Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 95 doc

30.1 Approximating a Function by a Polynomial 921 y y x x y = x y = x f(x) = sin x f(x) = sin x –2π 2π–1 –1 1 .1 .1 1 –ππ (a) (b) sin .1 Figure 30.1 Refining the tangent line approximation: Any polynomial approximation, P k (x), of sin x about x = 0 certainly ought to be as good a local approximation as is the tangent line approximation, P 1 (x) = x. Therefore, like sin x the graph of P k (x) must pass through (0, 0) and must have a slope of 1 at x = 0. This means that  P k (0) = 0 and P  k (0) = 1. P k (x) = a 0 + a 1 x + a 2 x 2 +···+a k x k so P k (0) = a 0 = 0. P  k (x) = a 1 + 2a 2 x + 3a 3 x 2 +···+ka k x k−1 so P  k (0) = a 1 = 1. Therefore, P k (x) is of the form x + a 2 x 2 + a 3 x 3 +···+a k x k . Note that sin x lies below the tangent line for x>0and above the tangent line for x<0. Therefore the approximation sin x ≈ x must be decreased for x>0and increased for x<0 in order to improve upon it. Second Degree Approximation P 2 (x) must be of the form x + a 2 x 2 , where a 2 is a constant. But a 2 x 2 cannot be negative for x>0and positive for x<0,as required; we cannot improve upon the tangent line approximation by using a second degree polynomial. We need at least a third degree polynomial to improve upon the tangent line approximation. 922 CHAPTER 30 Series Before moving on, let’s look at the second degree polynomial approximation from a geometric viewpoint. If P 2 (x) = a 0 + a 1 x + a 2 x 2 is to be the best parabolic approximation to f(x)=sin x about x = 0, then it must satisfy the following three conditions. It has the same value as sin x at x = 0. P 2 (0) = f(0) It has the same slope as sin x at x = 0. P  2 (0) = f  (0) It has the same concavity as sin x at x = 0. P  2 (0) = f  (0) Each one of these conditions determines the value of one coefficient of P 2 (x). The first two result in a 0 = 0 and a 1 = 1, respectively. The second derivative of sin x at x = 0 is zero. d 2 dx 2 sin x     x=0 = d dx cos x     x=0 =−sin x     x=0 = 0 The “parabola” must have a second derivative of zero; consequently, it is not a parabola at all. Third Degree Approximation To determine the coefficients of the third degree polynomial of best fit, we require that the polynomial, P 3 (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 , and f(x)= sin x agree at x = 0 and that each nonzero derivative of the polynomial is equal to the corresponding derivative of sin x at x = 0. These four conditions determine the four coefficients.        P 3 (0) = f(0) P  3 (0)=f  (0) P  3 (0) = f  (0) P  3 (0) = f  (0) We have already demonstrated that the first two conditions result in a 0 = 0, a 1 = 1. As an exercise, show that the third and fourth conditions require that a 2 = 0 and a 3 =− 1 6 , respectively. P 3 (x) = 0 + x + 0x 2 − 1 6 x 3 = x − 1 6 x 3 . Notice that the − 1 6 x 3 term is negative for x>0and positive for x<0,providing an appropriate adjustment to the tangent line approximation. (See Figure 30.2.) y x 2π–2π P 1 (x) = x P 3 (x) = x – f(x) = sin x x 3 6 Figure 30.2 EXERCISE 30.1 Let f(x)=sin x and P 3 (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 . Calculate P  (x), P  (x), P  (x), f  (x), f  (x), f  (x), and evaluate each at x = 0. Show that the four conditions given above 30.1 Approximating a Function by a Polynomial 923 determine a 0 , a 1 , a 2 , and a 3 , respectively and that a 0 = 0, a 1 = 1, a 2 = 0, and a 3 =− 1 6 . Using the third degree polynomial to approximate sin 0.1 gives sin 0.1 ≈ P 3 (0.1) = 0.1 − (0.1) 3 6 = 1 10 − 1 6000 = 0.0998 3. This matches the calculator estimate of sin 0.1 to six decimal places. The actual value is a bit larger than P 3 (0.1). Higher Degree Approximations To fi nd the kth degree polynomial approximation we require that P k (x) and sin x agree at x = 0 and each nonzero derivative of the polynomial matches that of sin x. The condition that the fourth derivatives agree ends up meaning that a 4 = 0, so we will proceed directly with the fifth degree polynomial. Let P 5 (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 . Requiring that all nonzero deriva- tives of P 5 (x) match the derivatives of f(x)=sin x at x = 0 means the following conditions must be satisfied.                  P 5 (0) = f(0) P  5 (0)=f  (0) P  5 (0) = f  (0) P  5 (0) = f  (0) P (4) 5 (0) = f (4) (0) where f (p) denotes the pth derivative of f P (5) 5 (0) = f (5) (0) As an exercise, show that these conditions determine a 0 , a 1 , a 2 , a 3 , a 4 , and a 5 , respectively, and that P 5 (x) = x − 1 6 x 3 + 1 120 x 5 . EXERCISE 30.2 Let f(x)=sin x and P 5 (x) be the fifth degree polynomial given above. Show that the six conditions stated mean that a 0 = f(0), a 1 =f  (0), a 2 = f  (0) 2! , a 3 = f  (0) 3! , a 4 = f (4) (0) 4! , a 5 = f (5) (0) 5! , where n! = n · (n − 1) ···3·2·1. Conclude that a 0 = 0, a 1 = 1, a 2 = 0, a 3 = −1 3! =− 1 6 , a 4 = 0 and a 5 = 1 5! = 1 120 . Using the fifth degree polynomial approximation to sin x to approximate sin(0.1) gives sin 0.1 ≈ P 5 (0.1) = 0.1 − (0.1) 3 6 + (0.1) 5 120 = 1 10 − 1 6000 + 1 12 · 10 6 ≈ 0.0998334166. This agrees with the 10 decimal places given for sin 0.1. 924 CHAPTER 30 Series y x 2π–2π P 1 (x) = x P 3 (x) = x – f(x) = sin x x 3 6 P 5 (x) = x – + x 3 6 x 5 120 Figure 30.3 The graphs of sin x, P 1 (x), P 3 (x), and P 5 (x) are given in Figure 30.3. ◆ EXERCISE 30.3 Using a computer or graphing calculator, graph f(x)=sin x, P 1 (x) = x, P 3 (x) = x − x 3 3! , and P 5 (x) = x − x 3 3! + x 5 5! . Zoom in around x = 0 and observe how well each polynomial approximates the values of sin x near x = 0. Try to guess formulas for P 7 (x), P 9 (x), and P 11 (x). Graph these as well and decide how much confidence you have in your answers. Below is a table of values given to 10 decimal places. x sin xP 1 (x) P 3 (x) P 5 (x) −0.2 −0.1986693308 −0.2 −0.1986666667 −0.1986693333 −0.1 −0.0998334166 −0.1 −0.0998333333 −0.0998334167 00 0 0 0 0.1 0.0998334166 0.1 0.0998333333 0.0998334167 0.5 0.4794255386 0.5 0.4791666667 0.4794270833 1 0.8414709848 1 0.8333333333 0.8416666667 2 0.9092974268 2 0.6666666667 0.9333333333 2.5 0.5984721441 2.5 −0.1041666667 0.7096354167 OBSERVATIONS From the graphical and numerical data gathered we observe that i. for a fixed x near zero, the higher the degree of the polynomial approximation the better its value approximates that of sin x, and ii. the higher the degree of the polynomial, the further away from zero the approximation is reasonable. NOTE In all the work we’ve done with sin x, x must be in radians, not in degrees. d dx sin x = cos x only for x in radians. Taylor Polynomial Approximations In the previous example we constructed polynomial approximations to f(x)=sin x around x = 0 by choosing the coefficients of the polynomial such that the polynomial and all its nonzero derivatives matched f and its corresponding derivatives at x = 0. This method of constructing polynomial approximations to a function f about a number x = b in its domain is remarkably useful. 30.1 Approximating a Function by a Polynomial 925 Let f be a function whose first n derivatives exist at x = b. For the sake of simplicity, we begin with the case b = 0. Definition The nth degree polynomial, P n (x), that is equal to f(0)when evaluated at x = 0 and whose first n derivatives are equal to those of f(x)when evaluated at x = 0 is called the nth degree Taylor polynomial generated by f at x = 0. The polynomial is said to be centered at x = 0, or expanded about x = 0. More generally, we can expand a function about x = b using a polynomial in powers of (x − b). P n (x) = a 0 + a 1 (x − b) + a 2 (x − b) 2 + a 3 (x − b) 3 +···+a n (x − b) n Definition The nth degree polynomial in powers of (x − b) that is equal to f(b)when evaluated at x = b and whose first n derivatives match those of f(x)at x = b is called the nth degree Taylor polynomial generated by f at x = b. We refer to b as the center of the polynomial. When evaluated at its center, a Taylor polynomial is equal to the value of its generating function. Our hope is that for x near the center the value of the polynomial is close to the value of the function. 1 We now turn our attention to computing Taylor polynomials. In the next section we will look at the accuracy of Taylor polynomial approximations, and subsequently will see what we get by allowing the degree of the Taylor polynomial to increase without bound. Computing a Taylor Polynomial Centered at x = 0 Suppose f and its first n derivatives exist at x = 0. We want to find constants a 0 , a 1 , a 2 , , a n such that P n (x) = a 0 + a 1 x + a 2 x 2 +···+a n x n is the Taylor polynomial generated by f about x = 0. We impose the following (n + 1) conditions; each enables us to solve for one coeffi- cient.                P n (0) = f(0) P  n (0)=f  (0) P  n (0) = f  (0) P  n (0) = f  (0) . . . P (n) n (0) = f (n) (0) (30.1) 1 While this is not always the case, often we will find it true. 926 CHAPTER 30 Series In short, P (k) n (0) = f (k) (0) for k = 0, 1, ,n. 2 Webegin by finding the first n derivatives of P n (x). P n (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 +···+ a n x n P  n (x) = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 +···+ na n x n−1 P  n (x) = 2 · a 2 + 3 · 2 · a 3 x + 4 · 3 · a 4 x 2 +···+ n(n − 1) · a n x n−2 P  n (x) = 3 · 2 · a 3 + 4 · 3 · 2 · a 4 x +···+ n(n − 1)(n − 2) · a n x n−3 P (4) n (x) = 4 · 3 · 2 · a 4 +···+ n(n − 1)(n − 2)(n − 3) · a n x n−4 . . . P (n−1) n (x) = (n − 1)(n − 2)(n − 3) ···3·2·a n−1 +n(n − 1)(n − 2) ···3·2·a n x P (n) n (x) = n(n − 1)(n − 2) ···3·2·a n Next we evaluate each expression at x = 0. P n (0) = a 0 P  n (0) = a 1 P  n (0) = 2 · a 2 P  n (0) = 3 · 2 · a 3 = 3!a 3 P (4) n (0) = 4!a 4 . . . P (n−1) n (0) = (n − 1)!a n−1 P (n) n (0) = n!a n We summarize: P (k) n (0) = k!a k for k = 0, 1, ,n.Returning to (30.1), the original (n + 1) conditions, we obtain 3 k!a k = f (k) (0) for k = 0, 1, ,n. Solving for a k , the coefficient of x k , we obtain a k = f (k) (0) k! for k = 0, 1, ,n.Wesumma- rize our result. The nth degree Taylor polynomial generated by f(x)at x = 0 is given by P n (x) = f(0)+f  (0)x + f  (0) 2! x 2 + f  (0) 3! +···+ f (n) (0) n! x n . That is, P n (x) = n  k=0 f (k) (0) k! x k . This work behind us, we compute the nth degree Taylor polynomial generated by f about x = 0 as follows. 2 Here we use the convention that P 0 (x) = P(x). 3 Recall: 0! = 1 and f (0) (x) = f(x). 30.1 Approximating a Function by a Polynomial 927 1. Compute the first n derivatives of f . Be alert to the possibility of patterns emerging. You improve your chances of noticing patterns by not multiplying out. For instance, 5 · 4 · 3 · 2 is easier to recognize as 5! than is 120. 2. Evaluate f and each of its derivatives at x = 0. 3. The coefficient of x k is the constant f (k) (0) k! . ◆ EXAMPLE 30.2 Find the nth degree Taylor polynomial generated by e x at x = 0. SOLUTION P n (x) = f(0)+f  (0)x + f  (0) 2! x 2 +···+ f (n) (0) n! x n The derivative of e x is e x ; therefore f (k) (0) = e 0 = 1 for k = 0, 1, 2, ,n.Thus P n (x) = 1 + x + x 2 2! + x 3 3! +···+ x n n! . ◆ Graphs of e x and several of its Taylor polynomials are shown in Figure 30.4. Note that P 1 (x) = 1 + x is simply the tangent line approximation to e x at x = 0. y y x x P 4 P 2 P 3 P 4 P 2 P 1 P 1 (x) P 1 (x) P 1 (x) = 1 + x P 2 (x) = 1 + x + P 3 (x) = 1 + x + P 4 (x) = 1 + x + P 2 (x) P 3 (x) P 4 (x) e x P 3 f(x) = e x f(x) = e x 15 10 5 5 –5 1 1–12 234 (a) (b) Magnification around x = 0 x 2 2! x 2 2! x 3 3! x 2 2! + x 3 3! + x 4 4! + Figure 30.4 On the following page is a table of values produced using Taylor polynomials for e x . Values are given to nine decimal places. 928 CHAPTER 30 Series xe x P 1 (x) P 2 (x) P 3 (x) P 4 (x) P 5 (x) 0.1 1.105170918 1.1 1.105 1.105166666 1.105170833 1.105170917 0.2 1.221402758 1.2 1.22 1.221333333 1.221400000 1.221402667 0.5 1.648721271 1.5 1.625 1.645833333 1.6484375 1.648697917 1 2.718281828 2 2.5 2.66666666 2.708333333 2.716666666 ◆ EXAMPLE 30.3 Find the 8th degree Taylor polynomial generated by f(x)=cos x about x = 0. SOLUTION P 8 (x) = f(0)+f  (0)x + f  (0) 2! x 2 +···+ f (8) (0) 8! x 8 f(x) =cos xf(0)=1f  (x) =−sin xf  (0)=0 f  (x) =−cos xf  (0) =−1 f  (x) = sin xf  (0) = 0 f (4) (x) = cos xf (4) (0)=1f (5) (x) =−sin xf (5) (0)=0 f (6) (x) =−cos xf (6) (0)=−1 f (7) (x) = sin xf (7) (0)=0 f (8) (x) = cos xf (8) (0)=1 P 8 (x) = 1 − x 2 2! + x 4 4! − x 6 6! + x 8 8! Notice that the coefficients of all the odd power terms are zero. This makes sense; cosine is an even function. Analogously, the coefficients of all even power terms in the expansion of sin x about x = 0 are zero since sin x is an odd function. ◆ The graph of cos x and some of its Taylor polynomials centered at x = 0 are given in Figure 30.5. (Graph them yourself and you can zoom in around x = 0.) P 4 (x) f(x) = cos x P 2 (x) = 1 – P 4 (x) = 1 – P 6 (x) = 1 – P 2 (x) P 6 (x) f(x) = cos x P 8 (x) x 2 2! x 2 2! x 4 4! x 2 2! + x 4 4! + x 6 6! – P 8 (x) = 1 – x 2 2! x 4 4! + x 6 6! + x 8 8! + 2π–2π y x Figure 30.5 Computing a Taylor Polynomial Centered at x = b Suppose we want to approximate ln 1.2 using a Taylor polynomial. We can’t use a Taylor polynomial for f(x)=ln x expanded about x = 0 because neither f nor any of its deriva- tives exist at x = 0. We can, however, either center the Taylor polynomial for ln(1 + x) at x = 0 or work with the Taylor polynomial for ln x expanded about x = 1. We will do the latter. First we will look at how to compute a Taylor polynomial centered at x = b. Recall that the nth degree Taylor polynomial for f(x) at x = b is an nth degree polynomial in powers of x − b, P n (x) = a 0 + a 1 (x − b) + a 2 (x − b) 2 +···+a n (x − b) n , 30.1 Approximating a Function by a Polynomial 929 such that the values of P n (x) and its nonzero derivatives are equal to those of f(x) when evaluated at x = b. That is, the coefficients a 0 , a 1 , a 2 , ,a n are determined by the conditions P (k) n (b) = f (k) (b) for k = 0, 1, 2, ,n. (30.2) As an exercise, calculate the first n derivatives of P n (x). (Don’t multiply out (x − b) k ; use the Chain Rule.) Evaluating these derivatives at a = b, conclude that P (k) n (b) = k!a k . This result, together with (30.2) enables us to solve for a k , k = 0, 1, 2, ,n. a k = f (k) (b) k! The nth degree Taylor polynomial generated by f(x)around x = b is given by P n (x) = f(b)+f  (b)(x − b) + f  (b) 2! (x − b) 2 +···+ f (n) (b) n! (x − b) n . That is, 4 P n (x) = n  k=0 f (k) (b) k! (x − b) k . Given a particular function, f , and center, b, we compute the first n derivatives of f , evaluate each at x = b, and use f (k) (b) k! as the coefficient of (x − b) k . Note: The equation of the line tangent to f at x = b is of the form y − y 1 = m(x − x 1 ), where (x 1 , y 1 ) = (b, f(b))and m = f  (b). The equation is therefore y = f(b)+f  (b)(x − b); this is P 1 (x), as expected. When b = 0we’re back to the Taylor polynomial centered at x = 0. ◆ EXAMPLE 30.4 (a) Find the nth degree Taylor polynomial for ln x centered at x = 1. (b) Use P 5 to estimate ln(1.2). SOLUTION (a) P n (x) = f(1)+f  (1)(x − 1) + f  (1) 2! (x − 1) 2 + f  (1) 3! (x − 1) 3 +···+ f (n) (1) n! (x − 1) n 4 In order to use this summation notation we must adopt the convention that (x − b) 0 = 1evenifx=b. 930 CHAPTER 30 Series Compute derivatives of ln x, looking for a pattern. f(x) =ln xf(1)=0 f  (x) = 1 x = x −1 f  (1) = 1 f  (x) =−x −2 f  (1) =−1=−1! f  (x) = 2 · x −3 f  (1) = 2 = 2! f (4) (x) =−3·2·x −4 f (4) (1)=−3·2=−3! f (5) (x) = 4 · 3 · 2 · x −5 f (5) (1) = 4 · 3 · 2 = 4! . . . . . . f (n) (x) = (−1) n+1 (n − 1)!x −n f (n) (1) = (−1) n+1 (n − 1)! P n (x) = 0 + 1(x − 1) + −1 2! (x − 1) 2 + 2! 3! (x − 1) 3 + −3! 4! (x − 1) 4 + ···+ (−1) n+1 (n − 1)! n! (x − 1) n P n (x) = (x − 1) − (x − 1) 2 2 + (x − 1) 3 3 − (x − 1) 4 4 +···+(−1) n+1 (x − 1) n n . More compactly, P n (x) = n  k=0 f (k) (1) k! (x − 1) k = n  k=0 (−1) k+1 (k − 1)! k! (x − 1) k = n  k=0 (−1) k+1 ( k − 1)! k · (k − 1)! (x − 1) k = n  k=0 (−1) k+1 (x − 1) k k . (b) ln x ≈ P 5 (x) = (x − 1) − (x − 1) 2 2 + (x − 1) 3 3 − (x − 1) 4 4 + (x − 1) 5 5 ln 1.2 ≈ P 5 (1.2) = 0.2 − (0.2) 2 2 + (0.2) 3 3 − (0.2) 4 4 + (0.2) 5 5 = 0.182330 6 Compare this with the actual value of ln 1.2; it matches for the first four decimal places. ◆ Aside: Dealing with Factorials and Alternating Signs Factorials: Parentheses are important. (2n)! = (2n) · (2n − 1) · (2n − 2) ···3·2·1 =2n·(2n−1)! On the other hand, 2n! = 2 · n! = 2[n · (n − 1) ···3·2·1]. Similarly, (2n + 1)! = 2n + 1! = 2n + 1. . x> 0and increased for x<0 in order to improve upon it. Second Degree Approximation P 2 (x) must be of the form x + a 2 x 2 , where a 2 is a constant. But a 2 x 2 cannot be negative for x> 0and. in around x = 0 and observe how well each polynomial approximates the values of sin x near x = 0. Try to guess formulas for P 7 (x), P 9 (x), and P 11 (x). Graph these as well and decide how much. multiplying out. For instance, 5 · 4 · 3 · 2 is easier to recognize as 5! than is 120. 2. Evaluate f and each of its derivatives at x = 0. 3. The coefficient of x k is the constant f (k) (0) k! . ◆ EXAMPLE