents
of a thought is meaningless if there are *no* *individual*
*thoughts*.At any rate either conciousness moves in time
whilewhat is perceived is static, or what is perceivedchanges
while conciousness remains (paradoxically) 'xed.claims that
conciousness (i.e. me and you) is absolutely 'xedand I
consider that to be a likely proposition.(i.e. the prime
uni'er = you, now and forever).I apologise to the readers of
sci.math for havingstarted this crankery long ago. I am cross
posting this postto alt.philosophy and I would appreciateit if
all replies to this post were posted there(and not to
sci.math), since this subject isnot mathematical in
nature.Sorry Doug for not reading and digesting your own
reply.It really isnOt up my alley right now, and I do 
notmean
to imply that your ideas are uninterestingor not worthy of
consideration.
=X is the number of aces in a set of 13
cards drawn.I would say its Poisson distribution because,
foreach card drawn not being ace, the probability thatthe
next card is an ace increases.On the other hand the
probability of getting more thanfour aces is zero, no matter
how many cards we willdraw. Any suggestions?Please note that
Im not looking for an actual answerbut the resoning why it
should or should not be aPoisson (or any other)
distribution.--
KindlyKonrad-
places;their souls be chased by demons in Gehenna from one
room toanother for all eternity and more.Sleep - thing used
by ineffective people as a substitute for coffeeAmbition - a
poor excuse for not having enough sence to be
lazy
X is the number of aces in a set of 13 cards drawn.> .... You
may 'nd the sci.stat.math news group a better place to ask.
Ken Pledger.
X is the number of aces in a set of 13 cards
drawn.>I would say its Poisson distribution because, for>each
card drawn not being ace, the probability that>the next card
is an ace increases.Why would that indicate itOs Poisson?>On
the other hand the probability of getting more than>four aces
is zero, no matter how many cards we will>draw. Any
suggestions?Then how could it be Poisson?| Please note that
Im not looking for an actual answer| but the resoning why it
should or should not be a| Poisson (or any other)
distribution.You stated the reason, you just donOt realize
that you did.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
> I would
say its Poisson distribution because, for>> each card drawn
not being ace, the probability that>> the next card is an ace
increases.>> Why would that indicate itOs Poisson?No calls 
to
a larm central for 5 minutes gives lower chancethat a call
will occure within next minute than 10 minuteswithout a call.
ThatOs how i 'gure. Where do i step wrong?> 
...the
probability of getting more than four aces is zero...> Then
how could it be Poisson?According to my teacher, the number
of spelling errors ona page can be Poisson distributed, even
if the chance ofthem being, say 1.000.000, is zero. ThatOs
how. :)> You stated the reason, you just donOt realize that
you did.If not Poisson, then which one?--
KindlyKonrad-
places;their souls be chased by demons in Gehenna from one
room toanother for all eternity and more.Sleep - thing used
by ineffective people as a substitute for coffeeAmbition - a
poor excuse for not having enough sence to be
lazy
 I would say its Poisson distribution because, for> each
card drawn not being ace, the probability that> the next
card is an ace increases.>> Why would that indicate itOs
Poisson?>No calls to a larm central for 5 minutes gives lower
chance>that a call will occure within next minute than 10
minutes>without a call. ThatOs how i 'gure. 
Where do i step
wrong?YouOre wrong to think that this makes it a Poisson
distribution.If youOre talking about a Poisson process 
(where
the number ofoccurrences of something over any given time
interval is a random variable with Poisson distribution),
then the numbers ofoccurrences over disjoint intervals are
supposed to be independent.So if having no calls over the
last 10 minutes increases the probability of a call in the
next minute, this is _not_ a Poissonprocess and the total
number of calls will probably _not_ havea Poisson
distribution.> ...the probability of getting more than four
aces is zero...>> Then how could it be Poisson?>According to
my teacher, the number of spelling errors on>a page can be
Poisson distributed, even if the chance of>them being, say
1.000.000, is zero. ThatOs how. :)Well, either your teacher
is confused or you have misunderstoodwhat (s)he said. The
Poisson distribution with parameter lambdasays Pr(X=x) =
exp(-lambda) lambda^x/x! for _all_ nonnegative integers x. If
Pr(X=1000000) = 0, itOs not Poisson. Now it mightbe
_approximated_ by a Poisson distribution, but thatOs not 
what
you asked about.>If not Poisson, then which
one?Hypergeometric.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
=--
KindlyKonrad-
places;their souls be chased by demons in Gehenna from one
room toanother for all eternity and more.Sleep - thing used
by ineffective people as a substitute for coffeeAmbition - a
poor excuse for not having enough sence to be
lazy
=
which tools can do tests of randomness? which one is the
best? and rank data available!
 > which tools can do
tests of randomness? http://csrc.nist.gov/rng/
=[...]|>
However, Professor McKenzie began claiming my work is
algebraic|> geometry which is out of his 'eld. So I explained
the entire thing|> over more than an hour, having driven up
four hours from Atlanta|> metro, and he tells me itOs out of
his 'eld.[...]In other words, you talk stuff that is
orthogonal to real mathematics,not even wrong, and you were
wasting the guyOs time and so he sent 
youon.IOd have done the
same... much like the Indians who 'gured out theycould get 
rid
of the Spaniards by telling them the next village had lotsof
gold.If he tells you youOre wrong or talking nonsense, 
youOll
argue. If hetells you it isnOt his 'eld and he 
doesnOt know
much about it, but youcan go talk to that guy over there who
might be interested, he is ridof you.IOll only tell you this
once...-- cu,Brucedrift wave turbulence:
http://www.rzg.mpg.de/~bds/
 > [...]> > |> However,
Professor McKenzie began claiming my work is algebraic> |>
geometry which is out of his 'eld. So I explained the entire
thing> |> over more than an hour, having driven up four hours
from Atlanta> |> metro, and he tells me itOs out of his 
'eld. [...]> > In other words, you talk stuff that is orthogonal
to real mathematics,> not even wrong, and you were wasting
the guyOs time and so he sent you> on.Which 
de'es the fact
that I explained it point-by-point. Theprofessor challenged
me repeatedly on points and I handled allchallenges.The
discussion was over an hour in two sessions.What amazes me is
how clearly some of you are dedicated to falsebeliefs. Given
the evidence that matters you instead turn to
*social*pronouncements as if simply *saying* something
changes the truth.ItOs like something is broken in your 
heads
and your rationalfunctions are impaired or absent.
Fascinating.James Harris
 > [.snip.]> > >Well,
rationally, one would suppose that in explaining a math
argument> >a person has to go step-by-step, which reveals all
the logical links.> > You would think so. IOve explained 
step
by step why your claims about> the roots of x^3+3x-2 are
false. Did you look at it?> > [.snip.]No. > >However, youOve
given the information that you know of him, and he IS> >your
senior.> > Surely thatOs irrelevant? I thought it 
wasnOt
about who had what> degree, but about the math?> > (Yes, he
is my better, by far. HeOs a world-renowned expert in>
Universal Algebra, he solved the Tarski Problem a few years
back, he> invented Tame Congruence Theory, and has a
signi'cant number of> accomplishments under his bel)And he
liked me, and I must admit that the more I think about it
themore I wonder if I didnOt make a mistake taking the path 
I
have.But then IOve spent so much time wasted with a liar 
like
you Magidin.Maybe mathematicians in general arenOt so bad,
but you are.And I think back now on what I gave him. > >He
dismissed the objection youOve used so often, and in fact,
thatOs> >not surprising as it never made any sense to think
that f divided off> >P(m) as some function of m or dependent
on m, as itOs just not> >mathematics.> > Apparently, he
dismissed what you think or what you told him was my>
objection. Your track record is clear: you have great
dif'culty in> correctly paraphrasing other 
peopleOs
objections.Bull. He trashed your objection. ThatOs it.>
We have no way of knowing if what you told him was an
accurate> representation of my view, other than your say so.
And you have been> wrong on this subject pretty much every
time youOve tried to state> what my objection is.> > But,
tell you what: state my objection in full context, and
provide> a link to a post where I made it.> > ItOs not that 
f
divided off P(m) as some function of m or dependent> on m
because I have never used the words divided off, so thatOs>
your paraphrase. And it depends very much on just what the
heck f and> P(m) are supposed to be.> > So, go ahead. Provide
the complete statement, weOll see how accurate> you were in
reporting it.Come on Magidin, thatOs only necessary for all
those losers who donOtreally know 
mathematics.IOve hung out
with Professor McKenzie. I know where I stand.> >IOve called
it voodoo math.> > Yes. YouOve also said that saying that
something is a parameter is> rejecting algebra. Did this
professor also agree with you on that> point?> > [.snip.]The
professor was quite gracious in many ways, and IOve been
ratherobnoxious.I hate you Magidin. YouOve twisted things,
lied, and challenged me atevery point even when IOve been
right.You jealous bastard. So what if IOm a better
mathematician. Is thatany reason to 
7ï
WEcf
!.856[Eth]èÿÿ
objection> >Magidin has given for months, quickly dismissed
it.> > Non sequitur.> > By your own standards, seniority is
irrelevant, but let that be as it> may. I have absolutely no
problems admitting that Ralph McKenzie is> way smarter than I
am.> > However, there is not an iota of evidence that what you
presented to> this professor was an accurate report of my
objection.I presented your objection which relies on the
assertion that factorsof f can distribute as functions of m,
or dependent on m, and theprofessor dismissed it.IOve talked
to you too much Magidin. YouOve convinced me that
othermathematicians are corrupt and Professor McKenzie at my
own alma materwas nice to me and I responded in anger when he
didnOt give me exactlywhat I wanted: vindication.> >Rather
than simply tell the truth, Magidin now backs away from hisoriginal position, > > What was my original position?That my
work is wrong.> >I guess now claiming that Professor McKenzie
didnOt> >understand what I was discussing.> > No, I did not
make that claim. What I said, as shoudl be clear from a>
simple reading of the above, is that your Advanced
Polynomial> Factorization is unclear, and many statements are
ambiguous. That> certain interpretations of those statements
lead to absolutely correct> statements, which are not
applicable in the situation you are trying> to apply them
(your FLT argument); and that certain interpretations of> the
ambiguous statements lead to absolutely false statements. > >
Which interpretation did you present?I explained the argument
point-by-point you sick asshole.Professor McKenzie was
skeptical but I answered his objections.You sick twisted
bastard, IOve argued with you for years, and youstill refuse
to accept the math.> >The conclusion of the paper Advanced
Polynomial Factorization is that> >for the factorization> > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)> >where the aOs are all algebraic integers, one of them is
coprime to 5.> > And that conclusion is false. WeOve gone
over it, in detail. But here> wrong. (Original calculations
done by Dale Hall):> > 1. Let > > q1 = 8 (a_1)^2 - 76 (a_1) -
185> r1 = 8 (a_1)^2 - 4 (a_1) - 45> s1 = 4 (a_1)^2 - 37 (a_1)
- 104> > Since a_1 is an algebraic integer, each of q1, r1,
s1 are algebraic> integers.> > 2. Likewise, let> > q2 = 8
(a_2)^2 - 76 (a_2) - 185> r2 = 8 (a_2)^2 - 4 (a_2) - 45> s2 =
4 (a_2)^2 - 37 (a_2) - 104> > and> > q3 = 8 (a_3)^2 - 76 (a_3)
- 185> r3 = 8 (a_3)^2 - 4 (a_3) - 45> s3 = 4 (a_3)^2 - 37
(a_3) - 104> > Each of q2, r2, s2, q3, r3, s3 are algebraic
integers.> > 3. We have that> > q1*r1 = [8(a_1)^2 - 76(a_1) -
185][8(a_1)^2 - 4(a_1)-45]> = 64(a_1)^4 - 32(a_1)^3 -
360(a_1)^2> -608(a_1)^3 + 304(a_1)^2 + 3420(a_1)>
-1480(a_1)^2 + 740(a_1) + 8325> > = 64(a_1)^4 - 640(a_1)^3 -
1536(a_1)^2 + 4160(a_1) + 8325> > > Since (a_1)^3 - 12(a_1)^2
+ 65 = 0, we have that> > (a_1)^3 = 12(a_1)^2 - 65> > (a_1)^4
= 12(a_1)^3 - 65(a_1)> = 12( 12(a_1)^2 - 65) - 65(a_1)> =
144(a_1)^2 - 780 - 65(a_1)> = 144(a_1)^2 - 65(a_1) - 780,> >
so> > q1*r1 = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 +
4160(a_1) + 8325> = 64 [144(a_1)^2 - 65(a_1) - 780] > - 640
[12(a_1)^2 - 65]> - 1536(a_1)^2 + 4160(a_1) + 8325> =
9216(a_1)^2 - 4160(a_1) - 49920 - 7680(a_1)^2 + 41600>
-1536(a_1)^2 + 4160(a_1) + 8325> = 41600+8325-49920> = 5.> >
Since (a_2)^3 - 12(a_2)^2 + 65 = 0 and (a_3)^3 - 12(a_3)^2 +
65 = 0,> we also have> > q2*r2 = 5.> q3*r3 = 5.> > > > 4.
Using the same de'nitions, we have:> > r1*s1 = ( 8(a_1)^2 -
4(a_1) - 45) * (4(a_1)^2 - 37(a_1) - 104)> = 32(a_1)^4 -
296(a_1)^3 - 832(a_1)^2> - 16(a_1)^3 + 148(a_1)^2 + 416(a_1)>
- 180(a_1)^2 +1665(a_1) + 4680> > = 32(a_1)^4 - 312(a_1)^3 -
864(a_1)^2 + 2081(a_1) + 4680> = 32( 144(a_1)^2 - 65(a_1) -
780) - 312( 12(a_1)^2 - 65)> - 864(a_1)^2 + 2081(a_1) + 4680 = 4608(a_1)^2 - 2080(a_1) - 24960 - 3744(a_1)^2 + 20280> -
864(a_1)^2 + 2081(a_1) + 4680> = (a_1) + 20280 + 4680 -
24960> = a_1> > And so we also have> > r2*s2 = a_2> r3*s3 =
a_3.> > 5. Since r1, s1, q1 are algebraic integers, r1*q1 = 5
and r1*s1 = a_1,> it follows that r1 is a common algebraic
integer factor of a_1 and> 5.> > 6. Since r2, s2, q2 are
algebraic integers, r2*q2 = 5 and r2*s2 = a_2,> it follows
that r2 is a common algebraic integer factor of a_2 and> 5. 7. Since r3, s2, q3 are algebraic integers, r3*q3 = 5 and
r3*q3 = a_3,> it follows that r3 is a common algebraic
integer factor of a_3 and> 5.> > 8. We claim that r1, r2, and
r3 are roots of the polynomial:> > f(x) = x^3 - 969 x^2 + 315
x + 5.> > To verify this, plug in the value of r1, and use
the following> identities:> > > (a_1)^3 = 12(a_1)^2 - 65.>
(a_1)^4 = 144(a_1)^2 - 65(a_1) - 780.> > (a_1)^5 =
(a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2> = 12(a_1)^4 -
65(a_1)^2> = 12(144(a_1)^2 - 65(a_1)-780) - 65(a_1)^2> =
1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2> = 1663(a_1)^2 -
780(a_1) - 9360.> > (a_1)^6 = (a_1)^4 (a_1)^2> = (144(a_1)^2
- 65(a_1) - 780) (a_1)^2> = 144(a_1)^4 - 65(a_1)^3 -
780(a_1)^2> = 144 (144(a_1)^2 - 65(a_1) - 780) > -
65(12(a_1)^2 - 65)> - 780(a_1)^2> = 20736(a_1)^2 - 9360(a_1)
- 112320> -780(a_1)^2 + 4225> -780(a_1)^2> = 19176(a_1)^2 -
9360(a_1) - 108095.> > Same for r2 and r3, replacing a_1 for
a_2 and a_3, respectively> (omitted for space).> > 9. f(x) is
monic, primitive, and irreducible over Q. For the latter,> the
polynomial is reducible over Q if and only if it has a root>
over Q, since it is degree 3. The only possible rational
roots, by> the p/q test, are 1, -1, 5, and -5, and> f(1) =
-648> f(-1)=-1280> f(5) = -22520> f(-5)=-25920.> > > 10. r1
is an algebraic integer unit if and only if 1/r1 (its>
multiplicative inverse) is also an algebraic integer. But
1/r1 is> a root of the polynomial we obtain from > f(x) = x^3
- 969 x^2 + 315 x + 5. > by plugging in 1/x, setting equal to
0, and solving, that is, 1/r1> is a root of> > g(x) = 5x^3 +
315x^2 - 969x + 1> > which is a primitive, non-monic,
irreducible polynomial over> Q. Therefore, none of its roots
are algebraic integers. So 1/r1, a> root, is not an algebraic
integer. So r1 is not an algebraic> integer unit.Which is the
break in the logical chain you anti-mathematician.What is
wrong with you Magidin?Now why donOt you 'll in 
the gap?My
point is that an algebraic integer can be coprime to ALL
primeinteger and yet its multiplicative inverse can NOT be an
algebraicinteger.IOve proven my point repeatedly, and given
the *math* but you refuseto accept it, so how can you be a
mathematician?Consider
http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759where
*basic* algebra refutes you. ItOs *algebra* that refutes
youMagidin, but you refuse to accept the math.You have
hounded me for years, convinced people of falsehoods, andnow,
even when repudiated by Professor McKenzie, whom you know,
youSTILL attack the math.WHAT IS WRONG WITH YOU??!!!James
Harris
=Uh oh...Your worst fears con'rmed... they really
are lying to you.Of course this brings up a sticky issue: who
to believe? Thise that agreewith you, or those that donOt.As
to thebottom of this affair. At the very least, theyOll take
away his slide ruleand perhaps his library card. Maybe you
could write to the NYT, I mean whoknows what other things
heOs lied about. Is he really Madigan (maybe 
heOsSuddam
trying to undermine America)? Does he _really_ come from
Berkley? Andof course the list of questions could go on and
on and on and on....Thanb you for alerting us to this grave
danger!
 the NYT, I mean who knows what other things heOs
lied about. Is he> really Madigan (maybe heOs Suddam trying 
to
undermine America)? Does> he _really_ come from Berkley? And
of course the list of questionsYou mispelt Amercia.--
=
[.snip.]>> And that conclusion is false. WeOve gone over it,
in detail. But here>> wrong. (Original calculations done by
Dale Hall):>> >> 1. Let >> >> q1 = 8 (a_1)^2 - 76 (a_1) -
185>> r1 = 8 (a_1)^2 - 4 (a_1) - 45>> s1 = 4 (a_1)^2 - 37
(a_1) - 104>> >> Since a_1 is an algebraic integer, each of
q1, r1, s1 are algebraic>> integers.>> >> 2. Likewise, let>> q2 = 8 (a_2)^2 - 76 (a_2) - 185>> r2 = 8 (a_2)^2 - 4 (a_2)
- 45>> s2 = 4 (a_2)^2 - 37 (a_2) - 104>> >> and>> >> q3 = 8
(a_3)^2 - 76 (a_3) - 185>> r3 = 8 (a_3)^2 - 4 (a_3) - 45>> s3
= 4 (a_3)^2 - 37 (a_3) - 104>> >> Each of q2, r2, s2, q3, r3,
s3 are algebraic integers.>> >> 3. We have that>> >> q1*r1 =
[8(a_1)^2 - 76(a_1) - 185][8(a_1)^2 - 4(a_1)-45]>> =
64(a_1)^4 - 32(a_1)^3 - 360(a_1)^2>> -608(a_1)^3 + 304(a_1)^2
+ 3420(a_1)>> -1480(a_1)^2 + 740(a_1) + 8325>> >> = 64(a_1)^4
- 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> >> >> Since
(a_1)^3 - 12(a_1)^2 + 65 = 0, we have that>> >> (a_1)^3 =
12(a_1)^2 - 65>> >> (a_1)^4 = 12(a_1)^3 - 65(a_1)>> = 12(
12(a_1)^2 - 65) - 65(a_1)>> = 144(a_1)^2 - 780 - 65(a_1)>> =
144(a_1)^2 - 65(a_1) - 780,>> >> so>> >> q1*r1 = 64(a_1)^4 -
640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> = 64
[144(a_1)^2 - 65(a_1) - 780] >> - 640 [12(a_1)^2 - 65]>> -
1536(a_1)^2 + 4160(a_1) + 8325>> = 9216(a_1)^2 - 4160(a_1) -
49920 - 7680(a_1)^2 + 41600>> -1536(a_1)^2 + 4160(a_1) +
8325>> = 41600+8325-49920>> = 5.>> >> Since (a_2)^3 -
12(a_2)^2 + 65 = 0 and (a_3)^3 - 12(a_3)^2 + 65 = 0,>> we
also have>> >> q2*r2 = 5.>> q3*r3 = 5.>> >> >> >> 4. Using
the same de'nitions, we have:>> >> r1*s1 = ( 8(a_1)^2 -
4(a_1) - 45) * (4(a_1)^2 - 37(a_1) - 104)>> = 32(a_1)^4 -
296(a_1)^3 - 832(a_1)^2>> - 16(a_1)^3 + 148(a_1)^2 +
416(a_1)>> - 180(a_1)^2 +1665(a_1) + 4680>> >> = 32(a_1)^4 -
312(a_1)^3 - 864(a_1)^2 + 2081(a_1) + 4680>> = 32( 144(a_1)^2
- 65(a_1) - 780) - 312( 12(a_1)^2 - 65)>> - 864(a_1)^2 +
2081(a_1) + 4680>> >> = 4608(a_1)^2 - 2080(a_1) - 24960 -
3744(a_1)^2 + 20280>> - 864(a_1)^2 + 2081(a_1) + 4680>> =
(a_1) + 20280 + 4680 - 24960>> = a_1>> >> And so we also
have>> >> r2*s2 = a_2>> r3*s3 = a_3.>> >> 5. Since r1, s1, q1
are algebraic integers, r1*q1 = 5 and r1*s1 = a_1,>> it
follows that r1 is a common algebraic integer factor of a_1
and>> 5.>> >> 6. Since r2, s2, q2 are algebraic integers,
r2*q2 = 5 and r2*s2 = a_2,>> it follows that r2 is a common
algebraic integer factor of a_2 and>> 5.>> >> 7. Since r3,
s2, q3 are algebraic integers, r3*q3 = 5 and r3*q3 = a_3,>>
it follows that r3 is a common algebraic integer factor of
a_3 and>> 5.>> >> 8. We claim that r1, r2, and r3 are roots
of the polynomial:>> >> f(x) = x^3 - 969 x^2 + 315 x + 5.>> To verify this, plug in the value of r1, and use the
following>> identities:>> >> >> (a_1)^3 = 12(a_1)^2 - 65.>>
(a_1)^4 = 144(a_1)^2 - 65(a_1) - 780.>> >> (a_1)^5 =
(a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2>> = 12(a_1)^4 -
65(a_1)^2>> = 12(144(a_1)^2 - 65(a_1)-780) - 65(a_1)^2>> =
1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2>> = 1663(a_1)^2 -
780(a_1) - 9360.>> >> (a_1)^6 = (a_1)^4 (a_1)^2>> =
(144(a_1)^2 - 65(a_1) - 780) (a_1)^2>> = 144(a_1)^4 -
65(a_1)^3 - 780(a_1)^2>> = 144 (144(a_1)^2 - 65(a_1) - 780)
>> - 65(12(a_1)^2 - 65)>> - 780(a_1)^2>> = 20736(a_1)^2 -
9360(a_1) - 112320>> -780(a_1)^2 + 4225>> -780(a_1)^2>> =
19176(a_1)^2 - 9360(a_1) - 108095.>> >> Same for r2 and r3,
replacing a_1 for a_2 and a_3, respectively>> (omitted for
space).>> >> 9. f(x) is monic, primitive, and irreducible
over Q. For the latter,>> the polynomial is reducible over Q
if and only if it has a root>> over Q, since it is degree 3.
The only possible rational roots, by>> the p/q test, are 1,
-1, 5, and -5, and>> f(1) = -648>> f(-1)=-1280>> f(5) =
-22520>> f(-5)=-25920.>> >> >> 10. r1 is an algebraic integer
unit if and only if 1/r1 (its>> multiplicative inverse) is
also an algebraic integer. But 1/r1 is>> a root of the
polynomial we obtain from >> f(x) = x^3 - 969 x^2 + 315 x +
5. >> by plugging in 1/x, setting equal to 0, and solving,
that is, 1/r1>> is a root of>> >> g(x) = 5x^3 + 315x^2 - 969x
+ 1>> >> which is a primitive, non-monic, irreducible
polynomial over>> Q. Therefore, none of its roots are
algebraic integers. So 1/r1, a>> root, is not an algebraic
integer. So r1 is not an algebraic>> integer unit.>>Which is
the break in the logical chain you anti-mathematician.There
is no break. r1 is an algebraic integer unit if and only if
(a)it is an algebraic integer; and (b) its multiplicative
inverse is alsoan algebraic integer. You tried using this
result with your failedpolynomial x^3 -3x + 2, remember?We
know r1 is an algebraic integer. And weOve seen that
itsmultiplicative inverse is not an algebraic integer.
Therefore, r1 isnot an algebraic integer unit.>What is wrong
with you Magidin?>>Now why donOt you 'll in the 
gap?BecausetI
can do nothing about the empty space between your ears.>My
point is that an algebraic integer can be coprime to ALL
prime>integer and yet its multiplicative inverse can NOT be
an algebraic>integer.Your point is wrong, and is proven wrong
above. IOve producedEXPLICITLY a non-unit common factor of 
a1
and 5. Your claim is
wrong.
Why do you take so much trouble to expose such
a reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A manOs capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
'gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde'nite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
Arturo Magidinmagidin@math.berkeley.edu
[.snip.]>> Apparently, he dismissed what you think or what
you told him was my>> objection. Your track record is clear:
you have great dif'culty in>> correctly paraphrasing other
peopleOs objections.>>Bull. He trashed your objection.
ThatOs it.So you claim. Looks, from your report, that he
trashed what youreported to be my objection. Let me guess,
you said What he isbasically saying is... and then you
simpli'ed the situation so hecould see the gist of what you
said my claim was? [.snip.]>> So, go ahead. Provide the
complete statement, weOll see how accurate>> you were in
reporting it.>>Come on Magidin, thatOs only necessary for 
all
those losers who donOt>really know mathematics.Let me quote
you on this.>> >Facts not in evidence.>> >For those who 
donOt
know, and in case I messed up with my usage, that>> >means
that the poster has said something that requires outside>veri'cation. That is, to 'nd out if 
heOs telling the truth,
you>> >have to go do research.>ItOs a nasty little trick
because who is going to go do all that>research to
8)Is it a nasty little trick when you make claims not in
evidence? Minewere available to check by anyone who wanted to
do so. Your claims
areuncheckable.
= [Gabriele Rossetti] has left a vast
body of writings... in which he has attempted to prove the
truth of his unorthodox interpre- tation of medieval
literature. They present a formidable record of unsystematic
research in which we see an enthusiast plunging farther and
farther and farther from the logic of facts and good sense
until truth is lost in the dreadful nightmare of an idee 'xe.
There is no real evolution of the Theory although it grows and
expands until it embraces ever wider horizons. The numerous
inaccuracies of deduction, mis-statements of historical fact,
and self-contradictions...have caused critics to turn away
from them in disgust... [...] It is impossible to read far...
without realizing that we have to deal with a work of faith
and imagination rather than of reasoning. There is an
appearance of reason, for the author is set on proving by
logic the truth of what he already believes by intuition. The
truth is plain to him and he cannot comprehend why others do
not immediately accept it, but as they desire demonstration
he has multiplied his proofs. It is the redundancy and
confusion of a prophet expounding by a familiar method the
truth revealed to his own simple soul in a §ash of
inspiration... In such work as this... it is idle to look for
the calm reasoning of a scholar; we do not 'nd it, and there
is little or no advantage in attacking the obvious
inconsistencies and absurdities that abound. -- E.R. Vincent,
_Gabriele Rossetti in England_, quoted in _The Shakespearan
Ciphers Examined_, by William F. Friedman and Elizebeth S.
Friedman
==
==Arturo
Magidinmagidin@math.berkeley.edu
> >> [...]>>But then
IOve spent so much time wasted with a liar like you
Magidin.>>Maybe mathematicians in general arenOt so bad, but
you are.>And I think back now on what I gave him.>[...]>>The professor was quite gracious in many ways, and
IOve been rather>obnoxious.>>I hate you Magidin. 
YouOve
twisted things, lied, and challenged me at>every point even
when IOve been right.>>You jealous bastard. So what if 
IOm a
better mathematician. Is that>any reason to trash everything
you were taught?> >> [...]>>IOve talked to you too much
Magidin. YouOve convinced me that other>mathematicians are
corrupt and Professor McKenzie at my own alma mater>was nice
to me and I responded in anger when he didnOt give me
exactly>what I wanted: vindication.> [...]>>I explained the
argument point-by-point you sick asshole.>>Professor McKenzie
was skeptical but I answered his objections.>>You sick
twisted bastard, IOve argued with you for years, and
you>still refuse to accept the math.>[...]>>You have
hounded me for years, convinced people of falsehoods,
and>now, even when repudiated by Professor McKenzie, whom you
know, you>STILL attack the math.>>WHAT IS WRONG WITH
YOU??!!!>James HarrisDavid C.
Ullrich**************************As far as IOm concerend
youOre trying to wait until I die, so I 
'guremaybe you should
die instead. How about that, eh? WouldnOt that be abetter
twist?You refuse to follow the math, so the great Powers that
controlreality and *speak* in mathematics decide to kill you
instead of me.So what do you think about that, eh? Oh, canOt
hear Them talking?Well, I guess thatOs because you 
donOt
really understand Mathematics,the true language, which is THE
language.TheyOre talking about you now, and They agree with 
my
assessment, andwill not penalize me as They allowed the others
like Galois and Abelto be penalized.They will kill you
instead.James Harris speaking on Weird factorization,
genius
 >> ... stuff deleted ...> >The conclusion of
the paper Advanced Polynomial Factorization is that>for the
factorization>> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x +
1)(a_3 x + 1)>>where the aOs are all algebraic integers,
one of them is coprime to 5.>>And that conclusion is false.
WeOve gone over it, in detail. But here>>wrong. (Original
calculations done by Dale Hall):>>1. Let >> q1 = 8
(a_1)^2 - 76 (a_1) - 185>> r1 = 8 (a_1)^2 - 4 (a_1) - 45>> s1
= 4 (a_1)^2 - 37 (a_1) - 104>> Since a_1 is an algebraic
integer, each of q1, r1, s1 are algebraic>> integers.>>2.
Likewise, let>> q2 = 8 (a_2)^2 - 76 (a_2) - 185>> r2 = 8
(a_2)^2 - 4 (a_2) - 45>> s2 = 4 (a_2)^2 - 37 (a_2) -
104>>and>> q3 = 8 (a_3)^2 - 76 (a_3) - 185>> r3 = 8
(a_3)^2 - 4 (a_3) - 45>> s3 = 4 (a_3)^2 - 37 (a_3) - 104>>
Each of q2, r2, s2, q3, r3, s3 are algebraic integers.>>3.
We have that>> q1*r1 = [8(a_1)^2 - 76(a_1) - 185][8(a_1)^2
- 4(a_1)-45]>> = 64(a_1)^4 - 32(a_1)^3 - 360(a_1)^2>>
-608(a_1)^3 + 304(a_1)^2 + 3420(a_1)>> -1480(a_1)^2 +
740(a_1) + 8325>> = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 +
4160(a_1) + 8325>> >>Since (a_1)^3 - 12(a_1)^2 + 65 = 0, we
have that>> (a_1)^3 = 12(a_1)^2 - 65>> (a_1)^4 = 12(a_1)^3
- 65(a_1)>> = 12( 12(a_1)^2 - 65) - 65(a_1)>> = 144(a_1)^2 -
780 - 65(a_1)>> = 144(a_1)^2 - 65(a_1) - 780,>>so>> q1*r1
= 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> =
64 [144(a_1)^2 - 65(a_1) - 780] >> - 640 [12(a_1)^2 - 65]>> -
1536(a_1)^2 + 4160(a_1) + 8325>> = 9216(a_1)^2 - 4160(a_1) -
49920 - 7680(a_1)^2 + 41600>> -1536(a_1)^2 + 4160(a_1) +
8325>> = 41600+8325-49920>> = 5.>>Since (a_2)^3 - 12(a_2)^2
+ 65 = 0 and (a_3)^3 - 12(a_3)^2 + 65 = 0,>>we also
have>>q2*r2 = 5.>>q3*r3 = 5.>>4. Using the same
de'nitions, we have:>> r1*s1 = ( 8(a_1)^2 - 4(a_1) - 45) *
(4(a_1)^2 - 37(a_1) - 104)>> = 32(a_1)^4 - 296(a_1)^3 -
832(a_1)^2>> - 16(a_1)^3 + 148(a_1)^2 + 416(a_1)>> -
180(a_1)^2 +1665(a_1) + 4680>> = 32(a_1)^4 - 312(a_1)^3 -
864(a_1)^2 + 2081(a_1) + 4680>> = 32( 144(a_1)^2 - 65(a_1) -
780) - 312( 12(a_1)^2 - 65)>> - 864(a_1)^2 + 2081(a_1) +
4680>> >> = 4608(a_1)^2 - 2080(a_1) - 24960 - 3744(a_1)^2 +
20280>> - 864(a_1)^2 + 2081(a_1) + 4680>> = (a_1) + 20280 +
4680 - 24960>> = a_1>> And so we also have>> r2*s2 =
a_2>> r3*s3 = a_3.>>5. Since r1, s1, q1 are algebraic
integers, r1*q1 = 5 and r1*s1 = a_1,>> it follows that r1 is
a common algebraic integer factor of a_1 and>> 5.>>6. Since
r2, s2, q2 are algebraic integers, r2*q2 = 5 and r2*s2 =
a_2,>> it follows that r2 is a common algebraic integer
factor of a_2 and>> 5.>>7. Since r3, s2, q3 are algebraic
integers, r3*q3 = 5 and r3*q3 = a_3,>> it follows that r3 is
a common algebraic integer factor of a_3 and>> 5.>>8. We
claim that r1, r2, and r3 are roots of the polynomial:>> >>
f(x) = x^3 - 969 x^2 + 315 x + 5.>> To verify this, plug in
the value of r1, and use the following>> identities:>>
(a_1)^3 = 12(a_1)^2 - 65.>> (a_1)^4 = 144(a_1)^2 - 65(a_1) -
780.>> (a_1)^5 = (a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2>> =
12(a_1)^4 - 65(a_1)^2>> = 12(144(a_1)^2 - 65(a_1)-780) -
65(a_1)^2>> = 1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2>> =
1663(a_1)^2 - 780(a_1) - 9360.>> (a_1)^6 = (a_1)^4
(a_1)^2>> = (144(a_1)^2 - 65(a_1) - 780) (a_1)^2>> =
144(a_1)^4 - 65(a_1)^3 - 780(a_1)^2>> = 144 (144(a_1)^2 -
65(a_1) - 780) >> - 65(12(a_1)^2 - 65)>> - 780(a_1)^2>> =
20736(a_1)^2 - 9360(a_1) - 112320>> -780(a_1)^2 + 4225>>
-780(a_1)^2>> = 19176(a_1)^2 - 9360(a_1) - 108095.>> Same
for r2 and r3, replacing a_1 for a_2 and a_3, respectively>>
(omitted for space).>>9. f(x) is monic, primitive, and
irreducible over Q. For the latter,>> the polynomial is
reducible over Q if and only if it has a root>> over Q, since
it is degree 3. The only possible rational roots, by>> the p/q
test, are 1, -1, 5, and -5, and>> f(1) = -648>> f(-1)=-1280>>
f(5) = -22520>> f(-5)=-25920.>> >>10. r1 is an algebraic
integer unit if and only if 1/r1 (its>> multiplicative
inverse) is also an algebraic integer. But 1/r1 is>> a root
of the polynomial we obtain from >> f(x) = x^3 - 969 x^2 +
315 x + 5. >> by plugging in 1/x, setting equal to 0, and
solving, that is, 1/r1>> is a root of>> g(x) = 5x^3 +
315x^2 - 969x + 1>> which is a primitive, non-monic,
irreducible polynomial over>> Q. Therefore, none of its roots
are algebraic integers. So 1/r1, a>> root, is not an algebraic
integer. So r1 is not an algebraic>> integer unit.> > > Which
is the break in the logical chain you anti-mathematician.> Do
you claim that this step is incorrect?Why?I suggest, since you
apparently have a friendly, trustingrelationship with
Professor McKenzie, that you cut and pastethe text (steps 1 -
17, together with your preceding paragraphof the trouble
youOre having with some of the readers here, andask him for 
a
judgment on the correctness (or error, since youbelieve this
argument to be §awed) of the argument. I maintainthat the
argument is correct, as has every other reader who
hasexamined the steps.This is a perfectly simple task, and if
youOre correct, wouldbuy you a modicum of credibility. ...
stuff deleted ...> Now why donOt you 'll in the 
gap?> > My
point is that an algebraic integer can be coprime to ALL
prime> integer and yet its multiplicative inverse can NOT be
an algebraic> integer.> Either this statement is demonstrably
wrong, or I fail to understandit. Taken literally, youOre
suggesting that there is an algebraicinteger r, such that r
is coprime to r^2. You must realize that thiscannot be true:
Let us suppose that r is coprime to r^2 in the ring of
algebraic integers. Let m,n be algebraic integers, for which
mr + nr^2 = 1. Factoring, one has r(m + nr) = 1, giving (m +
nr), the multiplicative inverse of r, as an algebraic
integer. Thus, r must be a unit, for its multiplicative
inverse is an algebraic integer.> IOve proven my point
repeatedly, and given the *math* but you refuse> to accept
it, so how can you be a mathematician?> You will no doubt
claim that the above argument is §awed, as well.What is the
error in that argument? ... stuff deleted ...> You have
hounded me for years, convinced people of falsehoods, and>
now, even when repudiated by Professor McKenzie, whom you
know, you> STILL attack the math.> If you wish to count
Professor McKenzie as a proponent of the claimthat the above
argument, as well as MagidinOs re-phrasing of my
earlierargument, you are doing him a disservice by not
allowing him to expresshis independent opinion (which, by
your accounting, will surely supportyour position).It canOt
be made any simpler. Let Professor McKenzie see the
arguments.Let him speak for himself. I have been tempted to
address him myself,but have chosen not to do so out of
respect for your relationship withhim. If you also have
respect for that relationship, you will not asserthis
position, but will invite him to weigh in on it
himself.invite him to respond to you directly. Be sure to
obtain his permissionto quote his response to those
arguments, and report back to sci.math.> > James
HarrisDale.
=Just 'xing a poorly-formed sentence. My fault
for posting too lateat night, no doubt.> ... mumble ...> If
you wish to count Professor McKenzie as a proponent of the
claim> that the above argument, as well as MagidinOs
re-phrasing of my earlier> argument, ... (insert) ... are
incorrect ... > ... , you are doing him a disservice by not
allowing him to express> his independent opinion (which, by
your accounting, will surely support> your position).> ...
and so forth ...> > Dale.> Sorry for the sloppy
wording.Sorrier to have failed to follow it myself.Dale.
But then IOve spent so much time wasted with a liar like you
Magidin.> Prof. Magadin has spent countless hours of his time
carefully and meticulously understanding, explaining, and
'nding the errors in your arguments. He is unfailingly polite
and always sticks to the math. I donOt understand your anger
towards him. You should thank him.
> Tee-hee.>Nothing
makes a pig happier than a roll in the wallow, as this
tittering porker knows!
http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&
width=314&height=400#
 I thought I was set. With a high
I.Q. group set to publish my paper> on factoring polynomials
into non-polynomial factors I 'gured that> now certainly I
could push my agenda to fame and fortune. I was> wrong.If a
society is founded on the idea of not being accountable to
others because of superior intelligence, it is not likely to
be a source of respected papers.> So here I am back again,
humbled yet again, as thereOs no escaping it,> there is no
way IOll get anywhere pissing off mathematicians.You need to
remember this throughout the week, at least, before weOll
believe that *you* believe that.> So here are some
concessions.> [concessions deleted]> > So quit the lying you
dark evil people as IOm not out here claiming to> have a
proof of FermatOs Last Theorem and IOm not out 
here claiming
to> have found THE prime counting function.Umm, how are
insults supposed to constitute not pissing off
mathematicians?> > Take down all those webpages attacking me,
and quit with the posts> calling me a crank. IOm 
'nally tired
of being called a crank.IOm not one of them, but I suspect
theyOll be taken down when you *prove* that you are willing
to listen to people who are trying to explain the math to
you.> > I want to go legit.> > Um, there is that little
problem with algebraic integers to discuss> though; however,
IOm open-minded and willing to consider *proof* that> 
IOm
there.> > LetOs get back to it folks. No FLT. NO ING
PRIME COUNTING!!!I got that the 'rst time. No need to shout. But 'nally I want some straight answers on the ring of
algebraic> integers.Ok, how about starting with what you
understand the de'nitions of ring and algebraic integers to
be.> And thereOs no website of mine, so no way to claim that
IOm NOT> dropping FLT and THE prime counting you evil
bastards.Again, I thought you didnOt want to piss of
mathematicians. What part of this is not intended to be at
least irritating?-- Will Twentyman
=well, what has the IBF
... I mean,the Homeland Security out't found, thus far? your
terminology of correct math is unde'ned, althoughmany of us
would say that Liebniz criteriumof iff is what you really
mean; is it?... some folks say,bijective binary operation, or
some thing, an implicationthat can be read either way,
dependingupon how one uses the wording for necessity &
suf'ciency. the same lack of de'nition applies, 
it seems,to
your veri'cation or proof of the lying liarsO 
lies,who are so
tedious in their belittling of your efforts,even ham-handed
... but that doesnOt apply to Magidin,nor to Nora Baboobda
Baron. I mean,they seem fairly honest (and nice, butthatOs
beside the point, with an ingracious character,like Yusef) to
me. what have you proven about them,again?is prof. Mackenzie
going to deliver his judgement,soon?well, IOm not coming 
back
to this til *at least* after Oct.7;after all, itOs polls 
from
AOL-TimeWarner-HBO-CNN that I haveto deal with, plus Warren
BuffetOs hulking puppet,while we account the actual
polling-place ordeal. at least,heOs not taking steroids any
more!... see my sig. (I mean,youOve never replied to me,
before, so,thereOs not really any reason ... 
but,IOm hoping
to see some action from those high school wonders .-) > I
could meet each and every such challenge and show how each
step> followed logically, and therefore correctly. > ItOs 
not
about believing any of you but about your willingness or>
unwillingness to accept correct mathematics. What I veri'ed
is my > What my meeting with Professor McKenzie also con'rmed
for me is that> a professional mathematician wouldnOt accept
their primary objection,> which con'rms to me that they were
deliberately lying, as itOs> nonsensical to believe that
factors of f vary as functions or as> dependents on m, which
Professor McKenzie didnOt even seriously> consider as a
possibility. > What I have clearly veri'ed is that certain
posters who *are*> mathematicians, like Magidin, have been
lying about the mathematics,> and lying repeatedly with
objections that a professional mathematician> quickly
rejected.--les ducs de
Buffet!http://larouchepub.com
=Hey,Could I ask for some
homework help? Apologies beforehand if this isthe wrong
group; I saw alt.algebra.help but this isnOt algebra...d =
deltaSo, I want to prove lim (x approaches 3) of f(x) =
1/((x-3)^2) = +inf.Where the de'nition of limit is that, if
lim (x->a) f(x) = +inf, thenfor all N, there is a d > 0 s.t.,
for all x, if 0 < |x-a| < d, then wehave f(x) > N.Obviously,
the problem revolves around choosing appropriate N.
Anysuggestions would be welcomed.Also, does anyone have a
book recommendation for a 4th semestercalculus class? We did
the basic 3 semesters of calculus (plug & chugw/ Thomas), and
now are using Calculus by Spivak as a 1 semesterreview of
7ï
WEcf
!.8574èÿÿ
Theorem.If you
believe people arguing with me thereÍs only ONE small
thingthat they fight.Think about it. Basic algebra proves my
point. The argument isstraightforward and simple, but the
consequences are huge.ItÍs revolutionary for *social*
reasons. And you can see society,your society, fighting as it
has before. Learn the lessons from thebattles over irrational
and imaginary, and try to let the math beyour guide--not your
comfort zone.The work itself is conveniently available to you
24 hours a day, sevendays a week at
http://groups.msn.com/AmateurMathand hey, you can even play
with an applet (if you join the group).James
Harris
=documentstyle{report}begin{document}pagestyle{plain
}raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the following integral:$$ int_{-half a}^{+half
a} x^2 , dx / a = left[ frac{1}{3} x^3 right]_{-half
a}^{+half a} / a = frac{2.a^3}{3.8.a} = frac{a^2}{12}$$Now
consider:$$ lim_{sigma rightarrow infty} frac{ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }$$We
would expect that this more difficult integral nevertheless
convergesto the same result as before: $a^2 / 12$ ,
because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividing the integralin the denominator
by $a$ should be equivalent to differentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
finding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ... ;$ in
theirdenominators, so they should converge to zero. Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they do not match. WhatÍs going wrong ? 
Please
help !%end{document}-- * Han de Bruijn; Systems for Research A
little bit of Physics would be (
)* and Education (DTO/SOO),
Mekelweg 6 NO Idleness in Mathematics(HdB) @-O^O-@* 2628 CD
Delft, The Netherlands http://huizen.dto.tudelft.nl/deBruijn
#/_#
=>....................> So the two outcomes have the
same order of magnitude, but they do not match. > WhatÍs
going wrong ? Please help !> %following .tex
file,
documentstyle[leqno]{report}newcommand{hsp}{hspace*{0.5cm}}
newcommand{Erf}{{rm Erf}}begin{document}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the problem to
determinebegin{equation}label{eu11} fbox{$displaystyle
L(a):=lim_{sigma rightarrowinfty}I(a,sigma)hsp mbox{rm with}
hsp I(a,sigma)= frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } $}end{equation}However:$$ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2
int_{-half a}^{+half a} x , e^{-half x^2 /sigma^2} ,
dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half
a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = -
sigma^2 left[ x.e^{-half x^2 / sigma^2}
right]_{-halfa}^{+half a} + sigma^2 int_{-half a}^{+half a}
e^{-half x^2 / sigma^2} ,dx$$ $$ hieruit hsp I(a,sigma)=frac{
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } =
sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}right)$$Let us define $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,;
h to0; .; $ Then problem (ref{eu11}) is equivalent
withbegin{equation}label{eu12}fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto0}frac{1}{h^2}left(1-frac{he^{
-h^2}}{Erf(h)}right) hspmbox{rm where}hsp
Erf(h):=intlimits_{0}^h e^{-t^2}; dt$}hsp .end{equation}But
$displaystyle
Erf(h)=he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $
where[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k
}cdotfrac{z^k}{k!}hsp,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]Therefore
[L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}{h^
2cdotsumlimits_{k=0}^infty
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=0}^inftyfrac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k
}}{cdotsumlimits_{k=0}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}=frac{
a^2}{8}cdotfrac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12};
.]%end{document}
=> >....................> So the two
outcomes have the same order of magnitude, but they do not
match.> WhatÍs going wrong ? Please help !> %> following 
.tex
file,> 
documentstyle[leqno]{report}>
newcommand{hsp}{hspace*{0.5cm}}> newcommand{Erf}{{rm Erf}}>
begin{document}> pagestyle{plain} raggedbottom> def half
{frac{1}{2}}> def hieruit {quad Longrightarrow quad}> %>
subsection*{Should be the same but alas ...}> %> Consider the
problem to determine> begin{equation}> label{eu11}
fbox{$displaystyle L(a):=lim_{sigma rightarrow>
infty}I(a,sigma)hsp mbox{rm with} hsp I(a,sigma)= frac{>
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx }>
{ int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } $}>
end{equation}> However:> $$> int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx => - sigma^2 int_{-half
a}^{+half a} x , e^{-half x^2 /> sigma^2} ,> dleft(-half x^2
/ sigma^2right) => $$ $$> - sigma^2 int_{-half a}^{+half a} x
,> dleft(e^{-half x^2 / sigma^2}right) => - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half> a}^{+half a}> +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,>
dx> $$ $$> hieruit hsp I(a,sigma)=frac{> int_{-half a}^{+half
a} x^2 , e^{-half x^2 / sigma^2} , dx }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx } => sigma^2 left(
1 - frac{ a.e^{-half (a/2)^2 / sigma^2} }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx }> right)> $$> Let
us define $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,; h to> 0; .; $
Then problem (ref{eu11}) is equivalent with>
begin{equation}label{eu12}> fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto>
0}frac{1}{h^2}left(1-frac{he^{-h^2}}{Erf(h)}right) hsp>
mbox{rm where}hsp Erf(h):=intlimits_{0}^h e^{-t^2}; dt> $}hsp
.> end{equation}> But $displaystyle Erf(h)=>
he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $ where>
[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k}
cdotfrac{z^k}{k!}hsp> ,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]>
Therefore> [L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^infty> frac{(1)_k}{(3/2)_k}h^{2k}}>
{h^2cdotsumlimits_{k=0}^infty frac{(1)_k}{(3/2)_k}h^{2k}}=>
frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=0}^infty>
frac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k}}>
{cdotsumlimits_{k=0}^infty>
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}cdot>
frac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12}; .> ]> %>
end{document}I would like to read LaTex files. Is there some
Windows software out therethat works immediately without
having to run numerous fonts setup programscomputer.
= [
deleted ]Here is your answer, as simple as I can make
it:setlength{mathindent}{1.0cm}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}begin{document}%subsection*{Should be the
same but alas ...}%Consider the following integral:$$
int_{-half a}^{+half a} x^2 , dx / a = left[ frac{1}{3} x^3
right]_{-half a}^{+half a} / a = frac{2.a^3}{3.8.a} =
frac{a^2}{12}$$Now consider:$$ lim_{sigma rightarrow infty}
frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} ,
dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}$$We would expect that this more difficult integral
neverthelessconverges to the same result as before: $a^2 /
12$ , because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividingthe integral in the denominator
by $a$ should be equivalent todifferentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
finding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ...;$ in
their denominators, so they should converge to zero.Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they donot match. WhatÍs going wrong ? Please
help !%subsection*{The easy answer}Let[frac{a}{{2sigma }} =
lambda]and rewrite the desired integral ratio
asbegin{eqnarray*}frac{{int_{ - {a mathord{left/ {vphantom {a
2}} right. kern-nulldelimiterspace} 2}}^{ + {a mathord{left/
{vphantom {a 2}} right. kern-nulldelimiterspace} 2}} {dx,x^2
e^{ - {{x^2 } mathord{left/ {vphantom {{x^2 } {sigma ^2 }}}
right. kern-nulldelimiterspace} {sigma ^2 }}} } }}{{int_{ -
{a mathord{left/ {vphantom {a 2}} right.
kern-nulldelimiterspace} 2}}^{ + {a mathord{left/ {vphantom
{a 2}} right. kern-nulldelimiterspace} 2}} {dx,e^{ - {{x^2 }
mathord{left/ {vphantom {{x^2 } {sigma ^2 }}} right.
kern-nulldelimiterspace} {sigma ^2 }}} } }} &equiv& sigma ^2
frac{{int_0^lambda {du,u^2 e^{ - u^2 } } }} {{int_0^lambda
{du,e^{ - u^2 } } }} &=& sigma ^2 left( {frac{{frac{{lambda
^3 }}{{3 cdot 0!}} - frac{{lambda ^5 }}{{5 cdot 1!}} +
frac{{lambda ^7 }}{{7 cdot 2!}} - + ldots }}{{frac{{lambda ^1
}}{{1 cdot 0!}} - frac{{lambda ^3 }} {{3 cdot 1!}} +
frac{{lambda ^5 }}{{5 cdot 2!}} - + ldots }}} right) &=&
frac{{sigma ^2 lambda ^2 }}{3}left( {frac{{1 - frac{{3lambda
^2 }}{5} + - ldots }}{{1 - frac{{lambda ^2 }}{3} + - ldots
}}} right);mathop to limits_{sigma to infty }
;frac{{a^2}}{{12}}end{eqnarray*}end{document}-- Julian V.
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.
=> > I would like to read LaTex files. Is
there some Windows software out there> that works immediately
without having to run numerous fonts setup programs>
computer.Subscribe to the newsgroup comp.text.texand see what
people there are asking about and recommending.You will
probably want to install the MikTeX distribution, available
from http://ftp.uci.agh.edu.pl/pub/tex/systems/win32/miktex/I
strongly also recommend getting WinEdt, available from
http://www.winedt.com/It is shareware, but worth the money
($40, or so). It is a LaTeX andTeX -aware editor that makes
using these systems easy (I suggest LaTeXfor your own
work).The book A Guide to LaTeX by Kopka & Daly is extremely
helpful.Good luck.-- Julian V. NobleProfessor Emeritus of
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.

reals, no number has a next door neighbor on either side, so
what do you mean by neighbors of a real number? And what do
you mean by in'nitely far away on the sequence of neighbors?
No two real numbers can be in'nitely far away from each
other.Irrelevancies snipped.
 > > > > > Half of the
irrationals means that, for example, of the irrationals on> >
the interval [0,2] that half of them are in [0,1] and the
other half> > in [0,2], > > And here I thought that both
halves of [0,2] were in [0,2]. Just > shows how wrong I can
be, I guess!> > > > N+1 s the set of each element of N union
{N}, id est, N U {N}. > > If N+1 is N union {N}, then what
are N+2, and N+3, etc.?> > > P(N) is a superset of N U {N}. Wrong!> > 1, which is a member of N U {N}, is also a member
of many members of > P(N) but is not itself a member of P(N),
so P(N) is NOT a superset > of N U {N}!> > > About the
in'nite summation, in some respects the limit concept is> >
easily discarded, > > Not by those who know how to use it. It
may be elided, but not > discarded.> > > > WhatOs 
continuity?
I think itOs a construction on a point set> > topology.> > 
In
what context are you wanting to use continuity? The most
common > one is in the context of real functions, which do
not seem relevant > here. > > > > About the alternation of
rationals and irrationals in the reals, itOs> > something
along the lines of saying hereOs a rational, 
itOs neighbors are irrationals, their neighbors are rationals, for
in'nitely many> > neighbors in either direction the value is
inde'nite, and any other> > rational or irrational selected
(named except via inde'nite neighbor> > relation to the
de'nite point) is in'nitely far away on the 
sequence> > of
neighbors.> > Within the ordered 'eld of reals, no number has
a next door > neighbor on either side, so what do you mean by
neighbors of a > real number? > > And what do you mean by
in'nitely far away on the sequence of > neighbors? No two
real numbers can be in'nitely far away from each > other.>
Virgil, N U {N} is a subset of P(N). Represent each natural
as an ordinal. Then, for example, the subset {0} is an
element of P(N), {0} = 1. Theempty set, zero, is an element
of P(N), as is each singleton {n} for nin N, as is N. Thus
P(N) contains each element 0, 1, 2, ..., as wellas N. Thus N
U {N} is a subset of P(N).N+1 U {N+1} is N+2, etcetera, the
order type of N is N.The limit concept can be discarded when
its result is the same as i't were considered discarded.
Calculate the area of a square: itOse
7ï
WEcf
!.85;dèÿÿ
a sumof 16 terms !BUT if you compute the variance on a moving
windows, thereis a very fast algorithm, that you can 'gure
out
pretty easily.Your just have to perform :1. compute mean m and
replace your image I by M=(I-m)^22. compute moving average
along X direction. This can be doneby a recursive algorithm,
because if S(k) is the moving averageand F(k) is the initial
function (e.g. a row of M), you haveS(k) = S(k-1)+F(k)-F(k-9)
(here I assume window of size 8)3. compute moving average
along Y direction.All this stuff needs about O(n) operations
where n is the number of pixelsin the original image and
itOs
independant of the size of the window.Gabriel
You just
have 16 samples of a signal and you want the variance ?Sorry,
I meant 64 samples.> Obviously, there is no fast algorithm for
computing a sum> of 16 terms !Sorrrry once again 8x8=64 not 16
...G.
=my'le.> DON02.
Calc.Factors.FermatMers.Mersenne.CARMP163.SPMP163subject:
Mersenne numbers, Mp163 >
D.Calc.Factors.FermatMers.Mersenne.CARMP163.SPMP163We look at
the usual long multiplication of, for example, 123*48. 123 x
48 ------- 984 +492------- 5904. However, I wish to
concentrate just on the tens and units 'gures of the
resulting product. This is called Omultiplying modulo
100.O
That is, any whole number of 100s can be thrown away in the
process. Then, (100+23)*48 == 48*100+23*48 (is congruent
to)== (20+3)*(40+8) == 2*4*100+3*40+8*20+3*8 == 120+160+24 ==
104 == 04 modulo 100. In calculating the remainder of product
(integers w*z) after dividing by positive integer, t, it can
often be made easier by taking away convenient multiples of
the modulus, t, from w or z respectively before executing the
multiplication. Let w = at+b, say, and z = ct+d, say. Then, wz
= (at+b)(ct+d) = at(ct+d) +b(ct+d) = t(act +ad +bc) +bd == bd
modulo t. Bd should usually be less than w*z, but it may
require to be reduced further. I photographed car number
plate OMP163O at Hanson St, Newtown, Wellington
on today,
26-5-2002. This number plate could be shorthand for Mersenne
numbers, 2^prime-1. (Figure 2 raised to a prime index,
subtract 1.) The worldOs largest known prime number in 2001
often is2^13.4million-1. (Greater than 13 million 2s
(Dominion Post, NZ Herald, GIMPS greatinternet mersenne prime
search.)By the way, the 40th known Mersenne prime exponent isa
twin prime, 10x(23x63)^2 +/-1. And a previous recordholder was
virtually a palindrome, Table Mountain,
increasingdecreasing.Exponent (1*2*34*44432+1)= 3021377.
[sic.] _____ / A theorem of mathematics number theory says
if(a is a positive integer )..a^prime-1 has factor/s they
must be of the form 2*k*p+1. However, the Penguin Dictionary
of Curious and Interesting Numbers (1997), entry 28, table of
perfect numbers, does not give M_163 as a Mersenne prime.
Therefore, I have written myprogram Opowabc1O in
BBC
interpreted Basic64, which tests for just such factors.(On
Acorn A5000 computer, UK 1990, RISC OS-reduced instructionset
several programs found 150287 and 704161, etc. divide M_163.
Or 2^163 == 1 modulo 150287. Have I found a factor of M_163?
The following, I believe, shows that I have.Check. 2^163=
2^3*(2^20)^8 (is congruent to)==
8*1048576^82^20==1048576-6*150287 == 146854 mod 150287 ==
-3433. Squaring 2^40= 11785489== 63103 Squaring
2^80==3,981,988,609 ==134544== -15743 Squaring
2^160==247,842,049 ==18786 2^163==8*18786 ==150288==1 mod
150287. Q.e.d.By the way, my methods found (the least prime)
factor1580,187,223 of (10^9)^(10^9)+3. (sci.math sensation
2001.)This is, digit 1 followed by 9 billion zeroes.
Surpassedonly by GrahamOs number. I suspect gigaplex
=10^billion. (There is a typo error in Penguin Dictionaryof
Curious and Interesting Numbers, 1997.) yours sincerely, /
Donald S. McDonald (Wellington, New Zealand)
=There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
in La Jolla CA at UCSD during time Greg Benford describes
inTimescape living on Bonair Street Wind an Sea Beach near
UnicornTheater with Ken KeseyOs Merry Prankster Bus parked
nearby frequently.I was also teaching at San Diego State with
Fred Alan Wolf....PZ: I canOt imagine anything more
wrong-headed. And you say Rovelli is a bigshot?That is why I
say RovelliOs position is incoherent.JS: Is coherence in the
mind of the beholder?PZ: If you want to treat g_uv as a
physical 'eld, based on its dynamical character(i.e.
matter-dependence), then the natural thing to do is separate
the backgroundgeneralized Minkowski kinematical metric (which
is NOT matter dependent) aschronogeometric, and treat the
gravitational g_uv alone as physical -- byRovelliOs own
argument.JS: That will not work. It will not tip the light
cones. It will not give the correct bending of light.Also you
are too vague on what you mean by matter on the RHS
ofGuv(Einstein) = -(alpha)(alphaO)Tuv(Matter)alpha = e^2/hc
~
1/137alphaO = 8pi(WittenOs reciprocal string
tension)alphaO ~
(10^-32 cm)^2 in a common convention.I also include on
RHStuv(Vacuum) ~ [(alpha)(alphaO)]^-1/zpfguv/zpf =
Lp^-1[Lp^3|Vacuum Coherence|^2 - 1]Repulsive dark energy is
/zpf > 0Attractive dark matter is /zpf < 0FRW Omega(Dark
Energy + Dark Matter) ~ 0.96With Omega(Total) = 1 i.e. FLAT
SPACEPreferred foliation is where CMB is maximally isotropic
to ~ 10^-5.This allows accurate navigation with weightless
warp drive and traversable wormholes both supported by
con'gurations of dark energy and dark matter exotic w = -1
vacua.PZ: Yet he doesnOt even mention this.JS: There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
World...JS: Also at quantum level RovellimentionOs
DiracOs
insight that the Heisenberg Picture is better thanSchrodinger
Picture and that time as in the §ow of timeOs arrowis
not
dynamical time but is a statistical thermodynamic
construct.PZ: Sure, if you want to quantize everything and
abandon time.UnruhOs for example, and some of the
others.Unruh has some important ideas however.PZ: But then I
fail to see any material distinction between relabelingthe
bare unindividuatedspacetime points (passive diffeomorphism),
on the one hand, andshifting all physical'elds (including the
GR metric 'eld) with respect to such a rawmanifold
(activediffeomorphism), on the other. KretschmanOs point
appears to be fullyvalid in both cases:how can physics in
general -- any sensible physics -- possibly dependon a mere
re-labelingof raw unindividuated spacetime points; or, for
that matter, on acommon shift of all physicalsystems,
including the *physical* metric 'eld g_uv, with respect
tosuch a set of unindividuatedpoints?JS: Admittedly a sticky
wicket that I also need to understand moredeeply.PZ: ItOs an
arti'cial model that has nothing to do with physical
relativity IMO.JS: I found this remark by Goldstein
helpful:In the ADM formulation 4-diffeomorphism invariance
amounts to the requirement that one ends up with the same
space-time, up to coordinate transformations, regardless of
which path in multi-'ngered space-time is followed, i.e.
which lapse function N is uses. p.278OK so this idea goes
back to the archetypal notions of classical thermodynamics
with a state function, to holonomic integrability of equality
of mixed partial derivatives with not multiply-connected
manifolds, to a closed Cartan exterior differential form on a
cycle (no boundary), no topological defects and all that.The
passive coordinate transformations are like EM gauge
transformation (e/c)Au -> (e/c)Au + hChi,u in a 'xed gauge
constraint.PZ: Rotating a system in isotropic space is
connected with a true physical symmetryof the system
including the vacuum in which it is embedded. Yes, the
symmetryof the system Hamiltonian under such a transformation
is an active transformationalsymmetry that is characteristic
of the particular system -- unlike the invariance of
theproper formal expression of any physical law under a
*mere* coordinate transformation.JS: Note also 12.2.2 p.279
alluding to your digging up Kretchmann from Dr.
FrankensteinOs favorite graveyard. ;-)The fundamental
symmetry at the heart of general relativity is its invariance
under general coordinate transformations of spacetime. It is
important to stress that almost any theory can be formulated
in such a 4-diffeomorphism invariant manner by adding further
structure to the theory (e.g. a preferred foliation of
spacetime as a dynamical object). General relativity has what
is sometimes called serious diffeomorphism invariance, meaning
that it involves no spacetime structure beyond the 4-metric
and, in particular, singles out no special foliation of
spacetime.Goldstein and Teufel then knock standard QGrav
including even Ashtekar -> Loop Spin Foams that are perhaps
near to being falsi'ed by NASAOs EINSTEIN
inJack better check
this story out:Gary S. Bekkum...How to getlocalization in
space and the §ow of time as we experience it in ourimmediate
inner consciousnesshas nothing to do with the particular local
coordinate representationlike r and t in, for example,K =
e^2GM/c^2rdr/dt = c/Kfor null geodesicand in his Tables
generally in the context of potentially practicalmetric
engineering of the guv 'eld using the EM Au 'eld
in spite
ofthe enormous gravity string tension ~ c^4/G ~ 10^19 Gev per
10^-33 cm.PZ: You are blocking his actual de'nition of r.
Classic operationalism does notapply to a theory of this
type. That is a critical point. Miss that and of
coursenothing makes sense.JS: I am not ready to renounce PW
BridgmanOs Operationalism. Indeed, nothing Hal Puthoff says
about the foundations of his PV makes any sense to my mind.
If it ainOt broke, donOt 'x it.
Of course I am not a
doctrinaire positivist like Stephen Hawking proudly proclaims
he is....JS: Correct, with the proviso that matter includes
both real, i.e.on mass shell, sources as well as virtual,
i.e. off mass shellsources. The virtual sources divide into
two classes:I. Non-exotic near EM 'eld Fuv giant coherent
quantum states ofvirtual photons that contribute to
Omega(Matter) of the FRW metric andto Tuv in EinsteinOs
local
'eld equation.II. Exotic vacuum w = -1 zero point
stress-energy density local tensor~ (String Tension)/zpfguv
for both repulsive dark energy /zpf > 0 ofnegative pressure
and attractive dark matter /zpf < 0 of positivepressure.These
exotic vacuum virtual sources contribute Omega(Exotic Vacua)
~0.96 to Omega(Total) = 1 in our large-scale spatially
§atpost-in§ationary local Level I Hubble sphere brane 
world
as inLenny SusskindOs megalopolis Landscape subject to the
naturalselection of the Weak Anthropic Principle (WAP)OK, I
think I made an error above including brane worlds in sense
of parallel worlds?Here is why I think I made an error (If I
did so did Hawking and Scienti'c American in their popular
science reports):D-Branes are extended surfaces without
edges. In order that the black hole be a localized object, it
is assumed that our ordinary four dimensions (three space and
one time) are all orthogonal to these D-Brane surfaces ...
Thus to us these D-Branes would look as though they were
located at a point (or at least a very small region) of our
observable three dimensions of space. W. G. Unruh p. 168
Black holes, dumb holes, and entropy, i.e. the D-Branes are
in the compacti'ed Calabi-Yau space. I have to look again at
HawkingOs The Universe in a Nutshell that seems to give the
wrong idea here? Perhaps I mis-remember?Note also Ed
WittenOs
formula generalizing HeisenbergOs quantum uncertainty
principle, i.e. eq. (5.9) p. 136Delta X > h/DeltaP +
alphaO(DeltaP)/hThe second gravity-string source of
uncertainty should give the irreversible statistical arrow of
time not found when alphaO = 0 i.e. in'nite
string tension, or
in'nite space-time stiffness of action without reaction as is
also found in the signal locality of orthodox quantum theory
in sense of Antony ValentiniOs papers.Mass without mass, but
with a strong micro-gravityG* ~ 10^40G on 1 fermi scale. The
wormhole has an attractive dark matter exotic vacuum core
whereVacuum Coherence --> 0just like inside a quantized
vortex of circulation in super§uid HeII.Therefore, in the
core/zpf ~ - (1 fermi)^-2w = -1Therefore the quantum pressure
is positive and the exotic vacuum core gravitates
asGrad^2V(Exotic Vacuum) ~ -c^2(1 fermi)^-2this prevents the
spread out electric charge from exploding and also
compensates any quantized rotation centrifugal forces. For
now keep charge and rotation zero for simplicity. That is I
here only model a spin 0 neutral micro-geon. That means I
would need a high-power graviton laser as the Heisenberg
uncertainty scattering probe microscope. ThatOs OK since
this
is only a gedankenexperiment.The SSS metric then has the
factor1 - 2GM/c2rWhere G(mass density) is replaced by
c^2/zpfGM is replaced by ~ c^2/zpfR^3for a micro-geon of size
Rr is DeltaX as in WittenOs formula above with DeltaP as the
scattering momentum transfer between gravitons and the
micro-geon.Therefore, the Schwarzschild factor is (neglecting
factors of 2, pi etc all lumped into dimensionless parameter
b1 - bc^2/zpfR^3/DeltaXThe critical value of Delta X is then
the event horizon where the Schwarzschild factor vanishes -
can it be reached?This micro-geon is a solution of the exotic
vacuum local 'eld equationGuv(Einstein) + /zpfguv = 0At
critical Delta X, the geon looks like a POINT PARTICLE from
the huge space-warp induced by the probeOs momentum transfer
Delta P.C is the circumferencedC/dR = 2pi (1 -
bc^2/zpfR^3/DeltaX)^1/2Where one considers the radial size
'xed at R the scale of the throat of the wormhole.So when can
we haveDeltaX = bc^2/zpfR^3 ?Note that alphaO may be large
of
order (10^-11 cm)^2 not (10^-32 cm)^2 as in WittenOs
idea.
=Interesting!Jack,Superstring theory does not view
the Planck scale of spacetime as a quantumfoam but rather as
the Planck scale is approached the dimensionality ofspacetime
goes to 10-d. In string theory the spacetime does §uctuate
butthis §uctuation is harmonic rather than chaotic. In fact
the assumptionthat the harmonics of the strings absorbs all
the quantum §uctuation can beused to derive the
dimensionality of spacetime. Also Lorentz invariance
isassumed in this calculation. This was 'rst done by L. Brink
and H.B.Nielsen in 1973 (A Simple Physical Interpretation of
the Critical Dimensionof Space-time in Dual Models, *Physics
Letters* 45B:4 (1973) 332-336. Thispaper is also included in
the anthology edited by John Schwartz,*Superstrings: the
First 15 Years of Superstring Theory, Vol. 1*
(WorldScienti'c, 1985). [Note that Dual Models is the old
name for stringtheory, when it was still evolving away from
the terminology of s-matrixtheory. However, the Mandelstam
labels of S, T, and U duality haveresurfaced in membrane
theory!] I mentioned the Brink-Nielsen view of string theory
in my paper Notes onpdf.] Also on page 6 of F.W. SteckerOs
pdf paper [referred to in the NASAreport], he says We note
that there are variants of quantum gravity andlarge extra
dimension models which do not violate Lorentz invarianmm-133
=I have the following eigenvalue problem:Let A and B
be symmetric positive semi definite matrices.The underlying
graphs of A and B are trees. (Actually the
diagonalelementsare positive and the element A_ij= 0 (B_ij=
0) if vertices i and j arenotadjacent. A_ij= -1 (B_ij= -1) if
the vertices i and j are adjacent)Let a, b be the smallest
eigenvalues of the matrices A and B,respectively. a and b are
simple eigenvalues and the correspondingeigenvectors x and y
are positive (Perron-Frobenius Theorem).Ax=axBy=by,We know
that 0 < a <= b, y^tAy=b and x^tBx>a.For which matrices A and
B hold that a<b (a=b)?Any hints or references?Tuerker
Biyikoglu
=What I am looking for is an iterative algorithm,
like
:a[n+1]=f(a[n],...,b[n],...,c[n],...);b[n+1]=g(a[n],...,b[n],
...,c[n],...);c[n+1]=h(a[n],...,b[n],...,c[n],...);...such
thatF(a[k],b[k],c[k],...) for some (not large) k is
EllipticE(z,k).Chris
= Chris> What I am looking for is an
iterative algorithm, like : Chris>
a[n+1]=f(a[n],...,b[n],...,c[n],...); Chris>
b[n+1]=g(a[n],...,b[n],...,c[n],...); Chris>
c[n+1]=h(a[n],...,b[n],...,c[n],...); Chris> ... Chris> such
that Chris> F(a[k],b[k],c[k],...) for some (not large) k is
EllipticE(z,k).Look in netlib for CarlsonOs elliptic
integrals. There should be adescription of such an algorithm
for Elliptic E.Ray
=Correction : What I am looking for is
an iterative algorithm, like
:a[n+1]=f(a[n],...,b[n],...,c[n],...);b[n+1]=g(a[n],...,b[n],
...,c[n],...);c[n+1]=h(a[n],...,b[n],...,c[n],...);...such
thatF(a[m],b[m],c[m],...) for some (not large) m is
EllipticE(z,k).Chris
=Given a matrix OAO, IOm looking
to
find (numerically) a matrixOBO
such that OA * B = 1O, where
O1O is the appropriate unitmatrix.I had expected
that one
could treat the equation as a set o§inear equations OA * Bj
=
EjO. According to Numerical Recipes,if one has such a system
of equations one can obtain the SVD ofthe matrix
OAO: A = U *
W * Vtwhere OVtO is the tranpose of
OVO. Then the smallest
solutionis: Bj = V * WO * Ut * Ej(where the entries in
WO are
either the inverse of thecorresponding entries in W, or zero
if those entries are small).It seems clear to me (which
doesnOt mean that itOs true) thatthis means
that: B = V * WO
* Utis a good choice. In practice, however, where
OAO has
morecolumns than rows (ie, is singular) IOm
finding that OA *
BO isnot the unit matrix, though itOs close on
some rows and
columns.This makes no sense to me. IOve
confirmed that the
decompositionI get does, in fact, yield the matrix
OAO.Is
there a better way to go about this? Or should this work
(inwhich case I should try another SVD algorithm or debug the
oneIOve
got)?--
. . . Except when they donOt, Because sometimes
they wonOt. - Dr.
Seuss--
Jason Cooper jcooper@acs.ucalgary.ca
= >Given a
matrix OAO, IOm looking to
find (numerically) a matrix >OBO
such that OA * B = 1O, where
O1O is the appropriate unit
>matrix. > >I had expected that one could treat the equation
as a set of >linear equations OA * Bj = EjO.
According to
Numerical Recipes, >if one has such a system of equations one
can obtain the SVD of >the matrix OAO: > > A = U
* W * Vt where OVtO is the tranpose of
OVO. Then the smallest
solution >is: > > Bj = V * WO * Ut * Ej > >(where the
entries
in WO are either the inverse of the >corresponding entries
in
W, or zero if those entries are small). >It seems clear to me
(which doesnOt mean that itOs true) that >this
means that: > >
B = V * WO * Ut > >is a good choice. In practice, however,
where OAO has more >columns than rows (ie, is
singular) IOm
finding that OA * BO is >not the
unit matrix, though itOs
close on some rows and columns. >This makes no sense to me.
IOve confirmed that the decomposition >I get
does, in fact,
yield the matrix OAO. > >Is there a better way
to go about
this? Or should this work (in >which case I should try
another SVD algorithm or debug the one >IOve got)? 
-- > . . . Except when they donOt, > Because sometimes
they wonOt. - Dr. Seuss
>
-- >Jason Cooper jcooper@acs.ucalgary.ca if A=U*W*Vt, then
W has the same shape as A, hence in your case more columns
than rows.then , if A is of full rank, W will consist of an
invertiblediagonal matrix with a zero block appended to
right.then B = V*W#*Ut where W# has the same shape as AO and
consists of the inverted diagonal on the top and a zero block
appended at the bottom. then by the rules obviously A*B=I
where I has dimension=number of rwos of A.If A is not of full
rank, you will get a projector (some diagonalelements in I
change to zero) and in W# the reciprocal of zero is set to
zero. (B is the Moore-Penrose-pseudoinverse). hence you were
right.the problem with this approach is the decision which of
the singularvalues (the nonzero elements of W) to consider as
small. and if you set anything below eps to zero the
deviation of the product from I can be large ifyour decision
on the rank of A (implicit with this) was false.hence I
assume the problem here and not in the software.hope this
helpspeter
=This is a question on the relationship between,
the zero and non-zeroelements of the transition probability
matrix and the correspondingtransient and absorbing
states.Let T_a and T_b be the transition probability matrices
for two finite statediscrete Markov chains A and B,
respectively.We are given thata) i(T_a) = i(T_b), where i(X)
is the indicator matrix of Xi.e., i(X) has 1.0 entries
wherever X has non-zero entries and has 0.0entries wherever X
has zero entries.b) T_a^n converges as n -> infinity. Let the
limiting matrix be Ta^{inf}Now, I think I can show that T_b^n
also converges. However, I am trying tofind out if i(Ta^inf) =
i(Tb^inf).I have gotten this far:Since i(T_a) = i(T_b), we
have i(T_a^2) = i(T_b^2) and so on ...I can show that for any
finite k, i(T_a^k) = i(T_b^k),but does this property hold in
the limit?In other words, if i(T_a) = i(T_b), then do A and B
have the same transientand absorbing states?Kumar
went looking for a short proof of FermatOs Last Theorem
using
basic> algebra following my physics training. I expected
failures, false> starts, false positives, but mathematicians
apparently arenOt> scientists, as they attack me for having
them.Wrong! They refuted your arguments. You responded to
discovery of your errors by attacking them.> Worse, theyOd
use my own admissions of §awed approaches against me,> as if
that proved I couldnOt succeed. And worst of all, when I
did>
succeed they kept up a campaign of personal attacks and
smears.Succeed? At what? The only success you have had to
date is in establishing that you are a crank and mostof the
personal attacks and smears posted in this newsgroup were
authored by you.> And itOs not just with the polynomial
factorization as thereOs my> prime counting research as
well.>> ItOs too much to be believable given what you can
see
with your own> eyes that none of my work is worth mention,
when people can even see> the heat generated on several
newsgroups. These people are expending> a LOT of energy, and
you think itOs for nothing?The energy others are expending
has been to keep the record straight. Your lies and
distortions deservechallenges, and will continue to face
them.> My work is available 24 hours a day, seven days a week
at>> http://groups.msn.com/AmateurMath>> and if it were really
so wrong, why do these people keep fighting?Obviously, as
previously noted, to keep the record straight. DonOt you
expect scientists andmathematicians to resist error? Take
down that ridiculous treatise. It isnOt fit to
wrap the
garbage in.> What would happen if mathematicians even
acknowledged the> *possibility* that IOm right? They know.
Given people following and> keeping up with the details itOs
clear IOm right.Clear to whom? You are wrong. Many peoply
follow your posts for comic relief. Personally, I liken
youto:1) Monty PythonOs Black Knight -- who, having been cut
to pieces, insists ItOs only a §esh wound!2) An insect in
a
collectorOs jar, waving his ineffectual pinchers at the
world,3) A stupidly grinning in§atable clown who keeps
popping up whenever he is knocked down.> Scientists behave a
certain way. Mathematicians are behaving another> way....and
you are behaving like an insolent, arrogant and paranoid
megalomaniac.> The proof is out there.>> James HarrisYeah,
but not on your website. Let us know when you get back from
the Twilight Zone with your latestX-File.--It takes a village
to raise an idiot.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
=IOm dealing with a
problem of nonlinear optimization with boundconstraints. I
canOt use commercial software for this. So farL-BFGS-B from
Netlib has produced satisfactory results. Now Ineed to know
the value of the Hessian matrix at the end of thecomputation.
L-BFGS-B doesnOt seem to have this information.(The L stands
for Limited Memory, meaning that it stores onlyupdates to the
Hessian.)Currently IOm looking at DRMNGB, also from Netlib.
It
seemsas if it stores the Cholesky decomposition of the
Hessian.However, IOm having trouble figuring out
how to find it
andtransform it back into the full Hessian matrix. On my
firstattempt the numerical values donOt seem
right. Some
thingswhich should be positive are not. As far as I can
tell,it should be stored beginning at V(IV(42)), but itOs
notclear in what order the terms are stored.Does anyone know
about this? Or do you have another favoritesubroutine?-- *
Patrick L. Nolan ** W. W. Hansen Experimental Physics
Laboratory (HEPL) * * Stanford University *
= >IOm dealing
with a problem of nonlinear optimization with bound
>constraints. I canOt use commercial software for this. So
far >L-BFGS-B from Netlib has produced satisfactory results.
Now I >need to know the value of the Hessian matrix at the
end of the >computation. L-BFGS-B doesnOt seem to have this
information. >(The L stands for Limited Memory, meaning that
it stores only >updates to the Hessian.) > >Currently IOm
looking at DRMNGB, also from Netlib. It seems > as if it
stores the Cholesky decomposition of the Hessian. >However,
IOm having trouble figuring out how to
find it and >transform
it back into the full Hessian matrix. On my first >attempt the
numerical values donOt seem right. Some things >which should
be positive are not. As far as I can tell, >it should be
stored beginning at V(IV(42)), but itOs not >clear in what
order the terms are stored. > >Does anyone know about this?
Or do you have another favorite >subroutine? > >-- >* Patrick
L. Nolan * >* W. W. Hansen Experimental Physics Laboratory
(HEPL) * >* Stanford University *drmngb is a bfgs based code
and will hardly deliver a good approximation to the true
Hessian, although as a minimizer it is o.k.why not this one:
toms/636 keywords: estimating sparse hessian matrices,
difference of gradients gams: G4f title: DSSM and FDHS for:
estimating sparse Hessian matrices by: T.F. Coleman, B.S.
Garbow, and J.J. More ref: ACM TOMS 11 (1985) 363-377 and 378
Score: 100%hope that helpspeter
=are you saying you are
feeding the optimizer at best first derivativesand want to get
out second derivatives? One way to do it is toevaluate your
own with the final solution vector. Any Hessian of
theoptimizer will be an approximation at best (finite
difference,quasi-Newton). Only exact or automatic
differentiation can give youthe Hessian with sufficient
acuracy.Hans
Mittelmann
==> IOm dealing with a problem of
nonlinear optimization with bound> constraints. I canOt use
commercial software for this. So far> L-BFGS-B from Netlib
has produced satisfactory results. Now I> need to know the
value of the Hessian matrix at the end of the> computation.
L-BFGS-B doesnOt seem to have this information.> (The L
stands for Limited Memory, meaning that it stores only>
updates to the Hessian.)> > Currently IOm looking at DRMNGB,
also from Netlib. It seems> as if it stores the Cholesky
decomposition of the Hessian.> However, IOm having trouble
figuring out how to find it and> transform it back
into the
full Hessian matrix. On my first> attempt the numerical values
donOt seem right. Some things> which should be positive are
not. As far as I can tell,> it should be stored beginning at
V(IV(42)), but itOs not> clear in what order the terms are
stored.> > Does anyone know about this? Or do you have
another favorite> subroutine?
=>Try the family:> (x/a)^n +
(y/b)^n = 1>with n > 2 . Of course, n=2 is an ellipse
centered at the origin.I forgot to mention one important
thing. I need an ellipse-like curvethat is not symmetric in
the way an ellipse (or the curve above) is.
=>>Try the
family:> (x/a)^n + (y/b)^n = 1>with n > 2 . Of course,
n=2 is an ellipse centered at the origin.>>I forgot to
mention one important thing. I need an ellipse-like
curve>that is not symmetric in the way an ellipse (or the
curve above) is.Like a longitudinal cut of an egg, or an
etc,likehttp://www.mathematische-basteleien.de/eggcurves.htm=
IOve done my part with my mathematical research by working
hard toexplain it, arguing with people over details, writing a
paper andsending it to math journals. And now I realize that
part of theproblem is that the math is revolutionary.I use
methods to factor polynomials into non-polynomial factors,
whichis such a powerful approach that it leads to a short
proof of FermatOsLast Theorem that is just a few pages.The
math is mostly basic algebra.IOve given the information.
IOve
explained in detail butmathematicians can resist for a while,
just as they resisted otherrevolutionary mathematics, but
IOm
hopeful that here at least *some*of you recognize that itOs
not to the benefit of math society to actin that way.What does
it take to consider the mathematics?Reviewing the math at my
website, or considering the many posts whereI talk about the
factorization (v^3+1)x^3 - 3vx + 1 = (a_1 x + 1)(a_2 x +
1)(a_3 x + 1)or variations on it.Now you can actually look at
the factorization for v=1, and see thatas IOve stated for
some
time, ONLY TWO of the aOs can have a factor of2, and that
factor is sqrt(2).The polynomial of course is 2x^3 - 3x + 1,
and itOs easily factorable.My work gave a
specific prediction.
I can show you where thatprediction is the reality with a
reducible polynomial. The methodsthemselves DO NOT CARE about
reducibility, so they work when v givespolynomials not
reducible over Q.many of you realize that if mathematicians
refuse to do their part,then my task is difficult, even though
IOm right.Factoring polynomials in a special way such that
even the FLT equationitself can be factored, as weird as that
may sound, is a great ideawhose time has come. Your task now
is simply to accept the future,just as those before you
accepted imaginary numbers, and thosebefore them accepted
irrational numbers.You may be puzzled. You may not understand
why thereOs so much drama. But then now maybe you can
understand other math history a littlebetter. People like
comfort. Thinking you know all you need to knowin certain
areas is comforting. And then some guy comes along andstarts
a revolution.I am that guy.James Harris
=LOL.>
LOL.Yeah, it may be funny to you, but what IOm talking about
is a modernsituation where a mathematical argument given is
simply ignored orlied about by mathematicians.2x^3 - 3x + 1 =
(x-1)(2x^2 +2x -1) =(a_1 x + 1)(a_2 x + 1)(a_3 x + 1)My work
has specific predictions, which are the mathematical
reality.Mathematicians fighting work that is simple,
demonstrable, and correctcanOt be good for the
discipline.And
itOs not funny.James Harris
work that is simple, demonstrable, and correct> canOt be
good
for the discipline.>> And itOs not funny.>> James HarrisNo
one
is laughing at Mathematicians fighting work that is simple
demonstrable, and correct. Thesubject of discussion is your
error-ridden, false, oft-refuted, proof of FLT. They are
laughing at*you*, idiot.--It takes a village to raise an
idiot.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
Mathematicians fighting work that is simple, demonstrable, and
correct> canOt be good for the discipline.>> And
itOs not
funny.>> James Harris> > No one is laughing at Mathematicians
fighting work that is simple demonstrable, and correct. The>
subject of discussion is your error-ridden, false,
oft-refuted, proof of FLT. They are laughing at> *you*,
idiot.Is that the math world? Is that what all of you see
when you think ifmathematicians? People calling others
idiot?IOm on whatOs supposed to be *your*
turf. Basic algebra
and astep-by-step argument which proves what IOm
saying.Ultimately people attacking me are attacking
mathematics itself.Now think about it. ThereOs my prime
counting work, then thereOs mypaper, and also not least
thereOs the actual short proof of FermatOsLast
Theorem.If you
believe people arguing with me thereOs only ONE small
thingthat they fight.Think about it. Basic algebra proves my
point. The argument isstraightforward and simple, but the
consequences are huge.ItOs revolutionary for *social*
reasons. And you can see society,your society, fighting as it
has before. Learn the lessons from thebattles over irrational
and imaginary, and try to let the math beyour guide--not your
comfort zone.The work itself is conveniently available to you
24 hours a day, sevendays a week at
http://groups.msn.com/AmateurMathand hey, you can even play
with an applet (if you join the group).James
Harris
=documentstyle{report}begin{document}pagestyle{plain
}raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the following integral:$$ int_{-half a}^{+half
a} x^2 , dx / a = left[ frac{1}{3} x^3 right]_{-half
a}^{+half a} / a = frac{2.a^3}{3.8.a} = frac{a^2}{12}$$Now
consider:$$ lim_{sigma rightarrow infty} frac{ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }$$We
would expect that this more difficult integral nevertheless
convergesto the same result as before: $a^2 / 12$ ,
because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividing the integralin the denominator
by $a$ should be equivalent to differentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
finding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ... ;$ in
theirdenominators, so they should converge to zero. Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they do not match. WhatOs going wrong ?
Please
help !%end{document}-- * Han de Bruijn; Systems for Research A
little bit of Physics would be (
)* and Education (DTO/SOO),
Mekelweg 6 NO Idleness in Mathematics(HdB) @-O^O-@* 2628 CD
Delft, The Netherlands http://huizen.dto.tudelft.nl/deBruijn
#/_#
=>....................> So the two outcomes have the
same order of magnitude, but they do not match. > WhatOs
going wrong ? Please help !> %following .tex
file,
documentstyle[leqno]{report}newcommand{hsp}{hspace*{0.5cm}}
newcommand{Erf}{{rm Erf}}begin{document}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the problem to
determinebegin{equation}label{eu11} fbox{$displaystyle
L(a):=lim_{sigma rightarrowinfty}I(a,sigma)hsp mbox{rm with}
hsp I(a,sigma)= frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } $}end{equation}However:$$ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2
int_{-half a}^{+half a} x , e^{-half x^2 /sigma^2} ,
dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half
a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = -
sigma^2 left[ x.e^{-half x^2 / sigma^2}
right]_{-halfa}^{+half a} + sigma^2 int_{-half a}^{+half a}
e^{-half x^2 / sigma^2} ,dx$$ $$ hieruit hsp I(a,sigma)=frac{
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } =
sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}right)$$Let us define $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,;
h to0; .; $ Then problem (ref{eu11}) is equivalent
withbegin{equation}label{eu12}fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto0}frac{1}{h^2}left(1-frac{he^{
-h^2}}{Erf(h)}right) hspmbox{rm where}hsp
Erf(h):=intlimits_{0}^h e^{-t^2}; dt$}hsp .end{equation}But
$displaystyle
Erf(h)=he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $
where[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k
}cdotfrac{z^k}{k!}hsp,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]Therefore
[L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}{h^
2cdotsumlimits_{k=0}^infty
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=0}^inftyfrac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k
}}{cdotsumlimits_{k=0}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}=frac{
a^2}{8}cdotfrac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12};
.]%end{document}
=> >....................> So the two
outcomes have the same order of magnitude, but they do not
match.> WhatOs going wrong ? Please help !> %> following
.tex
file,>
documentstyle[leqno]{report}>
newcommand{hsp}{hspace*{0.5cm}}> newcommand{Erf}{{rm Erf}}>
begin{document}> pagestyle{plain} raggedbottom> def half
{frac{1}{2}}> def hieruit {quad Longrightarrow quad}> %>
subsection*{Should be the same but alas ...}> %> Consider the
problem to determine> begin{equation}> label{eu11}
fbox{$displaystyle L(a):=lim_{sigma rightarrow>
infty}I(a,sigma)hsp mbox{rm with} hsp I(a,sigma)= frac{>
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx }>
{ int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } $}>
end{equation}> However:> $$> int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx => - sigma^2 int_{-half
a}^{+half a} x , e^{-half x^2 /> sigma^2} ,> dleft(-half x^2
/ sigma^2right) => $$ $$> - sigma^2 int_{-half a}^{+half a} x
,> dleft(e^{-half x^2 / sigma^2}right) => - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half> a}^{+half a}> +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,>
dx> $$ $$> hieruit hsp I(a,sigma)=frac{> int_{-half a}^{+half
a} x^2 , e^{-half x^2 / sigma^2} , dx }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx } => sigma^2 left(
1 - frac{ a.e^{-half (a/2)^2 / sigma^2} }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx }> right)> $$> Let
us define $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,; h to> 0; .; $
Then problem (ref{eu11}) is equivalent with>
begin{equation}label{eu12}> fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto>
0}frac{1}{h^2}left(1-frac{he^{-h^2}}{Erf(h)}right) hsp>
mbox{rm where}hsp Erf(h):=intlimits_{0}^h e^{-t^2}; dt> $}hsp
.> end{equation}> But $displaystyle Erf(h)=>
he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $ where>
[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k}
cdotfrac{z^k}{k!}hsp> ,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]>
Therefore> [L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^infty> frac{(1)_k}{(3/2)_k}h^{2k}}>
{h^2cdotsumlimits_{k=0}^infty frac{(1)_k}{(3/2)_k}h^{2k}}=>
frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=0}^infty>
frac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k}}>
{cdotsumlimits_{k=0}^infty>
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}cdot>
frac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12}; .> ]> %>
end{document}I would like to read LaTex files. Is there some
Windows software out therethat works immediately without
having to run numerous fonts setup programscomputer.
= [
deleted ]Here is your answer, as simple as I can make
it:setlength{mathindent}{1.0cm}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}begin{document}%subsection*{Should be the
same but alas ...}%Consider the following integral:$$
int_{-half a}^{+half a} x^2 , dx / a = left[ frac{1}{3} x^3
right]_{-half a}^{+half a} / a = frac{2.a^3}{3.8.a} =
frac{a^2}{12}$$Now consider:$$ lim_{sigma rightarrow infty}
frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} ,
dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}$$We would expect that this more difficult integral
neverthelessconverges to the same result as before: $a^2 /
12$ , because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividingthe integral in the denominator
by $a$ should be equivalent todifferentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
finding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ...;$ in
their denominators, so they should converge to zero.Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they donot match. WhatOs going wrong ? Please
help !%subsection*{The easy answer}Let[frac{a}{{2sigma }} =
lambda]and rewrite the desired integral ratio
asbegin{eqnarray*}frac{{int_{ - {a mathord{left/ {vphantom {a
2}} right. kern-nulldelimiterspace} 2}}^{ + {a mathord{left/
{vphantom {a 2}} right. kern-nulldelimiterspace} 2}} {dx,x^2
e^{ - {{x^2 } mathord{left/ {vphantom {{x^2 } {sigma ^2 }}}
right. kern-nulldelimiterspace} {sigma ^2 }}} } }}{{int_{ -
{a mathord{left/ {vphantom {a 2}} right.
kern-nulldelimiterspace} 2}}^{ + {a mathord{left/ {vphantom
{a 2}} right. kern-nulldelimiterspace} 2}} {dx,e^{ - {{x^2 }
mathord{left/ {vphantom {{x^2 } {sigma ^2 }}} right.
kern-nulldelimiterspace} {sigma ^2 }}} } }} &equiv& sigma ^2
frac{{int_0^lambda {du,u^2 e^{ - u^2 } } }} {{int_0^lambda
{du,e^{ - u^2 } } }} &=& sigma ^2 left( {frac{{frac{{lambda
^3 }}{{3 cdot 0!}} - frac{{lambda ^5 }}{{5 cdot 1!}} +
frac{{lambda ^7 }}{{7 cdot 2!}} - + ldots }}{{frac{{lambda ^1
}}{{1 cdot 0!}} - frac{{lambda ^3 }} {{3 cdot 1!}} +
frac{{lambda ^5 }}{{5 cdot 2!}} - + ldots }}} right) &=&
frac{{sigma ^2 lambda ^2 }}{3}left( {frac{{1 - frac{{3lambda
^2 }}{5} + - ldots }}{{1 - frac{{lambda ^2 }}{3} + - ldots
}}} right);mathop to limits_{sigma to infty }
;frac{{a^2}}{{12}}end{eqnarray*}end{document}-- Julian V.
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.
=> > I would like to read LaTex files. Is
there some Windows software out there> that works immediately
without having to run numerous fonts setup programs>
computer.Subscribe to the newsgroup comp.text.texand see what
people there are asking about and recommending.You will
probably want to install the MikTeX distribution, available
from http://ftp.uci.agh.edu.pl/pub/tex/systems/win32/miktex/I
strongly also recommend getting WinEdt, available from
http://www.winedt.com/It is shareware, but worth the money
($40, or so). It is a LaTeX andTeX -aware editor that makes
using these systems easy (I suggest LaTeXfor your own
work).The book A Guide to LaTeX by Kopka & Daly is extremely
helpful.Good luck.-- Julian V. NobleProfessor Emeritus of
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.

integers with an> > >actual, real-live mathematician, and not
a possibly fake one that just> > >posts a lot on Usenet!!!> > > >Yes I used chalk and went over it all on the chalkboard
and even> > >talked over some of the math history like talking
about Dedekind.> > >> > >What I veri'ed is that oddity that
you CAN talk to a mathematician,> > >give a correct math
argument, go over each and every point> > >step-by-step, yet
have that mathematician simply dismiss you. > > > >
Fascinating. He did in fact dismiss your work, but
nonetheless> > this 'lls you with glee.> > However, the
imporant point is that he couldnOt 'nd an 
error. Since there
is no such thing as an error in *Number Theory*, you probably
should be thankful he even let you keep your computer.> There
are people who *dismiss* the idea that man landed on the
moon. They should, since it wasnOt an idea.> The key issue
here is mathematical correctness. My point is that I> went
through the math point-by-point and the professor could not
'nd> an error. Then you sould post your proof to someplace
other than alt.math.undegrad, Since the only thing anybody
does in undergrad is give money to moron *physicists*.> > In this> > >case the professor claimed it was out of his
area, and gee, guess> > >what? According to him no one else
at Vanderbilt University had the> > >necessary expertise.> > > >Basically he was blowing me off, but not because I was
wrong. > > > > Of course not. No matter how many people,
usenet posters,> > mathematicians you visit in person, no
matter _how_ many of> > them say youOre wrong, 
itOs simply
not possible that the reason> > theyOre all saying that is
that youOre _wrong_.> > But youOre now lying 
David Ullrich as
my point is that Professor> McKenzie did NOT say that I was
wrong.> It is such an obvious falsehood that I feel con'dent
in calling you> out here as a liar.> > The issue is
mathematical correctness. ThatOs impossible. Since
mathematical correctness is isomorphic to gravity
correctness, neither of which exist.
=:> this 'lls you with
glee.: However, the imporant point is that he couldnOt 
'nd an
error.Does this mean that he accepted your logic at each
step? Orcould he 'nd nothing lucid enough to call an
error?
 IOve seen, for JJ, sets have no properties, set
mappings are a> pipe dream, in'nity is ineffable, and
nonesense is a virtue.Ha Ha. Lovely typo that nonesense,
given its ambiguity between nonsenseand nonessence. ;)Since
some of JJ remarks seem loosely Wittgensteinian, perhaps he
c
7ï
WEcf
!.856¨èÿÿ
by the statement that one of them is> >coprime to 2.> > The
three roots are 1 (twice) and -2; 1 is coprime to 2, in the>
trivial way. The reason this is not a problem and not
something wacky> is that the polynomial is reducible over Q: x^3-3x+2 = (x-1)(x-1)(x+2).> > James later stated that he
should have written x^3+3x-2, which is> irreducible over Q.
In that case, none of the roots are units. all of> them are
factors of 2 and of themselves, and therefore you would be>
well to be surprised at the statement that one of them is
coprime to> 2.ThatOs actually a useful statement to show how
the wacky error withalgebraic integers can lead to a false
proof for those of you whothink it doesnOt matter to have
this particular error in core.Here Magidin IS correct in that
none of the roots can be units, butthatOs because *in the 
ring
of algebraic integers* itOs provably truethat any unit is a
root of a monic polynomial with integercoef'cients and a last
coef'cient of 1 or -1.But that does NOT prove that none of 
the
roots are coprime to 2.Here Magidin is using one thing, but
switching to another, as if youjust believe that to be
coprime to 2, one of the roots has to be aunit, then you
accept his claim as true. Notice he says well to
besurprised.To catch Magidin here you have to force him to
*prove*non-coprimeness, and he cannot. Various posters have
tried thisparticular trick, and IOd catch them on it, and
theyOd start going incircles.James Harris
 >>It
turns out that the de'nition for algebraic integers where 
they
are>>roots of *monic* polynomials leaves off certain numbers
that should be>>included, which is how I found a way to show
some wacky things.>>ItOs easy enough to explain as
consider>> x^3 - 3x + 2>>which as it turns out
has a root that is coprime to 2. That is, it>>doesnOt 
share
any non-unit factors with 2 in the ring of
algebraic>>integers.>>The roots of this cubic are three
algebraic integers, and their product>For this reason, I am
surprised by the statement that one of them is>coprime to
2.>>The three roots are 1 (twice) and -2; 1 is coprime to
2, in the>>trivial way. The reason this is not a problem and
not something wacky>>is that the polynomial is reducible over
Q:>>x^3-3x+2 = (x-1)(x-1)(x+2).>>James later stated that
he should have written x^3+3x-2, which is>>irreducible over
Q. In that case, none of the roots are units. all of>>them
are factors of 2 and of themselves, and therefore you would
be>>well to be surprised at the statement that one of them is
coprime to>>2.> > > ThatOs actually a useful statement to 
show
how the wacky error with> algebraic integers can lead to a
false proof for those of you who> think it doesnOt matter to
have this particular error in core.> > Here Magidin IS
correct in that none of the roots can be units, but> thatOs
because *in the ring of algebraic integers* itOs provably
true> that any unit is a root of a monic polynomial with
integer> coef'cients and a last coef'cient of 1 
or -1.> > But
that does NOT prove that none of the roots are coprime to 2. Here Magidin is using one thing, but switching to another,
as if you> just believe that to be coprime to 2, one of the
roots has to be a> unit, then you accept his claim as true.
Notice he says well to be> surprised.> > To catch Magidin
here you have to force him to *prove*> non-coprimeness, and
he cannot. Various posters have tried this> particular trick,
and IOd catch them on it, and theyOd start 
going in> circles.>
Okay, IOll take a stab at it:x^3+3x-2 factors as
(x-a)(x-b)(x-c) where a,b,c are roots.Multiplying this out,
we get that:-a-b-c=0ab+ac+bc=3abc=2Looking at the last
equation, we see three factors of 2: a,b,c
since2=a(bc),2=b(ac),2=c(ab).These factors are each factors
of themselves: a=1*a, b=1*b, c=1*cSo, a is a factor of a and
2,b is a factor of b and 2,c is a factor of c and 2.Since you
admit that a,b,c are not units, and you have de'ned two
numbers to be coprime if they have no non-unit factors in
common,a is not coprime to 2, because they have a common
non-unit factor: ab is not coprime to 2, because they have a
common non-unit factor: bc is not coprime to 2, because they
have a common non-unit factor: cTherefor, *none of the roots
are coprime to 2*. QED.Now, having said that, I expect you
will 'nd an objection somewhere.-- Will Twentyman
<snip>James later stated that he should have written
x^3+3x-2, which is>>irreducible over Q. In that case, none of
the roots are units. all of>>them are factors of 2 and of
themselves, and therefore you would be>>well to be surprised
at the statement that one of them is coprime to>>2.>><snip> Here Magidin IS correct in that none of the roots can be
units, but> thatOs because *in the ring of algebraic
integers* itOs provably true> that any unit is a root of a
monic polynomial with integer> coef'cients and a last
coef'cient of 1 or -1.> > But that does NOT prove that none
of the roots are coprime to 2.Of course not, nor did he claim
that as a proof. However itOs trivialto see that none of the
roots are coprime to 2. Denote the roots by r1,r2, and r3, so
thatx^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)Then r1*r2*r3 = 2,
so r1 divides 2 in the ring of algebraic integers(in fact 2 /
r1 = r2*r3). Thus r1 and 2 share a common nonunit
factor,namely r1. The same holds for r2 and r3, so none of
the roots arecoprime to 2.Rick
 > > > <snip>> > >>James
later stated that he should have written x^3+3x-2, which is>irreducible over Q. In that case, none of the roots are
units. all of> >>them are factors of 2 and of themselves, and
therefore you would be> >>well to be surprised at the
statement that one of them is coprime to> >>2.> > <snip>>  > Here Magidin IS correct in that none of the roots can be
units, but> > thatOs because *in the ring of algebraic
integers* itOs provably true> > that any unit is a root of a
monic polynomial with integer> > coef'cients and a last
coef'cient of 1 or -1.> > > > But that does NOT prove that
none of the roots are coprime to 2.> > > Of course not, nor
did he claim that as a proof. However itOs trivial> to see
that none of the roots are coprime to 2. Denote the roots by
r1,> r2, and r3, so that> > x^3 + 3x - 2 = (x - r1)(x - r2)(x
- r3)> > Then r1*r2*r3 = 2, so r1 divides 2 in the ring of
algebraic integers> (in fact 2 / r1 = r2*r3). Thus r1 and 2
share a common nonunit factor,> namely r1. The same holds for
r2 and r3, so none of the roots are> coprime to 2.That does
not prove that there does not exist algebraic integers 
OaOand
ObO such that r_1 a + 2b = 1 in the ring of 
algebraic
integers,and in fact for one of the rOs it is true that they
*do* exist as bothyou and Magidin are wrong.Again, the trick
is to try and use the result that it is not a unit,when in
fact, my point is that the ring of algebraic integers
isscrewed up in that it is not a unit.Mistakes in core
mathematics can be great fun to play with, so posterslike
Rick Decker can run themselves in knots, if they donOt 
follow
themath.Remember an error in core is a big deal. People can
prove all kindsof things with such an error, which is why it
should be handled.James Harris
> >> > >> <snip> >>James later stated that he should have written x^3+3x-2,
which is>> >>irreducible over Q. In that case, none of the
roots are units. all of>> >>them are factors of 2 and of
themselves, and therefore you would be>> >>well to be
surprised at the statement that one of them is coprime to>>2.>> >> <snip> > >> > Here Magidin IS correct in that
none of the roots can be units, but>> > thatOs because *in
the ring of algebraic integers* itOs provably true>> > that
any unit is a root of a monic polynomial with integer>> >
coef'cients and a last coef'cient of 1 or -1.>> 
> >> > But
that does NOT prove that none of the roots are coprime to
2.>> >> >> Of course not, nor did he claim that as a proof.
However itOs trivial>> to see that none of the roots are
coprime to 2. Denote the roots by r1,>> r2, and r3, so that>> x^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)>> >> Then r1*r2*r3
= 2, so r1 divides 2 in the ring of algebraic integers>> (in
fact 2 / r1 = r2*r3). Thus r1 and 2 share a common nonunit
factor,>> namely r1. The same holds for r2 and r3, so none of
the roots are>> coprime to 2.>>That does not prove that there
does not exist algebraic integers OaO>and 
ObO such that r_1 a
+ 2b = 1 in the ring of algebraic integers,Yes, it does. For
rewriting 2 as r1*(r2*r3), we have that if such an aand b
existed, then:1 = r_1*a + = r_1*a + r1*r2*r3*b =
r_1*(a+r2*r3*b)Since r2, r3, and b are algebraic integers, so
is r2*r3*b. Since a isalso an algebraic integer by assumption,
so is a+r2*r2*b.So if a and b existed, then there would be an
algebraic integer x(namely, x=a+r2*r3*b) such that r_1*x =
1.Therefore, r_1 would be an algebraic integer unit.But the
ONLY complex number x with the property that r_1*x = 1
isx=1/r_1.Therefore, you are claiming that a+r2*r3*b =
1/r1.But we know that a+r2*r3*b is an algebraic integer. And
we know that1/r1 is NOT an algebraic integer.This is a
contradiciton. The contradiction arises from assuming that
aand b exist. Therefore, no such a and no such b exist.The
same argument, word for word, exchanging r1 and r2, shows the
samefor r2; and exchanging r1 and r3, shows it for r3.But this
is all a red herring.>and in fact for one of the rOs it is
true that they *do* exist as both>you and Magidin are
wrong.Please state explicilty(a) which r;(b) what is the
value of a; and(c) what is the value of b.Otherwise, your
claim is unfounded. But this is all a red herring.>Red
herring.We used YOUR de'nition of coprime. It is using YOUR
de'nition ofcoprime that you conclude that one of r1, r2, or
r3 is coprime to2. But using THAT de'nition, it turns out
that NONE of r1, r2, or r3are coprime to 2. So your claim is
wrong.>Again, the trick is to try and use the result that it
is not a unit,>when in fact, my point is that the ring of
algebraic integers is>screwed up in that it is not a unit.So:
the trick is to use that it is not a unit, when in fact it is
nota unit.Why is it a trick to use something that is
true?>>Mistakes in core mathematics can be great fun to play
with, so posters>like Rick Decker can run themselves in
knots, if they donOt follow the>math.You seem to be tying
yourself in knots. You claim it is a trick (or anerror) to
use a true fact.
Why?
==
== [Gabriele Rossetti] has left a vast body of
writings... in which he has attempted to prove the truth of
his unorthodox interpre- tation of medieval literature. They
present a formidable record of unsystematic research in which
we see an enthusiast plunging farther and farther and farther
from the logic of facts and good sense until truth is lost in
the dreadful nightmare of an idee 'xe. There is no real
evolution of the Theory although it grows and expands until
it embraces ever wider horizons. The numerous inaccuracies of
deduction, mis-statements of historical fact, and
self-contradictions...have caused critics to turn away from
them in disgust... [...] It is impossible to read far...
without realizing that we have to deal with a work of faith
and imagination rather than of reasoning. There is an
appearance of reason, for the author is set on proving by
logic the truth of what he already believes by intuition. The
truth is plain to him and he cannot comprehend why others do
not immediately accept it, but as they desire demonstration
he has multiplied his proofs. It is the redundancy and
confusion of a prophet expounding by a familiar method the
truth revealed to his own simple soul in a §ash of
inspiration... In such work as this... it is idle to look for
the calm reasoning of a scholar; we do not 'nd it, and there
is little or no advantage in attacking the obvious
inconsistencies and absurdities that abound. -- E.R. Vincent,
_Gabriele Rossetti in England_, quoted in _The Shakespearan
Ciphers Examined_, by William F. Friedman and Elizebeth S.
Friedman
==
==Arturo
Magidinmagidin@math.berkeley.edu
mathematics can be great fun to play with, so posters> like
Rick Decker can run themselves in knots, if they donOt 
follow
the> math.>> Remember an error in core is a big deal. People
can prove all kinds> of things with such an error, which is
why it should be handled.>> James HarrisThe error is in your
argument, not in core mathematics. The only big deal here is
the magnitude of your egoand its reciprocal: your vanishingly
small intelligence.--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
= [.snip.]>> x^3-3x+2 =
(x-1)(x-1)(x+2).>> >> James later stated that he should have
written x^3+3x-2, which is>> irreducible over Q. In that
case, none of the roots are units. all of>> them are factors
of 2 and of themselves, and therefore you would be>> well to
be surprised at the statement that one of them is coprime
to>> 2.>>ThatOs actually a useful statement to show how the
wacky error with>algebraic integers can lead to a false proof
for those of you who>think it doesnOt matter to have this
particular error in core.>>Here Magidin IS correct in that
none of the roots can be units, but>thatOs because *in the
ring of algebraic integers* itOs provably true>that any unit
is a root of a monic polynomial with integer>coef'cients and
a last coef'cient of 1 or -1.Excellent! Last time I claimed
(and proved) this, you stated this
241%40agate.berkeley.eduTHEOREM. Let f(x) be a monic
polynomial with integer coef'cients,which is irreducible over
Q. Then a root of f(x) is an algebraicinteger unit if and only
if the constant term of f(x) is 1 or -1.Proof. The de'nition
of algebraic integer unit is an algebraicinteger u such that
1/u is also an algebraic integer.Let r be a root. If the
constant term is 1 or -1, then r divides 1 or-1 (in the ring
of algebraic integers), so it is a unit.Conversely, suppose
that r is an algebraic integer unit. Lets=1/r. Then s is an
algebraic integer. Writef(x) = x^n + ... + a_0.0 = s^nf(r) =
s^n(r^n+...+a_0) = 1 + a_{n-1}s + ... + a_0s^n.So s is a root
of g(y) = a_0x^n+...+1; since g(y) is the polynomial weget by
taking f(1/y) and multiplying through by y^n, it follows
fromirreducibility of f(x) that g(x) is irreducible.Since s
is the root of the irreducible primitive polynomial g(y)
withinteger coef'cients, it follows that the leading
coef'cient of g(y)is either 1 or -1. So a_0=1 or a_0=-1.
Therefore, if r is an algebraicinteger unit, then the
constant term of f(x) is either 1 or -1. QEDYour reply
it:-- Begin Insert --> THEOREM. Let f(x) be a monic
polynomial with integer coef'cients,> which is irreducible
over Q. Then a root of f(x) is an algebraic> integer unit if
and only if the constant term of f(x) is 1 or -1.> > Proof.
The de'nition of algebraic integer unit is an algebraic>
integer u such that 1/u is also an algebraic integer.> > Let
r be a root. If the constant term is 1 or -1, then r divides
1> or> -1 (in the ring of algebraic integers), sce and
forwhich the constraints considered here do not apply. BTW:
In the Acknowledgments on page 8, Serge Rudaz is one of
fourpeople thanked for helpful discussions. I remember
meeting Serge Rudaz inseminars. He was then a young physics
student (at Cornell?) who was an oldacquaintance of yours. He
told us about instantons -- which was a new ideathen. I have
not heard of him since 1976 -- until seeing
thisacknowledgement! Nuff said! Saul-PaulJS: Yes, Serge was
with me at UCSC in Summer 1973 when I went to see Jean
CocteauOs Orphee on campus with Helen Quinn (who was close
to
having he baby at that time) and I think Serge and a few
others. This was days before I went to SRI to meet with
Puthoff and Targ on the tape you have. The story is in the
book Destiny Matrix.----------Therefore quantum foam is
suspect. Also BohmOs quantum potentialview of vacuum
§uctuations is relevant in context of the recentpaper from
Teheran.I need to follow the experiment below more
carefully.Jack better check this story out:Gary S.
Bekkum
=i have realised a few mathematical functions?,
properties?, lately andwas wondering about them, such as
whether they are already known, orare even important.if
anyone is interested, hit me up,AIM:heartxenocide
=I really
theMAA, and IOve posted here a few, brief times, however
since IOm stilla lower division undergraduate most of the
subject matter is far abovemy head, so IOm not a regular
(yet!)With that said, I have created some math t-shirt
designs that justmake me smile and IOd like to share them
with the math community, withthe hopes that they delight
someone else as much as they delight me.the url is
http://www.cafeshops.com/subjective and the designs areunder
Math and Science (click on the Einstein picture)All comments,
critiques, commendations, cat-calls and creative
input
=lurking travis> I really hope no one takes this as
IOve posted here
a few, brief times, however since IOm still> a lower
division
undergraduate most of the subject matter is far above> my
head, so IOm not a regular (yet!)...> the url is
http://www.cafeshops.com/subjective and the designs are>
under Math and Science (click on the Einstein
picture)Baseball is ninety percent mental. The other half is
physical.-- Yogi BerraLH
=I am stuck in high school maths
mode, and canOt seem to get into university level maths.
This
might be because I am entirely self taught, but I donOt
know.
Does anyone else have this problem?I am at a level where I
understand most high school maths, and IOve studied Calculus
Made Easy. It would be helpful to have some kind of way to
check my knowledge.Does anyone know of any good textbooks
that cover high school maths with worked exercises, and any
texts that help the transition from high school maths to the
more exciting stuff at university level? A book that takes
time to explain things, point out applications and say why
rather than just how. I am trying to understand maths, not
just learn some techniques or shortcuts.By the way, I would
prefer internet resources and small paperback books. ThereOs
no way I can afford college textbooks
unfortunately.Johnathan
I am stuck in high school maths
mode, and canOt seem to get into> university level maths.
This might be because I am entirely self> taught, but I
donOt
know. Does anyone else have this problem?Most people who went
to high school have this problem. You have mysympathies.> I
am at a level where I understand most high school maths, and
IOve> studied Calculus Made Easy. It would be helpful to
have
some kind of> way to check my knowledge. Does anyone know of
any good textbooks that cover high school maths> with worked
exercises,There are lots of books like that. Several of the
SchaumOs outlines bookscover what is (really) high school
math with lots and lots of workedexamples. The explanations
in most of the SchaumOs books are not half bad aswell.> and
any texts that help the transition from> high school maths to
the more exciting stuff at university level? A> book that
takes time to explain things, point out applications and say>
why rather than just how.The best book I have seen of this
genre you desire is Basic Mathematics bySerge Lang
(Springer-Verlag), which teaches all of secondary
mathematicsfrom a university mathematics viewpoint. It has
answers to selectedexercises, some worked examples, and an
excellent overall perspective on howconcepts 't together. I
am enjoying using LangOs book with my son as hestudies math
at home.Hope this helps! Good luck.
= I am stuck in high
school maths mode, and canOt seem to get into> university
level maths. This might be because I am entirely self>
taught, but I donOt know. Does anyone else have this
problem?
I am at a level where I understand most high school maths, and
IOve> studied Calculus Made Easy. It would be helpful to
have
some kind of> way to check my knowledge. Does anyone know of
any good textbooks that cover high school maths> with worked
exercises, and any texts that help the transition from> high
school maths to the more exciting stuff at university level?
A> book that takes time to explain things, point out
applications and say> why rather than just how. I am trying
to understand maths, not just> learn some techniques or
shortcuts. By the way, I would prefer internet resources and
small paperback> books. ThereOs no way I can afford college
textbooks unfortunately. JohnathanYou could check
out:http://store.doverpublications.com/
by-subject-science-and-mathematics-mathematics-calculus.htmlth
ey are cheap and useful. You could also try SchaumOs
outlines; they arealso very good.For web content,
search, you will 'nd a plethora of math relatedwebsites.
bestfriend. You can 'nd anything you could possibly want to
> I am stuck
in high school maths mode, and canOt seem to get into> >
university level maths. This might be because I am entirely
self> > taught, but I donOt know. Does anyone else have this
problem?> > I am at a level where I understand most high
school maths, and IOve> > studied Calculus Made Easy. It
would be helpful to have some kind of> > way to check my
knowledge.> > Does anyone know of any good textbooks that
cover high school maths> > with worked exercises, and any
texts that help the transition from> > high school maths to
the more exciting stuff at university level? A> > book that
takes time to explain things, point out applications and say why rather than just how. I am trying to understand maths,
not just> > learn some techniques or shortcuts.> > By the
way, I would prefer internet resources and small paperback> >
books. ThereOs no way I can afford college textbooks
unfortunately.> > Johnathan> > You could check out:>
http://store.doverpublications.com/
by-subject-science-and-mathematics-mathematics-calculus.html they are cheap and useful. You could also try SchaumOs
outlines; they are> also very good.> > For web content, try:>
search, you will 'nd a plethora of math related> websites.
best> friend. You can 'nd anything you could possibly want to
include ext:pdf site:.edu in your searches. YouOlljust get
postscript notes, usually written up by
professors.Ocid
=<SNIP> In particular, include ext:pdf
site:.edu in your searches. YouOll> just get postscript
notes, usually written up by professors. OcidThat will
actually give you adobe acrobat 'les. If you want postscript,
tryext:ps site:.edu. Or donOt, unless you know a plug-in for
IE on the PCwhich allows it to read postscript 'les (I
donOt).
<SNIP> > In particular, include ext:pdf
site:.edu in your searches. YouOll> > just get postscript
notes, usually written up by professors.> > Ocid That will
actually give you adobe acrobat 'les. If you want
postscript,try> ext:ps site:.edu. Or donOt, unless you know
a
plug-in for IE on the PC> which allows it to read postscript
'les (I donOt).>Try Ghostscript (Post-script
viewer) for IE
and windows.
=There was a time when people spoke their
minds, and were not afraid to offend -and that since then,
too many truths have been buried.Mark Kurlansky, 1968: The
LauriaIn 1968 was in La Jolla CA at UCSD during time Greg
Benford describes inTimescape living on Bonair Street Wind an
Sea Beach near UnicornTheater with Ken KeseyOs Merry
Prankster
Bus parked nearby frequently.I was also teaching at San Diego
State with Fred Alan Wolf.OK, now I have looked at the rest
of RovelliOs argument, IOll work throughthis
one more
time.PZ: RovelliOs position is very odd. First he says:...Of
course, nothing [in GR] prevents us... from singling out
thegravitational'eld as Othe more equal among
equalsO, and
declaring that location isabsolute inGR, because it can be
de'ned with respect to it. (p 108)JS: But he rejects that
Paul. He says doing that misses the greatEinsteinian
insight.The great Einsteinian insight being the conditional
and physical nature ofthe gravitational metric 'eld -- which
con§icts with the great Einsteinian insightstrict
equivalence.The opposite of a profound truth is another
profound truth. -- Niels BohrWhat Rovelli doesnOt seem to
understand is that this all makes perfect sense onceyou give
up strict equivalence and distinguish the background and
physical metrics.JS: I do not understand this distinction.
Please give more details what you mean.Have you read pp. 112
- 114 that completely demolishes Hal PuthoffO suse ofdr/dt =
cO = c/K radial null geodesicin his Tables.PZ: It does no
such thing. I would not even characterize pp 112-114 as an
argument.It is simply a sketch of a model in which
*everything* is quantized except the rawmanifold.JS: It shows
no intrinsic meaning to PuthoffOs r and t as he means it in
his Tables.PZ: He wants to throw away time in order to keep a
uni'ed g_uv.Why do you think this is an argument against
cO =
c/K? As far as I can see it is simplya different
theory.Rovelli brie§y considers an alternative approach in
which we retain the Minkowskibackground of standard QFT. But
there he hasg_uv = n_uv + §uctuationswhich makes no sense to
me. He does not seem to understand the distinction
betweenkinematical g_uv and dynamic gravitational g_uv. He
cannot get his head over the uni'edmetric.What does he mean
by §uctuations?JS: What do you mean by kinematical g_uv and
dynamic gravitational g_uv apart from Ruvwl = 0 in the former
and not in the latter.JZ: Sounds like sheer nonsense to me.
Nonsense because it is divorced from the
physicalfundamentals.JS: No argument from me on that one.PZ:
This of course is exactly what the classic
Einsteinchronogeometric model does,going all the way back to
special relativity.JS: I think you are misreading Rovelli.PZ:
The Einsteinian model is a chronogeometric model, in which the
metric g_uv re§ects thefundamental structure of spacetime.JS:
If you mean, for example,dT = goo^1/2dtdR = grr^1/2drThen I
agree that dT and dR are physical and dt and dr are
not.However, to get a gravity shift of light frequency
dodTO/gooO^1/2 =
dT/goo^1O2treating dt as a kind of nonlocal
invariant.If you mean more than this, then explain with
detailed examples.PZ: That is the great Einsteinian insight
-- which is.unfortunately, based on strict Einstein
equivalence, which is 'ctitious.JS: Again I really do not
understand what you mean by this sentence.PZ: Now Rovelli
wants to pretend that the great Einsteinian insight is
something else entirely-- that the gravitational 'eld is a
physical 'eld.Loony tunes. He is tying himself up in
knots.JS: Yes, in global specialrelativity, NO in local
general relativity.PZ: In Einstein general relativity, the
uni'ed metric g_uv represents the fundamental structureof
spacetime, and in inseparable from it. ThatOs Einstein.It
doesnOt even make sense to me to say that this
chronogeometric model holds only locally.PZ: That is
precisely what distinguishes theEinsteinian from the
Lorentzian model. The transformational andmetric structureis
not, in the classic Einsteinian model, separable from
spacetimeitself.But then he says:There is no absolute
referent of motion in GR; the dynamical 'eldsmove withrespect
to each other. (p 108)JS: Again you are misreading. There is
no contradiction here inRovelliOs argument.PZ: This is like
arguing that everything is relative because everything is
relative to theabsolute, including the absolute.Sounds like
Lewis Carroll.JS: ;-)PZ: None of this has anything to do with
Einstein general relativity, which treats theinertial 'eld as
real. The uni'ed metric derives its physical meaning from
thisequivalence.Rovelli ignores all this and simply takes the
uni'ed Einstein g_uv as given.PZ: This second assertion seems
to approach the metric 'eld of GR asjust another
'eld,
whichis very close to and even indistinguishable from the
physical rubberrod and clock modelof PV and Yilmaz -- just
another physical 'eld which happens to havemetric
properties.And of course such a 'eld is fully relational with
respect tounindividuated spacetime pointson a raw manifold
stripped of all coordinate systems, transformationproperties,
and metrics.JS: Yes on just another 'eld. But NO that
itOs
like PV and Yilmaz. Nottrue at all because,at least in PV,
Hal uses an absolute non-dynamical background globalMinkowski
spacePZ: That is what Rovelli *should* be doing, but he
doesnOt even consider thispossibility. He seems to think you
can treat uni'ed g_uv as a physical 'eld.JS: Why
do you think
you cannot?PZ: I canOt imagine anything more wrong-headed.
And
you say Rovelli is a bigshot?That is why I say RovelliOs
position is incoherent.JS: Is coherence in the mind of the
beholder?PZ: If you want to treat g_uv as a physical 'eld,
based on its dynamical character(i.e. matter-dependence),
then the natural thing to do is separate the
backgroundgeneralized Minkowski kinematical metric (which is
NOT matter dependent) aschronogeometric, and treat the
gravitational g_uv alone as physical -- byRovelliOs own
argument.JS: That will not work. It will not tip the light
cones. It will not give the correct bending of light.Also you
are too vague on what you mean by matter on the RHS
ofGuv(Einstein) = -(alpha)(alphaO)Tuv(Matter)alpha = e^2/hc
~
1/137alphaO = 8piWittenOs reciprocal string
tension)alphaO ~
(10^-32 cm)^2 in a common convention.I also include on
RHStuv(Vacuum) ~ [(alpha)(alphaO)]^-1/zpfguv/zpf =
Lp^-1[Lp^3|Vacuum Coherence|^2 - 1]Repulsive dark energy is
/zpf > 0Attractive dark matter is /zpf < 0FRW Omega(Dark
Energy + Dark Matter) ~ 0.96With Omega(Total) = 1 i.e. FLAT
SPACEPreferred foliation is where CMB is maximally isotropic
to ~ 10^-5.This allows accurate navigation with weightless
warp drive and traversable wormholes both supported by
con'gurations of dark energy and dark matter exotic w = -1
vacua.PZ: Yet he doesnOt even mention this.JS: There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
WorldPZ: So he has not made any argument at all against a
bimetric approach. He has simplyignored it, even though his
own argument points to it.JS: with a literal meaning for
coordinates r & t, otherwise his formuladr/dt = c/Kis
physically meaningless, which it is in GR as explained by
Rovelli onpp. 112 - 114.PZ: Only if you insist exclusively on
operational meaning -- which would also knock outF = ma, which
by a similar argument is empirically meaningless.JS: Yes,
until you also add things likeF = - GMmr/r^3F = eE +
e(v/c)xBetc.That is, HalOs use of engineering is precisely
missing the thirdstep on RovelliOs p. 108.PZ:
RovelliOs third
step is merely a sketch of a proposal to quantize everything,
whileretaining a uni'ed g_uv treated in its entirety as a
physical 'eld.This is not an argument.RovelliOs
argument for
his proposal, such as it is, is to be found on p 109:In my
*opinion*, this is the right way to go.Well, in my *opinion*,
itOs fundamentally wrong-headed. In a
7ï
WEcf
!.85<Pèÿÿ
is contained in S_kfor all k > N), and if lim S_n is
countably in[Caply r1).Same thing with r2 and 2, and r3 and 2.The 3 is a
typo.>Actually IOm glad you presented it that way Magidin as
it shows how>youOve fooled so many by getting them to take 
an
illogical step.What step is illogical and y?>Here it IS true
that x and y are NOT coprime if there exists a>non-unit
factor common to both x and y, but the converse, is NOT
true>in the ring of algebraic integers as it is screwy.Point
the zeroth. so in fact you are contesting point 15, not
pointreally read more carefully. Point 15 was:>> 15. In the
ring of all algebraic integers, x and y are not coprime if>>
and only if there exists a non-unit common factor of x and y
(in>> the ring of all algebraic integers).So you are saying
that this is not true. It is not if and only if,you claim
that only only if holds. Fine.Replace 15 with:15O. In the
ring of all algebraic integers, if there exists a non-unit
common factor of x and y (in the ring of all algebraic
integers) then x and y are not coprime.That one you claim to
agree to.Point the 'rst: I never used the converse: there
exists a non-unitfactor common to both r1 and 2 (point 10),
and so I conclude that r1and 2 are not coprime. There exists
a non-unit factor common to bothr2 and 2 (point 12), so I
conclude that r2 and 2 are notcoprime. There exists a
non-unit factor common to both r3 and 2 (point14), so I
conclude that r3 and 2 are not coprime.Where did I supposedly
use the converse? The converse would be If r1and 2 are not
coprime, then there exists a non-unit factor common toboth r1
and 2. I never invoked that.Point the second: Actually, the
converse IS true in the ring ofalgebraic integers. I already
gave you a
40agate.berkeley.eduDedekindOs Theorem on the 
'niteness of
the class number.>You have a gap.Nope. I never used the
converse.>For those readers confused consider that in the
ring of *evens* 2 and>6 donOt share non-unit factors, but by
MagidinOs de'nition they are>NOT coprime.This 
is a red
herring. We are working in the ring of all algebraicintegers,
and I did NOT use the converse of the statement, I only
usedthe statement in its direct form.You have not correctly
pointed any gap. You INSINUATE that I use theconverse of a
result, when I used the result itself.Please read again
points 15 and 16 carefully. Replace point 15 bypoint 15O, 
and
you will see that point 16 follows from using ONLYpoint 15O,
not the
converse.
=
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A manOs capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
'gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde'nite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
Arturo Magidinmagidin@math.berkeley.edu>>Oh yeah, youOre right. That should be x^3 + 3x -
2.>>Okay, this one is irreducible over Q: if it were
reducible it would>>have a root, and the only possible
rational roots are 1, -1, 2, and>>-2. Checking each shows
that it does not have rational roots. > > > Well here we go
again. I noticed a poster trying to make a big deal> about
the lateness of my reply but hey, I reply through Google and
it> takes it a while to put up messages.> > The math proving
my point is basically high school stuff, so itOs> easy, 
which
makes it fascinating to me that so many of you *trust*>
Magidin. Now I understand, I think, why he keeps lying, as
what> choice does he have?> > But why do any of you believe
him and his voodoo math?> Why do you continually claim that
people are lying? Is your worldso delicate that there is no
possibility that you are in error?I, for one, 'nd 
MagidinOs
arguments to be carefully formulated,with none of the sham
informality that your arguments cloak themselvesin, and when
I study them carefully, I 'nd them to be correct. Thetimes
IOve done the same with regard to your arguments, I 
'nd
themfestooned with holes, in fact I 'nd them comprising more
hole thansubstance.> >>So, do you agree or disagree with each
of the following?>>1. If r1, r2, r3 are the three roots of
x^3+3x-2, then >> x^3+3x-2 = (x-r1)(x-r2)(x-r3).> > > Agree. >>2. Therefore, r1*r2*r3*(-1) = 2.> > > Agree.> > >>3.
Therefore, r1 is a factor of 2 in the ring of all algebraic>>
integers, since (-1)*r2*r3 is an algebraic integer, and>> r1*
( (-1)*r2*r3 ) = 2.> > > Trivially, yes, so I agree.> > >>4.
Likewise, r2 is a factor of 2 in the ring of all algebraic>>
integers, since (-1)*r1*r3 is an algebraic integer, and>> r2*
( (-1)*r1*r3 ) = 2.> > > Again, trivially, yes, so I agree.> >5. Likewise, r3 is a factor of 2 in the ring of all
algebraic>> integers, since (-1)*r2*r3 is an algebraic
integer, and>> r3* ( (-1)*r2*r3 ) = 2.> > > Yet again,
trivially, yes, so I agree.> > >>6. In the ring of all
algebraic integers, if x and y are algebraic>> integers, then
x is a common factor of x and x*y.> > > Agree.> > >>7. In the
ring of all algebraic integers, if x and y are algebraic>>
integers and x is not an algebraic integer unit, then x is
a>> non-unit common factor of x and x*y.> > > > Hmmm...what
does being a unit have to do with anything?> > If I have xy =
2, where y=2, isnOt x STILL a factor of 2?> > Then again,
yeah, sounds ok to me.> > Agree.> > >>8. r1 is a common
factor of r1 and 2; r2 is a common factor of r2 and>> 2; r3
is a common factor of r3 and 2.> > > Agree.> > >>9. r1 is not
a unit in the ring of algebraic integers, because 1/r1>>
satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>>
irreducible, primitive polynomial with integer coef'cients,
and>> therefore its roots are not algebraic integers.> > >
Agree.> > >> and 9).> > > Agree.> > >>11. r2 is not a unit in
the ring of algebraic integers. Because 1/r2>> satis'es the
polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible,
primitive polynomial with integer coef'cients, and>>
therefore its roots are not algebraic integers.> > > Agree. >> and 11).> > > Agree.> > >>13. r3 is not a unit in the
ring of algebraic integers. Because 1/r3>> satis'es the
polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible,
primitive polynomial with integer coef'cients, and>>
therefore its roots are not algebraic integers.> > > Agree. >> and 13).> > > Agree.> > >>15. In the ring of all
algebraic integers, x and y are not coprime if>> and only if
there exists a non-unit common factor of x and y (in>> the
ring of all algebraic integers).>>[This is what you use as
a de'nition; it is equivalent to the>>standard 
de'nition for
the ring of all algebraic integers, so it does>>not matter in
->that<- context]> > > Agree.> > >>16. r1 is not coprime to 2.
r2 is not coprime to 2. r3 is not coprime> > > Disagree.> >
Actually IOm glad you presented it that way Magidin as it
shows how> youOve fooled so many by getting them to take an
illogical step.> > Here it IS true that x and y are NOT
coprime if there exists a> non-unit factor common to both x
and y, but the converse, is NOT true> in the ring of
algebraic integers as it is screwy.> You state: x and y are
NOT coprime if there exists a non-unit factor common to x and
yMagidin has shown, and you have agreed, to the following: r1
is a non-unit factor common to r1 and 2. (step 10) r2 is a
non-unit factor common to r2 and 2. (step 12) r3 is a
non-unit factor common to r3 and 2. (step 14).Yet, in step
16, modulo a typo (Magidin surely intended to typer3 is not
coprime to 2), you disagree with the combined statement.Or,
perhaps you didnOt read his statement as containing the 
clear
typo-graphical error (typing 2 when he clearly [referencing
step 14] meantto type 3).> You have a gap.> > For those
readers confused consider that in the ring of *evens* 2 and>
6 donOt share non-unit factors, but by 
MagidinOs de'nition
they are> NOT coprime.> No. Magidin uses the de'nition that r
and s are coprime in the ring Rif the minimal ideal in R
containing r and s is R itself. If the ringcontains a
multiplicative identity, this is equivalent to the
existenceof members m and n for which mr + ns = 1.Given the
fact that 2Z doesnOt have a multiplicative identity, it
isnecessary to deal with the previous form of the de'nition
(involvingideals, rather than a decomposition of 1 into the
prospective coprimeelements).Given that de'nition, and the
ring 2Z, the members 2 and 6 arecoprime, since the minimal
ideal containing 2 and 6 is 2Z. How do wesee this? Let I be
an ideal (an additive subgroup of the ring, which is closed
under multiplication by ring elements) of 2Z, containing both
2 and 6. Since I is closed under multiplication by elements of
2Z, and 2 is already in I, then -2*2 = -4 is also in I. Since
6 is also in I, and since I is an additive subgroup of 2Z, we
then know that -4 + 6 = 2 is in I (I know, I already knew
that, but we continue). Multiplying by ring elements, we
obtain all other elements of 2Z. Actually, the minimal ideal
of 2Z containing 2 is already 2Z. Thus, 2 is coprime to every
other element, including 6, of the ring 2Z.> Why?> > Because 3
is not even, but 2(3) = 6, so by MagidinOs 
de'nition of>
coprime they are NOT coprime. Now the problem with algebraic
integers> is more complicated, but that example should help
you to see that> Magidin is lying to you.> No, youOre
mis-representing his position. Part of honest
argumentationinvolves attempting to attack the position that
the person *actually*holds, rather than purposely misreading
it and attacking that mistakenposition.> No, you only produce
gaps by taking advantage of typographical orother
inconsequential errors. When you attempt to address his
*actual*arguments, you consistently fail.> Further, since his
position has been repudiated by Professor McKenzie> who shot
down his key objection he is against the discipline of>
mathematics as are his supporters.> Hey, if you want to keep
holding Professor McKenzie up as a person whosupports your
claim, you should at least offer Professor McKenzie
thecourtesy of speaking for himself. It is a sign of profound
disrespectto this person who has befriended you, without any
apparent recompenseto himself, to act otherwise.> Again see
http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > >
James HarrisDale.
=Some corrections: [.snip.]>No. Magidin
uses the de'nition that r and s are coprime in the ring R>if
the minimal ideal in R containing r and s is R itself.This is
not the de'nition I was using, either. That would becomaximal
in a ring with 1, and in rings without 1 (or in theabsence of
the Axiom of Choice) it need not be equivalent to 
coprime.IOll
give the de'nitions in a second, but let me just point out
thatJamesOs objection is moot. I used only HIS 
de'nition in
theargument I gave (which is really just copying some results
that you,Dale, had gotten and posted, with a few more
multiplications ofpolynomials made explicitly.Anyway, the
de'nitions:Let R be a ring, which may or may not have a
multiplicativeidentity. An ideal I of R is a prime ideal if
and only if it is notequal to R, and whenever x and y are
elements of R such thatx*y is in I, then either x is in I or
y is in I.An ideal J of R is a maximal ideal if and only if
it is an ideal,properly contained in R, and there does not
exist any ideal I whichproperly contains J and is properly
contained in R.Two ideals A and B are coprime if and only if
there does not exist aprime ideal containing both. Two
elements r and s of R are coprime if and only if the
principalideals they generate, (r) and (s), are coprime.Two
ideals A and B are coprime if and only if there does not
exist amaximal ideal containing both.Two elements r and s of
R are comaximal if and only if the principalideals they
generate, (r) and (s), are comaximal. For rings with 1,this
is equivalent to the minimal ideal in R that contains r and
sbeing R itself.If the ring is commutative and has a 1, then
every maximal idealis prime. So coprime -> comaximal.If the
ring has a 1, and you assume the Axiom of Choice
(ZornOsLemma), then every proper ideal is contained in a
maximal ideal, socomaximal->coprime.So, if your ring is
commutative, has a 1, and you assume the Axiom ofChoice, then
coprime is equivalent to comaximal. Your de'nition isthe
de'nition of comaximal.> If the ring>contains a
multiplicative identity, this is equivalent to the
existence>of members m and n for which>> mr + ns = 1.This is
true for comaximal.>Given the fact that 2Z doesnOt have a
multiplicative identity, it is>necessary to deal with the
previous form of the de'nition (involving>ideals, rather than
a decomposition of 1 into the prospective
coprime>elements).>>Given that de'nition, and the ring 2Z,
the members 2 and 6 are>coprime, since the minimal ideal
containing 2 and 6 is 2Z. How do we>see this?>> Let I be an
ideal (an additive subgroup of the ring, which is> closed
under multiplication by ring elements) of 2Z, containing>
both 2 and 6.>> Since I is closed under multiplication by
elements of 2Z, and> 2 is already in I, then -2*2 = -4 is
also in I. Since 6 is> also in I, and since I is an additive
subgroup of 2Z, we then> know that -4 + 6 = 2 is in I (I
know, I already knew that, but> we continue). Multiplying by
ring elements, we obtain all other> elements of 2Z.Ehr, no.
Multiplying by ring elements once you have 2 gives you allthe
multiples of 4. However, an ideal is also closed under
differences,since you have 2 you have 0=2-2; and you have -4
so you have4=0-(-4); and you have 6 so you have 4-6 = -2.It
is also closed under sums, so you have 2+2, 2+2+2,
2+2+2+2,....,,so you get all positive multiples of 2; and you
also have -2, -2+(-2),-2+(-2)+(-2),.... so you get all
negative multiples of 2, so you getall of 2Z.> Actually, the
minimal ideal of 2Z containing> 2 is already 2Z.Yes. The
minimal ideal containing 2n is {2n*k : k an integer}. And
theideal generated by 2n is {4n*k: k an integer}.Using the
de'nition on principal ideals, note that (2)={4k: k is
aninteger} and (6) = {12k: k is an integer}. The former
contains thelatter, but (2) is not prime: 2*2 is in (2), but
2 is not. Therefore,if there were a prime ideal containing
both, then it would properlycontain {4k: k an integer}. That
means that there is an even integer awith a = 2 (mod 4) in
the ideal. But that means that a-2=4k, so a-4k =2; therefore,
the ideal contains 2, and as you show above, that meansthat
the ideal is all of 2Z, and hence is not prime. Thus, (2) and
(6)are coprime, so 2 and 6 are coprime.They are not, however,
comaximal: (2) = {4k: k is an integer} is amaximal ideal, as
just shown above. And it contains (6); so (2) and(6) are both
contained in the maximal ideal (2), hence they are
notcomaximal.James is using the de'nition a and b are coprime
in R if and only ifevery common divisor of a and b in R is a
unit. For this de'nitionto even ->begin<- to make sense, you
need to mm-132
=I have the following eigenvalue problem:Let A and B
be symmetric positive semi de'nite matrices.The underlying
graphs of A and B are trees. (Actually the
diagonalelementsare positive and the element A_ij= 0 (B_ij=
0) if vertices i and j arenotadjacent. A_ij= -1 (B_ij= -1) if
the vertices i and j are adjacent)Let a, b be the smallest
eigenvalues of the matrices A and B,respectively. a and b are
simple eigenvalues and the correspondingeigenvectors x and y
are positive (Perron-Frobenius Theorem).Ax=axBy=by,We know
that 0 < a <= b, y^tAy=b and x^tBx>a.For which matrices A and
B hold that a<b (a=b)?Any hints or references?Tuerker
Biyikoglu
=What I am looking for is an iterative algorithm, like
:a[n+1]=f(a[n],...,b[n],...,c[n],...);b[n+1]=g(a[n],...,b[n],
...,c[n],...);c[n+1]=h(a[n],...,b[n],...,c[n],...);...such
thatF(a[k],b[k],c[k],...) for some (not large) k is
EllipticE(z,k).Chris
= Chris> What I am looking for is an
iterative algorithm, like : Chris>
a[n+1]=f(a[n],...,b[n],...,c[n],...); Chris>
b[n+1]=g(a[n],...,b[n],...,c[n],...); Chris>
c[n+1]=h(a[n],...,b[n],...,c[n],...); Chris> ... Chris> such
that Chris> F(a[k],b[k],c[k],...) for some (not large) k is
EllipticE(z,k).Look in netlib for CarlsonÍs elliptic
integrals. There should be adescription of such an algorithm
for Elliptic E.Ray
=Correction : What I am looking for is
an iterative algorithm, like
:a[n+1]=f(a[n],...,b[n],...,c[n],...);b[n+1]=g(a[n],...,b[n],
...,c[n],...);c[n+1]=h(a[n],...,b[n],...,c[n],...);...such
thatF(a[m],b[m],c[m],...) for some (not large) m is
EllipticE(z,k).Chris
=Given a matrix ïAÍ, IÍm looking 
to
'nd (numerically) a matrixÍBÍ 
such that ïA * B = 1Í, where
ï1Í is the appropriate unitmatrix.I had expected 

that one
could treat the equation as a set o§inear equations ïA 
* Bj 
=
EjÍ. According to Numerical Recipes,if one has such a system
of equations one can obtain the SVD ofthe matrix 
ïAÍ: A = U *
W * Vtwhere ïVtÍ is the tranpose of 
ïVÍ. Then the smallest
solutionis: Bj = V * WÍ * Ut * Ej(where the entries in 
WÍ are
either the inverse of thecorresponding entries in W, or zero
if those entries are small).It seems clear to me (which
doesnÍt mean that itÍs true) thatthis means 
that: B = V * WÍ
* Utis a good choice. In practice, however, where 
ïAÍ has
morecolumns than rows (ie, is singular) IÍm 
'nding that ïA *
BÍ isnot the unit matrix, though itÍs close on 

some rows and
columns.This makes no sense to me. IÍve 
con'rmed that the
decompositionI get does, in fact, yield the matrix 
ïAÍ.Is
there a better way to go about this? Or should this work
(inwhich case I should try another SVD algorithm or debug the
oneIÍve
got)?--
 . . . Except when they donÍt, Because sometimes
they wonÍt. - Dr.
Seuss--
Jason Cooper jcooper@acs.ucalgary.ca
= >Given a
matrix ïAÍ, IÍm looking to 
'nd (numerically) a matrix >ÍBÍ
such that ïA * B = 1Í, where 
ï1Í is the appropriate unit
>matrix. > >I had expected that one could treat the equation
as a set of >linear equations ïA * Bj = EjÍ. 
According to
Numerical Recipes, >if one has such a system of equations one
can obtain the SVD of >the matrix ïAÍ: > > A = U 

* W * Vt where ïVtÍ is the tranpose of 
ïVÍ. Then the smallest
solution >is: > > Bj = V * WÍ * Ut * Ej > >(where the 
entries
in WÍ are either the inverse of the >corresponding entries 
in
W, or zero if those entries are small). >It seems clear to me
(which doesnÍt mean that itÍs true) that >this 

means that: > >
B = V * WÍ * Ut > >is a good choice. In practice, however,
where ïAÍ has more >columns than rows (ie, is 
singular) IÍm
'nding that ïA * BÍ is >not the 
unit matrix, though itÍs
close on some rows and columns. >This makes no sense to me.
IÍve con'rmed that the decomposition >I get 
does, in fact,
yield the matrix ïAÍ. > >Is there a better way 
to go about
this? Or should this work (in >which case I should try
another SVD algorithm or debug the one >IÍve got)? 
-- > . . . Except when they donÍt, > Because sometimes
they wonÍt. - Dr. Seuss
>
-- >Jason Cooper jcooper@acs.ucalgary.ca if A=U*W*Vt, then
W has the same shape as A, hence in your case more columns
than rows.then , if A is of full rank, W will consist of an
invertiblediagonal matrix with a zero block appended to
right.then B = V*W#*Ut where W# has the same shape as AÍ and
consists of the inverted diagonal on the top and a zero block
appended at the bottom. then by the rules obviously A*B=I
where I has dimension=number of rwos of A.If A is not of full
rank, you will get a projector (some diagonalelements in I
change to zero) and in W# the reciprocal of zero is set to
zero. (B is the Moore-Penrose-pseudoinverse). hence you were
right.the problem with this approach is the decision which of
the singularvalues (the nonzero elements of W) to consider as
small. and if you set anything below eps to zero the
deviation of the product from I can be large ifyour decision
on the rank of A (implicit with this) was false.hence I
assume the problem here and not in the software.hope this
helpspeter
=This is a question on the relationship between,
the zero and non-zeroelements of the transition probability
matrix and the correspondingtransient and absorbing
states.Let T_a and T_b be the transition probability matrices
for two 'nite statediscrete Markov chains A and B,
respectively.We are given thata) i(T_a) = i(T_b), where i(X)
is the indicator matrix of Xi.e., i(X) has 1.0 entries
wherever X has non-zero entries and has 0.0entries wherever X
has zero entries.b) T_a^n converges as n -> in'nity. Let the
limiting matrix be Ta^{inf}Now, I think I can show that T_b^n
also converges. However, I am trying to'nd out if i(Ta^inf) =
i(Tb^inf).I have gotten this far:Since i(T_a) = i(T_b), we
have i(T_a^2) = i(T_b^2) and so on ...I can show that for any
'nite k, i(T_a^k) = i(T_b^k),but does this property hold in
the limit?In other words, if i(T_a) = i(T_b), then do A and B
have the same transientand absorbing states?Kumar
went looking for a short proof of FermatÍs Last Theorem 
using
basic> algebra following my physics training. I expected
failures, false> starts, false positives, but mathematicians
apparently arenÍt> scientists, as they attack me for having
them.Wrong! They refuted your arguments. You responded to
discovery of your errors by attacking them.> Worse, theyÍd
use my own admissions of §awed approaches against me,> as if
that proved I couldnÍt succeed. And worst of all, when I 
did>
succeed they kept up a campaign of personal attacks and
smears.Succeed? At what? The only success you have had to
date is in establishing that you are a crank and mostof the
personal attacks and smears posted in this newsgroup were
authored by you.> And itÍs not just with the polynomial
factorization as thereÍs my> prime counting research as
well.>> ItÍs too much to be believable given what you can 
see
with your own> eyes that none of my work is worth mention,
when people can even see> the heat generated on several
newsgroups. These people are expending> a LOT of energy, and
you think itÍs for nothing?The energy others are expending
has been to keep the record straight. Your lies and
distortions deservechallenges, and will continue to face
them.> My work is available 24 hours a day, seven days a week
at>> http://groups.msn.com/AmateurMath>> and if it were really
so wrong, why do these people keep 'ghting?Obviously, as
previously noted, to keep the record straight. DonÍt you
expect scientists andmathematicians to resist error? Take
down that ridiculous treatise. It isnÍt 't to 
wrap the
garbage in.> What would happen if mathematicians even
acknowledged the> *possibility* that IÍm right? They know.
Given people following and> keeping up with the details itÍs
clear IÍm right.Clear to whom? You are wrong. Many peoply
follow your posts for comic relief. Personally, I liken
youto:1) Monty PythonÍs Black Knight -- who, having been cut
to pieces, insists ItÍs only a §esh wound!2) An 
insect in 
a
collectorÍs jar, waving his ineffectual pinchers at the
world,3) A stupidly grinning in§atable clown who keeps
popping up whenever he is knocked down.> Scientists behave a
certain way. Mathematicians are behaving another> way....and
you are behaving like an insolent, arrogant and paranoid
megalomaniac.> The proof is out there.>> James HarrisYeah,
but not on your website. Let us know when you get back from
the Twilight Zone with your latestX-File.--It takes a village
to raise an idiot.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
=IÍm dealing with a
problem of nonlinear optimization with boundconstraints. I
canÍt use commercial software for this. So farL-BFGS-B from
Netlib has produced satisfactory results. Now Ineed to know
the value of the Hessian matrix at the end of thecomputation.
L-BFGS-B doesnÍt seem to have this information.(The L stands
for Limited Memory, meaning that it stores onlyupdates to the
Hessian.)Currently IÍm looking at DRMNGB, also from Netlib. 
It
seemsas if it stores the Cholesky decomposition of the
Hessian.However, IÍm having trouble 'guring out 
how to 'nd it
andtransform it back into the full Hessian matrix. On my
'rstattempt the numerical values donÍt seem 
right. Some
thingswhich should be positive are not. As far as I can
tell,it should be stored beginning at V(IV(42)), but itÍs
notclear in what order the terms are stored.Does anyone know
about this? Or do you have another favoritesubroutine?-- *
Patrick L. Nolan ** W. W. Hansen Experimental Physics
Laboratory (HEPL) * * Stanford University *
= >IÍm dealing
with a problem of nonlinear optimization with bound
>constraints. I canÍt use commercial software for this. So
far >L-BFGS-B from Netlib has produced satisfactory results.
Now I >need to know the value of the Hessian matrix at the
end of the >computation. L-BFGS-B doesnÍt seem to have this
information. >(The L stands for Limited Memory, meaning that
it stores only >updates to the Hessian.) > >Currently IÍm
looking at DRMNGB, also from Netlib. It seems > as if it
stores the Cholesky decomposition of the Hessian. >However,
IÍm having trouble 'guring out how to 
'nd it and >transform
it back into the full Hessian matrix. On my 'rst >attempt the
numerical values donÍt seem right. Some things >which should
be positive are not. As far as I can tell, >it should be
stored beginning at V(IV(42)), but itÍs not >clear in what
order the terms are stored. > >Does anyone know about this?
Or do you have another favorite >subroutine? > >-- >* Patrick
L. Nolan * >* W. W. Hansen Experimental Physics Laboratory
(HEPL) * >* Stanford University *drmngb is a bfgs based code
and will hardly deliver a good approximation to the true
Hessian, although as a minimizer it is o.k.why not this one:
toms/636 keywords: estimating sparse hessian matrices,
difference of gradients gams: G4f title: DSSM and FDHS for:
estimating sparse Hessian matrices by: T.F. Coleman, B.S.
Garbow, and J.J. More ref: ACM TOMS 11 (1985) 363-377 and 378
Score: 100%hope that helpspeter 
=are you saying you are
feeding the optimizer at best 'rst derivativesand want to get
out second derivatives? One way to do it is toevaluate your
own with the 'nal solution vector. Any Hessian of
theoptimizer will be an approximation at best ('nite
difference,quasi-Newton). Only exact or automatic
differentiation can give youthe Hessian with suf'cient
acuracy.Hans
Mittelmann
==> IÍm dealing with a problem of
nonlinear optimization with bound> constraints. I canÍt use
commercial software for this. So far> L-BFGS-B from Netlib
has produced satisfactory results. Now I> need to know the
value of the Hessian matrix at the end of the> computation.
L-BFGS-B doesnÍt seem to have this information.> (The L
stands for Limited Memory, meaning that it stores only>
updates to the Hessian.)> > Currently IÍm looking at DRMNGB,
also from Netlib. It seems> as if it stores the Cholesky
decomposition of the Hessian.> However, IÍm having trouble
'guring out how to 'nd it and> transform it back 
into the
full Hessian matrix. On my 'rst> attempt the numerical values
donÍt seem right. Some things> which should be positive are
not. As far as I can tell,> it should be stored beginning at
V(IV(42)), but itÍs not> clear in what order the terms are
stored.> > Does anyone know about this? Or do you have
another favorite> subroutine?
=>Try the family:> (x/a)^n +
(y/b)^n = 1>with n > 2 . Of course, n=2 is an ellipse
centered at the origin.I forgot to mention one important
thing. I need an ellipse-like curvethat is not symmetric in
the way an ellipse (or the curve above) is.
=>>Try the
family:> (x/a)^n + (y/b)^n = 1>with n > 2 . Of course,
n=2 is an ellipse centered at the origin.>>I forgot to
mention one important thing. I need an ellipse-like
curve>that is not symmetric in the way an ellipse (or the
curve above) is.Like a longitudinal cut of an egg, or an
etc,likehttp://www.mathematische-basteleien.de/eggcurves.htm=
IÍve done my part with my mathematical research by working
hard toexplain it, arguing with people over details, writing a
paper andsending it to math journals. And now I realize that
part of theproblem is that the math is revolutionary.I use
methods to factor polynomials into non-polynomial factors,
whichis such a powerful approach that it leads to a short
proof of FermatÍsLast Theorem that is just a few pages.The
math is mostly basic algebra.IÍve given the information. 
IÍve
explained in detail butmathematicians can resist for a while,
just as they resisted otherrevolutionary mathematics, but 
IÍm
hopeful that here at least *some*of you recognize that itÍs
not to the bene't of math society to actin that way.What does
it take to consider the mathematics?Reviewing the math at my
website, or considering the many posts whereI talk about the
factorization (v^3+1)x^3 - 3vx + 1 = (a_1 x + 1)(a_2 x +
1)(a_3 x + 1)or variations on it.Now you can actually look at
the factorization for v=1, and see thatas IÍve stated for 
some
time, ONLY TWO of the aÍs can have a factor of2, and that
factor is sqrt(2).The polynomial of course is 2x^3 - 3x + 1,
and itÍs easily factorable.My work gave a 
speci'c prediction.
I can show you where thatprediction is the reality with a
reducible polynomial. The methodsthemselves DO NOT CARE about
reducibility, so they work when v givespolynomials not
reducible over Q.many of you realize that if mathematicians
refuse to do their part,then my task is dif'cult, even though
IÍm right.Factoring polynomials in a special way such that
even the FLT equationitself can be factored, as weird as that
may sound, is a great ideawhose time has come. Your task now
is simply to accept the future,just as those before you
accepted imaginary numbers, and thosebefore them accepted
irrational numbers.You may be puzzled. You may not understand
why thereÍs so much drama. But then now maybe you can
understand other math history a littlebetter. People like
comfort. Thinking you know all you need to knowin certain
areas is comforting. And then some guy comes along andstarts
a revolution.I am that guy.James Harris
=LOL.>
LOL.Yeah, it may be funny to you, but what IÍm talking about
is a modernsituation where a mathematical argument given is
simply ignored orlied about by mathematicians.2x^3 - 3x + 1 =
(x-1)(2x^2 +2x -1) =(a_1 x + 1)(a_2 x + 1)(a_3 x + 1)My work
has speci'c predictions, which are the mathematical
reality.Mathematicians 'ghting work that is simple,
demonstrable, and correctcanÍt be good for the 
discipline.And
itÍs not funny.James Harris
work that is simple, demonstrable, and correct> canÍt be 
good
for the discipline.>> And itÍs not funny.>> James HarrisNo 
one
is laughing at Mathematicians 'ghting work that is simple
demonstrable, and correct. Thesubject of discussion is your
error-ridden, false, oft-refuted, proof of FLT. They are
laughing at*you*, idiot.--It takes a village to raise an
idiot.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
Mathematicians 'ghting work that is simple, demonstrable, and
correct> canÍt be good for the discipline.>> And 
itÍs not
funny.>> James Harris> > No one is laughing at Mathematicians
'ghting work that is simple demonstrable, and correct. The>
subject of discussion is your error-ridden, false,
oft-refuted, proof of FLT. They are laughing at> *you*,
idiot.Is that the math world? Is that what all of you see
when you think ifmathematicians? People calling others
idiot?IÍm on whatÍs supposed to be *your* 
turf. Basic algebra
and astep-by-step argument which proves what IÍm
saying.Ultimately people attacking me are attacking
mathematics itself.Now think about it. ThereÍs my prime
counting work, then thereÍs mypaper, and also not least
thereÍs the actual short proof of FermatÍsLast 

Theorem.If you
believe people arguing with me thereÍs only ONE small
thingthat they 'ght.Think about it. Basic algebra proves my
point. The argument isstraightforward and simple, but the
consequences are huge.ItÍs revolutionary for *social*
reasons. And you can see society,your society, 'ghting as it
has before. Learn the lessons from thebattles over irrational
and imaginary, and try to let the math beyour guide--not your
comfort zone.The work itself is conveniently available to you
24 hours a day, sevendays a week at
http://groups.msn.com/AmateurMathand hey, you can even play
with an applet (if you join the group).James
Harris
=documentstyle{report}begin{document}pagestyle{plain
}raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the following integral:$$ int_{-half a}^{+half
a} x^2 , dx / a = left[ frac{1}{3} x^3 right]_{-half
a}^{+half a} / a = frac{2.a^3}{3.8.a} = frac{a^2}{12}$$Now
consider:$$ lim_{sigma rightarrow infty} frac{ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }$$We
would expect that this more dif'cult integral nevertheless
convergesto the same result as before: $a^2 / 12$ ,
because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividing the integralin the denominator
by $a$ should be equivalent to differentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
'nding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ... ;$ in
theirdenominators, so they should converge to zero. Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they do not match. WhatÍs going wrong ? 
Please
help !%end{document}-- * Han de Bruijn; Systems for Research A
little bit of Physics would be (
)* and Education (DTO/SOO),
Mekelweg 6 NO Idleness in Mathematics(HdB) @-O^O-@* 2628 CD
Delft, The Netherlands http://huizen.dto.tudelft.nl/deBruijn
#/_#
=>....................> So the two outcomes have the
same order of magnitude, but they do not match. > WhatÍs
going wrong ? Please help !> %following .tex
'le,
documentstyle[leqno]{report}newcommand{hsp}{hspace*{0.5cm}}
newcommand{Erf}{{rm Erf}}begin{document}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}%subsection*{Should be the same but alas
...}%Consider the problem to
determinebegin{equation}label{eu11} fbox{$displaystyle
L(a):=lim_{sigma rightarrowinfty}I(a,sigma)hsp mbox{rm with}
hsp I(a,sigma)= frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } $}end{equation}However:$$ int_{-half
a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2
int_{-half a}^{+half a} x , e^{-half x^2 /sigma^2} ,
dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half
a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = -
sigma^2 left[ x.e^{-half x^2 / sigma^2}
right]_{-halfa}^{+half a} + sigma^2 int_{-half a}^{+half a}
e^{-half x^2 / sigma^2} ,dx$$ $$ hieruit hsp I(a,sigma)=frac{
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } =
sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } {
int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}right)$$Let us de'ne $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,;
h to0; .; $ Then problem (ref{eu11}) is equivalent
withbegin{equation}label{eu12}fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto0}frac{1}{h^2}left(1-frac{he^{
-h^2}}{Erf(h)}right) hspmbox{rm where}hsp
Erf(h):=intlimits_{0}^h e^{-t^2}; dt$}hsp .end{equation}But
$displaystyle
Erf(h)=he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $
where[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k
}cdotfrac{z^k}{k!}hsp,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]Therefore
[L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}{h^
2cdotsumlimits_{k=0}^infty
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=0}^inftyfrac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k
}}{cdotsumlimits_{k=0}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}=frac{
a^2}{8}cdotfrac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12};
.]%end{document}
=> >....................> So the two
outcomes have the same order of magnitude, but they do not
match.> WhatÍs going wrong ? Please help !> %> following 
.tex
'le,> 
documentstyle[leqno]{report}>
newcommand{hsp}{hspace*{0.5cm}}> newcommand{Erf}{{rm Erf}}>
begin{document}> pagestyle{plain} raggedbottom> def half
{frac{1}{2}}> def hieruit {quad Longrightarrow quad}> %>
subsection*{Should be the same but alas ...}> %> Consider the
problem to determine> begin{equation}> label{eu11}
fbox{$displaystyle L(a):=lim_{sigma rightarrow>
infty}I(a,sigma)hsp mbox{rm with} hsp I(a,sigma)= frac{>
int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx }>
{ int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } $}>
end{equation}> However:> $$> int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx => - sigma^2 int_{-half
a}^{+half a} x , e^{-half x^2 /> sigma^2} ,> dleft(-half x^2
/ sigma^2right) => $$ $$> - sigma^2 int_{-half a}^{+half a} x
,> dleft(e^{-half x^2 / sigma^2}right) => - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half> a}^{+half a}> +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,>
dx> $$ $$> hieruit hsp I(a,sigma)=frac{> int_{-half a}^{+half
a} x^2 , e^{-half x^2 / sigma^2} , dx }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx } => sigma^2 left(
1 - frac{ a.e^{-half (a/2)^2 / sigma^2} }> { int_{-half
a}^{+half a} e^{-half x^2 / sigma^2} , dx }> right)> $$> Let
us de'ne $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,; h to> 0; .; $
Then problem (ref{eu11}) is equivalent with>
begin{equation}label{eu12}> fbox{$displaystyle
L(a)=frac{a^2}{8}limlimits_{hto>
0}frac{1}{h^2}left(1-frac{he^{-h^2}}{Erf(h)}right) hsp>
mbox{rm where}hsp Erf(h):=intlimits_{0}^h e^{-t^2}; dt> $}hsp
.> end{equation}> But $displaystyle Erf(h)=>
he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $ where>
[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k}
cdotfrac{z^k}{k!}hsp> ,hsp
(a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]>
Therefore> [L(a)=frac{a^2}{8}limlimits_{hto
0}frac{sumlimits_{k=1}^infty> frac{(1)_k}{(3/2)_k}h^{2k}}>
{h^2cdotsumlimits_{k=0}^infty frac{(1)_k}{(3/2)_k}h^{2k}}=>
frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=0}^infty>
frac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k}}>
{cdotsumlimits_{k=0}^infty>
frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}cdot>
frac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12}; .> ]> %>
end{document}I would like to read LaTex 'les. Is there some
Windows software out therethat works immediately without
having to run numerous fonts setup programscomputer.
= [
deleted ]Here is your answer, as simple as I can make
it:setlength{mathindent}{1.0cm}pagestyle{plain}
raggedbottomdef half {frac{1}{2}}def hieruit {quad
Longrightarrow quad}begin{document}%subsection*{Should be the
same but alas ...}%Consider the following integral:$$
int_{-half a}^{+half a} x^2 , dx / a = left[ frac{1}{3} x^3
right]_{-half a}^{+half a} / a = frac{2.a^3}{3.8.a} =
frac{a^2}{12}$$Now consider:$$ lim_{sigma rightarrow infty}
frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} ,
dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx
}$$We would expect that this more dif'cult integral
neverthelessconverges to the same result as before: $a^2 /
12$ , because:$$ lim_{sigma rightarrow infty} e^{-half x^2 /
sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 ,
e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half
a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 /
sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x ,
dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[
x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} +
sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,
dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half
x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2
/ sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half
(a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 /
sigma^2} , dx } right)$$It should be correct to anticipate
that $a<<sigma$. Then dividingthe integral in the denominator
by $a$ should be equivalent todifferentiation in $x = 0$ :$$
frac{int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx}{a}
= e^{-half 0^2 / sigma^2} = 1$$So the problem reduces to
'nding the outcome of:$$ lim_{sigma rightarrow infty} sigma^2
left[ 1 - e^{-half (a/2)^2 / sigma^2} right] = lim_{sigma
rightarrow infty} sigma^2 left[ 1 - left{ 1 - half (a/2)^2 /
sigma^2 + ; mbox{h.o.t} ; ... ; right} right]$$Where the
higher order terms contain $sigma^4 , sigma^6 , ; ...;$ in
their denominators, so they should converge to zero.Leaving
us with:$$ lim_{sigma rightarrow infty} sigma^2 half (a/2)^2
/ sigma^2 = half (a/2)^2 = frac{a^2}{8} quad ne frac{a^2}{12}
quad mbox{!!}$$So the two outcomes have the same order of
magnitude, but they donot match. WhatÍs going wrong ? Please
help !%subsection*{The easy answer}Let[frac{a}{{2sigma }} =
lambda]and rewrite the desired integral ratio
asbegin{eqnarray*}frac{{int_{ - {a mathord{left/ {vphantom {a
2}} right. kern-nulldelimiterspace} 2}}^{ + {a mathord{left/
{vphantom {a 2}} right. kern-nulldelimiterspace} 2}} {dx,x^2
e^{ - {{x^2 } mathord{left/ {vphantom {{x^2 } {sigma ^2 }}}
right. kern-nulldelimiterspace} {sigma ^2 }}} } }}{{int_{ -
{a mathord{left/ {vphantom {a 2}} right.
kern-nulldelimiterspace} 2}}^{ + {a mathord{left/ {vphantom
{a 2}} right. kern-nulldelimiterspace} 2}} {dx,e^{ - {{x^2 }
mathord{left/ {vphantom {{x^2 } {sigma ^2 }}} right.
kern-nulldelimiterspace} {sigma ^2 }}} } }} &equiv& sigma ^2
frac{{int_0^lambda {du,u^2 e^{ - u^2 } } }} {{int_0^lambda
{du,e^{ - u^2 } } }} &=& sigma ^2 left( {frac{{frac{{lambda
^3 }}{{3 cdot 0!}} - frac{{lambda ^5 }}{{5 cdot 1!}} +
frac{{lambda ^7 }}{{7 cdot 2!}} - + ldots }}{{frac{{lambda ^1
}}{{1 cdot 0!}} - frac{{lambda ^3 }} {{3 cdot 1!}} +
frac{{lambda ^5 }}{{5 cdot 2!}} - + ldots }}} right) &=&
frac{{sigma ^2 lambda ^2 }}{3}left( {frac{{1 - frac{{3lambda
^2 }}{5} + - ldots }}{{1 - frac{{lambda ^2 }}{3} + - ldots
}}} right);mathop to limits_{sigma to infty }
;frac{{a^2}}{{12}}end{eqnarray*}end{document}-- Julian V.
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.
=> > I would like to read LaTex 'les. Is
there some Windows software out there> that works immediately
without having to run numerous fonts setup programs>
computer.Subscribe to the newsgroup comp.text.texand see what
people there are asking about and recommending.You will
probably want to install the MikTeX distribution, available
from http://ftp.uci.agh.edu.pl/pub/tex/systems/win32/miktex/I
strongly also recommend getting WinEdt, available from
http://www.winedt.com/It is shareware, but worth the money
($40, or so). It is a LaTeX andTeX -aware editor that makes
using these systems easy (I suggest LaTeXfor your own
work).The book A Guide to LaTeX by Kopka & Daly is extremely
helpful.Good luck.-- Julian V. NobleProfessor Emeritus of
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows
only one commandment: contribute to science. -- Bertolt
Brecht, Galileo.
0311~1   

Desktop DB
is f?Non sequitur. I do not know what it was you told Prof.
McKenzie wasmy claim, but you are not to be trusted to
accurately report mywords. Unless you provided him with a
transcript of what you said andwhat I replied, your claims
about what Prof. McKenzie may or may nothave said about what
you think I said are worthless.This above is yet another
piece of evidence in my favor: You aremisrepresenting my
claims about this matter. Since you are doing sowhile having
my precise words before your very eyes, it is clear thatwe
cannot assume that you can accurately report my claims at
all, letalone to Prof. McKenzie from
memory.
= [Gabriele Rossetti] has left a vast body of
writings... in which he has attempted to prove the truth of
his unorthodox interpre- tation of medieval literature. They
present a formidable record of unsystematic research in which
we see an enthusiast plunging farther and farther and farther
from the logic of facts and good sense until truth is lost in
the dreadful nightmare of an idee 'xe. There is no real
evolution of the Theory although it grows and expands until
it embraces ever wider horizons. The numerous inaccuracies of
deduction, mis-statements of historical fact, and
self-contradictions...have caused critics to turn away from
them in disgust... [...] It is impossible to read far...
without realizing that we have to deal with a work of faith
and imagination rather than of reasoning. There is an
appearance of reason, for the author is set on proving by
logic the truth of what he already believes by intuition. The
truth is plain to him and he cannot comprehend why others do
not immediately accept it, but as they desire demonstration
he has multiplied his proofs. It is the redundancy and
confusion of a prophet expounding by a familiar method the
truth revealed to his own simple soul in a §ash of
inspiration... In such work as this... it is idle to look for
the calm reasoning of a scholar; we do not 'nd it, and there
is little or no advantage in attacking the obvious
inconsistencies and absurdities that abound. -- E.R. Vincent,
_Gabriele Rossetti in England_, quoted in _The Shakespearan
Ciphers Examined_, by William F. Friedman and Elizebeth S.
Friedman
==
==Arturo
Magidinmagidin@math.berkeley.edu
> >> Because you have a
track record of just not getting what the objection>> is. It
is interesting that I do not see anything in your>>
correspondences below with Ralph McKenzie or anyone else that
even>> mentions me, though.> Professor McKenzie dismissed your
objection claiming that the aOs can> have varying factors of 
f
dependent on m.What were the *exact words* you used in
describing the objection?Because unless you quoted Prof.
MagidinOs argument word-for-word,it wasnOt his 
objection that
was dismissed, it was your distortedmisunderstanding of it
that was dismissed.-- Wayne Brown | When your tailOs in a
crack, you improvisefwbrown@bellsouth.net | if youOre good
enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
ddition, Rovelli
hasgrossly mischaracterized the Feynman-type alternative --
which shows that he hasnot understood it; although he is
right that under this alternative, the GR
conceptualrevolution is indeed dis-valued, but to a much
greater extent than he seems to realize.JS: Also p. 112Recall
that Einstein described his great intellectual struggle to
'ndGR as Ounderstanding the meaning of
coordinatesO.This is
my key objection to PuthoffOs and IbisonOs use
of r ant tin
their PV modeling.They do not IMHO understand the meaning of
coordinates: in GR. Neitherdoes Yilmaz apparently.PZ: I have
no idea of what you are getting at here. EinsteinOs point
about coordinatesdepends entirely on his strict equivalence
thesis, which is untenable.JS: I guess I have never
understood your idea here. Can you try to explain it again
withcomplete clarity?PZ: Once you abandon strict equivalence,
the uni'cation of gravitational and inertialg_uv is formal
and
arbitrary. It has no deep physical meaning. Then we
recoverchronogeometric background coordinates together with a
physical rubber-rod-and-clock non-inertial g_uv.Rovelli
doesnOt seem to understand that.JS: Neither do I.pp 112 -
114
make this clear IMHO.PZ: This is not an argument -- itOs
merely a proposal: In my opinion....JS: Also at quantum level
RovellimentionOs DiracOs insight that the
Heisenberg Picture
is better thanSchrodinger Picture and that time as in the §ow
of timeOs arrowis not dynamical time but is a statistical
thermodynamic construct.PZ: Sure, if you want to quantize
everything and abandon time.UnruhOs for example, and some of
the others.Unruh has some important ideas however.PZ: But
then I fail to see any material distinction between
relabelingthe bare unindividuatedspacetime points (passive
diffeomorphism), on the one hand, andshifting all
physical'elds (including the GR metric 'eld)
with respect to
such a rawmanifold (activediffeomorphism), on the other.
KretschmanOs point appears to be fullyvalid in both
cases:how
can physics in general -- any sensible physics -- possibly
dependon a mere re-labelingof raw unindividuated spacetime
points; or, for that matter, on acommon shift of all
physicalsystems, including the *physical* metric 'eld g_uv,
with respect tosuch a set of unindividuatedpoints?JS:
Admittedly a sticky wicket that I also need to understand
moredeeply.PZ: ItOs an arti'cial model that has
nothing to do
with physical relativity IMO.JS: I found this remark by
Goldstein helpful:In the ADM formulation 4-diffeomorphism
invariance amounts to the requirement that one ends up with
the same space-time, up to coordinate transformations,
regardless of which path in multi-'ngered space-time is
followed, i.e. which lapse function N is uses. p.278OK so
this idea goes back to the archetypal notions of classical
thermodynamics with a state function, to holonomic
integrability of equality of mixed partial derivatives with
not multiply-connected manifolds, to a closed Cartan exterior
differential form on a cycle (no boundary), no topological
defects and all that.The passive coordinate transformations
are like EM gauge transformation (e/c)Au -> (e/c)Au + hChi,u
in a 'xed gauge constraint.PZ: Rotating a system in isotropic
space is connected with a true physical symmetryof the system
including the vacuum in which it is embedded. Yes, the
symmetryof the system Hamiltonian under such a transformation
is an active transformationalsymmetry that is characteristic
of the particular system -- unlike the invariance of
theproper formal expression of any physical law under a
*mere* coordinate transformation.JS: Note also 12.2.2 p.279
alluding to your digging up Kretchmann from Dr.
FrankensteinOs favorite graveyard. ;-)The fundamental
symmetry at the heart of general relativity is its invariance
under general coordinate transformations of spacetime. It is
important to stress that almost any theory can be formulated
in such a 4-diffeomorphism invariant manner by adding further
structure to the theory (e.g. a preferred foliation of
spacetime as a dynamical object). General relativity has what
is sometimes called serious diffeomorphism invariance, meaning
that it involves no spacetime structure beyond the 4-metric
and, in particular, singles out no special foliation of
spacetime.Goldstein and Teufel then knock standard QGrav
including even Ashtekar -> Loop Spin Foams that are perhaps
near to being falsi'ed by NASAOs EINSTEIN
inJack better check
this story out:Gary S. BekkumPZ: However, moving *everything
physical* -- including the uni'ed g_uv -- along the
rawspacetime manifold is quite another kettle of 'sh. It is
devoid of physical content IMHO.But again, a space-time frame
of reference is not *merely* a coordinate transformation,since
it represents the *motion* of a possible observer. Thus there
is no *a priori* reasonwhy physical laws should take the same
form in different frames of reference, contraryto Einsteinian
doctrine. Of course they *may* as a matter of fact, but that
is very differentfrom insisting a priori that they always
*should*.ThatOs yet another Einsteinian red herring.PZ: What
is physically signi'cant here is not this abstract
andarti'cial de'nition of diffeomorphismwith
respect to a raw
manifold conceived as an unindividuated pointset, but a shift
of physical'elds with respect to the metric-transformational
structure ofspacetime, which of course in GRdepends on the
distribution of matter.JS: Precisely, yes that is the idea I
think. ItOs a return to I thinkLeibnizOs
relationism?PZ:
Descartes. He is almost literally putting Descartes before
the horse ofphysical relativity.JS: ;-)How to getlocalization
in space and the §ow of time as we experience it in
ourimmediate inner consciousnesshas nothing to do with the
particular local coordinate representationlike r and t in,
for example,K = e^2GM/c^2rdr/dt = c/Kfor null geodesicand in
his Tables generally in the context of potentially
practicalmetric engineering of the guv 'eld using the EM Au
'eld in spite ofthe enormous gravity string tension ~ c^4/G ~
10^19 Gev per 10^-33 cm.PZ: You are blocking his actual
de'nition of r. Classic operationalism does notapply to a
theory of this type. That is a critical point. Miss that and
of coursenothing makes sense.JS: I am not ready to renounce
PW BridgmanOs Operationalism. Indeed, nothing Hal Puthoff
says about the foundations of his PV makes any sense to my
mind. If it ainOt broke, donOt
'x it. Of course I am not a
doctrinaire positivist like Stephen Hawking proudly proclaims
he is.PZ: So it appears that Rovelli ends up in a position
where he isessentially arguing that the matter-dependence of
the metric 'eld implies that the GR metric is really
aphysical metric, and the GRmetric 'eld is to be regarded as
a physical 'eld like any otherphysical 'eld.That
is, the
*uni'ed* metric 'eld. That is
RovelliOs tacit, yet arbitrary,
assumptionIMO.Then, the physicsdepends only on the relative
disposition of the various physical'elds with respect to each
other.JS: Correct, with the proviso that matter includes both
real, i.e.on mass shell, sources as well as virtual, i.e. off
mass shellsources. The virtual sources divide into two
classes:I. Non-exotic near EM 'eld Fuv giant coherent quantum
states ofvirtual photons that contribute to Omega(Matter) of
the FRW metric andto Tuv in EinsteinOs local
'eld
equation.II. Exotic vacuum w = -1 zero point stress-energy
density local tensor~ (String Tension)/zpfguv for both
repulsive dark energy /zpf > 0 ofnegative pressure and
attractive dark matter /zpf < 0 of positivepressure.These
exotic vacuum virtual sources contribute Omega(Exotic Vacua)
~0.96 to Omega(Total) = 1 in our large-scale spatially
§atpost-in§ationary local Level I Hubble sphere brane 
world
as inLenny SusskindOs megalopolis Landscape subject to the
naturalselection of the Weak Anthropic Principle (WAP)OK, I
think I made an error above including brane worlds in sense
of parallel worlds?Here is why I think I made an error (If I
did so did Hawking and Scienti'c American in their popular
science reports):D-Branes are extended surfaces without
edges. In order that the black hole be a localized object, it
is assumed that our ordinary four dimensions (three space and
one time) are all orthogonal to these D-Brane surfaces ...
Thus to us these D-Branes would look as though they were
located at a point (or at least a very small region) of our
observable three dimensions of space. W. G. Unruh p. 168
Black holes, dumb holes, and entropy, i.e. the D-Branes are
in the compacti'ed Calabi-Yau space. I have to look again at
HawkingOs The Universe in a Nutshell that seems to give the
wrong idea here? Perhaps I mis-remember?Note also Ed
WittenOs
formula generalizing HeisenbergOs quantum uncertainty
principle, i.e. eq. (5.9) p. 136Delta X > h/DeltaP +
alphaO(DeltaP)/hThe second gravity-string source of
uncertainty should give the irreversible statistical arrow of
time not found when alphaO = 0 i.e. in'nite
string tension, or
in'nite space-time stiffness of action without reaction as is
also found in the signal locality of orthodox quantum theory
in sense of Antony ValentiniOs papers.PZ: But while this may
allow or even imply a relational view of theraw spacetime
manifold, in theCartesian sense, it is clearly not physical
general relativity, inthe classic Einsteinian sense,which
requires fundamental identi'cation of the inertial
andgravitational metrics.JS: I do not understand what you
just said. EEP in every signi'cantsense still holds IMHO.PZ:
You are in a loop.JS: Ground Hog Day. Help let me out of my
bottle Oh Thief of Baghdad. The Magic Flying Carpet is
obviously the weightless Alcubierre timelike geodesic faster
than light warp drive powered by dark energy and dark matter
exotic vacua con'gurations. See my animated picture of this
in http://qedcorp.com/APS/StarGate1.movPZ: EEP is NOT
Einstein equivalence. EEP is merely a correspondence
principle.Einstein strict equivalence is NOT valid. And even
EEP is not strictly valid asadvertised (see e.g. Ohanian and
Ruf'ni Ch. 1).given guv 'eld still works for
example.PZ: This
simply re§ects weak equivalence -- equality of gravitational
and inertial mass.That does not imply Einstein equivalence.
It works whether we use a uni'ed g_uvor go to a bimetric
formalism, and regardless of whether we consider inertial
forcesas real or 'ctitious.JS: The so-called
'ctitious or
inertial forces, e.g. G-Force when a jet takes off (0 to 120
mph in 2 sec) from an aircraft carrier are physically real. I
can tell you that from 'rst-hand experience. The use of
'ctitious like the use of hidden variable are unfortunate
choices of words. Extra variable is better for the latter.PZ:
So this is not a valid argument IMO. It is an example of the
classical logical fallacyaf'rming the antecedent.JS: The
tidal stretch-squeezeworks etc.PZ: Of course. So?The point is
that you can always tell the difference between a real gravity
'eld andan inertial 'eld. They are NOT
completely physically
equivalent.JS: If you mean Ruvwl =/= 0 locally for a real
gravity 'eld, and is zero for an inertial 'eld,
I
agree.However, I do not see how you can writeguv =
guv(inertial) + guv(real)That is, apart fromguv = Globally
Flat Metric + du,v + dv,uI do not claim we can physically
distinguish the two terms on RHS nor is thesecond term small
compared to 'rst.The Bohm pilot constraint for the extra
variable, i.e. actual distortion of Hagen KleinertOs
elastic-plastic 4D world crystal latticei.e. density of
vortex line topological defects of curvature disclination and
possibly torsion dislocations isdu = Lp^2(Goldstone Phase of
Vacuum Coherence),u,u is ordinary partial derivative.Compare
my equation here to GoldsteinOs eq. 12.4 on p. 282.PZ:
Consequently,uni'ed g_uv has no deep physical meaning or
necessity -- as Feynman argued.PZ: Which all seems to
contradict his statement,...Of course, nothing [in GR] to
prevents us... from singling out thegravitational 'eld as
Othemore equal among equalsO, and declaring that
location is
absolute inGR, because it can bede'ned with respect to it.JS:
No, you have simply misread the context of RovelliOs remark.
Readit again more carefully. This is not a valid objection at
all toRovelliOs argument IMHO. The way I read him, his
argument issplendidly globally self-consistent.PZ: The way I
read him, he doesnOt actually *have* an argument. He simply
*prefers*to quantize space and time intervals rather than
unpack EinsteinOs uni'ed g_uv.PZ: So IMO
RovelliOs position
is incoherent. If anything, he isarguing for an
*alternative*non-Einsteinian interpretation of GR in which
matter-dependent g_uvis to be treated as a*physical 'eld*,
which means it can in principle be distinguishedfrom the
kinematical inertial'eld with its trivially valid
transformable metric tensor.JS: We are back to Square One.
The Rock has rolled back down onSisyphus. I cannot pinpoint
the precise misconception that is drivingyour position
here.PZ: It is no wonder you are having this dif'culty, since
it is *RovelliOs* misconceptionthat is causing the
problem.Rovelli doesnOt seem to understand that recognition
of the dynamical character ofnon-inertial g_uv fundamentally
distinguishes it from the kinematical g_uv, whichlatter
doesnOt depend on the distribution of matter. So the natural
thing to do isto split uni'ed g_uv and thus clearly separate
the dynamical and kinematical metric'elds.JS: Show me an
algorithm to do that in any problem.PZ: Yet he doesnOt even
consider this.In other words, Rovelli is really a
*Yilmazian*.JS: Not in my book.PZ: Actually, you are right --
because he does not follow the natural developmentof his own
argument to its logical conclusions. If he did, his 'rst
alternative (p 109)would fall squarely in the Feynman-Yilmaz
category, with rubber rods and clocksand a §at background
inertial metric.JS: It seems to me the objective is to
eliminate any §at background non-dynamical
metric.Nondynamical means ACTION WITHOUT REACTION.Note, I
have it formally in my equations but it is not physically
measurable separately.Indeed, the pre-in§ationary vacuum is
the perfectly §at world crystal with no vortex string
topologicaldefects of effective multiple-connectivity in the
space of the vacuum coherence order parameter whichis zero
everywhere prior to the in§ationary vacuum phase transition
like a normal metal going super-conducting.PZ: Obviously it
is a 'eld that depends on the distribution ofmatter, if
thatOs what hemeans. But it is the only 'eld
that is de'ned
in terms of aspacetime metric, so itis obviously unique in
that sense.JS: Read RovelliOs paper in the book. Then tell
me
what you think.PZ: I read it. Several times. IOve been
scratching my head for severaldays now.See above.
I
canOt 'nd a proof of the (multivariable) chain
rule that
makes sense.Assume g is differentiable at x and f is
differentiable at g(x), where differentiable means in the
total derivative sense. Let dg(x) and df(g(x)) denote the
associated linear differential approximants. Then f(g(x+h)) -
f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) =
df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) =
df(g(x))[dg(x)(h)] + df(g(x))[o(h)] + o(g(x+h)) - g(x)).The
'rst term in the last sum is what we want to see, namely the
composition of df(g(x)) and dg(x) evaluated at h. So weOre
done if we can show both df(g(x))[o(h)] and o(g(x+h)) - g(x))
are o(h), which is not too hard.
I canOt 'nd a proof of
the (multivariable) chain rule that makes sense.Assume g is
differentiable at x and f is differentiable at g(x), where
>differentiable means in the total derivative sense. Let
dg(x) and >df(g(x)) denote the associated linear differential
approximants. Then f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) -
g(x)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h) + o(h)] +
o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] +
o(g(x+h)) - g(x)).The 'rst term in the last sum is what we
want to see, namely the >composition of df(g(x)) and dg(x)
evaluated at h. So weOre done if we can >show both
df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not
too >hard.Fine, but she said I want to see a real proof that
uses de'nitions and little greek letters...I suppose we can
leave it to the reader to change a few lettersto greek to
make things easier to follow. But if sheOs never seenthis
proof then possibly sheOs never seen the
de'nition:(Def: o(h)
means some function e(h) with the propertythat e(h)/||h|| -> 0
as h -> 0.)Def: f: R^n -> R^m is _differentiable_ at x if
there exists alinear map T: R^n -> R^m such that f(x + h) =
f(x) + Th + o(h);if so then T = df(x).(Then we should also
point out how the notation she maybe looking at follows from
this: If df(x) exists then the matrixcorresponding to that
linear map has the partial derivativesof the components of f
for its entries, and the formula yousee for the chain rule in
some calculus books is justgiving the product of two
matrices.)************************David C. Ullrich
> I
canOt 'nd a proof of the (multivariable) chain
rule that makes
sense.>>Assume g is differentiable at x and f is
differentiable at g(x), where >>differentiable means in the
total derivative sense. Let dg(x) and >>df(g(x)) denote the
associated linear differential approximants. Then>> f(g(x+h))
- f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x))>> =
df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x))>> =
df(g(x))[dg(x)(h)] + df(g(x))[o(h)]>> + o(g(x+h)) -
g(x)).>>The 'rst term in the last sum is what we want to see,
namely the >>composition of df(g(x)) and dg(x) evaluated at h.
So weOre done if we can >>show both df(g(x))[o(h)] and
o(g(x+h)) - g(x)) are o(h), which is not too >>hard.> > Fine,
but she said I want to see a real proof that uses >
de'nitions
and little greek letters...> How many students have you found
whoOve said that? Not many?It reads to me as though this
person is taking a second year course at anAmerican
University. The text books taught from can be pretty dire
when itcomes to formal proof. Indeed one teaches three
OdifferentO (at least) chainrules for special
cases, rather
than just _the_ chain rule.IOm sure that the course
instructor would be happyto provide some way for the OP to
satisfy her curiosity, even if itOspointing her towards the
library and a relevant book. But just in case, asyouOre the
Analyst round here, which ones would be useful in
demonstratingthat actually you donOt want too many greek
letters §oating around? IsRudin any good? LotOs of grad
programs seem to prefer it.> I suppose we can leave it to the
reader to change a few letters> to greek to make things easier
to follow. But if sheOs never seen> this proof then possibly
sheOs never seen the de'nition:> > (Def: o(h)
means some
function e(h) with the property> that e(h)/||h|| -> 0 as h ->
0.)> > Def: f: R^n -> R^m is _differentiable_ at x if there
exists a> linear map T: R^n -> R^m such that> > f(x + h) =
f(x) + Th + o(h);> > if so then T = df(x).> > (Then we should
also point out how the notation she may> be looking at follows
from this: If df(x) exists then the matrix> corresponding to
that linear map has the partial derivatives> of the
components of f for its entries, and the formula you> see for
the chain rule in some calculus books is just> giving the
product of two matrices.)> > > ************************> >
David C. Ullrich
> I canOt 'nd a proof of the
(multivariable) chain rule that makes sense.>>Assume g is
differentiable at x and f is differentiable at g(x), where
>differentiable means in the total derivative sense. Let
dg(x) and >df(g(x)) denote the associated linear
differential approximants. Then>> f(g(x+h)) - f(g(x)) =
df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x))>> =
df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x))>> =
df(g(x))[dg(x)(h)] + df(g(x))[o(h)]>> + o(g(x+h)) -
g(x)).>>The 'rst term in the last sum is what we want to
see, namely the >composition of df(g(x)) and dg(x)
evaluated at h. So weOre done if we can >show both
df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not
too >hard.>> >> Fine, but she said I want to see a real
proof that uses >> de'nitions and little greek letters...>How many students have you found whoOve said that? Not
many?It reads to me as though this person is taking a second
year course at an>American University. The text books taught
from can be pretty dire when it>comes to formal proof. Indeed
one teaches three OdifferentO (at least)
chain>rules for
special cases, rather than just _the_ chain rule.IOm sure
that the course instructor would be happy>to provide some way
for the OP to satisfy her curiosity, even if itOs>pointing
her
towards the library and a relevant book. But just in case,
as>youOre the Analyst round here, which ones would be useful
in demonstrating>that actually you donOt want too many greek
letters §oating around? Um, all the comments about greek
letters have been tongue in cheek.I donOt know what text
would be useful for demonstrating that youdonOt want too
many
greek letters §oating around, because theidea that you
donOt
want too many greek letters §oating aroundis news to me.I
suspect that you were being facetious, actually asking
aboutdemonstrating something else. But I canOt
'gure out what
thesomething else might be.>Is>Rudin any good? LotOs of grad
programs seem to prefer it.>> I suppose we can leave it to
the reader to change a few letters>> to greek to make things
easier to follow. But if sheOs never seen>> this proof then
possibly sheOs never seen the de'nition:>> >>
(Def: o(h)
means some function e(h) with the property>> that e(h)/||h||
-> 0 as h -> 0.)>> >> Def: f: R^n -> R^m is _differentiable_
at x if there exists a>> linear map T: R^n -> R^m such that>> f(x + h) = f(x) + Th + o(h);>> >> if so then T = df(x).>> (Then we should also point out how the notation she may>>
be looking at follows from this: If df(x) exists then the
matrix>> corresponding to that linear map has the partial
derivatives>> of the components of f for its entries, and the
formula you>> see for the chain rule in some calculus books is
just>> giving the product of two matrices.)>> >> >>
************************>> >> David C.
Ullrich************************David C. Ullrich
=Need help
on math, visit www.helptosolve.com Just try.Want to help
others with math, visit www.helptosolve.com and join the
team.
>>Dean said that Northeasterners donOt talk about
religion. >>ThatOs not true, because I know of
Northeasterners who spend almost>>all day talking about The
Virgin Mary, Hearing Confession, whoOs Rabbi>>is where, what
Temple this person goes to, I have another Bar
Mitzvah,>>Episcopal Parishoners this, American Baptists
convention that.>>What Northeast is Dean from ??> > > What
planet are you from?We have people from every planet on Earth
in California. --- Formergovernor Gray Davis--
http://hertzlinger.blogspot.com
Many of the problems
with BushOs agenda is the dishonest> reinterpretation of
reality that is done so often by the> Democrats.Citations,
pleaslean transform (xO-x.cO)=(x-x.c), insures
that it is an
error to imagine there is anydifference between the data and
law in one frame and in another;the usual, convenient rest
frame is the best frame and only framerequired for universal
analysis. [Well, (xO<>x, x,cO<>x.c,
but(xO-x.cO)=(x-x.c).]Second, given that you
decide
unnecessarily to adapt a law tozz

admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oft
Dean said
that Northeasterners donOt talk about religion. K3wl? How is
that on topic for sci.math, sci.skeptic, alt.atheism
oralt.christnet? Of the 5 groups that you posted to,
onlyalt.politics.democrats is relevant; clearly your intent
was to trollfor a crss-posted §ame war.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
> > > Dean said that
Northeasterners donOt talk about religion. > > > > Maybe you
should furnish an actual quote before you build a castle in
the > > air upon what you think he might have said. > > >
OOMy father used to tell us how much strength he
got from
religion, but > we didnOt have Bible readings. There are
traditions where people do > that. We
didnOt,OO he said.
OOPeople in the Northeast donOt
talk about > their religion.
ItOs a very personal private matter, and
thatOs the >
tradition I was brought up in.OO Howard Dean,
to Boston Globe
last > week. Quoted in countless other papers.You gotta
interpret that in context. It means northeastern
politiciansdonOt talk to the papers about their religion
(they leave that to preachers). It was de'nitely more polite
than what I would have said, which would be, I didnOt come
here today to talk about religion, so whatOs your next
question?
>> > > Dean said that Northeasterners donOt
talk about religion. >> > >> > Maybe you should furnish an
actual quote before you build a castle in the >> > air upon
what you think he might have said. >> > >> OOMy
father used
to tell us how much strength he got from religion, but >> we
didnOt have Bible readings. There are traditions where
people
do >> that. We didnOt,OO he
said. OOPeople in the Northeast
donOt talk about >> their religion. ItOs a
very personal
private matter, and thatOs the >> tradition I was brought up
in.OO Howard Dean, to Boston Globe last >>
week. Quoted in
countless other papers.You gotta interpret that in context.
It means northeastern politicians>donOt talk to the papers
about their religion (they leave that to >preachers). It was
de'nitely more polite than what I would have said, >which
would be, I didnOt come here today to talk about religion,
so
>whatOs your next question?By the same logic the previous
poster used, Bush has also told lies.President BushOs
compassionate agenda resonates with the people ofboth New
York and California, -Tracey Schmitt,spokeswoman for the
Bush-Cheney O04campaignThatOs not true, because
I know of
Californians who are not impressedwith BushOs compassionate
agenda. -----Yang a.a. #28a.a. pastor #-273.15, the most
frigid church of Celcius nee Kelvin EAC Econometric Forecast
and Socerey DivisionProudly plonked by Lani Girl and
CrazyalecThe Bush ObalancedO budget: -525
billion and
worseningThe Bush OeconomicO policy: -3 million
jobs and
countingThe Bush Iraq lie: -472 GIs, one friendOs
co-workerOs
son and mountingHaving Bush  up my country: Worthless
Dean said that Northeasterners donOt talk about religion.We
donOt, certainly not when compared to other parts ofthe
country.Most people I know consider it an impolite subject,
atleast in public & amongst strangers.
= > > Dean said that
Northeasterners donOt talk about religion.> > Maybe you
should
furnish an actual quote before you build a castle inthe> > air
upon what you think he might have said.> OOMy
father used to
tell us how much strength he got from religion, but> we
didnOt have Bible readings. There are traditions where
people
do> that. We didnOt,OO he said.
OOPeople in the Northeast
donOt talk about> their religion. ItOs a very
personal
private matter, and thatOs the> tradition I was brought up
in.OO Howard Dean, to Boston Globe last> week.
Quoted in
countless other papers.I heard my father once say that people
in the South arevery conservative religiously. Well I live in
Atlanta nowand am friends with some folk at the Unitarian
Church.My father. What a dirty, rotten liar.
=|Did you see
my reply to the OP?Yes, although the contents were not still
clear in mind when I lastAmandaOs acquaintance objected to
her proof based on its being aproof by contradiction. I guess
I think the distinction between aproof by proving the
contrapositive and a proof by contradiction isfuzzy enough
not to make this strictly incorrect. For the questionof
whether the proof is valid, I think youOd agree
itOs also not
adistinction that could be relevant. People consider the proof
okaynot because itOs crossed this fuzzy line; people
consider
the proofokay because weOre considering a context where
proof
by contradictionand the whole complex of related forms of
reasoning are all accepted.You had two main reasons why proof
by contradiction was consideredworse:|(i) itOs easy to give
a
wrong proof by contradiction, where |the contradiction arises
from some error ||(ii) if you give a direct proof that A
implies B, by assuming|A and then deducing B, the steps in
the proof can give|some insight into what A really entails -
you show A implies|B by showing A implies C and C implies B,
and along the|way youOve shown two facts that might be
interesting and|useful elsewhere, that A implies C and that C
implies B.|You donOt get this sort of bonus from a proof by
contradiction,|since in the course of the proof youOre
assuming things|which it turns out never actually hold.I
think the practice of constructive proof may shed some light
onthis. The conclusion of a constructive proof tells you more
aboutthe nature of the proof than the conclusion of a
classical proof,generally. In particular, when the conclusion
is negative, the proofwas a proof by contradiction
(essentially) and when the conclusionis positive it wasnOt.I
canOt tell whether the circumstances reducing the risk of
(i)
andenhancing the prospects of (ii) are different from the
circumstanceswhere the theorem (as it would be stated in
constructive terms) ismore solidly stated.In constructive
mathematics one has the general advice of Bishopto cut down
on negations. If we seek to tighten up our theorems bygetting
rid of §abby negations, then we also as a side-effectrefrain
from proofs by contradiction.Lots of negations and
implications can be upgraded by supplying themwith more solid
content. Bishop has a paper which uses GoedelOsDialectica
paper for this. Goedel has a method of rewritingstatements
which puts them into a form (there exists w) (for all t)
P(w,t)where w ranges over a set of possible witnesses to the
truth ofthe statement, and t ranges over a set of possible
cases. If thisstatement is false, one would like to do more
than just negate it;one would like to say something about the
function f:w->t for whichP(w, f(w)) fails for all w. A proof
of the negative statementimplicitly provides such an f, as
witness. Even when a statementis negative, then, there can be
implicit solid content in theconstruction demonstrating the
fact, in terms of information onand around this f.I get the
impression that in cases where a theorem is negative,but has
relatively tangible constructive content, one is not
soespecially prone to err in the way described in (i),
because oneis regarding the construction in the proof (of f,
say), whichactually *does* exist, and trying to see whether
*it* is asadvertised. In the proof that there are in'nitely
many primes,to pick a simple example, the engine is the
method for gettinga new prime out of a 'nite set of old
primes. An argument whichfocusses on this kind of concrete
construction doesnOt seem tosuffer from the kind of problem
that some attempts at proof bycontradiction do.IOm going on
impressions, though, and it seems like it wouldtake somewhat
tricky casework to get a more solid sense for whatkinds of
proof do and do not tend to suffer from these problems.Keith
Ramsay
=|Did you see my reply to the OP?Yes, although the
contents were not still clear in mind when I lastAmandaOs
acquaintance objected to her proof based on its being a>proof
by contradiction. Although I donOt think that even he was
saying the proof wastherefore _wrong_, just that it was not
the best possible proof.(Cf for example the title of the
thread; itOs about 'nding abeautiful proof, not
a _valid_
proof...)>I guess I think the distinction between a>proof by
proving the contrapositive and a proof by contradiction
is>fuzzy enough not to make this strictly incorrect. IOm
honestly not sure what you mean by that sentence,
butregarding the minor premise, I donOt see
whatOs fuzzy
aboutthe distinction. To prove A implies B:(a) Assume ~B.
Prove ~Aor(b) Assume A and ~B. Prove P and ~P for some P.Some
proofs of A implies B have the form (a); somehave the form
(b). WhatOs fuzzy here?Oh. YouOre claiming
that the
distinction is fuzzy enoughthat those of us who are saying he
was wrong instating the proof was a proof by contradiction are
noton solid ground? I disagree. Um, lemme put that alittle
more strongly: thatOs not so. The proof Amandagave is of the
form (a), not of the form (b).I mean thatOs just a
mathematical _fact_. You can sayyou donOt see why it
matters,
'ne. But you say aboveyou think the distinction between (a)
and (b) isfuzzy - I donOt see any fuzzine
7ï
WEcf
!.85Ĭèÿÿ
A implies C and C implies B, and along the>|way youOve shown
two facts that might be interesting and>|useful elsewhere,
that A implies C and that C implies B.>|You donOt get this
sort of bonus from a proof by contradiction,>|since in the
course of the proof youOre assuming things>|which it turns
out never actually hold.I think the practice of constructive
proof may shed some light on>this. The conclusion of a
constructive proof tells you more about>the nature of the
proof than the conclusion of a classical proof,>generally. In
particular, when the conclusion is negative, the proof>was a
proof by contradiction (essentially) and when the
conclusion>is positive it wasnOt.I canOt tell
whether the
circumstances reducing the risk of (i) and>enhancing the
prospects of (ii) are different from the circumstances>where
the theorem (as it would be stated in constructive terms)
is>more solidly stated.In constructive mathematics one has
the general advice of Bishop>to cut down on negations. If we
seek to tighten up our theorems by>getting rid of §abby
negations, then we also as a side-effect>refrain from proofs
by contradiction.Lots of negations and implications can be
upgraded by supplying them>with more solid content. Bishop
has a paper which uses GoedelOs>Dialectica paper for this.
Goedel has a method of rewriting>statements which puts them
into a form (there exists w) (for all t) P(w,t)where w ranges
over a set of possible witnesses to the truth of>the
statement, and t ranges over a set of possible cases. If
this>statement is false, one would like to do more than just
negate it;>one would like to say something about the function
f:w->t for which>P(w, f(w)) fails for all w. A proof of the
negative statement>implicitly provides such an f, as witness.
Even when a statement>is negative, then, there can be implicit
solid content in the>construction demonstrating the fact, in
terms of information on>and around this f.I get the
impression that in cases where a theorem is negative,>but has
relatively tangible constructive content, one is not
so>especially prone to err in the way described in (i),
because one>is regarding the construction in the proof (of f,
say), which>actually *does* exist, and trying to see whether
*it* is as>advertised. In the proof that there are in'nitely
many primes,>to pick a simple example, the engine is the
method for getting>a new prime out of a 'nite set of old
primes. An argument which>focusses on this kind of concrete
construction doesnOt seem to>suffer from the kind of problem
that some attempts at proof by>contradiction do.IOm going on
impressions, though, and it seems like it would>take somewhat
tricky casework to get a more solid sense for what>kinds of
proof do and do not tend to suffer from these problems.The
context here, or at least the context I had in mind,
wasproofs by students in classes where half the point to the
classis learning to read and write proofs (beginning
algebra,analysis, topology classes are often in this
category).In that context I can tell you from experience that
(i)is a real danger. Of course itOs much less of a
problemwhen
weOre talking about proofs in the real world writtenby
grownups - one has an idea what _sort_ of contradictionto
expect, so when one gets a totally irrelevant
contradictionone looks for the error instead of saying
qed.>Keith Ramsay************************David C.
Ullrich
> For the greater cogency and obviousness in
your paper THERE> > SHOULD BE a TABLE with componentries of
system, which one are> > sorted out on their in§uence to
precision of system GPS as a whole.> > Such a table was
posted by Sam Wormley on the 22nd, see> > > Then any layman
can see that we can neglect neglible small> > relativistic
corrections as contrasted to by other factors> > de'ning and
restricting limiting accuracies GPS as a whole.> > > > VI.
Summary> > Excluding the deliberate degradation of SA, the
dominant error source> > for satellite ranging with single
frequency receivers is usually the> > ionosphere. It is on
the order of four meters, depending on the> > quality of the
single-frequency model. For dual-frequency (P/Y-code)> >
receivers (which eliminate SA) the Standard Error Model of
Table I> > has one principal change (in addition to the
elimination of the SA> > error). The ionospheric error is
reduced from four meters to about> > one meter.> > > The GR
correction is about 44 microseconds or 13km per day> and
would be cumulative. You can hardly call that negligible.1.
What periodicity of corrections of parameters in GPS? 2. What
parameters are adjusted in GPS? 3. What medial - statistical
values have parameters adjusted in GPS in each session of
corrections?--Aleksandr
> > > For the greater cogency
and obviousness in your paper THERE> > > SHOULD BE a TABLE
with componentries of system, which one are> > > sorted out
on their in§uence to precision of system GPS as a whole.> >
Such a table was posted by Sam Wormley on the 22nd, see> > >
Then any layman can see that we can neglect neglible small>  relativistic corrections as contrasted to by other factors > de'ning and restricting limiting accuracies GPS as a
whole.> > > VI. Summary> > > > Excluding the deliberate
degradation of SA, the dominant errorsource> > > for
satellite ranging with single frequency receivers is usually
the> > > ionosphere. It is on the order of four meters,
depending on the> > > quality of the single-frequency model.
For dual-frequency (P/Y-code)> > > receivers (which eliminate
SA) the Standard Error Model of Table I> > > has one principal
change (in addition to the elimination of the SA> > > error).
The ionospheric error is reduced from four meters to about>  one meter.> > The GR correction is about 44 microseconds or
13km per day> > and would be cumulative. You can hardly call
that negligible. 1. What periodicity of corrections of
parameters in GPS?The corrections are of the order of
nanoseconds per day socon'rm GR to roughly one part in 10,000
by that simplisticcomparison. There is much more information
available in webpages if you want to look for more details.>
2. What parameters are adjusted in GPS?> 3. What medial -
statistical values have parameters adjusted> in GPS in each
session of corrections?Happy New YearGeorge
=To base
10:STEP 1: For the characteristic: carry out repeated
division of the number by10 until the result falls below 10
and count the number of times.STEP 2: For the mantissa: raise
the result to the power of 10 usingmultiplication only, then
divide by 10 again as in Step 1. Continue to Step2 for
further digits.This process generates the logarithm digit by
digit, to the same accuracy asthe other arithmetical
functions. Also works ef'ciently in binary.This has almost
certainly been tried before. I would be interested in
any
=To base 10:>STEP 1: For the characteristic: carry out
repeated division of the number by>10 until the result falls
below 10 and count the number of times.>STEP 2: For the
mantissa: raise the result to the power of 10
using>multiplication only, then divide by 10 again as in Step
1. Continue to Step>2 for further digits.>This process
generates the logarithm digit by digit, to the same accuracy
as>the other arithmetical functions. Also works ef'ciently in
binary.>This has almost certainly been tried before. I would
be interested in any>See Textbook of Algebra v1 by Chrystal
pgs 513 to 519 where some of the moreelementary, although at
the same time more laborious, approximative methodsthat might
be used to compute logarithms are described. Yours seems to be
oneof them. rich
=equation.The equation isAx=0A is a very
large sparse matrix which have 1473-by-1473.Sum(x_i) is
1.0All x are not zero.Please, would you like to let me know
how to solve the simultaneousequation by
fortran.Additionally, My major is not mathmatics. Thus, it is
dif'cult tosolve the equation in a method of pseudoinverse,
SVD, etc.. Please,let me know that, easily and in
detail.
equation. The equation is Ax=0 A is a very large
sparse matrix which have 1473-by-1473.> Sum(x_i) is 1.0> All x
are not zero. Please, would you like to let me know how to
solve the simultaneous> equation by fortran. Additionally, My
major is not mathmatics. Thus, it is dif'cult to> solve the
equation in a method of pseudoinverse, SVD, etc.. Please,>
let me know that, easily and in detail.>Well, I havenOt
dealt
with matrices that large, but maybe you could tryCramerOs
rule. As for the code, you are on your own, I donOt know
Fortran.You could
try:http://www.library.cornell.edu/nr/bookfpdf.htmlLurch
The equation is>> Ax=0>> A is a very large sparse matrix which
have 1473-by-1473.>Well, I havenOt dealt with matrices that
large, but maybe you could try>CramerOs rule. You know, I
think all of us have a tendency to think we can answermore
questions well than we really can. I donOt know a whole
heckuva lotabout numerical analysis, but this strikes me as a
really, really badanswer. For one thing, CramerOs rule gives
an explicit solution toAx=b when A is invertible; the OPOs
matrix A is surely notinvertible, lest the only solution be x
= 0. For another thing,CramerOs rule expresses the answer in
terms of (here) 1474 determinantsof 1473x1473 matrices. How
do you propose that the determinants beevaluated? A naive
algorithm sums 1473! products and introduces thepossibility
of tremendous quantities of round-off error and
faster, is stilltime consuming and prone to error if the
entries are not, say, small integers. As useful as CramerOs
rule can be for theoretical work,I imagine itOs essentially
_never_ used for computing numerical solutionsto linear
equations with more than a handful of variables.So what
response might be better for the OP?ItOs probably true that
any good matrix package can handle yourmatrix (a couple
thousand rows is not considered huge these days)but since you
say your matrix is sparse, you would probably bene'tfrom
using
routines speci'cally written for this common special
case.Here
is a link which was posted the other day in
sci.math.num-analysis(which is a much more appropriate group
for technical questionsof this type):
http://vlsicad.cs.ucla.edu/sparse.htmlOf course the equation
Ax=0 has no solution with sum(x_i)=1 if A isnonsingular, so
it would be good to think about how you know thatyour problem
has a solution. More generally, it is likely to helpthose who
are helping you if you can indicate where your matrix
comesfrom, since that may signal the use of special routines
e.g. forsolving differential equations numerically.dave
>> A is a very large sparse matrix which have
1473-by-1473.ThatOs actually pretty small.> >Well, I
havenOt
dealt with matrices that large, but maybe you could tryCramerOs rule.*splorf*ThatOs the punchline to
some jokes I
know.Really bad idea. It has exponential complexity, while
most numericalalgorithms are n^3 or thereabouts.> Of course
the equation Ax=0 has no solution with sum(x_i)=1 if A is>
nonsingular, So I guess the matrix is singular, but you can
add that sum condition asa last row. Introduce a dummy extra
variable, and you have a squaresystem again. (Or am I
overlooking something?)> More generally, it is likely to
help> those who are helping you if you can indicate where
your matrix comes> from, since that may signal the use of
special routines e.g. for> solving differential equations
numerically.Yep. And with that information post on
sci.math.num-analysis where IOmsure someone will have seen
your problem before.V.-- homepage: cs utk edu tilde
lastname
> >> The equation is> > >> Ax=0> > >> A is a
very large sparse matrix which have 1473-by-1473. >Well, I
havenOt dealt with matrices that large, but maybe you could
try> >CramerOs rule. You know, I think all of us have a
tendency to think we can answer> more questions well than we
really can. I donOt know a whole heckuva lot> about
numerical
analysis, but this strikes me as a really, really bad> answer.
For one thing, CramerOs rule gives an explicit solution to>
Ax=b when A is invertible; the OPOs matrix A is surely not>
invertible, lest the only solution be x = 0. For another
thing,> CramerOs rule expresses the answer in terms of
(here)
1474 determinants> of 1473x1473 matrices. How do you propose
that the determinants be> evaluated? A naive algorithm sums
1473! products and introduces the> possibility of tremendous
Even using row operations, while much faster, is still> time
consuming and prone to error if the entries are not, say,>
small integers. As useful as CramerOs rule can be for
theoretical work,> I imagine itOs essentially _never_ used
for computing numerical solutions> to linear equations with
more than a handful of variables. So what response might be
better for the OP? ItOs probably true that any good matrix
package can handle your> matrix (a couple thousand rows is
not considered huge these days)> but since you say your
matrix is sparse, you would probably bene't> from using
routines speci'cally written for this common special case.>
Here is a link which was posted the other day in
sci.math.num-analysis> (which is a much more appropriate
group for technical questions> of this type):>
http://vlsicad.cs.ucla.edu/sparse.html Of course the equation
Ax=0 has no solution with sum(x_i)=1 if A is> nonsingular, so
it would be good to think about how you know that> your
problem has a solution. More generally, it is likely to help>
those who are helping you if you can indicate where your
matrix comes> from, since that may signal the use of special
routines e.g. for> solving differential equations
numerically. daveHey, itOs an open forum! I consider this NG
like an online math club. Ioffer suggestions, and they
arenOt
always useful, or correct. None of thisstuff is being
published, so relax. He isnOt paying me for my help; so,
heis
free to take my advice, or not. It is up to him. I also
provided a linkto an online book of Numerical analysis in
Fortran, which should answer hisquestion.Lurch
> The
equation is>> Ax=0>> A is a very large sparse matrix
which have 1473-by-1473.>Well, I havenOt dealt with matrices
that large, but maybe you could try>>CramerOs rule. You
know,
I think all of us have a tendency to think we can answer>more
questions well than we really can. [...]I have an explanation
for that... hmm, no I donOt.************************David C.
Ullrich
=I am looking for a formula for the calculation of
the midpoint for thesmallest, an area A circumscribing
circle. Is there like for trianglesa connection between the
center of gravity and the center of thatcircle? Are any
formulas for such a circle known?Carolin Hauner
=Caro> I am
looking for a formula for the calculation of the midpoint for
the> smallest, an area A circumscribing circle. Is there like
for triangles> a connection between the center of gravity and
the center of that> circle? Are any formulas for such a
circle known?The circle which meets the three corners of a
triangle is called thecircumcircle of the triangle, and the
center of that circle is thecircumcenter of the
trinangle.http://mathworld.wolfram.com/Circumcenter.htmlThe
circumcenter may be outside the triangle. But see
alsohttp://mathworld.wolfram.com/Incenter.htmlTo get formulas
for the cartesian coordinates of the circumcenter, writedown
two linear equations, one for each of two perpendicular
bisectors. Theincenter lies on both of those
lines.LH
=hiwikipedia needs our help to carry
allbyemax
=WhatOs happened to you ?I get up early every
morning to read your diatribes and I havenOt seen anything
for a few days now.Does this mean I have to learn maths if I
want to be a part of this newsgroup ?Does this mean I have to
rely on comics for my cackles each morning ?I hope the FBI
didnOt turn on you and that the Evil Mathematical
Establishment (tm) havenOt imprisioned you in an
in'nite
series !Please hurry back, my humour is suffering and my
education lacking - at least I was being *forced* to learn
maths while you posted for no other reason than trying to
understand the people who replied to you!Ivan.So this poster
whoOs opinion, by his own admissions in terms ofmathematics
is worthless, still feels that his opinion is of
value,clearly because heOs at least learned the true nature
of the mathcommunity. JSH writing about me in the Focus on
point of dispute, more math thread ... IOve been
villi'ed by
James, does this make me famous?
=the application of
nonlinear systems (particularly catastrophetheory) to human
behavior. Through the years I have struggled withbalancing
the criticisms of catastrophe methodology and the
heuristicmerits of the theory. IOve read most of the
catastrophe literature andtheory and singularities.Although
the proponents of catastrophe theory are also
somewhatconvincing, I question its applicability and
implementation inpsychological research. IOm at a point
where
I need to make adecision:Do I drop this line of research or do
I continue working toward thedevelopment of catastrophe
applications (e.g., write grants to developsoftware for
analyzing catastrophes/bifurcations in
psychologicaldata)?Kate
> the application of nonlinear
systems (particularly catastrophe> theory) to human behavior.
Through the years I have struggled with> balancing the
criticisms of catastrophe methodology and the heuristic>
merits of the theory. IOve read most of the catastrophe
literature and> theory and singularities.> > Although the
proponents of catastrophe theory are also somewhat>
convincing, I question its applicability and implementation
in> psychological research. IOm at a point where I need to
make a> decision:> > Do I drop this line of research or do I
continue working toward the> development of catastrophe
applications (e.g., write grants to develop> software for
analyzing catastrophes/bifurcations in psychological> data)? KateIt would seem that this is more a question for your
advisor/committee/faculty to help you answer. IMO, as a grad
student it is generallyadvisable to do something that gives
you a high probability ofproducing a respectable dissertation
ASAP. A speculative 'shingexpedition (i.e. one where
youOre
not sure there are any 'sh in thelake as opposed to the
standard risk of just not catching any of 'shthere are) is
not what you should undertake early in your career.However,
if you still want to pursue it, ask around about
fundingprospects, that is is anyone supporing this type of
work? Maybethere is all kinds of homeland securtiy money
being thrown atthis area in an attempt to 'gure out
terrorists or something (donOtlaugh, theyOre
spending
billions on a missle defense system thathasnOt worked yet).
This is where faculty should be able, throughtheir
connections, to help you with information on research
andfunding trends. But unless your advisor knows about this
area andis supportive of the idea, youOre asking for
problems, IMO.Good luck,Russell
> the application of
nonlinear systems (particularly catastrophe> > theory) to
human behavior. Through the years I have struggled with> >
balancing the criticisms of catastrophe methodology and the
heuristic> > merits of the theory. IOve read most of the
catastrophe literature and> > theory and singularities.> >
Although the proponents of catastrophe theory are also
somewhat> > convincing, I question its applicability and
implementation in> > psychological research. IOm at a point
where I need to make a> > decision:> > Do I drop this line of
research or do I continue working toward the> > development of
catastrophe applications (e.g., write grants to develop> >
software for analyzing catastrophes/bifurcations in
psychological> > data)?> > Kate It would seem that this is
more a question for your advisor/committee/> faculty to help
you answer. IMO, as a grad student it is generally> advisable
to do something that gives you a high probability of>
producing a respectable dissertation ASAP. A speculative
'shing> expedition (i.e. one where youOre not
sure there are
any 'sh in the> lake as opposed to the standard risk of just
not catching any of 'sh> there are) is not what you should
undertake early in your career.> However, if you still want
to pursue it, ask around about funding> prospects, that is is
anyone supporing this type of work? Maybe> there is all kinds
of homeland securtiy money being thrown at> this area in an
attempt to 'gure out terrorists or something
(donOt> laugh,
theyOre spending billions on a missle defense system that>
hasnOt worked yet). This is where faculty should be able,
through> their connections, to help you with information on
research and> funding trends. But unless your advisor knows
about this area and> is supportive of the idea, youOre
asking
for problems, IMO.OTOH, if youOre really interested in
'nding
the truth, regardless offunding, then do what you know is
right.Bob Muncaster (http://www.math.uiuc.edu/~muncast/) has
in the past studiedsome applications of math to political
science and sociology, and I assumepsychology is not too
different. I know heOs into dynamical systems, so heshould
be
able to answer your questions.Jon Miller
the application
of nonlinear systems (particularly catastrophe> theory) to
human behavior. Through the years I have struggled with>
balancing the criticisms of catastrophe methodology and the
heuristic> merits of the theory. IOve read most of the
catastrophe literature and> theory and singularities.
Although the proponents of catastrophe theory are also
somewhat> convincing, I question its applicability and
implementation in> psychological research. IOm at a point
where I need to make a> decision: Do I drop this line of
research or do I continue working toward the> development of
catastrophe applications (e.g., write grants to develop>
software for analyzing catastrophes/bifurcations in
psychological> data)? KateWhy study pseudo-science
anyway?
=} Why study pseudo-science anyway?All the better
to refute asinine one-liners, my dear.
} Why study
pseudo-science anyway? All the better to refute asinine
one-liners, my dear.>Are you claiming it isnOt a
pseudo-science?Lurch
> > the application of nonlinear
systems (particularly catastrophe> > theory) to human
behavior. Through the years I have struggled with> >
balancing the criticisms of catastrophe methodology and the
heuristic> > merits of the theory. IOve read most of the
catastrophe literature and> > theory and singularities.> >
Although the proponents of catastrophe theory are also
somewhat> > convincing, I question its applicability and
implementation in> > psychological research. IOm at a point
where I need to make a> > decision:> > Do I drop this line of
research or do I continue working toward the> > development of
catastrophe applications (e.g., write grants to develop> >
software for analyzing catastrophes/bifurcations in
psychological> > data)?> > Kate> > Why study pseudo-science
anyway?Which one? ;-)Russell
Does the natural density of
all positive integers that arethe sum of two odd> primes =
1/2? How about the sum of two squares? The sum oftwo
squarefree>
integers?**************************************************If
the Goldbach conjecture is true, then the answer to your'rst
question is yes.If a number is the sum of two squares then it
canOt have aprime factor that is 3 modulo 4 and whose
highest
power isodd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + .
. .diverges, this would imply that the natural density of
thenumbers that are the sum of two squares is 0. ( [
(1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) =
0.)_________________________________________________________
Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics
and StatisticsYoungstown State UniversityOne University
PlazaYoungstown, OH 44555-0001330-941-1817
> Does the
natural density of all positive integers that are>the sum of
two odd>> primes = 1/2? How about the sum of two squares? The
sum of>two squarefree>>
integers?**************************************************>
If the Goldbach conjecture is true, then the answer to
your>'rst question is yes.If a number is the sum of two
squares then it canOt have a>prime factor that is 3 modulo 4
and whose highest power is>odd. Since the series 1 + 1/3 +
1/7 + 1/11 + 1/19 + . . .>diverges, this would imply that the
natural density of the>numbers that are the sum of two squares
is 0.> ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)I give
up. Supposing that everything you say is true, whichI imagine
it is, how does it follow that the sums of two squareshave
density 0?(My guess is that you deduce somehow that the sum
of thereciprocals of the sums of two squares is 'nite. If so
thatOsinteresting because the sum of 1/(j^2 + k^2) is
in'nite...)__________________________________________________
_______Eric J. Wingler (wingler@math.ysu.edu)>Dept. of
Mathematics and Statistics>Youngstown State University>One
University Plaza>Youngstown, OH
44555-0001>330-941-1817>************************David C.
Ullrich
=Let T be the set of integers that can be written
as a sum of twosquares. I donOt think that the sum of 1/n
for
n in T converges. WhatI get from a rough L-series calculation
(which could probably be maderigorous if it were earlier in
the evening) is that the DirichletseriesD(T,s) = sum over n
in T of 1/n^sdiverges as s --> 1+, but that it blows up
more-or-less like1/sqrt(s-1). This implies that the Dirichlet
density of T isDirichDen(T) := (lim as s-->1+ of D(T,s)*(s-1))
= 0.Presumably the natural density will also be zero, though
since D(T,s)wonOt have an analytic continuation in a
neighborhood of s=1, onecanOt simply apply a standard
Tauberian theorem to get the result.Joe Silverman> > >> Does
the natural density of all positive integers that are> the
sum of two odd> >> primes = 1/2? How about the sum of two
squares? The sum of> two squarefree> >> integers?**************************************************> >If the
Goldbach conjecture is true, then the answer to your> >'rst
question is yes.> >If a number is the sum of two squares then
it canOt have a> >prime factor that is 3 modulo 4 and whose
highest power is> >odd. Since the series 1 + 1/3 + 1/7 + 1/11
+ 1/19 + . . .> >diverges, this would imply that the natural
density of the> >numbers that are the sum of two squares is
0.> > ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)> > I
give up. Supposing that everything you say is true, which> I
imagine it is, how does it follow that the sums of two
squares> have density 0?> > (My guess is that you deduce
somehow that the sum of the> reciprocals of the sums of two
squares is 'nite. If so thatOs> interesting
because the sum
of 1/(j^2 + k^2) is in'nite...)> >
_________________________________________________________Eric J. Wingler (wingler@math.ysu.edu)> >Dept. of
Mathematics and Statistics> >Youngstown State UniversityOne University Plaza> >Youngstown, OH 44555-0001330-941-1817> > > ************************> > David C.
Ullrich
>If a number is the sum of two squares then it
canOt have a>>prime factor that is 3 modulo 4 and whose
highest power is>>odd. Since the series 1 + 1/3 + 1/7 + 1/11
+ 1/19 + . . .>>diverges, this would imply that the natural
density of the>>numbers that are the sum of two squares is
0.>> ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)>I give
up. Supposing that everything you say is true, which>I
imagine it is, how does it follow that the sums of two
squares>have density 0?Let S(p) be the set of positive
integers that are either not divisible by por are divisible
by p^2. The natural density of S is 1 - 1/p + 1/p^2.Any
positive integer that is the sum of two squares must be in
S(p) forall primes p = 3 mod 4. If these primes are p_1, p_2,
p_3, ..., thenatural density of intersection_{j=1}^n S(p_j) is
product_{j=1}^n (1 - 1/p_j + 1/p_j^2) = product_{j=1}^n
exp(-1/p_j) (1+O(1/p_j^2)) <= C exp(-sum_{j=1}^n 1/p_j))which
goes to 0 as n -> in'nity because the series sum_{j=1}^infty
1/p_j diverges.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
>If a
number is the sum of two squares then it canOt have
a>prime
factor that is 3 modulo 4 and whose highest power is>odd.
Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . .
.>diverges, this would imply that the natural density of
the>numbers that are the sum of two squares is 0.> ( [
(1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)>I give up.
Supposing that everything you say is true, which>>I imagine
it is, how does it follow that the sums of two squares>>have
density 0?Let S(p) be the set of positive integers that are
either not divisible by p>or are divisible by p^2. The
natural density of S is 1 - 1/p + 1/p^2.Ah. ThatOs the bit
that hadnOt clicked - couldnOt see where the
minussigns in
the expansion of 1/(1+1/p) were going to come in, but thereit
is. Duh.>Any positive integer that is the sum of two squares
must be in S(p) for>all primes p = 3 mod 4. If these primes
are p_1, p_2, p_3, ..., the>natural density of
intersection_{j=1}^n S(p_j) is >product_{j=1}^n (1 - 1/p_j +
1/p_j^2) > = product_{j=1}^n exp(-1/p_j) (1+O(1/p_j^2))> <= C
exp(-sum_{j=1}^n 1/p_j))>which goes to 0 as n -> in'nity
because the series sum_{j=1}^infty 1/p_j >diverges.Robert
Israel israel@math.ubc.ca>Department of Mathematics
http://www.math.ubc.ca/~israel >University of British
Columbia >Vancouver, BC, Canada V6T
1Z2************************David C. Ullrich
=THe cost of a
long-distance telphone call is determined by a§at fee for teh
'rst 5 minutes and a 'xed amount for each
additional minute.
If a 15-minute telephone call costs $3.25 and a 23-minute
call costs $5.17, 'nd the cost of a 30-minute call. I was
thinking of setting it up as a system of equations but im not
really sure, can someone explain how i would set this problem
up, cause alwyas get stumped on these type of questions.
TIA---= 19 East/West-Coast Specialized Servers - Total
Privacy via Encryption =---
THe cost of a long-distance
telphone call is determined by a§at fee for> teh 'rst 5
minutes and a 'xed amount for each additional minute. If
a15-> minute telephone call costs $3.25 and a 23-minute call
costs $5.17, 'nd> the cost of a 30-minute call.> I was
thinking of setting it up as a system of equations but im
not> really sure, can someone explain how i would set this
problem up, cause> alwyas get stumped on these type of
questions. TIANews==----> http://www.newsfeed.com The #1
Newsgroup Service in the World! >100,000> ---= 19
East/West-Coast Specialized Servers - Total Privacy via
Encryption=---Let x = the §at feeand y = the 'xed amount
after the 'rst 've,then5x + 18y = 5.175x + 10y =
3.25You get
x = .17 and y = .24.Lurch
THe cost of a long-distance
telphone call is determined by a§at fee for>> teh 'rst 5
minutes and a 'xed amount for each additional minute. If
a>15->> minute telephone call costs $3.25 and a 23-minute
call costs $5.17, 'nd>> the cost of a 30-minute call.>>Let x
= the §at fee>and y = the 'xed amount after the
'rst
've>,then5x + 18y = 5.17>5x + 10y = 3.25You get x = .17 and y
= .24.Lurch>No, a §at fee is 'xed and pays for anything up
to
including 5minutes, not a per minute fee. You should have x
in the equationsinstead of 5x. The 'rst 've
minutes cost .85
(the §at fee) and theadditional minutes are .24 each. A 2
minute call would also cost .85because you donOt apply the
§at fee on a per minute basis.--Lynn--Lynn
= >> THe cost of
a long-distance telphone call is determined by a§at feefor> >>
7ï
WEcf
!{.99lèÿÿ
>> boxes about
>> the pc can't communicate with the on problem in
people> >who donOt have any formal training in psychology,
and perhaps> >understandably so.> > *mmmph*> *snort*> *gggg*>
*whaaaaahaaaahaaaaaaahaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aa*> > youOve never> > *heee*> > been> >
*heeeeeeeeeeeeeeeeeeeeeeeeeeeee*> > to> > *mmmph*> >
sci.psychology.psychotherapy> > *heee*> > have you?> > --
astri> *hahahahhahahahaha*> *phew*Well, it made me laugh.But
then almost everything makes me laugh.Larry
 -- astri>
*hahahahhahahahaha*> *phew*Remind yourself you didnOt want 
to
read this,and scolded me for bringing it here.These cutups are
hilarious, like marionettesworked by asylum residents. The
lips move,sounds come out, they dance their ritual jigs,but
the heads are still made all of wood, thedances betray no
meaning, and the sounds tellless than wind moving through the
trees.xanthian.And the best part is, they perform for free.--
Rickert: If LongleyOs views are compatible with credible
scientists in>psychology related 'elds, why is he not posting
in the>sci.psychology newsgroups?>GS: Perhaps you need
remedial reading lessons, Rickert. Behaviorists are in>the
minority, here and elsewhere.Dogmatic behaviorists are in a
minority -- as they deserve to be.However, there are plenty
sensible non-dogmatic behaviorists in AIareas.>Rickert:
Longley regularly takes *interpretations* of experimental
results,>and declares them to be evidence. He seems to not
even be aware that>the interpretations were made on the
assumption of the folk>psychological views that Longley
decries.>GS: He reinterprets the results of experiments
conducted in the mentalist>tradition.Fine. There is no
problem with that.The problem comes when he asserts that his
interepretation is theevidence, and when he insists that
anyone who disagrees with hisinterpretation is ignoring the
evidence.> The interaction between mentalistic concepts and
the experimental>procedures are not as straightforward as you
are making them out to be.Presumably that utterance is a
mechanical response to stimuli youreceived. We can ignore it.
Any as you are making them out to bein your statement is pure
'ction.>Rickert: For example, he cites work of Kahneman and
Tversky. I donOt have>any>problem with their experimental
work. It is sound research. But>what Longley asserts, is that
this provides evidence of>irrationality. But he can only claim
this based on>folk-psychological accounts of rationality. If
he wants to reject>folk psychology, he should equally reject
those accounts of>rationality.>GS: As I have told you many
times (and you have apparently ignored) one must>mention
folk-psychological terms sometimes in order to point to
the>BEHAVIORAL PHENOMENA said to require them. The phenomena
are real enough,>but the mentalistic accounts are rubbish.It
is up to Longley to provide a behaviorist account of
rationality(if he can). But he doesnOt. He takes the
mentalist account, andasserts that it is
behaviorist.>Rickert: He regularly cites Quine. But QuineOs
accounts of science (as in>folk-psychological assumptions.
Likewise, much of QuineOs writing on>language is derivative
of folk-psychological assumptions.>Strictly speaking, radical
behaviorism also is heavily dependent on>folk psychological
assumptions. If you removed all such assumptions>from radical
behaviorism, little of use would remain.>GS: On the contrary,
radical behaviorism is useful precisely because it>treats the
phenomena said to require mentalistic notions without pointing
to>such entities as causes.Then I guess we should say that
physics, chemistry, geology are allmentalistic. For they
certainly deal with causes.
<2eb3320723dad25fefbb4e1a8bf9b934@news.teranews.com>
<_qreb.23585$ev2.4804540@newssrv26.news.prodigy.com>
Rickert: He regularly cites Quine. But QuineOs accounts of
science (as in>>folk-psychological assumptions. Likewise,
much of QuineOs writing on>>language is derivative of
folk-psychological assumptions.This really does reveal how
little you have understood. You confuse your own
misunderstandings with what I have written and as you slowly
start to grasp what IOve been saying you have the audacity
and arrogance to think itOs your own idea!. read Fragments
properly. Why do you think experimental psychology studies
folk psychology? What is the relationship between Anomalous
Monism and Eliminative Materialism?IOve made this point 
about
your muddled, solipsistic thinking time and time again but
like some of the other ignoramuses now frequenting this
newsgroup the point doesnOt sink in because you 
donOt listen
and you wonOt read! Either you have inadequate understanding
of what behaviour science is all about or you are guilty of
perhaps one of the most egregious sins one can 'nd in an
academic - time will tell.>Strictly speaking, radical
behaviorism also is heavily dependent on>>folk psychological
assumptions. If you removed all such assumptions>>from
radical behaviorism, little of use would remain.>GS: On the
contrary, radical behaviorism is useful precisely because
it>>treats the phenomena said to require mentalistic notions
without pointing to>>such entities as causes.>>Then I guess
we should say that physics, chemistry, geology are
all>mentalistic. For they certainly deal with causes.>He said
entities !-- David Longley
<2eb3320723dad25fefbb4e1a8bf9b934@news.teranews.com>
<_qreb.23585$ev2.4804540@newssrv26.news.prodigy.com>
Rickert: If LongleyOs views are compatible with credible
scientists in>>psychology related 'elds, why is he not
posting in the>>sci.psychology newsgroups?>GS: Perhaps you
need remedial reading lessons, Rickert. Behaviorists are
in>>the minority, here and elsewhere.>>Dogmatic behaviorists
are in a minority -- as they deserve to be.>However, there
are plenty sensible non-dogmatic behaviorists in
AI>areas.>Rickert: Longley regularly takes
*interpretations* of experimental results,>>and declares them
to be evidence. He seems to not even be aware that>>the
interpretations were made on the assumption of the
folk>>psychological views that Longley decries.>GS: He
reinterprets the results of experiments conducted in the
mentalist>>tradition.>>Fine. There is no problem with
that.>>The problem comes when he asserts that his
interepretation is the>evidence, and when he insists that
anyone who disagrees with his>interpretation is ignoring the
evidence.> The interaction between mentalistic concepts and
the experimental>>procedures are not as straightforward as you
are making them out to be.>>Presumably that utterance is a
mechanical response to stimuli you>received. We can ignore
it. Any as you are making them out to be>in your statement is
pure 'ction.>Rickert: For example, he cites work of 
Kahneman
and Tversky. I donOt have>>any>>problem with their
experimental work. It is sound research. But>>what Longley
asserts, is that this provides evidence of>>irrationality.
But he can only claim this based on>>folk-psychological
accounts of rationality. If he wants to reject>>folk
psychology, he should equally reject those accounts
of>>rationality.>GS: As I have told you many times (and you
have apparently ignored) one must>>mention folk-psychological
terms sometimes in order to point to the>>BEHAVIORAL
PHENOMENA said to require them. The phenomena are real
enough,>>but the mentalistic accounts are rubbish.>>It is up
to Longley to provide a behaviorist account of
rationality>(if he can). But he doesnOt. He takes the
mentalist account, and>asserts that it is behaviorist.>Are
you really this stupid? Go and re-read what I have written in
Fragments - *carefully*.-- David Longley
=My favorite
fragments:You said this: >The two dogmas listed above, are,
it will be argued, a consequence of a>simple, but quite
radical misinterpretation of the data which inspired>this
revolution and which has subsequently adversely changed the
direction>of much of contemporary research in experimental
psychology. The most important item in that sentence is where
you saidmisinterpretation. This means itOs your OPINION. NOT
fact. Opinionsdiffer. Are you going to call me stupid now?You
would prefer human behaviour to be as regular as clockwork, a
result ofpurely quantitative processes, and you say so,
constantly. You explicitlyignore all qualitative factors as
irrelevant (or should Isay...intensional?) and repeatedly
expound on the glories of thequantitative, pure numbers, in
which you apparently hope to reducebehaviour to mere
formulae.Hence, you said: >IF such_and_such_behaviour THEN
food_pops_out_at_X >IF so_and_so_behaviour THEN NOT
food_pops_out_at_X.as applied to a study of the behaviour of
rats. Rats are not intelligent.They lack the cognitive
abilities you so despise. They lack, dare I say,INUTITION.For
the sheer fun of it, letOs apply that to you, using your own
rules.IF Longley_Behaviour_Is_Challenged THEN
Accuse_challenger_of_irrationalityIF Longley_Is_Agreed_with
THEN Praise_person_in_agreementI think that works for me. How
do you feel about it?bp
=[another copy, including trailing
replicated copiesof the orginals to which he was responding
and whichhe had already cut and pasted into the main body
ofhis discourse]Having no clue as to the public norms of
Usenet,Mr. Sizemore manages to post most of what he hasto say
in quadruplicate.Sigh.xanthian.--
=This is all you can
muster? Somehow I 'gured. Anyway, when most of thepeople
around here deserve some respect, IOll take the time to trim
posts.BTW, it is Dr. Sizemore to lamers like you.> [another
copy, including trailing replicated copies> of the orginals
to which he was responding and which> he had already cut and
pasted into the main body of> his discourse]>> Having no clue
as to the public norms of Usenet,> Mr. Sizemore manages to
post most of what he has> to say in quadruplicate.>> Sigh.>>
xanthian.> --
 [another copy, including trailing
replicated copies> of the orginals to which he was responding
and which> he had already cut and pasted into the main body
of> his discourse]Style notwithstanding, youOd think some
common-sense editing would come intoplay...oh wait...that
would be folk psychology! Evil!bp
 The German words ACHT
(8) and EINS (1) represent numbers and have> their letters in
alphabetical order. I can think of only> one de'nite English
example, excluding one-letter answers (e, i)> (but have two
other English examples if we get pushy).> Roman numerals such
as CCIV (204) are not acceptable,> but a letter may be
duplicated if all occurrences are adjacent.> DonOt forget
that some numbers have multiple names, > such as FOURTH or
QUARTER for 1/4.> > For readers whose native language is not
English, > how many examples can you 'nd in your
language?LetOs try some more exotic languages. (Accuracy not
guaranteed sincenone of these languages is my native
tongue.)Chinese (Mandarin)Well in its own script the question
cannot be asked since it does notrepresent the sound and there
is one character for a whole word. Butthere is a standard
romanization called pin yin. In that:0 ling No1 yi No2 er
Yes3 san No4 si No5 wu No6 liu No7 qi No8 ba No9 jiu No10 shi
NoNo further words are required to get up to 99 but all of 11
to 99include shi and hence do not qualify.100 bai No1000 qian
No10000 wan NoI forget any higher than that but these words
get you up to 99,999. So Chinese manages 2 out of 100,000.
Ahead of Finnish but nonethelessa low score.All these words
should have tone marks but I have no idea how to writethose
in a newsgroup.Tagalog (Filipino)1 isa No2 dalawa No3 tatlo
No4 apat No5 lima No6 anim No7 pito No8 walo No9 siyam No10
sampu NoAll of 11 to 19 start labi and hence are no20, 30, up
to 90 are compounds of 2 to 9 and hence are no100 daan no1,000
libu noI donOt know any bigger numbers. They are never used.
If the subjectwas money, Spanish numbers would be used. If
the subject wastechnical then English would be used.So
another zero langauge. MalayI would need to check but I would
expect a similar success rate toTagalog. They are related
languages are both are very fond of theletter a which rules
out most words.ThaiI would like to do this but it has its own
alphabet and I would haveto revise it. It has loads of
letters, with lots of redundancy. Thealphabet starts with a
letter which is between English g and k. It isthen followed
by four ks.JapaneseAs for Chinese there is the script issue
but again there is a standardromanization called romanji. I
would need to refresh my memory. ichi, ni, san.J
=I am
offering free basic website monitoring, looking for some
peopleinterested in trying it
out.http://www.mjcsoftware.com/main.asp?page=servicesMike
Curry
=The unfortunate reality that *professional*
mathematicians have beenworking to instill false mathematics
into many of you has made my joba lot more dif'cult. However,
my hope is that this latest expositionwill work to foil these
anti-mathematicians.ConsiderP(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)where as I demonstrate
with P(x)=11x+123=11^2 + 11x + 2 the specialgrouping is to
factor into non-polynomial factors. In my example Iused x but
now I use m where the letter change shouldnOt matter but 
Inote
it in case some 'nd it confusing.That factorization isP(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)so the aOs are roots of
the cubic a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)and
I have three factorsg_1=(a_1 x + uf), g_2=(a_2 x +
uf),g_3=(a_3 x + uf)(notice the *symmetry* here)where P(0)
gives the constant term, of course, and P(0)=f^2(3x u^2 + u^3
f), so two of the aOs go to 0 when m=0 which is also seen 
from
the cubicde'ning the aOs.Then arbitrarily 
picking a_1 and a_2
as the ones that go to 0 at m=0you haveg_1=uf, g_2=uf,
g_3=3x+uf.(Notice now though that the symmetry is broken.
ItOs that brokensymmetry which helps show why Galois Theory
canOt be used to attack myargument.)But P(0)/f^2=3x u^2 + 
u^3
f as f divides off only two of the gOs whilewith the third 
it
is blocked, as long as it is coprime to 3 and x, soassume it
is, and assume as well that f is coprime to u.Then it follows
from the constant terms that g_1 and g_2 each have afactor
that is f.Many have disputed that conclusion, so IOll
elaborate further.Now consider that asa_1 a_2 a_3 = f^2(m^3
f^4 - 3m^2 f^2 + 3m) = f^2 m(m^2 f^4 - 3m f^2 + 3)and as they
go to 0 when m=0, two of the aOs have a factor of m,
andconsider the possibility that factor is sqrt(m).But a_3
does NOT go to 0 when m=0, but consider the possibility
thatat m=0, a_3 - 3 has a factor that is m.Those choices are
consistent witha_1 a_2 a_3 = f^2 m(m^2 f^4 - 3m f^2 + 3) so
now let a_1 = sqrt(m)h_1(m), a_2 = sqrt(m) h_2(m), and a_3 =
mh_3(m) + 3where I introduce the hOs as functions of m.Then
from g_1=(a_1 x + uf), g_2=(a_2 x + uf),g_3=(a_3 x + uf)I
haveg_1=(sqrt(m)h_1(m) x + uf), g_2=(sqrt(m)h_2(m) x +
uf),g_3=((mh_3(m) + 3) x + uf)which works for m=0.Now then
P(m) = g_1 g_2 g_3, but P(m) has a factor that is f^2 asP(m)
= f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3
f)and dividing off f^2 gives a result with a constant term
that iscoprime to f, which requires that g_1 and g_2 have a
factor that is f,while g_3 does not.Notice here that the
non-polynomial nature of the factorization is notterribly
complicated with that possibility.So then, is it possible
that both a_1 and a_2 go mm-134
I have a funciton:F(u,v) = Integrate ( 0, 2*pi )
Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dta=constant>0 i=sqrt(-1)we
note thatF(u,0) = 2*pi*J_0(u*a)F(0,v) = 2*pi*J_0(v*a)where
J_0(x) = bessel function of 'rst kind; order zero.also if you
assume that F(u,v) is continuous w.r.t u and v,
anddifferentiate under the integral sign i *think* you can
show thatu*dF/dv - v*dF/du = 0 where the above are taken as
partial derivitives: not sure about thatidentity yet - might
have slipped with my algebra. dont know if knowingthat will
help resolve exactly what F is.i guess a good starting point
would be knowing how one calculates theintegral of say F(u,0)
to get the bessel function - however i cant recallthat at the
moment...im assuming it might possibly be by series, and
willwork on this.any ideas most appreciated.cheersmoth
F(u,v) = Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) )
dt>> a=constant>0 i=sqrt(-1)The OP has already solved the
problem him(her)self, but I would like tosuggest an alternate
solution that is much simpler:Rewrite u*Cos(t) + v*Sin(t) =
r*Cos(t-phi), for some values r and phi.Then we have r =
Sqrt(u^2 + v^2). The value of phi is irrelevant, since
theintegral is over a complete period. The function F(u,v) is
independent ofphi, and we may therefore just as well set phi
equal to zero. The integralthen becomes:Integrate ( 0, 2*pi )
Exp(-i*a*(r*Cos(t)) ) dt,which just happens to be the original
integral with u=r and v=0.-Michael.
=Some more:>> have a
funciton:>> F(u,v) = Integrate ( 0, 2*pi )
Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dt>> a=constant>0 i=sqrt(-1)>>
we note that>> F(u,0) = 2*pi*J_0(u*a)> F(0,v) =
2*pi*J_0(v*a)>> where J_0(x) = bessel function of 'rst kind;
order zero.>> also if you assume that F(u,v) is continuous
w.r.t u and v, and> differentiate under the integral sign i
*think* you can show that>> u*dF/dv - v*dF/du = 0>assuming
the above is correct you can show via the method of
characteristicsthat a function satisfying the above + the
intiial condition of say F(u,0) =2*pi*J_0(u*a) isF(u,v) =
2*pi*J_0(a*Sqrt( u^2 + v^2 ))however is this actually a
solution of the original integral as statedabove? (i think it
may be...but would like to prove it)cheersmoth
=some more
again ;-)> Some more:> have a funciton:>> F(u,v) =
Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dt>>
a=constant>0 i=sqrt(-1)>> we note that>> F(u,0) =
2*pi*J_0(u*a)> F(0,v) = 2*pi*J_0(v*a)>> where J_0(x) = bessel
function of 'rst kind; order zero.>> also if you assume that
F(u,v) is continuous w.r.t u and v, and> differentiate under
the integral sign i *think* you can show that>> u*dF/dv -
v*dF/du = 0> assuming the above is correct you can show via
the method ofcharacteristics> that a function satisfying the
above + the intiial condition of say F(u,0) 2*pi*J_0(u*a)
is>> F(u,v) = 2*pi*J_0(a*Sqrt( u^2 + v^2 ))>> however is this
actually a solution of the original integral as stated> above?
(i think it may be...but would like to prove it)>> cheers>
mothi am sure this is correct now as i can verify that
2*Integrate [ 0, oo] r*Exp(-r^2)*J_0(2*Sqrt(t*(u^2+v^2))
)drequates toExp(-t*(u^2+v^2))which is what i started with.
looks like i have answered my own problem ;-)sorry to waste
board space.cheersmoth
Yes, . means multiplication. I
thought that that was a kind of universal>notation.No,
because the . is too small and too low. A higher, bigger dot
meansmultiplication.John
Savardhttp://home.ecn.ab.ca/~jsavard/index.html
Yes, .
means multiplication. I thought that that was a kind of
universal>notation.No, because the . is too small and too
low. A higher, bigger dot means> multiplication.Not
universally. English typographical preference used to be
forrepresenting the decimal separator by such a raised
dot.Phil-- Unpatched IE vulnerability: DNSError folder
disclosureDescription: Gaining access to local security
zonesReference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html==
=Yes, . means multiplication. I thought that that was a
kind of universal>notation.No, because the . is too small and
too low. A higher, bigger dot means> multiplication.Yes, but
AFAIK that cannot be obtained in ASCII. The closest thing
there isis ..Jose Carlos Santos
>Yes, . means
multiplication. I thought that that was a kind of universalnotation.> No, because the . is too small and too low. A
higher, bigger dot means> multiplication.Yes, but AFAIK that
cannot be obtained in ASCII. The closest thing there is> is
..> Jose Carlos SantosSo many computer languages use *, that
it has almost become the newsgroup defacto standard for
multiplications. It is certainly less liable to be
misinterpreted than ..
>You seem to forget that the
Canadians kicked our ass in two con§icts....>Really? Which
two?One of them was vaguely contemporary with the War of
1812, as Irecall, but I donOt remember *another*
one.However,
I doubt very much that, given the *current* ratio of
militarymight between our two countries, that Canadians are
optimistic aboutbeing able to defend themselves against a
determined attack from thesouth. Rather, they have faith in
the good, peaceful, and democraticintentions of that
nation.Perhaps, unfortunately, unlike some of its brothers to
the south.But just today in the National Post, I found a
*reason* for the UnitedStates to annex Canada.It seems that
the sensible ordinary people in both countries, who wantthe
same things, are being forced to put up with governments
ofopposite kinds which are in many ways different from what
they wouldwant because of bloc voting groups. In the States,
they have the BibleBelt. In Canada, we have Quebec.So if we
eachother out!That means, of course, it would be Bill Clinton
and not George Bushwho would want to annex Canada.John
Savardhttp://home.ecn.ab.ca/~jsavard/index.html
>You
seem to forget that the Canadians kicked our ass in two
con§icts....>>Really? Which two?> One of them was vaguely
contemporary with the War of 1812, as I> recall, but I donOt
remember *another* one.HeOs probably referring to the
attempt
to move into the land north of the 49th. As some sticklers
here point out, there was no Canada then.However, I doubt
very much that, given the *current* ratio of military> might
between our two countries, that Canadians are optimistic
about> being able to defend themselves against a determined
attack from the> south. Rather, they have faith in the good,
peaceful, and democratic> intentions of that nation.Probably
not. IOm very lucky though, because if it ever came to that,
I have a right to work in the UK, and IOd be gone, in a
§ash.Perhaps, unfortunately, unlike some of its brothers to
the south.But just today in the National Post, I found a
*reason* for the United> States to annex Canada.It seems that
the sensible ordinary people in both countries, who want> the
same things, are being forced to put up with governments of>
opposite kinds which are in many ways different from what
they would> want because of bloc voting groups. In the
States, they have the Bible> Belt. In Canada, we have
Quebec.Interestingly though, Quebec in another one of those
things that the US has taken interest in and theyOve made it
clear, on more than one occasion, that they donOt want
Quebec
to leave, while some Canadians do. You had your civil war in
the US, weOve just been dealing with the Quebec issue, it
seems forever. Perhaps because we arenOt so anxious for
civil
war, never have been. Besides, the West has been making a lot
of noises about being unhappy about things too.So if we could
out!That means, of course, it would be Bill Clinton and not
George Bush> who would want to annex Canada.Eiterh way, IOd
leave, cross the pond, back to the land of my mother and my
fatherOs ancestors.DJohn Savard>
http://home.ecn.ab.ca/~jsavard/index.html
IOm searching a
conformal mapping that transform a rectangle in a ring>bounded
by two circles with the same center, i.e., a disc with
a>circle hole with the same center.>The horizontal edges of
the rectangle must transform in the two circle>(with the same
center), while the two vertical edges must coincide at>the end
of the mapping and they must be transformed in a
vertical>radius line going from the inner to the outer
circles.>MAy anybody help me?Elliptic
integrals.http://www.hypermaths.org/quadibloc/maps/mcf0703.
htmOh, you donOt want to transform just a *plain* rectangle.
I thinkyouOll have to go to the library and look this one
up.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html
>
Anyone know why Grad Schools will not accept your application
unless>> you have a BachelorOs? Is it not possible to have
the required>> knowledge and no degree? Is it so very hard to
verify that oneOs claim>> of knowledge.> Occasionally
someone
_is_ admitted without a bachelorOs degree. But> these are
rare exceptions. In mathematics, if an American applicant>
had a top score on the Graduate Record Exam subject exam,
then the math> department would probably be so eager to have
him/her that they would> intervene with the admissions people
to make an exception.Really? The GRE subject exam in math is
kind of trivial -- tests> perhaps what a sophomore math major
ought to know, and doesnOt> require any
dif'cult reasoning,
just techniques. IOd> think more weight would be given to
the
GRE Analytic score, which> I understand has now been or will
shortly be eliminated(?). what exactly does the GRE analytic
section (or for that matter theGMAT analytic section) really
measure? I achieved SAT scores above1300 (in 1970 before the
test was watered down) and GRE scores of720V/730M. The
results on these tests and IQ tests always placed mein the
98th percentile or so, yet my GRE analytic scores were
amediocre 60th percentile. What I remember of the test is
that none ofthe items were particularly dif'cult, just that
the items werelong-winded and highly detailed word problems;
while I probablyanswered everything I saw correctly, I
didnOt
get to complete or evenlook at several of the other items on
the test. Later on I read thatthis section (and the GMAT
section I mention above) seem to be highlysensitive to test
coaching techniques (Kaplan/Princeton Review). I.e.these
sections have less to do with aptitude and more to do with
testtaking knowledge: set up a matrix or table, eliminate
wrong choices,and donOt spend too much time on any one of
the
items. Withaptitude sections like the math and verbal sections
of the GRE andSAT you donOt have to keep these test taking
techniques in mind, ifyou know the answer or how to compute
it, you donOt have to waste alot of time getting to the
point
where you can write down your answer.
IOve actually seen
a trisection device with no moving parts at all.> ItOs
basically a strip with one end having square and circular>
attachments at the end. By placing the object in the right
position> relative to a given angle, the trisectors of the
angle are identi'ed.That rings a bell. Does it look like the
image on the following
page?http://www.math.cornell.edu/~dwh/books/eg00/prob14-4.
htmlAnd hereOs a page with a bunch of stuff (mostly cranky),
including onetrisection with a carpenterOs
square:http://www.jimloy.com/geometry/trisect.htm-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
cf(e^(1/x) = [1, x-1, 1 1,
3x-1, 1, 1, 5x-1, 1, ...]This can be obtained by a
transformation of LambertOs continued fraction for tan, or
equivalently(e^(1/x)-1)/(e^(1/x)+1) = tanh(1/(2x)) = [0, 2x,
6x, 10x, ... ]If this is z, then e^(1/x) = (1+z)/(1-z)Now if
y = (1+z)/(1-z), and z = 1/(2qx + z1), theny = 1 + 1/(qx-1 +
1/(1 + 1/y1)), then y1 = (1+z1)/(1-z1)and thuse^(1/x) = [1;
1768 or so.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
=Robert
5x-1, 1, ...]> This can be obtained by a transformation of
LambertOs continued fraction > for tan, or
equivalently(e^(1/x)-1)/(e^(1/x)+1) = tanh(1/(2x)) = [0, 2x,
6x, 10x, ... ]If this is z, then e^(1/x) = (1+z)/(1-z)Now if
y = (1+z)/(1-z), and z = 1/(2qx + z1), then> y = 1 + 1/(qx-1
+ 1/(1 + 1/y1)), then y1 = (1+z1)/(1-z1)> and thus> e^(1/x) =
[1; x-1, 1, 1, 3x-1, 1, 1, 5x-1, 1, ... ]This all dates back
to 1768 or so.> They had power and a superb elegance, those
days! Wow.So IOve found a red wine of already more than 200
years...a nice present just at the birthday-weekend
:-)Gottfried Helms
=let I=k[x_1,x_2,..,x_n] is polinomial
ring over 'eld of char=0 and> I_n - subspase gomogenius
polinomial of power n. Let sl_2 - 3 -> dimesional simple lie
algebra wich act at I_n in usual way. How 'nd a> irreducible
components of decomposition of this representation? Need> 'nd
something like as formulae of (Klebsh-Gordon)for tensor
product.You use n in two different ways. Let I_k be the
k-homogeneous part.What was the action of sl_2 again?. The
way I see it I_k is naturallyacted upon by sl_n for all k.
But these sl_n-modules are all simple(in characteristic zero)
so there is no decomposition. Obviouslysomething is
wrong.Jyrki Lahtonen, Turku, Finland
=f and abs:R -> R, x
-> |x|-
=Frederick S. Bustamante> can someone help me
with these practice problems?>> 1. Let f:X-->D for some
connected metric space X and some discrete> space D. Show f
is continuous iff f is constant.Any constant function is
continuous, of course, and if f is not constant,consider the
inverse images of two disjoint (perforce open) subsets of D.>
2. Let f:X-->R for some metric space X. If f is continuous,
then |f|> is continuous.|f| is the composition of two
continuous functions. (IOm trying to do thehinting
thing:)LH
=Is there a sequence S of sets such that lim card
S_n < card lim S_n, where card denotes cardinality, both
limits exist, and the inequality is strict? Here trans'nite
limits are considered to exist as long as they are
well-de'ned. Also, feel free to call on the axiom of
choice.If so, does there exist such a sequence with lim S_n
'nite?-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
Is there a sequence S of
sets such that lim card S_n < card lim S_n, > where card
denotes cardinality, both limits exist, and the inequality >
is strict? Here trans'nite limits are considered to exist as
long > as they are well-de'ned. Also, feel free to call on
the axiom of choice.If so, does there exist such a sequence
with lim S_n 'nite?Yes, I think it is possible.
LetOs try
this:Let S_n = all rationals in [0,1) whose decimal expansion
has onlyfindigits after the decimal point (not counting the
in'nite number of0Os which follow). This may
need to be
tidied up a bit to removealternative formats, like .5 =
.4999...Now card S_n = 10*n, thus lim card S_n = Aleph_0.
However lim S_n isthe set of all reals on [0,1), thus card
lim S_n is 2^Aleph_0.Thus lim card S_n < card lim S_n .Note:
I am a little unsure about the argument that lim card S_n
=Aleph_0, so maybe somebody can correct me if this is
invalid.
> Is there a sequence S of sets such that lim
card S_n < card lim S_n, >> where card denotes cardinality,
both limits exist, and the inequality >> is strict? Here
trans'nite limits are considered to exist as long >> as they
are well-de'ned. Also, feel free to call on the axiom of
choice.>> >> If so, does there exist such a sequence with lim
S_n 'nite?> Yes, I think it is possible. LetOs
try this:> Let
S_n = all rationals in [0,1) whose decimal expansion has only
n> digits after the decimal point (not counting the in'nite
number of> 0Os which follow). This may need to be tidied up
a
bit to remove> alternative formats, like .5 = .4999...> Now
card S_n = 10*n, thus lim card S_n = Aleph_0. However lim S_n
is> the set of all reals on [0,1), thus card lim S_n is
2^Aleph_0.No, because lim S_n is not the same thing as the
closure of the limit.The only numbers that are eventually in
S_n are the rationals with terminatingdecimals. You donOt
even get 1/3, for example.> Thus lim card S_n < card lim S_n
.No, card lim S_n = aleph_0 = lim card S_n < card (cl lim
S_n) = c.> Note: I am a little unsure about the argument that
lim card S_n  Aleph_0, so maybe somebody can correct me if
this is invalid.That part is ok.-- Dave SeamanJudge YohnOs
mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
Is there a sequence S of sets such that
lim card S_n < card lim S_n, >where card denotes cardinality,
both limits exist, and the inequality >is strict? Here
trans'nite limits are considered to exist as long >as they
are well-de'ned. Also, feel free to call on the axiom of
choice.If by sequence you mean an ordinary sequence indexed
by thepositive integers then no: This is clearly impossible
if lim S_n is'nite (in that case there exists N so that S_n
is contained in S_kfor all k > N), and if lim S_n is
countably in'nite then card S_ncannot be bounded by a
'nite
number - 'nally, if lim S_n isuncountable then the _union_ of
countably many sets ofsmaller cardinality has smaller
cardinality. (I believe thatlast bit needs AC.)If youOre
considering more general sequences, for exampleindexed by
ordinals, then the answer is yes - for examplethe countable
ordinals constitute a sequence where thelimit of the
cardinality is aleph_0 but the cardinality of thelimit is
aleph_1.>If so, does there exist such a sequence with lim S_n
'nite?************************David C. Ullrich
> >Is
there a sequence S of sets such that lim card S_n < card lim
S_n, >>where card denotes cardinality, both limits exist, and
the inequality >>is strict? Here trans'nite limits are
considered to exist as long >>as they are well-de'ned. Also,
feel free to call on the axiom of choice.>> >>If by
sequence you mean an ordinary sequence indexed by
the>positive integers>I do.> then no: This is clearly
impossible if lim S_n is>'nite (in that case there exists N
so that S_n is contained in S_k>for all k > N),>It is way too
early for me right now, but I believe this is false: E.g., let
S_n = {1,n}.> and if lim S_n is countably in'nite then card
S_n>cannot be bounded by a 'nite number - >How do we show
that, exactly?>'nally, if lim S_n is>uncountable then the
_union_ of countably many sets of>smaller cardinality has
smaller cardinality. (I believe that>last bit needs AC.)>I
see what you are getting at - that if card lim S_n > omega,
then it must be sup card S_n - but I need more sleep to
convince myself about the details of that one.-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
>> >Is
there a sequence S of sets such that lim card S_n < card lim
S_n, >where card denotes cardinality, both limits exist,
and the inequality >is strict? Here trans'nite limits are
considered to exist as long >as they are well-de'ned. Also,
feel free to call on the axiom of choice. >>If by sequence
you mean an ordinary sequence indexed by the>>positive
integers> I do.> >> then no: This is clearly impossible if
lim S_n is>>'nite (in that case there exists N so that S_n is
contained in S_k>>for all k > N),> It is way too early for
me right now, but I believe this is false: > E.g., let S_n =
{1,n}.> But what is the limit of this in your opinion? In
mine it would just beOtheO two point set
{*,&}, lim has a
Ouniversal propertyO categorical interpretation
which
IOmcertainly using, and I would think David is too, but
possibly withoutwanting to say so.here it means ifS_1 --> S_2
--> S_3 ...has a limit S, it is OtheO object
(set in this
case) that has mapsto it from every S_i, with the property
that the map S_j to S is the sameas mapping to S_k (k>i) and
then mapping to S.in your example, the maps between S_i and
S_(i+1) I assume takes 1 to 1but what does it do to i? if it
sends it to i+1, then youOre identifyingthem, and the limit
would be the two point set.ItOs not just the sets you need
to
consider but the map between sets.>> and if lim S_n is
countably in'nite then card S_n>>cannot be bounded by a
'nite
number - > How do we show that, exactly?> Other wise after a
certain point in the sequence you are just identifyingelements
of the set, thatOs what it means if there is a uniform bound
onthe size of the sets. The way the limit gets bigger is to
have bigger andbigger nested subsets.>>'nally, if lim S_n
is>>uncountable then the _union_ of countably many sets
of>>smaller cardinality has smaller cardinality. (I believe
that>>last bit needs AC.)> I see what you are getting at -
that if card lim S_n > omega, then it > must be sup card S_n
- but I need more sleep to convince myself about > the
details of that one.
>> >> >Is there a sequence
S of sets such that lim card S_n < card lim S_n, >>where
card denotes cardinality, both limits exist, and the
inequality >>is strict? Here trans'nite limits are
considered to exist as long >>as they are well-de'ned.
Also, feel free to call on the axiom of choice.> >If by
sequence you mean an ordinary sequence indexed by
the>positive integers> I do.>> > then no: This is
clearly impossible if lim S_n is>'nite (in that case there
exists N so that S_n is contained in S_k>for all k >
N),> It is way too early for me right now, but I believe
this is false: >> E.g., let S_n = {1,n}.>> >>But what is the
limit of this in your opinion? In mine it would just
be>OtheO
two point set {*,&}, ??? What he said _is_ a counterexample to
what IOd written (becausebut the limit of the sets {1,n} is
the one-point set {1}.>lim has a Ouniversal
propertyO
categorical interpretation which IOm>certainly using, and I
would think David is too, but possibly without>wanting to say
so.>>here it means if>>S_1 --> S_2 --> S_3 ...>>has a limit S,
it is OtheO object (set in this case) that has
maps>to it from
every S_i, with the property that the map S_j to S is the
same>as mapping to S_k (k>i) and then mapping to S.>>in your
example, the maps between S_i and S_(i+1) I assume takes 1 to
1>but what does it do to i? if it sends it to i+1, then
youOre
identifying>them, and the limit would be the two point
set.>>ItOs not just the sets you need to consider but the
map
between sets.This may well be a reasonable and standard
notion, but itOs certainlynot what I meant by lim S_n, and
not what I think he meant either.The notion of the limit of a
sequence of sets has at least one otherperfectly standard
meaning that has nothing to do with what youwrite here: S =
lim S_n means that every element of S is in S_nfor large n
and every non-element of S is a non-element of S_nfor large
n. (Ie, the indicator function of S_n converges pointwiseto
the indicator function of S.)> and if lim S_n is countably
in'nite then card S_n>cannot be bounded by a
'nite number -
> How do we show that, exactly?>> >>Other wise after a
certain point in the sequence you are just
identifying>elements of the set, thatOs what it means if
there is a uniform bound on>the size of the sets. The way the
limit gets bigger is to have bigger and>bigger nested
subsets.>'nally, if lim S_n is>uncountable then the
_union_ of countably many sets of>smaller cardinality has
smaller cardinality. (I believe that>last bit needs
AC.)> I see what you are getting at - that if card lim
S_n > omega, then it >> must be sup card S_n - but I need
more sleep to convince myself about >> the details of that
one.************************David C. Ullrich
>>But what
is the limit of this in your opinion? In mine it would just
be>>OtheO two point set {*,&}, ??? What he
said _is_ a
counterexample to what IOd written (because> but the limit
of
the sets {1,n} is the one-point set {1}.> Not if you were to
de'ne a map between {1,n} and {1,n+1} that, purely asa map of
sets, identi'ed 1 with 1 and n with n+1. As sets, the limit
isgoing to be a set with two elements, whih=ch i just denoted
by * and & forthe sheer hell of it.The problem is that my
notion of lim requires maps between the sets.If you were to
send both 1 and n to 1 in S_{n+1} then the limit would bethe
one point set.>>lim has a Ouniversal propertyO
categorical
interpretation which IOm>>certainly using, and I would think
David is too, but possibly without>>wanting to say
so.>>here it means if>>S_1 --> S_2 --> S_3 ...>>has a
limit S, it is OtheO object (set in this case)
that has
maps>>to it from every S_i, with the property that the map
S_j to S is the same>>as mapping to S_k (k>i) and then
mapping to S.>>in your example, the maps between S_i and
S_(i+1) I assume takes 1 to 1>>but what does it do to i? if
it sends it to i+1, then youOre identifying>>them, and the
limit would be the two point set.>>ItOs not just the sets
you need to consider but the map between sets.This may well
be a reasonable and standard notion, but itOs certainly> not
what I meant by lim S_n, and not what I think he meant
either.> The notion of the limit of a sequence of sets has at
least one other> perfectly standard meaning that has nothing
to do with what you> write here: S = lim S_n means that every
element of S is in S_n> for large n and every non-element of S
is a non-element of S_n> for large n. (Ie, the indicator
function of S_n converges pointwise> to the indicator
function of S.)> The category theory style one says that an
element of the limit is in oneof the sets, and then every one
after that too, and its images areidenti'ed. The natural Idea
of lim youOre using (and it is much more natural, andeasier
to think about on the surface) is better than mine because it
candeal with suggestive things like the S_i in the counter
example, where wekeep special meanings for the letters in the
set. I would have to tell youhow to OannihilateO
elements
which in general I canOt, because IOm
hidingsomething about
what I really mean by lim: you need an indexing categoryand
some functors lying around, but I suspect you either know
this alreadyor donOt care.I can force mine to be equivalent
by saying Osuppose I want to deal withsituations where S_i
OmapsO to S_{i+1} but i donOt
want to explicilty map ito
anything (for the 'rst time) (again keeping with the above
counter example). Then I simply remove i from the set, and
all its preimages from the previous sets.OAs it is, if one
is
allowed to just kill off elements at will, then it isalways
possible to create these examples of lim card not being
related tocard lim, necessarily.Take a vector space X of
abitrary dimension over a 'eld of abitrarycardinality,
de'ne
a chainX --> X --> X > ...with the zero map at every
arrow, then the lim is the zero vectorspace in my way of
thinking.If you want it to be X then use the identity map
between them.Oh, and IOve not apologized for presuming to
tell you what you werethinking, sorry.It wasnOt very clearly
explained, was it? sorry.>> and if lim S_n is countably
in'nite then card S_n>>cannot be bounded by a
'nite number
- > How do we show that, exactly?> >>Other wise
after a certain point in the sequence you are just
identifying>>elements of the set, thatOs what it means if
there is a uniform bound on>>the size of the sets. The way
the limit gets bigger is to have bigger and>>bigger nested
subsets.>>'nally, if lim S_n is>>uncountable then the
_union_ of countably many sets of>>smaller cardinality has
smaller cardinality. (I believe that>>last bit needs
AC.)> I see what you are getting at - that if card lim
S_n > omega, then it > must be sup card S_n - but I need
more sleep to convince myself about > the details of that
one.************************David C. Ullrich
>>But
what is the limit of this in your opinion? In mine it would
just be>OtheO two point set {*,&}, >> >> ???
What he said
_is_ a counterexample to what IOd written (because>> but the
limit of the sets {1,n} is the one-point set {1}.>> >>Not if
you were to de'ne a map between {1,n} and {1,n+1} that,
purely as>a map of sets, identi'ed 1 with 1 and n with n+1.
As sets, the limit is>going to be a set with two elements,
whih=ch i just denoted by * and & for>the sheer hell of
it.>>The problem is that my notion of lim requires maps
between the sets.ThatOs a problem, all
right.LetOs think
about this. Evidently there are at least two things thenotion
of the limit of a sequence of sets might mean. One of themis
in fact the limit of a sequence of sets - the other one is
_not_actually a notion of the limit of a sequence of sets,
itOs thelimit of a structure involving a bunch of sets
together withmaps between them.Now we have a problem about
the limit of a sequence of sets.With no maps given. DoesnOt
it seem plausible to assume thatthis was about the notion
that only involves a sequence ofsets?>If you were to send
both 1 and n to 1 in S_{n+1} then the limit would be>the one
point set.>lim has a Ouniversal propertyO
categorical
interpretation which IOm>certainly using, and I would
think
David is too, but possibly without>wanting to say
so.>>here it means if>>S_1 --> S_2 --> S_3
...>>has a limit S, it is OtheO object (set
in this case)
that has maps>to it from every S_i, with the property that
the map S_j to S is the same>as mapping to S_k (k>i) and
then mapping to S.>>in your example, the maps between S_i
and S_(i+1) I assume takes 1 to 1>but what does it do to i?
if it sends it to i+1, then youOre identifying>them, and
the limit would be the two point set.>>ItOs not just the
sets you need to consider but the map between sets.>> >> This
may well be a reasonable and standard notion, but itOs
certainly>> not what I meant by lim S_n, and not what I think
he meant either.>> The notion of the limit of a sequence of
sets has at least one other>> perfectly standard meaning that
has nothing to do with what you>> write here: S = lim S_n
means that every element of S is in S_n>> for large n and
every non-element of S is a non-element of S_n>> for large n.
(Ie, the indicator function of S_n converges pointwise>> to
the indicator function of S.)>> >>The category theory style
one says that an element of the limit is in one>of the sets,
and then every one after that too, and its images
are>identi'ed. >>The natural Idea of lim youOre
using (and it
is much more natural, and>easier to think about on the
surface) is better than mine because it can>deal with
suggestive things like the S_i in the counter example, where
we>keep special meanings for the letters in the set. I would
have to tell you>how to OannihilateO elements
which in
general I canOt, because IOm hiding>something
about what I
really mean by lim: you need an indexing category>and some
functors lying around, but I suspect you either know this
already>or donOt care.HereOs what
IOm curious about. Say we
de'ne a sequence of realsby letting x_n be sqrt(2) to n
decimals: x_1 = 1.4, x_2 = 1.41, etc.Can we say that lim x_n
= sqrt(2), or would that be keeping specialmeanings for the
digits in the decimal expansions?>I can force mine to be
equivalent by saying Osuppose I want to deal with>situations
where S_i OmapsO to S_{i+1} but i
donOt want to explicilty
map i>to anything (for the 'rst time) (again keeping with the
above counter >example). Then I simply remove i from the set,
and all its preimages from >the previous sets.O>>As it is,
if
one is allowed to just kill off elements at will, then it
is>always possible to create these examples of lim card not
being related to>card lim, necessarily.>>Take a vector space
X of abitrary dimension over a 'eld of abitrary>cardinality,
de'ne a chain>X --> X --> X > ...>>with the zero map at
every arrow, then the lim is the zero vector>space in my way
of thinking.>If you want it to be X then use the identity map
between them.>>Oh, and IOve not apologized for presuming to
tell you what you were>thinking, sorry.>>It wasnOt very
clearly explained, was it? sorry.> and if lim S_n is
countably in'nite then card S_n>cannot be bounded by a
'nite number - > How do we show that, exactly?>>Other wise after a certain point in the sequence you
are just identifying>elements of the set, thatOs what it
means if there is a uniform bound on>the size of the sets.
The way the limit gets bigger is to have bigger and>bigger
nested subsets.>'nally, if lim S_n
is>uncountable then the _union_ of countably many sets
of>smaller cardinality has smaller cardinality. (I
believe that>last bit needs AC.)> I see what you
are getting at - that if card lim S_n > omega, then it >>
must be sup card S_n - but I need more sleep to convince
myself about >> the details of that one.>> >>
************************>> >> David C.
Ullrich************************David C. Ullrich
>Is there a sequence S of sets such that lim card S_n <
card lim S_n, >where card denotes cardinality, both limits
exist, and the inequality >is strict? Here trans'nite
limits are considered to exist as long >as they are
well-de'ned. Also, feel free to call on the axiom of choice.
>>If by sequence you mean an ordinary sequence indexed by
the>>positive integers>I do.> then no: This is clearly
impossible if lim S_n is>>'nite (in that case there exists N
so that S_n is contained in S_k>>for all k > N),>It is way
too early for me right now, but I believe this is false:
>E.g., let S_n = {1,n}.There was a typo there - I meant to
say that if lim S_n is 'nitethen there exists N such that lim
S_n is contained in S_k forall k > N.>> and if lim S_n is
countably in'nite then card S_n>>cannot be bounded by a
'nite
number - >How do we show that, exactly?This almost seems
like either you havenOt thought aboutit or
weOre using
different de'nitions of lim S_N. Sojust for the record, the
de'nition I have in mind is this:lim S_n = S if (i) for every
x in S there exists N such thatx is in S_n for all n > N and
(ii) for every x not in S thereexists N such that x is in S_n
for no n > N.It follows that if S = lim S_n and A is a 'nite
subset ofS then there exists N such that A is contained in
S_nfor all n > N (this follows from the de'nition and thefact
that given any 'nite set of positive integers hasa largest
element.) Hence if S = lim S_n and S isin'nite then for every
positive integer k there existsN such that S_n has at least k
elements for all n > N.Hence if S_n has at most k elements
for all n andS = lim S_n then S has at most k
elements.>>'nally, if lim S_n is>>uncountable then the
_union_ of countably many sets of>>smaller cardinality has
smaller cardinality. This sentence is not true. ItOs also
not
what I really meant,and not what we need. WhatOs true, and
what we need here,is that if lim S_n is uncountable then the
union of countably many sets of cardinality bounded below
card lim S_nhas cardinality smaller than card lim S_n.>(I
believe that>>last bit needs AC.)>I see what you are
getting at - that if card lim S_n > omega, then it >must be
sup card S_n - but I need more sleep to convince myself about
>the details of that one.LetOs say S = lim S_n, and A = card
S. Suppose that lim card S_n = B < A. Then card S_n <= B for
all but'nitely many n; since lim S_n is unchanged by
altering'nitely many S_n, wlog card S_n <= B for all n.
Socard S <= card union S_n <= B.************************David
C. Ullrich
Is there a sequence S of sets such that lim
card S_n < card lim S_n, > where card denotes cardinality,
both limits exist, and the inequality > is strict? Here
trans'nite limits are considered to exist as long > as they
are well-de'ned. Also, feel free to call on the axiom of
choice.If so, does there exist such a sequence with lim S_n
'nite?I would imagine that this depends on what
youOre taking
your limit over.YouOve suggestively written n implying over
the naturals. Instead howabout taking the ('ltered direct)
limit indexed by some suitable category(or net, as I;ve been
reminded recently of for analysis). ItOs too nastyfor me to
consider before coffee. Any one any ideas on that?
=I am
looking for the volume of the solid of revolution obtained
byrotating the region bounded by y=x and y=x^2 around the
verify this?
I am looking for the volume of th
similar problem, and then she can just> 'nd the solution with