1- Using Newton's backward difference formula, construct an interpolating polynomial of degree 3 for the data: f (- 0.75) - - 0.0718125, f (- 0.5) -- 0.02475, f (- 0.25) = 0.3349375, f (0) = 1.10100. Hence find f (- 1/3); Correct your results to five decimal places.

1- Using Newton's backward difference formula, construct an interpolating polynomial of degree 3 for the data: f (- 0.75) - - 0.0718125, f (- 0.5) -- 0.02475, f (- 0.25) = 0.3349375, f (0) = 1.10100. Hence find f (- 1/3); Correct your results to five decimal places.

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

1- Using Newton's backward difference formula, construct an interpolating polynomial of
degree 3 for the data: f (- 0.75) = - 0.0718125, f (- 0.5) = - 0.02475, f (- 0.25) =
0.3349375, f (0) = 1.10100. Hence findf(- 13); Correct your results to five decimal
places.

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