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 assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | = \frac { 1 } { 2 ^ { n - 1 } }$. Deduce that $Q = 0$, then that $W = \frac { 1 } { 2 ^ { n - 1 } } T _ { n }$. One may consider the sum of the inequalities from the previous question and exploit question 6 applied to suitable data.
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 assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | = \frac { 1 } { 2 ^ { n - 1 } }$. Deduce that $Q = 0$, then that $W = \frac { 1 } { 2 ^ { n - 1 } } T _ { n }$.
One may consider the sum of the inequalities from the previous question and exploit question 6 applied to suitable data.