Calculate Resultant of $g(x)=X^3+pX+q$

[evinda]

Hi! (Smile)

I want to find the discriminant of $g(x)=X^3+pX+q$.

$$Res(g,g')=(-1)^{\frac{3(3-1)}{2}} D(g(X)) \Rightarrow Res(g,g')=-D(g(X))$$

$$Res(g,g')=det\begin{bmatrix}
q & p & 0 & 1\\
0 & q & p & 0\\
p & 0 & 3 &0 \\
0 & p & 0 & 3
\end{bmatrix}$$

How can we find the above determinant?

 

[topsquark]

Hi! (Smile)

I want to find the discriminant of $g(x)=X^3+pX+q$.

$$Res(g,g')=(-1)^{\frac{3(3-1)}{2}} D(g(X)) \Rightarrow Res(g,g')=-D(g(X))$$

$$Res(g,g')=det\begin{bmatrix}
q & p & 0 & 1\\
0 & q & p & 0\\
p & 0 & 3 &0 \\
0 & p & 0 & 3
\end{bmatrix}$$

How can we find the above determinant?

Expanding the determinant across the third row:
\(\displaystyle \left | \begin{matrix}q & p & 0 & 1\\ 0 & q & p & 0\\ p & 0 & 3 &0 \\ 0 & p & 0 & 3 \end{matrix} \right | = p \left | \begin{matrix} p & 0 & 1 \\ q & p & 0 \\ p & 0 & 3 \end{matrix} \right | + 3 \left | \begin{matrix} q & p & 1 \\ 0 & q & 0 \\ 0 & p & 3 \end{matrix} \right |\)
etc.

-Dan

 

[evinda]

Expanding the determinant across the third row:
\(\displaystyle \left | \begin{matrix}q & p & 0 & 1\\ 0 & q & p & 0\\ p & 0 & 3 &0 \\ 0 & p & 0 & 3 \end{matrix} \right | = p \left | \begin{matrix} p & 0 & 1 \\ q & p & 0 \\ p & 0 & 3 \end{matrix} \right | + 3 \left | \begin{matrix} q & p & 1 \\ 0 & q & 0 \\ 0 & p & 3 \end{matrix} \right |\)
etc.

-Dan

Calculating it, I found $9q^2+2p^3$. So, have I done something wrong? (Worried)

Because then it would be $D(g(X))=-9q^2-2p^3$ that does not stand... (Shake)

 

[topsquark]

Calculating it, I found $9q^2+2p^3$. So, have I done something wrong? (Worried)

Because then it would be $D(g(X))=-9q^2-2p^3$ that does not stand... (Shake)

(shrugs) I can't help with the rest of the problem, but you got the determinant of the given matrix correct.

-Dan

 

[I like Serena]

Hi! (Smile)

I want to find the discriminant of $g(x)=X^3+pX+q$.

$$Res(g,g')=(-1)^{\frac{3(3-1)}{2}} D(g(X)) \Rightarrow Res(g,g')=-D(g(X))$$

$$Res(g,g')=det\begin{bmatrix}
q & p & 0 & 1\\
0 & q & p & 0\\
p & 0 & 3 &0 \\
0 & p & 0 & 3
\end{bmatrix}$$

How can we find the above determinant?

Calculating it, I found $9q^2+2p^3$. So, have I done something wrong? (Worried)

Because then it would be $D(g(X))=-9q^2-2p^3$ that does not stand... (Shake)

Hey! (Wave)

I'm confused about what you want to find.

The discriminant of a cubic polynomial? (Wondering)
That should be something like:
$$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$$

What does $Res$ represent? (Wondering)
If it were a $Residue$, I'd expect a function and a point as parameters.
What does $D(g(X))$ represent? I'd expect a derivative, but apparently you're making it a determinant.

 

[Fallen Angel]

Hi,

I suppose $D(g(x))$ is the discriminant and $Res(g,g')$ is the resultant of this two polynomials, given by the determinant of the Sylvester matrix, but this matrix you have is not $Sylv(g,g')$. Sylvester matrix of two polynomials is always a square matrix of order the sum of the degrees of the polynomials.

In this case, you should have a $5\times 5$ matrix