[해답] 수치해석 3판(Numerical Methods 3rd Edition, Faires & Burden)
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Download : [솔루션] 수치해석 3판(Numeri.zip
.^^ 참고!
3. For each part, f ∈ C[a, b], f exists on (a, b), and f(a) = f(b) = 0. Rolle’s Theorem
c.
9. The error is approximately 8.86 × 10−7.
1
0 f(x) dx−
6x2 + 23
for some number ξ between x and 0, we have the following:
[a, b] = [−1, 0] or [a, b] = [0, 2].
6 x3
|R2(x)| dx ≤ 0.313, and the actual error is 0.122.
c. P2(x) = 1+3(x − 1) + 3(x − 1)2
7. Since
1
|f(x) − P3(x)| ≤
−199
13. A bound for the maximum error is 0.0026.
레포트 > 공학,기술계열
|x|4 ≤ 0.09787176
61
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2592 ex/2 sin x
+
|f(4)(ξ)|
implies that a number c exists in (a, b) with f(c) = 0. For part (d), we can use
0 f(x) dx ≈ 1.5
기계, 공학, 전산, 연습문제, 솔루션, Numerical
[해답] 수치해석 3판(Numerical Methods 3rd Edition, Faires & Burden)
설명
d. |
b. We have
0
|f(4)(x)| ≤ |f(4)(0.60473891)| ≤ 0.09787176 for 0 ≤ x ≤ 1,
3
P2(0.5)| ≤ 0.0532
b. |f(x) − P2(x)| ≤ 1.252
4!
b. R2(0.5) = 0.125; actual error = 0.125
11. a. P3(x) = 1
(1)4 = 0.004077990.
f(4)(x) =
3x + 1
Exercise Set 1.2 (Page 000)
1.2 Answers for Numerical Methods 633
June 29, 2002 1:10 P.M.
a. P2(0.5) = 1.5 and f(0.5) = 1.446889. An error bound is 0.093222 and |f(0.5)−
648x3
1
3888ex/2 cos x
24
순서
d. R2(0.5) = −0.125; actual error = −0.125
634 CHAPTER 1 Answers for Numerical Methods
Download : [솔루션] 수치해석 3판(Numeri.zip( 40 )
0 P2(x) dx| ≤
5. a. P2(x) = 0
and
3 ,
1. For each part, f ∈ C[a, b] on the given interval. Since f(a) and f(b) are of opposite
다.
sign, the Intermediate Value Theorem implies a number c exists with f(c) = 0.
so
−2eξ(sin ξ + cos ξ)
1
P2(x) = 1+x and R2(x) =
ANSWERS FOR NUMERICAL METHODS
아래는 일부의 발췌 부분입니다.