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In this article, we will be going to study the fixed point iteration method in numerical analysis and design using Python.
Algorithm
start
Read x,,epsilon(absolute error),x_before=0,i
Calculate f(x) and g(x) //f(x)=putting x in the fxn,g(x)=differentiation of f(x)
while (|x-x_before|>epsilon)
x_before=x
x=g(x)
print (i, "{0:.6f}".format(x), "{0:.6f}".format(abs(x - x_before)))
i=i+1
end loop
Flowchart
Program
def f(x):
return (x**3 - 9*x + 1 )
def g(x):
return (9*x-1)**(1/3)
epsilon = 0.0001
x = 2.7
x_before = 0
i = 0
while (abs(x - x_before) > epsilon):
x_before = x
x = g(x)
print(i, "{0:.6f}".format(x), "{0:.6f}".format(abs(x - x_before)))
i = i + 1Example
Q]Q] Find the real root of equation x^3-9x+1=0 by fixed-point iteration method?
Solution-
f(0)=1 , f(1)=-7 , f(2)=-9 , f(3)=1
Since root of f(x)=0 lies between 2 and 3 so taking x0=2.7
Assumption-
x= (9x-1)^1/3
F(x)= (9x-1)^1/3
F’(x)=3/(9x-1)^2/3
Since F’(2.7) coming less than 1,satisfy the condition . So taking
x(n+1)=(9x(n)-1)^1/3
n=0
x1=(9x0-1)^1/3=2.8562
x2=(9x1-1)^1/3=2.9125 -------------------------------------------
----------------------------------------
x7=2.9427
X8=2.9428
We get two approximate values having same decimal digits up to three places ,so the answer is 2.942
Happy Coding!
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