Calculate the discriminant of $f(x)= x^k + cx + t$

Question

Calculate the discriminant of $$f(x)= x^k + cx + t .$$

I tried with the definition of the discriminant, $$D(f) = a_0^{2n - 2} \prod_{1 \le i \lt j \le n} (a_i - a_j)^2$$ but I get $0$. Can you help me?

Answer 1

The given definition seems to confuse two expressions for the discriminant: The factors of the product should be $(x_i - x_j)^2$, where the $x_i$ vary over the roots of the polynomial. Since the roots of $f$ are hard to describe for general $k, c, t$, this formula won't be of much help here.

Instead, use that $$D(f) = \frac{(-1)^{n (n - 1) / 2}}{a_n} \textrm{Res}(f, f') .$$ Here, $\textrm{Res}(f, g)$ is the resultant of $f, g$, which can be written as the determinant of the corresponding Sylvester matrix, who nonzero entries are the suitably arranged coefficients of $f, g$ suitably arranged. In our case, $f'(x) = k x^{k - 1} + c$, and $$\textrm{Res}(f, f') = \det \left(\begin{array}{ccccc|ccc} 1 & 0 & \cdots & 0 & 0 & c & t & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0 & 0 & c & t & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 & 0 & 0 & 0 & \cdots & t & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 & 0 & 0 & \cdots & c & t \\ \hline k & 0 & \cdots & 0 & 0 & c & 0 & 0 & \cdots & 0 & 0 \\ 0 & k & \cdots & 0 & 0 & 0 & c & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & k & 0 & 0 & 0 & \cdots & c & 0 \\ 0 & 0 & \cdots & 0 & 0 & k & 0 & 0 & \cdots & 0 & c \end{array} \right) . $$ The matrix has size $(2 k - 1) \times (2 k - 1)$, and the block sizes are $k - 1, k$. I've marked horizontal and vertical lines to suggest a block decomposition of this matrix for which it's relatively easy to compute the determinant using the usual formula for the determinant of a block matrix.

Doing so gives (for $k > 1$) that $$\color{#bf0000}{\boxed{D(f) = (-1)^{(k - 1) (k - 2) / 2} (k - 1)^{k - 1} c^k + (-1)^{k (k - 1) / 2} k^k t^{k - 1}}} .$$

Answer 2

I would like to show that one can obtain this discriminant (up to a factor constant factor, see Remark 2 below) in a simple, heuristic, way. Here is how.

Let us recall that the discriminant of a polynomial $f$ involving parameters (here $c$ and $t$) is a polynomial expression in these parameters which is zero if and only if $f$ has a multiple root.

But (see figure below for the case $k=4$) :

$$x_0 \ \text{is a multiple root of} \ f(x)= x^k + cx + t \ \ \ \iff \ \ \text{curves of} \ \begin{cases}y&=&x^{k}\\y&=&-cx-t \end{cases} \ \text{have a double point}$$

Said otherwise, multiple roots occur as double roots each time the straight line $L$ with equation $y=-cx-t$ is tangent to the curve $(C_k)$ with equation $y=x^k$. In fact no other cases exist as we will check later on.

As the equation of the tangent to curve $(C_k)$ at point $P$ with abscissa $x_0$ is

$$\tag{1}y-x_0^k=kx_0^{k-1}(x-x_0) \ \ \ \ \iff \ \ \ \ y=kx_0^{k-1}x+(1-k)x_0^k,$$

it suffices now to identify the coefficients of (1) with those of $y=-cx-t$, giving

$$\begin{cases}-c&=&kx_0^{k-1}\\-t&=&(1-k)x_0^k \end{cases}\ \ \ \ \iff \ \ \ \ \begin{cases}x_0^{k-1}&=&-\dfrac{c}{k}& \ \ \ (a)\\x_0^k&=&\dfrac{t}{k-1} & \ \ \ (b) \end{cases} $$

Raising (a) to the power $k$ and (b) to the power $k-1$, and equating the RHSides, one gets:

$$(-1)^k\dfrac{c^k}{k^k}=\dfrac{t^{k-1}}{(k-1)^{k-1}},$$

which is equivalent to the condition "discriminant $ \ = \ 0$", up to a constant factor.

Remarks: 1) One could object that the case of multiple complex roots has not been considered. But the polynomial we have obtained has exactly the right degree for being eligible as the discriminant: there is no more place for supplementary factors! Moreover, remember that I propose here a heuristic method... :). In fact, one could use the irreducibility of the polynomial, but I had not in mind in writing this anwer to be perfectly rigorous.

2) The absence of constant factor in a discriminant is unimportant in applications because it is the fact that the discriminant is zero that is useful.