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$T_n(x)=\frac{1}{2^n}[(x+\sqrt{1-x^2})^n+(x-\sqrt{1-x^2})^n]

Source: 3-rd Taiwanese Mathematical Olympiad 1994

January 14, 2007
algebrapolynomialalgebra proposed

Problem Statement

For −1≤x≤1-1\leq x\leq 1 and n∈Nn\in\mathbb N define Tn(x)=12n[(x+1−x2)n+(x−1−x2)n]T_{n}(x)=\frac{1}{2^{n}}[(x+\sqrt{1-x^{2}})^{n}+(x-\sqrt{1-x^{2}})^{n}]. a)Prove that TnT_{n} is a monic polynomial of degree nn in xx and that the maximum value of ∣Tn(x)∣|T_{n}(x)| is 12n−1\frac{1}{2^{n-1}}. b)Suppose that p(x)=xn+an−1xn−1+...+a1x+a0∈R[x]p(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\in\mathbb{R}[x] is a monic polynomial of degree nn such that p(x)>−12n−1p(x)>-\frac{1}{2^{n-1}} forall xx, −1≤x≤1-1\leq x\leq 1. Prove that there exists x0x_{0}, −1≤x0≤1-1\leq x_{0}\leq 1 such that p(x0)≥12n−1p(x_{0})\geq\frac{1}{2^{n-1}}.
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