| Topic: |
Science > Physics |
| User: |
"James Harris" |
| Date: |
12 Nov 2003 11:19:34 AM |
| Object: |
Factorization, basic to advanced concepts |
If you saw
(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, I think few of you would have a
problem realizing that only two of the c's have 7 as a factor.
But, of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
so I could also write it as
(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.
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
For instance
(f_1(x) + 7) versus (f_3(x) + 1), where at x=0, both functions equal
0, and 1 and 7 are both factors of the constant term.
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1)
multiplies out to give
49(x^3 + 5 x^2 + 3x + 1).
so f_1(x) f_2(x) f_3(x) = 49, for instance.
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(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 I can see all the constant term factors.
What you have just seen is a major advance in mathematical thinking
where I've used a rather simple abstraction and a special polynomial
to analyze the *roots* of another *different* polynomial.
It is a powerful tool that is new to mathematical analysis, as I
discovered it only a few years ago.
Unfortunately the concepts seem advanced enough to attract posters who
react with fury when they can't quite get it, who refuse to
acknowledge the basic principles, who also post a LOT!!!
They've created a false picture that mathematicians as a group have
refuted the information provided here, when it's impossible to refute
mathematical logic.
Basically, a few people with an agenda, posting a lot, have created an
atmosphere of confusion and distrust which has fed upon itself.
But now I hope to reach other, more sophisticated and intelligent
people who can consider the actual facts.
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/
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| User: "Nora Baron" |
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| Title: Re: Factorization, basic to advanced concepts |
12 Nov 2003 07:21:24 PM |
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(James Harris) wrote in message news:<3c65f87.0311120919.271d4d70@posting.google.com>...
If you saw
(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, I think few of you would have a
problem realizing that only two of the c's have 7 as a factor.
But, of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
so I could also write it as
(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.
Nice little challenge ...
Let H(x) = x^3 + 5*x^2 + 3*x + 1.
Let f_1(x) = sqrt(abs(x)).
Let f_2(x) = sqrt(abs(x)).
Let f_3(x) = 49*H(x)/(sqrt(abs(x)) + 7)^2 - 1.
Then clearly (f_1(x) + 7)*(f_2(x) + 7)*(f_3(x) + 1) = 49*H(x).
Note also that f_1(0) = f_2(0) = f_3(x) = 0, since H(0) = 1.
This appears to be a *nonpolynomial* factorization that satisfies
all your requirements, for what it's worth.
But in what sense do f_1(x) and f_2(x) have "7" as a factor?
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
Above functions certainly do that.
For instance
(f_1(x) + 7) versus (f_3(x) + 1), where at x=0, both functions equal
0, and 1 and 7 are both factors of the constant term.
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1)
multiplies out to give
49(x^3 + 5 x^2 + 3x + 1).
Yes, certainly true for the functions above also.
so f_1(x) f_2(x) f_3(x) = 49, for instance.
No - that is NOT required. I think what you meant here
is (f_1(0) + 7)*(f_2(0) + 7)*(f_1(0) + 1) = 49.
Is the set of three functions that I gave above the kind
of thing you want? There are infinitely many ways to
construct them, of course. Maybe you have some other
restrictions in mind?
Nora B.
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(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 I can see all the constant term factors.
What you have just seen is a major advance in mathematical thinking
where I've used a rather simple abstraction and a special polynomial
to analyze the *roots* of another *different* polynomial.
It is a powerful tool that is new to mathematical analysis, as I
discovered it only a few years ago.
Unfortunately it doesn't work as you think it does.
Unfortunately the concepts seem advanced enough to attract posters who
react with fury when they can't quite get it, who refuse to
acknowledge the basic principles, who also post a LOT!!!
They've created a false picture that mathematicians as a group have
refuted the information provided here, when it's impossible to refute
mathematical logic.
It is no longer clear what your are claiming to have proved.
For what you were claiming previously, that the a's were
divisible in the algebraic integers by "f", we have given
examples that show rigorously that your claims were false.
You have given no refutations of those examples and in fact
you have recently used the same methods with regard to
show that coefficients in the factorizations of other
polynomials are not algebraic integers.
You need now to state your current claims unambiguously, to
clear the air. We need to start over in assessing what you
have done. It is a confusing mess at present. It was simpler
when you were maintaining a web page with a clear statement
of your current claims and arguments intended to back them
up. At present we don't know which of your recent statements
to believe, especially when they are likely to be retracted
the next day.
Basically, a few people with an agenda, posting a lot, have created an
atmosphere of confusion and distrust which has fed upon itself.
Good example: your posting in the last few days on the
polynomial with constant term "2". You argued at length on that
and started several different threads on the same idea. People
pointed out problems. You modified what you were saying. More
problems were pointed out. You ended up retracting the whole
thing. It wasn't a matter of "confusion and distrust which has
fed upon itself". It was simply a matter of your being wrong from
the start. You accused your critics of being cranks and implied
they were lying and incompetent. That wasn't it. You were simply
wrong. Now you want to act as though it was other people sowing
"confusion and distrust"! What hypocrisy.
But now I hope to reach other, more sophisticated and intelligent
people who can consider the actual facts.
Bull. You just want people to shut up and agree with you,
no matter what you say. Better stick with your agreeable friends
at the Mega Society if that is what you require.
Nora B.
James Harris
"My math discoveries, found for profit"
http://mathforprofit.blogspot.com/
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| User: "William Hughes" |
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| Title: Re: Factorization, basic to advanced concepts |
14 Nov 2003 04:08:35 PM |
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(James Harris) wrote in message news:<3c65f87.0311120919.271d4d70@posting.google.com>...
If you saw
(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, I think few of you would have a
problem realizing that only two of the c's have 7 as a factor.
But, of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
so I could also write it as
(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.
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
You are not "moving to functions", you are moving
from linear functions to general functions. You cannot assume
that stuff that is true for linear functions is also true for
general functions.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
For instance
(f_1(x) + 7) versus (f_3(x) + 1), where at x=0, both functions equal
0, and 1 and 7 are both factors of the constant term.
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1)
multiplies out to give
49(x^3 + 5 x^2 + 3x + 1).
so f_1(x) f_2(x) f_3(x) = 49, for instance.
Yes, but you cannot claim that in general, f_1(x) and f_2(x) are
divisible by 7. This is true if f_1(x) = c_1 x and f_2(x) = c_2 x.
This is not true in general.
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(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 I can see all the constant term factors.
The factors are not polynomial factors. Thus you cannot
determine anything about the divisiblity of the factors for
x not equal to 0.
- "William Hughes"
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| User: "Dik T. Winter" |
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| Title: Re: Factorization, basic to advanced concepts |
12 Nov 2003 07:07:24 PM |
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In article <3c65f87.0311120919.271d4d70@posting.google.com> (James Harris) writes:
If you saw
(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, I think few of you would have a
problem realizing that only two of the c's have 7 as a factor.
If they realise that by the two 7's in the factorisation they are too
much relying on intuition. Change the constants 1 to 2 and it is
no longer true in the algebraic integers. In fact with
(c1 x + 7)(c2 x + 7)(c3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)
there are *no* c's possible that are also algebraic integers.
However, in this case the roots of the cubic (r1, r2 and r3) are units,
meaning that 1/r1, 1/r2 and 1/r3 are also algebraic integers. And we
have: c1 = 7/r1, c2 = 7/r2 and c3 = 1/r3.
But, of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x,
This is an irrelevant red herring.
so I could also write it as
(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.
And this is another red herring. It is true in *every* ring that
contains the roots of x^3 + 5x^2 + 3x + 1.
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
My problem with your position is that you claim that is the *only* way
you can distribute 1/49 over the three factors while the factors remain
functions from the algebraic integers to the algebraic integers. There
are indeed other ways. Moreover, this example is a complete red herring
as the roots of the cubic are units.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
No, first you need to outline it in a generalised form where the roots
of the cubic are *not* units. Next you can get to the complication of
arbitrary functions from the algebraic integers to the algebraic integers.
But I will give an alternative factorisation:
Define:
w3(x) = gcd(c3 x + 1, 7) (and pray note that this is not always 1)
w2(x) = gcd(c2 x + 7, 7)/w3(x);
P(x)/7 = [(c1 x + 7)/7] [(c2 x + 7)/w2(x)] [(c3 x + 1)/w3(x)]
is *also* a valid factorisation.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
For instance
(f_1(x) + 7) versus (f_3(x) + 1), where at x=0, both functions equal
0, and 1 and 7 are both factors of the constant term.
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1)
multiplies out to give
49(x^3 + 5 x^2 + 3x + 1).
so f_1(x) f_2(x) f_3(x) = 49, for instance.
Eh? Is the product f1(x) f2(x) f3(x) *independent* of x? In your case
it si *not* 49, it is 49 x^3.
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(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 I can see all the constant term factors.
I never complained about your discussion upto this point. Nor have I seen
other people object to this. So I do not see what you mean with this. It
is your step 6 that is broken, and that is beyond this stage.
--
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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| User: "The Ghost In The Machine" |
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| Title: Re: Factorization, basic to advanced concepts |
13 Nov 2003 02:59:42 AM |
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In sci.physics, James Harris
<jstevh@msn.com>
wrote
on 12 Nov 2003 09:19:34 -0800
<3c65f87.0311120919.271d4d70@posting.google.com>:
JSH states logic similar to the following. I've changed
the equations slightly for various reasons, so am not using
the customary chevrons. However, as far as I can tell,
James is of the opinion that the equation doesn't matter;
it should work for equations. (I do not share this
opinion, mind you, but am wondering what the issues are.)
My comments are indented in (), within the emulated text.
[begin emulation/check]
If one sees
(c_1 x + 7)(c_2 x + 7)( c_3 x + 15) = 49(x^3 + 9*x^2 + 23*x + 15),
with the c's algebraic integers, I think few of you would have
a problem realizing that only two of the c's have 7 as a factor.
(Note that:
(x^3 + 9*x^2 + 23*x + 15) = (x + 1) * (x + 3) * (x + 5).
Since I happen to know the roots (by construction) of this
particular polynomial, let's see how far we get with this
logic. It is not clear whether JSH is arguing within
the space of all degree 3 polynomials in Z[x] with a_3 = 1,
or the space of all degree 3 polynomials in Z[x] with a_3 = 1
and a_0 = ±1, or even the rather esoteric space of all
degree 3 polynomials in A[x] with a_3 a unit, where A is the
ring of all algebraic integers.)
But of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, f_3(x) = c_3 x,
so I could also write it as
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 15) = 49(x^3 + 9*x^2 + 23*x + 15)
(Yeah, so?)
Notice that dividing both sides by 49 gives
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 9*x^2 + 23*x + 15
as long as you're in a ring where 7 is not a factor of 1.
(That is one representation of result, yes. Other representations
are possible, such as
(f_1(x)/49 + 1/7)(f_2(x) + 7)( f_3(x) + 1) = x^3 + 9*x^2 + 23*x + 15.
Also, see below.)
That's an important point and represents an issue over which I've
found a lot of people willing to argue, and it may seem vague to move
to functions, even though in the previous example they were actually
functions but they weren't being *called* functions.
So some rules need to be outlined in generalizing from the basic
polynomial factors with their simple functions like c_1 x, where c_1
is constant, to more complicated ones that we might not even have
imagined yet.
One thing that's clearly important is that the functions *must* equal
0 when x=0, as then you have factors of the constant term opposite
them.
(Not a problem here so far.)
For instance
(f_1(x) + 7) versus (f_3(x) + 15), where at x=0, both functions equal
0, and 15 and 7 are both factors of the constant term.
(Which they are.)
Next, the factorization must multiply out correctly, which just means
that
(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 15)
multiplies out to give
49(x^3 + 9*x^2 + 23*x + 15).
so f_1(x) f_2(x) f_3(x) = 49, for instance.
(Actually, 49 * x^3. And even that may not be the case;
one could, for example, set f_1(x) = 49(x^3 + 9*x^2 + 23*x + 15) - 7,
f_2(x) = -6, f_3(x) = -14. The product of these would
be a multiple of 49 * 12, not 49 * x^3. I'm not even certain
one can conclude that their product is divisible by 49;
consider f_1(x) = -6, f_2(x) = -6,
f_3(x) = 49(x^3 + 9*x^2 + 23*x + 15) - 14, The product there
is divisible by 7, that much I can tell you.)
I'm abstracting and generalizing to functions because I've faced
arguments with a much more complicated example, where the basic
principles are the same:
(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)
(I've left the above two equations intact. Unlike my
example above the polynomial in () is irreducible
in Z[x] -- and for that matter in Q[x].)
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 I can see all the constant term factors.
What you have just seen is a major advance in mathematical thinking
where I've used a rather simple abstraction and a special polynomial
to analyze the *roots* of another *different* polynomial.
(Looks more sloppy than advanced to me; JSH is leaping to
some strange conclusions. For example, given that
(c_1(x) + 7)*(c_2(x) + 7)*(c_3(x) + a_0)
= 49(a_3*x^3 + a_2*x^2 + a_1*x^1 + a_0)
I can conclude very little about the c_i() functions.
If one postulates that they are within Q[x] of degree 0 or 1,
one might fare slightly better -- but that's a postulate
JSH has currently not made.)
[end of emulation/check]
--
#191,
It's still legal to go .sigless.
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| User: "Uncle Al" |
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| Title: Re: Factorization, basic to advanced concepts |
12 Nov 2003 11:28:04 AM |
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James Harris wrote:
If you saw
(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) =
49(x^3 + 5x^2 + 3x + 1)
[snip]
you'd next see
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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| User: "Sam Wormley" |
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| Title: Re: Factorization, basic to advanced concepts |
12 Nov 2003 05:41:48 PM |
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James Harris wrote:
If you saw...... [snip]
Crank Information
http://www.crank.net/harris.html
http://www.crank.net/usenet.html
http://www.google.com/search?q=harris+site%3Awww.crank.net
http://www.google.com/search?q=%22james+harris%22+site%3Ausers.pandora.be
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