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By Carothers N.L.

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If (∗) represents an even function, then it can be written using only cosines. Corollary. The collection T , consisting of all trig polynomials, is both a subspace and a subring of C 2π (that is, T is closed under both linear combinations and products). In other words, T is a subalgebra of C 2π . It’s not hard to see that the procedure we’ve described above can be reversed; that is, each algebraic polynomial in cos x and sin x can be written in the form (∗). For example, 4 cos3 x = 3 cos x + cos 3x.

Proof. Exercise. [Hint: If p(x) is a polynomial of degree n with leading coefficient 1, then p˜(x) = p((2x − b − a)/(b − a)) is a polynomial of degree n with leading coefficient 2n /(b − a)n . ] a≤x≤b −1≤x≤1 Properties of the Chebyshev Polynomials As we’ve seen, the Chebyshev polynomial Tn (x) is the (unique, real) polynomial of degree n (having leading coefficient 1 if n = 0, and 2n−1 if n ≥ 1) such that Tn (cos θ) = cos nθ Best Approximation 57 for all θ. The Chebyshev polynomials have dozens of interesting properties and satisfy all sorts of curious equations.

31. Proof. Since all the zeros of Tn lie in (−1, 1), we know that Tn (x0 ) = 0. Thus, we may consider the polynomial q(x) = p(x0 ) Tn (x) − p(x) ∈ Pn . Tn (x0 ) If the claim is false, then p < p(x0 ) . Tn (x0 ) 61 Best Approximation Now at each of the points yk = cos(kπ/n), k = 0, 1, . . , n, we have Tn (yk ) = (−1)k and, hence, q(yk ) = (−1)k p(x0 ) − p(yk ). Tn (x0 ) Since |p(yk )| ≤ p , it follows that q alternates in sign at these n + 1 points. In particular, q must have at least n zeros in (−1, 1).