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cubicinterpbookex

# cubicinterpbookex - T(3 M0 = 1.2995 For the quartic(4-th...

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For the cubic interpolation to get the viscosity at T=22, select T1=0, T2=10,T3=20,T4=30. Construct divided differences table. (I used the code, but you are supposed to construct manually.) >> T=[0,10,20,30]; >> M=[10.60,3.810,1.492,0.629]; >> D = divDiffTable(T,M) D = 10.6000 0 0 0 %f[T1] % 3.8100 -0.6790 0 0 %f[T2] f[T1,T2] 1.4920 -0.2318 0.0224 0 %f[T3] f[T2,T3] f[T1,T2,T3] 0.6290 -0.0863 0.0073 -0.0005 %f[T4] f[T3,T4] f[T2,T3,T4] f[T1,T2,T3,T4] >> C=diag(D) C = 10.6000 %f[T1] -0.6790 %f[T1,T2] 0.0224 %f[T1,T2,T3] -0.0005 %f[T1,T2,T3,T4] >> T0=22; >> M0=C(1)+C(2)*(T0-T(1))+C(3)*(T0-T(1))*(T0-T(2))+ C(4)*(T0-T(1))*(T0-T(2))*(T0-
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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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