Interpolation and Approximation | California State University, Northridge

University:
California State University, Northridge
Course:
MATH 331 | Mathematical Explorations
Academic year:

2023
Views:

295

Pages:

20
Author:

Yung D

UNIT-I INTERPOLATION AND APPROXIMATION. LAGRANGIAN POLYNOMIALS O. Find the polynomial fex) by using Lagrange's formula and hence find Sch: By Lagrange's interpolation formerla, we have =(2-1) (2-2)(x-5) (2) + (0-1) (0-2) (0-5) (2-0) (2-2) (2-5) (3) + (2-(2-1)(x-5) ( (1-0) (1-2) (1-5) (12) -(2-0)(2-1) (2-5) + (2-0) (2-1)(x-2) (147) (5-0) (5-1) (5-2) (2-1) (-2)(x-5) (-10) (2) + ((x-2)(x-5) 4 (3)+(-) (2-5) (12) -6 = 6(3) = (3-1) (3-2) (3-5) -10 (2) + + 3(3-1) (3-2) (147) 60 3(3-2) (3-5) (3)+ 4 3(3-1) (3-5) (12) -6 (2-707(2) = 6 (2) + (-10) (3) ()(-2) (3) + 3 (2) (-2) 4 (-6) (12) + (3) (2) (1) (147) 60 =(2)-(3)+2(12)+(147) 18 =-1+24+. 24+147 35 Find the third degree polynomial dex) satisfying the following data. 34 120 336 720 y 24 The Lagrange's interpolation formula is y=f(x)= (2-4) (0) (0) (20~13) (210)(2-2)(x-3) +(-10) (3) (3-22) (x-3)(2-7) (336) (5-1) (5-3) (5-7) (7) (15) (17) (24) + (-) (35) (27) (120) + (x-5) (720) (7-1) (7-3) (7-5) [152²+712-105]+[x-132²+47x-35]-21 [2ー112²+312-21] =+15[x-9x²+23x-15] =[-/+-21+1]+[- +[-52+641-225] 3 2 = x²+6x²+11x+6 +231-135]2²++5-605+345] 614) = 4+6(42) + (4) +6 64+96+44+6 = 210 using Lagrange's interpolation formula find f(4) given that +6)=2, 6(1)=3, f(2)= 12, f(15)=3 Güven By Lagrango's formula Y= f (4) = (4-1) (4-2) (4-15) (2)+ (4-0) (4-2) (4~15) (3) (0-1) (0-2) (0-15) (1-0) (1-2)(1-15) (2-0) (2-1) (2-15) +(4-0) (4-1) (4-15) (12) + (4-0) (4-1) (4-2) 3587 (15-0) (15-1) (15-2) (3) (2) (-11) (2) + (-1) (-2) (-15) (4) (2) (-11) (3) + (4) (3) (-11) (12) (1)(-1)[-14) (2)(1) (-19) (3587) + (4) (3) (2) (15)84)(13) Find the missing tam in the following table using Lagrange's Interpolation. By Lagrange's interpolation formula