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Using the following data 20 = 0.3 21 = 0.43 22 = 0.56 23 = 0.69 24 = 0.82 25 = 0.95 f(x) 2.3185 2.4159 2.478 2.5 2.48 2.4197 1. Af(x0) = 2. A² f(x0) = 3. A³ f(xo) =

Using the following data 20 = 0.3 21 = 0.43 22 = 0.56 23 = 0.69 24 = 0.82 25 = 0.95 f(x) 2.3185 2.4159 2.478 2.5 2.48 2.4197 1. Af(x0) = 2. A² f(x0) = 3. A³ f(xo) =

Advanced Engineering Mathematics
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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Using the following data x0 = 0.3 0.43 0.56 = 0.69 0.82 0.95 22 = f(x) 2.3185 2.4159 2.478 2.5 2.48 2.4197 1. Af(x0) = 2. Δ'f(τ0) - 3. A° f(z0) = 4. A³ f(x2) = 5. Aʻf(xo) = 6. Aʻƒ(x1) = 7. Δf(z ) - 8. To approximate f(0.60333333333333) what value of s should be used in newton's forward difference formula? s = (:) -1 (:) - 9. 4 10. 11. Using Newton's forward difference formula, what is the approximation for f(0.60333333333333) ?
Transcribed Image Text:Using the following data x0 = 0.3 0.43 0.56 = 0.69 0.82 0.95 22 = f(x) 2.3185 2.4159 2.478 2.5 2.48 2.4197 1. Af(x0) = 2. Δ'f(τ0) - 3. A° f(z0) = 4. A³ f(x2) = 5. Aʻf(xo) = 6. Aʻƒ(x1) = 7. Δf(z ) - 8. To approximate f(0.60333333333333) what value of s should be used in newton's forward difference formula? s = (:) -1 (:) - 9. 4 10. 11. Using Newton's forward difference formula, what is the approximation for f(0.60333333333333) ?
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