Unformatted text preview: T(3)) M0 = 1.2995 For the quartic (4-th order) interpolation, add T5=40. >> TT=[T,40] TT = 0 10 20 30 40 >> MM=[M,0.2754] MM = 10.6000 3.8100 1.4920 0.6290 0.2754 >> D = divDiffTable(TT,MM) D = 10.6000 0 0 0 3.8100 -0.6790 0 0 1.4920 -0.2318 0.0224 0 0.6290 -0.0863 0.0073 -0.0005 0.2754 -0.0354 0.0025 -0.0002 0.0000 >> CC=diag(D) CC = C 10.6000-0.6790 0.0224-0.0005 0.0000 Since CC(5)=f(T1,t2,t3,t4,t5)=0, the additional contribution is zero. Thus the quartic interpolation is the same as the cubic interpolation....
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- Spring '11
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- Cubic function, Divided differences, cubic interpolation, 10.6000 0 0 0 %
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