а. i. Find the second Taylor polynomial PĄ(x) for the function f (x) = xe** about xo = 0 ii. Use P3(0.5) to approximate f (0.5). iii. Find an upper bound for the error |f (x) – PĄ(x)| for 0 < x< 0.4 b. Given is the elliptic integral of second kind E(t) = T/2__1 V1-tsin² 6 de Evaluate E(0.25) by applying appropriate interpolation formula E(0.20) = 1.6596, E(0.22) = 1.6698, E(0.24) = 1.6804, %3D E(0.26) = 1.6912, E (0.28) = 1.7024, E(0.30) = 1.7139 %3D %3D
а. i. Find the second Taylor polynomial PĄ(x) for the function f (x) = xe** about xo = 0 ii. Use P3(0.5) to approximate f (0.5). iii. Find an upper bound for the error |f (x) – PĄ(x)| for 0 < x< 0.4 b. Given is the elliptic integral of second kind E(t) = T/2__1 V1-tsin² 6 de Evaluate E(0.25) by applying appropriate interpolation formula E(0.20) = 1.6596, E(0.22) = 1.6698, E(0.24) = 1.6804, %3D E(0.26) = 1.6912, E (0.28) = 1.7024, E(0.30) = 1.7139 %3D %3D
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

Transcribed Image Text:а.
i. Find the second Taylor polynomial P3(x) for the function f (x) = xe** about x, = 0
ii. Use P3(0.5) to approximate f (0.5).
iii. Find an upper bound for the error |f (x) – PĄ(x)| for 0<x< 0.4
b. Given is the elliptic integral of second kind E (t) = [,"
1/2
5o Ti-tsin² e
de
Evaluate E(0.25) by applying appropriate interpolation formula
E (0.20) = 1.6596,
E (0.22) = 1.6698,
E(0.24) = 1.6804,
E(0.26) = 1.6912,
E(0.28) = 1.7024,
E (0.30) = 1.7139
Page 1 of 2
c. Validate value of E(0.20) and E in part b, on any one of the given points using Matlab
(Attach screenshot)
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