Substitution to get rid of cubic terms in a two-variable expression
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Consider the quadratic form
$$Q(u,v) = au^4 + bu^3v+cuv^3+du^2v^2+ev^4$$
If $a,b,c,d = 1$ then we can get rid of the cubic terms by substituting $u:=x-y, v:=x+y$. But what would be the general substitution for any value of $a,b,c,d$ to get rid of the cubic terms? I've experimented with $u:=x-frac{lambda}{x}$, $v:=y-frac{lambda}{y}$ (with $lambda$ to be determined later) but that dind't give me much result.
linear-algebra quadratic-forms substitution change-of-variable
asked Nov 23 at 1:59
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4,17131035
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Consider the quadratic form
$$Q(u,v) = au^4 + bu^3v+cuv^3+du^2v^2+ev^4$$
If $a,b,c,d = 1$ then we can get rid of the cubic terms by substituting $u:=x-y, v:=x+y$. But what would be the general substitution for any value of $a,b,c,d$ to get rid of the cubic terms? I've experimented with $u:=x-frac{lambda}{x}$, $v:=y-frac{lambda}{y}$ (with $lambda$ to be determined later) but that dind't give me much result.
linear-algebra quadratic-forms substitution change-of-variable
asked Nov 23 at 1:59
sequence
4,17131035
1
The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the quadratic form
$$Q(u,v) = au^4 + bu^3v+cuv^3+du^2v^2+ev^4$$
If $a,b,c,d = 1$ then we can get rid of the cubic terms by substituting $u:=x-y, v:=x+y$. But what would be the general substitution for any value of $a,b,c,d$ to get rid of the cubic terms? I've experimented with $u:=x-frac{lambda}{x}$, $v:=y-frac{lambda}{y}$ (with $lambda$ to be determined later) but that dind't give me much result.
linear-algebra quadratic-forms substitution change-of-variable
asked Nov 23 at 1:59
sequence
4,17131035
Consider the quadratic form
$$Q(u,v) = au^4 + bu^3v+cuv^3+du^2v^2+ev^4$$
If $a,b,c,d = 1$ then we can get rid of the cubic terms by substituting $u:=x-y, v:=x+y$. But what would be the general substitution for any value of $a,b,c,d$ to get rid of the cubic terms? I've experimented with $u:=x-frac{lambda}{x}$, $v:=y-frac{lambda}{y}$ (with $lambda$ to be determined later) but that dind't give me much result.
linear-algebra quadratic-forms substitution change-of-variable
linear-algebra quadratic-forms substitution change-of-variable
asked Nov 23 at 1:59
sequence
4,17131035
asked Nov 23 at 1:59
sequence
4,17131035
asked Nov 23 at 1:59
sequence
4,17131035
asked Nov 23 at 1:59
sequence
4,17131035
asked Nov 23 at 1:59
sequence
4,17131035
4,17131035
1
The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02
add a comment |
1
The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02
1
1
The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02
The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02
add a comment |
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The trick that worked for $a,b,c,d=1$ was linear combinations of $x$ and $y$. Have you tried setting $u = Ax +By, v = Cx + Dy$, summing up the various expressions, and setting the coefficients of the cubic terms to $0$ to solve for $A, B, C, D$ in terms of $a,b,c,d,e$? With 2 equations in 4 unknowns, you should have infinitely many solutions, unless the 2 equations are contradictory.
– Paul Sinclair
Nov 23 at 14:02