Question

class: numerical analysis

= COS Problem 1: Recall that the Chebyshev nodes x4, x1,...,xy are determined on the interval (-1,1] as the zeros of Tn+1(x) = cos((n + 1) arccos(x)) and are given by 2j +10 Xj j = 0,1, ... 1 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 x x 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 y E (-5,5] Ifn+1(c) If (y) – Pn(y)= |(y – y*)(y – yi)... (y – ym) (n + 1)! for some ce(-5,5). Part 1: Show that 5n+1 (y – y*)(y – yi)... (y – Y*) = -Tn+1(x), where x = 2n y/5 Part II: It can be shown that there exists R > 0 such that \f(n)(y)] < RN for all YE(-5,5). Assuming this, show that lim max{\f(y) - P(y), y € (-5,5)} = 0 no

I wish if it was written in block letter

Sorry I can't read cursive

Answer 1

Answer:- Given that The interval [1,1] as the zeros of Tnt, (x) = cos(n+1) arcos () (2jt IT j=0,i,...n x; COS ht.12 f(x): On the interval t = 5,5 * (its?) Y = 5xx; and let Pn (y) denote the polynomial at degree m that interpolates f at these polats. Then the error estimate for polynomial interpolation gives for ye (-5,5) If() – Pm (y)] = 18 ht.cc | Cy-yt ](y-y#) ---(y-yst) (n+1)! for some ce [-5,5]. (y-y*)(y-y*) (y- und f * * * = 5*,y, * =52, 5 Yn=5am y=50 (52-544) (58-5*) (5x -5**) (y-yo)(y-9")... (y-yht) = 51+!De-xo) (x -x;')... (x-x) -> To (x)=1, T,(x) = x Tn (x) = cos(ncosia int, (C) = 2x To(x) - Tn-; (3) for 2 E[1,1]. Tnti () = (x - Xest) (1x_x*).--. (x-x) from :- for so T, (x) = (x-x*) AD= xao = x o=1 m=0,J=0 * Xo - cos cos G

m=1:- 24 T> (x) = cos (2cos () cos (06) - O coso - Cos(20) = 2 cos 0-1 2 2x = 1 T₂ (*) = 2X-1 from T2(x) = A (-4) (x-x*) 1 jal, m1 j=0, mal - cos COS ( COS a get = 605 () xt = cos ( 2 1 - 2 cos (3 s(TT- a-cost ta Ty(x) = A (x-x+ () i AL for m=2 :- from 73(x) = (*) *-**)(x-2)62) j=o, m=2 j=1, ma m2 K xo - cos 3 시 -EOS cos(1) cosa Loft J= 2, m=2 5 X2 COS 2

- cost 51 6 cos ( = cos(o-ho) COS 5 T3 (3) = cos (3005'(x)) Cos00)=0 COSOX COS (30) = y coso - 3cOSE 433-34 42?_3x = (2-3) (3-0) (x+3 24(2-3 Az A A2=4 so , proceeding thus we have an an Tnti (*) - [(x-xf)(x-**) ....Cox -**)]2. part ② 2 So, from A we have ས་ལ་ (y-yes) (y-y* )-----(y-yht)- n+1 5 Thai (0) 2n where na Ills part: It can be shown that there exists R>0 such that LAN (4) <R for all ye [-5,5 and we know, error estimate Entl (c) if(y)- PM Pnly)) licy-yit )ex=y)...-{y-yout) (n+1)!

Rati santi < 2n 1 FC9) - Pr.Cu)) == CORD! Rnti 5nti an < R9+1 5nti 2n Inti (x) (n+1)! [by part o and Ifa (U) Isen] Cos (1) cos() (nt)! (cos(n+1) C051(c) si (n+1)! moo maxf(y) – Pn(y) RO+I5nti mo (m+1)! 2n n+1 (n+1)! 2h it maelf (g)-Pnly) 1:0 (se)" for YE [-5,5 (ntli 少f It ntl R n 5 R. 5R ulo p p 11 5R it ph (n+1)! ent as (n+1)! is increasing very fast as n grows large. it hence no ph (n+1)!