Solve this equation with fixed point iteration method in python

Question

f(x) = x^2- 2x - 3 = 0

How can I solve this equation non-linear, and used fixed point iteration method in Python ?

Answer 1

Fixed Point Iteration

f(x) = x^2-2x-3 = 0 ⇒ x(x-2) = 3 ⇒ x = 3/(x-2)

import math

def g(x):
   if 2 == x:
       return x + 1e-10
   return 3/(x-2)

def quadratic(ff,x=0):
    while abs(x-ff(x)) > 1e-10:
        x = ff(x);
    return x

print(quadratic(g))

-1

root finding formula

# -*- coding:utf8 -*-
import math
def quadratic(a, b, c):
    ''' ax² + bx + c = 0
        return
        True  : all real number
        Fasle :  no solutions
        (x1,x2)
    '''
    if not isinstance(a, (int, float)):
        raise TypeError('a is not a number')
    if not isinstance(b, (int, float)):
        raise TypeErrot('b is not a number')
    if not isinstance(c, (int, float)):
        raise TypeError('c is not a number')
    derta = b * b - 4 * a * c
    if a == 0:
        if b == 0:
            if c == 0:
                return True
            else:
               return False
        else:
            x1 = -c / b
            x2 = x1
            return x1, x2
    else:
        if derta < 0:
            return False
        else:
            x1 = (-b + math.sqrt(derta)) / (2 * a)
            x2 = (-b - math.sqrt(derta)) / (2 * a)
            return x1, x2

print(quadratic(1, -2, -3))

(3.0, -1.0)