grandes-ecoles 2023 Q20
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For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n } : \mathbb { R } \rightarrow \mathbb { R }$ is defined by $$\forall x \in ] - \infty , - \sqrt { n } - \frac { 1 } { \sqrt { n } } [ , \quad B _ { n } ( x ) = 0$$ $$\forall k \in \llbracket 0 , n \rrbracket , \forall x \in \left[ x _ { n , k } - \frac { 1 } { \sqrt { n } } , x _ { n , k } + \frac { 1 } { \sqrt { n } } [ , \right. \quad B _ { n } ( x ) = \frac { \sqrt { n } } { 2 } \binom { n } { k } \frac { 1 } { 2 ^ { n } }$$ $$\forall x \in \left[ \sqrt { n } + \frac { 1 } { \sqrt { n } } , + \infty [ , \right. \quad B _ { n } ( x ) = 0$$ and $\Delta _ { n } = \sup _ { x \in \mathbb { R } } \left| B _ { n } ( x ) - \varphi ( x ) \right|$ where $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.
Show that, for all $n \in \mathbb { N } ^ { * }$, we have the equality $$\Delta _ { n } = \sup _ { x \geqslant 0 } \left| B _ { n } ( x ) - \varphi ( x ) \right| .$$