Using Newton's Divided Difference, the polynomial that interpolates x + sinxat the points x, =1,x2 = 2,x3 = 3,x4 = 4 , and x5 = 5 can be written asP(x) = c, + c,(x – 1) + c2(x – 1)(x – 2) + c3 (x – 1)(x – 2)(x – 3)+C,(х — 1)(х — 2)(х — 3)(х — 4).What is the value of 2c, – c,?A1.71766B) 2.553662-0.8142851.485832

The objective is to find the value of 2c3-c2

Using Newton's Divided Difference, the polynomial that interpolates x + sinx at the points x, = 1,x2 = 2,x3 = 3,x4 = 4 , and x5 = 5 can be written as Р(x) 3 Со + с, (х — 1) + с,(х — 1)(х — 2) + с;(х — 1)(х — 2)(х — 3) +c,(х — 1)(х — 2)(х — 3)(х — 4).

Using Newton's Divided Difference, the polynomial that interpolates x + sinx at the points x, = 1,x2 = 2,x3 = 3,x4 = 4 , and x5 = 5 can be written as Р(x) 3 Со + с, (х — 1) + с,(х — 1)(х — 2) + с;(х — 1)(х — 2)(х — 3) +c,(х — 1)(х — 2)(х — 3)(х — 4).

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Using Newton's Divided Difference, the polynomial that interpolates x + sinx
at the points x, =1,x2 = 2,x3 = 3,x4 = 4 , and x5 = 5 can be written as
P(x) = c, + c,(x – 1) + c2(x – 1)(x – 2) + c3 (x – 1)(x – 2)(x – 3)
+C,(х — 1)(х — 2)(х — 3)(х — 4).
What is the value of 2c, – c,?
A
1.71766
B) 2.553662
-0.814285
1.485832

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