3.8 Newton’s Method 225 3.8 Newton’s Method Approximate a zero of a function using Newton’s Method. Newton’s Method In this section, you will study a technique for approximating the real zeros of a function. The technique is called Newton’s Method, and it uses tangent lines to approximate the graph of the function near its x-intercepts. To see how Newton’s Method works, consider a function f that is continuous on the y interval ͓a, b͔ and differentiable on the interval ͑a, b͒. If f͑a͒ and f͑b͒ differ in sign, then, by the Intermediate Value Theorem,f must have at least one zero in the interval ͑a, b͒. To estimate this zero, you choose (x , f(x )) 1 1 ϭ x x1 First estimate Tangent line as shown in Figure 3.60(a). Newton’s Method is based on the assumption that the graph ͑ ͑ ͒͒ of f and the tangent line at x1, f x1 both cross the x-axis at about the same point. Because you can easily calculate the x-intercept for this tangent line, you can use it as b acx x x a second (and, usually, better) estimate of the zero of f. The tangent line passes through 1 2 ͑ ͑ ͒͒ Ј͑ ͒ the point x1, f x1 with a slope of f x1 . In point-slope form, the equation of the tangent line is (a) Ϫ ͑ ͒ ϭ Ј͑ ͒͑ Ϫ ͒ y f x1 f x1 x x1 ϭ Ј͑ ͒͑ Ϫ ͒ ϩ ͑ ͒ y y f x1 x x1 f x1 . Letting y ϭ 0 and solving for x produces (x , f(x )) f͑x ͒ 1 1 ϭ Ϫ 1 x x1 Ј͑ ͒. Tangent line f x1 So, from the initial estimate x1, you obtain a new estimate ͑ ͒ x f x1 c 2 b x ϭ x Ϫ . Second estimate [See Figure 3.60(b).] x 2 1 fЈ͑x ͒ a x1 x3 1 You can improve on x2 and calculate yet a third estimate (b) ͑ ͒ f x2 x ϭ x Ϫ . Third estimate The x-intercept of the tangent line 3 2 fЈ͑x ͒ approximates the zero of f. 2 Figure 3.60 Repeated application of this process is called Newton’s Method. Newton’s Method for Approximating the Zeros of a Function NEWTON’S METHOD Let f͑c͒ ϭ 0, where f is differentiable on an open interval containing c. Then, Isaac Newton first described the method for approximating the real to approximate c, use these steps. zeros of a function in his text 1. Make an initial estimate x that is close to c. (A graph is helpful.) Method of Fluxions. Although the 1 book was written in 1671, it 2. Determine a new approximation was not published until 1736. f͑x ͒ Meanwhile, in 1690, Joseph ϭ Ϫ n xnϩ1 xn Ј͑ ͒. Raphson (1648–1715) published f xn a paper describing a method for Ϫ approximating the real zeros of 3. When Խxn xnϩ1Խ is within the desired accuracy, let xnϩ1 serve as the a function that was very similar final approximation. Otherwise, return to Step 2 and calculate a new to Newton’s. For this reason, the approximation. method is often referred to as the Newton-Raphson method. Each successive application of this procedure is called an iteration. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 226 Chapter 3 Applications of Differentiation Using Newton’s Method REMARK For many Calculate three iterations of Newton’s Method to approximate a zero of f͑x͒ ϭ x2 Ϫ 2. functions, just a few iterations Use x ϭ 1 as the initial guess. of Newton’s Method will 1 produce approximations having Solution Because f͑x͒ ϭ x2 Ϫ 2, you have fЈ͑x͒ ϭ 2x, and the iterative formula is very small errors, as shown in f͑x ͒ x 2 Ϫ ϭ Ϫ n ϭ Ϫ n 2 Example 1. xnϩ1 xn Ј͑ ͒ xn . f xn 2xn y The calculations for three iterations are shown in the table. x = 1 1 f͑x ͒ f͑x ͒ x ͑ ͒ Ј͑ ͒ n x Ϫ n x = 1.5 n xn f xn f xn Ј͑ ͒ n Ј͑ ͒ 2 f xn f xn 1 1.000000 Ϫ1.000000 2.000000 Ϫ0.500000 1.500000 −1 2 1.500000 0.250000 3.000000 0.083333 1.416667 3 1.416667 0.006945 2.833334 0.002451 1.414216 4 1.414216 f(x) = x2 − 2 Of course, in this case you know that the two zeros of the function are ±Ί2. To six decimal places,Ί2 ϭ 1.414214. So, after only three iterations of Newton’s Method, The first iteration of Newton’s Method you have obtained an approximation that is within 0.000002 of an actual root. The first Figure 3.61 iteration of this process is shown in Figure 3.61. Using Newton’s Method See LarsonCalculus.com for an interactive version of this type of example. Use Newton’s Method to approximate the zeros of f͑x͒ ϭ 2x3 ϩ x2 Ϫ x ϩ 1. Continue the iterations until two successive approximations differ by less than 0.0001. y Solution Begin by sketching a graph of f, as shown in Figure 3.62. From the graph, you can observe that the function has only one zero, which occurs near x ϭϪ1.2. Next, 3 2 f(x) = 2x + x − x + 1 2 differentiate f and form the iterative formula f͑x ͒ 2x 3 ϩ x 2 Ϫ x ϩ 1 x ϭ x Ϫ n ϭ x Ϫ n n n . nϩ1 n Ј͑ ͒ n 2 ϩ Ϫ f xn 6xn 2xn 1 1 The calculations are shown in the table. x − − f͑x ͒ f͑x ͒ 2 1 ͑ ͒ Ј͑ ͒ n x Ϫ n n xn f xn f xn Ј͑ ͒ n Ј͑ ͒ f xn f xn After three iterations of Newton’s Ϫ Ϫ Method, the zero of f is approximated 1 1.20000 0.18400 5.24000 0.03511 1.23511 to the desired accuracy. 2 Ϫ1.23511 Ϫ0.00771 5.68276 Ϫ0.00136 Ϫ1.23375 Figure 3.62 3 Ϫ1.23375 0.00001 5.66533 0.00000 Ϫ1.23375 4 Ϫ1.23375 Because two successive approximations differ by less than the required 0.0001, you can estimate the zero of f to be Ϫ1.23375. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 3.8 Newton’s Method 227 When, as in Examples 1 and 2, the approximations approach a limit, the sequence x1, x2, x3, . ., xn, . is said to converge. Moreover, when the limit is c, it can be shown that c must be a zero of f. FOR FURTHER INFORMATION Newton’s Method does not y For more on when Newton’s always yield a convergent sequence. Method fails, see the article One way it can fail to do so is “No Fooling! Newton’s Method shown in Figure 3.63. Because f ′(x ) = 0 Can Be Fooled” by Peter Horton Newton’s Method involves division 1 Ј͑ ͒ in Mathematics Magazine. To by f xn , it is clear that the method view this article, go to will fail when the derivative is zero x x MathArticles.com. for any xn in the sequence. When 1 you encounter this problem, you Ј͑ ͒ ϭ Newton’s Method fails to converge when f xn 0. can usually overcome it by choosing Figure 3.63 a different value for x1. Another way Newton’s Method can fail is shown in the next example. An Example in Which Newton’s Method Fails The function f ͑x͒ ϭ x1͞3 is not differentiable at x ϭ 0. Show that Newton’s Method ϭ fails to converge using x1 0.1. Ј͑ ͒ ϭ 1 Ϫ2͞3 Solution Because f x 3 x , the iterative formula is ͑ ͒ 1͞3 f xn xn x ϭ x Ϫ ϭ x Ϫ ϭ x Ϫ 3x ϭϪ2x . nϩ1 n Ј͑ ͒ n 1 Ϫ2͞3 n n n f xn 3xn The calculations are shown in the table. This table and Figure 3.64 indicate that xn continues to increase in magnitude as n → ϱ, and so the limit of the sequence does not exist. f͑x ͒ f͑x ͒ ͑ ͒ Ј͑ ͒ n x Ϫ n n xn f xn f xn Ј͑ ͒ n Ј͑ ͒ f xn f xn 1 0.10000 0.46416 1.54720 0.30000 Ϫ0.20000 REMARK In Example 3, 2 Ϫ0.20000 Ϫ0.58480 0.97467 Ϫ0.60000 0.40000 the initial estimate x ϭ 0.1 1 Ϫ fails to produce a convergent 3 0.40000 0.73681 0.61401 1.20000 0.80000 sequence. Try showing that 4 Ϫ0.80000 Ϫ0.92832 0.3680 Ϫ2.40000 1.60000 Newton’s Method also fails for every other choice of x1 (other than the actual zero).