Chapter 1 The solution of nonlinear Equations f(a)=0
Chapter 1 The Solution of Nonlinear Equations f(x)=0
1.1 Iteration for Solving x=g(x)
1.1 Iteration for Solving x=g(x)
ppp p1=9(0 9(P Pk=9(k-1) Pk+1=9(k)
Example 1. 1. The iterative rule po 1 and pk+1= 1.001pk for k=0, 1,..pro- duces a divergent sequence. The first 100 terms look as follows: P1=1.0170=(1.001010001.00100 p2=1011=(1001)(1.0000001 3=1012=(1001)(1.002011.00300 p100=1.0019(1.001)(1.104012)=1.105116
1. 1. 1 Finding Fixed Points
1.1.1 Finding Fixed Points
Definition 1. 1(Fixed Point). A ficed point of a function g(a)is a real number P such that P=g(P) Geometrically, the fixed points of a function y=g()are the points of intersection of y=g(a)and y=0 Definition 1. 2(Fixed-point Iteration). The iteration Pn+1=g(pn )for n=0,1 is called fi.aced-point iteration
Theorem 1.1. Assume that g is a continuous function and that ipn jooo is a se- quence generated by fixed-point iteration. If limn-ooPn=P, then P is a fixed point
Example 1.2. Consider the convergent iteration po=0.5 and ph+ for k=0.1 The first 10 terms are obtained by the calculations n1=e-0.5000000.606531 P2=e-0603=0.545239 P3=e-05239=0.579703 p=e-0.566409 0.567560 0.567560 p10=e =0.566907
Theorem 1.2 Assume that g ECla, b If the range of the mapping y=y(r) satisfies y∈,列 for all a∈[,,then g has a fixed point in a, b (1.3) Furthermore, suppose that g(a)is defined over(a, b) and that a positive constant K 1 exists with Ig()s K<1 for all c E(a, b), then g has a(1.4 unique fixed point P in a, 6
Example 1.3. Apply Theorem 1.2 to rigorously show that g(a)=cos( r) has a unique fixed point in