Use Newton's backward-difference formula to construct interpolating polynomials of degree one two, and three for the following data: f(0) = 1, f(0.25) = 1.64872, f(0.5) = 2.71828, f(0.75) = 4.48169. Approximate f(0.43) using each of the polynomials.
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25 in this case. So, we have: Δf(0.75) = f(0.75) - f(0.5) = 4.48169 - 2.71828 = 1.76341 Δf(0.5) = f(0.5) - f(0.25) = 2.71828 - 1.64872 = 1.06956 Δf(0.25) = f(0.25) - f(0) = 1.64872 - 1 = 0.64872 The second backward difference is calculated as: Δ²f(x) = Δf(x) - Show more…
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