5128_Ch04_pp186-260

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Section 4.5 Linearization and Newton’s Method 235 EXAMPLE 3 Approximating Binomial Powers Example 1 introduces a special case of a general linearization formula that applies to powers of 1 x for small values of x: (1 x) k 1 kx. If k is a positive integer this follows from the Binomial Theorem, but the formula actually holds for all real values of k. (We leave the justification to you as Exercise 7.) Use this formula to find polynomials that will approximate the following functions for values of x close to zero: (a) 3 1 1 x (b) (c) 1 5x 4 (d) 1 x 1 1x 2 SOLUTION We change each expression to the form (1 y) k , where k is a real number and y is a function of x that is close to 0 when x is close to zero. The approximation is then given by 1 ky. (a) 3 1 x (1 (x)) 1/3 1 1 3 (x) 1 x 3 1 (b) (1 (x)) 1 1 (1)(x) 1 x 1 x (c) 1 5x 4 ((1 5x 4 )) 1/2 1 1 2 (5x4 ) 1 5 2 x4 (d) 1 1 x 2 ((1 (x2 )) 1/2 1 1 2 (x2 ) 1 1 2 x2 Now try Exercise 9. EXAMPLE 4 Approximating Roots Use linearizations to approximate (a) 123 SOLUTION and (b) 3 123 Part of the analysis is to decide where to center the approximations. (a) Let f (x) x . The closest perfect square to 123 is 121, so we center the linearization at x 121. The tangent line at (121, 11) has slope f (121) 1 2 (121)1/2 1 2 • 1 121 1 . 22 So 1 121 L(121) 11 (123 121) 11.09. 2 2 (b) Let f (x) 3 x. The closest perfect cube to 123 is 125, so we center the linearization at x 125. The tangent line at (125, 5) has slope f (125) 1 3 (125)2/3 1 3 • 1 ( 3 1) 25 2 1 . 7 5 So 3 1 123 L(123) 5 (123 125) 4.973. 7 5 A calculator shows both approximations to be within 10 3 of the actual values. Newton’s Method . Now try Exercise 11. Newton’s method is a numerical technique for approximating a zero of a function with zeros of its linearizations. Under favorable circumstances, the zeros of the linearizations
236 Chapter 4 Applications of Derivatives 0 y Root sought (x 3 , f(x 3 )) x 4 x 3 Fourth Third (x 2 , f(x 2 )) APPROXIMATIONS (x 1 , f(x 1 )) x 2 Second y f(x) x 1 First x converge rapidly to an accurate approximation. Many calculators use the method because it applies to a wide range of functions and usually gets results in only a few steps. Here is how it works. To find a solution of an equation f x 0, we begin with an initial estimate x 1 , found either by looking at a graph or simply guessing. Then we use the tangent to the curve y f x at x 1 , f x 1 to approximate the curve (Figure 4.47). The point where the tangent crosses the x-axis is the next approximation x 2 . The number x 2 is usually a better approximation to the solution than is x 1 . The point where the tangent to the curve at x 2 , f x 2 crosses the x-axis is the next approximation x 3 . We continue on, using each approximation to generate the next, until we are close enough to the zero to stop. There is a formula for finding the n 1st approximation x n1 from the nth approximation x n . The point-slope equation for the tangent to the curve at x n , f x n is y f x n f x n x x n . We can find where it crosses the x-axis by setting y 0 (Figure 4.48). 0 f x n f x n x x n Figure 4.47 Usually the approximations rapidly approach an actual zero of y f x. 0 y y f(x) Point: (x n , f(x n )) Slope: f'(x n ) Equation: y – f(x n ) f'(x n )(x – x n ) Root sought (x n , f(x n )) f(x n ) x n+1 x n – ——– f'(x n ) Figure 4.48 From x n we go up to the curve and follow the tangent line down to find x n1 . x n Tangent line (graph of linearization of f at x n ) x f x n f x n • x f x n • x n f x n • x f x n • x n f x n f xn x x n If fx f x n 0 This value of x is the next approximation x n1 . Here is a summary of Newton’s method. Procedure for Newton’s Method 1. Guess a first approximation to a solution of the equation f x 0. A graph of y f x may help. 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula f xn x n1 x n . f x EXAMPLE 5 Using Newton’s Method Use Newton’s method to solve x 3 3x 1 0. SOLUTION Let f x x 3 3x 1, then f x 3x 2 3 and f xn x n1 x n x f x n n x n 3 3xn 1 3x 2. n 3 The graph of f in Figure 4.49 on the next page suggests that x 1 0.3 is a good first approximation to the zero of f in the interval 1 x 0. Then, x 1 0.3, x 2 0.322324159, x 3 0.3221853603, x 4 0.3221853546. The x n for n 5 all appear to equal x 4 on the calculator we used for our computations. We conclude that the solution to the equation x 3 3x 1 0 is about 0.3221853546. Now try Exercise 15. n n
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