CODEWITHSUNDEEP
pythonnumerical-analysisbisection
bbnn
bbnn

Reputation: 3602

Solving equation using bisection method

Is there a bisection method I can find online, specifically for python?

For example, given these equations how can I solve them using the bisection method?

x^3 = 9  
3 * x^3 + x^2 = x + 5  
cos^2x + 6 = x

Upvotes: 11

Views: 20209

Answers (2)

Shamanth Muroor
Shamanth Muroor

Reputation: 3

This could help you!

import numpy as np

def fn(x):
    # This the equation to find the root
    return (x**3 - x - 1) #x**2 - x - 1

def find_root_interval():
    for x in range(0, 1000):
        if fn(x) < 0:
            lower_interval = x
            if fn(x+1) > 0:
                higher_interval = x + 1
                return lower_interval, higher_interval
    return False

def bisection():
    a,b = find_root_interval()
    print("Interval: [{},{}]".format(a,b))
    # Create a 1000 equally spaced values between interval
    mid = 0
    while True:
        prev_mid = mid
        mid = (a+b)/2
        print("Mid value: "+str(mid))
        # 0.0005 is set as the error range
        if abs(mid-prev_mid) < 0.0005:
            return mid
        elif fn(mid) > 0:
            b = mid
        else:
            a = mid

root = bisection()
print(root)

Upvotes: 0

unutbu
unutbu

Reputation: 879181

Using scipy.optimize.bisect:

import scipy.optimize as optimize
import numpy as np

def func(x):
    return np.cos(x)**2 + 6 - x

# 0<=cos(x)**2<=1, so the root has to be between x=6 and x=7
print(optimize.bisect(func, 6, 7))
# 6.77609231632

optimize.bisect calls _zeros._bisect, which is implemented in C.

Upvotes: 16

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