# Polynomial maximizing its discriminant

Consider a polynomial $$p(x) = (x-x_0)(x-x_1)\dots (x-x_{n-1})(x-x_n)$$, with $$-1 = x_0 < x_1 < \dots < x_{n-1} < x_n = 1$$. What values for the roots $$x_1, \dots x_{n-1}$$ maximize the discriminant of $$p$$ or, equivalently, the absolute value of the Vandermonde determinant of the roots $$x_i$$ of $$p$$, i.e. $$|\det V(x_0,\dots, x_n)| = |\prod_{0 \le i < j \le n} (x_i - x_j)|$$?

It is clear that roots should be chosen symmetric to zero, so for $$n$$ even we have $$x_{n/2} = 0$$. For $$n = 3,4$$ this leads to polynomial optimization in one variable, which can be solved analytically, yielding $$n = 3 \rightarrow x_{1,2} = \mp \sqrt{\frac{1}{5}},\quad n = 4 \rightarrow x_{1,3} = \mp \sqrt{\frac{3}{7}}.$$

Beyond that the results of numerical optimization based on gradient descent revealed nothing particularly noteworthy.

• And I also got $x_{1,2,3,4} = \pm\sqrt{\frac{7\pm 2\sqrt{7}}{21}}$ for $n=5.$ Commented May 19 at 22:06
• I think I have found a proof that $n(n+1)p(x)=p''(x)(x^2-1),$ I still have to work out the details. This would give us a simple method to obtain $p$ by equating coefficients. The roots would probably have to be determined numerically in most cases. Commented May 19 at 23:46

For $$i\in\{1,\ldots,n-1\}$$, the partial derivatives of $$V(x_0,\ldots,x_n)$$ with respect to $$x_i$$ are the following $$\frac{\partial }{\partial x_i} V(x_0,\ldots,x_n) =\pm \prod_{\substack{j\neq i \\ k\neq i \\ j The first product is not $$0$$ and it does not depend on $$x_i.$$ Therefore, the necessary condition for getting a maximum of $$V$$ is $$\sum_{\substack{k=0 \\ k\neq i}}^{n} \prod_{\substack{j=0 \\ j\neq i \\ j\neq k}}^n (x_i-x_j) =0 \;\;\; \forall\; i\in\{1,\ldots,n-1\}$$ Now we take a look at the second derivative of $$p(x)$$ with respect to $$x.$$ We get $$p'(x) = \sum_{k=0}^{n}\prod_{\substack{j=0 \\ j\neq k}}^n (x-x_j)$$ and $$p''(x) = \sum_{k_1=0}^{n}\sum_{\substack{k_2=0 \\ k_2\neq k_1}}^{n} \prod_{\substack{j=0 \\ j\neq k_1 \\ j\neq k_2 }}^n (x-x_j)$$ Now we check what happens when we evaluate $$p''$$ at $$x_i.$$ The product in this expression avoids the factor $$(x_j-x_j)$$ only if $$i=k_1$$ or $$i=k_2.$$ Only one of $$i=k_1$$ and $$i=k_2$$ can be true at any given moment, because $$k_1\neq k_2.$$ We split the double sum into one part in which $$k_1\neq i = k_2$$ and one part in which $$k_1 = i \neq k_2$$ and we get the following $$p''(x_i) = \sum_{\substack{k_1=0 \\ k_1\neq i }}^{n} \prod_{\substack{j=0 \\ j\neq k_1 \\ j\neq i }}^n (x_i-x_j) + \sum_{\substack{k_2=0 \\ k_2\neq i }}^{n} \prod_{\substack{j=0 \\ j\neq i \\ j\neq k_2 }}^n (x_i-x_j) = 2 \sum_{\substack{k=0 \\ k\neq i }}^{n} \prod_{\substack{j=0 \\ j\neq i \\ j\neq k }}^n (x_i-x_j)$$ Setting this to $$0$$ is precisely the necessary condition for getting the maximum of $$V.$$ Therefore, the necessary condition can also be written as $$p''(x_i) = 0 \;\;\; \forall\; i\in\{1,\ldots,n-1\}$$ As the degree of $$p''$$ is two less than the degree of $$p,$$ we know that those are all roots of $$p''.$$ $$p''$$ cannot have any other roots. Therefore, $$p''$$ can be written as follows $$p''(x) = c(x-x_1)(x-x_2)\ldots (x-x_{n-1})$$ with a suitable constant $$c.$$ We also know this constant, because we know the degree of $$p.$$ As this degree is $$n+1$$, the constant $$c$$ must be $$n(n+1)$$. So we have $$\begin{eqnarray} n(n+1)p(x) & = & c(x-x_0)(x-x_1)\ldots (x-x_n) \\ & = & (x-x_0)(x-x_n)\;c(x-x_1)(x-x_2)\ldots (x-x_{n-1}) \\ & = & (x^2-1)p''(x) \end{eqnarray}$$ We can now write $$p$$ in its expanded form as $$p(x) = \sum_{k=0}^{n+1} a_kx^k$$ and the second derivative is $$p''(x) = \sum_{k=2}^{n+1} k(k-1)a_kx^{k-2}$$ The coefficients can easily be found by using $$n(n+1)p(x)=(x^2-1)p''(x)$$ and sorting by the powers of $$x.$$ Note that the result of equating the coefficients is unique only up to a constant factor, this can be fixed by setting $$a_{n+1}=1.$$
Once we have $$p$$, we can find $$x_1,\ldots x_{n-1}$$ using numerical methods. I am not aware of an analytical approach to get the roots in general.
However, the maximum value of $$V$$ can be determined by using the resultant of $$p$$ and $$p'$$ without knowing the roots.
• Very nice, thank you! If I see this right, this characterization leads to the Gegenbauer polynomials for $\alpha = -\frac{1}{2}$. Commented May 21 at 15:20