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?
- A.
13.125
- B.
8.0625
- C.
10.0625
- D.
11.125
Attempted by 4 students.
Show answer & explanation
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.