Explanation Given: The equation 4 e − x 2 sin x = x 2 − x + 1 . Theorem used: Newton’s Formula: The Newton’s formula is, x n + 1 = x n − f ( x n ) f ′ ( x n ) , where f ( x ) is the given function. Calculation: Consider the given equation in the form of a function rewritten as, f ( x ) = x 2 − x + 1 − 4 e − x 2 sin x (1) Use online graphing calculator to graph the function f ( x ) = x 2 − x + 1 − 4 e − x 2 sin x as shown below in Figure 1. Differentiate f ( x ) = x 2 − x + 1 − 4 e − x 2 sin x with respect to x as, d d x ( f ( x ) ) = d d x ( x 2 − x + 1 − 4 e − x 2 sin x ) = d d x ( x 2 ) − d d x ( x ) + d d x ( 1 ) − d d x ( 4 e − x 2 sin x ) = 2 x − 1 + 0 − 4 d d x ( e − x 2 sin x ) = 2 x − 1 − 4 d d x ( e − x 2 sin x ) On further simplifications, d d x ( f ( x ) ) = 2 x − 1 − 4 ( e − x 2 d d x ( sin x ) + sin x d d x ( e − x 2 ) ) = 2 x − 1 − 4 ( e − x 2 ( cos x ) + sin x ( − 2 x e − x 2 ) ) = 2 x − 1 − 4 e − x 2 cos x + 8 x e − x 2 sin x That is, f ′ ( x ) = 2 x − 1 − 4 e − x 2 cos x + 8 x e − x 2 sin x . (2) The first root is close to 0.2. Thus, the initial approximation from the graph is x 0 = 0.2 . By Newton’s Formula, x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) . That is, the next x value is found from, x − x 2 − x + 1 − 4 e − x 2 sin x 2 x − 1 − 4 e − x 2 cos x + 8 x e − x 2 sin x (3) Substitute x = 0.2 in formula (3) as, x 1 = 0.2 − ( 0.2 ) 2 − ( 0.2 ) + 1 − 4 e − ( 0.2 ) 2 sin ( 0.2 ) 2 ( 0.2 ) − 1 − 4 e − ( 0.2 ) 2 cos ( 0.2 ) + 8 ( 0.2 ) e − ( 0.2 ) 2 sin ( 0.2 ) = 0.21883273 Again apply for x = 0.21883273 in formula (3) and obtain x 2 as follows. x 2 = 0.21883273 − ( 0.21883273 ) 2 − ( 0.21883273 ) + 1 − 4 e − ( 0.21883273 ) 2 sin ( 0.21883273 ) { 2 ( 0.21883273 ) − 1 − 4 e − ( 0.21883273 ) 2 cos ( 0.21883273 ) + 8 ( 0.21883273 ) e − ( 0.21883273 ) 2 sin ( 0

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Author: James Stewart

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Chapter 6.3, Problem 64AE

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To find: An initial approximation to the root of the equation 4e−x2sinx=x2−x+1 with the graph and use Newton’s method to find the root correct to eight decimal places.

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Chapter 6.3, Problem 63AE

Chapter 6.3, Problem 65AE

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An initial approximation to the root of the equation 4 e − x 2 sin x = x 2 − x + 1 with the graph and use Newton’s method to find the root correct to eight decimal places.

Solution Summary: The author explains Newton's formula for finding the root of the equation 4e-x2mathrmsin.

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To find: An initial approximation to the root of the equation 4e−x2sinx=x2−x+1 with the graph and use Newton’s method to find the root correct to eight decimal places.

Want to see the full answer?

Chapter 6 Solutions