mm-176
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simplifed
Visiting Assistant Professor at the University of Montana.
[.snip.]
>So go back now and read posts from Arturo Magidin, Keith
Ramsay, and
>Nora Baron among others, and ask yourself: What do they think
of me,
>really?
What does it matter? Your arguments fall on their own.
Now, go back and read posts from James Harris, and ask
yourself: what
does he think of me, really? And why?
=
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 'gures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the inde'nite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
> Well I'm going to try and break it down even more to try 
and
see if
> y'all will accept the mathematics:
Ok I have
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 +
u^3 f)
where f is a prime integer other than 3, and u is coprime to
f, and
> looking at that x, I see the possibility for the
factorization
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
That's what I've been putting up a lot where 
you see a LOT of
symbols,
> which seem to confuse people. The ring is algebraic
integers, and let
> me get rid of as many of those symbols as I can:
Let x=2, f=5, u=1, so that I have
P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
which is what I put up earlier, but I'm wary about some of 
you
still
> 'nding that confusing, so I'll work it out 
more to get
P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and now you can see what the polynomial P(m) looks like
without so
> many symbols.
Now from before where I had x, I *still* have that
P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
So
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(5000m^3 - 600 m^2 - 126m + 11).
Now setting m=0 gives me
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
Now let's say you accept that any factor of a polynomial can 
be
> written like r+c, where r=0, or r varies as the polynomial
variable
> varies, while c remains constant and is a factor of the
constant term.
Notice that
a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 -
75m^2 + 3m)
> which equals 0, when m=0, so at least one of the a's must
equal 0,
> when m=0.
And to get that factor that is 25, you must have two a's 
that
go to 0,
> when m=0.
Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims to
the effect
that you think these functions are continuous, so they
certainly should
take on non-algebraic integer values.
Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
would make your work easier to read and much clearer.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
> would make your work easier to read and much clearer.
... and much easier to debunk, so he won't do it.
Dirk Vdm
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
>>would make your work easier to read and much clearer.
> ... and much easier to debunk, so he won't do it.
>
Maybe if he did it he could 'nd his own mistakes and not
bother us with
them. Or maybe it has something to do with his Object Math. I
looked
at his website but couldn't make much sense of it. Lack of
clear
de'nitions with examples, perhaps.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
Visiting Assistant Professor at the University of Montana.
>Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
>would make your work easier to read and much clearer.
>> ... and much easier to debunk, so he won't do it.
>>
Maybe if he did it he could 'nd his own mistakes and not
bother us with
>them. Or maybe it has something to do with his Object Math. I
looked
>at his website but couldn't make much sense of it. Lack of
clear
>de'nitions with examples, perhaps.
The shortcomings of his de'nition of object have been pointed
out
numerous times. While denying any problems repeatedly, he has
nonetheless made subtle changes to it; yet, it still remains a
problem. Since it is supposed to be the basis of his whole
enterprise,
that is enough to establish that it tumbles down.
At one point, I tried to explain exactly why his de'nition of
object
was faulty, and why even if one were to give it its most
liberal
reading, the de'nition is such that it only includes 
integers.
He
never replied.
40agate.berkeley.edu
James is not interested in 'nding his own mistakes. Never has
been. In fact, he will avoid confronting them as long as
humanly
possible, if not longer.
=
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 'gures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the inde'nite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
=
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>Also, why do you INSIST on not using the notation a_i(m)
or a_i(0)? It
>>would make your work easier to read and much clearer.
> ... and much easier to debunk, so he won't do it.
 Maybe if he did it he could 'nd his own mistakes and not
bother us with
> them. Or maybe it has something to do with his Object Math.
I looked
> at his website but couldn't make much sense of it. Lack of
clear
> de'nitions with examples, perhaps.
IMO the only thing someone who suffers from this kind of
illness is interested in, is attention - it doesn't matter 
if
it is
positive or negative. He's getting plenty of it, so I guess 
we
are all very kind to James :-)
Dirk Vdm
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>Also, why do you INSIST on not using the notation a_i(m)
or a_i(0)? It
>>would make your work easier to read and much clearer.
>... and much easier to debunk, so he won't do it.
>>Maybe if he did it he could 'nd his own mistakes and not
bother us with
>>them. Or maybe it has something to do with his Object Math.
I looked
>>at his website but couldn't make much sense of it. Lack of
clear
>>de'nitions with examples, perhaps.
> The shortcomings of his de'nition of object have been
pointed out
> numerous times. While denying any problems repeatedly, he has
> nonetheless made subtle changes to it; yet, it still remains
a
> problem. Since it is supposed to be the basis of his whole
enterprise,
> that is enough to establish that it tumbles down.
At one point, I tried to explain exactly why his de'nition of
object
> was faulty, and why even if one were to give it its most
liberal
> reading, the de'nition is such that it only includes
integers. He
> never replied.
40agate.berkeley.edu
James is not interested in 'nding his own mistakes. Never has
> been. In fact, he will avoid confronting them as long as
humanly
> possible, if not longer.
His habit of not responding to counter-proofs directly, along
with his
tendency to start new threads and simply reassert his claims is
something I've become all too familiar with in the past 
month.
I still
haven't 'gured out why he wants all of us to 
join MSN's
Amateur Math to
just post direct links there.
Right now I wonder how long it will be before I tire of
reading his
and the practice serves no purpose.
Do you think there's a chance of getting him to play around
with Set
Theory or Computability Theory?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>Also, why do you INSIST on not using the notation a_i(m)
or a_i(0)? It
>>would make your work easier to read and much clearer.
>... and much easier to debunk, so he won't do it.
>>Maybe if he did it he could 'nd his own mistakes and not
bother us with
>>them. Or maybe it has something to do with his Object Math.
I looked
>>at his website but couldn't make much sense of it. Lack of
clear
>>de'nitions with examples, perhaps.
> IMO the only thing someone who suffers from this kind of
> illness is interested in, is attention - it doesn't matter
if it is
> positive or negative. He's getting plenty of it, so I 
guess
we
> are all very kind to James :-)
Dirk Vdm
If only he would reciprocate and give each of us as much
attention as we
give him. Answers to objections might be nice. Minus the
generalized
attacks on character would be nicer.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>Also, why do you INSIST on not using the notation
a_i(m) or a_i(0)?
It
>>would make your work easier to read and much clearer.
>... and much easier to debunk, so he won't do it.
>>Maybe if he did it he could 'nd his own mistakes and not
bother us
with
>>them. Or maybe it has something to do with his Object Math.
I looked
>>at his website but couldn't make much sense of it. Lack of
clear
>>de'nitions with examples, perhaps.
> IMO the only thing someone who suffers from this kind of
> illness is interested in, is attention - it doesn't matter
if it is
> positive or negative. He's getting plenty of it, so I 
guess
we
> are all very kind to James :-)
> Dirk Vdm
If only he would reciprocate and give each of us as much
attention as we
> give him. Answers to objections might be nice. Minus the
generalized
> attacks on character would be nicer.
That would be nice indeed. It would also be nice if mental
disease would be effectively curable with proper medication.
But don't let me put you off - enjoy it while you can :-)
Dirk Vdm
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>Well I'm going to try and break it down even more to try 
and
see if
>y'all will accept the mathematics:
[...]
So go back now and read posts from Arturo Magidin, Keith
Ramsay, and
>Nora Baron among others, and ask yourself: What do they think
of me,
>really?
Fascinating argument: I know, I'll behave like a complete 
and
total
ass. Then people will hate me, and then when they say I'm 
wrong
about the math I can explain that it's just because they 
hate
me.
Too bad nobody's stupid enough to buy it.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Well I'm going to try and break it down even more to try 
and
see if
> y'all will accept the mathematics:
> Ok I have
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2
+ u^3 f)
> where f is a prime integer other than 3, and u is coprime to
f, and
> looking at that x, I see the possibility for the
factorization
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
> That's what I've been putting up a lot where 
you see a LOT
of symbols,
> which seem to confuse people. The ring is algebraic
integers, and let
> me get rid of as many of those symbols as I can:
> Let x=2, f=5, u=1, so that I have
> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
> which is what I put up earlier, but I'm wary about some of
you still
> 'nding that confusing, so I'll work it out 
more to get
> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> and now you can see what the polynomial P(m) looks like
without so
> many symbols.
> Now from before where I had x, I *still* have that
> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> So
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
> 25(5000m^3 - 600 m^2 - 126m + 11).
> Now setting m=0 gives me
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
> Now let's say you accept that any factor of a polynomial 
can
be
> written like r+c, where r=0, or r varies as the polynomial
variable
> varies, while c remains constant and is a factor of the
constant term.
> Notice that
> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 -
75m^2 + 3m)
That's like saying, if I write y=mx+b, that you think it
should be
y(x)=mx+b.
> which equals 0, when m=0, so at least one of the a's must
equal 0,
> when m=0.
> And to get that factor that is 25, you must have two a's
that go to 0,
> when m=0.
Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims to
the effect
> that you think these functions are continuous, so they
certainly should
> take on non-algebraic integer values.
The a's are roots of a monic polynomial with integer
coef'cients,
given that m is an integer.
Now if you wish to make m rational then you can explore
continuity.
> Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
> would make your work easier to read and much clearer.
Why oh why do people ever insist on writing y=mx+b instead of
y(x)=mx+b?
It seems to me that you need to begin a crusade to make
certain that
such a horror is no longer committed.
You can be a paragon of mathematical style on a quest to save
the
ignorant masses from this travesty of ever writing such a
thing as
y=mx+b as instead you make certain that people write y(x) = mx
+ b!!!
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
contributions.
> Well I'm going to try and break it down even more to try 
and
see if
> y'all will accept the mathematics:
Ok I have
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 +
u^3 f)
where f is a prime integer other than 3, and u is coprime to
f, and
> looking at that x, I see the possibility for the
factorization
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
That's what I've been putting up a lot where 
you see a LOT of
symbols,
> which seem to confuse people. The ring is algebraic
integers, and let
> me get rid of as many of those symbols as I can:
Let x=2, f=5, u=1, so that I have
P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
which is what I put up earlier, but I'm wary about some of 
you
still
> 'nding that confusing, so I'll work it out 
more to get
P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and now you can see what the polynomial P(m) looks like
without so
> many symbols.
Now from before where I had x, I *still* have that
P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
So
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(5000m^3 - 600 m^2 - 126m + 11).
This is greatly simpli'ed. I'm with you so 
far.
Now setting m=0 gives me
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
Now let's say you accept that any factor of a polynomial can 
be
> written like r+c, where r=0, or r varies as the polynomial
variable
> varies, while c remains constant and is a factor of the
constant term.
Notice that
a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
Well, no, I don't notice that. But I don't 
think it matters.
which equals 0, when m=0, so at least one of the a's must
equal 0,
> when m=0.
And to get that factor that is 25, you must have two a's 
that
go to 0,
> when m=0.
>
OK. So you've declared that two of the a's are 
0 at m=0.
Therefore, the third a=3 at m=0. I'll simplify if you 
don't
mind; makes it
easier for me:
a_3 = b_3 + 3
(2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(11).
OK. Still with you. And all of a_1, a_2, b_3 are 0 at m=0.
> (Note: Some posters have gotten a lot of mileage out of
calling that a
> degenerate case, but they were just fooling you into
forgetting your
> basic algebra and what you know about polynomials. I think
they did
> so deliberately as the math isn't complicated.)
Now comes the question of what happens when m is NOT 0, and
answering
> that question requires looking again at
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(5000m^3 - 600 m^2 - 126m + 11)
and noticing that the constant term is 25(11), but you can
divide off
> that 25 to get P(m)/25 which gives you a constant term
that's 11. And
> 11 and 5 are coprime. That's very important. In fact, 
that's
the
> *key* fact which should stick in your mind.
So now looking at
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 =
(5000m^3 - 600 m^2 - 126m + 11)
> I know that if 2 a_1 + 5 has a factor of 5, when m=0, then
that factor
> is a factor of the constant term, and in fact, it'd have a
factor of
> the constant term that is 5. So why would you think that the
factor
> of the constant term would move or change when m changes?
Well it can't.
I agree with you. I think I'm following. 5 is a factor of 5
for any value
of
m.
Given that with 2 a_1 + 5, that 5 in there is a factor of the
constant
> term of
25(5000m^3 - 600 m^2 - 126m + 11)
you *still* have a factor of the constant term when 25 is
separated
> off.
But now your constant term is 11.
That forces all the factors of 5 to go away from 2 a_1 + 5.
Oops. Think you just lost me. You've shown that 5 is a 
factor
of 2a_1 + 5
when a_1 is zero; not for a_1
non-zero. Could you clarify what you mean here?
Now you may wish for there to be someway for some factors to
remain,
> but what actuall happens is you get
(2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =
(5000m^3 - 600 m^2 - 126m + 11)
while posters have argued that *all* the a's would have some
factor of
> 5.
>
Well, not quite. They independently proved that *none* of the
a's were
coprime to 5 for a
completely different polynomial. I can't see the relevance.
> But let's let m=0 again and see what happens now, as that
gives
(2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11
and you'll notice I have 1 where I had 5 before, so it can 
all
work.
That is, the a's that went to 0 before will go to 0 again.
Some posters have claimed otherwise which I've called voodoo
math.
You're being far too polite. All your doing is dividing by 
5,
and asserting
that if a is zero then a/5 is
zero. If people are claiming otherwise they're complete 
liars.
Of course, if you think about it, you'll understand that 
they
probably
> knew the truth and just lied to you as they didn't think 
you
could
> 'gure it out, and probably didn't expect me 
to simplify in
this way.
>
Yep, they're liars.
> Those people aren't your friends. You may trust them. You
may admire
> them, but they clearly don't think much of you.
>
Yep, they're not my friends.
> So go back now and read posts from Arturo Magidin, Keith
Ramsay, and
> Nora Baron among others, and ask yourself: What do they
think of me,
> really?
>
It's not my place to opine what they think of you. I 
haven't
noticed any of
them claiming that a/5 is not zero
if a=0 however.
But. There's no conclusion here.
What are you saying?
Given:
(2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) =
25(5000m^3 - 600 m^2 - 126m + 11)
That some of the a's are coprime to 5?
How about
(2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) =
25(5000m^3 - 600 m^2 - 126m + 11)
By exactly the same reasoning, some of the c's are coprime 
to
7?
I think there's something missing in all this.
James Harris
Phil Nicholson
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Maybe if he did it he could 'nd his own mistakes and not
bother us with
them.
What you call bothering, he proudly calls modern problem
solving
techniques. Take a look -
3c65f87.0304261407.7bb46534@posting.goo
gle.com
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> That would be nice indeed. It would also be nice if mental
> disease would be effectively curable with proper medication.
> But don't let me put you off - enjoy it while you can :-)
I enjoy the entertainment JSH provides as much as anyone, but
... I have
a nagging feeling that this is a modern intellectual version
of the
medieval court enjoying the antics of the fool, who was a
cripple or a
hunchback or otherwise handicapped.
Gib
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> That would be nice indeed. It would also be nice if mental
> disease would be effectively curable with proper medication.
> But don't let me put you off - enjoy it while you can :-)
I enjoy the entertainment JSH provides as much as anyone, but
... I have
> a nagging feeling that this is a modern intellectual version
of the
> medieval court enjoying the antics of the fool, who was a
cripple or a
> hunchback or otherwise handicapped.
... or the village idiot providing the necessary
entertainment on a hot Sunday afternoon.
Of course, that's very obvious.
Dirk Vdm
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>> That would be nice indeed. It would also be nice if mental
>> disease would be effectively curable with proper medication.
>> But don't let me put you off - enjoy it while you can :-)
> I enjoy the entertainment JSH provides as much as anyone,
but ... I have
> a nagging feeling that this is a modern intellectual version
of the
> medieval court enjoying the antics of the fool, who was a
cripple or a
> hunchback or otherwise handicapped.
That does seem to be the case. The difference is: in medieval
courts,
the fool was aware of his situation, as well as the true
purpose of his
being there.
I think similar analogies could be made to people throwing
rocks in the
school yard. There are times when I have to wonder if we
wouldn't
perform a better service by responding only to his math and
ignoring/deleting everything else that comes from his mouth.
Maybe then
he would learn to focus on the math as well.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>Maybe if he did it he could 'nd his own mistakes and not
bother us with
them.
> What you call bothering, he proudly calls modern problem
solving
techniques. Take a look -
>
3c65f87.0304261407.7bb46534@posting.goog
le.com
>
His claim that he later gathers the ideas that make the grade
is...
disturbing.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>Well I'm going to try and break it down even more to try 
and
see if
>y'all will accept the mathematics:
>Ok I have
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2
)xu^2 + u^3 f)
>where f is a prime integer other than 3, and u is coprime
to f, and
>looking at that x, I see the possibility for the
factorization
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
>That's what I've been putting up a lot where 
you see a LOT
of symbols,
>which seem to confuse people. The ring is algebraic
integers, and let
>me get rid of as many of those symbols as I can:
>Let x=2, f=5, u=1, so that I have
> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
>which is what I put up earlier, but I'm wary about some of
you still
>'nding that confusing, so I'll work it out 
more to get
> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>and now you can see what the polynomial P(m) looks like
without so
>many symbols.
>Now from before where I had x, I *still* have that
> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
>So
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
> 25(5000m^3 - 600 m^2 - 126m + 11).
>Now setting m=0 gives me
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
>Now let's say you accept that any factor of a polynomial
can be
>written like r+c, where r=0, or r varies as the polynomial
variable
>varies, while c remains constant and is a factor of the
constant term.
>Notice that
> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
>I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 -
75m^2 + 3m)
> That's like saying, if I write y=mx+b, that you think it
should be
> y(x)=mx+b.
You say this as if y(x)=mx+b is uncommon. When teaching about
linear
functions in algebra, it's usually written as y=mx+b. In
calculus, I've
seen both. This is especially true when switching rapidly back
and
forth between y as a function of x and y evaluated at a
particular value
of x.
>which equals 0, when m=0, so at least one of the a's must
equal 0,
>when m=0.
>And to get that factor that is 25, you must have two a's
that go to 0,
>when m=0.
>Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims 
to
the effect
>>that you think these functions are continuous, so they
certainly should
>>take on non-algebraic integer values.
> The a's are roots of a monic polynomial with integer
coef'cients,
> given that m is an integer.
I missed that m is an integer. What does that have to do with
the
possible values of a_i(0) though? You've said the 
a's are
roots of a
monic polynomial, but also that they don't need to be
polynomials
themselves. As a result, I see no reason for assuming anything
about
the codomain of those functions.
>>Also, why do you INSIST on not using the notation a_i(m) or
a_i(0)? It
>>would make your work easier to read and much clearer.
> Why oh why do people ever insist on writing y=mx+b instead of
> y(x)=mx+b?
Because it's a shortcut AND then don't use the 
symbol y to
refer to
both mx+b and m*0+b with little warning as to which it will be
in the
next line.
> It seems to me that you need to begin a crusade to make
certain that
> such a horror is no longer committed.
I'll take that under advisement. Would you care to assist 
me?
> You can be a paragon of mathematical style on a quest to
save the
> ignorant masses from this travesty of ever writing such a
thing as
> y=mx+b as instead you make certain that people write y(x) =
mx + b!!!
Can I use you as a model of the increased clarity that
happens? Maybe a
before and after snapshot.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
contributions.
> Well I'm going to try and break it down even more to try 
and
see if
> y'all will accept the mathematics:
> Ok I have
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2
)xu^2 + u^3 f)
> where f is a prime integer other than 3, and u is coprime
to f, and
> looking at that x, I see the possibility for the
factorization
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
> That's what I've been putting up a lot where 
you see a LOT
of symbols,
> which seem to confuse people. The ring is algebraic
integers, and let
> me get rid of as many of those symbols as I can:
> Let x=2, f=5, u=1, so that I have
> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
> which is what I put up earlier, but I'm wary about some of
you still
> 'nding that confusing, so I'll work it out 
more to get
> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> and now you can see what the polynomial P(m) looks like
without so
> many symbols.
> Now from before where I had x, I *still* have that
> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> So
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
> 25(5000m^3 - 600 m^2 - 126m + 11).
This is greatly simpli'ed. I'm with you so 
far.
Good.
> Now setting m=0 gives me
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
> Now let's say you accept that any factor of a polynomial
can be
> written like r+c, where r=0, or r varies as the polynomial
variable
> varies, while c remains constant and is a factor of the
constant term.
> Notice that
> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
Well, no, I don't notice that. But I don't 
think it matters.
It's very important as it shows part of the relationship
between the
a's and m, which is fully de'ned by three 
expressions, but I'm
focusing on one.
Here the point is that at *least* one of the a's is equal to
0, when
m=0.
Remember the a's come from the factorization of P(m), where
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 +
u^3 f),
and I've put in some numbers for f, x, and u, as f=5, x=2, 
and
u=1.
The factorization is
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
Note to readers: I'm leaving everything in from the previous
post,
which makes things kind of bulky, but I'm hoping that in 
this
case
you'll be better served by having all the information 
readily
available. Though I do fear there may be a bit of overload for
some
of you.
> which equals 0, when m=0, so at least one of the a's must
equal 0,
> when m=0.
> And to get that factor that is 25, you must have two a's
that go to 0,
> when m=0.
> OK. So you've declared that two of the a's 
are 0 at m=0.
No that's required by the given expressions.
> Therefore, the third a=3 at m=0. I'll simplify if you 
don't
mind; makes
it
> easier for me:
> a_3 = b_3 + 3
Whatever works for you is 'ne with me, as long as it works,
and is
mathematically correct.
That looks ok as you're identifying part of the constant 
term
for the
factor
2a_3 + 5
as the full value is 11 for its constant term is 11.
Here you have b_3 as part of the varying portion.
> (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(11).
OK. Still with you. And all of a_1, a_2, b_3 are 0 at m=0.
Yup.
> (Note: Some posters have gotten a lot of mileage out of
calling that a
> degenerate case, but they were just fooling you into
forgetting
your
> basic algebra and what you know about polynomials. I think
they did
> so deliberately as the math isn't complicated.)
> Now comes the question of what happens when m is NOT 0,
and answering
> that question requires looking again at
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> and noticing that the constant term is 25(11), but you can
divide off
> that 25 to get P(m)/25 which gives you a constant term
that's 11. And
> 11 and 5 are coprime. That's very important. In fact, 
that's
the
> *key* fact which should stick in your mind.
> So now looking at
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 =
> (5000m^3 - 600 m^2 - 126m + 11)
> I know that if 2 a_1 + 5 has a factor of 5, when m=0, then
that factor
> is a factor of the constant term, and in fact, it'd have a
factor of
> the constant term that is 5. So why would you think that the
factor
> of the constant term would move or change when m changes?
> Well it can't.
I agree with you. I think I'm following. 5 is a factor of 5
for any value
of
> m.
Ok, let see what's next.
> Given that with 2 a_1 + 5, that 5 in there is a factor of
the constant
> term of
> 25(5000m^3 - 600 m^2 - 126m + 11)
> you *still* have a factor of the constant term when 25 is
separated
> off.
> But now your constant term is 11.
> That forces all the factors of 5 to go away from 2 a_1 + 5.
Oops. Think you just lost me. You've shown that 5 is a 
factor
of 2a_1 + 5
> when a_1 is zero; not for a_1
> non-zero. Could you clarify what you mean here?
Hmmm...how about this explanation?
Remember that the polynomial is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and you see that factor 25. Now imagine that I have the
polynomial
Q(m), where
P(m) = 25 Q(m)
so you have as a factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
and the question is, how does that factor of 25 divide out?
Well, checking at Q(0), gives 11, so how can those factors of
25
divide through Q(m) in such a way as to give 11, at m=0?
There's only one way, which gives you
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
If that bothers you, remember that at m=0, two of the a's
equal 0.
If it still bothers you, try and get everything to work some
other
way.
And yes, it can be shown rigorously, but right now there's 
the
problem
of getting that feel for the math.
Then you can go to the paper Advanced Polynomial Factorization.
> Now you may wish for there to be someway for some factors
to remain,
> but what actuall happens is you get
> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =
> (5000m^3 - 600 m^2 - 126m + 11)
> while posters have argued that *all* the a's would have
some factor of
> 5.
>
Well, not quite. They independently proved that *none* of the
a's were
> coprime to 5 for a
> completely different polynomial. I can't see the 
relevance.
Nope. It turns out that they're making their claim for the
polynomial
I've given you, but in the past there have been *symbols*
where I've
now given numbers.
> But let's let m=0 again and see what happens now, as that
gives
> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11
> and you'll notice I have 1 where I had 5 before, so it can
all work.
> That is, the a's that went to 0 before will go to 0 again.
> Some posters have claimed otherwise which I've called
voodoo math.
You're being far too polite. All your doing is dividing by 
5,
and
asserting
> that if a is zero then a/5 is
> zero. If people are claiming otherwise they're complete
liars.
Hmmm...looks like I screwed up in my statement, as what I
should have
said is that they've made that claim when m does NOT equal 
0.
My
apologies.
The voodoo math has come in when m doesn't equal 0, as if
suddenly
constant factors shift around.
> Of course, if you think about it, you'll understand that
they probably
> knew the truth and just lied to you as they didn't think 
you
could
> 'gure it out, and probably didn't expect me 
to simplify in
this way.
> Yep, they're liars.
Well that is true as can be seen by the replies from some of
them in
several threads I created yesterday, like this one.
My simpli'cation removes areas for doubts, and if they 
weren't
liars,
they'd simply acknowledge the truth at this point.
> Those people aren't your friends. You may trust them. You
may admire
> them, but they clearly don't think much of you.
> Yep, they're not my friends.
Hmmm...so much agreement puts me somewhat at a loss for words.
> So go back now and read posts from Arturo Magidin, Keith
Ramsay, and
> Nora Baron among others, and ask yourself: What do they
think of me,
> really?
> It's not my place to opine what they think of you. I 
haven't
noticed any
of
> them claiming that a/5 is not zero
> if a=0 however.
That is true.
> But. There's no conclusion here.
> What are you saying?
Given:
> (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
That some of the a's are coprime to 5?
It turns out that in the ring of algebraic integers they all
are,
which is why the ring has problems.
> How about
> (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
By exactly the same reasoning, some of the c's are coprime 
to
7?
No.
> I think there's something missing in all this.
That's ok. My suggestion is to consider my reply and see if
you have
that same feeling.
Remember what I'm doing is considering a polynomial
factorization,
with speci'c reference to its constant term *because* 
I'm
using that
as my anchor given that I'm using non-polynomial factors of
that
polynomial.
That is, see the constant term as a lighthouse in the darkness.
Or better yet, see the math here as requiring that you §y by
your
instruments, as otherwise you may spiral into the ground,
'guratively
speaking.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
Fair bit of snippage. I've got some work to do now to follow
this up
properly.
<Snip > Notice that
> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
> Well, no, I don't notice that. But I don't 
think it
matters.
Sorry, poorly worded on my part. I *DID* notice that a_1 a_2
a_3 *DOESN'T*
in fact equal
It
doesn't
affect
your main point.
It's very important as it shows part of the relationship
between the
> a's and m, which is fully de'ned by three 
expressions, but
I'm
> focusing on one.
Here the point is that at *least* one of the a's is equal to
0, when
> m=0.
Which was this. And agreed.
<Snip > And to get that factor that is 25, you must have two
a's that go to
0,
> when m=0.
> OK. So you've declared that two of the a's 
are 0 at m=0.
No that's required by the given expressions.
>
OK. I'm guessing that you're right. But 
I'll check this and
get back if
necessary.
> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =
> (5000m^3 - 600 m^2 - 126m + 11)
> while posters have argued that *all* the a's would have
some factor
of
> 5.
> Well, not quite. They independently proved that *none* of
the a's were
> coprime to 5 for a
> completely different polynomial. I can't see the 
relevance.
Nope. It turns out that they're making their claim for the
polynomial
> I've given you, but in the past there have been *symbols*
where I've
> now given numbers.
I don't recall any claims being made by others. I did see
proofs made by
others that
for the polynomial 65x^3 -12x +1, none of the a's are 
coprime
to 5.
The polynomial under discussion here is 5000m^3 -600m^2 -126m
+11.
<Snip > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> That some of the a's are coprime to 5?
It turns out that in the ring of algebraic integers they all
are,
> which is why the ring has problems.
Surely not. What about when m=5? What are the a's in that 
case?
> How about
> (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> By exactly the same reasoning, some of the c's are coprime
to 7?
No.
I believe you'd be right.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Fair bit of snippage. I've got some work to do now to 
follow
this up
> properly.
> <Snip > Notice that
> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
> Well, no, I don't notice that. But I don't 
think it
matters.
Sorry, poorly worded on my part. I *DID* notice that a_1 a_2
a_3
*DOESN'T*
> in fact equal
It doesn't
affect
> your main point.
Yeah, you're correct. That should be
a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m).
> It's very important as it shows part of the relationship
between the
> a's and m, which is fully de'ned by three 
expressions, but
I'm
> focusing on one.
> Here the point is that at *least* one of the a's is equal
to 0, when
> m=0.
Which was this. And agreed.
<Snip > And to get that factor that is 25, you must have two
a's that go to
0,
> when m=0.
> OK. So you've declared that two of the a's 
are 0 at m=0.
> No that's required by the given expressions.
> OK. I'm guessing that you're right. But 
I'll check this and
get back if
> necessary.
Well I guess you're thinking that I just picked the
possibility that
two of the a's are 0, and maybe it's possible 
for, say, only
one of
the a's to equal 0.
But if you do that then you can only get a factor of 5 that is
5, for
the result.
> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =
> (5000m^3 - 600 m^2 - 126m + 11)
> while posters have argued that *all* the a's would
have some factor
of
> 5.
> Well, not quite. They independently proved that *none*
of the a's
were
> coprime to 5 for a
> completely different polynomial. I can't see the 
relevance.
> Nope. It turns out that they're making their claim for the
polynomial
> I've given you, but in the past there have been *symbols*
where I've
> now given numbers.
I don't recall any claims being made by others. I did see
proofs made by
> others that
> for the polynomial 65x^3 -12x +1, none of the a's are
coprime to 5.
> The polynomial under discussion here is 5000m^3 -600m^2
-126m +11.
Then you haven't looked closely. The claim is against the
polynomial
factorization
(v^3+1)x^3 - 3v xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
where v=-1+mf^2, and readers can get from there to
25(5000m^3 -600m^2 -126m +11)
easily enough by plugging in f=5, x=2, y=5.
> <Snip > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> That some of the a's are coprime to 5?
> It turns out that in the ring of algebraic integers they
all are,
> which is why the ring has problems.
Surely not. What about when m=5? What are the a's in that 
case?
The problem is with the ring of algebraic integers as I've
explained.
Consider the following which you deleted out from my previous
post:
Hmmm...how about this explanation?
Remember that the polynomial is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and you see that factor 25. Now imagine that I have the
polynomial
Q(m), where
P(m) = 25 Q(m)
so you have as a factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
and the question is, how does that factor of 25 divide out?
Well, checking at Q(0), gives 11, so how can those factors of
25
divide through Q(m) in such a way as to give 11, at m=0?
There's only one way, which gives you
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
If that bothers you, remember that at m=0, two of the a's
equal 0.
If it still bothers you, try and get everything to work some
other
way.
> How about
> (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> By exactly the same reasoning, some of the c's are coprime
to 7?
> No.
> I believe you'd be right.
The math should be simple now that so many symbols have been
replaced
with numbers.
The issue is an error in taught mathematics revealed by what
I've
shown here.
My position is that mathematicians should be diligent in
teaching the
truth, and should acknowledge an error, so that it is no longer
taught.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Well I'm going to try and break it down even more to try 
and
see if
> y'all will accept the mathematics:
Ok I have
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 +
u^3 f)
where f is a prime integer other than 3, and u is coprime to
f, and
> looking at that x, I see the possibility for the
factorization
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
That's what I've been putting up a lot where 
you see a LOT of
symbols,
> which seem to confuse people. The ring is algebraic
integers, and let
> me get rid of as many of those symbols as I can:
Let x=2, f=5, u=1, so that I have
P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)
which is what I put up earlier, but I'm wary about some of 
you
still
> finding that confusing, so I'll work it out 
more to get
P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and now you can see what the polynomial P(m) looks like
without so
> many symbols.
Now from before where I had x, I *still* have that
P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
>
Interesting. When the x's were left 
unspeci'ed,
the form of the factorization was
P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5).
Now that you have substituted in x = 2, it is no
longer a factorization of a polynomial in x. It is
just a factorization as a product of three numbers.
So 2*a_1 + 5, for example, is just an algebraic
integer. The factorization is
[1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5).
So then what you are asserting below is that if you factor
the number P(m) as in [1], then one of a1, a2, or
a3 must be coprime to 5. Right?
Let's take m = 1. Then
P(m) = 25 * 4285.
This can be factored as
5 * 5 * 4285,
which yields a1 = 0, a2 = 0, and a3 = 2140.
None of these is coprime to 5. End of story.
Don't like a1 = a2 = 0 ? Other things work
too - e.g., a1 = -5, a2 = -5, a3 = 2140. None
coprime to 5, as before.
Read on, however -
> So
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(5000m^3 - 600 m^2 - 126m + 11).
Now setting m=0 gives me
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).
Now let's say you accept that any factor of a polynomial can 
be
> written like r+c, where r=0, or r varies as the polynomial
variable
> varies, while c remains constant and is a factor of the
constant term.
Notice that
a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
which equals 0, when m=0, so at least one of the a's must
equal 0,
> when m=0.
And to get that factor that is 25, you must have two a's 
that
go to 0,
> when m=0.
(Note: Some posters have gotten a lot of mileage out of
calling that a
> degenerate case, but they were just fooling you into
forgetting your
> basic algebra and what you know about polynomials. I think
they did
> so deliberately as the math isn't complicated.)
Now comes the question of what happens when m is NOT 0, and
answering
> that question requires looking again at
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(5000m^3 - 600 m^2 - 126m + 11)
and noticing that the constant term is 25(11), but you can
divide off
> that 25 to get P(m)/25 which gives you a constant term
that's 11. And
> 11 and 5 are coprime. That's very important. In fact, 
that's
the
> *key* fact which should stick in your mind.
So now looking at
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 =
(5000m^3 - 600 m^2 - 126m + 11)
> I know that if 2 a_1 + 5 has a factor of 5, when m=0, then
that factor
> is a factor of the constant term, and in fact, it'd have a
factor of
> the constant term that is 5. So why would you think that the
factor
> of the constant term would move or change when m changes?
Well it can't.
Given that with 2 a_1 + 5, that 5 in there is a factor of the
constant
> term of
25(5000m^3 - 600 m^2 - 126m + 11)
you *still* have a factor of the constant term when 25 is
separated
> off.
But now your constant term is 11.
That forces all the factors of 5 to go away from 2 a_1 + 5.
Now you may wish for there to be someway for some factors to
remain,
> but what actuall happens is you get
(2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =
(5000m^3 - 600 m^2 - 126m + 11)
while posters have argued that *all* the a's would have some
factor of
> 5.
>
Yep, sure. See above. With actual numbers!
> But let's let m=0 again and see what happens now, as that
gives
(2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11
and you'll notice I have 1 where I had 5 before, so it can 
all
work.
That is, the a's that went to 0 before will go to 0 again.
Some posters have claimed otherwise which I've called voodoo
math.
Of course, if you think about it, you'll understand that 
they
probably
> knew the truth and just lied to you as they didn't think 
you
could
> 'gure it out, and probably didn't expect me 
to simplify in
this way.
Those people aren't your friends. You may trust them. You 
may
admire
> them, but they clearly don't think much of you.
So go back now and read posts from Arturo Magidin, Keith
Ramsay, and
> Nora Baron among others, and ask yourself: What do they
think of me,
> really?

> James Harris
> I created several threads today where I went ahead and used
my ability
> to put in some actual numbers for symbols in some key
expressions, and
> looking over the newsgroup, I see a lot of replies.
My suggestion is that you consider those replies and notice
how few
> talked about the math presented by actually referencing it,
and how
> many just asserted things--after deleting it out.
It's almost funny until you realize why there was so much
frenetic
> energy, and why these people are so desperate to hide the
math that
> they continually delete it out:
The mathematics is rather simple, the algebra is basic, but the
> conclusion is dramatic.
So let's say you were facing people who needed to keep 
people
from
> looking closely, and every time you tried to simplify and
show
> details, they'd jump in and hide details and make things 
more
> convoluted?
Well you might do what I've done today and stretch them out.
>
What this means is: start a lot of new threads in the
hope that people won't notice that the problems in the
old ones were not answered. Mr. Harris imagines that he
is playing to a huge audience of silent lurkers, perhaps
even math undergrads. I am sure they will be impressed by
this clever evasive maneuver!
See below for what happens when people look closely and
show details ... be sure to note where I have hidden details
and made things more convoluted ... where I was so desperate
to hide the math that I continually deleted it out ...
> What I'm doing is not very complicated as 
I'm using the
extra symbols
> to factor a polynomial into non-polynomial factors.
Now that's a fascinating idea for factoring polynomials that 
I
guess
> is new to the math world.
And yes, bad apples can take advantage when there's 
something
new.
What is a surprise though is that mathematical society can be
taken in
> by people like Arturo Magidin or Nora Baron, when they're
posting so
> desperately, so quickly when stretched out, that they can
barely get
> to the math, or say things that are clearly false.
But what if they never really believed in mathematics?
What if for them it is a fashion show?
Remember mathematics can be about appearance for some people.
They
> might have gotten a lot of mileage out of being able to put
down math
> that looked good.
That is, you may have people who are cons in your midst who
are used
> to playing a game upon other mathematicians, and now 
they're
> trapped--stretched out--by a lot of threads and math where I
did the
> sudden move of putting in numbers where once there were
symbols.
It seems to me that in mathematics there is a certain respect
that is
> to be given to ideas, logic, and mathematical truth. Sure
being a
> rather large society it's not surprising that corrupt 
people
can slip
> in, and maybe get some stature, or get through a Ph.d
program.
But in society when the corrupt people reveal themselves
clearly by
> 'nally pushing beyond obvious limits--like denying basic
algebra--it
> becomes time to clean house. Look over those threads, look
at their
> desperation, and their contempt for algebra and your algebra
> knowledge, then please ask yourselves how it cannot now be
that time.
> James Harris
I looked a little more closely at Advanced Polynomial
Factorization.
It turns out that, in addition to basic conceptual problems,
it also
has careless mistakes in algebra.
Start with
P(m) = f^2 *((m^3*f^4 - 3*m^2*f^2 + 3*m)*x^3
- 3*(-1 + m*f^2)*x*u^2 + u^3*f),
as you give early in Section 2 of APF. You note that P(m) has
a factor of f^2.
Nothing wrong with this particular formula. However, on the
last
page, when you substitute in m = 1, f = sqrt(5), and u = 1,
you do
NOT obtain
65*x^3 - 12*x + 1.
Instead you get
65*x^3 - 60*x + 5*sqrt(5).
Now if this is factored in the form
(a1*x + 1)*(a2*x + 1)*(a3*x + 1),
it obviously doesn't work: 1*1*1 is not equal to 5*sqrt(5).
Back on the second page, in a parenthetical note at
the bottom, you say: Note: the a's are roots of a monic
polynomial with algebraic integer coef'cients so they
are algebraic integers. Not true at all. Note that
the polynomial P(m), as given above, if considered as a
polynomial in x, does not have constant term equal to 1
or -1. The constant term with respect to x is u^3*f^3.
Thus the polynomial of which the a's are roots is not monic.
I am not sure what you actually intended.
This is carelessly written, even independent
of the unsoundness of the ideas. Parts of your posts
from yesterday on this topic also contained similar
algebraic errors.
I think it was still your intention in APF however to
show that if
65*x^3 - 12*x + 1
were factored in the form
(a1*x + 1)*(a2*x + 1)*(a3*x + 1),
where a1, a2, and a3 are algebraic integers, then one
of a1, a2, or a3 is coprime to 5. No matter
how you cut it, that is still false, as shown by
the proofs given in other threads by me and W. Dale
Hall. Yes, you have started other several new threads
today, but you have explicitly avoided a direct response to
either me or Dale Hall on this.
Even if you are unable to perform the computations that
Dale proposes and are incapable of understanding the proof that
I presented, you cannot ignore the simple algebraic errors
in APF. Something there has to change.
Nora B.
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Fair bit of snippage. I've got some work to do now to
follow this up
> properly.
> And again.
 <Snip > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) =
> 25(5000m^3 - 600 m^2 - 126m + 11)
> That some of the a's are coprime to 5?
> It turns out that in the ring of algebraic integers
they all are,
> which is why the ring has problems.
>
a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not
coprime to
5.
Perhaps you mean that none of the a's are coprime to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
It's actually fascinating as it's a bizarre 
result, and I
think you
realize that I'm correct as you *again* deleted out the
following
wonderful simpli'cation which shows clearly that 
I'm right.
The problem is with the ring of algebraic integers as I've
explained.
Consider the following which you deleted yet again from my
previous
post:
Hmmm...how about this explanation?
Remember that the polynomial is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and you see that factor 25. Now imagine that I have the
polynomial
Q(m), where
P(m) = 25 Q(m)
so you have as a factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
and the question is, how does that factor of 25 divide out?
Well, checking at Q(0), gives 11, so how can those factors of
25
divide through Q(m) in such a way as to give 11, at m=0?
There's only one way, which gives you
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
If that bothers you, remember that at m=0, two of the a's
equal 0.
If it still bothers you, try and get everything to work some
other
way.
Readers who want some sense of what I'm up against should 
read
the
various posts in this thread from people trying to escape that
obvious
conclusion.
Above the *only* thing that works at m=0, is a_1=a_2=0, while
a_3 = 3.
That's trivial enough and has been admitted, but posters 
seem
to want
the constants to move around if m doesn't equal 0, which is
what I
call voodoo math.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
<deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not
coprime to
5.
> Perhaps you mean that none of the a's are coprime to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
It's actually fascinating as it's a bizarre 
result, and I
think you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm right.
Yuck. Unfortunately in my haste I neglected to refute the
claim, as
it is in fact NOT true that none of the a's are coprime to 
5,
as in
fact the fascinating as it's a bizarre result is that they 
ALL
are
coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers as I've
explained.
Consider the following which you deleted yet again from my
previous
> post:
Hmmm...how about this explanation?
Remember that the polynomial is
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
which gives the reader a chance to see P(m) with only the m
left as
a symbol
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
P(m) = 25 Q(m)
so you have as a factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
and the question is, how does that factor of 25 divide out?
I want readers to consider how many posters have apparently
attempted
to make them believe that the factor of f^2, here 25, was in
some
sense welded into the expression, when in fact it's a factor
of P(m)
such that I can write
P(m) = 25 Q(m)
as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those factors
of 25
> divide through Q(m) in such a way as to give 11, at m=0?
There's only one way, which gives you
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
If that bothers you, remember that at m=0, two of the a's
equal 0.
And remember that posters in trying to refute have
continuously called
m=0 a degenerate case when in fact they need you to ignore the
obvious.
That is, given that f^2 is this factor that's multipled 
times
P(m) as
it is, then of course, it can be separated off, and it's not 
so
complicated and extraordinary that you need Galois Theory or
any of
all that extra technicality.
> If it still bothers you, try and get everything to work some
other
> way.
Indeed.
> Readers who want some sense of what I'm up against should
read the
> various posts in this thread from people trying to escape
that obvious
> conclusion.
Above the *only* thing that works at m=0, is a_1=a_2=0, while
a_3 = 3.
That's trivial enough and has been admitted, but posters 
seem
to want
> the constants to move around if m doesn't equal 0, which 
is
what I
> call voodoo math.
But what's at stake is the *belief* that mathematicians 
could
not have
taught erroneous mathematics for over a hundred years.
And in fact the §awed mathematics is STILL in the textbooks.
It is rather weak-minded that mathematicians would continue to
fight
over this, but I'm unfortunately familiar with how weak 
people
often
are.
The truth can be a bitter pill, expecially for fakes 'ghting 
to
preserve the lie.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> The truth can be a bitter pill, expecially for fakes 
'ghting
to
> preserve the lie.
James Harris
Well, enjoy the pill! You've earned it. The argument you 
have
given is
erroneous. It leads to a false
conclusion. The problem is not any contradiction in math, or
incompleteness
of any ring, but in the errors
you made in your argument. You are now §ogging a dead horse.
Please mount
the beast and ride off into the
wilderness where you belong.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed

<deleted
>a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
>For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e.
not coprime to
5.
>Perhaps you mean that none of the a's are coprime to 5?
>Which is trivially obvious and uninteresting after all.
>Sorry to have bothered you.
>It's actually fascinating as it's a bizarre 
result, and I
think you
>>realize that I'm correct as you *again* deleted out the
following
>>wonderful simpli'cation which shows clearly that 
I'm right.
> Yuck. Unfortunately in my haste I neglected to refute the
claim, as
> it is in fact NOT true that none of the a's are coprime to
5, as in
> fact the fascinating as it's a bizarre result is that they
ALL are
> coprime to 5 in the ring of algebraic integers.
>
Pretty much like the a's in the factorization
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
are all coprime to 5?
Why should your argument work in the one case, and not in the
other?
Oh, in case you forgot: I have shown explicit factors that are
shared
by 5, and *EACH* of the a's in this example. As a result, 5
CANNOT be
coprime to any of these a's.
Why do you §ee from the case you should be 'xing?
>The problem is with the ring of algebraic integers as I've
explained.
>>
You have not proven a thing.
>>Consider the following which you deleted yet again from my
previous
>>post:
>Hmmm...how about this explanation?
>Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> which gives the reader a chance to see P(m) with only the m
left as
> a symbol
>>and you see that factor 25. Now imagine that I have the
polynomial
>>Q(m), where
> P(m) = 25 Q(m)
So Q(m) = 5000 m^3 - 600 m^2 - 126 m + 11?
>so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
Rather, a factorization:
P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)?
What result guarantees that such a factorization is possible?
>and the question is, how does that factor of 25 divide out?
>
What happened to the extra dollar?
I want readers to consider how many posters have apparently
attempted
> to make them believe that the factor of f^2, here 25, was in
some
> sense welded into the expression, when in fact it's a 
factor
of P(m)
> such that I can write
P(m) = 25 Q(m)
as I've done here, or P(m) = f^2 Q(m), in general.
>
I'm sorry, I was thinking about your other case (I mentioned
it above)
where you said all the a's were coprime to 5, yet I *showed*
you the
common factors.
I was wondering why you wouldn't even address that failure 
of
your
method, rather than moving to another (most likely incorrect)
case?
Why wouldn't a person address what can be computed directly,
rather
than running to another case where the computations are almost
surely
more complicated? Is it the very intractibility of the new
problem
that makes you safe? If no one can tell you're wrong, does
that make
you right?
What if you could be shown wrong with Galois theory? I see in
your paper
that you call such applications *overinterpretations* of
Galois theory,
although you have admitted that you know less about Galois
theory than
virtually any human, & you prove that point by pointing out
(as though
itwere some §aw) that Galois theory typically deals with
'elds, where
you're working with rings.
>Well, checking at Q(0), gives 11, so how can those factors
of 25
>>divide through Q(m) in such a way as to give 11, at m=0?
>There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
>If that bothers you, remember that at m=0, two of the a's
equal 0.
> And remember that posters in trying to refute have
continuously called
> m=0 a degenerate case when in fact they need you to ignore
the
> obvious.
>
The case is degenerate in that the degree of the polynomial
drops from 3
to 1 when m=0. Frequently, calculations change in degenerate
cases.
> That is, given that f^2 is this factor that's multipled
times P(m) as
> it is, then of course, it can be separated off, and it's 
not
so
> complicated and extraordinary that you need Galois Theory or
any of
> all that extra technicality.
>
Do you mean divided? Why say separated? Is it necessary to use
your own
private language?
>If it still bothers you, try and get everything to work some
other
>>way.
> Indeed.
>>Readers who want some sense of what I'm up against should
read the
>>various posts in this thread from people trying to escape
that obvious
>>conclusion.
>Above the *only* thing that works at m=0, is a_1=a_2=0, while
a_3 = 3.
>>
What do you mean? My PC works when m=0, and it has nothing
whatsoever
to do with those a's. My car works, and it is totally 
unaware
of the
values of m and the a's. I work, for that matter, and I
haven't worried
about your m and a's, not ever.
>>That's trivial enough and has been admitted, but posters
seem to want
>>the constants to move around if m doesn't equal 0, which 
is
what I
>>call voodoo math.
>
What has been admitted?
But what's at stake is the *belief* that mathematicians 
could
not have
> taught erroneous mathematics for over a hundred years.
>
Could you please address your failure to 'x the claim you had
about the
above polynomial factorization? Please make correct assertions
before
you claim that mathematicians have taught erroneous
mathematics.
> And in fact the §awed mathematics is STILL in the textbooks.
It is rather weak-minded that mathematicians would continue to
'ght
> over this, but I'm unfortunately familiar with how weak
people often
> are.
You *are* familiar with how weak a *certain person* often is.
You are
the source of dishonesty, error, and cynicism in these
threads, yet you
make claims of being entirely honest and forthright. Doesn't
your
practice of hypocrisy cause you any discomfort? What do your
parents
think of your hypocrisy? Are they hypocrites as well?
The truth can be a bitter pill, expecially for fakes 'ghting 
to
> preserve the lie.
>
Fighting to preserve the lie. That's how JSH refers to a 
person
pointing out that he [JSH] is wrong; pointing this fact out
for days
on end, and with algebra that is as simple and direct as
anyone could
imagine.
Fakes. That's who can show what JSH says 
doesn't exist, and
who can
point those non-existent things out in a simple fashion.
James Harris
Perhaps our esteemed Mister Harris is warning everyone about
himself.
Dale.
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
<deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e.
not coprime
to 5.
> Perhaps you mean that none of the a's are coprime to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
> It's actually fascinating as it's a bizarre 
result, and I
think you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm right.
Yuck. Unfortunately in my haste I neglected to refute the
claim, as
> it is in fact NOT true that none of the a's are coprime to
5, as in
> fact the fascinating as it's a bizarre result is that they
ALL are
> coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers as I've
explained.
> Consider the following which you deleted yet again from my
previous
> post:
> Hmmm...how about this explanation?
> Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
which gives the reader a chance to see P(m) with only the m
left as
> a symbol
>
Yes! I like this example. See below.
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
> P(m) = 25 Q(m)
> so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> and the question is, how does that factor of 25 divide out?
I want readers to consider how many posters have apparently
attempted
> to make them believe that the factor of f^2, here 25, was in
some
> sense welded into the expression, when in fact it's a 
factor
of P(m)
> such that I can write
P(m) = 25 Q(m)
as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those factors
of 25
> divide through Q(m) in such a way as to give 11, at m=0?
> There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
> If that bothers you, remember that at m=0, two of the a's
equal 0.
>
Proving what, exactly? Clearly a_1, a_2 and a_3 are functions
of m. When m = 0, two of them are 0. However when m is not
zero,
none of them are zero. So how do their values when m = 0
determine what they will be when m <> 0 ?
> And remember that posters in trying to refute have
continuously called
> m=0 a degenerate case when in fact they need you to ignore
the
> obvious.
That is, given that f^2 is this factor that's multipled 
times
P(m) as
> it is, then of course, it can be separated off, and it's 
not
so
> complicated and extraordinary that you need Galois Theory or
any of
> all that extra technicality.
> If it still bothers you, try and get everything to work some
other
> way.
Indeed.
> Readers who want some sense of what I'm up against should
read the
> various posts in this thread from people trying to escape
that obvious
> conclusion.
> Above the *only* thing that works at m=0, is a_1=a_2=0,
while a_3 = 3.
> That's trivial enough and has been admitted, but posters
seem to want
> the constants to move around if m doesn't equal 0, which 
is
what I
> call voodoo math.
But what's at stake is the *belief* that mathematicians 
could
not have
> taught erroneous mathematics for over a hundred years.
>
Let's go back to your original polynomial,
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and assume it is factored *as you propose* in the form
[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
Now instead of m = 0, let's try m = 1:
P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
This can be factored in the form [*] by letting
a_1 = -5, a_2 = -5, and a_3 = 2140.
Note that EACH ONE of these is divisible by 5:
NONE ARE COPRIME TO 5.
Clearly m = 0 is a special case. It is different
in an essential way from m = 1 and other nonzero
values of m.
Now, what's your explanation?
Nora B.
> And in fact the §awed mathematics is STILL in the textbooks.
It is rather weak-minded that mathematicians would continue to
'ght
> over this, but I'm unfortunately familiar with how weak
people often
> are.
The truth can be a bitter pill, expecially for fakes 'ghting 
to
> preserve the lie.
> James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> But what's at stake is the *belief* that mathematicians
could not have
> taught erroneous mathematics for over a hundred years.
No, it's not at stake. It's not even a serious 
consideration
for
anyone but you.
--
Wayne Brown | When your tail's in a crack, you improvise
fwbrown@bellsouth.net | if you're good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> <deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e.
not coprime
to 5.
> Perhaps you mean that none of the a's are coprime to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
> It's actually fascinating as it's a bizarre 
result, and I
think you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm right.
> Yuck. Unfortunately in my haste I neglected to refute the
claim, as
> it is in fact NOT true that none of the a's are coprime to
5, as in
> fact the fascinating as it's a bizarre result is that they
ALL are
> coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers as I've
explained.
> Consider the following which you deleted yet again from my
previous
> post:
> Hmmm...how about this explanation?
> Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> which gives the reader a chance to see P(m) with only the m
left as
> a symbol
 Yes! I like this example. See below.
Fascinating.
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
> P(m) = 25 Q(m)
> so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> and the question is, how does that factor of 25 divide out?
> I want readers to consider how many posters have apparently
attempted
> to make them believe that the factor of f^2, here 25, was in
some
> sense welded into the expression, when in fact it's a 
factor
of P(m)
> such that I can write
> P(m) = 25 Q(m)
> as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those factors
of 25
> divide through Q(m) in such a way as to give 11, at m=0?
> There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
> If that bothers you, remember that at m=0, two of the a's
equal 0.
 Proving what, exactly? Clearly a_1, a_2 and a_3 are functions
> of m. When m = 0, two of them are 0. However when m is not
zero,
> none of them are zero. So how do their values when m = 0
> determine what they will be when m <> 0 ?
Well you have P(m) = 25 Q(m) and that factor 25, which has
been the
focus of all the arguing. Now considering Q(m) it turns out
that
looking at the factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
the a's are in the way, and it's not clear how 
you divide out.
Now you *know* that the 25 divides through some way, and 
it's
clear
that whatever way that is would be true for any m, just like
with
2(x^2 + 2x + 1) = (x+1)(2x+2)
when you divide off 2, you do it for all x.
For readers, people like Nora Baron have basically been
arguing that
you can have something like
Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)
where h_1 h_2 h_3 = 5, and each is not a unit.
But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25,
when in
fact Q(0)=11.
Given that rigorous fact they claim that m=0 is a some kind of
special
case and try to cast doubt on the result, which is an attack on
algebra that I call voodoo math.
Surprisingly they've been quite successful from what 
I've
gathered in
convincing people, which makes you wonder about people's
understanding
of algebra.
Because from algebra, it turns out that you can just set m=0,
to
'gure out how that 25 divides out, as I've 
done.
> And remember that posters in trying to refute have
continuously called
> m=0 a degenerate case when in fact they need you to ignore
the
> obvious.
> That is, given that f^2 is this factor that's multipled
times P(m) as
> it is, then of course, it can be separated off, and it's 
not
so
> complicated and extraordinary that you need Galois Theory or
any of
> all that extra technicality.
> If it still bothers you, try and get everything to work
some other
> way.
> Indeed.
> Readers who want some sense of what I'm up against should
read the
> various posts in this thread from people trying to escape
that
obvious
> conclusion.
> Above the *only* thing that works at m=0, is a_1=a_2=0,
while a_3 =
3.
> That's trivial enough and has been admitted, but posters
seem to want
> the constants to move around if m doesn't equal 0, which
is what I
> call voodoo math.
> But what's at stake is the *belief* that mathematicians
could not have
> taught erroneous mathematics for over a hundred years.
 Let's go back to your original polynomial,
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and assume it is factored *as you propose* in the form
[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
Now instead of m = 0, let's try m = 1:
Ok.
> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
This can be factored in the form [*] by letting
a_1 = -5, a_2 = -5, and a_3 = 2140.
Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
seem to be just a tad bit strange to you?
> Note that EACH ONE of these is divisible by 5:
Ok, you *pick* values from the a's as if 
they're not de'ned by
a
cubic, which they are, and your picked values are supposed to
prove
something?
> NONE ARE COPRIME TO 5.
So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3
= 2140?
Remember you simply tossed those numbers out there, but there
*is* a
way to calculate the a's so your assertion can be tested, as
they are
roots to a cubic.
Now if you can do that Nora Baron then clearly I made some
kind of
mistake and there's no more room for me to argue.
So why don't you go back to the cubic which 
de'nes the a's,
stick in
all the values and see if the a's come out as you have 
above,
and then
it's over.
> Clearly m = 0 is a special case. It is different
> in an essential way from m = 1 and other nonzero
> values of m.
Clearly you haven't proven your assertion.
> Now, what's your explanation?
Nora B.
Well I've explained above, and again, of course, you 
can't
just *pick*
values for the a's for a particular m, as their values are 
set
rigorously. What I'm curious about are the people who 
believe
Nora
Baron had a valid point.
Speak up, and don't be shy as I suspect you may believe she
still has
a valid point and I need to understand why she's so 
effective
in
convincing people.
Come on, speak up, do you think I answered her objections? Do
you
believe they were valid in the 'rst place?
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
and you see that factor 25. Now imagine that I have the
polynomial
>>Q(m), where
> P(m) = 25 Q(m)
>so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
>and the question is, how does that factor of 25 divide
out?
>I want readers to consider how many posters have apparently
attempted
>to make them believe that the factor of f^2, here 25, was
in some
>sense welded into the expression, when in fact it's a
factor of P(m)
>such that I can write
> P(m) = 25 Q(m)
>as I've done here, or P(m) = f^2 Q(m), in general.
>>Well, checking at Q(0), gives 11, so how can those factors
of 25
>>divide through Q(m) in such a way as to give 11, at m=0?
>There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
>If that bothers you, remember that at m=0, two of the a's
equal 0.
>> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions
>>of m. When m = 0, two of them are 0. However when m is not
zero,
>>none of them are zero. So how do their values when m = 0
>>determine what they will be when m <> 0 ?
> Well you have P(m) = 25 Q(m) and that factor 25, which has
been the
> focus of all the arguing. Now considering Q(m) it turns out
that
> looking at the factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
the a's are in the way, and it's not clear how 
you divide out.
Now you *know* that the 25 divides through some way, and 
it's
clear
> that whatever way that is would be true for any m, just like
with
2(x^2 + 2x + 1) = (x+1)(2x+2)
when you divide off 2, you do it for all x.
For readers, people like Nora Baron have basically been
arguing that
> you can have something like
Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)
where h_1 h_2 h_3 = 5, and each is not a unit.
But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25,
when in
> fact Q(0)=11.
Given that rigorous fact they claim that m=0 is a some kind of
special
> case and try to cast doubt on the result, which is an attack
on
> algebra that I call voodoo math.
Surprisingly they've been quite successful from what 
I've
gathered in
> convincing people, which makes you wonder about people's
understanding
> of algebra.
Because from algebra, it turns out that you can just set m=0,
to
> 'gure out how that 25 divides out, as I've 
done.
>And remember that posters in trying to refute have
continuously called
>m=0 a degenerate case when in fact they need you to ignore
the
>obvious.
>That is, given that f^2 is this factor that's multipled
times P(m) as
>it is, then of course, it can be separated off, and it's
not so
>complicated and extraordinary that you need Galois Theory
or any of
>all that extra technicality.
>>If it still bothers you, try and get everything to work
some other
>>way.
>Indeed.
>>Readers who want some sense of what I'm up against should
read the
>>various posts in this thread from people trying to escape
that obvious
>>conclusion.
>Above the *only* thing that works at m=0, is a_1=a_2=0,
while a_3 = 3.
>That's trivial enough and has been admitted, but posters
seem to want
>>the constants to move around if m doesn't equal 0, which
is what I
>>call voodoo math.
>But what's at stake is the *belief* that mathematicians
could not have
>taught erroneous mathematics for over a hundred years.
>> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>and assume it is factored *as you propose* in the form
>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
> Ok.
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
> Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
> seem to be just a tad bit strange to you?
You haven't provided a way to de'ne what the 
a_i(m) are, so of
course
she can simply select values that work.
>> Note that EACH ONE of these is divisible by 5:
Ok, you *pick* values from the a's as if 
they're not de'ned by
a
> cubic, which they are, and your picked values are supposed
to prove
> something?
They AREN'T de'ned by a cubic. They are 
constrained by a
cubic. If
you want them to be de'ned by a cubic, you'll 
have to say more
about
how they are supposed to be constructed.
>> NONE ARE COPRIME TO 5.
> So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
Remember you simply tossed those numbers out there, but there
*is* a
> way to calculate the a's so your assertion can be tested, 
as
they are
> roots to a cubic.
Not that you've provided. If you have a way in mind, please
provide it
to us and include it in your paper. You've only claimed this
things
exist and have certain properties. If you want to change them,
so be
it. But that may cause more problems for you.
> Now if you can do that Nora Baron then clearly I made some
kind of
> mistake and there's no more room for me to argue.
There's
no more
room for you to argue.
> So why don't you go back to the cubic which 
de'nes the a's,
stick in
> all the values and see if the a's come out as you have
above, and then
> it's over.
For each value of m, she can de'ne a_i(m) to be whatever 
three
values
makes both sides equal. You've given nothing more 
speci'c
about the
values to require more.
>> Now, what's your explanation?
> Nora B.
> Well I've explained above, and again, of course, you 
can't
just *pick*
> values for the a's for a particular m, as their values are
set
> rigorously. What I'm curious about are the people who
believe Nora
> Baron had a valid point.
Where have you established how to rigorously compute them.
What are they?
> Speak up, and don't be shy as I suspect you may believe 
she
still has
> a valid point and I need to understand why she's so
effective in
> convincing people.
Because you have not stated all of the conditions you want to
exist.
Once you do that, you will lose the results you are trying to
achieve.
> Come on, speak up, do you think I answered her objections?
Do you
> believe they were valid in the 'rst place?
I think you didn't understand her objections.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
the only answer is,
JSH is always-and-only every where dense.
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and assume it is factored *as you propose* in the form
[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
Now instead of m = 0, let's try m = 1:
P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
This can be factored in the form [*] by letting
a_1 = -5, a_2 = -5, and a_3 = 2140.
Note that EACH ONE of these is divisible by 5:
NONE ARE COPRIME TO 5.
Clearly m = 0 is a special case. It is different
> in an essential way from m = 1 and other nonzero
> values of m.
Now, what's your explanation?
--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
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> <deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of 25.
i.e. not
coprime to 5.
> Perhaps you mean that none of the a's are coprime to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
> It's actually fascinating as it's a bizarre 
result, and
I think you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm right.
> Yuck. Unfortunately in my haste I neglected to refute the
claim, as
> it is in fact NOT true that none of the a's are coprime to
5, as in
> fact the fascinating as it's a bizarre result is that they
ALL
are
> coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers as
I've
explained.
> Consider the following which you deleted yet again from
my previous
> post:
> Hmmm...how about this explanation?
> Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> which gives the reader a chance to see P(m) with only the
m left
as
> a symbol
> Yes! I like this example. See below.
Fascinating.
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
> P(m) = 25 Q(m)
> so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> and the question is, how does that factor of 25 divide
out?
> I want readers to consider how many posters have
apparently attempted
> to make them believe that the factor of f^2, here 25, was
in some
> sense welded into the expression, when in fact it's a
factor of P(m)
> such that I can write
> P(m) = 25 Q(m)
> as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those
factors of 25
> divide through Q(m) in such a way as to give 11, at m=0?
> There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
> If that bothers you, remember that at m=0, two of the
a's equal 0.
> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions
> of m. When m = 0, two of them are 0. However when m is not
zero,
> none of them are zero. So how do their values when m = 0
> determine what they will be when m <> 0 ?
Well you have P(m) = 25 Q(m) and that factor 25, which has
been the
> focus of all the arguing. Now considering Q(m) it turns out
that
> looking at the factorization
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
the a's are in the way, and it's not clear how 
you divide out.
Now you *know* that the 25 divides through some way, and 
it's
clear
> that whatever way that is would be true for any m, just like
with
2(x^2 + 2x + 1) = (x+1)(2x+2)
when you divide off 2, you do it for all x.
For readers, people like Nora Baron have basically been
arguing that
> you can have something like
Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)
where h_1 h_2 h_3 = 5, and each is not a unit.
But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25,
when in
> fact Q(0)=11.
Given that rigorous fact they claim that m=0 is a some kind of
special
> case and try to cast doubt on the result, which is an attack
on
> algebra that I call voodoo math.
Surprisingly they've been quite successful from what 
I've
gathered in
> convincing people, which makes you wonder about people's
understanding
> of algebra.
Because from algebra, it turns out that you can just set m=0,
to
> 'gure out how that 25 divides out, as I've 
done.
>
No voodoo at all. What you would like to have here is that
the same a_1, a_2, and a_3 work when m = 0 as when m <> 0.
But you know, and everyone else knows, that that is totally
impossible. Why? Because when m = 0, two of the a's must be
zero. When m <> 0, NONE of the a's can be zero.
It's real simple.
The a's are functions of m. And I don't mean 
CONSTANT
functions.
Got that?
Different m's give different a's.
You believe that because a_1 is divisible by 5 when m = 0,
it must be divisible by 5 for other values of m also. It is
some kind of hunch that you have, based on intuition. It seems
like a nice parallel construction. You think it's obvious.
It isn't. In fact it is false. Your intuition this time is
wrong. Your hunch has been disproved.
What the a's are when m = 0 is a special case. It does NOT
determine what they are when m <> 0. Sure, when m = 0,
you can assume that a_1 and a_2 are zero. Can you can say
that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of course
you can - they are both 0! After all, ANY number divides 0.
For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are
algebraic integers also, since both are 0. Does that mean that
137 must
divide two of the a's, when m <> 0 ???? By your logic, YES!
Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS 
OF
M.
WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT BE
ASSUMED
TO BE TRUE WHEN M <> 0. *****
Why is this so hard for you to understand? Why do you keep
trying to
act like I am pulling some kind of trick when I say this, or
pretending
that I am lying?
> And remember that posters in trying to refute have
continuously
called
> m=0 a degenerate case when in fact they need you to ignore
the
> obvious.
> That is, given that f^2 is this factor that's multipled
times P(m) as
> it is, then of course, it can be separated off, and it's
not so
> complicated and extraordinary that you need Galois Theory
or any of
> all that extra technicality.
> If it still bothers you, try and get everything to work
some other
> way.
> Indeed.
> Readers who want some sense of what I'm up against
should read the
> various posts in this thread from people trying to escape
that
obvious
> conclusion.
> Above the *only* thing that works at m=0, is a_1=a_2=0,
while a_3 =
3.
> That's trivial enough and has been admitted, but posters
seem to
want
> the constants to move around if m doesn't equal 0, which
is what I
> call voodoo math.
> But what's at stake is the *belief* that mathematicians
could not
have
> taught erroneous mathematics for over a hundred years.
> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> and assume it is factored *as you propose* in the form
> [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
Ok.
> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
> seem to be just a tad bit strange to you?
>
Nope. It was your idea, not mine, to replace the x's
with a constant value of 2. When you did that you were
no longer factoring a polynomial. You are just factoring
an ordinary number. There are no hidden rules that say it
must be factored as a polynomial.
You can't have it both ways. You want it factored as
a polynomial, leave the x's in there (and see below). You
want it factored as an ordinary number (remember? VERY much
simpli'ed) then one must assume that any factorization is
legitimate, and you get my example above.
So let's go back to assuming you want it factored as a
polynomial. Of course you know that both I and W. Dale
Hall have proofs that that is not going to work. You have
not provided a valid objection to my proof, and you have
not provided any response whatsoever to what Dale did. I
don't see how you can. It's a simple 
computation. You
do the arithmetic, and Dale is either right or he is wrong.
You don't have to understand ANYTHING. There is no wool
to be pulled over anyone's eyes. Try it and see.
See also below.
> Note that EACH ONE of these is divisible by 5:
Ok, you *pick* values from the a's as if 
they're not de'ned by
a
> cubic, which they are, and your picked values are supposed
to prove
> something?
>
Yes. It proves what you said was wrong. You said that one
of the a's had to be coprime to 5. Clearly, totally,
unambiguously false.
> NONE ARE COPRIME TO 5.
So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3
= 2140?
Remember you simply tossed those numbers out there, but there
*is* a
> way to calculate the a's so your assertion can be tested, 
as
they are
> roots to a cubic.
>
They *would* be roots of a cubic in the variable x, but you
replaced x by 2. At that point there is no longer a polynomial.
The restriction that they be roots of a cubic is gone. It was
your idea, not mine, to oversimplify things and make the
substitution.
Remember: if you want to go back to assuming that a_1, a_2, and
a_3 are roots of a cubic in the variable x, to continue to
maintain what you think is true, you are going to have to 'nd
an error in the proof I have presented (and is appended
below).
> Now if you can do that Nora Baron then clearly I made some
kind of
> mistake and there's no more room for me to argue.
>
Yep. You should not have substituted x = 2. You want it
factored as a polynomial, you had better leave it a polynomial
and not try to oversimplify. You have shown confusion
previously
over what the difference between a polynomial and the
evaluation
of that polynomial when you choose a particular value for x.
The moral here is, this little example should show you pretty
convincingly that when people tried to get you to understand
that
difference, they were not just playing with semantics.
There is another more general moral here, and you have run
afoul of this one many a time in the past also: Examples are
not a substitute for proofs. And do NOT oversimplify when you
are trying to do mathematics.
> So why don't you go back to the cubic which 
de'nes the a's,
stick in
> all the values and see if the a's come out as you have
above, and then
> it's over.
>
I will do better than that. I will reproduce my proof that your
claim that one of the a's is coprime to 5 is wrong, but I 
will
do it in
more generality. That way I will be heading off any examples
you might try to come up with in the future. What you want is
a consequence of the following claim, which you have stated
elsewhere:
CLAIM: It is possible to 'nd a monic polynomial of degree
3, with integer coef'cients, irreducible over the rationals.
and with roots a1, a2, and a3, such that at least one of
a1, a2, or a3 is coprime, in the ring of algebraic
integers, to a prime factor of the constant term of
the polynomial.
Right? You have claimed as recently as yesterday.
DISPROOF OF CLAIM:
Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are
integers and c = p*v, where p is a prime and v is another
integer. Q(x) is clearly monic. Assume Q(x) is
irreducible. Let a1, a2, and a3 be roots of Q(x). Note
that by de'nition, a1, a2, and a3 are algebraic
integers.
The claim is that at least one of a1, a2, or a3 is
coprime to p. Assume a1 is coprime to p.
By standard theory***, there exists an automorphism F12 of
the 'eld of algebraic numbers such that:
1. F12 leaves the sub'eld of rational numbers 
'xed,
i.e., if q is rational, F12(q) = q.
2. F12(a1) = a2.
3. If t is an algebraic integer, F12(t) is also an
algebraic integer.
Now since a1 is relatively prime to p, there exist
algebraic integers r and s such that
[1] r*a1 + s*p = 1.
Now apply the automorphism F12 to both sides of [1]:
F12(r)*F12(a1) + F12(s)*F12(p) = F12(1).
By the properties above, F12(p) = p and F12(1) = 1.
Moreover, r' = F12(r) is an algebraic integer, and
s' = F12(s) is an algebraic integer. Finally,
F12(a1) = a2. Thus one obtains:
r'*u2 + s'*p = 1,
which says: a2 and p are coprime in the algebraic
integers.
Similarly one shows that a3 and p are coprime.
Therefore if one of a1, a2, or a3 is coprime
to p, then they all are.
But a1 * a2 * a3 = p * v. That is, p divides
the product of a1, a2, and a3. Therefore p cannot
be coprime to each of a1, a2, and a3.
Putting all this together, one concludes that
NONE of a1, a2, or a3 can be coprime to p.
This directly contradicts the CLAIM made above.
Please feel free to point out any errors in the
proof I just gave.
***: Reference on automorphisms:
http://www.math.niu.edu/~beachy/aaol/galois.html
or the excellent textbook Abstract Algebra, by
John Beachy and William D. Blair
> Clearly m = 0 is a special case. It is different
> in an essential way from m = 1 and other nonzero
> values of m.
Clearly you haven't proven your assertion.
> Now, what's your explanation?
> Nora B.
Well I've explained above, and again, of course, you 
can't
just *pick*
> values for the a's for a particular m, as their values are
set
> rigorously.
If you can just *pick* x = 2, changing the problem from
factoring
a polynomial to factoring an ordinary number, certainly I can
*pick* any factorization that works. Lesson here: be careful
what
you *pick*.
> What I'm curious about are the people who believe Nora
> Baron had a valid point.
Speak up, and don't be shy as I suspect you may believe she
still has
> a valid point and I need to understand why she's so
effective in
> convincing people.
Come on, speak up, do you think I answered her objections? Do
you
> believe they were valid in the 'rst place?
>
On this point I agree. Speak up, lurkers!
Nora B.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
<deleted >> Let's go back to your original polynomial,
>> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>>and assume it is factored *as you propose* in the form
>>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
>> Now instead of m = 0, let's try m = 1:
> Ok.
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
>> This can be factored in the form [*] by letting
>> a_1 = -5, a_2 = -5, and a_3 = 2140.
> Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
> seem to be just a tad bit strange to you?
You haven't provided a way to de'ne what the 
a_i(m) are, so of
course
> she can simply select values that work.
That's false as the a's are given by the 
factorization
(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 y=uf.
For my example I used x=2, f=5, u=1. And now letting m=1, to
test
your claim I have
13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
so the a's are the roots of the cubic
a^3 + 72a^2 - 13825 = 0.
And checking with a=-5, gives -12150, and not 0.
So Nora Baron's claim and yours is refuted.
>> Note that EACH ONE of these is divisible by 5:
That was Nora Baron's comment for her picked values for the
a's.
> Ok, you *pick* values from the a's as if 
they're not de'ned
by a
> cubic, which they are, and your picked values are supposed
to prove
> something?
They AREN'T de'ned by a cubic. They are 
constrained by a
cubic. If
> you want them to be de'ned by a cubic, you'll 
have to say
more about
> how they are supposed to be constructed.
That statement is false as I've shown by giving the cubic.
>> NONE ARE COPRIME TO 5.
> So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
That was my question to Nora Baron.
> Remember you simply tossed those numbers out there, but
there *is* a
> way to calculate the a's so your assertion can be tested, 
as
they are
> roots to a cubic.
Not that you've provided. If you have a way in mind, please
provide it
> to us and include it in your paper. You've only claimed 
this
things
> exist and have certain properties. If you want to change
them, so be
> it. But that may cause more problems for you.
I simplify and people like you come in and try to confuse, and
sadly,
I think that the readership of sci.math wants to be confused
by you.
> Now if you can do that Nora Baron then clearly I made some
kind of
> mistake and there's no more room for me to argue.
There's
no more
> room for you to argue.
Well now that I've given the cubic, I challenge you to admit
*you*
made a mistake.
> So why don't you go back to the cubic which 
de'nes the a's,
stick in
> all the values and see if the a's come out as you have
above, and then
> it's over.
For each value of m, she can de'ne a_i(m) to be whatever 
three
values
> makes both sides equal. You've given nothing more 
speci'c
about the
> values to require more.
That is false as I've repeatedly given the expression 
de'ning
the
a's.
>> Now, what's your explanation?
>> Nora B.
> Well I've explained above, and again, of course, you 
can't
just *pick*
> values for the a's for a particular m, as their values are
set
> rigorously. What I'm curious about are the people who
believe Nora
> Baron had a valid point.
Where have you established how to rigorously compute them.
What are
they?
For m=1, f=5, the a's are the roots of the cubic
a^3 + 72a^2 - 13825 = 0.
> Speak up, and don't be shy as I suspect you may believe 
she
still has
> a valid point and I need to understand why she's so
effective in
> convincing people.
Because you have not stated all of the conditions you want to
exist.
> Once you do that, you will lose the results you are trying
to achieve.
I've stated those conditions repeatedly. Are you now 
claiming
that
you have never seen the factorization
(v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)?
> Come on, speak up, do you think I answered her objections?
Do you
> believe they were valid in the 'rst place?
I think you didn't understand her objections.
I think you're lying.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
Visiting Assistant Professor at the University of Montana.
[.snip.]
>> What I'm curious about are the people who believe Nora
>> Baron had a valid point.
>> Speak up, and don't be shy as I suspect you may believe 
she
still has
>> a valid point and I need to understand why she's so
effective in
>> convincing people.
>> Come on, speak up, do you think I answered her objections?
Do you
>> believe they were valid in the 'rst place?
>>
On this point I agree. Speak up, lurkers!
Hmmm... Within the last 24 Hours, James incorrectly claimed
that Keith
Ramsay was doing an appeal to the gallery, and correctly noted
that
appealing to the gallery is a logical fallacy. Perhaps James
does not
realize that what he is doing here is a ->classic<- example of
appealing to the gallery? After all, he does not know what
appeal to
authority, ad hominem, and any of the other logical fallacies
he
claims others are committing are...
=
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 'gures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the inde'nite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
=
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
 <deleted
>> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>and assume it is factored *as you propose* in the form
>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
>Ok.
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
>Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
>seem to be just a tad bit strange to you?
>You haven't provided a way to de'ne what the 
a_i(m) are, so
of course
>>she can simply select values that work.
> That's false as the a's are given by the 
factorization
(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 y=uf.
For my example I used x=2, f=5, u=1. And now letting m=1, to
test
> your claim I have
13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
so the a's are the roots of the cubic
a^3 + 72a^2 - 13825 = 0.
And checking with a=-5, gives -12150, and not 0.
So Nora Baron's claim and yours is refuted.
I thought you said the a's vary with m. If they are roots of
the cubic,
then they are constants. Note that in your paper you use the
a's in two
different ways. Which way are they used in THIS case? I
believe Dale
Hall is the one who showed that if you view the a's as
constants, they
still don't have the properties you want, but I could easily
be mistaken.
>> Note that EACH ONE of these is divisible by 5:
> That was Nora Baron's comment for her picked values for 
the
a's.

>Ok, you *pick* values from the a's as if 
they're not de'ned
by a
>cubic, which they are, and your picked values are supposed
to prove
>something?
>They AREN'T de'ned by a cubic. They are 
constrained by a
cubic. If
>>you want them to be de'ned by a cubic, you'll 
have to say
more about
>>how they are supposed to be constructed.
> That statement is false as I've shown by giving the cubic.
What you've shown is that you are inconsistent in the use of
your
notation. Perhaps if you used a_1(m) when dealing with
functions and
a_1 when dealing with constants this confusion wouldn't 
arise.
>> NONE ARE COPRIME TO 5.
>So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
> That was my question to Nora Baron.
>Remember you simply tossed those numbers out there, but
there *is* a
>way to calculate the a's so your assertion can be tested,
as they are
>roots to a cubic.
>Not that you've provided. If you have a way in mind, please
provide it
>>to us and include it in your paper. You've only claimed 
this
things
>>exist and have certain properties. If you want to change
them, so be
>>it. But that may cause more problems for you.
> I simplify and people like you come in and try to confuse,
and sadly,
> I think that the readership of sci.math wants to be confused
by you.
I didn't use the same symbols to represent two different 
types
of
quantities.
>Now if you can do that Nora Baron then clearly I made some
kind of
>mistake and there's no more room for me to argue.
There's no more
>>room for you to argue.
> Well now that I've given the cubic, I challenge you to 
admit
*you*
> made a mistake.
Challenge declined. You want the a's to both vary with m and 
be
constants. You'll have to make up your mind which it is.
>So why don't you go back to the cubic which 
de'nes the a's,
stick in
>all the values and see if the a's come out as you have
above, and then
>it's over.
>For each value of m, she can de'ne a_i(m) to be whatever
three values
>>makes both sides equal. You've given nothing more 
speci'c
about the
>>values to require more.
> That is false as I've repeatedly given the expression
de'ning the
> a's.
If the a's are constants. If they are, then your expression
above fails
to be true when you vary m.
>> Now, what's your explanation?
> Nora B.
>Well I've explained above, and again, of course, you 
can't
just *pick*
>values for the a's for a particular m, as their values are
set
>rigorously. What I'm curious about are the people who
believe Nora
>Baron had a valid point.
>Where have you established how to rigorously compute them.
What are
they?
> For m=1, f=5, the a's are the roots of the cubic
a^3 + 72a^2 - 13825 = 0.
There's nothing in the nature of the problem that (to me)
makes such a
de'nition either obvious or necessary.
>Speak up, and don't be shy as I suspect you may believe she
still has
>a valid point and I need to understand why she's so
effective in
>convincing people.
>Because you have not stated all of the conditions you want to
exist.
>>Once you do that, you will lose the results you are trying
to achieve.
> I've stated those conditions repeatedly. Are you now
claiming that
> you have never seen the factorization
(v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)?
Is THIS where it's de'ned??? Ok... and do you 
want the choice
of your
a_i(m) to hold for all x and y? Your simpli'cation does not
make
that obvious. Worse, there is no good way of assigning the
values
generated for each choice of m to each a_i(m).
Note: when you CHOSE values for x and y, you changed the
nature of the
problem being looked at. The observations made by myself and (I
believe) Nora are based on the special case of x=2, y=5. The
irony is,
you've been defending a leap from speci'c to 
general in your
arguments,
but won't let us look at the speci'c while 
ignoring the
general.
Have you considered keeping everything in one thread so it's
easier to
follow things?
>Come on, speak up, do you think I answered her objections?
Do you
>believe they were valid in the 'rst place?
>I think you didn't understand her objections.
> I think you're lying.
That is your perogative.
> James Harris
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
> seem to be just a tad bit strange to you?
For strangeness, nothing equals JSH's attempts at 
mathematics.
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> I think you didn't understand her objections.
I think you're lying.
> James Harris
We doubt that you are thinking and we know you're lying!
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> <deleted >> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>and assume it is factored *as you propose* in the form
>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
>Ok.
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
>Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
>seem to be just a tad bit strange to you?
>>You haven't provided a way to de'ne what the 
a_i(m) are,
so of course
>>she can simply select values that work.
> That's false as the a's are given by the 
factorization
> (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 y=uf.
> For my example I used x=2, f=5, u=1. And now letting m=1, to
test
> your claim I have
> 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
> so the a's are the roots of the cubic
> a^3 + 72a^2 - 13825 = 0.
> And checking with a=-5, gives -12150, and not 0.
> So Nora Baron's claim and yours is refuted.
I thought you said the a's vary with m. If they are roots of
the cubic,
> then they are constants. Note that in your paper you use the
a's in two
> different ways. Which way are they used in THIS case? I
believe Dale
> Hall is the one who showed that if you view the a's as
constants, they
> still don't have the properties you want, but I could 
easily
be mistaken.
Are you being deliberately obtuse? Nora Baron's assertion is
with
m=1.
>> Note that EACH ONE of these is divisible by 5:
> That was Nora Baron's comment for her picked values for 
the
a's.
Ok, you *pick* values from the a's as if 
they're not de'ned
by a
>cubic, which they are, and your picked values are supposed
to prove
>something?
>>They AREN'T de'ned by a cubic. They are 
constrained by a
cubic. If
>>you want them to be de'ned by a cubic, you'll 
have to say
more about
>>how they are supposed to be constructed.
> That statement is false as I've shown by giving the cubic.
What you've shown is that you are inconsistent in the use of
your
> notation. Perhaps if you used a_1(m) when dealing with
functions and
> a_1 when dealing with constants this confusion wouldn't
arise.
The following should not be new to you:
(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 y=uf.
And what I did to *simplify* was to let x=2, f=5, u=1. Then
Nora
Baron made an assertion about the a's for m=1, and I gave 
the
de'ning
cubic.
>> NONE ARE COPRIME TO 5.
>So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
> That was my question to Nora Baron.
>Remember you simply tossed those numbers out there, but
there *is* a
>way to calculate the a's so your assertion can be tested,
as they are
>roots to a cubic.
>>Not that you've provided. If you have a way in mind,
please provide it
>>to us and include it in your paper. You've only claimed 
this
things
>>exist and have certain properties. If you want to change
them, so be
>>it. But that may cause more problems for you.
> I simplify and people like you come in and try to confuse,
and sadly,
> I think that the readership of sci.math wants to be confused
by you.
I didn't use the same symbols to represent two different 
types
of
> quantities.
If I have y=mx+b, and then a little later I have y=x+1, and
then later
talk of with x=1, y=2, is that using the same symbols to
represent
different types of quantities?
If you're having trouble keeping up, then you can work 
through
it
*yourself* putting in whatever symbols help you.
However, I've yet to be convinced that others would be more
helped by
my adding in extra than hurt by the additional level of
complexity.
For instance, if I have
P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x + y)(a_2(m) x +
y)(a_3(m)
x + y)
does that help more than it hurts?
And it's even busier with the FLT argument where there are
even MORE
symbols.
Are there people who will now get even more confused?
What kind of arguments might I face then from people who want
to drag
me into arguments about functions?
>Now if you can do that Nora Baron then clearly I made some
kind of
>mistake and there's no more room for me to argue.
There's no more
>>room for you to argue.
> Well now that I've given the cubic, I challenge you to 
admit
*you*
> made a mistake.
Challenge declined. You want the a's to both vary with m and 
be
> constants. You'll have to make up your mind which it is.
You lack intellectual honesty.
>So why don't you go back to the cubic which 
de'nes the a's,
stick in
>all the values and see if the a's come out as you have
above, and then
>it's over.
>>For each value of m, she can de'ne a_i(m) to be whatever
three values
>>makes both sides equal. You've given nothing more 
speci'c
about the
>>values to require more.
> That is false as I've repeatedly given the expression
de'ning the
> a's.
If the a's are constants. If they are, then your expression
above fails
> to be true when you vary m.
You can't be that lost. I simply 'nd it hard to 
believe that
you
don't know more as I think you do but are attempting to 
confuse
others.
>> Now, what's your explanation?
> Nora B.
>Well I've explained above, and again, of course, you
can't just *pick*
>values for the a's for a particular m, as their values are
set
>rigorously. What I'm curious about are the people who
believe Nora
>Baron had a valid point.
>>Where have you established how to rigorously compute them.
What are
they?
> For m=1, f=5, the a's are the roots of the cubic
> a^3 + 72a^2 - 13825 = 0.
There's nothing in the nature of the problem that (to me)
makes such a
> de'nition either obvious or necessary.
You mean like expressions that actually de'ne the 
a's?
You seem to think you don't need to bother with the math!
>Speak up, and don't be shy as I suspect you may believe she
still has
>a valid point and I need to understand why she's so
effective in
>convincing people.
>>Because you have not stated all of the conditions you want
to exist.
>>Once you do that, you will lose the results you are trying
to achieve.
> I've stated those conditions repeatedly. Are you now
claiming that
> you have never seen the factorization
> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)?
Is THIS where it's de'ned??? Ok... and do you 
want the choice
of your
> a_i(m) to hold for all x and y? Your simpli'cation does not
make
> that obvious. Worse, there is no good way of assigning the
values
> generated for each choice of m to each a_i(m).
That is b.s. as you can easily enough solve for the a's, 
even
with the
expression in that form by creating a cubic as I have done,
and then
solving it.
There'll be a LOT of symbols, but it can be done.
> Note: when you CHOSE values for x and y, you changed the
nature of the
> problem being looked at. The observations made by myself and
(I
> believe) Nora are based on the special case of x=2, y=5. The
irony is,
> you've been defending a leap from speci'c to 
general in your
arguments,
> but won't let us look at the speci'c while 
ignoring the
general.
Actually my work has always covered a family of values, which
is why I
can stick in values as I did in this thread. Previously, with
FLT, of
course, I couldn't stick in values for x, y and z.
> Have you considered keeping everything in one thread so 
it's
easier to
> follow things?
The threads become HUGE with hundreds of posts, and then 
it's
harder
to follow things.
>Come on, speak up, do you think I answered her objections?
Do you
>believe they were valid in the 'rst place?
>>I think you didn't understand her objections.
> I think you're lying.
That is your perogative.
You've either been making a horrendous number of mistakes in
replying
to me while continually ignoring information that has been
given, or
you're lying.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> <deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2
+ 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of 25.
i.e. not
coprime to 5.
> Perhaps you mean that none of the a's are coprime to
5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
> It's actually fascinating as it's a bizarre 
result,
and I think
you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm
right.
> Yuck. Unfortunately in my haste I neglected to refute
the claim,
as
> it is in fact NOT true that none of the a's are coprime to
5, as in
> fact the fascinating as it's a bizarre result is that they
ALL
are
> coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers as
I've
explained.
> Consider the following which you deleted yet again
from my
previous
> post:
> Hmmm...how about this explanation?
> Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> which gives the reader a chance to see P(m) with only
the m left
as
> a symbol
> Yes! I like this example. See below.
> Fascinating.
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
> P(m) = 25 Q(m)
> so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> and the question is, how does that factor of 25 divide
out?
> I want readers to consider how many posters have
apparently
attempted
> to make them believe that the factor of f^2, here 25, was
in some
> sense welded into the expression, when in fact it's a
factor of
P(m)
> such that I can write
> P(m) = 25 Q(m)
> as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those
factors of 25
> divide through Q(m) in such a way as to give 11, at m=0?
> There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
> If that bothers you, remember that at m=0, two of the
a's equal
0.
> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions
> of m. When m = 0, two of them are 0. However when m is not
zero,
> none of them are zero. So how do their values when m = 0
> determine what they will be when m <> 0 ?
> Well you have P(m) = 25 Q(m) and that factor 25, which has
been the
> focus of all the arguing. Now considering Q(m) it turns out
that
> looking at the factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> the a's are in the way, and it's not clear 
how you divide
out.
> Now you *know* that the 25 divides through some way, and
it's clear
> that whatever way that is would be true for any m, just like
with
> 2(x^2 + 2x + 1) = (x+1)(2x+2)
> when you divide off 2, you do it for all x.
> For readers, people like Nora Baron have basically been
arguing that
> you can have something like
> Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)
> where h_1 h_2 h_3 = 5, and each is not a unit.
> But notice then that setting m=0, gives (h_1 h_2 h_3)^2 =
25, when in
> fact Q(0)=11.
> Given that rigorous fact they claim that m=0 is a some kind
of special
> case and try to cast doubt on the result, which is an attack
on
> algebra that I call voodoo math.
> Surprisingly they've been quite successful from what 
I've
gathered in
> convincing people, which makes you wonder about people's
understanding
> of algebra.
> Because from algebra, it turns out that you can just set
m=0, to
> 'gure out how that 25 divides out, as I've 
done.
>
No voodoo at all. What you would like to have here is that
> the same a_1, a_2, and a_3 work when m = 0 as when m <> 0.
> But you know, and everyone else knows, that that is totally
> impossible. Why? Because when m = 0, two of the a's must 
be
> zero. When m <> 0, NONE of the a's can be zero.
That everyone else knows is an appeal to the gallery which is a
logical fallacy.
The a's are de'ned 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 I guessed that with so many symbols
people were
confused, so I stuck in some values to lessen the symbol load.
> It's real simple.
The a's are functions of m. And I don't mean 
CONSTANT
functions.
Got that?
Different m's give different a's.
You believe that because a_1 is divisible by 5 when m = 0,
> it must be divisible by 5 for other values of m also. It is
> some kind of hunch that you have, based on intuition. It
seems
> like a nice parallel construction. You think it's obvious.
> It isn't. In fact it is false. Your intuition this time is
> wrong. Your hunch has been disproved.
<deleted>
Here you claim that I'm acting on a hunch about the 
a's.
But, when I used x=2, f=5, y=5, I have
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and noticing that 25 I used P(m) = 25 Q(m), so I have
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
where the question is, how does that 25 divide through?
But since I know that Q(0)=11, I know that it must be
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
At Q(0) that is
Q(0) = (2 (0) + 1)(2 (0) + 1)(2 (3) + 5) = 11.
So in fact it's algebra and not a hunch.
It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can
see how
the 2 divides out, but if you couldn't see, and 
couldn't
guess, like
if you had
P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2),
with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,
you could have P(x) = 2Q(x), and set x=0, with Q(0) to get 1,
showing
that
Q(x) = (a_1 + 1)(a_2/2 + 1), is correct.
Of course, you can also just see that is the only way that
works.
But regardless there is only ONE way that will work, and it
doesn't
change as x changes, and it doesn't change above as m 
changes.
Assaulting algebra is a rather disturbing step taken by posters
continuing to argue with me. What is also troubling is how
uncaring
the newsgroup seems to be about their denial of basic algebra.
But it
is telling.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> Assaulting algebra is a rather disturbing step taken by
posters
> continuing to argue with me. What is also troubling is how
uncaring
> the newsgroup seems to be about their denial of basic
algebra. But it
> is telling.
No one is assaulting algebra. They are denying that you have
constructed
a valid proof. And they are correct. You
are wrong. What is telling here is your monumental
incompetence.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>and assume it is factored *as you propose* in the form
>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
>Now instead of m = 0, let's try m = 1:
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
>This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
>Um, Nora Baron, now you willy-nilly give the a's
values. Doesn't that
>seem to be just a tad bit strange to you?
>You haven't provided a way to de'ne what the 
a_i(m) are,
so of course
>>she can simply select values that work.
>That's false as the a's are given by the 
factorization
> (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 y=uf.
>For my example I used x=2, f=5, u=1. And now letting m=1,
to test
>your claim I have
> 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)
>so the a's are the roots of the cubic
> a^3 + 72a^2 - 13825 = 0.
>And checking with a=-5, gives -12150, and not 0.
>So Nora Baron's claim and yours is refuted.
>I thought you said the a's vary with m. If they are roots 
of
the cubic,
>>then they are constants. Note that in your paper you use the
a's in two
>>different ways. Which way are they used in THIS case? I
believe Dale
>>Hall is the one who showed that if you view the a's as
constants, they
>>still don't have the properties you want, but I could 
easily
be mistaken.
> Are you being deliberately obtuse? Nora Baron's assertion 
is
with
> m=1.
Correct. She was DEFINING the values of the a_i(1). The
discussion had
dealt with what the a_i(0) were, but she simply showed that
when m
changes from 0 to 1, it is possible for the a_i(m) to change
from having
the properties you wanted to ones that you did NOT want.
Nothing
obtuse. Now, do you understand what a function is? Do you
perhaps see
the importance of de'ning the characteristics of these
functions at all
values of m if you want them to have certain properties?
>>Note that EACH ONE of these is divisible by 5:
>That was Nora Baron's comment for her picked values for the
a's.
Ok, you *pick* values from the a's as if 
they're not
de'ned by a
>cubic, which they are, and your picked values are
supposed to prove
>something?
>They AREN'T de'ned by a cubic. They are 
constrained by a
cubic. If
>>you want them to be de'ned by a cubic, you'll 
have to say
more about
>>how they are supposed to be constructed.
>That statement is false as I've shown by giving the cubic.
>What you've shown is that you are inconsistent in the use 
of
your
>>notation. Perhaps if you used a_1(m) when dealing with
functions and
>>a_1 when dealing with constants this confusion wouldn't
arise.
The following should not be new to you:
(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 y=uf.
And what I did to *simplify* was to let x=2, f=5, u=1. Then
Nora
> Baron made an assertion about the a's for m=1, and I gave
the de'ning
> cubic.
When you set x=2,f=5,u=1, then you change the nature of the
problem. If
you would prefer to think of the a_i as varying over m,x,f,u,
we can
do that but then Nora is still able to choose the values she
did. If
you don't want people de'ning the 
a_i's in inconvenient ways,
you will
have to specify what is legal. Don't be surprised if the
speci'cations
you provide limit you to cases that don't give the broad
result you want
to achieve.
>>NONE ARE COPRIME TO 5.
>So are your claiming that at m=1, a_1 = -5, a_2 = -5,
and a_3 = 2140?
>That was my question to Nora Baron.
>Remember you simply tossed those numbers out there, but
there *is* a
>way to calculate the a's so your assertion can be tested,
as they are
>roots to a cubic.
>Not that you've provided. If you have a way in mind,
please provide it
>>to us and include it in your paper. You've only claimed
this things
>>exist and have certain properties. If you want to change
them, so be
>>it. But that may cause more problems for you.
>I simplify and people like you come in and try to confuse,
and sadly,
>I think that the readership of sci.math wants to be
confused by you.
>I didn't use the same symbols to represent two different
types of
>>quantities.
> If I have y=mx+b, and then a little later I have y=x+1, and
then later
> talk of with x=1, y=2, is that using the same symbols to
represent
> different types of quantities?
I have no problem with this.
> If you're having trouble keeping up, then you can work
through it
> *yourself* putting in whatever symbols help you.
I do. That's why you keeping seeing my posts with a_1(m).
> However, I've yet to be convinced that others would be 
more
helped by
> my adding in extra than hurt by the additional level of
complexity.
It's not for the others, it's to clearly 
de'ne what are the
independent
variables for the functions a_i. It's so that you know what
you're
talking about and can clearly communicate it to us.
> For instance, if I have
P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x + y)(a_2(m) x +
y)(a_3(m)
> x + y)
does that help more than it hurts?
It helps a great deal. How does it hurt?
> And it's even busier with the FLT argument where there are
even MORE
> symbols.
Are there people who will now get even more confused?
Unlikely. It may be busier but it is now precise. You have
clearly
indicated that the a_i depend only on m, and that when you
change x or
y, the a_i will not change. Of course, y and v are both 
de'ned
in
terms of f, so there's probably a little more going on. Your
introductions of v and y is hiding a relationship.
> What kind of arguments might I face then from people who
want to drag
> me into arguments about functions?
Ones that you can either clearly defend against or concede to
as
everyone will be discussing the same things. Right now there
are no
de'nitions so people are interpreting the same symbols
differently.
>Now if you can do that Nora Baron then clearly I made
some kind of
>mistake and there's no more room for me to argue.
There's no more
>>room for you to argue.
>Well now that I've given the cubic, I challenge you to
admit *you*
>made a mistake.
>Challenge declined. You want the a's to both vary with m 
and
be
>>constants. You'll have to make up your mind which it is.
You lack intellectual honesty.
You are the one that is being dragged into de'ning your
symbols.
>So why don't you go back to the cubic which 
de'nes the
a's, stick in
>all the values and see if the a's come out as you have
above, and then
>it's over.
>For each value of m, she can de'ne a_i(m) to be whatever
three values
>>makes both sides equal. You've given nothing more 
speci'c
about the
>>values to require more.
>That is false as I've repeatedly given the expression
de'ning the
>a's.
>If the a's are constants. If they are, then your expression
above fails
>>to be true when you vary m.
You can't be that lost. I simply 'nd it hard to 
believe that
you
> don't know more as I think you do but are attempting to
confuse
> others.
If the a's vary with m, then there must be a way to 
de'ne them
for each
choice of m. You have NOT done that. You have given what looks
like a
de'nition but would lead to the a's being 
constants. Perhaps
if you go
back and POST your paper HERE we can see if you've changed
things enough
to address the issues that you feel are valid. I'm working
with a copy
that is approximately a month old.
>>Now, what's your explanation?
>Nora B.
>Well I've explained above, and again, of course, you
can't just *pick*
>values for the a's for a particular m, as their values
are set
>rigorously. What I'm curious about are the people who
believe Nora
>Baron had a valid point.
>Where have you established how to rigorously compute
them. What are
they?
>For m=1, f=5, the a's are the roots of the cubic
> a^3 + 72a^2 - 13825 = 0.
>There's nothing in the nature of the problem that (to me)
makes such a
>>de'nition either obvious or necessary.
> You mean like expressions that actually de'ne the 
a's?
You seem to think you don't need to bother with the math!
Ok, I just noticed the below... That connection would have
been helpful.
Why do you berate me for not connecting something below with
that above?
>Speak up, and don't be shy as I suspect you may believe
she still has
>a valid point and I need to understand why she's so
effective in
>convincing people.
>Because you have not stated all of the conditions you
want to exist.
>>Once you do that, you will lose the results you are trying
to achieve.
>I've stated those conditions repeatedly. Are you now
claiming that
>you have never seen the factorization
> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x +
y)?
>Is THIS where it's de'ned??? Ok... and do you 
want the choice
of your
>>a_i(m) to hold for all x and y? Your simpli'cation does not
make
>>that obvious. Worse, there is no good way of assigning the
values
>>generated for each choice of m to each a_i(m).
That is b.s. as you can easily enough solve for the a's, 
even
with the
> expression in that form by creating a cubic as I have done,
and then
> solving it.
There'll be a LOT of symbols, but it can be done.
Here's the funny thing: Nora did it. It worked. You just 
don't
like
it. If she failed to do it correctly, point out the *speci'c*
§aw.
>>Note: when you CHOSE values for x and y, you changed the
nature of the
>>problem being looked at. The observations made by myself and
(I
>>believe) Nora are based on the special case of x=2, y=5. The
irony is,
>>you've been defending a leap from speci'c to 
general in your
arguments,
>>but won't let us look at the speci'c while 
ignoring the
general.
Actually my work has always covered a family of values, which
is why I
> can stick in values as I did in this thread. Previously,
with FLT, of
> course, I couldn't stick in values for x, y and z.
What family of values? It might help if you informed people
about that
in advance.
>>Have you considered keeping everything in one thread so 
it's
easier to
>>follow things?
> The threads become HUGE with hundreds of posts, and then
it's harder
> to follow things.
It's easier than trying to keep track of 20 short threads. I
can mark a
thread for future reference. It's harder to mark a lot of
threads, and
harder to follow the logic between the threads. A concession
in one
thread is not apparent in another thread.
> James Harris
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>
[deleted]
It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can
see how
> the 2 divides out, but if you couldn't see, and 
couldn't
guess, like
> if you had
P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2),
with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,
Correction 1: 2 a_1(x) + a_2(x) = 4x.
Correction 2: only if a_1(x) = x, a_2(x) = 2x.
If you want to allow anything else, your assertion is
incorrect. And
you consistently want to allow the a_i(x) to be
non-polynomials.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> <deleted > a_1 a_2 a_3 = 25 (625 m^3 - 75
m^2 + 3m)
> For any integer m, a_1 a_2 a_3 is a multiple of
25. i.e. not
coprime to 5.
> Perhaps you mean that none of the a's are coprime
to 5?
> Which is trivially obvious and uninteresting after all.
> Sorry to have bothered you.
> It's actually fascinating as it's a bizarre 
result,
and I think
you
> realize that I'm correct as you *again* deleted out the
following
> wonderful simpli'cation which shows clearly that 
I'm
right.
> Yuck. Unfortunately in my haste I neglected to refute
the claim,
as
> it is in fact NOT true that none of the a's are coprime
to 5, as
in
> fact the fascinating as it's a bizarre result is that
they ALL
are
> coprime to 5 in the ring of algebraic integers.
> The problem is with the ring of algebraic integers
as I've
explained.
> Consider the following which you deleted yet again
from my
previous
> post:
> Hmmm...how about this explanation?
> Remember that the polynomial is
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> which gives the reader a chance to see P(m) with only
the m
left as
> a symbol
> Yes! I like this example. See below.
> Fascinating.
> and you see that factor 25. Now imagine that I have the
polynomial
> Q(m), where
> P(m) = 25 Q(m)
> so you have as a factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> and the question is, how does that factor of 25
divide out?
> I want readers to consider how many posters have
apparently
attempted
> to make them believe that the factor of f^2, here 25,
was in some
> sense welded into the expression, when in fact it's a
factor of
P(m)
> such that I can write
> P(m) = 25 Q(m)
> as I've done here, or P(m) = f^2 Q(m), in general.
> Well, checking at Q(0), gives 11, so how can those
factors of
25
> divide through Q(m) in such a way as to give 11, at m=0?
> There's only one way, which gives you
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
> If that bothers you, remember that at m=0, two of
the a's equal
0.
> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions
> of m. When m = 0, two of them are 0. However when m is not
zero,
> none of them are zero. So how do their values when m = 0
> determine what they will be when m <> 0 ?
> Well you have P(m) = 25 Q(m) and that factor 25, which has
been the
> focus of all the arguing. Now considering Q(m) it turns
out that
> looking at the factorization
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
> the a's are in the way, and it's not clear 
how you divide
out.
> Now you *know* that the 25 divides through some way, and
it's clear
> that whatever way that is would be true for any m, just
like with
> 2(x^2 + 2x + 1) = (x+1)(2x+2)
> when you divide off 2, you do it for all x.
> For readers, people like Nora Baron have basically been
arguing that
> you can have something like
> Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3
+ h_1
h_2)
> where h_1 h_2 h_3 = 5, and each is not a unit.
> But notice then that setting m=0, gives (h_1 h_2 h_3)^2 =
25, when in
> fact Q(0)=11.
> Given that rigorous fact they claim that m=0 is a some
kind of
special
> case and try to cast doubt on the result, which is an
attack on
> algebra that I call voodoo math.
> Surprisingly they've been quite successful from what 
I've
gathered in
> convincing people, which makes you wonder about people's
understanding
> of algebra.
> Because from algebra, it turns out that you can just set
m=0, to
> 'gure out how that 25 divides out, as I've 
done.
> No voodoo at all. What you would like to have here is that
> the same a_1, a_2, and a_3 work when m = 0 as when m <> 0.
> But you know, and everyone else knows, that that is totally
> impossible. Why? Because when m = 0, two of the a's must 
be
> zero. When m <> 0, NONE of the a's can be zero.
That everyone else knows is an appeal to the gallery which is a
> logical fallacy.
>
Just a statement of fact, and anyway backed up immediately
by a speci'c justi'cation. As for an appeal to 
the gallery:
that seems like an odd complaint coming from you, with your
frequent indictments of mathematicians as corrupt liars who
want to deceive the innocent naive silent readers of sci.math,
alt.math, and alt.math.undergrad.
> The a's are de'ned 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 I guessed that with so many symbols
people were
> confused, so I stuck in some values to lessen the symbol
load.
>
You oversimpli'ed things and paid the price.
> It's real simple.
> The a's are functions of m. And I don't mean 
CONSTANT
functions.
> Got that?
> Different m's give different a's.
> You believe that because a_1 is divisible by 5 when m = 0,
> it must be divisible by 5 for other values of m also. It is
> some kind of hunch that you have, based on intuition. It
seems
> like a nice parallel construction. You think it's obvious.
> It isn't. In fact it is false. Your intuition this time is
> wrong. Your hunch has been disproved.
<deleted>
Here you claim that I'm acting on a hunch about the 
a's.
But, when I used x=2, f=5, y=5, I have
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
and noticing that 25 I used P(m) = 25 Q(m), so I have
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25
where the question is, how does that 25 divide through?
But since I know that Q(0)=11, I know that it must be
Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).
At Q(0) that is
Q(0) = (2 (0) + 1)(2 (0) + 1)(2 (3) + 5) = 11.
So in fact it's algebra and not a hunch.
>
What you have here is that when m = 0, a_1 = a_2 = 0,
and therefore a_1 and a_2 are divisible by 5, and
2*a_1 / 5 and 2*a_2 / 5
are both algebraic integers (because both are 0, and
conclude that a_1 and a_2 must be divisible by 5
when m takes on nonzero values also. You do not
prove this. You do not cite any mathematical principle
that justi'es it. It is clearly just your hunch. It
looks like a nice pattern, so it must be true. Of
course a_1 and a_2 are actually dependent on m. When
m = 0, they are both 0 and both divisible by 5, but
how does that tell you anything about either of them
when m <> 0 ?
You have not provided a justi'cation. You simply make the
statement.
And you deleted out the rest of my post which dealt with
this. Why?
Further, you have not found a valid objection to my proof that
your central claim related to this is incorrect. Nor have you
responded at all to Dale Hall's separate proof. To
show that Dale is wrong, all you would have to do would
be carry out a simple computation. You don't have to 
understand
anything about Galois theory or Gauss' lemma or anything
else from the algebraic theory of numbers. But you haven't
done anything at all. Why?
Since you deleted my proof that your central claim is
wrong, I am appending it again below. Feel free to point out
any errors in it, or post a question if there are parts
of it that you don't understand.
=
JSH CLAIM:
It is possible to 'nd a 3rd degree polynomial with integer
coef'cients, monic and irreducible over the rationals, such
that
if a1, a2, and a3 are the roots, then at least one of a1, a2,
or
a3 is coprime (in the algebraic integers) to a prime factor of
the constant term of the polynomial.
DISPROOF OF CLAIM:
Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are
integers and c = p*v, where p is a prime and v is another
integer. Q(x) is clearly monic. Assume Q(x) is
irreducible. Let a1, a2, and a3 be roots of Q(x). Note
that by de'nition, a1, a2, and a3 are algebraic
integers.
You are claiming that at least one of a1, a2, or a3 is
coprime to p. Assume a1 is coprime to p.
By standard theory, there exists an automorphism F12 of
the 'eld of algebraic numbers with the following
properties *** :
1. F12 leaves the sub'eld of rational numbers 
'xed,
i.e., if q is rational, F12(q) = q.
2. F12(a1) = a2.
3. If t is an algebraic integer, F12(t) is also an
algebraic integer.
Now since a1 is relatively prime to p, there exist
algebraic integers r and s such that
[1] r*a1 + s*p = 1.
Now apply the automorphism F12 to both sides of [1]:
F12(r)*F12(a1) + F12(s)*F12(p) = F12(1).
By the properties above, F12(p) = p and F12(1) = 1.
Moreover, r' = F12(r) is an algebraic integer, and
s' = F12(s) is an algebraic integer. Finally,
F12(a1) = a2. Thus one obtains:
r'*a2 + s'*p = 1,
which says: a2 and p are coprime in the algebraic
integers.
Similarly one shows that a3 and p are coprime.
Therefore if one of a1, a2, or a3 is coprime
to p, then they all are.
But a1 * a2 * a3 = p * v. That is, p divides
the product of a1, a2, and a3. Therefore p cannot
be coprime to each of a1, a2, and a3.
Putting all this together, one concludes that
NONE of a1, a2, or a3 can be coprime to p.
This directly contradicts your claim which
was quoted above.
Again, please feel free to point out any errors in the
proof I just gave.
*** Reference on automorphisms:
http://www.math.niu.edu/~beachy/aaol/galois.html
See especially Proposition 8.6.2 on that page. Or
see the excellent textbook, Abstract Algebra, by
John Beachy and William D. Blair.}
=
> It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you 
can
see how
> the 2 divides out, but if you couldn't see, and 
couldn't
guess, like
> if you had
P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2),
with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,
you could have P(x) = 2Q(x), and set x=0, with Q(0) to get 1,
showing
> that
Q(x) = (a_1 + 1)(a_2/2 + 1), is correct.
Of course, you can also just see that is the only way that
works.
But regardless there is only ONE way that will work, and it
doesn't
> change as x changes, and it doesn't change above as m
changes.
>
Oversimpli'ed examples are NOT a substitute for proofs. I
would think you would have learned that by now, if nothing
else.
Nora B.
> Assaulting algebra is a rather disturbing step taken by
posters
> continuing to argue with me. What is also troubling is how
uncaring
> the newsgroup seems to be about their denial of basic
algebra. But it
> is telling.
>
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
sorry for violating my promise, today;
tomorrow, it will hoepfully remain in full force.
anyway, I had to say that your statement is peculiar, because
an appeal to the gallery can be countered by anyone
who happens to be *in* the gallery (*I* don't know that 
e.g.);
the important part, apparently, was that
*you* should perhaps know it, given your extensive 10-year
mission
to prove la premiere theoreme de Fermat.
That everyone else knows is an appeal to the gallery which is a
logical fallacy.
The a's are de'ned 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 I guessed that with so many symbols
people were
confused, so I stuck in some values to lessen the symbol load.
--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
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
<deleted>
Previously I answered Nora Baron's claim that I was acting 
on
a hunch
as I refuted that claim mathematically, and then I deleted out
the
rest as I wanted to shorten the length of posts.
This time I'm deleting down to just past her hunch 
assertion.
Nora Baron's comments follow.
> What the a's are when m = 0 is a special case. It does NOT
> determine what they are when m <> 0. Sure, when m = 0,
> you can assume that a_1 and a_2 are zero. Can you can say
> that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of
course
> you can - they are both 0! After all, ANY number divides 0.
For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are
> algebraic integers also, since both are 0. Does that mean
that 137 must
> divide two of the a's, when m <> 0 ???? By your logic, 
YES!
Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS 
OF
M.
> WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT
BE ASSUMED
> TO BE TRUE WHEN M <> 0. *****
I'm leaving that in as part of the setup without saying much
more
here, as I handle everything below. What's fascinating is 
how
far she
goes in trying to push against algebra, as I'll show below.
> Why is this so hard for you to understand? Why do you keep
trying to
> act like I am pulling some kind of trick when I say this, or
pretending
> that I am lying?
Well either you're lying or you think you've 
refuted algebra,
as I
show below.
I'd think that you'd prefer that people think 
you're lying.
<deleted>
Nora Baron's comments follow.
> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> and assume it is factored *as you propose* in the form
> [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
> Ok.
> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
> Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
> seem to be just a tad bit strange to you?
>
Nope. It was your idea, not mine, to replace the x's
> with a constant value of 2. When you did that you were
> no longer factoring a polynomial. You are just factoring
> an ordinary number. There are no hidden rules that say it
> must be factored as a polynomial.
You can't have it both ways. You want it factored as
> a polynomial, leave the x's in there (and see below). You
> want it factored as an ordinary number (remember? VERY much
> simpli'ed) then one must assume that any factorization is
> legitimate, and you get my example above.
The factorization exists for all x so why would it go away for
a
particular x?
> So let's go back to assuming you want it factored as a
> polynomial. Of course you know that both I and W. Dale
> Hall have proofs that that is not going to work. You have
> not provided a valid objection to my proof, and you have
> not provided any response whatsoever to what Dale did. I
> don't see how you can. It's a simple 
computation. You
> do the arithmetic, and Dale is either right or he is wrong.
> You don't have to understand ANYTHING. There is no wool
> to be pulled over anyone's eyes. Try it and see.
See also below.
Well you're 'ghting algebra. Guess what? The 
algebra wins.
See below...
> Note that EACH ONE of these is divisible by 5:
> Ok, you *pick* values from the a's as if 
they're not de'ned
by a
> cubic, which they are, and your picked values are supposed
to prove
> something?
 Yes. It proves what you said was wrong. You said that one
> of the a's had to be coprime to 5. Clearly, totally,
> unambiguously false.
It can only be false if you wish to refute algebra Nora Baron.
See
below...
> NONE ARE COPRIME TO 5.
> So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
> Remember you simply tossed those numbers out there, but
there *is* a
> way to calculate the a's so your assertion can be tested, 
as
they are
> roots to a cubic.
>
They *would* be roots of a cubic in the variable x, but you
> replaced x by 2. At that point there is no longer a
polynomial.
> The restriction that they be roots of a cubic is gone. It was
> your idea, not mine, to oversimplify things and make the
> substitution.
Let me ask you a question, are you claiming that given
P(x) = (x+1)(x+2)
that if I stick in actual values for x, the factorization is
gone?
Yes, if I have P(2)=12, it is true that you just see a number,
but
notice that P(2) = 3(4).
The expression I use is
(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 y=uf, and now you are arguing that the
factorization goes away simply because I set x=2?
Um, that's not algebra Nora Baron; that's more 
voodoo math.
What I've done is put in actual numbers for x, f, and u, as 
I
set x=2,
f=5, and u=1, which gave me a polynomial. But the factorization
*still* exists, and your wish that it doesn't--no matter how
desperate--won't change that reality.
> Remember: if you want to go back to assuming that a_1, a_2,
and
> a_3 are roots of a cubic in the variable x, to continue to
> maintain what you think is true, you are going to have to 
'nd
> an error in the proof I have presented (and is appended
> below).
Why?
> Now if you can do that Nora Baron then clearly I made some
kind of
> mistake and there's no more room for me to argue.
>
Yep. You should not have substituted x = 2. You want it
> factored as a polynomial, you had better leave it a
polynomial
> and not try to oversimplify. You have shown confusion
previously
> over what the difference between a polynomial and the
evaluation
> of that polynomial when you choose a particular value for x.
So you think that factorizations go away if you stick in
values?
> The moral here is, this little example should show you pretty
> convincingly that when people tried to get you to understand
that
> difference, they were not just playing with semantics.
There is another more general moral here, and you have run
> afoul of this one many a time in the past also: Examples are
> not a substitute for proofs. And do NOT oversimplify when you
> are trying to do mathematics.
It doesn't seem that I can oversimplify for you Nora Baron.
> So why don't you go back to the cubic which 
de'nes the a's,
stick in
> all the values and see if the a's come out as you have
above, and then
> it's over.
>
I will do better than that. I will reproduce my proof that your
> claim that one of the a's is coprime to 5 is wrong, but I
will do it in
> more generality. That way I will be heading off any examples
> you might try to come up with in the future. What you want is
> a consequence of the following claim, which you have stated
> elsewhere:
CLAIM: It is possible to 'nd a monic polynomial of degree
> 3, with integer coef'cients, irreducible over the 
rationals.
> and with roots a1, a2, and a3, such that at least one of
> a1, a2, or a3 is coprime, in the ring of algebraic
> integers, to a prime factor of the constant term of
> the polynomial.
Right? You have claimed as recently as yesterday.
> DISPROOF OF CLAIM:
<deleted>
And Nora Baron rattles on a bit more. Where from before her
primary
position is that given a factorization I lose the
factorization by
actually putting in a value for x, as I used x=2. The key
expression
is
(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 y=uf, and I put in actual numbers for
several
values as a simpli'cation to remove room for bogus objections
like
Nora Baron's here. Yet Nora Baron in one reply simply put in
some
picked values for the a's for m=1 in what she thought of as 
a
refutation, and now seems to think that my saying x=2 takes
away the
factorization above. That's hardly even voodoo mathematics.
I think it's more simply: nonsensical.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
> What the a's are when m = 0 is a special case. It does 
NOT
>>determine what they are when m <> 0. Sure, when m = 0,
>>you can assume that a_1 and a_2 are zero. Can you can say
>>that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of
course
>>you can - they are both 0! After all, ANY number divides 0.
> For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are
>>algebraic integers also, since both are 0. Does that mean
that 137 must
>>divide two of the a's, when m <> 0 ???? By your logic, 
YES!
> Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS
OF M.
>>WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT
BE ASSUMED
>>TO BE TRUE WHEN M <> 0. *****
>> Why is this so hard for you to understand? Why do you keep
trying to
>>act like I am pulling some kind of trick when I say this, or
pretending
>>that I am lying?
> Well either you're lying or you think you've 
refuted
algebra, as I
> show below.
Or you don't understand.
> I'd think that you'd prefer that people 
think you're lying.
<deleted>
Nora Baron's comments follow.
>> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>>
Please note: you are factoring 25(5000m^3 - 600m^2 - 126m + 11)
Where is x? Answer: nowhere. Now stop talking about it, or stop
talking about this polynomial.
[deleted]
>Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't that
>seem to be just a tad bit strange to you?
> Nope. It was your idea, not mine, to replace the x's
>>with a constant value of 2. When you did that you were
>>no longer factoring a polynomial. You are just factoring
>>an ordinary number. There are no hidden rules that say it
>>must be factored as a polynomial.
> You can't have it both ways. You want it factored as
>>a polynomial, leave the x's in there (and see below). You
>>want it factored as an ordinary number (remember? VERY much
>>simpli'ed) then one must assume that any factorization is
>>legitimate, and you get my example above.
> The factorization exists for all x so why would it go away
for a
> particular x?
Because you got rid of x. If you see an x up there, get
glasses.
>> So let's go back to assuming you want it factored as a
>>polynomial. Of course you know that both I and W. Dale
>>Hall have proofs that that is not going to work. You have
>>not provided a valid objection to my proof, and you have
>>not provided any response whatsoever to what Dale did. I
>>don't see how you can. It's a simple 
computation. You
>>do the arithmetic, and Dale is either right or he is wrong.
>>You don't have to understand ANYTHING. There is no wool
>>to be pulled over anyone's eyes. Try it and see.
> See also below.
> Well you're 'ghting algebra. Guess what? The 
algebra wins.
See below...

>> Note that EACH ONE of these is divisible by 5:
>Ok, you *pick* values from the a's as if 
they're not de'ned
by a
>cubic, which they are, and your picked values are supposed
to prove
>something?
>> Yes. It proves what you said was wrong. You said that one
>>of the a's had to be coprime to 5. Clearly, totally,
>>unambiguously false.
> It can only be false if you wish to refute algebra Nora
Baron. See
> below...

>> NONE ARE COPRIME TO 5.
>So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
>Remember you simply tossed those numbers out there, but
there *is* a
>way to calculate the a's so your assertion can be tested,
as they are
>roots to a cubic.
> They *would* be roots of a cubic in the variable x, but you
>>replaced x by 2. At that point there is no longer a
polynomial.
>>The restriction that they be roots of a cubic is gone. It was
>>your idea, not mine, to oversimplify things and make the
>>substitution.
> Let me ask you a question, are you claiming that given
P(x) = (x+1)(x+2)
that if I stick in actual values for x, the factorization is
gone?
Yes, if I have P(2)=12, it is true that you just see a number,
but
> notice that P(2) = 3(4).
Notice: P(2)=2(6)=1(12)
The uniqueness of the factorization into 2 factors is gone.
The expression I use is
(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 y=uf, and now you are arguing that the
> factorization goes away simply because I set x=2?
Does this look anything like the polynomial above? Does it
have the
same number of variables?
> Um, that's not algebra Nora Baron; that's 
more voodoo math.
You substituted for x but want to keep x. That's voodoo
alright.
> What I've done is put in actual numbers for x, f, and u, 
as
I set x=2,
> f=5, and u=1, which gave me a polynomial. But the
factorization
> *still* exists, and your wish that it doesn't--no matter 
how
> desperate--won't change that reality.
The factorization still exists, but not uniquely. Perhaps 
I've
been
wrong all along and you meant a_i(m,x,f,u). If this is the
case, what
she did *still* works. She de'ned a_i(1,2,5,1). She just
didn't do it
the way you want.
>> Remember: if you want to go back to assuming that a_1, a_2,
and
>>a_3 are roots of a cubic in the variable x, to continue to
>>maintain what you think is true, you are going to have to 
'nd
>>an error in the proof I have presented (and is appended
>>below).
Why?
Because otherwise your paper is wrong.
>Now if you can do that Nora Baron then clearly I made some
kind of
>mistake and there's no more room for me to argue.
> Yep. You should not have substituted x = 2. You want it
>>factored as a polynomial, you had better leave it a
polynomial
>>and not try to oversimplify. You have shown confusion
previously
>>over what the difference between a polynomial and the
evaluation
>>of that polynomial when you choose a particular value for x.
> So you think that factorizations go away if you stick in
values?
See above with P(2)=12.
>So why don't you go back to the cubic which 
de'nes the a's,
stick in
>all the values and see if the a's come out as you have
above, and then
>it's over.
> I will do better than that. I will reproduce my proof that
your
>>claim that one of the a's is coprime to 5 is wrong, but I
will do it in
>>more generality. That way I will be heading off any examples
>>you might try to come up with in the future. What you want is
>>a consequence of the following claim, which you have stated
>>elsewhere:
> CLAIM: It is possible to 'nd a monic polynomial of degree
>> 3, with integer coef'cients, irreducible over the 
rationals.
>> and with roots a1, a2, and a3, such that at least one of
>> a1, a2, or a3 is coprime, in the ring of algebraic
>> integers, to a prime factor of the constant term of
>> the polynomial.
> Right? You have claimed as recently as yesterday.
> DISPROOF OF CLAIM:
> <deleted>
At this point, how are you dealing with her proof? I would
think that
of all people, you would know better than to delete material
and then
respond to it. You were very upset when I did that by
accident, yet you
make sure everyone knows you deleted it.
> And Nora Baron rattles on a bit more. Where from before her
primary
> position is that given a factorization I lose the
factorization by
> actually putting in a value for x, as I used x=2. The key
expression
> is
(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 y=uf, and I put in actual numbers for
several
> values as a simpli'cation to remove room for bogus
objections like
> Nora Baron's here. Yet Nora Baron in one reply simply put 
in
some
> picked values for the a's for m=1 in what she thought of as 
a
> refutation, and now seems to think that my saying x=2 takes
away the
> factorization above. That's hardly even voodoo 
mathematics.
I think it's more simply: nonsensical.
> James Harris
This does not refute her argument. It does not even deal with
it.
There is something you are welcome to own, however. voodoo
mathematics. I'm willing to let you have it. 
It's all yours.
Whatever it is.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>>
[deleted]
> It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you 
can
see how
> the 2 divides out, but if you couldn't see, and 
couldn't
guess, like
> if you had
> P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2),
> with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,
Correction 1: 2 a_1(x) + a_2(x) = 4x.
> Correction 2: only if a_1(x) = x, a_2(x) = 2x.
If you want to allow anything else, your assertion is
incorrect. And
> you consistently want to allow the a_i(x) to be
non-polynomials.
Hmmm...which might lead a *rational* reader to suppose 
you're
trying
to claim that it doesn't work with non-polynomial factors.
How about this?
p(x) = sqrt(2(x^2+2x+1)) = sqrt((a_1+1)(a_2+2))
with a_1 a_2 = 2x^2, 2a_1 + a_2 = 4x?
Mathematicians, what a crew. But you know what? They take
themselves
SO seriously, even when they're 'ghting basic 
algebra.
So why would *mathematicians* 'ght algebra? Because 
they're
actually
like English professors...really stuck-up English
professors...who
believe they're owed a certain amount of respect.
But you see, English professors can pooh-pooh a literary work,
so
mathematicians acting like really stuck-up English professors
think
they can pooh-pooh valid mathematics, which inevitably leads
to them
attacking the foundations of mathematics because math is funny
that
way.
James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>
... stuff deleted ...
>> So let's go back to assuming you want it factored as a
>>polynomial. Of course you know that both I and W. Dale
>>Hall have proofs that that is not going to work. You have
>>not provided a valid objection to my proof, and you have
>>not provided any response whatsoever to what Dale did. I
>>don't see how you can. It's a simple 
computation. You
>>do the arithmetic, and Dale is either right or he is wrong.
>>You don't have to understand ANYTHING. There is no wool
>>to be pulled over anyone's eyes. Try it and see.
> See also below.
> Well you're 'ghting algebra. Guess what? The 
algebra wins.
See below...
>
... stuff deleted ...
It can only be false if you wish to refute algebra Nora Baron.
See
> below...
>
... stuff deleted ...
> The expression I use is
(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 y=uf, and now you are arguing that the
> factorization goes away simply because I set x=2?
Um, that's not algebra Nora Baron; that's more 
voodoo math.
What I've done is put in actual numbers for x, f, and u, as 
I
set x=2,
> f=5, and u=1, which gave me a polynomial. But the
factorization
> *still* exists, and your wish that it doesn't--no matter 
how
> desperate--won't change that reality.
>
You obtained the polynomial
65 x^3 - 12 x + 1
by substituting v=4, y=1 into that expression, correct?
You claim that this factorization:
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
with the a's being algebraic integers, has the 
a's being
coprime to 5, correct?
What you say is wrong, and repeating it doesn't change the
facts. I've lost count of how many times I've 
posted this
particular argument, but I don't mind, since I can easily
I have proposed that these polynomials
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
have the property that, for any root z of the polynomial
p(x)= x^3 - 12 x^2 + 65,
we have
q(z)*r(z) = 5
r(z)*s(z) = z.
I've shown how this can be veri'ed, by doing 
the following
multiplications (courtesy of DOE Macsyma):
First, here are the products that I'm making claims about:
q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325
r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680
Next, a couple of products of p(x) = x^3 - 12 x^2 + 65 with
polynomials of degree 1:
(64 x + 128)*(x^3 - 12 x^2 + 65)
= 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320
(32 x + 72)*(x^3 - 12 x^2 + 65)
= 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680
Finally, we compare the results and see this:
q(x)*r(x) = (64 x + 128)*p(x) + 5
r(x)*s(x) = (32 x + 72)*p(x) + x,
Note that, for any value xo that makes p(xo) = 0,
that same value xo will make q(xo)r(xo) = 5, so
r(xo) is a factor of 5.
That value of xo also makes r(xo)*s(xo) = xo, so
r(xo) is a factor of xo.
In short, r(xo) becomes a factor of *both* xo and 5.
Since r(x) is a polynomial with integral coef'cients,
r(xo) is an algebraic integer whenever xo is.
It is similarly simple to demonstrate that, whenever
xo is a root of p(x), then r(xo) is a root of the
polynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5:
First, expand the polynomial mpr(r(x)):
mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5
= (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2
+ 315 (8 x^2 - 4 x - 45) + 5
= 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
Next, multiply these two polynomials:
p(x) = x^3 - 12 x^2 + 65,
and
w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808
to get this:
p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3
+ 731136 x^2 - 374400 x - 2067520
Notice the equality
mpr(r(x)) = p(x)*w(x).
That means for every value of x, the polynomial you get by
computing
r(x), then evaluating mpr(x) at that value, is equal to the
product of
p(x) and w(x).
If xo is a root of p(x), you have p(xo) = 0, so p(xo)*w(xo) =
0, and
therefore r(xo) is a root of mpr(x).
Note that there are three such roots of mpr(x), and (taking
the three
roots x1,x2,x3 of p(x)), three values
r1 = r(x1) ~ 968.67481
r2 = r(x2) ~ -0.01517
r3 = r(x3) ~ 0.34036
correspond to the three real roots of mpr(x). As such, their
product
must be -5.
Your earlier claim that the a's must be coprime to 5 implies
that the
r's must be units (since they're common 
factors of ai with 5,
in the
ring of algebraic integers).
If you multiply units, even you must realize that the product
is again
a unit.
Therefore, -5 (and equivalently, 5 itself) must be a unit in
the ring
of algebraic integers.
Do you agree or not?
If so, then say so. I'll make it easy; here's 
a form for you
to 'll out and post to your full array of newsgroups:
I, James S. Harris, af'rm my belief that, in
the ring of algebraic integers, the (rational)
integer 5 ['ve] is a unit.
James S. Harris.
If not, then show me (hey, don't worry about me, show your
public!)
where my error is. I've done all the multiplication; you can
verify
or refute all this very easily, given the ability to multiply
or expand polynomial expressions.
Show how highly you value algebra: Do some.
It's easy: ordinary polynomials, ordinary polynomial
multiplication,
nothing up my sleeves, no salesman will call. You don't even
need
to solve any equations. Just multiply, combine terms, show me
wrong!
Prove me wrong. Your mama says you can't.
... stuff deleted ...
And Nora Baron rattles on a bit more. Where from before her
primary
> position is that given a factorization I lose the
factorization by
> actually putting in a value for x, as I used x=2. The key
expression
> is
(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 y=uf, and I put in actual numbers for
several
> values as a simpli'cation to remove room for bogus
objections like
> Nora Baron's here. Yet Nora Baron in one reply simply put 
in
some
> picked values for the a's for m=1 in what she thought of as 
a
> refutation, and now seems to think that my saying x=2 takes
away the
> factorization above. That's hardly even voodoo 
mathematics.
I think it's more simply: nonsensical.
>
I haven't bothered with your evaluation at x=2, and 
don't care
about
whether you think anyone is doing anything.
What I do notice is the level of cowardice it must take to sit
there
and claim that people are, what was it, 'ghting algebra?
You don't even 'ght. The algebra 
I've presented is simple
enough
that you shouldn't evade it. Well, I know that if I had made 
an
actual mistake (as contrasted to the typos that Wayne Brown
picked
up), you would be all over it like §ies on a pile of poo, but
if
someone shows up with the actual goods, you're Brave Sir 
Robin:
Brave Sir Robin ran away.
Bravely ran away, away!
When danger reared its ugly head,
He bravely turned his tail and §ed.
Yes, brave Sir Robin turned about
And gallantly he chickened out.
Bravely taking to his feet
He beat a very brave retreat,
Bravest of the brave, Sir Robin!
He is packing it in and packing it up
And sneaking away and buggering up
And chickening out and pissing off home,
Yes, bravely he is throwing in the sponge...
You might think I'll give up on this.
James Harris
Dale.
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>[deleted]
>>It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you 
can
see how
>the 2 divides out, but if you couldn't see, and 
couldn't
guess, like
>if you had
> P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2),
>with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,
>Correction 1: 2 a_1(x) + a_2(x) = 4x.

>>Correction 2: only if a_1(x) = x, a_2(x) = 2x.
>If you want to allow anything else, your assertion is
incorrect. And
>>you consistently want to allow the a_i(x) to be
non-polynomials.
> Hmmm...which might lead a *rational* reader to suppose
you're trying
> to claim that it doesn't work with non-polynomial factors.
How about this?
p(x) = sqrt(2(x^2+2x+1)) = sqrt((a_1+1)(a_2+2))
with a_1 a_2 = 2x^2, 2a_1 + a_2 = 4x?
Mathematicians, what a crew. But you know what? They take
themselves
> SO seriously, even when they're 'ghting basic 
algebra.
So why would *mathematicians* 'ght algebra? Because 
they're
actually
> like English professors...really stuck-up English
professors...who
> believe they're owed a certain amount of respect.
But you see, English professors can pooh-pooh a literary work,
so
> mathematicians acting like really stuck-up English
professors think
> they can pooh-pooh valid mathematics, which inevitably leads
to them
> attacking the foundations of mathematics because math is
funny that
> way.
> James Harris
Let's stick with the original problem.
P(x)=2(x^2 + 2x + 1)
P(x)= (a_1(x) + 1)(a_2(x) + 2)
a_1(x)=sqrt(x)
a_2(x)=2x sqrt(x) - 2x + 6 sqrt(x) - 8 + 8/(sqrt(x)+1)
These are valid factorizations in the algebraic functions.
a_1(x) a_2(x) = 2x^2 - 2x sqrt(x) + 6x - 8 sqrt(x) + [8
sqrt(x)]/[sqrt(x)+1]
Therefor, a_1(x) a_2(x) =/= 2x^2.
Put bluntly: I *am* claiming that non-polynomial factors 
don't
have the
properties you want them to have. Simply put, non-polynomial
factorizations do not behave like polynomial factorizations.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
<deleted>
Previously I answered Nora Baron's claim that I was acting 
on
a hunch
> as I refuted that claim mathematically, and then I deleted
out the
> rest as I wanted to shorten the length of posts.
This time I'm deleting down to just past her hunch 
assertion.
Nora Baron's comments follow.
>
Some of!
> What the a's are when m = 0 is a special case. It does NOT
> determine what they are when m <> 0. Sure, when m = 0,
> you can assume that a_1 and a_2 are zero. Can you can say
> that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of
course
> you can - they are both 0! After all, ANY number divides 0.
> For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are
> algebraic integers also, since both are 0. Does that mean
that 137 must
> divide two of the a's, when m <> 0 ???? By your logic, 
YES!
> Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS
OF M.
> WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT
BE ASSUMED
> TO BE TRUE WHEN M <> 0. *****
I'm leaving that in as part of the setup without saying much
more
> here, as I handle everything below. What's fascinating is
how far she
> goes in trying to push against algebra, as I'll show 
below.
> Why is this so hard for you to understand? Why do you keep
trying to
> act like I am pulling some kind of trick when I say this, or
pretending
> that I am lying?
Well either you're lying or you think you've 
refuted algebra,
as I
> show below.
I'd think that you'd prefer that people think 
you're lying.
>
People should certainly judge for themselves, unless you
consider that playing to the gallery.
> <deleted>
Nora Baron's comments follow.
>
Again?
> Let's go back to your original polynomial,
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
> and assume it is factored *as you propose* in the form
> [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).
> Now instead of m = 0, let's try m = 1:
> Ok.
> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.
> This can be factored in the form [*] by letting
> a_1 = -5, a_2 = -5, and a_3 = 2140.
> Um, Nora Baron, now you willy-nilly give the a's values.
Doesn't
that
> seem to be just a tad bit strange to you?
> Nope. It was your idea, not mine, to replace the x's
> with a constant value of 2. When you did that you were
> no longer factoring a polynomial. You are just factoring
> an ordinary number. There are no hidden rules that say it
> must be factored as a polynomial.
> You can't have it both ways. You want it factored as
> a polynomial, leave the x's in there (and see below). You
> want it factored as an ordinary number (remember? VERY much
> simpli'ed) then one must assume that any factorization is
> legitimate, and you get my example above.
The factorization exists for all x so why would it go away for
a
> particular x?
>
My point in showing what happened when you substituted 2
for x was in part that it completely changes the problem.
You started with statements about factoring a polynomial.
You evaluated the polynomial at x = 2. The evaluation is
simply an integer. Your proposed factorization was the
factorization of an ordinary integer. If you wanted us to
continue regarding it as a polynomial factorization, you
should have left it as a polynomial.
This point is admittedly peripheral to the main argument
here, that is, statements you have made regarding
factorizations
of polynomials. However I thought it was worth making anyway,
to demonstrate how polynomials and their evaluations are
different things. Sorry you didn't get it.
> So let's go back to assuming you want it factored as a
> polynomial. Of course you know that both I and W. Dale
> Hall have proofs that that is not going to work. You have
> not provided a valid objection to my proof, and you have
> not provided any response whatsoever to what Dale did. I
> don't see how you can. It's a simple 
computation. You
> do the arithmetic, and Dale is either right or he is wrong.
> You don't have to understand ANYTHING. There is no wool
> to be pulled over anyone's eyes. Try it and see.
> See also below.
Well you're 'ghting algebra. Guess what? The 
algebra wins.
See below...
 Note that EACH ONE of these is divisible by 5:
> Ok, you *pick* values from the a's as if 
they're not
de'ned by a
> cubic, which they are, and your picked values are supposed
to prove
> something?
> Yes. It proves what you said was wrong. You said that one
> of the a's had to be coprime to 5. Clearly, totally,
> unambiguously false.
It can only be false if you wish to refute algebra Nora Baron.
See
> below...
>
Nope. You made a statement about a factorization of an ordinary
integer. You said it could not be of a certain form. I showed
directly and unambiguously that your statement was wrong. You
did NOT say that, after you had evaluated the polynomial and
obtained an ordinary integer, that you still wanted the
factorization
to be that associated with the polynomial. You were, literally,
wrong. You tried to get an oversimpli'ed example to prove 
your
point. I gave a perfectly valid counterexample. Now you want to
go back and say, No, I meant only polynomial-type
factorizations!
To which I say, Well, then, why didn't you leave it in the 
form
of a polynomial in the 'rst place? It doesn't 
make it any
simpler
to replace ïx' with 
ï2' if you are just going to consider the
ï2'
as a polynomial variable, rather than a constant. Why did
you evaluate to an ordinary integer anyway if you are not
going to take
advantage of it? What did you expect to gain?
Again, this is peripheral to the main argument. If you don't
get my point here and want to think of this as a minor
distraction,
it's 'ne with me, as long as you address the 
*real* problem as
discussed below.
> NONE ARE COPRIME TO 5.
> So are your claiming that at m=1, a_1 = -5, a_2 = -5, and
a_3 = 2140?
> Remember you simply tossed those numbers out there, but
there *is* a
> way to calculate the a's so your assertion can be tested,
as they are
> roots to a cubic.
> They *would* be roots of a cubic in the variable x, but you
> replaced x by 2. At that point there is no longer a
polynomial.
> The restriction that they be roots of a cubic is gone. It was
> your idea, not mine, to oversimplify things and make the
> substitution.
Let me ask you a question, are you claiming that given
P(x) = (x+1)(x+2)
that if I stick in actual values for x, the factorization is
gone?
>
No, it's just the opposite. If you stick in actual values
for x, you may get factorizations which are not consistent
with the polynomial factorization. For example, you
let x = 2. P(x) = 12. The factorization that is consistent
with the polynomial factorization that you just gave is
12 = 3 * 4 = (2 + 1)*(2 + 2).
There are, however, other factorizations: for example 2 * 6
or 1 * 12. These are NOT consistent with the polynomial
factorization.
> Yes, if I have P(2)=12, it is true that you just see a
number, but
> notice that P(2) = 3(4).
>
See above.
> The expression I use is
(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 y=uf, and now you are arguing that the
> factorization goes away simply because I set x=2?
>
Not at all. You don't lose any of the polynomial 
factorizations
when you evaluate. You gain some *new* ones. That is exactly
what
happened in what I posted.
> Um, that's not algebra Nora Baron; that's 
more voodoo math.
>
Nonsense. It's just boring arithmetic.
> What I've done is put in actual numbers for x, f, and u, 
as
I set x=2,
> f=5, and u=1, which gave me a polynomial. But the
factorization
> *still* exists, and your wish that it doesn't--no matter 
how
> desperate--won't change that reality.
>
You got it backwards. You don't lose any of the polynomial
factorizations. You just gain some other ones. That is why
your oversimpli'ed example didn't make any 
sense. It most
certainly did not prove your point. It went exactly the other
direction: it showed that factorizations existed which were
exactly of the form you said could not occur. You would
have been better off leaving it as a polynomial in x.
> Remember: if you want to go back to assuming that a_1, a_2,
and
> a_3 are roots of a cubic in the variable x, to continue to
> maintain what you think is true, you are going to have to 
'nd
> an error in the proof I have presented (and is appended
> below).
Why?
>
A truly strange question.
I have pointed out an explicit place in your argument where you
have made a mistake. I have pointed out exactly what went
wrong with your reasoning. You still apparently do not
understand it. That is unfortunate. But in addition to
identifying your error [which basically boils down to assuming
without a shred of proof that the form of a factorization is
the same when m <> 0 as it is in the degenerate case, when
m = 0], I have a proof that your main conclusion *** cannot be
correct ***.
It is not a long or dif'cult proof.
Most people would react by trying to 'nd an error in my
proof. You have failed to do so. It stands.
I have found your error. You have not found mine, if I
have one. It's that simple.
Worse yet: W. Dale Hall also has a proof, totally inde-
pendent from mine, that one of your main claims is false.
In his case, all you have to do to check whether he is
right or not is carry out a little computation. You have
not done it, or if you have, you are being mighty quiet
about it. Dale's proof stands also.
Yes, most people would not be happy to have two
independent unrefuted proofs that their beloved work was
wrong, hanging out there unrefuted. They would try
'nd where the proofs went wrong.
That's why.
> Now if you can do that Nora Baron then clearly I made some
kind of
> mistake and there's no more room for me to argue.
> Yep. You should not have substituted x = 2. You want it
> factored as a polynomial, you had better leave it a
polynomial
> and not try to oversimplify. You have shown confusion
previously
> over what the difference between a polynomial and the
evaluation
> of that polynomial when you choose a particular value for x.
So you think that factorizations go away if you stick in
values?
>
No, no, no, again: you don't lose factorizations which work
for the underlying polynomial when you stick in values. You
gain
new ones.
> The moral here is, this little example should show you pretty
> convincingly that when people tried to get you to understand
that
> difference, they were not just playing with semantics.
> There is another more general moral here, and you have run
> afoul of this one many a time in the past also: Examples are
> not a substitute for proofs. And do NOT oversimplify when you
> are trying to do mathematics.
It doesn't seem that I can oversimplify for you Nora Baron.
>
No, on the contrary. As has often happened with your
simplistic little toy examples, you went too far. It
back'red on you.
> So why don't you go back to the cubic which 
de'nes the
a's, stick in
> all the values and see if the a's come out as you have
above, and
then
> it's over.
> I will do better than that. I will reproduce my proof that
your
> claim that one of the a's is coprime to 5 is wrong, but I
will do it in
> more generality. That way I will be heading off any examples
> you might try to come up with in the future. What you want is
> a consequence of the following claim, which you have stated
> elsewhere:
> CLAIM: It is possible to 'nd a monic polynomial of degree
> 3, with integer coef'cients, irreducible over the 
rationals.
> and with roots a1, a2, and a3, such that at least one of
> a1, a2, or a3 is coprime, in the ring of algebraic
> integers, to a prime factor of the constant term of
> the polynomial.
> Right? You have claimed as recently as yesterday.
> DISPROOF OF CLAIM:
<deleted
Whoa here, pardner! Looks like I am going to have to repeat
that proof again! I don't want you to lose sight of the fact
that
this is really your main #1 problem here, and I am sure you
wouldn't
want other people to overlook it either. So I am putting it
back in.
Here it is, and again *please* feel free to point out any
errors!
DISPROOF OF CLAIM
Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are
integers, and assume Q(x) is irreducible over the rationals.
Assume that c = p * v, where p is a prime and v is an integer.
Let a1, a2, and a3 be the roots of Q(x). Note that by
de'nition, since Q(x) is monic, a1, a2, and a3 are algebraic
integers.
Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime
to p. This means that there exist algebraic integers r and s
such that
[1] r*a1 + s*p = 1.
Now, let F12 be an automorphism of the 'eld of algebraic
numbers with the following properties:
1. F12(a1) = a2.
2. F12 leaves the sub'eld of rationals invariant, that is,
if d is rational, F12(d) = d.
3. If t is an algebraic integer, F12(t) is also an algebraic
integer.
The existence of such an F12 is an elementary result in Galois
theory. See, e.g., Abstract Algebra, by Beachy & Blair.
Now apply F12 to both sides of equation [1]:
[2] F12(r) * F12(a1) + F12(s) * F12(p) = F12(1).
Note that F12(a1) = a2, F12(p) = p, and F12(1) = 1. Let
r' = F12(r) and s' = F12(s). Note that 
r' and s' are
algebraic integers. With these substitutions, [2] becomes
r' * a2 + s' * p = 1,
which immediately implies that a2 and p are coprime.
Similarly, use another automorphism F13 to show that a3
and p are also coprime.
Therefore: if a1 is coprime to p, so are a2 and a3. That
is, if one of the three is coprime to p, they all are.
But: you know that a1 * a2 * a3 = -c = -p * v.
Therefore p is a divisor of a1 * a2 * a3. Therefore at
least one of a1, a2 and a3 is NOT coprime to p.
This contradiction shows that the original assumption
was false. The original assumption was based on your
claim that one of a1, a2, or a3 must be coprime to p.
Therefore you were wrong.
Again, feel free to poke any holes in this that you can.
Nora B.
> And Nora Baron rattles on a bit more. Where from before her
primary
> position is that given a factorization I lose the
factorization by
> actually putting in a value for x, as I used x=2. The key
expression
> is
(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 y=uf, and I put in actual numbers for
several
> values as a simpli'cation to remove room for bogus
objections like
> Nora Baron's here. Yet Nora Baron in one reply simply put 
in
some
> picked values for the a's for m=1 in what she thought of as 
a
> refutation, and now seems to think that my saying x=2 takes
away the
> factorization above. That's hardly even voodoo 
mathematics.
I think it's more simply: nonsensical.
> James Harris
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
[...]
| DISPROOF OF CLAIM
|
| Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are
| integers, and assume Q(x) is irreducible over the rationals.
| Assume that c = p * v, where p is a prime and v is an
integer.
| Let a1, a2, and a3 be the roots of Q(x). Note that by
| de'nition, since Q(x) is monic, a1, a2, and a3 are 
algebraic
| integers.
|
| Now ASSUME, as you claim, that one of a1, a2, or a3 is
coprime
| to p.
It seems to me that it should be possible to prove this
without relying
upon the existence of the automorphism of the algebraic
numbers you used.
Keith Ramsay
===
Subject: Re: Advanced Polynomial Factorization: VERY much
simpli'ed
>| DISPROOF OF CLAIM
>|
>| Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are
>| integers, and assume Q(x) is irreducible over the rationals.
>| Assume that c = p * v, where p is a prime and v is an
integer.
>| Let a1, a2, and a3 be the roots of Q(x). Note that by
>| de'nition, since Q(x) is monic, a1, a2, and a3 are 
algebraic
>| integers.
>|
>| Now ASSUME, as you claim, that one of a1, a2, or a3 is
coprime
>| to p.
It seems to me that it should be possible to prove this without
> relying upon the existence of the automorphism of the
algebraic
> numbers you used.
Keith Ramsay
>
It seems to me that you are both wasting your time if you hope
to
educate JSH - I have been following the many JSH initiated
threads for
maybe all of 2 months or so now and I have not yet seen a
instance of him
refuting an apparently sensible and logical argument that is
contrary to
his position.
I can't follow the mathematics involved but the standard of
discussion
involved is simultaneously entertaining and educational: I
de'nitely learn
more as I attempt to follow the sensible discussion and I am
most
certainly amused by the mental acrobactics that JHS is
displaying for
public entertainment ... all this and I don't have to worry
about paying
for my degree (*laughs loudly*) and it comes complete with a
free pass
to
watch the monkey in the zoo!
As I have said previously, I seriously hope that you (Keith)
and Nora (and
Will Twentyman and Arturo Magidin and David Ulrich - sorry if
I don't have
all the names perfectly correct) will continue to respond
seriously and
sensibly to JSH as this is fantastic learning stuff for me ...
as you have
to make your argument and discussion more and more simple in
an attempt to
show JSH the many ways in which you believe him to be
incorrect I learn
more and more.
Keith and Nora,
has been in the area of mathematical proof - I always believed
that proofs
belonged solely to the realm of geometry (showing one triangle
to be the
same as the other) and now I have seen how proof pervades even
(especially ?) the most esoteric math (not that I believe,
looking at other
threads in this NG, that this stuff is particularly esoteric).
I wonder, in the spirit of JSH, whether that last sentence is
syntactically
and gramatically correct ?
I'm not sure if I have set the NG stuff correctly - if 
I've
done this right
then only sci.math will see me ... I don't understand why 
JSH
feels a need
to post to sci.*, alt.*, *.fr and so on ... if it's
mathematics then it
belongs here doesn't it ? (unless it's 
speci'cally related to
undergraduate studies, the de'nition of which varies by
country)
Ivan McDonagh.
===
Subject: algebra
Good morning,
Is the symmetric algebra generated by diagonal tensors?
Tern_
===
Subject: Algebraic element
headers otherwise
we will be unable to process your complaint properly.
Let E is a extension 'eld of a 'eld F.
Prove that: If u in E is algebraic over F
then, u^2 is algebraic over F
my pf>Let [F(u^2):F]=n, then
{1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F.
so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over F.
therefore, c_0,c_1,...,c_n are not all zero, where
c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0.
so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x), then
f(u^2)=/=0 and deg f(x) >=1 .
hence, u^2 is algebraic over F.
----------------------------------------------------------
I wonder whether this proof is really correct or not.
If not, please show me right way.
In addition, How can I prove that F(u)=F(u^2) ?
If anyone know this, please post reply.
===
Subject: Re: Algebraic element
in message <DJPLa.3020725$CK1.453343@news.easynews.com>:
> Let E is a extension 'eld of a 'eld F.
> Prove that: If u in E is algebraic over F
> then, u^2 is algebraic over F
[...]
> In addition, How can I prove that F(u)=F(u^2) ?
You can't, because it's not true in general.
Consider F = Q, u = sqrt(2).
--
Jim Heckman
===
Subject: Re: Algebraic element
headers otherwise
we will be unable to process your complaint properly.
If E is 'nite extension 'eld of a 
'eld F and [F(u):F] is odd,
then Is it true, in general ?
If then, Can you give me some hint for proving ...
> in message <DJPLa.3020725$CK1.453343@news.easynews.com>:
> Let E is a extension 'eld of a 'eld F.
> Prove that: If u in E is algebraic over F
> then, u^2 is algebraic over F
[...]
> In addition, How can I prove that F(u)=F(u^2) ?
You can't, because it's not true in general.
> Consider F = Q, u = sqrt(2).
--
> Jim Heckman
===
Subject: Re: Algebraic element
>Let E is a extension 'eld of a 'eld F.
>Prove that: If u in E is algebraic over F
> then, u^2 is algebraic over F
my pf>Let [F(u^2):F]=n,
>
Right here, you've assumed what you want to prove. You have 
to
prove
that [F(u^2):F] is 'nite.
What do you want to do to prove that u^2 is algebraic? I get
the idea
that you're trying to use a theorem (if E>K>F and E 
algebraic,
then K
algebraic and [E:F]=[E:K][K:F]). Not knowing what book 
you're
using (if
you are using a book), I don't know if the point of the
exercise is to
reinforce the theorem (and make you check the assumptions), or
if you're
just supposed to use the de'nition of algebraic to see if u^2
is
algebraic (which is what I do assume in the absence of any
other
information). That is, this exercise is setting up that nice
theorem on
the degrees of extensions.
> then
> {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F.
>
Why? (disclaimer below signature)
> so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over
F.
> therefore, c_0,c_1,...,c_n are not all zero, where
> c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0.
> so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x),
then
> f(u^2)=/=0 and deg f(x) >=1 .
hence, u^2 is algebraic over F.
>----------------------------------------------------------
>I wonder whether this proof is really correct or not.
>If not, please show me right way.
In addition, How can I prove that F(u)=F(u^2) ?

The basic idea here would be to to 'nd [F(u):F(u^2)]. 
You've
really
only got two choices here. But once you've done the 
'rst part
(u^2 is
algebraic), you know that [F(u):F] = [F(u):F(u^2)][F(u^2):F]
(if you
haven't proven this yet, now is a good time, because 
it's a
very
convenient tool), which tells you an awful lot.
Jon Miller
Disclaimer: Really, why? It's not true.
===
Subject: Re: Algebraic element
> Let E is a extension 'eld of a 'eld F.
> Prove that: If u in E is algebraic over F
> then, u^2 is algebraic over F
my pf>Let [F(u^2):F]=n, then
do you know such an n exists? this is pretty much what 
you're
trying to
prove...
> {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F.
> so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over
F.
> therefore, c_0,c_1,...,c_n are not all zero, where
> c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0.
> so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x),
then
> f(u^2)=/=0 and deg f(x) >=1 .
> hence, u^2 is algebraic over F.
this doesn't work as your starting point is wrong.
But it looks as though you've already shown that u is
algebraic over F iff
[F(u):F] is 'nite. (And in fact that [F(u):F] is the degree 
of
the
minimal
polynomial of u over F, but we don't need this for what
follows). If so,
you can make your approach work in a slightly modi'ed form...
You need to start from what is given... u is algebraic over F,
so F(u):F
is
of 'nite dimension?
Also u^2 is in F(u), and so powers of u^2 are all in F(u). But
there's a
limit to how many independent powers of u^2 there can be in
F(u):F (why?),
which is where your approach above can naturally come in...
HTH
Mike.
> ----------------------------------------------------------
> I wonder whether this proof is really correct or not.
> If not, please show me right way.
In addition, How can I prove that F(u)=F(u^2) ?
If anyone know this, please post reply.
===
Subject: Re: Algebraic element
[one top-posting 'xed for free, next one will not be 
answered]
in message <Z2SLa.3030834$CK1.454965@news.easynews.com>:
> in message <DJPLa.3020725$CK1.453343@news.easynews.com>:
> Let E is a extension 'eld of a 'eld F.
> Prove that: If u in E is algebraic over F
> then, u^2 is algebraic over F
> [...]
> In addition, How can I prove that F(u)=F(u^2) ?
> You can't, because it's not true in 
general.
> Consider F = Q, u = sqrt(2).
If E is 'nite extension 'eld of a 
'eld F and [F(u):F] is odd,
> then Is it true, in general ?
Yes.
> If then, Can you give me some hint for proving ...
Another poster already has. Consider that [F(u):F] =
[F(u):F(u^2)] [F(u^2):F].
--
Jim Heckman
===
Subject: Another Integration Problem
How do I integrate
e^ ((1/3200)x^2) between 0 and inf?
Havn't done integration by parts in many a year so might 
need
to be written
in Help for Dummys format :-)
===
Subject: Re: Another Integration Problem
Originator: mtx014@linux.services.coventry.ac.uk (Robert Low)
How do I integrate
e^ ((1/3200)x^2) between 0 and inf?
You don't... that function gets bigger and bigger as
x does, so the integral from 0 to in'nity 
doesn't
converge.
If it should have been e^ (-(1/3200)x^2) then consider
substituting some multiple of x and then looking
up a table for the integral of e^(-x^2), which is
a standard result.
--
Rob. http://www.mis.coventry.ac.uk/~mtx014/
===
Subject: Re: Another Integration Problem
Sorry should have been
e^ ((-1/3200)x^2) between 0 and inf?
How do I integrate
e^ ((1/3200)x^2) between 0 and inf?
Havn't done integration by parts in many a year so might 
need
to be
written
> in Help for Dummys format :-)
>
===
Subject: Re: Another Integration Problem
<33e026c6ab406b67a91411950a8f80b6@TeraNews>
containing
illegal or copyrighted material through this service be
tolerated!!
> Sorry should have been
> e^ ((-1/3200)x^2) between 0 and inf?
>
Look up the Gamma function
Gamma(n) = integral(0,oo) x^(n-1) e^-x dx
Prove
Gamma(n) Gamma(1-n) = pi/sin pi.n
Show
Gamma(1/2) = 2 integral(0,oo) e^(-t^2) dt
Conclude
integral(0,oo) e^(-t^2) dt = (sqr pi)/2
Transform variable
t = x(sqr 2)/80
> How do I integrate
> e^ ((1/3200)x^2) between 0 and inf?
===
Subject: Re: Another Integration Problem
>>Sorry should have been
>>e^ ((-1/3200)x^2) between 0 and inf?
>>
Look up the Gamma function
> Gamma(n) = integral(0,oo) x^(n-1) e^-x dx
There's an even easier function to look up - without wanting
to give it
away, it's relevant that this was posted to a stats NG.
Bob
--
Bob O'Hara
Rolf Nevanlinna Institute
P.O. Box 4 (Yliopistonkatu 5)
FIN-00014 University of Helsinki
Finland
Telephone: +358-9-191 23743
Mobile: +358 50 599 0540
Fax: +358-9-191 22 779
WWW: http://www.RNI.Helsinki.FI/~boh/
===
Subject: Re: Another Integration Problem
>>Sorry should have been
>>e^ ((-1/3200)x^2) between 0 and inf?
> Look up the Gamma function
> Gamma(n) = integral(0,oo) x^(n-1) e^-x dx
There's an even easier function to look up - without wanting
to give it
> away, it's relevant that this was posted to a stats NG.
Bob
--
> Bob O'Hara
Rolf Nevanlinna Institute
> P.O. Box 4 (Yliopistonkatu 5)
> FIN-00014 University of Helsinki
> Finland
> Telephone: +358-9-191 23743
> Mobile: +358 50 599 0540
> Fax: +358-9-191 22 779
> WWW: http://www.RNI.Helsinki.FI/~boh/
>
Now I am really confused....
===
Subject: Re: Another Integration Problem
: Sorry should have been
: e^ ((-1/3200)x^2) between 0 and inf?
There's a short way and a long way.
The short way is to think about what this integral is - it's
very
nearly the same as somethig which is easy to 'nd tabulated.
The long way is to 'nd the integral of exp(-x^2) over the 
same
range.
Which is a wee bit of a pain to do in the general case, but in
this
case is traditionally done by noting that if
I = int(exp(-x^2), x = 0 .. inf) dx
then
I^2 = int(exp(-x^2), x = 0 .. inf) dx int(exp(-y^2), y = 0 ..
inf) dy
And by not being too fussy about orders and limits
I^2 = int(exp(-(x^2 + y^2) dx dy
which is a surface integral. Put it into polar coordinates,
note that
dx dy = r dr dtheta, 'ddle around a bit, get another I to
appear and
orff yer goes.
Ian
--
===
Subject: Re: Another Integration Problem
<33e026c6ab406b67a91411950a8f80b6@TeraNews>
<cCUlhtvFIYkV-pn2-jUKSlLtd4mCi@localhost>
containing
illegal or copyrighted material through this service be
tolerated!!
: Sorry should have been
> : e^ ((-1/3200)x^2) between 0 and inf?
I = int(exp(-x^2), x = 0 .. inf) dx
> then
> I^2 = int(exp(-x^2), x = 0 .. inf) dx int(exp(-y^2), y = 0
.. inf) dy
And by not being too fussy about orders and limits
> I^2 = int(exp(-(x^2 + y^2) dx dy
which is a surface integral. Put it into polar coordinates,
note that
> dx dy = r dr dtheta, 'ddle around a bit, get another I to
appear and
>
Indeed, the easy way. I^2 =
integral(0,oo) integral(0,oo) e^-(x^2 + y^2) dx dy
= integral(0,pi/2) integral(0,oo) e^(-r^2) r dr dt
= 1/2 * integral(0,pi/2) integral(0,oo) e^-u du dt
= 1/2 * integral(0,pi/2) (-e^-u)|0,oo dt
= 1/2 * integral(0,pi/2) dt
= pi/4
I = (sqr pi)/2
integral(-oo,oo) e^(-t^2) dt = sqr pi = Gamma(1/2) = (-1/2)!
===
Subject: Re: Another Integration Problem
> How do I integrate
e^ ((1/3200)x^2) between 0 and inf?
OK, we've gone off track here. If you want an answer, it 
does
help
to ask the right question. It may no longer matter, but let me
see
if I can clear up the confusion.
To recap, we want the expected value of the density
f(x) = (x/1600) exp(-(1/3200) x^2).
See:
.0307072121.2142515%40posting.g
oogle.com
The expected value is integral(0,inf) x f(x) dx, i.e.,
integral(0,inf) (x^2)/1600 exp(-(1/3200) x^2) dx
For convenience, let a = 1/3200, so the integrand is
2 a x^2 exp(-a x^2). Let t = x sqrt(a). Then dt = dx sqrt(a),
so the integral (call it E) is
E = integral(0 sqrt(a), inf sqrt(a)) 2 t^2 exp(-t^2) dt/sqrt(a)
= (1/sqrt(a)) integral(0,inf) 2 t^2 exp(-t^2) dt
Let u = t, and let dv = 2 t exp(-t^2) dt. Then du = dt, and
v = -exp(-t^2). So putting the factor 1/sqrt(a) aside for the
moment, the integral is
u(inf) v(inf) - u(0) v(0) - integral(0,inf) v du
= 0 - 0 + integral(0,inf) exp(-t^2) dt
= sqrt(pi)/2
as pointed out elsewhere in this thread. So the overall result
is
E = (1/sqrt(a)) sqrt(pi)/2 = sqrt(3200) sqrt(pi)/2,
which is approximately 50.
Well, I'm pretty sure I'm answering the right 
question, and I
hope
I didn't make any elementary mistakes in working out the
solution.
For what it's worth,
Robert an integral a day Dodier
--
Science may be described as the art of systematic
over-simpli'cation.
-- Karl Popper
===
Subject: another proof
I need to show that limsup(s(n)+t(n))<=limsup s(n) + lim sup
t(n)
can I do this by proving that they can be equal or the lhs can
be smaller
than the rhs or do I need to prove these two ideas together?
===
Subject: Re: another proof
> I need to show that limsup(s(n)+t(n))<=limsup s(n) + lim sup
t(n)
can I do this by proving that they can be equal or the lhs can
be smaller
> than the rhs or do I need to prove these two ideas together?
Do both together - there is really not much choice. You might
want to
give an example where equality holds and one where it does not.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: associative property question
I have a solution manual that states this problem is
associative:
Determine whether the binary operation * defned is commutative
and
whether * is associative...
* de'ned on Z by a*b=a-b
I can clearly see how to check to see if this problem is
commutative
because I have 2 numbers.
1-2 != 2-1 therefore not commutative
My question is, how do you check if its associative without 3
numbers?
The solution manual said: 2=1-(2-3) != (1-2)-3=-4
...where did the 3rd number (or c) come from?
Marcos
===
Subject: Re: associative property question
containing
illegal or copyrighted material through this service be
tolerated!!
> I have a solution manual that states this problem is
associative:
>
Never heard of such, would you explain what an associative
problem is?
> Determine whether the binary operation * defned is
commutative and
> whether * is associative...
> * de'ned on Z by a*b=a-b
I can clearly see how to check to see if this problem is
commutative
> because I have 2 numbers.
> 1-2 != 2-1 therefore not commutative
> My question is, how do you check if its associative without
3 numbers?
The solution manual said: 2=1-(2-3) != (1-2)-3=-4
> ...where did the 3rd number (or c) come from?
>
I suppose it comes with 1 & 2 but until you tell us what set
the binary
operator operates upon, how are we to know?
Who's c? You've yet to introduce her to us.
As for a two number counterexample, presuming - is
substraction of
integers.
1-(1-1) = 1-0 = 1
(1-1)-1 = 0-1 = -1
===
Subject: Re: associative property question
> I have a solution manual that states this problem is
associative:
Determine whether the binary operation * defned is commutative
and
> whether * is associative...
> * de'ned on Z by a*b=a-b
I can clearly see how to check to see if this problem is
commutative
> because I have 2 numbers.
> 1-2 != 2-1 therefore not commutative
> My question is, how do you check if its associative without
3 numbers?
The solution manual said: 2=1-(2-3) != (1-2)-3=-4
> ...where did the 3rd number (or c) come from?
> Marcos
>
An operation * on set R is de'ned to be associative if it
satis'es the
following property:
For all numbers a,b,c in R, (a*b)*c = a*(b*c)
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: associative property question
> I have a solution manual that states this problem is
associative:
> Determine whether the binary operation * defned is
commutative and
>> whether * is associative...
>> * de'ned on Z by a*b=a-b
> I can clearly see how to check to see if this problem is
commutative
>> because I have 2 numbers.
>> 1-2 != 2-1 therefore not commutative
>> My question is, how do you check if its associative without
3 numbers?
> The solution manual said: 2=1-(2-3) != (1-2)-3=-4
>> ...where did the 3rd number (or c) come from?
>> Marcos
>>
An operation * on set R is de'ned to be associative if it
satis'es the
> following property:
For all numbers a,b,c in R, (a*b)*c = a*(b*c)
Sorry I wasn't speci'c enough. I 
didn't know how to ask the
question.
I understand what the associative property says. I was looking
for an
operational procedure to 'gure out if the problem was
associative.
Lucky for me, I've 'gured out how to work the 
problem.
Binary operation * de'ned on set Z by a*b=a-b where a, b are
members of
Z.
Commutative if and only if a*b=b*a:
a-b != b-a // therefore not commutative
Associative if (a*b)*c=a*(b*c):
(a-b)*c ?= a*(b-c) // where a-b and b-c represents a*b
(a-b)-c ?= a-(b-c) // where ()-c and a-() represents a*b
a-b-c != a-b+c // therefore not associative
Marcos
===
Subject: basic algebra
if x<y and z<q can I assert that x+z<y+q?
===
Subject: Re: basic algebra
>if x<y and z<q can I assert that x+z<y+q?
>
No.
My answer is based on intuition, and sometimes intuition by
itself could be
wrong.
G C
===
Subject: Re: basic algebra
containing
illegal or copyrighted material through this service be
tolerated!!
> if x<y and z<q can I assert that x+z<y+q?
>
Yes. Use theorem
x < y ==> x+a < y+a
twice. With that hint can you prove
x < y, r < s ==> x+r < y+s ?
===
Subject: Re: basic algebra
> if x<y and z<q can I assert that x+z<y+q?
Yes.
A quickie proof
x<y
x - y < y - y
x - y < 0
z<q
z - z < q - z
0 < q - z
Since x - y < 0 and 0 < q - z, then x - y < q - z
x - y < q - z
x - y + z < q - z + z
x - y + z < q
x - y + z + y < q + y
x + z < q + y
x + z < y + q QED
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: basic algebra
> if x<y and z<q can I assert that x+z<y+q?
Yes, you can even prove it.
By de'nition, x < y <=> y - x is positive.
Similarly z < q <=> q - z is positive.
So ( that the sum of positives is positive is part of the
de'nition of
positive), (y - x) + (q - z) = (y + q) - (x + z) is positive
so x + z < y + q.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: basic algebra
>Yes, you can even prove it.
By de'nition, x < y <=> y - x is positive.
>Similarly z < q <=> q - z is positive.
So ( that the sum of positives is positive is part of the
de'nition of
>positive), (y - x) + (q - z) = (y + q) - (x + z) is positive
>so x + z < y + q.
--
>Paul Sperry
>Columbia, SC (USA)
Brilliant and simple. Amazing how some of the proving had been
missing
from
some of the elementary algebra courses; only a few of them
asked in each
elementary course. Other aspects were 'ne, but proving was
usually
expected
only in Geometry and some in Pre-Calculus, and brie§y in
Trigonometry
(identities).
Questions like that are probably what separates the
Mathematicians from the
studied-math-as-prerequisite-for-certain-courses type.
Applying Math well
is
one thing; proving stuff within the subject is sometimes
another thing.
G C
===
Subject: Re: basic algebra
if x<y and z<q can I assert that x+z<y+q?
It depends on what rules govern < and +. In a ring a set P of
so-called positive elements has the property that x and y in P
implies x
+ y in P, and x <y is de'ned to mean y - x in P. So you have
y - x in P and q - z in P,
hence
(y - x) + (q - z) in P
Now ring manipulations give
(y + q) - (x + z) in P.
Though rings are the usual setting for such things, you can do
the same
in an additive group. But can it be done in _your_ system? I
don't
know.
GC
===
Subject: Re: basic algebra
>if x<y and z<q can I assert that x+z<y+q?
Yes, and what is more, your assertion will be correct. :-)
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/
Walrus meat as a diet is less repulsive than seal.
===
Subject: basis of Null Space
headers otherwise
we will be unable to process your complaint properly.
___________________________
Find a basis of Ker T,
where T:R^2 ->R^3, T(x_1,x_2)=(x_1, x_2, 2x_1 + x_2)
__________________________________________________
Ker T={(x_1,x_2) | T(x_1,x_2)=(0,0,0)}
={(x_1,x_2) | (x_1, x_2, 2x_1 + x_2)=(0,0,0)}
={(0,0)}
Clearly, Ker T = Null Space of T and,
I think it is impossible to knowing dimension of Ker T by
counting # of
basis.
And I think maybe there exist a de'nition about a basis &
dimension about
Null Space.
But, I don't think the de'nition of dimension 
and basis about
Null Space.
If anyone know the de'niton or have suggestion, please post
reply.
===
Subject: Re: basis of Null Space
> ___________________________
Find a basis of Ker T,
> where T:R^2 ->R^3, T(x_1,x_2)=(x_1, x_2, 2x_1 + x_2)
> __________________________________________________
Ker T={(x_1,x_2) | T(x_1,x_2)=(0,0,0)}
> ={(x_1,x_2) | (x_1, x_2, 2x_1 + x_2)=(0,0,0)}
> ={(0,0)}
Yes, so far so good.
> Clearly, Ker T = Null Space of T and,
> I think it is impossible to knowing dimension of Ker T by
counting # of
> basis.
What is bothering you I think is dim( {0} ). But, think of it
this way:
what is the size of the largest linearly independent set of
vectors?
Answer: 0. So, dim( {0} ) = 0.
> And I think maybe there exist a de'nition about a basis &
dimension
about
> Null Space.
> But, I don't think the de'nition of dimension 
and basis
about Null
Space.
> If anyone know the de'niton or have suggestion, please post
reply.
There is a Theorem which says that the dimiension of the null
space
plus the dimension of the range space (_not_ the codomain)
equals the
dimension of the domain but I don't think that is very 
useful
here.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: basis of Null Space
> I think it is impossible to knowing dimension of Ker T by
counting #
> of basis. And I think maybe there exist a de'nition about a
basis
> & dimension about Null Space. But, I don't think the
de'nition of
> dimension and basis about Null Space. If anyone know the
de'niton
> or have suggestion, please post reply.
Remember, the de'nition of a vector space is the number of
elements
of a maximal linearly independent set. (It is a standard
result that
all 'nite such have the same number of elements.) Can you 
'nd a
maximal linearly independent set inside the null space?
/olov
--
I'm no believer in a random massacre of 
theatreland's sacred
cows. I just
believe in chasing them down King's Parade on a bright 
orange
spacehopper.
-- Ed Richardson, The Cambridge Student, 14/2/2002
===
Subject: Bipartite graph?
I was just wondering if someone could this graph to verify
that it is
bipartite (answers/solutions aren't in the back of the
textbook for all
questions and I got the 'rst one wrong that it has a solution
for).
I've uploaded it to 
www.geocities.com/bjj_freak/bipartite.jpg
(might need
to
be copied & pasted into your browser). I chopped off e1 and
the start of v
of v5 in the scan.
Bon
===
Subject: Re: Bipartite graph?
I was just wondering if someone could this graph to verify
that it is
>bipartite (answers/solutions aren't in the back of the
textbook for all
>questions and I got the 'rst one wrong that it has a 
solution
for).
I've uploaded it to 
www.geocities.com/bjj_freak/bipartite.jpg
(might need
to
>be copied & pasted into your browser). I chopped off e1 and
the start of v
>of v5 in the scan.
>
I don't see how I can give hints to this.
What's a bipartite graph? It's got two sets of 
vertices, and
all the
edges go between vertices in opposite sets.
So, if your graph is bipartite, you ought to be able to list
the two
sets of vertices.
Alternatively, you could draw the graph so that there's a
dotted line
down the middle, and all the edges cross the dotted line.
(This will
force the two disjoint vertex sets to be on opposite sides of
the line.)
Jon Miller
===
Subject: Re: Bipartite graph?
>I was just wondering if someone could this graph to verify
that it is
>bipartite (answers/solutions aren't in the back of the
textbook for all
>questions and I got the 'rst one wrong that it has a 
solution
for).
>I've uploaded it to 
www.geocities.com/bjj_freak/bipartite.jpg
(might need
to
>be copied & pasted into your browser). I chopped off e1 and
the start of v
>of v5 in the scan.
It's 'ne.
Brian
===
Subject: Re: Bipartite graph?
[...]
>So, if your graph is bipartite, you ought to be able to list
the two
>sets of vertices.
(S)he did.
[...]
Brian
===
Subject: Re: Bipartite graph?
[...]
>So, if your graph is bipartite, you ought to be able to list
the two
>>sets of vertices.
>(S)he did.
[...]
>
I guess my slider bar doesn't work. Requires thinking enough
to click
on it and move it.
I apologize to group and OP for being snippy.
Jon Miller
===
Subject: Re: Bipartite graph?
Bon
> It's 'ne.
Brian
===
Subject: Books recommendations?
Hi all,
I have a 4-year B. Sc. degree in math. I obtained it in a
non-English
speaking country, but now I work in a predominantly
English-speaking
environment. Can anybody recommend some standard
undergrad textbooks (in US) for 1) linear algebra, 2)
mathematical
analysis (1-d and N-d), 3) discrete mathematics and 4)
probability and
statistics?
I am a predominantly geometrical thinker (I tend to visualize
everything
I read in terms of visual machinery), thus the books should be
structured
correspondingly. (Emphasis on geometrical
presentation/rationale [think
Leibnitz instead of Newton], clear layout, multicolor fonts,
rich
diagrams.)
Excessive rigor (proof-wise) is absolutely not required.Thx,
===
Subject: Re: Books recommendations?
Help! Standard US undergrad textbook recommendations
wanted:
1) linear algebra,
2) mathematical analysis (1-d and N-d),
3) discrete mathematics and
4) probability and statistics.
What do you guys use?
===
Subject: Re: Books recommendations?
What level of Linear algebra are you lookin for?
You could try Howard Anton's book Linear algebra
Analysis:
Principle of mathematical analysis by Walter Rudin
Discrete:
Kenneth Rosen's Discrete Math and its applications
Prob. & Stats:
Jay L. DeVore Probability and Statistics for Scientists and
Engineers
> Help! Standard US undergrad textbook recommendations
> wanted:
1) linear algebra,
> 2) mathematical analysis (1-d and N-d),
> 3) discrete mathematics and
> 4) probability and statistics.
> What do you guys use?
>
S 
===
Subject: Re: Conditional Probability
> Hi:
Could someone help me understand how to model the following
problem
with
> conditional probability?
Suppose a box has 10 balls, 6 black and 4 white. Find the
probability
that
> all three balls removed are black if at least one of the
removed balls is
> black.
>
I assume that three balls are removed without replacement.
Let the sample space, S, consist of all possible sets of three
balls removed (where each ball is taken to be distinct from
the rest).
Let E be the event that all three balls removed are black, and
let
F be the event that at least one of them is black. Then
the number of outcomes in E is
n(E) = C(6,3) = 20.
We also have
n(S) = C(10,3) = 120.
Let F' be the complement of F, i.e., F' is the 
event that all
three
balls are White. Then
n(F') = C(4,3) = 4.
So,
n(F) = n(S) - n(F') = 116.
So, the desired conditional probability is
n(E)/n(F) = 20/116.
> I attempted to model this problem letting
B_i = a black ball is removed on the i'th removal for i = 
1,2,3
The conditional probability I came up with is
> P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3)
) / P( B_1 U
> B_2 U B_3 )
Is this model right?
I don't understand your notation, but I believe that keeping
track
of which ball was removed 'rst, which second, and which third
is
unnecessarily complicated.
===
Subject: Re: Conditional Probability
> Could someone help me understand how to model the following
problem
with
>conditional probability?
Suppose a box has 10 balls, 6 black and 4 white. Find the
probability
that
>all three balls removed are black if at least one of the
removed balls is
>black.
I attempted to model this problem letting
B_i = a black ball is removed on the i'th removal for i = 
1,2,3
The conditional probability I came up with is
>P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) )
/ P( B_1 U
>B_2 U B_3 )
Is this model right?
I don't know, but it's unnecessarily 
complicated.
If you remove three balls, of which at least one is known to be
black, from a container of 6 black + 4 white, that is
identical to
removing two balls (about which you know nothing) from a
container
of 5 black + 4 white.
The probability of 2 black is 5/9 * 4/8 = 5/18, and therefore
that
is the probability of three black in the original problem.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/
You 'nd yourself amusing, Blackadder.
I try not to §y in the face of public opinion.
===
Subject: Re: Conditional Probability
>If you remove three balls, of which at least one is known to
be
>black, from a container of 6 black + 4 white, that is
identical to
>removing two balls (about which you know nothing) from a
container
>of 5 black + 4 white.
No it's not identical. That would be true only if you knew
_which_
ball was black.
Apologies for any confusion I created.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/
You 'nd yourself amusing, Blackadder.
I try not to §y in the face of public opinion.
===
Subject: Re: Conditional Probability
>Hi:
> Could someone help me understand how to model the following
problem
with
>conditional probability?
>Suppose a box has 10 balls, 6 black and 4 white. Find the
probability
that
>all three balls removed are black if at least one of the
removed balls is
>black.
>I attempted to model this problem letting
>B_i = a black ball is removed on the i'th removal for i =
1,2,3
>The conditional probability I came up with is
>P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) )
/ P( B_1 U
>B_2 U B_3 )
>Is this model right?
Yes. The numerator is then C(6, 3)/C(10, 3), where C(n, m) is
the binomial coef'cient n choose m, and the denominator is
1 - C(4, 3)/C(10, 3) = [C(10, 3) - C(4, 3)]/C(10, 3), making
the
desired probability C(6, 3)/[C(10, 3) - C(4, 3)] = 20/(120 -
4) =
20/116 = 5/29.
Brian
===
Subject: constructing some in'nite series, and think about 
how
you are
goint
to use it?
series that
converge uniformely to some functions. All of the examples of
the
in'nite series that I saw had very simple forms. For example,
1) s(x)=1+x^2+x^3+...+x^n+....
2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...
When I was reading the de'nition of what power series are, 
the
book said:
if all of the coef'cients in the in'nite series 
are
independent of
x, the series is called a power series.
If this is the case, how about the in'nite series like this
one?
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x
^6)/(2^4)}+...
The series above has the following properties.
1) whenever x^(even numbers, like 0,2,4..),then the
denominators of each
term is in the form: 2^(some number).
2) whenever x^(odd numbers), then the denominators of each
term is in
the form: 1^(some numbers)
Since all of the coef'cients of the terms in the 
in'nite
series is
independent of x, could I call this series a power series? If
so is
there any way for me to know what this power series converges
to?
The reason why I am asking this question is that it is
relatively easy
to make many kinds of in'nite series, but then I 
don't know
which
in'nite series I come up with is more useful than the others.
I also
would like to know if there is any general method to check
that the
in'nite series that I am interested in can be converged to 
some
function of a simple form? When I was looking at the way book
explained
how some of the in'nite series can be changed to a simpler
forms, it
seems that they strongly depended on the form of the in'nite
series.
However, if there is some general way to tackle problems of
simplifying
in'nite series, that would help me very much.
brackets
in my in'nite series. I could not use some math text editor 
to
write
more neatly than that.
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
In'nite series can be thought of as polynomials that go on
forever (no
in'nite zero coef'cients)
3x^5+2x^7-8x^16+.... is an in'nite series for example.
series that
> converge uniformely to some functions. All of the examples
of the
> in'nite series that I saw had very simple forms. For 
example,
1) s(x)=1+x^2+x^3+...+x^n+....
> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...
When I was reading the de'nition of what power series are, 
the
book
said:
> if all of the coef'cients in the in'nite 
series are
independent of
> x, the series is called a power series.
If this is the case, how about the in'nite series like this
one?
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x
^6)/(2^4)}+...
The series above has the following properties.
1) whenever x^(even numbers, like 0,2,4..),then the
denominators of each
> term is in the form: 2^(some number).
2) whenever x^(odd numbers), then the denominators of each
term is in
> the form: 1^(some numbers)
Since all of the coef'cients of the terms in the 
in'nite
series is
> independent of x, could I call this series a power series?
If so is
> there any way for me to know what this power series
converges to?
> The reason why I am asking this question is that it is
relatively easy
> to make many kinds of in'nite series, but then I 
don't know
which
> in'nite series I come up with is more useful than the
others. I also
> would like to know if there is any general method to check
that the
> in'nite series that I am interested in can be converged to
some
> function of a simple form? When I was looking at the way
book explained
> how some of the in'nite series can be changed to a simpler
forms, it
> seems that they strongly depended on the form of the 
in'nite
series.
> However, if there is some general way to tackle problems of
simplifying
> in'nite series, that would help me very much.
brackets
> in my in'nite series. I could not use some math text editor
to write
> more neatly than that.
>
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
Yes,
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x
^6)/(2^4)}+... is an in'nite series
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x
^6)/(2^4)}+...=
(1/2)+(1/1)x+(1/2^2)x^2+(1/1^2)x^3+(1/2^3)x^4+...
The point is, an in'nite series can be expressed as the sum
and difference
of an in'nte number of (combined) terms of the form-- a
constant times x
raised to a non-negative integer. If the meaning of integer,
constant
and/or
non-negative integer are not clear then you MUST look them up.
Good luck
series that
> converge uniformely to some functions. All of the examples
of the
> in'nite series that I saw had very simple forms. For 
example,
1) s(x)=1+x^2+x^3+...+x^n+....
> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...
When I was reading the de'nition of what power series are, 
the
book
said:
> if all of the coef'cients in the in'nite 
series are
independent of
> x, the series is called a power series.
If this is the case, how about the in'nite series like this
one?
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x
^6)/(2^4)}+...
The series above has the following properties.
1) whenever x^(even numbers, like 0,2,4..),then the
denominators of each
> term is in the form: 2^(some number).
2) whenever x^(odd numbers), then the denominators of each
term is in
> the form: 1^(some numbers)
Since all of the coef'cients of the terms in the 
in'nite
series is
> independent of x, could I call this series a power series?
If so is
> there any way for me to know what this power series
converges to?
> The reason why I am asking this question is that it is
relatively easy
> to make many kinds of in'nite series, but then I 
don't know
which
> in'nite series I come up with is more useful than the
others. I also
> would like to know if there is any general method to check
that the
> in'nite series that I am interested in can be converged to
some
> function of a simple form? When I was looking at the way
book explained
> how some of the in'nite series can be changed to a simpler
forms, it
> seems that they strongly depended on the form of the 
in'nite
series.
> However, if there is some general way to tackle problems of
simplifying
> in'nite series, that would help me very much.
brackets
> in my in'nite series. I could not use some math text editor
to write
> more neatly than that.
>
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
series that
> converge uniformely to some functions. All of the examples
of the
> in'nite series that I saw had very simple forms. For 
example,
1) s(x)=1+x^2+x^3+...+x^n+....
> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...
When I was reading the de'nition of what power series are, 
the
book
said:
> if all of the coef'cients in the in'nite 
series are
independent of
> x, the series is called a power series.
If this is the case, how about the in'nite series like this
one?
>
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x^
6)
> /(2^4)}+...
The series above has the following properties.
1) whenever x^(even numbers, like 0,2,4..),then the
denominators of each
> term is in the form: 2^(some number).
2) whenever x^(odd numbers), then the denominators of each
term is in
> the form: 1^(some numbers)
Since all of the coef'cients of the terms in the 
in'nite
series is
> independent of x, could I call this series a power series?
Yes
> If so is
> there any way for me to know what this power series
converges to?
Yes.
Note:
(1) 1/2 + x^2/2^2 + x^4/2^3 + x^6/2^4 +... =
1/2(1 + (x^2/2)^1 + (x^2/2)^2 + (x^2/2)^3 +...) The series is
geometric
in x^2/2.
Also
(2) x + x^3 + x^5 +....=
x(1 + (x^2) + (x^2)^2 +...). The series is also geometric in
x^2.
The two series both converge for -1 < x < 1, you know what they
converge to and your series is the sum of (1) and (2).
> The reason why I am asking this question is that it is
relatively easy
> to make many kinds of in'nite series, but then I 
don't know
which
> in'nite series I come up with is more useful than the 
others.
That's backwards. You don't make up a series 
and then 'gure
out what
it's good for; basically, you come up with a series when all
else
fails. (I may get some argument about that.)
> I also
> would like to know if there is any general method to check
that the
> in'nite series that I am interested in can be converged to
some
> function of a simple form?
Simple form? Well you know series for several common functions
so you
could try to turn your series into one of those. As a rule, if
you are
lucky, ingenuity might do it. In general, it is a non-trivial
problem
and even though the power series may converge with non-zero
radius of
convergence, it doesn't converge to an elementary function.
> When I was looking at the way book explained
> how some of the in'nite series can be changed to a simpler
forms, it
> seems that they strongly depended on the form of the 
in'nite
series.
Sure.
> However, if there is some general way to tackle problems of
simplifying
> in'nite series, that would help me very much.
I'm afraid there is none. Also, to make matters worse, not 
all
series
are power series.
[...]
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
containing
illegal or copyrighted material through this service be
tolerated!!
> When I was reading the de'nition of what power series are,
the book
said:
> if all of the coef'cients in the in'nite 
series are
independent of
> x, the series is called a power series.
If this is the case, how about the in'nite series like this
one?
>
s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x
^5)/(1^3)}+{(x^
6)/(2^4)}+...
The series above has the following properties.
1) whenever x^(even numbers, like 0,2,4..),then the
denominators of each
> term is in the form: 2^(some number).
2) whenever x^(odd numbers), then the denominators of each
term is in
> the form: 1^(some numbers)
Since all of the coef'cients of the terms in the 
in'nite
series is
> independent of x, could I call this series a power series?
If so is
> there any way for me to know what this power series
converges to?
Yes & yes.
Try seperating the two parts into two series and see if each
converges.
> The reason why I am asking this question is that it is
relatively easy
> to make many kinds of in'nite series, but then I 
don't know
which
> in'nite series I come up with is more useful than the
others. I also
> would like to know if there is any general method to check
that the
> in'nite series that I am interested in can be converged to
some
> function of a simple form?
Wishful thinking.
> When I was looking at the way book explained
> how some of the in'nite series can be changed to a simpler
forms, it
> seems that they strongly depended on the form of the 
in'nite
series.
> However, if there is some general way to tackle problems of
simplifying
> in'nite series, that would help me very much.
>
Yup, practice, experience, insight, repeat as necessary.
brackets
> in my in'nite series. I could not use some math text editor
to write
> more neatly than that.
>
Don't bother, often they do not show up as you intend.
x + (x^2)/(2^2) + ... is suf'cient
x + x^2 / 2^2 + ... is loose tho possible
Please use some spaces for easier reading. Compare:
x^2=3y+7=27+6z
x^2 = 3y+7 = 27+6z
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint
to use it?
> The reason why I am asking this question is that it is
relatively easy
>to make many kinds of in'nite series, but then I 
don't know
which
>in'nite series I come up with is more useful than the 
others.
I also
>would like to know if there is any general method to check
that the
>in'nite series that I am interested in can be converged to
some
>function of a simple form? When I was looking at the way book
explained
>how some of the in'nite series can be changed to a simpler
forms, it
>seems that they strongly depended on the form of the in'nite
series.
>However, if there is some general way to tackle problems of
simplifying
>in'nite series, that would help me very much.
To rephrase the question: given a convergent series, how can
you 'nd the
sum in closed form (if indeed a closed form exists)?
There is no completely general answer AFAIK. There are certain
tricks,
e.g.:
1) Check for telescoping series, i.e. series that can be
written in the
form sum_n (f(n+1)-f(n)) for some function f.
2) Write the series as a power series, or as a special case of
a power
series, i.e. if your series is sum_n f(n) look at sum_n f(n)
z^n. Then
a) recognize the series for some of the standard elementary
functions,
or even some of the non-elementary ones (e.g. hypergeometric
series
cover a lot of territory)
b) see if differentiation or integration will simplify the
form of the
series
c) see if the series satis'es some differential equation that
can be
solved
3) If the coef'cients are rational functions of n times an
n'th power,
use a partial fraction decomposition.
4) If the sum is numeric, evaluate it numerically to high
precision and
see if it is recognized by the Inverse Symbolic Calculator at
<http://www.cecm.sfu.ca/projects/ISC/ISCmain.html>
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
>> The reason why I am asking this question is that it is
relatively easy
>>to make many kinds of in'nite series, but then I 
don't know
which
>>in'nite series I come up with is more useful than the
others. I also
>>would like to know if there is any general method to check
that the
>>in'nite series that I am interested in can be converged to
some
>>function of a simple form? When I was looking at the way
book explained
>>how some of the in'nite series can be changed to a simpler
forms, it
>>seems that they strongly depended on the form of the 
in'nite
series.
>>However, if there is some general way to tackle problems of
simplifying
>>in'nite series, that would help me very much.
> To rephrase the question: given a convergent series, how can
you 'nd the
> sum in closed form (if indeed a closed form exists)?
There is no completely general answer AFAIK. There are certain
tricks,
> e.g.:
1) Check for telescoping series, i.e. series that can be
written in the
> form sum_n (f(n+1)-f(n)) for some function f.
2) Write the series as a power series, or as a special case of
a power
> series, i.e. if your series is sum_n f(n) look at sum_n f(n)
z^n. Then
> a) recognize the series for some of the standard elementary
functions,
> or even some of the non-elementary ones (e.g. hypergeometric
series
> cover a lot of territory)
> b) see if differentiation or integration will simplify the
form of the
series
> c) see if the series satis'es some differential equation
that can be
> solved
3) If the coef'cients are rational functions of n times an
n'th power,
> use a partial fraction decomposition.
4) If the sum is numeric, evaluate it numerically to high
precision and
> see if it is recognized by the Inverse Symbolic Calculator at
> <http://www.cecm.sfu.ca/projects/ISC/ISCmain.html>
Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
Also, take a look at A=B at
http://www.cis.upenn.edu/~wilf/AeqB.html
Absolutely amazing stuff.
Martin Cohen
===
Subject: Re: constructing some in'nite series, and think 
about
how you are
goint to use it?
[...]
> Also, take a look at A=B at
http://www.cis.upenn.edu/~wilf/AeqB.html
Absolutely amazing stuff.
Martin Cohen
>
Martin.
You can even download the whole book.
===
Subject: Correction: Error in NOTICE
My message NOTICE: Error in Currently Taught Mathematics
posted June
Here's the relevant portion from the post:
<Quote>
That de'nition depends on a polynomial P(x) of degree n with
integer
coef'cients being monic so that you have
P(x) = (x + a_1)...(x + a_n)
but that is unbalanced as a complete ring must handle all non
monic
polynomials of degree n with integer coef'cients to get
P(x) = (a_1 x + b_1)...(a_n x + b_n)
but in fact the a's and b's cannot here always 
be algebraic
integers
as I've shown in my paper Advanced Polynomial 
Factorization...
</Quote>
However the paper does not show that all the a's and 
b's
cannot be
algebraic integers for that factorization as it shows a
coprimeness
result for a particular family of polynomials, which then can
be used
to show a problem in the ring as explained in multiple
postings.
My apologies for the error and the lateness in issuing a
correction.
James Harris
===
Subject: Re: Correction: Error in NOTICE
> My message NOTICE: Error in Currently Taught Mathematics
posted June
Here's the relevant portion from the post:
<Quote That de'nition depends on a polynomial P(x) of degree 
n
with integer
> coef'cients being monic so that you have
P(x) = (x + a_1)...(x + a_n)
but that is unbalanced as a complete ring must handle all non
monic
> polynomials of degree n with integer coef'cients to get
For the sake of those of us who have not been following in
detail . . .
can you de'ne complete ring. please.
Also, what does it mean for a ring to handle a polynomial?
===
Subject: Re: Correction: Error in NOTICE
> My apologies for the error and the lateness in issuing a
correction.
James Harris
Please expand the apology to cover posting your error-ridden
attempt at a
proof in the 'rst place. Then take it down. It is toast.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: Correction: Error in NOTICE
Bait ïn' switch.
> My message NOTICE: Error in Currently Taught Mathematics
posted June
Here's the relevant portion from the post:
<Quote That de'nition depends on a polynomial P(x) of degree 
n
with integer
> coef'cients being monic so that you have
P(x) = (x + a_1)...(x + a_n)
but that is unbalanced as a complete ring must handle all non
monic
> polynomials of degree n with integer coef'cients to get
P(x) = (a_1 x + b_1)...(a_n x + b_n)
but in fact the a's and b's cannot here always 
be algebraic
integers
> as I've shown in my paper Advanced Polynomial
Factorization...
> </Quote>
However the paper does not show that all the a's and 
b's
cannot be
> algebraic integers for that factorization as it shows a
coprimeness
> result for a particular family of polynomials, which then
can be used
> to show a problem in the ring as explained in multiple
postings.
>
I may be misreading this, but it seems as though the JSHter
may be
attempting to distort my correction to his mistaken claim
regarding
divisibility properties of the coef'cients of a certain
factorization
the same impression as I, of the above paragraph) is that JSH
takes a
certain form of argument, to a manifestly incorrect conclusion
(the
divisibility result). I point out that his conclusion is
incorrect,
therefore his argument is incorrect. In the above paragraph,
JSH
appears (to me) to be stating that the multiple postings
highlight
a problem in the ring.
> My apologies for the error and the lateness in issuing a
correction.
> James Harris
Apologize all you like, but your claim to have proven that any
a's are
coprime to 5 in the factorization
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
is quite obviously mistaken, as I have shown on multiple
occasions.
Any claim that this represents a problem in the ring of
algebraic
integers, any attempt to return to the original family of
polynomials,
anyh other discussion is simply your attempt at diverting
attention from
your own inability.
This so-called apology is, as are virtually all of your
missives,
thoroughly misguided and irrelevant. If you can't even 
answer
my
example showing your so-called Primary argument is incorrect,
then you are admitting that your whole argument is a fraud.
It's really too bad you didn't have what it 
took to address a
mathematical argument. Perhaps that Java coding will earn you
the
fame you're so desperately seeking. Look what it did for Al
Hirt.
Dale
===
Subject: Re: Correction: Error in NOTICE
>My message NOTICE: Error in Currently Taught Mathematics
posted June
No, really? <chortle>
************************
David C. Ullrich
===
Subject: Re: Correction: Error in NOTICE
> My message NOTICE: Error in Currently Taught Mathematics
posted
June
> Here's the relevant portion from the post:
> <Quote > That de'nition depends on a polynomial P(x) of
degree n with integer
> coef'cients being monic so that you have
> P(x) = (x + a_1)...(x + a_n)
> but that is unbalanced as a complete ring must handle all
non monic
> polynomials of degree n with integer coef'cients to get
> For the sake of those of us who have not been following in
detail . . .
> can you de'ne complete ring. please.
Also, what does it mean for a ring to handle a polynomial?
I'll make one reply in this thread, which is this one.
The de'nition of algebraic integers allows one to 
'nd algebraic
integers a_1, a_2, and a_3, such that they are roots of a
monic cubic
irreducible over Q with integer coef'cients, where one is
provably
coprime to a prime factor of the last coef'cient. That
coprimeness
results leads inevitably to the conclusion that all must be
coprime to
that prime factor in the ring of algebraic integers, which is
the
contradictory result which proves the incompleteness of the
ring.
By incomplete I mean that elements that should be in the ring
are
not such that the contradiction arises.
By handle all non-monic polynomials above I meant that the
factorization for all non-monic polynomials should be in the
ring.
After that I asserted that my paper Advanced Polynomial
Factorization
proved they did not, which is the error corrected by this
post, as
actually I prove a coprimeness result.
James Harris
===
Subject: Re: Correction: Error in NOTICE
>My message NOTICE: Error in Currently Taught Mathematics
posted June
>Here's the relevant portion from the post:
><Quote>>That de'nition depends on a polynomial P(x) of
degree n with integer
>coef'cients being monic so that you have
> P(x) = (x + a_1)...(x + a_n)
>but that is unbalanced as a complete ring must handle all
non monic
>polynomials of degree n with integer coef'cients to get
>For the sake of those of us who have not been following in
detail . . .
>>can you de'ne complete ring. please.
>Also, what does it mean for a ring to handle a polynomial?
> I'll make one reply in this thread, which is this one.
The de'nition of algebraic integers allows one to 
'nd algebraic
> integers a_1, a_2, and a_3, such that they are roots of a
monic cubic
> irreducible over Q with integer coef'cients, where one is
provably
> coprime to a prime factor of the last coef'cient. That
coprimeness
> results leads inevitably to the conclusion that all must be
coprime to
> that prime factor in the ring of algebraic integers, which
is the
> contradictory result which proves the incompleteness of the
ring.
>
Not so. Your claim that one of the a's is provably coprime 
to
a prime
factor of the last coef'cient is misleading. You should state
that you
have developed an argument that purports to prove as much. The
fact of
the matter is that this argument that you continue to promote
is §awed.
If you had a proof, using standard techniques, you might have
something
worth talking about. As it is, you exercise a §awed method of
argument,
fail to see its inadequacy, and arrive at an incorrect
conclusion (your
so-called contradictory result). What is amazing is your
inability to
recognize that the 'rst obligation after 'nding 
results that
appear to
have been contradictory is *not* to attack the foundations of
the 'eld
[in this case, algebraic number theory], but to validate the
method used
to locate the contradictions. You refuse to do this, relying
instead on
a well-refuted sense of intuition.
> By incomplete I mean that elements that should be in the
ring are
> not such that the contradiction arises.
>
Elements that should be in the ring are not such that the
contradiction
arises? Let's be speci'c. In the case
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
where the a's are de'ned below (following Let 
...), please
state
*what elements* should be in the ring of algebraic integers,
but are
not.
Let
u = (63 + i*sqrt(12415))/2,
ubar = (63 - i*sqrt(12415))/2,
zeta = (-1 + i*sqrt(3))/2,
zetabar = (-1 - i*sqrt(3))/2.
Then the values a1,a2,a3 can be given as follows:
x1 = -(u^(1/3) + ubar^(1/3) + 4)
x2 = -(zeta*u^(1/3) + zetabar*ubar^(1/3) + 4)
x3 = -(zetabar*u^(1/3) + zeta*ubar^(1/3) + 4)
where the (...)^(1/3) above are the values with argument = 1/3
times
the argument of (...). Note that this convention forces the
three
roots to be real (as they must be, for this polynomial). The
values are
approximately:
a1 ~ -11.50930
a2 ~ 2.14375
a3 ~ -2.63445
> By handle all non-monic polynomials above I meant that the
> factorization for all non-monic polynomials should be in the
ring.
> After that I asserted that my paper Advanced Polynomial
Factorization
> proved they did not, which is the error corrected by this
post, as
> actually I prove a coprimeness result.
>
You seem to be back to your periodic denial that Magidin &
McKinnon were
incorrect in their rediscovery of the result of Cohn, McAdam,
and Rush.
Why not pick one side of that issue and remain on it? For your
sake you
might try to choose the side that has an actual proof going
for it.
As far as the claim of a coprimeness result, you should say
that you
have an argument that you claim proves such a result. I have
refuted
your claim of coprimeness in the case I cite above, but you
are just
ignoring me. That is your prerogative, of course, but ignoring
the truth
is not the same as proving your case.
What *is* true is that your argument is §awed.
Why is it §awed?
Who knows, maybe your mom didn't treat you right.
What is the evidence that it is §awed?
It produces incorrect conclusions.
Don't those conclusions really mean that mathematics itself
has a problem? I mean, mathematicians are human, aren't 
they?
Don't §atter yourself. It is overwhelmingly more likely
that you have made a mistake than hundreds of thousands,
if not hundreds of millions, of independent observations
have made *exactly* the same error in not seeing the
contradictions that you claim.
Why can't I see the errors in my argument?
Many people have asked the same question. The consensus of
opinion is that you refuse to believe that you can be wrong.
Your notion that mathematics is like some battle of wills,
where perseverence makes right, and ones critics are like
enemies that need to be denied endlessly, is a perversion.
Why can't anyone else see the errors in my argument?
Several reasons:
(1) it is impenetrable, and idiosyncratic
in the extreme,
(2) statements in the argument are not 'xed
in meaning, having one interpretation at
one stage and another interpretation later on
(3) hand-waving
(4) argument by example rather than construction
(5) people *have* pointed out errors, which you deny.
For the readers: I don't expect JSH to address these points,
or any
actual points. I'm not as dumb as I look. However, the 
present
thread
is an affront to the honesty and competence of a number of
people who
have participated in the periodic debunking of JSH and his
style of
argument [both mathematical and personal]. As such, it
shouldn't
stand unanswered.
James Harris
Dale.