Entrance examination Mines-Ponts MP 25: mathématiques 2.
I found the problem well done, the progression well thought out and the results interesting.
It all starts with the following definition: for \(p\in\mathbb{R}[X]\), a polynomial split over \(\mathbb{R}\), \[ \begin{alignat*}{1} p&=\sum_{i=0}^n a_iX^i\ \end{alignat*} \] we define the polynomial \(p_0\) by reversing the order of the coefficients as follows: \[ \begin{alignat*}{1} p_0&=\sum_{i=0}^n a_{n-i}X^i\ \end{alignat*} \]
We introduce the now well-known matrix \(S\)
We define the symmetrical matrix: \[ \begin{alignat*}{1} J(p)&= p_0(S)^Tp_0(S) - p(S)^Tp(S) \end{alignat*} \]
At the heart of the problem lies the notion of stable roots: \[ \begin{alignat*}{1} \alpha\text{ is a stable root of }p&\Leftrightarrow \alpha\neq0\land \alpha\text{ is a root of }p\land \frac{1}{\alpha}\text{ is a root of }p. \end{alignat*} \]
Finally, we define \(\pi(M)\) the number of positive real eigenvalues of \(M\) counted with their multiplicity, and \(\sigma(p)\) the number of real roots of \(p\) belonging to the interval \(]-1; 1[\), counted with their multiplicity.
The first remarkable result is obtained in question 18 ( Schur-Cohn criterion ): \[ \begin{alignat*}{1} J(p)\in GL_n(\mathbb{R})&\Leftrightarrow p\text{ does not have any stable roots and }\sigma(p)=\pi(J(p)) \end{alignat*} \]
Then, in the case where all the roots of \(p\) are stable and of multiplicity 1, we obtain \[ \begin{alignat*}{1} \sigma(p)&= n-1-\pi(J(p')) \end{alignat*} \]
Finally, in the last section, we construct a method for counting the roots of \(p\) in \(]-1,1[\) in the general case.
Answers to the Maths 2 MP test on April 24th in entrance examination Mines-Ponts 2025:
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