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Math Advanced Math ADVANCED ENGINEERING MATH W/ACCESS

The error bound of the Lagrange polynomial function. The error bound is 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 _ . Formula used: The general Lagrange interpolation polynomial function is p n ( x ) = ∑ k = 0 n L k ( x ) f k = ∑ k = 0 n l k ( x ) l k ( x k ) f k . Where L 0 ( x ) = ( x − x 1 ) ( x − x 2 ) ( x 0 − x 1 ) ( x 0 − x 2 ) , L 1 ( x ) = ( x − x 0 ) ( x − x 2 ) ( x 1 − x 0 ) ( x 1 − x 2 ) and L 2 ( x ) = ( x − x 0 ) ( x − x 1 ) ( x 2 − x 0 ) ( x 2 − x 1 ) . Calculation: The value of [ f ″ ( t ) ] ′ is calculated as: [ f ″ ( t ) ] ′ = d 3 d t 3 [ 2 π ∫ 0 x e − t 2 d t ] = d 2 d x 2 [ 2 π e − x 2 d x ] = d d x [ − 4 π x e − x 2 d x ] = 4 π ( 2 x 2 − 1 ) e − x 2 Obtain the fourth derivative of the function as follows. f iv ( x ) = d d x [ 4 π ( 2 x 2 − 1 ) e − x 2 ] = 8 π x ( 3 − 2 x ) e − x 2 From the above function, it can be observed that the expression x , ( 3 − 2 x ) and e − x 2 are positive on interval [ 0.25 , 1 ] . So, the third derivative is increasing on the interval. [ f ″ ( 0.25 ) ] ′ ≤ [ f ″ ( t ) ] ′ ≤ [ f ″ ( 1 ) ] ′ − 1 . 8 5 5 0 2 ≤ [ f ″ ( t ) ] ′ ≤ 0.83022 Substitute, x = 0.75 in the equation ε ( x ) = ( x − 0.25 ) ( x − 0.5 ) ( x − 1 ) [ f ″ ( t ) ] ′ 3 ! . ε ( 0.75 ) = ( 0.75 − 0.25 ) ( 0.75 − 0.5 ) ( 0.75 − 1 ) [ f ″ ( t ) ] ′ 3 ! ε ( 0.75 ) = − ( 0.5 ) ( 0.7 ) ( 0.25 ) [ f ″ ( t ) ] ′ 6 ε ( 0.75 ) = − 0.0052083 [ f ″ ( t ) ] ′ Therefore, the error bound can be written as: − 0.0052083 [ f ″ ( 1 ) ] ′ ≤ ε ( 0.75 ) ≤ − 0.0052083 [ f ″ ( 0.25 ) ] ′ − 0 . 0 0 4 3 2 4 0 ≤ ε ( 0.75 ) ≤ 0.0096615 − 0 . 0 0 4 3 2 4 0 ≤ f ( 0.75 ) − p 2 ( 0.75 ) ≤ 0.0096615 − 0 . 0 0 4 3 2 4 0 + p 2 ( 0.75 ) ≤ f ( 0.75 ) ≤ 0.0096615 + p 2 ( 0.75 ) On further simplification, the following is obtained. − 0 . 0 0 4 3 2 4 0 + 0 . 7 0 2 2 9 ≤ f ( 0.75 ) ≤ 0.0096615 + 0 . 7 0 2 2 9 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 Therefore, the value of error bond is 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 _ .

The error bound of the Lagrange polynomial function. The error bound is 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 _ . Formula used: The general Lagrange interpolation polynomial function is p n ( x ) = ∑ k = 0 n L k ( x ) f k = ∑ k = 0 n l k ( x ) l k ( x k ) f k . Where L 0 ( x ) = ( x − x 1 ) ( x − x 2 ) ( x 0 − x 1 ) ( x 0 − x 2 ) , L 1 ( x ) = ( x − x 0 ) ( x − x 2 ) ( x 1 − x 0 ) ( x 1 − x 2 ) and L 2 ( x ) = ( x − x 0 ) ( x − x 1 ) ( x 2 − x 0 ) ( x 2 − x 1 ) . Calculation: The value of [ f ″ ( t ) ] ′ is calculated as: [ f ″ ( t ) ] ′ = d 3 d t 3 [ 2 π ∫ 0 x e − t 2 d t ] = d 2 d x 2 [ 2 π e − x 2 d x ] = d d x [ − 4 π x e − x 2 d x ] = 4 π ( 2 x 2 − 1 ) e − x 2 Obtain the fourth derivative of the function as follows. f iv ( x ) = d d x [ 4 π ( 2 x 2 − 1 ) e − x 2 ] = 8 π x ( 3 − 2 x ) e − x 2 From the above function, it can be observed that the expression x , ( 3 − 2 x ) and e − x 2 are positive on interval [ 0.25 , 1 ] . So, the third derivative is increasing on the interval. [ f ″ ( 0.25 ) ] ′ ≤ [ f ″ ( t ) ] ′ ≤ [ f ″ ( 1 ) ] ′ − 1 . 8 5 5 0 2 ≤ [ f ″ ( t ) ] ′ ≤ 0.83022 Substitute, x = 0.75 in the equation ε ( x ) = ( x − 0.25 ) ( x − 0.5 ) ( x − 1 ) [ f ″ ( t ) ] ′ 3 ! . ε ( 0.75 ) = ( 0.75 − 0.25 ) ( 0.75 − 0.5 ) ( 0.75 − 1 ) [ f ″ ( t ) ] ′ 3 ! ε ( 0.75 ) = − ( 0.5 ) ( 0.7 ) ( 0.25 ) [ f ″ ( t ) ] ′ 6 ε ( 0.75 ) = − 0.0052083 [ f ″ ( t ) ] ′ Therefore, the error bound can be written as: − 0.0052083 [ f ″ ( 1 ) ] ′ ≤ ε ( 0.75 ) ≤ − 0.0052083 [ f ″ ( 0.25 ) ] ′ − 0 . 0 0 4 3 2 4 0 ≤ ε ( 0.75 ) ≤ 0.0096615 − 0 . 0 0 4 3 2 4 0 ≤ f ( 0.75 ) − p 2 ( 0.75 ) ≤ 0.0096615 − 0 . 0 0 4 3 2 4 0 + p 2 ( 0.75 ) ≤ f ( 0.75 ) ≤ 0.0096615 + p 2 ( 0.75 ) On further simplification, the following is obtained. − 0 . 0 0 4 3 2 4 0 + 0 . 7 0 2 2 9 ≤ f ( 0.75 ) ≤ 0.0096615 + 0 . 7 0 2 2 9 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 Therefore, the value of error bond is 0.697966 ≤ f ( 0.75 ) ≤ 0.71195 _ .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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ADVANCED ENGINEERING MATH W/ACCESS

Ch. 19.1 - Floating point. Write 84.175, −528.685,...Ch. 19.1 - Write −76.437125, 60100, and −0.00001 in...Ch. 19.1 - Prob. 3P Ch. 19.1 - Order of terms, in adding with a fixed number of...Ch. 19.1 - Prob. 5P Ch. 19.1 - Nested form. Evaluate at x = 3.94 using 3S...Ch. 19.1 - Quadratic equation. Solve x2 − 30x + 1 = 0 by (4)...Ch. 19.1 - Solve x2 − 40x + 2 = 0, using 4S-computation. Ch. 19.1 - Prob. 9P Ch. 19.1 - Instability. For small |a| the equation (x − k)2 =...

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