def newton(f,Df,x0,epsilon,max_iter):
    '''Approximate solution of f(x)=0 by Newton's method.

    Parameters
    ----------
    f : function
        Function for which we are searching for a solution f(x)=0.
    Df : function
        Derivative of f(x).
    x0 : number
        Initial guess for a solution f(x)=0.
    epsilon : number
        Stopping criteria is abs(f(x)) < epsilon.
    max_iter : integer
        Maximum number of iterations of Newton's method.

    Returns
    -------
    xn : number
        Implement Newton's method: compute the linear approximation
        of f(x) at xn and find x intercept by the formula
            x = xn - f(xn)/Df(xn)
        Continue until abs(f(xn)) < epsilon and return xn.
        If Df(xn) == 0, return None. If the number of iterations
        exceeds max_iter, then return None.

    Examples
    --------
    >>> f = lambda x: x**2 - x - 1
    >>> Df = lambda x: 2*x - 1
    >>> newton(f,Df,1,1e-8,10)
    Found solution after 5 iterations.
    1.618033988749989
    '''
    xn = x0
    for n in range(0,max_iter):
        fxn = f(xn)
        if abs(fxn) < epsilon:
            print('Found solution after',n,'iterations.')
            return xn
        Dfxn = Df(xn)
        if Dfxn == 0:
            print('Zero derivative. No solution found.')
            return None
        xn = xn - fxn/Dfxn
    print('Exceeded maximum iterations. No solution found.')
    return None


#test function f(x) = x^3 - x^2 - 1
print(" test function 1 ")
f = lambda x: x**3 - x**2 - 1
Df = lambda x: 3*x**2 - 2*x
approx = newton(f,Df,1.5,1e-10,10)
print(approx)

#test function 2: f(x)=x^{1/3}
print(" test function 2 ")
f = lambda x: x**(1/3)
Df = lambda x: (1/3)*x**(-2/3)
approx = newton(f,Df,0.5,1e-2,100)