Use Theorem 10.6 to show that G: D R3 R3 has a unique fixed point

Question:

Use Theorem 10.6 to show that G: D ⊂ R3 → R3 has a unique fixed point in D. Apply functional iteration to approximate the solution to within 10−5, using the l∞ norm.
a.

D = {(x1, x2, x3)t | −1 ≤ xi ≤ 1, i = 1, 2, 3 }
b.

D = {(x1, x2, x3)t | 0 ≤ x1 ≤ 1.5, i = 1, 2, 3 }
c. G(x1, x2, x3) = (1 − cos(x1 x2 x3), 1 − (1 − x1)1/4 − 0.05x23+ 0.15x3, x21
+ 0.1x22 − 0.01x2 + 1)t;
D = {(x1, x2, x3)t | − 0.1 ≤ x1 ≤ 0.1,− 0.1 ≤ x2 ≤ 0.3, 0.5 ≤ x3 ≤ 1.1 }
d.

D = {(x1, x2, x3)t | −1 ≤ xi ≤ 1, i = 1, 2, 3 }
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Numerical Analysis

ISBN: 978-0538733519

9th edition

Authors: Richard L. Burden, J. Douglas Faires

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