Find e 0 25 and e 0 75 by linear interpolation of e x with x 0 0, x 1 0 5 and x 0 0 5, x 1 1, respectively Then find p 2 (x) by quadratic interpolation of e x with x 0 0, x 1 0 5, x 2 1 and from it e 0 25 and e 0 75 Compare the errors Use 4S values of e x ...

For e025 and e075 e025 p2x e075 P2 x e025 p2x e075 p2x 0 p2xpq 0 p q 0 or 2131213343 0 or pq 01633 o ...

Find e -0.25 and e -0.75 by linear interpolation of e -x with x 0 = 0,

Question:

Find e^-0.25and e^-0.75by linear interpolation of e^-x with x₀ = 0, x₁ = 0.5 and x₀ = 0.5, x₁ = 1, respectively. Then find p₂(x) by quadratic interpolation of e^-x with x₀ = 0, x₁ = 0.5, x₂ = 1 and from it e^-0.25 and e^-0.75.Compare the errors. Use 4S-values of e^-x.