Answers for "secant method"

secant method

def secant_method(f, x0, x1, iterations):
    """Return the root calculated using the secant method."""
    for i in range(iterations):
        x2 = x1 - f(x1) * (x1 - x0) / float(f(x1) - f(x0))
        x0, x1 = x1, x2
    return x2

def f_example(x):
    return x ** 2 - 612

root = secant_method(f_example, 10, 30, 5)

print("Root: {}".format(root))  # Root: 24.738633748750722

secant method

x(i+1) = x(i-1)f(xi)-x(i)f(xi-1)/f(xi)-f(xi-1)

secant method

def secant(f,a,b,N):
    '''Approximate solution of f(x)=0 on interval [a,b] by the secant method.

    Parameters
    ----------
    f : function
        The function for which we are trying to approximate a solution f(x)=0.
    a,b : numbers
        The interval in which to search for a solution. The function returns
        None if f(a)*f(b) >= 0 since a solution is not guaranteed.
    N : (positive) integer
        The number of iterations to implement.

    Returns
    -------
    m_N : number
        The x intercept of the secant line on the the Nth interval
            m_n = a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))
        The initial interval [a_0,b_0] is given by [a,b]. If f(m_n) == 0
        for some intercept m_n then the function returns this solution.
        If all signs of values f(a_n), f(b_n) and f(m_n) are the same at any
        iterations, the secant method fails and return None.

    Examples
    --------
    >>> f = lambda x: x**2 - x - 1
    >>> secant(f,1,2,5)
    1.6180257510729614
    '''
    if f(a)*f(b) >= 0:
        print("Secant method fails.")
        return None
    a_n = a
    b_n = b
    for n in range(1,N+1):
        m_n = a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))
        f_m_n = f(m_n)
        if f(a_n)*f_m_n < 0:
            a_n = a_n
            b_n = m_n
        elif f(b_n)*f_m_n < 0:
            a_n = m_n
            b_n = b_n
        elif f_m_n == 0:
            print("Found exact solution.")
            return m_n
        else:
            print("Secant method fails.")
            return None
    return a_n - f(a_n)*(b_n - a_n)/(f(b_n) - f(a_n))