Solving two polynomials in two variables
Solving two polynomials in two variables
I had an algorithm for solving two polynomials in 2 variables by converting it into an equation in 1 variable. I'll see if I can remember it.Let us start with 2 polynomials. (solution: x=2 y=3). We try to eliminate x.
(1) x2+4xy+2y2+3x+4y - 64 = 0
(2) 3x2 -xy + y2+5x-y -22 = 0
First eliminate x2
(3)=-(2)+3(1): 13xy +5y2 +4x +13y -170 = 0
or (13y+4)x + (5y2+13y -170) = 0
Next use this equation to eliminate x2 again from (1).
(1).(13y+4) : (13y+4)x2+ (52y2+55y+12)x + (13y+4)(2y2+4y - 64) = 0
-(3).x: (13y+4)x2 + (5y2+13y -170)x = 0
=(4): (47y2+42y+182)x + (13y+4)(2y2+4y - 64) = 0
Now use these two linear equations to eliminate x:
(13y+4)(4) - (47y2+42y+182)(3) : (13y+4)(13y+4)(2y2+4y - 64) - (47y2+42y+182)(5y2+13y -170) =0
or (y-3) (103y3+372y2-2718y-9972)=0
Which tells us that one solution has y=3. (There are 3 other solutions too!)
Feeding this back into (3) gives:
43x -86 = 0
giving x=2. So one solution is (x=2,y=3)
The same method works for polynomials of any degree. Just eliminate the higher powers one by one.
It can also be used to find out if two equations have the same root, if A,B,C.. etc. don't depend on y:
(1) Ax2+Bx+C =0
(2) Dx2+Ex+F =0
D(1)-A(2) = (3): (BD-AE)x+(DC-FA)=0
(BD-AE)(1): A(BD-AE)x2+(BD-AE)Bx+(BD-AE)C =0
-Ax(3): A(BD-AE)x2+A(DC-FA)x = 0
=(4): ((BD-AE)B-A(DC-FA))x+(BD-AE)C = 0
((BD-AE)B-A(DC-FA))(3) - (BD-AE)(4)
=(5): ((BD-AE)B-A(DC-FA))(DC-FA)-(BD-AE)(BD-AE)C = 0
Which apart from a factor of A is the condition that both equations share a common root.
I haven't tried it with 3 polynomials in 3 variables!
Conclusion
To eliminate a variable, x, from a system of polynomials is equivalent to finding the condition that the polynomials in x have a common root. Which is the resultant of the polynomials.
General case
(1) ax2+2bxy+cy2+2dx+2ey +f = 0 <= xQx=0
(2) Ax2+2Bxy+Cy2+2Dx+2Ey +F = 0 <= xRx=0
A(1)-a(2):(3): 2((bA-2Ba)y+(dA-Da))x + ((cA-Ca)y2+2(eA-Ae)y+2(fA-Fa)) = 0
(4): (a((cA-Ca)y2+2(eA-Ae)y+2(fA-Fa)) -4((bA-2Ba)y+(dA-Da))(by+d))x
-2((bA-2Ba)y+(dA-Da))(cy2+2ey +f )= 0
((cA-Ca)y2+2(eA-Ae)y+2(fA-Fa)) (a((cA-Ca)y2+2(eA-Ae)y+2(fA-Fa)) -4((bA-2Ba)y+(dA-Da))(by+d)) + 4((bA-2Ba)y+(dA-Da))(cy2+2ey +f )((bA-2Ba)y+(dA-Da)) =0
Then find the discriminant!! of this quartic!
y4(a(cA-Ca)2+4c(bA-2Ba)2)+...
(Dividing by a?) Each coefficient has power 4 so the conditional polynomial will have power 12.
Q* has power 2
det(Q) has power 3 = (Tr(Q)3-3Tr(Q2)Tr(Q)+2Tr(Q3))/6
Tr(Q) has power 1
e.g. a det(Q)2det(R)2 + b det(Q)det(R)3+ c det(R)4+ d det(Q)3det(R) + e det(Q)4 = 0
It can all be built out of traces. But symmetry suggests should all be built out of determinants??
0 = Tr(Q6)Tr(Q)3Tr(R2)3+...etc.
Is this useful:
∫ (exp (-xQx) ) dx = 1/det(Q)^(1/2)
How to extend that to two formulae?
∫ (exp (-xQx) + exp (-xRx) ) dx = 1/det(Q)^(1/2) + 1/det(R)^(1/2)
Conditions:
det(R)2>det(Q)2 ??det(R)2>-det(Q)2
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