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Math 128A, Spring 2007
Homework 6 Solution
5.1.12 It’s enough to check exactness for 1, x, x^2 , x^3 , etc. The degree of precision is the first n for which our rule is not exact for xn+1. ∫ (^1)
0
1 dx = 1
0
xdx = 1/ 2
4 2 /3 = 1/^2
0
x^2 dx = 1/ 3 1 4
0 +^3
0
x^3 dx = 1/ 4
so the degree is 2.
5.1.14 We have
I(f ) =
− 1
f (x)
− 1
[f (− 1 /3)(3/2)(1/ 3 − x) + f (1/3)(3/2)(x + 1/3)]dx
= f (− 1 /3) + f (1/3)
and we can check the degree is 2 by the method of problem 12.
5.1.15 When f (x) = 1, we have
− 1 f^ (x)dx^ = 2 = 1 + 1 =^ f^ (−β) +^ f^ (β) for any β. When f (x) = x, we have
− 1 f^ (x)dx^ = 0 =^ −β^ +^ β^ =^ f^ (−β) +^ f^ (β) for any β. Hence the degree is at least 1. To make it work for f (x) = x^2 , we need
− 1 f^ (x)dx^ = 2/3 =^ f^ (−β) + f (β) = 2β^2 , i.e. β = ± 1 /
- This choice of β is also exact for f (x) = x^3 but not for f (x) = x^4. Hence the degree is 3.
5.2.1b By (5.23) with f (x) = e−x 2 and [a, b] = [0, 1],
I(f ) − Tn(f ) = − h^2 (b − a) 12 f ′′(cn) = (4c^2 n − 2)e−c (^2) n 12 n^2 with cn ∈ [0, 1]. Since |(4c^2 n − 2)e−c (^2) n | ≤ 2 we get a bound |I(f ) − Tn(f )| ≤
6 n^2
5.2.4 For the integral I(2)^ =
0
dx 1 + x^2 the error formula (5.32) says
I(2)^ − Tn(f ) ≈ − h
2 12 [f ′(b) − f ′(a)] = 32 867 n^2 Tabulating the error and error estimates,
n actual estimate 2 -1.33e-1 9.23e- 4 -3.59e-3 2.31e- 8 5.64e-4 5.77e- 16 1.44e-4 1.44e- 32 3.60e-5 3.60e- 64 9.01e-6 9.01e- 128 2.25e-6 2.25e-
5.2.5c As in the previous problem
I(f ) − Tn(f ) ≈ − h
2 12 [f ′(b) − f ′(a)] = − 4 e
− 4 3 n^2 We want |I(f ) − Tn(f )| ≤ 10 −^10 , so we solve 4 e−^4 3 n^2 ≤^10
− 10
and obtain n ≥
4 e−^4 3 · 10 −^10
≈ 1. 56 × 104
5.2.7 (a) By (5.23)
I(f ) − Tn(f ) = − h
(^2) (b − a) 12 f ′′(cn) = 8 12 n^2
c^2 n with cn ∈ [1, 3]. Since 1/c^2 n ≤ 1, we have
|I(f ) − Tn(f )| ≤ 2 3 n^2 By (5.32) I(f ) − Tn(f ) ≈ h^2 12 [f ′(b) − f ′(a)] = h^2 18 To insure that |I(f ) − Tn(f )| ≤ 5 × 10 −^8 it is enough to find n so that 2 3 n^2
≤ 5 × 10 −^8
