Question

Problem 1: Recall that the Chebyshev nodes 20, 21, ...,.are determined on the interval (-1,1) as the zeros of Tn+1(x) cos((n + 1) arccos(x)) and are given by 2; +17 Tj = COS , j = 0,1,...n. n+1 2 Consider now interpolating the function f(x) = 1/(1 + x2) on the interval (-5,5). We have seen in lecture that if equispaced nodes are used, the error grows unbound- edly as more points are used. The purpose of this problem is to show that if the Chebyshev nodes are used, this problem disapears. First denote by y; = 5 xx; and let Pn (y) denote the polynomial of degree n that interpolates f at these points. Then the error estimate for polynomial interpolation gives for ye 1-5,5) if(y) – Pn(y)] | fm+1 (0)||(4 – 97)(y – yt)... (y – ym) (n + 1)! for some ce [-5,5). Part 1: Show that 5n+1 (y – y*)(y - y1)... (y – ym) -Tn+1(2), where x = y/5 2n Part II: It can be shown that there exists R > 0 such that \f(n)(y)] < RN for all y E1-5,5). Assuming this, show that lim max{\f(y) - Pn(y)], y € (-5,5)} = 0 n>00

Answer 1

We have x n+ hoots or where, 2), joo, lyg zeros of ( (*) Inti(x) Inti(x) = cos((n + 1) cos(x)). the interval [-5,5 1 It x² and y 5 **; C. Pm ly) is the polynomial that interpolates f at these points, such that the shhor estimate is I cay - yot) ... (y yo If.ly) - Pm ly) ) = 1 + and some CE [-5,5 for де E [-5,5 1 0.18.g.) (9.8,") ... (y-yo) 5nti جا (where ula

* xi 1 2 V2 Note that, To(x) = 1 Tn (x) T> (x) = 2x=4 (x*- and in general, Inti (x) = 2n (x-xo) (x-x*) ... (x-xh (in whiting Tata), I use the fact that cos (20) = 2 cosao 1) eam. 0, could white : (y-go) ... (y- y) = 5n+ Inti (x) 16 From ul an (I have substituted for the product of (-27) in terms of Thr, (x)).

Now, 7 R > 0 11 such that, I f! (y) , < Rn e Rm for all ay € (-5,5] (me) R1+1 5nti :. || fly) - Pn (y) | (n+1)! an lintila) * The (where I have used relation in equ. and equ. ① from Part ③). abone inequality holds true for € (-5,5] By definition of Intl (x) (x = [1,1]) that, I Inti (x) 1 sl & MEIN all y € he see ole, n=0. 1 Thus, If (y) - Pn (y) | ₂ 2 2 ( 52 )** (n+1)!

let 5R > 0 By 2 Archimedean such that property I NE IN N> a Now, a a sa a a ok n! 11 no and as @) Therefore, from the squeeze theorem m+1 of limits, lim 2 (n+1)! Hence, lim max nax 31419) - Pm (9) || €15,5] noo