MATH 128A, SUMMER 2009: HOMEWORK 3 SOLUTIONS
3.1.5a. To minimize the error bound, we should take as our xithe (n+ 1) points nearest x= 8.4. When
n= 2 there’s a tie; both reasonable answers are shown below.
octave:1> format long
octave:2> neville(8.4,[8.3,8.6], [17.56492, 18.50515]) %n=1
ans = 17.8783300000000
octave:3> neville(8.4,[8.3,8.6,8.1], [17.56492, 18.50515, 16.94410]) %n=2
ans = 17.8771300000000
octave:4> neville(8.4,[8.3,8.6,8.7], [17.56492, 18.50515, 18.82091]) %n=2
ans = 17.8771550000000
octave:5> neville(8.4,[8.3,8.6,8.1,8.7], [17.56492, 18.50515, 16.94410, 18.82091]) %n=3
ans = 17.8771425000000
3.1.19c. There are many ways to find the polynomial. A correct answer, when expanded with a calculator,
should come out to P(x)=0.1970056667x3−1.06259055x2+ 2.532453189x−1.666868305. The
error, for x∈[1,1.4], is of the form ln(4) (ξ(x))
4! (x−1)(x−1.1)(x−1.3)(x−1.4). Since ln(4)(ξ) =
(−3)(−2)(−1)x−4=−6ξ−4, it’s at worst −6(1)−4=−6. |(x−1)(x−1.1)(x−1.3)(x−1.4)|is at
most 4 ·10−4, attained when x= 1.2. So P(x) approximates ln(x) to within 10−4on [1,1.4].
Extra. f[0,1,2] = f[1,2]−f[0,1]
2−0=
f[2]−f[1]
2−1−f[1]−f[0]
1−0
2=(8−1)−(1−0)
2= 3.
3.2.11. (a) Calculation shows P(x) = f(x) and Q(x) = f(x) for x=−2,−1,0,1,2.
(b) Both polynomials are equal (to x3−3x+ 1), just written in a different form.
3.3.1a. Construct a divided-difference table as follows.
x0= 8.3f[x0] = f(8.3) = 17.56492
f[x0, z0] = f0(8.3) = 3.116256
z0= 8.3f[z0] = f(8.3) = 17.56492 0.0594800
f[z0, x1] = · · · · · · = 3.134100 −0.002022222
x1= 8.6f[x1] = f(8.6) = 18.50515 0.0588733
f[x1, z1] = f0(8.3) = 3.151762
z1= 8.6f[z1] = f(8.6) = 18.50515
(See, e.g.,
Table 3.13 in the text.) The coefficients for the Newton form of the approximating polynomial are on the top
edge of this table: H(x) = 17.56492 +3.116256(x−8.3) +0.0594800(x−8.3)2−0.002022222(x−8.3)2(x−8.6).
Date: Due Thursday 7/09.
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