11.9. Convergence of Taylor Series; Error Estimates. Taylor s Theorem

Transcription

1 11.9 Convergence of Taylor Series; Error Estimates Convergence of Taylor Series; Error Estimates This section addresses the two uestions left unanswered by Section 11.8: 1. When does a Taylor series converge to its generating function? 2. How accurately do a function s Taylor polynomials approximate the function on a given interval? Taylor s Theorem We answer these uestions with the following theorem.

2 812 Chapter 11: Infinite Seuences and Series THEOREM 22 Taylor s Theorem If ƒ and its first n derivatives ƒ, ƒ, Á, ƒ snd are continuous on the closed interval between a and b, and ƒ snd is differentiable on the open interval between a and b, then there exists a number c between a and b such that ƒsbd = ƒsad + ƒ sadsb - ad + ƒ sad + ƒsnd sad n! sb - ad 2 + Á sb - ad n + ƒsn + 1d scd sb - adn + 1. Taylor s Theorem is a generalization of the Mean Value Theorem (Exercise 39). There is a proof of Taylor s Theorem at the end of this section. When we apply Taylor s Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor s formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change. Taylor s Formula If ƒ has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, ƒsxd = ƒsad + ƒ sadsx - ad + ƒ sad + ƒsnd sad n! sx - ad n + R n sxd, sx - ad 2 + Á (1) where R n sxd = f sn + 1d scd sx - adn + 1 for some c between a and x. (2) When we state Taylor s theorem this way, it says that for each x H I, The function R n sxd is determined by the value of the sn + 1dst derivative ƒ at a point c that depends on both a and x, and which lies somewhere between them. For any value of n we want, the euation gives both a polynomial approximation of ƒ of that order and a formula for the error involved in using that approximation over the interval I. Euation (1) is called Taylor s formula. The function R n sxd is called the remainder of order n or the error term for the approximation of ƒ by P n sxd over I. If R n sxd : 0 as n : for all x H I, we say that the Taylor series generated by ƒ at x = a converges to ƒ on I, and we write Often we can estimate R n ƒsxd = P n sxd + R n sxd. ƒsxd = a ƒ skd sad k! sx - ad k. sn + 1d without knowing the value of c, as the following example illustrates.

3 11.9 Convergence of Taylor Series; Error Estimates 813 EXAMPLE 1 The Taylor Series for Revisited e x Show that the Taylor series generated by ƒsxd = e x at x = 0 converges to ƒ(x) for every real value of x. Solution The function has derivatives of all orders throughout the interval s -, d. Euations (1) and (2) with ƒsxd = e x and a = 0 give I = e x = 1 + x + x2 + Á + xn n! + R nsxd Polynomial from Section 11.8, Example 2 and Since e x is an increasing function of x, e c lies between e 0 = 1 and e x. When x is negative, so is c, and e c 6 1. When x is zero, e x = 1 and R n sxd = 0. When x is positive, so is c, and e c 6 e x. Thus, and Finally, because lim R nsxd = 0, n: R n sxd = ƒ R n sxd ƒ 6 e x x n + 1 when x 7 0. lim n: and the series converges to e x = a e c xn + 1 for some c between 0 and x. ƒ R n sxd ƒ ƒ x ƒ n + 1 when x 0, x n + 1 = 0 for every x, e x for every x. Thus, x k k! = 1 + x + x2 + Á + xk k! + Á. Section 11.1 (3) Estimating the Remainder It is often possible to estimate R n sxd as we did in Example 1. This method of estimation is so convenient that we state it as a theorem for future reference. THEOREM 23 The Remainder Estimation Theorem If there is a positive constant M such that ƒ ƒ sn + 1d std ƒ M for all t between x and a, inclusive, then the remainder term R n sxd in Taylor s Theorem satisfies the ineuality ƒ R n sxd ƒ M ƒ n + 1 x - a ƒ. If this condition holds for every n and the other conditions of Taylor s Theorem are satisfied by ƒ, then the series converges to ƒ(x).

4 814 Chapter 11: Infinite Seuences and Series We are now ready to look at some examples of how the Remainder Estimation Theorem and Taylor s Theorem can be used together to settle uestions of convergence. As you will see, they can also be used to determine the accuracy with which a function is approximated by one of its Taylor polynomials. EXAMPLE 2 The Taylor Series for sin x at x = 0 Show that the Taylor series for sin x at x = 0 converges for all x. Solution The function and its derivatives are ƒsxd = sin x, ƒ sxd = cos x, ƒ sxd = o - sin x, ƒ sxd = - cos x, o so ƒ (2k) sxd = s -1d k sin x, ƒ (2k + 1) sxd = s -1d k cos x, f s2kd s0d = 0 and f s2k + 1d s0d = s -1d k. The series has only odd-powered terms and, for n = 2k + 1, Taylor s Theorem gives sin x = x - x3 3! + x5 5! - Á + s -1dk x 2k + 1 s2k + 1d! All the derivatives of sin x have absolute values less than or eual to 1, so we can apply the Remainder Estimation Theorem with M = 1 to obtain ƒ R 2k + 1 sxd ƒ 1 # ƒ x ƒ 2k + 2 s2k + 2d!. + R 2k + 1 sxd. Since sƒ x ƒ 2k + 2 >s2k + 2d!d : 0 as k :, whatever the value of x, R 2k + 1 sxd : 0, and the Maclaurin series for sin x converges to sin x for every x. Thus, sin x = a s -1d k x 2k + 1 s2k + 1d! = x - x3 3! + x5 5! - x7 7! + Á. (4) EXAMPLE 3 The Taylor Series for cos x at x = 0 Revisited Show that the Taylor series for cos x at x = 0 converges to cos x for every value of x. Solution We add the remainder term to the Taylor polynomial for cos x (Section 11.8, Example 3) to obtain Taylor s formula for cos x with n = 2k: cos x = 1 - x2 + x4 4! - Á + s -1d k x 2k s2kd! + R 2ksxd.

5 11.9 Convergence of Taylor Series; Error Estimates 815 Because the derivatives of the cosine have absolute value less than or eual to 1, the Remainder Estimation Theorem with M = 1 gives For every value of x, R 2k : 0 as k :. Therefore, the series converges to cos x for every value of x. Thus, cos x = a ƒ R 2k sxd ƒ 1 # ƒ x ƒ 2k + 1 s2k + 1d!. s -1d k x 2k s2kd! = 1 - x2 + x4 4! - x6 6! + Á. (5) EXAMPLE 4 Finding a Taylor Series by Substitution Find the Taylor series for cos 2x at x = 0. Solution We can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x: cos 2x = a = 1-22 x 2 = a s -1d k s2xd 2k s2kd! + 24 x 4 4! s -1d k 2 2k x 2k s2kd!. = 1 - s2xd2-26 x 6 6! + Á + s2xd4 4! - s2xd6 6! + Á Euation (5) with 2x for x Euation (5) holds for - 6 x 6, implying that it holds for - 6 2x 6, so the newly created series converges for all x. Exercise 45 explains why the series is in fact the Taylor series for cos 2x. EXAMPLE 5 Finding a Taylor Series by Multiplication Find the Taylor series for x sin x at x = 0. Solution We can find the Taylor series for x sin x by multiplying the Taylor series for sin x (Euation 4) by x: x sin x = x ax - x3 3! + x5 5! - x7 7! + Á b = x 2 - x4 3! + x6 5! - x8 7! + Á. The new series converges for all x because the series for sin x converges for all x. Exercise 45 explains why the series is the Taylor series for x sin x. Truncation Error The Taylor series for at x = 0 converges to for all x. But we still need to decide how many terms to use to approximate e x to a given degree of accuracy. We get this information from the Remainder Estimation Theorem. e x e x

6 816 Chapter 11: Infinite Seuences and Series EXAMPLE 6 Calculate e with an error of less than Solution We can use the result of Example 1 with x = 1 to write e = Á + 1 n! + R ns1d, with R n s1d = e c 1 for some c between 0 and 1. For the purposes of this example, we assume that we know that e 6 3. Hence, we are certain that 1 6 R ns1d 6 3 because 1 6 e c 6 3 for 0 6 c 6 1. By experiment we find that 1>9! , while 3>10! Thus we should take sn + 1d to be at least 10, or n to be at least 9. With an error of less than 10-6, e = ! + Á + 1 9! L EXAMPLE 7 For what values of x can we replace sin x by x - sx 3 >3!d with an error of magnitude no greater than 3 * 10-4? Solution Here we can take advantage of the fact that the Taylor series for sin x is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorem (Section 11.6), the error in truncating after sx 3 >3!d is no greater than sin x = x - x3 3! x5 x ` 5 5! ` = ƒ x ƒ ! - Á Therefore the error will be less than or eual to 3 * 10-4 if ƒ x ƒ * 10-4 or ƒ x ƒ * 10-4 L Rounded down, to be safe The Alternating Series Estimation Theorem tells us something that the Remainder Estimation Theorem does not: namely, that the estimate x - sx 3 >3!d for sin x is an underestimate when x is positive because then x 5 >120 is positive. Figure shows the graph of sin x, along with the graphs of a number of its approximating Taylor polynomials. The graph of P 3 sxd = x - sx 3 >3!d is almost indistinguishable from the sine curve when - 1 x 1.

7 11.9 Convergence of Taylor Series; Error Estimates 817 y 2 1 P 1 P 5 P 9 P 13 P 17 y sin x x 1 P 3 P 7 P 11 P 15 P 19 2 FIGURE The polynomials P 2n + 1 sxd = a n s -1d k x 2k + 1 s2k + 1d! converge to sin x as n :. Notice how closely P 3 sxd approximates the sine curve for x 6 1 (Example 7). You might wonder how the estimate given by the Remainder Estimation Theorem compares with the one just obtained from the Alternating Series Estimation Theorem. If we write sin x = x - x3 3! + R 3, then the Remainder Estimation Theorem gives ƒ R 3 ƒ 1 # ƒ 4 x ƒ 4! which is not as good. But if we recognize that x - sx 3 >3!d = 0 + x + 0x 2 - sx 3 /3!d + 0x 4 is the Taylor polynomial of order 4 as well as of order 3, then sin x = x - x3 3! R 4, and the Remainder Estimation Theorem with M = 1 gives ƒ R 4 ƒ 1 # ƒ x ƒ 5 This is what we had from the Alternating Series Estimation Theorem. 5! = ƒ 4 x ƒ 24, = ƒ x ƒ Combining Taylor Series On the intersection of their intervals of convergence, Taylor series can be added, subtracted, and multiplied by constants, and the results are once again Taylor series. The Taylor series for ƒsxd + gsxd is the sum of the Taylor series for ƒ(x) and g(x) because the nth derivative of f + g is f snd + g snd, and so on. Thus we obtain the Taylor series for s1 + cos 2xd>2 by adding 1 to the Taylor series for cos 2x and dividing the combined results by 2, and the Taylor series for sin x + cos x is the term-by-term sum of the Taylor series for sin x and cos x.

8 818 Chapter 11: Infinite Seuences and Series Euler s Identity As you may recall, a complex number is a number of the form a + bi, where a and b are real numbers and i = 2-1. If we substitute x = iu ( u real) in the Taylor series for e x and use the relations i 2 = -1, i 3 = i 2 i = -i, i 4 = i 2 i 2 = 1, i 5 = i 4 i = i, and so on, to simplify the result, we obtain e iu = 1 + iu 1! + i2 u 2 + i3 u 3 3! + i4 u 4 4! + i5 u 5 5! + i6 u 6 6! + Á = a1 - u2 + u4 4! - u6 6! + Á b + i au - u3 3! + u5 5! - Á b = cos u + i sin u. This does not prove that e iu = cos u + i sin u because we have not yet defined what it means to raise e to an imaginary power. Rather, it says how to define e iu to be consistent with other things we know. DEFINITION For any real number u, e iu = cos u + i sin u. (6) Euation (6), called Euler s identity, enables us to define to be e a # e bi e a + bi for any complex number a + bi. One conseuence of the identity is the euation e ip = -1. When written in the form e ip + 1 = 0, this euation combines five of the most important constants in mathematics. A Proof of Taylor s Theorem We prove Taylor s theorem assuming a 6 b. The proof for a 7 b is nearly the same. The Taylor polynomial P n sxd = ƒsad + ƒ sadsx - ad + ƒ sad sx - ad 2 + Á + f snd sad n! and its first n derivatives match the function ƒ and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form Ksx - ad n + 1, where K is any constant, because such a term and its first n derivatives are all eual to zero at x = a. The new function f n sxd = P n sxd + Ksx - ad n + 1 and its first n derivatives still agree with ƒ and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = f n sxd agree with the original curve y = ƒsxd at x = b. In symbols, ƒsbd = P n sbd + Ksb - ad n + 1, or K = ƒsbd - P nsbd sb - ad n + 1. sx - ad n (7)

9 11.9 Convergence of Taylor Series; Error Estimates 819 With K defined by Euation (7), the function measures the difference between the original function ƒ and the approximating function for each x in [a, b]. We now use Rolle s Theorem (Section 4.2). First, because Fsad = Fsbd = 0 and both F and F are continuous on [a, b], we know that Next, because F sad = F sc 1 d = 0 and both F and F are continuous on [a, c 1 ], we know that Rolle s Theorem, applied successively to F, F, Á, F implies the existence of c n in sa, c n - 1d such that F snd sc n d = 0. Finally, because F snd is continuous on [a, c n ] and differentiable on sa, c n d, and F snd sad = F snd sc n d = 0, Rolle s Theorem implies that there is a number c n + 1 in sa, c n d such that If we differentiate Fsxd = ƒsxd - P n sxd - Ksx - ad n + 1 a total of n + 1 times, we get Euations (8) and (9) together give Euations (7) and (10) give This concludes the proof. c 3 in sa, c 2 d such that F sc 3 d = 0, c 4 in sa, c 3 d such that F s4d sc 4 d = 0, o Fsxd = ƒsxd - f n sxd F sc 1 d = 0 for some c 1 in sa, bd. F sc 2 d = 0 for some c 2 in sa, c 1 d. F sn + 1d sc n + 1 d = 0. sn - 1d F sn + 1d sxd = ƒ sn + 1d sxd K. K = ƒsn + 1d scd for some number c = c n + 1 in sa, bd. ƒsbd = P n sbd + ƒsn + 1d scd sb - adn + 1. f n (8) (9) (10)