mm-82
===
Here m varies and you question that possibility for the a's
to be| continuous everywhere. Now what do you see in those
equations Keith| Ramsay?You say whether a1, a2, and a3 are
continuous is irrelevant, so if youwant to you, feel free to
skip the following explanation for why theycan't all three be
continuous and defined everywhere. The explanationcan be made
more precise if you don't believe me.It's called monodromy.
It's related to Galois theory! But it'smore geometric.At m=0,
two of the a's are the same. For m near 0, but not exactly
at0, two of them are close together, while the third is
farther away,staying close to the value it has at 0.If we
take m on a path around 0, going in small steps, because the
a'sare continuous they would have also to change in small
steps. What isinteresting is what happens to them when m is
brought back to where itstarted: those two a's are forced to
exchange places with each other.But that would contradict the
fact that they have the values they doat the original m close
to 0, and that they're continuous, and defined for all
algebraic integers m. (It's possible to step along such a
pathstepping only on algebraic integers, because there are
algebraicintegers close to any point on the complex
plane.)It's easier to see in the case of x^2-m =
(x-b1)(x-b2), so that b1 andb2 are the two complex square
roots of m (b2=-b1). If we take m in aloop around m=0, then
its two square roots exchange places with eachother. They
also go around 0 in the complex plane, but half as fast asm
does. So for b1 and b2 both to be continuous is impossible.
The usual? is to decide on a place to cut the plane, and
make them switch asthey cross the cut.I would have to do a
more careful calculation to be sure that the thirda in the
example from the paper can't be continuous either, becauseon
a different path it changes place with one of the first two,
but Ithink that's also true. One would have to look carefully
at the way thea's change going to one of the other values of m
where two a's coincide,and I don't see any way to make the
calculation simple yet.[...]| Hmmm...maybe it'd help to
consider the other g's now that you bring| them up.| | You
have to remember that I can consider P(m)/f^2, and then you
have| | P(m)/f^2 = g_1 g_2 g_3/f^2| | and I'm merely noting
the consequence of P(0)/f^2 being coprime to f.Namely, that
the factor of f in g_1 must go away. Really, the
mainquestion I had at this point was what you meant by go
away. Dividingg_1 g_2 g_3 by f^2 doesn't necessarily
correspond to any operation ong_1, g_2, and g_3 separately.
This concept of what factor of f istaken away from g_1 when
the product of g_1, g_2, and g_3 is dividedby f^2 hasn't been
de?ed.It sounds like (this is a guess) you mean that you want
to write acommon re?ement of the two factorizations of P,
namely f^2*(P/f^2)and g_1*g_2*g_3. This would be a way to
write P as a product ABCDEFwhere AB=g_1, CD=g_2, and EF=g_3,
and f^2=ACE, and P/f^2=BDF. Itfurther sounds like by the
factor of f in g_1 going away you mean torefer to the
factor A in it, which is a common factor of f^2 and g_1.If
this is what you mean, you would be considering the product
as beingdivided by f^2 by having the factors A,C,E taken out
of g_1, g_2, andg_3 separately.If this is so, then one of the
main questions is what values thiscommon factor has when m is
not 0. I don't see a problem with sayingit's f when m=0, but
I also don't see why it shouldn't be a varyingfactor of f as
m changes.[...]| > It's looking like the factor going away
is supposed to be a factor| > of a_1 in some sense or
another. If so, what sense?| | In considering P(m)/f^2 = g_1
g_2 g_3/f^2, I'm ?uring out how that| factor f^2 goes
through the g's.| | What I ?d is that it only makes sense if
only two of the g's have a| factor of f that is f, but then
you're forced out of the ring of| algebraic integers, though
it doesn't appear to have happened from| within the argument,
which is why that ring is ?I was thinking we knew that
two of the a's were 0 at m=0, so that thecorresponding g's
were uf there, and hence divisible by f. I don't seehow you
plan to apply this to the situation when m is not 0.| >
Perhaps it would help if you clari?d and/or con?med a few
things.| > When in the paper you write that g1 must have
that same factor (of| > f) in general, does in general
mean that you're saying the value| > of g1 for each allowed
value of m is divisible by f? | | It depends on context. It's
like if you have y=2x+1, and consider| various values. Now I
see pressure to write that as f(x)=2x+1, but I| don't think
it's worth the potential confusion.| | The problem is that
though you see m as the key variable, it may in| fact be a
*factor* of m that is the true independent variable, which|
I'm sure has confused quite a few people. Now it's a measure
of your| mathematical ability if you know what I mean there,
and let's just say| that I'm not interested in expanding out
in a way that leaves *more*| room for
confusion.Unfortunately, if the only way that the argument
actually works foryou is that in the background, you're
implicitly permitting yourselfthis kind of reparametrization,
then the thing you have written downisn't a complete
proof.Yes, there are clever techniques involving substituting
parameters,sometimes used to circumvent the presence of a
Galois group ormonodromy, but if you plan to go that route
you still need to say whatyou're doing.| >Or do you mean| >
something else? Is it a fair paraphrase of the step where
you| > introduce a1, a2, and a3, to say that they are three
functions from| > the algebraic integers to the algebraic
integers, a1(m), a2(m), and| > a3(m), having the property
that P(m) = (a1*x+uf)(a2*x+uf)(a3*x+uf) for| > every x?| |
They're not necessarily functions of m.At some point you need
actually to say what kind of thing a1, a2, anda3 are, without
leaving any special wriggle room. If treating them
asfunctions of m is not quite good enough, then you really
need to saywhat they are.| Think about it Keith Ramsay. Sit
down, go over what I've given and| think very carefully over
just that one point, and maybe, if you're| honestly confused,
it'll help you to break through to an| understanding.[...]|
Posts should get *shorter* and not longer.Keith Ramsay
===
>
> Here I'm right, and you're wrong, as the *de?ition* of
algebraic> integer leads to an incomplete ring, which I
prove mathematically.> The trouble is, I'm still not
following your logic, just to be able to> see what the great
contradiction is.> (a) I have two algebraic integers x and
y> (b) I claim that y is a factor of x in the algebraic
integersNope. There is no claim that y is a factor as it's
*provably* afactor.Rather than deal with the term factor as
people get confused, in mypaper Advanced Polynomial
Factorization, I have three numbers a_1,a_2, and a_3, and I
prove that a_3 is coprime to a factor I call f.You also have
that a_1 a_2 a_3 have f^2 as a factor, so proving thata_3 is
coprime to f, means that a_1 and a_2 shouldn't be.But there's
a contradiction because they don't technically have
*any*factors of f, so that means that the de?ition of
algebraic integers,with a strict interpretation of the word
factor would mean thatneither a_1, a_2, nor a_3 have f as a
factor when a_1 a_2 a_3 has f^2as a factor. > (c) But there is
no algebraic integer z (=x/y) such that y*z = x> This
leads inexorably to the conclusion that the claim in (b) is
simply> false. There are *lots* of possible algebraic
integers which I can> multiply by *something* to get x, but
there are only a few choices that> I can multiply by another
algebraic integer to get x, and those are the> factors.You're
running in circles.First consider that the de?ition of
algebraic integers as the*roots* of monic polynomials with
integer coef?ients does notspeci?ally give you any
protection from contradiction.The de?ition is over the ?ld
of rationals, and it's not clear injust looking at it, if it
will give a complete ring.Mathematicians apparently *assumed*
it did, and have gone on thatassumption for some time.People
seem ?ed on the *de?ition* with the assumption, whenthere's
no proof based on the de?ition that the ring is complete.And
there can't be such a proof because I've *proven* it's
notcomplete, as that de?ition leads to a contradiction.>
Basically, I do not see in what sense you can claim that y is
a factor> of x without demonstrating the properties of x/y...
just being able to> multiply y by something to get x isn't
enough to be able to claim that.> And without that claim, the
conclusion about incompleteness isn't there.Go to the
following link, ?d and read my paper Advanced
PolynomialFactorization: http://groups.msn.com/AmateurMathand
you will see how you can prove that x has y as a factor.>
Perhaps it could help if you could give me a rigorous
de?ition of> what you mean by incomplete ring, or
conversely complete ring. Is> this a property which is
distinct from being, say, closed under>
multiplication?Consider c=ab, where ?c' is an algebraic
integer, and ?a' is analgebraic integer, but ?b' is not.Now
if you include fractions or move to a ?ld like algebraic
numbersthat's ok. But I'm talking about ?b' that's not in any
way a fractionor fractional.It's like with the ring of evens,
and given 6 = 2(3), you have that 3is outside the ring, as in
the ring of evens, 2 is NOT a factor of 6.The situation is
analogous.Do you understand?James Harris
===
> [...]> | Here m
varies and you question that possibility for the a's to be> |
continuous everywhere. Now what do you see in those equations
Keith> | Ramsay?> You say whether a1, a2, and a3 are
continuous is irrelevant, so if you> want to you, feel free
to skip the following explanation for why they> can't all
three be continuous and de?ed everywhere. The explanation>
can be made more precise if you don't believe me.If I have
y=mx+b in the ring of algebraic integers, is y continuous?The
ring is algebraic integers, but you're trying to wander off
intosomething unrelated to the paper, which I guess is some
knowledgeyou're proud to have memorized.But do you
*understand* it? > It's called monodromy. It's related to
Galois theory! But it's> more geometric.> At m=0, two of
the a's are the same. For m near 0, but not exactly at> 0,
two of them are close together, while the third is farther
away,> staying close to the value it has at 0.> If we take
m on a path around 0, going in small steps, because the a's>
are continuous they would have also to change in small steps.
What is> interesting is what happens to them when m is brought
back to where it> started: those two a's are forced to
exchange places with each other.My guess is that you have
some way to expand on that, and if you do,and it depends on
some p-adic interpretation then I will rip you apartby
forcing you down to the rigorous and EXACT numbers.You see
Mathematics is an in?ite world where your brain can getquite
lost, but be proud of itself anyway as it can simply tell
itselfthat it's not lost.> But that would contradict the fact
that they have the values they do> at the original m close to
0, and that they're continuous, and de?ed> for all algebraic
integers m. (It's possible to step along such a path> stepping
only on algebraic integers, because there are algebraic>
integers close to any point on the complex plane.)Why don't
you just quit babbling, solve for the a's, and with
thatsolution try to prove your claim?And that's what I mean
readers about people who think mathematics is afashion
show.Mathematics is an in?ite subject, which forces
perfection. Humanbeings, however, have mathematics, which is
a ?ite subject made upof what people can discover in a
fascinatingly short period of time,and unfortunately, people
can also screw up and put errors inmathematics.> It's easier
to see in the case of x^2-m = (x-b1)(x-b2), so that b1 and>
b2 are the two complex square roots of m (b2=-b1). If we take
m in a> loop around m=0, then its two square roots exchange
places with each> other. They also go around 0 in the complex
plane, but half as fast as> m does. So for b1 and b2 both to
be continuous is impossible. The usual> ? is to decide on a
place to cut the plane, and make them switch as> they cross
the cut.That's an operator problem. That is, in trying to
evaluate a valuefor b1 or b2, you use the sqrt() operator,
and get lost on itsinherent ambiguity.However for the math
there is no such problem as it simply deals withthe value of
b1 and b2 at each point. It doesn't think to itself,take the
square, ?ure out if it's negative or positive.You, however,
can get confused.If you wish to see it in some sense the way
the math sees it, use onlygaussian integers for m in your
loop that are squares in the ring ofgaussian integers.Try
that and report back Keith Ramsay.> I would have to do a more
careful calculation to be sure that the third> a in the
example from the paper can't be continuous either, because>
on a different path it changes place with one of the ?st
two, but I> think that's also true. One would have to look
carefully at the way the> a's change going to one of the
other values of m where two a's coincide,> and I don't see
any way to make the calculation simple yet.And you've gone
off on a tangent. But as you mention it, I suggestyou go
ahead and be *careful* instead of possibly hoping that you
canget away with casting doubt.It's not a political
discussion Keith Ramsay. You're not on Fox News.> [...]> |
Hmmm...maybe it'd help to consider the other g's now that you
bring> | them up.> | > | You have to remember that I can
consider P(m)/f^2, and then you have> | > | P(m)/f^2 = g_1
g_2 g_3/f^2> | > | and I'm merely noting the consequence of
P(0)/f^2 being coprime to f.> Namely, that the factor of f
in g_1 must go away. Really, the main> question I had at
this point was what you meant by go away. Dividing> g_1 g_2
g_3 by f^2 doesn't necessarily correspond to any operation on>
g_1, g_2, and g_3 separately. This concept of what factor of f
is> taken away from g_1 when the product of g_1, g_2, and g_3
is divided> by f^2 hasn't been de?ed.That's nonsensical. Now
I guess you're questioning the assumptionthat given that g_1
g_2 g_3 have f^2 as a factor, some of the g's musthave some
non unit factor in common with f.Well gee, it turns out that
*in the ring of algebraic integers* exceptfor at m=0, NONE of
the g's share any non unit factors in common withf.It's
bizarre; it's wacky; and it shows that the ring of
algebraicintegers is incomplete.> It sounds like (this is a
guess) you mean that you want to write a> common re?ement of
the two factorizations of P, namely f^2*(P/f^2)> and
g_1*g_2*g_3. This would be a way to write P as a product
ABCDEF> where AB=g_1, CD=g_2, and EF=g_3, and f^2=ACE, and
P/f^2=BDF. It> further sounds like by the factor of f in g_1
going away you mean to> refer to the factor A in it,
which is a common factor of f^2 and g_1.> If this is what you
mean, you would be considering the product as being> divided
by f^2 by having the factors A,C,E taken out of g_1, g_2,
and> g_3 separately.> If this is so, then one of the main
questions is what values this> common factor has when m is
not 0. I don't see a problem with saying> it's f when m=0,
but I also don't see why it shouldn't be a varying> factor of
f as m changes.Yeah, and I gave an example yesterday which
should tell you why. P(x) = 2x^2 + 4x + 2 (a_1 + 1)(a_2 + 2)
= 2x^2 + 4x + 2 a_1 a_2 = 2x^2,and I note that when x=0, at
*least* one of the a's must equal 0, andfurther note that the
constant term of the factor a_1 + 1, which I'llcall g_1 is 1
at that point. Further the factor a_2 + 2 has a factorthat is
2 at that point, and I'll call it g_2.Now then Keith Ramsay,
do you question what g_2 will do when x doesnot equal 0?So
why do you do it above?Ultimately, you may try and claim that
irreducibility over Q is what'spertinent, if so, explain in
detail.> [...]> | > It's looking like the factor going away
is supposed to be a factor> | > of a_1 in some sense or
another. If so, what sense?> | > | In considering P(m)/f^2 =
g_1 g_2 g_3/f^2, I'm ?uring out how that> | factor f^2 goes
through the g's.> | > | What I ?d is that it only makes
sense if only two of the g's have a> | factor of f that is f,
but then you're forced out of the ring of> | algebraic
integers, though it doesn't appear to have happened from> |
within the argument, which is why that ring is ?> I
was thinking we knew that two of the a's were 0 at m=0, so
that the> corresponding g's were uf there, and hence
divisible by f. I don't see> how you plan to apply this to
the situation when m is not 0.I've generalized that factors
of polynomials can be written as r+c,where r=0, or varies
while the polynomial factor varies, and c isconstant and is a
factor of the constant term of the polynomial.That's easy to
see with something like x+2 as a factor of x^2+3x+2.In the
paper I have that as a lemma, with a proof.If anyone had an
actual error in the paper, they could simply showfault with
the lemma, as it's the key linchpin.Instead I get overlong
posts where people bring up tangents, and keepquestioning
things that are obvious.Mathematicians are such dweebs.> | >
Perhaps it would help if you clari?d and/or con?med a few
things.> | > When in the paper you write that g1 must have
that same factor (of> | > f) in general, does in
general mean that you're saying the value> | > of g1 for
each allowed value of m is divisible by f? > | > | It depends
on context. It's like if you have y=2x+1, and consider> |
various values. Now I see pressure to write that as
f(x)=2x+1, but I> | don't think it's worth the potential
confusion.> | > | The problem is that though you see m as the
key variable, it may in> | fact be a *factor* of m that is the
true independent variable, which> | I'm sure has confused
quite a few people. Now it's a measure of your> |
mathematical ability if you know what I mean there, and let's
just say> | that I'm not interested in expanding out in a way
that leaves *more*> | room for confusion.> Unfortunately,
if the only way that the argument actually works for> you is
that in the background, you're implicitly permitting
yourself> this kind of reparametrization, then the thing you
have written down> isn't a complete proof.Then prove it.There
are lots of things that I think about that aren't needed to
showa proof, and I don't put them into discussions unless it
suits me.Remember, a proof begins with a truth and proceeds
by logical steps toa conclusion which then must be true.I
only need give that truth in the beginning and proceed by
logicalsteps.That you can move deeper into portions in a way
that isn't material tothe argument makes the math just a
little more fun...well it does forme at least.> Yes, there
are clever techniques involving substituting parameters,>
sometimes used to circumvent the presence of a Galois group
or> monodromy, but if you plan to go that route you still
need to say what> you're doing.That is irrelevant. See? I
knew that bringing the subject up mightjust distract you.Why
don't you focus on the important path of the proof, and not
worryabout sideroads?> | >Or do you mean> | > something else?
Is it a fair paraphrase of the step where you> | > introduce
a1, a2, and a3, to say that they are three functions from> |
> the algebraic integers to the algebraic integers, a1(m),
a2(m), and> | > a3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) for> | > every x?> | > | They're
not necessarily functions of m.> At some point you need
actually to say what kind of thing a1, a2, and> a3 are,
without leaving any special wriggle room. If treating them
as> functions of m is not quite good enough, then you really
need to say> what they are.You didn't connect back with your
own statement, fascinating.I'm repeating myself with that
statement, but before you talked ofreparametrization in
reply.I fear you are a parrot. You look for key words and
then regurgitatebased on something you think is relevant, but
you need to THINK.> | Think about it Keith Ramsay. Sit down,
go over what I've given and> | think very carefully over just
that one point, and maybe, if you're> | honestly confused,
it'll help you to break through to an> | understanding.>
[...]> | Posts should get *shorter* and not longer.> Keith
RamsayAnd I'll reiterate that posts should get shorter and
not longer.It's called cutting to the chase, or getting to
the nitty-gritty.James Harris
===
| > [...]| > | Here m varies
and you question that possibility for the a's to be| > |
continuous everywhere. Now what do you see in those equations
Keith| > | Ramsay?| > | > You say whether a1, a2, and a3 are
continuous is irrelevant, so if you| > want to you, feel free
to skip the following explanation for why they| > can't all
three be continuous and de?ed everywhere. The explanation| >
can be made more precise if you don't believe me.| | If I have
y=mx+b in the ring of algebraic integers, is y
continuous?Assuming you mean that m and b are algebraic
integers, and y is beinggiven as a function of x,
yes.Remember (unlike e.g. the Gaussian integers) the
algebraic integersare dense in the complex plane, meaning
that any circle, no matterhow small, has algebraic integers
inside it. | The ring is algebraic integers, but you're
trying to wander off into| something unrelated to the paper,
which I guess is some knowledge| you're proud to have
memorized.My memory is decent, but nothing to be proud of.No,
my main point in bringing this stuff up was to try to
convinceyou that de?ing a1, a2, and a3 in such a way that
their values atone point are related to their values at other
points in a useful way.If my explanation isn't working, ?e,
someone will eventually ?da better way of explaining it to
you.| > If we take m on a path around 0, going in small
steps, because the a's| > are continuous they would have also
to change in small steps. What is| > interesting is what
happens to them when m is brought back to where it| >
started: those two a's are forced to exchange places with
each other.| | My guess is that you have some way to expand
on that, and if you do,| and it depends on some p-adic
interpretation then I will rip you apart| by forcing you down
to the rigorous and EXACT numbers.You have a way of being
melodramatic when it doesn't make sense.First of all, don't
try to hint that I'm prone to pretending that I'mdealing with
complex numbers when I'm actually sneakily dealing withp-adic
numbers. You're the one who's been dealing with a1, a2, anda3
having indeterminate type.Second, p-adic numbers ARE exact.
They're just a different system.It's easier in a lot of ways
to be rigorous with them than it isto be rigorous with the
real numbers or the complex numbers.Third, you're surely
aware that nobody will be ripping anybodyapart or forcing
anybody to do anything. People have tried hardto force you
down to rigor, for example by telling you what,allegedly,
the common factors between 5 and the roots of that cubicare.
Doesn't phase you a bit! And you've shown us all how easy it
isto be impervious to demands for rigor.| You see Mathematics
is an in?ite world where your brain can get| quite lost, but
be proud of itself anyway as it can simply tell itself| that
it's not lost.on the algebraic integers, when the property I
actually needed wascontinuity not only at the algebraic
integers, but at all complexnumbers. My mistake was to think
that the fact that the algebraicintegers are dense in the
plane would be good enough, but it isn't.You might say that
the discontinuities can be hidden atnon-algebraic-integer
points. There is, for example, a uniquefunction f(z) de?ed
on the complex numbers a+bi such thatb <> pi*a, satisfying
f(z)^2=z, f(1)=1, and f(-1)=i. It can'tbe extended to the
line b=pi*a because it approaches differentlimits on either
side of that line (it jumps there) on theside where a,b <
0. If we restrict it to the algebraic integers,then it's
continuous (because continuous would mean only continuouson
algebraic integers).I'm still saying that the situation with
a1, a2 and a3 is analogousto the situation with the two
so-called branches of the squareroot multiple-valued
function. It's not all that hard to give arigorous proof of
what I said about continuous square root functions:Suppose
that f(z) is a function on the complex plane having
complexvalues, and satisfying f(z)^2=z for all z. The goal is
to show thatthis leads to a contradiction. Actually, it's
enough to consider iton the circle, which consists of complex
numbers of the form e^{it}where t is a real number.Because for
each real t, f(e^{it})^2 - e^{it}=
(f(e^{it})-e^{it/2})(f(e^{it})+e^{it/2}), we also have
thateither f(e^{it}) = e^{it/2} or f(e^{it}) = -e^{it/2}.
Denote by Athe set of t satisfying f(e^{it}) = e^{it/2} and
by B the set oft satisfying f(e^{it}) = -e^{it/2}. Since
e^{it/2} <> 0 for anyreal t, the two sets A and B are
disjoint, and I've just shownthat their union is the whole
real line.The next thing is to prove that they are both
closed sets. A closedset is a set like A which has as an
element every complex number zhaving the property that for
every epsilon>0, there is an elementof A within epsilon of z.
But A and B are the sets of zeros of thecontinuous functions
f(e^{it})-e^{it/2} and f(e^{it})+e^{it/2}respectively, and
the set of zeros of a continuous function is alwaysclosed.
The proof of that is that if g is continuous, and g(z) <>
0,then there exists by the de?ition of continuity a delta >
0 suchthat |g(z')-g(z)|<|g(z)| when |z'-z|<delta, and
|g(z')-g(z)|<|g(z)|implies g(z')<>0 as well. So if g is
continuous and z isn't a zero,then there's a delta>0 such
that no other zeros of g come withindelta of z. Conversely,
if there are zeros within any delta>0 ofz, then z must also
be a zero.Finally, we use a standard result that the real
line can't bebroken into two disjoint nonempty sets which are
both closed.The standard proof is by the least upper bound
principle. If Aand B are the two nonempty closed sets,
suppose a is an elementof A and b is an element of B. Without
loss of generality, a < b;the proof for b < a is essentially
the same. Let a' be the leastupper bound of the elements of A
less than b. Because a' is anupper bound, there are no
elements of A between a' and b. So forevery delta>0,
delta<b-a', the element a'+delta is in B. So bythe fact that
B is closed, a' is an element of B. On the otherhand, by the
fact that A is closed, this means there is a delta>0such that
every number within delta of a' is also in B, includingthe
open segment from max(a, a'-delta) to a'. That meansmax(a,
a'-delta) is a lesser upper bound on A, contradicting thefact
that a' was the least upper bound.It follows that one of A or
B is empty, and the other is the wholereal line. But that's
impossible, since if f(1) = f(e^{0*i}) =e^{0*i/2} = 1 then
f(1) = f(e^{2*pi*i}) = -(-1) = -e^{pi*i}= -e^{2*pi*i/2}, and
if f(e^{0*i})=-e^{0*i/2} then f(e^{2*pi*i})= e^{2*pi*i/2}.
This is where the wraparound comes in; if 0 isin A then
2*pi is in B and vice-versa.So in fact the original
assumption that there was such an f wasfalse, Q.E.D.Now, a1
and a2 are not square roots, but they are similar enoughto a
constant times the square root of m to make the same type
ofargument apply.Keith Ramsay
===
> | > [...]> | > | Here m
varies and you question that possibility for the a's to be> |
> | continuous everywhere. Now what do you see in those
equations Keith> | > | Ramsay?> | > | > You say whether
a1, a2, and a3 are continuous is irrelevant, so if you> | >
want to you, feel free to skip the following explanation for
why they> | > can't all three be continuous and de?ed
everywhere. The explanation> | > can be made more precise if
you don't believe me.> | > | If I have y=mx+b in the ring of
algebraic integers, is y continuous?> Assuming you mean
that m and b are algebraic integers, and y is being> given as
a function of x, yes.> Remember (unlike e.g. the Gaussian
integers) the algebraic integers> are dense in the complex
plane, meaning that any circle, no matter> how small, has
algebraic integers inside it.Hmmm...interesting assertion,
but what about singularities? > | The ring is algebraic
integers, but you're trying to wander off into> | something
unrelated to the paper, which I guess is some knowledge> |
you're proud to have memorized.> My memory is decent, but
nothing to be proud of.That's not what I said, as I assume
that no matter what your level ofpride in your memory might
be you do actually remember some things.> No, my main point
in bringing this stuff up was to try to convince> you that
de?ing a1, a2, and a3 in such a way that their values at>
one point are related to their values at other points in a
useful way.Um, but that is an insinuation that they aren't
already rigorouslyde?ed, when in fact they are.> If my
explanation isn't working, ?e, someone will eventually ?d>
a better way of explaining it to you.That is an irrational
statement. Where do you suppose this someoneis?Possibly
your memory tells you they're in some potential ?ld?> | > If
we take m on a path around 0, going in small steps, because
the a's> | > are continuous they would have also to change
in small steps. What is> | > interesting is what happens to
them when m is brought back to where it> | > started: those
two a's are forced to exchange places with each other.> | > |
My guess is that you have some way to expand on that, and if
you do,> | and it depends on some p-adic interpretation then
I will rip you apart> | by forcing you down to the rigorous
and EXACT numbers.> You have a way of being melodramatic
when it doesn't make sense.> First of all, don't try to
hint that I'm prone to pretending that I'm> dealing with
complex numbers when I'm actually sneakily dealing with>
p-adic numbers. You're the one who's been dealing with a1,
a2, and> a3 having indeterminate type.That is false as the
a's are determined both in form and type.> Second, p-adic
numbers ARE exact. They're just a different system.> It's
easier in a lot of ways to be rigorous with them than it is>
to be rigorous with the real numbers or the complex
numbers.Is that something you have from memory Keith Ramsay?>
Third, you're surely aware that nobody will be ripping
anybody> apart or forcing anybody to do anything. People
have tried hard> to force you down to rigor, for example by
telling you what,> allegedly, the common factors between 5 and
the roots of that cubic> are. Doesn't phase you a bit! And
you've shown us all how easy it is> to be impervious to
demands for rigor.That's an appeal to the gallery, which is
a logical fallacy.> | You see Mathematics is an in?ite world
where your brain can get> | quite lost, but be proud of
itself anyway as it can simply tell itself> | that it's not
lost.> on the algebraic integers, when the property I
actually needed was> continuity not only at the algebraic
integers, but at all complex> numbers. My mistake was to
think that the fact that the algebraic> integers are dense in
the plane would be good enough, but it isn't.> You might say
that the discontinuities can be hidden at>
non-algebraic-integer points. There is, for example, a
unique> function f(z) de?ed on the complex numbers a+bi such
that> b <> pi*a, satisfying f(z)^2=z, f(1)=1, and f(-1)=i. It
can't> be extended to the line b=pi*a because it approaches
different> limits on either side of that line (it jumps
there) on the> side where a,b < 0. If we restrict it to the
algebraic integers,> then it's continuous (because continuous
would mean only continuous> on algebraic integers).>> I'm
still saying that the situation with a1, a2 and a3 is
analogous> to the situation with the two so-called branches
of the square> root multiple-valued function. It's not all
that hard to give a> rigorous proof of what I said about
continuous square root functions:> Suppose that f(z) is a
function on the complex plane having complex> values, and
satisfying f(z)^2=z for all z. The goal is to show that> this
leads to a contradiction. Actually, it's enough to consider
it> on the circle, which consists of complex numbers of the
form e^{it}> where t is a real number.> Because for each
real t, f(e^{it})^2 - e^{it}> =
(f(e^{it})-e^{it/2})(f(e^{it})+e^{it/2}), we also have that>
either f(e^{it}) = e^{it/2} or f(e^{it}) = -e^{it/2}. Denote
by A> the set of t satisfying f(e^{it}) = e^{it/2} and by B
the set of> t satisfying f(e^{it}) = -e^{it/2}. Since
e^{it/2} <> 0 for any> real t, the two sets A and B are
disjoint, and I've just shown> that their union is the whole
real line.Your limitation is in needing the sqrt() operator,
but being unable tosimply give the result in general. The
math doesn't have that problemas it doesn't need operators,
but simply sees the result as it is.For instance, consider
(1+sqrt(-3))/2, which is in fact a unit in thering of
algebraic integers, but it *looks* nothing like 1, or
i,because you have the two operators sqrt() and /.(Purists
may note that this algebraic integer cannot be expressedusing
only ring operators.)You have no meaningful means of further
resolution.But the math sees a number. > The next thing is to
prove that they are both closed sets. A closed> set is a set
like A which has as an element every complex number z> having
the property that for every epsilon>0, there is an element> of
A within epsilon of z. But A and B are the sets of zeros of
the> continuous functions f(e^{it})-e^{it/2} and
f(e^{it})+e^{it/2}> respectively, and the set of zeros of a
continuous function is always> closed. The proof of that is
that if g is continuous, and g(z) <> 0,> then there exists by
the de?ition of continuity a delta > 0 such> that
|g(z')-g(z)|<|g(z)| when |z'-z|<delta, and
|g(z')-g(z)|<|g(z)|> implies g(z')<>0 as well. So if g is
continuous and z isn't a zero,> then there's a delta>0 such
that no other zeros of g come within> delta of z. Conversely,
if there are zeros within any delta>0 of> z, then z must also
be a zero.> Finally, we use a standard result that the real
line can't be> broken into two disjoint nonempty sets which
are both closed.> The standard proof is by the least upper
bound principle. If A> and B are the two nonempty closed
sets, suppose a is an element> of A and b is an element of B.
Without loss of generality, a < b;> the proof for b < a is
essentially the same. Let a' be the least> upper bound of the
elements of A less than b. Because a' is an> upper bound,
there are no elements of A between a' and b. So for> every
delta>0, delta<b-a', the element a'+delta is in B. So by> the
fact that B is closed, a' is an element of B. On the other>
hand, by the fact that A is closed, this means there is a
delta>0> such that every number within delta of a' is also in
B, including> the open segment from max(a, a'-delta) to a'.
That means> max(a, a'-delta) is a lesser upper bound on A,
contradicting the> fact that a' was the least upper bound.>
It follows that one of A or B is empty, and the other is the
whole> real line. But that's impossible, since if f(1) =
f(e^{0*i}) => e^{0*i/2} = 1 then f(1) = f(e^{2*pi*i}) = -(-1)
= -e^{pi*i}> = -e^{2*pi*i/2}, and if f(e^{0*i})=-e^{0*i/2}
then f(e^{2*pi*i})> = e^{2*pi*i/2}. This is where the
wraparound comes in; if 0 is> in A then 2*pi is in B and
vice-versa.> So in fact the original assumption that there
was such an f was> false, Q.E.D.Can you switch to
congruences? What about something like f(e^{it})^2 - e^{it} =
0(mod 35)in the ring of algebraic integers? > Now, a1 and a2
are not square roots, but they are similar enough> to a
constant times the square root of m to make the same type of>
argument apply.> Keith RamsayWhy don't you just give the a's
explicitly and prove it then?James Harris
===
[...]|And you've
gone off on a tangent. But as you mention it, I suggest|you
go ahead and be *careful* instead of possibly hoping that you
can|get away with casting doubt.||It's not a political
discussion Keith Ramsay. You're not on Fox News.You're hardly
in any position to complain, either about gliblyexpressing
doubts about things you don't understand, or about
makingpolitical statements, or about being careful
either![...]| > If this is so, then one of the main questions
is what values this| > common factor has when m is not 0. I
don't see a problem with saying| > it's f when m=0, but I
also don't see why it shouldn't be a varying| > factor of f
as m changes.| | Yeah, and I gave an example yesterday which
should tell you why.| | P(x) = 2x^2 + 4x + 2| | (a_1 + 1)(a_2
+ 2) = 2x^2 + 4x + 2| | a_1 a_2 = 2x^2,Is this last equation
another assumption? Otherwise, what's to ruleout a_1=1, a_2 =
x^2+2x-1, for instance? | and I note that when x=0, at *least*
one of the a's must equal 0, and| further note that the
constant term of the factor a_1 + 1, which I'll| call g_1 is
1 at that point. Further the factor a_2 + 2 has a factor|
that is 2 at that point, and I'll call it g_2.|| Now then
Keith Ramsay, do you question what g_2 will do when x does|
not equal 0?| | So why do you do it above?Continually
expressing shock that I don't agree with you, when youhaven't
indicated a principle to justify your claim, is just a bunchof
posturing. What's the general principle?| Ultimately, you may
try and claim that irreducibility over Q is what's|
pertinent, if so, explain in detail.The two cases aren't
parallel. You really should get out of thehabit of thinking
that analogy and cajolery add up to a logicalargument. Giving
a bunch of examples which work the way you wantanother example
to work is not a proof.If I were to give an explanation for
why the examples that work,work, I'd give it in terms of the
following principle: (*) If a divides bc, then there exist m
and n such that a=mn, m divides b and n divides c.You give
examples where the product of two polynomials is divisibleby
an integer. Given that (*) is true in these cases, we can
thentry to ?ure out what m and n would work by considering
the constantterms and so on as you do. All of your simpli?d
examples are oneswhere (*) works.But (*) is not generally
true for functions having algebraic integervalues.[...]| > |
>Or do you mean| > | > something else? Is it a fair
paraphrase of the step where you| > | > introduce a1, a2, and
a3, to say that they are three functions from| > | > the
algebraic integers to the algebraic integers, a1(m), a2(m),
and| > | > a3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) for| > | > every x?| > | | > |
They're not necessarily functions of m.| > | > At some point
you need actually to say what kind of thing a1, a2, and| > a3
are, without leaving any special wriggle room. If treating
them as| > functions of m is not quite good enough, then you
really need to say| > what they are.| | You didn't connect
back with your own statement, fascinating.| | I'm repeating
myself with that statement, but before you talked of|
reparametrization in reply.| | I fear you are a parrot. You
look for key words and then regurgitate| based on something
you think is relevant, but you need to THINK.I didn't think I
would need to put in the previous quoted text inorder for you
to remember what you had just written:#> Perhaps it would
help if you clari?d and/or con?med a few things.#> When in
the paper you write that g1 must have that same factor
(of#> f) in general, does in general mean that you're
saying the value#> of g1 for each allowed value of m is
divisible by f? ##It depends on context. It's like if you
have y=2x+1, and consider#various values. Now I see pressure
to write that as f(x)=2x+1, but I#don't think it's worth the
potential confusion.##The problem is that though you see m as
the key variable, it may in#fact be a *factor* of m that is
the true independent variable, which#I'm sure has confused
quite a few people. Now it's a measure of your#mathematical
ability if you know what I mean there, and let's just
say#that I'm not interested in expanding out in a way that
leaves *more*#room for confusion.So you're telling us that m
might not be the true independentvariable. Do you realize
that switching from one independentvariable to another one is
known as reparametrizing?A reasonable reader, who had read
the paper but not this discussion,would naturally tend to
suspect that a1, a2 and a3, which aredescribed as having
values at given values of m, would tend tosuspect that they
are simply functions of m. If this is notnecessarily so, it's
because you've left out any explanation forwhat else you
intend for them to be.This stuff about how being more speci?
is bad because the rest ofus (being inferior to you in
mathematical ability of course) wouldbe liable to be
confused is just an excuse. You are, because ofyour lack of
practice in writing actual proofs, failing to bespeci?, and
as a result getting confused. I think you're just tryingto
bluff your way past the fact that you don't know how to
de?ethem in such a way as to get the argument in the paper
to work.Keith Ramsay
===
[...]|And you've gone off on a
tangent. But as you mention it, I suggest|you go ahead and be
*careful* instead of possibly hoping that you can|get away
with casting doubt.||It's not a political discussion Keith
Ramsay. You're not on Fox News.You're hardly in any position
to complain, either about gliblyexpressing doubts about
things you don't understand, or about makingpolitical
statements, or about being careful either![...]| > If this is
so, then one of the main questions is what values this| >
common factor has when m is not 0. I don't see a problem with
saying| > it's f when m=0, but I also don't see why it
shouldn't be a varying| > factor of f as m changes.| | Yeah,
and I gave an example yesterday which should tell you why.| |
P(x) = 2x^2 + 4x + 2| | (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2| |
a_1 a_2 = 2x^2,Is this last equation another assumption?
Otherwise, what's to ruleout a_1=1, a_2 = x^2+2x-1, for
instance? | and I note that when x=0, at *least* one of the
a's must equal 0, and| further note that the constant term of
the factor a_1 + 1, which I'll| call g_1 is 1 at that point.
Further the factor a_2 + 2 has a factor| that is 2 at that
point, and I'll call it g_2.|| Now then Keith Ramsay, do you
question what g_2 will do when x does| not equal 0?| | So why
do you do it above?Continually expressing shock that I don't
agree with you, when youhaven't indicated a principle to
justify your claim, is just a bunchof posturing. What's the
general principle?| Ultimately, you may try and claim that
irreducibility over Q is what's| pertinent, if so, explain in
detail.The two cases aren't parallel. You really should get
out of thehabit of thinking that analogy and cajolery add up
to a logicalargument. Giving a bunch of examples which work
the way you wantanother example to work is not a proof.If I
were to give an explanation for why the examples that
work,work, I'd give it in terms of the following principle:
(*) If a divides bc, then there exist m and n such that a=mn,
m divides b and n divides c.You give examples where the
product of two polynomials is divisibleby an integer. Given
that (*) is true in these cases, we can thentry to ?ure out
what m and n would work by considering the constantterms and
so on as you do. All of your simpli?d examples are oneswhere
(*) works.But (*) is not generally true for functions having
algebraic integervalues.[...]| > | >Or do you mean| > | >
something else? Is it a fair paraphrase of the step where
you| > | > introduce a1, a2, and a3, to say that they are
three functions from| > | > the algebraic integers to the
algebraic integers, a1(m), a2(m), and| > | > a3(m), having
the property that P(m) = (a1*x+uf)(a2*x+uf)(a3*x+uf) for| >
| > every x?| > | | > | They're not necessarily functions of
m.| > | > At some point you need actually to say what kind of
thing a1, a2, and| > a3 are, without leaving any special
wriggle room. If treating them as| > functions of m is not
quite good enough, then you really need to say| > what they
are.| | You didn't connect back with your own statement,
fascinating.| | I'm repeating myself with that statement, but
before you talked of| reparametrization in reply.| | I fear
you are a parrot. You look for key words and then
regurgitate| based on something you think is relevant, but
you need to THINK.I didn't think I would need to put in the
previous quoted text inorder for you to remember what you had
just written:#> Perhaps it would help if you clari?d and/or
con?med a few things.#> When in the paper you write that g1
must have that same factor (of#> f) in general, does in
general mean that you're saying the value#> of g1 for each
allowed value of m is divisible by f? ##It depends on
context. It's like if you have y=2x+1, and consider#various
values. Now I see pressure to write that as f(x)=2x+1, but
I#don't think it's worth the potential confusion.##The
problem is that though you see m as the key variable, it may
in#fact be a *factor* of m that is the true independent
variable, which#I'm sure has confused quite a few people. Now
it's a measure of your#mathematical ability if you know what I
mean there, and let's just say#that I'm not interested in
expanding out in a way that leaves *more*#room for
confusion.So you're telling us that m might not be the true
independentvariable. Do you realize that switching from one
independentvariable to another one is known as
reparametrizing?A reasonable reader, who had read the paper
but not this discussion,would naturally tend to suspect that
a1, a2 and a3, which aredescribed as having values at given
values of m, would tend tosuspect that they are simply
functions of m. If this is notnecessarily so, it's because
you've left out any explanation forwhat else you intend for
them to be.This stuff about how being more speci? is bad
because the rest ofus (being inferior to you in mathematical
ability of course) wouldbe liable to be confused is just an
excuse. You are, because ofyour lack of practice in writing
actual proofs, failing to bespeci?, and as a result getting
confused. I think you're just tryingto bluff your way past
the fact that you don't know how to de?ethem in such a way
as to get the argument in the paper to work.Keith
Ramsay
===
[...]|And you've gone off on a tangent. But as you
mention it, I suggest|you go ahead and be *careful* instead
of possibly hoping that you can|get away with casting
doubt.||It's not a political discussion Keith Ramsay. You're
not on Fox News.You're hardly in any position to complain,
either about gliblyexpressing doubts about things you don't
understand, or about makingpolitical statements, or about
being careful either![...]| > If this is so, then one of the
main questions is what values this| > common factor has when
m is not 0. I don't see a problem with saying| > it's f when
m=0, but I also don't see why it shouldn't be a varying| >
factor of f as m changes.| | Yeah, and I gave an example
yesterday which should tell you why.| | P(x) = 2x^2 + 4x + 2|
| (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2| | a_1 a_2 = 2x^2,Is this
last equation another assumption? Otherwise, what's to ruleout
a_1=1, a_2 = x^2+2x-1, for instance? | and I note that when
x=0, at *least* one of the a's must equal 0, and| further
note that the constant term of the factor a_1 + 1, which
I'll| call g_1 is 1 at that point. Further the factor a_2 + 2
has a factor| that is 2 at that point, and I'll call it g_2.||
Now then Keith Ramsay, do you question what g_2 will do when x
does| not equal 0?| | So why do you do it above?Continually
expressing shock that I don't agree with you, when youhaven't
indicated a principle to justify your claim, is just a bunchof
posturing. What's the general principle?| Ultimately, you may
try and claim that irreducibility over Q is what's|
pertinent, if so, explain in detail.The two cases aren't
parallel. You really should get out of thehabit of thinking
that analogy and cajolery add up to a logicalargument. Giving
a bunch of examples which work the way you wantanother example
to work is not a proof.If I were to give an explanation for
why the examples that work,work, I'd give it in terms of the
following principle: (*) If a divides bc, then there exist m
and n such that a=mn, m divides b and n divides c.You give
examples where the product of two polynomials is divisibleby
an integer. Given that (*) is true in these cases, we can
thentry to ?ure out what m and n would work by considering
the constantterms and so on as you do. All of your simpli?d
examples are oneswhere (*) works.But (*) is not generally
true for functions having algebraic integervalues.[...]| > |
>Or do you mean| > | > something else? Is it a fair
paraphrase of the step where you| > | > introduce a1, a2, and
a3, to say that they are three functions from| > | > the
algebraic integers to the algebraic integers, a1(m), a2(m),
and| > | > a3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) for| > | > every x?| > | | > |
They're not necessarily functions of m.| > | > At some point
you need actually to say what kind of thing a1, a2, and| > a3
are, without leaving any special wriggle room. If treating
them as| > functions of m is not quite good enough, then you
really need to say| > what they are.| | You didn't connect
back with your own statement, fascinating.| | I'm repeating
myself with that statement, but before you talked of|
reparametrization in reply.| | I fear you are a parrot. You
look for key words and then regurgitate| based on something
you think is relevant, but you need to THINK.I didn't think I
would need to put in the previous quoted text inorder for you
to remember what you had just written:#> Perhaps it would
help if you clari?d and/or con?med a few things.#> When in
the paper you write that g1 must have that same factor
(of#> f) in general, does in general mean that you're
saying the value#> of g1 for each allowed value of m is
divisible by f? ##It depends on context. It's like if you
have y=2x+1, and consider#various values. Now I see pressure
to write that as f(x)=2x+1, but I#don't think it's worth the
potential confusion.##The problem is that though you see m as
the key variable, it may in#fact be a *factor* of m that is
the true independent variable, which#I'm sure has confused
quite a few people. Now it's a measure of your#mathematical
ability if you know what I mean there, and let's just
say#that I'm not interested in expanding out in a way that
leaves *more*#room for confusion.So you're telling us that m
might not be the true independentvariable. Do you realize
that switching from one independentvariable to another one is
known as reparametrizing?A reasonable reader, who had read
the paper but not this discussion,would naturally tend to
suspect that a1, a2 and a3, which aredescribed as having
values at given values of m, would tend tosuspect that they
are simply functions of m. If this is notnecessarily so, it's
because you've left out any explanation forwhat else you
intend for them to be.This stuff about how being more speci?
is bad because the rest ofus (being inferior to you in
mathematical ability of course) wouldbe liable to be
confused is just an excuse. You are, because ofyour lack of
practice in writing actual proofs, failing to bespeci?, and
as a result getting confused. I think you're just tryingto
bluff your way past the fact that you don't know how to
de?ethem in such a way as to get the argument in the paper
to work.Keith Ramsay
===
[...]|And you've gone off on a
tangent. But as you mention it, I suggest|you go ahead and be
*careful* instead of possibly hoping that you can|get away
with casting doubt.||It's not a political discussion Keith
Ramsay. You're not on Fox News.You're hardly in any position
to complain, either about gliblyexpressing doubts about
things you don't understand, or about makingpolitical
statements, or about being careful either![...]| > If this is
so, then one of the main questions is what values this| >
common factor has when m is not 0. I don't see a problem with
saying| > it's f when m=0, but I also don't see why it
shouldn't be a varying| > factor of f as m changes.| | Yeah,
and I gave an example yesterday which should tell you why.| |
P(x) = 2x^2 + 4x + 2| | (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2| |
a_1 a_2 = 2x^2,Is this last equation another assumption?
Otherwise, what's to ruleout a_1=1, a_2 = x^2+2x-1, for
instance? | and I note that when x=0, at *least* one of the
a's must equal 0, and| further note that the constant term of
the factor a_1 + 1, which I'll| call g_1 is 1 at that point.
Further the factor a_2 + 2 has a factor| that is 2 at that
point, and I'll call it g_2.|| Now then Keith Ramsay, do you
question what g_2 will do when x does| not equal 0?| | So why
do you do it above?Continually expressing shock that I don't
agree with you, when youhaven't indicated what principle it
is that justi?s your claim,is just a bunch of posturing.
What's the general principle?| Ultimately, you may try and
claim that irreducibility over Q is what's| pertinent, if so,
explain in detail.The two cases aren't parallel. You really
should get out of thehabit of thinking that analogy and
cajolery add up to a logicalargument. Giving a bunch of
examples which work the way you wantanother example to work
is not a proof.If I were to give an explanation for why the
examples that work,work, I'd give it in terms of the
following principle: (*) If a divides bc, then there exist m
and n such that a=mn, m divides b and n divides c.You give
examples where the product of two polynomials is divisibleby
an integer. Given that (*) is true in these cases, we can
thentry to ?ure out what m and n would work by considering
the constantterms and so on as you do. All of your simpli?d
examples are oneswhere (*) works.But (*) is not generally
true for functions having algebraic integervalues.Another way
to look at it is that in order to show that a1 is divisibleby
f (say), you have either to show that all three of them are
divisibleby f, or you have to make some additional assumption
which somehow forces a1at each value of m not to be the value
or one of the values of the three(a1, a2, and a3) that isn't
divisible by f. If a1, a2 and a3 were simplythree functions
of m, which at each speci? m satisfy the equation
de?ingthem, then the order in which they're assigned to
those three values is stillarbitrary, leaving nothing to
keep a1 from being the wrong value.[...]| > | >Or do you
mean| > | > something else? Is it a fair paraphrase of the
step where you| > | > introduce a1, a2, and a3, to say that
they are three functions from| > | > the algebraic integers
to the algebraic integers, a1(m), a2(m), and| > | > a3(m),
having the property that P(m) = (a1*x+uf)(a2*x+uf)(a3*x+uf)
for| > | > every x?| > | | > | They're not necessarily
functions of m.| > | > At some point you need actually to say
what kind of thing a1, a2, and| > a3 are, without leaving any
special wriggle room. If treating them as| > functions of m
is not quite good enough, then you really need to say| > what
they are.| | You didn't connect back with your own statement,
fascinating.| | I'm repeating myself with that statement, but
before you talked of| reparametrization in reply.| | I fear
you are a parrot. You look for key words and then
regurgitate| based on something you think is relevant, but
you need to THINK.I didn't think I would need to put in the
previous quoted text inorder for you to remember what you had
just written:|> Perhaps it would help if you clari?d and/or
con?med a few things.|> When in the paper you write that g1
must have that same factor (of|> f) in general, does in
general mean that you're saying the value|> of g1 for each
allowed value of m is divisible by f? ||It depends on
context. It's like if you have y=2x+1, and consider|various
values. Now I see pressure to write that as f(x)=2x+1, but
I|don't think it's worth the potential confusion.||The
problem is that though you see m as the key variable, it may
in|fact be a *factor* of m that is the true independent
variable, which|I'm sure has confused quite a few people. Now
it's a measure of your|mathematical ability if you know what I
mean there, and let's just say|that I'm not interested in
expanding out in a way that leaves *more*|room for
confusion.So you're telling us that m might not be the true
independentvariable. Do you realize that switching from one
independentvariable to another one is known as
reparametrizing?A reasonable person reading the proof (and
not this discussion)would look at that formula supposedly
de?ing a1, a2, and a3,and since you don't say speci?ally
what they are, they'd have tomake a bit of a guess. That they
are at least functions of m wouldbe quite a reasonable
interpolation. Otherwise, you're leaving themin the dark.This
stuff about how being more speci? is bad because the rest
ofus (being inferior to you in mathematical ability of
course) wouldbe liable to be confused is just an excuse.
You are, because ofyour lack of practice in writing actual
proofs, failing to bespeci?, and as a result getting
confused.I think you're just trying to bluff your way past
the problem. Anybodywho actually had a proof would be able to
answer such simple questionsas these much more
speci?ally.Keith Ramsay
===
> If my explanation isn't
working, ?e, someone will eventually ?d> a better way of
explaining it to you.> That is an irrational statement.
Where do you suppose this someone> is?The someone who can
bring JSH to understanding does not exist.
===
>> Remember
(unlike e.g. the Gaussian integers) the algebraic integers>>
are dense in the complex plane, meaning that any circle, no
matter>> how small, has algebraic integers inside
it.>>Hmmm...interesting assertion, but what about
singularities?I'll bite. What about singularities?We aren't
talking about a function here, just about two sets,
thecomplex numbers and the algebraic integers. There aren't
anysingularities. And if there were, what does that have to
do withwhether there are in?itely many algebraic integers
everywhere youlook? - Randy
===
Sorry about the duplicate
postings; I got caught by this thing whereI get a connection
timed out page while trying to post with Google,but
apparently the posting goes out anyway.[...]| > Remember
(unlike e.g. the Gaussian integers) the algebraic integers| >
are dense in the complex plane, meaning that any circle, no
matter| > how small, has algebraic integers inside it.| |
Hmmm...interesting assertion, but what about singularities?As
usual by a circle I mean one having positive radius, which
doesn'thave singularities.[...]| > No, my main point in
bringing this stuff up was to try to convince| > you that
de?ing a1, a2, and a3 in such a way that their values at| >
one point are related to their values at other points in a
useful way.| | Um, but that is an insinuation that they
aren't already rigorously| de?ed, when in fact they are.I
don't see much sign of your knowing what it means to give a
rigorousBut an equation is only meaningful if one knows
somehow what kind ofthings the terms in it represent. The
paper refers to the value onehas at m=0, so there has to be
some sense in which one can substitutem=0 into one of them.
It's natural then to assume you mean for themto be functions
of m. But then you say that they're not necessarilyfunctions
of m, without saying speci?ally what else they might be.(You
hint that expanding on this point might cause
confusion.)Obviously you're not giving a rigorous de?ition.|
> If my explanation isn't working, ?e, someone will
eventually ?d| > a better way of explaining it to you.| |
That is an irrational statement. Where do you suppose this
someone| is?I see at least a couple of other people who
also understand this pointabout how the values of a1, a2, and
a3 at one point relate to theirvalues at another point, and
have been trying to explain it to you.I'm just assuming that
eventually one of us will come up with a betterway of
explaining it to you.| Possibly your memory tells you they're
in some potential ?ld??? [...]| That is false as the a's are
determined both in form and type.As in, sort of like
functions of m but not necessarily, and youwouldn't want to
confuse poor confused us by going into furtherdetails about
it.| > Second, p-adic numbers ARE exact. They're just a
different system.| > It's easier in a lot of ways to be
rigorous with them than it is| > to be rigorous with the real
numbers or the complex numbers.| | Is that something you have
from memory Keith Ramsay?No, from my own experience.| >
Third, you're surely aware that nobody will be ripping
anybody| > apart or forcing anybody to do anything. People
have tried hard| > to force you down to rigor, for example
by telling you what,| > allegedly, the common factors between
5 and the roots of that cubic| > are. Doesn't phase you a bit!
And you've shown us all how easy it is| > to be impervious to
demands for rigor.| | That's an appeal to the gallery,
which is a logical fallacy.I'm just answering this suggestion
you made that if I had a furtherexplanation in mind using
p-adic numbers, you were going to rip meapart and force me
to be rigorous. I'll be less likely to remind youof how poor
your rigor has been up until now, if you refrain frombold
predictions about what mine is going to be like in the
future.[...]| > So in fact the original assumption that there
was such an f was| > false, Q.E.D.| | Can you switch to
congruences? What about something like | | f(e^{it})^2 -
e^{it} = 0(mod 35)| | in the ring of algebraic
integers?That's an interesting question. The multiples of 35
in the algebraicintegers are also dense, which means that we
have a lot of freedomto adjust the values of f. I would guess
that this makes it possibleto have a function f which is
continuous on the whole complex planeand satis?s f(z)^2 = z
(mod 35) for all algebraic integers z, butI would have to
think some more to prove it.e^{it} is an algebraic integer
only for certain selected t, of course. | > Now, a1 and a2
are not square roots, but they are similar enough| > to a
constant times the square root of m to make the same type of|
> argument apply.| > | > Keith Ramsay| | Why don't you just
give the a's explicitly and prove it then?If you mean the a1,
a2 and a3 as I would de?e them, i.e. asfunctions on the
complex integers with complex integer values,it's just the
time required. Applying the cubic formula, as youseem to be
suggesting I do, is pretty tedious when the coef?ientsare
expressions involving m and another constant or two. We'll
seewhether I have enough spare time this week.If it's a1, a2
and a3 that might have a different independentvariable than
m, it would of course depend upon what that variableis.Keith
Ramsay
===
[...]>>Remember (unlike e.g. the Gaussian integers)
the algebraic integers>>are dense in the complex plane,
meaning that any circle, no matter>>how small, has algebraic
integers inside it.> Hmmm...interesting assertion, but
what about singularities?[...] 2> sqrt(2) > 1 so, 1> sqrt(2)
- 1 >0Also, sqrt(2) - 1 is an algebraic integer because
sqrt(2) and 1 are.In fact, 1/2 > sqrt(2) - 1 >0 .So for any
n>=1: (1/2)^n > (sqrt(2) - 1)^n > 0 .Note that, by repeated
multiplication, (sqrt(2) - 1)^n is also analgebraic
integer.Given any positive real number d, there exists an n
such that: d> (sqrt(2) - 1)^n > 0.Letting M and N vary
through all ordinary integers, we get a latticeor square grid
from the collection of algebraic integers M* (sqrt(2)-1)^n +
N*i*(sqrt(2)-1)^n .The basic , small squares in the grid have
a side of (sqrt(2) - 1)^n < d [ the n depends on d...]So we
can construct grids with d=0.1, 0.01, 0.001, .... d=
10^(-100) ... , and all the numbers in all the gridsare
algebraic integers.David Bernier
===
>> Remember (unlike
e.g. the Gaussian integers) the algebraic integers>> are
dense in the complex plane, meaning that any circle, no
matter>> how small, has algebraic integers inside it.>>
>Hmmm...interesting assertion, but what about singularities?>
> I'll bite. What about singularities?> We aren't talking
about a function here, just about two sets, the> complex
numbers and the algebraic integers. There aren't any>
singularities. And if there were, what does that have to do
with> whether there are in?itely many algebraic integers
everywhere you> look?> - RandyA singularity is a circle of
0 radius, and in the complex plane I canpick an in?ite
number of fractions.Fascinating how far the discussion is
from my paper, but maybe not sofar from the subject line,
eh?James Harris
===
> [...]> |And you've gone off on a
tangent. But as you mention it, I suggest> |you go ahead and
be *careful* instead of possibly hoping that you can> |get
away with casting doubt.> |> |It's not a political discussion
Keith Ramsay. You're not on Fox News.> You're hardly in any
position to complain, either about glibly> expressing doubts
about things you don't understand, or about making> political
statements, or about being careful either!Support what you say
mathematically. It's not a political discussionand you're not
on Fox News!> [...]> | > If this is so, then one of the main
questions is what values this> | > common factor has when m
is not 0. I don't see a problem with saying> | > it's f when
m=0, but I also don't see why it shouldn't be a varying> | >
factor of f as m changes.> | > | Yeah, and I gave an example
yesterday which should tell you why.> | > | P(x) = 2x^2 + 4x
+ 2> | > | (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2> | > | a_1 a_2
= 2x^2,> Is this last equation another assumption?
Otherwise, what's to rule> out a_1=1, a_2 = x^2+2x-1, for
instance?The last equation rules it out, which is why I gave
it.> | and I note that when x=0, at *least* one of the a's
must equal 0, and> | further note that the constant term of
the factor a_1 + 1, which I'll> | call g_1 is 1 at that
point. Further the factor a_2 + 2 has a factor> | that is 2
at that point, and I'll call it g_2.> |> | Now then Keith
Ramsay, do you question what g_2 will do when x does> | not
equal 0?> | > | So why do you do it above?> Continually
expressing shock that I don't agree with you, when you>
haven't indicated a principle to justify your claim, is just
a bunch> of posturing. What's the general principle?The
general principle is that factors of polynomials can be
written asr+c, where r varies or is 0, while c is
constant.That's it. Like I've said before that simple thing
is the linchpin ofmy position.To refute my mathematical
argument, you have to refute it.However, posters keep making
long-winded posts that manage to avoidit.> | Ultimately, you
may try and claim that irreducibility over Q is what's> |
pertinent, if so, explain in detail.> The two cases aren't
parallel. You really should get out of the> habit of thinking
that analogy and cajolery add up to a logical> argument.
Giving a bunch of examples which work the way you want>
another example to work is not a proof.I've given the proof.
> If I were to give an explanation for why the examples that
work,> work, I'd give it in terms of the following
principle:> (*) If a divides bc, then there exist m and n
such that a=mn,> m divides b and n divides c.> You give
examples where the product of two polynomials is divisible>
by an integer. Given that (*) is true in these cases, we can
then> try to ?ure out what m and n would work by considering
the constant> terms and so on as you do. All of your simpli?d
examples are ones> where (*) works.>> But (*) is not generally
true for functions having algebraic integer>
values.Irrelevant.> [...]> | > | >Or do you mean> | > | >
something else? Is it a fair paraphrase of the step where
you> | > | > introduce a1, a2, and a3, to say that they are
three functions from> | > | > the algebraic integers to the
algebraic integers, a1(m), a2(m), and> | > | > a3(m), having
the property that P(m) = (a1*x+uf)(a2*x+uf)(a3*x+uf) for> | >
| > every x?> | > | > | > | They're not necessarily functions
of m.> | > | > At some point you need actually to say what
kind of thing a1, a2, and> | > a3 are, without leaving any
special wriggle room. If treating them as> | > functions of m
is not quite good enough, then you really need to say> | >
what they are.> | > | You didn't connect back with your own
statement, fascinating.> | > | I'm repeating myself with that
statement, but before you talked of> | reparametrization in
reply.> | > | I fear you are a parrot. You look for key words
and then regurgitate> | based on something you think is
relevant, but you need to THINK.> I didn't think I would
need to put in the previous quoted text in> order for you to
remember what you had just written:It's not necessary that
they be functions of m. The question ofwhether or not the a's
for the speci? example I gave are functionsof m, keeps coming
up, but it's not really relevant.In general, there is no
necessity that given P(x), a factor of P(x)can be written as
a function of x plus a constant.> #> Perhaps it would help if
you clari?d and/or con?med a few things.> #> When in the
paper you write that g1 must have that same factor (of> #>
f) in general, does in general mean that you're saying
the value> #> of g1 for each allowed value of m is divisible
by f? > #> #It depends on context. It's like if you have
y=2x+1, and consider> #various values. Now I see pressure to
write that as f(x)=2x+1, but I> #don't think it's worth the
potential confusion.> #> #The problem is that though you see
m as the key variable, it may in> #fact be a *factor* of m
that is the true independent variable, which> #I'm sure has
confused quite a few people. Now it's a measure of your>
#mathematical ability if you know what I mean there, and
let's just say> #that I'm not interested in expanding out in
a way that leaves *more*> #room for confusion.> So you're
telling us that m might not be the true independent>
variable. Do you realize that switching from one independent>
variable to another one is known as reparametrizing?That's
not relevant.> A reasonable reader, who had read the paper
but not this discussion,> would naturally tend to suspect
that a1, a2 and a3, which are> described as having values at
given values of m, would tend to> suspect that they are
simply functions of m. If this is not> necessarily so, it's
because you've left out any explanation for> what else you
intend for them to be.Questions about whether or not they're
functions of m are notrelevant.> This stuff about how being
more speci? is bad because the rest of> us (being inferior
to you in mathematical ability of course) would> be liable to
be confused is just an excuse. You are, because of> your
lack of practice in writing actual proofs, failing to be>
speci?, and as a result getting confused. I think you're
just trying> to bluff your way past the fact that you don't
know how to de?e> them in such a way as to get the argument
in the paper to work.The paper has a lemma. Given that lemma
its conclusion follows easilyenough.If you have a problem
with the paper, you need to show itmathematically, instead of
talking on all these extraneous subjects.For instance, the
question of whether or not the a's are functions ofm, is
irrelevant. The lemma is in the paper for a reason. In
dodgingthe lemma and in continually bringing up irrelevant
issues, it seemsto me that you are deliberately being
intellectually dishonest, with aperverse need to try and
in? others to believe falsehoods.James Harris
===
[snip]>
The paper has a lemma. Given that lemma its conclusion follows
easily> enough.The lemma is wrong.> If you have a problem with
the paper, you need to show it> mathematically, instead of
talking on all these extraneous subjects.The errors have been
shown mathematically already. You have chosen to ignore the
truth.> For instance, the question of whether or not the a's
are functions of> m, is irrelevant. The lemma is in the paper
for a reason. In dodging> the lemma and in continually
bringing up irrelevant issues, it seems> to me that you are
deliberately being intellectually dishonest, with a> perverse
need to try and in? others to believe falsehoods.The
arguments you present in your paper have been conclusively
refuted. It is you who are beingintellectually dishonest,
with a perverse need to try and in? others to believe
falsehoods.You have consistently revealed yourself to be a
defender of lies and an enemy of truth. When confronted
withspeci? errors you ignore the facts and continue your
delirium.James Harris, you should be ashamed of your behavior
-- especially since you persist in document it in
publicrecords.--A man with integrity will identify,
acknowledge and correct his errors. A man without integrity
will ignore,deny or defend them.--Democracy: The triumph of
popularity over principle.--http://www.crbond.com
===
> Sorry
about the duplicate postings; I got caught by this thing
where> I get a connection timed out page while trying to
post with Google,> but apparently the posting goes out
anyway.> [...]> | > Remember (unlike e.g. the Gaussian
integers) the algebraic integers> | > are dense in the
complex plane, meaning that any circle, no matter> | > how
small, has algebraic integers inside it.> | > |
Hmmm...interesting assertion, but what about singularities?>
> As usual by a circle I mean one having positive radius,
which doesn't> have singularities.0 is a positive number. >
[...]> | > No, my main point in bringing this stuff up was to
try to convince> | > you that de?ing a1, a2, and a3 in such a
way that their values at> | > one point are related to their
values at other points in a useful way.> | > | Um, but that
is an insinuation that they aren't already rigorously> |
de?ed, when in fact they are.> I don't see much sign of
your knowing what it means to give a rigorous> But an
equation is only meaningful if one knows somehow what kind
of> things the terms in it represent. The paper refers to the
value one> has at m=0, so there has to be some sense in which
one can substitute> m=0 into one of them. It's natural then
to assume you mean for them> to be functions of m. But then
you say that they're not necessarily> functions of m, without
saying speci?ally what else they might be.> (You hint that
expanding on this point might cause confusion.)> Obviously
you're not giving a rigorous de?ition.Obviously?The a's are
de?ed by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x +
y)(a_3 x + y),where v=-1+mf^2.claims with mathematics. > | >
If my explanation isn't working, ?e, someone will eventually
?d> | > a better way of explaining it to you.> | > | That is
an irrational statement. Where do you suppose this someone>
| is?> I see at least a couple of other people who also
understand this point> about how the values of a1, a2, and a3
at one point relate to their> values at another point, and
have been trying to explain it to you.> I'm just assuming
that eventually one of us will come up with a better> way of
explaining it to you.Well I should have helped you and these
hypothetical others out withthe information above. Solve that
cubic and see what happens.> | Possibly your memory tells you
they're in some potential ?ld?> ??Don't worry about it.>
[...]> | That is false as the a's are determined both in form
and type.> As in, sort of like functions of m but not
necessarily, and you> wouldn't want to confuse poor confused
us by going into further> details about it.The a's are de?ed
by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x +
y),where v=-1+mf^2.And how do you know? Wouldn't it be from
your *memory* Keith Ramsay?That's why it's called experience,
as in it happened in the past, soyou have to use memory unless
possibly you're relying on writings? > | > Third, you're
surely aware that nobody will be ripping anybody> | > apart
or forcing anybody to do anything. People have tried hard> | >
to force you down to rigor, for example by telling you
what,> | > allegedly, the common factors between 5 and the
roots of that cubic> | > are. Doesn't phase you a bit! And
you've shown us all how easy it is> | > to be impervious to
demands for rigor.> | > | That's an appeal to the gallery,
which is a logical fallacy.> I'm just answering this
suggestion you made that if I had a further> explanation in
mind using p-adic numbers, you were going to rip me> apart
and force me to be rigorous. I'll be less likely to remind
you> of how poor your rigor has been up until now, if you
refrain from> bold predictions about what mine is going to be
like in the future.Your mathematical arguments should stand on
their own Keith Ramsay. And there's no need to shrink back if
you're correct. It's about themath, not your fears. > [...]>
| > So in fact the original assumption that there was such an
f was> | > false, Q.E.D.> | > | Can you switch to congruences?
What about something like > | > | f(e^{it})^2 - e^{it} = 0(mod
35)> | > | in the ring of algebraic integers?> That's an
interesting question. The multiples of 35 in the algebraic>
integers are also dense, which means that we have a lot of
freedom> to adjust the values of f. I would guess that this
makes it possible> to have a function f which is continuous
on the whole complex plane> and satis?s f(z)^2 = z (mod 35)
for all algebraic integers z, but> I would have to think some
more to prove it.> e^{it} is an algebraic integer only for
certain selected t, of course. Of course. Feel free to think
further on it.> | > Now, a1 and a2 are not square roots, but
they are similar enough> | > to a constant times the square
root of m to make the same type of> | > argument apply.> | >
> | > Keith Ramsay> | > | Why don't you just give the a's
explicitly and prove it then?> If you mean the a1, a2 and
a3 as I would de?e them, i.e. as> functions on the complex
integers with complex integer values,> it's just the time
required. Applying the cubic formula, as you> seem to be
suggesting I do, is pretty tedious when the coef?ients> are
expressions involving m and another constant or two. We'll
see> whether I have enough spare time this week.> If it's
a1, a2 and a3 that might have a different independent>
variable than m, it would of course depend upon what that
variable> is.> Keith RamsaySo you're admitting to bringing
up what I've called irrelevant issueswithout having actually
checked them, while refusing to fullyacknowledge the
mathematical argument in the paper including a ratherbasic
lemma that you simply don't mention.James Harris
===
...
stuff deleted ...> The paper has a lemma. Given that lemma
its conclusion follows easily> enough.> If you have a
problem with the paper, you need to show it> mathematically,
instead of talking on all these extraneous subjects.> I have
shown that your Primary Argument reaches a false
conclusion,by showing (what I'm sure you must have seen by
now, but are choosingto ignore) explicit formulas for factors
that each of your a's havein common with 5.Surely that cannot
be considered extraneous, as it has to do withwhat one must
assume is the primary result in the paper. Thatprimary
result is false, thus your argument must be incorrect.in
threads that you have started, and keep participating in.
Thearithmetic is simple, if tedious, and the argument, unlike
yours,can be veri?d, line by line.Yet, you persist in this
denial.the argument. Just say the word.> For instance, the
question of whether or not the a's are functions of> m, is
irrelevant. The lemma is in the paper for a reason. In
dodging> the lemma and in continually bringing up irrelevant
issues, it seems> to me that you are deliberately being
intellectually dishonest, with a> perverse need to try and
in? others to believe falsehoods.> Is it irrelevant to
point out that your conclusion (regarding the a'sbeing
coprime to 5) is false? Do you deny that I have done this?Is
it perverse to have demonstrated the falsity of your
conclusion?Is it perverse to mention that fact?Is it a
falsehood that I am perpetrating?If so, then why?If not, then
why do you insist on asserting that othersare lying, when you
knowingly insist that an incorrectresult is true?> James
HarrisDale.
===
>>[...]>>| > If this is so, then one of the
main questions is what values this>>| > common factor has
when m is not 0. I don't see a problem with saying>>| > it's
f when m=0, but I also don't see why it shouldn't be a
varying>>| > factor of f as m changes.>>| >>| Yeah, and I
gave an example yesterday which should tell you why.>>| >>|
P(x) = 2x^2 + 4x + 2>>| >>| (a_1 + 1)(a_2 + 2) = 2x^2 + 4x +
2>>| >>| a_1 a_2 = 2x^2,>>Is this last equation another
assumption? Otherwise, what's to rule>>out a_1=1, a_2 =
x^2+2x-1, for instance?> The last equation rules it out,
which is why I gave it.The last equation is not necessarily
true. What must be true is that a_1(x) a_2(x) + a_2(x) + 2
a_1(x) = 2x^2 + 4x. Because we do not know the a_i(x) are
polynomials, we cannot assume that they will behave exactly
the way terms of a polynomial would.>>| and I note that when
x=0, at *least* one of the a's must equal 0, and>>| further
note that the constant term of the factor a_1 + 1, which
I'll>>| call g_1 is 1 at that point. Further the factor a_2 +
2 has a factor>>| that is 2 at that point, and I'll call it
g_2.>>|>>| Now then Keith Ramsay, do you question what g_2
will do when x does>>| not equal 0?>>| >>| So why do you do
it above?>>Continually expressing shock that I don't agree
with you, when you>>haven't indicated a principle to justify
your claim, is just a bunch>>of posturing. What's the general
principle?> The general principle is that factors of
polynomials can be written as> r+c, where r varies or is 0,
while c is constant.This is trivially true, but because you
have failed to say anything further about the nature of r, it
also communicates virtually nothing.> That's it. Like I've
said before that simple thing is the linchpin of> my
position.It's trivially true, but you've done nothing
rigorous with it.> To refute my mathematical argument, you
have to refute it.No one is trying to refute that fact. It's
the conclusions you leap to from it that are the problem. If
you say something a little more precise about r, then you may
be able to ?l in the holes in your lemma.> However, posters
keep making long-winded posts that manage to avoid> it.No,
it's the only part of the paper that is true. It's the *rest*
of your paper that has problems.>>| Ultimately, you may try
and claim that irreducibility over Q is what's>>| pertinent,
if so, explain in detail.>>The two cases aren't parallel.
You really should get out of the>>habit of thinking that
analogy and cajolery add up to a logical>>argument. Giving a
bunch of examples which work the way you want>>another
example to work is not a proof.> I've given the proof.No,
you've given a claim that you have failed to defend. You also
tend to change the subject or start new threads rather than
deal with the problems that have been shown.>>If I were to
give an explanation for why the examples that work,>>work,
I'd give it in terms of the following principle:>> (*) If a
divides bc, then there exist m and n such that a=mn,>> m
divides b and n divides c.>>You give examples where the
product of two polynomials is divisible>>by an integer. Given
that (*) is true in these cases, we can then>>try to ?ure out
what m and n would work by considering the constant>>terms and
so on as you do. All of your simpli?d examples are
ones>>where (*) works.>>But (*) is not generally true for
functions having algebraic integer>>values.> Irrelevant.>
What you are trying to prove, and the reason your proof
doesn't work is irrelevant? How does that work?>>[...]>>| >
| >Or do you mean>>| > | > something else? Is it a fair
paraphrase of the step where you>>| > | > introduce a1, a2,
and a3, to say that they are three functions from>>| > | >
the algebraic integers to the algebraic integers, a1(m),
a2(m), and>>| > | > a3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) for>>| > | > every x?>>| > | >>|
> | They're not necessarily functions of m.>>| >>| > At some
point you need actually to say what kind of thing a1, a2,
and>>| > a3 are, without leaving any special wriggle room. If
treating them as>>| > functions of m is not quite good enough,
then you really need to say>>| > what they are.>>| >>| You
didn't connect back with your own statement, fascinating.>>|
>>| I'm repeating myself with that statement, but before you
talked of>>| reparametrization in reply.>>| >>| I fear you
are a parrot. You look for key words and then regurgitate>>|
based on something you think is relevant, but you need to
THINK.>>I didn't think I would need to put in the previous
quoted text in>>order for you to remember what you had just
written:> It's not necessary that they be functions of m.
The question of> whether or not the a's for the speci?
example I gave are functions> of m, keeps coming up, but it's
not really relevant.> In general, there is no necessity that
given P(x), a factor of P(x)> can be written as a function of
x plus a constant.They can be constant functions of x. Then
you have a constant (written as a function) plus a constant.
Also note that you've just said it's ok to do what you
objected to at the beginning of this post. Keith made a_1(x)
= 1, a_2(x)=x^2+2x-1. Here, a_1(x)+1 is in the form constant
+ constant. Is this ok or not?>>#> Perhaps it would help if
you clari?d and/or con?med a few things.>>#> When in the
paper you write that g1 must have that same factor (of>>#>
f) in general, does in general mean that you're saying
the value>>#> of g1 for each allowed value of m is divisible
by f? >>#>>#It depends on context. It's like if you have
y=2x+1, and consider>>#various values. Now I see pressure to
write that as f(x)=2x+1, but I>>#don't think it's worth the
potential confusion.>>#>>#The problem is that though you see
m as the key variable, it may in>>#fact be a *factor* of m
that is the true independent variable, which>>#I'm sure has
confused quite a few people. Now it's a measure of
your>>#mathematical ability if you know what I mean there,
and let's just say>>#that I'm not interested in expanding out
in a way that leaves *more*>>#room for confusion.>>So you're
telling us that m might not be the true
independent>>variable. Do you realize that switching from one
independent>>variable to another one is known as
reparametrizing?> That's not relevant.Are you saying it
doesn't matter what your independent variables are? If so,
then you aren't saying anything and can draw no conclusions.
It may be inconvenient, but is highly relevant.>>A reasonable
reader, who had read the paper but not this discussion,>>would
naturally tend to suspect that a1, a2 and a3, which
are>>described as having values at given values of m, would
tend to>>suspect that they are simply functions of m. If this
is not>>necessarily so, it's because you've left out any
explanation for>>what else you intend for them to be.>
Questions about whether or not they're functions of m are
not> relevant.If they cannot vary with m, they are constants.
If they are constants, they can be computed and we can easily
verify whether your paper is true. If they do vary with m,
they are functions of m. In that case, different types of
analysis can be done on them. If we don't know what is being
talked about, then there is no way to conclude anything. You
are refusing to address the key issue of de?ing what you are
talking about. When you do that, you are not saying anything
meaningful. Instead, you are babbling.>>This stuff about how
being more speci? is bad because the rest of>>us (being
inferior to you in mathematical ability of course) would>>be
liable to be confused is just an excuse. You are, because
of>>your lack of practice in writing actual proofs, failing
to be>>speci?, and as a result getting confused. I think
you're just trying>>to bluff your way past the fact that you
don't know how to de?e>>them in such a way as to get the
argument in the paper to work.> The paper has a lemma.
Given that lemma its conclusion follows easily> enough.Then
start by defending your lemma. There have been several
objections (and disproofs) that you have completely failed to
address.> If you have a problem with the paper, you need to
show it> mathematically, instead of talking on all these
extraneous subjects.> What do you think everyone's been
doing? You are the one who brings up Fox News. You are the
one who brings up politics. You are the one who brings up
moral character and the supposed ?n our characters.
Please deal with the math as we have been.> For instance, the
question of whether or not the a's are functions of> m, is
irrelevant. The lemma is in the paper for a reason. In
dodging> the lemma and in continually bringing up irrelevant
issues, it seems> to me that you are deliberately being
intellectually dishonest, with a> perverse need to try and
in? others to believe falsehoods.> The lemma has been
shown to be highly questionable at best. You have failed to
defend it. Who is dodging issues about the lemma?> James
Harris-- Will Twentyman
===
> 0 is a positive number.What is
your de?ition of a positive number?Can a number be both
positive and negative?If you have a non-standard notion of
positive,then the following argument may prove zero alsois
negative.Suppose +x = -x.Then 2x = 0and x = 0.So, -0 = 0 =
+0.
===
>> Sorry about the duplicate postings; I got caught by
this thing where>> I get a connection timed out page while
trying to post with Google,>> but apparently the posting goes
out anyway.> [...]>> | > Remember (unlike e.g. the
Gaussian integers) the algebraic integers>> | > are dense in
the complex plane, meaning that any circle, no matter>> | >
how small, has algebraic integers inside it.>> | >> |
Hmmm...interesting assertion, but what about singularities?>>
>> As usual by a circle I mean one having positive radius,
which doesn't>> have singularities.>>0 is a positive
number.Good one; about equivalent to your infamous integers
areirrationals.No, 0 is NOT a positive number. Positive
numbers are strictly largerthan 0. 0 is a non-negative
number, and a non-positive number; it isnot a positive
number.
===
======================================

===
=========Why do you take so much trouble to expose
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A man's capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
?ures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde?ite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
Augustus de
Morgan
===
=======================================

===
=========Arturo
Magidinmagidin@math.berkeley.edu
===
[snip]Give it up, James
Harris. Your so-called proof of FLT has gone down in
? Your pitiful attempts torescue it with diversions and
side shows is laughable. You have covered the territory from
gad?troll toloon to failure to fool to clown. Take down
your idiotic and failed attempt at a proof. It's polluting
theinternet.Recently you seem focussed on the notion that
mathematicians are liars. (See topic of this thread.) Why
notpost a list of those lies so others can observe them?--or
are *you* the liar?--There are two things you must never
attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
> 0 is a positive
number.> What is your de?ition of a positive number?>
Can a number be both positive and negative?> If you have a
non-standard notion of positive,> then the following argument
may prove zero also> is negative.> Suppose +x = -x.> Then 2x
= 0> and x = 0.> So, -0 = 0 = +0.Oh, ok, if you all agree that
0 isn't a positive number that's ?e with me.James
Harris
===
[snip....>> As usual by a circle I mean one having
positive radius, which doesn't>> have singularities.>>0 is a
positive number.> I think I have boiled all of JHs problems
down.....[snip...
===
> 0 is a positive number. > What
is your de?ition of a positive number?James must be a
follower of Bourbaki.-- dik t. winter, cwi, kruislaan 413,
1098 sj amsterdam, nederland, +31205924131home: bovenover
215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
> 0 is a positive number.>
> What is your de?ition of a positive number?> Can a
number be both positive and negative?> If you have a
non-standard notion of positive,> then the following
argument may prove zero also> is negative.> Suppose
+x = -x.> Then 2x = 0> and x = 0.> So, -0 = 0 = +0.>
Oh, ok, if you all agree that 0 isn't a positive number
that's ?e with me.> James HarrisJust to keep JSH
guessing, to the Bourbaki set, 0 is both positive and
negative, but not strictly either.
===
[...]| > Obviously
you're not giving a rigorous de?ition.| | Obviously?Yes.|
The a's are de?ed by| | (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x +
y)(a_2 x + y)(a_3 x + y),| | where v=-1+mf^2.You seem often
to have the attitude that equations speak forthemselves.
But if you had just left this equation by itself,it could
have been referring to quaternion-valued functionsa_1(m),
a_2(m) and a_3(m). Just writing an equation doesn't tellus
what the equation means.But then you denied that guess. m is
not necessarily the independentvariable. So evidently not
only does the de?ition not quite saywhat type of functions
or values the a_i are, it misleads the readerinto thinking
that they're something more obvious than they are.Don't try
to just skim past this issue of what the independentvariable
is in these functions. | | a^3 + 3va^2 -(v^3+1) = 0, where
again v=-1+mf^2, and the a's are its| roots.| | So you can
solve for the a's and get some result, instead of endlessly|
talking.The usual solution of the cubic starts by changing
the variable toget rid of the square term. So let b = a+v,
and we getb^3 - 3v^2b + (v^3-1) = 0. The discriminant is D =
(v^3-1)^2/4 - v^6 = -(3/4)v^6 - (1/2)v^3 + (1/4) = -(3v^3 -
1)(v^3 + 1)/4.(although in number theory we use a different
constant factor).This shows there are six values of v for
which two of the rootscoincide.By Cardan's formula: b =
((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3},or
a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3}
- v,| It might be especially interesting to see the result and
see what| happens at v=-1.Because D =
(v+1)(1-3v^3)(1+v+v^2)/4, near v=-1, D is approximatelyv+1.
The two values of sqrt(D) trade places with each other as
wesqrt(D) with -sqrt(D) wouldn't change the value of a. But
the twocube roots are chosen so that at v=-1, for example,
they are eitherboth 1, or one is (-1+sqrt(3)*i)/2 and the
other is (-1-sqrt(3)*i)/2,or the ?st is (-1-sqrt(3)*i)/2 and
the other is (-1+sqrt(3)*i)/2.So changing the sign on the
square root of D is equivalent to leavingthe ?st root (where
both cube roots are taken to be 1) alone, whileexchanging the
other two roots. As I said before, if you take a looparound
v=-1, the value of one of the roots comes back to the value
ofthe other one. | Reminder to readers, Keith Ramsay has
brought up a continuity| question, which actually is
irrelevant, but what the hell?I was trying to convince you
that it's unsafe to assume that thevalues of a_1, a_2, and
a_3 extrapolate in a natural way from onevalue of m to
another. This would be needed, for example, for a_2to be
always divisible by f or anything like that. Otherwise,
howcould it possibly be true that it's one particular one of
the threethat has a certain divisibility property, if the
manner in whichwe've decided which of the three roots to
assign to each of thethree variables in an arbitrary way
depending on the value of m?I'm sorry if the relevance
doesn't seem clear to you.The one obvious way to attempt to
extrapolate values of a_1, a_2and a_3 from one value of m to
another is by continuity. Given apath in the complex plane
which avoids all the points where twoof the roots coincide,
we can see what happens to the a_i if wecontinuously change m
along that path, while letting the a_i alsochange continuously
as we go. Unfortunately (or fortunately,actually) which a_i is
which when we reach the end of the pathwill depend upon which
path we took from one endpoint to the other.So if I have a
given value of a at m=0 (say a=0), the question ofwhich of
the three values at m=1 corresponds to it is
meaninglessunless I make an arbitrary convention.That's
essentially what prevents them from being
continuouseverywhere.I don't have the time today, but what
one can do is to show thatfor each epsilon>0, there exists a
delta>0 having the property thatif |v+1| < delta, then one of
the roots is within epsilon of thenonzero root at v=-1 (what
was it, a=1?), while the other two satisfy|a^2/(v+1) - C| <
epsilon where C is some constant (probably just 1).Then we
can consider the values v = -1+e^{it}. For small epsilon,the
inequality forces a/e^{it/2} to be close to one of the
squareroots of C. Then we can carry out the same argument as
last timebut with the two sets being the t for which
|a/e^{it/2} - sqrt(C)|is smaller and the t for which
|a/e^{it/2} + sqrt(C)| is smaller.[...]| So you're admitting
to bringing up what I've called irrelevant issues| without
having actually checked them, while refusing to fully|
acknowledge the mathematical argument in the paper including
a rather| basic lemma that you simply don't mention.I ?d the
lemma very uninteresting. I don't understand how youmanage to
think of it as important.Similar to a lot of arguments
written by undergraduate math majorsearly in their studies,
this argument plods its way slowly throughthe routine parts
of the argument, but condenses into a rapid sprintthe part
which is most in need of explanation, and also hardest
tobelieve.Keith Ramsay
===
> [...]> | > Obviously you're not
giving a rigorous de?ition.> | > | Obviously?> Yes.Let's
see if you support your accusation.> | The a's are de?ed by>
| > | (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x
+ y),> | > | where v=-1+mf^2.> You seem often to have the
attitude that equations speak for> themselves. But if you
had just left this equation by itself,> it could have been
referring to quaternion-valued functions> a_1(m), a_2(m) and
a_3(m). Just writing an equation doesn't tell> us what the
equation means.On the contrary, the equation clearly shows
that there is afactorization.> a_1, a_2, and a_3 to be
functions de?ed for algebraic integers m,> with algebraic
integer values, having the property that for each m> they
make the equation above hold for all x and y.Of course the
ring is algebraic integers as that's been the ringthroughout
the thread and is the ring in the paper Advanced
PolynomialFactorization.The factorization should hold for all
f, m, x and y, in the ring.There is no need to mention
functions. > If that guess had been correct, then I would
have said that the> de?ition was at least close to rigorous,
just leaving a couple of> things implicit that should be
explicit. All that it would take to> make it rigorous would
be a few clarifying words. (As I tried to> explain to you
before, you also couldn't be taking x to be a constant,> for
example.)That's false. There is no requirement on the
factorization that x notbe constant.Factorizations don't work
that way.> But then you denied that guess. m is not
necessarily the independent> variable. So evidently not only
does the de?ition not quite say> what type of functions or
values the a_i are, it misleads the reader> into thinking
that they're something more obvious than they
are.Factorizations are not dependent on the dependency of the
variables.> Don't try to just skim past this issue of what the
independent> variable is in these functions.Factorizations are
not dependent on functions.> | > | a^3 + 3va^2 -(v^3+1) = 0,
where again v=-1+mf^2, and the a's are its> | roots.> | > |
So you can solve for the a's and get some result, instead of
endlessly> | talking.> The usual solution of the cubic
starts by changing the variable to> get rid of the square
term. So let b = a+v, and we get> b^3 - 3v^2b + (v^3-1) = 0.
The discriminant is> D = (v^3-1)^2/4 - v^6> = -(3/4)v^6 -
(1/2)v^3 + (1/4)> = -(3v^3 - 1)(v^3 + 1)/4.> (although in
number theory we use a different constant factor).> This
shows there are six values of v for which two of the roots>
coincide.> By Cardan's formula:> b = ((1-v^3)/2 +
sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3},> or> a =
((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} -
v,> | It might be especially interesting to see the result
and see what> | happens at v=-1.Readers should note that
assuming the above is correct at v=-1, D=0,so you get a =
1^{1/3} + 1^{1/3} + 1which gives you the three a's, a_1=0,
a_2=0, a_3=3.Now consider Keith Ramsay's reply. > Because D =
(v+1)(1-3v^3)(1+v+v^2)/4, near v=-1, D is approximately> v+1.
The two values of sqrt(D) trade places with each other as we>
sqrt(D) with -sqrt(D) wouldn't change the value of a. But the
two> cube roots are chosen so that at v=-1, for example, they
are either> both 1, or one is (-1+sqrt(3)*i)/2 and the other
is (-1-sqrt(3)*i)/2,> or the ?st is (-1-sqrt(3)*i)/2 and the
other is (-1+sqrt(3)*i)/2.> So changing the sign on the
square root of D is equivalent to leaving> the ?st root
(where both cube roots are taken to be 1) alone, while>
exchanging the other two roots. As I said before, if you take
a loop> around v=-1, the value of one of the roots comes back
to the value of> the other one.Um, did *anyone* see an answer
in there?> | Reminder to readers, Keith Ramsay has brought up
a continuity> | question, which actually is irrelevant, but
what the hell?> I was trying to convince you that it's
unsafe to assume that the> values of a_1, a_2, and a_3
extrapolate in a natural way from one> value of m to another.
This would be needed, for example, for a_2> to be always
divisible by f or anything like that. Otherwise, how> could
it possibly be true that it's one particular one of the
three> that has a certain divisibility property, if the
manner in which> we've decided which of the three roots to
assign to each of the> three variables in an arbitrary way
depending on the value of m?> I'm sorry if the relevance
doesn't seem clear to you.There is no relevance. In
polynomial equations you *can* have itwhere certain roots
have a factor while others do not.> The one obvious way to
attempt to extrapolate values of a_1, a_2> and a_3 from one
value of m to another is by continuity. Given a> path in the
complex plane which avoids all the points where two> of the
roots coincide, we can see what happens to the a_i if we>
continuously change m along that path, while letting the a_i
also> change continuously as we go. Unfortunately (or
fortunately,> actually) which a_i is which when we reach the
end of the path> will depend upon which path we took from one
endpoint to the other.> So if I have a given value of a at m=0
(say a=0), the question of> which of the three values at m=1
corresponds to it is meaningless> unless I make an arbitrary
convention.Actually you simply take the three values of the
cuberoot operator,just like with square roots you take the
two values of the square rootoperator.For those who are
confused simply remember that x^3 - 1 = 0, gives youthree
values for x.> That's essentially what prevents them from
being continuous> everywhere.Looks to me like you put down a
lot of words Keith Ramsay, yet somehowmanaged to not even
refer back to a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 -
sqrt(D))^{1/3} - vfrom your work before.> I don't have the
time today, but what one can do is to show that> for each
epsilon>0, there exists a delta>0 having the property that>
if |v+1| < delta, then one of the roots is within epsilon of
the> nonzero root at v=-1 (what was it, a=1?), while the
other two satisfy> |a^2/(v+1) - C| < epsilon where C is some
constant (probably just 1).> Then we can consider the values
v = -1+e^{it}. For small epsilon,> the inequality forces
a/e^{it/2} to be close to one of the square> roots of C. Then
we can carry out the same argument as last time> but with the
two sets being the t for which |a/e^{it/2} - sqrt(C)|> is
smaller and the t for which |a/e^{it/2} + sqrt(C)| is
smaller.Well, if you don't have the time it's not really
worth mentioning, nowis it?After all, the discussion isn't a
political one, and you're NOT on FoxNews!!!> [...]> | So
you're admitting to bringing up what I've called irrelevant
issues> | without having actually checked them, while
refusing to fully> | acknowledge the mathematical argument in
the paper including a rather> | basic lemma that you simply
don't mention.> I ?d the lemma very uninteresting. I don't
understand how you> manage to think of it as important.That
lemma is the linchpin of the paper!!!> Similar to a lot of
arguments written by undergraduate math majors> early in
their studies, this argument plods its way slowly through>
the routine parts of the argument, but condenses into a rapid
sprint> the part which is most in need of explanation, and
also hardest to> believe.Your belief here is irrelevant as a
proof begins with a truth thenproceeds by logical steps to a
conclusion which then must be true. Challenging a paper
requires that you ?d that it did not begin witha truth, or
that you ?d a break in the logical chain.James Harris
===
>>
> Remember (unlike e.g. the Gaussian integers) the
algebraic integers> are dense in the complex plane,
meaning that any circle, no matter> how small, has
algebraic integers inside it.>>Hmmm...interesting
assertion, but what about singularities?> I'll bite. What
about singularities?> We aren't talking about a function
here, just about two sets, the>> complex numbers and the
algebraic integers. There aren't any>> singularities. And
if there were, what does that have to do with>> whether there
are in?itely many algebraic integers everywhere you>> look?>>
>>A singularity is a circle of 0 radius,No, it's not. A
singularity is a property of a function. And intopological
statements like a circle, no matter how small, it
isunderstood that the term circle refers to a ?ite
radius.> and in the complex plane I can>pick an in?ite
number of fractions.What does that have to do with the
statement? The statement is this:Pick a complex value z. Pick
a radius epsilon >0. The neighborhood {win C: |z - w| <
epsilon} has in?itely many algebraic integers in it.What the
heck does I can pick an in?ite number of fractions haveto
do with the density of algebraic integers?Within a radius of
0.01 of 1 + i there are an in?ite number ofalgebraic
integers.Within a radius of 0.00002 of -sqrt(3) + i*pi there
are an in?itenumber of algebraic integers.What does I can
pick an in?ite number of fractions have to do
withstatements like those? - Randy
===
yeah, but a circle of
no radius would have itselfas a singularity. did we uncover a
new property/axiom/pedagogical twist?otherwise, Buddy,a ring
is what I say it is, now; capiche? > No, it's not. A
singularity is a property of a function. And in> topological
statements like a circle, no matter how small, it is>
understood that the term circle refers to a ?ite
radius.--A church-school McCrusade (Blair's
ideals?):Harry-the-Mad-Potter want's US to kill Iraqis?...
http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
http://www.rwgrayprojects.com/synergetics/plates/plates.html
http://quincy4board.homestead.com/?es/curriculum/Cosmo.PCX==
==I'm trying to ?ure out how to correctly form the logical
negation ofthis sentence:For all real numbers satisfying
a<b, there exists a natural number n,such that a + 1/n <
b.Converting this sentence into symbolic form gives
something like:(forall a,b in R ):( a < b),(exists n
in N):( a+1/n <b )and then forming the negation, I
have(exists a,b in R):(a>= b),(forall n in N):(a+1/n
>= b)i.e. There exist real numbers satisfying a>=b, such
that for allnatural numbers n, a+1/n >= b.However, I am
obviously doing something wrong in the negation, sinceBOTH of
the above sentences appear to be true, and by the law
ofexcluded middle, P and not P cannot BOTH be true.So, what
am I doing wrong?thanks.
===
> I'm trying to ?ure out how to
correctly form the logical negation of> this sentence:> For
all real numbers satisfying a<b, there exists a natural
number n,> such that a + 1/n < b.> Converting this
sentence into symbolic form gives something like:> (forall
a,b in R ):( a < b),(exists n in N):( a+1/n <b )> and
then forming the negation, I have> (exists a,b in
R):(a>= b),(forall n in N):(a+1/n >= b)> i.e. There
exist real numbers satisfying a>=b, such that for all>
natural numbers n, a+1/n >= b.> However, I am obviously
doing something wrong in the negation, since> BOTH of the
above sentences appear to be true, and by the law of>
excluded middle, P and not P cannot BOTH be true.> So, what
am I doing wrong?Your original statement says that if there
are two reals, a and b with a < b, there exists n such that a
+ 1/n < b.The negation of that would be that there exist two
reals, a and b, with a < b, such that for all n, a + 1/n >=
b.I think you got into trouble when you swapped a < b for a
>= b in your attempted negation.
===
> For all real numbers
satisfying a<b, there exists a natural number n,> such that a
+ 1/n < b.>> Converting this sentence into symbolic form
gives something like:> (forall a,b in R ):( a <
b),(exists n in N):( a+1/n <b )>for all a,b in R, (a < b
==> some n in N with a+1/n < b)> and then forming the
negation, I have> (exists a,b in R):(a>= b),(forall n
in N):(a+1/n >= b)>some a,b in R with (a < b and for all n
in N, not a+1/n < b)> i.e. There exist real numbers
satisfying a>=b, such that for all> natural numbers n, a+1/n
>= b.>Archimedean principle for real numbers: for all r in
R, some n in N with r < n.if a < b, then b-a /= 0 some n in N
with 1/(b-a) < n 1/n < b-a as 0 < b-a a + 1/n < a + b-a = b>
So, what am I doing wrong?You converted wrong and you may
have used ~(p&q) <-> (~p&~q)instead of ~(p&q) <-> (~p v
~q)when forming the negation.
===
> I'm trying to ?ure out
how to correctly form the logical negation of> this
sentence:> For all real numbers satisfying a<b, there exists
a natural number n,> such that a + 1/n < b.The tricky part
here is the phrase satisfying a < b.Actually, the same
problem applies to the phrase real also,except that real
is a unary relation, while a < b is abinary relation.That
is, strictly, you can use only for all x and notfor all
real x, etc. That is, the phrase for all real xis an
abreviation for the phrase for all x,... (if x is real,then
...).Note that you can swap (for all real a)(for all real b)
toget (for all real b)(for all real a) with no problem.But,
you can not swap (for all real a)(for all real b with a <
b)to get (for all real b with a < b)(for all real a) sincethe
?st a is then left unquanti?d.> Converting this sentence
into symbolic form gives something like:> (forall a,b in
R ):( a < b),(exists n in N):( a+1/n <b )This is the ?st
time I have seen the notation :(a < b).This is what is
causing you a problem. You don't have acorrect rule on how to
handle the negation through it.I convert that sentence into
symbolic form as follows: (for all a in R)(for all b in
R)(for some n in N) (if a < b then a + 1/n < b)Since I am
going to form the negation, I transform theif ... then ...
into and and or and not termsas follows: (for all a in
R)(for all b in R)(for some n in N) (not(a < b) or (a + 1/n <
b))> and then forming the negation, I have> (exists a,b
in R):(a>= b),(forall n in N):(a+1/n >= b) (for some a
in R)(for some b in R)(for all n in N) ((a < b) and (a + 1/n
>= b))> i.e. There exist real numbers satisfying a>=b, such
that for all> natural numbers n, a+1/n >= b.You should not
have negated a<b, just like you did notnegate a in R.>
However, I am obviously doing something wrong in the
negation, since> BOTH of the above sentences appear to be
true, and by the law of> excluded middle, P and not P cannot
BOTH be true.> So, what am I doing wrong?Eliminating a in
R etc in the phrase for a in R, I have: (for all a)(for
all b)(for some n) (if a in R and b in R and n in N and a < b
then a + 1/n < b)Try negating that and see that a in R
remains and does notbecome a not in R.Or, even better,
since I don't like the phrase a < b unlessI de?itely
already know that a and b are real numbers, I supposethat the
above should be written: (for all a)(for all b)(for some n)
(if a in R and b in R and n in N, then if a < b then a + 1/n
< b)You could try and negate that also. I didn't do it
myself.I hope it works out ok.-- Bill Hale
===
>I'm trying to
?ure out how to correctly form the logical negation of>this
sentence:>For all real numbers satisfying a<b, there exists
a natural number n,>such that a + 1/n < b.>Converting this
sentence into symbolic form gives something like:>(forall
a,b in R ):( a < b),(exists n in N):( a+1/n <b )
(forall a,b in R) (a < b --> exists n in N (a + 1/n <
b))>and then forming the negation, I have>(exists a,b in
R):(a>= b),(forall n in N):(a+1/n >= b)Nope. The negation
is (exists a,b in R) ~(a < b --> exists n in N (a +
1/n < b)).Since ~(p --> q) is p & ~q, the desired negation is
(exists a,b in R) (a < b & ~exists n in N (a + 1/n <
b)),or, ?ally, (exists a,b in R) (a < b & forall n
in N (a + 1/n >= b)).[...]Brian
===
[cut]> Eliminating a in
R etc in the phrase for a in R, I have:> (for all a)(for
all b)(for some n)> (if a in R and b in R and n in N and a <
b then a + 1/n < b)I eliminated n in N from the phrase for
some n in N wrong.It should be: (for all a)(for all b)(for
some n) (n in N and (if a in R and b in R and a < b then a +
1/n < b))-- Bill Hale
===
>I'm trying to ?ure out how to
correctly form the logical negation of>this sentence:>For
all real numbers satisfying a<b, there exists a natural
number n,>such that a + 1/n < b.For some real numbers a<b,
there is no natural number n such that a+1/n < b.The negation
of For all X, Y is For some X, not Y. (Or: There exists
at least one X such that Y.)>i.e. There exist real numbers
satisfying a>=b, such that for all>natural numbers n, a+1/n
>= b.This is For some not X, Z. :-)-- Stan Brown, Oak Road
Systems, Cortland County, New York, USA
http://OakRoadSystems.com/You ?d yourself amusing,
Blackadder.I try not to ?the face of public
opinion.Mime-version: 1.0
===
> I'm trying to ?ure out how
to correctly form the logical negation of> this sentence:>
For all real numbers satisfying a<b, there exists a natural
number n,> such that a + 1/n < b.> Converting this
sentence into symbolic form gives something like:> (forall
a,b in R ):( a < b),(exists n in N):( a+1/n <b )> and
then forming the negation, I have> (exists a,b in
R):(a>= b),(forall n in N):(a+1/n >= b)> i.e. There
exist real numbers satisfying a>=b, such that for all>
natural numbers n, a+1/n >= b.> However, I am obviously
doing something wrong in the negation, since> BOTH of the
above sentences appear to be true, and by the law of>
excluded middle, P and not P cannot BOTH be true.> So, what
am I doing wrong?> thanks.Assuming that your N does not
contain 0, consider:(exists a,b in R)(forall n in
N):if(a >= b) then (a+1/n >= b)Which negates to:(forall a,b
in R)(exists n in N):(a>= b) and (a+1/n <
b)
===
Determine a>0 so that the equationsystem: x^2+y^2=18{
y=-x+ahas exactly one solution.How Do I solve a problem of
this kind?
===
>Determine a>0 so that the equationsystem:>
x^2+y^2=18> y=-x+a>has exactly one solution.>>How Do I solve
a problem of this kind?Solve the equations simultaneously.
Since one is a quadratic, you will end up with some +/-
radical expression, and the radicand will probably contain a.
Determine a so that the choice of + or - makes no difference,
i.e. so that the radicand is 0.-- Stan Brown, Oak Road
Systems, Cortland County, New York, USA
http://OakRoadSystems.com/You ?d yourself amusing,
Blackadder.I try not to ?the face of public
opinion.
===
> Determine a>0 so that the equationsystem:>
x^2+y^2=18> {> y=-x+a> has exactly one solution.> How Do
I solve a problem of this kind?> To rephrase another
answer, substitute a - x into x^2 + y^2 - 18 = 0.Square the a
- x and collect things up and you'll have a quadratic.Think of
the quadratic formula. The discriminant, the part under
thesquare root (b^2 - 4*a*c - this a is not _your_ a)
determines howmany solutions there are: if it is zero there
is only one so set itequal to zero and solve for a. I get a =
6 (and a = -6.)-- Paul SperryColumbia, SC (USA)
===
>
Determine a>0 so that the equationsystem:> x^2+y^2=18>
> {> y=-x+a> has exactly one solution.> How Do
I solve a problem of this kind?> To rephrase another
answer, substitute a - x into x^2 + y^2 - 18 = 0.> Square
the a - x and collect things up and you'll have a
quadratic.[snip]That'll all work, but another approach is a
little geometryand some trig.x^2 + y^2 = 18 is a circle
centered on (0,0) withradius 3sqrt(2). y=-x+a with a>0 is a
line with slope -1 thatpasses through quadrants I, II, and
IV. Having one solutionmeans the line intersects the circle
at only one point, whichmeans it must be tangent to the
circle. By the constraintsof the problem, we'll be in
quadrant I. In quadrant I, acircle only has slope -1 at the
point where x=y. Thatimmediately gives us the point (3,3).So
the desired tangent line has slope -1 and passes through(3,
3): y - 3 = (-1)(x - 3) y - 3 = -x + 3 y = -x + 6Thus a=6.--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
>Determine
a>0 so that the equationsystem:>> x^2+y^2=18>{>
y=-x+a>>has exactly one solution.>>How Do I solve a
problem of this kind?>To rephrase another answer,
substitute a - x into x^2 + y^2 - 18 = 0.>>Square the a - x
and collect things up and you'll have a quadratic.>>
>[snip]>>That'll all work, but another approach is a little
geometry>and some trig.>>x^2 + y^2 = 18 is a circle centered
on (0,0) with>radius 3sqrt(2). y=-x+a with a>0 is a line with
slope -1 that>passes through quadrants I, II, and IV. Having
one solution>means the line intersects the circle at only one
point, which>means it must be tangent to the circle. By the
constraints>of the problem, we'll be in quadrant I. In
quadrant I, a>circle only has slope -1 at the point where
x=y. That>immediately gives us the point (3,3).>>So the
desired tangent line has slope -1 and passes through>(3, 3):>
y - 3 = (-1)(x - 3)> y - 3 = -x + 3> y = -x + 6>>Thus a=6.>And
avoids the spurious a=-6 solution.This reminds me of one
de?ition of a mathematician as someone who will spend two
weeks looking for the way to reduce two hours of calculation
to 10 minutes. (Actually, I ?st heard that as trick rather
than way, but many of the tricks are actually rather
useful, more so than a method is a device, used twice would
lead you to believe.)My experience has been that very few
students appreciate the elegance of this and most view it as
a trick. In other words, it's totally inappropriate for
classroom use.I wonder what we'll ?d out about Mr. Aspen's
mathematical sense.Jon Miller
===
> Determine a>0 so that the
equationsystem:>> x^2+y^2=18> {> y=-x+a>> has exactly one
solution.>> How Do I solve a problem of this kind?>>I dont
know what more to say! =)Said Aspen
===
[skip]> Since I
explained it the implication that I didn't is false.Since you
didn't actually explained anything, the implication is
correct.I asked you to explain what *kind* of thing factor
g is, and your response was -Not even close. Read the
paper. That's not an explanation.
===
Does James Harris have
a degree in math?
===
> Does James Harris have a degree in
math?No. He claims to have an undergraduate degree in
physics.-- Wayne Brown | When your tail's in a crack, you
improvisefwbrown@bellsouth.net | if you're good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) =
-1 -- Euler | -- John Myers Myers, Silverlock
===
>
> I was going back over my work as I often do when I ?ally
realized> that my lemma in my paper>
Advanced Polynomial Factorization>
http://groups.msn.com/AmateurMath/Documents/FactProof.pdf>
> was wrong.> I had a factor g of the
polynomial P(x) where g = r+c, and r is> supposedly
proven to not be coprime to x, when in fact it can be>
coprime to x.> What I was actually using for
the paper is the fact that r varies with> x, while c
remains constant, so I made the simplest change.>
> It is possible to prove that r is not coprime to P(x)-P(0),
but not> worth the effort here.> My
apologies for the error.> In the ?st paragraph you
give the wrong roots for x^2+x-5=0. You> might want to
correct them, too.> James, I am new to your work. You
say: As an example consider> sqrt(x+1) which is a non
polynomial factor of x+1. What do you mean> by saying that
sqrt(x+1) is a factor of x+1?> John PetersThere are two
ways to look at it. Since the ring is algebraicintegers, you
can see it as that given x, which is going to be analgebraic
integer as that's the ring, sqrt(x+1) is a factor of x+1.For
instance with x=2, which is an algebraic integer, sqrt(3) is
afactor of 3.The other way to look at it is that x+1 can be
considered the resultof the multiplication sqrt(x+1) times
sqrt(x+1) equals x+1; therefore,sqrt(x+1) is a factor of x+1,
where ring operations allow thatexpression. But they don't
allow (x+1)/(y+1) in general as the / isnot allowed in
general as division is not a ring operation.So I can also say
that given the polynomial P(x)=x+1, sqrt(x+1) is afactor of
the polynomial, though you may wonder what to now
callsqrt(x+1). I don't care what you call it.Now sure, if you
qualify by saying y=x, then y+1 is a factor of x+1 inthe ring,
but not, in general, otherwise, if all you know is that y+1is
in the ring.Some may see that I don't have to talk
speci?ally about polynomialrings or something wackier to
handle sqrt(x+1) by taking thisapproach.If you ?d that makes
your head swim, you can consider values of x+1from the ?st
point of view.James Harris
===
> ...> sqrt(x+1) is a factor
of x+1, where ring operations allow that> ...In _what_ is
sqrt(x+1) a factor of x+1?GC
===
> ...>> sqrt(x+1) is a
factor of x+1, where ring operations allow that>> ...>In
_what_ is sqrt(x+1) a factor of x+1?I believe this is what
JSH means: If x is an algebraic integer, then so is x+1 and
sqrt(x+1). Moreover, there is an algebraic integer r (viz. r
= sqrt(x+1)) such that r*sqrt(x+1) = x+1. Therefore sqrt(x+1)
is a factor of x+1 in the ring of algebraic integers.If this
IS what he means, then he is, of course, quite correct. (Oh,
GOD! It's ?e o'clock in the morning of midsummer's eve and
JSH is starting to make sense!)The problem is that he
categorically denies that 2 is a factor of x^2 + x in the
ring of integers, so my interpretation may be wrong.--
Thomas Wasell | A great many people think they are thinking
when theywasell@bahnhof.se | are merely rearranging their
prejudices. | -- William James
===
>> ...>> sqrt(x+1)
is a factor of x+1, where ring operations allow that>> ...>
>>In _what_ is sqrt(x+1) a factor of x+1?>> I believe this
is what JSH means:>> If x is an algebraic integer, then so is
x+1 and sqrt(x+1).Moreover, there> is an algebraic integer r
(viz. r = sqrt(x+1)) such that r*sqrt(x+1)= x+1.> Therefore
sqrt(x+1) is a factor of x+1 in the ring of
algebraicintegers.>> If this IS what he means, then he is, of
course, quite correct. (Oh,GOD!> It's ?e o'clock in the
morning of midsummer's eve and JSH isstarting to> make
sense!)>> The problem is that he categorically denies that 2
is a factor ofx^2 + x> in the ring of integers, so my
interpretation may be wrong.James,Building on the insight
provided by Thomas, let me frame a possiblede?ition:1) Let S
be a set.2) Let f, g and h be [single-valued] functions de?ed
on S and takingvalues in S. That is, their domain and codomain
are S.3) Let * be a [single-valued] binary operator whose
operands areelements of S and which takes values in S.4)
Suppose, for all x in S, f(x)*g(x)=h(x).De?ition: In this
case we say that f is a factor of h. Or(equivalently) that h
is divisible by f.In the case you are interested in, S is the
set of algebraic integers,and * is (complex)
multiplication.(Note that our sqrt(x+1) example might have
some problems becausesqrt(x+1) might not be single-valued.)Is
the above de?ition correct?John Peters
===
> ...>>
sqrt(x+1) is a factor of x+1, where ring operations allow
that>> ...>In _what_ is sqrt(x+1) a factor of x+1?>
> I believe this is what JSH means: > If x is an algebraic
integer, then so is x+1 and sqrt(x+1). Moreover, there > is
an algebraic integer r (viz. r = sqrt(x+1)) such that
r*sqrt(x+1) = x+1. > Therefore sqrt(x+1) is a factor of x+1
in the ring of algebraic integers.> If this IS what he
means, then he is, of course, quite correct. (Oh, GOD! >
It's ?e o'clock in the morning of midsummer's eve and JSH is
starting to > make sense!)> The problem is that he
categorically denies that 2 is a factor of x^2 + x > in the
ring of integers, so my interpretation may be wrong.That's
not true. The original post had something about that being
inthe same sense if I remember correctly. So you left off
keyinformation in your claim.After all, in the ring of
algebraic integers, 2 is NOT in general afactor of x^2 +
x.For instance, with x=(1+sqrt(-3))/2, x^2 + x =
sqrt(-3).That it is in the ring of integers is, in my mind,
trivia, which isnot in the same sense as something like x
being a factor of x^2 + x,which IS true in the ring of
algebraic integers, so it's NOT in thesame sense.What I've
seen time and time again is a grasping at any little
commentof mine to try and see someway to make it not make
sense.Here when I talk of factorizations in what I think is a
rather obviousway, bringing up a trivial result in the ring of
integers as if it hasgreat sign?ance just makes me
wonder.Don't ANY of you care about what the truth is? Don't
bother me with trivia. James Harris
===
nevermind. let's start
another item!> I had a factor g of the polynomial P(x)
where g = r+c, and r is> supposedly proven to not be
coprime to x, when in fact it can be> coprime to x.> ...>
> Or r can be constant and 0.> What you fail to state
anywhere is how g is restricted. Or is it> unrestricted? If
unrestricted:> g(x) = def: 1 (if x = 0), x + 5 (elsewhere)>
g(x) is a factor of P(x) = x + 5 in the sense you state. What
are r and c?--les ducs d'Enron!
http://quincy4board.homestead.com/ Funny.html (schoolboard
stuf?')
===
> The problem is that he categorically denies
that 2 is a factor of x^2 + x > in the ring of integers,
so my interpretation may be wrong.> That's not true. The
original post had something about that being in> the same
sense if I remember correctly. So you left off key>
information in your claim.> After all, in the ring of
algebraic integers, 2 is NOT in general a> factor of x^2 +
x.> For instance, with x=(1+sqrt(-3))/2, x^2 + x =
sqrt(-3).> But, 2 IS a factor of sqrt(-3), the other factor
being sqrt(-3)/2,which is a root of 4x^2+3, hence
algebraic.
===
[snip]>But, 2 IS a factor of sqrt(-3), the
other factor being sqrt(-3)/2,>which is a root of 4x^2+3,
hence algebraic.But it's not an algebraic *integer*. I.e.
it's not the root of a *monic* polynomial.-- Thomas Wasell |
For a man to truly understand rejection, he must
?stwasell@bahnhof.se | be ignored by a cat.
===
> [snip]>
>But, 2 IS a factor of sqrt(-3), the other factor being
sqrt(-3)/2,>which is a root of 4x^2+3, hence algebraic.>
But it's not an algebraic *integer*. I.e. it's not the root
of a *monic* > polynomial.I'm just playing by the same rules
that James does --a polynomial isan algebraic integer,a
function,an integer,a square root,a ring of polynomials,a
real number,a ring of algebraic integers,a monic polynomial,a
complex number,a transcendental function,a Laplace Transform,
a ...Hell, no wonder math is full of contradictions!
===

... a buncha crap ...> What I've seen time and time again is
a grasping at any little comment> of mine to try and see
someway to make it not make sense.> Here when I talk of
factorizations in what I think is a rather obvious> way,
bringing up a trivial result in the ring of integers as if it
has> great sign?ance just makes me wonder.> Don't ANY of
you care about what the truth is? > Apparently more than you
do. Note this:De?e the polynomials q(x) = 8 x^2 - 76 x -185
r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104Then for z
any root of the polynomial p(x) = x^3 - 12 x^2 + 65,we have
q(z)*r(z) = 5 r(z)*s(z) = z.By virtue of the polynomials
q,r,s having integer coef?ients, andof z being a root of
p(x), we know that z, q(z), r(z), and s(z) are allalgebraic
integers.Finally, the minimal polynomial for r(z) is
MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5The divisibility
properties were veri?d in Maxima; the minimalpolynomial was
provided by KASH, and veri?d in KASH.Your claim regarding
the properties of the coef?ients ai inthe factorization: 65
x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)namely, that
ONE of the ai's is coprime to 5, is clearlyincorrect.
Therefore, your argument is ?Once again, your methods
are shown to produce incorrectresults, despite your con?ence
in your own ability tospot correct logic. Your logic, your
methods, your graspof mathematics are *all* unsound.> Don't
bother me with trivia. > How trivial was that?Score: sci.math
omega JSH nullset> James HarrisDale.
===
John Peters
<tafelnder@yahoo.com> skrev i melding> ...>>
sqrt(x+1) is a factor of x+1, where ring operations allow
that>> ...>>In _what_ is sqrt(x+1) a factor of
x+1?>> I believe this is what JSH means:>> If x is an
algebraic integer, then so is x+1 and sqrt(x+1).> Moreover,
there> is an algebraic integer r (viz. r = sqrt(x+1)) such
that r*sqrt(x+1)> = x+1.> Therefore sqrt(x+1) is a factor
of x+1 in the ring of algebraic> integers.>> If this IS
what he means, then he is, of course, quite correct. (Oh,>
GOD!> It's ?e o'clock in the morning of midsummer's eve
and JSH is> starting to> make sense!)>> The problem is
that he categorically denies that 2 is a factor of> x^2 + x>
> in the ring of integers, so my interpretation may be
wrong.>> James,>> Building on the insight provided by Thomas,
let me frame a possible> de?ition:>> 1) Let S be a set.>> 2)
Let f, g and h be [single-valued] functions de?ed on S and
taking> values in S. That is, their domain and codomain are
S.>> 3) Let * be a [single-valued] binary operator whose
operands are> elements of S and which takes values in S.>> 4)
Suppose, for all x in S, f(x)*g(x)=h(x).>> De?ition: In this
case we say that f is a factor of h. Or> (equivalently) that
h is divisible by f.>> In the case you are interested in, S
is the set of algebraic integers,> and * is (complex)
multiplication.>> (Note that our sqrt(x+1) example might
have some problems because> sqrt(x+1) might not be
single-valued.)Isn't sqrt(a) single-valued? Don't mistake it
to be +/- ,because it isn't.Sqrt is allways positive! It
follows from the de?ition of Sqrt.>> Is the above de?ition
correct?>> John Peters>>
===
don't bother me with mathematics,
jerk. now, let's start another item! > By virtue of the
polynomials q,r,s having integer coef?ients, and> of z being
a root of p(x), we know that z, q(z), r(z), and s(z) are all>
algebraic integers.> Finally, the minimal polynomial for
r(z) is> MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5> The
divisibility properties were veri?d in Maxima; the minimal>
polynomial was provided by KASH, and veri?d in KASH.> Your
claim regarding the properties of the coef?ients ai in> the
factorization:> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3
x + 1)> namely, that ONE of the ai's is coprime to 5, is
clearly--Dec.2000 ?WAND' Chairman Paul O'Neill, reelectedto
Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
===
I'm
living in HK. We had one of the biggest rally against the HK
gov't since the June 4th ?89 incident. The press reported
that there are 500K peoplewent to the street for the
protest.I'm very interested with the reported ?ures. Is
there really a 'reasonable' method based on statistical
model to approximate the total no. ofprotesters with
acceptable degree of con?ence? Can anyone show me the
concept.
===
>I'm living in HK. We had one of the biggest
rally against the HK gov't since the June 4th ?89 incident.
The press reported that there are 500K people>went to the
street for the protest.>>I'm very interested with the
reported ?ures. Is there really a 'reasonable' method based
on statistical model to approximate the total no.
of>protesters with acceptable degree of con?ence? Can
anyone show me the concept.>I Have Been Told (tm) that no
one estimating crowd sizes uses any reasonable method to
estimate them. I don't remember the reference.The way to
estimate crowd size is the same as the one astronomers use to
estimate the number of visible stars. You pick a small area of
surveillance, ?d the volume of space that area views,
actually count the number of people/stars/whatever. Then pick
another area and repeat. You need to use statistical sampling
methodology to assure that your samples are representative of
the total population. You can get quite sophisticated with
this, so that your efforts are concentrated in the areas with
the most importance/dif?ulty/whatever. So for crowds, you'd
probably concentrate your efforts at dense points, rather
than along the edges. Then you average (possibly with
weights) your observations to get an average density, and
then multiply that by the volume to get the total number. Of
course, for crowds you'd use surface area instead of volume,
but you get the same idea.I have done this, but if your
interest isn't in the results (that is, if BigNum crowds in
headlines sell newspapers/support your political point of
view/etc. and accuracy isn't really relevant) it's probably
not worth the effort.At some time, someone will probably add
a fairly simple routine to the camera analysis software to
have crowd size printed out in the automatically generated
report, but that's at least a couple of years down the road.
http://www.ohio.com/mld/ohio/news/6211959.htm I'm guessing
that deployment of the CTS software/hardware will take a
couple of years to implement, and crowd size estimation will
be low enough priority that they just don't get around to
it.So they won't be able to tell you how many people are in
the crowd, but they'll be able to give you all their names.
Well, there's another method: just count the number of names
in the list of attendees. Maybe it won't take a couple of
years.Jon Miller
===
Jon/Mary,The methodology you mentioned,
does it make a assumption that the crowd in question is
stationary just like the stars in the clear sky? My scenario
isthe crowd to be measured has varied mobility (e.g. the
crowd is moving at non-constant speed at one direction).
When the crowd moves forward, newin-coming crowd comes in and
it took 7 hours to travel from Point A to Point B.I'm just
thinking of differential equations to solve it. Does it make
sense?James>I'm living in HK. We had one of the biggest
rally against the HK gov't since the June 4th ?89 incident.
The press reported that there are 500K people>>went to the
street for the protest.>>I'm very interested with the
reported ?ures. Is there really a 'reasonable' method based
on statistical model to approximate the total no.
of>>protesters with acceptable degree of con?ence? Can
anyone show me the concept.>I Have Been Told (tm) that no
one estimating crowd sizes uses any >reasonable method to
estimate them. I don't remember the reference.>>The way to
estimate crowd size is the same as the one astronomers use to
>estimate the number of visible stars. You pick a small area
of >surveillance, ?d the volume of space that area views,
actually count >the number of people/stars/whatever. Then
pick another area and repeat. > You need to use statistical
sampling methodology to assure that your >samples are
representative of the total population. You can get quite
>sophisticated with this, so that your efforts are
concentrated in the >areas with the most
importance/dif?ulty/whatever. So for crowds, >you'd probably
concentrate your efforts at dense points, rather than >along
the edges. Then you average (possibly with weights) your
>observations to get an average density, and then multiply
that by the >volume to get the total number. Of course, for
crowds you'd use surface >area instead of volume, but you get
the same idea.>>I have done this, but if your interest isn't
in the results (that is, if >BigNum crowds in headlines sell
newspapers/support your political point >of view/etc. and
accuracy isn't really relevant) it's probably not worth >the
effort.>>At some time, someone will probably add a fairly
simple routine to the >camera analysis software to have crowd
size printed out in the >automatically generated report, but
that's at least a couple of years >down the road.
http://www.ohio.com/mld/ohio/news/6211959.htm I'm >guessing
that deployment of the CTS software/hardware will take a
couple >of years to implement, and crowd size estimation will
be low enough >priority that they just don't get around to
it.>>So they won't be able to tell you how many people are in
the crowd, but >they'll be able to give you all their names.
Well, there's another >method: just count the number of names
in the list of attendees. Maybe >it won't take a couple of
years.>>Jon Miller
===
> The methodology you mentioned, does
it make a assumption that the crowd> in question is
stationary just like the stars in the clear sky? My> scenario
is the crowd to be measured has varied mobility (e.g. the
crowd> is moving at non-constant speed at one direction).
When the crowd moves> forward, new in-coming crowd comes in
and it took 7 hours to travel from> Point A to Point B.>No,
it makes sense to photograph crowd from satellite, plane
orhellicopter perhaps several times to assure estimation of
peak event.However, these times, as protests are ignored,
crowd estimates aren'tneeded nor event and attendance deemed
news wirthy.> I'm just thinking of differential equations to
solve it. Does it make sense?No, renting a plane and an
aerial survey camara does.
===
>In alt.math.undergrad you
write:>At some time, someone will probably add a fairly
simple routine to the >>camera analysis software to have
crowd size printed out in the >>automatically generated
report, but that's at least a couple of years >>down the
road.>I recall some press release from a company that I
think was dealing with>designing places for selling tickets,
handling crowds, that sort of thing, >and if I'm not
confused, I think they had done substantial work on
counting>automatically with cameras. But it has been a while
and I'm not able to>think of any of the keywords that might
turn this up.>Could be. Now that I think about it, I think
the discussion I read was about the size of the crowd at the
Million Man March, so my info is certainly dated.I guess I'll
post this to alt.math.undergrad since it's relevant, and I
don't mind appearing stupid (I have children, so it happens
all the time Jon Miller
===
Market research for a new mobile
phone suggests that the potential markethas a ceiling of 40
million units and that 0 sales will be achieved
withoutmarketing. After spending 1 million on marketing 20
million sales wererecorded.Can anyone model for sales (S) in
terms of advertising expenditure (x) ofthe form:S =
a+(be)POWER-cx
===
> Market research for a new mobile phone
suggests that the potential market> has a ceiling of 40
million units and that 0 sales will be achieved without>
marketing. After spending 1 million on marketing 20 million
sales were> recorded.> Can anyone model for sales (S) in
terms of advertising expenditure (x) of> the form:> S =
a+(be)POWER-cxI assume you mean S = a + b e^(-cx)Well, you
have some known values, so use them: (0, 0) (+oo, 40mil)
(1mil, 20 mil)Plug those (x, S) pairs into the equation for S
and see what theyimply about constants a, b, and c.-- Rich
Carreiro rlcarr@animato.arlington.ma.us
===
>>Market
research for a new mobile phone suggests that the potential
market>>has a ceiling of 40 million units and that 0 sales
will be achieved without>>marketing. After spending 1
million on marketing 20 million sales were>>recorded.>>Can
anyone model for sales (S) in terms of advertising
expenditure (x) of>>the form:>>S = a+(be)POWER-cx> I
assume you mean> S = a + b e^(-cx)> Well, you have some
known values, so use them:> (0, 0)> (+oo, 40mil)> (1mil, 20
mil)> Plug those (x, S) pairs into the equation for S and
see what they> imply about constants a, b, and c.>
===
Let's
say I have a function, f(x), whose inverse for some reason I
can'tcalculate. Can I use the formula:g(x) = f^(-1)(x)g'(x) =
1 / [ f'(g(x)) ]so ?d f^(-1)(x) explicitly? If so, can I
always ?d it?Also, does every function whose inverse can be
expressed in terms of a?ite number of elementary operations
be expressed in terms of a ?itenumber of elementary
operations? Also, is it possible to have a functionthat can't
be expressed in terms of a ?ite number of elementary
operationshave an inverse that can be?Jeremy
===
> Let's say
I have a function, f(x), whose inverse for some reason I
can't> calculate. Can I use the formula:>> g(x) = f^(-1)(x)>
g'(x) = 1 / [ f'(g(x)) ]>g(x) = integral g'(x) dx = integral
1/f'(g(x)) dxlet u = g(x) = integral du/f'(u)g'(x) = integral
du = u = g(x)> so ?d f^(-1)(x) explicitly? If so, can I
always ?d it?>Let's test it out. For n > 5 letf(x) =
sum(i=0..n) x^i restricted to some interval where it's
1-1g'(x) = 1/sum(i=1..n) i.g(x)^(i-1)What's used for g(x) to
calculate this sum?> Also, does every function whose inverse
can be expressed in terms of a> ?ite number of elementary
operations be expressed in terms of a ?ite> number of
elementary operations?> Also, is it possible to have a
function that can't be expressed in terms> of a ?ite number
of elementary operations have an inverse that can
be?Basically these two questions are the same.Let g be the
inverse of f as given above.Can g be expressed in a ?ite
number of elementary operations?
===
> Let's say I have a
function, f(x), whose inverse for some reason I can't>
calculate. Can I use the formula:> g(x) = f^(-1)(x)>
g'(x) = 1 / [ f'(g(x)) ]> so ?d f^(-1)(x) explicitly? If
so, can I always ?d it? [...]Do you mean, by integrating 1 /
(f'(g(x)) explicitly? I don't see howyou can do the
integration without knowing what g(x) is, unless you wantto
solve an integral equation. (And the way to solve that
particularintegral equation will probably involve ?ding the
inverse of f(x) ?st.) -- Christopher Heckman
===
> Do you
mean, by integrating 1 / (f'(g(x)) explicitly? I don't see
how> you can do the integration without knowing what g(x) is,
unless you want> to solve an integral equation. (And the way
to solve that particular> integral equation will probably
involve ?ding the inverse of f(x)?st.)Not necessarily by
doing it explicitly, but maybe by doing something moreclever
than I can think of, or by ?ding a method to solve the
differentialequation, or at least do something good to aid
in the ?ding of f^(-1)(x).- Jeremy
===
> so ?d f^(-1)(x)
explicitly? If so, can I always ?d it?>> Let's test it
out. For n > 5 let> f(x) = sum(i=0..n) x^i restricted to some
interval where it's 1-1> g'(x) = 1/sum(i=1..n) i.g(x)^(i-1)>
What's used for g(x) to calculate this sum?I'm confused on
what you're doing here.> Also, does every function whose
inverse can be expressed in terms of a> ?ite number of
elementary operations be expressed in terms of a ?ite>
number of elementary operations?>> Also, is it possible to
have a function that can't be expressed in terms> of a
?ite number of elementary operations have an inverse that
can be?>> Basically these two questions are the same.> Let g
be the inverse of f as given above.> Can g be expressed in a
?ite number of elementary operations?I would guess so, but
I'm confused as how to prove it.- Jeremy
===
> Let's say I
have a function, f(x), whose inverse for some reason I can't>
calculate. Can I use the formula:>> g(x) = f^(-1)(x)>> g'(x) =
1 / [ f'(g(x)) ]>> so ?d f^(-1)(x) explicitly? If so, can I
always ?d it?>> Also, does every function whose inverse can
be expressed in terms of a> ?ite number of elementary
operations be expressed in terms of a ?ite> number of
elementary operations? Also, is it possible to have a
function> that can't be expressed in terms of a ?ite number
of elementaryoperations> have an inverse that can be?>>
Jeremy>>Suppose I want to ?d the inverse off(x) = e^x + x -
1x = e^y + y - 1 swap variables1 = (e^y)y' + y' take the
derivative of both sidesy' = 1/(e^y + 1) solve for y'y'' =
.....y^(p) = ...note that when y=1, x = e so y(e) = 1(if
y=g(x), then y(e)= g(e) = 1)The Taylor expression becomes,y =
1 + y'(e) (x-e)+y''(e) (x-e)^2 /2! + ...swapping back
variables leads to the inverse,so yes, implicit
differentation can be used.
===
>> so ?d f^(-1)(x)
explicitly? If so, can I always ?d it?>> Let's test it
out. For n > 5 let> f(x) = sum(i=0..n) x^i restricted to
some interval where it's 1-1> g'(x) = 1/sum(i=1..n)
i.g(x)^(i-1)> What's used for g(x) to calculate this sum?>
I'm confused on what you're doing here.As you clipped the
orginal problem, I'm limited to general description.How would
you like it if I clipped down to ?What's used ... ?I'm taking
the function f(x) as I described and applying your
clippedformula to it to point out the circularity of your
attempt and to giveexample for later.> Also, does every
function whose inverse can be expressed in terms of a>
?ite number of elementary operations be expressed in terms
of a ?ite> number of elementary operations?>>
Also, is it possible to have a function that can't be
expressed in terms> of a ?ite number of elementary
operations have an inverse that can be?>> Basically
these two questions are the same.> Let g be the inverse of
f as given above.> Can g be expressed in a ?ite number of
elementary operations?>> I would guess so, but I'm confused
as how to prove it.>It depends upon what you mean as
elementary. For starters there's Abel'sfamous theorem that a
polynomial of degree greater than 4 can't be solvedby
radicals. Indeed prior to his time, the solution for quartics
was thebig ado.
===
> Let's say I have a function, f(x), whose
inverse for some reason I can't> calculate. Can I use the
formula:> g(x) = f^(-1)(x)> g'(x) = 1 / [ f'(g(x)) ]>
so ?d f^(-1)(x) explicitly? If so, can I always ?d it?>
Also, does every function whose inverse can be expressed in
terms of a> ?ite number of elementary operations be
expressed in terms of a ?ite> number of elementary
operations? Also, is it possible to have a function> that
can't be expressed in terms of a ?ite number of elementary
operations> have an inverse that can be?> JeremyMore on
the subject that you want to know is contained is the book
byfamous British mathematician G.H. Hardy The integration of
functionsof a single variable Hafner, New York, 1971A formula
due to J. L. Liouville uses inverse function to ?d aprimitive
function of a function.Say that you want to integrate f(x)
Int[f(x)]dx = Int[x'*f(x)]dx {Since x is 1} (A)Now since
f^-1(f(x)) = x, then x' = (f^-1(f(x)))' then (A) is equal to:
Int[f(x)]dx = Int[(f^-1(f(x)))'*f(x)]dx {using partial
integration} = f^-1(f(x))*f(x) - Int[f^-1(f(x))*f'(x)]dx =
x*f(x) - Int[(F^-1(f(x)))'dx] = x*f(x) - F^-1(f(x))where F^-1
is primitive of f^-1This is because f^-1(f(x))*f'(x) =
[F^-1(f(x))]' f^-1(f(x))*f'(x)For example if f(x) = ln(x), x
> 0, the above forumula gives:Int[ln(x)]dx = x*ln(x) -
exp(ln(x)) + C = x[ln(x) - 1] + CSince primitive of exp(x) is
exp(x) + C/Marko
===
Wow, thanks, I'll have to look into that
further!
===
A quick question: Is every ?ite 2-generated
abelian group, i.e. G =< x, y >, isomorphic to Z_a x Z_b for
some integers a and b? It seemsso (since such a G can be
displayed in a lattice just like one woulduse to visually
represent Z_a x Z_b, and there must be relations ofthe form
mx=0 for some m, ny=0 for some n), but I'm not 100%
certainone way or the other.BDG
===
> A quick question: Is
every ?ite 2-generated abelian group, i.e. G => < x, y >,
isomorphic to Z_a x Z_b for some integers a and b? It seems>
so (since such a G can be displayed in a lattice just like
one would> use to visually represent Z_a x Z_b, and there
must be relations of> the form mx=0 for some m, ny=0 for some
n), but I'm not 100% certain> one way or the other.> BDGIf
you mean G is isomorphic to the direct sum of at most
twonon-trivial cyclics the answer is yes.If you mean G is
isomorphic to the direct sum of _exactly_ twonon-trivial
cyclics then the answer is no: Z_5 = < 2, 3 >.One way to
prove the former is to note that G is the homorphic image ofZ
x Z and the kernel is of the form K x H with K <= Z x {0} and
H <= {0} x Z.-- Paul SperryColumbia, SC (USA)
===
> A
quick question: Is every ?ite 2-generated abelian group,
i.e. G => < x, y >, isomorphic to Z_a x Z_b for some
integers a and b? It seems> so (since such a G can be
displayed in a lattice just like one would> use to visually
represent Z_a x Z_b, and there must be relations of> the
form mx=0 for some m, ny=0 for some n), but I'm not 100%
certain> one way or the other.> BDG> If you mean G
is isomorphic to the direct sum of at most two> non-trivial
cyclics the answer is yes.> If you mean G is isomorphic to
the direct sum of _exactly_ two> non-trivial cyclics then the
answer is no: Z_5 = < 2, 3 >.> One way to prove the former
is to note that G is the homorphic image of> Z x Z and the
kernel is of the form K x H with K <= Z x {0} and > H <= {0}
x Z.If we adopt Z_1={0} then apparently everything is OK,
Isn't it?Alireza Abdollahi
===
> A quick question: Is every
?ite 2-generated abelian group, i.e. G =>> < x, y >,
isomorphic to Z_a x Z_b for some integers a and b? It seems>>
so (since such a G can be displayed in a lattice just like one
would>> use to visually represent Z_a x Z_b, and there must be
relations of>> the form mx=0 for some m, ny=0 for some n), but
I'm not 100% certain>> one way or the other.> BDG>>If you
mean G is isomorphic to the direct sum of at most
two>non-trivial cyclics the answer is yes.>>If you mean G is
isomorphic to the direct sum of _exactly_ two>non-trivial
cyclics then the answer is no: Z_5 = < 2, 3 >.But Z_5 = Z_5 x
Z_1, so the answer to his question is still yes.An even more
trivial special case is < 1, 1 > = Z_1 x Z_1.Derek Holt.>One
way to prove the former is to note that G is the homorphic
image of>Z x Z and the kernel is of the form K x H with K <=
Z x {0} and >H <= {0} x Z.>>-- >Paul Sperry>Columbia, SC
(USA)
===
> A quick question: Is every ?ite 2-generated
abelian group, i.e. G =>> < x, y >, isomorphic to Z_a x Z_b
for some integers a and b? It seems>> so (since such a G
can be displayed in a lattice just like one would>> use to
visually represent Z_a x Z_b, and there must be relations of>
>> the form mx=0 for some m, ny=0 for some n), but I'm not
100% certain>> one way or the other.> BDG>>If
you mean G is isomorphic to the direct sum of at most two>
>non-trivial cyclics the answer is yes.>>If you mean G is
isomorphic to the direct sum of _exactly_ two>non-trivial
cyclics then the answer is no: Z_5 = < 2, 3 >.> But Z_5 =
Z_5 x Z_1, so the answer to his question is still yes.Or no.
(As far as I'm concerned, Z_1 is trivial.)[...]-- Paul
SperryColumbia, SC (USA)
===
> A quick question: Is every
?ite 2-generated abelian group, i.e. G => < x, y >,
isomorphic to Z_a x Z_b for some integers a and b? It seems>>
>> so (since such a G can be displayed in a lattice just like
one would> use to visually represent Z_a x Z_b, and there
must be relations of> the form mx=0 for some m, ny=0 for
some n), but I'm not 100% certain> one way or the
other.>>If you mean G is isomorphic to the direct sum of at
most two>>non-trivial cyclics the answer is yes.>>If you
mean G is isomorphic to the direct sum of _exactly_ two>>
>non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.>>
But Z_5 = Z_5 x Z_1, so the answer to his question is still
yes.>Or no. (As far as I'm concerned, Z_1 is trivial.)The
answer to his question as he asked it is ?yes'; you're
addinga condition that wasn't part of the
question.Brian
===
>> A quick question: Is every ?ite
2-generated abelian group, i.e. G => < x, y >, isomorphic
to Z_a x Z_b for some integers a and b? It seems> so
(since such a G can be displayed in a lattice just like one
would> use to visually represent Z_a x Z_b, and there
must be relations of> the form mx=0 for some m, ny=0 for
some n), but I'm not 100% certain> one way or the
other.> BDG>>If you mean G is isomorphic to
the direct sum of at most two>>non-trivial cyclics the
answer is yes.>>If you mean G is isomorphic to the
direct sum of _exactly_ two>>non-trivial cyclics then the
answer is no: Z_5 = < 2, 3 >.> But Z_5 = Z_5 x Z_1, so
the answer to his question is still yes.>Or no. (As far as
I'm concerned, Z_1 is trivial.)But the answer to *his*
question is yes (since 1 is an integer).Derek Holt.
===
Let A
and B be nonempty subsets of R which are bounded above.
Alsode?e the following sets of upper bounds for A and B:S_A
: = {s in R : forall a in A, a <= s}S_B : = {s in R : forall
b in B, b <= s}I want to ?d formal (rigorous) justi?ation
for the following: B subseteq A --> S_A subseteq S_BIn
words: If B is a subset of A, then the set of upper bounds
for Ais a subset of the set of upper bounds for B.I can
easily see that this is true by drawing a picture, but I
wouldlike to have formal justi?ation, starting from the
de?tions ofsubset and upper bound. Any takers? thanks.
===
>
Let A and B be nonempty subsets of R which are bounded above.
Also> de?e the following sets of upper bounds for A and B:>
S_A : = {s in R : forall a in A, a <= s}> S_B : = {s in R :
forall b in B, b <= s}> I want to ?d formal (rigorous)
justi?ation for the following:> B subseteq A --> S_A
subseteq S_B> In words: If B is a subset of A, then the
set of upper bounds for A> is a subset of the set of upper
bounds for B.> I can easily see that this is true by
drawing a picture, but I would> like to have formal
justi?ation, starting from the de?tions of> subset and
upper bound. Any takers? thanks.Let x be a member of S_A. By
de?ition of S_A we have a <= x for all elements a of A. As B
is a subset of A, every element b of B is also an element of
A, and therefore b <= x. So for every element b of B we have
b <= x, which means by de?ition of S_B that x is an element
of S_B. As x was an arbitrary member of S_A, we have shown
that every element of S_A is also an element of S_B, which
means that S_A is a subset of S_B. By the way, the upper
bound is irrelevant. Every relation between a and s in the
de?itions of S_A and S_B would work exactly in the same
way.
===
I have a question concerning ?ite-state machines
(FSM). I need toshow that no FSM with input and output set
{0, 1} outputs a 1 wheneverthe number of ones and zeros input
are equal and 0 otherwise.In my attempt of constructing such
an FSM, I ran into a problem. Itseems that the number of
states required for a particular input stringis not the same
for all input strings. I then ?ured the proof wouldbe by
contradiction involving the size of the set of states of
theFSM. However, I haven't had luck in developing any
rigorous argumentto convince myself of such. Any clues or
suggestions?Bernd
===
> I have a question concerning
?ite-state machines (FSM). I need to> show that no FSM with
input and output set {0, 1} outputs a 1 whenever> the number
of ones and zeros input are equal and 0 otherwise.> In my
attempt of constructing such an FSM, I ran into a problem.
It> seems that the number of states required for a particular
input string> is not the same for all input strings. I then
?ured the proof would> be by contradiction involving the
size of the set of states of the> FSM. However, I haven't had
luck in developing any rigorous argument> to convince myself
of such. Any clues or suggestions?The important thing to
know is how many ahead (or behind) the 1s areon the 0s. At
every stage of reading a string, this is some numbern. Can
the machine be in the same state for two different values
ofn?/olov-- Supplicant: I don't know if I'm alive or
dead.Oracle: So much for Descartes. -- Internet Oracularity
#1151-05
===
> The important thing to know is how many ahead
(or behind) the 1s are> on the 0s. At every stage of reading a
string, this is some number> n. Can the machine be in the same
state for two different values of> n?> /olovApparently not
since each value of n requires a seperate state.However, I
can't entirely convince myself of this. I think theargument
that surges is that the number of states changes fordifferent
values of n. Am I thinking in the right direction?Bernd
===
>
>>The important thing to know is how many ahead (or behind)
the 1s are>>on the 0s. At every stage of reading a string,
this is some number>>n. Can the machine be in the same state
for two different values of>>n?>>/olov> Apparently not
since each value of n requires a seperate state.> However, I
can't entirely convince myself of this. I think the> argument
that surges is that the number of states changes for>
different values of n. Am I thinking in the right direction?>
> BerndThis post is relatively old, so nobody may still be
interested in it. You are indeed thinking in the right
direction. The problem can be proved using the Pumping Lemma
for regular languages. A google search is fruitful,
especially since the ?st entry (for me) was a delightful
poem-turned-lemma!Robert
===
> ...> The
gist of it is that I have a *proof* that an algebraic
integer> which I called x in the original post has
another algebraic integer> factor that I called y, but
x/y is NOT an algebraic integer.> That's it. That's
the contradiction which I say proves that the ring> of
algebraic integers is incomplete as x/y should be included.>
> ...> Why? Why if x and y are in a ring,
should x/y be in it as well? > Consider the ring of
rational integers.> GC> In a special case, I prove
that a number I'll call x has a factor I'll> here call y,
where both are in the ring of algebraic integers.> Given
that y is a factor x, by de?ition x/y is in the ring of>
algebraic integers, but in the special case it provably is
not.> That's the contradiction.<deleted>It's been hard to
explain.Later I realized that y can't be a factor of x in the
ring ofalgebraic integers, so I'd been going to a higher ring,
like talkingof 2 and 6 being coprime in the ring of evens, and
then saying that 2is a factor of 6, which it is, but in the
ring of integers.Thinking about it this afternoon it seems to
me that the gist of theargument is that given xyz=w, where w
is an integer, you can ?d x, yand z, algebraic integers,
such that neither xy, xz, nor yz, is analgebraic integer.More
speci?ally considering the expressions I've often used,
youhave a_1 a_2 a_3 = 65where neither a_1 a_2, a_2 a_3, nor
a_1 a_3 is an algebraic integer.The a's here given by, from
the expression used in my paper AdvancedPolynomial
Factorization: a^3 + 12a^2 - 65 = 0.Given that you have the
product of two algebraic integers NOT being analgebraic
integer, the ring is incomplete.Note: Any proof that a_1 a_2,
a_2 a_3, or a_1 a_3 is an algebraicinteger would de?itely
cause me problems. And I'm not sayingproof but a proof,
which means that the mathematical argument istrue, not just
that there's some yahoo claiming they have a proof,when
actually they just have a ?argument.James Harris
===
>
Thinking about it this afternoon it seems to me that the gist
of the> argument is that given xyz=w, where w is an integer,
you can ?d x, y> and z, algebraic integers, such that
neither xy, xz, nor yz, is an> algebraic integer.You may want
to check
http://www.jmilne.org/math/CourseNotes/math594f.pdfJ.S.Milne,
Fields and Galois Theory, Page 9, Lemma 1.15: The algebraic
integers form a subring of the complex numbers.
===

[.snip.]>Thinking about it this afternoon it seems to me that
the gist of the>argument is that given xyz=w, where w is an
integer, you can ?d x, y>and z, algebraic integers, such
that neither xy, xz, nor yz, is an>algebraic integer.In
short, you are saying that the algebraic integers do not form
aring, as they are not closed under multiplication.That will
come as a great surprise to Eisenstein, Gauss's
favoritepupil, who proved that if x and y are algebraic
integers, then so arex+y and x*y...>Note: Any proof that a_1
a_2, a_2 a_3, or a_1 a_3 is an algebraic>integer would
de?itely cause me problems. And I'm not saying>proof but a
proof, which means that the mathematical argument is>true, not
just that there's some yahoo claiming they have a proof,>when
actually they just have a ?argument.Amazing. You have
just described perfectly most of your arguments(that's some
yahoo claiming they have a proof, when actually theyjust have
a ?argument).Recall: a complex number c is an ALGEBRAIC
INTEGER if and only ifthere exists a monic polynomial f(x)
with integer coef?ients suchthat f(c)=0.Recall also that by
subring of C we mean a nonempty set R of complexnumbers
with the following properties:(i) 0 is in R.(ii) 1 is in
R.(iii) If a and b are in R, then a-b is in R (usual complex
subtraction).(iv) If a and b are in R, then a*b is in R
(usual complex multiplication).Finally, let R be a subring of
C. We say that the additive group of Ris ?itely generated
to mean that there exist a ?ite number ofelements of R,
a_1,...,a_m, such that every element of R may bewritten in at
least one way asc_1*a_1 + ... + c_m*a_m, where c_1,...,c_m are
(rational) integers.THEOREM. The following are equivalent for
a complex number a (that is,if one of them is true, then they
are all true; if one of them isfalse, then they are all
false): (i) a is an algebraic integer. (ii) The additive
group of the ring Z[a] (the smallest ring contained in the
complex numbers which contains all integers and contains a;
it consists of all polynomial expressions in a with
coef?ients in Z) is ?itely generated.(iii) a is a member of
some subring of the complex numbers whose additive group is
?itely generated. (iv) There exists a ?itely generated
additive subgroup of C, called A, with the property that aA
is contained in A.Proof. To prove the equivalence, we show
that if (i) holds, then (ii)holds; that if (ii) holds, then
(iii) holds; that if (iii) holds, then(iv) holds; and that if
(iv) holds, then (i) holds. That is, we provethat
(i)->(ii)->(iii)->(iv)->(i).(i)->(ii) Assume that a is an
algebraic integer. Then thereexists a monic polynomial with
integer coef?ients, f(x), such thatf(a)=0; letf(x) = x^n +
a_{n-1}*x^{n-1} + ... + a_1*x + a_0,with a_i an integer. By
substituting a^n = -(a_{n-1}*a^{n-1} + ... + a_1*a + a_0)
into anyexpression of the formb_m*a^m + ... b_1*a + b_0, with
b_i integers, we see that everyelement of Z[a] may be written
as a polynomial expression in a ofdegree strictly less than n
and integer coef?ients. That is,Z[a] = {c_{n-1}*a^{n-1} + ...
+ c_1*a + c_0 : c_i integers}.As a group, Z[a] is generated by
1, a, a^2, ... , a^{n-1}. Therefore,Z[a] is ?itely generated,
proving (ii).(ii) -> (iii). Assume that Z[a] has ?itely
generated additivegroup. Then Z[a] is a subring of C which
contains a, and with theproperty that its additive group is
?itely generated. Thus, if (ii)holds, then (iii) must hold
as well.(iii)->(iv) Assume there is a subring R of the
complex numbers, withthe properties that a is in R, and the
additive group of R is ?itelygenerated. Since R is a ring,
the product of any two elements of R must be inR. Therefore,
aR = {a*r : r in R} is a subset of R (since both a and rare
in R); thus, A=R shows that (iv) is true.Finally, we prove
that if (iv) holds, then (i) holds.Assume that (iv) holds.
Let A be a ?itely generated additivesubgroup of C such that
a*A is contained in A. Let x_1,...,x_m beelements of A such
that every element of A may be written asc_1*x_1 + ... +
c_m*x_m, with c_1,...,c_m integers.Furthermore, we may assume
that none of the x_i are equal to 0.Our assumption is that for
every choice of integers c_1,...,c_m,(c_1*x_1 + ... +
c_m*x_m)*a is in A.For each i, a*x_i lies in A; therefore,
a*x_i may be written asabove. We write:a*x_1 = b_{11}*x_1 +
b_{21}*x_2 + ... + b_{m1}*x_ma*x_2 = b_{12}*x_1 + b_{22}*x_2
+ ... + b_{m2}*x_m . . .a*x_m = b_{1m}*x_1 + b_{2m}*x_2 + ...
+ b_{mm}*x_m.Let M be the matrix ( b_{11} b_{21} ... b_{m1} )
( b_{12} b_{22} ... b_{m2} ) ( . . . . ) M= ( . . . . ) ( . .
. . ) ( b_{1m} b_{2m} ... b_{mm} )this is an m x m matrix with
integer coef?ients, such that (x_1) (x_1) (x_2) (x_2) ( . ) (
. )a*( . ) = M( . ) ( . ) ( . ) (x_m) (x_m)If I is the m x m
identity matrix, this means that (x_1) (x_2) ( . )(a*I - M)(
. ) ( . ) (x_m)is the zero vector. Since none of the x_i are
equal to 0, it followsthat a is an eigenvalue of M and the
vector of the x_i is aneigenvector of M. That means that
det(aI-M) = 0. Equivalently, thecharacteristic polynomial of
M, det(xI-M), has a as a root. Since Mhas integer
coef?ients, the characteristic polynomial of M is monicand
has integer coef?ients. Therefore, a is the root of a
monicpolynomial with integer coef?ients. Therefore, a is an
algebraicinteger, proving that (i) holds whenever (iv) holds.
QEDCorollary. Let a and b be algebraic integers. Then a+b,
a-b, and a*bare all algebraic integers.Proof. Consider the
ring Z[a,b], the smallest subring of C whichcontains a, b,
and all the integers. Note that it contains a+b, a-b,and a*b,
by de?ition of subring.By clause (iii) of the Theorem, in
order to prove that a+b, a-b, anda*b are algebraic integers
it is enough to prove that Z[a,b] has?itely generated
additive group.Since a is an algebraic integer, we know that
a satis?s a monicpolynomial with integer coef?ients,
say:f(x) = x^n + a_{n-1}*x^{n-1} + ... + a_1*x +
a_0;likewise, b satis?s a monic polynomial with integer
coef?ients,say:g(x) = x^m + b_{m-1}*x^{m-1} + ... + b_1*x +
b_0.CLAIM: Z[a,b] is generated, as an additive group, by1, a,
a^2, ..., a^{n-1}, b, ba, ba^2, ..., ba^{n-1},
b^2,....,b^{m-1}a^{n-1}.Note that establishing the claim will
prove the corollary.Every element of Z[a,b] may be written as
a polynomial expression in aand b with integer coef?ients.
Sincea^n = -(a_{n-1}*a^{n-1} + ... + a_1*a + a_0)b^m =
-(b_{m-1}*b^{m-1} + ... + b_1*b + b_0),we may substitute any
occurance of any power of a greater thana^{n-1}, and any
power of b greater than b^{m-1}, by an expressioninvolving
only lesser powers. Therefore, by repeating thesubstitution,
we may write every element of Z[a,b]
as:c_{n-1,m-1}*a^{n-1}*b^{m-1} + c_{n-1,m-2}*a^{n-1}*b^{m-2}
+ ... + + c_{n-1,0}*a^{n-1} + c_{n-2,m-1}*a^{n-2}b^{m-1} +
... + a_{0,0}*1.This proves the claim. This proves the
Corollary. QEDThis proves that if a and b are algebraic
integers, then so is a*b.This de?itely cause[s you]
problems.
===
====================================

===
============Why do you take so much trouble to expose
such a reasoner as Mr. Smith? I answer as a deceased friend of
mine used to answer on like occasions - A man's capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
?ures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde?ite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
Augustus de
Morgan
===
=======================================

===
=========Arturo
Magidinmagidin@math.berkeley.edu
===
[snip latest sideshow]>
Note: Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an
algebraic> integer would de?itely cause me problems. And I'm
not saying> proof but a proof, which means that the
mathematical argument is> true, not just that there's some
yahoo claiming they have a proof,> when actually they just
have a ?argument.It's already been done by several
posters (not yahoos). Wake up and smell the coffee you
moron. You arehereby banned from the sandbox. Go climb back
under the rock whence you came.--There are two things you
must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
> ...>
> The gist of it is that I have a *proof* that an
algebraic integer> which I called x in the original
post has another algebraic integer> factor that I
called y, but x/y is NOT an algebraic integer.>
That's it. That's the contradiction which I say proves that
the ring> of algebraic integers is incomplete as x/y
should be included.> ...> Why?
Why if x and y are in a ring, should x/y be in it as well? >
> Consider the ring of rational integers.> GC>
> In a special case, I prove that a number I'll call x has a
factor I'll> here call y, where both are in the ring of
algebraic integers.> Given that y is a factor x, by
de?ition x/y is in the ring of> algebraic integers, but in
the special case it provably is not.> That's the
contradiction.> <deleted>> It's been hard to explain.>
Later I realized that y can't be a factor of x in the ring
of> algebraic integers, so I'd been going to a higher ring,
like talking> of 2 and 6 being coprime in the ring of evens,
and then saying that 2> is a factor of 6, which it is, but in
the ring of integers.> Thinking about it this afternoon it
seems to me that the gist of the> argument is that given
xyz=w, where w is an integer, you can ?d x, y> and z,
algebraic integers, such that neither xy, xz, nor yz, is an>
algebraic integer.Nope. That's not it. > More speci?ally
considering the expressions I've often used, you> have> a_1
a_2 a_3 = 65> where neither a_1 a_2, a_2 a_3, nor a_1 a_3 is
an algebraic integer.They are as is easily shown.> The a's
here given by, from the expression used in my paper Advanced>
Polynomial Factorization:> a^3 + 12a^2 - 65 = 0.> Given
that you have the product of two algebraic integers NOT being
an> algebraic integer, the ring is incomplete.> Note: Any
proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic>
integer would de?itely cause me problems. And I'm not
saying> proof but a proof, which means that the
mathematical argument is> true, not just that there's some
yahoo claiming they have a proof,> when actually they just
have a ?argument.Oh well, that was wrong.However, I ?d
it interesting that my earlier post received some replies.
James Harris
===
>> [...]>><deleted>It's been hard to
explain.>>[...]>>Thinking about it this afternoon it seems to
me that the gist of the>argument is that given xyz=w, where w
is an integer, you can ?d x, y>and z, algebraic integers,
such that neither xy, xz, nor yz, is an>algebraic
integer.Yes, that _is_ hard to explain. Why? Because it's
simply _false_. Easily shown to be false, alsowell-known to
be false. When you say this is hard to explain it's_exactly_
like if you'd said you'd had a hard time explaining
yourtheory, then you realized the gist of it was that 2 + 2 =
5.(Well, not exactly like that - the difference is that _you_
knowenough math to see that 2 + 2 = 5 is false. But to
someonewho actually knows something about this stuff it's
exactlythe same. You talk a lot about how we all lie about
thesethings. When you do that the impression you make
isexactly like someone complaining about people lying,saying
that 2 + 2 = 4. Honest.)>More speci?ally considering the
expressions I've often used, you>have>> a_1 a_2 a_3 =
65>>where neither a_1 a_2, a_2 a_3, nor a_1 a_3 is an
algebraic integer.>>The a's here given by, from the
expression used in my paper Advanced>Polynomial
Factorization:>> a^3 + 12a^2 - 65 = 0.>>Given that you have
the product of two algebraic integers NOT being an>algebraic
integer, the ring is incomplete.>>Note: Any proof that a_1
a_2, a_2 a_3, or a_1 a_3 is an algebraic>integer would
de?itely cause me problems. And I'm not saying>proof but a
proof, which means that the mathematical argument is>true, not
just that there's some yahoo claiming they have a proof,>when
actually they just have a ?argument.>James
Harris************************David C. Ullrich
===
with this
informal admission,I hereby (again) declare my solemn
intentionnot to participate in heckling monsieur Harris, any
more, ifI can possibly help it. even if he can't help
instigating it. of course, there could be an ulterior motive
to all of this,as it's so hard to believe! > Note: Any
proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic>
integer would de?itely cause me problems. And I'm not
saying> proof but a proof, which means that the
mathematical argument is> true, not just that there's some
yahoo claiming they have a proof,> when actually they just
have a ?argument.> Oh well, that was wrong.--UN
HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...La
Troi Phases d'Exploitation de la Protocols des Grises de
Kyoto:(FOSSILISATION [McCainanites?]
(TM/sic))/BORE/GUSH/NADIR @
http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/
bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de
COUP D'ETAT, 3/30/81
===
> And I'm not saying> proof but a
proof, which means that the mathematical argument is> true,
not just that there's some yahoo claiming they have a proof,>
when actually they just have a ?argument.> James
HarrisJSH has lots of experience in this area.
===
I was
leading my class through related rates yesterday, and one of
the students stunned me by asking for a good qualitative
description of what a derivative means. (He asked it in a
more basic form, which is why I was stunned.)Our textbook,
Larson/Hostetler/Edwards, really doesn't do a very good job
of explaining what a derivative _means_, now that I look at
it with fresh eyes.I pointed him at /Calculus Made Easy/,
which I think is pretty good at that. But is there a Web
resource I can give to all my students? Once again, the issue
is not how to ?d derivatives but how to understand them.--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/You ?d yourself amusing,
Blackadder.I try not to ?the face of public
opinion.
===
A derivative is the rate of change of a
function. If you graph the functionthe derivative is the
slope of the graph.A good example is that the derivative of
position is velocity, thederivative of velocity is
acceleration.>> I was leading my class through related rates
yesterday, and one of> the students stunned me by asking for
a good qualitative description> of what a derivative means.
(He asked it in a more basic form, which> is why I was
stunned.)>> Our textbook, Larson/Hostetler/Edwards, really
doesn't do a very> good job of explaining what a derivative
_means_, now that I look at> it with fresh eyes.>> I pointed
him at /Calculus Made Easy/, which I think is pretty good> at
that. But is there a Web resource I can give to all my
students?>> Once again, the issue is not how to ?d
derivatives but how to> understand them.>> -- > Stan Brown,
Oak Road Systems, Cortland County, New York, USA>
http://OakRoadSystems.com/> You ?d yourself amusing,
Blackadder.> I try not to ?the face of public
opinion.
===
> A derivative is the rate of change of a
function. If you graph thefunction> the derivative is the
slope of the graph.> A good example is that the derivative of
position is velocity, the> derivative of velocity is
acceleration.derivative is often interpreted as a rate of
change. You mean well, weknow, and I may be wrong but I think
what Stan has in mind is he is havingtrouble explaining,
satisfactorilly, and at a certain intuitive/qualitativelevel,
what a derivative _means_. Textbooks are good at giving
veryunambiguous de?itions, but not so good many times at
explaining thesethings at a more basic level.Although I don't
have the most current edition of Larson, et al, I do havethe
5th ed. handy and if the one Stan has is anything like it, I
understandexactly what he is talking about. In my edition,
the way the authorsattempt to put a derivative in layman's
terms, so to speak, is fairlyconfusing. They begin this
attempt somewhere in ch. 1 with a sectionentitled the
tangent line problem then they drop that discussion
entirely,moving on to other things, and later in chapter 2 I
believe, they pick backup with that discussion. At face value,
this seems not that unreasonablesince the problem of ?ding a
derivitive *is* essentially the same as?dling the slope of a
tangent line (hence, the tangent line itself) but ifyou have
had to experience reading the actual material, in the
mannerpresented, you may agree that the authors were not very
effective in thisattempt to explain a derivative. Yeah, they
do a good job of explaing howthe secant line approaches the
tangent line, etc, etc, and how the limitingposition of the
secant line is the tangent line, etc. but this reallydoesn't
address what a derivative _means_. It just addresses what one
_is_.so why _are_ we interested in tangent lines? What do
_they_ mean?All that said, it's not the easiest thing in the
world to explain what aderivative _means_ as Stan has
acknowledged (and Larson, et al.demonstrates). But here are a
couple of places you may want to look, if youhaven't
already:http://karlscalculus.org/calculus.htmlKarl goes into
extreme vivid detailed explanations, in layman's
terms,(borderline children's talk, in some places, if that
kind of stuff doesn'tbother you or your students). All the
rigor is there, but buried deepwithin lengthy, informal
explanations in story format. It may be necessaryto start his
story with the limit
chapter.http://www.sosmath.com/calculus/diff/der00/der00.
htmlThis is an excellent site that is much more concise than
karl's, geared tomore of a grown up audience, if you will.
They discuss both commoninterpretations, that of a physical
rate of change and that of the(geometrical) slope of a
tangent line. but they do a much better job at it,IMO, than
larson, et. al.HTH,-- Darrell> I was leading my class
through related rates yesterday, and one of> the students
stunned me by asking for a good qualitative description> of
what a derivative means. (He asked it in a more basic form,
which> is why I was stunned.)>> Our textbook,
Larson/Hostetler/Edwards, really doesn't do a very> good
job of explaining what a derivative _means_, now that I look
at> it with fresh eyes.>> I pointed him at /Calculus
Made Easy/, which I think is pretty good> at that. But is
there a Web resource I can give to all my students?>>
Once again, the issue is not how to ?d derivatives but how
to> understand them.>> -- > Stan Brown, Oak Road
Systems, Cortland County, New York, USA>
http://OakRoadSystems.com/> You ?d yourself amusing,
Blackadder.> I try not to ?the face of public
opinion.>>
===
>> A derivative is the rate of change of a
function. If you graph the>function>> the derivative is the
slope of the graph.>> A good example is that the derivative
of position is velocity, the>> derivative of velocity is
acceleration.>>derivative is often interpreted as a rate of
change. You mean well, we>know, and I may be wrong but I
think what Stan has in mind is he is having>trouble
explaining, satisfactorilly, and at a certain
intuitive/qualitative>level, what a derivative _means_. You
interpret rightly. Since I'm teaching calculus, I should
certainly hope I know what a derivative is!
:-)>http://karlscalculus.org/calculus.html>Karl goes into
extreme vivid detailed explanations, in layman's
terms,>(borderline children's talk, in some places, if that
kind of stuff doesn't>bother you or your students).Nope. When
the student was explaining his dif?ulty, he said, Talk to me
like I'm
4.>http://www.sosmath.com/calculus/diff/der00/der00.html>
This is an excellent site that is much more concise than
karl's, geared to>more of a grown up audience, if you will.
Every time I look at sosmath I block it out mentally because I
don't like the ad for calc101.com, which is essentially Pay
us and we'll do your homework. But the page itself does look
good.-- Stan Brown, Oak Road Systems, Cortland County, New
York, USA http://OakRoadSystems.com/You ?d yourself
amusing, Blackadder.I try not to ?the face of public
opinion.
===
I'm trying to derive the properties of
determinants, as opposed to just provingthem. Hence I'm
pretending ignorance of determinants, and trying to
computeeigenvalues by primitive methods. In case that kind of
thing doesn't interestyou, you've been warned. This is where
I'm at:Suppose A is a square matrix and I is the identity
matrix of the same order. LetX_0 be (A-xI), where x is a
variable. Assuming X_0[1,1]<>0, there are rowoperations that
convert X_0 into a matrix X_1 with X_1[1,j]=0 for
j>1.Continuing in the same way you eventually obtain a matrix
X_n that can beconverted into I provided none of the entries
on its main diagonal is zero.Hence every eigenvalue of A is a
root of numer(X_n[1,1])*...*numer(X_n[n,n]),where numer(r)
means the polynomial p with r=p/q for p and q relatively
prime.Now, it turns out that numer(X_n[n,n]) is the
characteristic polynomial of A(recall that I don't of?ially
know about the characteristic polynomial at thispoint), so
that none of the numer(X_n[i,i])'s with i<n has an eigenvalue
of Afor a root.Can anybody explain this fact without invoking
a bunch of involved theory?--Peace,EJ
===
> I'm trying to
derive the properties of determinants, as opposed to just
proving> them. Hence I'm pretending ignorance of
determinants, and trying to compute> eigenvalues by
primitive methods.I don't think computing eigenvalues is a
property of determinants.So this might not be the best way
to investigate determinants.I think of the properties of a
determinant as being multilinear andskew symmetric.One
approach at viewing determinants is to consider it as
thevolume of the parallelpiped formed by the row vectors of
the matrix.Another approach is to consider it as a
differential n-form.A third approach is to see how
determinants pop-up when tryingto solve n equations in n
unknowns using Kramer's rule. I liketo use Kramer's rule in
the form D * x_i = c_i since it istrue even when D is zero
(here D is the determinant).> In case that kind of thing
doesn't interest> you, you've been warned. This is where I'm
at:> Suppose A is a square matrix and I is the identity
matrix of the same order. Let> X_0 be (A-xI), where x is a
variable. Assuming X_0[1,1]<>0, there are row> operations
that convert X_0 into a matrix X_1 with X_1[1,j]=0 for j>1.>
Continuing in the same way you eventually obtain a matrix X_n
that can be> converted into I provided none of the entries on
its main diagonal is zero.> Hence every eigenvalue of A is a
root of numer(X_n[1,1])*...*numer(X_n[n,n]),> where numer(r)
means the polynomial p with r=p/q for p and q relatively
prime.> Now, it turns out that numer(X_n[n,n]) is the
characteristic polynomial of A> (recall that I don't
of?ially know about the characteristic polynomial at this>
point), so that none of the numer(X_n[i,i])'s with i<n has an
eigenvalue of A> for a root.> Can anybody explain this fact
without invoking a bunch of involved theory?Let A be the
square matrix [a,b;c,d].Let u = (x,y) be an eigenvector with
eigenvalue k.Then, A*u = k*u.Thus, a*x + b*y = k*x c*x + d*y
= k*y.That is, (a-k)*x + b*y = 0 c*x + (d-k)*y = 0Note that
(0,0) is not considered an eigenvector.Thus, either x or y is
non-zero.I have two equations in two unknowns x and y, with
paramenter k.Solve for the non-zero unknown, say x, using the
method youlearned in high school.You get: (a-k)*(d-k)*x -
b*c*x = 0.Since x is not zero, you get (a-k)*(d-k) - b*c =
0.This is just the determinant since the determinantpops up
when you solve linear equations for the unknowns.This doesn't
seem to be what you are looking for though.-- Bill Hale
===
>
I'm trying to derive the properties of determinants, as
opposed to just proving> them. Hence I'm pretending
ignorance of determinants, and trying to compute>
eigenvalues by primitive methods. In case that kind of thing
doesn't interest> you, you've been warned. This is where I'm
at:> Suppose A is a square matrix and I is the identity
matrix of the same order. Let> X_0 be (A-xI), where x is a
variable. Assuming X_0[1,1]<>0, there are row> operations
that convert X_0 into a matrix X_1 with X_1[1,j]=0 for j>1.>
Continuing in the same way you eventually obtain a matrix X_n
that can be> converted into I provided none of the entries on
its main diagonal is zero.> Hence every eigenvalue of A is a
root of numer(X_n[1,1])*...*numer(X_n[n,n]),> where numer(r)
means the polynomial p with r=p/q for p and q relatively
prime.> Now, it turns out that numer(X_n[n,n]) is the
characteristic polynomial of A> (recall that I don't
of?ially know about the characteristic polynomial at this>
point), so that none of the numer(X_n[i,i])'s with i<n has an
eigenvalue of A> for a root.> Can anybody explain this fact
without invoking a bunch of involved theory?Have a look at
the book Linear Algebra Done Right by Sheldon Axler,for a
consistently determinant-free approach to basic linear
algebra. The book's homepage is
at:http://math.sfsu.edu/axler/LADR.htmlHTH,Felix.P.S: Note
that Axler allows the zero vector to be an eigenvector,which
is quite non-standard usage. Apart from that, the book is
anexcellent presentation of classic material in an elegant
way.P.P.S[Full Disclousre]: Personally, I think that
determinants are justok.
===
>URGENT PLEASE READ>Our friend
recently suffered a heart attack at 46 years old. >He is a
regularly working contracted employee (self employed) who
will be unable to work for 6 or 7 weeks from the time of the
heart attack with absolutely no income.>He has paid taxes all
of his life but is not eligible for government assistance>of
any type as his wife is working earning $1400.00 monthly.
This obviously does not support their family including their
2 kids but does exclude them from any agency help. >Mrs.
Sheffer was even told that if she were a crack head or
unemployed, they could then offer assistance.>They have now
missed their rent, had to give up their car, had to add $300
monthly in medications and the quality of their life has
sunken to new lows. >This is a kind and considerate hard
working family who has fallen on temporary hard times and
could sure use a kindly hand before it's too late.>We are
trying to prove that people DO still care even if our
government doesn't.>We have set up a donation PO Box at the
address below and could sure use your help!>>Donna Sheffer,
3216 Eglinton ave. east, Scarborough, Ontario, Canada, M1J
2H6>Your generous donation $1, $2, $5, $10, $20 whatever
would be greatly appreciated.>Help restore a little faith
and save a real family.>God Bless>See the angioplasty image
here. http://www.otwebdesign.com/apimage.jpg>>This is no
scam>Please help!! I faxed you a $100 bill ! Do you really
think we're all idiots?
===
[snip by Stan]>>This is no
scam>>Please help!!> I faxed you a $100 bill !>> Do you
really think we're all idiots?Some of we apparently are.
You found it necessary to repost the whole scam, including
the Web site and the postal address, thus giving the spammer
another shot at our eyes.-- Stan Brown, Oak Road Systems,
Cortland County, New York, USA http://OakRoadSystems.com/You
?d yourself amusing, Blackadder.I try not to ?the face
of public opinion.
===
>> [snip by Stan]>>This is no scam>
>>Please help!!> I faxed you a $100 bill !>> Do
you really think we're all idiots?>> Some of we apparently
are. You found it necessary to repost the> whole scam,
including the Web site and the postal address, thus> giving
the spammer another shot at our eyes.However, the tragedy may
be real, and the arrogantreplies thus all the more hurtful.
But we cannotknow for sure: hence paranoia
prevails...Christian--In the paranoid-schizoid position ...
there is no concernfor the consequences of actions, and
certainly not for theireffects on others. At most there is
the operation of a talionmorality, a cold exchange of goods
and evils.- Eugene Victor Wolfenstein:
?Psychoanalytic-Marxism, Groundwork'
===
I need help with
proving the following statement: When the vertex of an angle
is in the exterior of a circle, the measureof the angle is
one-half the difference of the measures of the
interceptedarcs.Any hints or directions to look in would be
appreciated.
===
>I need help with proving the following
statement:>> When the vertex of an angle is in the exterior
of a circle, the measure>of the angle is one-half the
difference of the measures of the intercepted>arcs.>>Any
hints or directions to look in would be appreciated.Are we to
assume that the sides of the angle are tangent to the
circle?If so, try drawing radii to the two points of
tangency. You will then have two right angles; you will also
have an interior angle of the circle (which has a simple
relationship to both arc lengths). If you draw the diagonal
of the quadrilateral from the center of the circle to the
vertex of the external angle, you will have two congruent
right triangles.I think if you try that and do some playing
around you should see a method of attack. But I should
disclose that I haven't completed the proof myself, so I'm
telling you how I _would_ start and not how I _did_ start.--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/Walrus meat as a diet is less
repulsive than seal.
===
> When the vertex of an angle is in
the exterior of a circle, the measure> of the angle is
one-half the difference of the measures of the intercepted>
arcs.>> Any hints or directions to look in would be
appreciated.>What happens if the angle's line miss the circle
or just one passes thruor touches the circle?
===
In a
particular retail store the average price of a typical
two-seater sofahas been 200 and average sales were 1000 per
year. This year the store hasrun a half price sale on such
sofas and found that sales increased by 50%.Question: Find a
hyperbolic model for demand in terms of price.
===
Suppose
that in a triangle ABC a,b,c are the sides ,2s=a+b+c , r=
radius of incircle , R= radius of circumcircle.Denote D=
sqrt{s^2-3r(4R+r)} .Prove that(1) D =< max{a,b,c} -
min{a,b,c} =< 2D/sqrt{3} .If a =< b =< c , prove or disprove
the inequalities: (2) min{b-a,c-b} =< D/sqrt{3} ,(3) D =<
2c-(a+b) =< 2D , |2b-(a+c)| =< D , D=< (b+c)-2a =< 2D
.
===
=To: geometry-college@moderators.isc.org===Suppose that
in a triangle ABC a,b,c are the sides , r= radius of incircle
, R= radius of circumcircle.Denote D= sqrt{s^2-3r(4R+r)}
.Prove that(1) D =< max{a,b,c} - min{a,b,c} =< 2D/sqrt{3} .If
a =< b =< c , prove or disprove the inequalities: (2)
min{b-a,c-b} =< D/sqrt{3} ,(3) D =< 2c-(a+b) =< 2D ,
|2b-(a+c)| =< D , D=< (b+c)-2a =< 2D .
===
====Let G be an
in?ite abelian group such that every proper subgroup ofG is
?ite. Show that G is isomorphic to Z(p^infty),
whereZ(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }I've
been looking at this one for a week and nothing. Can
somebodyhelp! It's driving me crazy.
===
>Let G be an in?ite
abelian group such that every proper subgroup of>G is ?ite.
Show that G is isomorphic to Z(p^infty), where>>Z(p^infty) =
{ [m/p^n] in Q/Z | m in Z, n in N }>>I've been looking at
this one for a week and nothing. Can somebody>help! It's
driving me crazy.You say you have gotten nothing, so some
comments, which may helpgetting you started; I haven't
thought it through in detail, so Icould be wrong in some
things. But it seems right to me:First, note that G is a
torsion group; as such, it can be decomposedinto a direct sum
of its p-parts, that isG = G_2 oplus G_3 oplus G_5 oplus
...where G_p = {g in G; g^{p^m}=e for some m>0}.Now, let's
consider only the non-trivial p-parts; so we haveG = G_{p_1}
oplus G_{p_2} oplus ...Note that there cannot be an
in?ite number of non-trivial p-parts;if there were, then the
subgroup G_{p_2}oplus ... would be a properin?ite subgroup.
SoG= G_{p_1} oplus G_{p_2} oplus ... oplus G_{p_m};since
G is in?ite, at least one of the G_{p_i} is in?ite. Butthen,
G_{p_i} itself is an in?ite subgroup of G, nontrivial.
HenceG=G_{p_i}. That is, G is an in?ite p-group.Now consider
the quotients G/pG, pG/p^2G, p^2G/p^3G,...these quotients are
vector spaces over F_p, hence they have awell-de?ed
dimension. Prove that the dimensions are non-increasing,able
to conclude (a) that there exists m>0 such that
p^mG/p^{m+1}/G iscyclic; and then that (b) all of them are
cyclic; using the fact thatG does not have in?ite proper
subgroups.Once you have that all of them are cyclic, it
should not be hard towrite down an
isomorphism.
===
==================================

===
=============It's not denial. I'm just very
selective about what I accept as reality. --- Calvin
(Calvin and
Hobbes)
===
=====================================

===
==========Arturo
Magidinmagidin@math.berkeley.edu
===
Hmmm...I think I follow
you, but I still don't see the actual isomorphism...>Let G
be an in?ite abelian group such that every proper subgroup
of>G is ?ite. Show that G is isomorphic to Z(p^infty),
where>>Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }>
>>I've been looking at this one for a week and nothing. Can
somebody>help! It's driving me crazy.> You say you have
gotten nothing, so some comments, which may help> getting you
started; I haven't thought it through in detail, so I> could
be wrong in some things. But it seems right to me:> First,
note that G is a torsion group; as such, it can be
decomposed> into a direct sum of its p-parts, that is> G =
G_2 oplus G_3 oplus G_5 oplus ...> where G_p = {g in
G; g^{p^m}=e for some m>0}.> Now, let's consider only the
non-trivial p-parts; so we have> G = G_{p_1} oplus
G_{p_2} oplus ...> Note that there cannot be an in?ite
number of non-trivial p-parts;> if there were, then the
subgroup G_{p_2}oplus ... would be a proper> in?ite
subgroup. So> G= G_{p_1} oplus G_{p_2} oplus ...
oplus G_{p_m};> since G is in?ite, at least one of the
G_{p_i} is in?ite. But> then, G_{p_i} itself is an in?ite
subgroup of G, nontrivial. Hence> G=G_{p_i}. That is, G is an
in?ite p-group.> Now consider the quotients G/pG, pG/p^2G,
p^2G/p^3G,...> these quotients are vector spaces over F_p,
hence they have a> well-de?ed dimension. Prove that the
dimensions are non-increasing,> able to conclude (a) that
there exists m>0 such that p^mG/p^{m+1}/G is> cyclic; and
then that (b) all of them are cyclic; using the fact that> G
does not have in?ite proper subgroups.> Once you have that
all of them are cyclic, it should not be hard to> write down
an isomorphism.>

===
===========================================

===
====> It's not denial. I'm just very selective about>
what I accept as reality.> --- Calvin (Calvin and Hobbes)>

===
============================================

===
===> Arturo Magidin> magidin@math.berkeley.edu===>
Let G be an in?ite abelian group such that every proper
subgroup of> G is ?ite. Show that G is isomorphic to
Z(p^infty), where> Z(p^infty) = { [m/p^n] in Q/Z | m in Z,
n in N }> I've been looking at this one for a week and
nothing. Can somebody> help! It's driving me crazy.This may
involve some machinery that you don't have available. You can
easily see that G must be a p-group. There is a
theorem[Kulikov] that says G has a direct summand of the form
Z_(p^n), 1 <= n <= oo. That summand can't be ?ite or else its
complementarysummand would also be ?ite and hence G would be
?ite. So, thesummand is Z_(p^oo). The complementary summand
must be 0 or else G hasan in?ite proper subgroup. Hence G is
(isomorphic to) Z_(p^oo).-- Paul SperryColumbia, SC (USA)
===
>
Let G be an in?ite abelian group such that every proper
subgroup of> G is ?ite. Show that G is isomorphic to
Z(p^infty), where> Z(p^infty) = { [m/p^n] in Q/Z | m in Z,
n in N }But is every proper subgroup of this ?ite?Perhaps
the question was: every proper subgroup of ?ite index?--
Timothy Murphy tel: +353-86-233 6090
===
>>Let G be an
in?ite abelian group such that every proper subgroup of>>G
is ?ite. Show that G is isomorphic to Z(p^infty), where>>
>Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }>>I've
been looking at this one for a week and nothing. Can
somebody>>help! It's driving me crazy.> You say you
have gotten nothing, so some comments, which may help>>
getting you started; I haven't thought it through in detail,
so I>> could be wrong in some things. But it seems right to
me:> First, note that G is a torsion group; as such, it
can be decomposed>> into a direct sum of its p-parts, that
is> G = G_2 oplus G_3 oplus G_5 oplus ...>
where G_p = {g in G; g^{p^m}=e for some m>0}.> Now, let's
consider only the non-trivial p-parts; so we have> G =
G_{p_1} oplus G_{p_2} oplus ...> Note that there
cannot be an in?ite number of non-trivial p-parts;>> if
there were, then the subgroup G_{p_2}oplus ... would be a
proper>> in?ite subgroup. So> G= G_{p_1} oplus G_{p_2}
oplus ... oplus G_{p_m};> since G is in?ite, at least
one of the G_{p_i} is in?ite. But>> then, G_{p_i} itself is
an in?ite subgroup of G, nontrivial. Hence>> G=G_{p_i}. That
is, G is an in?ite p-group.> Now consider the quotients
G/pG, pG/p^2G, p^2G/p^3G,...>> these quotients are vector
spaces over F_p, hence they have a>> well-de?ed dimension.
Prove that the dimensions are non-increasing,>> able to
conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G
is>> cyclic; and then that (b) all of them are cyclic; using
the fact that>> G does not have in?ite proper subgroups.>>
>> Once you have that all of them are cyclic, it should not
be hard to>> write down an isomorphism.>Hmmm...I think I
follow you, but I still don't see the actual
isomorphism...Well, assuming everything I said was correct,
and you have shown thatG is a p-group, and moreover, that
each of p^iG/p^{i+1}G is cyclic,what does that mean? Can you
show that G/p^iG is cyclic for all i>0?Now, take G/pG. This
is cyclic of order p, say with generatorx+pG. Since G/p^2G is
also cyclic, you should be able to see that itis generated by
an element which, when multiplied by p, is equal tox. Call it
x/p. Then you need to ?d one which when multiplied by pis
x/p; call it x/p^2.
Etc.
===
========================================

===
=======It's not denial. I'm just very selective about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=====================================

===
==========Arturo
Magidinmagidin@math.berkeley.edu
===
>Let G be an in?ite
abelian group such that every proper subgroup of>>G is
?ite. Show that G is isomorphic to Z(p^infty), where>>
>Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }>>I've
been looking at this one for a week and nothing. Can
somebody>>help! It's driving me crazy.> You say you
have gotten nothing, so some comments, which may help>>
getting you started; I haven't thought it through in detail,
so I>> could be wrong in some things. But it seems right to
me:> First, note that G is a torsion group; as such, it
can be decomposed>> into a direct sum of its p-parts, that
is> G = G_2 oplus G_3 oplus G_5 oplus ...>
where G_p = {g in G; g^{p^m}=e for some m>0}.> Now, let's
consider only the non-trivial p-parts; so we have> G =
G_{p_1} oplus G_{p_2} oplus ...> Note that there
cannot be an in?ite number of non-trivial p-parts;>> if
there were, then the subgroup G_{p_2}oplus ... would be a
proper>> in?ite subgroup. So> G= G_{p_1} oplus G_{p_2}
oplus ... oplus G_{p_m};> since G is in?ite, at least
one of the G_{p_i} is in?ite. But>> then, G_{p_i} itself is
an in?ite subgroup of G, nontrivial. Hence>> G=G_{p_i}. That
is, G is an in?ite p-group.> Now consider the quotients
G/pG, pG/p^2G, p^2G/p^3G,...>> these quotients are vector
spaces over F_p, hence they have a>> well-de?ed dimension.
Prove that the dimensions are non-increasing,>> able to
conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G
is>> cyclic; and then that (b) all of them are cyclic; using
the fact that>> G does not have in?ite proper subgroups.>>
>> Once you have that all of them are cyclic, it should not
be hard to>> write down an isomorphism.>Hmmm...I think I
follow you, but I still don't see the actual
isomorphism...Probably because the ?al two paragraphs are
complete bollocks... Allthe quotients given will be trivial.I
screwed up; I used the p-powers subgroups instead of the
subgroupsof elements of exponent the p-powers...Here's a
better way to proceed; we know that G is an in?itegroup.
First, consider, pG. If pG were trivial, then G would be
ofexponent p, hence an in?ite dimensional vector space over
F_p, whichclearly has in?ite proper subgroups; so pG is
nontrivial. If pGis ?ite, then G/pG is an in?ite
dimensional vector space over F_p,and you can lift an in?ite
proper subgroup to G. So pG is in?ite,hence pG=G. So G is
divisible (being a p-group and p-divisible).At this point you
could invoke classi?ation theorems for divisible
abeliangroups, which will tell you that G must be a bunch of
copies ofZ_p^{infty}, and thereby conclude that it must be
equal to one ofthem. But here's a way to proceed without
invoking thoseclassi?ation theorems:For each n>0, let G[p^n]
= {g in G : g^{p^n}=e}, the subgroup ofelements of epxonent
p^n. G[p] is nontrivial; let x in G[p] be different from e1.
Since G isp-divisible, there is an element x_2 such that px_2
= x; an elementx_3 such that px_3=x_2; ...; an element x_{n+1}
such that px_{n+1} =x_n.Let H = < x, x_2, x_3,...>It is now
easy to verify that H is nontrivial and in?ite, so H=G. Itis
now also easy to given an isomorphism from Z_{p^{infty}} to G:
sent1/p + Q/Z to x, and 1/p^n +Q/Z to x_n,
n>1.
===
========================================

===
=======It's not denial. I'm just very selective about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=====================================

===
==========Arturo
Magidinmagidin@math.berkeley.edu
===
Can someone explain why
the set of all integers is not a ?ld?
===
> Can someone
explain why the set of all integers is not a ?ld?>What's the
multiplicative inverse of 2? Is it an integer?
===
>Can someone
explain why the set of all integers is not a ?ld?A ?ld is a
set F with two binary operations, usually denoted + and *(sum
and multiplication, respectively). The fact that they
arebinary operations means they are functions from FxF to F
(pairs ofelements to elements). However, it is usual in most
places to usein? notation, and write x+y instead of
+(x,y).The set F, together with the two operations + and *,
must satisfycertain properties before it is called a ?ld:F
must be an abelian group under the operation +. That
means:(1) + must be associative: for all x,y,z in F, (x+y)+z
= x+ (y+z).(2) There exists an element 0 in F such that for
all x in F, x+0 = 0+x = x.(3) For every x in F there exists y
in F such that x+y = y+x = 0. y is called the inverse of x
and is often denoted -x; in that case, x-y means
x+(-y).(4) + is commutative: for all x,y in F, x+y = y+x.In
addition to these properties, F must be a commutative ring
underthe operations + and *. That means:(5) * must be
associative: for all x,y,z in F, (x*y)*z = z*(y*z).(6) * must
distribute over +: for all x,y,z in F, x*(y+z) = (x*y) + (x*z)
(y+z)*x = (y*x) + (z*x).(7) * must be commutative: for all
x,y, in F, x*y = y*x.In addition to these properties, F must
have a one:(8) There exists an element 1 in F such that for
all x in F, 1*x = x*1 = x.And ?ally, in order to be a ?ld,
every nonzero element of F musthave a multiplicative
inverse:(9) For every x in F, if x is not equal to 0 (the
same 0 in property (2) above), then there exists y in F such
that x*y = y*x = 1.Now, are the integers a ?ld? First you
must specify which twooperations you are considering. If you
are using the usual additionand multiplication, then the
answer is no because the integers failproperty (9): for
example, if x=2, then x is not equal to 0, but theredoes not
exist any integer y such that 2*y =
1.
===
==========================================

===
======It's not denial. I'm just very selective about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=====================================

===
==========Arturo
Magidinmagidin@math.berkeley.edu
===
alt.math.undergrad,
Cieslak Family <cieslakr002@hawaii.rr.com>Can someone
explain why the set of all integers is not a
?ld?http://mathworld.wolfram.com/FieldAxioms.htmlCan you see
which ?ld axiom is _not_ satis?d by the integers?(spoiler
below)What is the multiplicative inverse of 6? Is it a member
of the integers?-- Stan Brown, Oak Road Systems, Cortland
County, New York, USA http://OakRoadSystems.com/Walrus meat
as a diet is less repulsive than seal.
===
> Can someone
explain why the set of all integers is not a ?ld?>>
===
>
>Can someone explain why the set of all integers is not a
?ld?>> A ?ld is a set F with two binary operations, usually
denoted + and *> (sum and multiplication, respectively). The
fact that they are> binary operations means they are
functions from FxF to F (pairs of> elements to elements).
However, it is usual in most places to use> in? notation,
and write x+y instead of +(x,y).>> The set F, together
with the two operations + and *, must satisfy> certain
properties before it is called a ?ld:>> F must be an
abelian group under the operation +. That means:>> (1) +
must be associative: for all x,y,z in F,> (x+y)+z = x+
(y+z).> (2) There exists an element 0 in F such that for all
x in F,> x+0 = 0+x = x.> (3) For every x in F there exists y
in F such that x+y = y+x = 0. y is> called the inverse of x
and is often denoted -x; in that> case, x-y means
x+(-y).> (4) + is commutative: for all x,y in F,> x+y =
y+x.>> In addition to these properties, F must be a
commutative ring under> the operations + and *. That
means:>> (5) * must be associative: for all x,y,z in F,>
(x*y)*z = z*(y*z).>> (6) * must distribute over +: for all
x,y,z in F,>> x*(y+z) = (x*y) + (x*z)> (y+z)*x = (y*x) +
(z*x).>> (7) * must be commutative: for all x,y, in F,>> x*y
= y*x.>> In addition to these properties, F must have a
one:>> (8) There exists an element 1 in F such that for all
x in F,>> 1*x = x*1 = x.>> And ?ally, in order to be a ?ld,
every nonzero element of F must> have a multiplicative
inverse:>> (9) For every x in F, if x is not equal to 0 (the
same 0 in property> (2) above), then there exists y in F such
that>> x*y = y*x = 1.> Now, are the integers a ?ld? First
you must specify which two> operations you are considering.
If you are using the usual addition> and multiplication, then
the answer is no because the integers fail> property (9): for
example, if x=2, then x is not equal to 0, but there> does
not exist any integer y such that 2*y = 1.>>

===
===========================================

===
====> It's not denial. I'm just very selective about>
what I accept as reality.> --- Calvin (Calvin and Hobbes)>

===
============================================

===
===>> Arturo Magidin> magidin@math.berkeley.edu>===I am
currently learning to do integration by parts. I'm
havingdif?ulty on one problem, for which I get an answer
that seems to beright to me, but which disagrees with the
answer in the book. I'velooked for the errata for the book,
but can't ?d it. Since I can'tdo the integral sign, I've
made an image which can be found
athttp://www.geocities.com/venture_free/Integration.jpg which
displaysmy problem. I'll wait here while you go look at
it.Now, the answer I keep coming up with is:xln2x - 1/2x -
ln2 +1/2The answer in the back of the book, however, is:xln2x
- x - ln2 + 1I think that I am assuming something different
than the book is, and Idon't know if I'm wrong, or the book
is. Basically, all else beingequal, I get my answer if:d/dt
ln2t = 1/2tand I get the books answer if:d/dt ln2t = 1/t.
Which of us is right, and why?Also, since I have your
attention, is there some easier way to ?dbook errata than by
just looking on Google?-----------------------------
===
[...]>
I get my answer if:> d/dt ln2t = 1/2t> and I get the books
answer if:> d/dt ln2t = 1/t. > Which of us is right, and
why?Sorry, your book is right: d/dt(ln(2t)) = 1/(2t)*d/dt(2t)
= 2/(2t) = 1/t.Or: ln(2t) = ln(2) + ln(t) so d/dt(ln(2t)) =
d/dt(ln(t) = 1/t.For future reference, your problem was
int(ln(2t), t=1..x).> Also, since I have your attention, is
there some easier way to ?d> book errata than by just
looking on Google?It never would have occurred to me to look
_there_.-- Paul SperryColumbia, SC (USA)
===
> I am currently
learning to do integration by parts. I'm having> dif?ulty on
one problem, for which I get an answer that seems to be> right
to me, but which disagrees with the answer in the book. I've>
looked for the errata for the book, but can't ?d it. Since I
can't> do the integral sign, I've made an image which can be
found at>
http://www.geocities.com/venture_free/Integration.jpg which
displays> my problem. I'll wait here while you go look at
it.Bah, boo.> Now, the answer I keep coming up with is:>
xln2x - 1/2x - ln2 +1/2integeral log x dx = x log x - x> The
answer in the back of the book, however, is:> xln2x - x - ln2
+ 1>> I get my answer if:> d/dt ln2t = 1/2t>d/dt log 2t = 1/2t
* 2 = 1/t> and I get the books answer if:> d/dt ln2t = 1/t.>
Which of us is right, and why?>The book, because you made a
mistake.> Also, since I have your attention, is there some
easier way to ?d> book errata than by just looking on
Google?>Go googling by title, publisher and/or author.
===
I'm
trying to do a Fourier Series, but it's been too long since I
tookCalculus. I'm starting with:Integral [limits 0 to pi]
x*sin(nx)dxIntegrating by parts:u = x du = dxdv = sin(nx)dxv
= -cos(nx) Integral x*sin(nx)dx = -x*cos(nx) - [integral]
(-1)cos(nx)dx = -x*cos(nx) + [integral] cos(nx)dx
===
> I'm
trying to do a Fourier Series, but it's been too long since I
took> Calculus. > I'm starting with:> Integral [limits 0
to pi] x*sin(nx)dx> Integrating by parts:> u = x > du =
dx> dv = sin(nx)dx> v = -cos(nx) > Integral x*sin(nx)dx =
-x*cos(nx) - [integral] (-1)cos(nx)dx> = -x*cos(nx) +
[integral] cos(nx)dx> constant (n). Odd, since you
presumably (incorrectly) integrated sin(nx) to get v.In both
cases substitute for nx.-- Paul SperryColumbia, SC
(USA)
===
>I'm trying to do a Fourier Series, but it's been
too long since I took>Calculus. >>I'm starting
with:>>Integral [limits 0 to pi] x*sin(nx)dx>>Integrating by
parts:>>u = x >du = dx>dv = sin(nx)dx>v = -cos(nx) >>Integral
x*sin(nx)dx = -x*cos(nx) - [integral] (-1)cos(nx)dx> =
-x*cos(nx) + [integral] cos(nx)dx>>constant (n). Is this part
going to be equal to sin(nx) or something>else?> DUH! THINK!
what's the derivative of sin(nx) ????
===
I am trying to model
two different savings plans here in the US, andseem to be
running into contradictory results.There are two plans, an
unmatched plan and a matched plan.You can add money to the
unmatched plan like so:2004: 16k2005: 18k (it remains at 18k
from here on)Then there is a matched plan (the employer adds
4k)2004: 20k2005: 22k (it remains at 22k from here on)When
you withdraw your money, the unmatched plan is taxed at
15%,while the matched plan is taxed at 35% (this tax is only
applied once,when withdrawing all the money). Assume 10%
interest compoundedanually.It initially appears that after
several years, the extra 4k added overand over again will
make the matched plan more attractive. Apparentlythe latest
BusinessWeek mentioned something to this effect.me
contradictory results.Approaching the problem mathematically,
I solved for U_n and M_n (forn > 2)U_n = unmatched funds after
n yearsM_n = matched funds after n yearsI determined the
formulas to be:U_n = 14000 * (1.1)^n + 16000 * (1.1)^(n-1) +
18000* (1.1^(n-1) -1.1)/(.1)M_n = 18000 * (1.1)^n + 20000 *
(1.1)^(n-1) + 22000* (1.1^(n-1) -1.1)/(.1)The last terms of
each equation are a simpli?d geometric seriesI want n, such
that .85 * U_n = .65 * M_n. How do I solve for n?If you see
any obvious problems with my program, feel free to pointthem
out.If the funds never equal each other (which is what my
program hintsat), how could I go about proving this?This
problem has been bugging me. I would greatly appreciate any
help.ChrisBegin Python Program---------------rate = .1years =
50Over50 = [14000, 16000, 18000]Over50M = [18000, 20000,
22000]def GetSum(funds): result = [] total = 0.0 for x in
range(years): if x < 2: total += funds[x] else: total +=
funds[-1] total *= (1.0 + rate) result.append(total) return
resultdef PrintResults(a, b): print ?Year', ?t',
?Unmatched', ?t', ?Matched', ?t', 'Unmatched -matched'
for x in range(years): unmat = a[x] * .85 mat = b[x] * .65
print (x+1), ?t', unmat, ?t', mat, ?t', unmat -
matresultOver50 = GetSum(Over50)resultOver50M =
GetSum(Over50M)PrintResults(resultOver50, resultOver50M)
===
>
I am trying to model two different savings plans here in the
US, and> seem to be running into contradictory results.>
There are two plans, an unmatched plan and a matched plan.>
You can add money to the unmatched plan like so:> 2004: 16k>
2005: 18k (it remains at 18k from here on)> Then there is a
matched plan (the employer adds 4k)> 2004: 20k> 2005: 22k (it
remains at 22k from here on)> When you withdraw your money,
the unmatched plan is taxed at 15%,> while the matched plan is
taxed at 35% (this tax is only applied once,> when withdrawing
all the money). Assume 10% interest compounded> anually.> It
initially appears that after several years, the extra 4k added
over> and over again will make the matched plan more
attractive. Apparently> the latest BusinessWeek mentioned
something to this effect.> me contradictory results.>
Approaching the problem mathematically, I solved for U_n and
M_n (for> n > 2)> U_n = unmatched funds after n years> M_n
= matched funds after n years> I determined the formulas to
be:> U_n = 14000 * (1.1)^n + 16000 * (1.1)^(n-1) + 18000*
(1.1^(n-1) -> 1.1)/(.1)> M_n = 18000 * (1.1)^n + 20000 *
(1.1)^(n-1) + 22000* (1.1^(n-1) -> 1.1)/(.1)> The last
terms of each equation are a simpli?d geometric series> I
want n, such that .85 * U_n = .65 * M_n. How do I solve for
n?You don't. As you observed, they are never equal.[...]U_n =
14000*A + 16000*B + 18000*K;M_n = 18000*A + 20000*B +
22000*K;0.85*U_n = 11900*A + 13600*B + 15300*K;0.65*M_n =
11700*A + 13000*B + 14300*K.And A, B and K are positive (I
don't agree with your K but it doesn'tmatter).So unmatched,
taxed at 15% is always better that matched at 35%.-- Paul
SperryColumbia, SC (USA)
===
>I am trying to model two
different savings plans here in the US, and>seem to be
running into contradictory results.>>There are two plans, an
unmatched plan and a matched plan.>>You can add money to the
unmatched plan like so:>2004: 16k>2005: 18k (it remains at
18k from here on)>>Then there is a matched plan (the employer
adds 4k)>2004: 20k>2005: 22k (it remains at 22k from here
on)>>When you withdraw your money, the unmatched plan is
taxed at 15%,>while the matched plan is taxed at 35% (this
tax is only applied once,>when withdrawing all the money).
Assume 10% interest compounded>anually.>Where do you get
these tax rates? I'll bet the difference isn't the
calculations, it's the assumptions.Is it possible that the
unmatched plan paid with after-tax dollars, so that some
percentage is going to taxes (or you're really making bigger
contributions)? Maybe the best thing to do is give us the
generic names of the plans, like Roth IRA vs.
employer-sponsered 401(k) plan.The rates also depend on
income. You need to use your (foreknown) marginal tax rate
for the years you're going to be withdrawing the
money.Keeping the rest because downthread comments may refer
to them.Jon Miller>It initially appears that after several
years, the extra 4k added over>and over again will make the
matched plan more attractive. Apparently>the latest
BusinessWeek mentioned something to this effect.>>me
contradictory results.>>Approaching the problem
mathematically, I solved for U_n and M_n (for>n > 2)>>U_n =
unmatched funds after n years>M_n = matched funds after n
years>>I determined the formulas to be:>>U_n = 14000 *
(1.1)^n + 16000 * (1.1)^(n-1) + 18000* (1.1^(n-1)
->1.1)/(.1)>M_n = 18000 * (1.1)^n + 20000 * (1.1)^(n-1) +
22000* (1.1^(n-1) ->1.1)/(.1)>>The last terms of each
equation are a simpli?d geometric series>>I want n, such
that .85 * U_n = .65 * M_n. How do I solve for n?>>If you see
any obvious problems with my program, feel free to point>them
out.>>If the funds never equal each other (which is what my
program hints>at), how could I go about proving this?>>This
problem has been bugging me. I would greatly appreciate any
help.>>Chris>>Begin Python Program>--------------->>rate =
.1>years = 50>>Over50 = [14000, 16000, 18000]>Over50M =
[18000, 20000, 22000]>>def GetSum(funds):> result = []> total
= 0.0> for x in range(years):> if x < 2:> total += funds[x]>
else:> total += funds[-1]> total *= (1.0 + rate)>
result.append(total)> return result>>def PrintResults(a, b):>
print ?Year', ?t', ?Unmatched', ?t', ?Matched', ?t',
'Unmatched ->matched'> for x in range(years):> unmat = a[x]
* .85> mat = b[x] * .65> print (x+1), ?t', unmat, ?t',
mat, ?t', unmat - mat>>resultOver50 =
GetSum(Over50)>resultOver50M =
GetSum(Over50M)>>PrintResults(resultOver50,
resultOver50M)>
===
Can anyone help with the following....I've
been given a function h(x) = 1/1600 x, x>=0and told that the
function Q(x) = exp (- integral between x and 0 h(u) du)I've
evaluated this to e ^ -1/3200x^2Is this right - my
integration is VERY rusty.I've then worked out that the CDF
of X isF(x) = 1 -Q(x) = 1 - e ^ -1/3200x^2Hence the pdf
isf(x) = 1/3200 e ^-1/3200x^2Which has given me an
exponential distribution with lambda = square root1/3200 (
I've square rooted cos of the ^2 term in e above) giving a
mean of56.56854249 (1/lambda)Can someone just check my
calculations
===
> Can anyone help with the following....>>
I've been given a function h(x) = 1/1600 x, x>=0> and told
that the function Q(x) = exp (- integral between x and 0 h(u)
du)>> I've evaluated this to e ^ -1/3200x^2> Is this right -
my integration is VERY rusty.>integral(0,x) 1/1600u du = (log
u)/1600 |0,x which gives problem at u = 0integral(0,x)
(1/1600)u du = (1/2)(1/1600)u^2 |0,x = x^2 / 3200 =
(x^2)/3200Becareful about 1/2 u, 1/2u, (1/2)u, 1/(2u)> I've
then worked out that the CDF of X is> F(x) = 1 -Q(x)> = 1 - e
^ -1/3200x^21 - e^[(-1/3200)x^2]It appeared you said 1 -
e^[-1/(3200x^2)]Beware e^a+b, e^a + b, e^(a+b), (e^a) +
b.[clip]>> Can someone just check my calculations>The problem
I see is lack of clarity writing formulas in one line
ascii.
===
Debbie Sewell <debbie.sewell@virgin.net>:> I've
been given a function h(x) = 1/1600 x, x>=0> and told that
the function Q(x) = exp (- integral between x and 0 h(u)
du)> I've evaluated this to e ^ -1/3200x^2> Is this right
- my integration is VERY rusty.> I've then worked out that
the CDF of X is> F(x) = 1 -Q(x)> = 1 - e ^ -1/3200x^2> Hence
the pdf is> f(x) = 1/3200 e ^-1/3200x^2> Which has given me
an exponential distribution with lambda = square root> 1/3200
( I've square rooted cos of the ^2 term in e above) giving a
mean of> 56.56854249 (1/lambda)Close, but be careful with
the details. This problem is formally identical to ?ding the
event density givena hazard function in survival analysis.
That doesn't help solve integrals,It looks to me like you're
OK up to and including the calculation ofF(x) = 1 -
exp(-(1/3200) x^2). Recall that the derivative of exp(y(x))is
y'(x) exp(y(x)) where y(x) is some function of x. So we have
f(x) = -(d/dx) (-(1/3200) x^2) exp(-(1/3200) x^2) = (x/1600)
exp(-(1/3200) x^2)de?itions given, this is always the case:
the hazard function is the ratio of the event density to the
survival function.Note that this density f(x) is not an
exponential density; an exponential event density has a
constant (not linear) hazard function.The expected value of
the event density is integral(0,inf) x f(x) dx,which you can
solve by integration by parts. Recall that (d/dx) exp(x^2)is
just 2 x exp(x^2).Hope this helps,Robert Dodier--If I have
not seen as far as others, it is because giants werestanding
on my shoulders. -- Hal Abelson
===
> Debbie Sewell
<debbie.sewell@virgin.net>:>> I've been given a function
h(x) = 1/1600 x, x>=0>> and told that the function Q(x) =
exp (- integral between x and 0 h(u)du)>> I've evaluated
this to e ^ -1/3200x^2> Is this right - my integration is
VERY rusty.>> I've then worked out that the CDF of X is>
> F(x) = 1 -Q(x)> = 1 - e ^ -1/3200x^2> Hence the pdf is>
> f(x) = 1/3200 e ^-1/3200x^2>> Which has given me an
exponential distribution with lambda = square root> 1/3200
( I've square rooted cos of the ^2 term in e above) giving a
meanof> 56.56854249 (1/lambda)>> Close, but be careful
with the details.>> This problem is formally identical to
?ding the event density given> a hazard function in survival
analysis. That doesn't help solve integrals,>> It looks to me
like you're OK up to and including the calculation of> F(x) =
1 - exp(-(1/3200) x^2). Recall that the derivative of
exp(y(x))> is y'(x) exp(y(x)) where y(x) is some function of
x. So we have>> f(x) = -(d/dx) (-(1/3200) x^2) exp(-(1/3200)
x^2)> = (x/1600) exp(-(1/3200) x^2)>> de?itions given, this
is always the case: the hazard function> is the ratio of the
event density to the survival function.>> Note that this
density f(x) is not an exponential density; an> exponential
event density has a constant (not linear) hazard function.>>
The expected value of the event density is integral(0,inf) x
f(x) dx,> which you can solve by integration by parts. Recall
that (d/dx) exp(x^2)> is just 2 x exp(x^2).>> Hope this
helps,> Robert Dodier> --> If I have not seen as far as
others, it is because giants were> standing on my shoulders.
-- Hal Abelsonintegration above x^2 etc.. !!So, can you help
me with the step by step way of using integration by partsfor
integral(0,inf) exp(-(1/3200) x^2). ??Debbie
===
Is there a way
to completely describe the isomorphisms from Z_p x Z_q(where p
and q are any integers, not necessarily relatively prime)
toitself? What are these automorphisms?
===
>Is there a way to
completely describe the isomorphisms from Z_p x Z_q>(where p
and q are any integers, not necessarily relatively prime)
to>itself? What are these automorphisms?Well, since any ?ite
abelian group G is the direct product of its Sylowsubgroups
P_1,...,P_r, and Aut(G) = Aut(P_1) x ... x Aut(P_r), this
problemimmediately reduces to the case when p amnd q are both
powers of the sameprime - so let's change notation, and
consider A = Aut(G), withG = Z_{p^a} x Z_{p^b}, where p is
prime and 0 <= a <= b.In all cases, A has a normal p-subgroup
N which acts trivially on G/G^p,and A/N is the induced action
of A on G/G^p. I did a quick calculation,(and may have made
mistakes!), and gotCase 1. a = 0, G is cyclic. This case is
well known. |N| = p^(b-1) and |A/N| = p-1.Case 2. 0 < a < b.
|N| = p^(3a+b-2), |A/N| = (p-1)^2.Case 3. 0 < a = b. |N| =
p^(4(a-1)), |A/N| = |GL(2,p)| = (p^2-1)(p^2-p).I am sure the
structure could be described more precisely if necessary.I
believe that I have seen it in the literature.Derek
Holt.
===
> Is there a way to completely describe the
isomorphisms from Z_p x Z_q> (where p and q are any integers,
not necessarily relatively prime) to> itself? What are these
automorphisms?I know of no nice description for arbitrary p
and q.Hom(Z_p x Z_q, Z_p x Z_q) is isomorphic toHom(Z_p, Z_p)
x Hom(Z_p, Z_q) x Hom(Z_q, Z_p) x Hom(Z_q, Z_q).The ?st and
last Hom's are isomorphic to Z_p and Z_q respectively.Thetwo
mixed Hom's depend strongly on how p and q are related.
Moreover,it is not obvious (to me, anyway) how to identify
the images of theautomorphisms of Z_p x Z_q in the product of
the Hom's.-- Paul SperryColumbia, SC (USA)
===
> Is there a way
to completely describe the isomorphisms from Z_p x Z_q> (where
p and q are any integers, not necessarily relatively prime)
to> itself? What are these automorphisms?(I shall use the
notation Z/p and Z/q instead of Z_p and Z_q)Further, and in
contrast to Derek Holt's more enlightened approach, Ihave an
elementary prime-free description of the situation.Let g =
gcd(p, q), and p' = p/g, q' = q/g.Let e_1 and e_2 be
generators of Z/p x Z/q, with orders p and qrespectively.Then
one can represent the endomorphism ring of Z/p x Z/q as the
ringof 2x2 matricesa b = Ac dsuch thata belongs to Z/p,b
belongs to Z/p and is a multiple of p'c belongs to Z/q and is
a multiple of q'd belongs to Z/q.These matrices act on the
group in a natural way, and also multiplynaturally to
represent composition.Necessary conditions for such a matrix
A to represent an automorphism(i.e. to be invertible) are(1)
gcd(det A, p, q) = 1.(2) gcd(a, p') = 1(3) gcd(d, q') =
1Conversely, suppose that A satis?s all three conditions.
Let A' bethe reduction of A mod g. Then (1) tells us that A'
is invertible,with inverse B', say. We want to construct a
matrix B of the aboveform which reduces to B' mod g. Well,
the non-diagonal entries of B'determine those of B. Then
apply the Chinese remainder theorem to getthe diagonal
entries of B, and hopefully a simple calculation shouldshow
that the matrix we construct is indeed the inverse of A.
===
T
H O M A S20 8 15 13 1 19 = 76 I met Joshua in Bessborough Park
beside the drinking fountain.76+ Mom 16 7 53 197/168 +1312289
Joshua 4 8 81 216/149 8934Joshua 74 Christopher 139 Thomas 76
Joshua is a 6 lettered name, Joshua is Bible Book 6, Joshua
adds to74 (a factor of 666). Joshua begins with the 10th
letter of thealphabet (6th non-prime). The vowels and
consonants in Joshua both addnon-prime). Joshua was born on
the 216th (6x6x6th) day of the year, itis the 13x13th
non-prime (while 13 in turn is the 6th prime). Joshua(Book 6)
contains 24 (6+6+6+6) chapters, I am 24 years older
thanJoshua.1-50 - Genesis51-90 - Exodus91-117 -
Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 -
Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 -
John1019-1046 - Acts1047-1062 - Romans 188 <-the opening
chapter of Book 6 is 6x6x6 short of the 404 verses of Bible
Book 66, it is the 6th prime squared (13x13) short of the 357
verses of Daniel (also in part about 666) 193 <-Book 6 chapter
6 is the 44th prime, while 44 is in turn 66.666...% of 66 211
<-the terminating chapter of Book 6 is approximately 66.6% of
the 66th prime (317) 357 <-the opening chapter of Book 6 plus
the 6th prime squared is the 357 verses of Daniel (in part
about 666) 404 <-the 6th prime squared (13x13) plus the 6th
prime squared (13x13) plus 66 adds to the 404 verses of Bible
Book 661062 <-666 plus 6x66 is a combination of the 658 verses
of Bible Book 6 plus the 404 verses of Bible Book 66, and is
the terminating chapter of New Testament Book 61070 <-666
plus the 404 verses of Book 66 is the 1070 verses of Job
(Book 6+6+6)1213 <-Exodus terminates at chapter 90 (66th non-
prime) with 1213 verses (the 198th or the 66+66+66th
prime)1292 <-the 658 verses of Book 6 plus twice the 66th
prime (317) is the 1292 verses of Isaiah (the Book contains
66 chapters) Mom was born on the 16th day of the month,
perhaps in 53 (16thprime), and she gave birth on the 216th
day of the year. She gavebirth with 149 days remaining in the
year, it's the length of BibleBook 48 (16+16+16). Joshua's
name adds to 289, corresponding to the(166th
non-prime).Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12
19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61
27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107
40 109 42 113 44 127 45 131 46 137 48 139 49 149 50 151 51 157
52 163 54 167 55 173 56 179 57 181 58 191 60 193 62 197 63 199
64 211 65 223 <-48th-> 66---- ---4661 1710 Joshua was born
with 149 days remaining in the year, it's thethe 48th prime
(223) and the 48th non-prime (66). Primes In Prime Primes
Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <-
17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 41 --- 108 Mom and
Joshua were born on days of the year adding to 413, it
isexactly 59 weeks (the 17th prime). Joshua was born in 81
(59thnon-prime while 59 is the 17th prime). He was born a
multiple of 17days into the century (29801=17x1753). Mom and
Joshua were togetherborn a multiple of 108 days into the
century (49356=108x457), thereare 108 verses in Bible Book 59
(the 17th prime, it's the 17th primeday, month and year of
birth adds to 93 (Leviticus 3 with 17 verses).plus the 17th
prime, chapter 76 is Exodus 26 (17th non-prime). Todaymom and
Joshua are together 71.77 years old. Primes In Prime Primes
Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <-
17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317
59 <- 59 --- 167 Esther Book 17 Joshua was born 19 days after
mom's birthday. Mom was born on day197, it's the number of
verses in Bible Book 28 (19th non-prime).Mom's day, month and
year of birth adds to 76 and she married Thomas(76=4x19).
Joshua was born 67 days closer to the end of the year thanto
the beginning of the year (19th prime), and keep in mind that
the2460 verses of Bible Book 19
is19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus the 19th
prime (67).unrepeated letters add to 77 (the primes up to
19). Mom was born onthe 19555th day of the century and
married Thomas (76, the 55thnon-prime). The Samuels are Books
9 and 10 (for 19) with 55 chaptersand 1505 verses. I
criticized churches (I said they had Egyptian penises on
theirroofs, and this was in opposition to God's Second
Commandment),Protestants and Catholics lobbied my abusive
parents to have metreated, my abusive parents quickly
complied and I was repeatedlyProtestants and Catholics then
sat on psychiatric appeal paneldetain and torture me. I
begged and begged for assistance to get outof the country to
no avail, people would tell me that I was there
(inpsychiatric facilities) because I was supposed to be
there. And so itis with Joshua, he now ?ds himself posted on
the usenet becuase heis supposed to be here. I showed Joshua
gems, and like everybody else,he was so cheap and ignorant
that he did not even have the decency totime out of my life
to do so. But come December, he will please hisparents (or
his girlfriend's parents) and attend their church, he
willcomment on the beauty of their decorated trees or their
steeple (arepresntation of a penis), and will give them money
(the churches thatteach you people to turn pagan fertility
symbols into decorated idolsare supposedly worthy of your
support and respect). In my effort toget out of the brutal
cycle of psychiatric abuuse, I told theProtestants and
Catholics seated in judgment against me that the
Biblerepeatedly condemns turning trees into idols, and that
they in factbow to their decorated trees via the placement
and retieval ofpresents at the base of their trees. At one
psychiatric appeal panelhearing Dr. Gene Marcoux heard me say
this, he responded by sayingthat I was religiously deluded to
believe this, and then all theProtestants and Catholics
nodded in agreement and gave him permissonto detain and
torture me for the following three weeks. Year afteryear you
people collectively spend billions of dollars on turningtrees
into decorated idols, you have spent millions of dollars
havingme tortured in an attempt to make me shut up about this
and othertraditions being taught by your churches, and you are
so cheap andignorant that you can't even offer to pay me
minimum wage for my workso I could attempt to get out of this
country on my own dime. I wasfrantic to leave the country as I
lost summer after summer aftersummer after summer after summer
after summer after summer topsychiatric torture, and the
subsequent summers were effectly lost aswell as I remained
frantic to leave the country (I expected to bearrested and
tortured in the years that I was not), and there isn't
asingle person who has the compassion to assist me to get out
of thiscountry for one single winter!!! You are the shit of
the earth JoshuaChristopher Thomas, you are an
incompassionate turd, and now I beg Godto honor Exodus 20:5
and Hosea 4:6 as promises so that He wouldterminate your
life, or at least turn away from you in your hour ofneed. If
I hear of your death I will cheer with utter glee, it will
bein accordance to Scripture (Psalm 137:9), and I will post
your statsagain. God is about to spread you people out like
dung over thesurface of the earth, and in this I rejoice. You
have billions ofdollars to spend on turning trees into idols,
you have millions ofand so ignorant that you can't even buy a
postage stamp in order tosend me a cheap letter expressing
thanx for showing you evidence thatyour very name is a gift
from God.Daryl Shawn KabatoffBox 7134Saskatoon
SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God
gives you your name!!!
===
T H O M A S20 8 15 13 1 19 = 76 I
met Joshua in Bessborough Park beside the drinking
fountain.76+ Mom 16 7 53 197/168 +1312289 Joshua 4 8 81
216/149 8934Joshua 74 Christopher 139 Thomas 76 Joshua is a 6
lettered name, Joshua is Bible Book 6, Joshua adds to74 (a
factor of 666). Joshua begins with the 10th letter of
thealphabet (6th non-prime). The vowels and consonants in
Joshua both addnon-prime). Joshua was born on the 216th
(6x6x6th) day of the year, itis the 13x13th non-prime (while
13 in turn is the 6th prime). Joshua(Book 6) contains 24
(6+6+6+6) chapters, I am 24 years older thanJoshua.1-50 -
Genesis51-90 - Exodus91-117 - Leviticus118-153 -
Numbers154-187 - Deuteronomy188-211 - Joshua930-957 -
Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 -
Acts1047-1062 - Romans 188 <-the opening chapter of Book 6 is
6x6x6 short of the 404 verses of Bible Book 66, it is the 6th
prime squared (13x13) short of the 357 verses of Daniel (also
in part about 666) 193 <-Book 6 chapter 6 is the 44th prime,
while 44 is in turn 66.666...% of 66 211 <-the terminating
chapter of Book 6 is approximately 66.6% of the 66th prime
(317) 357 <-the opening chapter of Book 6 plus the 6th prime
squared is the 357 verses of Daniel (in part about 666) 404
<-the 6th prime squared (13x13) plus the 6th prime squared
(13x13) plus 66 adds to the 404 verses of Bible Book 661062
<-666 plus 6x66 is a combination of the 658 verses of Bible
Book 6 plus the 404 verses of Bible Book 66, and is the
terminating chapter of New Testament Book 61070 <-666 plus
the 404 verses of Book 66 is the 1070 verses of Job (Book
6+6+6)1213 <-Exodus terminates at chapter 90 (66th non-
prime) with 1213 verses (the 198th or the 66+66+66th
prime)1292 <-the 658 verses of Book 6 plus twice the 66th
prime (317) is the 1292 verses of Isaiah (the Book contains
66 chapters) Mom was born on the 16th day of the month,
perhaps in 53 (16thprime), and she gave birth on the 216th
day of the year. She gavebirth with 149 days remaining in the
year, it's the length of BibleBook 48 (16+16+16). Joshua's
name adds to 289, corresponding to the(166th
non-prime).Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12
19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61
27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107
40 109 42 113 44 127 45 131 46 137 48 139 49 149 50 151 51 157
52 163 54 167 55 173 56 179 57 181 58 191 60 193 62 197 63 199
64 211 65 223 <-48th-> 66---- ---4661 1710 Joshua was born
with 149 days remaining in the year, it's thethe 48th prime
(223) and the 48th non-prime (66). Primes In Prime Primes
Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <-
17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 41 --- 108 Mom and
Joshua were born on days of the year adding to 413, it
isexactly 59 weeks (the 17th prime). Joshua was born in 81
(59thnon-prime while 59 is the 17th prime). He was born a
multiple of 17days into the century (29801=17x1753). Mom and
Joshua were togetherborn a multiple of 108 days into the
century (49356=108x457), thereare 108 verses in Bible Book 59
(the 17th prime, it's the 17th primeday, month and year of
birth adds to 93 (Leviticus 3 with 17 verses).plus the 17th
prime, chapter 76 is Exodus 26 (17th non-prime). Todaymom and
Joshua are together 71.77 years old. Primes In Prime Primes
Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <-
17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317
59 <- 59 --- 167 Esther Book 17 Joshua was born 19 days after
mom's birthday. Mom was born on day197, it's the number of
verses in Bible Book 28 (19th non-prime).Mom's day, month and
year of birth adds to 76 and she married Thomas(76=4x19).
Joshua was born 67 days closer to the end of the year thanto
the beginning of the year (19th prime), and keep in mind that
the2460 verses of Bible Book 19
is19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus the 19th
prime (67).unrepeated letters add to 77 (the primes up to
19). Mom was born onthe 19555th day of the century and
married Thomas (76, the 55thnon-prime). The Samuels are Books
9 and 10 (for 19) with 55 chaptersand 1505 verses. I
criticized churches (I said they had Egyptian penises on
theirroofs, and this was in opposition to God's Second
Commandment),Protestants and Catholics lobbied my abusive
parents to have metreated, my abusive parents quickly
complied and I was repeatedlyProtestants and Catholics then
sat on psychiatric appeal paneldetain and torture me. I
begged and begged for assistance to get outof the country to
no avail, people would tell me that I was there
(inpsychiatric facilities) because I was supposed to be
there. And so itis with Joshua, he now ?ds himself posted on
the usenet because heis supposed to be here. I showed Joshua
gems, and like everybody else,he was so cheap and ignorant
that he did not even have the decency totime out of my life
to do so. But come December, he will please hisparents (or
his girlfriend's parents) and attend their church, he
willcomment on the beauty of their decorated trees or their
steeple (arepresentation of a penis), and will give them
money (the churchesthat teach you people to turn pagan
fertility symbols into decoratedidols are supposedly worthy
of your support and respect). In my effortto get out of the
brutal cycle of psychiatric abuse, I told theProtestants and
Catholics seated in judgment against me that the
Biblerepeatedly condemns turning trees into idols, and that
they in factbow to their decorated trees via the placement
and retrieval ofpresents at the base of their trees. At one
psychiatric appeal panelhearing Dr. Gene Marcoux heard me say
this, he responded by sayingthat I was religiously deluded to
believe this, and then all theProtestants and Catholics
nodded in agreement and gave him permissionto detain and
torture me for the following three weeks. Year afteryear you
people collectively spend billions of dollars on turningtrees
into decorated idols, you have spent millions of dollars
havingme tortured in an attempt to make me shut up about this
and othertraditions being taught by your churches, and you are
so cheap andignorant that you can't even offer to pay me
minimum wage for my workso I could attempt to get out of this
country on my own dime. I wasfrantic to leave the country as I
lost summer after summer aftersummer after summer after summer
after summer after summer topsychiatric torture, and the
subsequent summers were effectively lostas well as I remained
frantic to leave the country (I expected to bearrested and
tortured in the years that I was not), and there isn't
asingle person who has the compassion to assist me to get out
of thiscountry for one single winter!!! You are the shit of
the earth JoshuaChristopher Thomas, you are an
incompassionate turd, and now I beg Godto honor Exodus 20:5
and Hosea 4:6 as promises so that He wouldterminate your
life, or at least turn away from you in your hour ofneed. If
I hear of your death I will cheer with utter glee, it will
bein accordance to Scripture (Psalm 137:9), and I will post
your statsagain. God is about to spread you people out like
dung over thesurface of the earth, and in this I rejoice. You
have billions ofdollars to spend on turning trees into idols,
you have millions ofand so ignorant that you can't even buy a
postage stamp in order tosend me a cheap letter expressing
thanx for showing you evidence thatyour very name is a gift
from God.Daryl Shawn KabatoffBox 7134Saskatoon
SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God
gives you your name!!!
===
Lately I've been explaining how I
found an error in taught mathematicsas I prove that in the
ring of algebraic integers a number I'll herecall x has a
factor I'll here call y, but x/y is not an
algebraicinteger.Now at least *some* of you should be
wondering how that works giventhat division is not a ring
operation.Part of the problem for those of you who wonder
about that is that themathematicians among you probably know
how that all plays out as I'vebeen arguing for a while now,
but a poster named Randy Poe posted somethings that lead me
to consider that all of you aren't necessarilymath experts
i.e. mathematicians.Ok, here it goes. I'm going to go kind of
fast but you should be ableto keep up as it's simple.First
thing is my paper Advanced Polynomial Factorization which is
asusual found at http://groups.msn.com/AmateurMathshows that
given these three numbers two of them have this factor Icall
f and the ring is algebraic integers.That's all you get
there, and there's no mention of anything likex/y, which
I've talked about a lot.And there's no mention because the
ring operations are addition andmultiplication.Division is
not a ring operation. Division is de?ed for ?lds,
notrings.So then, it's *provable* that a number I'll call x
has a factor I'llcall y in the ring of algebraic integers, is
what the paper covers. There you have the a's like before and
two of the a's have a factor f,while the third is coprime to
f, which is proven in the paper.That is, given the de?ition
for algebraic integer I can prove thatthere's this number
I'll call x, which is itself an algebraic integer,that has
this number I'll call y, which is an algebraic integer, as
afactor.That's it for the paper.Now then, taking that result
in that ring, I can now move to the ?ldof algebraic
numbers.Now I have division, and in that ?ld, I take x/y and
notice thatit's NOT an algebraic integer, which is a
contradiction.That's how it's done.Now for the
mathematicians, what I just said shouldn't be news, but alot
of you are basically, um, what's the right word? Well
thereprobably isn't a right word so I'll leave it
there.Notice I'm only posting to sci.math and
alt.math.undergrad here, as Idon't see the point in bothering
anyone else with such trivialdetails.James Harris
===
[...]>
Now then, taking that result in that ring, I can now move to
the ?ld> of algebraic numbers.> Now I have division, and
in that ?ld, I take x/y and notice that> it's NOT an
algebraic integer, which is a contradiction.> That's how
it's done.The ?ld of algebraic numbers over the rationals Q
contains the ringof algebraic integers over Q, but the two
are not equal.So if x and y are algebraic integers over Q,
and y is not zero,then surely x/y is an algebraic number. By
no means doesx, y algebraic integers imply x/y is an
algebraic integer.David Bernier
===
> Lately I've been
explaining how I found an error in taught mathematics> as I
prove that in the ring of algebraic integers a number I'll
here> call x has a factor I'll here call y, but x/y is not an
algebraic> integer.> Can you explain what you mean by factor
in this instance?For example, 3 and 5 are algebraic integers,
but 3/5 is not an algebraic integer. That's because the
algebraic integers are a ring but not a ?ld. What do you
mean that y is a factor of x? If x and y are algebraic
integers, the accepted de?ition of factor is that there is
some algebraic integer z such that yz = x. If so, then we say
that y is a factor of x.If we consider the example above in
the ring of the usual integers, 5 is not a factor of 3. But
if we extend the ring of integers to the ?ld of rationals,
it turns out that there is a rational, namely 3/5, with the
property that 5 * 3/5 = 3. The same example works in the ring
of algebraic integers.So all you seem to be saying is that
you've determined that the algebraic integers are a ring but
not a ?ld. This is already well known.Is this what you're
talking about?
===
>Lately I've been explaining how I found an
error in taught mathematics>as I prove that in the ring of
algebraic integers a number I'll here>call x has a factor
I'll here call y, but x/y is not an
algebraic>integer.>[...]>>Now then, taking that result in
that ring, I can now move to the ?ld>of algebraic
numbers.>>Now I have division, and in that ?ld, I take x/y
and notice that>it's NOT an algebraic integer, which is a
contradiction.No, that's not a contradiction. Because you get
a factor in the?ld of algebraic numbers, _not_ in the
algebraic integers.The following looks like one of those
nasty parodies peoplesometimes post, but actually it's an
illustration of _exactly_the error you're making, translated
to a more familiar context:Lately I've been explaining how I
found an error in taughtmathematics as I prove that in the
ring of integers a number I'll here call x has a factor I'll
here call y, but x/y is not aninteger.Let x = 2, y = 3.Now
then, taking that result in that ring, I can now move to the
?ldof rational numbers.Now I have division, and in that
?ld, I take x/y and notice thatit's NOT an integer, which is
a contradiction.Honest. What you're saying here makes
_exactly_ as muchsense as saying that 3 is a factor of 2
(because 2/3 existsin the rationals) although 2/3 is not an
integer, contradiction.There's no contradiction, just two
true facts:(i) 3 is _not_ a factor of 2, _in_ the
integers.(ii) 3 _is_ a factor of 2 _in_ the rationals (and
luckily 2/3 is rational.)************************David C.
Ullrich
===
> Lately I've been explaining how I found an
error in taught mathematics> as I prove that in the ring of
algebraic integers a number I'll here> call x has a factor
I'll here call y, but x/y is not an algebraic> integer.>
> Can you explain what you mean by factor in this
instance?> For example, 3 and 5 are algebraic integers, but
3/5 is not an algebraic > integer. That's because the
algebraic integers are a ring but not a > ?ld. > What do
you mean that y is a factor of x? If x and y are algebraic >
integers, the accepted de?ition of factor is that there is
some > algebraic integer z such that yz = x. If so, then we
say that y is a > factor of x.> If we consider the example
above in the ring of the usual integers, 5 is > not a factor
of 3. But if we extend the ring of integers to the ?ld > of
rationals, it turns out that there is a rational, namely 3/5,
with > the property that 5 * 3/5 = 3. > The same example
works in the ring of algebraic integers.> So all you seem
to be saying is that you've determined that the > algebraic
integers are a ring but not a ?ld. This is already well >
known.> Is this what you're talking about?This is going to
seem rather odd. There are times when I thinkwe put too much
stress on form and not enough on substance. Allowing for the
fact that James Harris activelyresists the use of standard
terminology and wants to remainignorant of conventions and
de?itions, is there any waythat his statement about x and y
could make sense?I think so. If x and y are algebraic
integers and y isa factor of x in the algebraic integers,
anyone wouldnaturally conclude that x/y must be an algebraic
integer simply from the de?ition of factor. But there is
still the possibility that x/y could be a root of a non-monic
polynomial with integer coef?ients, primitive and irreducible
over the rationals: that is, that possibility would exist if
it were not for the following known theorem: Theorem. If r is
a root of a non-monic polynomial with integer coef?ients that
is primitive and irreducible over the rationals, then r cannot
be an algebraic integer.Now if Harris could show that x/y is
such a root, onewould have to conclude that either (1) the
theorem aboveis wrong, or (2) something else is wrong with
coremathematics, i.e., the mathematics that is
currentlytaught.Harris argued for many months that the
Theorem is wrong.He tried over and over again to come up with
a counter-example. He ?ally consented to read a proof of
thetheorem, *only after* someone that he regarded as
anauthority, Robert Israel, stated that the theorem
wasvalid. He then said that he accepted the theorem. Whether
he ever actually read the proof I don't know.However it
appears now that he has slid back to grinding thisold axe in
a new form: he no longer says the theoremis wrong; instead he
says there is something wrong withcore mathematics and in
particular something wrong with the de?ition of algebraic
integers, or that thering of algebraic integers is
incomplete.The point here is, there is possible content in
whathe says. I totally disagree with his conclusion, butit
could make a loony kind of sense. If he actually did?d x and
y as above, and a non-monic polynomial with integer
coef?ients, primitive and irreducibleover the rationals with
the algebraic integer x/y as a root, he would indeed have
found something wrong with mathematics. Of course he hasn't,
but theoretically it could make sense.Thus: quit taking him
to task for the super?ial problemof not knowing the
de?ition of ?factor'. Take him to task for yet again trying
to show that the Theorem above, which he has accepted, is
false. Ask him to specify the non-monic polynomial of which
x/y is a root.Andrzej
===
> What do you mean that y is a
factor of x? If x and y are algebraic > integers, the
accepted de?ition of factor is that there is some >
algebraic integer z such that yz = x. If so, then we say that
y is a > factor of x.[snip]> This is going to seem rather
odd. There are times when I think> we put too much stress on
form and not enough on > substance. Allowing for the fact
that James Harris actively> resists the use of standard
terminology and wants to remain> ignorant of conventions and
de?itions, is there any way> that his statement about x and
y could make sense?> I think so. If x and y are algebraic
integers and y is> a factor of x in the algebraic integers,
anyone would> naturally conclude that x/y must be an
algebraic > integer simply from the de?ition of factor.Uh,
yeah.> But > there is still the possibility that x/y could be
a root > of a non-monic polynomial with integer coef?ients,
> primitive and irreducible over the rationals: that is, >
that possibility would exist if it were not for the >
following known theorem:> Theorem. If r is a root of a
non-monic polynomial> with integer coef?ients that is
primitive and > irreducible over the rationals, then r cannot
be > an algebraic integer.Snip. You haven't clari?d the
problem as far as I cansee. If the relationship between x and
y is notx/y is an algebraic integer, then what is it?You
seem to be saying that the only thing known aboutx/y is that
it is the root of such a polynomial. That'seasy.> The point
here is, there is possible content in what> he says.And what
is that? I don't get what you're trying tosay. It seems to be
that: (1) There exists x and y algebraic integers such that
(2) r = x/y is not an algebraic integer, and (3) r is the
root of a non-monic polynomial with integer coef?ients, that
is primitive and irreducible over the rationals.Is that it? If
so, isn't that just a trivial restatementof the theorem? -
Randy
===
> What do you mean that y is a factor of x? If
x and y are algebraic > integers, the accepted de?ition
of factor is that there is some > algebraic integer z
such that yz = x. If so, then we say that y is a >
factor of x.> [snip]> This is going to seem rather odd.
There are times when I think> we put too much stress on
form and not enough on > substance. Allowing for the fact
that James Harris actively> resists the use of standard
terminology and wants to remain> ignorant of conventions
and de?itions, is there any way> that his statement about
x and y could make sense?> I think so. If x and y are
algebraic integers and y is> a factor of x in the algebraic
integers, anyone would> naturally conclude that x/y must be
an algebraic > integer simply from the de?ition of
factor.> Uh, yeah.> But > there is still the
possibility that x/y could be a root > of a non-monic
polynomial with integer coef?ients, > primitive and
irreducible over the rationals: that is, > that possibility
would exist if it were not for the > following known
theorem:> Theorem. If r is a root of a non-monic
polynomial> with integer coef?ients that is primitive and
> irreducible over the rationals, then r cannot be > an
algebraic integer.> Snip. You haven't clari?d the problem
as far as I can> see. If the relationship between x and y is
not> x/y is an algebraic integer, then what is it?>
Obviously it is. What Harris may be trying to say is,I have
found an algebraic integer which is a root of a*non-monic*
polynomial etc., thus defying a well-knowntheorem. Therefore
there is something wrong withmathematics: either the theorem
is wrong (as he haslong contended) or the de?ition of
algebraic integeris wrong, or something else. I am not sure
why this needs explaining. I am veryde?itely not saying that
I agree with Harris: simplythat instead of continuing to
criticize him for refusingto recognize a tautology, consider
that he may actuallyhave had something more in mind -
something equally wrong,to be sure, but at a different level.
I actually don't know how he goes from his main
AdvancedPolynomial Factorization result (which has now
beenproved false by two other posters) to obtaining x andy as
he describes. I am not sure he remembers how either - not that
it matters at all anyway. > You seem to be saying that the
only thing known about> x/y is that it is the root of such a
polynomial. That's> easy.> What??? Basically the monic
criterion for algebraicintegers is if and only if - more
precisely, an algebraicnumber is an algebraic integer if and
only if its minimalpolynomial with integer coef?ients is
monic.> The point here is, there is possible content in
what> he says.> And what is that? I don't get what you're
trying to> say. Yes, I see that.> It seems to be that:> (1)
There exists x and y algebraic integers such that> (2) r =
x/y is not an algebraic integer, and No - just the opposite -
r = x/y IS an algebraic integer.> (3) r is the root of a
non-monic polynomial with> integer coef?ients, that is
primitive and irreducible> over the rationals.> Is that it?
If so, isn't that just a trivial restatement> of the theorem?>
Essentially he may be saying yet again, for the nth time,that
he has found a counterexample to the theorem. The obvious
question he should answer is: what is the polynomial? Andrzej
> - Randy
===
> Obviously it is. What Harris may be trying to
say is,> I have found an algebraic integer which is a root
of a> *non-monic* polynomial etc., thus defying a well-known>
theorem. Therefore there is something wrong with> mathematics:
either the theorem is wrong (as he has> long contended) or the
de?ition of algebraic integer> is wrong, or something
else.OK, if he's made that statement. I thought you
wereinterpreting the remarks about I have proved that x is
afactor of y in the algebraic integers, but x/y is notan
algebraic integer.Or perhaps you are. Is that his proof
that x is afactor of y? I've never yet been able to ?ure
outwhich part of James' writings are supposed to be theproof
part, but every once in a while he startstalking as if a
proof went by when nobody was looking. - Randy
===
>
Obviously it is. What Harris may be trying to say is,> I
have found an algebraic integer which is a root of a>
*non-monic* polynomial etc., thus defying a well-known>
theorem. Therefore there is something wrong with>
mathematics: either the theorem is wrong (as he has> long
contended) or the de?ition of algebraic integer> is wrong,
or something else.> OK, if he's made that statement. I
thought you were> interpreting the remarks about I have
proved that x is a> factor of y in the algebraic integers,
but x/y is not> an algebraic integer.> Or perhaps you are.
Is that his proof that x is a> factor of y? I've never yet
been able to ?ure out> which part of James' writings are
supposed to be the> proof part, but every once in a while he
starts> talking as if a proof went by when nobody was
looking.> - Randy In connection of the x/y thing, he says
he has foundproblems in core mathematics and taught
mathematics -he has said the algebraic integers are
incomplete(presumably as opposed to objects, where he has
acircular de?ition and he has yet to identify even one object
which is not an integer). My reply was none too clear, partly
because I am notcertain of what he is implying. I <think> he
could be saying,I can prove that x/y is an algebraic integer
but alsothe theorem implies that it's *not* an algebraic
integer.Therefore something must be wrong with math. With
regardto where the proof is in this case, as I said before,
Idon't know and I am not sure he now knows. If it dependson
his Advanced Polynomial Factorization paper, it isnowhere,
per Hall and Baron. Andrzej
===
Ullrich, you're busted;you
have a *lot* of explaining to do,to the Order of the Undead
Mathematicians!> Lately I've been explaining how I found an
error in taught> mathematics as I prove that in the ring of
integers a number > I'll here call x has a factor I'll here
call y, but x/y is not an> integer.> Let x = 2, y = 3.>
Now then, taking that result in that ring, I can now move to
the ?ld> of rational numbers.> Now I have division, and in
that ?ld, I take x/y and notice that> it's NOT an integer,
which is a contradiction.--A church-school McCrusade (Blair's
ideals?):Harry-the-Mad-Potter want's US to kill Iraqis?...For
a 1000-year anglo-american hegemony?HEY, JIMMY; LET'S US
and SU FIGHT -then-PM of England &
Zbiggyhttp://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
http://www.rwgrayprojects.com/synergetics/plates/plates.html
http://quincy4board.homestead.com/?es/curriculum/Cosmo.PCX==
==Q. There is a Chinese steel company that produces steel
rods for local construction company. The plant manager knew
from past statistics that the rodsproduced are normally
distributed with a mean length of 300cm. The standard
deviation is 8cm. If 22% of the sample means are more than
speci? lengthK, what is the value of K?Solution:Given: u =
300cm, sigma = 8cm, n = 14, p(>K)= 0.22Let K = be the length
of the steel rodSince u (population) = u (sample) = 300,
sigma (sample) = sigma /squareroot(14) = 2.138099P(bar X >
K) = [K - u(sample)]/2.138099 = Zvalue(0.22)(K - 300) /
2.138089935 = Zvalue(0.22)(K - 300) / 2.138089935 = 0.58K =
2.138089935(0.58) + 300K = 301.2401cmTherefore, if 22% of the
sample means are more than a speci? length K, then K is =
301.2401cm.Is my answer correct?
===
I am a cabinetmaker and a
novice programmer. I have a interest in writinga plywood
optimization application. This app takes a number of piecesof
plywood and ?ds the best way to orient the pieces on a
standardsheet size i.e. 4 x 8. I`m looking for a mathematical
algorithm that dealswith this type of problem. I`m wondering
if you might able to guide mein the right direction.Here is a
link of a commercial product that provides a optimization
ofplywood. http://www.sheetlayout.com/Best Wishes, Barry
Golash
===
I tried to post this message before. I hope I'm not
just being impatient and posting the same thing twice. Anyway,
this is my question. Lets say I have y=f[x1(t),x2(t)] and I
want to plot y(t) after removing the effect of x2 so that,
in reality, I can have an idea of what x1(t) looks like. The
data I have is y(t) (in which x1 and x2 are both taken into
account) and x2(t). Ok, I hope I have made myself clear.
===
>
I have y=f[x1(t),x2(t)] and I want to plot y(t) after
removing the> effect of x2 so that, in reality, I can have
an idea of what x1(t) looks> like. The data I have is y(t)
(in which x1 and x2 are both taken into> account) and x2(t).
Ok, I hope I have made myself clear.No, hardly at all. You
want to plot, y = f(x1(t), x2(a)) to see whatx1(t) looks
like? Oh, you have y(t) = f(x1(t), x2(t)) and want to
?dx1(t) from y(t) and x2(t) ?What if f(x,y) = g(y) ? Then
y(t) = f(x1(t), x2(t)) = g(x2(t))and x1(t) could be anything
and do anything it wanted as alongas y(t) and x2(t) were
arbitrarly related as y(t) = g(x2(t)).
===
> I tried to post
this message before. I hope I'm not just being impatient >
and posting the same thing twice. Anyway, this is my
question. Lets say > I have y=f[x1(t),x2(t)] and I want to
plot y(t) after removing the > effect of x2 so that, in
reality, I can have an idea of what x1(t) looks > like. The
data I have is y(t) (in which x1 and x2 are both taken into >
account) and x2(t). Ok, I hope I have made myself clear.In
order to solve such a problem, you need to assume something
aboutthe relationship between y, x1 and x2 (e.g. that x1 and
x2 areindependent, and that y is dependent on both
variables). In otherwords, you need to specify the problem
more precisely. In particular,you must make an assumption
about the functional relationship. Forexample, a common
problem in linear system theory is that ofdeconvolution. i.e.
assume y = x1 * x2, where * denotes convolution. The goal in
that case would be to solve for x1 given x2 and y. Generally,
this sort of problem falls under the category of aninverse
problem. This is a very complicated subject with
manypossible approaches. The chosen approach depends very
much on yourparticular problem, and what you may assume about
the relationshipbetween your variables x1, x2, and y. I hope
this helps.
===
>>I tried to post this message before. I
hope I'm not just being impatient >>and posting the same
thing twice. Anyway, this is my question. Lets say >> I have
y=f[x1(t),x2(t)] and I want to plot y(t) after removing the
>>effect of x2 so that, in reality, I can have an idea of
what x1(t) looks >>like. The data I have is y(t) (in which x1
and x2 are both taken into >>account) and x2(t). Ok, I hope I
have made myself clear.> In order to solve such a
problem, you need to assume something about> the relationship
between y, x1 and x2 (e.g. that x1 and x2 are> independent,
and that y is dependent on both variables). In other> words,
you need to specify the problem more precisely. In
particular,> you must make an assumption about the functional
relationship. For> example, a common problem in linear system
theory is that of> deconvolution. i.e. assume y = x1 * x2,
where * denotes convolution.> The goal in that case would
be to solve for x1 given x2 and y. > Generally, this sort of
problem falls under the category of an> inverse problem.
This is a very complicated subject with many> possible
approaches. The chosen approach depends very much on your>
particular problem, and what you may assume about the
relationship> between your variables x1, x2, and y. I hope
this helps.It does help. If nothing else, it lets me know
that it wasn't as simple
===
By the way, I do know that y is
dependent on both variables and that x1 and x2 are
independent but that's about it. Oh, and I also have the
following data: y vs t and x2 vs t. Anyway, like I said, I'll
have to look into it some more.>>I tried to post this
message before. I hope I'm not just being impatient >>and
posting the same thing twice. Anyway, this is my question.
Lets say >> I have y=f[x1(t),x2(t)] and I want to plot y(t)
after removing the >>effect of x2 so that, in reality, I
can have an idea of what x1(t) looks >>like. The data I have
is y(t) (in which x1 and x2 are both taken into >>account)
and x2(t). Ok, I hope I have made myself clear.> In order
to solve such a problem, you need to assume something about>
the relationship between y, x1 and x2 (e.g. that x1 and x2
are> independent, and that y is dependent on both variables).
In other> words, you need to specify the problem more
precisely. In particular,> you must make an assumption about
the functional relationship. For> example, a common problem
in linear system theory is that of> deconvolution. i.e.
assume y = x1 * x2, where * denotes convolution.> The goal
in that case would be to solve for x1 given x2 and y. >
Generally, this sort of problem falls under the category of
an> inverse problem. This is a very complicated subject
with many> possible approaches. The chosen approach depends
very much on your> particular problem, and what you may
assume about the relationship> between your variables x1, x2,
and y. I hope this helps.
===
> Lately I've been explaining how
I found an error in taughtmathematics> as I prove that in the
ring of algebraic integers a number I'll here> call x has a
factor I'll here call y, but x/y is not an algebraic>
integer.James,I tried to frame a de?ition of ?factor' in the
thread ?Error in mypaper Advanced Polynomial Factorization'. I
asked if my proposedde?ition was correct, but you did not
respond. Is that de?itioncorrect?John Peters
===
>> Lately
I've been explaining how I found an error in taught>
mathematics> as I prove that in the ring of algebraic
integers a number I'llhere> call x has a factor I'll here
call y, but x/y is not an algebraic> integer.>> James,>> I
tried to frame a de?ition of ?factor' in the thread ?Error
in my> paper Advanced Polynomial Factorization'. I asked if
my proposed> de?ition was correct, but you did not respond.
Is that de?ition> correct?>> John PetersOops, posting under
wrong pseudonym.Clive
Tooth
===
<3c65f87.0306171917.10e891a3@posting.google.com>,>
Oh yeah, it'd help if you know mathematics.> James
HarrisIf who knew mathematics?