But q is a nonzero polynomial of degree 2 n 2 exactly Therefore x 1 and x n 2

But q is a nonzero polynomial of degree 2 n 2 exactly

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But q is a nonzero polynomial of degree 2 n + 2 exactly. Therefore, x 1 and x n +2 have to be simple zeros and so x 1 = - 1 and x n +1 = 1. Note that the polynomial p ( x ) = (1 - x 2 )[ e 0 n ( x )] 2 P 2 n +2 has the same zeros as q and so p = cq , for some constant c . Comparing the coefficient of the leading order term of p and q it follows that c = ( n + 1) 2 . Therefore, e n satisfies the ordinary differential equation (1 - x 2 )[ e 0 n ( x )] 2 = ( n + 1) 2 k e n k 2 - e 2 n ( x ) . (2.61) We know e 0 n P n and its n zeros are the interior points x 2 , . . . , x n +1 . There- fore, e 0 n cannot change sign in [ - 1 , x 2 ]. Suppose it is nonnegative for x [ - 1 , x 2 ] (we reach the same conclusion if we assume e 0 n ( x ) 0) then, taking square roots in (2.61) we get e 0 n ( x ) p k e n k 2 - e 2 n ( x ) = n + 1 1 - x 2 , for x [ - 1 , x 2 ] . (2.62) Using the trigonometric substitution x = cos θ , we can integrate to obtain e n ( x ) = k e n k cos[( n + 1) θ ] , (2.63) for x = cos θ [ - 1 , x 2 ] with 0 < θ π , where we have chosen the constant of integration to be zero so that e n (1) = k e n k . Recall that e n is a polynomial of degree n + 1 then so is cos[( n + 1) cos - 1 x ]. Since these two polynomials agree in [ - 1 , x 2 ], (2.63) must also hold for all x in [ - 1 , 1].
2.4. CHEBYSHEV POLYNOMIALS 33 Definition 5. The Chebyshev polynomial (of the first kind) of degree n , T n is defined by T n ( x ) = cos nθ, x = cos θ, 0 θ π. (2.64) Note that (2.64) only defines T n for x [ - 1 , 1]. However, once the coefficients of this polynomial are determined we can define it for any real (or complex) x . Using the trigonometry identity cos[( n + 1) θ ] + cos[( n - 1) θ ] = 2 cos cos θ, (2.65) we immediately get T n +1 (cos θ ) + T n - 1 (cos θ ) = 2 T n (cos θ ) · cos θ (2.66) and going back to the x variable we obtain the recursion formula T 0 ( x ) = 1 , T 1 ( x ) = x, T n +1 ( x ) = 2 xT n ( x ) - T n - 1 ( x ) , n 1 , (2.67) which makes it more evident the T n for n = 0 , 1 , . . . are indeed polynomials of exactly degree n . Let us generate a few of them. T 0 ( x ) = 1 , T 1 ( x ) = x, T 2 ( x ) = 2 x · x - 1 = 2 x 2 - 1 , T 3 ( x ) = 2 x · (2 x 2 - 1) - x = 4 x 3 - 3 x, T 4 ( x ) = 2 x (4 x 3 - 3 x ) - (2 x 2 - 1) = 8 x 4 - 8 x 2 + 1 T 5 ( x ) = 2 x (8 x 4 - 8 x 2 + 1) - (4 x 3 - 3 x ) = 16 x 5 - 20 x 3 + 5 x. (2.68) From these few Chebyshev polynomials, and from (2.67), we see that T n ( x ) = 2 n - 1 x n + lower order terms (2.69) and that T n is an even (odd) function of x if n is even (odd), i.e. T n ( - x ) = ( - 1) n T n ( x ) . (2.70)
34 CHAPTER 2. FUNCTION APPROXIMATION Going back to (2.63), since the leading order coefficient of e n is 1 and that of T n +1 is 2 n , it follows that k e n k = 2 - n . Therefore p * n ( x ) = x n +1 - 1 2 n T n +1 ( x ) (2.71) is the best uniform approximation of x n +1 in [ - 1 , 1] by polynomials of degree at most n . Equivalently, as noted in the beginning of this section, the monic polynomial of degree n with smallest infinity norm in [ - 1 , 1] is ˜ T n ( x ) = 1 2 n - 1 T n ( x ) . (2.72) Hence, for any other monic polynomial p of degree n max x [ - 1 , 1] | p ( x ) | > 1 2 n - 1 . (2.73) The zeros and extrema of T n are easy to find. Because T n ( x ) = cos and 0 θ π , the zeros occur when θ is an odd multiple of π/ 2. Therefore, ¯ x j = cos (2 j + 1) n π 2 j = 0 , . . . , n - 1 . (2.74) The extrema of T n (the points x where T n ( x ) = ± 1) correspond to = for j = 0 , 1 , . . . , n , that is x j = cos n , j = 0 , 1 , . . . , n. (2.75)

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