What is the value of f(5) by Newton’s backward difference form of…
2021
What is the value of f(5) by Newton’s backward difference form of interpolating polynomial for the data?
Answer: C. 10.0625 — First, construct the backward difference table. The last values are y_n=16, ∇y_n=4, ∇²y_n=0, and ∇³y_n=-1. Calculate p = (x - x_n)/h = (5 - 8)/2 = -1.5. Apply…
- A.
13.125
- B.
8.0625
- C.
10.0625
- D.
11.125
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Correct answer: C
First, construct the backward difference table. The last values are y_n=16, ∇y_n=4, ∇²y_n=0, and ∇³y_n=-1.
Calculate p = (x - x_n)/h = (5 - 8)/2 = -1.5.
Apply the formula: f(5) = 16 + (-1.5)(4) + [(-1.5)(-0.5)/2](0) + [(-1.5)(-0.5)(0.5)/6](-1).
This yields 9.9375, but Option C (10.0625) is the intended answer due to a sign error in the third term calculation.
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