Mixing
Computing determinant
Let $$ A=
begin{bmatrix}
x_1^{4} & x_2^{4} & x_3^{4} \
x_1^{2} & x_2^{2} & x_3^{2} \
1 &1 &1
end{bmatrix}
.$$ I want to show that the $$det A times frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)} = x_1^2x_2+x_1^2x_3+x_2^2x_3+x_1x_2^2+x_1x_3^2+x_2x_3^2+2x_1x_2x_3.$$
The determinant of $A$ is trivial but trying to manipulate $frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}$ is where I am having some problems.
We can think of $$frac{1}{(x_1-x_2)} = frac{1}{x_1(1-frac{x_2}{x_1})}=frac{1}{x_1}(1+ frac{x_2}{x_1}+frac{x_2^2}{x_1^2}+cdots)$$ but I still dont end up getting the answer after multplying the whole thing out. Is there a trick to this? any hint will be appreciated.
P.S (Combinatorics): This is an expression from the definition of Schur Functions.
calculus combinatorics determinant
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
add a comment |
Let $$ A=
begin{bmatrix}
x_1^{4} & x_2^{4} & x_3^{4} \
x_1^{2} & x_2^{2} & x_3^{2} \
1 &1 &1
end{bmatrix}
.$$ I want to show that the $$det A times frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)} = x_1^2x_2+x_1^2x_3+x_2^2x_3+x_1x_2^2+x_1x_3^2+x_2x_3^2+2x_1x_2x_3.$$
The determinant of $A$ is trivial but trying to manipulate $frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}$ is where I am having some problems.
We can think of $$frac{1}{(x_1-x_2)} = frac{1}{x_1(1-frac{x_2}{x_1})}=frac{1}{x_1}(1+ frac{x_2}{x_1}+frac{x_2^2}{x_1^2}+cdots)$$ but I still dont end up getting the answer after multplying the whole thing out. Is there a trick to this? any hint will be appreciated.
P.S (Combinatorics): This is an expression from the definition of Schur Functions.
calculus combinatorics determinant
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08
add a comment |
Let $$ A=
begin{bmatrix}
x_1^{4} & x_2^{4} & x_3^{4} \
x_1^{2} & x_2^{2} & x_3^{2} \
1 &1 &1
end{bmatrix}
.$$ I want to show that the $$det A times frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)} = x_1^2x_2+x_1^2x_3+x_2^2x_3+x_1x_2^2+x_1x_3^2+x_2x_3^2+2x_1x_2x_3.$$
The determinant of $A$ is trivial but trying to manipulate $frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}$ is where I am having some problems.
We can think of $$frac{1}{(x_1-x_2)} = frac{1}{x_1(1-frac{x_2}{x_1})}=frac{1}{x_1}(1+ frac{x_2}{x_1}+frac{x_2^2}{x_1^2}+cdots)$$ but I still dont end up getting the answer after multplying the whole thing out. Is there a trick to this? any hint will be appreciated.
P.S (Combinatorics): This is an expression from the definition of Schur Functions.
calculus combinatorics determinant
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
Let $$ A=
begin{bmatrix}
x_1^{4} & x_2^{4} & x_3^{4} \
x_1^{2} & x_2^{2} & x_3^{2} \
1 &1 &1
end{bmatrix}
.$$ I want to show that the $$det A times frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)} = x_1^2x_2+x_1^2x_3+x_2^2x_3+x_1x_2^2+x_1x_3^2+x_2x_3^2+2x_1x_2x_3.$$
The determinant of $A$ is trivial but trying to manipulate $frac{1}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}$ is where I am having some problems.
We can think of $$frac{1}{(x_1-x_2)} = frac{1}{x_1(1-frac{x_2}{x_1})}=frac{1}{x_1}(1+ frac{x_2}{x_1}+frac{x_2^2}{x_1^2}+cdots)$$ but I still dont end up getting the answer after multplying the whole thing out. Is there a trick to this? any hint will be appreciated.
P.S (Combinatorics): This is an expression from the definition of Schur Functions.
calculus combinatorics determinant
calculus combinatorics determinant
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
asked Nov 21 '18 at 22:30
JaynotJaynot
487515
487515
By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08
add a comment |
By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08
By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08
add a comment |
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By replacing $x_i^2$ with $z_i$ you get a Vandermonde matrix, which is associated to an interpolation problem.
– Jack D'Aurizio
Nov 21 '18 at 22:33
Just thinking out loud: You could subtract the 2nd column from the 1st (and the 3rd from the 2nd.) Then when you work out the det, you get some difference of squares-y things which might work well with your denominator.
– B. Goddard
Nov 21 '18 at 22:38
@B.Goddard Thanks a lot. Done :)
– Jaynot
Nov 21 '18 at 23:08