MathDB

1

Part of 1972 Vietnam National Olympiad

Problems(1)

T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x), polynomial, cosn\alpha=T_n(x)

Source: Vietnamese MO (VMO) 1972

8/24/2018
Let α\alpha be an arbitrary angle and let x=cosα,y=cosnαx = cos\alpha, y = cosn\alpha (n∈Zn \in Z). i) Prove that to each value x∈[−1,1]x \in [-1, 1] corresponds one and only one value of yy. Thus we can write yy as a function of x,y=Tn(x)x, y = T_n(x). Compute T1(x),T2(x)T_1(x), T_2(x) and prove that Tn+1(x)=2xTn(x)−Tn−1(x)T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x). From this it follows that Tn(x)T_n(x) is a polynomial of degree nn. ii) Prove that the polynomial Tn(xT_n(x) has nn distinct roots in [−1,1][-1, 1].
algebrapolynomialtrigonometryfunctional equation