JSH: Basic tautological spaces example
JSH
spaces. Partly as a commemoration of the anniversary this month here
is a demonstration post of a basic tautological spaces example.
What are tautological spaces? Most simply they are identities.
A simple tautological space: x+y+vz = x+y+vz
Identities are so dismissed that I also have the bonus of being
someone who found a surprisingly simple use for them--probing
equations for analysis. I invented the term "tautological space" as
well.
The 'v' variable is free, so you always have one extra degree of
freedom.
My innovation moving forward from Gauss was to use "mod" with
identities:
x+y+vz = 0(mod x+y+vz), which is the equivalent of x+y+vz = x+y+vz
You use tautological spaces with non-identities which I call
conditionals.
Here's using tautological spaces with x^2 + y^2 = z^2.
x+y+vz = 0(mod x+y+vz), so x+y = -vz(mod x+y+vz), squaring both
sides:
x^2 + 2xy + y^2 = v^2 z^2 (mod x + y + vz), now subtract out
x^2 + y^2 = z^2, giving
2xy = (v^2 - 1) z^2 (mod x+y+vz), which is
(v^2 - 1) z^2 = 2xy (mod x+y +vz), so:
(v^2 - 1)z^2 - 2xy = 0 (mod x+y+vz)
and I can get rid of x, with x = -y-vz (mod x+y+vz), so I have:
(v^2 - 1)z^2 + 2vyz + 2y^2 = 0 (mod x+y+vz)
which just says that for ANY v you choose, you have that x+y+vz, will
be a factor of what's on the left hand side which is the residue.
And that is not really ring specific, as if you don't use "mod" you
can do it all explicitly. But it's most meaningful in rings where
"factor" is meaningful.
So like with v=1, I have 2yz + 2y^2 = 0 (mod x+y+z), and trivially
with x=3, y=4, z=5, you have:
2(4)(5) + 2(16) = 72 has 12 as a factor.
So you get the result that if x^2 + y^2 = z^2, then 2y(z+y) has x+y+z
as a factor.
Notice that is true over infinity. Tautological spaces give answers
over infinity in general. So their use always encapsulates an
infinite set.
They add on to existing knowledge, as notice you can analyze x^2 +
y^2 = z^2 separately, or throw anything you want at its residue using
the variable 'v' which is your control variable.
Now this month ten years of tautological spaces! A fascinatingly
simple idea using identities, which just so happens to capture
infinity.
James Harris
Junoexpress
> So you get the result that if x^2 + y^2 = z^2, then 2y(z+y) has x+y+z
> as a factor.
>
This is high school algebra James:
(x+y+z)^2 = x^2+y^2+z^2 + 2xy+2xz+2yz
where you substitute x^2+y^2 in for z^2 on the RHS and regroup
(x+y+z)^2 = 2x[x+ y+z]+2y(y+z)
so
2y(y+z) =(x+y+z) [y+z - x]
WOW! I'm sure Gauss *never* saw that acomin'
Happy Anniversary: hope your next ten years are as fruitful and
productive (or at least as funny)!
M
JSH
> On Dec 19, 3:34 pm, JSH <jst...@gmail.com> wrote:
>
> > So you get the result that if x^2 + y^2 = z^2, then 2y(z+y) has x+y+z
> > as a factor.
>
> This is high school algebra James:
Yup, meant to be a simple example.
> (x+y+z)^2 = x^2+y^2+z^2 + 2xy+2xz+2yz
> where you substitute x^2+y^2 in for z^2 on the RHS and regroup
> (x+y+z)^2 = 2x[x+ y+z]+2y(y+z)
> so
> 2y(y+z) =(x+y+z) [y+z - x]
Yeah, I gave a basic example which is meant to be easy to understand.
Given your taunts I thought it worth tossing in something a little
more complicated at the end.
> WOW! I'm sure Gauss *never* saw that acomin'
>
> Happy Anniversary: hope your next ten years are as fruitful and
> productive (or at least as funny)!
Well, I also used tautological spaces on binary quadratic Diophantine
equations. The result then is:
Given
c_1*x^2 + c_2*xy + c_3*y^2 = c_4 + c_5*x + c_6*y
you have a simple form of
(2A(x+y) - B)^2 - 4As^2 = B^2 - 4AC
where A, B and C are defined by the c's and are:
A = (c_2 - 2c_1)^2 + 4c_1(c_2 - c_1 - c_3)
B = 2(c_2 - 2c1)(c_6 - c_5) + 4c_5(c_2 - c_1 - c_3)
and
C = (c_6 - c_5)^2 - 4c_4(c_2 - c_1 - c_3).
x+y and s are the new unknowns where s is actually a function of x
and
y. which can be solved for if you wish by hand or by using a math
program.
If you substitute everything and simplify then you just get back:
c_1*x^2 + c_2*xy + c_3*y^2 = c_4 + c_5*x + c_6*y
Easily derived with tautological spaces.
Care to try and show your own derivation here? Not so easy then, eh?
Students who wish to understand how impressive it truly is need only
search Google (has to be Google).
Search: binary quadratic Diophantine equations
James Harris
Junoexpress
You just do not get it. Anything you could derive of any importance
using simple algebraic identities would have been done a long time
ago. The thing it takes to make a *significant* advance in math is a
new idea, of which you have none (and no, a simple algebraic identity
is not a new idea, no matter how ridiculously pretentious a name you
may call it.) As has been pointed out to you time and time and time
again, most of what you think is new (just because it isn't in written
on Google somewhere) is well known to most other people who actually
work in the field. But you keep shukin and jivin though, hopin you'll
hit the jackpot, like some whino scratching off lottery tickets
outside the liquor store, 'cuz you'll never realize there are no
shortcuts to success.
HTH,
M
JSH
> James,
>
> You just do not get it. Anything you could derive of any importance
> using simple algebraic identities would have been done a long time
> ago. The thing it takes to make a *significant* advance in math is a
> new idea, of which you have none (and no, a simple algebraic identity
> is not a new idea, no matter how ridiculously pretentious a name you
> may call it.) As has been pointed out to you time and time and time
> again, most of what you think is new (just because it isn't in written
> on Google somewhere) is well known to most other people who actually
Then provide another and equal relation for binary quadratic
Diophantine equations.
I'm not even asking for a better one, just an EQUAL one.
So that it's clear what is being discussed, given the binary quadratic
Diophantine form:
c_1*x^2 + c_2*xy + c_3*y^2 = c_4 + c_5*x + c_6*y
provide an equation which reduces to the simpler form of u^2 + Dv^2 =
C.
To reiterate as you deleted it out in your reply, my form is:
(2A(x+y) - B)^2 - 4As^2 = B^2 - 4AC
where A, B and C are defined by the c's and are:
A = (c_2 - 2c_1)^2 + 4c_1(c_2 - c_1 - c_3)
B = 2(c_2 - 2c1)(c_6 - c_5) + 4c_5(c_2 - c_1 - c_3)
and
C = (c_6 - c_5)^2 - 4c_4(c_2 - c_1 - c_3).
What's interesting here is that IS the best in the world. There
actually is no other known better for that task!!!
But more remarkably when I note that (as I have noted it before)
posters simply glide past that reality.
The answer is to provide a better form. If one exists you should be
able to copy and paste from some source.
Students should note that people CAN lie about mathematics--if they
are allowed to just say what they wish!
My math is the best in the world.
There is none better. So yeah, I can toss out equations at will when
challenged, but that doesn't matter if you can just accept it when
others ignore that math, and Usenet isn't the full story.
Math professors at universities around the world should pick up this
research!
Why? Because it is the best in the world.
The best. The best the human species has found to this date. The
best in the world.
Now then, I can find the best. Can show the best. But if you refuse
the best then what can I do?
Proverbially, you can lead a horse to water but you can't make him (or
her) drink.
Now you should really understand that proverb.
Because now, it describes many of you.
James Harris
MichaelW
ax^2+bxy+cx^2+dx+ey+f=0
Set D=4ac-b^2
Set A=2cd-be
Set F=D*(4cf-e^2)-A^2
u=(Dx+A)
v=(bx+2cy+e)
u^2+D*v^2+F=0
Restrictions: D<>0, c<>0
Pretty easy to do with algebra although using matrices makes it
easier.
Merry Christmas! Michael W.