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• Manisha Sahu

# Root Finding Techniques | Fixed Point Iteration Method

In this article, we will be going to study the fixed point iteration method in numerical analysis and design using Python.

### Algorithm

• start

• 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

### 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 + 1```

### 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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