mm-51
===
I am working on a project in Macromedia Flash MX using its
drawing APIto construct Bezier curves
parametrically.Ultimately, I need to examine a set of curves
and ?d all theintersections of those curves.I am currently
scouring the web for different approaches to solvingthis
problem.Any code, algorithms, references, etc. would be
greatly appreciated.Luke
===
> I am working on a project in
Macromedia Flash MX using its drawing API> to construct
Bezier curves parametrically.> Ultimately, I need to
examine a set of curves and ?d all the> intersections of
those curves.> I am currently scouring the web for
different approaches to solving> this problem.> Any code,
algorithms, references, etc. would be greatly appreciated.You
could look at Knuth.a6s Metafont (mf.web) where he tackleswhat
seems to be the same question.-- Timothy Murphy tel:
+353-86-233 6090
===
>
ucdavis!newsfeed.berkeley.edu!ucberkeley!
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l.syr.edu!sn-xit-03!sn-xit-01!sn-post-01!supernews.com!
news.supernews.com!w> aderameyxiii>
<3F46BE3A.E057DE36@mdli.com> For convenience, I.a6m
going to change this relative form of the paradox> to an
absolute form. > Achilles starts at spot A and is
trying to get to spot B. When he is> halfway between A and
B, he needs to get to halfway between where he is> and B.
Even when he gets to that spot, he still has to get to the
point> halfway between where he is and B. Ad in?itum.
Therefore, he can> never get to B.> Therefore? What
process of reasoning gets us to he can never get to B? > I
don.a6t see the paradox.> I didn.a6t mean therefore to 
indicate
a logical deduction. That.a6s justthe way it.a6s usually 
stated,
at least the versions I.a6ve seen. Of course,with any kind of
would-be paradox, there.a6s an element of trickeryinvolved.The
basic issue seems to be that Achilles at any point of this
in?iteprocess is never there, but he gets there in the
end. If you canexplain this, I.a6ll accept there.a6s no 
paradox.
But good luck trying toexplain it just in terms of
mathematics.> The common thing to do at this point is to
say, But of course, we know> from Calculus that in?ite
series of positive numbers can add up to some> ?ite
number. How silly of Zeno not to realize this!> This
kind of thinking misses two major points. The setting of
the> paradox relies on the fact that in?ite series of
positive numbers can> addup to some ?ite number.> In
particular, it relies on the fact that> you can divide up a
geometric segment in half, and then in half, and then> in
half, ad in?itum. Knowing this fact is part of the *setting*
of the> paradox. It in no way resolves anything.> I agree
that the in?ite subdivisibility of time and distance are in
play > - it.a6s mentioned explicitly - but that is not the
same as understanding the > convergence of in?ite series.
The ?st concept is easy to grasp; you can > convince a
child of its plausibility. The second is a much more >
sophisticated idea. (Not that I.a6m claiming in?ite series
resolve the > paradox, because I still don.a6t see what the
paradox is.)> I never said you have to understand all the
subtleties of in?ite seriesto understand in?ite
subdivisibility. But understanding the former willnot resolve
the paradox.> The other point is that Zeno, as a Greek
philosopher, was well aware of> this kind of calculus-type
reasoning. He lived during the ?th and fourth> centuries
BC, but the Pythagoreans, for example, had been around
before> him. The kinds of things they studied, for example,
constructing the> golden rectangle, and creating the spiral
formed by it, would have, I> believe, been known to Zeno.>
Well, the truth is we don.a6t know what Zeno knew. But 
it.a6s
beside the point. > The paradox should be stated as clearly
as possible. If it quali?s as a > genuine paradox, it
should need no defense based on allusions to ideas > extant
in Zeno.a6s time that he may or may not have known.> I.a6m 
sorry
you don.a6t care to know what Zeno may have known. As you
sayit.a6s irrelevant (in some sense), but others may feel
broadening theirunderstanding of the historical context is a
good thing.> To better see that calculus does not resolve
the Achilles paradox,> consider the following addition to
the above absolute form of the paradox:> as Achilles
travels in smaller intervals of space, he alternates
opening> and closing his mouth, i.e. When he is traveling to
the ?st midpoint, he> opens his mouth, when he is traveling
from the ?st midpoint to the> second midpoint, he closes his
mouth, ad in?itum. What is the position> of his mouth when
he arrives at his destination? Open, or closed? Or>
neither?> Come now. This is not a serious example.> Well, I
say it is. Should we continue this back and forth?> The
typical objection to this question is to say, there is no
last term in> a sequence, so it doesn.a6t make sense to ask
this question. You can.a6t> assign some state to each event
in this in?ite sequence of events, and> then ask what is
the ?al state, since that is contradictory to the> idea
of an in?ite sequence. > There are better objections: You
can certainly talk about a ?al state in > the presence of
continuity.> For centuries linear motion has been modeled
as a smooth function f : [a,b] > -> R, where [a,b] is to be
thought of as a time interval and f(t) is the > location
along a line. This model has been a resounding success.
Zeno.a6s > paradox seems to be this: Achilles can.a6t 
reach f(b),
because for each t < > b, f(t) < f(b)!This is falling into
the same kind of trap as saying in?ite seriesresolves the
Achilles paradox.If you model the situation as you have, then
that is a different situationthan that outlined in the
paradox. Look at it this way: does your f correspond to
anything in the paradox? Are you seriously suggesting that
there is an uncountable number ofpositions that Achilles is
reaching before he reaches f(b)?The key point here is that
such a model of linear motion sidesteps theparadoxes of
Zeno.a6s much like how the epsilon delta de?ition of
limitavoids the kinds of paradoxes the founders of analysis
had to tangle with. Stating that this model has been a
resounding success doesn.a6t resolveany of the issues; it just
shows that to a large extent Zeno.a6s paradoxesdon.a6t 
have to be
resolved in order to do physics. 
===
>
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news.supernews.com!w> aderameyxiii>
<3F46BE3A.E057DE36@mdli.com> For centuries
linear motion has been modeled as a smooth function f :
[a,b] > -> R, where [a,b] is to be thought of as a time
interval and f(t) is the > location along a line. This
model has been a resounding success. Zeno.a6s > paradox
seems to be this: Achilles can.a6t reach f(b), because for each
t < > b, f(t) < f(b)!> This is falling into the same kind
of trap as saying in?ite series> resolves the Achilles
paradox.> If you model the situation as you have, then that
is a different situation> than that outlined in the paradox.
But maybe it isn.a6t. More on this below. However, 
there.a6s a
solution to this particular paradox that ishidden in the
continuous motion model. Consider:Zeno does allow a ?st step
in this particular paradox, does he not?This ?st step itself
is the limit point of a sequence that, justlike the original
sequence in question, cannot be actually completed,right? And
so this ?st step is a limit point that, just like theoriginal
limit point in question, cannot be actually reached,
right?Zeno is trying to have it both ways, and so we have a
solution tothe paradox in that it cannot even be
consistently stated.You referred to another form of the
paradox:Achilles can never move. For, before he can get to
some spot, heneeds to move through a spot in between that and
his starting spot.And of course, I can divide even that
further. Since there is no lastterm in a sequence, Achilles
can never get started.What was stated here is actually
called the Dichotomy Paradox. Here isa fuller statement of
the Dichotomy Paradox:If we are at 0 in 0, . . ., 1/4, 1/2,
1, then how can we get startedin traversing all the steps if
we can.a6t even take a ?st step?Answer: We can.a6t, and 
so
motion is not possible - we are forever stuckat 0.Zeno.a6s aim
with his Paradoxes was to prove motion impossible - thatmotion
is an illusion. This Dichotomy Paradox seems trickier than
theAchilles Paradox, because completing an in?ite
step-by-step processwithout taking a last step (the Achilles
Paradox) seems lesspsychologically troublesome than starting
an in?ite step-by-stepprocess without taking a ?st step
(the Dichotomy Paradox). This isprobably because with the
latter, the problem is right at one.a6sdoorstep but with the
former, the problem is in?itely removed.Note: Actually
reaching a limit point of a sequence is equivalent
tocompleting an in?ite step-by-step process without taking
a laststep and it not equivalent to taking a last step in
an in?itestep-by-step process (because there is no such
thing as such a laststep). If one considers completing an
in?ite step-by-step processwithout taking a last step
(Achilles) and especially starting anin?ite step-by-step
process without taking a ?st step (Dichotomy)to be
nonsensical, them maybe one is seeing what Zeno was trying
toget at.If we restrict ourselves to only the sequences given
by Zeno, thenthere do seem to be problems. But why should we
do this? Zeno may nothave required us to do this. You also
stated:A clari?ation: Zeno.a6s paradoxes are basically
objections to thiskind of model of motion. To accept the
model is to disregard anyobjections, such as the
paradoxes.In other words, Zeno seems to not have accepted
the existence o??ite sets, in that he was trying to derive
absurdities from thepostulated existence of motion over a
continuous interval (an in?iteset) by looking at certain
subsets - sequences and their limit points.But the
absurdities he derived are not mathematical, but
merelypsychological. Many thinkers in the past did and some
in the presentdid not make the distinction between these
types of absurdity withregard to in?ite sets.For example, it
was argued that in?ite sets could not exist becauseif they
did, some proper subsets would have the same cardinality
asthe whole set. Some just couldn.a6t psychologically handle
this, and soderiving this absurdity was deemed a proof that
there could not bein?ite sets.Zeno is doing the same type of
thing. By looking at certain subsets ofan interval, sequences
and their limit points, he is derivingabsurdities of motion
along a continuous interval.In response to this ?st
absurdity about in?ite sets, that someproper subsets would
have the same cardinality as the whole set, somehave taken
this absurdity to be a de?ing property of an objectcalled
in?ite sets.So why not give the same response to those of
Zeno.a6s paradoxes thatcan be consistently stated? Just take
these derived absurdities ofsuch as Achilles and Dichotomy
as some de?ing properties of someparts of an object called
motion along a continuous interval? Somemight not be able
to psychologically handle this, but why letpsychology get in
the way?And if one objects to embedding the sequences of the
paradoxes inother sets, then the same points can still be
made: The derivedabsurdities are still not mathematical,
and they are still merelypsychological. One can still take
these absurd consequences asde?ing properties of objects.
Again, why let psychology get in theway?And also, what if one
claims that Achilles and Dichotomy are notequally problematic
(or non-problematic) on mathematical grounds, thatthe latter
is more problematic? Answer: Motion along a
continuousinterval covers both. But if that doesn.a6t satisfy,
look intonon-well-founded set theory and in?ite descending
chains for apossible answer to the latter.Again, as I said in
the other post, I think it all boils down to thephilosophical
bias one brings to the table with regard to the questionof
potential vs. actual in?ity: Does one accept the existence
ofactual in?ity - of in?ite sets, AND THE CONSEQUENCES OF
THEIREXISTENCE? (Prior to Cantor, many did not, and some
still today donot.)Paul
===
> ...[snippage]> To better see
that calculus does not resolve the Achilles paradox,>
consider the following addition to the above absolute form of
the paradox:> as Achilles travels in smaller intervals of
space, he alternates opening> and closing his mouth, i.e.
When he is traveling to the ?st midpoint, he> opens his
mouth, when he is traveling from the ?st midpoint to the>
second midpoint, he closes his mouth, ad in?itum. What is the
position> of his mouth when he arrives at his destination?
Open, or closed? Or> neither?> Come now. This is not a
serious example.> Well, I say it is. Should we continue
this back and forth?The position of his mouth is whatever you
say it should be,and you haven.a6t said. You haven.a6t 
given any
rule for whathe does with his mouth at the destination and
beyond. Sinceapparently he can change the position of his
mouth in zero time,it.a6s no problem for Achilles to arrange it
either way. Just tellhim what to do.Btw if he can.a6t change it
in zero time, then there.a6s no problemat all as I.a6m 
sure you
can see.> The typical objection to this question is to
say, there is no last term in> a sequence, so it doesn.a6t
make sense to ask this question. You can.a6t> assign some
state to each event in this in?ite sequence of events,
and> then ask what is the ?al state, since that is
contradictory to the> idea of an in?ite sequence.>
There are better objections: You can certainly talk about a
?al state in> the presence of continuity.> For
centuries linear motion has been modeled as a smooth function
f : [a,b]> -> R, where [a,b] is to be thought of as a time
interval and f(t) is the> location along a line. This
model has been a resounding success. Zeno.a6s> paradox seems
to be this: Achilles can.a6t reach f(b), because for each t <>
> b, f(t) < f(b)!> This is falling into the same kind of
trap as saying in?ite series> resolves the Achilles
paradox.> If you model the situation as you have, then that
is a different situation> than that outlined in the paradox.>
> Look at it this way: does your f correspond to anything in
the paradox?> Are you seriously suggesting that there is an
uncountable number of> positions that Achilles is reaching
before he reaches f(b)?Why not?(Of course, Zeno only
considers a denumerable subset of them; infact that is
probably what causes all the brouhaha, since it.a6s
notimmediately clear that a denumerable set of points can be
dense.)> The key point here is that such a model of linear
motion sidesteps the> paradoxes of Zeno.a6s much like how the
epsilon delta de?ition of limit> avoids the kinds of
paradoxes the founders of analysis had to tangle with.>
Stating that this model has been a resounding success
doesn.a6t resolve> any of the issues; it just shows that to a
large extent Zeno.a6s paradoxes> don.a6t have to be 
resolved in
order to do physics.If you put it that way, I guess I see
your point; however, asyou yourself said, the paradox relies
on trickery and (I wouldadd) vagueness of one.a6s de?itions,
so I.a6m not sure how muchthat point is worth.
===
> For
centuries linear motion has been modeled as a smooth function
f : [a,b] > -> R, where [a,b] is to be thought of as a time
interval and f(t) is the > location along a line. This
model has been a resounding success. Zeno.a6s > paradox seems
to be this: Achilles can.a6t reach f(b), because for each t <
> b, f(t) < f(b)!> This is falling into the same kind of
trap as saying in?ite series> resolves the Achilles
paradox.> If you model the situation as you have, then that
is a different situation> than that outlined in the paradox.
> Look at it this way: does your f correspond to anything
in the paradox? > Are you seriously suggesting that there is
an uncountable number of> positions that Achilles is reaching
before he reaches f(b)?> The key point here is that such a
model of linear motion sidesteps the> paradoxes of Zeno.a6s
much like how the epsilon delta de?ition of limit> avoids
the kinds of paradoxes the founders of analysis had to tangle
with.> Stating that this model has been a resounding
success doesn.a6t resolve> any of the issues; it just shows
that to a large extent Zeno.a6s paradoxes> don.a6t have to 
be
resolved in order to do physics.A clari?ation: Zeno.a6s
paradoxes are basically objections to this kind ofmodel of
motion. To accept the model is to disregard any objections,
suchas the paradoxes.
===
> How to compute this de?ite
integral?> Integrate[LegendreP[15, x]
ChebyshevT[11, x], {x, 0, 1}]> Excuse me, I didn.a6t make
myself very clear.> I want to compute this integral without
using the computer.> If it were true that>
Integrate[LegendreP[15, x] x^k , {x, 0, 1}]> is equal to zero
for all odd k, then the problem> wouldn.a6t be very dif?ult.On
the other hand, a bit of experimenting suggests that
Integrate[LegendreP[2m+1,x]ChebyshevT[2n+1,x],{x,0,1}] ==
0whenever 0 <= n < m.-- Dave SeamanJudge Yohn.a6s mistakes
revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
In alt.anagrams,>David James Polewka used
the Gregg Method to post ...> Typing Monkeys Don.a6t Write
Shakespeare>=>Spanking the keys typed, No worries
mate.>Tyke.a6s gay peep show is rotten in Denmark.>Yon
thirteen-toed yakkers spewing SPAM>Mork Went To See The Dark
Gypsy In Spain.>We paid something rotten. Press any key ...
?KNorth Korea: Peking swept my seat,
Sidney!
===
================ Endeavor to
persevere
===
===================> Let > h(0,m) = 1,
for all positive integers m.> Let h(n,m) = sum{k=1 to m}
h(n-1,k)/k,> for all positive integers m and n.> Then,>
(1/n) sum{k=1 to oo} h(k-1,n) /(k 2^k)> => sum{k=1 to n}
binomial(n,k) ln(binomial(2k,k)) (-1)^k> As ascii-image:>
oo > ---> 1  h(k-1,n)> --- / --------- = > n --- k 2^k> k=1>
> n> --- / n  / 2k  k>  | | ln(| |) (-1)> /  k /  k
/> ---> k=1> Even if true, this is pretty pointless...>
This is ONLY true if n is >= 2.I think that the identity is
correct for these n.a6s. But now I amstarting to have my doubts
about the truth of the above for EVERYinteger n >= 2,
generally. Why would there be an exception to thisidentity at
n = 1?[note: I have editted the ascii-image in the copied
original post sothat the upper limit of the lower sum is ?n.a6
instead of ?oo.a6, eventhough the earlier version was correct
as well.]Leroy Quet
===
> Relax, I.a6m just yanking your
chain.> Yeah I want to share.> I.a6m pondering a closed
group to discuss an aspect.> Like anyone I have but bits
and pieces.> I.a6m waiting also for my partner to decide to
drop out of get with> it.. If he drops I am free to share .>
Anyway yes .. I have a something to suggest. > So far my
experience with this is once one door is open I see several>
more that need to be..> You should know I actually respect
your abilities. > Ok, the ball.a6s in your court. If you have
something to discuss, postit here or in the Collatz
Community.
===
> Relax, I.a6m just yanking your
chain.> Yeah I want to share.> I.a6m pondering
a closed group to discuss an aspect.> Like anyone I
have but bits and pieces.> I.a6m waiting also for my
partner to decide to drop out of get with> it.. If he
drops I am free to share .> Anyway yes .. I have a
something to suggest. > So far my experience with this
is once one door is open I see several> more that need to
be..> You should know I actually respect your
abilities. > Ok, the ball.a6s in your court. If you
have something to discuss, post> it here or in the Collatz
Community. Alright. I.a6ll get it together this weekend. 
I.a6ll
go over there and log in.I.a6m working overtime this week so
I.a6m dog tired. This idea I.a6m thinking of needs to 
be kicked
around a bit so, Ihope Collatz people will give it a kick or
two before it gets postedto the world on Google. My partner
is into higher math studies so he would like to knowhow it
goes but doesn.a6t have the time at this time to work with me
onit. Ernst
===
> This does not directly solve the original
poster.a6s question, but those > interested in this topic
should take, if they like, a look at this old > post of
mine: >
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=j4wtyvj> i98vr%40forum.mathforum.com&rnum=3&prev=>
> That link is broken for me.> If my result is correct,
it should be noted that MANY A1.a6s (a(1).a6s = x in > 
my
post) lead to later unde?ed A.a6s (a(k).a6s).> I believe 
that
the A1.a6s that you refer to are simply the roots of> (f@@n)(z)
= 1 where f(z) = z^2/(z-1) and `@@` indicates iterated>
function composition.> (For example, if a(1) = 1, I get
that a(2) is 1/0. But a(3) is 1.> {because 1 + oo + 1 = 1
*oo *1....snicker})> I think it.a6s best to avoid that
silliness by ignoring the original> posing of the problem in
terms of sums and products of arbitrary> length and rather
just using the recurrence p[n+1] = p[n]^2/(p[n]-1)There has
appeared to be a techno-glitch which only allowed the
?stpart of the URL to actually be linked to.Here is the
link
again:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe
=off&threadm=j4wtyvji98vr%40forum.mathforum.com&rnum=3&prev=
The recursion I got APPEARS to be different from yours, but I
am mostcertainly not understanding what you mean
exactly.Anyway, in case the link still does not link,here is
my post copy/pasted anyway:>One may wonder what sequence of
a[m].a6s is such that the ?st n terms>summed equals the ?st n
terms multiplied.>i.e.: >sum{k=1 to n} a[k] =>product{k=1 to
n} a[k] ,>for every positive integer n.>This is how to
calculate the terms:>a[1] = x;>a[2] = x /(x-1);>and for m
>=3,>a[m] = a[m-1]^2 /(a[m-1]^2 -a[m-1] +1) .>These are the
?st few a[m].a6s:>x, x /(x-1), x^2 /(x^2 -x +1), x^4 /(x^4
-x^3 +2x^2 -2x +1),>x^8 /(x^8 -x^7 +3x^6 -6x^5 +9x^4 -10x^3
+8x^2 -4x +1),...>The m_th (m >= 2) term is of the
form>x^(2^(m-2)) /Q_m(x) ,>where Q_m(x) is a 2^(m-2)-order
polynomial,>where Q_2(x) = x-1;>and for m >= 3,>Q_m(x) =
x^(2^(m-2)) - Q_{m-1}(x) *x^(2^(m-3)) +(Q_{m-1}(x))^2.>Is
there a closed form for the Q_m(x).a6s ?>This whole thing may
be well studied. Any additional info?>Leroy Quet
===
> I
think it.a6s best to avoid that silliness by ignoring the
original> posing of the problem in terms of sums and
products of arbitrary> length and rather just using the
recurrence p[n+1] = p[n]^2/(p[n]-1)> The recursion I got
APPEARS to be different from yours, but I am most> certainly
not understanding what you mean exactly.Your recurrence is
a[n+1] = a[n]/(a[n]-1). It generates the A.a6s, theterms of the
sum. My recurrence generates the sum (or product -- samething)
of all the terms.
===
I.a6ve noticed that there has been a
preponderance of JSH threads that donot have JSH in the
subject lines. Especially recently it seems. I.a6ve isolated
one reason for this. Many of the sci.math regulars
whoparticipate in JSH threads are no longer actively using
the JSH pre? intheir responses.Here is a (partial) list of
names of regulars who have participated in JSHthreads without
the JSH pre?:James HarrisDavid C. UllrichRandy PoeLarry
HammickWill TwentymanArturo MagidinEd HookBrian Quincy
HutchingsNora BaronVirgilRobin ChapmanGeorge CoxBill
HaleClive ToothI got these names from just a cursory glance
through the last couplemonths worth of postings on google
groups. I think anybody on this listposts enough to sci.math,
and in particular, to JSH related threads, thatthe cumulative
effect of all these people not using the JSH pre? iscausing
an increasingly large jump in my news traf?.I encourage
people (especially those on the list, and in particular
JamesHarris) to use the JSH pre? in their future posts to
JSH relatedthreads. The non-JSH part of sci.math will be very
grateful.
===
> I.a6ve noticed that there has been a
preponderance of JSH threads that do> not have JSH in the
subject lines. Especially recently it seems. > I.a6ve
isolated one reason for this. Many of the sci.math regulars
who> participate in JSH threads are no longer actively using
the JSH pre? in> their responses.> Here is a (partial)
list of names of regulars who have participated in JSH>
threads without the JSH pre?:> James Harris> David C.
Ullrich> Randy Poe> Larry Hammick> Will Twentyman> Arturo
Magidin> Ed Hook> Brian Quincy Hutchings> Nora Baron> Virgil>
Robin Chapman> George Cox> Bill Hale> Clive Tooth> I got
these names from just a cursory glance through the last
couple> months worth of postings on google groups. I think
anybody on this list> posts enough to sci.math, and in
particular, to JSH related threads, that> the cumulative
effect of all these people not using the JSH pre? is>
causing an increasingly large jump in my news traf?.how
large is that jump?> I encourage people (especially those on
the list, and in particular James> Harris) to use the JSH
pre? in their future posts to JSH related> threads. The
non-JSH part of sci.math will be very grateful.
===
 > I
encourage people (especially those on the list, and in
particular James> Harris) to use the JSH pre? in their
future posts to JSH related> threads. The non-JSH part of
sci.math will be very grateful.Another attempt to suppress
Harris.a6s work. Typical. I can almost hear thebleating,
Outsiiide the baaahx! Baaaad!The funny thing is, it *would*
be a good idea for us all to use JSH in thesubject line.
James.a6s true audience is the future. It may be 1000
yearsbefore the human race has evolved enough to understand
James.a6s ideas. Whynot make it easy for them to collect his
work, and the pitiful whining ofhis detractors, by
distinguishing it from the general noise on sci.math withthe
simple tag, JSH?I imagine a day, when the world is almost
exclusively digital, when peoplewill buy and sell original
electronic documents the way we would tradeStarry Night or an
original copy of the Declaration of Independence. Oneof the
most coveted e-documents will be the Harris originals. (You
laugh,but your great^n-grandchildren won.a6t.) And you know
what the most frequentlament of the lucky owners will be?He
signed it, but didn.a6t initial it!I think James should sign
and initial all his work, if it.a6s not too muchtrouble. For
the sake of posterity.
===
>I.a6ve noticed that there has been a
preponderance of JSH threads that do>not have JSH in the
subject lines. Especially recently it seems. >I.a6ve isolated
one reason for this. Many of the sci.math regulars
who>participate in JSH threads are no longer actively using
the JSH pre? in>their responses.You are correct. Bad habit.
I was never very reliable at rememberingthe pre?. I need to
develop the habit.That said, I have to say I would really
miss JSH threads were they tostop. Though for his own sake I
do wish he.a6d get some help. In myopinion his postings are
escalating in a number of ways. I think thatclaiming
de?itions to be wrong is a relatively recent thing,
forinstance. - Randy
===
>I.a6ve noticed that there has
been a preponderance of JSH threads that do>not have JSH
in the subject lines. Especially recently it seems. >I.a6ve
isolated one reason for this. Many of the sci.math regulars
who>participate in JSH threads are no longer actively using
the JSH pre? in>their responses.> You are correct. Bad
habit. I was never very reliable at remembering> the pre?. I
need to develop the habit.> That said, I have to say I would
really miss JSH threads were they to> stop. Though for his own
sake I do wish he.a6d get some help. In my> opinion his postings
are escalating in a number of ways. I think that> claiming
de?itions to be wrong is a relatively recent thing, for>
instance.> - Randy> What I ?d odd is that he seems to
think rational numbers were designed to be ?lds, and then
the de?ition made to ? that. That.a6s the only way I can make
sense of his claims about them. He doesn.a6t appear to
understand that the de?ition comes ?st, then we see what
category they ? into (Ring, Division Ring, Field, etc).--
Will Twentyman
===
>I.a6ve noticed that there has been a
preponderance of JSH threads that do>not have JSH in the
subject lines. Especially recently it seems. >I.a6ve isolated
one reason for this. Many of the sci.math regulars
who>participate in JSH threads are no longer actively using
the JSH pre? in>their responses.>Here is a (partial) list
of names of regulars who have participated in JSH>threads
without the JSH pre?:>James Harris>David C. Ullrich>Randy
Poe>Larry Hammick>Will Twentyman>Arturo Magidin>Ed Hook>Brian
Quincy Hutchings>Nora Baron>Virgil>Robin Chapman>George
Cox>Bill Hale>Clive ToothWow. Probably posting this list of
names will shame us allinto abiding by your wishes.If it
doesn.a6t work I.a6d suggest sending the same list to the 
FBI.>I
got these names from just a cursory glance through the last
couple>months worth of postings on google groups. I think
anybody on this list>posts enough to sci.math, and in
particular, to JSH related threads, that>the cumulative
effect of all these people not using the JSH pre? is>causing
an increasingly large jump in my news traf?.>I encourage
people (especially those on the list, and in particular
James>Harris) to use the JSH pre? in their future posts to
JSH related>threads. The non-JSH part of sci.math will be
very grateful.************************David C. Ullrich
===
>
Does anybody know of a pointer to the Levinson Time Eqns.?
What are> these eqns. supposed to describe/model?I.a6ve heard
this come up in relation to a certain Montauk Project.
Anybodyknow??
===
>Could anyone point to me some references or
give me some help with>following optimal control problem
?>Problem :minimize integral of f(x,u)dt over [t1, t2] >
dx/dt=g(n,lambda,u)Do you mean g(x,lambda,u)?Is lambda a
constant? If so, you could try including it as a
statevariable with d(lambda)/dt = 0.> subject to state
inequality constraints : dx/dt >= -E ; lambda <= 1> and state
equality constraints : (dx/dt + E)*(lambda-1)=0Here you have
only three possibilities: either dx/dt= -E, lambda = 1,or
both. If E is a constant, then the ?st case restricts x to
be alinear function of time, in fact, g = - E, etc.--John E.
PrussingUniversity of Illinois at Urbana-ChampaignDepartment
of Aerospace
Engineeringhttp://www.uiuc.edu/~prussing
===
>..........
[lots snipped] > I might as well add that,> If f(x) happens
to be such that> if> a[j] (the j_th derivative of f(x) at
x = 0)> is an integer multiple of (j-1) for all j.a6s >= 1,
(and if a[0] = 0)> then:> b[j] is, for *ALL* integer
j.a6s >= 1, an integer multiple of (j-1),> where b[j] is
(as noted in original post) the j_th derivative of exp(f(x))
at x = 0.> I might as well add again that:(with a little
rede?ing of {a(k)} and {b(k)}.)If (for n = ?ed nonnegative
integer)B(x) = exp(integral{0 to x} A(y) y^n dy),where A(x) =
sum{k=0 to oo} a(k) x^k/k!,and B(x) = sum{k=0 to oo} b(k)
x^k/k!,and integral{0 to x} A(y) y^n dy = sum{k=0 to oo} a(k)
x^(k+n+1)/(k!(k+n+1))(if the above is all kosher, converging
uniformly),then:b(0) = 1;b(m) = 0 for 1<=m<=n; b(m+n+1) =
((m+n)!/m!) sum{k=0 to m} b(m-k) a(k).(hopefully true.)So, if
{a(k)} is an integer sequence, b(m+n+1) is divisible by
((m+n)!/m!),for m = all nonegative integers.Leroy Quet
===
>
YOUCH! That.a6s a heavy condition. So the prime factorization of
n> has no lone primes. I would call that powerful
rather than squareful.> Yes, the term powerful is also
used. But the thing about squareful > (or, squarefull) is
that one can then use cubefull for a number n > with the
property that p divides n implies p^3 divides n, and k-full
> to mean p divides n implies p^k divides n.Likewise, we can
similarly use the term cubeful to mean it has at leastone
prime factor to power 3 or more.I mentioned these once in
class, maybe in a problem:- to prove that everynatural can be
uniquely expressed as a product of a perfect cube anda
cube-free number. Cubeful numbers arose naturally along the
way.----------------------------------------------------------
-------------------- Bill Taylor
W.Taylor@math.canterbury.ac.nz-------------------------------
----------------------------------------------- The sum of
the cubes of the ?st n natch, Is square the sum of the
batch!------------------------------------------------------
------------------------
===
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WongCo-Editor-In-Chief
===
>[...]>I.a6d like to put the
paper on the main math preprint server but have>been blocked
by an automatic message saying that some established>
>mathematician has to vouch for me, as it apparently checks
my domain>and notices I.a6m not using a *.edu account.> You
should thank them. Eventually you.a6ll agree that the current
Proof> is wrong, just like you eventually agreed that the
previous ones were> wrong.In case you hadn.a6t noticed 
I.a6ve had
each of my *three* major resultsthoroughly critiqued now,
which is why I.a6m moving to the ?al phase.I realized a while
back that my FLT proof was just complicated enoughthat
mathematicians could block me inde?itely.Then I thought my
prime counting function would break things open, butsaw a
strong and successful resistance.I realized then that I
needed a third result, and ?ding the error inthe ring of
algebraic integers gave it to me, with an important
twist:With the previous results people could stay quiet, but
whenmathematicians teach the old ?mathematics I can make
a case forfraud, and make that case on the federal
level.Individual mathematicians may ?d themselves at best
having to payback federal funds, and also entire universities
may ?d themselvesdisgraced as well as facing penalties.And,
it takes out an entire cross-section of universities from
toptier ones like Harvard and Yale all the way to the
bottom.I will, of course, personally protect Vanderbilt
University. >As the academic year begins here in the United
States, remember that>the error in taught mathematics can
potentially be part of fraud cases>against universities and
individual mathematicians.> People really really wish you.a6d
_do_ something about these bizarre> threats of lawsuits.
Really.I don.a6t intend on suing anyone. If things go according
to plan, atleast some of the suits will come from the federal
government.>Some of you need to read the ?e print on those
federal funds you>receive.Basically now I can shift between
my three results and hammer atmathematicians endlessly,
continually shifting until I see a break. When I see the
break I.a6ll push with overwhelming force.When the story ?ally
breaks, then the investigations will de?itelybegin in
earnest, and meanwhile, mathematicians build the case
againstthemselves by teaching the ?mathematics, thus
removing plausibledeniability.And believe me, if the United
States even just tells you to pay backfederal funds that you
received for math research, you won.a6t like it.And your
universities will bail on you when even a whiff of
thepossibility gets to them.But that.a6s the big plan. For now,
I look for weak spots, while inAmerica, many of you begin
teaching, building my case against you withyour own energy.As
I continue to probe, I will ?d a break, and then it will be
overquickly.James Harris
===
>[...]>I.a6d like to
put the paper on the main math preprint server but have>
>been blocked by an automatic message saying that some
established>mathematician has to vouch for me, as it
apparently checks my domain>and notices I.a6m not using a
*.edu account.> You should thank them. Eventually you.a6ll
agree that the current Proof> is wrong, just like you
eventually agreed that the previous ones were> wrong.>In
case you hadn.a6t noticed I.a6ve had each of my *three* 
major
results>thoroughly critiqued now, which is why I.a6m moving to
the ?al phase.That.a6s very funny. The critique of your
proofs of FLT and APF haveconsisted of pointing out errors in
the parts that were statedcoherently enough that this was
possible. Given that, most peoplewould think that the ?al
stage would be to give up and startover (or give up and not
start over.)>I realized a while back that my FLT proof was
just complicated enough>that mathematicians could block me
inde?itely.>Then I thought my prime counting function would
break things open, but>saw a strong and successful
resistance.>I realized then that I needed a third result,
and ?ding the error in>the ring of algebraic integers gave
it to me, with an important twist:Tee-hee. A _ring_ contains
an _error_? Doesn.a6t make muchsense.>With the previous results
people could stay quiet, but when>mathematicians teach the old
?mathematics I can make a case for>fraud, and make that
case on the federal level.>Individual mathematicians may ?d
themselves at best having to pay>back federal funds, and also
entire universities may ?d themselves>disgraced as well as
facing penalties.>And, it takes out an entire cross-section
of universities from top>tier ones like Harvard and Yale all
the way to the bottom.>I will, of course, personally protect
Vanderbilt University.>As the academic year begins here
in the United States, remember that>the error in taught
mathematics can potentially be part of fraud cases>against
universities and individual mathematicians.> People really
really wish you.a6d _do_ something about these bizarre> threats
of lawsuits. Really.>I don.a6t intend on suing anyone. If
things go according to plan, at>least some of the suits will
come from the federal government.>Some of you need to read
the ?e print on those federal funds you>
>receive.>Basically now I can shift between my three results
and hammer at>mathematicians endlessly, continually shifting
until I see a break. >When I see the break I.a6ll push with
overwhelming force.>When the story ?ally breaks, then the
investigations will de?itely>begin in earnest, and
meanwhile, mathematicians build the case against>themselves
by teaching the ?mathematics, thus removing
plausible>deniability.>And believe me, if the United States
even just tells you to pay back>federal funds that you
received for math research, you won.a6t like it.>And your
universities will bail on you when even a whiff of
the>possibility gets to them.>But that.a6s the big plan. For
now, I look for weak spots, while in>America, many of you
begin teaching, building my case against you with>your own
energy.>As I continue to probe, I will ?d a break, and then
it will be over>quickly.What sort of break do you have in
mind here? I mean, assumingfor a second that your bizarre
fantasies about how everyone onsci.math is lying, every
journal editor in the world seems to beimmediately seem to
realize that your work is going to overthrowthe establishment
so they need to reply with lies about howit.a6s wrong or
trivial, etc are true - given that the entire worldseems to
be united against you here it.a6s hard to see what sort of
break you.a6re waiting for - if some renegade were to boldly
speak the Truth about your work he.a6d be suppressed just as
you.a6ve been.Giggle.We all really do wish you.a6d post 
the text
of the complaintyou sent to the FBI and the text of their
reply. I meanthese will be important documents some day when
theTruth ?ally comes out - we should get them on therecord
as soon as possible.>James
Harris************************David C. Ullrich
===
> In case
you hadn.a6t noticed I.a6ve had each of my *three* major 
results>
thoroughly critiqued now, which is why I.a6m moving to the ?al
phase.But I think your 4th result, the Z[1/2] = R is much
more important.The logic is so simple that even undergrads
can follow it, butall mathematitians in sci.math still reject
it for some reason.I.a6m sure it.a6s the weakest spot.
===
> In
case you hadn.a6t noticed I.a6ve had each of my *three* 
major
results> thoroughly critiqued now, which is why I.a6m moving
to the ?al phase.> But I think your 4th result, the
Z[1/2] = R is much more important.> The logic is so simple
that even undergrads can follow it, but> all mathematitians
in sci.math still reject it for some reason.> I.a6m sure 
it.a6s
the weakest spot.I doubt that you.a6re sincere.My original post
in this thread points out a result which should
raisesuspicions about other posters, as I can show with 2x^3
- 3xy^2 + y^3that my predictions are correct as you can *see*
the answer.What some of you may fail to realize is that
posters who dispute myargument are claiming that
mathematicians are blind in that area.Their argument is that
NO ONE can ?ure out the factors that I saycan be determined,
and I.a6ve read some fascinating posts on thesubject.For those
of you who don.a6t understand, I.a6ve proven that 2x^3 - 
3xy^2 +
y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)where *only two* of
the a.a6s have a factor that is sqrt(2).That.a6s easily
checkable.Posters have tried to claim that the reducibility
over Q allows that,so that they can try and claim that there
is NO WAY to ?ure out thefactors if it.a6s irreducible.For
instance with 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x +
y)(a_3 x + y)these posters claim that NO ONE can say how the
65, and in particular,the factor 5 of 65 separates out into
the a.a6s, while I say I can proveonly two of the a.a6s 
have a
factor that is sqrt(5).The difference they cite is
reducibility over Q, lack of which theyargue blinds humanity,
blocking the determination that I say ispossible.They claim
intellectual weakness for mathematics; I prove
greaterstrength.James Harris
===
>In case you hadn.a6t
noticed I.a6ve had each of my *three* major
results>thoroughly critiqued now, which is why I.a6m moving
to the ?al phase.>But I think your 4th result, the
Z[1/2] = R is much more important.>The logic is so simple
that even undergrads can follow it, but>all mathematitians
in sci.math still reject it for some reason.>I.a6m sure 
it.a6s
the weakest spot.> I doubt that you.a6re sincere.Why should
sincerity have anything to do with the truth? You
apparentlyhave not done as I requested, which is: 1. Provide
the de?ition of the ring Z[x]. This will establish for us
that you know what you.a6re talking about (or not, which is
really the case). 2. Provide the de?ition of the ring
Z[1/2], and verify that it is the particular case of applying
the de?ition you gave in 1. above to the situation x = 1/2.
3. Prove that the result of the de?ition in 2. above
actually yields the ring of real numbers.I.a6ve said before
that I can prove 1 and 2, and can prove that 3 isincorrect;
furthermore, most of your mathematical critics can do
thesame. If you care to be other than a curiosity, a sideshow
freak, thenyou.a6ll take the time and effort to provide
responses to these threeitems. As it is, you appear to be
content as the butt of every availablejoke on sci.math.
That.a6s your prerogative, to be sure, but it isn.a6t 
theway to
get your case heard, either here or at the FBI or CIA or NSA
orNSC or whatever 3-letter agency strikes your fancy.> My
original post in this thread points out a result which should
raise> suspicions about other posters, as I can show with>
2x^3 - 3xy^2 + y^3> that my predictions are correct as you
can *see* the answer.> What some of you may fail to realize
is that posters who dispute my> argument are claiming that
mathematicians are blind in that area.> Their argument is
that NO ONE can ?ure out the factors that I say> can be
determined, and I.a6ve read some fascinating posts on the>
subject.> For those of you who don.a6t understand, 
I.a6ve
proven that> 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x +
y)(a_3 x + y)> where *only two* of the a.a6s have a factor
that is sqrt(2).It is easily veri?d that, up to order, the
a.a6s are thesenumbers: a_1 = -1 a_2 = -1+sqrt(3) a_3 =
-1-sqrt(3)As the person making the claim that two and only
two of thesenumbers are divisible by sqrt(2), please indicate
which, andthen show the factorization.> That.a6s easily
checkable.> So, what.a6s keeping you? I.a6ve shown you 
the
values, now it.a6s upto you to make good on your implicit
promise to factor the particulara.a6s to make that factor
evident.> Posters have tried to claim that the reducibility
over Q allows that,> so that they can try and claim that
there is NO WAY to ?ure out the> factors if it.a6s
irreducible.> For instance with> 65x^3 - 12xy^2 + y^3 =
(a_1 x + y)(a_2 x + y)(a_3 x + y)> these posters claim that
NO ONE can say how the 65, and in particular,> the factor 5 of
65 separates out into the a.a6s, while I say I can prove> only
two of the a.a6s have a factor that is sqrt(5).> Untrue, and
you are aware of that. Knowing that I have shown
you*PRECISELY* the factorization of the a.a6s that exhibits the
commonfactor, your statement that these posters claim that NO
ONE can say how the 65, and in particular, the factor 5 of 65
separates out into the a.a6sis clearly a lie. As a person who
claims to be interested in theTRUTH, and who claims that
mathematicians are notorious liars, I?d it suspicious that
you have no problem indulging in lying here.BTW, none of the
a.a6s appears divisible by sqrt(5).> The difference they cite
is reducibility over Q, lack of which they> argue blinds
humanity, blocking the determination that I say is>
possible.> They claim intellectual weakness for
mathematics; I prove greater> strength.> explicitly claiming
intellectual weakness for mathematics. Also,I haven.a6t seen
any proofs coming from you. I.a6ve seen more than myshare of
*claims* of proof, but [ain.a6t it amazin.a6?] an equal 
numberof
fatal errors.> James HarrisDale.
===
> <snippage>
My original post in this thread points out a result which
should raise> suspicions about other posters, as I can show
with> 2x^3 - 3xy^2 + y^3> that my predictions are
correct as you can *see* the answer.> What some of you may
fail to realize is that posters who dispute my> argument are
claiming that mathematicians are blind in that area.>
Their argument is that NO ONE can ?ure out the factors that
I say> can be determined, and I.a6ve read some fascinating
posts on the> subject.> For those of you who don.a6t
understand, I.a6ve proven that> 2x^3 - 3xy^2 + y^3 = (a_1 x
+ y)(a_2 x + y)(a_3 x + y)> where *only two* of the a.a6s
have a factor that is sqrt(2).> It is easily veri?d
that, up to order, the a.a6s are these> numbers:> a_1 = -1>
a_2 = -1+sqrt(3)> a_3 = -1-sqrt(3)> As the person making
the claim that two and only two of these> numbers are
divisible by sqrt(2), please indicate which, and> then show
the factorization.>James is correct here, since sqrt(2) times
the algebraic integer unit(sqrt(2)/2)(-1 + sqrt(3)) is indeed
equal to -1 + sqrt(3),for example.However, this does
absolutely nothing to make his larger point,since he.a6s chosen
a polynomial that.a6s _reducible_ over Z.In fact, if one divides
off the linear factor over Z, his bumfoggeryactually provides
an example of what you and others have beensaying all
along.Rick
===
 > For those of you who don.a6t understand, I.a6ve
proven that> 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x +
y)(a_3 x + y)> where *only two* of the a.a6s have a factor
that is sqrt(2).> That.a6s easily checkable.some elemantary
calculations yield thata_1 = -1a_2 = -1+sqrt(3)a_3 =
-1-sqrt(3)solves the problem. But I can.a6t see any sqrt(2)s
here!!! The same argument shows that this is the only
solution up to permutation of the a.a6s. So what?I would really
be happy to see how this contradicts to what I learned about
polynomial factorization.Imre Polik
===
>In case you
hadn.a6t noticed I.a6ve had each of my *three* major
results>thoroughly critiqued now, which is why I.a6m moving
to the ?al phase.>But I think your 4th result, the
Z[1/2] = R is much more important.>The logic is so simple
that even undergrads can follow it, but>all mathematitians
in sci.math still reject it for some reason.>I.a6m sure 
it.a6s
the weakest spot.> I doubt that you.a6re sincere.> My
original post in this thread points out a result which should
raise> suspicions about other posters, as I can show with>
2x^3 - 3xy^2 + y^3> that my predictions are correct as you
can *see* the answer.> What some of you may fail to realize
is that posters who dispute my> argument are claiming that
mathematicians are blind in that area.> Their argument is
that NO ONE can ?ure out the factors that I say> can be
determined, and I.a6ve read some fascinating posts on the>
subject.> For those of you who don.a6t understand, 
I.a6ve
proven that> 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x +
y)(a_3 x + y)> where *only two* of the a.a6s have a factor
that is sqrt(2).> That.a6s easily checkable.> Posters have
tried to claim that the reducibility over Q allows that,> so
that they can try and claim that there is NO WAY to ?ure out
the> factors if it.a6s irreducible.> For instance with>
65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)>
these posters claim that NO ONE can say how the 65, and in
particular,> the factor 5 of 65 separates out into the a.a6s,
while I say I can prove> only two of the a.a6s have a factor
that is sqrt(5).If you would like, I.a6ll be happy to calculate
the a.a6s for you, calculate the common root of each a and 5
(using W. Dale Hall.a6s equations), and examine wither or not
sqrt(5) is actually a factor of your a.a6s. Would that be
suf?ient to cause you to give up?-- Will Twentyman
===
>
> In case you hadn.a6t noticed I.a6ve had each of my 
*three*
major results> thoroughly critiqued now, which is why I.a6m
moving to the ?al phase.> But I think your 4th result,
the Z[1/2] = R is much more important.> The logic is so
simple that even undergrads can follow it, but> all
mathematitians in sci.math still reject it for some reason.>
> I.a6m sure it.a6s the weakest spot.>I doubt that 
you.a6re
sincere.Why would you doubt that? It may or may not be the
weakestspot in mathematical terms - hard to say whether one
bit ofnonsense is more or less nonsensical than another. But
itseems clear that it.a6s the weakest spot, in the sense that
it.a6sthe spot that.a6s most obviously nonsense to 
readers
withthe least knowledge of mathematics.Which of course makes
it a very welcome addition tothe canon, since it clari?s
things for lots of readers whomight not follow the arguments
regarding FLT or APF.>My original post in this thread points
out a result which should raise>suspicions about other
posters, as I can show with> 2x^3 - 3xy^2 + y^3>that my
predictions are correct as you can *see* the answer.>What
some of you may fail to realize is that posters who dispute
my>argument are claiming that mathematicians are blind in
that area.>Their argument is that NO ONE can ?ure out the
factors that I say>can be determined, and I.a6ve read some
fascinating posts on the>subject.>For those of you who don.a6t
understand, I.a6ve proven that> 2x^3 - 3xy^2 + y^3 = (a_1 x +
y)(a_2 x + y)(a_3 x + y)>where *only two* of the a.a6s have a
factor that is sqrt(2).>That.a6s easily checkable.>Posters
have tried to claim that the reducibility over Q allows
that,>so that they can try and claim that there is NO WAY to
?ure out the>factors if it.a6s irreducible.>For instance
with> 65x^3 - 12xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x +
y)>these posters claim that NO ONE can say how the 65, and
in particular,>the factor 5 of 65 separates out into the a.a6s,
while I say I can prove>only two of the a.a6s have a factor that
is sqrt(5).>The difference they cite is reducibility over Q,
lack of which they>argue blinds humanity, blocking the
determination that I say is>possible.Uh, right. Exactly where
has anyone claimed that irreducibilityover Q blinds
humanity?>They claim intellectual weakness for mathematics; I
prove greater>strength.All you prove is that if you ignore
what words mean you canspout a lot of nonsense. This does
show you can do thingsthat actual mathematicians cannot, but
they.a6re things anactual mathematician would not want to
do.>James Harris************************David C. Ullrich
===
>
In my previous thread I talked about some simple experiments
with> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x
u^2 + u^3 f)> which show a problem with the ring of
algebraic integers, which some> posters seem to be *wishing*
doesn.a6t exist as they reply back with> dedication, and wrong
arguments.You forgot to mention W. Dale Hall.a6s concrete
counter-example for 65x^3-12x+1. If you.a6d like I.a6ll 
work on
the explicit form of the numbers in his counter-example.>
I.a6m posting here to remind you of a case where you can check
my> conclusion directly as using f=sqrt(2), m=1, y=sqrt(2) u,
you get> 2x^3 - 3xy^2 + y^3> which factors over Q.> My
simple experiments would indicate that you have> 2x^3 -
3xy^2 + y^3 = (sqrt(2)a_1 x + y)(sqrt(2)a_2 x + y)(a_3 x +
y)> and in fact that is the case.> Posters have claimed
that reducibility is why my methods work here,> when in fact,
reducibility only allows a_1 and a_2 to be in the ring> of
algebraic integers.Part of the scienti? method is exploring
how certain situations may be special cases. If I drop an
unfolded sheet of paper its acceleration is *not* 9.8 m/s/s.
Is that a special case or a counter-example to acceleration
due to gravity *in a vacuum* at the Earth.a6s surface is 9.8
m/s/s? Reducibility could be a special case.> That.a6s it.>
Why don.a6t any of you wonder why these posters themselves 
can.a6t
predict> where factors must go? Instead they simply challenge
my work, unable> to do better themselves, as they claim that
all is hidden.> Yup, think carefully, and you.a6ll realize
that posters are claiming> that NO ONE can ?ure out where
factors distribute, which is a claim> of forced ignorance.>
I.a6m saying that the mathematics gives tools to see, where they
claim> humanity is blind.> Beyond experiments I have
rigorously proven my case and written a> paper Advanced
Polynomial Factorization.> That paper is linked to on my
primary website> http://groups.msn.com/AmateurMath> where
though you both have to have an MSN Passport and join the
group.I.a6ve joined the group twice and been removed twice,
apparently by the moderator. Any ideas who that might be? In
the mean time, being removed makes it very dif?ult to check
your paper for any changes that may occur.-- Will
Twentyman
===
> [...]> I.a6m saying that the mathematics
gives tools to see, where they claim> humanity is blind.>
> Beyond experiments I have rigorously proven my case and
written a> paper Advanced Polynomial Factorization.>
That paper is linked to on my primary website>
http://groups.msn.com/AmateurMath> where though you both
have to have an MSN Passport and join the group.>I.a6ve joined
the group twice and been removed twice, apparently by the
>moderator. Any ideas who that might be? In the mean time,
being >removed makes it very dif?ult to check your paper for
any changes that >may occur.This is really quite curious. The
math is supposed to speak for itself; publishing it but
only allowing true believers to view thepublication certainly
gives the lie to his claims about howanyone can check the
proof for themselves.************************David C.
Ullrich
===
Good news??....(for once)Leroy===> Good
news??....(for once)> Yeah. But weren.a6t the scores
renormalized a few years back?--Ron BruckApproved: Daniel
Grayson, dan@math.uiuc.edu, moderator for
sci.math.research
===
> I wanted too but the only copy our
library has is on loan till 2015.This seems excessive even
for a university librarian.-- Timothy Murphy tel: +353-86-233
6090
===
> My suggestion would be to keep track of the
smallest prime factor> for 48^p - 47^p for the many primes
which have already been tried.> Perhaps a pattern in the
smallest of prime factors will pop out.> Actually, there
is at least one thing that can be said about factors of>
48^p-47^p (with p prime) : they have the form 2*k*p+1 for
some integer k...>BTW, this is true for any integer n >= 1
that factors of (n+1)^p-n^p havetheform 2*k*p+1; not just for
the special value n=47.
===
>Let f(n) = 48^n - 47^n. Apparently
[ie, based on looking at factors >of a few hundred f(n)] for
n>0, 5|f(2n), 7|f(3n), 11|f(5n), 337|f(7n), >23|f(11n),
79|f(13n+13), 443|f(17n), 7937|f(31n), etc. Well, yes: 3^2 =
2^2 mod 5 so f(2n) = 0 mod 5; 6^3 = 5^3 mod 7 so f(3n) = 0
mod 7;and in general for each prime p, if m is the least
positive integersuch that 48^m = 47^m mod p, then f(mn) = 0
mod p. And m will alwaysbe a divisor of p-1 (by Fermat.a6s
Little Theorem).Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
===
>Let
f(n) = 48^n - 47^n. Apparently [ie, based on looking at
factors>of a few hundred f(n)] for n>0, 5|f(2n), 7|f(3n),
11|f(5n), 337|f(7n),>23|f(11n), 79|f(13n+13), 443|f(17n),
7937|f(31n), etc.> Well, yes: 3^2 = 2^2 mod 5 so f(2n) = 0
mod 5;> 6^3 = 5^3 mod 7 so f(3n) = 0 mod 7;> and in general
for each prime p, if m is the least positive integer> such
that 48^m = 47^m mod p, then f(mn) = 0 mod p. And m will
always> be a divisor of p-1 (by Fermat.a6s Little Theorem).I
usually see the obvious after I post, not before. :)Before my
earlier post I reduced 48^n and 47^n mod 47to see there is no
k such that f(kn)=0 mod 47, but didn.a6t think to reduce 48^n
and 47^n separately for other moduli.-jiw
===
The three-signed
plane is converted to the cartesian plane by thefollowing: x =
s - 1/2( p + m ) y = sqrt(3)/2( m - p ) where x and y are the
traditional cartesian coordinates and s, mand p are the star,
minus, and plus components of a value in threesigned space.I
have written some code that performs the conversion and I
havechecked products against their x + iy form and get
complete agreement!For two values y1, y2 in Y the product
y1y2 is the same as if Iconvert them to z1, z2 in C and
compute the product z1z2.This means that angles add in the
product.I believe that this is quite a discovery.Simply put,
the complex plane is equivalent to three-signed space as
Ihave de?ed it. <bidipl$gcs$1@Xenon.Stanford.EDU>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
===
 at 05:59 PM, amorgan@Xenon.Stanford.EDU (Alan
Morgan) said:>At what level? Any?>Would you suggest teaching
Cauchy sequences to>junior high school students?Would you
suggest sprinkling them with pixie dust? That has as
muchwouldn.a6t suggest introducing real numbers in terms of
rational numbersat all; I.a6d suggest introducing real numbers
geometrically at an earlyage and only teaching constructions
much later.When it was time to introduce them to more
rigorous methods, I.a6d startthem on an axiomatic approach well
before I did a construction fromthe integers. And if I did get
around to teaching a construction fromPeano.a6s Postulates, it
most certainly wouldn.a6t be rdecimals[1]expansions, because
that is by far the *most* complex approach underdiscussion.
Look at all the asinine threads on the claim that .9 ...is
not equal to 1.>but most classes don.a6t do thatThen why
introduce the complications inheren.a6t in using
repeatingdecimals as a de?ition? Do it geometrically, and
teach decimals assimle nomenclature.>Nice. It.a6s almost
impossible to ?d quali?d math teachers for>public school
now. You.a6ll only make it harder.No. I.a6d make it easier 
to ?d
*quali?d* teachers. It doesn.a6t botherme that I.a6d make 
it
harder to ?d unquali?d teachers.[1] Why base 10, anyway?--
Shmuel (Seymour J.) Metz, SysProg and JOATReply to domain
Patriot dot net user shmuel+news to contact me. Do not
replyto spamtrap@library.lspace.org
===
I wonder if Conway.a6s
surreal numberswill ever be the standard way of constructing
the rationals, the reals and more?Logically, it seems much
simpler than any other method.It.a6s surprising (to me) 
it.a6s
made no headway,even at university level.It seems so much
neater than non-standard analysiswhich has (used to have?) a
few fanatical supporters.-- Timothy Murphy tel: +353-86-233
6090
===
Fred Galvin> Congratulations, it looks like you.a6ve
rediscovered Ramsey.a6s theorem.> More precisely, the in?ite
version (there is also a ?ite version)> of Ramsey.a6s theorem,
Theorem A in F. P. Ramsey.a6s 1930 paper On a> problem of
formal logic, which says in part:> If the increasing pairs
of natural numbers [ordered pairs (i,j) where> i and j are
natural numbers and i < j] are partitioned arbitrarily> into
a ?ite number of classes, then there is an in?ite set M of>
natural numbers such that all pairs (i,j) with i, j in M
belong to the> same class.> Actually, Ramsey proved the
analogous statement for r-tuples; what I> have stated above
is the case r = 2.Nice. I new the ?ite Ramsey theorem but
didn.a6t spot any connection. Andthanks G.A.E.
also.LH
===
>Original Format>Don.a6t let the harping from the
king.a6s men of science [KMS]>discourage you. You are working on
the frontiers of knowledge>with what corresponds to the
Hamilton Visualization [HV] of>spin [they call it Lie
algebras and Clifford algebras and>spinors and twistors and
such today but it.a6s all the same HV>with the names changed
and the pictures destroyed] from the>early part of the 1800s.
I.a6ve described it as grade school level>geometry when 
it.a6s
done right as Hamilton did with his giant>science
breakthroughs. I.a6ve explained why I don.a6t agree 
with>some of
your specialized techniques but the KMS carping with>their
phony baloney fancy word games which they don.a6t understand>as
you note, is just a diversion. The French changed all
the>Hamilton pictures and made it into king.a6s secret symbol
manipulation>[symbman] without real meaning but ?ally their
cover up is being>exposed worldwide and your work is helpful.
As you have recognized>they suppressed the HV for over 150
years and your methods are a>restoration of the truth
geometric meanings.>Good seeing. JD>I found him JD.Your KMS
mastermind. (at least for the 1800s)I have less doubts
now.pmjhttp://www.hypercomplex.com
===
I hope you don.a6t mind
if I tag this thread from here.It is the only post I can see
in my newsreader.Was at google for the
rest._______________Quote:The spinors for a sub algebra of
geometric algebra. The algebra of spinors was discovered
independent of the full geometric algebra by Hamalton, who
gave it the name quaternion algebra. Some readers may want to
say, rather, that we have here two isomorphic algebras, but
there is no call for any such distinction. A quaternion is a
spinor. The identi?ation of quaternions with spinors is
fully justi?d not only because they have equivalent
algebraic properties, but more important, because they have
the same geometric signi?ance.It is unfortunate that
Hamilton chose the name quaternion, for this name merely
refers to the comparatively insigni?ant fact that the
quaternions compose a linear space of four dimensions. The
name quaternion diverts attention from the key fact that
Hamilton had invented a geometric algebra. Hamilton.a6s work
itself shows clearly the crucial form of geometry in his
invention. Hamilton was consciously looking for a system of
numbers to represent rotations in three dimensions. He was
looking for a way to describe a geometry by algebra, and so
he found a geometric algebra.In the twentieth century
quaternions seem to be generally regarded as something of a
mathematical museum piece, a curious artifact from the
nineteenth century or a diversion from the mainstream of
modern mathematics. One reason for this unfortunate
circumstance might be found in the fact that Frobenius proved
in 1878 that the quaternions constitute the largest
associative divisions algebra. Thus, if division is a
characteristic feature of quaternions, then quaternion
algebra cannot be generalized. This is a typical diversion
from the geometrical spirit of Hamiltons inquiry. The
predominance of such diversion from the geometric to the
algebraic features of quaternions goes far to explain why
quaternions have not yet found the central place in
mathematics which Hamilton had foreseen.The treatment of
re?ns and rotations given above originated with
Hamilton....a6Vetors, Spinors, And Complex Numbers in Classical
and Quantum Physics.a6David Hestenes, ?American Journal of
Physics.a6, Sept. 1971_____________Best, Dan.-- if( this ==
NULL ) return that;
===
----- Original Message -----> invented
a geometric algebra.Airy fought against the teaching of
Algebraic Geometry, GeometricAlgebrais just another
modern twist on the use of algebra to represent
geometry,here.a6s a quote,QUOTE: He was a practical man, with
his work directed at application, not theory.tripos
examinations for Cambridge in 1859:I have looked very
carefully over the Examination Papers and think them onthe
whole very bad. They are utterly perverted by the insane love
ofProblems, and by the foolish importance given to wholly
useless parts ofuseful application of pure mathematics are
cut down or not
mentioned.http://coldrain.net/lucas/airy.htmlEND QUOTEHe
helps destroy Hamilton.a6s career --QUOTE:So witches some
enchanted WAND bestride, And think they through the AIRY
regions ride.> -
Oldham-------------------------------------------------------
----------------1. AIRY.a6s taunt during a dinner at the estate
of Lord Rosse induced thesensitive William Rowan Hamilton
(1805-1865), then two years on thewagon, to resume drinking,
thus darkening the latter half of a careerwhose ?st half put
Hamilton in the front rank of mathematicians inthat
century.http://groups.google.com/groups?selm=
3B5CC257.48D4975D%40erols.com&oe=UTF-8&output=gplainEND
QUOTEHamilton lost the woman he loved to another because of
?ances,while Airy fought his poverty to win his woman, and
so while Hamiltonstarts drinking to drown out his sadness,
Airy is a ?hting man winingall his lifes battles and
challlenges. But Hamilton gives up drinking witha new
resoution to change, and actually is alcohol free for 2 years
beforethis social dinner where Airy unkindly reminds him of
the difference betweenthe two of them, taunts Hamilton about
drinking water and being unableto hold his liquor and using
this quote from Oldham to refer to Hamiltonas a WAND whose
creativity must be the result of some witches, whilehe AIRY
is a warrior who makes things happen by his own efforts,
etc.from this point on Hamilton went downhill,END QUOTEAiry.a6s
battles are not scienti?, but using whatever he can to
win.Not only did he try to stop the teaching of Algebraic
Geometrybut also Michael Faraday.a6s Electromagnetic
theoryQUOTEKnown for his sarcasm and caustic personality,
Airy had an ongoing battlewith Charles Babbage in which he
prevailed professionally and ?ancially tothe detriment of
science. Airy.a6s infamy ranges from ignoring John
Adams.a6discovery of Neptune to dismissing Michael 
Faraday.a6s
?ld
theory.http://micro.magnet.fsu.edu/optics/timeline/people/
airy.htmlEND QUOTEWhat would have happened to J.C. Maxwell.a6s
career then? Good thing forscience Airy.a6s views 
didn.a6t
always prevail.But since he didn.a6t restrict himself to
science in his methods of battle,evenwhen ?hting
scienti? wars, but used whatever tactic was nearest at
hand,being a practical man, many question some of his
scienti? claims, likeQUOTE:Nick Kollerstrom, an
independent researcher and SHA council member,followed with a
remarkable paper given under a very unremarkable
title.Presenting his work as simply ?archiving the British
Neptune correspondence(1837-1848).a6 Nick began with the
curious story of where these papers hadbeen prior to their
recovery in Brazil in 1999 when he started to studythem. He
then went on to tell the story that they told, namely of the
coverup, engineered by such great names in 19th century
astronomy, as theAstronomer Royal, George Biddell Airy, the
Cambridge Professor JamesChallis, and the Gentleman amateur,
John Herschel, of turning the discoveryof Neptune into a
British triumph. In fact, as Nick pointed out, it wasDennis
Rawlins, in 1993 that ?st came up with the theory, but
Nick.a6sresearch con?ms his ?dings. According to Rawlins,
these Britishastronomers manufactured their own hero in the
shape of John Couch Adamsfollowing the Berlin Observatory.a6s
discovery of Neptune, based on the Frenchmathematician
Leverrier.a6s very accurate predictions. In doing so Airy
andChallis had agreed to take the blame for failing to
respond to Adams.a6mathematical predictions of the planet,
thereby allowing the French to beatthem to it but at the same
time saving the reputation of British astronomyand mathematics
as being technically able to compete with the French.
Nick.a6s?dings do con?m this version of events, the
manuscripts alleged to beAdams.a6 predictions show for example
they were not on a par with Leverrier.a6s.Similarly his diary
shows just how close Adams and Challis were, suggestingit
would have been unlikely that Adams would have approached
Airy ?stinstead of Challis as the of?ial story has
it.http://www.shastro.org.uk/report100403.htmEND QUOTE>
spirit of Hamiltons inquiry. The predominance of such
diversion from> the geometric to the algebraic features of
quaternions goes far to> explain why quaternions have not yet
found the central place in> mathematics which Hamilton had
foreseen.>I think after the quaternion battle was lost to the
vector camp, sciencesimply went down another path, and since
we always tend to look backand think that past knowledge is
in some sense less valuable than currentideas, because
progress is always marching forward, it becomes moredif?ult
for scientists to adopt an open mind and explore things we
believewe once examined thoroughly and discarded. If it was
discarded by menof the past, it must have been useless then.
The idea that it was discardedbecause it was too useful,
would not be accepted by scientists today.> The treatment of
re?ns and rotations given above originated with>
Hamilton...> ?Vetors, Spinors, And Complex Numbers in
Classical and Quantum Physics.a6> David Hestenes, ?American
Journal of Physics.a6, Sept.
1971>pmjhttp://www.hypercomplex.com
===
math is this
?TIA,Lurch
===
>math is this ?It.a6s an English letter, used in
various words like math, etc. It.a6salso often used as a
variable, denoting some mathematical objectin an equation.
What object it denotes varies from context to context;if the
thing was written properly you can ?ure out what the h
meansby looking for phrases like let h = ....(An h with a
funny slash through it is a standard symbol in _physics_,I
think for Plank.a6s
constant.)>TIA,>Lurch>************************David C.
Ullrich
===
> (An h with a funny slash through it is a
standard symbol in _physics_,> I think for Plank.a6s
constant.)> Actually Planck.a6s constant is usually noted h; h
with a funny slash throughis h/(2*pi)Sam-- People sometimes
ask me if it is a sin in the Church of Emacs to use vi. Using
a free version of vi is not a sin; it.a6s a penance. - Richard
Stallman
===
> (An h with a funny slash through it is a
standard symbol in _physics_,> I think for Plank.a6s
constant.)Also h without the slash. The slash means to divide
by 2 Pi.
===
> (An h with a funny slash through it is a
standard symbol in _physics_,> I think for Plank.a6s
constant.)>Also h without the slash. The slash means to
divide by 2 Pi.Oh. Right, I knew
that...************************David C. Ullrich
===
Charlie
Johnsonof> math is this ?The typsetting program called TeX,
and its variations such as Latex, use$...$ around all kinds
of mathematical notation. That.a6s the only usage of$...$ that
I know of.LH
===
Two motivating examples:- One can construct a
?itary computational model of the integers inmany ways. One
way is to represent them as a string of binary digits,with
the classic two.a6s-completement algorithms for binary
addition,subtraction, multiplication, division, and
comparison. If therepresentation is constrained to have no
redundent leading digits,then the mapping from integers to
representations is bijective.- Given a computational model of
the integers, one can construct a?itary model of the rational
numbers, by representing each rationalnumber as a pair
(numerator,denominator). The elementary operationscan be
de?ed in the obvious way, i.e. (a,b)+(c,d)=(a*d+b*c,b*d).
Ifthe representation is constrained to be in lowest terms
with positivedenominator, the mapping from rational numbers
to representations isbijective.Now a few questions:- Does
there exist a ?itary computational model of the
algebraicnumbers, such that there exists a bijection between
mathematicalalgebraic numbers and their model, and there
exist terminatingalgorithms for addition, subtraction,
multiplication, division, andcomparison? To clarify ?itary
here, I mean a representation suchthat if you start with a
?ite set of rational numbers and perform a?ite sequence of
?ld operations and polynomial-solving operationson them,
then the result is guaranteed to be representable in
?itespace.- Does there exist a halting algorithm that, for
every ?ite set ofpolynomial coef?ients
(a0+a1*x+a2*x^2+..=0), can determine all o?s solutions in
the algebraic numbers in such a representation?- Any
references to computer implementations of the above?
===
>Two
motivating examples:>- One can construct a ?itary
computational model of the integers in>many ways. One way is
to represent them as a string of binary digits,>with the
classic two.a6s-completement algorithms for binary
addition,>subtraction, multiplication, division, and
comparison. If the>representation is constrained to have no
redundent leading digits,>then the mapping from integers to
representations is bijective.>- Given a computational model
of the integers, one can construct a>?itary model of the
rational numbers, by representing each rational>number as a
pair (numerator,denominator). The elementary operations>can
be de?ed in the obvious way, i.e. (a,b)+(c,d)=(a*d+b*c,b*d).
If>the representation is constrained to be in lowest terms
with positive>denominator, the mapping from rational numbers
to representations is>bijective.>Now a few questions:>-
Does there exist a ?itary computational model of the
algebraic>numbers, such that there exists a bijection between
mathematical>algebraic numbers and their model, and there
exist terminating>algorithms for addition, subtraction,
multiplication, division, and>comparison? To clarify
?itary here, I mean a representation such>that if you
start with a ?ite set of rational numbers and perform
a>?ite sequence of ?ld operations and polynomial-solving
operations>on them, then the result is guaranteed to be
representable in ?ite>space.>- Does there exist a halting
algorithm that, for every ?ite set of>polynomial coef?ients
(a0+a1*x+a2*x^2+..=0), can determine all of>its solutions in
the algebraic numbers in such a representation?>I think the
answer to all of the above questions is yes.The KANT system,
which can be accessed through Magma, allows you tocreate and
compute in algebraic number ?lds (and also within theirrings
of integers). You have to bear in mind that the roots of an
irreducible polynomial,like x^2 -5 over Q are conjugate under
the automorphism group ofthe algebraic numbers A, and hence
are indistinguishable over A.A computation within KANT would
always take place within a particular?ite extension of Q,
which might be constructed as a sequence ofsimple
extensions.To do exactly what you are asking for, I think you
would need to ?stdecide on a particular enumeration of the
irreducible polynomials over Q.You would then insist that you
were going to always contruct your?ite algebraic extensions
of Q by adjoining the roots of these equationsin the order
speci?d. That way, you could perform any
calcualtioninvolving algebraic numbers in a well-de?ed
manner.Derek Holt.
===
> decide on a particular enumeration of
the irreducible polynomials over Q.do you have one in
mind?er.. one that is reasonable to compute with?That is
you -could- just enumerate tuples of integer coeffs
(reasonablyeasy) reduce the poly (also reasonable) and throw
out ones you.a6vealready seen (-not- reasonable).Is there a
method for enumertaing them similar to enumerating
therationals via Farey sequences (or the related Stern-Brocot
tree)?-- Mitch HarrisLehrstuhl fuer Automatentheorie,
Fakultaet InformatikTechnische Universitaet Dresden,
Deutschlandhttp://tcs.inf.tu-dresden.de/~harris
===
> decide
on a particular enumeration of the irreducible polynomials
over Q.>do you have one in mind?>er.. one that is
reasonable to compute with?>That is you -could- just
enumerate tuples of integer coeffs (reasonably>easy) reduce
the poly (also reasonable) and throw out ones you.a6ve>already
seen (-not- reasonable).I don.a6t think the particular
enumeration is very important in terms ofpractical
computation.The way I described things, for every irreducible
polynomialencountered, you explicitly construct its splitting
?ld over theextension of Q that you have constructed so far.
As far aspractical computation is concerned, this extension
building processwill grind to a halt long before the problem
of discarding irreduciblesyou have seen already becomes
noticeable.In practice is it only possible to compute using
extensions of reasonablysmall degree over Q. So I guess this
procedure for computing in A isnotional, and is not
recommended for real computation. Real computationswould
always be carried out in the smallest possible extension of
Qin which the particular algebraic numbers lie.>Is there a
method for enumertaing them similar to enumerating
the>rationals via Farey sequences (or the related
Stern-Brocot tree)?I don.a6t know, but probably not!Derek
Holt.>-- >Mitch Harris>Lehrstuhl fuer Automatentheorie,
Fakultaet Informatik>Technische Universitaet Dresden,
Deutschland>http://tcs.inf.tu-dresden.de/~harris>
===
>
decide on a particular enumeration of the irreducible
polynomials over Q.>do you have one in mind?>er.. one
that is reasonable to compute with?>That is you -could-
just enumerate tuples of integer coeffs (reasonably>easy)
reduce the poly (also reasonable) and throw out ones you.a6ve>
>already seen (-not- reasonable).> I don.a6t think the
particular enumeration is very important in terms of>
practical computation.The enumeration is needed only when you
want to model *all* of thealgebraic numbers at once. Is that
correct?> The way I described things, for every irreducible
polynomial> encountered, you explicitly construct its
splitting ?ld over the> extension of Q that you have
constructed so far.The hard part is to ?d the irreducible
factors of a polynomialwith coef?ients in a ?ite extension
of Q.Newton knew how to ?d the irreducible factors of a
polynomialwith coef?nts in Q (although not with an ef?ient
method).In the 1980.a6s there appeared several papers on how to
do themore general problem (Zassenhaus, Berkelamp,
others).Note that when you have an irreducible polynomial
p(X) andadjoin a root r to the base ?ld, then you need to
?dthe irreducible factors of p(X) / (X - r) to continue on.>
As far as> practical computation is concerned, this extension
building process> will grind to a halt long before the
problem of discarding irreducibles> you have seen already
becomes noticeable.> In practice is it only possible to
compute using extensions of reasonably> small degree over
Q.You didn.a6t specify how small small is. But certain
polynomials areeasier than others. For example, X^p - 1 might
be reasonalbe up toprimes less than or equal to 257. This is
so because such X^p - 1split completely when you add just one
non-trivial root.> So I guess this procedure for computing in
A is> notional, and is not recommended for real computation.
Real computations> would always be carried out in the
smallest possible extension of Q> in which the particular
algebraic numbers lie.poster.a6s idea of starting with just the
integers and building upeverything from there, even the
rational numbers as being orderedpairs of integers. The
Scheme interpreter was implemented in Java,so the program was
not too fast. I recall that X^19 - 1 took severalminutes to
computer the model of the algebraic integers for
thespitting ?ld of X^19 - 1. My goal was to solve equations
likeX^7 - 1 in terms of radicals.-- Bill Hale
===
I didn.a6t
know what to write for the subject, I hope it.a6s
understandable.My problem ist, that I have 8 points Mn(x,y,z)
that describe a cuboid. Andthen the 8 points get shifted each
in another direction, thus I get thepoints Rn(x,y,z). These
Points all are known.Now, I want to translate any point Pm
inside the cuboid Mn into the space ofRn using a quanti?d
linear combination. (Don.a6t know how to say this inEnglish.)
What I mean is, that a point of Mn that is closer to Pm has
morein? than one that.a6s further away.Can anyone help?--
-GernotIn order to reply, revert my
forename.________________________________________Looking for
a good game? Do it yourself!GLBasic - you can
dowww.GLBasic.com
===
Not sure that I have understood the
question, and I am sure there are bettersolutions to the one
I offer, but just to get the ball rolling....Each corner of
the cuboid (x_i,y_i,_z_i) get shifted in their own
separateway so the end shape is no longer cuboid. so new
corners are given by x.a6_i =x_i + b_i, i=1 to 8 and same for
y and z. So there are 24 b_i slet d_i be the distances from
some point (X,Y,Z) in original space tocorners of cuboid. Let
D be their average.Then new X.a6=X + B and also for Y and Z.Need
some algorithm for BCould take B=Sum (i= 1 to 8) (b_i * (1 -
d_i/D))This at least has the property that a point near a
corner stays near acorner, but there are other properties
that it may lack like the pointstaying inside the shape.hope
this helps> I didn.a6t know what to write for the subject, I
hope it.a6s understandable.> My problem ist, that I have 8
points Mn(x,y,z) that describe a cuboid. And> then the 8
points get shifted each in another direction, thus I get the>
points Rn(x,y,z). These Points all are known.> Now, I want to
translate any point Pm inside the cuboid Mn into the spaceof>
Rn using a quanti?d linear combination. (Don.a6t know how to
say this in> English.) What I mean is, that a point of Mn
that is closer to Pm has more> in? than one that.a6s
further away.> Can anyone help?> --> -Gernot> In order to
reply, revert my forename.>
________________________________________> Looking for a good
game? Do it yourself!> GLBasic - you can do>
www.GLBasic.com>
===
>If S=set and P(S)=collection of all
subsets of S,>then: there is no 1-1 corespondence between S
and P(S).> This is Cantor.a6s Theorem. The usual proof is to
take an arbitrary> function f:S->P(S), and show it is not
surjective; this works for> ?ite and in?ite sets.>Here.a6s a
generalization. surjection f:S -> A^S ==> |A| <= 1> Given f,
consider A = {x in S : x is not a member of f(x)}> Note that
f(x) is a subset of S, so it makes sense to ask whether x is>
a member of f(x) or not.> (1) Prove that A is a subset of S
(easy).> (2) Prove that for every y in S, f(y) is not equal
to A (hint: try> (arguing by contradiction) [hardest part of
the proof]> (3) Explain why (1) & (2) prove that there can be
no 1-1> correspondence between S and P(S).>
===
> at 09:50 AM,
Aatu Koskensilta <aatu.koskensilta@xortec.? said:>No,
because if it did you could well-order any uncountable set.>
> How? Remember, I is countable.Oh. Sorry. What I said was
incorrect.-- Aatu Koskensilta
(aatu.koskensilta@xortec.?Wovon man nicht sprechen kann,
daruber muss man schweigen - Ludwig Wittgenstein, Tractatus
Logico-Philosophicus
===
A few years ago, I programmed a
Matrix QR decomposition algorithm to solvesystems of linear
equations. It worked for complex coef?ients as well,until I
came across this matrix:[1 0][i 1]which cannot be decomposed
although the description of QR algorithm saysit.a6s always
possible.Since then, I sometimes wander if there is any
trick to overcome this?Quick check tells that the matrix is
invertible and LU decomposition ispossible. The problem cannot
be detected by eigenvalues, since the secondcolumn can be of
your choice and thus eigenvalues can be any.What is the most
numerically stable way of inverting a matrix that is*always*
possible if the inverse exists? (I tried SVD as well but I
don.a6thave any math program to run it directly and the source
codes on theinternet are too complicated..., but it seems to
have problems as well). Milan Vandrovec
===
Ahoj Milane,I
don.a6t understand the details of a complex QR algorithm, but
here goes what I obtained in Octave (and the same result in
Matlab):octave:5> A = [1 0;i 1]A = 1 + 0i 0 + 0i 0 + 1i 1 +
0ioctave:6> [Q,R,P] = qr(A)Q = -0.70711 + 0.00000i 0.00000 +
0.70711i 0.00000 - 0.70711i 0.70711 + 0.00000iR = -1.41421 +
0.00000i 0.00000 + 0.70711i 0.00000 + 0.00000i 0.70711 +
0.00000iP = 1 0 0 1Zdenek Hurak> A few years ago, I
programmed a Matrix QR decomposition algorithm to solve>
systems of linear equations. It worked for complex
coef?ients as well,> until I came across this matrix:> [1
0]> [i 1]> which cannot be decomposed although the
description of QR algorithm says> it.a6s always possible.>
Since then, I sometimes wander if there is any trick to
overcome this?> Quick check tells that the matrix is
invertible and LU decomposition is> possible. The problem
cannot be detected by eigenvalues, since the second> column
can be of your choice and thus eigenvalues can be any.>
What is the most numerically stable way of inverting a matrix
that is> *always* possible if the inverse exists? (I tried SVD
as well but I don.a6t> have any math program to run it directly
and the source codes on the> internet are too complicated...,
but it seems to have problems as well).> Milan
Vandrovec
===
> Q => -0.70711 + 0.00000i 0.00000 + 0.70711i>
0.00000 - 0.70711i 0.70711 + 0.00000iI have forgotten to
replace transposition in R with transposition-conjugatein C.
I believe I will be able to ? the algorithm now.Milan
Vandrovec
===
all helpful for my problem.BestAxel===it turns
out indeed, that the problem can be solvedfor arbitrary p.
Applying Eulers formula,one gets e.g.sin(p*arctan(x))=1/(2i)
* ( (1+ik)^p - (1-ik^p) ) / (1+k^2)^{p/2}where i denotes the
imaginary unit. Similar resultsare obtained for the tan- and
cos-case.Axel
===
> sin(p*arctan(x))=1/(2i) * ( (1+ix)^p -
(1-ix)^p ) / (1+x^2)^{p/2} But if one wants to do an actual
computation with non integer p, one will still have to resort
to trigonometric functions.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
> it turns out indeed, that
the problem can be solved> for arbitrary p. Applying Eulers
formula,> one gets e.g.Just for the record, there are two
types of typographical errors below.1) On the right, k should
be x instead.2) A power is misplaced: Instead of (1-ix^p), it
should be (1-ix)^p.> sin(p*arctan(x))=1/(2i) * ( (1+ik)^p -
(1-ik^p) ) / (1+k^2)^{p/2}David
===
> Just for the record,
there are two types of typographical errors below.> 1) On the
right, k should be x instead.> 2) A power is misplaced:>
Instead of (1-ix^p), it should be (1-ix)^p.>
>sin(p*arctan(x))=1/(2i) * ( (1+ik)^p - (1-ik^p) ) /
(1+k^2)^{p/2}yes, thanks:sin(p*arctan(x))=1/(2i) * ( (1+ix)^p
- (1-ix)^p ) / (1+x^2)^{p/2}
===
>I.a6m not much up on the
current rage of category theory.><snip> To understand
this, consider the following...<snip of cogent answer to
question>Ooops, forgot to say - thanks for the answer!!
<iSJ2b.14151$_5.322454@news1.telusplanet.net>
===
 > M proper
subset M / { x | xMa }x{ y | bMy } in P, which cannot be
>Since a,b are arbitrary incomparable elements, we can
conclude that >any order is the intersection of the total
orders that extend it. >That.a6s so slick it goes over my head.
Would you clarify?Don.a6t we have to show the intersection of
total orders isn.a6t empty?----
===
I.a6ll call the empty set E,
since I don.a6t know a better way torepresent it in ASCII and
I.a6ll call the power set of a set A, P(A).Is the following
correct?P(E) = {E}P(P(E)) = { E, {E} }P(P(P(E))) = { E, {E},
{{E}}, {E, {E}} }P(P(P(P(E)))) = { ... } will have 2^4 = 16
members (?)P(P(P(P(P(E))))) = { ... } will have 2^(2^4) =
2^16 = 65536 members (?)My questions are related to an
exercise in Elements of Set Theory,Chapter 1. Question 7:
List all the members of V_3. List all the members of V_4.
(It is to be assumed here that there are no atoms.)V_n is
de?ed to be V_n+1 = P(V_n). And V_0 = E.If my observations
about the number of members resulting fromrepeated
applications of power set function are correct, then Ican
conclude V_4 has over 65,000 members...Ok - so it.a6s obvious
I.a6ve gone wrong somewhere.
===
> I.a6ll call the empty set E,
since I don.a6t know a better way to> represent it in ASCII and
I.a6ll call the power set of a set A, P(A).> Is the following
correct?> P(E) = {E}> P(P(E)) = { E, {E} }> P(P(P(E))) = {
E, {E}, {{E}}, {E, {E}} }> P(P(P(P(E)))) = { ... } will have
2^4 = 16 members (?)> P(P(P(P(P(E))))) = { ... } will have
2^(2^4) = 2^16 = 65536 members (?)But this is V_5 (using the
notation from below).> My questions are related to an
exercise in Elements of Set Theory,> Chapter 1. Question 7:>
List all the members of V_3. List all the members of V_4.>
(It is to be assumed here that there are no atoms.)> V_n is
de?ed to be V_n+1 = P(V_n). And V_0 = E.> If my observations
about the number of members resulting from> repeated
applications of power set function are correct, then I> can
conclude V_4 has over 65,000 members...> Ok - so it.a6s
obvious I.a6ve gone wrong somewhere.>V_0 has 0 elements.V_1 has
2^0=1 element.V_2 has 2^1=2 elements.V_3 has 2^2=4
elements.V_4 has 2^4=16 elements.-- Clive
Toothhttp://www.clivetooth.dk
===
> I.a6ll call the empty set
E, since I don.a6t know a better way to> represent it in ASCII
and I.a6ll call the power set of a set A, P(A).> Is the
following correct?> P(E) = {E}> P(P(E)) = { E, {E} }>
> P(P(P(E))) = { E, {E}, {{E}}, {E, {E}} }> P(P(P(P(E)))) =
{ ... } will have 2^4 = 16 members (?)> P(P(P(P(P(E))))) = {
... } will have 2^(2^4) = 2^16 = 65536 members (?)> But this
is V_5 (using the notation from below).> My questions are
related to an exercise in Elements of Set Theory,> Chapter
1. Question 7:> List all the members of V_3. List all
the members of V_4.> (It is to be assumed here that there
are no atoms.)> V_n is de?ed to be V_n+1 = P(V_n). And
V_0 = E.> If my observations about the number of members
resulting from> repeated applications of power set function
are correct, then I> can conclude V_4 has over 65,000
members...> Ok - so it.a6s obvious I.a6ve gone wrong
somewhere.> V_0 has 0 elements.> V_1 has 2^0=1 element.>
V_2 has 2^1=2 elements.> V_3 has 2^2=4 elements.> V_4 has
2^4=16 elements.> -- > Clive Tooth>
http://www.clivetooth.dk
===
> Oops... the equation didn.a6t
come out right. Let me retype it out in words:> dX =
Kappa*(Miu - X)dt + sigma*X^gammadWThis process is mean
reverting to Miu. If Miu is zerothen it will be absorbing as
the stochastic term will dissapear.If Miu is not zero then
its not absorbing (which I guess means re?g ?)as the
stochastic term will knock it away from the
mean.hthPeter.
===
>Oops... the equation didn.a6t come out
right. Let me retype it out in words:>dX = Kappa*(Miu -
X)dt + sigma*X^gammadW> This process is mean reverting to
Miu. If Miu is zero> then it will be absorbing as the
stochastic term will dissapear.> If Miu is not zero then its
not absorbing (which I guess means re?g ?)> as the
stochastic term will knock it away from the mean.Could you
maybe explain your statements some more? I.a6m interested :-)>
> hth> Peter.Nijmegen, Netherlands
===
2nd appeal... please
try to help.
===
> 2nd appeal... please try to help.OK, some
thoughts again then... Note that it.a6s a bit of handwaving and
I haven.a6t checked the calculations.If you look at the process
I provided and you write the SDE in integral form you can
obtain the expectation and the variance of r(t)-r(0) (by
de?ing m(t)=E(r(t)-r(0)) en solving the equation for m(t)).
The increments of r(t) have a normal distribution, so then
you have the complete distribution. And you can see that the
mean tends to b/a.term made me wonder whether the increments
of the process are now still look it up in a textbook if it
would make any difference, but I haven.a6t gotten around to
that yet. *If* it is the case that the resulting process
still has normal increments, than you could *maybe* apply the
same technique as I did above to determine the mean and
variance of this normal distribution. But I think the X^gamma
will give rise to computational problems again. If you can do
it (determine that the increments are normal, what the
expectation and the variance are) though, then you can
determine the level to which the process is reverting.
Another possibility is to determine the moment-generating
function of X(t), but here also the X^gamma gives a problem I
think.Maybe someone who knows by heart what effect the X^gamma
in the diffusion term has, could tell it here. Otherwise I.a6ll
try to look it up.Another thought I have on this is maybe to
make a substitution. A certain conditional expectation of the
solution of the SDE you mention is the solution to a
particular partial differential equation. These are linked
via the Feynman-Kac theorem. Maybe the PDE is more easy to
handle in terms of computability. Maybe a substitution
(something like exp{X^gamma} or so) in this PDE to get rid of
the X^gamma (which will come into the PDE somehow) will work.
But I wouldn.a6t know how and I have bad hopes for a
substitution to work. But in any case the PDE might be
numerically tractable.A ?al thought is to apply numerical
procedures to the SDE. I have no experience in that area, so
just a mere thought.Nijmegen, Netherlands
===
> Forget
math/science types, what about the whole population of
humans. I have never met a single person who does not derive
a lot of pleasure from at least some kind of music.False.
Professor Andrei Zavrotzky, told me that for him music was
mere noise.Luis Rodriguez
===
> Forget math/science
types, what about the whole population of humans. I> have
never met a single person who does not derive a lot of
pleasure from> at least some kind of music.> False.Not
necessarily, if the previous poster had never met Prof
Zavrotzky.> Professor Andrei Zavrotzky, told me that for him
music was mere> noise. Luis Rodriguez-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
===
>I can agree with mathematics being a science,
or a tool of science, but >the standard methods don.a6t always
line up. My objections is not to math >being a science, but
to the notion that math can be done by the >scienti? method.
JSH appears to use experimental evidence as his >proof,
because he doesn.a6t see how math formulas can display widely
>varying behaviors after very small changes in the
parameters. He isn.a6t saying so in this round, but in the past
he has claimed thatall the fallacious proofs ARE his
experiments, and are his version ofthe scienti? method. You
try a proof, you see if it succeeds, youtry another
proof. If sooner or later you get a proof accepted,you.a6ve
done math by the scienti? method. - Randy
===
>I can
agree with mathematics being a science, or a tool of science,
but >the standard methods don.a6t always line up. My objections
is not to math >being a science, but to the notion that math
can be done by the >scienti? method. JSH appears to use
experimental evidence as his >proof, because he doesn.a6t see
how math formulas can display widely >varying behaviors
after very small changes in the parameters. > He isn.a6t
saying so in this round, but in the past he has claimed that>
all the fallacious proofs ARE his experiments, and are his
version of> the scienti? method. You try a proof, you see
if it succeeds, you> try another proof. If sooner or later
you get a proof accepted,> you.a6ve done math by the scienti?
method.> - Randy> If that.a6s what he means, he 
isn.a6t doing
anything new. Attempt a proof, look for holes, patch holes,
look for more holes, patch, etc until there are no holes.--
Will Twentyman
===
> I can agree with mathematics being
a science, or a tool of science, > but the standard methods
don.a6t always line up. My objections is not > to math being a
science, but to the notion that math can be done by > the
scienti? method. JSH appears to use experimental evidence as
> his proof, because he doesn.a6t see how math formulas can
display > widely varying behaviors after very small changes
in the parameters. > He isn.a6t saying so in this round,
but in the past he has claimed that> all the fallacious
proofs ARE his experiments, and are his version of> the
scienti? method. You try a proof, you see if it succeeds,
you> try another proof. If sooner or later you get a proof
accepted,> you.a6ve done math by the scienti? method.> -
Randy> If that.a6s what he means, he isn.a6t doing 
anything
new. Attempt a proof, > look for holes, patch holes, look for
more holes, patch, etc until there > are no holes.> But you
forgot the most important steps: 1. Insist that those
claiming holes are not to be believed: they are lying,
confused, inept, corrupt, or engaged in a conspiracy to
silence him. 2. Threaten lawsuit, congressional subcommittee
meetings, the downfall of the corrupt mathematical
establishment. 3. Invoke The Hammer. 4. Inform the FBI. 5.
Oh, well. It wasn.a6t important anyway.Dale.
===
>
> I can agree with mathematics being a science, or a
tool of science, > but the standard methods don.a6t always
line up. My objections is not > to math being a science,
but to the notion that math can be done by > the scienti?
method. JSH appears to use experimental evidence as > his
proof, because he doesn.a6t see how math formulas can display
> widely varying behaviors after very small changes in the
parameters. > He isn.a6t saying so in this round,
but in the past he has claimed that> all the fallacious
proofs ARE his experiments, and are his version of> the
scienti? method. You try a proof, you see if it succeeds,
you> try another proof. If sooner or later you get a
proof accepted,> you.a6ve done math by the scienti?
method.> - Randy> If that.a6s what he means, he
isn.a6t doing anything new. Attempt a > proof, look for holes,
patch holes, look for more holes, patch, etc > until there
are no holes.> But you forgot the most important steps:>
> 1. Insist that those claiming holes are not to be
believed:> they are lying, confused, inept, corrupt, or
engaged in> a conspiracy to silence him.> 2. Threaten
lawsuit, congressional subcommittee meetings,> the downfall
of the corrupt mathematical establishment.> 3. Invoke The
Hammer.> 4. Inform the FBI.> 5. Oh, well. It wasn.a6t
important anyway.> Dale.> Are you suggesting that there
are new steps in the Scienti? Method? Does James own that
now too?-- Will Twentyman
===
>Facing vigorous
opposition to my warnings about a problem with an>esoteric
area in mathematics, I make this post to show how
the>scienti? method can be applied to what turns out to be
a question in>a complex system.>The scienti? method is
not math. Period. If you use the scienti? >method to draw
conclusions in math, any correct conclusions will be a
>result of pure luck.> I wonder how the latter
statement can be reconciled with the uncertain> foundations
of mathematics cf G.9adel.a6s undecidability results. Uhm.
What.a6s so uncertain here? G.9adel.a6s 1. 
incompleteness
theorem simplys prove that the set of arithmetical truths is
not recursively enumerable and that there is a primitive
recursive function G, s.t. for any recursively enumerable
axiomatisation A of a theory capable of representing certain
primitive recursive functions, the sentence G(A) is not
provable from A, yet N |= G(A) (G(A) is true about the
natural numbers). > If the> axioms of arithemetic are
uncertain surely results derived therefrom are> also
uncertain. I.a6m not attempting to support the original poster,
but> it seems wrong to be so dogmatic about an un-provable
system. The axioms of arithmetics are not uncertain. There
is a second order axiomatisation of arithmetics that has the
natural numbers as its only model (modulo isomorphism).--
Aatu Koskensilta (aatu.koskensilta@xortec.?Wovon man nicht
sprechen kann, daruber muss man schweigen - Ludwig
Wittgenstein, Tractatus Logico-Philosophicus
<3f49662f$1_2@newsfeed>
<CFex2jAEaxS$EwEo@jessikat.fsnet.co.uk>
<585ab5d8.0308260809.18ae3351@posting.google.com>
===
......>
At the level of a talk for undergraduates the result is that
you >can propose a proposition that you can neither prove nor
disprove.>At that point you are free to have a larger system
with that >proposition as a new axiom which may be either
asserted or denied>by the new axiom. The axiom of choice
was eventually proven to>be independent of the normal
Zormelo-Frankel axioms and lead to>systems which either had
the axiom of choice true or had it denied.>problems was
answered negatively by the Godel proof. Most >interesting
mathematics had assumed AoC so whether it was a>theorem under
Z-F or an axiom was a minor technical issue at best.>I rather
naively imagined that the ?problem.a6 was slightly deeper so
thatit applied to all such systems ie that even after adding
each new.a6axiom.a6 there would remain undecidable 
propositions.
If that were thecase there could be in?itely many ways to
construct particular proofs.Hopefully (Z-F+A)+B is identical
to (Z-F+B)+A, but I.a6m a bit stupid anddon.a6t 
understand
axiomatic independence.I have heard of the experimental
mathematical approach has it arisenbecause of computers or
because of the dif?ulty at the cutting edge?>In fact I
would characterize the certainty of the axioms>as one of the
most essential differences between science>and mathematics.
In science, your axioms are only guesses>and there is always
Nature out there who might tell you that>you guessed wrong.
In mathematics, there is no out there.>The universe is
built entirely from your axioms. The axioms>can.a6t possibly
be wrong so long as they.a6re self-consistent.>Some would say
that all axioms are permissible except some>will lead to
systems with no content, a situation which might>be
summarized as not self-consistent axioms.> - Randy>--
Robin Becker
===
>......>At the level of a talk for
undergraduates the result is that you >can propose a
proposition that you can neither prove nor disprove.>At that
point you are free to have a larger system with that
>proposition as a new axiom which may be either asserted or
denied>by the new axiom. The axiom of choice was
eventually proven to>be independent of the normal
Zormelo-Frankel axioms and lead to>systems which either had
the axiom of choice true or had it denied.>problems was
answered negatively by the Godel proof. Most >interesting
mathematics had assumed AoC so whether it was a>theorem
under Z-F or an axiom was a minor technical issue at
best.>I rather naively imagined that the ?problem.a6 was
slightly deeper so that>it applied to all such systems ie
that even after adding each new>.a6axiom.a6 there would 
remain
undecidable propositions. Precisely! In the new system there
will be some new proposition thatcan be neither proven nor
disproved. And so on and so on ...> If that were the>case
there could be in?itely many ways to construct particular
proofs.>Hopefully (Z-F+A)+B is identical to (Z-F+B)+A, but
I.a6m a bit stupid and>don.a6t understand axiomatic
independence.There would be ?itely many axioms, the
propositions would involve a ?ite number of symbols and be
proven in a ?ite number of stepsso ini?ities would not be
present. There are always alternate proofsbut they are also
?ite. Mechanical proofs are often not very understandable to
humans. Technical questions about what is a miinimalproof etc
are the guist of topics in either Logic or Computer
Science.>I have heard of the experimental mathematical
approach has it arisen>because of computers or because of the
dif?ulty at the cutting edge?>In fact I would characterize
the certainty of the axioms>as one of the most essential
differences between science>and mathematics. In science,
your axioms are only guesses>and there is always Nature out
there who might tell you that>you guessed wrong. In
mathematics, there is no out there.>The universe is built
entirely from your axioms. The axioms>can.a6t possibly be
wrong so long as they.a6re self-consistent.>Some would say
that all axioms are permissible except some>will lead to
systems with no content, a situation which might>be
summarized as not self-consistent axioms.> -
Randy>-- >Robin Becker
===
After a discussion of the
signi?ance of incrementalincreases in athletic prowess I
checked out the followingURL for world records in the men.a6s
100m
eventhttp://multimedia.olympic.org/pdf/en_report_70.
pdfLooking at the data it.a6s clear that despite progressin
health/diet/sports science in postwar years theamount by
which the record is changing is reducing.It.a6s an interesting
question, I think at least,to say, is there a time below
which human beingswill not be able to run the 100m?The ?st
check I thought of doing was a Padeapproximation, (courtesy
of Numerical Recipes.a6 routine)and true enough there seems to
be a singularity in the date at which the world record will
be under 9seconds(although breaking the 9.5s barrier will
take anothercentury or so!).However, my question is, is there
a statistically rigorous way to determine a lower bound on
data of this type? I guess it would have a range of very
dubious applications (populations, technical advancesetc.)
and looks likely to have been studied at some
point.Tom
===
Laurent,For numerical solutions, I would suggest
LAPACK (from www.netlib.org) andLAPACK95. If you prefer C, you
can use CLAPACK.You can obtain the explicit inverse using
LAPACK by solving the system ofequations A*X=I, where A is
your Hermitian matrix.However, are you sure you really need
an explicit inverse? Usually, youend up multiplying the
inverse by something else, so you.a6re better off(numerically)
solving a system of equations.If you need the answer
symbolically (for low order matrices), you can tryMaple,
Mathematica or MuPad (all cost a lot of money).Good
luck,John> Is there any method that quickly inverts an
hermitian matrix?> Laurent>
===
I started writing a program
to enumerate the k-subsets of a set {1,..., n} and it.a6s
taking a little longer than I thought it would.I notice that
a subset of a set of n elements can be represented as ann-bit
word, with an on bit for that the element at that index is
inthe subset, and an off bit for that it is not. For example,
for aneight bit word, that can represent a subset of a set of
eightelements, in this case {1, ..., 8}. The set {1, ..., 8}
is 11111111or in hexadecimal notation 0xFF. The set {1} would
be 10000000, {8}would be 00000001, with the empty set {} being
00000000.Anyways, I have started writing a recursive method to
enumerate thesubsets, and it.a6s a hassle to write and has quite
a few expressions ofn and k. In frustration I started typing
patterns and noticedsomething about how the subsets may be
generated with a seeminglysimpler method, at least
computationally.Consider a set of say, six elements, {1, 2,
3, 4, 5, 6}, and the3-subsets of that, each subset with three
elements. Start with {1, 2,3} in the binary notation, and then
shift right, carrying over theright bit back to the beginning
of the sequence.1, 1, 1, 0, 0, 00, 1, 1, 1, 0, 00, 0, 1, 1,
1, 00, 0, 0, 1, 1, 11, 0, 0, 0, 1, 11, 1, 0, 0, 0, 1Then, I
start with {1, 3, 4}, having shifted the bits after the
?stone bit right. Then I repeat the right-shift with
wraparound n=6times again.1, 0, 1, 1, 0, 00, 1, 0, 1, 1, 00,
0, 1, 0, 1, 11, 0, 0, 1, 0, 11, 1, 0, 0, 1, 00, 1, 1, 0, 0,
1Next is {1, 4, 5}, shifted six times.1, 0, 0, 1, 1, 00, 1,
0, 0, 1, 11, 0, 1, 0, 0, 11, 1, 0, 1, 0, 00, 1, 1, 0, 1, 00,
0, 1, 1, 0, 1Finally there is {1, 3, 5}, there have been
eighteen sequencesrepresenting different subsets thus far,
two more will make theremaining 3-subsets.1, 0, 1, 0, 1, 00,
1, 0, 1, 0, 1I thought I had stumbled upon some incredibly
simple method toenumerate the k-subsets of a set. Yet, it
wasn.a6t immediately clearhow it would work with other values
of n and k, besides the fact thatit is very easy to implement
as a machine.The 1-subsets of {1, ..., 6} are simple enough:1,
0, 0, 0, 0, 00, 1, 0, 0, 0, 00, 0, 1, 0, 0, 00, 0, 0, 1, 0,
00, 0, 0, 0, 1, 00, 0, 0, 0, 0, 1That is {1}, {2}, {3}, {4},
{5}, {6}. The 2-subsets of {1, ..., 6}are not quite so
simple. There are 6!/4!2! many of them, or6*5/2*1=15:1, 1, 0,
0, 0, 00, 1, 1, 0, 0, 00, 0, 1, 1, 0, 00, 0, 0, 1, 1, 00, 0,
0, 0, 1, 11, 0, 0, 0, 0, 11, 0, 1, 0, 0, 00, 1, 0, 1, 0, 00,
0, 1, 0, 1, 00, 0, 0, 1, 0, 11, 0, 0, 0, 1, 00, 1, 0, 0, 0,
11, 0, 0, 1, 0, 00, 1, 0, 0, 1, 00, 0, 1, 0, 0, 1Very good.
It seems that there is a simple pattern to generatesequences
representing the subset elements. Yet, this is so far onlyfor
values of n and k such that n is around 2k, and that k is a
factorof n, for small n. I guess I should implement this and
see if itworks.Consider the 4-subsets of {1, ..., 6}. There
are just as many4-subsets of six as there are 2-subsets of
six, 6!/2!4! = 6!/4!2!. Those sequences above are the
2-subsets of 6, their complements arethe 4-subsets of six.
Thus I can consider only the cases where k <=n/2. The process
generates the k-subsets and (n-k)-subsets at aboutthe same
time, it is a matter of a complement operation on each.Of
course, my goal here is to sum the products of those subsets,
the4-subsets would have higher products than the
2-subsets.Given one of these sequences, I wonder if there.a6s a
way to quicklycalculate the product of the elements that it
represents. The bitsequence up to word size is typically
stored on a machine register ina computer program.I can see
some problems with this method for higher values of n.
Thesize of n isn.a6t a problem, it.a6s when k is small 
compared
to n,generating the starter sequences requires its own
algorithm. It.a6swhen k is relatively prime compared to n that
this is more complex,when k is a factor of n it is more
simple. For example, consider the3-subsets of {1, ..., 8}.
There would be 8!/5!3! of them, 8*7*6/3*2 =56. These would be
the starter sequences with 8 many each:1, 1, 1, 0, 0, 0, 0,
01, 0, 1, 1, 0, 0, 0, 01, 0, 0, 1, 1, 0, 0, 01, 0, 0, 0, 1,
1, 0, 01, 0, 0, 0, 0, 1, 1, 0That.a6s 40 of them, then there
are 16 more. 1, 0, 1, 0, 1, 0, 0, 01, 0, 0, 1, 0, 0, 1,
0Hmm..., well there they are, those two sequences each
generate eightsubsets. That.a6s different from the case where
the sequence 0, 1, 0,1, ... generates only two subsets.I
wonder how it is different for odd n. How many starter
sequences are there for k many sequences apiece? I.a6ll call
them the blocked starter sequences. Let.a6s look at that: for
n=6, k=1 there is one blocked sequence, for n=6, k=2 there
are twoblocked sequences, for n=6, k=3 there are 3. For n=8,
k=1 there isone, n=8, k=2, there could be 6 except {1, 3} is
considered under thealternating starter sequence, it would
have to be not considered. Itshould be a blocked starter
sequence and not repeated in thealternating starter sequence.
For n=8, k=3 there are 5, and for n=8,k=4, there is four.Then
there are the alternating starter sequences, there is a
givennumber of them based upon n and k and whether k is a
factor of n, andthen they generate a given number of
sequences based upon n!/k!(n-k)!- k(n-k) and the number of
sequences. Here I.a6m going to work with k <n/2, so the
function there is about the greatest integer x thus thatk/n <
1/x. For example for n=8, k=3, 1/3 < 3/8 < 1/2, x =2, there
are2 alternating starter sequences. I guess it.a6s easier to
compare n/kto an integer, 8/3 in integer math will give the
answer of x=2. Then,to generate the alternating starter
sequences there is 1, 0, 1, 0,..., spacing the on bits by one
off bit, then 1, 0, 0, 1, 0, 0,etctera, the sequence yn+1 for
postive integer y <= x, i.e. {1, 3, 5,...}, {1, 4, 7, ...},
{1, 5, 9, ...}, each of those. For example inthe above
description for k=3 and n=8, x =2 and thus for y=1 and y=2the
starter sequences are for {1, 3, 5} and {1, 4, 7}.Why is it
that there are k many subsets from alternating
startersequences where k is not a factor of n and n/k many
for k being afactor of n? Maybe it is more apt to say that as
k divides into nthat there are only n/k many subsets from the
alternating startersequence.In a way the k-subsets relate
back to the concept of the canonicalsequence. This is where
the canonical sequence for k=n/2 is101010..., and the other
k-subset.a6s sequences canonicalize to that.For n=8, there are
more starting sequences. This is where there areenough
sequence elements to separate blocks of multiple elements.
For n=10, k=5, there are 10!/5!5! or 10*9*8*7*6/5*4*3*2*1
=2*3*2*7*3=252 5-subsets of {1, ..., 10}.1, 1, 1, 1, 1, 0, 0,
0, 0, 01, 0, 1, 1, 1, 1, 0, 0, 0, 01, 0, 0, 1, 1, 1, 1, 0, 0,
01, 0, 0, 0, 1, 1, 1, 1, 0, 01, 0, 0, 0, 0, 1, 1, 1, 1, 01,
1, 0, 1, 1, 1, 0, 0, 0, 01, 1, 0, 0, 1, 1, 1, 0, 0, 01, 1, 0,
0, 0, 1, 1, 1, 0, 01, 1, 0, 0, 0, 0, 1, 1, 1, 01, 1, 0, 1, 1,
0, 1, 0, 0, 01, 1, 0, 1, 1, 0, 0, 1, 0, 01, 1, 0, 1, 1, 0, 0,
0, 1, 01, 1, 0, 1, 0, 1, 1, 0, 0, 01, 1, 0, 1, 0, 1, 0, 1, 0,
01, 1, 0, 1, 0, 1, 0, 0, 1, 01, 1, 0, 1, 0, 0, 1, 1, 0, 01,
1, 0, 0, 1, 1, 0, 1, 0, 01, 1, 0, 0, 1, 1, 0, 0, 1, 01, 1, 0,
0, 0, 1, 1, 0, 1, 01, 0, 1, 0, 1, 1, 1, 0, 0, 01, 0, 1, 0, 0,
1, 1, 1, 0, 01, 0, 1, 0, 0, 0, 1, 1, 1, 01, 0, 0, 1, 0, 1, 1,
1, 0, 01, 0, 0, 1, 0, 0, 1, 1, 1, 01, 0, 0, 0, 1, 0, 1, 1, 1,
0Those 25 sequences, if correct, rotate to each form ten
sequencesforming a total of 250 sequences representing
5-subsets of 10.With those blocked starting sequences there
is the alternatingsequence, it generates 10/5 = 2
5-subsets:1, 0, 1, 0, 1, 0, 1, 0, 1, 0At that point again it
is about as not simple as the regular recursivemethod to
enumerate the subsets. Yet, it is much more
computationallyef?ient.Figuring this out would probably help
me determine how the productlists as discussed earlier would
progress.I thought to look at discussion of Gray codes, they
are progressionsof sequences, but their use is not
immediately apparent here as thesequences here have ?ed
numbers of on and off bits. The Hammingdistance is the
minimum number of bits to be changed to convert onesequence
to another. What.a6s the regular way to enumerate thesequences
with a given number of on bits that are each different
onrotation?I ?ure the easiest way to generate the byte
arrays of each rotationof a starter sequence will be to put
two copies of the startersequence into a sequence of bits,
and then to extract subsequences. For example, for the
starter sequence1, 0, 0put into memory 1, 0, 0, 1, 0, 0and
then extract the three subsequences of length three starting
atthe index from one to three:[ 1, 0, 0, ] 1, 0, 0 -> 1, 0,
01, [ 0, 0, 1, ] 0, 0 -> 0, 0, 11, 0, [ 0, 1, 0, ] 0 -> 0, 1,
0This is similar to having the sequence being written about a
circle,and then reading off a sequence of that length from
each element,beads on a string. Some machine architectures
might have instructionsto rotate a register.I was reading
MathWorld and notice that Mathematica has a KSubsetsmethod to
list the k-subsets of a set, I.a6m still in the process
o?plementing it. The other day I was reading the
documentation ofHYPERG, a Hypergeometric function package for
Maple, based on HYP. Today I.a6ll read the documentation of
HYP.Heh, bait.Ross
===
> I started writing a program to
enumerate the k-subsets of a set {1,> ..., n} and it.a6s taking
a little longer than I thought it would.If Ross means subsets
with k elements, he should try the binomial coef?ient,
C(n,k).If he does not mean that, he should explain up front
what he does mean.
===
>it seems to me that a quaternion is
nothing more than a simple>ordered quadruplet. And that
multiplication/addition is simply>extended in such a way that
it is de?ed over such ordered>quadruplets. It.a6s as if I were
to write a geometry book where I>called points binarinions
and write them as a#{(b}) instead of>(a,b).. the point
being that it would be arbitrary and accomplish>nothing and,
in fact, greatly obfuscate something for no reason. >This, it
seems, is the entire nature of quaternions.IIRC, I used
quaternions once in a computer graphics program thatneeded to
do rotations of objects in 3D, and it simpli?d the
programnicely.
===
>is a group of transformations of a vector
space over the quaternions:>in your notation>Sp(2n) = {T
in GL(n,H): T*T = I}>Symplectic structure. it seems to me,
arises if and only if>one has an underlying vector space over
the quaternions.> Only if one also demands that the matrices
are unitary.> The above group is automatically constrained to
be compact.> And compactness and the fact that the group is
quaternionic> *together* imply that the group is
symplectic.You seem to enjoy making life more complex (pun
intended).It seems to me much simpler to use the term
symplectic groupfor the groups Sp(n) (or Sp(2n)),just as it
is simpler to use the term unitary groupfor the groups
U(n).You might as well say that unitary has nothing to do
with complex numbersbecause there are quasi-unitary groups
over ?ite ?lds.As far as I can see, whenever symplectic
structure is important(as in symplectic geometry)one can in
fact express the structure in terms of quaternions.-- Timothy
Murphy tel: +353-86-233 6090
===
>is a group of
transformations of a vector space over the quaternions:>in
your notation>Sp(2n) = {T in GL(n,H): T*T = I}>
>Symplectic structure. it seems to me, arises if and only
if>one has an underlying vector space over the
quaternions.> Only if one also demands that the matrices
are unitary.> The above group is automatically constrained
to be compact.> And compactness and the fact that the group
is quaternionic> *together* imply that the group is
symplectic.>You seem to enjoy making life more complex (pun
intended).>It seems to me much simpler to use the term
symplectic group>for the groups Sp(n) (or Sp(2n)),>just as
it is simpler to use the term unitary group>for the groups
U(n).But U(p,q) are the only unitary groups. Sp(2p,2q) are
not theonly symplectic groups. Similarly, SO(p,q) and O(p,q)
(and connected components, etc, thereof) are not the only
orthogonal groups.>You might as well say that unitary has
nothing to do with complex numbers>because there are
quasi-unitary groups over ?ite ?lds.>As far as I can see,
whenever symplectic structure is important>(as in symplectic
geometry)>one can in fact express the structure in terms of
quaternions.With respect to symplectic geometry and
symplectic manifolds, the appropriate symplectic group is
generally noncompact, and is in fact generally equal to
Sp(2n,R), whose elements are most simply described by
matrices over the real numbers.David McAnally--------------
<biahmt$gh$1@bunyip.cc.uq.edu.au>
<bie3np$i6j$1@bunyip.cc.uq.edu.au>
<EBw2b.29046$pK2.45750@news.indigo.ie>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
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===
 at 12:02 AM, Timothy Murphy
<tim@birdsnest.maths.tcd.ie> said:>Surely nobody interested
in quantum theory>would worry about the Axiom of Choice?Why
not, if they can.a6t derive what they need without it?>In
practice, the terms quaternionic and symplectic>seem more
or less interchangeable to me.?Complex I could understand and
even sympathize with, but Quaternionic?No way. The
dimensionality of a symplectic manifold need not be amultiple
of 4.-- Shmuel (Seymour J.) Metz, SysProg and JOATReply to
domain Patriot dot net user shmuel+news to contact me. Do not
replyto spamtrap@library.lspace.org
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tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
===
 at 04:58 AM, doug_protocols@yahoo.com (Doug B)
said:>What I was saying>was that these results could all be
proven just as easily with only>the most trivial
modi?ations, if you skipped the whole quaternion>detour
and just de?ed multiplication and addition of
ordered>quadruplets.Sure, it.a6s easy to see in hindsight the
proper way to de?emultiplication of OF THE 4-touples of
reals. It.a6s not so easy to dowhen you.a6re the ?st to 
try it
and don.a6t have the advantage of modernVector Analysis, Group
Theory, etc. For his time and place it was amajor
breakthrough.>would be just as applicable if the entire
theory simply used ordered>quadruplets. Correct.>So what I
meant to say is that it seems>that the quaternion apparatus
merely makes the results unnecessarily>opaque.Doing it the
way you want might take more machinery than you realize,and
Hamilton would probably have found it to be more opaque than
hisway.-- Shmuel (Seymour J.) Metz, SysProg and JOATReply to
domain Patriot dot net user shmuel+news to contact me. Do not
replyto spamtrap@library.lspace.org
===
>2) given a
formula in +,-,*,/, and radicals, and its computed minimal>
>polynomial (for which the formula is a solution), are the
other real>solutions expressible similarly (as formulas
in +,-,+,/,radicals)?> Yes, the other roots (ie the
conjugates of the given root)> are obtained by replacing
each nth root by the other nth roots,> eg replacing
a^{1/n} by wa^{1/n} where w is an nth root of 1.>but is
that really in the given syntax? w (=e^(2pi i/n) )>isn.a6t
really allowed here. none of e, i, or pi are de?able (I
also>left out the restriction that powers can only be
rationals).> This is really a religious question.> You can
either regard the algebraic numbers > as a sub?ld of the
complex numbers> in which case you can set w = exp(2pi i/n)
as you say;> or one can regard the algebraic numbers> as
constructed from the rationals,> in which case w is one of
the roots of x^n - 1.OK, maybe this is allowable
symbolically. But my thoughts about it are not clear enough
really to say yes or no.> In any case, if you want to look at
all the roots of the equation> you will have to go outside the
?ld Q(r) where r is your root.> But I feel I.a6m just saying
what you know perfectly well anyway!Maybe not perfectly. I
was under the impression that, by de?ition,algebraics were a
subset of the reals. but looking around the web itseems that
sometimes it is speci?d as such but mostly not. And maaybe
now I realize I don.a6t care (i.e. algebraics that are not
reals are OK).If any spirituality is involved, it is in my
desire to have only symbolic operations involved, no
numerical ones (yes, I realize that is problematic
distinction, let.a6s say no bit counting?).I think my problem
may have been better stated in other threads (Whatis an
algebraic integer? and Computational algebraic
integers).-- Mitch HarrisLehrstuhl fuer Automatentheorie,
Fakultaet InformatikTechnische Universitaet Dresden,
Deutschlandhttp://tcs.inf.tu-dresden.de/~harris
===
> A
question please: In the phrase if and only if is the ?st
if> necessary?> IOW, A only if B is not the same
as A if and only if B?> Yes, both if.a6s are
signi?ant.> The following statements all mean exactly
the same thing:> 1. A only if B.> 2. B if A.> 3. If
A, then B.> 4. A is suf?ient for B.> 5. B is necessary
for A.> My dif?ulty is with the statement 1 and I would
never imagine that it is> equivalent to A implies B. I
would rather say that what it means is onlyif B> is true
is A true but I am not sure that this is a meaningful phrase
in> English.> That is precisely what it means. I don.a6t
understand why you think there> is a difference.Because I
thought that the statement Only if a teacher has given
permission isa student allowed to enter this room (from an
Oxford dictionary explaining themeaning of only if)
implies the statement If a teacher has given permission,a
student is allowed to enter this room. It seems I was wrong
and myEnglish is worse than I thought :-)> To be speci?,
the following statements all are equivalent:> 1. A only if
B.> 2. A implies B.> 3. Only if B is true is A true.> 4. A is
not true unless B is true.> 5. B or not A.> -- > Dave Seaman
dseaman@purdue.edu> Manager, Advanced Applications YONG 517>
http://www-rcd.cc.purdue.edu/~dseaman/ FAX: (765)
494-0566
===
Repeatedly I.a6ve given out a de?ition for
counting prime numbers, butI.a6m afraid that something has not
been communicated. So ?st I.a6mgoing to give an example,
counting the number of prime numbers up to100, and afterwards
I.a6ll give the function. In that way I hope topoint clearly to
a key feature of my research.What my function does is count
the number of composites for eachinteger up to the square
root of your base number. So for countingprimes up to 100 the
base number is 100 and its square root is 10.Here are the
composite counts: dS(100,2) = 49; dS(100,3) = 16; dS(100,4) =
0; dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0;
dS(100,10)= 0;and summing the dS values gives you 74. Notice
when dS equals 0.Now you add 1 for the number 1, as its not
considered prime, andsubtract from 100 to get 25, which is
the count of primes up to 100.And those prime numbers are2,
3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61,67, 71, 73, 79, 83, 89, and 97.Now a key feature of dS
that I want to emphasize is that for dS(x,y)if y is not prime
dS equals 0. I want to repeat that for emphasis,for dS(x,y) if
y is not prime dS equals 0.Amazingly enough, mathematicians
never discovered such a function asmy dS throughout their
entire history.Some posters have tried to convince you
otherwise. While someapparently have tried to convince you
that it doesn.a6t matter if that.a6strue. Basically I 
think many
of them have just worked to confuse you.Now here.a6s that
de?ition I.a6ve given so many times before:dS(x,y) = [pi(x/y,
y-1) - pi(y-1, sqrt(y-1))] [pi(y, sqrt(y)) - pi(y-1,
sqrt(y-1))],S(x,1) = 0. And pi(x, y) = ?) - S(x, y) - 1,
and you get S as the sum of dSfrom dS(x,1) to dS(x,y). Note:
pi(x,sqrt(x)) here gives the same value as the
traditionalpi(x).For faster calculations you need to use
dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] [pi(y,
sqrt(y)) - pi(y-1, sqrt(y-1))]when sqrt(x/y) < y-1.See
http://groups.msn.com/AmateurMath/
primecountingfunction.msnwIf your eyes glazed over, or you
found yourself wishing to quickly runaway from what looks
very complicated then you might understand howsome posters
could so easily manipulate the discussion away from
thetruth.The truth is that the dS function does the amazing,
as you see acomplete de?ition above, where you don.a6t see any
prime numbersexcept 2 but *somehow* as you noticed above
dS(x,y) equals 0 when y isa prime number.You see, the
function *knows* when a number is prime, making
itunique.Those values from above were dS(100,2) = 49;
dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,7) = 3;
dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0;and you can see
the de?ition I gave above at work, as it picks itsway
through the numbers dropping to 0, whenever y is not
prime.That.a6s so huge that I can.a6t emphasize it enough. 
It.a6s
also somethingthat should give you a certain sadness because
what should havehappened is that mathematicians should have
excitedly embraced such aresult.And I think for some of you,
disbelieving *me* is easier thanconsidering even the
possibility that a discipline like mathematicscould have
enough corruption that mathematicians would ?ht what
theyshould be celebrating.But focusing on me is a mistake.If
you really wish to consider human achievement as some
personalissue, where you wish to make me that powerful as a
single human beingthat the accomplishments of thousands of
years of effort can bereduced to just being about the
individuals involved, then you shatterthe foundations of
human civilization, and make it just some fashionshow.The
truth is that the discoveries are more important than
thediscoverers, and in letting it turn into some celebrity
issue, youdemean human society and its accomplishments.For
some of you I fear that the idea that I might become a
celebrityis all that matters to you. And I fear you think
that.a6s all that evermattered through all of human history, as
if the truth were just aword, and life has always just been a
big popularity contest.The truth is not just a word and
emphasizing truth over social issueshas helped humanity to
make the technological achievments which makemy communication
through this medium possible.Where would we be if engineers
had paused to ?ure out who they likedand disliked before
they bothered to use the information
discovered?Mathematicians aren.a6t doing you any favors. I
suggest that all of youfocus less on me than on the
information, and forget about whatbene?s telling the truth
will give me, as you consider that thevalue to society is
far, far greater.So how can mathematicians behave as they
have? That.a6s an important question, and I suggest we ?d out
why popularityhas so much importance in their world.James
Harris
===
> Repeatedly I.a6ve given out a de?ition for
counting prime numbers, [snuip]and been made an utter fool by
competent folks who trivially refuteyour psychotic spews with
concrete counterexamples that you ignore.
http://w0rli.home.att.net/youare.swfhttp://www.mazepath.com/
uncleal/sunshine.jpg--Uncle
Alhttp://www.mazepath.com/uncleal/eotvos.htm (Do something
naughty to physics)
===
<...> Now a key feature of dS that I
want to emphasize is that for dS(x,y)> if y is not prime dS
equals 0. I want to repeat that for emphasis,> for dS(x,y) if
y is not prime dS equals 0.> Amazingly enough, mathematicians
never discovered such a function as> my dS throughout their
entire history.Function.<...>
===
> <...> Now a key
feature of dS that I want to emphasize is that for dS(x,y)>
if y is not prime dS equals 0. I want to repeat that for
emphasis,> for dS(x,y) if y is not prime dS equals 0.>
Amazingly enough, mathematicians never discovered such a
function as> my dS throughout their entire history.>Can you
_prove_ this negative for me please?(bait = never + entire
history)
===
> <...> Now a key feature of dS that
I want to emphasize is that for dS(x,y)> if y is not prime
dS equals 0. I want to repeat that for emphasis,> for
dS(x,y) if y is not prime dS equals 0.> Amazingly
enough, mathematicians never discovered such a function as>
> my dS throughout their entire history.> Can you
_prove_ this negative for me please?> (bait = never +
entire history)Well for context it helps if the information
you deleted out is given.What follows is what the poster
deleted out, up to the statement onwhich focus was placed by
that deletion.What my function does is count the number of
composites for eachinteger up to the square root of your base
number. So for countingprimes up to 100 the base number is 100
and its square root is 10.Here are the composite counts:
dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6;
dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)= 0;and
summing the dS values gives you 74. Notice when dS equals
0.Now you add 1 for the number 1, as its not considered
prime, andsubtract from 100 to get 25, which is the count of
primes up to 100.And those prime numbers are2, 3, 5, 7, 11,
13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,67, 71,
73, 79, 83, 89, and 97.Now a key feature of dS that I want to
emphasize is that for dS(x,y)if y is not prime dS equals 0. I
want to repeat that for emphasis,for dS(x,y) if y is not
prime dS equals 0.Amazingly enough, mathematicians never
discovered such a function asmy dS throughout their entire
history.Ok then, now to that statement, it.a6s amazing but
true. Throughouttheir *entire* history mathematicians never
discovered such afunction.Obviously, if that were false the
easiest thing would be for someposter to show you something
that mathematicians have that does behavethe same way.But the
closest thing they might try works by brute force
andexploiting rounding. I.a6ll see if any try to put it up
before I gointo details.Now notice that the poster deleted
out the data, and focused on astatement which said poster
probably ?ured most of you would simplyreject as being
impossible.But that depends on your faith in the very people
I.a6m explaining toyou aren.a6t living up to that 
faith.Remember,
math people know that many of you ?d math eitherunappealing
or dif?ult, after all, math people *teach* mathematics. By
pushing you to depend on your trust, deeply ingrained, this
postershowed a fascinating disdain for some of you.James
Harris
===
><...>Now a key feature of dS that
I want to emphasize is that for dS(x,y)>if y is not prime
dS equals 0. I want to repeat that for emphasis,>for
dS(x,y) if y is not prime dS equals 0.>Amazingly
enough, mathematicians never discovered such a function
as>my dS throughout their entire history.>Can you
_prove_ this negative for me please?>(bait = never +
entire history)> Well for context it helps if the
information you deleted out is given.Ok, so you are going to
answer his question about proving mathematicians never
discovered something like dS...> What follows is what the
poster deleted out, up to the statement on> which focus was
placed by that deletion.> What my function does is count
the number of composites for each> integer up to the square
root of your base number. So for counting> primes up to 100
the base number is 100 and its square root is 10.> Here are
the composite counts:> dS(100,2) = 49; dS(100,3) = 16;
dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3; dS(100,8) =
0;> dS(100,9) = 0; dS(100,10)= 0;> and summing the dS
values gives you 74. Notice when dS equals 0.> Now you add
1 for the number 1, as its not considered prime, and>
subtract from 100 to get 25, which is the count of primes up
to 100.> And those prime numbers are> 2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,> 67, 71, 73,
79, 83, 89, and 97.> Now a key feature of dS that I want to
emphasize is that for dS(x,y)> if y is not prime dS equals 0.
I want to repeat that for emphasis,> for dS(x,y) if y is not
prime dS equals 0.> Amazingly enough, mathematicians never
discovered such a function as> my dS throughout their entire
history.> Ok then, now to that statement, it.a6s amazing but
true. Throughout> their *entire* history mathematicians never
discovered such a> function.> Obviously, if that were false
the easiest thing would be for some> poster to show you
something that mathematicians have that does behave> the same
way.But that doesn.a6t show that it.a6s true. Only that 
it.a6s
(presumably) easy for us to show it false.> But the closest
thing they might try works by brute force and> exploiting
rounding. I.a6ll see if any try to put it up before I go> into
details.> Now notice that the poster deleted out the data,
and focused on a> statement which said poster probably ?ured
most of you would simply> reject as being impossible.> But
that depends on your faith in the very people I.a6m explaining
to> you aren.a6t living up to that faith.> Remember, math
people know that many of you ?d math either> unappealing or
dif?ult, after all, math people *teach* mathematics. > By
pushing you to depend on your trust, deeply ingrained, this
poster> showed a fascinating disdain for some of you.>
James HarrisBut, can you, for that matter *will* you, prove
that no mathematician came up with anything like your dS? You
didn.a6t answer the question, just showed us what the function
is and talked about methods of disproof.-- Will
Twentyman
===
> Repeatedly I.a6ve given out a de?ition for
counting prime numbers, but> I.a6m afraid that something has
not been communicated. So ?st I.a6m> going to give an example,
counting the number of prime numbers up to> 100, and
afterwards I.a6ll give the function. In that way I hope to>
point clearly to a key feature of my research.> What my
function does is count the number of composites for each>
integer up to the square root of your base number. So for
counting> primes up to 100 the base number is 100 and its
square root is 10.> Here are the composite counts:>
dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) =
6; dS(100,7) = 3; dS(100,8) = 0;> dS(100,9) = 0; dS(100,10)=
0;> and summing the dS values gives you 74. Notice when dS
equals 0.> Now you add 1 for the number 1, as its not
considered prime, and> subtract from 100 to get 25, which is
the count of primes up to 100.> And those prime numbers
are> 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61,> 67, 71, 73, 79, 83, 89, and 97.> Now a key
feature of dS that I want to emphasize is that for dS(x,y)> if
y is not prime dS equals 0. I want to repeat that for
emphasis,> for dS(x,y) if y is not prime dS equals 0.>
Amazingly enough, mathematicians never discovered such a
function as> my dS throughout their entire history.> Some
posters have tried to convince you otherwise. While some>
apparently have tried to convince you that it doesn.a6t matter
if that.a6s> true. Basically I think many of them have just
worked to confuse you.> Now here.a6s that de?ition 
I.a6ve
given so many times before:> dS(x,y) = [pi(x/y, y-1) -
pi(y-1, sqrt(y-1))]> [pi(y, sqrt(y)) - pi(y-1,
sqrt(y-1))],> S(x,1) = 0.> And pi(x, y) = ?) - S(x,
y) - 1, and you get S as the sum of dS> from dS(x,1) to
dS(x,y).> Note: pi(x,sqrt(x)) here gives the same value as
the traditional> pi(x).> For faster calculations you need
to use> dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)]>
> [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))]> when sqrt(x/y) <
y-1.> See
http://groups.msn.com/AmateurMath/
primecountingfunction.msnwOk, let.a6s step through the
calculation of dS(100,2)=49.dS(100,2)= [pi(100/2,2-1) -
pi(2-1,sqrt(2-1))] * [ pi(2,sqrt(2)) - pi(2-1,
sqrt(2-1)]dS(100,2) = [pi(50,1) - pi(1,1)] [pi(2,sqrt(2)) -
pi(1,1)]Evaluating the pi.a6s:pi(50,1)= ?0)- S(50,1)- 1 =
50 - 0 - 1 = 49pi(1,1) = ?) - S(1,1) - 1 = 1 - 0 - 1 =
0pi(2,sqrt(2))=?) - S(2,sqrt(2)) - 1 = 1 -
S(2,sqrt(2))pi(1,1) = ?) - S(1,1) - 1 = 1 - 0 - 1 =
0Evaluating the odd S(2,sqrt(2)):S(2,sqrt(2)) = dS(2,1) +
dS(2,sqrt(2)) [Is this correct?]dS(2,1) = [pi(1,0) - pi(0,0)]
[pi(1,1) - pi(0,0)]pi(1,0) = ?) - S(1,0) - 1 = -S(1,0)
which I don.a6t see how to calculate.At this point I know that
S(2,sqrt(2)) should be 1 to give the answer you arrived at,
but I.a6m not sure how you are getting there.-- Will
Twentyman
===
> Repeatedly I.a6ve given out a de?ition for
counting prime numbers, but> I.a6m afraid that something has
not been communicated. So ?st I.a6m> going to give an example,
counting the number of prime numbers up to> 100, and
afterwards I.a6ll give the function. In that way I hope to>
point clearly to a key feature of my research.> What my
function does is count the number of composites for each>
integer up to the square root of your base number. So for
counting> primes up to 100 the base number is 100 and its
square root is 10.> Here are the composite counts:>
dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) =
6; dS(100,7) = 3; dS(100,8) = 0;> dS(100,9) = 0; dS(100,10)=
0;> and summing the dS values gives you 74. Notice when dS
equals 0.> Now you add 1 for the number 1, as its not
considered prime, and> subtract from 100 to get 25, which is
the count of primes up to 100.> And those prime numbers
are> 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61,> 67, 71, 73, 79, 83, 89, and 97.> Now a key
feature of dS that I want to emphasize is that for dS(x,y)> if
y is not prime dS equals 0. I want to repeat that for
emphasis,> for dS(x,y) if y is not prime dS equals 0.>
Amazingly enough, mathematicians never discovered such a
function as> my dS throughout their entire history.> Some
posters have tried to convince you otherwise. While some>
apparently have tried to convince you that it doesn.a6t matter
if that.a6s> true. Basically I think many of them have just
worked to confuse you.> Now here.a6s that de?ition 
I.a6ve
given so many times before:> dS(x,y) = [pi(x/y, y-1) -
pi(y-1, sqrt(y-1))]> [pi(y, sqrt(y)) - pi(y-1,
sqrt(y-1))],> S(x,1) = 0.> And pi(x, y) = ?) - S(x,
y) - 1, and you get S as the sum of dS> from dS(x,1) to
dS(x,y).> Note: pi(x,sqrt(x)) here gives the same value as
the traditional> pi(x).> For faster calculations you need
to use> dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)]>
> [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))]> when sqrt(x/y) <
y-1.> I am not going to argue with the correctness of your
results. A point of curiousity about this method is, how
ef?ient is it (in terms of big-O notation)? If your new
algorithm is considerably slower than existing algorithms, it
is only a curiousity. If it is faster, then you have done
something worthy of note.> See
http://groups.msn.com/AmateurMath/
primecountingfunction.msnwThis would be the site I can only
see when I.a6m not logged into my msn passport.> If your eyes
glazed over, or you found yourself wishing to quickly run>
away from what looks very complicated then you might
understand how> some posters could so easily manipulate the
discussion away from the> truth.> The truth is that the dS
function does the amazing, as you see a> complete de?ition
above, where you don.a6t see any prime numbers> except 2 but
*somehow* as you noticed above dS(x,y) equals 0 when y is> a
prime number.> You see, the function *knows* when a number
is prime, making it> unique.> Those values from above were
> dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; >
dS(100,5) = 6; dS(100,7) = 3; dS(100,8) = 0;> dS(100,9) = 0;
dS(100,10)= 0;> and you can see the de?ition I gave above
at work, as it picks its> way through the numbers dropping to
0, whenever y is not prime.> That.a6s so huge that I 
can.a6t
emphasize it enough. It.a6s also something> that should give
you a certain sadness because what should have> happened is
that mathematicians should have excitedly embraced such a>
result.You haven.a6t stated *why* it.a6s exciting. You 
have a new
method of counting primes. That is not, in itself, exciting.
How fast is your method? Does dS return 0 for composites
faster than checking whether a number is composite by other
methods?Note: these are not questions I have the answer to,
these are the questions that will help determine the value of
this work.[non mathematics deleted.]-- Will Twentyman
===
>
Repeatedly I.a6ve given out a de?ition for counting prime
numbers, but> I.a6m afraid that something has not been
communicated. So ?st I.a6m> going to give an example,
counting the number of prime numbers up to> 100, and
afterwards I.a6ll give the function. In that way I hope to>
point clearly to a key feature of my research.> What my
function does is count the number of composites for each>
integer up to the square root of your base number. So for
counting> primes up to 100 the base number is 100 and its
square root is 10.> Here are the composite counts:>
> dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5)
= 6; dS(100,7) = 3; dS(100,8) = 0;> dS(100,9) = 0;
dS(100,10)= 0;> and summing the dS values gives you 74.
Notice when dS equals 0.> Now you add 1 for the number
1, as its not considered prime, and> subtract from 100 to
get 25, which is the count of primes up to 100.> And
those prime numbers are> 2, 3, 5, 7, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61,> 67, 71, 73, 79,
83, 89, and 97.> Now a key feature of dS that I want to
emphasize is that for dS(x,y)> if y is not prime dS equals
0. I want to repeat that for emphasis,> for dS(x,y) if y is
not prime dS equals 0.> Amazingly enough, mathematicians
never discovered such a function as> my dS throughout their
entire history.> Some posters have tried to convince
you otherwise. While some> apparently have tried to
convince you that it doesn.a6t matter if that.a6s> 
true.
Basically I think many of them have just worked to confuse
you.> Now here.a6s that de?ition I.a6ve given so 
many
times before:> dS(x,y) = [pi(x/y, y-1) - pi(y-1,
sqrt(y-1))]> [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))],>
> S(x,1) = 0.> And pi(x, y) = ?) - S(x, y) - 1,
and you get S as the sum of dS> from dS(x,1) to dS(x,y).>
> Note: pi(x,sqrt(x)) here gives the same value as the
traditional> pi(x).> For faster calculations you need
to use> dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1,
sqrt(y-1)]> [pi(y, sqrt(y)) - pi(y-1, sqrt(y-1))]>
> when sqrt(x/y) < y-1.> I am not going to argue with
the correctness of your results. A point > of curiousity about
this method is, how ef?ient is it (in terms of > big-O
notation)? If your new algorithm is considerably slower than
> existing algorithms, it is only a curiousity. If it is
faster, then you > have done something worthy of note.It is
just old Legendre.a6s formula, slightly repacked. Legendre
de?ed phi (x, k) := number of positive integers <= x which
are not divisible by any of the ?st k primes. You get phi
(x, 0) = x for all x, and if k > 0 then phi (x, k) = 0 if x =
0 phi (x, k) = 1 if 1 <= x < p (k+1), where p(k+1) is the
(k+1)st prime phi (x, k) = pi (x) - k + 1 if p(k) <= x <
(p(k+1))^2and from the last equation it follows that with k =
pi (x^(1/2)) pi (x) = phi (x, k) + k - 1And then there is the
recursion formula phi (x, k) = phi (x, k-1) - phi (x/p(k),
k-1). That one is quite obvious because the difference
between phi (x, k) and phi (x, k-1) are exactly those
integers <= x with p(k) as the smallest prime factor, and
these integers can be written as y*p(k), where y has no
divisor <= p(k) and y <= x / p(k). I think you can ?ure out
easily how phi (x, k) and his function dS (x, y) are related.
Execution time is O (N). Meissel.a6s algorithm is a trivial
variation of this; if you take the ?st recursion level of
Legendre.a6s formula then you will see that you end up
calculating many values pi (x.a6) where x.a6 <= x^(2/3). 
You can
get all those by using a sieve in O (N^(2/3)), and the
remaining work is done in O (N / log^3 N), I think; so this
is quite simple and much much faster. Btw. I wonder if he
ever ?ured out the cute trick how to calculate phi (x, k)
for lets say k <= 6 and arbitrarily large x using just one
division and two multiplications. I don.a6t know if Meissel
knew it, but Lehmer certainly did.
===
>Repeatedly
I.a6ve given out a de?ition for counting prime numbers,
but>I.a6m afraid that something has not been communicated. So
?st I.a6m>going to give an example, counting the number of
prime numbers up to>100, and afterwards I.a6ll give the
function. In that way I hope to>point clearly to a key
feature of my research.>What my function does is count
the number of composites for each>integer up to the square
root of your base number. So for counting>primes up to 100
the base number is 100 and its square root is 10.>Here
are the composite counts:> dS(100,2) = 49; dS(100,3) =
16; dS(100,4) = 0; > dS(100,5) = 6; dS(100,7) = 3;
dS(100,8) = 0;> dS(100,9) = 0; dS(100,10)= 0;>and
summing the dS values gives you 74. Notice when dS equals
0.>Now you add 1 for the number 1, as its not considered
prime, and>subtract from 100 to get 25, which is the count
of primes up to 100.>And those prime numbers are>2,
3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61,>67, 71, 73, 79, 83, 89, and 97.>Now a key feature
of dS that I want to emphasize is that for dS(x,y)>if y is
not prime dS equals 0. I want to repeat that for
emphasis,>for dS(x,y) if y is not prime dS equals
0.>Amazingly enough, mathematicians never discovered
such a function as>my dS throughout their entire
history.>Some posters have tried to convince you
otherwise. While some>apparently have tried to convince you
that it doesn.a6t matter if that.a6s>true. Basically I 
think
many of them have just worked to confuse you.>Now here.a6s
that de?ition I.a6ve given so many times before:>dS(x,y) =
[pi(x/y, y-1) - pi(y-1, sqrt(y-1))]> [pi(y, sqrt(y)) -
pi(y-1, sqrt(y-1))],>S(x,1) = 0.>And pi(x, y) =
?) - S(x, y) - 1, and you get S as the sum of dS>from
dS(x,1) to dS(x,y).>Note: pi(x,sqrt(x)) here gives the
same value as the traditional>pi(x).>For faster
calculations you need to use> dS(x,y) = [pi(x/y,
sqrt(x/y)) - pi(y-1, sqrt(y-1)]> [pi(y, sqrt(y)) -
pi(y-1, sqrt(y-1))]>when sqrt(x/y) < y-1.>I am
not going to argue with the correctness of your results. A
point >of curiousity about this method is, how ef?ient is
it (in terms of >big-O notation)? If your new algorithm is
considerably slower than >existing algorithms, it is only a
curiousity. If it is faster, then you >have done something
worthy of note.> It is just old Legendre.a6s formula,
slightly repacked. > Legendre de?ed> phi (x, k) :=
number of positive integers <= x which are not > divisible by
any of the ?st k primes. > You get phi (x, 0) = x for all
x, and if k > 0 then> phi (x, k) = 0 if x = 0> phi (x, k) =
1 if 1 <= x < p (k+1), where p(k+1) is the (k+1)st prime> phi
(x, k) = pi (x) - k + 1 if p(k) <= x < (p(k+1))^2> and from
the last equation it follows that with k = pi (x^(1/2))> pi
(x) = phi (x, k) + k - 1> And then there is the recursion
formula> phi (x, k) = phi (x, k-1) - phi (x/p(k), k-1). >
That one is quite obvious because the difference between phi
(x, k) and > phi (x, k-1) are exactly those integers <= x
with p(k) as the smallest > prime factor, and these integers
can be written as y*p(k), where y has > no divisor <= p(k)
and y <= x / p(k). > I think you can ?ure out easily how
phi (x, k) and his function dS (x, > y) are related.
Execution time is O (N). Meissel.a6s algorithm is a > trivial
variation of this; if you take the ?st recursion level of >
Legendre.a6s formula then you will see that you end up
calculating many > values pi (x.a6) where x.a6 <= x^(2/3). 
You
can get all those by using a > sieve in O (N^(2/3)), and the
remaining work is done in O (N / log^3 N), > I think; so this
is quite simple and much much faster. And with this, I
believe, we arrive at why James is not getting the primes any
faster. It is slower than existing algorithms, and therefore a
curiousity.> Btw. I wonder if he ever ?ured out the cute
trick how to calculate phi > (x, k) for lets say k <= 6 and
arbitrarily large x using just one > division and two
multiplications. I don.a6t know if Meissel knew it, but >
Lehmer certainly did.-- Will Twentyman
===
Let f be an
holomorphic function on an B(0,R) = {z, |z|<R}Let M(r) =
max{|f(z)| / |z|=r} for r < RWe know that there exists a
complex number z_0=r*exp(it) with |f(z_0)| =M(r)and
therefore, C_r = {z, |f(z)| = M(r) and |z|=r} is not
empty.Can we have C_r = { re^(it) / t in [t_0,t_1]} where
t_0<>t_1 and C_r is notthe entire circle of radius
r.Olivier.
===
> Let f be an holomorphic function on an B(0,R)
= {z, |z|<R}> Let M(r) = max{|f(z)| / |z|=r} for r < R> We
know that there exists a complex number z_0=r*exp(it) with
|f(z_0)| => M(r)> and therefore, C_r = {z, |f(z)| = M(r) and
|z|=r} is not empty.> Can we have C_r = { re^(it) / t in
[t_0,t_1]} where t_0<>t_1 and C_r is not> the entire circle
of radius r.No, in fact |f| cannot be constant on any subarc
of a circle without being constant on the full circle; it
doesn.a6t matter if the constant is the maximum value. Proof:
|f|^2 is real analytic on the circle in question. If a real
analytic function on a circle is constant on any subarc of a
circle, it is constant on the circle. (In fact, if it
constant on any in?ite subset of the circle, it is constant
on the circle.)
===
The World Wide Wade
<waderameyxiii@attbi.remove13.com> a .8ecrit dans le>
Let f be an holomorphic function on an B(0,R) = {z, |z|<R}>
> Let M(r) = max{|f(z)| / |z|=r} for r < R> We know
that there exists a complex number z_0=r*exp(it) with
|f(z_0)| => M(r)> and therefore, C_r = {z, |f(z)| = M(r)
and |z|=r} is not empty.> Can we have C_r = { re^(it) / t
in [t_0,t_1]} where t_0<>t_1 and C_r isnot> the entire
circle of radius r.> No, in fact |f| cannot be constant on
any subarc of a circle without being> constant on the full
circle; it doesn.a6t matter if the constant is the> maximum
value. Proof: |f|^2 is real analytic on the circle in
question. If> a real analytic function on a circle is
constant on any subarc of acircle,> it is constant on the
circle. (In fact, if it constant on any in?ite> subset of
the circle, it is constant on the circle.)Ok, but can you
give me a proof with more details....
===
> No, in fact |f|
cannot be constant on any subarc of a circle without being>
> constant on the full circle; it doesn.a6t matter if the
constant is the> maximum value. Proof: |f|^2 is real
analytic on the circle in question. If> a real analytic
function on a circle is constant on any subarc of a>
circle,> it is constant on the circle. (In fact, if it
constant on any in?ite> subset of the circle, it is
constant on the circle.)> Ok, but can you give me a proof
with more details...De?ition: Suppose g: I -> R, where I is
an interval. We say g is real analytic on I if g is locally
expressible on I as a power series.Fact 1. If g is real
analytic on [a,b] and g is constant on an in?ite subset of
[a,b], then g is constant on [a,b]. Fact 2: Let f = u + iv be
holomorphic as in your problem. Then the map t ->
[u(r*exp(it))]^2 + [v(r*exp(it))]^2 is a real analytic
function of t in R.
===
> For this to be true you need H to
be the Borel ?ld generated by F (or, > more generally, for H
to be contained in the u-completion of the Borel > ?ld
generated by F).> It seems you are worrying over the proof
of the seocnd part of Theorem 4 > (pages 382-383) in Chung.a6s
book. To your credit it must be noted that> it is implicit in
Chung.a6s argument that the sequence you call (Q_n)> (and that
Chung labels (Omega_n) is *increasing*. This sequence,
which > bears witness to the sigma-?iteness of u, can
always be taken to be > increasing if desired. And in the
assertion on page 383 that (in your > notation)> liminf
u(Q_n intersect A) = u(A) > the reference to property (e)
of the measure u makes it clear that the > author *is*
assuming (Q_n) to be monotone increasing. [The same problem>
occurs in the proof of Theorem 3 on page 381; to salvage
things one > should simply added the hypothesis (Omega_n)
increasing to De?ition > 5 on page 381.]This is, indeed,
my problem and the reason for my posts. Now, re.Theorem 3, p.
381 I believe I can prove that for every nu_1(A Union (1 to
n)Omega_n)= u_2(A Union (1 to n)Omega_n)It would appear that
this would eliminate the need to assume thatOmega_n is
monotonic increasing in proving u_1(A) = u_2(A). One canthen
assume sigma-?iteness of u without quali?ations. Is
thiscorrect?If true, then maybe there is a similar approach
to avoiding theadditional hypothesis for the second part of
Theorem 4.It.a6s hard enough for a beginner to learn this stuff
without textsmaking it even harder by their mistakes and
omissions, and that.a6s whyyour help is extremely useful to
me.
===
> For this to be true you need H to be the Borel
?ld generated by F (or, > more generally, for H to be
contained in the u-completion of the Borel > ?ld generated
by F).> It seems you are worrying over the proof of the
seocnd part of Theorem 4 > (pages 382-383) in Chung.a6s
book. To your credit it must be noted that> it is implicit
in Chung.a6s argument that the sequence you call (Q_n)> (and
that Chung labels (Omega_n) is *increasing*. This sequence,
which > bears witness to the sigma-?iteness of u, can
always be taken to be > increasing if desired. And in the
assertion on page 383 that (in your > notation)>
liminf u(Q_n intersect A) = u(A) > the reference to
property (e) of the measure u makes it clear that the >
author *is* assuming (Q_n) to be monotone increasing. [The
same problem> occurs in the proof of Theorem 3 on page
381; to salvage things one > should simply added the
hypothesis (Omega_n) increasing to De?ition > 5 on
page 381.]> This is, indeed, my problem and the reason for
my posts. Now, re.> Theorem 3, p. 381 I believe I can prove
that for every n> u_1(A Union (1 to n)Omega_n)= u_2(A Union
(1 to n)Omega_n)> It would appear that this would eliminate
the need to assume that> Omega_n is monotonic increasing in
proving u_1(A) = u_2(A). One can> then assume sigma-?iteness
of u without quali?ations. Is this> correct?Yes > If true,
then maybe there is a similar approach to avoiding the>
additional hypothesis for the second part of Theorem 4.>
It.a6s hard enough for a beginner to learn this stuff without
texts> making it even harder by their mistakes and omissions,
and that.a6s why> your help is extremely useful to me.This is
not a matter of additional hypotheses: a measure u on a
sigma-?ld H is sigma-?ite if and only if there is an
*increasing*sequence (Q_n) of elements of H such that
u(Omega_n) is ?ite for each n. [For if u is sigma-?ite in
the sense of de?ition 5, page 381, of Chung, with u(Omega_n)
?ite and union(Omega_n)=S, then Q_n = Omega_1 U Omega_2 U ...
U Omega_n , n=1,2,3,...de?es an increasing sequence of
elements of H with u(Q_n) <= sum_{k=1}^n u(Omega_n) ?ite.]
The problem is that Chung did not include the word
increasing in his De?ition 5 (though he might as well
have, with no loss of generality) but he argues in the proofs
of Theroems 3 and 4 as if he had.-- A.
===
> Think about TWO
axiom systems S1 and S2.> How can they be related? (uselessly
vague possibilities snipped)lol - a full understanding of the
following might answer the question:Boolos & Jeffrey, Logic &
Computability (or something like that)Chang & Keisler, Model
Theory (ditto)Hartley Rogers, Recursive Function Theory (one
more time)have fun,(ro?
===
> Under the assumption that
both S1 and S2 are consistent? Yes, *and* that (see above)
one isn.a6t a subset of the other,which this... > If 
that.a6s
what > you.a6re asking the answer is no. As a trivial example,
consider > Presburger arithmetic (a decidable theory of
arithmetic having axioms > concerning only + and successor)
and Presburger arithmetic + 1*2=2. ...looks to fall
under.Maybe if one is Pr+1*2=2 and the other Pr+1*3=3? --
Hauke Reddmann <:-EX8 For our chemistry workgroup,remove
math from the addressFor spamming, remove anything
else
===
Repeatedly I.a6ve given out a de?ition for counting
prime numbers, butI.a6m afraid that something has not been
communicated. So ?st I.a6mgoing to give an example, counting
the number of prime numbers up to100, and afterwards I.a6ll
give the function. In that way I hope topoint clearly to a
key feature of my research.What my function does is count the
number of composites for eachinteger up to the square root of
your base number. So for countingprimes up to 100 the base
number is 100 and its square root is 10.Here are the
composite counts: dS(100,2) = 49; dS(100,3) = 16; dS(100,4) =
0; dS(100,5) = 6; dS(100,6) = 0; dS(100,7) = 3; dS(100,8) = 0;
dS(100,9) = 0; dS(100,10)= 0;and summing the dS values gives
you 74. Notice when dS equals 0.Now you add 1 for the number
1, as its not considered prime, andsubtract from 100 to get
25, which is the count of primes up to 100.And those prime
numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61,67, 71, 73, 79, 83, 89, and 97.Now a key
feature of dS that I want to emphasize is that for dS(x,y)if
y is not prime dS equals 0. I want to repeat that for
emphasis,for dS(x,y) if y is not prime dS equals 0.Amazingly
enough, mathematicians never discovered such a function asmy
dS throughout their entire history.Some posters have tried to
convince you otherwise. While someapparently have tried to
convince you that it doesn.a6t matter if that.a6strue. 
Basically
I think many of them have just worked to confuse you.Now
here.a6s that de?ition I.a6ve given so many times 
before:dS(x,y)
= [pi(x/y, y-1) - pi(y-1, sqrt(y-1))] [pi(y, sqrt(y)) -
pi(y-1, sqrt(y-1))],S(x,1) = 0. And pi(x, y) = ?) - S(x,
y) - 1, and you get S as the sum of dSfrom dS(x,2) to dS(x,y).
Note: pi(x,sqrt(x)) here gives the same value as the
traditionalpi(x).For faster calculations you need to use
dS(x,y) = [pi(x/y, sqrt(x/y)) - pi(y-1, sqrt(y-1)] [pi(y,
sqrt(y)) - pi(y-1, sqrt(y-1))]when sqrt(x/y) < y-1.See
http://groups.msn.com/AmateurMath/
primecountingfunction.msnwIf your eyes glazed over, or you
found yourself wishing to quickly runaway from what looks
very complicated then you might understand howsome posters
could so easily manipulate the discussion away from
thetruth.The truth is that the dS function does the amazing,
as you see acomplete de?ition above, where you don.a6t see any
prime numbersexcept 2 but *somehow* as you noticed above
dS(x,y) equals 0 when y isa prime number.You see, the
function *knows* when a number is prime, making
itunique.Those values from above were dS(100,2) = 49;
dS(100,3) = 16; dS(100,4) = 0; dS(100,5) = 6; dS(100,6) = 0;
dS(100,7) = 3; dS(100,8) = 0; dS(100,9) = 0; dS(100,10)=
0;and you can see the de?ition I gave above at work, as it
picks itsway through the numbers dropping to 0, whenever y is
not prime.That.a6s so huge that I can.a6t emphasize it 
enough.
It.a6s also somethingthat should give you a certain sadness
because what should havehappened is that mathematicians
should have excitedly embraced such aresult.And I think for
some of you, disbelieving *me* is easier thanconsidering even
the possibility that a discipline like mathematicscould have
enough corruption that mathematicians would ?ht what
theyshould be celebrating.But focusing on me is a mistake.If
you really wish to consider human achievement as some
personalissue, where you wish to make me that powerful as a
single human beingthat the accomplishments of thousands of
years of effort can bereduced to just being about the
individuals involved, then you shatterthe foundations of
human civilization, and make it just some fashionshow.The
truth is that the discoveries are more important than
thediscoverers, and in letting it turn into some celebrity
issue, youdemean human society and its accomplishments.For
some of you I fear that the idea that I might become a
celebrityis all that matters to you. And I fear you think
that.a6s all that evermattered through all of human history, as
if the truth were just aword, and life has always just been a
big popularity contest.The truth is not just a word and
emphasizing truth over social issueshas helped humanity to
make the technological achievments which makemy communication
through this medium possible.Where would we be if engineers
had paused to ?ure out who they likedand disliked before
they bothered to use the information
discovered?Mathematicians aren.a6t doing you any favors. I
suggest that all of youfocus less on me than on the
information, and forget about whatbene?s telling the truth
will give me, as you consider that thevalue to society is
far, far greater.So how can mathematicians behave as they
have? That.a6s an important question, and I suggest we ?d out
why popularityhas so much importance in their world.James
Harris
===
> Repeatedly I.a6ve given out a de?ition for
counting prime numbers, but> I.a6m afraid that something has
not been communicated. So ?st I.a6m> going to give an example,
counting the number of prime numbers up to> 100, and
afterwards I.a6ll give the function. In that way I hope to>
point clearly to a key feature of my research.> What my
function does is count the number of composites for each>
integer up to the square root of your base number. So for
counting> primes up to 100 the base number is 100 and its
square root is 10.> Here are the composite counts:>
dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) =
6; dS(100,6) = 0; dS(100,7) = 3; > dS(100,8) = 0; dS(100,9) =
0; dS(100,10)= 0;> and summing the dS values gives you 74.
Notice when dS equals 0.> Now you add 1 for the number 1,
as its not considered prime, and> subtract from 100 to get
25, which is the count of primes up to 100.> And those
prime numbers are> 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
37, 41, 43, 47, 53, 59, 61,> 67, 71, 73, 79, 83, 89, and 97.>
> Now a key feature of dS that I want to emphasize is that for
dS(x,y)> if y is not prime dS equals 0. I want to repeat that
for emphasis,> for dS(x,y) if y is not prime dS equals 0.>
Amazingly enough, mathematicians never discovered such a
function as> my dS throughout their entire history.<deleted>I
know that for some my claims of uniqueness for my function
thatcounts primes just by itself seems impossible, as you
probably havesome exalted view of mathematicians or
something.But do the research. They missed it.What
mathematicians found were approximations to what I have, and
it.a6seasy for me to show.Then they took off chasing faster and
faster counting algorithms,never having found the full
function, which eluded them for quite sometime.And in fact, I
can get to some of their work from mine, but they can.a6tget to
mine from theirs unless they exploit the same principle I
use.Various posters have claimed that they do, but here.a6s
whatmathematicians actually exploit.Given a positive integer
x, a positive integer y, divides into it x/ytimes.Yup, that.a6s
the big principle behind what they call Legendre.a6sMethod.For
instance, take 11/2 and drop the remainder which gives you
5,which correspondes to 2, 4, 6, 8 and 10, then subtract 1
for the 2 asits prime to get you 4. Now take 11/3 and drop
the remainder and youget 3, which corresponds to 3, 6 and 9.
Mathematicians would againsubtract 1 to handle 3, to get the
composite count including 6 and 9,so you have 2. But THEN
they.a6d go to 11/6 which gives 1 and*subtract* it to handle 6,
as it.a6s been counted *twice* from the 2count and the 3
count.That way you get the entire composite count using that
simpleprinciple.Then you subtract that count from your x, and
subtract 1 and you havethe number of primes. For instance
above you have 4 + 2 - 1 = 5 and11-5-1 = 5, and those primes
are 2, 3, 5, 7, and 11.But mathematicians found that method
ever harder to work out as youexpanded as you have to keep
adding and subtracting thesecombinations.Over a hundred years
they found various tricks to speed things up, andactually got
close to what I discovered but apparently never made theleap
to considering a pi(x,y) function versus a pi(x) function.So
what I.a6m facing now may be a massive ego tantrum
frommathematicians.James Harris
===
> Repeatedly I.a6ve given
out a de?ition for counting prime numbers, but> I.a6m afraid
that something has not been communicated. So ?st I.a6m> going
to give an example, counting the number of prime numbers up
to> 100, and afterwards I.a6ll give the function. In that way I
hope to> point clearly to a key feature of my research.>
What my function does is count the number of composites for
each> integer up to the square root of your base number. So
for counting> primes up to 100 the base number is 100 and its
square root is 10.> Here are the composite counts:>
dS(100,2) = 49; dS(100,3) = 16; dS(100,4) = 0; > dS(100,5) =
6; dS(100,6) = 0; dS(100,7) = 3; > dS(100,8) = 0; dS(100,9) =
0; dS(100,10)= 0;> and summing the dS values gives you 74.
Notice when dS equals 0.> Yup, on the composites.> If your
eyes glazed over, or you found yourself wishing to quickly
run> away from what looks very complicated then you might
understand how> some posters could so easily manipulate the
discussion away from the> truth.> The truth is that the dS
function does the amazing, as you see a> complete de?ition
above, where you don.a6t see any prime numbers> except 2 but
*somehow* as you noticed above dS(x,y) equals 0 when y is> a
prime number.No, it.a6s 0 when y is a composite.> You see,
the function *knows* when a number is prime, making it>
unique.> Now, can you explain *how* it knows when a number is
prime? Also, this is far from unique, there are functions
(often called isPrime(n)) that return true/false, 1/0
depending on whether the number is prime.-- Will
Twentyman
===
> There are always *more* divisions than
multiplications.> always? That assumes the very thing
you are trying to> prove.> I can run this on my grid
http://www.gridontap.com and let you know... I am> 99.999%
sure that there will be no counter examples that show there
are more> multiplications then divisions.Since every
multiplication step is immediately followed by a division,
isn.a6tit trivial that there aren.a6t more 
multiplications than
divisions (other thanif you happen to stop counting right
after a multiplication step in a sequencethat has alternated
multiplication and divisions)?-- Evidence Eliminator is
worthless: www.evidence-eliminator-sucks.com--Tim
Smith
===
>Ok, I am at 3,000,000 and I have not yet found a
counter example to There>are always *more* divions then
multiplications. I will keep it going...>and post updates on
the list?According to a note on this
page:http://ksl.stanford.edu/people/bmac/papers/peim.html,the
Collatz conjecture has been veri?d up to about 3
trillion.Wikipedia says 1.2 trillion.
http://www.wikipedia.org/wiki/ConjectureMathworld says that
as of 1999 the record is up to 3*2^53,which is in excess of
27 quadrillion in USA notation
(2.7*10^16).http://mathworld.wolfram.com/
CollatzProblem.htmlSo you have a little competition. -
Randy
===
>Man, sometimes I hate math... ;) I am a
programmer... (so be easy on me)>hehe.>This collatz thing is
making me bash my head off of my keyboard. Is this>problem
unsolveable like the NP problem? I mean, you would have to
check>every integer (and that is not possible as you
say)Right. So the truth has to be determined by proof, not by
simplychecking out integers.In?ite loops are possible. I was
just in another Collatz thread afew days ago, and somebody
pointed out that if you use 3x + 3 insteadof 3x + 1, you get
lots of loops, including a short one at 19.So it.a6s clearly
not true that all such expressions eventually take youto 1
for all integers. Then the question becomes whether
there.a6ssomething special about 3x + 1. - Randy
===
> In?ite
loops are possible. I was just in another Collatz thread a>
few days ago, and somebody pointed out that if you use 3x + 3
instead> of 3x + 1, you get lots of loops, including a short
one at 19.My memory was wrong. If you use 3x+3, you do get a
loopstarting at 19, but it.a6s a rather uninteresting one:12,
6, 3, 12. The whole sequence is:19, 60, 30, 15, 48, 24, 12,
6, 3, 12, ...3x+5 is the one that gives a somewhat more
interestingloop at 19:19, 62, 31, 98, 49, 152, 76, 38, 19,
... - Randy
===
> In?ite loops are possible. I was just in
another Collatz thread a> few days ago, and somebody pointed
out that if you use 3x + 3 instead> of 3x + 1, you get lots
of loops, including a short one at 19.>My memory was wrong.
If you use 3x+3, you do get a loop>starting at 19, but it.a6s a
rather uninteresting one:>12, 6, 3, 12. The whole sequence
is:>19, 60, 30, 15, 48, 24, 12, 6, 3, 12, ...>3x+5 is the
one that gives a somewhat more interesting>loop at 19:>19,
62, 31, 98, 49, 152, 76, 38, 19, ...> - Randy3x+3 is not
very interesting because you only get multiples of 3 andthe
successors of 3n with the 3x+3 problem are 3 times the
successorsof n with the 3x+1 problem, because n is even iff
3n is even (3n)/2 = 3(n/2) 3(3n) + 3 = 3 (3n+1)-- Wim
Benthem
===
>Man, sometimes I hate math... ;) I am a
programmer... (so be easy on me)>hehe.>This collatz
thing is making me bash my head off of my keyboard. Is this>
>problem unsolveable like the NP problem? I mean, you would
have to check>every integer (and that is not possible as
you say)> Right. So the truth has to be determined by
proof, not by simply> checking out integers.> In?ite loops
are possible. I was just in another Collatz thread a> few days
ago, and somebody Me.>pointed out that if you use 3x + 3
insteadThat was 3x + 5.> of 3x + 1, you get lots of loops,
including a short one at 19.> So it.a6s clearly not true that
all such expressions eventually take you> to 1 for all
integers. Then the question becomes whether there.a6s>
something special about 3x + 1.There is absolutely something
special about 3x+1.Generalize the problem to 3x+C, where C is
an odd integer constant.For every possible sequence of
iterations of x/2 and 3x+C, a Hailstonefunction can be
written in terms of the last number in the sequence:f_e d_c b
af = (8*a - 5*C)/9Every Hailstone function can be expressed
as(X*a - Z*C)/Ywhere X is always a power of 2, Y is always a
power of 3, and Z is amixture based on how the sequence
zig-zags. z y x w v u_t s r q p o n m_l k_j i h g f_e d c_b
athe Crossover Point works out to
be(35343985*C)/33552245which factors
to(5*23*307339*C)/(5*6710449)which reduces
to(7068797*C)/6710449So the system 3x + 6710449 has a loop at
7068797, which is easilyveri?d7068797 27916840 13958420
6979210 3489605 17179264 8589632 4294816 2147408 1073704
536852 268426 134213 7113088 3556544 1778272 889136 444568
222284 111142 55571 6877162 3438581 17026192 8513096 4256548
2128274 1064137 9902860 4951430 2475715 14137594
7068797What.a6s special about 3x + 1 is that with C=1, the
Crossover functionisZ/(X-Y)which means the only way to get an
integer is if the factors of Zcancel all the factors of (X-Y).
And apparently, that is a dauntingtask. There are many
candidate sequences (where Z > (X-Y)), but?ding one with
even one common factor, let alone all, is dif?ult.Very often
X-Y is prime. A survey of all sequences of length 4 anddepth 8
ended up with 28 prime and 41 composite X-Y values. And
oftenthe composites have large prime factors.So it.a6s possible
that if you looked at all sequences up to a millioniterations,
you might ?d one where the factors cancel when C=1. ButI
can.a6t prove it one way or the other.> - Randy
===
> Man,
sometimes I hate math... ;) I am a programmer... (so be easy
on me)> hehe.> This collatz thing is making me bash my head
off of my keyboard. Is this> problem unsolveable like the NP
problem? I mean, you would have to check> every integer (and
that is not possible as you say)... can it be said that>
there are more divisions than multiplications, thus it always
ends up at 1?Nobody knows if it is solvable or unsolvable.
You might ?d a start value x other than 1, 2 or 4 that
eventually returns to x again. That would solve it. You might
?d a start value x that seems to go off to larger and larger
values. That would indicate the conjecture is false. However,
you would have to prove that these values get larger and
larger, which might be dif?ult. You might ?d a proof that
every value has to return back to 1. It took a few hundred
years for Fourier.a6s Last Theorem, so there is still hope. I
haven.a6t got the slightest idea how to start such a proof, so
I don.a6t think ?ding a proof will be easy. (There might be a
simple proof that everyone has overlooked. )
===
...> You
might ?d a proof that every value has to return back to 1.
It took> a few hundred years for Fourier.a6s Last Theorem, so
there is still hope.> I haven.a6t got the slightest idea how to
start such a proof, so I don.a6t> think ?ding a proof will be
easy. (There might be a simple proof that> everyone has
overlooked. )And what was Fourier.a6s Last Theorem about, that
made it dif?ult?-jiw
===
> There are always *more*
divisions than multiplications.> always? That assumes
the very thing you are trying to> prove.> I can run
this on my grid http://www.gridontap.com and let you know...
I am> 99.999% sure that there will be no counter examples
that show there are more> multiplications then divisions.Do
you realize that if you have more multiplications than
divisions itmeans that somewhere in the sequence you have at
least two consecutivemutiplications?And do you further
realize that you cannot have consecutivemultiplications
because every iteration of 3x+1 returns an evennumber?The
best you could possibly achieve is parity between
multiplicationsand divisons. But you can.a6t even do that
because the divisions willwin when you reach 16.
===
>
There are always *more* divisions than multiplications.>
always? That assumes the very thing you are trying to>
prove.> I can run this on my grid http://www.gridontap.com
and> let you know... I am 99.999% sure that there will be>
no counter examples that show there are more>
multiplications then divisions.> I.a6m 100% sure you 
won.a6t
check all the integers.True statement, but I don.a6t think the
idea that you don.a6tprove things in math by testing a subset
of the possibilitiesand calling that a proof has soaked
into his mind yet.To save him a little time, I checked the
?st six billion orso integers (got to love 2GHz processors)
using the code
here:http://www.well.com/user/xanthian/java/Hailstone.javanot
with any misapprehension that I was going to do a proof
byexhausting all possibilities, but just to look at the
extremecases to try to ?d a pattern from which a proof might
be begun(to no avail, but it was fun trying, and perhaps half
a dozenfolks in the comp.lang.java.* groups got involved in a
bit of acompetition, which my code very much did not
win).xanthian.BTW, _this_ is an interesting thing to
encounter:http://books.lulu.com/content/9828Anybody got
opinions on whether that.a6s legit, orknow if it.a6s been 
peer
reviewed and con?med?It was this page that convinced me the
problem hadlong outstripped my math skills and yet
remainedunsolved by those much smarter than me, and waslikely
to remain so for considerably
longer:http://www.cecm.sfu.ca/organics/papers/lagarias/--
===
>There are always *more* divisions than
multiplications.>always? That assumes the very thing
you are trying to>prove.>I can run this on my grid
http://www.gridontap.com and>let you know... I am 99.999%
sure that there will be>no counter examples that show there
are more>multiplications then divisions.>I.a6m 100% sure
you won.a6t check all the integers.> True statement, but I
don.a6t think the idea that you don.a6t> prove things in 
math by
testing a subset of the possibilities> and calling that a
proof has soaked into his mind yet.> To save him a little
time, I checked the ?st six billion or> so integers (got to
love 2GHz processors) using the code here:>
http://www.well.com/user/xanthian/java/Hailstone.java> not
with any misapprehension that I was going to do a proof by>
exhausting all possibilities, but just to look at the
extreme> cases to try to ?d a pattern from which a proof
might be begun> (to no avail, but it was fun trying, and
perhaps half a dozen> folks in the comp.lang.java.* groups
got involved in a bit of a> competition, which my code very
much did not win).> xanthian.> BTW, _this_ is an
interesting thing to encounter:>
http://books.lulu.com/content/9828> Anybody got opinions on
whether that.a6s legit, or> know if it.a6s been peer 
reviewed and
con?med?Couldn.a6t tell you, but he.a6s also claiming to 
have
proved the Goldbach Conjecture in a seperate book. I ?d it
interesting that neither is available at amazon.com.> It
was this page that convinced me the problem had> long
outstripped my math skills and yet remained> unsolved by
those much smarter than me, and was> likely to remain so for
considerably longer:>
http://www.cecm.sfu.ca/organics/papers/lagarias/> -- Will
Twentyman
===
>There are always
*more* divisions than multiplications.>always?
That assumes the very thing you are trying to>prove.>
>I can run this on my grid http://www.gridontap.com and>
>let you know... I am 99.999% sure that there will be>no
counter examples that show there are more>multiplications
then divisions.>I.a6m 100% sure you won.a6t 
check all
the integers.> True statement, but I don.a6t think
the idea that you don.a6t> prove things in math by testing a
subset of the possibilities> and calling that a proof has
soaked into his mind yet.> To save him a little time, I
checked the ?st six billion or> so integers (got to love
2GHz processors) using the code here:>
http://www.well.com/user/xanthian/java/Hailstone.java>
not with any misapprehension that I was going to do a proof
by> exhausting all possibilities, but just to look at the
extreme> cases to try to ?d a pattern from which a proof
might be begun> (to no avail, but it was fun trying, and
perhaps half a dozen> folks in the comp.lang.java.* groups
got involved in a bit of a> competition, which my code very
much did not win).> xanthian.> BTW, _this_ is an
interesting thing to encounter:>
http://books.lulu.com/content/9828> Anybody got
opinions on whether that.a6s legit, or> know if it.a6s 
been
peer reviewed and con?med?> Couldn.a6t tell you, but 
he.a6s
also claiming to have proved the Goldbach > Conjecture in a
seperate book. I ?d it interesting that neither is >
available at amazon.com.> I just read the preview and I
assume from that blurb he has shownthat any integer will get
smaller and that a relationship exists withthree digit ( base
10 I assume ) numbers. Or maybe it has something to do with
his brother or sister. I.a6m notsure.Ernst> It was this
page that convinced me the problem had> long outstripped my
math skills and yet remained> unsolved by those much smarter
than me, and was> likely to remain so for considerably
longer:>
http://www.cecm.sfu.ca/organics/papers/lagarias/>
>
===
>I.a6m 100% sure you won.a6t check all the integers.> 
>
Is this a conjecture or can you prove it ? ;-))> J-L.>
There are countably in?itely many integers.There is only
?ite matter to use for building computers for the
computations.Therefor, in ?ite time with ?ite resources he
can only check ?itely many integers he can check.QEDHmm...
wait a sec, what if the universe doesn.a6t expire? I.a6ll 
have to
rethink my proof.-- Will Twentyman
===
In sci.math, Will
Twentyman<wtwentyman@read.my.sig><3f4ab69c$1_3@newsfeed>:>
I.a6m 100% sure you won.a6t check all the integers.> 
> Is
this a conjecture or can you prove it ? ;-))> J-L.>
> There are countably in?itely many integers.> There is only
?ite matter to use for building computers for the >
computations.> Therefor, in ?ite time with ?ite resources
he can only check > ?itely many integers he can check.>
QED> Hmm... wait a sec, what if the universe doesn.a6t
expire? I.a6ll have to > rethink my proof.> Don.a6t worry. 
Those
computers also consume energy and the energydensity of any
given section of the Universe is ?ite. :-)-- #191,
ewill3@earthlink.netIt.a6s still legal to go .sigless.
===
> In
sci.math, Will Twentyman> <wtwentyman@read.my.sig>
<3f4ab69c$1_3@newsfeed>:>I.a6m 100% sure you 
won.a6t check
all the integers.>Is this a conjecture or can you
prove it ? ;-))>J-L.>There are countably
in?itely many integers.>There is only ?ite matter to use
for building computers for the >computations.>Therefor,
in ?ite time with ?ite resources he can only check
>?itely many integers he can check.>QED>Hmm... wait a
sec, what if the universe doesn.a6t expire? I.a6ll have 
to
>rethink my proof.> Don.a6t worry. Those computers also
consume energy and the energy> density of any given section
of the Universe is ?ite. :-)> Whew! I was worried that I
might have to start invoking Newton.a6s laws or something. Or
prove that the universe will collapse.-- Will Twentyman
===
>
> [snip]>I don.a6t think humans made it to the top of the
food chain by>out-nice-ing the other animals - I think
we out-predated them,>and we bear the mark of that
evolutionary history today. > No. We got to where we
are _precisely_ by out-nice-ing other animals:> the
cooperation in a modern human society dwarfs that in a
beehive.>This is true, but if you factor out generic
behavioral complexity>and look at a ratio of
cooperative/altruistic behavior to sel?h>behavior, I
would think that human behavior is far less cooperative>
>than that of bees.> I doubt that very much. Factor out
sexual and sex-moderated> competition, which is not possible
for worker bees, and how muchRight--it.a6s not possible. So in
that sense, any comparison betweenhumans and bees is apples
to oranges.Given the centrol role sex plays in biology, it.a6s
dif?ult to factorout sex. Everything---every biological
feature exhibiting complexdesign, that is---stems from
differential reproductive advantage, andhence is tied to sex.
Including reciprocal altruism.> uncooperative human behavior
is left? Even when we compete, we> cooperate (e.g., team
sports, business competition, war).>Put another way, I
would venture that most ethnologists would consider>bees to
be a more cooperative species (well, group of species...)
than>humans.> IMO this just means human behavior is far
more complex than bee> behavior.Right, which is why it.a6s
apples to oranges---which species is morecooperative depends
on the de?ition and metric used. > -- Roy L
===
>
[snip]>I don.a6t think humans made it to the top of the
food chain by>out-nice-ing the other animals - I think
we out-predated them,>and we bear the mark of that
evolutionary history today. > No. We got to where
we are _precisely_ by out-nice-ing other animals:> the
cooperation in a modern human society dwarfs that in a
beehive.>This is true, but if you factor out generic
behavioral complexity>and look at a ratio of
cooperative/altruistic behavior to sel?h>behavior, I
would think that human behavior is far less cooperative>
>than that of bees.>I doubt that very much. Factor out
sexual and sex-moderated>competition, which is not possible
for worker bees, and how much>uncooperative human behavior
is left? Even when we compete, we>cooperate (e.g., team
sports, business competition, war).>Put another way, I
would venture that most ethnologists would consider>bees
to be a more cooperative species (well, group of species...)
than>humans.>IMO this just means human behavior is
far more complex than bee>behavior.> Humans are the most
co-operative species by far, which is an essential> part of
human behavior being more complex.It depends how you de?e
cooperative.> Pack in millions of total strangers (not
sisters) of any other speciesnot sisters---that.a6s a
reasonable constraint, but note that thediscussion is then
constrained to e.g. non-kin reciprocal altruism(not that
there.a6s anything wrong with that).> at a density equivalent
of the NYC or Tokyo subways, and see how well> they cooperate
in each getting to their different destinations.> Or to
borrow an example of Milton Friedman.a6s: can one imagine>
thousands of members of any other species, all total
strangers to one> another, located all around the world and
not even knowing that each> other exist as individuals,
cooperating to manufacture a pencil?But this example isn.a6t
species-typical---modern industrialcivilization is a ?iolo
gically speaking. Though it.a6s of coursea good example of
cooperation. (Of course, even the cooperativeefforts within a
small tribe are pretty complex.)While humans are one of the
few advanced species marked by altruismdirected at non-kin,
we can also be pretty nasty---as seen by the factthat murder
(even murder of kin) appears to be species-typical.This to me
is what makes social life interesting---the tension
betweensel?h and cooperative urges.
===
> 3+1>
1.. Pick any positive integer n.> 2.. If n is even divide
by 2 else add 1.> 3.. If n = 1, stop; else go back to step
2.> If 3x+1 is disproved, then 3+1 will be disproved at
the same time.> Your conjecture is true. Easy proof by
induction. Does that mean> you.a6ve proven the Collatz
conjecture? Cool. (hehe)> lol, I wish, you know, I bet the
answer is in the Beal Ciphers... hehe Interesting idea :) As a
average guy I spend way too much time on the Collatz
conjectureand also have worked on the Beal ciphers. I know
what you mean :)Good luck Ernst Gametheory how to play
StockMarket
===
I need something from either biology, or
chemistry or physics that says something thatif a entity
increases in a day time frame of above 5% or above8% that the
likelihood if it increasing in the near future above that
initialrise is remote.And I saw it in action just a few days
or a week ago when Intel had a daily rise of 8%but has since
then fallen back to previous prices.I need something tangible
in the world of science to anchor the idea that whenever
yousee a 8% rise or more in one day, that the Theory of
Probabilitywould say the likelihood of further rises is too
remote.Perhaps the maximum of a living organism is a 8%
increase per day. Perhapssomething in chemistry says that a
8% or more is the limit of a parameter. Perhapssomething in
physics says that 8% or more of a physical parameteris tops
to be expected because tomorrow it will have fallen
back.whole entire Universe is just one big atom where dotsof
the electron-dot-cloud are galaxies
<3F43D5C6.2C7909DE@dtgnet.com> <3F43D99C.E8303B33@dtgnet.com>
<3F446C78.8C784A85@dtgnet.com> <3F4654AA.9DE4FF4B@dtgnet.com>
<3F4CEAE9.CF6A3D6A@dtgnet.com>
===
In message
<3F4CEAE9.CF6A3D6A@dtgnet.com>, Archimedes Plutonium >I need
something from either biology, or chemistry or physics that
says >something that>if a entity increases in a day time
frame of above 5% or above>8% that the likelihood if it
increasing in the near future above that initial>rise is
remote.>And I saw it in action just a few days or a week ago
when Intel had a >daily rise of 8%>but has since then fallen
back to previous prices.>I need something tangible in the
world of science to anchor the idea >that whenever you>see a
8% rise or more in one day, that the Theory of
Probability>would say the likelihood of further rises is too
remote.>Perhaps the maximum of a living organism is a 8%
increase per day. Perhaps>something in chemistry says that a
8% or more is the limit of a >parameter. Perhaps>something in
physics says that 8% or more of a physical parameter>is tops
to be expected because tomorrow it will have fallen back.>But
what if...There is a hostile take-over situation and a stock
rises 10% and this is followed by a bidding competition from
multiple companies leading to lucrative gains for your good
self?-- Jeremy Boden
===
> Hypothetical situation:> I hire a
man to go and get grapes for me. We agree that I will pay> him
1 cent for each grape he brings me (irrespective of source),
with> my payment to be rounded to the nearest $1. After a
time, he brings> me 30 grapes. I quickly devour them and give
him zero payment.> Under common rounding rules, x.3 rounds
down (e.g. 1.3 becomes 1). > Hence my payment of zero (0.3
becomes 0). However, is there any> legitimate argument for a
payment of $1, instead of zero in my> hypothetical? Stated
another way, are there any reasons why positive> values ought
not be rounded to zero, when common rounding rules> suggest
they should?> Would the answer be any different if I were
the only source of income> available to the man and I
dictated the terms of employment, rather> than reaching arm.a6s
length agreement.> I appreciate any constructive responses.
I use a hypothetical to> spare you the boring details of my
real life application.Under that arrangement, what possible
incentive would the man have forbringing you less than 50
grapes? Assuming that grapes aren.a6tperishable in the short
run and that they have market value to others,the only
logical thing for him to do in the event that he picked
lessthan 50 would be to try to sell them to someone else
instead.
===
> Under that arrangement, what possible
incentive would the man have for> bringing you less than 50
grapes? Assuming that grapes aren.a6t> perishable in the short
run and that they have market value to others,> the only
logical thing for him to do in the event that he picked less>
than 50 would be to try to sell them to someone else instead.I
agree, but in my application, it.a6s not really a question
o?centive, but rather a question of if the grape provider
has anargument that his payment should be non-zero. The buyer
is the onlygame in town.
===
> Hypothetical situation:> I
hire a man to go and get grapes for me. We agree that I will
pay> him 1 cent for each grape he brings me (irrespective of
source), with> my payment to be rounded to the nearest $1.
After a time, he brings> me 30 grapes. I quickly devour them
and give him zero payment.> Under common rounding rules,
x.3 rounds down (e.g. 1.3 becomes 1). > Hence my payment of
zero (0.3 becomes 0). However, is there any> legitimate
argument for a payment of $1, instead of zero in my>
hypothetical? No; the contract sounds clear - you pay him for
(and therefore keeptrack of) ) each and every grape he brings
you, but you pay only evendollars. So you pay him nothing at
this time, and when later hebrings you (app.) 70 more grapes,
you pay him $1.The only reasons for rounding off the payment
being (1) he has areason to keep bringing grapes, at this
point, (2) it.a6s easier for youto pay, (3) if the contract
ends at some point, you don.a6t have tosettle up and ?ure the
exact amount left owing by either party.Now, if you.a6d claimed
that, because you had eaten the ?st 30 grapes,they no longer
counted and he had to start over from zero, you.a6d betrying to
cheat him. And if, had you done that and he.a6d objected, youhad
cancelled the contract at that point, you would have cheated
him -it would not be fair for you to on one hand cancel the
contract, andon the other insist that he was still bound by
it and so deservednothing for the grapes he had already
brought you.> Stated another way, are there any reasons why
positive> values ought not be rounded to zero, when common
rounding rules> suggest they should?> Would the answer be
any different if I were the only source of income> available
to the man and I dictated the terms of employment, rather>
than reaching arm.a6s length agreement.> I appreciate any
constructive responses. I use a hypothetical to> spare you
the boring details of my real life application.
===
my actual
application, I should state that this was a one-time
event(i.e. no carryovers of unpaid balances, etc) and the
terms arenon-negotiable. Under the hypothetical, I have
received something (30grapes), my counterparty has received
nothing ($0). While thisseems intuitively unfair, it may be
perfectly legitimate. I guess Iwas looking for some rule or
logic that suggests that rounding to zerois inappropriate for
payment in a commercial transaction, when thereis positive,
albeit small, contribution by the counterparty.This is
important in my application as, by ?t, a zero
paymenttriggers one unrelated, but very signi?ant, outcome,
while anon-zero, positive payment triggers a different
outcome.Another possible argument, might be that there is a
difference betweenzero dollars and $0, as suggested in an
earlier post. In thecontract controlling the unrelated
outcome, the application trigger isbased on zero payment in
the ?st contract, while the ?stcontract itself (e.g. payment
for the grapes), calls for paymentrounded to the nearest $1,
which in this case is $0.B Stephens> Hypothetical
situation:> I hire a man to go and get grapes for me.
We agree that I will pay> him 1 cent for each grape he
brings me (irrespective of source), with> my payment to be
rounded to the nearest $1. After a time, he brings> me 30
grapes. I quickly devour them and give him zero payment.>
> Under common rounding rules, x.3 rounds down (e.g. 1.3
becomes 1). > Hence my payment of zero (0.3 becomes 0).
However, is there any> legitimate argument for a payment of
$1, instead of zero in my> hypothetical? > No; the
contract sounds clear - you pay him for (and therefore keep>
track of) ) each and every grape he brings you, but you pay
only even> dollars. So you pay him nothing at this time, and
when later he> brings you (app.) 70 more grapes, you pay him
$1.> The only reasons for rounding off the payment being
(1) he has a> reason to keep bringing grapes, at this point,
(2) it.a6s easier for you> to pay, (3) if the contract ends at
some point, you don.a6t have to> settle up and ?ure the exact
amount left owing by either party.> Now, if you.a6d claimed
that, because you had eaten the ?st 30 grapes,> they no
longer counted and he had to start over from zero, you.a6d be>
trying to cheat him. And if, had you done that and he.a6d
objected, you> had cancelled the contract at that point, you
would have cheated him -> it would not be fair for you to on
one hand cancel the contract, and> on the other insist that
he was still bound by it and so deserved> nothing for the
grapes he had already brought you.> Stated another way,
are there any reasons why positive> values ought not be
rounded to zero, when common rounding rules> suggest they
should?> Would the answer be any different if I were
the only source of income> available to the man and I
dictated the terms of employment, rather> than reaching
arm.a6s length agreement.> I appreciate any constructive
responses. I use a hypothetical to> spare you the boring
details of my real life application.
===
> my actual
application, I should state that this was a one-time event>
(i.e. no carryovers of unpaid balances, etc) and the terms
are> non-negotiable. Under the hypothetical, I have received
something (30> grapes), my counterparty has received
nothing ($0). While this> seems intuitively unfair, it may
be perfectly legitimate. I guess I> was looking for some rule
or logic that suggests that rounding to zero> is inappropriate
for payment in a commercial transaction, when there> is
positive, albeit small, contribution by the counterparty.>
This is important in my application as, by ?t, a zero
payment> triggers one unrelated, but very signi?ant,
outcome, while a> non-zero, positive payment triggers a
different outcome.> Another possible argument, might be that
there is a difference between> zero dollars and $0, as
suggested in an earlier post. In the> contract controlling
the unrelated outcome, the application trigger is> based on
zero payment in the ?st contract, while the ?st> contract
itself (e.g. payment for the grapes), calls for payment>
rounded to the nearest $1, which in this case is $0.Are you
asking whether such an arrangement can be fair? Certainly it
can be customary. For example, you can make a one-time
contract with a performance clause. If the contractor
produces at least 50 grapes he gets $1, if he produces at
least 150 grapes he gets $2, etc. There could even be
penalties for nonperformance, if he produces less than 20
grapes he pays you. People do sign such agreements on
occasion. It.a6s generally considered to be legal and
enforceable. It.a6s generally considered proper to make sure
both parties understand the agreement ?st.If you have made
such an agreement and now are disputing what it means then
there.a6s a communication gap that should have been cleared up
before the contract was signed.
===
nearest $1, which in this
case is $0.> Are you asking whether such an arrangement can
be fair?Not really, I am simply asking if there is any way the
employee canlegitimately claim he is entitled to payment
greater than zero. I amstruggling to come up with anything
defendable. Certainly it > can be customary. For example, you
can make a one-time contract with a > performance clause. If
the contractor produces at least 50 grapes he > gets $1, if
he produces at least 150 grapes he gets $2, etc. There >
could even be penalties for nonperformance, if he produces
less than 20 > grapes he pays you. People do sign such
agreements on occasion. It.a6s > generally considered to be
legal and enforceable. It.a6s generally > considered proper to
make sure both parties understand the agreement ?st.> If
you have made such an agreement and now are disputing what it
means > then there.a6s a communication gap that should have been
cleared up before > the contract was signed.The agreement in
my actual case is a monopoly provider tariff, not astandard
arms-length agreement. Disputing what it means is
exactlywhat.a6s at stake, as the circumstances at hand, a tiny
amount thatrounds to 0, likely have not occurred before in
the application of thetariff. To state it in the context of
my grape provider, all othergrape providers before him
brought either none (zero) or thousands ofgrapes. Rounding
precision was not an issue.
===
Dear Maplesoft: I would like
to be able to recommend your software to the members of
thecoalitions I build and help build. These coalitions work
on issues of interestto me (any one of about 125 inventions
I am preparing for patent), or to thepublic. I facilitate
the purchase of Mathcad software for these coalitions
andassist the members in preparing documentation to justify
the academic licenseprice when appropriate. I also
facilitate the use of coalition member.a6s airmiles for travel
by their and other coalition.a6s members to
meetings,conferences, and symposia. I am an expert and
licensed Mathcad user and have asked MathSoft pay me
acommission when facilitated purchases are made. I would like
to be able torecommend Maple to my coalition members when
appropriate and would like you topay me a commission when
facilitated purchases are made. I intend to make this same
request to Wolfram, the makers of Mathematica. Onecoalition
member, and possibly others, is a licensed and expert
Mathematicauser. I am examining Scienti? Notebook right now,
as well. I.a6d like to share a little something I noticed on
Usenet. In Messaged-ID:
Xns93E1C99B8D4DAcam0@202.4.251.50----------------------------
begin quoted textyou can just download maple from kazaa. its
only 40MB. the program is great. iuse it for large
computations and numerical analysis
problems.---------------------------------------------end
quoted textwhich is a clear violation of the provisions of
DCMA. Read it carefully: I ambeing attributed with
recommending violation of your copyright. I respect,create,
and license intellectual property. Please let me know if you
are ableto prosecute this user for downloading your
copyrighted software from Kazaa, based on my tip.Until then,
I.a6d just like to hear what it takes to enroll with you to
receivecommisions on facilitated sales. That should be an
easy question to answer.Maybe I just have to ?l out a
form.transmission to you at bizdev@maplesoft.com.Yours,Doug
GonczReplikon ResearchPO Box 4394 (preferred)6187 Greenwood
Dr. Apt. 102Seven Corners, VA 22044-0394 (or
2514)703-536-2367 voice, message703-536-5469 faxYours,Doug
GonczReplikon Research (via aol.com)Nuclear weapons are just
Pu.a6s way of ensuring that plenty of Pu will beavailable for
The Next Big Experiment, outlined in a post
tosci.physics.research at Google Groups under
supercritical
===
Dear Dr. Sarfatti,only some of what I can
understand.own fashion. You say dark energy is an exotic
vacuum phase ofnegative quantum zero point ?tions, and
dark matter is thepositive quantum /zpf. But dark energy
is a positive energy formleft over from the in? of
space. The dark energy ?an accounting gimmic to explain
the negative energy (work) requiredto in?pace (and have
the universe remain ?In fact,I don.a6t see any reason to
suspect that dark energy is necessarilyany different than
regular matter.No you do not understand the basic physics
here and have not properly described above what I am talking
about.Part of the basic physics is on pages 25-26 of John
Peacock.a6s Cosmological Physics showing why
covariancerequires w = -1 for zero point ?tions of all
quantum ?lds.Dark energy has / zpf > 0, i.e. positive zero
point energy density with equal and opposite negative zero
point pressure.What matters in Einstein.a6s GR is energy
density + 3pressure in the Poisson equation for the system,
in this case, the exotic quantum vacuum.Dark matter is simply
/zpf < 0, i.e. negative zero point energy density with equal
and opposite positive zero point pressure.This explains the
dark matter not only in large scale but also in micro-scale
i.e.1. Why self-charge does not cause electron to explode,
i.e. the old Abraham-Becker self-energy problem.scattering
yet e/mc^2 ~ 1 fermi and h/mc ~ 137e^2/mc^2e^2|/zpf|^1/2
where microscale dark matter cores of lepto-quarks have
/zpf < 0 and |/zpf|~ (1 fermi)^-24. This same micro
/zpf gives the correct universal Regge slope of hadronic
resonances alpha.a6 ~ (1Gev)^-25. The universal micro dark
matter exotic vacuum core of the lepto-quarks is|/zpf| ~
Lp^-4/3(c/Ho)^-2/3 from Susskind.a6s world hologram
model.Lp^2 = hG/c^3For more details go to
http://qedcorp.com/APS/Ukraine.dochttp://qedcorp.com/APS/
Vigier4.pdfMy theory predicts null results for all
conventional dark matter I explain both large scale and small
scale structure of universe with one simple mainstream idea
from P.W. Anderson More is different (ODLRO) - Occam.a6s
Razor.If I haven.a6t offended your scienti? sensibilities yet,
I wouldspace. Does it exist? What are its properties? Is there
any roomfor Mach.a6s principle in this picture? You will recall
that Machclaimed that inertial frames are determined by the
disposition ofmatter in the surrounding universe. Does your
concept of space allowfor the succint derivation of the
fundamental equation, F=ma?Yes, of course. Read my papers
above. :-)F = ma is basically timelike geodesic equation for
neutral test ?lds.I derive Einstein.a6s GR from the modulation
of the local Goldstone phase ?ld of the broken symmetry
macro-quantum vacuum, hence the geodesic equation is an
automatic result.Global special relativity and quantum ?ld
theory is already built into the unstable Minkowski
space.Basic force laws come from the quantum action principle
is the standard way.In other words F = ma is built into the
Feynman path quantum action Mach.a6s principle is inherent in
the Feynman path-Hoyle-Narlikar in? functional
response of the universe approach to cosmology from the
quantum action principle. That is consistent with my theory.
Aharonov.a6s advanced destiny quantum state vector is
automatically built in with the Costa De Beauregard/John
CramerFeynman Zig Zag explanation of micro-quantum EPR
nonlocal correlations violating Bell.a6s locality
inequality.
===
>A question I sometimes ask...>Give an
example of an object, de?ed in the language of ZF,>such
that (1) it is provable in ZF that it is a countably in?ite>
>set of pairs, but that (2) there is no obvious choice
function.>When you omit countably in?ite I can do it:
the set of>all pairs of subsets of R. But of course that is
not countably>in?ite.> Consider the Galois ?lds F_n =
G(p^2^(n-1)), n = 1, ... , p odd.> There are two isomorphic
mappings from F_n to F_{n-1}; how can one> choose which
one?Can we be more speci? here? Like this? ... -> G(p^16) ->
GF(p^8) -> G(p^4) -> G(p^2) -> G(p)What sort of maps are we
talking about? Are the ?lds G(p^2^(n-1))presented to us
explicitly in some way? Or do we (before we even
start)arbitrarily choose one representative of each
isomorphism class?-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
> I realize
that to you my claims sound miraculous and that my answers>
seem incomprehensibly mysterious (like hocus-pocus), but try
to> question your programming and understand what I.a6m
saying. SR boasts> that there is no absolute time order,
no absolute frame of> reference, and that motion faster
than light results in some> observer interpreting the
motion as time-travel into the past. > Never into the
past... ever... The arrow of time only> points
forward.Paul,I.a6m asking you to set aside your own opinions of
the world and selectone of two SR type models of the universe.
Who do you think inventedthe time-travel industry? I.a6ll give
you a hint. It wasn.a6t Rod Serling,Gene Roddenberry or any
travel agency. It comes from
SR.http://www.everythingimportant.org/viewtopic.php?t=605http
://www.everythingimportant.org/relativity/
simultaneity.htmEugene Shubert
===
> I realize that to
you my claims sound miraculous and that my answers> seem
incomprehensibly mysterious (like hocus-pocus), but try to>
> question your programming and understand what I.a6m saying. SR
boasts> that there is no absolute time order, no
absolute frame of> reference, and that motion faster
than light results in some> observer interpreting the
motion as time-travel into the past.> Never into the
past... ever... The arrow of time only> points forward.>
Paul,> I.a6m asking you to set aside your own opinions of the
world and select> one of two SR type models of the universe.
Who do you think invented> the time-travel industry? I.a6ll
give you a hint. It wasn.a6t Rod Serling,> Gene Roddenberry or
any travel agency. It comes from SR.>
http://www.everythingimportant.org/viewtopic.php?t=605>
http://www.everythingimportant.org/relativity/
simultaneity.htm> Eugene ShubertSince you asked...I cannot
set aside my own opinions just as you cannot set asideyours.
Any view I have is based on my world view biasjust as yours
is ... My understanding of the observed universeis different
from yours and yours is different from SmFarts (
welleveryone.a6s is different from smFarts) ( Sorry I had
too)Given that both views consider only the effect aspect of
matterin motion in relation to other matter in motion there
are missingrules and assumption used in both that give me
more issues thanthe differences between your hypothesis and
the standard modelsof SR.I tend to think that the existing
rules, to the point of failure dueto scale, have worked
pretty good and I also tend to give thestandard models of
relativistic interplay. Yours has merit forthe effort and I
would never denigrate a works that hasintellectual content,,,
even if shown to be incorrect,,,,As for running the numbers I
have made it clear in the past thatmy interest are more
generalized and I suck at Math so I leavethat to others to
pick at and I will read the explanation of themathematical
expressions and then may have a better view...But so as to
not miss lead you if you have not read my lunacyin the
past... I.a6m an engineer, not a physicist and tend tolook for
ways to poke at the horned rimmed glasses andpocket protector
crowd ( with great respect) to see if maybe anew idea my pop
into homebodies head ...mine ,,, theirs... yours..which
ever....I will say that I do not think there are any paradox
in physics... itsmore of a philosophical issue and crops up
when the existingrules fail (almost always due to scale) .. I
also say the samefor several aspects of physics... In?ity is
an illusion that occurswhen physical models fail due to
scale, c is a variable thatvaries at the rate of separation
of all matter universally ( avariance so out of scale as to
be undetectable as other than aconstant on human scale)
...the UGC is a variable thatvaries at the rate of separation
of all matter universally ( avariance so out of scale as to be
undetectable as other than aobservable aspect of duality are
only EM waves sampledby a method and concept biased as to
lead to interpretation
<c45b45b3.0308260922.4f2dd3df@posting.google.com>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
===
 at 10:22 AM, perfectlyInnocent@as-if.com
(Perfectly Innocent) said:>The special theory of relativity
is a mathematical subject.The. SR is a [hysical model, and
the issue of its validity is Physics,not Mathematics. If you
don.a6t understand a basic point like that, thenit woiuld be a
waste of time to read your web site.>I have proposed a simple
mathematical riddle: Do logically>consistent, alternative, SR
type theories exist?You.a6re a century too late, and the answer
is irrelevant. The properquestions are: 1. Is an aether theory
with identical predictions easier to use? 2. Can an
alternative theory be extended to QFT without
addingcomplexities? 3. Can an alternative theory be extended
to gravitation withoutadding complexities?>The question has
been submitted in terms of axioms.The applicability of those
axioms to the real world, and their utilityor lack thereof,
are questions for the physicists, not themathematicians.>The
solicited proof for or against my proposed solutionWhat a
mathematician means by proof is very different from what
aphysicist means by it. For a physical theory, the proof
isimperical, and is a matter for physicists. A mathematician
can saywhether your predicitions follow from your axioms, but
can havenothing to say about whether they match the real
world.Let me give an analogy. There is a lot of Physics
involved in abaseball game. But were I to post a play-by-play
description of a gameis sci.physics, the reader would, quite
properly, chastise me for anoff topic post, and would not buy
my claim that it was Physics.-- Shmuel (Seymour J.) Metz,
SysProg and JOATReply to domain Patriot dot net user
shmuel+news to contact me. Do not replyto
spamtrap@library.lspace.org
===
> at 10:22 AM, Perfectly
Innocent said:>The special theory of relativity is a
mathematical subject.> SR is a physical model, and the issue
of its validity is Physics, not> Mathematics. If you don.a6t
understand a basic point like that, then> it would be a waste
of time to read your web site.SR is Minkowski geometry.
Geometry is mathematics. SR is also aphysical model but I.a6m
not raising the issue of its validity. It.a6sobvious that you
misunderstand my question, do logically
consistent,alternative, SR type theories exist? By exist I
mean: exist asmathematical models, not as real physical
realities in a paralleluniverse.>I have proposed a simple
mathematical riddle: Do logically>consistent, alternative,
SR type theories exist?> You.a6re a century too late, and the
answer is irrelevant. So some mathematical questions are
irrelevant? Please de?emathematical irrelevance.>The
question has been submitted in terms of axioms.> The
applicability of those axioms to the real world, and their
utility> or lack thereof, are questions for the physicists,
not the> mathematicians.Where in the world do you get the
idea that my view of those axiomshas to do with applicability
and utility?>The solicited proof for or against my proposed
solution > What a mathematician means by proof is very
different from what > a physicist means by it. For a physical
theory, the proof is> imperical, and is a matter for
physicists. I.a6m using the word proof in its strictest
mathematical sense. Thequestion I.a6ve raised is strictly
mathematical and has nothing to dowith practical measuring by
experimental physicists.> A mathematician can say> whether
your predicitions follow from your axioms, but can have>
nothing to say about whether they match the real
world.Exactly! And as a mathematician, I.a6m saying that all
the predictionsof my model are identical to the standard
model within our galaxy butdiffer on a cosmological scale.I.a6m
only asking that mathematicians con?m or refute my claim
interms of mathematics, which you just said they could
do.already answered that question. Physics is too dif?ult
for
thephysicists.http://www.everythingimportant.org/
viewtopic.php?t=605Eugene Shubert
===
> Exactly! And as a
mathematician, I.a6m saying that all the predictions> of my
model are identical to the standard model within our galaxy
but> differ on a cosmological scale.That.a6s physics, not
mathematics.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
===
Portfolio of PAF as of 27AUG0350 BCE 21.64
$1,082.0050 BLS 25.25 $1,262.50100 BMY 25.37 $2,537.0050 DT
14.57 $728.5051,000 Q 4.50 $229,500.0010,250 SBC 22.73
$232,982.50380 VZ 35.10 $13,338.0080 WYE 42.64
$3,411.20realestate land 3APR03 of 3 lots $19,000realestate
land 30JUL03 another lot $11,500art of science-lithographs &
porcelain JAN-JUN03 for $12,000Well, today I sold 950 shares
of BMY at 25.35 in order to buy 1050shares of SBC at 22.75
because of the Crossover.BMY.I expect in the near term future
that SBC will climb considerably higherthan BMY and I can
thence re-play the above all over again and laughingall the
wayto the whatever.The Stockmarket as an entity is like a
gigantic fruit orchard withthousands and millions of free
cherries to pick. Absolutely free becausethey are Crossover
entities. If I had bought SBC only back in 3 OCT2002, then
the PAF portfoliowould be 20,900 shares of SBC and only that.
Collecting the dividendswould be the only income. But with
Crossover, PAF has gainedconsiderably more in wealth. Even
though PAF ran into a dismalexperience with SGP.Expectation:
I expect in the next 3 months that SBC will be
considerablyhigher than BMY and that the fat dividend that
SBC pays will also bevery nice. Perhaps SBC will do another
one of those extra dividends of10 cents on top of its
regular dividend something that BMY is surelynot capable of
at this time.The VonNeumann Gametheory applied to the
Stockmarket is a picture ofCrossover applications where
someone continues to pick free cherries,freepeaches, free
whatever as they continue to apply Crossover
phenomenon.Whereas a person who buys a stock and sits on it
for years with theexpectancy of a gain plus collecting of
dividends. The Crossover expectsnot only the dividends but
also the future stock rise but the addeddimension of free
pickings of the Crossover itself.So a regular investor has
only 2 dimensional gain expectancy. TheCrossover player has 3
dimensional gain expectancy.Good Question: can the Crossover
technique have avoided me of the dogthat was SGP? I think
not. I think, though, that a Crossover player isbetter able
to detect dogs in a portfolio than nonCrossover players,
inthat the zest orvitality of a company when compared with
others is more easily observedand pronounced. But as to
whether the Crossover player can avoid dogscompletely is
skeptical. It may turn out that SGP is bought in the
nextweeks at a price of nice proportions, but it is also
likely that SGPwill fall to the price of Elan of$5 a share
and just ?r for 2 years in the bucket.whole entire
Universe is just one big atom where dotsof the
electron-dot-cloud are galaxies
===
I am having a problem
linking to the CLU library from the examples using
http://www.perwass.de/cbup/cludownload.htmlI had to make
changes like making class members public, it wasn.a6t
recognizing the calls marked as friend.At this point, the
library compiles ?e but linking has unresolved symbols like,
BladeList<double>, MultiV<double>, Blade<?in all those
functions that were declared friend.
===
typo-corrections &
additions 2nd draftDear Dr. Sarfatti,only some of what I can
understand.own fashion. You say dark energy is an exotic
vacuum phase ofnegative quantum zero point ?tions, and
dark matter is thepositive quantum /zpf. But dark energy
is a positive energy formleft over from the in? of
space. The dark energy ?an accounting gimmic to explain
the negative energy (work) requiredto in?pace (and have
the universe remain ?In fact,I don.a6t see any reason to
suspect that dark energy is necessarilyany different than
regular matter.No you do not understand the basic physics
here and have not properly described above what I am talking
about.Part of the basic physics is on pages 25-26 of John
Peacock.a6s Cosmological Physics showing why
covariancerequires w = -1 for zero point ?tions of all
quantum ?lds.Dark energy has / zpf > 0, i.e. positive zero
point energy density with equal and opposite negative zero
point pressure.What matters in Einstein.a6s GR is energy
density + 3pressure in the Poisson equation for the system,
in this case, the exotic quantum vacuum.Dark matter is simply
/zpf < 0, i.e. negative zero point energy density with equal
and opposite positive zero point pressure.This explains the
dark matter not only in large scale but also in micro-scale
i.e.1. Why self-charge does not cause electron to explode,
i.e. the old Abraham-Becker self-energy problem.scattering
yet e/mc^2 ~ 1 fermi and h/mc ~ 137e^2/mc^2e^2|/zpf|^1/2
where microscale dark matter cores of lepto-quarks have
/zpf < 0 and |/zpf|~ (1 fermi)^-24. This same micro
/zpf gives the correct universal Regge slope of hadronic
resonances alpha.a6 ~ (1Gev)^-25. The universal micro dark
matter exotic vacuum core of the lepto-quarks is|/zpf| ~
Lp^-4/3(c/Ho)^-2/3 from Susskind.a6s world hologram
model.Lp^2 = hG/c^3For more details go to
http://qedcorp.com/APS/Ukraine.dochttp://qedcorp.com/APS/
Vigier4.pdfMy theory predicts null results for all
conventional dark matter I explain both large scale and small
scale structure of universe with one simple mainstream idea
from P.W. Anderson More is different (ODLRO from
spontaneously self-organizing BCS pairing of virtual
lepto-quarks with virtual anti lepto-quarks inside the
physical vacuum creating the complex scalar in? ?ld.) -
Occam.a6s Razor.If I haven.a6t offended your scienti?
sensibilities yet, I wouldspace. Does it exist? What are its
properties? Is there any roomfor Mach.a6s principle in this
picture? You will recall that Machclaimed that inertial
frames are determined by the disposition ofmatter in the
surrounding universe. Does your concept of space allowfor the
succint derivation of the fundamental equation, F=ma?Yes, of
course. Read my papers above. :-)F = ma is basically timelike
geodesic equation for neutral test ?lds.I derive Einstein.a6s
GR from the modulation of the local Goldstone phase ?ld of
the broken symmetry macro-quantum vacuum, hence the geodesic
equation is an automatic result.Global special relativity and
quantum ?ld theory is already built into the unstable
Minkowski space.Basic force laws come from the quantum action
principle in the standard way.In other words F = ma is
built into the Feynman path quantum action Mach.a6s principle
is inherent in the Feynman path-Hoyle-Narlikar in?
functional response of the universe approach to cosmology
from the quantum action principle. That is consistent with my
theory. Aharonov.a6s advanced destiny quantum state vector is
automatically built in with the Costa De Beauregard/John
CramerFeynman Zig Zag explanation of micro-quantum EPR
nonlocal correlations violating Bell.a6s locality
inequality.I.a6m having a hard time understanding the use of
epsilon and delta in proofs. Are there any books that just
focus on this rather than going through it prettyfast and
moving on to the rest of analysis? I would appreciate any
help atthis point. I.a6m really frustrated.
===
>I.a6m having a
hard time understanding the use of epsilon and delta in
proofs. >Are there any books that just focus on this rather
than going through it pretty>fast and moving on to the rest
of analysis? I would appreciate any help at>this point. I.a6m
really frustrated. Let.a6s consider a simple proof. A function
f : R -> R is saidto be _continuous_ at a point x0 if given
any delta > 0,there exists an epsilon > 0 such that | f(x) -
f(x0) | < deltawhenever | x - x0 | < epsilon. Your job is to
prove that if f1 and f2 are continuous at x0, thenf1 + f2 is
also continuous at x0, where (f1 + f2)(x) is de?ed to be
f1(x) + f2(x). An outsider picks delta, subject only to delta
> 0.To distinguish this speci? number from the generic
rolesof delta and epsilon, I.a6ll call his input delta0.You,
the prover, are told the value of delta0 and need to supply
epsilon0.Your epsilon0, which may depend on delta0, must
ensure that |(f1 + f2)(x) - (f1 + f2)(x0) | < delta0 whenever
|x - x0| < epsilon0 First use the de?ition of f1 + f2 to
translate your requirement into |f1(x) + f2(x) - f1(x0) -
f2(x0) | < delta0 whenever |x - x0| < epsilon0 Variants of
the triangle inequality come up frequently inanalysis. Your
requirement will be satis?d if |f1(x) - f1(x0)| + |f2(x) -
f2(x0)| < delta0whenever |x - x0| < epsilon0. You have not
speci?d epsilon0 yet.One way to achieve your revised goal
requires both |f1(x) - f1(x0)| < delta0/2and |f2(x) - f2(x0)|
< delta0/2whenever |x - x0| < epsilon0. This looks simpler
than the earlier goal because youhave separated the roles of
f1 and of f2. The delta you.a6ve been handed (namely delta0)is
guaranteed to be positive; hence delta0/2 > 0 too. Somebody
else is guaranteeing that f1 and f2are continuous at x0 --
this time you can require him to supplyan epsilon > 0 for
each delta > 0 you send him. |f1(x) - f1(x0)| < delta0/2
whenever |x - x0| < epsilon1and an epsilon2 > 0 such that
|f2(x) - f2(x0)| < delta0/2 whenever |x - x0| < epsilon2.
Observe that you are allowed to submit thesame positive value
delta0/2 for both ?d me an epsilon queriesbut the returned
values of epsilon may be different.They have been given
different names, epsilon1 and epsilon2.Both are guaranteed to
be positive. Now you try to supply a suitable epsilon (namely
epsilon0) such that both | x - x0 | < epsilon1and | x - x0 |
< epsilon2are achieved whenever |x - x0| < epsilon0. You
choose epsilon0 = min(epsilon1, epsilon2),the smaller of the
two values returned by queries about f1, f2.Your epsilon0 is
guaranteed to be positive since bothepsilon1 and epsilon2 are
positive. The actual proof would be written up in backwards
order,with much of the material abbreviated (and usingdelta,
epsilon in place of delta0, epsilon0) Let delta > 0. Using
the continuity of f1 and f2 at x0, select epsilon1 > 0 and
epsilon2 > 0 such that |f1(x) - f1(x0)| < delta/2 whenever |x
- x0| < epsilon1 |f2(x) - f2(x0)| < delta/2 whenever |x - x0|
< epsilon2 . Let epsilon = min(epsilon1, epsilon2) > 0. If |x
- x0| < epsilon, then |x - x0| < epsilon1 and |x - x0| <
epsilon2. Hence | (f1 + f2)(x) - (f1 + f2)(x0)) | = | f1(x) +
f2(x) - f1(x0) - f2(x0) | <= |f1(x) - f1(x0)| + |f2(x) -
f2(x0)| < delta/2 + delta/2 = delta Most of the proof
consists of understanding the de?itionsof continuous and of
f1 + f2. The strategic parts were i) Deciding where and how
to employ the triangle inequality. ii) Replacing one
inequality with bound delta by two inequalities each with
bound delta/2. iii) Realizing that we could use epsilon =
min(epsilon1, epsilon2) in order to simultaneously satisfy
both inequalities in ii) after we had satis?d them
individually.These techniques will occur repeatedly in your
analysis course.Understand de?itions carefully. Do your
homework to practice thetechniques. If you master these
steps, you.a6ll master the course.-- Wanted: Experts at
choosing the best of 100+ applicants for a position.Register
as a California voter by September 22, and vote on October 7.
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
===
Katie88265 a dit :> I.a6m having
a hard time understanding the use of epsilon and delta in>
proofs. Are there any books that just focus on this rather
than going> through it pretty> fast and moving on to the rest
of analysis? I would appreciate any> help at> this point. I.a6m
really frustrated.Epsilons and deltas are easy to understand
and to visualize, provided you : - draw what you want to
prove - understand what depends on what (does this delta
depends on x or does it not ?) : if you try to understand
this (for instance with the notation delta with a x
subscripted, you know that delta depends on x) you will
understand the proofs easily.-- Alexandre CharitopoulosEm6 /
Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7
===
> I.a6m having a hard
time understanding the use of epsilon and delta in> proofs.
Are there any books that just focus on this rather than
going> through it pretty fast and moving on to the rest of
analysis? I would> appreciate any help at this point. I.a6m
really frustrated.Many Calculus texts have special sections,
often in the backin an appendix, where delta/epsilon proofs
for limits and suchare carefully delineated. (The darned
things are 1200 pagesthick now, so they ought to have
_something_ about rigorouslimits in them.)Bart
===
> I.a6m
having a hard time understanding the use of epsilon and delta
in proofs. > Are there any books that just focus on this
rather than going through it > pretty> fast and moving on to
the rest of analysis? I would appreciate any help at> this
point. I.a6m really frustrated.You could take a look at
Principles of Mathematical Analysis by Rudin. There are many
e/d proofs there, in the context of metric spaces.If you give
an example of a particular proof you.a6re having trouble with,
perhaps readers here could help.
===
I saw a poster mention
the Twin Primes Conjecture and it got me tothinking.What I
know from my prime counting function is that given X a
primenumber, it is required for X+2 to be a prime number if
pi(X/2,1) = pi((X+2)/2,1)and pi(X/3, 2) = pi((X+2)/3,2).For
instance, with 17 and 19, you have pi(17/2,1) = 4, and
pi(19/2, 1)=4,while pi(17/3, 2) = 3 and pi(19/3) = 3.That.a6s
because with such a short difference as 2 between X and
X+2,only two dS values can be different.So proving the Twin
Prime Conjecture is just a matter of proving thatas you go
out to in?ity you can always ?d an X, such that pi(X/2,1) =
pi((X+2)/2,1)and pi(X/3, 2) = pi((X+2)/3,2).Oh, you need my
prime counting function to evaluate, of course.It seems to me
that giving that away might speed things up, so one ofyou can
just go out, see if you can prove it, and post so then you
cantake some credit for proving the conjecture.If you decide
to be silly, I.a6ll come in later and prove it myself,
andprobably have yet another thing that mathematicians want
to ?ht meover.But if you ever wanted to be famous, here.a6s
your chance to put yourname up there, and get in the history
books.James Harris
===
I think you forget a very fundamental
effect in your works : you do not takein account the effects
of Einstein.a6s relativity.Let me explain what I mean : twin
primes are primes that are distant of 2;right ?But it is well
known from relativity that the fastest you go, the
moredistances shrink !So you must admit that as soon as you
are in motion, the distance betweenany 2 twinprimes is no
longer 2 but a bit less that 2, add thus this two primes
arenot twin any longer !!!So my question is : are you sure
that each time you try to make ademonstrationdealing with
twin primes your speed is 0 ?;-))J-L.
===
> I saw a poster
mention the Twin Primes Conjecture and it got me to>
thinking.> What I know from my prime counting function is
that given X a prime> number, it is required for X+2 to be a
prime number if> pi(X/2,1) = pi((X+2)/2,1)> and>
pi(X/3, 2) = pi((X+2)/3,2).> For instance, with 17 and 19,
you have > pi(17/2,1) = 4, and pi(19/2, 1)=4,According to
your de?ition elsewhere, pi(x,y)= ?) - S(x,y) - 1,
where S(x,1) = 0pi(17/2,1) = ?.5) -1 = 7pi(19/2,1) =
?.5) -1 = 8Have you changed the de?ition of pi(x,y) in
the past few hours?-- Will Twentyman
===
> I saw a poster
mention the Twin Primes Conjecture and it got me to>
thinking.> ...> If you decide to be silly, I.a6ll come in
later and prove it myself, andBut what if it.a6s false?--
G.C.
===
> I saw a poster mention the Twin Primes Conjecture
and it got me to> thinking.> What I know from my prime
counting function is that given X a prime> number, it is
required for X+2 to be a prime number if> pi(X/2,1) =
pi((X+2)/2,1)> and> pi(X/3, 2) = pi((X+2)/3,2).> For
instance, with 17 and 19, you have> pi(17/2,1) = 4, and
pi(19/2, 1)=4,> while pi(17/3, 2) = 3 and pi(19/3) = 3.>
That.a6s because with such a short difference as 2 between X
and X+2,> only two dS values can be different.> So proving
the Twin Prime Conjecture is just a matter of proving that>
as you go out to in?ity you can always ?d an X, such that>
pi(X/2,1) = pi((X+2)/2,1)> and> pi(X/3, 2) =
pi((X+2)/3,2).Seems like a piece of cake. Just plug in
?in?ity.a6 for X and solve.> Oh, you need my prime counting
function to evaluate, of course.Naturally. Otherwise there is
no way to perform these calculations.> It seems to me that
giving that away might speed things up, so one of> you can
just go out, see if you can prove it, and post so then you
can> take some credit for proving the conjecture.> If you
decide to be silly, I.a6ll come in later and prove it myself,
and> probably have yet another thing that mathematicians want
to ?ht me> over.Why wait? Go ahead and prove it yourself
now.> But if you ever wanted to be famous, here.a6s your chance
to put your> name up there, and get in the history books.>
James HarrisIsn.a6t that what you.a6ve been working 
towards? Why
stop with all yourprevious achievements. Add a new one to
your trophy case.--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
===
> working on it for a
year and I.a6m looking for an intelligent and> thoughtful
evaluation and feedback.> Evaluating the paper requires
knowledge of posets and Dilworth.a6s> Theorem, as this is the
key insight in the proof (In the paper, we> look at the
subset-sum problem and think of the set of all subset-sums>
as a poset). I welcome and take seriously any feedback.>
Craigyou gotta some funny ideas about modern rigor, buddy.
but there areprobably amatuers out there who would ?d that
proof apealing. goodluck.
===
> the norm topology, T_B is the
weak topology? Or is that backwards?>a Baire space with
respect to the weak topology.It isn.a6t. Let
{e_n}_{n=1}^in?ity be an orthonormal basis of yourThis is
open and dense in the weak topology. But the intersection of
the G_n is empty. Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
===
>
> programs. How do you de?e programming?>
Writing:> FAC(x,I)^LT(x,I)^~(exists
A)FAC(A,I)^LT(A,I)^LT(x,A)> Is programming, no
matter what system translates that input into a program.>
> Sam> 1. Do you consider Predicate Calculus wffs
to be programs?>(If it is> written in a logic textbook, it
is not a program; if it is passed to a> Prolog interpreter,
it is.)What would the Prolog interpreter do? Could you use
the above wff or~LT(x,2)^~(exists A)PPFAC(A,x) as an
example and describe what itwould do and produce?> 3. Do
you know of a better way to sepcify the largest proper
factor> of a given number?> Yes: the largest proper
factor of a given number.So if a system took that as input
and produced a computer program,then it would constitute a
program synthesis system? And has anyoneever accomplished
that, to your knowledge? Have they tried? Is it aworthwhile
problem?> 4. Do you know of a way to specify it that isn.a6t
programming?> Yes: the largest proper factor of a given
number.It would be pretty straightforward to translate
between that phraseand the predicate calculus wff. We need
only de?e the constructionof each term such as largest [
P(a) -> P(a)^~(exists A)P(A)^LT(a,A)] and factor [ (exists
A)MUL(A,a) ].How could the wff be programming and the English
equivalent not, whenthe semantics are the same - they only
difffer in the syntax used?> 6. How do you think that a
Mathematician would specify it?> The largest proper factor
of a given number.You don.a6t think there is a formal
mathematical expression for thatEnglish phrase?> 7. Did you
know that the state-of-the-art in Program Synthesis is to>
specify the program requirement as a Predicate Calculus wff?>
> Not surprising. I doubt it.a6s really state-of-the-art
though; I> expect people having doing that for decades.What
makes you think that? Have you ever seen or heard of a
systemthat succeeds at translating between predicate calculus
wffs thatrefer to the natural numbers (Number Theory) and
computer programs?> 8. Do you see a qualitative difference
between a Predicate Calculus> wff and a computer program?>
> Yes; a wff is an assertion, while a program is a
command.Agreed. That is my whole point.> 9. How about the
fact that a Predicate Calculus wff has no assignment,>
conditional execution, loops or the possibility of not
terminating?> Many programming languages don.a6t have any of
these things, except the> last. And what would it mean for a
wff to not terminate anyway?It would only have meaning if a
wff were somehow executed as a programis.> 10. Do you think
that Predicate Calculus wffs and computer programs> are at
the same level of abstraction?> No, wffs are at a higher
level of abstraction, because they need not> be associated
with any sort of computer.Exactly.> 12. Do you think that
computer programs have to be analyzed to> determine what
function they compute, and in general you cannot do> that?>
> No. The (partial) function determined by program x is
already> determinate, independently of any analysis.But is it
discernible in general?> 18. Do you know of any system other
than mine that determines programs> that compute a given
wff?> Yes; Prolog springs to mind immediately.How would
Prolog determine programs that compute wff~LT(x,2)^~(exists
A)PPFAC(A,x), that is, list the prime numbers? OrFAC(I,J)
determine if one number is a factor of another?[My
understanding of Prolog is that it is just a database that
youcreate and query, having no relationship to the study of
in?itemathematical objects such as the natural numbers that
are the basis ofNumber Theory.]Charlie VolkstorfCambridge,
MA
===
> In my case, after the programming language
is axiomatized, nobody> system (using the de?itions,
axioms and rules.)>The formalism is still programing. >
> I.a6ve been wondering whether anyone else would point this
out...Do you see a difference between programs and predicate
calculus wffs? Are they at the same level of abstraction? In
what sense is apredicate calculus wff a program?Is there
value in being able to map back and forth between a
computerprogram and a predicate calculus wff that de?es the
result ofrunning that program? Is this a problem that has
been activelypursued by researchers?>Just because you
disguise it in >axiomatic form does not change the
underlying realities. In the last >step all that you write
will be translated into machine language, if and >when it
actually runs.But what input is required to producre that
program?Isn.a6t that by de?ition of the problem of program
synthesis itself? You produce a program without programming.
Your output is a computerprogram. Yes, you can translate or
execute or print it if you wish. That.a6s what a program is
supposed to do.That is, your assertion applies to any program
synthesis system byde?ition of the fact that we have a system
that creates computerprograms.Is the notion of program
synthesis a valid one? Is there anyconceivable scenario (not
that it is necessarily possible) which wouldconstitute
program synthesis? Then what would the input be? Can
youexpress it in formal terms?>You either are kidding us,
or you believe in symbolic juju and magicI give detailed
algorithms in my papers as to how the programs areproduced,
including several examples from Number Theory, DatabaseQuery
Processing and the Theory of Computation.Charlie
VolkstorfCambridge, MA>Bob Kolker>Sam>
************************> David C. Ullrich
===
> Is there
value in being able to map back and forth between a computer>
program and a predicate calculus wff that de?es the result
of> running that program? Is this a problem that has been
actively> pursued by researchers?how is that question
relevant to the thread? AFAIK, no one has criticized
theusefulness of your proposed system yet; the main critic
is the fact thatdespite what you say, using PC wff to specify
tasks is programming.Sam-- [...] but the delight and pride of
Aule is in the deed of making, and in thething made, and
neither in possession nor in his own mastery; wherefore
hegives and hoards not, and is free from care, passing ever
on to some newwork. - J.R.R. Tolkien, Ainulindale
(Silmarillion)
===
> 1. Do you consider Predicate Calculus
wffs to be programs?> Some Predicate Calculus wffs are
programs.> 2. How do you de?e a program?> A program
is an machine-readable speci?ation of acceptable (computer)>
behavior.How would a computer behave given a Predicate
Calculus wff?> 3. Do you know of a better way to sepcify
the largest proper factor> of a given number?> No.>
4. Do you know of a way to specify it that isn.a6t
programming?> No.Then how would a program synthesis
system generate a program todetermine the largest proper
factor of a given number without beingmerely programming?>
10. Do you think that Predicate Calculus wffs and computer
programs> are at the same level of abstraction?>
Predicate Calculus.a6s heavy use of quanti?rs makes it more
abstract at the> language level than many of today.a6s
computer programming languages.That.a6s right - it is at a
higher level of abstraction. PC de?es aset. A program de?es
an algorithm that performs a particularoperation on a set.>
16. Do you see value in being able to determine computer
programs that> implement a given predicate calculus wff?>
Yes.> 17. Does my system determmine programs that compute
a given wff?> Not if those programs are required to halt
for all inputs.Why and who said it did?> 18. Do you know of
any system other than mine that determines programs> that
compute a given wff?> False premise.No premise
intended.Charlie VolkstorfCambridge, MA
===
>The
formalism is still programing.> I.a6ve been wondering
whether anyone else would point this out...> That.a6s what
I.a6ve been trying to explain from the beginning. PC is a>
programming language, though it does not have conditional
branching and other> common features. I have also noted
that, according to Charlie.a6s paper, it is> translated to APC
(which is similar to pseudo-code in its syntax). Therefore,>
the system is actually a translator from a high-level
non-procedural language> to a procedural one. That does not
contradict the fact that it is a program> generator (Charlie
does not seem to accept that idea, though), provided it>
actually works (since Charlie has not accpted to make his
implementation> available, even though he supposedly has used
it in replying to one of my> post)> And again, almost any
syntax that speci?s a task to be performed is a> programming
language. In fact, English can be one (check out the
Shakespeare> programming language)> SamThis is deja vue
all over again, so I will proceed with what I noticedawhile
ago. One of the strongest ways to attack something is to do
soat a very high, general level of abstraction. That is,
declare in themost general terms how it violates required
principles. There is noalternative than to throw out the
entire attempt.The problem with this approach is that it is
so broad that it doesn.a6tattack the proposed solution at all.
It attacks the problem itself. It paints a condemnation so
broad that the problem itself takes thecondemned form.In the
case with program synthesis, the attack might be that
thesystem for representing the program requirement is no
good. And theattack continues with as broad a paint brush as
possible - thuscondemning the notion of program sythesis
itself.There is input and a program is created. Nothing is
said about thesyntax or semantics of the input. But it is
still condemned as beingprogramming and thus is not program
synthesis. How could a programsynthesis system work without
having input to de?e the programmingrequirement?What would
be a better system than the Predcate Calculus? This iswhat
mathematicians have been using for 100 years and is the
ultimatecase of Occam.a6s Razor: we are using existing,
established mathematicsthat is here to stay regardless - so
if we use that, then it is forfree. We minimize the system by
not adding anything new.Is the concept of a program synthesis
system a valid one? If yes,then describe how it would work.
What would be a good input? If yourepeat your claim that the
LISP programming system is programsynthesis, I ask only: Do
you (plural) have to input new LISP programsas you use the
system? Are you not merely entering in a series ofprograms,
and ?ally a single line that calls these progams in acertain
order? Then you are merely typing in a program in LISP. Where
is the program that is supposed to be created and output by
thesystem? There is none. You input a program rather than
outputtingone.If a person could type in a short, free-form
description of what hewanted the new program to do, and out
popped one or more programs thatdid just that, would that
constitute program synthesis?But at the same time, you have
condemened program synthesis itself: any syntax that
speci?s a task to be performed is a programming language.
English can be one.To perform program synthesis, we must
have a syntax to specify thetask to be performed, no? But any
such syntax constitutesprogramming. So how could a program
synthesis system ever exist?The Predicate Calculus is the (or
at least a) standard mathematicalway to represent sets and
function. It is very general. It issimple. It has only a
handful of concepts: not, and, or, exists, forall,
paremtheses. It doesn.a6t require you to specify an algorithm
toperform any particular task. You can construct multiple
differentalgorithms for it. It has no assignment, if
commands, in?ite loops.These are some of the reasons that I
use the Predicate Calculus. Isthere a better way? Is anything
else even close? What is next best?PREDICATE CALCULUS VS
PROGRAMMING LANGUAGESThe predicate calculus is a system for
specifying predicates that areeither true or false. It is
declarative - nonprocedural. We statewhat the condition is,
not how to get it in any particular way.A program is a series
of functions that are executed until they run tocompletion or
get into an in?ite loop. The author must develop aseries of
commands that will be executed in order to actually
output(list) a set, or calculate a single value, or decide if
a givenpredicate is true or false.The Predicate Calculus
doesn.a6t require you to give a program thatcalculates the set
or function being de?ed.The Predicate Calculus is
declarative like English. How would youde?e what a prime
number is? Someone must tell the system what itmeans to be a
prime number. That is the lowest level of abstraction. What
is a prime number?I would de?e a prime number as a number
that is not less than 2 andwhich has no proper factors. I
would de?e a proper factor of N as anumber that is a factor
of N and is not equal to 1 or N. I wouldde?e a factor of N
as a number such that there is a number that whenmultiplied
by the factor equals N.How would you?And how do we express
this in the Predicate Calculus? Well, sinceboth English and
the Predicate Calculus are declarative, the semanticsare the
same. We have a predicate written in English that we wish
toexpress in the Predicate Calculus.a number that is not
less than 2 : ~LT(x,2)and : ^which has no : ~(exists
A)proper factors : PPFAC(A,x)Concatonate it verbatim and we
get:a number that is not less than 2 and which has no proper
factorsis:~LT(x,2)^~(exists A)PPFAC(A,x)The semantics are
the same. It is a matter of translating the syntaxof English
into the syntax of the Predicate Calculus. What could
beeasier or simpler?A lot of people have spent a lot of time
studying the relationshipbetween the Predicate Callculus and
computer programs. Some have eventried very hard to translate
between Predicate Calculus wffs andcomputer programs.How would
you represent the functionality provided by a
computerprogram?You said that a Predicate Calculus wff is a
program and a program issomething that a computer can
execute. So I asked you how a computerwould execute
~LT(x,2)^~(exists A)PPFAC(A,x) and you said you didn.a6tknow
my system.But this is not a question about my system per se.
I make no mentionof my system in the question. I am only
asking how this program canbe executed.The answer is, it
can.a6t be executed in the normal sense of the word. Yes, it
can be input into a system. Yes, there can be output fromthat
systen. And if you want to call that a translation, then
goahead, although a de?ition should relate something unknown
tosomething known.But rather than being a series of simple
(recursive) functions likeconstant zero, or add one, or
multiply two numbers as in an actualprogram that are executed
in series, a Predicate Calculus wff doesn.a6tsupply a series of
functions for us to execute. Instead, it is astatement - a
predicate - that is true when a given value is in a setwe are
de?ing. How can we execute the declaration that x is notless
than two and has no proper factors?A Predicate Calculus wff
de?es a set. A program is an algorithm tocarry some
particular process concerning a set, such as listing
itselements or deciding if a given value is an element. They
are atdifferent levels of abstraction.Programs de?e
(processes on) sets and functions that can be provento have
their particular values. Predicate Calculus wffs de?e
setsand functions in terms of wffs that are true when the set
or functionis stated to have a correct value.The distinction
is that between the provable wffs and the true wffs. (Godel
showed they are not the same.)It is much easier to declare
something by giving a statement that istrue or false
depending on the condition (that is called talking)than it
is to give a wff that is provable (aka programed) when
thecorrect values for that set or function are given. In some
cases, thelatter is impossible while the former is
possible.Charlie VolkstorfCambridge, MA
===
I.a6ve added
sci.math, I understand they know these sorts of things.>
> Afken says>
> div V = 1/sqrt(g) @(sqrt(g)V^k)/@q^k>
where g is the determinant of the metric. So for
spherical>coordinates> I.a6m getting
g=r*sqrt(sin(theta)) and> sqrt(g) = r *
sin(theta)>shouldn.a6t that be> sqrt(g) =
r^2 * sin(theta) ?>;-)> Should be. I had a
little bit of idiocy going back there, but I tried it>
that way and got the wrong answer, so I managed to convince
myself that I> had too many squares in the determinant,
and still got the wrong answer.>Hehe, that.a6s what
happens when you run around>in Minkowski-circles for too
long ;-)> So I started doing imaginary math. I guess I
should have expected that,> under the circumstances.>
Maybe someone can ?d an error if I go through this step by
step. All> derivatives are partial, so I.a6ll just use 
d.a6s
instead of trying to> symbolize partials with a-holes or
something.> Arfken says:> div V = sum_k 1/sqrt(g)
d(sqrt(g) V^k)/dq^k> Spherical polar coordinates has
sqrt(g)=r^2 sin(theta). Go through the> k.a6s one by one.>
k=r: 1/r^2 sin(theta) d/dr (r^2 sin(theta) V^r)> = 1/r^2
d/dr (r^2 V^r)> Beautiful.> k=theta: 1/r^2 sin(theta)
d/d(theta) (r^2 sin(theta) V^theta)> = 1/sin(theta)
d/d(theta) (sin(theta) V^theta)> Wrong. The derivative
should be multiplied by 1/r sin(theta). But there> was an
r^2 in the top and an r^2 in the bottom and no d/dr, so
they> canceled on me.> k=phi: 1/r^2 sin(theta) d/d(phi)
(r^2 sin(theta) V^phi)> = d/d(phi) V^phi> Wrong, the
derivative part should be multiplied by 1/r sin(theta). But>
the r^2 sin(theta) just passed right through the derivative
and canceled.> I.a6m doing something wrong here, but I
don.a6t know what.>You probably are expecting the physical
components V_(k)>in stead of the contravariant components V^k
of the vector.>They are related by> V_(k) = sqrt(g_kk) V^k (no
summation)>i.o.w.> V^r = V_(r)> V^t = 1/r V_(t)> V^f = 1/[r
sin(t)] V_(f)>When you ?l in these ones in the above
result, you get>what you are looking forYes, I.a6ve noticed
that, but I couldn.a6t ?l in the missing steps to prove it.
The books I.a6ve looked at have been kind of vague on that sort
of thing. Like when you have vector components multiplied by a
basis, do the indices of the components have any particular
meaning, or just the basis vectors? E.g. V^i e_i and V_i e^i?
Does V_i e_i make sense? It would be tempting to raise one of
them like g_ij V^j e_j, but that.a6s not sqrt(g_ij).I get the
same thing as above if I try to compute div V with the
Christoffel symbol method, div V = V^i_;iand I.a6ve checked my
Christoffel symbols again and again, so I thought I must be
missing some relation between what I get and what.a6s on the
inside cover of Jackson.>I have been wondering about this too
when I was working>through my Schaum.a6s tensors last year.>See
also example 1 of>
>http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/
Maths-Stuff/tensorcalc/part7.PDF-- A nice adaptation of
conditions will make almost any hypothesis agreewith the
phenomena. This will please the imagination but does not
advanceour knowledge. -- J. Black, 1803.
===
Dear Jack,I
have sent yesterday the abstract to Richard Amoroso
(attached)....I am glad to see, that the good correlation
between some of our ideas is existing.The interplay and
feedback [IT <
===
> BIT] is one of my Uni?d model feature
also, if under IT to understand a matter (including
biosystems) and under BIT its virtual replica (VR) in
Bivacuum, containing the information about IT. The [IT + BIT]
represents in general case complex macroscopic (star systems)
levels, able to self-organization.I do not understand what
you mean by VR and Bivacuum?In Bohm.a6s realism this is
very simple. BIT is simply the deBroglie qubit according tov
= (h/m)Grad(argPsi)Self-organization means the BIT Psi has as
a source its own IT hidden variable trajectory making the
fragile Bohm quantum potential robust.IT is a system point
rolling in gradient ? the BIT landscape.rigid BIT
landscape with a fragile quantum potential where BIT has no
This is IT FROM BIT without BIT FROM IT.Post-quantum theory
with signal nonlocality means that IT is no longer a This is
spontaneous self-organization pumped from environmental
stochastic inputs from BOTH gauge forces making dynamical
phases and also variable boundaries making Berry topological
phases from paths in parameter space.I like your idea, that
just a speci? (exotic) properties of vacuum (bivacuum) with
/ < 0 and / > 0 are responsible for attractive dark
matter and repulsive dark energy, correspondingly. However, I
did not ?d in your materials the clear explanation of
physical mechanisms, responsible for such properties of
vacuum and the ways of their regulation (vacuum engineering
;-) ).Details are in
http://qedcorp.com/APS/Ukraine.dochttp://qedcorp.com/APS/
Vigier4.pdfAs you can see from my Uni?d Model (UM), it
contains the mechanism of Bivacuum ?contraction.a6 due to
spin-spin interaction between introduced Bivacuum fermions
(BVF) of opposite spins and Bivacuum ?expansion.a6 due to Pauli
repulsion between BVF of parallel spins in the huge volumes of
3D domains of virtual BC, formed by BVF.I do not understand
that as yet. Why do you need to introduce new ideas when the
old ones seem to work nicely? More with less is always
better. I do not see the need to introduce any new
non-orthodox ideas to explain any of the fundamental data in
both the new precision cosmology and also easily explained
and understood needing only mainstream ideas:1. Einstein.a6s
general theory of relativity supplemented by Hagen Kleinert.a6s
world crystal lattice analogy and Wheeler.a6s geometrodynamics
of Mass without mass etc.2. Standard quantum ?ld theory in
globally ?acetime.3. P.W. Anderson.a6s More is different
general theory of emergent complexity in spontaneous broken
symmetry ground states.4. Bohm.a6s quantum realism
interpretation of Wheeler.a6s IT and BIT.5. Lenny
Susskind.a6s world hologram generalization of Jacob
Bekenstein.a6s black hole thermodynamics.6.
Feynman-Wheeler/Costa De
Beauregard/Hoyle-Narklikar/Cramer/Aharonov notions of
advanced in?s from the future.The other possible
explanation of attraction/repulsion between gravitation in
UM. It could be a result of interference of uncompensated
Bivacuum virtual pressure waves, radiated by two or more
interacting bodies, excited by [C - W] pulsation of their
elementary two pulsing in liquid bodies the attraction force
has (1/r^2) dependence and may change to repulsion when the
distance between bodies exceeds the wave-length of density
waves in liquid medium (i.e. super?ivacuum). For details
see Chapter 11 in my paper:
http://arxiv.org/abs/physics/0207027 (see full PDF
version).I don.a6t understand this.In my theory/zpf =
Lp*^-2[1 - Lp*^3|PSI|^2]The ordinary non-gravitating vacuum
has/zpf = 0This corresponds to the critical vacuum
condensate density<|PSI|^2> = Lp*^-3whereLp* =
Lp^2/3(c/Ho)^1/3 ~ 1 fermiAnti-gravitating zero point dark
energy density causing the large scale 3D space geometry to
accelerate in its expansion from the initial singularity
vortex core where PSI = 0 has /zpf > 0 for FRW Omega(Dark
Energy) ~ 0.73. This implies a low density super?cro
-quantum vacuum condensate density|PSI|^2 < Lp*^-3Lp* is
the equilibrium Holographic World Crystal Lattice spacing
(combining Kleinert with Susskind). Therefore the low BEC
density antigravitating dark energy phase of exotic vacuum
has dilated or expanded unit cells away from equilibrium.
Note these are 4D unit cells with generalized ?ite groups of
different kinds of broken symmetries.Similarly the gravitating
dark matter phase of exotic vacuum is high density vacuum
condensate with the unit cells compressed smaller than the
equilibrium value where FRW Omega(Dark Matter) ~ 0.23.Now if
you can connect your picture in yourIt is shown in
hydrodynamic Bjorkness theory, that between two pulsing in
liquid bodies the attraction force has (1/r^2) dependence and
may change to repulsion when the distance between bodies
exceeds the wave-length of density waves in liquid medium
(i.e. super?ivacuum)to what I just said, that would be
interesting. There seems to be an analogy or similarity if
the equilibrium wavelength of your density waves is my Lp*
and your bodies correspond to the vertices of the unit cell
in the world crystal lattice?
===
My previous post on Twin
Primes had a mistake. Figuring out thatmistake (it was just a
goof, but oh well) is crucial to getting apiece of the glory
of proving the Twin Primes Conjecture.Here.a6s the Twin Primes
Formula:Given X a prime, then iff ?/3) - ?/6) =
?X+2)/3) - ?X+2)/6)then X+2 is a prime as
well.
===
> My previous post on Twin Primes had a
mistake.Flashback to 8 years ago. Post #1: incorrect proof.
Post #2: correction.I think I know how it goes from
there.V.-- 
===
>My previous post on Twin Primes had a
mistake. Figuring out that>mistake (it was just a goof, but
oh well) is crucial to getting a>piece of the glory of
proving the Twin Primes Conjecture.>Here.a6s the Twin Primes
Formula:>Given X a prime, then iff> ?/3) - ?/6) =
?X+2)/3) - ?X+2)/6)>then X+2 is a prime as well.Let
X=47.?7/3) - ?7/6) = 15 - 7 = 8?9/3) -
?9/6) = 16 - 8 = 8Your Twin Primes Formula concludes
that 49 is prime.
===
> Given X a prime, then iff> ?/3)
- ?/6) = ?X+2)/3) - ?X+2)/6)> then X+2 is a
prime as well.Cool! I suggest calling 2, 3 and 6 Harris
Numbers.
===
> Given X a prime, then iff> ?/3) -
?/6) = ?X+2)/3) - ?X+2)/6)> then X+2 is a
prime as well.> Cool! I suggest calling 2, 3 and 6 Harris
Numbers.Alas, the formula proved to be wrong. But just
imagine math students startedlearning McLarren series asf(x)
= f(0) + f.a6(0)*x + f(0)*x^2/H2 + f[Paragraph](0)*x^3/H3 
+...where H1
and H3 are the ?st and the third Harris numbers.
===
>
>Given X a prime, then iff> ?/3) - ?/6) =
?X+2)/3) - ?X+2)/6)>then X+2 is a prime as
well.> Cool! I suggest calling 2, 3 and 6 Harris
Numbers.> Why stop there? By the above conjecture, all
multiples of 3 are prime!Proof:For n a whole number, consider
X=3n?/3) - ?/6) = ?) - ?/2)?x+2)/3) -
?x+2)/6) = ?+2/3) - ?/2+1/3)This gives us two
cases:Case 1: If n is even, there exists whole number m such
that n=2m?m)-?m/2) = 2m-m =
m?m+2/3)-?m/2+1/3)=2m-m=mCase 2: if n is odd, there
is a whole number m such that n=2m+1?m+1)-?2m+1)/2)
= (2m+1) - m = m+1?m+1+2/3) - ?2m+1)/2+1/3) =
(2m+1) - m = m+1So, if the Harris conjecture is true, then
for all whole numbers n, 3n is a prime number.Therefore,
since (by de?ition) 3n is NOT a prime for all n <> 1, the
Harris conjecture is false.Obvious Harrisian Conclusion: the
de?ition of prime is broken.-- Will Twentyman
===
>
Given X a prime, then iff> ?/3) - ?/6) =
?X+2)/3) - ?X+2)/6)> then X+2 is a prime as
well.> Cool! I suggest calling 2, 3 and 6 Harris
Numbers.> Why stop there? By the above conjecture, all
multiples of 3 are prime!> Proof:> For n a whole number,
consider X=3n> ?/3) - ?/6) = ?) - ?/2)>
?x+2)/3) - ?x+2)/6) = ?+2/3) - ?/2+1/3)>
This gives us two cases:> Case 1: If n is even, there exists
whole number m such that n=2m> ?m)-?m/2) = 2m-m =
m> ?m+2/3)-?m/2+1/3)=2m-m=m> Case 2: if n is odd,
there is a whole number m such that n=2m+1>
?m+1)-?2m+1)/2) = (2m+1) - m = m+1> ?m+1+2/3)
- ?2m+1)/2+1/3) = (2m+1) - m = m+1> So, if the Harris
conjecture is true, then for all whole numbers n, 3n > is a
prime number.> Therefore, since (by de?ition) 3n is NOT a
prime for all n <> 1, the > Harris conjecture is false.>
Obvious Harrisian Conclusion: the de?ition of prime is
broken.I think you are becoming cynical.Gib
===
>
Given X a prime, then iff> ?/3) - ?/6) =
?X+2)/3) - ?X+2)/6)> then X+2 is a prime as
well.> Cool! I suggest calling 2, 3 and 6
Harris Numbers.> Why stop there? By the above
conjecture, all multiples of 3 are prime!> Proof:> For
n a whole number, consider X=3n> ?/3) - ?/6) =
?) - ?/2)> ?x+2)/3) - ?x+2)/6) =
?+2/3) - ?/2+1/3)> This gives us two cases:>
Case 1: If n is even, there exists whole number m such that
n=2m> ?m)-?m/2) = 2m-m = m>
?m+2/3)-?m/2+1/3)=2m-m=m> Case 2: if n is odd,
there is a whole number m such that n=2m+1>
?m+1)-?2m+1)/2) = (2m+1) - m = m+1> ?m+1+2/3)
- ?2m+1)/2+1/3) = (2m+1) - m = m+1> So, if the Harris
conjecture is true, then for all whole numbers n, 3n > is a
prime number.> Therefore, since (by de?ition) 3n is NOT a
prime for all n <> 1, the > Harris conjecture is false.>
Obvious Harrisian Conclusion: the de?ition of prime is
broken.> I think you are becoming cynical.> Gib> Nope,
not cynical... realistic. I.a6ve seen James claim that
de?itions are broken far more often than I.a6ve seen him admit
to error. Of course, I won.a6t be entirely disappointed if he
admits to making a mistake. What would be more useful is if
he had posted his logic behind the claim. Then we might have
helped him re?e it to a correct equivalence.Ok, maybe a
little bit cynical, but it.a6s hard to tell when dealing with
James.-- Will Twentyman
===
Hash: SHA1Exactly what part of
math is the study of nomenclature notation likeGF(p^q), Z/pZ,
etc..I know they teach it through the course of usage in
number theory booksbut is there a speci? book [or subject]
that teaches how to read
thatnotation?TomiD8DBQE/TRhjsP+
tEsHHY0ARAj2pAJ9B7OIQgpB1WOGlICGnFroAsW6rOACfZd5TMiQooZIHGdwfR
tLnwk0D3rg==epqM
===
Tom St Denis a dit :> Hash: SHA1>
Exactly what part of math is the study of nomenclature
notation like> GF(p^q), Z/pZ, etc..The Book-prefacing
theory part of math ...All this notation stuff is written at
the beginning of each book.As notation is a part of what it
represents, you will ?d Z/pZ and GF(p^q) in an algebra book
(ring theory for the former, group theory for the latter)>
I know they teach it through the course of usage in number
theory> books but is there a speci? book [or subject] that
teaches how to> read that notation?I don.a6t think so.--
Alexandre CharitopoulosEm6 / Eb7(5b) / Dm7 / Db7(5b, 9b) /
Cmaj7
===
Sorry but I thought I saw a quick way to prove the
Twin PrimesConjecture. On further thought I.a6m not so
sure.While it is true that given X a prime, ?/3) -
?/6) = ?X+2)/3) - ?X+2)/6)if X+2 is prime,
it.a6s not suf?ient to force X+2 to be prime.My apologies, but
I thought I had something there for a while.James Harris
===
>
Sorry but I thought I saw a quick way to prove the Twin
Primes> Conjecture. On further thought I.a6m not so sure.>
While it is true that given X a prime,> ?/3) -
?/6) = ?X+2)/3) - ?X+2)/6)> if X+2 is prime,
it.a6s not suf?ient to force X+2 to be prime.> My apologies,
but I thought I had something there for a while.No apology is
required, as history knows many theorems being valid for
ashort period of time. It.a6s the US: for a success you 
don.a6t
have to becorrect, you just have to be loud.
===
: It.a6s the
US: for a success you don.a6t have to be correct, you just
have: to be loud.Kindly explain this rude
comment.-Justin
===
> : It.a6s the US: for a success you don.a6t
have to be correct, you just have> : to be loud.> Kindly
explain this rude comment.Oh please, do I have to be that
speci??- In politics they were shouting he has WMD!-
Enviromentalist are screaming Watch for Global Warming!-
You can put the most stupid show on TV and you are a
sensation.- You can sue anybody for the most ridiculous
reason. Watch for idiot.a6srights movement.
===
> Sorry but I
thought I saw a quick way to prove > the Twin Primes
Conjecture. However, you managed to get your name into the
history books. -
jb----------------------------------------------------------
Dr.
Frankensteinhttp://seattlepi.nwsource.com/horsey/
viewbydate.asp?id=870----------------------------------------
------------------
===
> As I.a6m sure most know, the rhombic
dodecahedral numbers and the> tesseract /8-cell numbers are,
respectively, the 3rd and 4th> dimensional nexus numbers. The
cuboctahedron, of course, is the dual> of the rhombic
dodecahedron and the hyperoctahedron/16-cell is the> dual of
the tesseract/8-cell. What I.a6m hoping is that someone can>
tell me what the formulae are, if they exist, for computing
what I> guess one would call the cuboctahedral and
hyperoctahedral numbers. > Any help would be appreciated.>
RossTotally slipped my mind to check
Sloane.a6s(http://www.research.att.com/projects/OEIS?Anum=
A005901). So I foundthe cuboctahedral. Still looking for the
hyperoctahedral...Ross
===
I think that you.a6d ?d it in
Coxeter.a6s _Regular Polytopes_,using quaternions. actually,
I.a6m not sure what you mean by nexus numbers.> As 
I.a6m sure
most know, the rhombic dodecahedral numbers and the>
tesseract /8-cell numbers are, respectively, the 3rd and 4th>
dimensional nexus numbers. The cuboctahedron, of course, is
the dual> of the rhombic dodecahedron and the
hyperoctahedron/16-cell is the> dual of the tesseract/8-cell.
What I.a6m hoping is that someone can> tell me what the formulae
are, if they exist, for computing what I> guess one would call
the cuboctahedral and hyperoctahedral numbers. --les ducs
d.a6Enron!http://members.tripod.com/~american_almanac
===
> I
think that you.a6d ?d it in Coxeter.a6s _Regular 
Polytopes_,>
using quaternions.> actually, I.a6m not sure what you mean by
nexus numbers.> As I.a6m sure most know, the rhombic
dodecahedral numbers and the> tesseract /8-cell numbers
are, respectively, the 3rd and 4th> dimensional nexus
numbers. The cuboctahedron, of course, is the dual> of the
rhombic dodecahedron and the hyperoctahedron/16-cell is the>
> dual of the tesseract/8-cell. What I.a6m hoping is that
someone can> tell me what the formulae are, if they exist,
for computing what I> guess one would call the
cuboctahedral and hyperoctahedral numbers. > --les ducs
d.a6Enron!>
http://members.tripod.com/~american_almanachttp://
mathworld.wolfram.com/NexusNumber.html. Conway and Guy
de?edthem in The Book of Numbers. The (n+1)th rhombic
dodecahedral number= 1 + 4n + 6n^2 + 4n^3. The nth = n^4 -
(n-1)^4. Likewise, the(n+1)th tesseract number = 1 + 5n +
10n^2 + 10n^3 + 5n^4. The nth =n^5 - (n-1)^5.Ross
===
>
For instance (-1+sqrt(3)i)/2 is an algebraic integer.>
Yes. I know that since I know it.a6s a root of x^2 + x + 1.>
So, are you excluding it from your ring? Yes or no? > No.>
> Suppose we leave the land of radicals. Let a be the
largest of the> three real roots of x^5-3125x-5. Is a
included in your ring? Do you> consider it to be an
integer? Yes or no?> It.a6s an algebraic integer, so 
it.a6s
included.> The polynomial is irreducible over Q, so it.a6s
not an integer.> If it.a6s an algebraic integer, it.a6s
included, just like algebraic> integers include gaussian
integers, and I remind that I extended from> Dedekind.Very
well then. Then since your ring (whatever it is) contains all
thealgebraic integers, it contains all the units that the ring
ofalgebraic integers has. For you to assert that 1 and -1 are
the onlyunits in your ring is to be in ?t denial of the
existence ofthings like (to use your example of an algebraic
integer)(-1+sqrt(3)i)/2, whose reciprocal is (-1-sqrt(3)i)/2,
which is also analgebraic integer. Learn the properties of
your own ring before youmake assertions about it.----
David
===
>For instance (-1+sqrt(3)i)/2 is an algebraic
integer.>Yes. I know that since I know it.a6s a root of
x^2 + x + 1.>So, are you excluding it from your ring? Yes
or no? >No.>Suppose we leave the land of radicals.
Let a be the largest of the>three real roots of
x^5-3125x-5. Is a included in your ring? Do you>consider it
to be an integer? Yes or no?>It.a6s an algebraic integer, so
it.a6s included.>The polynomial is irreducible over Q, so
it.a6s not an integer.>If it.a6s an algebraic integer, 
it.a6s
included, just like algebraic>integers include gaussian
integers, and I remind that I extended from>Dedekind.>
Very well then. Then since your ring (whatever it is)
contains all the> algebraic integers, it contains all the
units that the ring of> algebraic integers has. For you to
assert that 1 and -1 are the only> units in your ring is to
be in ?t denial of the existence of> things like (to use
your example of an algebraic integer)> (-1+sqrt(3)i)/2, whose
reciprocal is (-1-sqrt(3)i)/2, which is also an> algebraic
integer. Learn the properties of your own ring before you>
make assertions about it.> ---- DavidI got caught on this
too. 1,-1 are the only integer units in the ring.-- Will
Twentyman
===
I.a6m stuck with just part of a problem (just the
parts that I.a6m stuck on arelisted below) and was looking for
some helpPoor Bobby wants to gamble, but he has no money.
Luckily, his benevolentfather offers to give him ?e dollars
to play quarter-dollar slot machines,saying if you ever go
bust, you can return to me for another ?e-dollarsworth of
quarters, but if your total holdings ever exceeds ten
dollars, youmay keep ?e dollars worth of quarters but return
everything else to me.Suppose Bobby.a6s slot machines only take
quarter-dollar bets (i.e. onequarter per spin) and only two
outcomes are possible for each spin: (i) areturn of nine
quarters with probability .1 and (ii) no return
withprobability .9.i. Compute the steady state probabilities
for each recurrentclass above.ii. What is the expected number
of quarters in Bobby.a6s stash in steadystate?I easily got the
following equations but have no idea what to do next to
getthe solutions to i and ii.Let N = {1,2,3...,47} be the
number of quarters that Bob has.For 0 < n < 40P(n , n+8) =
.1P(n , n-1) = .9For n >= 40P( n , 20 ) = 1For n = 0P ( n ,
20 ) = 1.
===
> I want to make a table that will be level on
an unlevel ground> surface. Assuming 3 or 4 vertical legs,
how many legs will need to be> adjustable?> Thinking about
this prompted another question: Is there a function z =>
f(x,y) in which no two points (x,y) map to the same value of
z?> In particular, could I ever be unlucky enough to need
to adjust all> but one of the legs? Let the table be as large
as you like, if that> helps.> John Carter> 25-8-03See
previous postings.So I take it you think:there is no
continuous surface where no two points have the sameheight
?above deck.a6i.e.there does not exist a continuous function
z = f(x,y) for which z isdifferent for all values of x and
yIf that.a6s true, then it must be true for every region of
the surface.Then in any region, no matter how small, there
must be at least twopoints with the same ?height.a6 z. It
follows that in any region, theremust be an unlimited number
of points with the same height (so tablesize doesn.a6t matter).
It follows that in any region there must be Npoints with the
same height, where N is the number of legs on thetable. So
provided the legs are long enough (to accommodate
peaksbetween the legs), they can all be the same length, at
least.John Carter27-8-03
<Pine.LNX.4.55L-032.0308251842440.14025@unix47.andrew.cmu.edu
>
===
> I want to make a table that will be level on an
unlevel ground> surface. Assuming 3 or 4 vertical legs, how
many legs will need to be> adjustable?> Two out of three,
or three out of four.> If you.a6re allowed to rotate the table
any way you like, then> I think 1 out of 3, or 2 out of 4 (or
N-2 out of N) is enough,> but I don.a6t have a proof.Another
puzzle, sort of in line with John.a6s question below: Does
there exist a continuous [in?ite] surface (or patch of
ground) such that no two points on the surface are both (a)
at the same height, and (b) exactly 1 unit
apart?Equivalently, is it possible, given any table with two
?ed legs,to construct a patch of ground on which that table
*cannot* be leveled?Any answers/proofs will be welcome. (I.a6m
almost certain the answeris no, no such patches of ground
exist, but can.a6t be sure of it.)> In particular, could I
ever be unlucky enough to need to adjust all> but one of
the legs? Let the table be as large as you like, if that>
helps.> Why would the size of the table matter? Unless
you.a6re saying that> the table.a6s size can be varied, 
too --
but I don.a6t *think* that would> help.[And then you replied,
something like: Yes, the table.a6s size varies.]That 
doesn.a6t
explain much better. Do you mean: Given a table of size X and
?ed legs F, and a patch Z, level the table. You can vary the
un?ed legs L.or: Given a table of ?ed legs F, and a patch
Z, level the table. You can vary the un?ed legs L and the
table size X.or something else? And what, exactly, does it
mean to vary thesize of a table? Is stretching allowed?
Do the ?ed legs getlonger proportionally, or stay the same
lengths? (Frankly, I thinkthe original problem is much more
interesting. Fewer variables.)-Arthur
===
> I
want to make a table that will be level on an unlevel ground>
> surface. Assuming 3 or 4 vertical legs, how many legs will
need to be> adjustable?> Two out of three, or three
out of four.> If you.a6re allowed to rotate the table any
way you like, then> I think 1 out of 3, or 2 out of 4 (or
N-2 out of N) is enough,> but I don.a6t have a 
proof.You.a6re
right, at least for the case where the ground iscontinuous
and nowhere vertical (or where there is a patchof suf?ient
size satisfying these constraints). And withone other caveat:
we need to make the table top transparentto bumps, otherwise
we can easily create an array of thin,very high spikes such
that the ?ite-length legs of a giventable can never reach
the level ground below.> Another puzzle, sort of in line
with John.a6s question below:> Does there exist a continuous
[in?ite] surface (or patch of> ground) such that no two
points on the surface are both (a) at> the same height, and
(b) exactly 1 unit apart?> Equivalently, is it possible,
given any table with two ?ed legs,> to construct a patch of
ground on which that table *cannot* be leveled?> Any
answers/proofs will be welcome. (I.a6m almost certain the
answer> is no, no such patches of ground exist, but can.a6t
be sure of it.)No follows immediately from the intermediate
value theorem:Put the table down anywhere; if leg A is on
higher ground thanleg B, obviously there is a 180-degree
rotation that swaps thetwo positions so that leg B is on
higher ground. It followsfrom continuity that there is some
intermediate angle of rotation0 <= theta <= pi such that the
height difference between A and Bis zero.> In
particular, could I ever be unlucky enough to need to adjust
all> but one of the legs? Let the table be as large as
you like, if that> helps.> Why would the size of the
table matter? Unless you.a6re saying that> the table.a6s 
size
can be varied, too -- but I don.a6t *think* that would>
help.It doesn.a6t matter at all, as long as it ?s completely
onthe surface in any orientation.I don.a6t know the answer if
we allow vertical patches; you wouldcertainly (in some cases)
have to put both legs on vertical ground,which is a rather odd
thing to do with a table. My intuition isthat there are spiky
or oddly reticulated surfaces that won.a6t permiteven that, but
I really don.a6t know.
===
> I want to make a table that
will be level on an unlevel ground> surface. Assuming 3 or
4 vertical legs, how many legs will need to be>
adjustable?> I think you have a Goedel Escher Bach record
player problem: for any> ?ed table size and shape, there is
a ground surface on which that> table cannot be leveled
unless all but one leg is adjustable. For any> ?ed ground
shape, you can probably scale the table to make it> levelable
with more than one ?ed leg length.> Consider the table with
three legs, one adjustable, and for simplicity,> assume the
legs are attached to the table in an equilateral triangle>
with unit side length.> Fix the location of the adjustable
leg.> Consider the right cylinder formed by the trace of
the extention of the> other two legs as the table is rotated
in a circle around the ?ed leg.> Consider that cylinder.a6s
intersection with the ground, forming a closed> curve around
the cylinder.> Since the table legs will ? six times
around that closed curve, make> the curve a repeat seven sine
wave; then at no point will the two legs> be on the same
phase, thus height,Same phase requires same height, but the
converse is *not*true. For a sine curve of any frequency, all
you have to dois position a peak (or trough) at the exact
midpoint betweenthe two legs, and the height at each leg will
be equal, sincesine is symmetrical about any of its peaks (or
troughs).
===
> Same phase requires same height, but the
converse is *not*> true. For a sine curve of any frequency,
all you have to do> is position a peak (or trough) at the
exact midpoint between> the two legs, and the height at each
leg will be equal, since> sine is symmetrical about any of
its peaks (or troughs).Ah, right you are. That.a6s what I get
for trying to argue fromintuition.xanthian.-- 
===
>
Same phase requires same height, but the converse is *not*>
true. For a sine curve of any frequency, all you have to do>
is position a peak (or trough) at the exact midpoint between>
> the two legs, and the height at each leg will be equal,
since> sine is symmetrical about any of its peaks (or
troughs).> Ah, right you are. That.a6s what I get for trying
to argue from> intuition.But I shamelessly used your setup
anyway, for a proof ofwhat you were trying to disprove. So,
thanks!
===
See below. I don.a6t see why, in the three-leg case,
you say there arejust six positions to rotate the table around
the adjustable leg.And what about my more general question,
which amounts to: ** Is there any surface where no two points
are the same height.a6above the deck.a6? **> I want to 
make a
table that will be level on an unlevel ground> surface.
Assuming 3 or 4 vertical legs, how many legs will need to be>
> adjustable?> I think you have a Goedel Escher Bach record
player problem: for any> ?ed table size and shape, there is
a ground surface on which that> table cannot be leveled
unless all but one leg is adjustable. For any> ?ed ground
shape, you can probably scale the table to make it> levelable
with more than one ?ed leg length.> Consider the table with
three legs, one adjustable, and for simplicity,> assume the
legs are attached to the table in an equilateral triangle>
with unit side length.> Fix the location of the adjustable
leg.> Consider the right cylinder formed by the trace of
the extention of the> other two legs as the table is rotated
in a circle around the ?ed leg.> Consider that cylinder.a6s
intersection with the ground, forming a closed> curve around
the cylinder.> Since the table legs will ? six times
around that closed curve, make> the curve a repeat seven sine
wave; then at no point will the two legs> be on the same
phase, thus height, of the sine wave, so a second leg> must
be adjustable to make the table levelable.> Further analysis
would un? the ?ed leg to allow the table to move in> three
degrees of freedom: x, y, and theta, instead of just theta.>
> That is a bit beyond my skills, but I suspect a similar
trick can be> extended to that case.> xanthian.
===
> See
below. I don.a6t see why, in the three-leg case, you say there
are> just six positions to rotate the table around the
adjustable leg.I think he just meant that you can ? 6
equilateral trianglesaround a point before you start
repeating yourself; so hetried to solve the problem with a
function whose periodabout the circle thus de?ed is
relatively prime to 6. Butas I already posted, that turns out
not to work.> And what about my more general question, which
amounts to:> ** Is there any surface where no two points are
the same height> ?above the deck.a6? **No, not if the height
function is continuous in any region.(And if it isn.a6t, I
don.a6t think you.a6d want to call it a 
surface.Additionally,
there are of course surfaces for which height isnot a
function at all, e.g. a wall; I assume you are ruling
theseout from the get-go.)Pick any two points (A and B) in
the region of continuity; thenthere exists a simple closed
curve in the surface connectingthose points. If z(A)=z(B) we
are done; otherwise, by the meanvalue theorem there is some
point C on the clockwise leg of thecurve from A to B, and
some point D on the counterclockwise leg,C and D distinct,
that both have the same height (z(A)+z(B))/2.We can say more.
Starting from Kent Paul Dolan.a6s picture ofthe intersection of
cylinder with ground, and let A_0 and B_0be the positions of
the two (non-adjustable) legs at rotationangle theta=0.
Suppose that z(A_0) > z(B_0). Rotate 60 degreesand it may
still be the case that z(A_60) > z(B_60); but by thetime you
have done 4 more rotations it cannot be the case everytime,
because A_300 is the same as B_0 and therefore z(A_300)is
less than z(A_0). Therefore the difference function
z(A)-z(B)at any angle theta is a continuous function with
both positiveand negative values; we are therefore assured
that it is zero atsome angle theta. So, this solves your
special case.(Of course, speaking practically, you might have
a case wherethese two legs are at a height that differs from
that of theadjustable leg by more than the range of
adjustability.)As Arthur J. O.a6Dwyer already responded, there
*are* discontinuousfunctions that have no two values the
same.
===
...> Additionally, there are of course surfaces for
which height is> not a function at all, e.g. a wall; I assume
you are ruling these> out from the get-go.)Actually you don.a6t
really have to rule these out at all,but I wanted to keep
things simple.... Therefore the difference function
z(A)-z(B)> at any angle theta is a continuous function with
both positive> and negative values;Lousy writing; I meant
that the difference z(A)-z(B) isa continuous function of
theta with both positive andnegative values over the domain 0
<= theta < 2*pi we are therefore assured that it is zero at>
some angle theta. So, this solves your special case.
===
The
following matrix arises when doing cubic spline interpolation
of aclosed curve of N=5 points:4 1 0 0 11 4 1 0 00 1 4 1 00 0
1 4 11 0 0 1 4In view of the symmetry of this matrix is it
possible to get an explicitformulae for the inverse of this
matrix (for the general NxN case)?Obviously only the ?st row
of the inverse need be given since the restfollow by
sysmmetry.
===
>I thought it relevant to inform that I
noti?d the FBI a couple of>months ago about some of the
math issues I.a6ve brought up here. I>received a single reply
that agents were looking into it, as I cited>national
security, given that mathematicians are so important in the>
>defense of this nation.> Whee! This made my day, thanks.>
>It was not a form letter reply. I.a6ve followed up but have
not gotten>further information from the FBI.>I have
also informed a couple of senators, but did not receive>
>anything other than form letter replies.>The senators
were McCain of Arizona and Graham of Florida.>Some of
you may be angered by my contacting important agencies like>
>the FBI who have VERY important work to do in defense of
this nation.> I doubt anyone.a6s angry about this. You
consumed a teensy bit> of FBI resources, and provided a few
minutes of entertainment> for sci.math readers around the
world - seems like an excellent> bargain.>However, I think
it very important if mathematicians are as adept at>lying as
I.a6ve seen, and the federal government needed to be noti?d.>
Yeah, there.a6s that too.>I will also suggest that those of
you who receive federal funds>carefully review the terms
and conditions you agreed to in order to>get them.>I
am not saying that I know of any investigations into
mathematicians>resulting from my contacts with the United
States Government. I would>suspect that I was simply
ignored as a crank, and that they referred>to
mathematicians who may have lied to them.> I doubt that they
needed to ask a mathematician.a6s opinion to> determine 
you.a6re a
crank. Your stuff just oozes crankitude.>However, it was my
duty to inform, and possibly at some future date,>if some
mathematicians did lie to the FBI or senators, they may face>
>further questions.>If I was mostly ignored by the FBI
and those senators, which is>probable, then, of course,
they didn.a6t ask anyone.>At a later date I will probably
make higher level contacts, hoping to>get feedback from
members of the National Security Council.>James
HarrisI bet the FBI has a Crank File.But anyone bothering to
pester them is goingto get some attention, and possibly
pro?ed in a database for furtherattention , if needed.RJ
P
===
> It.a6s comedy!> It.a6s drama!!> It.a6s 
tragedy!!!> It.a6s>
THE POWERS THAT BE!!!!!!!> WHOOHOOOOOOO!!!!!> [snipped for
brevity but strongly encourage reading the full post
forexcellent insight and humor]> That.a6s right. You go get
?em, tiger.> DaleThis was brilliant - thank you!
===
>
>However, I think it very important if
mathematicians are as adept at>lying as I.a6ve seen, and
the federal government needed to be noti?d.>I will
also suggest that those of you who receive federal funds>
>carefully review the terms and conditions you agreed to in
order to>get them.>Sigh. It.a6s about so much more
than federal funding. Fermat.a6s Last>Theorem, or 
Wiles.a6s
Theorem (if his proof holds up), or Harris.a6s Theorem>(if
the Wiles proof is found to be ?or fraudulent), is a
monumental>piece of number theory (by any name). I don.a6t
think the government cares>much about it, though.>
> fraudulent - such an ugly word.> If the government
could prove that the Wiles proof was fraudulent, that>
would be just the high-pro?e lever that they need. With
Wiles in prison> the time would be ripe to push through
legislation mandating Object Oriented> Math and banning
the ?rings and ?lds that mathematicians have>
conspired to use for so many shame-?led years.> Things
seem to be moving very quickly now. I think that we will see
more and> more mathematicians espousing Harrisian
mathematics. The ?rings and> ?lds will probably be
explained away as a useful ?tion. A sort of>
luminiferous ether upon which the various algebraic
structures (now> revealed by Harris as Objects) ? >
Over the next few months we will witness a paradigm shift in
mathematics.> Those who shift early will be in the
vanguard of the new movement.> ALL HAIL, Jesus Harris,
OUR BLESSED SAVIOUR.> Your sarcasm comes through loud and
clear. The last line indicates that it.a6s> aimed more at me
than at James. Yes, I called James a mathematical messiah.>
I called for people to run out in the streets shouting James
Harris is God!> I admit that this does sound a little
overboard.> But put yourself in my shoes for a moment. I
grew up believing the party line.> When I realized one day
that everything I had been taught was false, and that> the
person whom I thought was a total joke was really a profound
genius, I went> into a state of shock. My world view was
thrown into upheaval. If I was so> wrong about this, I could
be wrong about everything. In this mental chaos, the> only
thing that was blindingly clear was James Harris.a6s oppressive
brilliance.> I squirmed under it, unable to understand his
proof, yet convinced of its truth> (a guy that smart doesn.a6t
make mistakes). I fell to worshipping James and> posted some
things that seem a little over the top now.> I don.a6t see
James as a messiah any longer. He.a6s not here to save the
human race.> He.a6s merely one light that turned on in a vast
black plain. And the creatures of> darkness cringed at the
sight and are trying, even now, to extinguish it.> It
grieves me to see the cruel mockery of this great intellect.
It grieves me> all the more that I once so mocked. Your
vision of a new Harrisian mathematics> could come to be in a
perfect world, but never in one as ?as ours.> I have to
go now. I can.a6t type through the bitter tears I.a6m 
weeping for
the> human race.Priceless exchange!Dirk Vdm
===
In sci.math,
James
Harris<jstevh@msn.com><3c65f87.0308260745.7ae17601@
posting.google.com>:> I thought it relevant to inform that I
noti?d the FBI a couple of> months ago about some of the
math issues I.a6ve brought up here. I> received a single reply
that agents were looking into it, as I cited> national
security, given that mathematicians are so important in the>
defense of this nation.OK. Someone please walk me through
this *very* slowly.Why would the Federal Bureau of
Investigation, which is primarilyinterested in looking for
criminals, both federal and state,be remotely interested in
your mathematical theories?The FBI can do simple stuff like 1
bullet + 1 corpse =1 murder :-), [*] and probably has a rather
comprehensiveset of lab setups and such for ?uring where that
?ercame from and whether it was rubbed onto that piece
ofgrass or blew thereonto, but I fail to see how pure
mathwould even interest them, apart from the rather
obviousarea of cryptography, which is more likely covered
bythe NSA but could fall under their jurisdiction if thecase
hinges on who did what to whom when and what theysaid about
it to whomever.If one of course wishes to claim the existence
of a deepunderground laboratory where Agent J assigns work
forthe day to ensure that the visiting aliens aren.a6t
commitingcrimes, don.a6t call us, they.a6ll call you. :-)> 
> It
was not a form letter reply. I.a6ve followed up but have not
gotten> further information from the FBI.> I have also
informed a couple of senators, but did not receive> anything
other than form letter replies.> The senators were McCain
of Arizona and Graham of Florida.I don.a6t see how Graham would
be interested unless itinvolved spending tax dollars :-) (in
which case he.a6dbe vehemently against it). Dunno about McCain
althoughhis big buzz was straight, honest talk about
variousissues IIRC.(No, LGMs, UFOs, and math equations
weren.a6t among them.Maybe Social Security.)> Some of you may
be angered by my contacting important agencies like> the FBI
who have VERY important work to do in defense of this
nation.Yes, they do. How that work is augmented by your
interestingideas on math equations is far from clear.>
However, I think it very important if mathematicians are as
adept at> lying as I.a6ve seen, and the federal government
needed to be noti?d.Oh no. Mathematicians are terrorists!I
see it all now: the accountant.a6s cap, the pencil
newlysharpened, the math book laid open with a ruler on the
page --it.a6s all part of the sinister plan of Al
Mathla[+].Wooooooooooooooo..... Better inform the...oh, wait,
you already did.> I will also suggest that those of you who
receive federal funds> carefully review the terms and
conditions you agreed to in order to> get them.You mean I
have to give back my tax refund?!> I am not saying that I
know of any investigations into mathematicians> resulting
from my contacts with the United States Government. I would>
suspect that I was simply ignored as a crank, and that they
referred> to mathematicians who may have lied to them.Wow. A
shred of sanity ?mongst the silliness.> However, it was my
duty to inform, and possibly at some future date,> if some
mathematicians did lie to the FBI or senators, they may face>
further questions.> If I was mostly ignored by the FBI and
those senators, which is> probable, then, of course, they
didn.a6t ask anyone.> At a later date I will probably make
higher level contacts, hoping to> get feedback from members
of the National Security Council.> Well, whatever ?your
boat. I personally would preferto be strapped into an F-15
interceptor or a space shuttleand puke all over myself as he
does various hair-raisingmaneuvers, but to each his own.:-)>
> James Harris[*] I should note here that murder is a state
crime, not a federal one, unless the victim was serving the
Federal Government in some capacity (e.g., a U.S. Marshall or
an IRS bureaucrat). It is not clear whether the individual
would have had to be serving in that capacity at the time of
the crime, or not; there are a number of issues here but
they.a6re all peripheral to the discussion.[+] not to be
confused with a certain terrorist group in the Middle East.--
#191, ewill3@earthlink.netIt.a6s still legal to go
.sigless.
===
>In sci.math, James
Harris><jstevh@msn.com><3c65f87.0308260745.7ae17601@
posting.google.com>:> I thought it relevant to inform that I
noti?d the FBI a couple of> months ago about some of the
math issues I.a6ve brought up here. I> received a single reply
that agents were looking into it, as I cited> national
security, given that mathematicians are so important in the>
defense of this nation.>OK. Someone please walk me through
this *very* slowly.>Why would the Federal Bureau of
Investigation, which is primarily>interested in looking for
criminals, both federal and state,>be remotely interested in
your mathematical theories?I.a6m surprised he didn.a6t 
mention
the NSA. Thrillers have been full ofNSA agents lately. It
doesn.a6t seem to bother the writers that theNSA has a mission
which doesn.a6t involve anything remotely like chasingpeople
(or aliens) around with guns. More like sitting in rooms
withheadphones on, which lacks that James Bond quality.I
guess it.a6s that they really don.a6t know what the NSA 
does, so
thatmakes it de facto a mysterious 3-letter government
agency.And of course there are the sekrit agents who work for
an agency soshadowy that being secret FBI/NSA agents is
their COVER STORY.James should contact them. - Randy
===
> And
of course there are the sekrit agents who work for an agency
so> shadowy that being secret FBI/NSA agents is their COVER
STORY.> James should contact them.Ah, those must be the agents
from Scarecrow and Mrs. King, who workedfor an agency so
secret that it had no name but The Agency. (And no,it
wasn.a6t the CIA; in at least one episode a CIA agent and an
Agencyagent worked together.)-- Wayne Brown | When your
tail.a6s in a crack, you improvisefwbrown@bellsouth.net | if
you.a6re good enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
In sci.math, Randy
Poe<rpoePA@yahoo.com><7v5pkv414v50k320lupievbuu5lmic50pd@
4ax.com>:>In sci.math, James
Harris><jstevh@msn.com><3c65f87.0308260745.7ae17601@
posting.google.com>:> I thought it relevant to inform that
I noti?d the FBI a couple of> months ago about some of the
math issues I.a6ve brought up here. I> received a single reply
that agents were looking into it, as I cited> national
security, given that mathematicians are so important in
the> defense of this nation.>OK. Someone please walk me
through this *very* slowly.>Why would the Federal Bureau
of Investigation, which is primarily>interested in looking
for criminals, both federal and state,>be remotely
interested in your mathematical theories?> I.a6m surprised he
didn.a6t mention the NSA. Thrillers have been full of> NSA
agents lately. It doesn.a6t seem to bother the writers that
the> NSA has a mission which doesn.a6t involve anything
remotely like chasing> people (or aliens) around with guns.
More like sitting in rooms with> headphones on, which lacks
that James Bond quality.> I guess it.a6s that they really
don.a6t know what the NSA does, so that> makes it de facto a
mysterious 3-letter government agency.> And of course there
are the sekrit agents who work for an agency so> shadowy that
being secret FBI/NSA agents is their COVER STORY.> James
should contact them.If he can ?d them. I suspect they.a6re so
secret even theydon.a6t know. :-)> - Randy> -- #191,
ewill3@earthlink.netIt.a6s still legal to go .sigless.
===
In
sci.math,
a<a@shell3.shore.net><3JQ2b.14373$Kj3.5517@
nntp-post.primus.ca>:> I thought it relevant to inform that
I noti?d the FBI a couple of> months ago about some of the
math issues I.a6ve brought up here. >Could you post a copy
of the letter you sent? I for one would>love to see it.>
So would I. I.a6d also like the see the FBI.a6s reply, 
which of
course> was not a form letter.> Well, it probably had the
form of a letter, anyway... :-)-- #191,
ewill3@earthlink.netIt.a6s still legal to go .sigless.
===
>
MAMA: C.a6mere, boy, and bring me that switch! JSH: Get it
yourself, I ain.a6t no in.a6 electrician.--Mensanator2 of
Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
=J. Edgar Harris> I thought it relevant to inform that I
noti?d the FBI a couple of> months ago about some of the
math issues I.a6ve brought up here. I> received a single reply
that agents were looking into it, as I cited> national
security, given that mathematicians are so important in the>
defense of this nation....I.a6ve seen these symptoms before. I
expect the FBI will add only one name toits databases, and
its initials will be JSH.
===
The only thing I don.a6t
understand is why you post this here. Wouldn.a6t it help the
FBI to have the element of surprise?Gib
===
>The only thing I
don.a6t understand is why you post this here. Wouldn.a6t 
>it help
the FBI to have the element of surprise?>Gib>It.a6s hard to
intimidate all those wacky math people that don.a6t recognize
his genious if he doesn.a6t tell you about it.I somehow think
he.a6d be slapping a lot of people with frivolous lawsuits if
he had enough money to spend on it. Right or wrong doesn.a6t
matter with that tactic, if the instigator has deep enough
pockets to drag it through the courts long enough to break
the little guy.-- A nice adaptation of conditions will make
almost any hypothesis agreewith the phenomena. This will
please the imagination but does not advanceour knowledge. --
J. Black, 1803.
===
>The only thing I don.a6t understand is why
you post this here. Wouldn.a6t >it help the FBI to have the
element of surprise?>Gib> It.a6s hard to intimidate
all those wacky math people that don.a6t recognize > his genious
if he doesn.a6t tell you about it.I.a6m not interested in 
people
breaking the law, so the warning is thesame as the previous
warnings, which is--don.a6t break it.Mathematicians may believe
they have a right to continue teachingalgebraic integers as if
there isn.a6t a problem with them, but I saythat.a6s 
fraud.There
are students who believe that they.a6re getting value for
theirtime, and money (or their parents.a6 money), as well as
the UnitedStates itself, which promotes math research and
math studies for the*bene?* of society. > I somehow think
he.a6d be slapping a lot of people with frivolous lawsuits > if
he had enough money to spend on it. Right or wrong doesn.a6t
matter with > that tactic, if the instigator has deep enough
pockets to drag it through > the courts long enough to break
the little guy.And I think many of you fail to see that
blocking math results, evenif only by silence, can mean that
you.a6re on the outside of societyworking against it.As
lawbreakers you will be like other criminals, and in
facingpunishment, it won.a6t be like worrying about
lawsuits.Make no mistake, if the United States Government
gets convinced tocome after mathematicians, individuals and
math departments who areresponsible for them, who teach the
?mathematics, then you willbe paying attention from that
time forward.And none of you will be joking about it.I.a6d like
to remind that I.a6ve contacted major universities all over
theUnited States including top tier ones like Harvard and
Yale.My strategy is deliberate. Another bene? of warning
those of you inacademia is that it removes some excuses, and
will help take away thepossibility of trying to rely on
public sympathy.As far as I.a6m concerned many of you are
criminals or are about to becriminals as you go into the
classroom to give young minds ?nformation, and I don.a6t
want the public to get distracted from thatfact.James
Harris
===
> As far as I.a6m concerned many of you are criminals
or are about to be> criminals as you go into the classroom to
give young minds ? information, and I don.a6t want the
public to get distracted from that> fact.Applying scienti?
method. 2 hypotheses are:1. There are so many criminals2.
There is just one conspiracy theoristMany would prefer #2 as
the *simpler* one.
===
God that is so ed up> Believe it
or not, this guy is dead serious.> For some background on
James, you can go here:>
http://groups.google.com/groups?selm=EWVpa.3442$z15.3317@
news.primus.ca>I thought it relevant to inform that I
noti?d the FBI a couple of>months ago about some of the
math issues I.a6ve brought up here. I>received a single reply
that agents were looking into it, as I cited>national
security, given that mathematicians are so important in the>
>defense of this nation.>It was not a form letter reply.
I.a6ve followed up but have not gotten>further information
from the FBI.>I have also informed a couple of senators,
but did not receive>anything other than form letter
replies.>The senators were McCain of Arizona and Graham
of Florida.>Some of you may be angered by my contacting
important agencies like>the FBI who have VERY important
work to do in defense of this nation.>However, I think
it very important if mathematicians are as adept at>lying
as I.a6ve seen, and the federal government needed to be
noti?d.>I will also suggest that those of you who
receive federal funds>carefully review the terms and
conditions you agreed to in order to>get them.>I am
not saying that I know of any investigations into
mathematicians>resulting from my contacts with the United
States Government. I would>suspect that I was simply
ignored as a crank, and that they referred>to
mathematicians who may have lied to them.>However, it
was my duty to inform, and possibly at some future date,>if
some mathematicians did lie to the FBI or senators, they may
face>further questions.>If I was mostly ignored by the
FBI and those senators, which is>probable, then, of course,
they didn.a6t ask anyone.>At a later date I will probably
make higher level contacts, hoping to>get feedback from
members of the National Security Council.>James
Harris>
===
Will Twentyman <wtwentyman@read.my.sig> skrev i
melding> I thought it relevant to inform that I noti?d the
FBI a couple of> months ago about some of the math issues
I.a6ve brought up here. I> received a single reply that
agents were looking into it, as I cited> national security,
given that mathematicians are so important in the> defense
of this nation.> It was not a form letter reply. I.a6ve
followed up but have not gotten> further information from
the FBI.> I have also informed a couple of senators, but
did not receive> anything other than form letter replies.>
> The senators were McCain of Arizona and Graham of
Florida.> Some of you may be angered by my contacting
important agencies like> the FBI who have VERY important
work to do in defense of this nation.> However, I think
it very important if mathematicians are as adept at> lying
as I.a6ve seen, and the federal government needed to be
noti?d.> I will also suggest that those of you who
receive federal funds> carefully review the terms and
conditions you agreed to in order to> get them.> I am
not saying that I know of any investigations into
mathematicians> resulting from my contacts with the United
States Government. I would> suspect that I was simply
ignored as a crank, and that they referred> to
mathematicians who may have lied to them.> However, it
was my duty to inform, and possibly at some future date,>
if some mathematicians did lie to the FBI or senators, they
may face> further questions.> If I was mostly ignored
by the FBI and those senators, which is> probable, then, of
course, they didn.a6t ask anyone.> At a later date I will
probably make higher level contacts, hoping to> get
feedback from members of the National Security Council.>
> James Harris> Interesting followup note to this action:
I joined his AmateurMath page> so I could look at his proof
of the Advanced Polynomial Factorization,> and he has now
blocked me out of the entire site, even the pages that> you
do *not* have to register to view.> -- > Will
Twentyman>That.a6s my experience too. I have had a passport
since January 2002 and theonly site I can.a6t watch is 
James.a6
site. I am not allowed. I think that hereads the post
carefully and ban those he doesn.a6t like!Karl-Olav Nyberg
===
>
Will Twentyman <wtwentyman@read.my.sig> skrev i melding>
>I thought it relevant to inform that I noti?d the FBI a
couple of>months ago about some of the math issues I.a6ve
brought up here. I>received a single reply that agents were
looking into it, as I cited>national security, given that
mathematicians are so important in the>defense of this
nation.>It was not a form letter reply. I.a6ve followed up
but have not gotten>further information from the FBI.>I
have also informed a couple of senators, but did not
receive>anything other than form letter replies.>The
senators were McCain of Arizona and Graham of
Florida.>Some of you may be angered by my contacting
important agencies like>the FBI who have VERY important
work to do in defense of this nation.>However, I think
it very important if mathematicians are as adept at>lying
as I.a6ve seen, and the federal government needed to be
noti?d.>I will also suggest that those of you who
receive federal funds>carefully review the terms and
conditions you agreed to in order to>get them.>I am
not saying that I know of any investigations into
mathematicians>resulting from my contacts with the United
States Government. I would>suspect that I was simply
ignored as a crank, and that they referred>to
mathematicians who may have lied to them.>However, it
was my duty to inform, and possibly at some future date,>if
some mathematicians did lie to the FBI or senators, they may
face>further questions.>If I was mostly ignored by the
FBI and those senators, which is>probable, then, of course,
they didn.a6t ask anyone.>At a later date I will probably
make higher level contacts, hoping to>get feedback from
members of the National Security Council.>James
Harris>Interesting followup note to this action: I joined
his AmateurMath page>so I could look at his proof of the
Advanced Polynomial Factorization,>and he has now blocked
me out of the entire site, even the pages that>you do *not*
have to register to view.>-- >Will Twentyman>
That.a6s my experience too. I have had a passport since January
2002 and the> only site I can.a6t watch is James.a6 
site. I am
not allowed. I think that he> reads the post carefully and
ban those he doesn.a6t like!> Karl-Olav Nyberg> I 
haven.a6t
even tried to post anything. I just want to be able to get at
his .pdf in case he updates it. The last time he made a change
I was unaware of it until looking again. I found I was arguing
against a signi?antly different paper than I had originally
found.-- Will Twentyman
===
Will Twentyman
<wtwentyman@read.my.sig> skrev i melding> Will Twentyman
<wtwentyman@read.my.sig> skrev i melding>I thought
it relevant to inform that I noti?d the FBI a couple of>
>months ago about some of the math issues I.a6ve brought up
here. I>received a single reply that agents were looking
into it, as I cited>national security, given that
mathematicians are so important in the>defense of this
nation.>It was not a form letter reply. I.a6ve
followed up but have not gotten>further information from
the FBI.>I have also informed a couple of senators,
but did not receive>anything other than form letter
replies.>The senators were McCain of Arizona and
Graham of Florida.>Some of you may be angered by my
contacting important agencies like>the FBI who have VERY
important work to do in defense of this nation.>
>However, I think it very important if mathematicians are
as adept at>lying as I.a6ve seen, and the federal
government needed to be noti?d.>I will also suggest
that those of you who receive federal funds>carefully
review the terms and conditions you agreed to in order to>
>get them.>I am not saying that I know of any
investigations into mathematicians>resulting from my
contacts with the United States Government. I would>
>suspect that I was simply ignored as a crank, and that
they referred>to mathematicians who may have lied to
them.>However, it was my duty to inform, and
possibly at some future date,>if some mathematicians did
lie to the FBI or senators, they may face>further
questions.>If I was mostly ignored by the FBI and
those senators, which is>probable, then, of course, they
didn.a6t ask anyone.>At a later date I will probably
make higher level contacts, hoping to>get feedback from
members of the National Security Council.>James
Harris>Interesting followup note to this action: I
joined his AmateurMath page>so I could look at his proof
of the Advanced Polynomial Factorization,>and he has now
blocked me out of the entire site, even the pages that>you
do *not* have to register to view.>-- >Will
Twentyman> That.a6s my experience too. I have had a
passport since January 2002 andthe> only site I can.a6t watch
is James.a6 site. I am not allowed. I think thathe> reads the
post carefully and ban those he doesn.a6t like!> Karl-Olav
Nyberg> I haven.a6t even tried to post anything. I just
want to be able to get at> his .pdf in case he updates it.
The last time he made a change I was> unaware of it until
looking again. I found I was arguing against a> signi?antly
different paper than I had originally found.> -- > Will
Twentyman>I meant posting (commenting) to mr. Harris.a6 treads
here on sci. math.Karl-Olav Nyberg
===
> Will Twentyman
<wtwentyman@read.my.sig> skrev i melding>Will
Twentyman <wtwentyman@read.my.sig> skrev i
melding>Interesting followup note to this action: I
joined his AmateurMath page>so I could look at his proof
of the Advanced Polynomial Factorization,>and he has now
blocked me out of the entire site, even the pages that>you
do *not* have to register to view.>That.a6s my
experience too. I have had a passport since January 2002 and>
> the>only site I can.a6t watch is James.a6 site. I am 
not
allowed. I think that> he>reads the post carefully and
ban those he doesn.a6t like!>Karl-Olav Nyberg>I
haven.a6t even tried to post anything. I just want to be able
to get at>his .pdf in case he updates it. The last time he
made a change I was>unaware of it until looking again. I
found I was arguing against a>signi?antly different paper
than I had originally found.> I meant posting
(commenting) to mr. Harris.a6 treads here on sci. math.>
Karl-Olav Nyberg> That makes sense. It would be a true
tragedy if we started posting in his space, rather than let
it be a bastion of Pro-JSH.-- Will Twentyman
===
James
Harris <jstevh@msn.com> skrev i melding> I thought it
relevant to inform that I noti?d the FBI a couple of> months
ago about some of the math issues I.a6ve brought up here. I>
received a single reply that agents were looking into it, as
I cited> national security, given that mathematicians are so
important in the> defense of this nation.> It was not a form
letter reply. I.a6ve followed up but have not gotten> further
information from the FBI.> I have also informed a couple of
senators, but did not receive> anything other than form
letter replies.> The senators were McCain of Arizona and
Graham of Florida.> Some of you may be angered by my
contacting important agencies like> the FBI who have VERY
important work to do in defense of this nation.> However, I
think it very important if mathematicians are as adept at>
lying as I.a6ve seen, and the federal government needed to be
noti?d.> I will also suggest that those of you who receive
federal funds> carefully review the terms and conditions you
agreed to in order to> get them.> I am not saying that I
know of any investigations into mathematicians> resulting
from my contacts with the United States Government. I would>
suspect that I was simply ignored as a crank, and that they
referred> to mathematicians who may have lied to them.>
However, it was my duty to inform, and possibly at some
future date,> if some mathematicians did lie to the FBI or
senators, they may face> further questions.> If I was mostly
ignored by the FBI and those senators, which is> probable,
then, of course, they didn.a6t ask anyone.> At a later date I
will probably make higher level contacts, hoping to> get
feedback from members of the National Security Council.>
James HarrisMr. HarrisYou are quite a guy! When your
arguments fail, you call in the heavy guys.Is it like that
you will press your arguments on us? If we dont like them,you
will force them?Grow up! It will not work in Norway!And aside
this post, can you explain to me how the scienti? method
withtrials and fails can be used in mathematics. I will
repeat that thescienti? method doesn.a6t prove a thing, 
it.a6s
only conjuctures things. Wehave conjuctures in mathemathics
too, but thats why we struggle to provethem or disprove them.
We never say that veri?ation through examples is aproof!The
method of problem solving (Polya) often use tools that looks
like thescienti? method, but it.a6s fare from it, because the
result is either trueor false.The scienti? method doesn.a6t
work in mathematics. We want proof, notveri?ations. And
veri?ations aren.a6t that good in the scienti? methodeither.
One falsi?ation breaks and distroys the theory.Consider
Newton.a6s rule of force. He almost had it, but he 
didn.a6t
belive thatmass or time could differ with the distance
between two bodies. Einstein didshow that this could happen.
Newton set this to 0 (zero), but Einstein showthat this was
not so.We use Newton.a6s laws here, but have to use
Einstein.a6s laws when it comesto the extremes. And what
happens when thing are more extreme? Masses movingabove the
speed of light? Einstein doesn.a6t apply.Mr. Harris, you say
that you have a degree in physics. Well, then you musthave
learned that the laws of physics are not theoremes, they are
(in amathematical meaning conjuctures). They are conjuctures
to be prooved ordisaprooved. No trial and veri?ation can
proove them. It only needs onecounterexample to falsify
them.In mathematics we do this in another way. We build the
mathematics on someour theoremes. They can.a6t be disputed.
Through your work you are questioningset-theory, be aware;
you are attacking the axiomes! It is meaningless!Theoremes
are prooved by these axiomes! You can.a6t change that!We can
construct some sets that make problems for the set-theory,
but noneof the sets you treat are among these.The basic thing
that I will say to you is that (I have said it before)
inmathemathics we are dealig with theoremes when we will
proove something, notconjuctures and examples.However if you
could give an example of a number that should be an
algebraicinteger and is excluded from (what I know of) the
set of algebraic integersI would really like to know! Can.a6t
you just give me one?Karl-Olav Nyberg
===
> You are quite a
guy! When your arguments fail, you call in the heavy guys.>
Is it like that you will press your arguments on us? If we
dont like them,> you will force them?> Grow up! It will not
work in Norway!Don.a6t think you are safe in Norway! James
plansto get the US Army involved, too. As he said
in3c65f87.0304191552.511ad5b4%40posting.google.com on April
19th:> Yup, you guessed it. If worse comes to worse, I *will*
turn to the> Army to help me with mathematicians. And then
mathematicians don.a6t> think the NSA or CIA can save your
asses, as generals LIKE me.So you see, you can expect troops
to invade Norway any day now...-- Wayne Brown | When your
tail.a6s in a crack, you improvisefwbrown@bellsouth.net | if
you.a6re good enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
>Grow up! It will not work in Norway!Ah, you
don.a6t see this, but you.a6re mistaken. Unitentional 
humour, as
providedby Mr. Harris, works in any country.Doug
===
Doug
Norris <norrisdt@rintintin.colorado.edu> skrev i melding>
>Grow up! It will not work in Norway!> Ah, you don.a6t see
this, but you.a6re mistaken. Unitentional humour, asprovided>
by Mr. Harris, works in any country.> DougI see the
humourous side of it, but I don.a6t think Harris is trying to
befunny. He really thinks that FBI will slay us. :)Karl-Olav
Nyberg
===
>Doug Norris <norrisdt@rintintin.colorado.edu>
skrev i melding>Grow up! It will not work in Norway!>
Ah, you don.a6t see this, but you.a6re mistaken. 
Unitentional
humour, as>provided> by Mr. Harris, works in any country.>I
see the humourous side of it, but I don.a6t think Harris is
trying to be>funny. He really thinks that FBI will slay us.
:)That.a6s why I called it unintentional humour.Doug
===
> I
thought it relevant to inform that I noti?d the FBI a couple
of> months ago about some of the math issues I.a6ve brought up
here. I> received a single reply that agents were looking
into it, as I cited> national security, given that
mathematicians are so important in the> defense of this
nation.>Did the FBI have any comments related to m = 0 ?
Like,Agent Smith: Look here, I was investigating this m = 0
deal.Harris: Yeah? What of it ?Agent Smith: I.a6ll do the
questioning here. Looks to me like there.a6s some kinda coverup
here. You know. m = 0 is kind of a, uh, special case. You get
my drift?Harris: Um ... no ... you mean, where I say 5 is a
factor of a_1 because when m = 0, 5 is a factor of zero
because anything divides 0 therefore 5 is a factor of a_1 no
matter what m is, even though a_1 isn.a6t 0 when m is not 0,
but still it must always be divisible by 5 because it was in
that one special case where it happens to be divisible by
everything and 5 just happens to be a special case of
everything but still it obviously works in that one case so
it must work all the time. Is that what you mean?Agent Smith:
Yeah. Somethin.a6 like that.Harris: Well, clearly you have been
talking to a bunch of lying thieving low-down cheating
federal-fund-grubbing (ugh!) mathematicians.Agent Smith:
Yeah. I mean, I don.a6t know if they were thieving. They mighta
been though. I wouldn.a6t put anything past ?em. Look, kid. 
I.a6m
gonna go easy on ya. Just write down a little proof for f =
3. Yeah, I know, it.a6s gonna be trivial and all. But the Boss,
he says proving this thing for f = 3 ain.a6t enough. He says
what you get with 3 is reducible. He says ...Harris: I see
what you.a6re doing. It.a6s the old good-cop bad- cop 
routine.
He.a6s feeding you this crap about f = 3 probably from some
low-down lying thieving cheating mathematician who can.a6t
stand that an amateur can come along and after only 8 years
of total bull come out with another load of total
bull ... no, wait, that didn.a6t come out right ... I mean,
comes along with the Best Proof Ever Written. Now about f = 3.
Again here by SCIENTIFIC EXPERIMEN- TATION I ?d that when f =
3, a.k.a. 2 + 1, a bona ?e element of a ring if there ever
was one and a number whose inverse, like all real numbers, is
in the ring generated by Z and 1/2, I say when f = 3 (and
mathmaticians REFUSE TO ACKNOWLEDGE THIS), even though it is
also a special case because of the irreducibility, I say,
mathematicians cannot ?d *how* the factors of f split out
into the roots, all they have is some cowpoop of a proof that
all the roots have some f in them - I say, by scienti?
experimen- tation, when f = 3, again the numbers factor just
like I say they should. And obviously when you prove a thing
for m = 0 and you prove it for f = 3, it MUST be true for all
other m.a6s and f.a6s, RIGHT? I mean, take a typical math
statement, say, the statement that all numbers of the form
m*f are divisible by 3. Well, obviously it.a6s true when m = 0,
right? And obviously it.a6s true when f = 3, right? SO IT MUST
BE TRUE WHEN m = anything and f = anything, right? Couldn.a6t
possibly be just a special case. That.a6s what any normal
person would accept as a valid proof, right? But that.a6s just
the kind of thing that these putrid, fetid, lying cheating
mathema- ticians deny right and left. They WILL NOT ACCEPT MY
GREATNESS!!! They keep giving these asshole PROOFS and
COUNTEREXAMPLES and they keep covering up their stupidity by
asking for  like DEFINITIONS and for WHAT RING AM I IN? I
mean for Christ.a6s sake, what possible difference does it make
what ring I am in? Agent Smith: Yup.Harris: I mean, if it
doesn.a6t ?, you must acquit. Right?Agent Smith:
Yessir.Harris: You.a6re a smart guy, Smith. But you.a6re 
not
convinced.Agent Smith: I didn.a6t say that.Harris: Look, 
I.a6ll
go over it again. You got it when m = 0. True, it.a6s a
degenerate case, and true, the polynomial only has degree 1
and I want it to be true when it has degree p, and true,
everything divides 0. So you got it without the slightest
shadow of a doubt when m = 0. Right?Agent Smith: Uh -
right.Harris: And you got it when f = 3, even though f = 3 is
a special case that I don.a6t even care about, and no one is
squawking that I am wrong about f = 3 because I don.a6t have
irreducibility. Right?Agent Smith: Uh - right. No
irredoucheability or whatever.Harris: So you sort of got a
great big square with an f side and an m side, and you
got one side of this big square, where m = 0, sort of, and
you got another side of the square when f = 3, and even
though neither one of these is what I want or need, I got
both of them, sort of, and so it.a6s pretty goddamn obvious,
isn.a6t it?Agent Smith: Maybe you better spell it out for
me.Harris: Well, you got it for m = 0 and f = 3. Therefore
you got it for all m and f. Right?Agent Smith: Somethin.a6
missing there. Can.a6t put my ?ger on it.Harris: Look, wienie.
I say you got the factorization I want when m = 0 and f = 3.
So it must be true for all the other m.a6s and f.a6s.Agent 
Smith:
You mean, say, if m = 7 and f = 13?Harris: Yeah, good example.
See, I showed it (sort of) for m = 0 and f = 3. So it must be
true when m = 7 and f = 13. Only an idiot would fail to see
the logical connection. Listen to the Math!Agent Smith: Yup.
I.a6m listening to the Math. The Math knows.Harris: And 
that.a6s
what these lying cheating bull artist mathematicians are
trying to tell you. That the argument I gave is not a
proof.Agent Smith: Yup. > It was not a form letter reply.
I.a6ve followed up but have not gotten> further information
from the FBI.> Dear Mr. Harris, This is not a form letter. We
will inform you when we have further information. Your
friend, J. Edgar Hoover, Jr.> I have also informed a couple
of senators, but did not receive> anything other than form
letter replies.> Dear [boxholder] one question. When you say
m = 0 or f = 3, why is that suf?ient? This is a form letter.
Please contribute to my next campaign. Sincerely, Amos S.
Senator, Washington, DC> The senators were McCain of Arizona
and Graham of Florida.> Some of you may be angered by my
contacting important agencies like> the FBI who have VERY
important work to do in defense of this nation.> I.a6m madder
than hell and I.a6m not going to take it any more.> However, I
think it very important if mathematicians are as adept at>
lying as I.a6ve seen, and the federal government needed to be
noti?d.> But ... it.a6s the only thing they are good at,
lying. Take thataway, what do they have left? They.a6ll go on
welfare.> I will also suggest that those of you who receive
federal funds> carefully review the terms and conditions you
agreed to in order to> get them.> Whew! Nothing in there
about not lying. I.a6m in the clear. > I am not saying that I
know of any investigations into mathematicians> resulting
from my contacts with the United States Government. > I
would> suspect that I was simply ignored as a crank, and
that they referred> to mathematicians who may have lied to
them.> A crank ??? How could that be. Those dastards. They
have in?-trated the innermost inner sanctums of the
Goverment. How dare theycall you a crank. You can start a car
with a crank. Do peoplestart cars with you? No. You are not a
crank. You are more likean ashtray, just under the dash.>
However, it was my duty to inform, and possibly at some
future date,> if some mathematicians did lie to the FBI or
senators, they may face> further questions.> Questions like,
What about m = 17 and f = 37?> If I was mostly ignored by
the FBI and those senators, which is> probable, then, of
course, they didn.a6t ask anyone.> Ask James Harris, you fools.
Or ask Rummie. He.a6ll vouchfor old ex-soldier James. > At a
later date I will probably make higher level contacts, hoping
to> get feedback from members of the National Security
Council.> Fat chance. The mathematicians have stuffed the
Council withtheir own men, none of whom are going to admit in
a millionyears that proving whatever for m = 0 and f = 3
should be enough for anybody. They are such dishonest crooks,
plusright now they are busy planning a pre-emptive attack on
Paraguay.Andrzej.> James Harris
===
What does the
J(x)=Li(x)-sum Li(x^p)-log(2)+integral formula look ifwe are
looking at 1 mod 4 primes? We could modify Pi(x) to
Pi4,1(x).How do you de?e J and can we get a formula
involving L(chi(4))? Ithink the correct de?itions would
let us calculate the differencebetween 4 mod 3 and mod 1
primes. Do you have a reference?
===
Marriage mathematicsThe
Deccan HeraldEverything on this earth can be modelled
mathematically,they say. A professor from the University of
Washingtonhas proven this statement once again, by
formulating amodel that can predict the chances of the
success of aproposed marriage.He claims data from a couple.a6s
conversations, convertedinto algebra, is 94% accurate in
determining how long acouple will remain wed. Professor James
Murray, the man behind this theory,creates a graph of their
conversations, and believes thatit could eventually be used
to predict the likelihood ofdivorce.In this approach, he
gives a numerical value for thephrases used by a couple in
their conversation, with therange being from minus four to
plus four. So, a reference to a partner ?being stupid.a6 would
bescored as minus four, but a joke which results inlaughter
from a partner would be scored as plus two. Foreach
conversation slot, he gets a number and over time agraph
showing how the couple are interacting is obtained.Prof
Murray says that the information for the graph iscollated
when couples, who are planning to get married,are brought
into a lab and there conversation is assessedfor 15 minutes.
He uses a scoring system on their reactions andrepresents
their reactions with algebraic terms. With thedata collected,
he works out whether they are likely tostay together or if
they are more likely to divorce. ?.a6Maths provides a language
for interpreting the humaninteraction. It quanti?s one
person.a6s effect on theother, and it is not dif?ult. The
maths we are usingcould be done by secondary school pupils
with basicalgebra. Once they are shown the basic model, they
inserttheir data and make the simple calculation,.a6.a6 
says
theProfessor. T G Srinidhi Read the complete news
at:http://deccanherald.com/deccanherald/aug26/snt1.aspJai
Maharajhttp://www.mantra.com/jaiOm ShantiShubhanu Nama
Samvatsare Dakshinaya Jivana Ritau Singh Mase Shukl Pakshe
Buddh Vasara YuktayamMagh-Poorvaphalguni Nakshatr Shiv-Siddha
Yog Naag-Kinstughn Karan Amavasya-Pratham Yam
Tithauhttp://www.mantra.com/holocausthttp://www.hindu.orghttp
://www.hindunet.orgThe truth about Islam and
Muslimshttp://www.?m/~jai/satyamevajayate o Not for
commercial use. Solely to be fairly used for theeducational
purposes of research and open discussion. The contents ofthis
post may not have been authored by, and do not necessarily
representthe opinion of the poster. The contents are
protected by copyright lawand the exemption for fair use of
copyrighted works.considered or answered if it does not
contain your full legal name,are not necessarily those of the
poster.
===
No Im not a student, just an old dude who hasn.a6t
had to do this basicalgebra for a looooong time...But I need
to solve the following quadratic for x:a*(off + x * width)^2
+ b * (off + x * width) + cCan anyone take pity and help
out?
===
> No Im not a student, just an old dude who hasn.a6t
had to do this basic> algebra for a looooong time...> But I
need to solve the following quadratic for x:> a*(off + x *
width)^2 + b * (off + x * width) + c> Can anyone take pity
and help out?> If you set it equal to 0, and replace off
+ x * width with y, then you have a*y^2+b*y+c=0. This is
easily solved with the quadratic equation.Then, since
off+x*width=y, subtract off and divide by width on both
sides.-- Will Twentyman
===
> Nobody wins. You.a6re using
include in two different ways:> According to standard
set theory, no set includes itself,> meaning that no set
is an element of itself. But it _is_> basic that every set
includes itself, meaning that every> set is a subset of
itself.> @Christine:> In symbols:> For any set x:>
> x c x.> But (at least in Zermelo-Fraenkel.a6s ZF):> For
any set x:> not (x e x) ,> which is equivalent with>
There is no set x such that:> x e x.That.a6s because they
weren.a6t able to ?ure out how to de?e a setthat contains
itself. But I solved that problem a few months ago:
Letf(x)={x(x)}. Then f(f)={f(f)} so that f(f) is an element
of itself.Just think about recursion theory in which a
program can outputitself. set=program, element=output.If we
think in general terms, then sets, functions, predicates,
lists,trees, graphs etc. are each arbitrary mathematical
domains that hasits own paradox and incompleteness and needs
to be de?ed in relationto the others.Using my notation, it
is easy to generate Godel.a6s 1st IncompletenessTheorem (based
on soundness) and dozens of other theorems related toit. Some
of these theorems have been published (e.g. Smullyan.a6s
DualForm theorem) while others seem to be new.Compare this to
books and papers that claim to be a system thatgenerates
exactly one theorem!Charlie VolkstorfCambridge,
MA
===
-------------------------------------------------------
--------------I can do question #1 but I.a6m not sure about #2.
Can someone please check my answer for me?M means LamdaDave
is taking a multiple-choice exam. You may assume that
thenumber of questions is in?ite. Simultaneously, but
independently, his conscious andconscious and subconscious
are always working on different questions.) Conscious
responsesare generated at a rate of Mc responses per minute.
Subconscious responses aregenerated at a rate of Ms
responses per minute. Each conscious response is an
independentBernoulli trial with probability pc of being
correct. Similarly, each subconsciousresponse is an
independent Bernoulli trial with probability ps of being
correct. Daveresponds only once to each question, and you
can assume that his time for recordingthese conscious and
subconscious responses is negligible.(1) The papers are to be
collected as soon as Dave has completed exactly n
responses.Determine:(i) The expected number of questions he
will answer correctlyAnswer: n*Mc*Pc/(Mc + Ms) + n*Ms*Ms/(Mc
+ Ms) where M means Lamda(ii) The probability mass function
for L, the number of questions he answerscorrectly.Answer:
Each question is answered correctly with probability P =
Mc*Pc/(Mc + Ms) + Ms*Ms/(Mc + Ms). So, P( L = K ) = ( n C k
)* P^k ( 1 - P)^ (n - k)(2) Repeat part (f) for the case in
which the exam papers are to be collected at theend of a
?ed interval of t minutes.Answer?: P( L = K ) = summation
from n = k to oo [e^(-Mt)*Mt^n/n! * P^k( 1-p)^(n - k)]