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Faires&Burden : ¨ùöÄ¢®Ç¨ª¨ù¢ç 3¨¡Ç ¨ùÖ¡¤ç¨ùÇ (Numerical Methods 3rd Edition , Faires & Burden)

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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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