mm-331
===
Subject: Re: Final Rout of Synchronization Clocks in
Relativity
>The only parameters used in the calculation is the
gravitational
>field of the Earth, and the speed and altitude of the clocks.
>> So what is the actual connection between a few maths
equations and
the physical
> change in clock rates that is observed when they are in
free fall?
The connection is that the cause of the proper times of the
clocks
>>is to be found in the gravitaional field of the Earth and
the speed
>>and altitude of the clocks, and that GR correctly accounts
for these.
Does that mean that the same clocks would not work if they
were not in
the
>> Earth's field?
When cornered, talk nonsense, eh?
> Are you jotting down your own thoughts, Paul?
Indeed I am.
My thought's are::
When Henry Wilson is cornered, he talk nonsense.
He does of course talk nonsense when he isn't cornered too.
>> Or does it mean that one free fall is as good as any
other, as far as
clocks
>> are concerned?
Gravity probe A?
>HahA!
>So there!
> It doesn't work.
OBVIOUSLY not.
HahA!
Can't refute that!
>>Kick and scream all you want, Henry.
Claiming that a correction calcula by GR is
>> NOT a relativistic correction and that the cause
>>of how the clocks behave is something not accoun
>>for in the equations that predict the behaviour correctly
>>to a precision better than 10^-12 is a stupidity beyond
belief.
If that is true, which I very much doubt because it was never
tes
properly,
>> it is purely coincidental. The prediction is wrong for
other orbits
anyway.
For instance, it MUST be wrong for an orbit 1 centimetre
above the
Earth's
>> surface since the gravitational potential is the same
there as o the
ground and
>> since we know that clock rates are not affec by movement.
(to
discourage
>> smartarse replies from the idiots here - you know who I
mean - assume
the earth
>> is perfectly round and has no atmosphere)
So a clock circling the Earth at that height would not run at
a
different rate
>> from one at rest and so it woud remain always in synch
with the GC.
Beautiful, Henry.
>Thanks for yet another great demo!
> Oh thank you, Paul. Yes, I thought it was quite clever
myself.
> It completely annihilates your argument and brings down
your religion.
Of course you assertion of what the outcome of an impossible
experiment would be is fatal to GR.
After all, you KNOW!
>Henry Wilson knows that the outcome of an unfeasible
experiment
>never done MUST support Henry Wilson's idea of reality,
>and thus will falsify GR.
> Do you not agre that the gravitational potential 1cm
(better still, 1 um)
ABOVE
> the surface is near enough to that ON the surface, for the
purpose of
this
> experiment?
Sure, Henry.
As I said, since you know that the outcome of this experiment
MUST falsify GR, there is nothing more to discuss.
This is simply too stupid to be funny, Henry.
I am bored by this game.
Paul
===
Subject: Re: Polynomials and the AC Method
Adjunct Assistant Professor at the University of Montana.
>I am moving forward with my math studies. I am working on a
text
>called Intermediate Algebra. I recently was working on
systems of
>three equations. I encountered for the first time matrix
operations.
>The operations themselves are pretty easy. I think you just
have to
>memorize them. The hard part is understanding why they work
(which I
>don't really intend to do).
>Otherwise, I have a math question:
>There is a method to solve polynomials called the ac method.
If you
>have a polynomial of the form ax^2+bx+c, you can factor this
by
>substituting m and n for b, such that m+n=b and mn=ac.
I don't understand what substituting m and n for b means!
My first guess would be to take ax^2+bx+c and replace it with
ax^2+mx+c and ax^2+nx+c; that's the obvious meaning of
substituting
m and n for b in my mind. But this makes no sense if you are
talking
about factoring this polynomial.
>Here is my question: the method is pretty easy, and works.
But I can't
>figure out why it works. I see why m+n must equal b. This is
obvious.
>But where does the requirement that mn=ac come in? There is
no
>obvious reason that I can see.
If I knew what you meant, perhaps I could help...
--
===
Subject: Re: Automobile Speeds.
>Does anyone know why automobile speeds are measures in miles
per hour
>>rather than <say> feet/sec or perhaps meters/sec?
> They are not measured in miles per hour.
> They are measured in kilometers per hour.
> But YMMV.
Dirk Vdm
Don't you mean YKMV?
btw, what is the form of kilometer that corresponds to
mileage?
Kilometerage?
===
Subject: Re: JSH: Limi intellects
> I've concluded that the problem I'm facing 
is that the
human brain
> isn't built to handle Mathematics. Human beings have done
well with
> some simple mathematics and basic concepts, but it only
goes so far,
> and what have, is beyond those limits.
>
I thought of myself as calm and cool about that assessment.
Still I'm facing an incredibly strange situation which 
defies
easy
explanation.
Consider, mathematicians *supposedly* care about pure math
and the
importance of mathematical knowledge, but have for close to
two years
now obstinately refused to accept some rather dramatic
results.
Rather than face intrigued interest and help in further
research I've
been ignored and insul, as if mathematicians *hate* me for
finding
what I have.
It's like there's some kind of odd war, where 
they will fight
no
matter what, against all reason, against the very values they
claimed
to have, and for what?
That's what I guess I need to figure out now.
===
Subject: Re: Action Device Tragedy
http://www.geocities.com/inertial_propulsion
17 years ago, in 1986, when I was 17 year old, I ran away
from one of
>the most prestigious engineering college in India, VJTI
Bombay. Reason
>was, I don't want to be Engineer. I want to become Pilot.
through which we can hang the things in air. We can build
air vehicles
>and any one can become Pilot.
But I am back to square 1. I will have to become Engineer to
build
>this action device. There are no mechanical engineers in any
workshop
>in this town. All you people in world are forcing me to
build this
>device myself even though you are aware that at least one
person has
>built one of the parts of this device and it works. Shame!
Shame!
>Shame!
I don't want to be Engineer. I want to be Pilot, you 
Morons!
I am not going to build this action device. That spring in
hardware
>shop struck to my finger very hard.
-Abhi.
> What little is left of your mind is missing again.
This individual is permit to ßy aircraft?
Now thats a comforting thought.
-Pragmatist -
===
Subject: Re: Action Device Tragedy
> http://www.geocities.com/inertial_propulsion
17 years ago, in 1986, when I was 17 year old, I ran away
from one of
> the most prestigious engineering college in India, VJTI
Bombay. Reason
> was, I don't want to be Engineer. I want to become Pilot.
through which we can hang the things in air. We can build air
vehicles
> and any one can become Pilot.
But I am back to square 1. I will have to become Engineer to
build
> this action device. There are no mechanical engineers in
any workshop
> in this town. All you people in world are forcing me to
build this
> device myself even though you are aware that at least one
person has
> built one of the parts of this device and it works. Shame!
Shame!
> Shame!
I don't want to be Engineer. I want to be Pilot, you Morons!
I am not going to build this action device. That spring in
hardware
> shop struck to my finger very hard.
-Abhi.
Abhi,
Can you not see the hand of fate here?
You were obviously destined to become the greatest and most
innovative
of all the engineers that ever were, - and you gave it up.
Now the task is forced upon you. - Kismet!
Good luck with your new career. Write if you get work.
- Pragmatist
(The above passage is carefully calcula to deprave the cultiva
reader.)
===
Subject: Re: Automobile Speeds.
>Does anyone know why automobile speeds are measures in
miles per hour
>>rather than <say> feet/sec or perhaps meters/sec?
>
They are not measured in miles per hour.
> They are measured in kilometers per hour.
> But YMMV.
Dirk Vdm
>
Don't you mean YKMV?
Hardly, afaik there's only one kilometer ;-)
> btw, what is the form of kilometer that corresponds to
mileage?
> Kilometerage?
In French and ÔFlemish': kilometrage.
In Dutch: kilometerstand.
Dirk Vdm
===
Subject: Factorials
Here is a little problem i do not know answer of.
(I am preparing for Programming Contest).
Find the rightmost non-zero digit on m! followed by
the number of zeros after that digit
(1 <= m <= 10^6)
Could anyone provide a mathematical solution to it?
P.Krumins
===
Subject: Re: Monotonic function
> Situation:
> Let y = a^n + b^n (1) where (a, b) = 1, b > a > 1 and odd n
3.
> Let a1, a2 are the two values of a and b1,b2 are the two
values of b
This, then, doesn't make much sense to me. If 
I'm reading you
> incorrectly and they aren't integewhat are they?
Here a2 > a1 and b2 > b1
> y1 = a1^n + b1^n (2) and y2 = a2^n + b2^n (3)
> Therefore, y2 > y1
> Statement:
> y in (1) also increases monotonically for a given odd n.
This is another problem with viewing a and b as integers. How
does
> their sum increase linearly?
Another way to look at this problem is that choose a1,a2,a3,
.... and
b1, b2, b3, .... so that a1 + b1 < a2 + b2 < a3 + b3 < ....
Then y1<y2<y3 < ..... etc etc.
> Any comment upon the correctness of the above statement
will be
> apprecia.
===
Subject: Re: JSH: My quick math quide
Adjunct Assistant Professor at the University of Montana.
>> [.snip.]
> Proofs that your definition of coprime and 
Virgil's definition
>are equivalent in the algebraic integers have been known
since the
>time of Dedekind.
> You have never shown the slightest awareness of the
Euclidean
>algorithm, one of the absolutely basic tools in algebraic
number
>theory and the real reason that the definition that Virgil
uses
>for coprime is the standard accep definition.
>>No, the Euclidean algorithm is not really at issue here.
> Yes I know that. The point was, Harris has never shown any
>awareness of it or of the fact that in the integethe E.
algorithm
>is used to show that c = GCD(a,b) can be written as a*s +
b*t = c.
>He has seen this often enough in many posts but has never
>shown any sign of appreciating it, and my guess is it just
>went over his head every time. As you may recall he never
>understood the correct proof of Area One. His view of coprime
>is strictly in terms of factors. His view of Virgil's 
second
>step may have been that it had nothing to do with coprimeness
>as he understands it, so Virgil must have been assuming what
>he wan to prove.
I would say that the difficulties may arise even earlier than
the
Euclidean algorithm, at the division algorithm itself. See
for example
his first reply to the proof about minimal polynomials of
algebraic
integeback in December 2002:
http://groups.google.com/groups?selm=
3c65f87.0212201947.1587a25%40posting.go
ogle.com
and similar replies along the same lines which he had given
to earlier
attempts at showing him the proof.
> Yes, I completely agree on that. In fact it seems like
>incredible luck that gcd's can be found at all in the
algebraic
>integers. I wonder if something even deeper, not yet
discovered,
>might be another explanation.
Here I agree with that it seems likely the underlying
reason here is the existence of a Hilbert Class Field.
--
===
Subject: Re: Would you recommend a school of math for a
soon-to-be college
student
> If you think you have the qualifications, you should go for
the
>> absolute top schools.
Don't necessarily agree. U.C. Berkeley has the one of the
top-ra math
> departments in the world, but freshman and sophomore
classes are still
> taught as huge lectures. If you went to a community college
for two
> years and transferred in as a junior you'd be just as well
off. Being a
> freshperson at Berkeley has its pleasures and benefits, but
academically
> you're just another mad cow in a real big herd. Same goes
for any other
> large state university.
I disagree. If the OP has Talent, a school with a
research-orien math
department is a much better choice, even as a freshman. A CC
is likely to
have no math courses beyond first-year calculus, and certainly
no proof-
orien courses. A talen undergrad should be in courses that
the CC
doesn't offer certainly as a sophomore, probably as a
freshman.
===
Subject: Re: Dishonesty
I note that dishonesty has become the normal thing for
ÔBilge'.
> [snip] < stupid typical response from a moron.
> [unsnip]
> He makes the return address something other than
sci.physics.relativity,
> such as alt.troll or alt.moron.
> Obviously such a cheap trick can be considered amusing, but
wears thin
when
> repea.
> He claims to be able to Ôderive' 
sqrt(1-v^2/c^2) as though
it were
something
> special that he alone can do, using the trigonometric
functions of a
conic
> section.
You have me there. What are the trigonometric functions of a
conic
section?
> Obviously this is the full extent of his mathematical
capabilities,
which,
> when exhaus, causes him to resort to personal abuse and
cheap
practical
> jokes as his only counter argument when his absurd claims
are challenged.
It is not a cheap practical joke. It is a highly amusing
thing to do with
the likes of you. Just look at the lovely response it drew
from you!
[snip]
Franz
Franz
===
Subject: Re: area of a spherical right triangle
In-reply-to: Michael JÀrgensen <ingen@ukendt.dk> following formula for the area of a spherical right
triangle is derived:
>> tan(E/2) = tan(a/2) tan(b/2)
>> Where E is the area (spherical excess) and a and b are the
legs of the
>> right spherical triangle.
>> I have not seen this formula before and I have been unable
to find it
>> on the web or in the books that I have searched. Does
anyone have a
>> reference for it? I can't imagine that such a simple
formula has not
>> been recorded before.
>It's a nice formula, but it can be derived from the other
well-known
>formulas in spherical trigonometry:
>The half-side formulas are
><http://mathworld.wolfram.com/SphericalTrigonometry.htmltan(a/2) = K * cos(S-A) and
>tan(b/2) = K * cos(S-B),
>where K^2 = - cos(S) / [cos(S-A) * cos(S-B) * cos(S-C)], and
2S = A+B+C =
E
>+ pi.
>Multiplying the two equations and inserting C = pi/2 leads
to the above
>result.
>Hope that helps.
Thanks for the reply.
I don't doubt that the formula is true or is provable from
other
above; I was seeing where the polar form for area would lead.
When I
saw how nice the formula was, I was surprised that I had not
seen it
before; I was curious whether anyone else had seen it before.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: a base for R?
> For instance, as the union of all (a,b) where -pi < a < b <
pi.
a, b rational of course.
===
Subject: Re: Explicit prime counting formula, complexity
> The way I calculate explicit prime formulas is
straightforward, as I
> star with
dS(N,3) = ßoor(N-4)/6. with even N
and itera.
So
N/2 - ßoor((N-4)/6) - ßoor((N-16)/10) + ßoor((N-16)/30) -
> ßoor((N-36)/14) + ßoor((N-22)/42) + ßoor((N-106)/70) -
> ßoor((N-106)/210) + 2.
is the result of two iterations covering 5 and 7.
I'm curious now about whether or not there are smaller
expressions
> that do what it does out to infinity as I sat down and
figured out how
> terms get added.
All these problems that are so hard on your poor little brain
are nicely
solved on my webpage, together with (shall I mention it? ) an
implementation of a prime counting algorithm that is about
1000 times
faster than anything you've ever crea...
> In general, with x=sqrt(N) there are approximately 2^{x/ln
x}/ln N
> terms.
Wrong. I mean massively, completely wrong.
===
Subject: Re: JSH: Limi intellects
> Consider, mathematicians *supposedly* care about pure math
and the
> importance of mathematical knowledge, but have for close to
two years
> now obstinately refused to accept some rather dramatic
results.
Mathematicians _do_ care about mathematics. What you produce
is pure
bull and has nothing to do with mathematics whatsoever, and
there
have been definitely no results, whether dramatic or
undramatic.
> Rather than face intrigued interest and help in further
research I've
> been ignored and insul, as if mathematicians *hate* me for
finding
> what I have.
Nobody could possibly hate you for finding what you have,
because you
have never found anything. People hate you because you are a
prick,
that's all.
> It's like there's some kind of odd war, 
where they will
fight no
> matter what, against all reason, against the very values
they claimed
> to have, and for what?
That's the problem with having a condition that needs
psychiatric
attention: You have lost all connections to reality, and you
can't see
things the way they are.
> That's what I guess I need to figure out now.
Good luck.
===
Subject: Re: JSH: Focusing on counting prime numbers
I've never seen Shaq attack, and
I've never seen I Spy. I also don't know any 
marketing!...
But, it looks like they're about to converge
into one, big, middle digit!
> Isn't replying to your own posts considered trolling?
Sure, [my Usenet presence is] like Shaq playing against you
in your
backyard, but that has its perks, as I find ways to have my
fun *and*
I can send messages to certain people in the Uni States
Government
without concern that the rest of you understand them. --
--YOU don't *have* to be a rocket-scientist -- I beg you!
(my highschool career counselor's last day)
===
Subject: Re: Factorials
Use Congruence.
 Here is a little problem i do not know answer of.
> (I am preparing for Programming Contest).
> Find the rightmost non-zero digit on m! followed by
> the number of zeros after that digit
> (1 <= m <= 10^6)
> Could anyone provide a mathematical solution to it?
 P.Krumins
===
Subject: Re: Polynomials and the AC Method
X-ID: XjSi8vZbgeRl1uvwzpf6gry2W9YrIT2kqQ+1CGf1JOnqlLt7lLPXg8
I am moving forward with my math studies. I am working on a
text
> called Intermediate Algebra. I recently was working on
systems of
> three equations. I encountered for the first time matrix
operations.
The operations themselves are pretty easy. I think you just
have to
> memorize them. The hard part is understanding why they work
(which I
> don't really intend to do).
Otherwise, I have a math question:
There is a method to solve polynomials called the ac method.
If you
> have a polynomial of the form ax^2+bx+c, you can factor
this by
> substituting m and n for b, such that m+n=b and mn=ac.
>
Assuming the form ax^2+bx+c = a*(x-n)*(x-m), we get on the
right-hand-side
ax^2 - a(n+m)x + anm
So n and m such that n+m = -b/a and nm = c/a is the right
choice.
> Here is my question: the method is pretty easy, and works.
But I can't
> figure out why it works. I see why m+n must equal b. This is
obvious.
> But where does the requirement that mn=ac come in? There is
no
> obvious reason that I can see.
--
Just because you're paranoid
Don't mean they're not after you
reverse my forename for mail!
===
Subject: Re: Riemann Surfaces in Analysis
As posts to the thread by others seem to have ended I would
like to
say I am grateful to my critics for helping me to clarify my
ideas and
to remark that the useful part of what I had to say is collec
on my
website www.riemannsurfaces.info.
===
Subject: Logs, bases and Casio calculators
for my calculations is not allowed for an exam I am sitting
shortly.
Therefore I have been forced to borrow a less complica Casio
FX-83WA.
I'm familiar with converting from base 10 logs to other
bases, but does
anyone know wether the Casio family of characters has a
button or syntax
for specifying the base? Surely I don't have to perform 
manual
calculations each time I want a base other than 10.
Many thanks in advance.
Jeff
===
Subject: Re: Dishonesty
>
I note that dishonesty has become the normal thing for
ÔBilge'.
[snip] < stupid typical response from a moron.
[unsnip]
> He makes the return address something other than
sci.physics.relativity,
> such as alt.troll or alt.moron.
> Obviously such a cheap trick can be considered amusing, but
wears thin
> when
> repea.
> He claims to be able to Ôderive' 
sqrt(1-v^2/c^2) as though
it were
> something
> special that he alone can do, using the trigonometric
functions of a
conic
> section.
> You have me there. What are the trigonometric functions of
a conic
section?
I would suggest you refer that question to Bilge, he's the
advocate of
applying them to special relativity. I am not responsible for
your
ignorance.
> Obviously this is the full extent of his mathematical
capabilities,
which,
> when exhaus, causes him to resort to personal abuse and
cheap
practical
> jokes as his only counter argument when his absurd claims
are
challenged.
> It is not a cheap practical joke. It is a highly amusing
thing to do
with
> the likes of you. Just look at the lovely response it drew
from you!
As I said, it can be considered amusing once and I can accept
it as a joke.
When it becomes repetitive it simply becomes boring, childish
and pathetic,
which aptly describes you as well.
Androcles
===
Subject: Re: Dishonesty
> I note that dishonesty has become the normal thing for
ÔBilge'.
> <dubious@radioactivex.lebesque-al.netHe makes the return
address something other than
sci.physics.relativity,
> such as alt.troll or alt.moron.
> He's always done that. Nothing new for bilge. It just 
means
that he
> either doesn't want to discuss it or he's 
unable to figure
something
> out so he doesn't want to be anymore embarassed than he
alread is.
I find your analysis to be accurate and concur.
Androcles
===
Subject: Re: Dishonesty
> Being unable to correspond with a fool that indulges in
such childish
> behaviour and dishonest claims, I merely ignore any further
communiqu.8es
he
> may post.
> Promise?
No.
Androcles
===
Subject: Re: Automobile Speeds.
> Does anyone know why automobile speeds are measures in
miles per hour
> rather than <say> feet/sec or perhaps meters/sec?
Because the distances on road signs are given in miles. So if
the
sign says
Chicago 125 mi
and you're driving at 55 mph, your travel time is a little
over
2 hours.
Try this in your head:
Madison 633,600 ft
and your speed is
88 ft/sec
what's your travel time?
===
Subject: delayed boolean equations
Hi:
I was wondering if someone could give me a hint on the
following
delayed Boolean equation. Actually, this WAS a homework
assignment
from a long time ago, but I've never been able to 
figure it
out. Now that
I got time to kill, I want to figure this out.
Suppose
_ _
V(t) = V(t-t1) & V(t-t2) & V(t-t3) (1.1)
t1 > t2 > t3
overbar means the opposite state of V and & means AND
operation.
Is there a way to figure out the period of the system in
general in
the ASYMPTOTIC limit?
I tried using (1.1) to figure out what V(t-t2) is:
_ _
V(t-t2) = V(t-t1-t2) & V(t-2t2) & V(t-t3-t2) (1.2)
If we subsitute (1.2) into (1.1) and keep repeating in
subsituting for
V(t-n*t2) where n is an integer (depending on the
number of (n-1) times the subsitution you did for V,
sooner or later, you get a recursive relation, quite ugly:
_ _ _
V(t) = V(t-t1) & V(t-t1-t2) & .... & V(t-t1-n*t2) & V(t-n*t2)
&
_ _ _
V(t-t3) & V(t-t3-t2) & .... & V(t-t3-n*t2)
I argued that in the asymptotic limit, the period of the
system
is t2 because the above has to be true for all n.
My TA marked me wrong, but I never find out the answer.
Could somebody solve the mystery for me? Or provide a hint?
Thanks. --- Angel
===
Subject: Re: Final Rout of Synchronization Clocks in
Relativity
Expires: 28 days
> For instance, it MUST be wrong for an orbit 1 centimetre
above the
Earth's
> surface since the gravitational potential is the same there
as o the
ground and
> since we know that clock rates are not affec by movement.
(to
discourage
> smartarse replies from the idiots here - you know who I
mean - assume
the earth
> is perfectly round and has no atmosphere)
>> So a clock circling the Earth at that height would not run
at a
different rate
> from one at rest and so it woud remain always in synch with
the GC.
Beautiful, Henry.
>>Thanks for yet another great demo!
>> Oh thank you, Paul. Yes, I thought it was quite clever
myself.
>> It completely annihilates your argument and brings down
your religion.
>Of course you assertion of what the outcome of an impossible
>experiment would be is fatal to GR.
>After all, you KNOW!
>>Henry Wilson knows that the outcome of an unfeasible
experiment
>>never done MUST support Henry Wilson's idea of reality,
>>and thus will falsify GR.
>> Do you not agre that the gravitational potential 1cm
(better still, 1 um)
ABOVE
>> the surface is near enough to that ON the surface, for the
purpose of
this
>> experiment?
>Sure, Henry.
>As I said, since you know that the outcome of this experiment
>MUST falsify GR, there is nothing more to discuss.
>This is simply too stupid to be funny, Henry.
>I am bored by this game.
>Paul
This message typifies your tactics when you find 
yourself well
and truly
cornered. You resort to ridicule, jokes, anything.... that
will divert
attention from the question you cannot answer.
Paul, I'll ask you another question.
Is the drum of your washing machine at rest wrt your house
when it spinning
dry?
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
===
Subject: Re: Logs, bases and Casio calculators
> for my calculations is not allowed for an exam I am sitting
shortly.
> Therefore I have been forced to borrow a less complica
Casio FX-83WA.
> I'm familiar with converting from base 10 logs to other
bases, but does
> anyone know wether the Casio family of characters has a
button or syntax
> for specifying the base? Surely I don't have to perform
manual
> calculations each time I want a base other than 10.
> Many thanks in advance.
> Jeff
As far as I know, if you want a different base, you need to
apply the
change-of-base formula. If you don't know what this is, any
good College
Algebra book will have it or Google for it.
===
Subject: Re: Automobile Speeds.
> Does anyone know why automobile speeds are measures in
miles per hour
> rather than <say> feet/sec or perhaps meters/sec?
Units are for the convenience of humans, and the SI is not
Procrustes'
bed. If I set out to drive 100 kilometers at a speed of 100
kilometers
per hour, I know I will arrive at my destination in 1 hour --
and I
can determine that with only minimal arithmetic. But if I set
out to
drive 100 kilometers at a speed of 30 meters per second, then
the same
calculation now requires the sort of arithmetic that it is
important
not to perform while traveling at such speeds.
On the other hand, if I'm doing crash test research, it may
be much
more convenient for me to express speed in meters per second.
--
Chris Green
===
Subject: Re: Proposal for responding to idiots
> idiot
> one word: killfile
another word: filter
Use whichever one your non-Googled newsreader has available.
The problem is when otherwise highly valuable contributors to
the
group lower themselves (in my view) to engaging the idiots.
I cannot bring myself to killfile those from which I could
learn
who have the nasty habit of engaging the idiots, which means
I'm
left with a 50% portion of the threads I'd rather not see.
(who believes that the JSH-bot was a good idea)
--
Unpatched IE vulnerability: NavigateAndFind file proxy
Description: c-domain scripting, cookie/data/identity
theft, command execution
Reference:
http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ
PU-Content.HTM
Exploit:
http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ
PU-MyPage.htm
===
Subject: Re: Final Rout of Synchronization Clocks in
Relativity
Expires: 28 days
>The MMX confirms SR.
>> Well, indirectly, yes. SR relies on source dependency. It
clearly
requires that
> light travels at c from any source. Any observer at rest
with that
source will
> receive the same light at c.
> If the light is returned to the source by a mirror at rest
wrt the
source that
> the return travel time of that light will be constant,
irespective of
the speed
> of the system.
It is an indisputable fact that the MMX confirms SR.
>>That's why it isn't dispu.
>> The MMX proves source dependency, fair and square. No
doubt about it.
>Amen.
QED
>The MMX confirms source dependent light theory
>> Absolutely correct. No doubt about that.
It is an indisputable fact that the MMX falsifies Michelsons
ether
theory.
>>That's why it isn't dispu.
>> The reason it isn't dispu is that its NULL result was
easily
manipulable to
>> give the answer that the physics mafia wan.
>So in 1886 Michelson was part of a physics mafia that wan
>to falsify the ether to prove that the establishment were
wrong.
>So that's why they manipula the highly unwan NULL result
>to give the answer they didn't want, but which the 
mafia wan.
>Right?
Wrong Paul. Physics was an honourable profession in the 19th
century.
>> Null results prove very litle if anything.
>> They usually reveal a ßaw in the experiment.
>> My demo shows that ßaw. IF the michelson aether exis, then
the 45
mirror
>> would not remain at 45 after its length was contrac. The c
beam
would be
>> bent backwards as shown, thus exactly compensating for the
presumed
Ôdiagonal
>> effect'..
>AH! Length contraction in Michelson's ether!.
>So THAT's what your demo now shows.
Where is the joke? In aether theories, lengths DO physically
contract. If
the
45 mirror contracts in the parallel direction, it obviously
ends up sloping
greater than 45.
Even though my demo was not based on that fact, it clearly
shows that it
happens.
>You are SO funny, Henry!
If one length contracts, then ALL lengths must contract. eh?
Where is the
joke?
Every experiment MUST support [my idea of] reality, ie, source
dependency.
>>Thus even the experiments falsifying source dependency MUST
support it!
>>So there!
>> There are NO experiments that falsify source dependency.
De Sitter was
wrong
>> and there are no others.
>Every experiment MUST support reality, ie, source dependency
>So the experiments that falisfies source dependency
>simply CANNOT exist.
>So there!
There is not one believeable bit of experimental evidence
that refutes
source
dependency.
Even your own theory is based on it.
>Paul
Henri Wilson.
www.users.bigpond.com/hewn/index.htm
===
Subject: Re: Mathematics forums
Pos-And-Mailed: yes
===
> Are there any other mathematics forums? I just want to be a
part of a
> couple and this is the only one I can find. I am interes in
> tricky competion style problems or advancements in pure
mathematics.
> Thanks for any help....
Here's my list of Usenet news groups concerning mathematics.
They're lis roughly in order of activity, so sci.math comes
first.
sci.math
sci.logic
alt.math.undergrad
de.sci.mathematik (German)
aus.mathematics
alt.math
uk.education.maths
alt.math.recreational
sci.math.symbolic
sci.math.research (advanced)
fr.education.entraide.maths (French)
fr.sci.maths (French)
alt.algebra.help
k12.ed.math
alt.sci.math.combinatorics
soc.history.science
schule.mathe (German)
geometry.puzzles
geometry.research
geometry.pre-college
geometry.college
z-netz.wissenschaft.mathematik (German)
geometry.software.dynamic
alt.math.alonline
alt.sci.math.galois_fields
alt.algebra
Also some statistics and probability groups:
sci.stat.math
sci.stat.edu
sci.stat.consult
alt.sci.math.probability
alt.sci.math.statistics.prediction
HTH
Ken Pledger.
===
Subject: Generators of Prime
In a group, Z_p, where p is a prime, is it true that EVERY
number other
than
p is a generator of Z_p?
If so, can someone give me a proof?
===
Subject: Re: C^n proper inclusions problem
> The function f(x) = |x|*x^n is in C^n(R) but not in
C^(n+1)(R). If a is
in
> U, an open subset of R^k, then the function x -> f(x1 - a1)
is in C^n(U)
> but not in C^(n+1)(U).
I don't understand your notation :
What is x1 and a1? R denotes the reals?
===
Subject: Bijection between (0,1) and [0,1)?
I can't think of a one to one onto mapping between (0,1) and
[0,1). I
don't
expect it to be continuous. Plus is it only possible to
assert the
existence thereof as opposed to giving a specification for
such a mapping?
Am I missing something obvious?
===
Subject: Re: rank and eigenvalue
Suppose A is 3x3, and the 3 eigenvalues are 0,1,2.
Then AX=0X=0.
Suppose the geometric multiplicity of the eigenvalue 0 is 1.
This means AX=0 has a nonzero solution X.
If rank(A)=3, AX=0 must have unique solution X=0.
Hence we know rank(A)<=2.
Is this all info we can get?
eigenvectors are independent. This implies A is
diagonalizable. But
does diagonalizable have relationship with rank?
Thanks
>Given the eigenvalues, can we know the rank of a matrix?
I assume it's a square matrix...
Given the eigenvalues _with their algebraic multiplicities_,
we can
> (of course we only need the multiplicity of 0).
Oops: I meant geometric multiplicities.
Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Google's rankings
(1) http://www.nakedteens.com ==> 4/10
(2) http://www.datashaping.com ==> 0/10
(1) has multiple invading pop-ups and crudly displays poor
quality
pornographic material without asking your age
(2) has none of the above -- it's just a website for
mathematicians
(1) is better than (2) according to Google's page ranking
algorithm.
Visit (1) and (2) and judge by yourself.
> http://www.scientology.com ==> 4/10
> http://www.datashaping.com ==> 0/10
> This gives you an idea on how biased Google's ranking is,
favoring
> criminals over legitimate businesses.
===
Subject: Re: Google's rankings
> ...
> (1) is better than (2) according to Google's page ranking
algorithm.
> Visit (1) and (2) and judge by yourself.
Google PR technology has nothing to do with 
Ôbetter' or any
other value
judgement as you well know.
--
William Tasso - http://WilliamTasso.com
===
Subject: Re: Bijection between (0,1) and [0,1)?
>I can't think of a one to one onto mapping between (0,1) 
and
[0,1). I
don't
>expect it to be continuous. Plus is it only possible to
assert the
>existence thereof as opposed to giving a specification for
such a mapping?
>Am I missing something obvious?
Yes. For (0,1) -> [0,1) map 1/2 to 0, 1/4 to 1/2, 1/8 to 1/4
etc. and
everything else to itself.
--Lynn
===
Subject: Re: Bijection between (0,1) and [0,1)?
> I can't think of a one to one onto mapping between (0,1)
and [0,1). I
don't
> expect it to be continuous. Plus is it only possible to
assert the
> existence thereof as opposed to giving a specification for
such a
mapping?
Map (0, .9] linearly to [0, .9).
Map (.9, .99] linearly to [.9, .99).
Map (.99, .999] linearly to [.99, .999).
Etc.
--
===
Subject: Re: Bijection between (0,1) and [0,1)?
> I can't think of a one to one onto mapping between (0,1)
and [0,1).
Hint: If E is a countably infinite subset of (0,1), 
can't we
map E
bijectively onto E U {0}?
===
Subject: Re: Multiple-Sum Congruence
> I am betting that this is easily proved somehow,...maybe ...
(I wonder if I pos it already a long while back...)
Let each S(0,m) = a(m), which is any rational you like, as
long as
> a(m) is defined for every positive integer m, and each
{m!*a(m)} is an
> integer.
> Let S(n,m) = sum{j=1 to m} S(n-1,j)
for every positive integer n.
> then (perhaps), if q and r are fixed integers where 0<= q <
r:
m! *S(q,m)
is always congruent to
m! *S(r,m) (mod {r-q})
I may very well have made a mistake, since I was careless;
> but this result seems as if it *must* be a trivial
consequence of the
> fact that:
S(n,m) =
 sum{k=1 to m} binomial(m+n-k-1,n-1) * a(k).
But I myself did not use this particular sum to get the
result.
 Here is a proof using immediately previous sum.
> S(n,m) =
sum{k=1 to m} sum{j=0 to m-k} n^j *S(m-k,j) a(k) /(m-k)!,
where S() is an unsigned Stirling number of the 1st kind.
> (All that is important here is that S() is an integer.)
> So,
m!(S(r,m) - S(q,m)) =
sum sum S(m-k,j) *a(k) *k! *binomial(m,k) *(r^j -q^j).
So, if {a(k) *k!} = integer, each term of double sum is
integer multiplied
by
> (r^j - q^j), and (r^j - q^j) is divisible by (r-q).
QED
> Leroy
> Quet
UGG! I made, if not a mistake in my math, a mistake in my
notation...
I reused ÔS()' for 2 different things above in 
my proof.
I will rewrite the proof with each unsigned Stirling number
represen as T(m,n) instead.
S(n,m) =
sum{k=1 to m} sum{j=0 to m-k} n^j *T(m-k,j) a(k) /(m-k)!,
where T() is an unsigned Stirling number of the 1st kind.
(All that is important here is that T() is an integer.)
So,
m!(S(r,m) - S(q,m)) =
sum sum T(m-k,j) *a(k) *k! *binomial(m,k) *(r^j -q^j).
So, if {a(k) *k!} = integer, each term of double sum is
integer multiplied
by
(r^j - q^j), and (r^j - q^j) is divisible by (r-q).
QED
Leroy
Quet
===
Subject: Re: Dishonesty
> Gauge:
>> I note that dishonesty has become the normal thing for
ÔBilge'.
>> <dubious@radioactivex.lebesque-al.net> He makes the return address something other than
sci.physics.relativity,
>> such as alt.troll or alt.moron.
He's always done that. Nothing new for bilge. It just means
that he
>either doesn't want to discuss it or he's 
unable to figure
something
>out so he doesn't want to be anymore embarassed than he
alread is.
 Or your ranting is off-topic so follow-ups are set
appropriately.
> You always semm to overlook the obvious.
Incorrect as usual
===
Subject: Re: Would you recommend a school of math for a
soon-to-be college
student
>> If you think you have the qualifications, you should go for
the
>> absolute top schools.
Don't necessarily agree. U.C. Berkeley has the one of the
top-ra math
> departments in the world, but freshman and sophomore
classes are still
> taught as huge lectures. If you went to a community college
for two
> years and transferred in as a junior you'd be just as well
off. Being a
> freshperson at Berkeley has its pleasures and benefits, but
academically
> you're just another mad cow in a real big herd. Same goes
for any other
> large state university.
I disagree. If the OP has Talent, a school with a
research-orien math
> department is a much better choice, even as a freshman. A
CC is likely
to
> have no math courses beyond first-year calculus, and
certainly no proof-
> orien courses. A talen undergrad should be in courses that
the CC
> doesn't offer certainly as a sophomore, probably as a
freshman.
>
I agree with Mike that the CC and transfer plan doesn't 
sound
too good;
however, Fishfry makes a good point regarding the size of
classes.
While Berkeley is well-known to all mathematicians as a first
class
mathematical institution, I don't think that translates to 
it
being a
great undergrad math school. I think it's much better to go
to a
smaller school that is still ranked highly in mathematics,
and I would
suggest a university over a college (there are advantages and
disadvantages to going to a small college, but that's for
another
discussion). The smallness of a school can contribute heavily
to
familiarity with the faculty and lead to opportunities for
even a shy,
not so agressive, student. Of course, I agree that a student
with
talent can be very aggressive and make the most of
opportunities even
at a very large school, but I personally wouldn't have felt
comfortable
doing that.
===
Subject: Re: Rationals are Uncountable
http://en2.wikipedia.org/wiki/Cantor%27s_first_uncountability_
proof
Replace R wth Q (the rationals), or any other set dense in
the reals.
It would show the rationals uncountable, by the same logic
that the
reals were, of the logic of Cantor's first 
proof.
If Q is countable, as has been sugges by the existence of
functions
between N and Q, then what's the ßaw in the mentioned 
proof?
If Q is uncountable, yet countable, are countability and
uncountability not mutually contradictory?
I would think that somebody else has already mentioned this
point.
Where this is so, we could examine its contemporary arguments.
What is the meaning of the more general result of this proof
with
regards to any set dense in the reals?
===
Subject: Re: envelope stuffing probability problem (question)
There's your problem: For example, for P(E2' | 
E1'), you need
to
> consider whether envelope 1 has letter 2 or not.
P(E2' | E1 Ô) = P(E1' 
E2') / PE1'
For E1' E2', either envelope 1 has letter 2 
and envelope 2
has any of
> n-1 letteor envelope 1 has neither letter 1 nor 2 and
envelope 2 has
> any of n-2 letters. Thus,
P(E1' E2') = [n-1 + (n-2) (n-2)] / [n(n-1)] = 
(n^2 - 3n + 3) /
[n(n-1)]
> PE1' = (n-1) / n
> P(E1' | E2') = (n^2 - 3n + 3) / (n-1)^2
stephen,
doo'h (slapping forehead).
correct. thanks. this really did help.
j.
===
Subject: Re: Bijection Thanks for the quick answers?
Why didn't I think of that?
(0,1/2){1/2}(1/2,1/4){1/4}...--->{0}(0,1/2){1/2}...
> I can't think of a one to one onto mapping between (0,1)
and [0,1). I
don't
> expect it to be continuous. Plus is it only possible to
assert the
> existence thereof as opposed to giving a specification for
such a
mapping?
> Am I missing something obvious?
===
Subject: Re: Skeptickal Inquirer UFO
Sorry Borg fool, this time it's not me, it's 
you.
One perfectly good use for the moon, besides obtaining
he3/3he, or of
those super terrific VLA-SAR imaging, is for implementing an
affordable strategy upon interplanetary communications.
Though Venus nighttime atmosphere is still relatively hot and
nasty:
Unless you're a nocturnal lizard sort of folk, perhaps 
Cathar
and not
about to die off simply because your world is situa next door
to
the dumbest and most pathetic other planet in the universe.
Though as
downright embarrassing as that might be, it's certainly not
worth
dying over.
At least our inferior DNA/RNA might have to take the back
seat on this
one, staying within a protective cacoon in order to survive
their
environment, though since no one has been invi, I don't
understand
what all the fuss is about.
Apparently there's been well established results that only
further
proves my previous guestimates, such as the nearly 68 bar
working
environments crea by this following research lab, obviously
upon
humankind that were given every benefit of the doubt, 
gradually
exposed to this horrific pressure, sustained for a 
sufficient
period
of time while accomplishing various tasks, then thoughtfully
brought
back into Earth's surface environment, again with all due
respect for
the what ifs and of avoiding all the nasty issues of otherwise
dying. http://www.comex.fr/Pages1/Page6.html
Again, this is not me suggesting that we should be planning
upon going
out of our way to be visiting those Venus lizard folks, at
least not
before attempting a few long-distance but local
interplanetary area
code phone calls via laser packets, or perhaps a composite
analog
laser beam hosting a few teabytes worth of quanntum binary
packets, so
that the widest possible range of species and technologies
can detect
and hopefully devise something as crude and primitive in
order to
transmitt their reply, something like; please go away, or
else!
This notion of a laser beam capable of sufficiently
illuminating
through those relatively cool nighttime clouds is not rocket
science,
it's more or less like initially utilizing a near UV 
spectrum
(400~450
nm) as a pointer, though configured so as to obtaining the 
best
penetration. Although, once a data link is established (even
1 bps is
good for go), obviously depending upon the mutual technology
interface
or weakest link, which well probably be on our side of this
equation,
that spectrum may become almost anything.
Even crude radio can be implemen, although I don't know why
they'd
bother coming down to that level, unless we supplied the
interactive
kiosk for them to utilize, as radio might be somewhat of an
antique
road show as far as they're concerned, somewhat like if you
were
handed a 78 LP or perhaps an older dicta-phone cylinder, and
of those
giving you this analog record made of hard wax were waiting
to see you
smile, as though you were somehow impressed by the level of
their
technology.
OK, at this point I'd certainly take whatever I could get,
even if it
were a couple of soup cans using string, especially if at the
other
end was a Venus ET trying to respond, even though for decades
we've
all been too snookered by folks having the right stuff, so
much so
that we didn't even bother to take Osama bin Laden 
seriously,
nor of a
few dozen other clues, starting off with the USS LIBERTY
fiasco, nor
of GW Bush and of his Salem bin Laden business dealings, not
to
mention his preemptive tactics upon overthrowing Iraq from
the very
get-go.
If you think you can contribute to any of these issues, or to
this
novel of life and consequences, I'm all eathough 
I've been
told
that there's not all that much between them ears because, my
Borg
collective interface has been broken down for at least the
past three
years and counting, though your's is probably still fully
functioning
and synchronized to the collective.
http://mittymax.com/Archive/0085-
SaddamHusseinAndTheSandPirates.htm
The latest round of insults to this Mars/Moon/Venus class
action
injury:
http://guthvenus.tripod.com/gv-what-if.htm
Some other recent file updates:
http://guthvenus.tripod.com/gv-gwb-moon.htm
http://guthvenus.tripod.com/gv-interplanetary.htm
http://guthvenus.tripod.com/gv-illumination.htm
http://guthvenus.tripod.com/gv-moon-02.htm
http://guthvenus.tripod.com/moon-04.htm
Brad Guth / IEIS~GASA
===
Subject: Re: rank and eigenvalue
>Suppose A is 3x3, and the 3 eigenvalues are 0,1,2.
>Then AX=0X=0.
>Suppose the geometric multiplicity of the eigenvalue 0 is 1.
It would have to be (each eigenvalue has multiplicity at
least 1,
and the total is 3).
>This means AX=0 has a nonzero solution X.
>If rank(A)=3, AX=0 must have unique solution X=0.
>Hence we know rank(A)<=2.
>Is this all info we can get?
No, we know the rank is exactly 2: the geometric multiplicity
of the
eigenvalue 0 is the nullity, i.e. the dimension of the
nullspace, and
rank = column dimension - nullity.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: u is eig vector associa with eig value zero. Ax=u
has no
solution?
Given a matrix A, suppose u is an eig vector associa with eig
value
zero. Can we prove that Ax=u has no solution?
Thanks
===
Subject: Re: u is eig vector associa with eig value zero.
Ax=u has no
solution?
>Given a matrix A, suppose u is an eig vector associa with
eig value
>zero. Can we prove that Ax=u has no solution?
Hint: try
[ 0 1 ]
A = [ 0 0 ]
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: DC Proof: Rationale for a more User-Friendly
Notation
Richard,
I do list the order of precedence in the user manual. (Look
up Order of
precedence in the index). If, however, you have a look at the
sample
proofs
in the Tutorial, I think you will agree that that order of
precedence is
quite natural. I don't think the beginner would even be 
aware
of it
initially. The alternative would be a confusing tangle of
brackets that
would obscure the meaning of statements. It makes quite a
difference.
Dan
Visit DC Proof Online at http://www.dcproof.com -- FREE
Download (new
release today was temporarily unavailable)
> You may think that precedence is natural, but it is not,
once
> you get beyond + and *. In fact, you don't even need to go
that
> far in tex, sum_{0le i le n} f + g .
> either sums f from 0 to n , and then adds g
> or sums f+g from 0 to n.
> the precedence of all() and exists() would hardly be
clearcut
> to someone just learning the subject.
> Good luck.
> RJF
> (fixed typo)
>
Last month here, I announced my new proof-writing program, DC
Proof.
Some
> users have commen on its use of nonstandard notation.
Following is a
> rationale for the notation I have adop (also pos at my
homepage).
Dan
Visit DC Proof Online at http://www.dcproof.com -- FREE
Download (new
> release yesterday)

DC Proof: Rationale for a more User-Friendly Notation
While it may not be the standard notation of logicians, in
most cases,
the
> notation used by DC Proof does not vary significantly from
it. It was
> designed primarily for the non-specialist mathematics
student in first
or
> second year university or in advanced senior high school
courses.
To give just one example, most university algebra and calculus
(analysis)
> textbooks, even the advanced ones it seems, do not make use
of
universal
and
> existential quantifiers (the logician's inver 
A's and
backwards E's
> respectively). They usually spell out for all x or there
exists a
y
such
> that etc. Students should be used to working with such
notation. It
was
> felt that the use of ALL and EXIST would be a more natural
and
> suggestive notation.
By the way, users need not type out these words or commit the
notation
to
> memory. To minimize keystrokes and to help students learn
the notation,
> there is a convenient, pull-down menu of all symbols and
notation used
in DC
> Proof (click Notation on the Premise form, etc.).
Also, in a effort to make variable and predicate names more
meaningful,
DC
> Proof allows names of any length. This, of course,
necessitates the use
of
> some extra delimiters (round and square brackets) for
readability. DC
Proof
> automatically minimizes bracketing, however, with its
built-in, order
of
> precedence (or operations). This makes it easier for the
user to
recognize
> the occurrence of structures like double negations, for
example.
Generally speaking, the DC Proof notation was designed to be
as
familiar
as
> possible to students, and to make it easy to learn and
master.
>
Return

>
===
Subject: Re: DC Proof: Rationale for a more User-Friendly
Notation
You may think that precedence is natural, but it is not, once
you get beyond + and *. In fact, you don't even need to go
that
far in tex, sum_{0le i le n} f + g .
either sums f from 0 to n , and then adds g
or sums f+g from 0 to n.
the precedence of all() and exists() would hardly be clearcut
to someone just learning the subject.
Good luck.
RJF
> (fixed typo)
> Last month here, I announced my new proof-writing program,
DC Proof. Some
> users have commen on its use of nonstandard notation.
Following is a
> rationale for the notation I have adop (also pos at my
homepage).
Dan
Visit DC Proof Online at http://www.dcproof.com -- FREE
Download (new
> release yesterday)
DC Proof: Rationale for a more User-Friendly Notation
While it may not be the standard notation of logicians, in
most cases,
the
> notation used by DC Proof does not vary significantly from
it. It was
> designed primarily for the non-specialist mathematics
student in first or
> second year university or in advanced senior high school
courses.
To give just one example, most university algebra and
calculus (analysis)
> textbooks, even the advanced ones it seems, do not make use
of universal
and
> existential quantifiers (the logician's inver 
A's and
backwards E's
> respectively). They usually spell out for all x or there
exists a y
such
> that etc. Students should be used to working with such
notation. It
was
> felt that the use of ALL and EXIST would be a more natural
and
> suggestive notation.
By the way, users need not type out these words or commit the
notation to
> memory. To minimize keystrokes and to help students learn
the notation,
> there is a convenient, pull-down menu of all symbols and
notation used in
DC
> Proof (click Notation on the Premise form, etc.).
Also, in a effort to make variable and predicate names more
meaningful,
DC
> Proof allows names of any length. This, of course,
necessitates the use
of
> some extra delimiters (round and square brackets) for
readability. DC
Proof
> automatically minimizes bracketing, however, with its
built-in, order of
> precedence (or operations). This makes it easier for the
user to
recognize
> the occurrence of structures like double negations, for
example.
Generally speaking, the DC Proof notation was designed to be
as familiar
as
> possible to students, and to make it easy to learn and
master.
> Return
>
===
Subject: Re: help!!!!!!! indicials
> hi guys,
2^(x+1)+2^(3-x)=17
logging gives
(x+1)lg2+(3-x)lg2=lg17
expanding gives
lg2x+lg2+3lg2-lg2x=lg17
but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but
can't get
there
> numerically.....please help i know i'm missing something
simple
best regards
> shaun
Another poster has provided a good solution, probably better
than this
one; what I notice from the equation
2^(x+1)+2^(3-x)=17
that the left side is a sum of powers of 2, but the right
side is odd.
To me, that says that one of the terms on the left must be 1
= 2^0.
So, let's look at the possibilities:
Case 1:
x+1 = 0, so x = -1, 3-x = 4.
The equation becomes
2^(x+1) + 2^(3-x) = 2^0 + 2^4 = 17.
so, x=-1 works.
Case 2:
3-x = 0, so x = 3, and x+1 = 4.
The equation becomes
2^(x+1) + 2^(3-x) = 2^4 + 2^0 = 17
so, x=3 works.
Your equation has two solutions: x = -1, or x = 3.
Dale.
===
Subject: Re: Google's rankings
[snip endless whining]
Vincent, you were bus for referrer log spamming. Google is
providing a
service as a business, and if they want to weed out creeps
like you than
they are 100% within their rights to do so. It doesn't 
matter
if your
site is the most legitimate piece of net real estate on the
planet. You
did something Google thought was naughty, and now you whine
and complain
because somehow it's your *right* to be lis.
This has nothing to do with inefficiencies in the Google
PageRank
algorithm, and has *everything* to do with
www.datashaping.com being
banned from the Google database. Comparing your site to
others with
higher PageRank than yours means nothing except that they are
lis and
you are not.
Everyone else can prove it for themselves:
http://www.google.com/search?q=site%3Awww.datashaping.com+-
sdsdfsdf
Perhaps if you cleaned up your act and asked nicely, you'd 
be
given your
precious PageRank back again. On the other hand, it's 
awfully
amusing
watching you humiliate yourself by bearing this grudge like a
grade-schooler who had his pencilcase stolen. Because despite
all your
degrees (which you lis in a previous rant a few months ago)
you still
haven't grown up.
Grey
--
The technical axiom that nothing is impossible sinisterly
implies the
pitfall corollory that nothing is ridiculous.
- http://www.greywyvern.com - ORCA - Camoußaged PHP Web
Scripts.
===
Subject: Re: Google's rankings
How can you tell he was banned by google and what is referrer
log
spamming?...How did you find out this info on his site?
> [snip endless whining]
> Vincent, you were bus for referrer log spamming. Google is
providing
a
> service as a business, and if they want to weed out creeps
like you than
> they are 100% within their rights to do so. It doesn't
matter if your
> site is the most legitimate piece of net real estate on the
planet. You
> did something Google thought was naughty, and now you whine
and complain
> because somehow it's your *right* to be lis.
> This has nothing to do with inefficiencies in the Google
PageRank
> algorithm, and has *everything* to do with
www.datashaping.com being
> banned from the Google database. Comparing your site to
others with
> higher PageRank than yours means nothing except that they
are lis and
> you are not.
> Everyone else can prove it for themselves:
>
http://www.google.com/search?q=site%3Awww.datashaping.com+-
sdsdfsdf
> Perhaps if you cleaned up your act and asked nicely, you'd
be given your
> precious PageRank back again. On the other hand, it's
awfully amusing
> watching you humiliate yourself by bearing this grudge like
a
> grade-schooler who had his pencilcase stolen. Because
despite all your
> degrees (which you lis in a previous rant a few months ago)
you still
> haven't grown up.
> Grey
> --
> The technical axiom that nothing is impossible sinisterly
implies the
> pitfall corollory that nothing is ridiculous.
> - http://www.greywyvern.com - ORCA - Camoußaged PHP Web
Scripts.
===
Subject: Re: Google's rankings
> How can you tell he was banned by google and what is
referrer log
> spamming?...How did you find out this info on his site?
STFW
Grey
--
The technical axiom that nothing is impossible sinisterly
implies the
pitfall corollory that nothing is ridiculous.
- http://www.greywyvern.com - Orca RingMaker: PHP web ring
creation and
management
===
Subject: Re: Google's rankings
(1) http://www.nakedteens.com ==> 4/10
> (2) http://www.datashaping.com ==> 0/10
(1) has multiple invading pop-ups and crudly displays poor
quality
> pornographic material without asking your age
> (2) has none of the above -- it's just a website for
mathematicians
(1) is better than (2) according to Google's page ranking
algorithm.
> Visit (1) and (2) and judge by yourself.
> http://www.scientology.com ==> 4/10
> http://www.datashaping.com ==> 0/10
This gives you an idea on how biased Google's ranking is,
favoring
> criminals over legitimate businesses.
Hey stooopid - Google ranking is decided by software based on
empirically valida statistical elements. Google uses their
model
because it works. Google favors nobody, and better is
orthognal to
their analysis. If you were looking for a whole lot of
strange and
ended up at the math site, you'd be holding your little
wee-wee in
your palsied hand and screaming that Google had chea you.
Tell ya what, stooopid, Google Amanda Peet. One is an actress
who
takes off her clothes a lot, the other is a respec string
theorist. Why don't you get up on your hind legs and tell us
which
one is the ripoff?
--
Uncle Al
http://www.mazepath.com/uncleal/qz.pdf
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
===
Subject: Re: New question
> Suppose we have a group G and a homomorphism h exists taking
> elements of G into a ring (R, *, +) such that h(x)*h(y) =
h(x*y).
>
It's not clear what purpose the ring structure serves, other
than
to imply that not all elements of R have multiplicative
inverses.
> Next, suppose that for a given ring, the number of such
> non-isomorphic G's could be vast or infinite 
and that,
because
> of the complexity of the ring, we are initially only aware
> of a few.
>
You're looking for different groups that have non-trivial
homomorphisms
into the group of units of R? Have you tried the set of
groups that
have non-trivial homomorphisms into G? That would give you
something,
but it wouldn't really have anything to say about R.
> Given one such G and a homomorphism, surely one way to
> try ones luck at finding at least some of the other 
G's is to
> systematically look for elements x, y of R such that there
> exist m, n in naturals: x^m, y^n in h(G). Then we
automatically
> know the multiplicative inverses of x and y exist, and we
> also know that the set
> G(x,y) = {z | exists m, n >= 0: z = (x^m)*(y^n) }
I don't see how this is a group. The multiplication is not
necessarily commutative, and G is not necessarily finite,
so how do you deduce that (for instance) z = xy has
an inverse; that is, how do you show that the element
z^(-1) = y^(-1)x^(-1)
which certainly exists in R (since y^(-1), x^(-1) exist) is
of the form x^m y^n ? I can imagine that it might not. I mean,
this difficulty might be avoided by taking the group genera
by the above set G(x,y), or by taking all words in x,y subject
to the constraints imposed by R itself, but that could make
the
topic cumbersome.
> -with x^0 defined as h(1)- is a group. In particular
G(x) = {z | exists m >= 0: z = x^m } is a subgroup of G(x,y).
>
Why? What makes x^(-1) an element of this set? G isn't 
finite,
is it?
> Does this procedure have a name?
> p.s. I'm *not* asking this question as a hypothetical
(that
> would be really quite boring!), since I have used it
> a few times to generate a string of new groups branching
> out all over the place on a ring and, at the moment,
> I don't particularly see and end to these groups in
sight.
> At first glance, the method would appear 
inefficient:
> how the hell does one keep finding new x and 
y's (and
z's
> etc. !)? It seems that at least for the ring at hand, there
> is a generic method for guessing new ones based on looking
> for especially symmetric elements of R.
>
What specific examples have you been working with? What is the
ring
at hand, anyhow?
Note that for any unit u of R, you can produce a homomorphism
from Z
to R by n |--> u^n. Given k units u_1,..., u_k, there is a
homomorphism
from the free group on k generators to the subgroup genera by
the
u's. Modulo it's a joke and 
it's not a joke) the relations
imposed by
the group R* of units of R, that appears to be the full set
of things
available, but I'm no algebraist, nor do I play one on
sci.math.
That's all I've been able to 
figure out. Others may be more
helpful.
> Please remember that I am unfortunately still a novice in
your
> answers.
> C. Dement
>
Dale
===
Subject: Re: Left Truncatable Primes Derived from Fibonacci
> While playing with the Fibonacci sequence in a spreadsheet,
the
> following interesting pattern appeared.
> When the Fibonacci number is divided by its position in the
list, the
> result is a whole number when the position corresponds to a
left
> truncatable prime*. I ran out of accuracy to test this past
the 67th
> number and I know little about left truncatable primes.
Does anyone
> know if this property is anything worth digging into? A
google
search
> on fibonacci and primes numbers returns a few things but
nothing
of
> great depth.
> If anyone has more info about Fibonacci and primes I would
appreciate
> anything you care to share.
> The numbers from the spreadsheet are below. Primes(*) Column
> A=Fibonacci Column B=Position Column C=A/B Also notice the
primes
> appear the intervals of 6,4,6,14,6,4,6,14... wonder if that
would
> continue? Do I have too much time on my hands?
> Many
> Bob Carlson
> Fibonacci
> Sequence Position
> A B A/B
> 1 0
> 1 1 1.000000
> 2 2 1.000000
> 3 3* 1.000000 -----
> 5 4 1.250000
> 8 5 1.600000
> 13 6 2.166667
> 21 7* 3.000000-------
> 34 8 4.250000
> 55 9 6.111111
> 89 10 8.900000
> 144 11 13.090909
> 233 12 19.416667
> 377 13* 29.000000------
> 610 14 43.571429
> 987 15 65.800000
> 1597 16 99.812500
> 2584 17* 152.000000-----
> 4181 18 232.277778
> 6765 19 356.052632
> 10946 20 547.300000
> 17711 21 843.380952
> 28657 22 1302.590909
> 46368 23* 2016.000000-----
> 75025 24 3126.041667
> 121393 25 4855.720000
> 196418 26 7554.538462
> 317811 27 11770.777778
> 514229 28 18365.321429
> 832040 29 28691.034483
> 1346269 30 44875.633333
> 2178309 31 70268.032258
> 3524578 32 110143.062500
> 5702887 33 172814.757576
> 9227465 34 271396.029412
> 14930352 35 426581.485714
> 24157817 36 671050.472222
> 39088169 37* 1056437.000000-----
> 63245986 38 1664368.052632
> 102334155 39 2623952.692308
> 165580141 40 4139503.525000
> 267914296 41 6534495.024390
> 433494437 42 10321296.119048
> 701408733 43* 16311831.000000-----
> 1134903170 44 25793253.863636
> 1836311903 45 40806931.177778
> 2971215073 46 64591632.021739
> 4807526976 47* 102287808.000000----
> 7778742049 48 162057126.020833
> 12586269025 49 256862633.163265
> 20365011074 50 407300221.480000
> 32951280099 51 646103531.352941
> 53316291173 52 1025313291.788460
> 86267571272 53* 1627690024.000000----
> 139583862445 54 2584886341.574070
> 225851433717 55 4106389703.945450
> 365435296162 56 6525630288.607140
> 591286729879 57 10373451401.386000
> 956722026041 58 16495207345.534500
> 1548008755920 59 26237436541.016900
> 2504730781961 60 41745513032.683300
> 4052739537881 61 66438353080.016400
> 6557470319842 62 105765650320.032000
> 10610209857723 63 168416029487.667000
> 17167680177565 64 268245002774.453000
> 27777890035288 65 427352154389.046000
> 44945570212853 66 680993488073.530000
> 72723460248141 67* 1085424779823.000000 ----
> 117669030460994 68 1730426918544.030000
Bob,
Using your same list above, where A is the Fibonacci sequence
> in the first column and B is the second column of positions
then
> where if any B is (ODD) then B = prime if A == 0 or 1 (mod
B).
Correction!
Except of course, prime (5) which I overlooked.
 At least up to B = 67. ;-)
I don't know if this fails for any higher (ODD) values of B.
Dan
Bob,
In my haste it should have read above ---
All (ODD) integers in your B position where B > 9, then B = a
prime
if A == 0 or 1 (mod B).
I know your index is off by one, but this is the only way my
discovery works for all primes = > 11 and at this point = <
163
I have checked all odd integers (B) > 9 up to B = 163 and all
B's that are primes to this point are either a 0 or 1 (mod 
B)
all
other odd integecomposites, are A == >1 (mod B).
I am now only checking B > 163 where all odd B's THAT ARE 
NOT
congruent to 0 (mod 3) or 0 (mod 5).
This is very interesting if this does not fail for higher odd
values
of B because this simple summation of integers in the
Fibonnaci
sequence could be so closely linked to the primes!
There are other links to primes like if ((p-1)! +1) == 0 (mod
p)
then p = prime. The problem here is these factorials are so
large
to find a small p it is not a practical function.
The ratio of A/B in this example using Fibonnaci numbers (A)
is
much smaller, thus more interesting if true! Also more
interesting
because it is the Fibonacci sequence with so many other
interesting
properties.
Thanks for your OP, I would not have discovered this!
Dan
===
Subject: Re: Left Truncatable Primes Derived from Fibonacci
> While playing with the Fibonacci sequence in a spreadsheet,
the
> following interesting pattern appeared.
> When the Fibonacci number is divided by its position in the
list,
the
> result is a whole number when the position corresponds to a
left
> truncatable prime*. I ran out of accuracy to test this past
the
67th
> number and I know little about left truncatable primes. Does
anyone
> know if this property is anything worth digging into? A
google
search
> on fibonacci and primes numbers returns a few things but
nothing
of
> great depth.
> If anyone has more info about Fibonacci and primes I would
appreciate
> anything you care to share.
> The numbers from the spreadsheet are below. Primes(*) Column
> A=Fibonacci Column B=Position Column C=A/B Also notice the
primes
> appear the intervals of 6,4,6,14,6,4,6,14... wonder if that
would
> continue? Do I have too much time on my hands?
> Many
> Bob Carlson
> Fibonacci
> Sequence Position
> A B A/B
> 1 0
> 1 1 1.000000
> 2 2 1.000000
> 3 3* 1.000000 -----
> 5 4 1.250000
> 8 5 1.600000
> 13 6 2.166667
> 21 7* 3.000000-------
> 34 8 4.250000
> 55 9 6.111111
> 89 10 8.900000
> 144 11 13.090909
> 233 12 19.416667
> 377 13* 29.000000------
> 610 14 43.571429
> 987 15 65.800000
> 1597 16 99.812500
> 2584 17* 152.000000-----
> 4181 18 232.277778
> 6765 19 356.052632
> 10946 20 547.300000
> 17711 21 843.380952
> 28657 22 1302.590909
> 46368 23* 2016.000000-----
> 75025 24 3126.041667
> 121393 25 4855.720000
> 196418 26 7554.538462
> 317811 27 11770.777778
> 514229 28 18365.321429
> 832040 29 28691.034483
> 1346269 30 44875.633333
> 2178309 31 70268.032258
> 3524578 32 110143.062500
> 5702887 33 172814.757576
> 9227465 34 271396.029412
> 14930352 35 426581.485714
> 24157817 36 671050.472222
> 39088169 37* 1056437.000000-----
> 63245986 38 1664368.052632
> 102334155 39 2623952.692308
> 165580141 40 4139503.525000
> 267914296 41 6534495.024390
> 433494437 42 10321296.119048
> 701408733 43* 16311831.000000-----
> 1134903170 44 25793253.863636
> 1836311903 45 40806931.177778
> 2971215073 46 64591632.021739
> 4807526976 47* 102287808.000000----
> 7778742049 48 162057126.020833
> 12586269025 49 256862633.163265
> 20365011074 50 407300221.480000
> 32951280099 51 646103531.352941
> 53316291173 52 1025313291.788460
> 86267571272 53* 1627690024.000000----
> 139583862445 54 2584886341.574070
> 225851433717 55 4106389703.945450
> 365435296162 56 6525630288.607140
> 591286729879 57 10373451401.386000
> 956722026041 58 16495207345.534500
> 1548008755920 59 26237436541.016900
> 2504730781961 60 41745513032.683300
> 4052739537881 61 66438353080.016400
> 6557470319842 62 105765650320.032000
> 10610209857723 63 168416029487.667000
> 17167680177565 64 268245002774.453000
> 27777890035288 65 427352154389.046000
> 44945570212853 66 680993488073.530000
> 72723460248141 67* 1085424779823.000000 ----
> 117669030460994 68 1730426918544.030000
> Bob,
> Using your same list above, where A is the Fibonacci
sequence
> in the first column and B is the second column of positions
then
> where if any B is (ODD) then B = prime if A == 0 or 1 (mod
B).
Correction!
Except of course, prime (5) which I overlooked.
 At least up to B = 67. ;-)
> I don't know if this fails for any higher (ODD) values of 
B.
> Dan
Bob,
In my haste it should have read above ---
All (ODD) integers in your B position where B > 9, then B = a
prime
> if A == 0 or 1 (mod B).
I know your index is off by one, but this is the only way my
> discovery works for all primes = > 11 and at this point = <
163
 Thanks for your OP, I would not have discovered this!
Dan
Dan, I moved the index off by one because I thought the
starting point
in Fibonacci is one not zero. I have no formal training in
math, but
I am consumed by numeric sequences and the patterns in those
sequences. Starting at zero just didn't look right. I see
patterns
in PI, e, and PHI by laying those numbers in columns next to
each
other with a 4th column in the base 6 numbering system (only
allows
digits 0,1,2,3,4,5). The number 5 and sqrt(5) do interesting
things
to PHI that make it have the same pattern as PI.
Good luck!
===
Subject: Hints for simulation techniques?
Hello sci.math people.
I'm presently attempting to produce a model of a system as 
it
evolves
over time. It involves the interaction of four components. The
derivatives are:
dydt[0] = B * y[0] * (A *y[3] - N_1) + seed * y[3] * gamma;
dydt[1] = B * y[1] * (A * y[3] - N_1) + seed * y[3] * gamma;
dydt[2] = C * y[2] * (y[3] - N_0);
dydt[3] = -B * (y[0] + y[1]) * (A * y[3] - N_1) -
C * y[2] * (y[3] - N_0) - y[3] * D * gamma;
where A and D are ~1. B and C are ~ 10^-12. N_0 and N_1 are
~10-^26. Gamma is a decay rate (~10^3 per sec) and seed is a
small
number (I've used 10^-8). If you're interes, 
this is supposed
to
be the interaction of two lasers (y[0] and y[1]) being
amplified by
some gain medium (y[3]), pumped by another laser (y[2]). The
above is
for the interaction at a single position - I repeat this many
times
over the length of the medium.
Computer simulations are not really my area, but I've spent
time
going through Numerical Recipes to try to clue up. I tried to
model
the above system with a 4th order Runge-Kutta method, but
found that
when (A *y[3] - N_1) approached zero (y[0] and y[1] being
~10^20 at
this point) the values of y[0] and y[1] rapidly became
unphysical
(lots of NaNs). I tried implementing step controls, but the
model
showed no sign of progressing even after allowing very small
timesteps
(I try to take an initial step of about 10^-12 s, and allowed
the step
control to do steps 1/1000 of this).
I thought perhaps the model was too stiff, so I have also
tried the
Rosenbrock recipe. This produces similar results but y[0] and
y[1]
grow exponentially once (A *y[3] - N_1) converges on zero. I
expect
y[0] and y[1] to converge to their own steady states, since
there
should be a limit where the energy going in to the system
equals the
energy
===
Subject: Re: Hints for simulation techniques?
> Hello sci.math people.
> I'm presently attempting to produce a model of a system as
it evolves
> over time. It involves the interaction of four components.
The
> derivatives are:
> dydt[0] = B * y[0] * (A *y[3] - N_1) + seed * y[3] * gamma;
> dydt[1] = B * y[1] * (A * y[3] - N_1) + seed * y[3] * gamma;
> dydt[2] = C * y[2] * (y[3] - N_0);
> dydt[3] = -B * (y[0] + y[1]) * (A * y[3] - N_1) -
> C * y[2] * (y[3] - N_0) - y[3] * D * gamma;
> where A and D are ~1. B and C are ~ 10^-12. N_0 and N_1 are
> ~10-^26. Gamma is a decay rate (~10^3 per sec) and seed is
a small
> number (I've used 10^-8). If you're interes, 
this is
supposed to
> be the interaction of two lasers (y[0] and y[1]) being
amplified by
> some gain medium (y[3]), pumped by another laser (y[2]).
The above is
> for the interaction at a single position - I repeat this
many times
> over the length of the medium.
> Computer simulations are not really my area, but I've 
spent
time
> going through Numerical Recipes to try to clue up. I tried
to model
> the above system with a 4th order Runge-Kutta method, but
found that
> when (A *y[3] - N_1) approached zero (y[0] and y[1] being
~10^20 at
> this point) the values of y[0] and y[1] rapidly became
unphysical
> (lots of NaNs). I tried implementing step controls, but the
model
> showed no sign of progressing even after allowing very
small timesteps
> (I try to take an initial step of about 10^-12 s, and
allowed the step
> control to do steps 1/1000 of this).
> I thought perhaps the model was too stiff, so I have also
tried the
> Rosenbrock recipe. This produces similar results but y[0]
and y[1]
> grow exponentially once (A *y[3] - N_1) converges on zero.
I expect
> y[0] and y[1] to converge to their own steady states, since
there
> should be a limit where the energy going in to the system
equals the
> energy
It sounds strange that the values for y[0] and y[1] should
start to grow
exponentially, when A * y[3] - N_1 is close to zero. What
happens with y[2]
and y[3]; are they approximately constant? What happens with
the step size;
is it decreasing to smaller and smaller values?
Are you sure you have typed in the correct formulae for the
derivatives? It
seems safe to use Runge Kutta, so I suspect some programming
error.
Before diving into the numerics, I would try to obtain some
analytical
results. Perhaps you've already done this. For instance, 
when
gamma is zero
there is supposedly no decay. Is it possible to write down an
expression
for
the energy E(y[0], y[1], y[2], y[3]) such that E is constant
over time. In
other words, the time derivative of E should be zero when
gamma is zero.
Perhaps you can estimate the dynamics of E as function of
time for nonzero
gamma. You can then compare with the numerical evaluation of
E as function
of time.
Are there any steady state solutions? A rela question, when A
* y[3] -
N_1 is close to zero, will it remain close to zero? This
could probably be
answered analytically.
Another suggestion. Could you perhaps scale your parameters
so they are
close to 1. This includes scaling time, which I suppose is of
the order
10^-12. Doing the scaling will enable you to more easily
determine which
parameters are large and which are small. For instance, if A
* y[3]
-
N_1 happens to be positive and large, then that would give
you an
exponential growth in y[0] and y[1].
Perhaps there is a physical mechanism not included in the
equations that
prevents A * y[3] - N_1 from being positive.
Hope that helps.
-Michael.
Subject: Re: Hints for simulation techniques?
===
> Hello sci.math people.
I'm presently attempting to produce a model of a system as 
it
evolves
> over time. It involves the interaction of four components.
The
> derivatives are:
dydt[0] = B * y[0] * (A *y[3] - N_1) + seed * y[3] * gamma;
> dydt[1] = B * y[1] * (A * y[3] - N_1) + seed * y[3] * gamma;
> dydt[2] = C * y[2] * (y[3] - N_0);
> dydt[3] = -B * (y[0] + y[1]) * (A * y[3] - N_1) -
> C * y[2] * (y[3] - N_0) - y[3] * D * gamma;
where A and D are ~1. B and C are ~ 10^-12. N_0 and N_1 are
> ~10-^26. Gamma is a decay rate (~10^3 per sec) and seed is
a small
> number (I've used 10^-8). If you're interes, 
this is
supposed to
> be the interaction of two lasers (y[0] and y[1]) being
amplified by
> some gain medium (y[3]), pumped by another laser (y[2]).
The above is
> for the interaction at a single position - I repeat this
many times
> over the length of the medium.
Computer simulations are not really my area, but I've spent
time
> going through Numerical Recipes to try to clue up. I tried
to model
> the above system with a 4th order Runge-Kutta method, but
found that
> when (A *y[3] - N_1) approached zero (y[0] and y[1] being
~10^20 at
> this point) the values of y[0] and y[1] rapidly became
unphysical
> (lots of NaNs). I tried implementing step controls, but the
model
> showed no sign of progressing even after allowing very
small timesteps
> (I try to take an initial step of about 10^-12 s, and
allowed the step
> control to do steps 1/1000 of this).
--snip--
Your problem seems to be one of scales, since at first glance
your
system of equations is not too badly behaved. Before trying
numerical
simulation you should scale your variables such that they are
all
approximately of the same order. So, if y[0] andd y[1] ~
10^15 (say)
but y[2] and y[3] ~1, then introduce new variables x[0] =
10^(-15) y[0]
and x[1] = 10^(-15) y[2]. This will also scale the parameters
in your
equations. This should help with numerics.
If this does not work, report back. Also give the ranges
where you
want your y's to be.
HTH,
Michael.
--
&&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
Dr. Michael Ulm
FB Mathematik, Universitaet Rostock
michael.ulm@mathematik.uni-rostock.de
===
Subject: Re: Hints for simulation techniques?
Huge apologies. I meant to finish what I was writing...
> Hello sci.math people.
I'm presently attempting to produce a model of a system as 
it
evolves
> over time. It involves the interaction of four components.
The
> derivatives are:
dydt[0] = B * y[0] * (A *y[3] - N_1) + seed * y[3] * gamma;
> dydt[1] = B * y[1] * (A * y[3] - N_1) + seed * y[3] * gamma;
> dydt[2] = C * y[2] * (y[3] - N_0);
> dydt[3] = -B * (y[0] + y[1]) * (A * y[3] - N_1) -
> C * y[2] * (y[3] - N_0) - y[3] * D * gamma;
where A and D are ~1. B and C are ~ 10^-12. N_0 and N_1 are
> ~10-^26. Gamma is a decay rate (~10^3 per sec) and seed is
a small
> number (I've used 10^-8). If you're interes, 
this is
supposed to
> be the interaction of two lasers (y[0] and y[1]) being
amplified by
> some gain medium (y[3]), pumped by another laser (y[2]).
The above is
> for the interaction at a single position - I repeat this
many times
> over the length of the medium.
Computer simulations are not really my area, but I've spent
time
> going through Numerical Recipes to try to clue up. I tried
to model
> the above system with a 4th order Runge-Kutta method, but
found that
> when (A *y[3] - N_1) approached zero (y[0] and y[1] being
~10^20 at
> this point) the values of y[0] and y[1] rapidly became
unphysical
> (lots of NaNs). I tried implementing step controls, but the
model
> showed no sign of progressing even after allowing very
small timesteps
> (I try to take an initial step of about 10^-12 s, and
allowed the step
> control to do steps 1/1000 of this).
I thought perhaps the model was too stiff, so I have also
tried the
> Rosenbrock recipe. This produces similar results but y[0]
and y[1]
> grow exponentially once (A *y[3] - N_1) converges on zero.
I expect
> y[0] and y[1] to converge to their own steady states, since
there
> should be a limit where the energy going in to the system
equals the
> energy
... lost from the system (each iteration about 5% is lost).
If anyone has any experience with using such numerical models
and is
able to point me in the direction of a better approach I'd 
be
very
grateful.
Bryn.
===
Subject: Re: Integer problem and dynamic programming
>I encountered the following problem. It's an integer 
problem
from the
>following type:
>sum(Min(4a11+6a21,1),Min(4a12+6a22+3a31,4),Min(6a23+5a41,10)
,Min(5a42,4))
>S.T:
>a11+ a12+....=1 //we can just choose one to be 1 at the same
time
>a21+ a22+....=1
>a31+ a32+....=1
...
>am1+ am2+....=1
>for all i,j: aij={0,1} //the solution must be 1/0
>I managed to solve it in n^2(log n) using dynamic
programming.( n is
>the number of Min operation that I have to do, in this
example-4).
>Does anyone know a better solution (complexity) to solve
this problem?
>
Do you want to minimixe the sum? Find all possible solutions?
What?
Probably better pos to sci.op-research, where I have c-pos
this post.
--
===
Subject: Integer problem and dynamic programming
I encountered the following problem. It's an integer problem
from the
following type:
sum(Min(4a11+6a21,1),Min(4a12+6a22+3a31,4),Min(6a23+5a41,10),
Min(5a42,4))
S.T:
a11+ a12+....=1 //we can just choose one to be 1 at the same
time
a21+ a22+....=1
a31+ a32+....=1
......
am1+ am2+....=1
for all i,j: aij={0,1} //the solution must be 1/0
I managed to solve it in n^2(log n) using dynamic
programming.( n is
the number of Min operation that I have to do, in this
example-4).
Does anyone know a better solution (complexity) to solve this
problem?
thanks in advance,
Michal
===
Subject: what is squeeze theorem?
Is there anyone who can tell what is squeeze theorem and how
to use
squeeze theorem?
Someone thinks the squeeze theorem not to exist and wrong, Is
this
kind of view correct?
===
Subject: Re: what is squeeze theorem?
> Is there anyone who can tell what is squeeze theorem and
how to use
> squeeze theorem?
It is about limits of sequences and functions. Also called
squeeze rule
and squeeze test.
http://planetmath.org/encyclopedia/SqueezeRule.html
LH
===
Subject: Re: what is squeeze theorem?
===
> Someone thinks the squeeze theorem not to exist and wrong,
Is this
> kind of view correct?
>
Is his name Garry Denke? :)
Tim Wouters
===
Subject: Re: what is squeeze theorem?
> Someone thinks the squeeze theorem not to exist and wrong,
Is this
> kind of view correct?
> Is his name Garry Denke? :)
Tim Wouters
who is Garry Denke?
what did he say?
Is squeeze theorem another expression of Continuous functions?
Someone told me that f(x) is g(x-0) and h(x) is g(x-0).Is
that right?
Is f(x) lower limit of g(x)?
Is h(x) upper limit of g(x)?
===
Subject: Re: what is squeeze theorem?
Injector-Info: news.mailgate.org;
posting-host=adsl-66-126-134-202.dsl.frsn01.pacbell.net;
posting-account=48257; posting-date=1074589859
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
6f149fe987cfd80ea15a79b7a22be5
07.48257%40mygate.mailgate.org
> who is Garry Denke?
> what did he say?
> Is squeeze theorem another expression of Continuous
functions?
> Someone told me that f(x) is g(x-0) and h(x) is g(x-0).Is
that right?
> Is f(x) lower limit of g(x)?
> Is h(x) upper limit of g(x)?
Slow down, take a deep breath.
This is Usenet, not all answers are serious ones.
Not all responses are helpful.
No question should be pos that needs an answer before it is
asked.
Usenet is not a real time communications medium, though it
gets closer
with each improvment.
No one owes you an answer, patience and polite manners are
useful.
Read all the answethen follow the links sugges, then still
don't panic. It's still all only 1s and 0s, as 
a wise man,
Gene
Spafford, once explained.
xanthian, don't take life too seriously, it 
doesn't last long.
--
Pos via Mailgate.ORG Server - http://www.Mailgate.ORG
===
Subject: Re: what is squeeze theorem?
>Is there anyone who can tell what is squeeze theorem and how
to use
>squeeze theorem?
If f(x) <= g(x) <= h(x) for all x and lim f(x) = lim h(x) = b
for
some finite b, then lim g(x) = b as well. (Some niceties 
ommit,
since they probably would confuse you.)
--
===
Subject: Re: what is squeeze theorem?
>>Is there anyone who can tell what is squeeze theorem and
how to use
>>squeeze theorem?
>If f(x) <= g(x) <= h(x) for all x and lim f(x) = lim h(x) =
b for
>some finite b, then lim g(x) = b as well.
There's no need for finite here.
>(Some niceties ommit,
>since they probably would confuse you.)
??? What sort of nicety is missing here?
(There's a little missing from the _notation_ -
we need to understand lim as lim_{x->a}.
But that seems clear enough, and doesn't
seem like something that's likely to confuse
him...?)
************************
===
Subject: Re: what is squeeze theorem?
Is there anyone who can tell what is squeeze theorem and how
to use
>squeeze theorem?
>If f(x) <= g(x) <= h(x) for all x and lim f(x) = lim h(x) =
b for
>>some finite b, then lim g(x) = b as well.
>There's no need for finite here.
>(Some niceties ommit,
>>since they probably would confuse you.)
>??? What sort of nicety is missing here?
>(There's a little missing from the _notation_ -
>we need to understand lim as lim_{x->a}.
>But that seems clear enough, and doesn't
>seem like something that's likely to confuse
>him...?)
That the inequalities need only hold in a punctured
neighborhood of the
limit point.
--
===
Subject: 2D Determinants.
Given three points in 2D:
a = (xa, ya)
b = (xb, yb)
c = (xb, yc)
and the determinant
| xa xb xc |
| ya yb yc |
| 1 1 1 |
How could you calculate the area of a triangle formed by the
three points a, b and c (1/2bh) by using the determinant?
===
Subject: Re: 2D Determinants.
> Given three points in 2D:
a = (xa, ya)
> b = (xb, yb)
> c = (xb, yc)
and the determinant
| xa xb xc |
> | ya yb yc |
> | 1 1 1 |
How could you calculate the area of a triangle formed by the
> three points a, b and c (1/2bh) by using the determinant?
As is well-known, it's half the absolute value of this
determinant.
As a followup exercise, given four points in 3-space, find
a determinantal formula for the volume of the tetrahedron
with them as vertices.
--
===
Subject: Re: 2D Determinants.
>As a followup exercise, given four points in 3-space, find
>a determinantal formula for the volume of the tetrahedron
>with them as vertices.
I followed your advice for the triangle in 2D, and worked
it out finally. But I can't seem to solve this 
other problem
you've sugges. Could you tell me how to do it?
===
Subject: Re: 2D Determinants.
>>As a followup exercise, given four points in 3-space, find
>>a determinantal formula for the volume of the tetrahedron
>>with them as vertices.
I followed your advice for the triangle in 2D, and worked
> it out finally. But I can't seem to solve this 
other problem
> you've sugges. Could you tell me how to do it?
Yes.
--
===
Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A
Smith)
> John Schoenfeld:
My friend Bilge, d|v|/dt is not a vector,
 So, what you're saying is that all the nonsense about 
vector
> functions was just that - nonsense and you are simply taking
> the the derivative of v^2 = v.v like I said. Sheeesh....
it is a scalar - it is the rate of change of speed.
So much for your argument about vector functions.
[THREAD DISCONTINUED]
JS
===
Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A
Smith)
> John Schoenfeld:
>>I find it unbelievable that you persist with your ridiculous
position
>>that
>>v(t)=sin(t) is somehow a counter-example to v.a<0 being
deceleration.
>>Listen idiot, something decelerates when d|v|/dt < 0
>> I find it unbelievable that you are attempting to use an
absolute
>> value to define a vector quantity, especially after I told
how
>> to define it as thescalar v^2 avoid that problem.
The absolute quantity that you are refering to is |v|, which
is the
>speed.
 You are clearly a complete crackpot.
[THREAD DISCONTINUED]
JS
===
Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A
Smith)
> John Schoenfeld:
>>I find it unbelievable that you persist with your ridiculous
position
>>that
>>v(t)=sin(t) is somehow a counter-example to v.a<0 being
deceleration.
>>Listen idiot, something decelerates when d|v|/dt < 0
>> I find it unbelievable that you are attempting to use an
absolute
>> value to define a vector quantity, especially after I told
how
>> to define it as thescalar v^2 avoid that problem.
The absolute quantity that you are refering to is |v|, which
is the
>speed.
You are clearly a complete crackpot.
> [THREAD DISCONTINUED]
[Translation: John is unable to defend his crackpot position,
so he runs
away with his tail tucked between his legs]
===
Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A
Smith)
Decreasing velocity (i.e. dv/dt < 0) does not mean
decreasing speed
>(i.e. d|v|/dt < 0).
>
what does it mean for a vector to be less than zero? or for
that matter it
to decrease? i can think of one possibility, but it's the
same as its
magnitude decreasing
===
Subject: Re: Logs, bases and Casio calculators
> for my calculations is not allowed for an exam I am sitting
shortly.
> Therefore I have been forced to borrow a less complica
Casio FX-83WA.
I'm familiar with converting from base 10 logs to other
bases, but does
> anyone know wether the Casio family of characters has a
button or syntax
> for specifying the base? Surely I don't have to perform
manual
> calculations each time I want a base other than 10.
For heaven's sake.... Real Mathematicians never use base 10
logs.
If you do have to calculate log_b a, it's (log a)/(log b).
Anyway the Casio I occasionally used to use had a log button
for calculating base 10 logs (groan) and a ln button (groan)
for calulating logs.
--
===
Subject: Re: Logs, bases and Casio calculators
> For heaven's sake.... Real Mathematicians never use base 
10
logs.
So what bases do real mathematicians use? e of course. Logs
to base 2 are
useful in many cases. Any others that come up in real math?
--
===
Subject: Re: Logs, bases and Casio calculators
>> For heaven's sake.... Real Mathematicians never use base
10 logs.
So what bases do real mathematicians use? e of course. Logs
to base 2
> are
> useful in many cases. Any others that come up in real math?
What's this math that you and others talk about?
Anyway Real Mathematicians use logarithms, but never bases.
--
===
Subject: Re: Logs, bases and Casio calculators
> For heaven's sake.... Real Mathematicians never use base 
10
logs.
So what bases do real mathematicians use? e of course. Logs
to base 2
> are
> useful in many cases. Any others that come up in real math?
What's this math that you and others talk about?
Anyway Real Mathematicians use logarithms, but never bases.
All your base are belong to us.
===
Subject: Re: Logs, bases and Casio calculators
>> For heaven's sake.... Real Mathematicians never use base 
10
logs.
> So what bases do real mathematicians use? e of course. Logs
to base
2
> are
> useful in many cases. Any others that come up in real math?
What's this math that you and others talk about?
Anyway Real Mathematicians use logarithms, but never bases.
All your base are belong to us.
One leg is both the same.
===
Subject: Re: Logs, bases and Casio calculators
>For heaven's sake.... Real Mathematicians never use base 10
logs.
So what bases do real mathematicians use? e of course. Logs
to base
2
>> are
>> useful in many cases. Any others that come up in real math?
What's this math that you and others talk about?
Anyway Real Mathematicians use logarithms, but never bases.
All your base are belong to us.
Que?
--
===
Subject: Re: Logs, bases and Casio calculators
>> All your base are belong to us.
Que?
It's Engrish (Japanese transla badly into English). It comes
from the
introductory cut scene for the Sega Genesis version of Zero
Wing. Here
is a transcript:
In A.D. 2101
War was beginning
Captain: What happen?
Mechanic: Somebody set up us the bomb.
Operator: We get signal.
Captain: What!
Operator: Main screen turn on.
Captain: It's You!!
Cats: How are you gentlemen!!
Cats: All your base are belong to us.
Cats: You are on the way to destruction.
Captain: What you say!!
Cats: You have no chance to survive make your time.
Cats: Ha Ha Ha Ha ....
Captain: Take off every ÔZig'!!
Operator: You know what you doing.
Captain: Move ÔZig'.
Captain: For great justice.
The phrase all your base are belong to us has since shown up
all over
the
place. Many many many many people have Photoshoped it into
photos and
pos
them. These range from as dumb as Ô math to as funny as James
Harris' attempts to do math.
You can see the cut scene in an anima GIF here:
http://www.planettribes.com/allyourbase/story.shtml
--
===
Subject: Re: Logs, bases and Casio calculators
$m2a$3@news7.svr.pol.co.uk:
> Que?
>
If you really want to know, google for the phrase. But let me
give you a
piece of advice: you *don't* really want to know.
--
My email address has an extra @ (spell it out) and an extra
invalid. Please
remove them if you are not a spammer or list broker and want
to reply.
===
Subject: Re: Logs, bases and Casio calculators
> For heaven's sake.... Real Mathematicians never use base 
10
logs.
> So what bases do real mathematicians use? e of course. Logs
to base 2
are
> useful in many cases. Any others that come up in real math?
> --
>
As a meteorologist, I've only used Natural Logs and Common
Logs; Natural
Logs more since I deal with calculus quite a bit. I can't
remember the last
time I used any other base.
===
Subject: Re: Logs, bases and Casio calculators
> for my calculations is not allowed for an exam I am sitting
shortly.
> Therefore I have been forced to borrow a less complica
Casio FX-83WA.
I'm familiar with converting from base 10 logs to other
bases, but does
> anyone know wether the Casio family of characters has a
button or syntax
> for specifying the base? Surely I don't have to perform
manual
> calculations each time I want a base other than 10.
Many thanks in advance.
Jeff
Well, I cannot tell you about the casio calculator, but if it
has any
log button (which is probable), you should use the change of
base
formula. If you forget it, its on most reference sheets.
Where the log
base a of b = log any base of b divided by log same base of a.
===
Subject: Re: Factorials
Peteris Krumins <pkruminsREMOVETHIS@inbox.lv> escribi.97 en
el mensaje
 Here is a little problem i do not know answer of.
> (I am preparing for Programming Contest).
> Find the rightmost non-zero digit on m! followed by
> the number of zeros after that digit
> (1 <= m <= 10^6)
> Could anyone provide a mathematical solution to it?
The problem is the number of new zeros crea in each
multiplication. If
you disregard the trailing zeros in both factothe maximum of
new zeros
is the maximum exponent of a power of 2 less than the limit.
You need a
digit more. Then, tracking the 20 rightmost nonzero digits
must be enough
to
m <= 10^6, although possibly it be excesive. It is only
necessary if the
product, without trailing zeros, is sometimes multiple of
5^19.
With that DERIVE function
ultciffact(n) :=
Prog
k := 1
c := 1
Loop
If k >= n
RETURN MOD(c, 10)
k :+ 1
q := k
Loop
If MOD(q, 10) = 0
q :/ 10
exit
c :* q
Loop
If MOD(c, 10) = 0
c :/ 10
exit
c := MOD(c, 10^20)
I get 4 for the last non-zero digit of 1000000! in 300s on a
P3-450.
--
Best
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Factorials
In-reply-to: Ignacio Larrosa Ca.96estro
<ilarrosaQUITARMAYUSCULAS@mundo-r.com>, Peteris Krumins
<pkruminsREMOVETHIS@inbox.lvPeteris Krumins <pkruminsREMOVETHIS@inbox.lv> escribi.97 en
el mensaje
>> Here is a little problem i do not know answer of.
>> (I am preparing for Programming Contest).
>> Find the rightmost non-zero digit on m! followed by
>> the number of zeros after that digit
>> (1 <= m <= 10^6)
>> Could anyone provide a mathematical solution to it?
>The problem is the number of new zeros crea in each
multiplication. If
>you disregard the trailing zeros in both factothe maximum of
new zeros
>is the maximum exponent of a power of 2 less than the limit.
You need a
>digit more. Then, tracking the 20 rightmost nonzero digits
must be enough
to
>m <= 10^6, although possibly it be excesive. It is only
necessary if the
>product, without trailing zeros, is sometimes multiple of
5^19.
>With that DERIVE function
>ultciffact(n) :=
> Prog
> k := 1
> c := 1
> Loop
> If k >= n
> RETURN MOD(c, 10)
> k :+ 1
> q := k
> Loop
> If MOD(q, 10) = 0
> q :/ 10
> exit
> c :* q
> Loop
> If MOD(c, 10) = 0
> c :/ 10
> exit
> c := MOD(c, 10^20)
>I get 4 for the last non-zero digit of 1000000! in 300s on a
P3-450.
This agrees with algorithm that I gave in
http://groups.google.com/groups?selm=b1606r$329$1@
zinnia.noc.ucla.edu
the Mathematica coding of the algorithm there gives 4 in no
perceptible
time whatsoever; it runs in (log(n))^2 time.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: Factorials
Rob Johnson <rob@trash.whim.org> escribi.97 en el mensaje
>> I get 4 for the last non-zero digit of 1000000! in 300s on
a P3-450.
> This agrees with algorithm that I gave in
>
http://groups.google.com/groups?selm=b1606r$329$1@
zinnia.noc.ucla.edu
> the Mathematica coding of the algorithm there gives 4 in no
> perceptible
> time whatsoever; it runs in (log(n))^2 time.
> Rob Johnson <rob@trash.whim.orgtake out the trash before
replying
Yes, that is a relly good approach. Thanks.
--
Best
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Factorials
> Here is a little problem i do not know answer of.
> (I am preparing for Programming Contest).
> Find the rightmost non-zero digit on m! followed by
> the number of zeros after that digit
> (1 <= m <= 10^6)
> Could anyone provide a mathematical solution to it?
It has been discussed at some length in various newsgroups,
including this one. See the thread Tough FACTORIAL math
problem...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Generators of Prime
>In a group, Z_p, where p is a prime, is it true that EVERY
number other
than
>p is a generator of Z_p?
>If so, can someone give me a proof?
No, 2p is not a generator - and neither is np for any integer
n.
Derek Holt.
===
Subject: Re: Generators of Prime
>In a group, Z_p, where p is a prime, is it true that EVERY
number other
than
>p is a generator of Z_p?
>If so, can someone give me a proof?
No, 2p is not a generator - and neither is np for any integer
n.
But we're not in the integewe're in Z_p. There 
is no distinct
2p.
Having said that, OP mentioned p, which would more usually
just be
called 0.
--
Unpatched IE vulnerability: Security zone transfer
Description: Automatically opening IE + Executing attachments
Published: March 22nd 2002
Reference: http://security.greymagic.com/adv/gm002-ie/
===
Subject: Re: Generators of Prime
>>In a group, Z_p, where p is a prime, is it true that EVERY
number other
than
>>p is a generator of Z_p?
>>If so, can someone give me a proof?
No, 2p is not a generator - and neither is np for any integer
n.
>But we're not in the integewe're in Z_p. 
There is no
distinct 2p.
>Having said that, OP mentioned p, which would more usually
just be
>called 0.
But he said `number'. Since the elements of Z_p are not
numbers at all,
I interpre number as meaning integer, and the question as
meaning
`Is the image in Z_p of every integer other than p a
generator of Z_p?',
and decided that the answer was no.
Derek Holt.
===
Subject: Re: Generators of Prime
>>In a group, Z_p, where p is a prime, is it true that EVERY
number other
>>than p is a generator of Z_p?
>>If so, can someone give me a proof?
No, 2p is not a generator - and neither is np for any integer
n.
Is 2p other than p?
--
===
Subject: Re: Generators of Prime
>In a group, Z_p, where p is a prime, is it true that EVERY
number other
>than p is a generator of Z_p?
>If so, can someone give me a proof?
No, 2p is not a generator - and neither is np for any integer
n.
>Is 2p other than p?
He said `number'. Is 6 the same number as 3 ? :-{
Derek Holt.
===
Subject: Re: Generators of Prime
> In a group, Z_p, where p is a prime, is it true that EVERY
number other
> than p is a generator of Z_p?
> If so, can someone give me a proof?
Apply Lagranges theorem to <a> where a is a nonzero element
of Z_p.
--
===
Subject: Re: Generators of Prime
>> In a group, Z_p, where p is a prime, is it true that EVERY
number other
>> than p is a generator of Z_p?
>> If so, can someone give me a proof?
>Apply Lagranges theorem to <a> where a is a nonzero element
of Z_p.
Surely this is just a trivial corollary of Lagrange's
theorem, so the
OP would need to first prove Lagrange's 
theorem.
Theorem:
Assume the group Z_p (p prime) and g some element of Z_p
which does
not generate it. Then g == 0.
Proof:
Let g be any element of Z_p which does not generate it. Then
g^m == 0
for some 1 <= m < p. For all n in Z_p we also know that n^p
== 0. This
gives:
g^m == g^p == 0
which is only possible if p = nm for some n in Z. But since p
is
prime, the only possibilities are n = 1 or m = 1. Since 1 <=
m < p the
only possibility is that m = 1. Now:
g^1 == 0 => g == 0
QED.
===
Subject: Re: Generators of Prime
 In a group, Z_p, where p is a prime, is it true that EVERY
number other
> than p is a generator of Z_p?
> If so, can someone give me a proof?
>>Apply Lagranges theorem to <a> where a is a nonzero element
of Z_p.
Surely this is just a trivial corollary of Lagrange's
theorem, so the
> OP would need to first prove Lagrange's 
theorem.
Theorem:
Assume the group Z_p (p prime) and g some element of Z_p
which does
> not generate it. Then g == 0.
Proof:
Let g be any element of Z_p which does not generate it. Then
g^m == 0
?
Is g^m equal to g m ?
?
> for some 1 <= m < p. For all n in Z_p we also know that n^p
== 0. This
> gives:
g^m == g^p == 0
so gm = gp in Z_p.
> which is only possible if p = nm for some n in Z.
Now I don't get this .... you must be using g =/= 0 in Z_p 
for
if g = 0 this is always true, and then m needn't be a
multiple of p.
--
===
Subject: Re: Generators of Prime
<buiq6n$1mjktl$1@athena.ex.ac.uk>
<hmfq009r2fqg33uco24pi9ar1sas7uje2a@no.spam>
<bujtjo$5u8$2@news5.svr.pol.co.uk Theorem:
Assume the group Z_p (p prime) and g some element of Z_p
which does
> not generate it. Then g == 0.
Proof:
Let g be any element of Z_p which does not generate it. Then
g^m == 0
> ?
> Is g^m equal to g m ?
Yes. I use ^-notation as a general repea operation of an
element
with itself. In case of Z_p, g^m = gm, where m is the
smallest integer
for which g^m == 0 holds.
> for some 1 <= m < p. For all n in Z_p we also know that n^p
== 0. This
> gives:
g^m == g^p == 0
> so gm = gp in Z_p.
> which is only possible if p = nm for some n in Z.
> Now I don't get this .... you must be using g =/= 0 in Z_p
for
> if g = 0 this is always true, and then m needn't be a
multiple of p.
Nowhere did I say that m be a multiple of p, but that p must
be a
multiple of m. Why is this? We know that g^m == 0 and g^p ==
0.
Then g^(m+1) == g, so (g^(m+1))^r == 0 for some r. Since m is
the
smallest such integer by definition of g we have
g^((m+1)m) = g(m^2+m) == 0
Repeating the same process as previously, we have:
g^(m^2+m+1) == g => (g^(m^2+m+1))^m == 0 => g^(m^3+m^2+m) == 0
ad infinitum.
And we see that for any r such that g^r == 0 it must be that
r == 0 mod m, and since especially for p we have that it has
no factors other than 1 and p then either m = 1 or m = p.
Since m < p the only option is m == 1, and if g^1 == 0 then
g = 0.
===
Subject: Re: Generators of Prime
> Theorem:
Assume the group Z_p (p prime) and g some element of Z_p
which does
>> not generate it. Then g == 0.
Proof:
Let g be any element of Z_p which does not generate it. Then
g^m == 0
>> ?
>> Is g^m equal to g m ?
Yes. I use ^-notation as a general repea operation of an
element
> with itself. In case of Z_p, g^m = gm, where m is the
smallest integer
> for which g^m == 0 holds.
> for some 1 <= m < p. For all n in Z_p we also know that n^p
== 0. This
>> gives:
g^m == g^p == 0
>> so gm = gp in Z_p.
>> which is only possible if p = nm for some n in Z.
>> Now I don't get this .... you must be using g =/= 0 in 
Z_p
for
>> if g = 0 this is always true, and then m needn't be a
multiple of p.
Nowhere did I say that m be a multiple of p, but that p must
be a
> multiple of m. Why is this? We know that g^m == 0 and g^p
== 0.
> Then g^(m+1) == g, so (g^(m+1))^r == 0 for some r. Since m
is the
> smallest such integer by definition of g we have
Hold on. What's all this about m the the smallest whatnot?
You didn't mention that before :-(
> g^((m+1)m) = g(m^2+m) == 0
Repeating the same process as previously, we have:
g^(m^2+m+1) == g => (g^(m^2+m+1))^m == 0 => g^(m^3+m^2+m) == 0
ad infinitum.
All of this looks fairly pointless.
If it was relevant to prove that (m^2+m)g = 0, then the swift
way
is to tnote that (m^2 + m)g = (m+1) m g = 0 (as m g = 0).
> And we see that for any r such that g^r == 0 it must be that
> r == 0 mod m,
Nonsense. If m g = r g = 0 in an additive group, it does not
follow
that m|r .
--
===
Subject: Re: Generators of Prime
> In a group, Z_p, where p is a prime, is it true that EVERY
number other
than
> p is a generator of Z_p?
> If so, can someone give me a proof?
Pick a (non-zero) number. What is the subgroup genera by it?
Subject: re:Generators of Prime
===
Thank you, Bob for that. :)
Sorry, what is a coset? Sorry, i'm really new to this :oops:
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===
Subject: Re: Generators of Prime
> Thank you, Bob for that. :)
> Sorry, what is a coset? Sorry, i'm really new to this 
:oops:
Let G be a group and H a subgroup of G. If g is in G then the
products
genera by gx for all x in H (or {gx|x /in H})is a coset of H
in G.
(With arbitrary groups, you have to consider left- and
right-cosets
separately, by Z_p is cummutative, so we can ignore that
nicety.)
If you think about it, you will realize that the cosets are
disjoint
and, taken together, partition G into a set of equivalence
classes.
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===
Subject: Re: Generators of Prime
I will assume that you really mean the additive
cyclic group Z_p (as written) and not Z_p*.
Yes, it is (almost) true. 0 is not a generator.
Here is a hint for the proof:
every integer in the group coprime to p has a
unique multiplicative inverse.
Now consider any coset.
You can lead a horse's ass to knowledge, but you 
can't make
him think.
===
Subject: Re: Matrix trace question
> I recently read, Recall that, if (matrix) A = A**2, then
Tr(A) =
> Rank(A). Actually, I don't recall that, though I have no
doubt it's
> true. (My linear algebra course did not dwell long on the
trace.)
If someone had a pointer to a proof of this, I would
appreciate seeing
> it.
If A is idempotent (A^2 = A) then A is diagonalizable.
As its only possible eigenvalues are 0 and 1, the diagonalized
matrix is diag(1,1,1,...,1,0,0,0,...,0). Diagonalization
preserves rank and trace.
--
===
Subject: Re: Matrix trace question
>I recently read, Recall that, if (matrix) A = A**2, then
Tr(A) =
>Rank(A). Actually, I don't recall that, though I have no
doubt it's
>true. (My linear algebra course did not dwell long on the
trace.)
>If someone had a pointer to a proof of this, I would
appreciate seeing
>it.
Let V be the null space of A and W the null space of A - I.
You can write any vector v = (v - A v) + A v, where v - A v
is in V
and A v is in W. Moreover, V intersect W = {0}. Taking a basis
of V and a basis of W, you get a basis of R^n in which the
matrix A
is diagonal, with only 1's and 0's on the 
diagonal. The rank
is the
number of 1's, which is also the trace (noting that a change
of basis
corresponds to a similarity transformation A -> S^(-1) A S,
which
leaves the trace unchanged since
Tr(S^(-1) A S) = Tr(A S S^(-1)) = Tr(A)).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Matrix trace question
I recently read, Recall that, if (matrix) A = A**2, then
Tr(A) =
Rank(A). Actually, I don't recall that, though I have no
doubt it's
true. (My linear algebra course did not dwell long on the
trace.)
If someone had a pointer to a proof of this, I would
appreciate seeing
it.
TIA,
George
===
Subject: Change of coordinates in complex plane differentials
In the complex plane, with z, z* gran as independent
variables, and z in
polar reprensation z = rho e^{i theta}, I would like to know
the partial
differential d/dz (THOSE are partials, not straight d).
The chain rule gives d/dz = d/dr (dr/dz) + d/dtheta
(dtheta/dz).
Now if I do dr/dz = 1/(dz/dr) = e^{-i theta}.
BUT
if I do dr/dz straight from r=sqrt(zz*), I get e^{-i theta}/2.
So is the complex partial d/dz = e^{-i theta}(d/dr - (i/r)
d/d theta) or
half of it?
I surmise that this is linked is some way I do not understand
with the fact
that z and z* are independent.
Strangely enough, this is nowhere to be found on the web, and
very few
litterature evokes this subject, so if you have some
references along with
an explanations, they are welcomed.
Thanks.
===
Subject: Re: Change of coordinates in complex plane
differentials
In-reply-to: Iwonder <yssual@yahoo.comIn the complex plane, with z, z* gran as independent
variables, and z
in
>polar reprensation z = rho e^{i theta}, I would like to know
the partial
>differential d/dz (THOSE are partials, not straight d).
>The chain rule gives d/dz = d/dr (dr/dz) + d/dtheta
(dtheta/dz).
>Now if I do dr/dz = 1/(dz/dr) = e^{-i theta}.
>BUT
>if I do dr/dz straight from r=sqrt(zz*), I get e^{-i
theta}/2.
>So is the complex partial d/dz = e^{-i theta}(d/dr - (i/r)
d/d theta) or
>half of it?
>I surmise that this is linked is some way I do not
understand with the
fact
>that z and z* are independent.
>Strangely enough, this is nowhere to be found on the web,
and very few
>litterature evokes this subject, so if you have some
references along with
>an explanations, they are welcomed.
Since you are using dr/dz to mean the partial derivatives,
you cannot
invoke
a rule for total derivatives and say that dz/dr = 1/(dr/dz).
When dealing
with partials, you have to use the Jacobian matrix to change
variables.
The chain rule, written in matrix form, is simply
+- -+
| dz/dr dz*/dr |
[ dz dz* ] = [ dr d@ ] | | [1]
| dz/d@ dz*/d@ |
+- -+
or, equivalently,
+- -+
| dr/dz d@/dz |
[ dz dz* ] | | = [ dr d@ ] [2]
| dr/dz* d@/dz* |
+- -+
The matrices in [1] and [2] are obviously inverses, so we see
that dr/dz
is not the reciprocal of dz/dr, but simply an element in the
inverse of
the Jacobian.
+- -+
| dr/dz d@/dz |
| |
| dr/dz* d@/dz* |
+- -+
+- -+ -1
| dz/dr dz*/dr |
= | |
| dz/d@ dz*/d@ |
+- -+
+- -+ -1
| z/r z*/r |
= | |
| iz -iz* |
+- -+
+- -+
1 | z* -iz*/r |
= -- | |
2r | z iz/r |
+- -+
Thus, dr/dz = z*/2r = e^{-i@}/2, as you compu from r^2 = zz*.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: Axiom exploring software?
> Surely there are general programs for the purpose of
> shortening strings with some given simple axioms
> (with the usual problem that some strings must be
> heavily lengthened before they can be shortened)?
It seems that you just want to do some rewriting.
Use ELAN or Maude or LP:
http://www.loria.fr/~vigneron/RewritingHP/systems.html
There are other commercial packages you can use (the obvious
captial M computer algebra packages)
There's a whole bunch of automa reasoning software available
that does other kinds of reasoning (first
order/tableau/resolution/etc):
http://www-formal.stanford.edu/clt/ARS/systems.html
--
Mitch Harris
(remove q to reply)
===
Subject: Re: Axiom exploring software?
> What you want is called a theorem prover. There are several
> available, but mainly as parts of larger programs that use
them to
> avoid non-worthwhile calculations. Check on google is my
best advice;
> I don't use them myself very often.
> On the other hand, if you want to prove that you CAN'T
reduce one
> string to another, you'll probably need a program based
that searches
> for invariants; I'm not sure if these are around, but if
the axioms
> are simple enough, it'd be doable, if long and ugly.
In fact, I have only one atom A, brackets, - (unary) and +
(binary
operator) and about 3 axioms (+duals). But complexity is in
the
smallest hut, to paraphrase a German proverb
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
>> [snip] For me to stand with a small
>> silver object in my hand and communicate, was something
that, as
little
>> as 200 years ago, would have had one burned at the stake.
Now it's
>> commonplace and easily understood.

>Could you explain what you mean?
> i think pianowow means a cellphone...either that or a magic
spoon.
I think he was definitely referring to cutlery.
Shaun aRe, born with a magic spoon in his mouth.
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
On a supercalifragilisticexpialidocious day, after dancing
about singing
Bibbety bobbety boo!, Shaun Rimmer ishkabibbled:
^
^>
^>> [snip] For me to stand with a small
^>> silver object in my hand and communicate, was something
that, as
little
^>> as 200 years ago, would have had one burned at the stake.
Now it's
^>> commonplace and easily understood.
^>
^>
^>Could you explain what you mean?
^>
^> i think pianowow means a cellphone...either that or a
magic spoon.
^
^I think he was definitely referring to cutlery.
^
^
^
^Shaun aRe, born with a magic spoon in his mouth.
^
^
^
^
Actually, I think that was MY quote, and I was referring to
technological advances that would have been seen as
witchcraft a few
short centuries ago.
And I was born with a plastic spoon in my mouth. My parents
couldn't
afford silver till much later.
--
The Queen of DXeas well as
Queen of the Commonwealth of Virginia, as well as
The Ruler of A.D.P., as well as
Saint Debbe, as well as
Our Lady of the Black Hole Exploratory Input Services
as OhFishAlly Appoin by the Psychedelic Pope, a/k/a
Saint Isidore of Seville
An Oin Minister of the Universal Life Church
Reverant of the Church of the SubGenius, UnOrthodox
Superior Mutha Superior of the Little Sistahs of the
Politically Incorrect
Worshipper of Eris, Goddess of Discord
I WON'T grow up!! -- Peter Pan
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
> On a supercalifragilisticexpialidocious day, after dancing
about singing
> Bibbety bobbety boo!, Shaun Rimmer ishkabibbled:
> ^
> ^^>> [snip] For me to stand with a small
> ^>> silver object in my hand and communicate, was something
that, as
little
> ^>> as 200 years ago, would have had one burned at the
stake. Now it's
> ^>> commonplace and easily understood.
> ^^^>Could you explain what you mean?
> ^^> i think pianowow means a cellphone...either that or a
magic spoon.
> ^
> ^I think he was definitely referring to cutlery.
> ^
> ^
> ^
> ^Shaun aRe, born with a magic spoon in his mouth.
> ^
> ^
> ^
> ^
> Actually, I think that was MY quote, and I was referring to
> technological advances that would have been seen as
witchcraft a few
> short centuries ago.
Sorry for propagating the misatribution there! Oh, knew
excatly what you
were referring too though Ô,;~}~
> And I was born with a plastic spoon in my mouth. My parents
couldn't
> afford silver till much later.
They shoulda saved up soon as they knew you were on the way!
Shaun aRe
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
X-Who-Cares: Who cares?
> On a supercalifragilisticexpialidocious day, after dancing
> about singing Bibbety bobbety boo!, Shaun Rimmer
> ishkabibbled: ^
> ^>> [snip] For me to stand with a small
> ^>> silver object in my hand and communicate, was
> something that, as little ^>> as 200 years ago, would
> have had one burned at the stake. Now it's ^>commonplace
and easily understood. ^^^>Could you explain what you mean?
> ^^> i think pianowow means a cellphone...either that or a
> magic spoon. ^
> ^I think he was definitely referring to cutlery.
> ^
> ^
> ^
> ^Shaun aRe, born with a magic spoon in his mouth.
> ^
> ^
> ^
> ^
> Actually, I think that was MY quote, and I was referring to
> technological advances that would have been seen as
> witchcraft a few short centuries ago.
All centuries are the same length, bimbo. There are no short
ones. Don't take my word for it. Ask your mommy.
> And I was born with a plastic spoon in my mouth. My parents
> couldn't afford silver till much later.
You can take it out now.
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
> On a supercalifragilisticexpialidocious day, after dancing
> about singing Bibbety bobbety boo!, Shaun Rimmer
> ishkabibbled: ^
> ^>> [snip] For me to stand with a small
> ^>> silver object in my hand and communicate, was
> something that, as little ^>> as 200 years ago, would
> have had one burned at the stake. Now it's ^>commonplace
and easily understood. ^^^>Could you explain what you mean?
> ^^> i think pianowow means a cellphone...either that or a
> magic spoon. ^
> ^I think he was definitely referring to cutlery.
> ^
> ^
> ^
> ^Shaun aRe, born with a magic spoon in his mouth.
> ^
> ^
> ^
> ^
> Actually, I think that was MY quote, and I was referring to
> technological advances that would have been seen as
> witchcraft a few short centuries ago.
> All centuries are the same length, bimbo.
Heheheh! Ya hear that Queenie? You're a bimbo now! LOL, FFS!
> There are no short
> ones. Don't take my word for it. Ask your mommy.
> And I was born with a plastic spoon in my mouth. My parents
> couldn't afford silver till much later.
> You can take it out now.
Silly little idiot!
Shaun aRe
===
Subject: Re: wooooooooooosssssssssshhhhhhhhhhh.............
>> [snip] For me to stand with a small
>> silver object in my hand and communicate, was something
that, as little
>> as 200 years ago, would have had one burned at the stake.
Now it's
>> commonplace and easily understood.
>Could you explain what you mean?
i think pianowow means a cellphone...either that or a magic
spoon.
===
Subject: Re: C^n proper inclusions problem
>> The function f(x) = |x|*x^n is in C^n(R) but not in
C^(n+1)(R). If a is
in
>> U, an open subset of R^k, then the function x -> f(x1 -
a1) is in C^n(U)
>> but not in C^(n+1)(U).
>I don't understand your notation :
>What is x1 and a1?
x = (x1, x2, ..., xk).
>R denotes the reals?
Yes.
************************
===
Subject: Re: C^n proper inclusions problem
Show in general that the inclusions
. C^{n}(U) subset of C^{n-1}(U) subset of ... subset of
C^{2}(U)
subset
of C^{1}(U)
>are proper. (Clearly if I dele are proper this would hold)
>But otherwise? Can I look for a function that is in C^{1}(U)
and not in
>C^{2}(U) and then somehow extend it in general to C^{n}(U)?
>C^n denotes functions whose partial derivatives up to order
n are all
>continuous and exist.
>Let a be any value in the first projection of U. Let f_0(x) 
=
>>|x-a|, f_(n+1) (x) = integral(a..x, f_n). Let g_n(x) =
f_n(x1). Then
>>g_n is in C^n, not in C^(n+1)
>What do you mean by the first projection of U? Also, as a
notation
>confusion, what do you mean by x1?
x1 = the first coordinate of the vector x. The 
first projection
of U is {x1: x in U}. I assume U is an open subset of R^m for
some m.
--
===
Subject: Re: C^n proper inclusions problem
>Show in general that the inclusions
>
. C^{n}(U) subset of C^{n-1}(U) subset of ... subset of
C^{2}(U)
subset
>of C^{1}(U)
are proper. (Clearly if I dele are proper this would hold)
But otherwise? Can I look for a function that is in C^{1}(U)
and not in
>C^{2}(U) and then somehow extend it in general to C^{n}(U)?
>C^n denotes functions whose partial derivatives up to order
n are all
>continuous and exist.
Let a be any value in the first projection of U. Let f_0(x) =
> |x-a|, f_(n+1) (x) = integral(a..x, f_n). Let g_n(x) =
f_n(x1). Then
> g_n is in C^n, not in C^(n+1)
What do you mean by the first projection of U? Also, as a
notation
confusion, what do you mean by x1?
> --
>
===
Subject: Madamard matrix
I read about Hadamard matrices and error-correction codes on
MathWorld.
I don't quite understand how to go from one iteration to the
next in the
given examples, but I inven the following simple system:
1.
start with
[ 1 -1 ]
[ -1 -1 ]
2.
now replace each number by a new 2x2 matrix
[ 1 -1 ]
[ -1 -1 ]
if the number is 1,
[ -1 1 ]
[ 1 1 ]
if the number is -1.
3.
repeat step 2
I found that the Hamming distance between each 2 row vectors
in a 2^n x 2^n
matrix is 2^(n-1). I checked this for up to n=12; I can't
prove it.
Now I can't believe I would be the first person 
to construct
this, so my
question is: does this type of matrix have a specific name? Is
it used as
error correction code?
BTW, I used the following recursive Delphi procedure to
construct the
matrix:
procedure TForm1.IterateMat(ALevel, AX, AY: Word; AValue:
ShortInt);
var
Offset: Word;
begin
Mat[AX, AY] := AValue;
if ALevel > 0 then begin
Offset := 1 shl (ALevel - 1);
IterateMat(ALevel - 1, AX, AY, AValue); // remove for
Sierpinski
triangle
IterateMat(ALevel - 1, AX, AY + Offset, -AValue);
IterateMat(ALevel - 1, AX + Offset, AY, -AValue);
IterateMat(ALevel - 1, AX + Offset, AY + Offset, -AValue);
end;
end;
which I called by
IterateMat(12, 0, 0, 1);
As commen in the procedure, if you remove the first recursive
call, you
get an isosceles right Sierpinski triangle, but I guess that
was known
also.
--
Steven
I'm not a mathematician, so please type slowly.
Remove pants if you want to reply by e-mail.
===
Subject: Re: Madamard matrix
This construction was first given (as far as I know) by
Sylvester in
J. J. Sylvester, Thoughts on inverse orthogonal matrices,
simultaneous
sign successions, and tessela pavements in two or more
colours,
with applications to Newton's rule, ornamental tile-work, 
and
the
theory of numbeLondon Edinburgh and Dublin os. Mag. and J.
Sci. 34 (1867) 461-475.
Its use in error-correction is described in many places,
including the
book
by F. J. MacWilliams and N. J. A. Sloane (see
http://www.research.att.com/~njas/doc/ms77.html).
===
Subject: Re: Madamard matrix
Injector-Info: news.mailgate.org;
posting-host=adsl-66-126-134-202.dsl.frsn01.pacbell.net;
posting-account=48257; posting-date=1074589859
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
ddc0da153f0ebb2b67f77a9aa5ddca
0c.48257%40mygate.mailgate.org
> 1.
> start with
> [ 1 -1 ]
> [ -1 -1 ]
2.
> now replace each number by a new 2x2 matrix
> [ 1 -1 ]
> [ -1 -1 ]
> if the number is 1,
> [ -1 1 ]
> [ 1 1 ]
> if the number is -1.
That looks quite a bit like the mechanism used to produce
the redundent chip coding of Code Domain Multiple Access
technology, except the pair are [0,1] rather than [-1,1].
See if this helps:
http://www.google.com/search?q=walsh.code
> I'm not a mathematician, so please type slowly.
I hope that sufficed, touch typists have problems with their
throttles sometimes. I used to be a mathematician, back about
1968. It didn't stick.
xanthian, former Qualcomm employee.
--
Pos via Mailgate.ORG Server - http://www.Mailgate.ORG
===
Subject: Re: Madamard matrix
> I read about Hadamard matrices and error-correction codes
on MathWorld.
> I don't quite understand how to go from one iteration to
the next in the
> given examples, but I inven the following simple system:
1.
> start with
> [ 1 -1 ]
> [ -1 -1 ]
2.
> now replace each number by a new 2x2 matrix
> [ 1 -1 ]
> [ -1 -1 ]
> if the number is 1,
> [ -1 1 ]
> [ 1 1 ]
> if the number is -1.
That is, replace the number a by
a ( 1 -1)
(-1 -1).
This is a special case of the well-known fact that the
Kronecker product of two Hadamard matrices is Hadamard.
See e.g., Hughes/Piper, _Design Theory_, CUP, 1985, Theorem
3.23.
--
===
Subject: Re: need help in understanding Torkel's ZFC comment
>
I'd love to see a couple of specific examples of 
formal proofs
of
>>consistency of subsets of ZFC in ZFC. (Actually, I just
want to test
>>out my simplification of ZFC and formal proofs within ZFC
are so hard
>>to find.)
> Actually ZFC has infinitely many axioms. Those 
infinitely
many
>> axioms can be described by finitely many axiom _schemata_.
>> (Schema? Oh hell, schemes?)
************************
Ok. I have always seen them lis as 10 (or less) axioms,
People use words in different ways in different contexts -
> those things that you've seen called the ten axioms are
> more correctly called axioms schemes. The author didn't
> think the difference mattered, but in the present context
> it does.
How do you know what the author thought? I didn't say where 
I
saw it.
How do you know that none of these contexts weren't similar
to the
present context and thus the difference matters there as well?
>but I
>agree with the interpretation given by you and others - to a
degree.
Giggle. The interpretations that have been given have just
been
> clarifications of what the words mean.
I'm at a loss as to the point of writing Giggle, and would 
be
glad
to hear an explanation.
I'm not talking about the interpretation of the words. 
I'm
talking
about the interpretation of the intuitive concepts that ZFC is
attempting to formalize. ZFC interprets them as all being
axioms (or
the illformed axiom schema.) I am saying that these various
concepts are in fact definitions, axioms, rules of inference
and
theorems, as I detailed, and could (should) be formalized as
such.
> You can agree to a degree,
> or if you want to actually get it right you could simply
agree - I
> mean it's not like there's anything to 
actually _agree_ to
here...
I agree that some of the axioms are sets of axioms, but they
can
better be formalized as rules of inference, as I explained.
>I have always felt that axiom schemas are just rules of
inference.
Giggle. I've always felt that automobiles were actually 
pieces
> of fruit. I can't understand why nobody seems interes in
> what I've always believed.
Maybe it's because you haven't explained or 
justified it as I
have.
So what are the Rules of Inference of ZFC then?
> ************************
===
Subject: Re: need help in understanding Torkel's ZFC comment
Charlie-Boo says...
>I'm not talking about the interpretation of the words. 
I'm
talking
>about the interpretation of the intuitive concepts that ZFC
is
>attempting to formalize. ZFC interprets them as all being
axioms (or
>the illformed axiom schema.) I am saying that these various
>concepts are in fact definitions, axioms, rules of inference
and
>theorems, as I detailed, and could (should) be formalized as
such.
Well, there are different ways to formalize a first-order
theory.
One extreme is to have only rules of inference. The other
extreme
is to have only a single rule of inference, modus-ponens, and
all
the content of the theory is in the axioms. This latter,
which is
called a Hilbert-style formal system, has the advantage that
it
leads to a very simple notion of what a valid proof is:
A proof is a sequence Phi_0, Phi_1, ..., Phi_n of formulas
such
that each formula is either an axiom, or follows by
modus-ponens
from earlier formulas.
The drawback is that for many theories, this requires an
infinite
number of axioms. The only restriction is that for what is
called
an axiomatizable theory, the set of axioms must be r.e. That
is,
there must be a computable function nthAxiom(n) such that
nthAxiom(n) = the nth axiom. An alternative characterization,
which turns out to be equivalent, is to require that there be
a
computable function isAnAxiom(x) which returns 1 if x is an
axiom,
and 0 otherwise.
Anyway, if you formalize ZFC in this way, it requires an
infinite
number of axioms. But that doesn't really matter, as long as
the
functions nthAxiom or isAnAxiom can be finitely 
specified. The
axiom schemas give an implicit rule for generating (or
recognizing)
the axioms of ZFC.
When people say that this or that theory is or is not finitely
axiomatizable, they are talking about its formalization as a
Hilbert-style formal system. If you consider axiom schemas to
be rules of inference, then you can always get away with a
finite
number of axioms (or even no axioms at all, only rules of
inference).
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: need help in understanding Torkel's ZFC comment
> Charlie-Boo says...
I'm not talking about the interpretation of the words. 
I'm
talking
>about the interpretation of the intuitive concepts that ZFC
is
>attempting to formalize. ZFC interprets them as all being
axioms (or
>the illformed axiom schema.) I am saying that these various
>concepts are in fact definitions, axioms, rules of inference
and
>theorems, as I detailed, and could (should) be formalized as
such.
Well, there are different ways to formalize a first-order
theory.
> One extreme is to have only rules of inference. The other
extreme
> is to have only a single rule of inference, modus-ponens,
and all
> the content of the theory is in the axioms.
How about the case where you formalize each concept according
to what
works best? In particular, has anyone every formalized ZFC so
that it
is nonempty sets of definitions, axioms, rules of inference 
and
theorems? (In the case of theorems, I am saying that some of
ZFC is
dropped altogether because it is derivable from the rest.)
And do they always use the combination of Predicate Calculus
(logical
connectives and quantifiers) and Set Theory (sets and set
membership)
syntax and semantics in which ZFC is expressed?
The third improvment that I make is to abstract out and make
into a
primitive the notion of expressing a collection, which
appears in
almost every ZFC axiom and also in the context of both
formulas and
sets. Who has done that before?
I have never seen ZFC formalized much differently than the
original
system.
> This latter, which is
> called a Hilbert-style formal system, has the advantage
that it
> leads to a very simple notion of what a valid proof is:
> A proof is a sequence Phi_0, Phi_1, ..., Phi_n of formulas
such
> that each formula is either an axiom, or follows by
modus-ponens
> from earlier formulas.
How is that different if we have a mixture of axioms and
rules?
> the set of axioms must be r.e.
> An alternative characterization,
> which turns out to be equivalent, is to require that there
be a
> computable function isAnAxiom(x) which returns 1 if x is an
axiom,
> and 0 otherwise.
Recursively enumerable is equivalemnt to recursive??
===
Subject: Re: need help in understanding Torkel's ZFC comment
Charlie-Boo says...
>> A proof is a sequence Phi_0, Phi_1, ..., Phi_n of formulas
such
>> that each formula is either an axiom, or follows by
modus-ponens
>> from earlier formulas.
>How is that different if we have a mixture of axioms and
rules?
It depends on what the rules are.
>> the set of axioms must be r.e.
>> An alternative characterization,
>> which turns out to be equivalent, is to require that there
be a
>> computable function isAnAxiom(x) which returns 1 if x is
an axiom,
>> and 0 otherwise.
>Recursively enumerable is equivalemnt to recursive??
No, but any theory with an r.e. set of axioms also can be
formalized with a recursive set of axioms.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: need help in understanding Torkel's ZFC comment
> (snip)
No, but any theory with an r.e. set of axioms also can be
> formalized with a recursive set of axioms.
This strikes me as implausible. What if you have an r.e. but
not
recursive set S.Define first order theory T with a 
constant 0, a
unary operation s, and a unary predicate R. Let 3 be
shorthand for
s(s(s(0))) etc. Let the *only* axioms of T be of the form
R(n), for
each n in S. You could doubtless expand T to something like T
+ PA in
such a way that you could axiomatize the recursive function
that
enumerates S, but in what sense could the resulting theory be
considered just another formalization of the original theory?
It seems
that there is an unavoidable overhead in passing from an r.e.
axiomatization to a recursive one. (Its been over a decade
since I've
studied logic - but I suspect that some notion of
interpretation of
one theory in another must lurk here).
-still scattered after all these years (you should see my
desk)
===
Subject: Re: need help in understanding Torkel's ZFC comment
still@scattered@yahoo.com says...
>> No, but any theory with an r.e. set of axioms also can be
>> formalized with a recursive set of axioms.
>This strikes me as implausible. What if you have an r.e. but
not
>recursive set S.Define first order theory T with 
a constant 0,
a
>unary operation s, and a unary predicate R. Let 3 be
shorthand for
>s(s(s(0))) etc. Let the *only* axioms of T be of the form
R(n), for
>each n in S.
Okay. Here's a fact about r.e. sets: every r.e. set is the
range
of some total recursive function (primitive recursive,
actually).
So let f be some total recursive function such that
S = { f(0), f(1), f(2), etc. }
Then we can axiomatize your theory T as follows:
Let A_0 = R(m_0)
Let A_{n+1} = A_n & R(m_n)
where m_n is the numeral representing the number f(n). It
isn't
necessary that f be definable in the theory, it is only
necessary
that f be used to determine the numerals m_0, m_1, etc.
The set of axioms A = { A_0, A_1, A_2, ... } is recursive, and
it has the same theorems as the set { R(0), R(1), R(2), ... }.
To see that A is recursive, here is a decision procedure to
see whether a formula Phi is in A:
Start generating the axioms A_0, A_1, A_2, ... in order.
When you come to some axiom A_n whose size (in number of
characters) is greater than or equal to Phi, then stop.
If the last axiom genera is equal to Phi, then Phi is
an axiom. Otherwise, Phi is not an axiom.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: need help in understanding Torkel's ZFC comment
>Charlie-Boo says...
>>I'm not talking about the interpretation of the words. 
I'm
talking
>>about the interpretation of the intuitive concepts that ZFC
is
>>attempting to formalize. ZFC interprets them as all being
axioms (or
>>the illformed axiom schema.) I am saying that these various
>>concepts are in fact definitions, axioms, rules of inference
and
>>theorems, as I detailed, and could (should) be formalized
as such.
>Well, there are different ways to formalize a first-order
theory.
>One extreme is to have only rules of inference. The other
extreme
>is to have only a single rule of inference, modus-ponens,
and all
>the content of the theory is in the axioms. This latter,
which is
>called a Hilbert-style formal system, has the advantage that
it
>leads to a very simple notion of what a valid proof is:
> A proof is a sequence Phi_0, Phi_1, ..., Phi_n of formulas
such
> that each formula is either an axiom, or follows by
modus-ponens
> from earlier formulas.
>The drawback is that for many theories, this requires an
infinite
>number of axioms. The only restriction is that for what is
called
>an axiomatizable theory, the set of axioms must be r.e. That
is,
>there must be a computable function nthAxiom(n) such that
>nthAxiom(n) = the nth axiom. An alternative characterization,
>which turns out to be equivalent, is to require that there
be a
>computable function isAnAxiom(x) which returns 1 if x is an
axiom,
>and 0 otherwise.
??? The second requirement seems to me to say that the
set of axioms is _recursive_, which is stronger than just
being recursively enumerable. (I would have thought that
the set of axioms would typically be required to be recursive,
since if we can't recognize an axiom when we see one 
we're
going to have a hard time verifying the correctness of a
proof...)
Why are these two equivalent?
>Anyway, if you formalize ZFC in this way, it requires an
infinite
>number of axioms. But that doesn't really matter, as long 
as
the
>functions nthAxiom or isAnAxiom can be finitely 
specified. The
>axiom schemas give an implicit rule for generating (or
recognizing)
>the axioms of ZFC.
>When people say that this or that theory is or is not 
finitely
>axiomatizable, they are talking about its formalization as a
>Hilbert-style formal system. If you consider axiom schemas to
>be rules of inference, then you can always get away with a
finite
>number of axioms (or even no axioms at all, only rules of
inference).
************************
===
Subject: Re: need help in understanding Torkel's ZFC comment
says...
>>The drawback is that for many theories, this requires an
infinite
>>number of axioms. The only restriction is that for what is
called
>>an axiomatizable theory, the set of axioms must be r.e.
That is,
>>there must be a computable function nthAxiom(n) such that
>>nthAxiom(n) = the nth axiom. An alternative
characterization,
>>which turns out to be equivalent, is to require that there
be a
>>computable function isAnAxiom(x) which returns 1 if x is an
axiom,
>>and 0 otherwise.
>??? The second requirement seems to me to say that the
>set of axioms is _recursive_, which is stronger than just
>being recursively enumerable. (I would have thought that
>the set of axioms would typically be required to be
recursive,
>since if we can't recognize an axiom when we see one 
we're
>going to have a hard time verifying the correctness of a
proof...)
>Why are these two equivalent?
I could swear that we've had exactly this discussion before.
I said turns out to be equivalent because it's not 
immediately
obvious that they are equivalent, but they are.
Let A = { A1, A2, ... } be an r.e. set of axioms that is
*not* recursive.
Then let B = { B1, B2, ... } be defined as follows:
B1 = A1
B2 = A1 & A2
B3 = A1 & A2 & A3
etc.
Then the set { B1, B2, B3, ... } is a recursive set of axioms
that
are equivalent to the r.e. set { A1, A2, A3, ... }
(equivalent in
the sense that they have the same theorems).
Do you want me to prove that B is recursive, or is that
obvious?
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: need help in understanding Torkel's ZFC comment
> says...
>The drawback is that for many theories, this requires an
infinite
>number of axioms. The only restriction is that for what is
called
>an axiomatizable theory, the set of axioms must be r.e. That
is,
>there must be a computable function nthAxiom(n) such that
>nthAxiom(n) = the nth axiom. An alternative characterization,
>which turns out to be equivalent, is to require that there
be a
>computable function isAnAxiom(x) which returns 1 if x is an
axiom,
>and 0 otherwise.
>>??? The second requirement seems to me to say that the
>>set of axioms is _recursive_, which is stronger than just
>>being recursively enumerable. (I would have thought that
>>the set of axioms would typically be required to be
recursive,
>>since if we can't recognize an axiom when we see one 
we're
>>going to have a hard time verifying the correctness of a
proof...)
>>Why are these two equivalent?
>I could swear that we've had exactly this discussion 
before.
Hmm, probably we have, what's below rings a bell. Sorry.
(To be fair to myself, the above looks like you're saying 
that
the two conditions on a set of axioms are equivalent, which
of course they're not. If you'd said that _the 
existence of_
an r.e. set of axioms was equivalent to _the existence
of_ an r. set of axioms I might have had a better chance
recalling the argument, because I wouldn't have been
distrac by the fact that what you seemed to be saying
was clearly false...)
>I said turns out to be equivalent because it's not
immediately
>obvious that they are equivalent, but they are.
>Let A = { A1, A2, ... } be an r.e. set of axioms that is
*not* recursive.
>Then let B = { B1, B2, ... } be defined as follows:
> B1 = A1
> B2 = A1 & A2
> B3 = A1 & A2 & A3
> etc.
>Then the set { B1, B2, B3, ... } is a recursive set of
axioms that
>are equivalent to the r.e. set { A1, A2, A3, ... }
(equivalent in
>the sense that they have the same theorems).
>Do you want me to prove that B is recursive, or is that
obvious?
No, that's fine.
************************
===
Subject: Re: need help in understanding Torkel's ZFC comment
> The other extreme
> is to have only a single rule of inference, modus-ponens,
and all
> the content of the theory is in the axioms. This latter,
which is
> called a Hilbert-style formal system,
Most Hilbert-style formalizations also have a generalization
rule.
> An alternative characterization,
> which turns out to be equivalent, is to require that there
be a
> computable function isAnAxiom(x) which returns 1 if x is an
axiom,
> and 0 otherwise.
Or rather, the corresponding axiomatizability requirements are
equivalent.
> If you consider axiom schemas to
> be rules of inference, then you can always get away with a
finite
> number of axioms (or even no axioms at all, only rules of
inference).
Or you can formulate a finitely axiomatizable conservative
extension
of the theory.
An exercise (from a logic book) which touches on this: where
does
the schematic proof of A->A has model break down for the
finitely
axiomatizable theory NBG?
===
Subject: Re: need help in understanding Torkel's ZFC comment
I'd love to see a couple of specific examples of 
formal proofs
of
>consistency of subsets of ZFC in ZFC. (Actually, I just want
to
test
>out my simplification of ZFC and formal proofs within ZFC are
so
hard
>to find.)
> Actually ZFC has infinitely many axioms. Those 
infinitely many
> axioms can be described by finitely many axiom _schemata_.
> (Schema? Oh hell, schemes?)
>> ************************
Ok. I have always seen them lis as 10 (or less) axioms,
People use words in different ways in different contexts -
>> those things that you've seen called the ten axioms are
>> more correctly called axioms schemes. The author didn't
>> think the difference mattered, but in the present context
>> it does.
>How do you know what the author thought? I didn't say where
I saw it.
>How do you know that none of these contexts weren't similar
to the
>present context and thus the difference matters there as
well?
You'd probably have more fruitful discussions with people if
you
didn't argue just for the sake of arguing, asking questions
that
you know the answer to very well. (For example, here, assuming
you're not simply stupid, it's clear that you 
know that you're
asking silly questions, because it's clear that I 
wasn't
actually
claiming to know what some author thought, just offering a
likely explanation for how it could happen that you saw
those things lis as axioms in a source written by
someone who knew what he was talking about, even
though they're not _actually_ axioms, in the strict
sense of the word being used elsewhere in this
thread.)
>>but I
>>agree with the interpretation given by you and others - to
a degree.
Giggle. The interpretations that have been given have just
been
>> clarifications of what the words mean.
>I'm at a loss as to the point of writing Giggle, and would
be glad
>to hear an explanation.
>I'm not talking about the interpretation of the words. 
I'm
talking
>about the interpretation of the intuitive concepts that ZFC
is
>attempting to formalize.
Also it would work better if you didn't pull this sort of
swithcheroo
- you referred to the interpretations given by me and others.
I've explained a little bit about what certain words mean, 
but
I haven't said _anything_ here about the interpretation of 
the
intuitive concepts that ZFC is attempting to formalize.
previous things you said indicate you haven't been paying
attention to who said what in you replies - whichever the
problem is, things would work better if you avoided that
behavior.
> ZFC interprets them as all being axioms (or
>the illformed axiom schema.)
What in the world is illformed here?
>I am saying that these various
>concepts are in fact definitions, axioms, rules of inference
and
>theorems, as I detailed, and could (should) be formalized as
such.
>> You can agree to a degree,
>> or if you want to actually get it right you could simply
agree - I
>> mean it's not like there's anything to 
actually _agree_ to
here...
>I agree that some of the axioms are sets of axioms,
I imagine you also agree that triangles have three sides?
Great.
(That's an attempt at explaning what I find 
amusing about what
you say you agree to...)
>but they can
>better be formalized as rules of inference, as I explained.
This is just silly. _Any_ axiom or axiom scheme can be
formalized
as a rule of inference. (i) this has nothing to do with the
distinction between axioms and axioms schemes, (ii) you
haven't
given any evidence for the better part (saying that this
formalization is better or worse than that one, _when_ 
wer're
talking about this level of unimportant detail, seems
exquisitely
in ZFC when they're doing mathematics _anyway_.)
>>I have always felt that axiom schemas are just rules of
inference.
Giggle. I've always felt that automobiles were actually 
pieces
>> of fruit. I can't understand why nobody seems interes in
>> what I've always believed.
>Maybe it's because you haven't explained or 
justified it as I
have.
>So what are the Rules of Inference of ZFC then?
You're lecturing us on this stuff and you really 
don't know?
Wow.
Hint: there's really no such thing as the rules of inference
of
ZFC per se. In what seems to me to be the most common
formalizations there are a small number of purely logical
rules of inference - sometimes just modus ponens, sometimes
modus ponens plus universal generalization, depending
on how one prefers to set up ones formal system.
> ************************
************************
===
Subject: Re: need help in understanding Torkel's ZFC comment
>>I'd love to see a couple of specific examples 
of formal
proofs of
>>consistency of subsets of ZFC in ZFC. (Actually, I just
want to test
>>out my simplification of ZFC and formal proofs within ZFC
are so hard
>>to find.)
>> Actually ZFC has infinitely many axioms. Those 
infinitely
many
>> axioms can be described by finitely many axiom _schemata_.
>> (Schema? Oh hell, schemes?)
>> ************************
>Ok. I have always seen them lis as 10 (or less) axioms,
People use words in different ways in different contexts -
those things that you've seen called the ten axioms are
more correctly called axioms schemes. The author didn't
think the difference mattered, but in the present context
it does.
>but I
>agree with the interpretation given by you and others - to a
degree.
Giggle. The interpretations that have been given have just
been
clarifications of what the words mean. You can agree to a
degree,
or if you want to actually get it right you could simply
agree - I
mean it's not like there's anything to 
actually _agree_ to
here...
>I have always felt that axiom schemas are just rules of
inference.
Giggle. I've always felt that automobiles were actually 
pieces
of fruit. I can't understand why nobody seems interes in
what I've always believed.
************************
===
Subject: Diophant equation
How do I solve Diophant equations like these:
x^2-2y^2=5
x^2-17y^2=1 ?
===
Subject: Re: Diophant equation
x^2-2y^2=5 --> x=2a+1 --> 4a^2+4a-2y^2=4 --> y=2b
-->4a^2+4a-8b^2=4 -->
a^2+a -2b^2=1, or a(a+1)-2b^2=1 --> 2 divides 1, impossible.
In fact,
x^2-2y^2=8k+5 has no solutions, via the same argument. HTH.
Robert
> How do I solve Diophant equations like these:
> x^2-2y^2=5
> x^2-17y^2=1 ?
===
Subject: Re: Diophant equation
al.mauri@hznet.hr a .8ecrit:
> How do I solve Diophant equations like these:
> x^2-2y^2=5
> x^2-17y^2=1 ?
I suggest to Google pell equation, and generalized pell
equation
You get there detailed methods for solving the pell equation
(x^2-D*y^2=1) and the generalized pell equation (x^2-D*y^2=K)
There are even on-line solvers for that.
My own solver gives x^2-17*y^2=1 fundamental's solution is
x_0=33 y_0=8
There are infinite number of solutions given by
x_(n+1) = 33*x_(n)+17*8*y_(n)
y_(n+1) = 8*x_(n)+33*y_(n)
x^2-2y^2=5 seems to have no solutions
as said by the on line solver at
http://www.alpertron.com.ar/QUAD.HTM
regards
--
===
Subject: Re: Diophant equation
En el mensaje:400D2C5A.2050301@freefr.invalid,
ippe che <anti-chephip-spam@freefr.invalid> escribi.97:
 al.mauri@hznet.hr a .8ecrit:
>> How do I solve Diophant equations like these:
>> x^2-2y^2=5
>> x^2-17y^2=1 ?
> I suggest to Google pell equation, and generalized pell
equation
> You get there detailed methods for solving the pell equation
> (x^2-D*y^2=1) and the generalized pell equation
(x^2-D*y^2=K)
> There are even on-line solvers for that.
> My own solver gives x^2-17*y^2=1 fundamental's solution is
x_0=33
> y_0=8 There are infinite number of solutions given by
> x_(n+1) = 33*x_(n)+17*8*y_(n)
> y_(n+1) = 8*x_(n)+33*y_(n)
> x^2-2y^2=5 seems to have no solutions
> as said by the on line solver at
> http://www.alpertron.com.ar/QUAD.HTM
That equation hasn't integer solutions, as can be sawn 
easily:
x must be odd ===> x^2 = 1 (mod 8) ===>
2y^2 = 4 (mod 8)
But this is imposible, because if y is odd, then y^2 = 1 (mod
8), and 2y^2
=
2 (mod 8); and if y is even, then y^2 = 0 or 4 (mod 8), and
2y^2 = 0 (mod
8)
--
Saludos,
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
Subject: Re: JSH: Limi intellects
===
>I've concluded that the problem I'm facing is 
that the human
brain
>>isn't built to handle Mathematics. Human beings have done
well with
>>some simple mathematics and basic concepts, but it only
goes so far,
>>and what have, is beyond those limits.
> I thought of myself as calm and cool about that assessment.
I had assumed you were simply frustra with the lack of
acceptance
your ideas were receiving. That isn't necessarily a bad 
mood.
Still I'm facing an incredibly strange situation which 
defies
easy
> explanation.
The explanation has been offered. We honestly and truly
believe that
you are wrong. What makes it seems strange is that you
honestly and
truly believe that you are right.
Consider, mathematicians *supposedly* care about pure math
and the
> importance of mathematical knowledge, but have for close to
two years
> now obstinately refused to accept some rather dramatic
results.
Because we believe they are wrong, and have provided the
appropriate
evidence to back up that position.
It doesn't help when your conclusion is that an error lies 
in
a definition.
Rather than face intrigued interest and help in further
research I've
> been ignored and insul, as if mathematicians *hate* me for
finding
> what I have.
No, there is no hate. Amusement, and what could be considered
cruel
sport by some, but no hate. Who resor to personal attacks
first?
It's like there's some kind of odd war, where 
they will fight
no
> matter what, against all reason, against the very values
they claimed
> to have, and for what?
Then why is your math rebut with math, instead of personal
attacks?
Why is our math rebut with personal attacks as often as with
math?
That's what I guess I need to figure out now.
You need to drop one of your axioms: I, , am right.
--
===
Subject: Re: JSH: Limi intellects
how's the marketing going, monsieur Harris?
> You need to drop one of your axioms: I, , am right.
as you know, if you've read anyhting that I typed,
there was a conspiracy of states that DID vote Gore -- and
the Supremes sealed that conspiracy on March 27, 2000,
by refusing to hear the appeal in LaRouche v. Fowler
(Don Fowler was the DNC Chair in Ô96;
Sentelle's 3-judge panel made the Voting Rights Act
unconsitutitonal.
but, hey; it's up for re-auhtorization in Ô07 
!-)
I am frankly scared of the touchscreen mentality. just like
the Supremes' abrogation of the USA and Florida 
constitutions
foments
a pop-culture hatred of the electoral college
on the part of some rabid Democrats. ah, so;
imagine if North Dakota ... rather, imagine if
Wyoming had less than one electoral college vote for
president.
anyway,
every one of us in this debate knows
about the Texas cirterium for chad --
it ain't just missing confetti!
thus saith:
but, at the national level,
will you be able to maintain the DNC's Any One But George --
unless it's Lyndon! media glut?... are you on the board
of GOPpers For Howie Troisieme?
http://www.wlym.com/pdf/iclc/communism.pdf
--Give the Gift of Dick Cheeny -- out of office, at last!
http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: JSH: Limi intellects
says...
>> I've concluded that the problem I'm facing 
is that the
human brain
>> isn't built to handle Mathematics. Human beings have done
well with
>> some simple mathematics and basic concepts, but it only
goes so far,
>> and what have, is beyond those limits.
>I thought of myself as calm and cool about that assessment.
>Still I'm facing an incredibly strange situation which 
defies
easy
>explanation.
>Consider, mathematicians *supposedly* care about pure math
and the
>importance of mathematical knowledge, but have for close to
two years
>now obstinately refused to accept some rather dramatic
results.
>Rather than face intrigued interest and help in further
research I've
>been ignored and insul, as if mathematicians *hate* me for
finding
>what I have.
>It's like there's some kind of odd war, where 
they will fight
no
>matter what, against all reason, against the very values
they claimed
>to have, and for what?
>That's what I guess I need to figure out now.
Consider this:
1) How would the mathematicians' behaviour differ from that
described above
if
you were wrong, if there was a genuine error in your work?
Might they not
be
expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
In other words - the behaviour looks strange if your work is
definitely
correct.
If you have made an error then it's not so strange is it?
2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
and later discovered them and admit them. In some of these
cases you
were
absolutely convinced that your work was correct right up
until the point
where
you discovered the mistake.
Given this, and given that *no-one* other than yourself
believes your
current
proof to be correct - shouldn't you at least consider the
possibility that
you
might have made a mistake here too?
===
Subject: Re: JSH: Limi intellects
> says...
> I've concluded that the problem I'm facing 
is that the
human brain
>> isn't built to handle Mathematics. Human beings have done
well with
>> some simple mathematics and basic concepts, but it only
goes so far,
>> and what have, is beyond those limits.
I thought of myself as calm and cool about that assessment.
Still I'm facing an incredibly strange situation which 
defies
easy
>explanation.
Consider, mathematicians *supposedly* care about pure math
and the
>importance of mathematical knowledge, but have for close to
two years
>now obstinately refused to accept some rather dramatic
results.
Rather than face intrigued interest and help in further
research I've
>been ignored and insul, as if mathematicians *hate* me for
finding
>what I have.
It's like there's some kind of odd war, where 
they will fight
no
>matter what, against all reason, against the very values
they claimed
>to have, and for what?
That's what I guess I need to figure out now.
>
Consider this:
1) How would the mathematicians' behaviour differ from that
described
above if
> you were wrong, if there was a genuine error in your work?
Might they not
be
> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
Certainly.
> In other words - the behaviour looks strange if your work
is definitely
correct.
> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't be.
> 2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
> and later discovered them and admit them. In some of these
cases you
were
> absolutely convinced that your work was correct right up
until the point
where
> you discovered the mistake.
Yup.
> Given this, and given that *no-one* other than yourself
believes your
current
> proof to be correct - shouldn't you at least consider the
possibility that
you
> might have made a mistake here too?
What proof are you talking about?
I think it fascinating to see how much you post based on
assumptions.
So then, what work of mine do you have in mind Paul Mason?
===
Subject: Re: JSH: Limi intellects
says...
>> says...
> I've concluded that the problem I'm facing 
is that the
human brain
> isn't built to handle Mathematics. Human beings have done
well with
> some simple mathematics and basic concepts, but it only
goes so far,
> and what have, is beyond those limits.
>
I thought of myself as calm and cool about that assessment.
Still I'm facing an incredibly strange situation which 
defies
easy
>>explanation.
Consider, mathematicians *supposedly* care about pure math
and the
>>importance of mathematical knowledge, but have for close to
two years
>>now obstinately refused to accept some rather dramatic
results.
Rather than face intrigued interest and help in further
research I've
>>been ignored and insul, as if mathematicians *hate* me for
finding
>>what I have.
It's like there's some kind of odd war, where 
they will fight
no
>>matter what, against all reason, against the very values
they claimed
>>to have, and for what?
That's what I guess I need to figure out now.
>> Consider this:
1) How would the mathematicians' behaviour differ from that
described
above if
>> you were wrong, if there was a genuine error in your work?
Might they not
be
>> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
>Certainly.
In which case you're not Ôfacing an incredibly 
strange
situation which
defies
easy explanation' are you? There is an easy explanation - 
you
simply find
it
unpalatable.
>In other words - the behaviour looks strange if your work is
definitely
correct.
>> If you have made an error then it's not so strange is it?
>Well I think the insults are strange, while being ignored
wouldn't be.
Usenet is a place where insults ßy freely. Add to that your
own tendency
to
insult others and I don't find this at all 
strange.
And you're not being ignored. You get many more replies than
any other
single
poster. You're being disagreed with, not ignored.
> 2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
>> and later discovered them and admit them. In some of these
cases you
were
>>absolutely convinced that your work was correct right up
until the point
where
>> you discovered the mistake.
>Yup.
> Given this, and given that *no-one* other than yourself
believes your
current
>>proof to be correct - shouldn't you at least consider the
possibility that
you
>> might have made a mistake here too?
>What proof are you talking about?
I was referring to your proof of a problem with the definition
of algebraic
integers. Were you *not* referring to that when you mentioned
Ôdramatic
results'
above?
>I think it fascinating to see how much you post based on
assumptions.
Which assumptions are you thinking of? Feel free to correct
me.
>So then, what work of mine do you have in mind Paul Mason?
See above.
I notice you've not answered my final question - 
are you able
to now that
I've
answered yours? I repeat - shouldn't you consider the
possibility that
you've
made a mistake and that's why your work hasn't 
been accep?
===
Subject: Re: JSH: Limi intellects
> says...
>
<dele> Consider this:
1) How would the mathematicians' behaviour differ from that
described
above if
>> you were wrong, if there was a genuine error in your work?
Might they
not be
>> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
Certainly.
>
In which case you're not Ôfacing an incredibly 
strange
situation which
defies
> easy explanation' are you? There is an easy explanation -
you simply find
it
> unpalatable.
>
I've considered that explanation carefully. It 
doesn't apply
here.
>In other words - the behaviour looks strange if your work is
definitely
correct.
>> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't be.
>
Usenet is a place where insults ßy freely. Add to that your
own tendency
to
> insult others and I don't find this at all 
strange.
Well that makes sense on the one hand, but the sheer volume
and level
of hostility seems to be off the scale, even for Usenet.
> And you're not being ignored. You get many more replies
than any other
single
> poster. You're being disagreed with, not ignored.
>
Mainstream mathematicians are more important from what I've
gathered
than Usenet readedespite various claims I've heard at times
that
mainstream mathematicians read Usenet.
My experiences with actual contacts has been that they don't
seem to
know anything about what happens on the sci.math newsgroup.
Some grad
students do, but not professors.
> 2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
>> and later discovered them and admit them. In some of these
cases you
were
>>absolutely convinced that your work was correct right up
until the point
where
>> you discovered the mistake.
Yup.
> Given this, and given that *no-one* other than yourself
believes your
current
>>proof to be correct - shouldn't you at least consider the
possibility
that you
>> might have made a mistake here too?
What proof are you talking about?
>
I was referring to your proof of a problem with the definition
of
algebraic
> integers. Were you *not* referring to that when you
mentioned Ôdramatic
results'
> above?
My prime counting function uses a partial difference equation
to count
prime numbers--a first in math history--which makes it not
only a
dramatic result, but one readily understandable to a wide
audience.
Rather than acknowledge the unique features of my find,
posters on
sci.math have tried to claim that it's just 
Legendre's
Method, or if
they're more sophistica they may claim it's 
some variant on
Meissel's formula.
I can repeat, repeat, repeat that it uses a partial difference
equation, explain why that's important, and they just keep
repeating
their own positions, while mainstream mathematicians, like
Lagarias
whom I recently contrac, are busy.
I think it fascinating to see how much you post based on
assumptions.
>
Which assumptions are you thinking of? Feel free to correct
me.
>
I have a fairly wide body of research at this point, so I find
it
interesting when some poster comes along, like yourself, with
various
assertions showing they have some particular bit of research
in mind,
as if that's all there is.
It's a fascinating bit of narrowmindedness, a tunnel vision,
which I
guess helps people like yourself believe that they're in the
right.
>So then, what work of mine do you have in mind Paul Mason?
See above.
I notice you've not answered my final question - 
are you able
to now that
I've
> answered yours? I repeat - shouldn't you consider the
possibility that
you've
> made a mistake and that's why your work 
hasn't been accep?
That was my early assumption. And in fact I've continually 
tes
that assumption with work that's been challenged.
Given my *YEARS* of mistakes, it's not a leap to consider
that I might
just be wrong, so yes, I have considered that possibility
with ALL of
my work, including my prime counting research.
Life is good. I had a lot of fun chasing after new math
results, and
now I get to kick back, and be a lot less frenetic about it
all.
It's cool.
===
Subject: Re: JSH: Limi intellects
says...
>> says...
><dele Consider this:
>1) How would the mathematicians' behaviour differ from that
described
above if
>> you were wrong, if there was a genuine error in your work?
Might they
not be
> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
Certainly.
>> In which case you're not Ôfacing an 
incredibly strange
situation which
defies
>> easy explanation' are you? There is an easy explanation -
you simply find
it
>> unpalatable.
>I've considered that explanation carefully. It 
doesn't apply
here.
Forgive me, but I don't believe you. I believe that if you
had truly
considered
that explanation, especially if you had considered it from
the point of view
of
your detractors - you would not find the opposition so
incredibly strange as
you
do.
After all you don't have to believe that they are correct to
not find their
disagreement strange do you?
In other words - the behaviour looks strange if your work is
definitely
>>correct.
> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't be.
>> Usenet is a place where insults ßy freely. Add to that
your own tendency
to
>> insult others and I don't find this at all 
strange.
>Well that makes sense on the one hand, but the sheer volume
and level
>of hostility seems to be off the scale, even for Usenet.
>
The volume is directly rela to your own posting volume.
And with all possible respect, the level of hostility in this
group to
anyone,
is tame compared to much of usenet.
>> And you're not being ignored. You get many more replies
than any other
single
>> poster. You're being disagreed with, not ignored.
>Mainstream mathematicians are more important from what I've
gathered
>than Usenet readedespite various claims I've heard at times
that
>mainstream mathematicians read Usenet.
>My experiences with actual contacts has been that they 
don't
seem to
>know anything about what happens on the sci.math newsgroup.
Some grad
>students do, but not professors.
So your complaint is not that you're being ignored in
sci.math but that
you're
being ignored by mainstream mathematicians? Ok, thanks for the
clarification.
2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
>> and later discovered them and admit them. In some of these
cases you
were
>>absolutely convinced that your work was correct right up
until the point
where
> you discovered the mistake.
Yup.
Given this, and given that *no-one* other than yourself
believes your
current
>>proof to be correct - shouldn't you at least consider the
possibility
that you
> might have made a mistake here too?
What proof are you talking about?
>> I was referring to your proof of a problem with the
definition of
algebraic
>>integers. Were you *not* referring to that when you
mentioned Ôdramatic
results'
>> above?
>My prime counting function uses a partial difference
equation to count
>prime numbers--a first in math history--which makes it not
only a
>dramatic result, but one readily understandable to a wide
audience.
Have you really reviewed all of the research on prime
counting up to this
point
in order to be so definite that it's a 
first?
Also isn't there a performance issue with your function?
Isn't it the case
that
other algorithms already exist with orders of magnitude
faster calculation
times? If that's the case then I can understand why it might
not appear so
dramatic to those reviewing it.
>Rather than acknowledge the unique features of my find,
posters on
>sci.math have tried to claim that it's just 
Legendre's
Method, or if
>they're more sophistica they may claim it's 
some variant on
>Meissel's formula.
Again, is that
a) because they simply don't want to accept your work?
or
b) because they genuinely believe that it's a variant of one
of those other
methods?
My problem is not so much that you assume a) - it's that you
seem to never
even
consider b).
>I can repeat, repeat, repeat that it uses a partial
difference
>equation, explain why that's important, and they just keep
repeating
>their own positions,
So they're only doing what you're doing - 
repeating a
position they firmly
believe to be true.
>while mainstream mathematicians, like Lagarias
>whom I recently contrac, are busy.
I think it fascinating to see how much you post based on
assumptions.
>> Which assumptions are you thinking of? Feel free to
correct me.
>I have a fairly wide body of research at this point, so I
find it
>interesting when some poster comes along, like yourself,
with various
>assertions showing they have some particular bit of research
in mind,
>as if that's all there is.
I'm not trying to claim that this is all of your work.
However I don't
think
it's unreasonable to assume that it was one of the 
Ôdramatic
results' you
were
referring to, since it's been one you've been 
posting about
until very
recently.
>It's a fascinating bit of narrowmindedness, a tunnel 
vision,
which I
>guess helps people like yourself believe that they're in 
the
right.
This is an odd statement. Look at my posts carefully. I'm
asking questions.
I'm
asking you to consider things from another point of view. 
I'm
not making
assertions about your work. So given that - what is it you
think I'm so
anxious
to be right about?
On a wider level I think it's a little sad that you see it 
as
a question of
being in the right at all - as if it's a battle with sides.
If you saw
it as a
co-operative process with the aim of furthering your own and
other
mathematical
knowledge - you might get further.
>>So then, what work of mine do you have in mind Paul Mason?
See above.
I notice you've not answered my final question - 
are you able
to now that
I've
>> answered yours? I repeat - shouldn't you consider the
possibility that
you've
>> made a mistake and that's why your work 
hasn't been accep?
>That was my early assumption. And in fact I've continually
tes
>that assumption with work that's been challenged.
>Given my *YEARS* of mistakes, it's not a leap to consider
that I might
>just be wrong, so yes, I have considered that possibility
with ALL of
>my work, including my prime counting research.
And yet there's little evidence of that in your posts.
Consider this
scenario,
which I believe is common,
- you post something
- some others post replies, pointing out what they believe
are ßaws in
your
argument
- in your replies to them, you are dismissive, casually
insulting and
accusing
them of having malicious motives for disagreeing with you.
Now if you are continually considering the posibility that
you may be
wrong,
then I would not expect you to doubt their motives and insult
them - *merely
for
disagreeing with you*
Furthermore - as I mentioned - you do occasionally admit your
mistakes -
but
I've never seen you apologise to those who helped you 
realise
it. Those who,
at
the time, you accused of being Ôsocially' motiva 
or whatever.
>Life is good. I had a lot of fun chasing after new math
results, and
>now I get to kick back, and be a lot less frenetic about it
all.
What's brought on the change? Have you found someone to
accept one of your
dramatic results? Or have you found other interests?
===
Subject: Re: JSH: Limi intellects
> says...
> says...
>
><dele>
Consider this:
>1) How would the mathematicians' behaviour differ from that
described
above if
>> you were wrong, if there was a genuine error in your work?
Might they
not be
> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
Certainly.
>> In which case you're not Ôfacing an 
incredibly strange
situation which
defies
>> easy explanation' are you? There is an easy explanation -
you simply
find it
>> unpalatable.
I've considered that explanation carefully. It 
doesn't apply
here.
>
Forgive me, but I don't believe you. I believe that if you
had truly
considered
> that explanation, especially if you had considered it from
the point of
view of
> your detractors - you would not find the opposition so
incredibly strange
as you
> do.
After all you don't have to believe that they are correct to
not find
their
> disagreement strange do you?
>
Yes I do. There are any number of people with wacky ideas
that I feel
are just wrong. At times I may even try to communicate that
to such a
person.
It's not like I get very exci about their reaction though.
It's like when NASA decided to try and convince certain
people that
men had really landed on the moon.
There was lots of criticism about the effort. Why bother?
In other words - the behaviour looks strange if your work is
definitely
> correct.
> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't
be.
>> Usenet is a place where insults ßy freely. Add to that
your own
tendency to
>> insult others and I don't find this at all 
strange.
Well that makes sense on the one hand, but the sheer volume
and level
>of hostility seems to be off the scale, even for Usenet.
> The volume is directly rela to your own posting volume.
>
Really? It seems to be a lot more.
> And with all possible respect, the level of hostility in
this group to
anyone,
> is tame compared to much of usenet.
>
That's not worth arguing over.
>> And you're not being ignored. You get many more replies
than any other
single
>> poster. You're being disagreed with, not ignored.
Mainstream mathematicians are more important from what I've
gathered
>than Usenet readedespite various claims I've heard at times
that
>mainstream mathematicians read Usenet.
My experiences with actual contacts has been that they 
don't
seem to
>know anything about what happens on the sci.math newsgroup.
Some grad
>students do, but not professors.
>
So your complaint is not that you're being ignored in
sci.math but that
you're
> being ignored by mainstream mathematicians? Ok, thanks for
the
clarification.
>
Yeah, I've used the sci.math newsgroup mainly for drafts. I
toss out
ideas, and see if they ßy.
Later I make contact with mainstream mathematicians, who in my
experience, usually haven't heard of me.
2) Everyone makes mistakes from time to time. You yourself
have made
mistakes
>> and later discovered them and admit them. In some of these
cases
you were
>>absolutely convinced that your work was correct right up
until the
point where
> you discovered the mistake.
Yup.
Given this, and given that *no-one* other than yourself
believes your
current
>>proof to be correct - shouldn't you at least consider the
possibility
that you
> might have made a mistake here too?
What proof are you talking about?
>> I was referring to your proof of a problem with the
definition of
algebraic
>>integers. Were you *not* referring to that when you
mentioned Ôdramatic
results'
>> above?
My prime counting function uses a partial difference
equation to count
>prime numbers--a first in math history--which makes it not
only a
>dramatic result, but one readily understandable to a wide
audience.
>
Have you really reviewed all of the research on prime
counting up to this
point
> in order to be so definite that it's a 
first?
I've checked the reliable sources.
My work stands out as no one else in recorded history has
found a way
to count prime numbers using a partial difference equation.
Also isn't there a performance issue with your function?
Isn't it the case
that
> other algorithms already exist with orders of magnitude
faster
calculation
> times? If that's the case then I can understand why it
might not appear
so
> dramatic to those reviewing it.
>
That's a red herring. A partial difference equation method 
is
significant because it leads to a partial difference equation.
I've
made some effort to check that partial difference equation,
and it is
closer to the prime distribution than li(x), which is
significant.
Besides, simply the existence of a partial difference
equation method,
where none was found before should be enough for people who
supposedly
care about pure math, but I've seen how much rhetoric has
differed
from reality with my own work!
And finally what makes the speed issue so disgusting is the
fact that
starting from the base mathematics with the partial difference
equation I found faster and faster prime counting methods,
making a
*Java* program that could compete with Mathematica, with only
a few
months effort!!!
Reasonably one might suppose that mathematicians actually
interes
in the truth with more expertise might be able to push even
further.
But at a minimum, my work even explains current methods, and
why they
work they way they do, as in fact, I can derive a piece of
Meissel's
function from my partial difference equation, but you can't
go the
other way.
It's just a HUGE area where I can go on and on with details
about how
my work is unique and relevant.
>Rather than acknowledge the unique features of my find,
posters on
>sci.math have tried to claim that it's just 
Legendre's
Method, or if
>they're more sophistica they may claim it's 
some variant on
>Meissel's formula.
>
Again, is that
a) because they simply don't want to accept your work?
> or
> b) because they genuinely believe that it's a variant of
one of those
other
> methods?
>
There is NO OTHER KNOWN METHOD for counting primes using a
partial
difference equation!!!
That means uniqueness at the start.
The partial difference equation leads to a partial
differential
equation, which when integra numerically is closer to the
prime
distribution, for the values I checked, than li(x).
There's more than enough for *reasonable* mathematicians to
show at
least a modicum of interest.
> My problem is not so much that you assume a) - it's that
you seem to never
even
> consider b).
>
I've derived a piece of Meissel's formula from 
my partial
difference
equation.
It seems to me that you have an idea in your head and either
don't
have all the facts, or couldn't care less about the facts.
Now I've researched what was known about counting prime
numbers before
my discovery, and clearly what I found is fundamental
mathematics,
from which you can not only derive what's currently known in
prime
counting, but you can go beyond.
Piddling around using Java I managed to find fast algorithms,
in just
a few months.
Rather than make the obvious logical step, posters like
yourself act
as if nothing matters unless I make the fastest prime counting
possible on my own efforts as if without doing that you have
proof
that my work is not important.
That's naive.
>I can repeat, repeat, repeat that it uses a partial
difference
>equation, explain why that's important, and they just keep
repeating
>their own positions,
So they're only doing what you're doing - 
repeating a
position they
firmly
> believe to be true.
>
They're annoying me by tracking me making posts like 
hecklers.
If they really think I'm wrong they can just say their piece
and
wonder off.
But what motivates these people to reply, and reply, and
reply, and
reply, and reply, and reply, over a period of YEARS?
It's just wacky.
Now these hecklers clearly are motiva by something. At times I
think they're just out to try and hurt my feelings, other
times I've
thought maybe they're trying to protect the status quo.
But something is going on with these people.
>while mainstream mathematicians, like Lagarias
>whom I recently contrac, are busy.
>
I think it fascinating to see how much you post based on
assumptions.
>> Which assumptions are you thinking of? Feel free to
correct me.
I have a fairly wide body of research at this point, so I
find it
>interesting when some poster comes along, like yourself,
with various
>assertions showing they have some particular bit of research
in mind,
>as if that's all there is.
>
I'm not trying to claim that this is all of your work.
However I don't
think
> it's unreasonable to assume that it was one of the
Ôdramatic results' you
were
> referring to, since it's been one you've 
been posting about
until very
recently.
>
You're working to justify your assumption.
>It's a fascinating bit of narrowmindedness, a tunnel 
vision,
which I
>guess helps people like yourself believe that they're in 
the
right.
This is an odd statement. Look at my posts carefully. I'm
asking
questions. I'm
> asking you to consider things from another point of view.
I'm not making
> assertions about your work. So given that - what is it you
think I'm so
anxious
> to be right about?
>
Your questions show your assumptions.
For instance, my prime counting work is neat for me to use in
answering you because there's no room to call it wrong.
After all, the count of primes is constant so a prime counting
function either works or it doesn't.
So in looking over your questions, and seeing your tone, I
focus on
that part of my research which most effectively and easily
refutes
your position.
Why should I work harder because you haven't?
If you're just ignorant of the breadth of my research, why
shouldn't I
simply make you aware rather than worrying about arguing over
narrow
areas?
I'm trying to help you see that a *rational* discourse
doesn't depend
on your own point of view, as if necessary, you need to be
able to
*expand* your point of view given the facts.
Now if you want to focus on one bit of my research, the onus
is on
your to be specific.
It's not on me.
> On a wider level I think it's a little sad that you see it
as a question
of
> being in the right at all - as if it's a battle with 
sides.
If you saw
it as a
> co-operative process with the aim of furthering your own
and other
mathematical
> knowledge - you might get further.
>
You asked questions which showed a biased view, so I went to
that
research which I felt best answered your questions and your
view.
My prime counting research is an excellent area to consider
your
questions from, and I haven't heard anything from you that
changes
that fact.
>So then, what work of mine do you have in mind Paul Mason?
See above.
I notice you've not answered my final question - 
are you able
to now
that I've
>> answered yours? I repeat - shouldn't you consider the
possibility that
you've
>> made a mistake and that's why your work 
hasn't been accep?
That was my early assumption. And in fact I've continually
tes
>that assumption with work that's been challenged.
Given my *YEARS* of mistakes, it's not a leap to consider
that I might
>just be wrong, so yes, I have considered that possibility
with ALL of
>my work, including my prime counting research.
>
And yet there's little evidence of that in your posts.
Consider this
scenario,
> which I believe is common,
- you post something
> - some others post replies, pointing out what they believe
are ßaws in
your
> argument
> - in your replies to them, you are dismissive, casually
insulting and
accusing
> them of having malicious motives for disagreeing with you.
Now if you are continually considering the posibility that
you may be
wrong,
> then I would not expect you to doubt their motives and
insult them -
*merely for
> disagreeing with you*
Furthermore - as I mentioned - you do occasionally admit your
mistakes -
but
> I've never seen you apologise to those who helped you
realise it. Those
who, at
> the time, you accused of being Ôsocially' 
motiva or
whatever.
>
So? My purpose in posting is to evaluate ideas. Others may
choose to
comment, and I tend to look for objective replies, especially
anyone
finding an error!!!
Now, if I find that a certain poster has some agenda, refuses
to
acknowledge *logical* positions which refute their claims,
and also
likes to find ways to insult me, then I discount such posters.
It's part of why I tend to look for new names when it comes 
to
replies, like I haven't seen your name much, and you seem to
be trying
to be rational, so it's worth it to me to reply to you.
But when it comes to the hecklewho've spent YEARS or many
months
obsessively replying to my posts, who cares?
They're already not very bright just for doing that, and 
I've
often
seen from trying to talk to them specifically where they 
don't
have a
clue, so I reply to them at a whim.
>Life is good. I had a lot of fun chasing after new math
results, and
>now I get to kick back, and be a lot less frenetic about it
all.
>
What's brought on the change? Have you found someone to
accept one of
your
> dramatic results? Or have you found other interests?
Research is difficult. When I'm pushing the 
mathematical
envelope I
get cranky.
I've decided I've discovered enough, and can 
settle down and
enjoy the
fruits of my labors.
So I'm no longer into extreme math.
It makes me a much nicer guy.
===
Subject: Re: JSH: Limi intellects
: Yeah, I've used the sci.math newsgroup mainly for drafts. 
I
toss out
: ideas, and see if they ßy.
Kindly explain what you mean. Historically you have used
sci.math for two
things:
1) Vigilant defense of things that other people have shown to
be false.
2) Vague ideas that you haven't bothered to look into
seriously yourself.
: Later I make contact with mainstream mathematicians, who in
my
: experience, usually haven't heard of me.
So give me the name of a single mainstream mathematician
who's perked up
and been interes enough in what you've done to...to...to do
anything...
: They're annoying me by tracking me making posts like
hecklers.
If you define heckling as pointing out where your mathematics
is wrong,
even in the face of your insults and stupidity then, well,
yes.
: If they really think I'm wrong they can just say their
piece and
: wonder off.
Actually I think that people here have gained real
intellectual
stimulation from discussion of your mistakes. I know from
reading
Ullrich, Arturo and Nora's posts (among others) 
I've learned
some
things. That's why I continue to follow along.
: But what motivates these people to reply, and reply, and
reply, and
: reply, and reply, and reply, over a period of YEARS?
See above.
: Why should I work harder because you haven't?
Right, so when you post a conjecture and don't bother to
write a simple
program to test it out, why should *we* work harder because
*you* haven't?
<rest snipped>
Best,
Justin
===
Subject: Re: JSH: Limi intellects
> Actually I think that people here have gained real
intellectual
> stimulation from discussion of your mistakes. I know from
reading
> Ullrich, Arturo and Nora's posts (among others) 
I've
learned some
> things. That's why I continue to follow along.
For example I learned that:
Not only are the algebraic numbers dense in R.
Not only are the algebraic integers dense in R.
Not only are the algebraic integer units dense in R.
But even the algebraic integer units which are the roots of
polynomials
of degree three are dense in R !
Someone with the limi intellect of wouldn't be able to
find a correct proof of this, obviously, even though it is
quite simple.
===
Subject: Re: JSH: Limi intellects
> My work stands out as no one else in recorded history has
found a way
> to count prime numbers using a partial difference equation.
To the best of my knowledge, no one has done it standing on
his head
either, but unless a new method shows a noticeable
improvement over
existing methods, what's the point?
Your method has been shown to be slower than a number of
methods already
available.
So your invention is on a par with what would make a talking
dog
notable.
After the record of your long unbroken sequence of
unsuccesful forays
into mathematics, it's not that your algorithm counts primes
well, it's
that it can count them at all that is notable.
===
Subject: Re: JSH: Limi intellects
> says...
>
<dele> Consider this:
>1) How would the mathematicians' behaviour differ from that
described
>>above if
>> you were wrong, if there was a genuine error in your work?
Might they
>> not be
>> expec to Ôobstinately refuse to accept' a 
genuinely ßawed
proof?
Certainly.
>
In which case you're not Ôfacing an incredibly 
strange
situation which
> defies
> easy explanation' are you? There is an easy explanation -
you simply
find
> it
> unpalatable.
> I've considered that explanation carefully. It 
doesn't
apply here.
How would you, with your ever increasing oops number, know
that?
>>In other words - the behaviour looks strange if your work is
definitely
>>correct.
>> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't be.
>
Usenet is a place where insults ßy freely. Add to that your
own
tendency
> to
> insult others and I don't find this at all 
strange.
Well that makes sense on the one hand, but the sheer volume
and level
> of hostility seems to be off the scale, even for Usenet.
 And you're not being ignored. You get many more replies
than any other
> single
> poster. You're being disagreed with, not ignored.
> Mainstream mathematicians are more important from what 
I've
gathered
> than Usenet readedespite various claims I've heard at 
times
that
> mainstream mathematicians read Usenet.
The really mainstream ones may lurk, but are are a good deal
more
careful about posting. I have seen an occasional name of
considerable
mathematical repute appearing here.
My experiences with actual contacts has been that they don't
seem to
> know anything about what happens on the sci.math newsgroup.
Some grad
> students do, but not professors.
>> 2) Everyone makes mistakes from time to time. You yourself
have made
>> mistakes
>> and later discovered them and admit them. In some of these
cases
you
>> were
>>absolutely convinced that your work was correct right up
until the
point
>>where
>> you discovered the mistake.
Yup.
>> Given this, and given that *no-one* other than yourself
believes your
>> current
>>proof to be correct - shouldn't you at least consider the
possibility
>>that you
>> might have made a mistake here too?
What proof are you talking about?
>
I was referring to your proof of a problem with the definition
of
algebraic
> integers. Were you *not* referring to that when you
mentioned Ôdramatic
> results'
> above?
My prime counting function uses a partial difference equation
to count
> prime numbers--a first in math history--which makes it not
only a
> dramatic result, but one readily understandable to a wide
audience.
Rather than acknowledge the unique features of my find,
posters on
> sci.math have tried to claim that it's just 
Legendre's
Method, or if
> they're more sophistica they may claim it's 
some variant on
> Meissel's formula.
I can repeat, repeat, repeat that it uses a partial difference
> equation, explain why that's important, and they just keep
repeating
> their own positions, while mainstream mathematicians, like
Lagarias
> whom I recently contrac, are busy.
Why is it important to anyone but you? There are hosts of
methods of
counting primes. A major increase in ease or speed of
computation might
be important, but your method seems to have neither of these.
That you
claim to have inven it is of importanve only to you.
===
Subject: Re: JSH: Limi intellects
[snip]
> My prime counting function uses a partial difference
equation to count
> prime numbers--a first in math history--which makes it not
only a
> dramatic result, but one readily understandable to a wide
audience.
It appears that you are less interes in finding your way into
the annals
of math then into
the Guinness Book of Records. But consider this (according to
Guinness):
Charles Osborne of Iowa had the hiccups for more than 66
years!
That's far more dramatic and understandable than anything 
you
have pos.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
===
Subject: Re: JSH: Limi intellects
>> says...
>> [...]
In which case you're not Ôfacing an incredibly 
strange
situation which
defies
>> easy explanation' are you? There is an easy explanation -
you simply find
it
>> unpalatable.
>I've considered that explanation carefully. It 
doesn't apply
here.
What was that phrase you kept repeating a few weeks ago? Oh
yeah:
think about it some more.
>In other words - the behaviour looks strange if your work is
definitely
correct.
> If you have made an error then it's not so strange is it?
Well I think the insults are strange, while being ignored
wouldn't be.
>> Usenet is a place where insults ßy freely. Add to that
your own tendency
to
>> insult others and I don't find this at all 
strange.
>Well that makes sense on the one hand, but the sheer volume
and level
>of hostility seems to be off the scale, even for Usenet.
Giggle. Like when I said
The only person I know who thinks this thing you call the
conventional thinking is _you_.
and you replied
You stupid head!!! What the is wrong with you Ullrich?
No matter how many ing times I tell you to off, you keep
replying to me!!!
What the is your problem you head?
You Ullrich are a stupid piece of dumb who refuses to get the
message when someone does NOT want to talk to you, you stupid
ing
ty asshole.
You are an ASSHOLE Ullrich!!! Now why don't you take your
dumb ass
stupid self somewhere to GET A ING CLUE and QUIT ING REPLYING
TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!
OFF!!!!
Can't you get it through your stupid head?
OFF!!!!!!!!!!!!!!!!!
Yes, that does seem a little much.
(Oh, you were talking about the level of hostility direc
towards
you? Never mind...)
>>[...]
>My prime counting function uses a partial difference
equation to count
>prime numbers--a first in math history--which makes it not
only a
>dramatic result, but one readily understandable to a wide
audience.
>Rather than acknowledge the unique features of my find,
posters on
>sci.math have tried to claim that it's just 
Legendre's
Method, or if
>they're more sophistica they may claim it's 
some variant on
>Meissel's formula.
>I can repeat, repeat, repeat that it uses a partial
difference
>equation, explain why that's important,
I don't recall you ever explaining why this is important.
I mean I recall plenty of times when you've explained that 
the
reason your algorithm is important is that it's the First in
History
to use a partial difference equation (which btw is not true)
but I don't recall _any_ explanation of why the fact that it
used a partial difference equation _made_ it important -
that's always seemed like something that you assume is
obvious (it's not). What's the reason, again?
>and they just keep repeating
>their own positions, while mainstream mathematicians, like
Lagarias
>whom I recently contrac, are busy.
[...]
I notice you've not answered my final question - 
are you able
to now that
I've
>> answered yours? I repeat - shouldn't you consider the
possibility that
you've
>> made a mistake and that's why your work 
hasn't been accep?
>That was my early assumption. And in fact I've continually
tes
>that assumption with work that's been challenged.
>Given my *YEARS* of mistakes, it's not a leap to consider
that I might
>just be wrong, so yes, I have considered that possibility
with ALL of
>my work, including my prime counting research.
You've considered the possibility. Great.
Think about it some more.
>Life is good. I had a lot of fun chasing after new math
results, and
>now I get to kick back, and be a lot less frenetic about it
all.
>It's cool.
>
**************************
As far as I'm concerend you're trying to wait 
until I die, so
I figure
maybe you should die instead. How about that, eh? Wouldn't
that be a
better twist?
You refuse to follow the math, so the great Powers that
control
reality and *speak* in mathematics decide to kill you instead
of me.
So what do you think about that, eh? Oh, can't hear Them
talking?
Well, I guess that's because you don't really 
understand
Mathematics,
the true language, which is THE language.
They're talking about you now, and They agree with my
assessment, and
will not penalize me as They allowed the others like Galois
and Abel
to be penalized.
They will kill you instead.
speaking on Weird factorization, genius
===
Subject: Re: JSH: Limi intellects
<buj08302o3a@drn.newsguy.com>
<3c65f87.0401200831.335b35cc@posting.google.com>
<bujp5502ei9@drn.newsguy.com>
<3c65f87.0401201514.73221396@posting.google.com>
Discussion, linux)
Cancel-Lock: sha1:rxcRiKvmImpSxaROc0LhtqprCSc=
> Mainstream mathematicians are more important from what 
I've
gathered
> than Usenet readedespite various claims I've heard at 
times
that
> mainstream mathematicians read Usenet.
*Where* have you heard such various claims? I've *never* 
seen
anyone
that's said so in this forum --- aside from you, of course,
at least
implicitly.
You've obviously believed that mainstream mathematicians 
read
sci.math, since shortly after your great new method for
counting
primes, you whined about how it wasn't in the math texts 
yet.
Evidently, you thought that enough mathematicians read
sci.math that
they would force a recall of those now out-of-date texts so
that
James's new discovery could be given all due attention, what
with
being the only method involving differential engines and
positrack
steering and all.
But aside from your own posts here, where have you seen
anything that
sugges that most mathematicians read Usenet (and,
particularly,
many or most of them read sci.math)?
--
That's what's annoying about Usenet as some 
loser will state
a case,
get their ass kicked, but STILL keep coming back as if nothing
happened. -- explains his strategy.
===
Subject: Re: JSH: Limi intellects
> Well that makes sense on the one hand, but the sheer volume
and level
> of hostility seems to be off the scale, even for Usenet.
The sheer volume and level of your stupidity is off scale,
even for
Usenet. On top of that comes behaviour that reminds me of the
lowest of
pond life. Does that explain it?
===
Subject: Re: JSH: Limi intellects
> I've concluded that the problem I'm facing 
is that the
human brain
> isn't built to handle Mathematics. Human beings have done
well with
> some simple mathematics and basic concepts, but it only
goes so far,
> and what have, is beyond those limits.
> I thought of myself as calm and cool about that assessment.
Still I'm facing an incredibly strange situation which 
defies
easy
> explanation.
Consider, mathematicians *supposedly* care about pure math
and the
> importance of mathematical knowledge, but have for close to
two years
> now obstinately refused to accept some rather dramatic
results.
Actually, since those results have been shown false, it would
be much
more dramatic if any serious mathematician had actually accep
tham.
Rather than face intrigued interest and help in further
research I've
> been ignored and insul, as if mathematicians *hate* me for
finding
> what I have.
You have hardly been ignored, since I dare say you have
genera more
responses than any other two posters combined, And the
insults you have
received are certainly no greater than those you have issued.
And mathematicians, just as other people, do not bother to
hate what
they do not bother to respect. Sneer, yes, hate, not worth it.
It's like there's some kind of odd war, where 
they will fight
no
> matter what, against all reason, against the very values
they claimed
> to have, and for what?
The only person in these discussions that has fought against
reason in
these discussions is JSH. And for what?
That's what I guess I need to figure out now.
>
===
Subject: Re: JSH: Limi intellects
>> I've concluded that the problem I'm facing 
is that the
human brain
>> isn't built to handle Mathematics. Human beings have done
well with
>> some simple mathematics and basic concepts, but it only
goes so far,
>> and what have, is beyond those limits.
>I thought of myself as calm and cool about that assessment.
>Still I'm facing an incredibly strange situation which 
defies
easy
>explanation.
>Consider, mathematicians *supposedly* care about pure math
and the
>importance of mathematical knowledge, but have for close to
two years
>now obstinately refused to accept some rather dramatic
results.
>Rather than face intrigued interest and help in further
research I've
>been ignored and insul, as if mathematicians *hate* me for
finding
>what I have.
>It's like there's some kind of odd war, where 
they will fight
no
>matter what, against all reason, against the very values
they claimed
>to have, and for what?
>That's what I guess I need to figure out now.
Fascinating. The typical line is about how we're all liars
engaged in
a great conspiracy to hide the truth. This thread you suddenly
switch to the idea that we're all morons. Now you seem to
realize
how crazy it sounds to suggest that the reason nobody says
you're
right is that nobody else on the planet is able to comprehend
your
work. Sounds for a second like the beginning of a hint of
sanity,
but no, instead of the simple explanation you fall back on the
global conspiracy.
Oh well.
>
************************
Subject: coupled boundary conditions for ode
===
i have n-1 3er order odes of the same form
F(y_i''',y_i'[C
apitalOTilde],y_i')=0, i=0...n-2
which must be solved with boundary conditions
y_i(X_i)=Y_i; y_i(X_{i+1})=Y_{i+1}; i=0...n-2;
and a coupled condition that is continuity under first
derivative in
every point.
Does this interpolating scheme have a solution?
Thanks in advance,
-Eduardo Su.87rez
===
Subject: dx/dy = 1/(dy/dx)
When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.
===
Subject: Re: dx/dy = 1/(dy/dx)
> When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.
It's somewhat sloppy when we do this evalution, for we 
should
be
stating conditions when dx/dy = 1/(dy/dx) at a point or over
an
interval.
OK, if x = x(y) is a smooth, invertible function from R to R,
then
x = x(y(x))
then differentiating both sides by x gives
dx/dx = 1 = (dx/dy)(dy/dx)
So,
dx/dy = 1/(dy/dx)
However, if x and y are functionallly independent, then
dx/dy = dy/dx = 0
Now, as an example (which uses matrix algebra), consider the
transformation (Galilean)
x' = x-vt
t' = t
where x and t are mutually independent, and v is a constant,
so, with
delta as the total derivative operator,
delta x delta t
-------- = -------- = 0
delta t delta x
Now, let F be a differentiable real function, then
_ _
| partial F partial F |
| --------- , ---------- | =
| partial x partial t |
- -
- -
|delta x' delta x' |
_ _ |--------- --------- |
| partial F partial F | |delta x delta t |
| ---------- , ---------- | | |
| partial x' partial t' | |delta 
t' delta t' |
- - |--------- --------- |
|delta x delta t |
- -
=
- - - -
| partial F partial F | | 1 -v |
| ---------- , ---------- | | |
| partial x' partial t' | | 0 1 |
- - - -
So, from all this we get that the partials transform as:
partial / partial x = partial / partial x'
and
partial / partial t
= -v(partial / partial x') + partial / partial 
t'
For more on this see the pdf files at
http://ajnpx.com/html/SD.html
===
Subject: Re: dx/dy = 1/(dy/dx)
Pos-And-Mailed: yes
===
> When is it true that one can write dx/dy = 1/(dy/dx) ?
> ....
Here is the best elementary version I know of that theorem,
for a
function of one variable (no partials).
If a function f has inverse f^(-1) which is continuous at b =
f(a),
and if f itself is differentiable at a, with f'(a) not zero,
then f^(-1) also is differentiable at b, with derivative
1/(f'(a)).
Is that the sort of thing you wan?
Ken Pledger.
===
Subject: Re: dx/dy = 1/(dy/dx)
> When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.
When dx/dy refers to a derivative, OK.
differential, well you will first have to tell me what the
quotient
means.
partial derivatives ... big trouble if you don't know what
you are
doing.
===
Subject: Re: dx/dy = 1/(dy/dx)
> When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.

> When dx/dy refers to a derivative, OK.
> differential, well you will first have to tell me what the
quotient
> means.
> partial derivatives ... big trouble if you don't know what
you are
> doing.
A nice example with partial derivatives:
the special relativity Lorentz transformation:
{ x = g(x'-vt')
{ t = g(t'-vx')
with g = 1/sqrt(1-v^2) (this is gamma)
and its inverse:
{ x' = g(x+vt)
{ t' = g(t+vx)
and the partial derivatives
@t/@t' = g
@t'/@t = g
but obviously
@t/@t' is not 1 / @t'/@t
... and someone who didn't know what he was doing:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/
PartialDiff.html
;-)
Dirk Vdm
===
Subject: Re: dx/dy = 1/(dy/dx)
A N Niel <pbai7uo02@sneakemail.com> a .8ecrit dans le message
de
> When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.

> When dx/dy refers to a derivative, OK.
> differential, well you will first have to tell me what the
quotient
> means.
Does it not mean a derivative?
If not, what do you mean by ok when it refers to a derivative
? how
does
the formula makes sense for y' ? (derivative of y function 
of
x)
> partial derivatives ... big trouble if you don't know what
you are
> doing.
Okay, thanks.
===
Subject: Re: dx/dy = 1/(dy/dx)
> A N Niel <pbai7uo02@sneakemail.com> a .8ecrit dans le
message de
When is it true that one can write dx/dy = 1/(dy/dx) ?
dx, dy refers either to a differential or to a partial
differential.

When dx/dy refers to a derivative, OK.
differential, well you will first have to tell me what the
quotient
> means.
Does it not mean a derivative?
> If not, what do you mean by ok when it refers to a
derivative ? how
does
> the formula makes sense for y' ? (derivative of y function
of x)
Let y=f(x) be a function, and let x=g(y) be the inverse
function. Then g'(y) = 1/f'(x). 
That's what I mean OK for
derivative.
That equation could be written dx/dy = 1/(dy/dx) if you like.
Differential would be something like dx that makes sense by
itself. If you don't know about that, then 
don't worry about
it.
partial derivatives ... big trouble if you don't know what
you are
> doing.
Okay, thanks.
>
===
Subject: Socrates' Nothing is Everything
(x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
-Doctor Garry Whilhelm Denke, 1655
Solves every equation.
===
Subject: Re: Socrates' Nothing is Everything
> (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
> -Doctor Garry Whilhelm Denke, 1655
Solves every equation.
Nonsense.
Bob Kolker
===
Subject: Re: Socrates' Nothing is Everything
>> (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
>> -Doctor Garry Whilhelm Denke, 1655
Solves every equation.
>Nonsense.
If _I_ of all people said you should not feed the trolls then
people would start looking up in the air for ßying pigs...
But it happens a lot that posts that people interpret as
trollery seem to me to be possible instances of actual
confusion instead. Otoh with Denke isn't it clear that
he _knows_ that the stuff he's posting is nonsense?
>Bob Kolker
************************
===
Subject: Re: Socrates' Nothing is Everything

>> (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
>> -Doctor Garry Whilhelm Denke, 1655
Solves every equation.
Nonsense.
If _I_ of all people said you should not feed the trolls then
> people would start looking up in the air for ßying pigs...
But it happens a lot that posts that people interpret as
> trollery seem to me to be possible instances of actual
> confusion instead. Otoh with Denke isn't it clear that
> he _knows_ that the stuff he's posting is nonsense?
>
Well, there's a grain of truth to what he is saying. If one
assumes
that x^2 + 1 = 0, then there is no logical reason not to
substitute
x^2 + 1 into the expression 0/0 = 0/0. This is just a
substitution of
co-referring terms. The trouble, however, is that 0/0 isn't
defined
in the reals, or the rationals, or the complex field. So
calling 0/0
a solution to the equation in the context of mathematics as
practiced
conventionally (that is, over a field, in this case) is false.
I think he's just a misguided person who equates 0--the
number--with
nothing. That is, nothingness. Either that or he's a troll.
In the
first case, I'm not going to try to 
fix his metaphysical
mistakes--too
bloody. In the second, I'll laugh when people try to correct
him.
Either way, hardly worth my time. (Not that it's worth much)
Ôcid Ôooh
===
Subject: Re: Socrates' Nothing is Everything

>> (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
>> -Doctor Garry Whilhelm Denke, 1655
Solves every equation.
Nonsense.
If _I_ of all people said you should not feed the trolls then
> people would start looking up in the air for ßying pigs...
But it happens a lot that posts that people interpret as
> trollery seem to me to be possible instances of actual
> confusion instead. Otoh with Denke isn't it clear that
> he _knows_ that the stuff he's posting is nonsense?
Bob Kolker
> ************************
It's not my fault dividing by 0 equals every number,
including -1.
It's only a mathematical fact, David, don't 
let it bother you
so.
===
Subject: Re: Socrates' Nothing is Everything

> (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
>> -Doctor Garry Whilhelm Denke, 1655
> Solves every equation.
Nonsense.
If _I_ of all people said you should not feed the trolls then
> people would start looking up in the air for ßying pigs...
But it happens a lot that posts that people interpret as
> trollery seem to me to be possible instances of actual
> confusion instead. Otoh with Denke isn't it clear that
> he _knows_ that the stuff he's posting is nonsense?
>Bob Kolker
> ************************
It's not my fault dividing by 0 equals every number,
including -1.
> It's only a mathematical fact, David, don't 
let it bother
you so.
Division by 0 is a mathematical operation. (Well, not really,
since
its undefined, but I'll grant that its an 
operation.) A number
is not
a mathematical operation. They are completely different sorts
of
objects. Hence, division by 0 is not a number. I suggest you
try
learning some mathematical facts before preaching here.
Ôcid Ôooh
===
Subject: Re: Socrates' Nothing is Everything
> If _I_ of all people said you should not feed the trolls
then
> people would start looking up in the air for ßying pigs...
> But it happens a lot that posts that people interpret as
> trollery seem to me to be possible instances of actual
> confusion instead. Otoh with Denke isn't it clear that
> he _knows_ that the stuff he's posting is nonsense?
Sadly yes, that seems clear even to me with his last post.
Never mind, I could do with plenty of lessons in patience.
Love and respect
Chris
===
Subject: Re: Socrates' Nothing is Everything
(x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
> -Doctor Garry Whilhelm Denke, 1655
Solves every equation.
Nonsense.
Bob Kolker
I agree. Zero sense.
===
Subject: Re: Sum of two beta distributions
> The pdf of the sum of 2 Beta rv's will generally NOT be
close to
Normal.
> For example, here is a quick 2-line derivation and pdf plot
using
> mathStatica: [...]
> Robert Dodier <robert_dodier@yahoo.com> replied:
> Well, if NUMERICAL approximations are allowed, here's a
short
> construction in Octave which amounts to the same thing [
snip ]
> Neat code. Since you have raised the stakes, I shall see
your
> Numerical solution, and raise you an EXACT SYMBOLIC solution
> (given parameter values). See:
> http://www.mathStatica.com/Sumof2Betas/
> Method 2: Transform Method
> Exact Symbolic solutions for sum of 2 Betas
> This yields some rather scrumptuous pdf plots.
Now that's neat indeed. I'll call your exact 
symbolic
solution and raise
with the following:
1. Sum of two Betas with different parameters - I guess this
is easy.
2. Sum of two Betas with different parameters and scaled
diffently, ie, min
and max not tied to zero and one respectively (this is the
scenario used
with Pert analysis).
(I have more to raise with but I'm keeping that in hand for
the moment ;-)
BTW: MathStatica is about to enter my list of want-to-have
toys - now I
just
need a business case for getting it so that my boss will pay
for it.
/Torben
--
P.S. The views expressed above are my own.
P.P.S. Remove dashes in mail address when replying.
===
Subject: Condition for absolute convergence of a series
Good morning
I have a question about the root test for absolute
convergence of
infinite series and will appreciate any comments.
Supposing the terms of Sum a_n don't vanish, the ratio 
test
is
usually sta like that:
If lim sup |a_n+1|/|a_n|, <1 then the series converges
If lim inf |a_n+1|/|a_n| >1, then the series diverges
If lim inf |a_n+1|/|a_n| <=1 <= lim sup |a_n+1|/|a_n|, then
nothing
can be affirmed.
series is not absolute convergent. That's my question. It
looks to me
that L = lim sup |a_n+1|/|a_n| >1 does imply divergence of
Sum |a_n|.
If this condition is satisfied, then (|a_n+1|/|a_n|) has a
subsequence
that converges to L and, therefore, for sufficient large k we
have
|a_n_k+1/|a_n_k| >1. Therefore, (|a_n_k|) and (|a_n|) itself
cannot
converge to 0 (they are unbounded), which makes impossible the
convergence of Sum |a_n|. Why is this argument wrong? Isn't 
it
similar to the argument used to prove the root test? If lim
sup
|a_n|^(1/n) >1, then Sum |a_n| diverges because there's a
subsequence
of
|a_n|^(1/n) that converges to a number >1, and this implies
the
existence of p>1 such that |a_n| > p^n for infinitely many n
(which
turns |a_n| unbounded), so that |a_n| cannot converge to 0.
Thank you
Artur
===
Subject: Re: Condition for absolute convergence of a series
Content-transfer-encoding: 8bit
> Good morning
> I have a question about the root test for absolute
convergence of
> infinite series and will appreciate any comments.
> Supposing the terms of Sum a_n don't vanish, the ratio
test is
> usually sta like that:
> If lim sup |a_n+1|/|a_n|, <1 then the series converges
> If lim inf |a_n+1|/|a_n| >1, then the series diverges
> If lim inf |a_n+1|/|a_n| <=1 <= lim sup |a_n+1|/|a_n|, then
nothing
> can be affirmed.
series is not absolute convergent. That's my question. It
looks to me
> that L = lim sup |a_n+1|/|a_n| >1 does imply divergence of
Sum |a_n|.
No. Think about it: take any two positive-term summable
series sum_n
a_n and sum_n b_n. Begin choosing terms from the two series,
starting
with the a's,
a(1), a(2), ..., a(n1)
so that b(1)/a(n1) > 2. (You can do this since b(1) > 0 and
a(n) -->
0.) Insert b(1) and continue taking a's,
a(1), a(2), ..., a(n1), b(1), a(n1+1), ..., a(n2)
until you have b(2)/a(n2) > 4; insert b2 and continue,
a(1),a(2),...,a(n1),b(1),a(n1+1),...,a(n2),b(2),a(n2+1),...,a
(n3)
so that b(3)/a(n3) > 8; etc. etc.
The new sequence of a's, with the b's inser in 
this fashion,
is
still summable, but the lim sup of the ratios is +infinity.
The lim
inf is <= 1, of course, since the series IS summable. With a
little
more care you can make sure the lim inf is 0.
David Ullrich has given an explicit counterexample. I prefer
to think
in the terms I've shown here (which, I suspect, are the same
terms
David thought in, too); and to leave the exposition that way.
--Ron Bruck
===
Subject: Re: Condition for absolute convergence of a series
>Good morning
>I have a question about the root test for absolute
convergence of
>infinite series and will appreciate any comments.
>Supposing the terms of Sum a_n don't vanish, the ratio
test is
>usually sta like that:
>If lim sup |a_n+1|/|a_n|, <1 then the series converges
>If lim inf |a_n+1|/|a_n| >1, then the series diverges
>If lim inf |a_n+1|/|a_n| <=1 <= lim sup |a_n+1|/|a_n|, then
nothing
>can be affirmed.
>series is not absolute convergent. That's my question. It
looks to me
>that L = lim sup |a_n+1|/|a_n| >1 does imply divergence of
Sum |a_n|.
It doesn't. Counterexample:
1 + 1/4 + 1/2 + 1/8 + 1/4 + 1/16 + ...
(the next term is obtained from the previous one by
alternately
multiplying by 2 or dividing by 4.)
>If this condition is satisfied, then (|a_n+1|/|a_n|) has a
subsequence
>that converges to L and, therefore, for sufficient large k we
have
>|a_n_k+1/|a_n_k| >1.
Yes, assuming you mean |a_{n_k}+1/|a_n_k| >1 and not
|a_n_{k+1}/|a_n_k| >1. (Actually what you really meant to
say was that there exists L > 1 such that for sufficiently
large k we have |a_n_k+1/|a_n_k| >L.)
>Therefore, (|a_n_k|) and (|a_n|) itself cannot
>converge to 0 (they are unbounded),
Doesn't follow. It _would_ follow if you had
|a_n_{k+1}/|a_n_k| > L > 1, but it doesn't follow from
what you actually have, |a_{n_k}+1/|a_n_k| > L > 1.
If you don't see why it doesn't follow, take 
your
argument and see what it comes to if you apply it
to the counterexample above: There you have
|a_{n+1}/a_n| = 2 for all even n, but a_n is not
unbounded.
>which makes impossible the
>convergence of Sum |a_n|. Why is this argument wrong? Isn't
it
>similar to the argument used to prove the root test? If lim
sup
>|a_n|^(1/n) >1, then Sum |a_n| diverges because there's a
subsequence
>of
>|a_n|^(1/n) that converges to a number >1, and this implies
the
>existence of p>1 such that |a_n| > p^n for infinitely many n
(which
>turns |a_n| unbounded), so that |a_n| cannot converge to 0.
>Thank you
>Artur
************************
===
Subject: Re: Condition for absolute convergence of a series
Good morning
>I have a question about the root test for absolute
convergence of
>infinite series and will appreciate any comments.
>Supposing the terms of Sum a_n don't vanish, the ratio
test is
>usually sta like that:
>If lim sup |a_n+1|/|a_n|, <1 then the series converges
>If lim inf |a_n+1|/|a_n| >1, then the series diverges
>If lim inf |a_n+1|/|a_n| <=1 <= lim sup |a_n+1|/|a_n|, then
nothing
>can be affirmed.
series is not absolute convergent. That's my question. It
looks to me
>that L = lim sup |a_n+1|/|a_n| >1 does imply divergence of
Sum |a_n|.
It doesn't. Counterexample:
1 + 1/4 + 1/2 + 1/8 + 1/4 + 1/16 + ...
(the next term is obtained from the previous one by
alternately
> multiplying by 2 or dividing by 4.)
If this condition is satisfied, then (|a_n+1|/|a_n|) has a
subsequence
>that converges to L and, therefore, for sufficient large k we
have
>|a_n_k+1/|a_n_k| >1.
Yes, assuming you mean |a_{n_k}+1/|a_n_k| >1 and not
> |a_n_{k+1}/|a_n_k| >1. (Actually what you really meant to
> say was that there exists L > 1 such that for sufficiently
> large k we have |a_n_k+1/|a_n_k| >L.)
Therefore, (|a_n_k|) and (|a_n|) itself cannot
>converge to 0 (they are unbounded),
Doesn't follow. It _would_ follow if you had
> |a_n_{k+1}/|a_n_k| > L > 1, but it doesn't follow from
> what you actually have, |a_{n_k}+1/|a_n_k| > L > 1.
If you don't see why it doesn't follow, take 
your
> argument and see what it comes to if you apply it
> to the counterexample above: There you have
> |a_{n+1}/a_n| = 2 for all even n, but a_n is not
> unbounded.
>
Thank you, I got the point.
We also have lim inf |a_n+1|/|a_n| <= lim sup |a_n|^(1/n) <=
lim sup
|a_n+1|/|a_n|. I've never seen this proof, all the books I
have leave
it to the reader...I'll try to work it out myself. Let L =
lim sup
|a_n+1|/|a_n|. For every r>L, there's a natural k such 
that
|a_n+1|/|a_n| < r for n>=k. Then, for n>=k if follows that
|a_n| <
|a_k|*r^(n-k) and |a_n|^(1/n) <r * s^(1/n), where s =
(|a_k|/r^k)
(independent of n). Since this inequality holds for every
n>=k, it
follows that lim sup |a_n|^(1/n) <= lim sup r * s^(1/n). But
the
sequence on the right hand side converges to r, which shows
lim sup
|a_n|^(1/n)<=r. We conclude that, for every r>L, we have lim
sup
|a_n|^(1/n)<=r. Therefore, we must have lim sup |a_n|^(1/n)<=
L =lim
sup |a_n+1|/|a_n|. I got a bit confused about the proof of the
inequality on the left hand size. I was about to say the
proof is
similar, but it's not. I'll think a bit 
more
about it.
Based on this conclusion, it gets easy to prove that a_n =
(n!)^(1/n)
-> inf. I was having difficulties to prove this directly, but
we
readily see that (a_n+1)/(a_n) = n+1 -> inf. Therefore, lim
inf (a_n)
= inf, which gives the conclusion. This argument may be
somewhat
artificial, but it's correct.
Artur
===
Subject: Re: Exponent operator laws - Can anyone name these
formulas?
> show FOUR Operator formulas for a^m, where m-n, m+n, m/n
and m*n are
> used
Oh, it boils down to homework problem
a^(m-n) = a^m / a^n
a^(m+n) = a^m a^n
a^(m/n) = (a^m)^(1/n)
solve for a^m
As that last one might be hard, I do it for you.
[a^(m/n)]^n = a^m
> a^m = a(m+n) / a^n
Huh? It's in error.
> Set a = 2, n = 0, m = 3.
> when n > or = 1 is what you missed
a^m = a*(m+n) / a^n
You mean a^m = a^(m+n) / a^n
n,m may be any real numbea must be positive real.
--
As for your ensuing lengthy and unbefitting drivel
I suggest you present your laments to alt.therapy.
----
===
Subject: Re: Exponent operator laws - Can anyone name these
formulas?
> Oh, it boils down to homework problem
Don't be surprised if you can't convince him. 
There's a
thread in
sci.crypt
concerning his discovery of the following:
N^2 = 1 + (N-1) * (N+1)
N^3 = N + (N-1)*(N)*(N+1)
N^4 = N^2 + (N^2)*(N-1)*(N+1)
N^5 = N + (N^2+1)*(N-1)*(N)*(N+1)
N^6 = N^2 + (N^3+N)*(N-1)*(N)*(N+1)
N^7 = N^3 + (N^4+N^2)*(N-1)*(N)*(N+1)
He doesn't believe the various people who've 
told him this is
trivial
stuff any grade school algebra student can do.
--
===
Subject: Re: Exponent operator laws - Can anyone name these
formulas?
Can anyone name these four formulas for operators of
exponents, they
> cover the basic rule for addition, subtraction, division and
> multiplication of exponents
A bunch of blah. Stick with
> a^(m+n) = a^m * a^n
> (a^n)^m = a^(nm)
> a^(-m) = 1/a^m
> and the stuff below will come quickly if perhaps it's
needed.
Your opinion
The following formula is a division formula for exponents.
a^m = a^y(^(m/y))
Correction: a^m = (a^y)^(m/y)
> Toss it out, it's redundant to
> (a^n)^m = a^(nm)
Your opinion, the point of the four formula's is to show an
exponent
equivalency formula for a^m to be equal to a formula using
m/n, so
your formula does not fit what the person is putting forth,
you need
to be thrown out
The following formula is a subtraction formula used to find
exponent
> factors for any exponent, the number of exponent factor
sets within
> this equation is equal to m/2
a^m = a^(m-n) x a^n
> (a^n)^m = a^(nm)
Don't use x, it could be confused with the variable x.
> Toss it out, it's redundant to
> a^(m+n) = a^m * a^n
minor point, the x is understandable, however, once again you
do not
put forth a substraction formula to equal a^m, so once again
you
should be thrown out since you do not comprehend what the
formula's
represent
show FOUR Operator formulas for a^m, where m-n, m+n, m/n and
m*n are
used
this post has nothing to do with better than etc., it is an
attempt to
show how a formula can be crea for any basic operator
(+,-,/,*) to
equal a^m
Your so called formula's are not better than etc., and most
do not
satisfy the point of the paper
The following formula is an addition formula used to find 
equal
> exponents
a^m = a(m+n) / a^n
Huh? It's in error.
> Set a = 2, n = 0, m = 3.
when n > or = 1 is what you missed
The following formula is a multiplication formula.
a^m = a(m*n) / a^((m*n) -n)
Error. Set n = 0, a = 1 = m.
> Does a(m*n) = a*(m*n) or something else?
How about
> a^m = a^(m+n) / a^n ?
> and
> a^m = a^(mn) / a^(mn - n) ?
But that's wrong, it's
> a^m = a^(mn) / a^(mn - m)
> Here's a generalization
> a^p = a^(mn) / a^(mn - p)
> Both are of minimal to negligible value and superceeded by
> a^p = a^m / a^(m-p)
once again WHEN n is > than or = to ONE
Exercise: quickly proof the generalization from the three
equations I
> gave at start.
I consider that referenced web page to be of dubious value.
> Frankly, there's much better stuff to be found on the web.
> Look around
> http://mathforum.org
> http://www.mathpages.com
> and forget the page you dredged up.
You're the dredge, your Opinions did nothing but show you
MISSED the
whole point of the page, put forth simple operator formulas
for
subtraction, addition, multiplication and division to a^m
As to the site, you must be a xtian since you wish to point
people
away from the author,
Everyone should be going to his site, if they did, THEY WOULD
KNOW THE
FUTURE
stupidity and remarks that had nothing to do with his premise
put this
WARNING out before 911
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&
selm=6tb7rp%24gv
r%241%40winter.news.erols.com
warner sta MAJOR EMERGENCY in DC on 911
Now if people followed your idiotic advice, no one would have
gotten
the warning
Well guess what, that warning has made that person who you
say should
be stayed away from ONE OF THE MOST DISCUSSED TOPICS on the
net
The same person has put out a simple riddle the past 10 years
for the
super bowl, HE HIT EVERY ONE with the exact score
The same person has put out exact date and location quake
warnings
where rare 7.0+ quakes then hit
So stick you head in the sand and ignore the writings of a
famous
person that is one of the most mentioned topics in the world
due to
his warnings about quakes, acts of terrorism, etc
The same person proofed Fermat in one page
The same person has a prime number algorithm named after him
The same person has the brains at NASA studying his PDF
(Planetary
Distance Formula) Formula based on an expanded version of
Titius-Bodes
Yet you spout off your mouth that some formula that is
SUPPOSE TO HAVE
a certain operator in it is inferior to your formula THAT
DOESN'T HAVE
THE CORRECT OPERATOR in it
SHUT UP
You need to be tossed for ignorance
You are nothing but another defective DNA program that would
be better
off if it expired
===
Subject: Re: Integral
The answer is certainly incorrect for some values of the
parameters. Try,
> for example, substituting 1 and 2 for p and q, resp.
>
Maple 9:
> f := sqrt(p + q*cos(t) + cos(t)^2);
(1/2)
/ 2
f := p + q cos(t) + cos(t) /
> f2 := subs({p=1,q=2},f);
(1/2)
/ 2
f2 := 1 + 2 cos(t) + cos(t) /
> int(f2,t);
/
1 |
- ------- (-sin(t) + cos(t) sin(t) - t + t cos(t))
2
sin(t)
(1/2)
/ 2 |
1 + 2 cos(t) + cos(t) / /
> diff(%,t);
...
> simplify(%);
csgn(cos(t) + 1) (cos(t) + 1)
CHECKED!
> fint := int(f,t):length(%);
33180
> subs({p=1,q=2},fint):
Error, numeric exception: division by zero
> fint1 := simplify(subs({p=1},fint)):length(%);
16992
> subs(q=2,fint1):
Error, numeric exception: division by zero
> limit(fint1,q=2);
... goes on for hours ...
--
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Mathematics forums
charset=iso-8859-1
Are there any other mathematics forums? I just want to be a
part of a
> couple and this is the only one I can find. I am interes in
> tricky competion style problems or advancements in pure
mathematics.
> Thanks for any help....
Here's my list of Usenet news groups concerning mathematics.
> They're lis roughly in order of activity, so sci.math 
comes
first.
sci.math
> sci.logic
> alt.math.undergrad
> de.sci.mathematik (German)
> aus.mathematics
> alt.math
> uk.education.maths
> alt.math.recreational
> sci.math.symbolic
> sci.math.research (advanced)
> fr.education.entraide.maths (French)
> fr.sci.maths (French)
> alt.algebra.help
> k12.ed.math
> alt.sci.math.combinatorics
> soc.history.science
> schule.mathe (German)
> geometry.puzzles
> geometry.research
> geometry.pre-college
> geometry.college
> z-netz.wissenschaft.mathematik (German)
> geometry.software.dynamic
> alt.math.alonline
> alt.sci.math.galois_fields
> alt.algebra
Also some statistics and probability groups:
sci.stat.math
> sci.stat.edu
> sci.stat.consult
> alt.sci.math.probability
> alt.sci.math.statistics.prediction
HTH
Ken Pledger.
Also:
sci.math.num-analysis
comp.soft-sys.math.maple
comp.soft-sys.math.mathematica
Jim Buddenhagen
--
To reply copy jbuddenh@REMOVEtexas.net to address bar and
edit out REMOVE
===
Subject: Re: Mathematics forums
> Maybe I am just looking in the wrong places, but it appears
to me that
> none of the webpages you provided led to post-reply style
forums like
> this one. If anyone has a link to a forum where you can
post your
> math problems i would appreciate if you provided it here
for me.
> Thanks for the help so far. If any of those links was what
I was
> looking for please tell me quickly how to post something
there.
>
There's a Hong Kong group that likes to solve problems:
http://mathdb.math.cuhk.edu.hk/forum/e_forum.php
Hey, and get this, you can even use LaTeX for your posts so
they look
a LOT better.
Oh and by the way, notice the replies you got here before
mine, as
sci.math is domina by a few people who are control freaks.
I'm not
surprised they didn't answer your question to your
satisfaction.
===
Subject: Re: Would you recommend a school of math for a
soon-to-be college
student
> If you think you have the qualifications, you should go for
the
> absolute top schools.
>> Don't necessarily agree. U.C. Berkeley has the one of the
top-ra math
>> departments in the world, but freshman and sophomore
classes are still
>> taught as huge lectures. If you went to a community
college for two
>> years and transferred in as a junior you'd be just as 
well
off. Being a
>> freshperson at Berkeley has its pleasures and benefits, but
academically
>> you're just another mad cow in a real big herd. Same goes
for any other
>> large state university.
>I disagree. If the OP has Talent, a school with a
research-orien math
>department is a much better choice, even as a freshman. A CC
is likely to
>have no math courses beyond first-year calculus, and
certainly no proof-
>orien courses. A talen undergrad should be in courses that
the CC
>doesn't offer certainly as a sophomore, probably as a
freshman.
It is proof-orien courses which are the important ones.
In fact, I would suggest not taking any more computational
courses than are required to get into those; I believe that
they are likely to do more harm than good.
See if you can get into the honors courses; many of the others
have been so weakened by the use of student evaluations, from
students who want to be given methods without understanding.
This understanding does NOT come easily later, unless you are
in the genius class, and even then there are problems.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Would you recommend a school of math for a
soon-to-be college
student
>Being a
>freshperson at Berkeley has its pleasures and benefits, but
academically
>you're just another mad cow in a real big herd. Same goes
for any other
>large state university.
At our 25,000 student state university, the largest
introductory math
classes
are capped at about 40 students.
--irascible since 1957
Subject: Re: Proposal for responding to idiots
===
>idiot
>one word: killfile
>>another word: filter
>>Use whichever one your non-Googled newsreader has available.
> The problem is when otherwise highly valuable contributors
to the
> group lower themselves (in my view) to engaging the idiots.
> I cannot bring myself to killfile those from which I could
learn
> who have the nasty habit of engaging the idiots, which
means I'm
> left with a 50% portion of the threads I'd rather not see.
> (who believes that the JSH-bot was a good idea)
What I would do is (and for the most part DO do), is ignore
the threads
star by those I don't wish to see. For example, most any
thread that
involves non-trivial (to me) statistics I will ignore. If you
don't
have a threaded news-reader, this can be a problem, of course.
I use Mozilla, but as I recall, pine works nicely as well.
--
===
Subject: Re: Proposal for responding to idiots
Solutions:
1. Join to the FAQ and create the ambient of looking it up
first.
There are many people that are habitual to this.
Or:
2. Create sci.math.beginners
I think it is very egoistic from you not to help newbies. So,
don't
you help your students? There are many beginners that study by
themselves and sci.math is the only place where their
questions can be
answered.
===
Subject: Re: Proposal for responding to idiots
Injector-Info: news.mailgate.org;
posting-host=adsl-66-126-134-202.dsl.frsn01.pacbell.net;
posting-account=48257; posting-date=1074589070
X-URL:
http://mygate.mailgate.org/mynews/sci/sci.math/
fe69fe464623ca282a5989004ca4d9
35.48257%40mygate.mailgate.org
> My proposal is that, instead of replying immediately
[everyone can fill in the rest from memory]
If there were a positive correlation between intelligence
and self-control, wars would be over quickly, because only
stupid, easily defea people would start them.
Right after you solve that problem, Usenet would really
welcome your input as a proven master of human relations.
Until then, we've all tried, in fact the attempts are old
enough to be of drinking age, and it just isn't happening.
The attempts just become more of the noise, hiding the
signal all the more thoroughly.
HTH.
xanthian, inventor of *.advocacy, which at least admits
that human nature exists and should be considered in
constructing Usenet group hierarchies.
OTOH, you are more than welcome to lead by example. People
who do that tend to end up with labels like elder god on
Usenet.
[On the gripping hand, some of us get labeled elder god
with no evidence whatever of having gained wisdom, but from
simple longevity, so check bona-fides, trust no one, beginning
at the top of this posting, obviously.]
--
Pos via Mailgate.ORG Server - http://www.Mailgate.ORG
===
Subject: Re: Proposal for responding to idiots
<fe69fe464623ca282a5989004ca4d935.48257@mygate.mailgate.org>:
>If there were a positive correlation between intelligence
>and self-control, wars would be over quickly, because only
>stupid, easily defea people would start them.
If anyone truly wants world peace they will find it only by
commencing a war that wipes out mankind completely with no
hope of a
return of the species.
However even without mankind nearly all aspects of the
universe are
in turmoil. Destruction and decay create new beginnings. New
beginnings bring hope. Decay creeps in to the point where
destruction
once again creates a new beginning.......
... and this (IMO) is how the universe came to be and how it
will
end where to be and to end are neither a beginning nor an
ending
as what we perceive as the universe neither begins nor does it
end....
it simply transforms...
===
Subject: Re: Proposal for responding to idiots
>Subject: Re: Proposal for responding to idiots
>Message-id: <87isj75uq5.fsf@nonospaz.fat.org> idiot
> one word: killfile
another word: filter
Use whichever one your non-Googled newsreader has available.
>The problem is when otherwise highly valuable contributors
to the
>group lower themselves (in my view) to engaging the idiots.
As someone who knows nothing, reading the replies of those who
can't resist the engaging of idiots is not always a problem.
>I cannot bring myself to killfile those from which I could
learn
>who have the nasty habit of engaging the idiots,
One does occaisionally learn something from the replies to
idiots,
although whether it's cost effective in terms of bandwith is
debatable.
>which means I'm
>left with a 50% portion of the threads I'd rather not see.
(who believes that the JSH-bot was a good idea)
Mensanator
(who believes that a bad idea is a bad idea, even in a good
cause)
>--
>Unpatched IE vulnerability: NavigateAndFind file proxy
>Description: c-domain scripting, cookie/data/identity
> theft, command execution
>Reference:
http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ
PU-Content.HTM
>Exploit:
http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ
PU-MyPage.htm
===
Subject: Re: Proposal for responding to idiots
Daniel McLaury
> Daniel McLaury
> Many newsgroup readers show the threads which have had the
most
recent
> responses at the top of the list. This leads to a problem
when some
> idiot comes on this board and says that .9999.. != 1, or
that he's
> squared the circle, or whatever, and like a million people
post
> replies to messages like this.
> ...
> But that's the problem -- such threads will get responses.
As I'm sure
the
> late C. N. Parkinson would say, the length of a thread is
inversely
> proportional to its value.
> LH
> Indeed, this is the problem; that's why I'm 
trying to help
fix it. If
> there's no problem, there's no need to come 
up with a
solution. I'm
> requesting that such threads do NOT get responses. If
enough people
> vow not to answer such threads in a way that takes up the
bandwidth of
> the rest of us, we can at least ameliorate the problem to
some extent.
I agree, but we're up against human nature. On the bright
side, I've
personally succeeded, after many failed attempts, at kicking
the habit of
responding to JSH.
LH
===
Subject: double integral question
I know that the following integral has no analytical solution:
+00
/ -x^2
f(phi) = | e dx
/
sin(phi)
What I have is:
2pi
/
g = | f(phi) d phi
/
0
Is there any trick that I can get rid of the double integral?
Or is this
just Ôas is'.
Thanx,
Jeroen
===
Subject: Re: double integral question
>I know that the following integral has no analytical
solution:
> +00
> / -x^2
>f(phi) = | e dx
> /
> sin(phi)
>What I have is:
> 2pi
> /
>g = | f(phi) d phi
> /
> 0
>Is there any trick that I can get rid of the double
integral? Or is this
>just Ôas is'.
It is easy to get rid of the double integral. Suppose one
integrates on phi first; this gives 0 for x < -1, 2pi for
x > 1, and (pi + 2arcsin(x)) for -1 < x < 1.
However, it is even easier, as
(pi + 2arcsin(x)) + (pi + 2arcsin(-x)) = 2pi
for all x between -1 and 1. So g = 2pi*int_0 exp(-x^2) dx,
which is pi^1.5.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: double integral question
>I know that the following integral has no analytical
solution:
+00
> / -x^2
>f(phi) = | e dx
> /
> sin(phi)
>What I have is:
2pi
> /
>g = | f(phi) d phi
> /
> 0
Is there any trick that I can get rid of the double integral?
Or is this
>just Ôas is'.
It is easy to get rid of the double integral. Suppose one
> integrates on phi first; this gives 0 for x < -1, 2pi for
> x > 1, and (pi + 2arcsin(x)) for -1 < x < 1.
However, it is even easier, as
(pi + 2arcsin(x)) + (pi + 2arcsin(-x)) = 2pi
for all x between -1 and 1. So g = 2pi*int_0 exp(-x^2) dx,
> which is pi^1.5.
Looks like for any continuous function g(phi) that is odd
about Pi, the
evenness of exp(-x^2) gives
int [0,2pi] int (g(phi),oo) exp(-x^2) dx d phi
= int [0,2pi] int [0,oo) exp(-x^2) dx d phi
= Pi^(1.5)
===
Subject: Re: Dishonesty
I note that dishonesty has become the normal thing for
ÔBilge'.
> [snip]
Having a hard time with objectively demonstra competence, are
you?
> Why don't you slit open your abdomen and augur the future
from your
> splayed guts?
We note that in ÔMichelson's Interference 
Experiment',
H.A.Lorentz
> [snip]
> Phys. Rev. Lett. 88(1) 010401 (2002)
> Phys. Rev. Lett. 42(9) 549 (1979)
> Phys. Bull. 21 255 (1970)
> Europhysics Lett. 56(2) 170 (2001)
> Gen. Rel. Grav. 34(9) 1371 (2002)
Try education rather than psychosis and personal snit. Try
emulating
> adult behavior. Here's a hint: God doesn't 
speak to you.
Do you know why your paper will never be accep,because it is
not
good enough to steal.
http://xxx.lanl.gov/PS_cache/physics/pdf/0310/0310134.pdf
There is no dishonesty really in sci.physics for ultimately
you are
cannon fodder for your peers and when you become irrelevent
you can be
as honest and dishonest as you wish.
===
Subject: Re: Dishonesty
Oriel36 gives a live demonstration
> There is no dishonesty really in sci.physics for ultimately
you are
> cannon fodder for your peers and when you become irrelevent
you can be
> as honest and dishonest as you wish.
what a written rendition of a total disconnect from
reality looks like.
===
Subject: Re: Dishonesty
> I note that dishonesty has become the normal thing for
ÔBilge'.
><dubious@radioactivex.lebesque-al.netHe makes the return
address something other than
sci.physics.relativity,
> such as alt.troll or alt.moron. Obviously such a cheap
trick can be
> considered amusing, but wears thin when repea. He claims to
be able
> to Ôderive' sqrt(1-v^2/c^2) as though it were 
something
special that he
> alone can do, using the trigonometric functions of a conic
section.
> Obviously this is the full extent of his mathematical
capabilities,
which,
> when exhaus, causes him to resort to personal abuse and
cheap
practical
> jokes as his only counter argument when his absurd claims
are
challenged.
We note that in ÔMichelson's Interference 
Experiment',
H.A.Lorentz (p.
7,
> The Principle of Relativity' Dover publications, SBN
486-60081-5) uses
this
> in a supposed Ô...shortening in the direction of motion in
the
proportion
> of 1 to sqrt(1-v^2/c^2)...' by which method Lorentz 
assumes
a spherical
> Earth becomes an ellipsoidal Earth pressed upon by aether.
That this ratio is such is hardly surprising, since the ratio
of the
minor
> axis of an ellipse to the major axis is given by
sqrt(1-e^2): 1, e
being
> the eccentricity.
Being unable to correspond with a fool that indulges in such
childish
> behaviour and dishonest claims, I merely ignore any further
communiqu.8es
he
> may post.
> That is probably your best tack... There have been only two
people in
> all of my years (nearly 9 now) in these groups that I can
say have
> demonstra a pathological behavior of overt dishonesty.
Sadly, Bilge
> is one of these.
Bilge has his strengths and hisweaknesses. But, those, like
you, who
accuse
him falsely of dishonesty are simply specimens whose esential
shallowness
he
has revealed.
Franz
===
Subject: Re: Dishonesty
> I note that dishonesty has become the normal thing for
ÔBilge'.
><dubious@radioactivex.lebesque-al.netHe makes the return
address something other than
sci.physics.relativity,
> such as alt.troll or alt.moron. Obviously such a cheap
trick can be
> considered amusing, but wears thin when repea. He claims to
be able
> to Ôderive' sqrt(1-v^2/c^2) as though it were 
something
special that he
> alone can do, using the trigonometric functions of a conic
section.
> Obviously this is the full extent of his mathematical
capabilities,
which,
> when exhaus, causes him to resort to personal abuse and
cheap
practical
> jokes as his only counter argument when his absurd claims
are
challenged.
We note that in ÔMichelson's Interference 
Experiment',
H.A.Lorentz (p.
7,
> The Principle of Relativity' Dover publications, SBN
486-60081-5) uses
this
> in a supposed Ô...shortening in the direction of motion in
the
proportion
> of 1 to sqrt(1-v^2/c^2)...' by which method Lorentz 
assumes
a spherical
> Earth becomes an ellipsoidal Earth pressed upon by aether.
That this ratio is such is hardly surprising, since the ratio
of the
minor
> axis of an ellipse to the major axis is given by
sqrt(1-e^2): 1, e
being
> the eccentricity.
Being unable to correspond with a fool that indulges in such
childish
> behaviour and dishonest claims, I merely ignore any further
communiqu.8es
he
> may post.
> That is probably your best tack... There have been only two
people in
> all of my years (nearly 9 now) in these groups that I can
say have
> demonstra a pathological behavior of overt dishonesty.
Sadly, Bilge
> is one of these.
David Evens must be the other, unless you overlooked him, and
Paul Andersen
is borderline when pressed.
And Androcles concludes that they will not meet each other
periodically!
> Many others have demonstra arrogance, egocentrism,
contempt, and
> blunt rudeness. However, at least, all of these others have
demonstra
> the common decency an integrity not specifically &
deliberately
represent
> others pressen material and/or positions in a manner that
is an
> intentional misrepresentation and distortion.
> I think that those that Ôsurvive' their 
first six months can
deal with
> derision & contempt as long as they know that what they
have presen
> has not been altered and distor. Basic disagreement are
bound to
> occur, perspectives & views may never be compatible, but,
at least, one
> should have the expectation the respondents will not alter
or
misrepresent
> their material or position.
> You can however, as recent events have demonstra, expect
him to
> persist this behavior, EVEN AFTER you make your intention
clear that
> you intend to NOT to respond to his posting.
> Oh well, these newsgroups are unmodera, so you have to
Ôlive with'
the
> uncouth along with the crude, rude, socially maladjus, as
well as the
> civilized, well mannered, and polite. But you most
certainly do not
have
> to respond or talk to them.
> Paul Stowe
That is quite correct, Paul. However, I do have a perverse
satisfaction in
trading
and upgrading insults with those that insult me. Until it
gets boring, that
is.
Androcles
===
Subject: Re: Dishonesty
Paul Stowe:
> I note that dishonesty has become the normal thing for
ÔBilge'.
>><dubious@radioactivex.lebesque-al.net> He makes the return address something other than
sci.physics.relativity,
>> such as alt.troll or alt.moron. Obviously such a cheap
trick can be
>> considered amusing, but wears thin when repea. He claims
to be able
>> to Ôderive' sqrt(1-v^2/c^2) as though it were 
something
special that he
>> alone can do, using the trigonometric functions of a conic
section.
>> Obviously this is the full extent of his mathematical
capabilities,
which,
>> when exhaus, causes him to resort to personal abuse and
cheap
practical
>> jokes as his only counter argument when his absurd claims
are
challenged.
>> We note that in ÔMichelson's Interference 
Experiment',
H.A.Lorentz (p.
7,
>> The Principle of Relativity' Dover publications, SBN
486-60081-5) uses
this
>> in a supposed Ô...shortening in the direction of motion in
the
proportion
>> of 1 to sqrt(1-v^2/c^2)...' by which method Lorentz
assumes a spherical
>> Earth becomes an ellipsoidal Earth pressed upon by aether.
>> That this ratio is such is hardly surprising, since the
ratio of the
minor
>> axis of an ellipse to the major axis is given by
sqrt(1-e^2): 1, e
being
>> the eccentricity.
>> Being unable to correspond with a fool that indulges in
such childish
>> behaviour and dishonest claims, I merely ignore any further
communiqu.8es he
>> may post.
 That is probably your best tack... There have been only two
people in
> all of my years (nearly 9 now) in these groups that I can
say have
> demonstra a pathological behavior of overt dishonesty.
Sadly, Bilge
> is one of these.
You only think that because I don't put up with your double
standard.
Provide an example.
 Many others have demonstra arrogance, egocentrism,
contempt, and
> blunt rudeness. However, at least, all of these others have
demonstra
> the common decency an integrity not specifically &
deliberately
represent
> others pressen material and/or positions in a manner that
is an
> intentional misrepresentation and distortion.
Give an example.
 I think that those that Ôsurvive' their 
first six months can
deal with
> derision & contempt as long as they know that what they
have presen
> has not been altered and distor. Basic disagreement are
bound to
> occur, perspectives & views may never be compatible, but,
at least, one
> should have the expectation the respondents will not alter
or
misrepresent
> their material or position.
It's not simply a disagreement over a viewpoint. You want to
taken seriously, yet you post nothing serious. I really find 
it
hard to believe that you consider what you call a theory to be
a competing theory to anything.
===
Subject: Re: Dishonesty
> I note that dishonesty has become the normal thing for
ÔBilge'.
><dubious@radioactivex.lebesque-al.netHe makes the return
address something other than sci.physics.relativity,
> such as alt.troll or alt.moron. Obviously such a cheap
trick can be
> considered amusing, but wears thin when repea. He claims to
be able
> to Ôderive' sqrt(1-v^2/c^2) as though it were 
something
special that he
> alone can do, using the trigonometric functions of a conic
section.
> Obviously this is the full extent of his mathematical
capabilities,
which,
> when exhaus, causes him to resort to personal abuse and
cheap
practical
> jokes as his only counter argument when his absurd claims
are challenged.
> We note that in ÔMichelson's Interference 
Experiment',
H.A.Lorentz (p. 7,
> The Principle of Relativity' Dover publications, SBN
486-60081-5) uses
this
> in a supposed Ô...shortening in the direction of motion in
the proportion
> of 1 to sqrt(1-v^2/c^2)...' by which method Lorentz 
assumes
a spherical
> Earth becomes an ellipsoidal Earth pressed upon by aether.
> That this ratio is such is hardly surprising, since the
ratio of the
minor
> axis of an ellipse to the major axis is given by
sqrt(1-e^2): 1, e being
> the eccentricity.
> Being unable to correspond with a fool that indulges in
such childish
> behaviour and dishonest claims, I merely ignore any further
communiqu.8es he
> may post.
That is probably your best tack... There have been only two
people in
all of my years (nearly 9 now) in these groups that I can say
have
demonstra a pathological behavior of overt dishonesty. Sadly,
Bilge
is one of these.
Many others have demonstra arrogance, egocentrism, contempt,
and
blunt rudeness. However, at least, all of these others have
demonstra
the common decency an integrity not specifically &
deliberately represent
others pressen material and/or positions in a manner that is
an
intentional misrepresentation and distortion.
I think that those that Ôsurvive' their 
first six months can
deal with
derision & contempt as long as they know that what they have
presen
has not been altered and distor. Basic disagreement are bound
to
occur, perspectives & views may never be compatible, but, at
least, one
should have the expectation the respondents will not alter or
misrepresent
their material or position.
You can however, as recent events have demonstra, expect him
to
persist this behavior, EVEN AFTER you make your intention
clear that
you intend to NOT to respond to his posting.
Oh well, these newsgroups are unmodera, so you have to Ôlive
with' the
uncouth along with the crude, rude, socially maladjus, as
well as the
civilized, well mannered, and polite. But you most certainly
do not have
to respond or talk to them.
Paul Stowe
===
Subject: Re: problem from gallian's contemporary abstract
algebra
> (this is not homework by the way.)
> ch. 16 problem 38,
let R be a commutative ring with unity,
> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
This seems to be incredibly hard.
> i have tried showing a(x)*b(x) in I[x] => a(x) in I[x] or
b(x) in
> I[x], i could not make this work so i tried using the
contrapositive,
if a(x) and b(x) are NOT in I[x] => a(x)*b(x) is NOT in I[x].
> This seems easier, but I can't seem to isolate the
coefficient of a*b
> that is not in I. If the first or last coeff. of a and b are
not in I
> then it is easy, but when the not in I coeff is lost in the
middle
> of a and b, I have a hard time determining where the not in
I coeff
> is in a*b, since so many things can happen when you add
terms not in
> I.
I agree with Robin, that you could produce a much simpler
proof
by exhibiting I[x] as the kernel of a suitable Homomorphism
from
R[x] to some integral domain.
However, if you want to complete your proof above, I suggest
the following.
You could already prove f,g not in I[x] ==> f*g not in I[x]
in the special case of f and g having their lowest coefficents
outside I.
Now, for f,g not in I[x], write f = f1 + f0 and g = g1 + g0
where
f0 and g0 have _all_ coefficients in I
f1 and g1 have their _lowest_ coefficients outside I.
Now apply the special case to f1 and g1 to obtain
(f - f0) * (g - g0) = f1 * g1 not in I
Marc
===
Subject: Re: problem from gallian's contemporary abstract
algebra
> (this is not homework by the way.)
> ch. 16 problem 38,
let R be a commutative ring with unity,
> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
This seems to be incredibly hard.
> i have tried showing a(x)*b(x) in I[x] => a(x) in I[x] or
b(x) in
> I[x], i could not make this work so i tried using the
contrapositive,
if a(x) and b(x) are NOT in I[x] => a(x)*b(x) is NOT in I[x].
> This seems easier, but I can't seem to isolate the
coefficient of a*b
> that is not in I. If the first or last coeff. of a and b are
not in I
> then it is easy, but when the not in I coeff is lost in the
middle
> of a and b, I have a hard time determining where the not in
I
coeff
> is in a*b, since so many things can happen when you add
terms not in
> I.
I agree with Robin, that you could produce a much simpler
proof
> by exhibiting I[x] as the kernel of a suitable Homomorphism
from
> R[x] to some integral domain.
However, if you want to complete your proof above, I suggest
> the following.
You could already prove f,g not in I[x] ==> f*g not in I[x]
> in the special case of f and g having their lowest
coefficents
> outside I.
Now, for f,g not in I[x], write f = f1 + f0 and g = g1 + g0
where
> f0 and g0 have _all_ coefficients in I
> f1 and g1 have their _lowest_ coefficients outside I.
Now apply the special case to f1 and g1 to obtain
(f - f0) * (g - g0) = f1 * g1 not in I
Marc
it seems i was too hasty.
if f0 and g0 are in I[x] then x*f0 and y*g0 are in I[x] for
ALL x,y in
R[x], which gets me the final step,
thanks for the suggestion.
===
Subject: Re: problem from gallian's contemporary abstract
algebra
> (this is not homework by the way.)
> ch. 16 problem 38,
let R be a commutative ring with unity,
> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
To prove this, I would write down a homomorphism
f: R[x] -> S where S is an integral domain (suitably chosen
of course!) having kernel I[x].
--
===
Subject: Re: problem from gallian's contemporary abstract
algebra
>However, if you want to complete your proof above, I suggest
>the following.
>You could already prove f,g not in I[x] ==> f*g not in I[x]
>in the special case of f and g having their lowest 
coefficents
>outside I.
Yes.
>Now, for f,g not in I[x], write f = f1 + f0 and g = g1 + g0
where
>f0 and g0 have _all_ coefficients in I
>f1 and g1 have their _lowest_ coefficients outside I.
>Now apply the special case to f1 and g1 to obtain
>(f - f0) * (g - g0) = f1 * g1 not in I[x]
i will use I for I[x] below:
so now, f*g - f0*g - f*g0 + f0*g0 is not in I
=> f*g - f0*g - f*g0 is not in I
=> f*g is not in I or -(f0*g + f*g0) is not in I
-a in I <=> a in I
=> f*g is not in I or (f0*g + f*g0) is not in I
=> f*g is not in I or { f0*g is not in I or f*g0 is not in I)
and...i'm stuck here.
i can't tell what happens to f0*g or f*g0, since f and g are
not in I.
>Marc
===
Subject: Re: problem from gallian's contemporary abstract
algebra
>That is an integral domain, of course, but ISN'T my choice
>for S .... we are dealing with polynomials, aren't we
Yes polynomials....
Can you give me some hint for an integral domain to use?
I'm out of ideas.
===
Subject: Re: problem from gallian's contemporary abstract
algebra
>That is an integral domain, of course, but ISN'T my choice
>>for S .... we are dealing with polynomials, aren't we
Yes polynomials....
> Can you give me some hint for an integral domain to use?
> I'm out of ideas.
Remember (or prove) the following
If S is an integral domain, then so is S[X].
Marc
===
Subject: Re: problem from gallian's contemporary abstract
algebra
(this is not homework by the way.)
> ch. 16 problem 38,
let R be a commutative ring with unity,
> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
To prove this, I would write down a homomorphism
> f: R[x] -> S where S is an integral domain (suitably chosen
> of course!) having kernel I[x].
The only obvious integral domain to use would be R/I.
If we want f: R[x]->R/I to have kernel I[x] then we need to
map
f(g) = 0 + I if g is in I[x]
and all the g's not in I[x] need to NOT map to 0+I. Figuring
out what
to do with the g's not in I[x] to create a homomorphism 
seems
equivalent to the problems I was having using my original
method.
Can you suggest a way to map a g not in I[x] to R/I that will
lead to
a homomorphism?
or is there another integral domain I am not thinking of here?
===
Subject: Re: problem from gallian's contemporary abstract
algebra
(this is not homework by the way.)
>> ch. 16 problem 38,
let R be a commutative ring with unity,
>> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
To prove this, I would write down a homomorphism
>> f: R[x] -> S where S is an integral domain (suitably chosen
>> of course!) having kernel I[x].
The only obvious integral domain to use would be R/I.
That is an integral domain, of course, but ISN'T my choice
for S .... we are dealing with polynomials, aren't we
--
===
Subject: Re: problem from gallian's contemporary abstract
algebra
>
(this is not homework by the way.)
>> ch. 16 problem 38,
let R be a commutative ring with unity,
>> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
To prove this, I would write down a homomorphism
>> f: R[x] -> S where S is an integral domain (suitably chosen
>> of course!) having kernel I[x].
The only obvious integral domain to use would be R/I.
That is an integral domain, of course, but ISN'T my choice
> for S .... we are dealing with polynomials, aren't we
please, at least give me a hint, i'm out of ideas.
===
Subject: Re: problem from gallian's contemporary abstract
algebra
(this is not homework by the way.)
> ch. 16 problem 38,
let R be a commutative ring with unity,
> if I is a prime ideal in R then I[x] is a prime ideal in
R[x].
To prove this, I would write down a homomorphism
> f: R[x] -> S where S is an integral domain (suitably chosen
> of course!) having kernel I[x].
The only obvious integral domain to use would be R/I.
That is an integral domain, of course, but ISN'T my choice
>> for S .... we are dealing with polynomials, aren't we
>please, at least give me a hint, i'm out of ideas.
Well, what rings can you define using R, I, and x, to which
R[x]/I[x]
might conceivably be isomorphic?
There are not many possibilities.
Derek Holt.
===
Subject: Re: problem from gallian's contemporary abstract
algebra
> if a(x) and b(x) are NOT in I[x] => a(x)*b(x) is NOT in
I[x].
> This seems easier, but I can't seem to isolate the
coefficient of a*b
> that is not in I. If the first or last coeff. of a and b are
not in I
> then it is easy, but when the not in I coeff is lost in the
middle
> of a and b, I have a hard time determining where the not in
I coeff
> is in a*b, since so many things can happen when you add
terms not in
> I.
ie, it is possible for x or y to NOT be in I, but x+y is in I.
In a(x) and b(x), consider the respective terms of lowest
degree with a
coefficient not in I.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
Subject: Re: Polynomials and the AC Method
===
>I am moving forward with my math studies. I am working on a
text
>>called Intermediate Algebra. I recently was working on
systems of
>>three equations. I encountered for the first time matrix
operations.
>>The operations themselves are pretty easy. I think you just
have to
>>memorize them. The hard part is understanding why they work
(which I
>>don't really intend to do).
>>Otherwise, I have a math question:
>>There is a method to solve polynomials called the ac
method. If you
>>have a polynomial of the form ax^2+bx+c, you can factor
this by
>>substituting m and n for b, such that m+n=b and mn=ac.
> I don't understand what substituting m and n for b means!
My first guess would be to take ax^2+bx+c and replace it with
> ax^2+mx+c and ax^2+nx+c; that's the obvious meaning of
substituting
> m and n for b in my mind. But this makes no sense if you
are talking
> about factoring this polynomial.
>Here is my question: the method is pretty easy, and works.
But I can't
>>figure out why it works. I see why m+n must equal b. This is
obvious.
>>But where does the requirement that mn=ac come in? There is
no
>>obvious reason that I can see.
> If I knew what you meant, perhaps I could help...
>
It's a technique for rewriting the polynomial in a form that
allows
factoring by grouping.
ax^2 + bx + c = ax^2 + mx + nx + c where mn = ac.
--
Subject: Re: Polynomials and the AC Method
===
I am moving forward with my math studies. I am working on a
text
> called Intermediate Algebra. I recently was working on
systems of
> three equations. I encountered for the first time matrix
operations.
The operations themselves are pretty easy. I think you just
have to
> memorize them. The hard part is understanding why they work
(which I
> don't really intend to do).
Otherwise, I have a math question:
There is a method to solve polynomials called the ac method.
If you
> have a polynomial of the form ax^2+bx+c, you can factor
this by
> substituting m and n for b, such that m+n=b and mn=ac.
Here is my question: the method is pretty easy, and works.
But I can't
> figure out why it works. I see why m+n must equal b. This is
obvious.
> But where does the requirement that mn=ac come in? There is
no
> obvious reason that I can see.
Look at your factorization: (ex+f)(gx+h)
Foiling, you get: (eg)x^2 + eh x + fg x + fh
or: (eg) x^2 + (eh+fg) x + fh
eh = m, fg = n, ac = egfh = ehfg = mn
HTH
--
===
Subject: Re: Polynomials and the AC Method
I am moving forward with my math studies. I am working on a
text
> called Intermediate Algebra. I recently was working on
systems of
> three equations. I encountered for the first time matrix
operations.
The operations themselves are pretty easy. I think you just
have to
> memorize them. The hard part is understanding why they work
(which I
> don't really intend to do).
Otherwise, I have a math question:
There is a method to solve polynomials called the ac method.
If you
> have a polynomial of the form ax^2+bx+c, you can factor
this by
> substituting m and n for b, such that m+n=b and mn=ac.
Here is my question: the method is pretty easy, and works.
But I can't
> figure out why it works. I see why m+n must equal b. This is
obvious.
> But where does the requirement that mn=ac come in? There is
no
> obvious reason that I can see.
Note that I leave a space between factors for multiplications.
Start with a x^2 + b c + c = 0, with a not being zero.
Then multiplying by a gives an equivalent equation
a^2 x^2 + a b x + a c = 0
And a^2 x^2 + a b x + a c = (a x + m)(a x + n)
provided a b x = a x m + a x n = a x ( m+ n), i.e., b = m + n,
and also, a c = m n as the constant term.
Does this help?
===
Subject: Re: Polynomials and the AC Method
I am moving forward with my math studies. I am working on a
text
> called Intermediate Algebra. I recently was working on
systems of
> three equations. I encountered for the first time matrix
operations.
The operations themselves are pretty easy. I think you just
have to
> memorize them. The hard part is understanding why they work
(which I
> don't really intend to do).
If you're memorizing something w/o understanding why it
works, you're
doing something wrong. It's not like you'll 
ever need to be
able to
solve a system of n equations in n variables for any
practical reason
(in the way you need to know, say, to brush your teeth and
drive a
car), so the entire point is for the math itself.
Indeed, the entire point of mathematics is understanding why
things
work, not just knowing how to get some answer or something.
The
reason that row operations work is that you have a matrix
equation:
M v = B
and you can multiply each side on the left by a matrix. Say
that
these are 3x3 matrices and you want to interchange the first
and third
rows. Then the matrix for this is
[ 0 0 1 ]
[ 0 1 0 ] = A
[ 1 0 0 ]
So we multiply each side by A on the left:
A M v = A B
The operation of, say, multiplying the 2nd row by 5 can be
represen
by:
[ 1 0 0 ]
[ 0 5 0 ]
[ 0 0 1 ]
And the operation of adding the second row to the first looks
like
this:
[ 1 1 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
Otherwise, I have a math question:
There is a method to solve polynomials called the ac method.
If you
> have a polynomial of the form ax^2+bx+c, you can factor
this by
> substituting m and n for b, such that m+n=b and mn=ac.
Here is my question: the method is pretty easy, and works.
But I can't
> figure out why it works. I see why m+n must equal b. This is
obvious.
> But where does the requirement that mn=ac come in? There is
no
> obvious reason that I can see.
Well, start with a factorization. Let's take
(sx + t)(ux + v) = us x^2 + (ut + sv) x + tv
Then a = us and c = tv, so ac = (us)(tv) = (ut)(sv). Also b =
ut +
sv. So you are basically just letting m = ut and n = sv.
Substituting these in, you get ac = mn and m + n = b.
===
> Ok, I'm starting to feel a little old and getting tired of
arguing, so
> it's time to shift to a less antagonistic strategy.
What I can say in explanation is that *finding* new
mathematical ideas
> isn't what's hard, it's 
critiquing the damn things.
>
Real mathematicians find it the other way round. Recently a
Russian
mathematician has written a proposed proof of the Poincare
conjecture.
It is being widely circula and critiqued. So far it is
holding up to
scrutiny. The author is somewhat of a recluse and has not even
submit his papers for publication. He appears to be making no
efforts to defend his work. There is a $1M prize for a
solution,
but he appears not to be interes.
He put his efforts into the math. If he has a solution it is
deep and difficult and requires a mastery of existing
knowledge.
You, on the other hand, spend most of your time in promotion
and
Ômarketing', You defame and insult your 
opponents rather than
deal with the math they present. You lie and evade. You are
doing it all for the fame and money. You fantasize yourself as
a superhuman brain, and at the same time fail to refute
arguments
that disprove what you claim. You make no effort to learn
basic
theory.
And now you pontificate on how easy *finding* new 
ideas is
compared
to Ôcritiquing' the Ôdamn 
things'.
> You can come up with idea after idea after idea that's 
just
crap, but
> for any number of reasons you can get attached to it.
>
True before, and still true.
> It's fun to get out there and argue and fight 
passionately
for certain
> ideas, and see if they survive.
Most of the ideas that I came up with over a period of more
than 8
> years of active research fell to the side, useless.
Now that was a glorious time.
Your anguished and hate-filled interchanges with other poste
your posts about mental illness, your obscene screaming posts
to , David Ullrich, me, and othe
your letters to Congressmen and the FBI, your threats of
invoking
the military and personal violence, all suggest that it was
something
less than glorious ...
> I'll miss it, but eventually you just
> get old and tired.
>
Good.
> Now I get to refine my ideas, mostly out of sight, and make
them part
> of established mathematics, mostly using some high-powered
people with
> advanced degrees, since I just have a B.S. in physics.
Meanwhile I'll check in here, and probably post from time to
time, as
> I seem to really like posting.
>
In the last two weeks you have run away from arguments put
forward
by Rick Decker and ÔVirgil'. Both struck at the 
heart of your
methodology and revealed its ßaws. In the case of Rick
Decker's
examples, you admit you were wrong but evidently that made no
difference: you still claim your method is right. In the case
of Virgil's objection, you made a minor comment 
(Virgil's
proof
is airtight, BTW), and then simply ran away.
But yeah, you really like posting.
> But I'm not interes in getting really inves in arguments
like in
> the past.
>
Especially when you keep being on the losing end.
> You see, the heavy lifting is done. Now it's time for
academic
> politics.
>
Yes, the Ôheavy lifting is done' ... 
you've used that phrase
a lot -
a couple of examples from some time ago -
=============================================================
===============
===
>Ok, the heavy lifting is done in that I now have the
factorization of
>2x^3 - 3x - 2 as a counterexample in my corner to refute
various
>claims of counterexamples from others.
>It's now important that you understand the argument that
proves that
>factorization, so I'm quite anxious to hear questions, and
provide as
>much help as necessary.
>(For those who find the back-and-forth tiresome 
there's light
at the
>end of the tunnel, as the difficulties up to now may have
something to
>do with mathematical knowledge at the limits, and the good
news is
>cool, startling mathematics.)
>Some of you may have noticed some sign errors and a typo in
an earlier
>post giving the factorization, and those have been fixed in 
my
>follow-up replies, but what's more important to me is that
you
>understand the how of the answers.
>I'll talk about it some here, so you understand 
what's going
on at a
>high level.
>The factorization is
> 2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)
>and I've been talking about it a lot because it turns out
that it's
>relatively easy to calculate the a's and b's 
where they are
all
>algebraic integers.
>It's relatively easy because of the coefficient 
-3, while the
leading
>and last coefficients are both 2, as that forces the 
a's and
b's for
>each factor to be coprime. For instance, a1 can't have non
unit
>factors of 2 shared with b1, as then that would simply
distribute out,
>and have to be in the third coefficient, but 3 is coprime to
non unit
>factors of 2.
>It turns out that means that I have a factorization that
looks like
> (g1 g2 x + g3^2)(g1 g3 x + g2^2)(g2 g3 x + g1^2)
>where the g's are algebraic integers and g1 g2 g3 = 
sqrt(2).
>If that is confusing then reply asking for more information,
and I'll
>go into more detail.
>Having that factorization is how I can go forward and find a
monic
>polynomial which is a solution for the g's, which is
> g^9 - 3g^3 - 2sqrt(2)
>and if that bothers you reply here and I'll go into more
detail.
>Getting the factorization is the easy part, and intriguingly
enough
>the g's are defined by unit factoand those 
factors are
> (1+sqrt(2))^{1/3} and (-1+sqrt(2))^{1/3}
>which helps out tremendously!!!
>The reason they are unit factors is that they multiply
together to
>give
> 1^{1/3}
>and I use that information in my analysis to actually find
out what
>factors of 2 are in each of the g's, which is fascinating.
>Well I think it's fascinating.
>It was also gratifying as now there's a direct example,
which as I
>mentioned above is my own counterexample to those who made a
claim of
>a counterexample to the key result I've called Area One.
>That is, exactly two of the a's have a factor that is
sqrt(2), and I
>can show it with my own example. As that's the really
important area,
>I'd prefer questions in detail about the argument that 
shows
that be
>made in the thread Challenge answer: Factorization of 2x^3 -
3x - 2,
>but I'd like replies here as to whether or not you
understand it or
>not.
>Though, of course, there's a mathematical proof that works
well for
>those of you who don't need examples and various versions
are all over
>the place but I'm highlighting the thread Polynomial
factorization
>analysis, while also remember there's a general proof that
covers p
>odd prime, specific to FLT, at my main website.
>The factorization is an easy area where there shouldn't be
much
>difficulty, but any people wishing to dispute my important
claims have
>to come into the area where I analyze the g's.
>I'm sure you wouldn't be surprised to hear 
that I hope the
process is
>a short one. After all, the gist of it all is now currently
pos
>and it's not that complica I'd think for 
professional
>mathematicians.
>What I want to emphasize now though is that things have
gotten bigger
>than the question of whether or not I have a proof of FLT,
as my
>analysis shows numbers which are roots of monic polynomials
with
>integer coefficients irreducible over Q, which are *not*
algebraic
>integers.
>That's the kind of information that should be of interest 
to
>mathematicians, and especially number theorists, all over
the world.
>
=============================================================
===============
====
>I must admit to a certain feeling of immense satisfaction at
the
>completion of the most important stage of a formidable task.
>Some of you may realize now that it's not enough to 
find a
spectacular
>proof--you must also be able to defend it.
>Of course, that's not odd as the people who 
didn't find the
proof, who
>are already established in the field, tend to get pissed off
at
>spectacular finds from outsiders.
>Which is why I've used the newsgroups to find 
objections.
>Thanks for the help.
>Now that all the heavy lifting is done, I can focus on
acceptance of
>my mathematical work.
>One of the benefits of such a long period taking place
between when I
>discovered the proof and when it finally gets accep is that I
have
>few acknowledgments necessary.
>After all, those of you who had no faith in my work, who
spent so much
>time attacking it, shouldn't expect to gain anything by its
>recognition.
>Neither should those of you who didn't dig deep enough to
find out the
>truth for yourselves and depended on trusting people who
told you
>things that were false, but you believed them anyway.
>Obviously, you should not gain for your failures, though I
hope you
>learn from them.
>Mathematics isn't about trust. It's about 
truth.
>It does make things much simpler this way.
>
=============================================================
===============
====
Ah yes, those glorious days when the heavy lifting was done.
Now
on to the politics, right?
===
> Ok, I'm starting to feel a little old and getting tired of
arguing, so
> it's time to shift to a less antagonistic strategy.
What I can say in explanation is that *finding* new
mathematical ideas
> isn't what's hard, it's 
critiquing the damn things.
> Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding up to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
You're living in a fantasy world, and my evaluation of you
still
holds.
You lack the ability to reason, but instead get an idea in
your head
and hold on to it against all evidence.
I don't find you credible, and reject your 
claims, as coming
from a
ßawed source.
Why don't you try to produce some facts that are checkable 
to
support
your claims?
I'm curious, about how you'll react to a 
simple challenge:
1. How do you know he is making no efforts to defend his work?
2. How do you know that he's not interes in a monetary 
prize?
I think you're simply NOT a rational person, you have some
idea in
your head that it'd sound great to talk up this particular
person, and
you couldn't care less what the facts actually are because
you're a
social animal.
My guess is that you're just a propagandist, desperate to
push a
political view without concern for the facts!
Now then, how do you answer?
===
> Ok, I'm starting to feel a little old and getting tired of
arguing,
so
> it's time to shift to a less antagonistic strategy.
> What I can say in explanation is that *finding* new
mathematical
ideas
> isn't what's hard, it's 
critiquing the damn things.
Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding up
to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
You're living in a fantasy world, and my evaluation of you
still
> holds.
You lack the ability to reason, but instead get an idea in
your head
> and hold on to it against all evidence.
I don't find you credible, and reject your 
claims, as coming
from a
> ßawed source.
Why don't you try to produce some facts that are checkable 
to
support
> your claims?
I'm curious, about how you'll react to a 
simple challenge:
1. How do you know he is making no efforts to defend his work?
2. How do you know that he's not interes in a monetary 
prize?
>
First, stick your challenge in whatever of your own bodily
orifices
is most convenient. You can find this information imediately
with a web search on Poincare Conjecture and Perelman. The
$1M prize requires that the proof be published in a journal
and
stand up to 2 years of scrutiny. If Perelman were interes
in the money he would NOT have put his papers out on a web
site;
he would have submit them to the top math journal, Annals of
Mathematics. The papers have been pos, so far as I know
without modification, since Nov. 2002. Perelman is described 
as
a recluse and a loner who does not grant interviews.
You, on the other hand, seek out interviews, write to your
Congressman, to magazines, journals [BTW, what is happening
with
the Southwest Journal ???], editoexperts, the FBI, the
President, etc.. You threaten. You write to employers. You
bluster
and whine. You ignore valid refutations of your work.
Recently you simply RAN AWAY from a counterexample by Rick
Decker, and again you RAN AWAY from an airtight proof by
ÔVirgil'. You lie, you evade, you attack people 
personally,
you pretend to be some kind of supergenius mutant -
Like YESTERDAY: the episode involving x^j + y^j = M = f_1*f_2
...
You claim you have a proof. Someone immediately submits a
counterexample.
You modify your claim slightly to evade the counterexample
and again
claim to have a proof. *Another* immediate counterexample.
You modify
your claim *again*. And ANOTHER immediate counterexample.
ANYONE
can see the pattern, even if they never heard of you till
January 20,
claiming one as a challenge or test for the rest of us! In
this case,
the refutation was strictly a matter of arithmetic and you
could not
possibly misunderstand it or deny it. Not exactly supergenius
mutant
behavior, eh? Do you think that is how Perelman or Wiles
behave?
It's only when we get into a little bit more complex math 
that
you reach your ceiling and deny perfectly sound mathematics
only
because you keep yourself deliberately ignorant of the theory
required to understand it. That is what has been going on for
years.
My motivation? I resent greatly the fact that you keep trying
to
discredit mathematicians with no real understanding of what
they do.
You represent a yahoo-ignoramus anti-intellectual approach to
thinking,
an uninformed arrogance toward rigorous logic and
mathematics, too common
and increasing, that should not go unanswered.
> I think you're simply NOT a rational person, you have some
idea in
> your head that it'd sound great to talk up this particular
person, and
> you couldn't care less what the facts actually are because
you're a
> social animal.
My guess is that you're just a propagandist, desperate to
push a
> political view without concern for the facts!
Now then, how do you answer?
>
How about responding to the MATH in my previous posts,
rather than deleting and making up excuses and foisting your
own
propaganda ad nauseum ?
Your pal,
===
> Ok, I'm starting to feel a little old and getting tired of
arguing,
so
> it's time to shift to a less antagonistic strategy.
> What I can say in explanation is that *finding* new
mathematical
ideas
> isn't what's hard, it's 
critiquing the damn things.
Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding up
to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
[sour grapes snipped]
===
> Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding up
to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
> You're living in a fantasy world, and my evaluation of you
still
> holds.
> You lack the ability to reason, but instead get an idea in
your head
> and hold on to it against all evidence.
> I don't find you credible, and reject your 
claims, as coming
from a
> ßawed source.
> Why don't you try to produce some facts that are checkable
to support
> your claims?
> I'm curious, about how you'll react to a 
simple challenge:
> 1. How do you know he is making no efforts to defend his
work?
> 2. How do you know that he's not interes in a monetary
prize?
> I think you're simply NOT a rational person, you have some
idea in
> your head that it'd sound great to talk up this particular
person, and
> you couldn't care less what the facts actually are because
you're a
> social animal.
> My guess is that you're just a propagandist, desperate to
push a
> political view without concern for the facts!
> Now then, how do you answer?
>
Read this James, and then apologize to
http://www.hpcwire.com/freehpc/106742.html
===
> Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding up
to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
You're living in a fantasy world, and my evaluation of you
still
> holds.
You lack the ability to reason, but instead get an idea in
your head
> and hold on to it against all evidence.
I don't find you credible, and reject your 
claims, as coming
from a
> ßawed source.
Why don't you try to produce some facts that are checkable 
to
support
> your claims?
I'm curious, about how you'll react to a 
simple challenge:
1. How do you know he is making no efforts to defend his work?
2. How do you know that he's not interes in a monetary 
prize?
I think you're simply NOT a rational person, you have some
idea in
> your head that it'd sound great to talk up this particular
person, and
> you couldn't care less what the facts actually are because
you're a
> social animal.
My guess is that you're just a propagandist, desperate to
push a
> political view without concern for the facts!
Now then, how do you answer?
>
Read this James, and then apologize to
> http://www.hpcwire.com/freehpc/106742.html
I very much dislike Nora Baron and will not apologize. But
thanks for
the link!
===
> Read this James, and then apologize to
> http://www.hpcwire.com/freehpc/106742.html
I very much dislike Nora Baron and will not apologize. But
thanks for
the
> link!
>
People tend to dislike those who so easily puncture their
overinßa
egos.
===
> Real mathematicians find it the other way round. Recently a
Russian
>>mathematician has written a proposed proof of the Poincare
conjecture.
>>It is being widely circula and critiqued. So far it is
holding up
to
>>scrutiny. The author is somewhat of a recluse and has not
even
>>submit his papers for publication. He appears to be making
no
>>efforts to defend his work. There is a $1M prize for a
solution,
>>but he appears not to be interes.
>You're living in a fantasy world, and my evaluation of you
still
>holds.
This is a personal attack by you.
>You lack the ability to reason, but instead get an idea in
your head
>and hold on to it against all evidence.
This is a personal attack by you.
>I don't find you credible, and reject your 
claims, as coming
from a
>ßawed source.
This is a personal attack by you and dismissal of claims.
>Why don't you try to produce some facts that are checkable
to support
>your claims?
A clear challenge by you.
>I'm curious, about how you'll react to a 
simple challenge:
>1. How do you know he is making no efforts to defend his
work?
See the link provided by Stan.
>2. How do you know that he's not interes in a monetary 
prize?
See the link provided by Stan.
>I think you're simply NOT a rational person, you have some
idea in
>your head that it'd sound great to talk up this particular
person, and
>you couldn't care less what the facts actually are because
you're a
>social animal.
Personal attack by you.
>My guess is that you're just a propagandist, desperate to
push a
>political view without concern for the facts!
Personal attack by you.
>Now then, how do you answer?
>Read this James, and then apologize to
>>http://www.hpcwire.com/freehpc/106742.html
> I very much dislike Nora Baron and will not apologize. But
thanks for
the link!
The link completely validates all of Nora's claims. It (or a
similar
one) has been around for a little while. So: you made at
least 4
personal attacks that were easily shown to be unjustified. You
questioned her assertions when they were public knowledge.
And you won't apologize because you don't like 
her?
Please keep this in mind the next time you complain about how
you are
trea. People who treat others with respect tend to receive
respect.
You have not shown respect to Nora, how much respect do you
think it
will inspire towards you?
--
===
[snip]
> I very much dislike Nora Baron and will not apologize.
>
That statement reveals much more about you than you think.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
===
> Real mathematicians find it the other way round. Recently a
Russian
> mathematician has written a proposed proof of the Poincare
conjecture.
> It is being widely circula and critiqued. So far it is
holding
up to
> scrutiny. The author is somewhat of a recluse and has not
even
> submit his papers for publication. He appears to be making
no
> efforts to defend his work. There is a $1M prize for a
solution,
> but he appears not to be interes.
You're living in a fantasy world, and my evaluation of you
still
> holds.
You lack the ability to reason, but instead get an idea in
your head
> and hold on to it against all evidence.
I don't find you credible, and reject your 
claims, as coming
from a
> ßawed source.
Why don't you try to produce some facts that are checkable 
to
support
> your claims?
I'm curious, about how you'll react to a 
simple challenge:
1. How do you know he is making no efforts to defend his work?
2. How do you know that he's not interes in a monetary 
prize?
I think you're simply NOT a rational person, you have some
idea in
> your head that it'd sound great to talk up this particular
person,
and
> you couldn't care less what the facts actually are because
you're a
> social animal.
My guess is that you're just a propagandist, desperate to
push a
> political view without concern for the facts!
Now then, how do you answer?
>
Read this James, and then apologize to
> http://www.hpcwire.com/freehpc/106742.html
> I very much dislike Nora Baron and will not apologize. But
thanks
for
the link!
>
James, You only dislike Nora Baron because she's proven you
wrong over and
over again. You need to grow up and act like an adult instead
of an
immature
child. I don't consider you an adult because you act like my
7 year old
brother. I've said it once and I'll say it 
again. GROW UP!
===
>My guess is that you're just a propagandist, desperate to
push a
>political view without concern for the facts!
Well, I have a proof that you're of this type. Can you prove
it?
===
[snip]
> Now I get to refine my ideas, mostly out of sight, and make
them part
> of established mathematics, mostly using some high-powered
people with
> advanced degrees,
I notice you didn't name them. At their request, perhaps?
> since I just have a B.S. in physics.
B.S. is the operative descriptor here.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
--
http://www.crbond.com
===
>Ok, I'm starting to feel a little old and getting tired of
arguing, so
>it's time to shift to a less antagonistic strategy.
>What I can say in explanation is that *finding* new
mathematical ideas
>isn't what's hard, it's 
critiquing the damn things.
Deciding whether a genuinely new concept is interesting or
useful
can be tricky. But that doesn't really apply to anything
you've
ever pos - most of what you've pos in the last few years
has been simply _erroneous_, and easily seen to be so. The
new things (like OOM) have not been defined coherently;
again, it's very easy to see that the 
definitions you give
simply make no sense.
>You can come up with idea after idea after idea that's just
crap, but
>for any number of reasons you can get attached to it.
Uh, once an idea has been shown to be crap _most_ of us have
no problem abandoning it.
>It's fun to get out there and argue and fight 
passionately
for certain
>ideas, and see if they survive.
Some people think so. Some people seem to think that it's 
fun
to make a complete and utter fool of oneself in front of the
entire planet, over and over. Tastes differ.
>Most of the ideas that I came up with over a period of more
than 8
>years of active research fell to the side, useless.
>Now that was a glorious time. I'll miss it, but eventually
you just
>get old and tired.
>Now I get to refine my ideas, mostly out of sight, and make
them part
>of established mathematics, mostly using some high-powered
people with
>advanced degrees, since I just have a B.S. in physics.
Pshaw. Do make certain to let us know when one of your ideas
has become part of established mathematics.
>Meanwhile I'll check in here, and probably post from time 
to
time, as
>I seem to really like posting.
>But I'm not interes in getting really inves in arguments
like in
>the past.
>You see, the heavy lifting is done. Now it's time for
academic
>politics.
>
************************
===
Ok, I'm starting to feel a little old and getting tired of
arguing, so
it's time to shift to a less antagonistic strategy.
What I can say in explanation is that *finding* new
mathematical ideas
isn't what's hard, it's 
critiquing the damn things.
You can come up with idea after idea after idea that's just
crap, but
for any number of reasons you can get attached to it.
It's fun to get out there and argue and fight 
passionately for
certain
ideas, and see if they survive.
Most of the ideas that I came up with over a period of more
than 8
years of active research fell to the side, useless.
Now that was a glorious time. I'll miss it, but eventually
you just
get old and tired.
Now I get to refine my ideas, mostly out of sight, and make
them part
of established mathematics, mostly using some high-powered
people with
advanced degrees, since I just have a B.S. in physics.
Meanwhile I'll check in here, and probably post from time to
time, as
I seem to really like posting.
But I'm not interes in getting really inves in arguments 
like
in
the past.
You see, the heavy lifting is done. Now it's time for 
academic
politics.