Question

Problem 2 Let f(x) = sin 2x and P() be the interpolation polynomial off with degree n at 20,***, Im Show that \,f(z) – P() Sin+1 – 20) (1 - 11). (I – In).

Answer 1

tnt1 +1)! 2 you can see any booke of numerical analysts and see that the errors terms of . Lagromgian interpolation with nginen points De 9 (fia) – Pola)) = CET (A-2)(x-»1) arak-an) where gore mincs) & <max(s) 392003---3 any Then put fex)=sinax 30 that fomu – pmix) = 29 sierpnia 2nt sin (et) + 28) (0-20) -- 11:24) ( +1) Taking modulus on both sides wehave X il fre) – Primal 3 207 11990) -- (**») | Ais one yorith can see my hand made more. CamScanner

pat (5) where MINU rrit Q18. Show that the remainder in approximating $ (.x) by the interpolating polynomial using distinct interpolating points Po, di,ata. I, is of the form /(x-xo)(x– x;)...(x– x, )(2+1)! Whi min {x, Xoy-1...., -,} <3 < max {1; Ag, d.;..., ;}. VH2010'2007 ☺. Let f (x) be a function which is continuously differentiable upto any desired number of times and $(x) be the interpolating polynomial of degree n which replaces f(x) on the set of (n+1) interpolating points x; (j = 0,1, 2,..., 1) i.e., $(x)) = $(x;), j = 0,1,2,..., 1 ... ... (1) and R,+1(x) is the remainder or error in approximating . f ( x) by 0(2). Scanned with CamScanner Or error 10

Numerical Analysis2 $(x)= $(x)+Rouy ()] or, Rou(x) = f (x)-°(x) \ -* ... (2) Thus by (1) and (2) We have Roxy(x) = 0, j = 0,1,2,..., . : . (3) av virtue of (3) we may assume R.: (x) = w()(x) ... ... (4) where w(x)= (-– xo)(x – *;)(x – x3 ...(x - 2,) ... ... (5) is a polynomial of degree (n+1) and y (:) is any function of x. wewe want to find the remainder at a particular x= , which is anecm x, (j=0,1,2,...,n). Thus, using (4) in (2) we have s(a)= v(a)+wla)y(a)... ... (6) let us consider an auxiliary function F(x) given by F(x) = f (x)-0(x) – w()y (a)... ... (7) Thus F(a)=0, by (6), F(x)=0, j = 0,1,2,...,17., by (2), (5) and (7) Thus, F(x) vanish at (n+ 2) points 6,1, (j = 0,1,2,...,.1) and F (x) is continuously differentiable up to any desired numbers of time, as f (x) is continuously differentiable up to any desired numbers of time. Therefore, by the repeated application of Rolle's theorem, we have F** (5) = 0 ...(8) where min{a, 1, ..., x}< < (11x4, X, Y,.., x, 7 Since $(x) is a polynomial of degree n and w(-1) is a polynomial of degree (1+1), we have $**' (x) = 0) and (o""(x= (1+1)! Now, differentiating (7) (12+1) times we have, F*** (x) = f*** (x)– (1+1)!y (a) Therefore from (8) we have, f'"'' (5)– (12+1)!: (a) = 0 , i.c., v (a) =%. (n+1)! (12-4-1)! Now from (4) we get, R., (a) = w(a)v (a)= w(a). Replacing a by x, we get R...(x)= w(s), 19) = (xr-x)(x = 1, ...(r = x) Lost where min{x, xo, d.,..,.x,} << < max {x,.', d.....\,}. Q19. If f**'(x) does not vary strongly in the region of interpolation and if h is also very small, then prove that h"'f" (5) - A"'y, where xo,;,,,.....I, are Sinterpolating points, ſ (ko) = yo, h is the step length. CamScanner ES

n In+1 inc you can see any booke of numerical analysis and see that the errors terms of Lagramgian interpolation with ngineen Points to main (41a) – Pora)) = CECO-w) (RmxW) and (aman) where one mincs) & <max(s) S - 9300,---, any Then put tex)=sinax 30 that fa) - Pm (2) = 28thesion tons : 2n+! Bi (671) + 2€ ) (-).- 1a-] (ott) paring modulus on both sides wehave l fron) – Pokrog 2"11020) --- (4-201)|| mti CS Scanned with CamScanner