THÔNG TIN TÀI LIỆU
G APPENDIX Newton’s Method: Using Derivatives to Approximate Roots Many problems involve finding roots of equations, and many equations we are unable to solve exactly. Consider for instance, finding the solutions to x 4 = 2x + 2. By graphing y = x 4 and y = 2x + 2wefind the graphs intersect at two points. The x-coordinates of the points correspond to the roots of f(x)=x 4 −2x −2. –2 –1 –1–2 –1 –2 –2 –3 12 1 1 2 3 4 5 2 y y x x 12 f(x) = x 4 –2x – 2 f(x) = x 4 f(x) = 2x + 2 Figure G.1 The roots appear to be in the vicinity of x =−0.8 and x = 1.5. Zooming in on the graphs would help us get more accurate answers, but the method we are about to present, Newton’s method, is more efficient. Newton’s method is a method of obtaining successive approximations to a root of a function. It uses the derivative to hone in on the root and generally produces a high degree of accuracy within a few iterations. The key notions are that it is simple to find the root of a linear equation, and the tangent line is a good linear approximation of a function about a point. 1121 1122 APPENDIX G Newton’s Method: Using Derivatives to Approximate Roots Let’s begin by finding the root of a linear equation. We will use the result of the next example repeatedly in applying Newton’s method. ◆ EXAMPLE G.1 Find the x-intercept of the line through (x k , y k ) with slope m, where m = 0. SOLUTION The equation of the line is y − y k = m(x − x k ). Set y = 0 and solve for x. mx − mx k =−y k mx = mx k − y k x = x k − y k m ◆ We can use this result to refine an approximation of the root, r, of an arbitrary differentiable function, f(x),asfollows. Approximate the root. Call this initial guess x 0 . For instance, in our example f(x)= x 4 −2x −2, one of the roots appears to be around 1.5, so let x 0 = 1.5. Find out where the tangent line to f at x 0 intersects the x-axis. Call this point x 1 . Generally 1 the point x 1 will be closer to r than was x 0 . y x f(x) (x 0 , y 0 ) x 1 x 0 r Figure G.2 From Example 2 we know x = x 0 − y 0 m , where y 0 = f(x 0 ) and m = the slope of the tangent line at x 0 = f (x 0 ).So x 1 =x 0 − f(x 0 ) f (x 0 ) . x 1 is our new guess. Repeat the process. x 2 is the x-intercept of the tangent line at x 1 . x 2 = x 1 − f(x 1 ) f (x 1 ) 1 If x 1 is not closer to r than x 0 , use another first guess. Newton’s Method: Using Derivatives to Approximate Roots 1123 x f(x) (x 0 , y 0 ) (x 1 , y 1 ) x 1 x 2 x 0 r Figure G.3 By repeating the process we can produce a sequence x 0 , x 1 , x 2 , of successive approxi- mations of r, where x k+1 = x k − f(x k ) f (x k ) . REMARK If, at any stage, f(x k )is zero, then we have found the root exactly. If, at any stage, f (x k ) is zero, then we’re stuck and must begin again. Newton’s Method for Approximating a Root of f i. Take an initial guess and call it x 0 . ii. Successive approximations x 1 , x 2 , x 3 , are found using x k+1 = x k − f(x k ) f (x k ) , f (x k ) = 0. Generally, the sequence of approximations will approach the root r as n increases, provided the initial guess was close enough to the root. ◆ EXAMPLE G.2 Use Newton’s method to approximate the larger of the two roots of f(x)=x 4 −2x −2. SOLUTION Our first guess is x 0 = 1.5, f(x 0 )=0.0625. f (x) = 4x 3 − 2 f (1.5) = 4(1.5) 3 − 2 = 11.5 x 1 = x 0 − f(x 0 ) f (x 0 ) = 1.5 − f(1.5) f (1.5) = 1.5 − 0.0625 11.5 ≈ 1.494565 f(1.494565) ≈ 3.9 × 10 −4 We have a high degree of accuracy in just one step. To get more accuracy repeat the process. ◆ 1124 APPENDIX G Newton’s Method: Using Derivatives to Approximate Roots Notice that a computer or programmable calculator can be programmed to perform iterations and arrive at a sequence of approximations with ease. Ways to be Led Astray We’ve already remarked that Newton’s method only makes sense when f (x k ) = 0. If f (x k ) = 0, then the tangent line is horizontal and does not intersect the x-axis. If this happens, start the process again. There are a few other problems one can run into, as illustrated in the figures below. y f(x) f(x) rr x x k x k x k + 1 x k + 1 y x The tangent line is too close to being horizontal. y f(x) f(x) r x y x x k x k x k + 1 x k + 1 The sequence converges to a different root of f. The lines tangent to f at x k and x k + 1 are parallel and Newton's method goes into a loop. The sequence of approximations does not converge. x k is too far from r. (a) (c) (b) (d) Figure G.4 When can we be guaranteed that Newton’s method won’t lead us astray? It can be shown that if x 0 is chosen in an interval around r over which f is monotonic and doesn’t change concavity, then the successive approximations will converge to r. In fact, the criteria can be weakened. Newton’s method will work if, in the interval between x 0 and r, the graph of f protrudes toward the x-axis, that is, f is concave up when f(x 0 )>0orconcave down when f(x 0 )<0.But even when the criteria do not hold, Newton’s method may still converge to the root. EXERCISE G.1 Verify for yourself that the graphs in Figure G.4 fail to meet these criteria. Newton’s Method: Using Derivatives to Approximate Roots 1125 ◆ EXAMPLE G.3 Approximate 3 √ 17 using Newton’s method, continuing until two successive approximations agree to five decimal places. SOLUTION This question is equivalent to finding the root x 3 − 17 = 0. f(x)=x 3 −17 f (x) = 3x 2 We know 2 < 3 √ 17 < 3sowe’ll let x 0 = 2.5. x 1 = x 0 − f(x 0 ) f (x 0 ) = 2.5 − f(2.5) f (2.5) = 2.57 3 x 2 = x 1 − f(x 1 ) f (x 1 ) = 2.57 3 − f(2.57 3) f (2.573) ≈ 2.571283226 x 3 = x 2 − f(x 2 ) f (x 2 ) = 2.571283226 − f(2.571283226) f (2.571283226) ≈ 2.57128591 x 3 and x 2 agree to five decimal places. Check: 2.57128 3 = 16.99996 and (2.571281591) 3 = 17.000085. ◆ PROBLEMS FOR APPENDIX G In Problems 1 through 5, approximate the quantity indicated. Use Newton’s method until two successive approximations agree to three decimal places. You are expected to use a calculator or computer. 1. The solution to ln x + x = 0. 2. The positive root of e x − x = 3. 3. The negative root of e x − x = 3. 4. 3 √ 20 5. Write a computer program to approximate roots using Newton’s method. 6. Approximate the negative root of x 4 − 3x 2 + 2. Use Newton’s Method until two successive approximations agree to five decimal places. 7. Approximate the largest root of x 3 − 3x 2 =−2. Use Newton’s method until two successive approximations agree to three decimal places. 8. Approximate the solution(s) to cos x = x. Use Newton’s method until two successive approximations agree to 5 decimal places. 9. Approximate the solution(s) to sin x =x 2 . Use Newton’s method until two successive approximations agree to four decimal places. H APPENDIX Proofs to Accompany Chapter 30, Series We begin this Appendix with a proof of Taylor’s Theorem. The proof rests on Rolle’s Theorem, which says that if f is continuous on [a, b] and differentiable on (a, b) and f(a)=f(b),then there exists a number c ∈ (a, b) such that f (c) = 0. Although the proof of Taylor’s Theorem will not seem natural, it gets the job done. Taylor’s Theorem Suppose f and all its derivatives exist in an open interval I centered at x = b. Then for each x in I , f(x)=f(b)+f (b)(x − b) + f (b) 2! (x − b) 2 +···+ f (n) (b) n! (x − b) n + R n (x), where R n (x) = f (n+1) (c) (n+1)! (x − b) n+1 , for some number c ∈ I , c between a and b. Proof Let a ∈ I . The function F given below is constructed to satisfy the conditions of Rolle’s Theorem. F(x)= (a − b) n+1 f(a)−f(x)−f (x)(a − x) − f (x) 2! (a − x) 2 −···− f (n) (x) n! (a − x) n − (a − x) n+1 f(a)−f(b)−f (b)(a − b) − f (b) 2! (a − b) 2 −···− f (n) (b) n! (a − b) n This is constant; denote it by k. 1127 1128 APPENDIX H Proofs to Accompany Chapter 30, Series F(a)=0and F(b)=0. By Rolle’s Theorem, there exists c between a and b such that F (c) = 0. Compute F (x). F (x) = (a − b) n+1 −f (x) + f (x) − f (x)(a − x) + f (x) − f (x) 2! (a − x) 2 + f (x) 2! (a − x) 2 −··· − f (n+1) (x) n! (a − x) n + (n + 1)k(a − x) n F (x) = (a − b) n+1 − f (n+1) (x) n! (a − x) n + (n + 1)k(a − x) n Factor out (a − x) n . F (x) = (a − x) n − f (n+1) (x) n! (a − b) n+1 + (n + 1)k There exists c between a and b such that F (c) = 0. For such c we have (a − c) n − f (n+1) (c) n! (a − b) n+1 + (n + 1)k = 0 k = f (n+1) (c) (n + 1)n! (a − b) n+1 f(a)−f(b)−f (b)(a − b) − f (b) 2! (a − b) 2 −···− f (n) (b) n! (a − b) n = f (n+1) (c) (n + 1)! (a − b) n+1 Solving for f(a)gives f(a)=f(b)+f (b)(a − b) + f (b) 2! (a − b) 2 +···+ f (n) (b) n! (a − b) n + f (n+1) (c) (n + 1)! (a − b) n+1 . Since a can be any number in I, we have completed the proof. Absolute Convergence Implies Convergence If ∞ k=1 |a k | converges, then ∞ k=1 a k converges. Proof Suppose ∞ k=1 |a k | converges, and ∞ k=1 |a k |=S.Then ∞ k=1 2|a k |=2S. |a k |is either a k or − a k ,so0≤a k +|a k |≤2|a k |. Proofs to Accompany Chapter 30, Series 1129 Therefore ∞ k=1 (a k +|a k |)is convergent; its sum is ≤ 2S. ∞ k=1 a k = ∞ k=1 (a k +|a k |)− ∞ k=1 |a k |, the difference of two convergent series. We conclude that ∞ k=1 a k converges. A Convergence Theorem for Power Series If a power series ∞ k=0 a k x k converges at x = c, c = 0 and |∝|<|c|,then the series converges absolutely for x =∝. Proof We assume a 0 + a 1 c + a 2 c 2 +···+a k c k +···converges. Therefore lim k→∞ a k c k = 0. Then there exists a positive constant M such that |a k c k | <Mfor all k, k = 0, 1, 2 0≤|a 0 |+|a 1 ∝|+|a 2 ∝ 2 |+|a 3 ∝ 3 |+···+|a n ∝ n | =|a 0 |+|a 1 c| ∝ c +|a 2 c 2 | ∝ c 2 +···+|a n c n | ∝ c n ≤ M + M ∝ c + M ∝ c 2 +···+M ∝ c n But this is a geometric sum with |r| < 1, so 0 ≤ n k=0 M ∝ c k = M 1 − ∝ c . Hence the sequence of partial sums of ∞ k=0 |a k ∝ k | is increasing (all terms are positive) and bounded below by zero and above by M 1− ∝ c . Therefore the sequence of partial sums converges and consequently ∞ k=0 |a k ∝ k | converges. ∞ k=0 a k x k converges absolutely at x =∝. The Integral Test for the Convergence of a Series Let ∞ k=1 a k be a series such that a k = f(k)for k = 1, 2, 3, where the function f is positive, continuous, and decreasing on [1, ∞). Then ∞ k=1 a k and ∞ 1 f(x)dx either both converge or both diverge. 1130 APPENDIX H Proofs to Accompany Chapter 30, Series Proof Given the hypotheses of the theorem, we can illustrate the relationship between the series and the integral in two ways, shown in Figure H.1 (with circumscribed rectangles) and Figure H.2 (with inscribed rectangles). y x 123456 The sum of the area of the shaded rectangles is a 1 + a 2 + a 3 + a 4 + a 5 a 1 a 2 a 3 a 4 a 5 y = f(x) (2, f(2)) Figure H.1 y x 123456 The sum of the area of the shaded rectangles is a 2 + a 3 + a 4 + a 5 + a 6 a 2 a 3 a 4 a 5 a 6 y = f(x) (2, f(2)) Figure H.2 Each rectangle has a base of 1, so its area corresponds to its height. In Figure H.1 the kth shaded rectangle has area = a k ; in Figure H.2 the kth shaded rectangle has area = a k+1 . Figure H.2 can be obtained from Figure H.1 by shifting the rectangles left one unit and dropping the rectangle corresponding to a 1 , from consideration. a 2 + a 3 +···+a n ≤ n 1 f(x)dx ≤a 1 +a 2 +···+a n−1 (H.1) We begin by demonstrating that the behavior of the series can be determined from that of the integral. The terms of the series are positive; therefore the sequence of partial sums is increasing. Suppose ∞ 1 f(x)dx converges. Denote ∞ 1 f(x)dx by A. S n = a 1 + a 2 + a 3 +···+a n ≤a 1 + n 1 f(x)dx . sequence of partial sums of ∞ k=0 |a k ∝ k | is increasing (all terms are positive) and bounded below by zero and above by M 1− ∝ c . Therefore the sequence of partial sums converges and. equation of the line is y − y k = m(x − x k ). Set y = 0 and solve for x. mx − mx k =−y k mx = mx k − y k x = x k − y k m ◆ We can use this result to refine an approximation of the root, r, of an arbitrary. 1 y x The tangent line is too close to being horizontal. y f(x) f(x) r x y x x k x k x k + 1 x k + 1 The sequence converges to a different root of f. The lines tangent to f at x k and x k +
Ngày đăng: 05/07/2014, 18:20
Xem thêm: Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 115 pps