Question

Problem 3: The following data is given | * | 12. 23.4 4.8 67 | 33. 245 2.8 a) Write the Lagrange polynomial that passes through the points; then use it calculate the interpolated value of y at x = 5.4. b) Write the polynomial in Newton's form that passes through the points; then use it calculate the interpolated value of y at x = 5.4.

Answer 1

3 solution 8 Given that From data the given problem, we get of the problem is the following the data. Y 2 | 2181 B . 13121 15 Lagrange Rolynomial ;- The value of a dare. xo=1 X =212 22 = 34 y = 4.8 X4 = 6 | Yo = 1 /4 = 218/42= 3 4 3 = 3:2 Yu=4 *5=7 45=5 l Now, (ex t x;) (x->a) (x-*y) (- xy) (xa As) 66-4) (%-*) (6-3) (%6-] [46-X5) Yo

© 4= (x7)(x-4) (x – Vg) (x=44] (x-x3] (Homo) (x-2) (Monty) (xo-tu) (26-25) = 60-80) (X-X) (4-13) (x-4) (0-05) (34) (-) (913-83] [84,- dy) (x2-45] too (x-²) (X-X) (x-12) Cx-tu) (x-xs) (*) (*-*) (*-*) (*-*) (-4) 35 ve (X-Xo) (x+ X6) (4=] [) (X-85) 43 [thumno) (heart) (421(1934)(x385) (-x) (1-x)(-12) (sada ] [4_xy) YJE (ay-xo) (15²) (xgmtz) (253) (alsty Now, Add all y' values y = y + y + y2 + ₃ +Yu+Y5

NOW, Put= soy substitute 'x' value in the Lyd equation. - X2 X 2.8 - X3 (5.4) - 2.2) (5.4–3.4 (5.4-4.8) (S.4-6) 5:42) (1-22) (1.3-4) (1.408) (116) (1.7) (5.4-1) (5:4–3.4) (5.4–48) (514-6) (5.4-7). (2.24) (2.2-3-4) (2.2 -48) (2.2-6) (2.2-7) +(514-) (5.4 -2.2) (Soy-48) 05:4-6) (5.4-7). (3-4-1) (3-4-2-2) (364=407) (34-6) (3.4-7 t (S.4-1) (5.4-2.8) (5.4-3-4[5:4-6)(5-4-7) (484) (4:8-213) (418 – 34) (4.9-6) (4.8 - +) * +(5:47) (5.4-22) (5.4-34) (5.4–4.8) (5:4-7) (6-1) (6-202)16–3.4) (6-4.8) (6-7) (504-1). (5.4-2.2) (5.4-34) (5.4 -4.8) (5:4-6) (711) (2-2.2) (7+3+4) (7-4.8) (7.6) 43:2 X4

y= (-01123X2) + ((0.074 224] X2184(0. suara) + 601740312x 32) +( OU5603244) +[Co. Dei vu) X5) --0.02246 +0.207827+ €0.644695+(2.36899) + (1182413) +(-0:22) y = 3.511589. Newton's polynomial method o or hoe T S I home-to = 2.2-1 =1.2 cleanlyn * 05.23164 | S, - L 1 h 112

The draw a table below. 0.81 con -0.6 Co2 2 2.2 2.8 -0 -0.2 2 0.6 -0.4 3.4 4.8 312 10. 80 2 The polynomial from is newton's given by to y =yotu. Ay+ U, (4-1) A²yot 2! in ucu-1) (4-2) 3. At 3.1

= 27 3.67 (0.8) - 3.67[3.67-1) 606)+3167 $ 3.67 8[3.67-1) (3•67-3) 367[3.670) (3-67-3) (3:17-3 6 SAI - ou- XO 2012 + 3.67(3-67-) (3-62-2) (3.67-3) (3,67-4) 120 = 1.996+(-0.1636) + (-0.63x10-3) 1Y = 0-3535)