# A short course on approximation theory (Math682) by Carothers N.L. By Carothers N.L.

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Extra info for A short course on approximation theory (Math682)

Example text

Clearly, each Pn is a polynomial. px for 0 x 1. Use Dini's theorem to (b) Check that 0 Pn(x) Pn+1(x) px on 0 1 ]. conclude that Pn(x) (c) Pn(x2 ) is also a polynomial, and Pn(x2 ) jxj on ;1 1 ]. 22. The result in problem 19 (or 20) shows that the polynomials are dense in C 0 1 ]. Using the results in 18, conclude that the polynomials are also dense in C a b ]. 23. How do we know that there are non-polynomial elements in C 0 1 ]? In other words, is it possible that every element of C 0 1 ] agrees with some polynomial on 0 1 ]?

As it happens, Tn(x) has the largest possible rate of growth outside of ;1 1 ] among all polynomials of degree n. Speci cally: Theorem. Let p 2 Pn and let kpk = max jp(x)j. Then, for any x0 with jx0j 1 and any k = 0 1 : : : n we have ;1 x 1 jp(k)(x0 )j kpk jTn(k)(x0 )j where p(k) is the k-th derivative of p. We'll prove only the case k = 0. In other words, we'll check that jp(x0 )j kpk jTn(x0 )j. 10, p. 31. Proof. Since all the zeros of Tn lie in (;1 1), we know that Tn (x0 ) 6= 0. Thus, we may consider the polynomial q(x) = Tp((xx0 )) Tn (x) ; p(x) 2 Pn: n 0 If the claim is false, then kpk < Tp((xx0 )) : n 0 Best Approximation 61 Now at each of the points yk = cos(k =n), k = 0 1 : : : n, we have Tn(yk ) = (;1)k and, hence, q(yk ) = (;1)k Tp((xx0 )) ; p(yk ): n 0 Since jp(yk )j kpk, it follows that q alternates in sign at these n + 1 points.

Bohman, Korovkin) Let (Tn) be a sequence of monotone linear operators on C 0 1 ] that is, each Tn is a linear map from C 0 1 ] into itself satisfying Tn(f ) Tn (g) whenever f g. Suppose also that Tn(f0 ) f0, Tn(f1 ) f1 , and Tn(f2 ) f2 . Prove that Tn(f ) f for every f 2 C 0 1 ]. ] 34. Find Bn(f ) for f (x) = x3 . ] The same method of calculation can be used to show that Bn(f ) 2 Pm whenever f 2 Pm and n > m. Polynomials 37 35. Let f be continuously di erentiable on a b ], and let " > 0. Show that there is a polynomial p such that kf ; pk < " and kf 0 ; p0 k < ".