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.
x=25;26;27;28;29;30y=0.42262;0.43837;0.45399;0.46947;0.48481;0.5Δy=−0.01575;−0.01562;−0.01548;−0.01534;−0.01519Δ2y=−0.00013;−0.00014;−0.00014;−0.00015Δ3y=0.0001;0;0.00001Δ4y=0.00001;−0.00001Δ5y=0.00002Pn(x)=y0+Δy1!h(x−x0)++Δ2y2!h2(x−x0)(x−x1)++…+Δnyn!hn(x−x0)…(x−xn)P5(x)=0.422262−0.01575q−0.00013∗q(q−1)2++0.00001∗q(q−1)(q−2)6++0.0001∗q(q−1)(q−2)(q−3)24++0.00002∗q(q−1)(q−2)(q−3)(q−4)120q=x−251=x−25P5(x)=0.422262−0.01575(x−25)−−0.00013∗(x−25)(x−26)2++0.00001∗(x−25)(x−26)(x−27)6++0.0001∗(x−25)(x−26)(x−27)(x−28)24++0.00002∗(x−25)(x−26)(x−27)(x−28)(x−29)120P5(x)=−16000000x5+11480000x4−−6350000x3+829372400000x2−−27379196000000x+64893250000x=28024′=28.40P5(28.4)=0.4756217876x = 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)=0.4756217876x=25;26;27;28;29;30y=0.42262;0.43837;0.45399;0.46947;0.48481;0.5Δy=−0.01575;−0.01562;−0.01548;−0.01534;−0.01519Δ2y=−0.00013;−0.00014;−0.00014;−0.00015Δ3y=0.0001;0;0.00001Δ4y=0.00001;−0.00001Δ5y=0.00002Pn(x)=y0+1!hΔy(x−x0)++2!h2Δ2y(x−x0)(x−x1)++…+n!hnΔny(x−x0)…(x−xn)P5(x)=0.422262−0.01575q−0.00013∗2q(q−1)++0.00001∗6q(q−1)(q−2)++0.0001∗24q(q−1)(q−2)(q−3)++0.00002∗120q(q−1)(q−2)(q−3)(q−4)q=1x−25=x−25P5(x)=0.422262−0.01575(x−25)−−0.00013∗2(x−25)(x−26)++0.00001∗6(x−25)(x−26)(x−27)++0.0001∗24(x−25)(x−26)(x−27)(x−28)++0.00002∗120(x−25)(x−26)(x−27)(x−28)(x−29)P5(x)=6000000−1x5+48000011x4−−5000063x3+240000082937x2−−60000002737919x+25000064893x=28024′=28.40P5(28.4)=0.4756217876
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