grandes-ecoles 2010 QII.B.1

grandes-ecoles · France · centrale-maths1__pc Proof Direct Proof of an Inequality

For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order.
Show that, for all $x \in [x_{n,1}, x_{n,n}]$, we have $$\sqrt{1 - x^2} \geq \frac{1}{n}$$

For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order.

Show that, for all $x \in [x_{n,1}, x_{n,n}]$, we have
$$\sqrt{1 - x^2} \geq \frac{1}{n}$$

Paper Questions

QI.A.1 QI.A.2 QI.A.3 QI.A.4 QI.A.5 QI.B.1 QI.B.2 QI.B.3 QI.C.1 QI.C.2 QI.C.3 QI.C.4 QII.A.1 QII.A.2 QII.A.3 QII.A.4 QII.A.5 QII.B.1 QII.B.2 QII.B.3 QII.C QIII.A.1 QIII.A.2 QIII.B.1 QIII.B.2 QIII.B.3 QIII.C.1 QIII.C.2 QIII.C.3 QIII.D.1 QIII.D.2