I need the answer as soon as possible 20. Derive interpolating polynomial for log,01-0, log₁, 2=0.3010, log, 3=0.4771, and hencecalculate log₁0 (15) using Newton divided difference formula.Ans.-0.06245x² +0.48835x-0.42590;log₁0 (1.5) = 0.16611250log₁0 (15) = log₁0 (10x1.5)= log₁0 (10)xlog (1.5)=1+0.16611250=1.1661125021. Approximate the exponential value e¹.³ by Newton divided difference polynomial from the dataset e = 1, e¹=2.7183, e² = 7.3891. What is the estimate for the error bound?Ans. e¹3 = 3.8095, Error bound = 0.3362

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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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Derive interpolating polynomial for log₁0 1=0, log₁, 2=0.3010, log, 3=0.4771, and hence calculate log₁0 (15) using Newton divided difference formula. Ans.-0.06245x² +0.48835x-0.42590; logo (1.5)= 0.16611250 (1.5)=1+0.16611250=1.16611250 logo (15)= log₁0 (10x1.5)= log₁ (10)xlog