Answer to Question #194844 in Quantitative Methods for ali khan

Question

Given sin250=0.42262, sin260=0.43837, sin270=0.45399, sin280=0.46947,

sin290=0.48481, sin300=0.5.

Using Newton interpolation formula find sin28024′. Estimate the error. 

Answer 1

$x = 25; 26; 27; 28; 29; 30 \\ y = 0.42262; 0.43837; 0.45399; 0.46947; 0.48481; 0.5 \\ \Delta y=-0.01575;-0.01562;-0.01548;-0.01534;-0.01519 \\ \Delta^2 y = -0.00013; -0.00014; -0.00014; -0.00015 \\ \Delta^3 y = 0.0001; 0; 0.00001 \\ \Delta^4 y = 0.00001; -0.00001 \\ \Delta^5 y = 0.00002 \\ P_n(x) = y_0 + {\Delta y \over 1!h}(x-x_0)+\\ +{\Delta^2 y \over 2!h^2}(x-x_0)(x-x_1)+\\ +\dotso + {\Delta^n y \over n!h^n}(x-x_0)\dotso(x-x_n) \\ P_5(x)=0.422262-0.01575q-0.00013*{q(q-1) \over 2}+\\ +0.00001*{q(q-1)(q-2) \over 6}+\\ +0.0001*{q(q-1)(q-2)(q-3) \over 24}+\\ +0.00002*{q(q-1)(q-2)(q-3)(q-4) \over 120}\\ q={x-25 \over 1}=x-25\\ P_5(x)=0.422262-0.01575(x-25)-\\ -0.00013*{(x-25)(x-26) \over 2}+\\ +0.00001*{(x-25)(x-26)(x-27) \over 6}+\\ +0.0001*{(x-25)(x-26)(x-27)(x-28) \over 24}+\\ +0.00002*{(x-25)(x-26)(x-27)(x-28)(x-29) \over 120} \\ P_5(x)={-1\over6000000}x^5+{11\over480000}x^4-\\ -{63 \over 50000}x^3+{82937\over2400000}x^2-\\ -{2737919\over6000000}x+{64893\over250000}\\ x=28^024'=28.4^0\\ P_5(28.4)=0.4756217876$