CODEWITHSUNDEEP
python
user123
user123

Reputation: 206

Bisection Method In Python

I'm coding the bisection method in Python. I have set a tolerance for the error in my routine. I have two questions:

  1. My function should be able to find the root of an arbitrary continuous scalar-valued function. How do I pass this to bisection_method and use it properly?
  2. How can I record how many iterations it takes to reach the specified tolerance?

Code:

def f(x):
    return(x**2 - 11)

def bisection_method(a, b, tol):
    if f(a)*f(b) > 0:
        #end function, no root.
        print("No root found.")
    else:
        while (b - a)/2.0 > tol:
            midpoint = (a + b)/2.0
            if f(midpoint) == 0:
                return(midpoint) #The midpoint is the x-intercept/root.
            elif f(a)*f(midpoint) < 0: # Increasing but below 0 case
                b = midpoint
            else:
                a = midpoint
        return(midpoint)

answer = bisection_method(-1, 5, 0.0001)
print("Answer:", answer)

which gives the following output:

Answer: 3.31671142578125

Upvotes: 1

Views: 5570

Answers (1)

Prune
Prune

Reputation: 77837

For the function, simply pass the function name as an argument. I've changed your function's name to root11 and made it the first argument to the bisection.

For the count ... you should have been able to look this up on line. Just count iterations as you would before you learned the for statement. Return this with the final answer.

Note that I removed your check for an exact answer: you'll find it on the next iteration, anyway.

def root11(x):
    return(x**2 - 11)

def bisection_method(f, a, b, tol):
    if f(a)*f(b) > 0:
        #end function, no root.
        print("No root found.")
    else:
        iter = 0
        while (b - a)/2.0 > tol:
            midpoint = (a + b)/2.0

            if f(a)*f(midpoint) < 0: # Increasing but below 0 case
                b = midpoint
            else:
                a = midpoint

            iter += 1
        return(midpoint, iter)

answer, iterations = bisection_method(root11, -1, 5, 0.0001)
print("Answer:", answer, "\nfound in", iterations, "iterations")

import math
answer, iterations = bisection_method(math.cos, 0, 2, 0.0001)
print("Answer:", answer, "\nfound in", iterations, "iterations")

Output:

Answer: 3.31671142578125 
found in 15 iterations
Answer: 1.5706787109375 
found in 14 iterations

Upvotes: 1

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