1. Fixed- Point Iteration
Sec:6.1
Fixed- Point Iteration

2. 1) Online Computing 2) MATLAB workshop 2) Exam-I and Exam-II
Exam-I Date: Oct 19
Exam-II Date: Oct 23
Time: two hours between 5PM-10PM

3. Sec:6.1 Fixed- Point Iteration
The number 𝑝 is a fixed point for a given function 𝑔 if.
𝑝=𝑔 𝑝
Example:
A fixed point for g occurs precisely when the graph of 𝑦 = 𝑔(𝑥) intersects the graph of 𝑦 = 𝑥
Determine any fixed points of the function 𝑔(𝑥) = 𝑥 2 − 2.
solution
𝑝=𝑔(𝑝)
𝑝= 𝑝 2 − 2
0= 𝑝 2 −𝑝− 2
𝑝=−1, 𝑝=2

4. Sec:6.1 Fixed- Point Iteration
The derivative mean-value theorem states that if a function 𝑔(𝑥) and its first derivative are continuous over an interval 𝑎 ≤ 𝑥 ≤ 𝑏, then there exists at least one value of 𝑥= 𝜉 within the interval such that
𝑔′(𝜉 ) = 𝑔(𝑏) − 𝑔(𝑎) 𝑏 − 𝑎
The right-hand side of this equation is the slope of the line joining 𝑔(𝑎) and 𝑔(𝑏).
𝑔 𝑏 − 𝑔 𝑎 = 𝑔 ′ 𝜉 (𝑏−𝑎)
Slope of the tangent line is
𝑔′(𝜉 )
Slope of this line is
𝑔(𝑏) − 𝑔(𝑎) 𝑏 − 𝑎 𝜉

5. Sec:6.1 Fixed- Point Iteration
Example1
𝑥 2 − 2𝑥 + 3 = 0
can be simply manipulated to yield
simple fixed-point iteration
Step 1
 𝑥= 𝑥
𝑥=𝑔(𝑥)
rearranging the function
𝒇 (𝒙) = 𝟎
so that x is on the left-hand side of the equation:
𝒙 = 𝒈(𝒙) (*)
Example2
sin⁡(𝑥)= 0
can be simply manipulated to yield
𝑥=𝑥+sin⁡(𝑥)
𝑥=𝑔(𝑥)
Step 2
given an initial guess at the root 𝑥 𝑖 , (*) can be used to compute a new estimate 𝑥 𝑖+1 as expressed by the iterative formula
𝒙 𝒊+𝟏 =𝒈( 𝒙 𝒊 )
Note
Convert the problem from root-finding to finding fixed-point

6. Sec:6.1 Fixed- Point Iteration
Step 1
Example1
Use simple fixed-point iteration to locate the root of
𝒇 𝒙 = 𝒆 −𝒙 −𝒙
rearranging the function
𝒙= 𝒆 −𝒙  
𝒙=𝒈(𝒙) 
𝒙 𝒏
Step 2 𝒏
𝒙 𝒊+𝟏 =𝒈( 𝒙 𝒊 )
𝒙 𝒊+𝟏 = 𝒆 − 𝒙 𝒊  
Starting with an initial guess of x0 = 0
𝒙 𝟏 =𝒈 𝒙 𝟎 = 𝒆 −𝟎 =𝟏
𝒙 𝟐 =𝒈 𝒙 𝟏 = 𝒆 −𝟏 =
𝒙 𝟑 =𝒈 𝒙 𝟐 = 𝒆 −𝟎.𝟑𝟔𝟕𝟖𝟕𝟗 = ⋮ ⋮ ⋮ ⋮
Thus, each iteration brings the estimate closer to the true value of the root:

7. Sec:6.1 Fixed- Point Iteration
𝜺 𝒕 = 𝒙 𝒏 − 𝒙 𝒓 𝒙 𝒓 ×𝟏𝟎𝟎%
clear; clc; format long
x(1) = 0;
g exp(-x);
f exp(-x) - x;
true_root = ;
for k=1:11
x(k+1) = g( x(k) );
end
e_t = 100 * abs( x - true_root ) ./ true_root;
res = [[1:12]' x' e_t'];
fprintf('%d %16.13f %16.13f\n', res');
𝒙 𝒏
𝜺 𝒕 (%) 𝒏
Notice that the true percent relative error for each iteration is roughly proportional (by a factor of about 0.5 to 0.6) to the error from the previous iteration. This property, called linear convergence

8. 𝑬 𝒕,𝟏𝟎 < 𝑬 𝒕,𝟗 <⋯< 𝑬 𝒕,𝟏 < 𝑬 𝒕,𝟎
Convergence
Sec:6.1 Fixed- Point Iteration
Suppose that the true solution is
𝑬 𝒕,𝒏+𝟏 = 𝒈′ 𝝃 ∙ 𝑬 𝒕,𝒏
𝒙 𝒓 =𝒈( 𝒙 𝒓 ) 
< 𝑬 𝒕,𝒏
the iterative equation is
( If 𝒈 𝝃 <𝟏 )
𝒙 𝒏+𝟏 =𝒈( 𝒙 𝒏 ) 
the errors decrease with each iteration
Subtracting these equations yields
𝑬 𝒕,𝒏+𝟏 < 𝑬 𝒕,𝒏
𝒙 𝒓 − 𝒙 𝒏+𝟏 =𝒈 𝒙 𝒓 −𝒈( 𝒙 𝒏 ) 
The derivative mean-value theorem gives
𝑬 𝒕,𝟏𝟎 < 𝑬 𝒕,𝟗 <⋯< 𝑬 𝒕,𝟏 < 𝑬 𝒕,𝟎
𝒙 𝒓 − 𝒙 𝒏+𝟏 =𝒈′ 𝝃 (  𝒙 𝒓 − 𝒙 𝒏 )
Step 1
where 𝜉 is somewhere between 𝑥 𝑟 and 𝑥 𝑛
rearranging the function 𝒇 (𝒙) = 𝟎
 so that x is on the left-hand side of the equation: 𝒙 = 𝒈(𝒙)
If the true error for iteration n is defined as
𝑬 𝒕,𝒏 = 𝒙 𝒓 − 𝒙 𝒏
𝑬 𝒕,𝒏+𝟏 =𝒈′ 𝝃 𝑬 𝒕,𝒏
Select 𝒈 𝒙 𝒔𝒐 𝒕𝒉𝒂𝒕
𝒈′ 𝝃 <𝟏
𝑬 𝒕,𝒏+𝟏 = 𝒈′ 𝝃 ∙ 𝑬 𝒕,𝒏