Example 7.4. 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). 718125, f(-0.5) (Anna, B.Tech., 2003)
Example 7.4. 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). 718125, f(-0.5) (Anna, B.Tech., 2003)
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
![Example 7.4. 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)
0.5) -
= 0.3349375, f(0) = 1.10100. Hence find f(-1/3).
(Anna, B.Tech., 2003)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F876195a2-6c5a-4875-b744-13eba8799255%2Fcd34cff8-814d-4adc-b96b-c0e6129320c8%2Fp17mdnc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 7.4. 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)
0.5) -
= 0.3349375, f(0) = 1.10100. Hence find f(-1/3).
(Anna, B.Tech., 2003)
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