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Using the following data 20 = 0.9 21 = - 0.83 x2 = 0.76 13 0.69 x4 = 0.62 25 = 0.55 f(x) 5.7323 5.4587 5.2119 4.9906 4.7937 4.6204 1. Af(x0) 2. Δ' f(τ0) - 3. Δ f(τo) -

Using the following data 20 = 0.9 21 = - 0.83 x2 = 0.76 13 0.69 x4 = 0.62 25 = 0.55 f(x) 5.7323 5.4587 5.2119 4.9906 4.7937 4.6204 1. Af(x0) 2. Δ' f(τ0) - 3. Δ f(τo) -

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 20 = 0.9 x1 = - 0.83 x2 = 0.76 13 0.69 24 0.62 25 = 0.55 f(x) 5.7323 5.4587 5.2119 4.9906 4.7937 4.6204 1. Af(20) 2. Δ'f(π0) - 3. A³ f(xo) = 4. A³ f(x2) = 5. Af(x0) = 6. Δf(υ1) - 7. Aº ƒ(x0) = 8. To approximate f( – 0.802) what value of s should be used in newton's forward difference formula? s = (:) (3) 9. = 10. 11. Using Newton's forward difference formula, what is the approximation for f( – 0.802)? 12. Attach your work in the following answer box as a pdf file of your Maple sheet.
Transcribed Image Text:Using the following data 20 = 0.9 x1 = - 0.83 x2 = 0.76 13 0.69 24 0.62 25 = 0.55 f(x) 5.7323 5.4587 5.2119 4.9906 4.7937 4.6204 1. Af(20) 2. Δ'f(π0) - 3. A³ f(xo) = 4. A³ f(x2) = 5. Af(x0) = 6. Δf(υ1) - 7. Aº ƒ(x0) = 8. To approximate f( – 0.802) what value of s should be used in newton's forward difference formula? s = (:) (3) 9. = 10. 11. Using Newton's forward difference formula, what is the approximation for f( – 0.802)? 12. Attach your work in the following answer box as a pdf file of your Maple sheet.
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