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mws_ind_inp_txt_lagrange_examples

# mws_ind_inp_txt_lagrange_examples - Chapter 05.04 Lagrange...

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05.04.1 Chapter 05.04 Lagrange Method of Interpolation – More Examples Industrial Engineering Example 1 The geometry of a cam is given in Figure 1. A curve needs to be fit through the seven points given in Table 1 to fabricate the cam. Figure 1 Schematic of cam profile. Table 1 Geometry of the cam. Point x in. y in. 1 2.20 0.00 2 1.28 0.88 3 0.66 1.14 4 0.00 1.20 5 –0.60 1.04 6 –1.04 0.60 7 –1.20 0.00 If the cam follows a straight line profile from 28 . 1 x to 66 . 0 x , what is the value of y at 10 . 1 x using a first order Lagrange polynomial? 1 2 3 5 6 7 x y 4

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05.04.2 Chapter 05.04 Solution For first order Lagrange polynomial interpolation (also called linear interpolation), the value of y is given by 1 0 ) ( ) ( ) ( i i i x y x L x y ) ( ) ( ) ( ) ( 1 1 0 0 x y x L x y x L Figure 2 Linear interpolation. Since we want to find the value of y at 10 . 1 x , using the two points 28 . 1 0 x and 66 . 0 1 x , then 88 . 0 , 28 . 1 0 0 x y x 14 . 1 , 66 . 0 1 1 x y x gives 1 0 0 0 0 ) ( j j j j x x x x x L 1 0 1 x x x x 1 1 0 1 1 ) ( j j j j x x x x x L 0 1 0 x x x x Hence ( x 0 , y 0 ) ( x 1 , y 1 ) f 1 ( x ) x y
Lagrange Method of Interpolation-More Examples: Industrial Engineering 05.04.3 ) ( ) ( ) ( 1 0 1 0 0 1 0 1 x y x x x x x y x x x x x y 28 . 1 66 . 0 ), 14 . 1 ( 28 . 1 66 . 0 28 . 1 ) 88 . 0 ( 66 . 0 28 . 1 66 . 0 x x x ) 14 . 1 ( 28 . 1 66 . 0 28 . 1 10 . 1 ) 88 . 0 ( 66 . 0 28 . 1 66 . 0 10 . 1 ) 10 . 1 ( y ) 14 . 1 ( 29032 . 0 ) 88 . 0 ( 70968 . 0 . in 95548 . 0 You can see that 70968 . 0 ) ( 0 x L and 29032 . 0 ) ( 1 x L are like weightages given to the values of y at 28 . 1 x and 66 . 0 x to calculate the value of y at 10 . 1 x . Example 2 The geometry of a cam is given in Figure 3. A curve needs to be fit through the seven points given in Table 2 to fabricate the cam. Figure 3 Schematic of cam profile. Table 2 Geometry of the cam. Point x in. y in. 1 2.20 0.00 2 1.28 0.88 3 0.66 1.14 4 0.00 1.20 5 –0.60 1.04 6 –1.04 0.60 7 –1.20 0.00 If the cam follows a quadratic profile from 20 . 2 x to 28 . 1 x to 66 . 0 x , what is the value of y at 10 . 1 x using a second order Lagrange polynomial? Find the absolute relative approximate error for the second order polynomial approximation. 1 2 3 5 6 7 x y 4

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05.04.4 Chapter 05.04 Solution For second order Lagrange polynomial interpolation (also called quadratic interpolation), the value of y given by 2 0 ) ( ) ( ) ( i i i x y x L x y ) ( ) ( ) ( ) ( ) ( ) ( 2 2 1 1 0 0 x y x L x y x L x y x L Figure 4 Quadratic interpolation. Since we want to find the value of y at 10 . 1 x , using the three points 20 . 2 0 x , 28 . 1 1 x , 66 . 0 2 x , then 00 . 0 , 20 . 2 0 0 x y x 88 . 0 , 28 . 1 1 1 x y x 14 . 1 , 66 . 0 2 2 x y x gives 2 0 0 0 0 ) ( j j j j x x x x x L 2 0 2 1 0 1 x x x x x x x x 2 1 0 1 1 ) ( j j j j x x x x x L 2 1 2 0 1 0 x x x x x x x x ( x 0 , y 0 ) ( x 1 , y 1 ) ( x 2 , y 2 ) f 2 ( x ) x y