grandes-ecoles 2022 Q14

grandes-ecoles · France · centrale-maths2__pc Proof Direct Proof of an Inequality
Let $n \in \mathbb{N}^*$, $W$ be a monic polynomial of degree $n$, $Q = \frac { 1 } { 2 ^ { n - 1 } } T _ { n } - W$, and for all $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. We now assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | = \frac { 1 } { 2 ^ { n - 1 } }$. Show that, for all $k \in \llbracket 0 , n \rrbracket$, $$\frac { Q \left( z _ { k } \right) } { \prod _ { \substack { j = 0 \\ j \neq k } } ^ { n } \left( z _ { k } - z _ { j } \right) } \geqslant 0.$$
Let $n \in \mathbb{N}^*$, $W$ be a monic polynomial of degree $n$, $Q = \frac { 1 } { 2 ^ { n - 1 } } T _ { n } - W$, and for all $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. We now assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | = \frac { 1 } { 2 ^ { n - 1 } }$. Show that, for all $k \in \llbracket 0 , n \rrbracket$,
$$\frac { Q \left( z _ { k } \right) } { \prod _ { \substack { j = 0 \\ j \neq k } } ^ { n } \left( z _ { k } - z _ { j } \right) } \geqslant 0.$$