unknown method

perigeu

New member
Joined
May 31, 2013
Messages
6
Can you please tell me which method of solving equations is this?


I have two equations:
1) a0*y^2+a1*y+a2=0
2) b0*y^4+b1*y^3+b2*y^2+b3*y+b4=0


And the professor solve this way (6x6 matrix form):


|a0 a1 a2 0 0 0 |
|0 a0 a1 a2 0 0 |
|0 0 a1 a2 a3 0 |
|0 0 0 a1 a2 a3 | =0
|b0 b1 b2 b3 b4 0|
|0 b0 b1 b2 b3 b4|


and the professor reach this equation, where a0....a2 and b0....b4 are in function of x:


po*x^12+p1*x^11+p2*x^10+p3*x^9+p4*x^8+p5*x^7+p6*x^6+p7*x^5+p8*x^4+p9*x^3+p10*x^2+p11*x+p12=0


Does anyone know which method is this?

Thanks in advance.
 
Last edited:
Can you please tell me which method of solving equations is this?


I have two equations:
1) a0*y^2+a1*y+a2=0
2) b0*y^4+b1*y^3+b2*y^2+b3*y+b4=0


And the professor solve this way (6x6 matrix form):


|a0 a1 a2 0 0 0 |
|0 a0 a1 a2 0 0 |
|0 0 a1 a2 a3 0 |
|0 0 0 a1 a2 a3 | =0
|b0 b1 b2 b3 b4 0|
|0 b0 b1 b2 b3 b4|


and the professor reach this equation, where a0....a2 and b0....b4 are in function of x:


po*x^12+p1*x^11+p2*x^10+p3*x^9+p4*x^8+p5*x^7+p6*x^6+p7*x^5+p8*x^4+p9*x^3+p10*x^2+p11*x+p12=0


Does anyone know which method is this?

Thanks in advance.

Basically is shifting columns until it reaches the end of the matrix, and each time you do this you move one column to the right.
 
1) a0*y^2+a1*y+a2=0
2) b0*y^4+b1*y^3+b2*y^2+b3*y+b4=0


And the professor solve this way

po*x^12+p1*x^11+p2*x^10+p3*x^9+p4*x^8+p5*x^7+p6*x^6+p7*x^5+p8*x^4+p9*x^3+p10*x^2+p11*x+p12=0

I do not understand how the 12th-degree polynomial in x represents a solution to the system of two equations in y.

I do not recognize the matrix method as anything that I have seen before. :(
 
I do not understand how the 12th-degree polynomial in x represents a solution to the system of two equations in y.

I do not recognize the matrix method as anything that I have seen before. :(

Thanks for your reply.

Forget about the x's, the x's comes from the determinant of the matrix because some values of a.n or b.n are in function of x and x^2.

I wonder if anybody knows about this method to join this two equations into one by this method.
The solution is not bordering me because is the determinant and I know how to do it.
 
Got no idea what you're asking or trying to do,
but if it helps, here's a solution where y = 2:
1) 3y^2 - 5y - 2 = 0
2) 1y^4 - 6y^3 + 4y^2 + 5y + 6 = 0

Changing variable names in your equations (yours are too confusing!):
1) ay^2 + by + c = 0
2) dy^4 + ey^3 + fy^2 + gy + h = 0
You can get rid of the "y^4" by manipulating, and end up with:
(bd - ae)y^3 + (cd - af)y^2 - agy - ah = 0

Hope that helps...

Edit: easy to solve(!); go see:
http://www.wolframalpha.com/input/?i=solve+(b*d-a*e)*y^3+++(c*d-a*f)*y^2+-+a*g*y+-+a*h+=+0+for+y


Thanks for your help, but I'm interested into know which method is this.
Solving it, I believe is not hard with any of the available tools as for example mathematica.

RGS
 
:idea: Try asking your professor whether this method goes by a specific name.

The professor didn't tell me, he says that I need to do my research, otherwise during the phd presentation they will eat me alive.

Anyway, after long hours studying old math books I find out which method is this. Search for Sylvester Matrix, is exactly this method that I was trying to learn about, now is easy ;)

Thanks for all your help guys.

Regards from Portugal
 
after long hours studying old math books I find out which method is this

This is what you ought to have done in the first place.

Asking people on the Internet for answers should not be accepted as research from doctoral students. :shock:
 
36009]This is what you ought to have done in the first place.

Asking people on the Internet for answers should not be accepted as research from doctoral students. :shock:[/QUOTE]

That was what I did in the first place, but in modern Engineering Algebra books (I search around 30 books), and finnaly found this method in a book of Signal Processing Mathematics (nothing to do with my research) from 1919.
 
Top