3.8 Newton's Method

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.

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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.

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 f͑x ͒ x 1͞3 x ϭ x Ϫ n ϭ x Ϫ n 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). y f(x) = x1/3

x1 x x x x −1 4 2 3 x5

−1

Newton’s Method fails to converge for every x-value other than the actual zero of f. Figure 3.64

It can be shown that a condition sufficient to produce convergence of Newton’s Method to a zero of f is that

f͑x͒ fЉ͑x͒ < 1 Condition for convergence Խ ͓ fЈ͑x͔͒2 Խ

on an open interval containing the zero. For instance, in Example 1, this test would yield f͑x͒ ϭ x2 Ϫ 2, fЈ͑x͒ ϭ 2x, fЉ͑x͒ ϭ 2, and f͑x͒ fЉ͑x͒ ͑x2 Ϫ 2͒͑2͒ 1 1 ϭ ϭ Ϫ . Example 1 Խ ͓ fЈ͑x͔͒2 Խ Խ 4x2 Խ Խ2 x2Խ On the interval ͑1, 3͒, this quantity is less than 1 and therefore the convergence of Newton’s Method is guaranteed. On the other hand, in Example 3, you have 1 2 f ͑x͒ ϭ x1͞3, fЈ͑x͒ ϭ xϪ2͞3, fЉ͑x͒ ϭϪ xϪ5͞3 3 9 and f͑x͒ f Љ͑x͒ x1͞3͑Ϫ2͞9͒͑xϪ5͞3͒ ϭ ϭ 2 Example 3 Խ ͓ fЈ͑x͔͒2 Խ Խ ͑1͞9͒͑xϪ4͞3͒ Խ which is not less than 1 for any value of x, so you cannot conclude that Newton’s Method will converge. You have learned several techniques for finding the zeros of functions. The zeros of some functions, such as f͑x͒ ϭ x3 Ϫ 2x2 Ϫ x ϩ 2 can be found by simple algebraic techniques, such as factoring. The zeros of other functions, such as ͑ ͒ ϭ 3 Ϫ ϩ NIELS HENRIK ABEL (1802–1829) f x x x 1 cannot be found by elementary algebraic methods. This particular function has only one real zero, and by using more advanced algebraic techniques, you can determine the zero to be 3 Ϫ Ί23͞3 3 ϩ Ί23͞3 x ϭϪΊ3 Ϫ Ί3 . 6 6 Because the exact solution is written in terms of square roots and cube roots, it is called a solution by radicals. The determination of radical solutions of a polynomial equation is one of the fundamental problems of algebra. The earliest such result is the Quadratic Formula, which dates back at least to Babylonian times. The general formula for the zeros of a cubic function was developed much later. In the sixteenth century, an Italian mathematician, Jerome Cardan, published a method for finding radical solutions to cubic and quartic equations. Then, for 300 years, the problem of finding a general EVARISTE GALOIS (1811–1832) quintic formula remained open. Finally, in the nineteenth century, the problem was Although the lives of both Abel answered independently by two young mathematicians. Niels Henrik Abel, a and Galois were brief, their work Norwegian mathematician, and Evariste Galois, a French mathematician, proved that it in the fields of analysis and abstract is not possible to solve a general fifth- (or higher-) degree polynomial equation by algebra was far-reaching. radicals. Of course, you can solve particular fifth-degree equations, such as See LarsonCalculus.com to read a biography about each of these x5 Ϫ 1 ϭ 0 mathematicians. but Abel and Galois were able to show that no general radical solution exists. The Granger Collection, New York

3.8 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Using Newton’s Method In Exercises 1–4, complete two 19. Mechanic’s Rule The Mechanic’s Rule for approximating iterations of Newton’s Method to approximate a zero of the Ίa, a > 0, is function using the given initial guess. ϭ 1 ϩ a ϭ ͑ ͒ ϭ 2 Ϫ ϭ xnϩ1 ΂xn ΃, n 1, 2, 3 . . . 1. f x x 5, x1 2.2 2 xn ͑ ͒ ϭ 3 Ϫ ϭ 2. f x x 3, x1 1.4 Ί where x1 is an approximation of a. ͑ ͒ ϭ ϭ 3. f x cos x, x1 1.6 (a) Use Newton’s Method and the function f ͑x͒ ϭ x2 Ϫ a to ͑ ͒ ϭ ϭ 4. f x tan x, x1 0.1 derive the Mechanic’s Rule. (b) Use the Mechanic’s Rule to approximate Ί5 and Ί7 to Using Newton’s Method In Exercises 5–14, approximate three decimal places. the zero(s) of the function. Use Newton’s Method and continue the process until two successive approximations differ by less 20. Approximating Radicals than 0.001. Then find the zero(s) using a graphing utility and (a) Use Newton’s Method and the function f ͑x͒ ϭ xn Ϫ a to compare the results. obtain a general rule for approximating x ϭΊn a. 5.f͑x͒ ϭ x3 ϩ 4 6. f͑x͒ ϭ 2 Ϫ x3 (b) Use the general rule found in part (a) to approximate Ί4 6 and Ί3 15 to three decimal places. 7.f͑x͒ ϭ x3 ϩ x Ϫ 1 8. f͑x͒ ϭ x5 ϩ x Ϫ 1 9.f͑x͒ ϭ 5Ίx Ϫ 1 Ϫ 2x 10. f ͑x͒ ϭ x Ϫ 2Ίx ϩ 1 Failure of Newton’s Method In Exercises 21 and 22, 11. f͑x͒ ϭ x3 Ϫ 3.9x2 ϩ 4.79x Ϫ 1.881 apply Newton’s Method using the given initial guess, and explain why the method fails. 12. f͑x͒ ϭ x 4 ϩ x3 Ϫ 1 ϭ 3 Ϫ 2 ϩ Ϫ ϭ 13.f͑x͒ ϭ 1 Ϫ x ϩ sin x 14. f͑x͒ ϭ x3 Ϫ cos x 21. y 2x 6x 6x 1, x1 1 y y Finding Point(s) of Intersection In Exercises 15–18,

apply Newton’s Method to approximate the x -value(s) of the 2 x indicated point(s) of intersection of the two graphs. Continue −121 the process until two successive approximations differ by less fͧxͨ ؊ gͧxͨ.] 1 ؍ than 0.001. [Hint: Let hͧxͨ −2 ͑ ͒ ϭ ϩ ͑ ͒ ϭ Ϫ 15.f x 2x 1 16. f x 3 x x x −3 1 1 2 g͑x͒ ϭ Ίx ϩ 4 g͑x͒ ϭ 2 ϩ x 1 Figure for 21 Figure for 22 y y 22. y ϭ x3 Ϫ 2x Ϫ 2, x ϭ 0 f 1 3 g 3 Fixed Point In Exercises 23 and 24, approximate the fixed f 2 point of the function to two decimal places. [ A fixed point x0 of ؍ ͨ ͧ a function f is a value of x such that f x0 x0.] 1 g 23. f͑x͒ ϭ cos x x x 24. f͑x) ϭ cot x, 0 < x < ␲ 123 123

17.f͑x͒ ϭ x 18. f͑x͒ ϭ x2 25. Approximating Reciprocals Use Newton’s Method to show that the equation g͑x͒ ϭ tan x g͑x͒ ϭ cos x ϭ ͑ Ϫ ͒ y y xnϩ1 xn 2 axn can be used to approximate 1͞a when x is an initial guess of 6 g 3 1 f the reciprocal of a. Note that this method of approximating 2 4 f reciprocals uses only the operations of multiplication and subtraction. (Hint: Consider 2 x 1 −ππ f ͑x͒ ϭ Ϫ a.΃ x x π 3π −1 g 2 2 26. Approximating Reciprocals Use the result of Exercise 1 1 25 to approximate (a) 3 and (b) 11 to three decimal places.

33. Minimum Time You are in a boat 2 miles from the WRITING ABOUT CONCEPTS nearest point on the coast (see figure). You are to go to a point 27. Using Newton’s Method Consider the function Q that is 3 miles down the coast and 1 mile inland. You can row f͑x͒ ϭ x3 Ϫ 3x2 ϩ 3. at 3 miles per hour and walk at 4 miles per hour. Toward what (a) Use a graphing utility to graph f. point on the coast should you row in order to reach Q in the least time? (b) Use Newton’s Method to approximate a zero with ϭ x1 1 as an initial guess. ϭ 1 (c) Repeat part (b) using x1 4 as an initial guess and observe that the result is different. 2 mi (d) To understand why the results in parts (b) and (c) are x 3 − x different, sketch the tangent lines to the graph of f at ͑ ͑ ͒͒ ͑1 ͑1͒͒ 1 mi the points 1, f 1 and 4, f 4 . Find the x -intercept of each tangent line and compare the intercepts with the 3 mi Q first iteration of Newton’s Method using the respective initial guesses. 34. Crime The total number of arrests T (in thousands) for all (e) Write a short paragraph summarizing how Newton’s males ages 15 to 24 in 2010 is approximated by the model Method works. Use the results of this exercise to ϭ 4 Ϫ 3 ϩ 2 Ϫ ϩ describe why it is important to select the initial guess T 0.2988x 22.625x 628.49x 7565.9x 33,478 carefully. for 15 Յ x Յ 24, where x is the age in years (see figure). 28. Using Newton’s Method Repeat the steps in Approximate the two ages that had total arrests of Exercise 27 for the function f͑x͒ ϭ sin x with initial 300 thousand. (Source: U.S. Department of Justice) guesses of x ϭ 1.8 and x ϭ 3. 1 1 T 29. Newton’s Method In your own words and using a sketch, describe Newton’s Method for approximating the 400 zeros of a function. 350 300 250 200 30. HOW DO YOU SEE IT? For what value(s) will 150

Newton’s Method fail to converge for the function Arrests (in thousands) 100 shown in the graph? Explain your x reasoning. 15 16 17 18 19 20 21 22 23 24 Age (in years) y

4 True or False? In Exercises 35–38, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

x p͑x͒ − − − 35. The zeros of f͑x͒ ϭ coincide with the zeros of p͑x͒. 6244 2 q͑x͒ −2 36. If the coefficients of a polynomial function are all positive, −4 then the polynomial has no positive zeros. 37. If f͑x͒ is a cubic polynomial such that fЈ͑x͒ is never zero, then any initial guess will force Newton’s Method to converge to the zero of f. Using Newton’s Method Exercises 31–33 present problems similar to exercises from the previous sections of this 38. The roots of Ίf͑x͒ ϭ 0 coincide with the roots of f͑x͒ ϭ 0. chapter. In each case, use Newton’s Method to approximate the ͑ ͒ ϭϪ solution. 39. Tangent Lines The graph of f x sin x has infinitely many tangent lines that pass through the origin. Use Newton’s 31. Minimum Distance Find the point on the graph of Method to approximate to three decimal places the slope of the f ͑x͒ ϭ 4 Ϫ x2 that is closest to the point ͑1, 0͒. tangent line having the greatest slope. 32. Medicine The concentration C of a chemical in the 40. Point of Tangency y The f(x) = cos x bloodstream t hours after injection into muscle tissue is given graph of f͑x͒ ϭ cos x and a by tangent line to f through the origin are shown. Find the 3t2 ϩ t C ϭ . coordinates of the point of x ϩ t3 π 2π 50 tangency to three decimal When is the concentration the greatest? places. −1

Section 3.8 (page 229) 29. Answers will vary. Sample answer:

If f is a function continuous on 1 In the answers for Exercises 1 and 3, the values in the tables have x f(x) ͓ ͔ ͑ ͒ 1 a, b and differentiable on a, b , x been rounded for convenience. Because a calculator and a computer 2 x where c ʦ ͓a, b͔ and f͑c͒ 0, c b program calculates internally using more digits than they display, −1 x 2 then Newton’s Method uses a 3 you may produce slightly different values from those shown in the −1 tables. tangent lines to approximate c. First, estimate an initial x1 close −2 1. to c. (See graph.) Then determine f ͑x ͒ f ͑x ͒ x using x x f ͑x ͒͞f͑x ͒. Calculate a third estimate x n x ͑ ͒ ͑ ͒ n n 2 2 1 1 1 3 n f xn f xn ͑ ͒ xn ͑ ͒ ͑ ͒͞ ͑ ͒ f xn f xn using x3 x2 f x2 f x2 . Continue this process until Խx x Խ is within the desired accuracy, and let x be the n n 1 n1 1 2.2000 0.1600 4.4000 0.0364 2.2364 final approximation of c. 2 2.2364 0.0015 4.4728 0.0003 2.2361 31. ͑1.939, 0.240͒ 33. x Ϸ 1.563 mi x2 1 35. False; let f͑x͒ . 37. True 39. 0.217 3. x 1 f ͑x ͒ f ͑x ͒ n x ͑ ͒ ͑ ͒ n n n f xn f xn ͑ ͒ xn ͑ ͒ f xn f xn 1 1.6 0.0292 0.9996 0.0292 1.5708

2 1.5708 0 1 0 1.5708

5. 1.587 7. 0.682 9. 1.250, 5.000 11. 0.900, 1.100, 1.900 13. 1.935 15. 0.569 17. 4.493 19. (a) Proof (b) Ί5 Ϸ 2.236; Ί7 Ϸ 2.646 ͑ ͒ 21. f x1 0 23. 0.74 25. Proof 27. (a) 4 (b) 1.347 (c) 2.532

−45

−2

4 − (d) y = 3x + 4 x-intercept of y 3x 4 is 3. y x-intercept of f y 1.313x 3.156 is approximately 2.404. 3

x −2 1 4 5

y = −1.313x + 3.156 (e) If the initial estimate x x1 is not sufficiently close to the desired zero of a function, then the x- intercept of the corresponding tangent line to the function may approximate a second zero of the function. y