Homework Assignment #2

Homework Assignment #2
2-1 Use the Bisection method to find p n , n = 1, 2, ..., 10 for f(x) = √ x − cos x on [0, 1]. (Algorithm 1)
2-2 Find an approximation to √ 3 correct to within 10 −4 using the Bisection Algorithm. [Hint: Consider
f(x) = x 2 − 3.]
2-3 Use the fixed-point iteration method to find an approximation to √ 3 that is accurate to within 10 −4 .
Compare your result and the number of iterations required with the answer obtained in the Bisection
Algorithm.
2-4 (i) Use the following algebraic manipulation to show that each has a fixed point at p precisely when
f(p) = 0, where f(x) = x 4 + 2x 2 − x − 3.
(a) g 1 (x) = (3 + x − 2x 2 ) 1/4 (b) g 2 (x) =
(c) g 3 (x) =
(
x + 3 − x
4
2
) 1/2
( ) x + 3 1/2
x 2 (d) g 4 (x) = 3x4 + 2x 2 + 3
+ 2
4x 3 + 4x − 1
(ii) Perform four iterations, if possible, on each of the function g i . Let p 0 = 1 and p n+1 = g(p n ), for
n = 0, 1, 2, 3. which function do you think gives the best approximation to the solution.
(Algorithm 2)
2-5 (i) Show that if A is any positive number then the sequence defined by
converges to √ A whenever x 0 > 0.
(ii) What happens if x 0 < 0 ?
x n = 1 2 x n−1 +
A
2x n−1
, for n ≥ 1,
2-6 Let f(x) = x 2 − 6 and p 0 = 1. Use Newton’s method to find p 2 .
2-7 Let f(x) = x 2 − 6 and p 0 = 3, p 1 = 2, find p 3 .
(a) Use the Secant method.
(b) Use the method of False Position.
(c) Which of (a) or (b) is closer to √ 6.
2-8 Suppose p is a zero of multiplicity m of f, where f ′′′ is continuous on an open interval containing p.
Show that the following fixed-point method has g ′ (p) = 0:
g(x) = x − mf(x)
f ′ (x)
2-9 Show that the Bisection Algorithm gives a sequence with an error bound that converges linearly to 0.
2-10 Use Steffensen’s method with p 0 = 2 to compute an approximation to √ 3 accurate to within 10 −4 .
Compare your result and the numbers of iterations required with the answer obtained in the Bisection
Algorithm and fixed-point iteration method.
2-11 The following sequences converge to 0. Use Aitken’s ∆ 2 method to generate {̂p n } until |̂p n | ≤
5 × 10 −2 : (a) p n = 1 n , n ≥ 1. (b) p n = 1 n 2 , n ≥ 1.
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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.
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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.
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