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Numerical interpolation using Stirling's formula
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Option :
1.  X 20 25 30 35 40 f(x) 49225 48316 47236 45926 44306
and x=28
2.  X 10 11 12 13 14 f(x) 0.23967 0.2806 0.31788 0.35209 0.38368
and x=12.2
3.  X 0 5 10 15 20 25 30 f(x) 0 0.0875 0.1763 0.2679 0.364 0.4663 0.5774
and x=16
 f(x) = x1 = and x2 = x = Step value (h) =  OR  Inverval (N) = Option : 1. Value f(2) 2. Interpolation table 3. Equation f(x) 4. Equation f(x) then value f(2) f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=x^3-x+1x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=x^3-x+1x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=x^3+x+2x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=x^3+x+2x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=sin(x)x1 = 0 and x2 = 1.57x = 2Step value (h) = 0.25 or N = 8f(x)=cos(x)x1 = 0 and x2 = 1.57x = 2Step value (h) = 0.25 or N = 8
 Decimal Place = 0 1 2 3 4 5 6 7 8
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