Homework Assignment #2

Solutions for Homework Assignment #2
2-1
n p n f(p n ) n p n f(p n )
1 0.5 −0.170476 6 0.640625 −0.00133182
2 0.75 0.134337 7 0.648438 0.00822774
3 0.625 −0.0203937 8 0.644531 0.00344555
4 0.6875 0.0563213 9 0.642578 0.00105626
5 0.65626 0.0178067 10 0.641602 −0.000137933
2-2 p 13 = 1.7320557, p = 1.732050807568877, and |p − p 13 | = 4.86 × 10 −6 .
2-4
(a) p n f(p n ) (b) p n f(p n )
0 1.18921 0.63922 0 1.22474 1.02526
1 1.08006 −0.386228 1 0.993666 −1.04402
2 1.14967 0.240826 2 1.22857 1.06842
3 1.10782 −0.147105 3 0.987506 −1.08621
(c) p n f(p n ) (d) p n f(p n )
0 1.1547 0.289744 0 1.14286 0.175344
1 1.11643 −0.0700688 1 1.12448 0.0032932
2 1.12605 0.0177429 2 1.1241231639 1.23208 × 10 −6
3 1.12364 −0.00444148 3 1.124123029704334 1.72659 × 10 −13
Hence the (d) is the best iterative function.
2-5 (i) Since A > 0 and x 0 > 0, x 1 = 1 2 x 0 + A
2x 0
> 0. Then x n > 0 for all n. Suppose x n converges to p ≥ 0, then
p = 1 2 p + A 2p ⇒ p = √ A.
(ii) If x 0 < 0, then x 1 = 1 2 x 0 + A
2x 0
< 0. Then x n < 0 for all n. Then x n converges to p ≤ 0, then
p = 1 2 p + A 2p ⇒ p = −√ A.
2-6 f(x) = x 2 − 6, f ′ (x) = 2x, and p 0 = 1, the Newton’s method p n+1 = p n − f(p n)
f ′ (p n ) . Then p 1 = 1 − f(1)
f ′ (1) = 3.5
and p 2 = 3.5 − f(3.5)
f ′ (3.5) = 2.607. (The exact solution √ 6 ≈ 2.4494897)
2-7 The Secant method and False Position is p n = p n−1 − f(p n−1)(p n−1 − p n−2 )
, p 0 = 3, p 1 = 2.
f(p n−1 ) − f(p n−2 )
(a) p 0 = 3, p 1 = 2, f(p 0 ) = 3 and f(p 1 ) = −2. Then p 2 = p 1 − f(p 1)(p 1 − p 0 )
f(p 1 ) − f(p 0 ) = 2.4, and f(p 2) = −0.24.
Then we have p 3 = p 2 − f(p 2)(p 2 − p 1 )
f(p 2 ) − f(p 1 ) = 2.45454.
(b) As in the Secant method. Since f(p 1 ) < 0 and f(p 2 ) < 0, then p 3 = p 2 − f(p 2)(p 2 − p 0 )
f(p 2 ) − f(p 0 ) = 2.44444.
The exact solution √ 6 ≈ 2.4494897. Hence (b) is closer to √ 6.
2-8 f(x) = (x − p) m q(x) where lim q(x) ≠ 0. Then
x→p
m(x − p) m q(x)
g(x) = x −
m(x − p) m−1 q(x) + (x − p) m q ′ (x) = x − m(x − p)q(x)
mq(x) + (x − p)q ′ (x) .
Therefore, g ′ (p) = 1 − mq(p)[mq(p)]
[mq(p)] 2
= 0. If f ′′′ is continuous, the sequence produces quadratic convergence.
p − p n+1 (b − a)/2 n+1
2-9 lim = lim
n→∞ p − p n n→∞ (b − a)/2 n = 1 . Then the Bisection Algorithm is linearly convergence.
2
2-10 For g(x) = 0.5(x + 3 x ) and p 0 = 0.5, then p 4 = 1.73205.
2-11 (a) ˆp 10 = 0.0 ¯45, (b) ˆp 2 = 0.0363.
2