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James Harris
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 Posted: Sun Nov 16, 2003 12:07 am Post subject: Advanced techniques, non-polynomial factorization |
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Polynomials are well-known in science and mathematics, but while
finding roots of polynomials is typically the aim of the average
researcher, polynomials themselves can be used as powerful tools for
analyzing the roots of *other* polynomials.
The concepts are advanced, but can be approached by first considering
a basic example.
The basic factorization to start is
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
49(x^3 + 5x^2 + 3x + 1)
with the c>s algebraic integers, notice that only two of the c>s have
7 as a factor.
It might help to go the *other* way, and start with
(d_1 x + 1)(d_2 x + 1)( d_3 x + 1) =
x^3 + 5x^2 + 3x + 1
and now multiply by 49.
In the first example you>re looking at a product and realizing that
from the distributive property a(b+c) = ab + ac, you know there>s
*one* way it could be produced, which is to multiply something like
the second example by 49.
The distributive property is key here. Understanding it thoroughly,
is of prime importance.
Now notice that you can abstract from here as you>re looking at
*functions* of x, as introducing
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
you have
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
Notice that dividing both sides by 49 gives
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
as long as you>re in a ring where 7 is not a factor of 1.
Which is consistent with what was found before, as only two of the
functions have the property that 7 is a factor.
Now I>ll move on to a more complicated example.
Let
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so they are functions of x, and since one of the roots equals 3 at
x=0, I have
b_3(x) = a_3(x) - 3,
so that all the functions in
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
equal 0, when x=0.
Those of you who find it hard to use the distributive property with
the *product* can imagine the factorization from *before* 49 being
multiplied.
It>s harder to show here as the polynomial which defines the function
in that factorization is not displayable in general.
So I started at the end, with 49 already multiplied because then I can
give
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
That slight change, starting at the end, means that you have to
understand the distributive property fully and *trust* it.
Now notice that I have the result that only two of the roots of the
cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
can have factors in common with 7, so the 49 splits between those two.
What>s so startling is that the result is for a *family* of
polynomials as it applies for any algebraic integer x.
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/ |
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Uncle Al
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| Posted: Sun Nov 16, 2003 12:22 am Post subject: Re: Advanced techniques, non-polynomial factorization |
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James Harris wrote:
[quote]
Polynomials are well-known in science and mathematics,
[snip][/quote]
but not to James Harris by disgustingly repeated explicit empirical
demonstration falsified by a legion of qualified mathematicians.
http://www.crank.net/harris.html
It>s not every braying jackass that gets a whole page at crank.net
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net! |
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Virgil
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| Posted: Sun Nov 16, 2003 1:58 am Post subject: Re: Advanced techniques, non-polynomial factorization |
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JSH still does not understand that examples are neither theorems nor
proofs. |
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Dik T. Winter
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| Posted: Sun Nov 16, 2003 7:39 am Post subject: Re: Advanced techniques, non-polynomial factorization |
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James, I think you presses a send button for an article you already
posted. I did comment on it already, but you did not answer. I think
you have not seen my response. So I will try again here.
In article <3c65f87.0311151007.3e1c8e66@posting.google.com> jstevh@msn.com (James Harris) writes:
[quote]Polynomials are well-known in science and mathematics, but while
finding roots of polynomials is typically the aim of the average
researcher, polynomials themselves can be used as powerful tools for
analyzing the roots of *other* polynomials.
[/quote]
Oh, well, It appears you have modified it a bit. Yes, polynomials are
powerful tools to analyse roots of other polynomials. Yup, this
paragraph was rewritten, as is the next.
[quote]The concepts are advanced, but can be approached by first considering
a basic example.
The basic factorization to start is
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
49(x^3 + 5x^2 + 3x + 1)
with the c>s algebraic integers, notice that only two of the c>s have
7 as a factor.
[/quote]
A comment I add this time and did not add in the last version. The c>s
can *only* be algabraic integers because the constant factor of the
polynomial is 1. When that constant factor is 2, there is *no* such
decomposition.
[quote]It might help to go the *other* way, and start with
(d_1 x + 1)(d_2 x + 1)( d_3 x + 1) =
x^3 + 5x^2 + 3x + 1
and now multiply by 49.
[/quote]
And again the same, newly added, comment: the d>s are *only* algebraic
integers because the constant term of the cubic is 1.
[quote]In the first example you>re looking at a product and realizing that
from the distributive property a(b+c) = ab + ac, you know there>s
*one* way it could be produced, which is to multiply something like
the second example by 49.
The distributive property is key here. Understanding it thoroughly,
is of prime importance.
[/quote]
Hrm, an added paragraph. Does not add much information.
[quote]Now notice that you can abstract from here as you>re looking at
*functions* of x, as introducing
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
you have
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
Notice that dividing both sides by 49 gives
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
as long as you>re in a ring where 7 is not a factor of 1.
[/quote]
A repeat:
This is independent of 7 being a factor of 1. Note however that it
is *not* the only way to distribute 49 amongst the three factors on
the left hand side. This is the only way *only* if you require that
the three factors on the left hand side are polynomials, if you have
not such an requirement it can be done differently. However, you can
have polynomials on the left hand side *only* because the roots of the
polynomial on the right hand side are units, i.e. divisors of 1. It
will *not* work if that is not the case.
An addition:
If you do not have the requirement that the factors on the left hand
side are polynomials, the following factorisation is also possible
(and gazillion others), and assuming x is integer:
Define:
w3(x) = gcd(f3(x) + 1, 7) { this can be 1 or some other
divisor of 7. }
w2(x) = 7 / w3(x).
Now:
[ (f1(x) + 7)/7 ][ (f2(x) + 7)/w2(x) ][ (f3(x) + 1)/w3(x) ]
is a perfectly valid factorisation in the algebraic integer valued
functions on the integers of:
x^3 + 5 x^2 + 3x + 1
Note that the distributive property dictates:
(a + b) * c = a * c + b * c
in a particular ring. Not that
(a + b) / c = a / c + b / c
in a ring. That is, that when (a + b)/c is in a ring, there is no
requirements that a/c and b/c are also in that ring. You keep on
assuming that.
[quote]Which is consistent with what was found before, as only two of the
functions have the property that 7 is a factor.
[/quote]
This is true *only* in some special cases...
[quote]Now I>ll move on to a more complicated example.
Let
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so they are functions of x, and since one of the roots equals 3 at
x=0, I have
[/quote]
Note, that they are *not* polynomials. So the situation is different
from the one above. And indeed, in general *none* of the factors is
divisible by 7. Moreover, you can not even have the requirement that
when you divide by 49 the factors on the left must be polynomials,
because you do not start with polynomials on the left. So there are
other ways to distribute 49 amongst the three factors, however the
way you distribute is a function of x.
Remainder of repetition skipped.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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William Hughes
Guest
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| Posted: Mon Nov 17, 2003 7:46 pm Post subject: Re: Advanced techniques, non-polynomial factorization |
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jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311151007.3e1c8e66@posting.google.com>...
[quote]
so that all the functions in
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
equal 0, when x=0.
[/quote]
We have been here many times. The problems is that a_1(x), a_2(x) and
b_3(x) are not polynomials. Therefore, we do not know that
the way in which the 49 distributes itself among the
three factors on the LHS is independent of x. Thus we cannot
conclude that 7 divides (5 a_1(x)+ 7) for all x. Thus we
cannot conclude that 7 divides a_1(x) for all x.
-William Hughes |
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Nora Baron
Guest
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| Posted: Mon Nov 17, 2003 10:17 pm Post subject: Re: Advanced techniques, non-polynomial factorization |
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jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311151007.3e1c8e66@posting.google.com>...
[quote]Polynomials are well-known in science and mathematics, but while
finding roots of polynomials is typically the aim of the average
researcher, polynomials themselves can be used as powerful tools for
analyzing the roots of *other* polynomials.
The concepts are advanced, but can be approached by first considering
a basic example.
The basic factorization to start is
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
49(x^3 + 5x^2 + 3x + 1)
with the c>s algebraic integers, notice that only two of the c>s have
7 as a factor.
It might help to go the *other* way, and start with
(d_1 x + 1)(d_2 x + 1)( d_3 x + 1) =
x^3 + 5x^2 + 3x + 1
and now multiply by 49.
In the first example you>re looking at a product and realizing that
from the distributive property a(b+c) = ab + ac, you know there>s
*one* way it could be produced, which is to multiply something like
the second example by 49.
The distributive property is key here. Understanding it thoroughly,
is of prime importance.
Now notice that you can abstract from here as you>re looking at
*functions* of x, as introducing
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
you have
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).
Notice that dividing both sides by 49 gives
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1
as long as you>re in a ring where 7 is not a factor of 1.
Which is consistent with what was found before, as only two of the
functions have the property that 7 is a factor.
[/quote]
This part so far is OK.
[quote]Now I>ll move on to a more complicated example.
Let
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where the a>s are roots of
[***] a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so they are functions of x, and since one of the roots equals 3 at
x=0, I have
b_3(x) = a_3(x) - 3,
so that all the functions in
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
equal 0, when x=0.
Those of you who find it hard to use the distributive property with
the *product* can imagine the factorization from *before* 49 being
multiplied.
[/quote]
As in the first example you gave above, the "49" can be
distributed among the three factors in several different
ways. There is no justification for your implied claim below
that two constant terms "7" should be cancelled completely.
This is where you are making your error.
[quote]It>s harder to show here as the polynomial which defines the function
in that factorization is not displayable in general.
So I started at the end, with 49 already multiplied because then I can
give
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
That slight change, starting at the end, means that you have to
understand the distributive property fully and *trust* it.
Now notice that I have the result that only two of the roots of the
cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
can have factors in common with 7, so the 49 splits between those two.
[/quote]
When you were discussing the simpler polynomial above, you
said that the coefficients c_1, c_2, c_3 were algebraic integers,
so presumably you are talking about the same thing here. That
is, you are saying the a>s are algebraic integers and two of them
are divisible by 7 in the ring of algebraic integers. Thus
assume a_1/7 is an algebraic integer; equivalently a_1 = 7*b_1,
where b1 is an algebraic integer. But since you are saying that
a_1 is a root of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
we must have that b_1 is a root of
7^3*b^3 + 3*(-1 + 49*x)*7^2*b^2 - 7^2*(2401*x^3 - 147*x^2 + 3*x) = 0.
Now divide through by 7^2:
7*b^3 + 3*(-1 + 49*x)*b^2 - (2401*x^3 - 147*x^2 + 3*x) = 0.
Finally, let x = 1:
7*b^3 + 144*b^2 - 2257 = 0.
This polynomial is primitive, irreducible, and non-monic. Therefore
none of its roots can be algebraic integers. Therefore a_1/7 is
not an algebraic integer. Therefore a_1 is not divisible by 7.
You continue to think that your proof that a_1 *is* divisible by
7 is valid. If it were, it would imply a mathematical contradiction.
Mathematics would be inconsistent. There is no sense in claiming
that the algebraic integers are "incomplete". They are a perfectly well-
defined subset of the real numbers. Something cannot both be in
that subset and not in that subset.
Please explain your own conclusions on this.
[quote]What>s so startling is that the result is for a *family* of
polynomials as it applies for any algebraic integer x.
[/quote]
For example, x = 1, which is what I used above.
Your "proof" implies a contradiction, so it is either
incorrect or math is inconsistent. I have identified exactly
above where you have made an unjustified assumption.
Your argument here is wrong.
Nora B.
[quote]
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/[/quote] |
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