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1.2 Answers for Numerical Methods 633 ANSWERS FOR NUMERICAL METHODS Exercise Set 1.2 (Page 000) 1. For each part, f ¢®ô C[a, b] on the given interval. Since f(a) and f(b) are of opposite sign, the Intermediate Value Theorem implies a number c exists with f(c) = 0. 3. For each part, f ¢®ô C[a, b], f exists on (a, b), and f(a) = f(b) = 0. Rolles Theorem implies that a number c exists in (a, b) with f(c) = 0. For part (d), we can use [a, b] = [−1, 0] or [a, b] = [0, 2]. 5. a. P2(x) = 0 b. R2(0.5) = 0.125; actual error = 0.125 c. P2(x) = 1+3(x − 1) + 3(x − 1)2 d. R2(0.5) = −0.125; actual error = −0.125 7. Since P2(x) = 1+x and R2(x) = −2e¥î(sin ¥î + cos ¥î) 6 x3 for some number ¥î between x and 0, we have the following: a. P2(0.5) = 1.5 and f(0.5) = 1.446889. An error bound is 0.093222 and |f(0.5)− P2(0.5)| ¢®Â 0.0532 b. |f(x) − P2(x)| ¢®Â 1.252 c. 1 0 f(x) dx ≈ 1.5 d. | 1 0 f(x) dx− 1 0 P2(x) dx| ¢®Â 1 0 |R2(x)| dx ¢®Â 0.313, and the actual error is 0.122. 9. The error is approximately 8.86 ¢®¢¯ 10−7. June 29, 2002 1:10 P.M. 634 CHAPTER 1 Answers for Numerical Methods 11. a. P3(x) = 1 3x + 1 6x2 + 23 648x3 b. We have f(4)(x) = −199 2592 ex/2 sin x 3 + 61 3888ex/2 cos x 3 , so |f(4)(x)| ¢®Â |f(4)(0.60473891)| ¢®Â 0.09787176 for 0 ¢®Â x ¢®Â 1, and |f(x) − P3(x)| ¢®Â |f(4)(¥î)| 4! |x|4 ¢®Â 0.09787176 24 (1)4 = 0.004077990. 13. A bound for the maximum error is 0.0026.
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