root of bessel function

import scipy.special

# Note that the num_steps being returned should be a list 
# of the number of steps being used in each method
def j0_zeros(x0, bracket_tolerance, tolerance):
        """Find the zeros of the bessel function using a combination of secant and newton's method
    
    :Input:
     - *x0* (list) - initia; bracket
     - *bracket_tolerance* (float) - tolerance for size of teh bracket
     - *tolerance* (float) - Stopping tolerance for iteration
     
    :Output:
    If the iteration was successful the return values are:
     - *x_l* (float) - Zero found via the given intial iterate.
     - *n* (int) - Number of iterations it took to achieve the specified
       tolerance.
  
    """
    #functions
    f= lambda y:scipy.special.jn(0,y)
    #parameters
    MAX_STEPS = 100
    
    
    x_km=x0[0]
    x_k=x0[1]
    # INSERT CODE HERE
    if numpy.abs(f(x_k)) < tolerance:
        return x,0
        
    for n in xrange(1,MAX_STEPS+1):
        if numpy.abs(x_k - x_km > bracket_tolerance): #do secant's method until bracket becomes small enough
            #print x_k-x_km +1000
            x_kp = x_k - f(x_k) * (x_k-x_km) / (f(x_k) - f(x_km))
            x_km = x_k
            x_k = x_kp
            continue
            
        x_k = x_k-f(x_k)/scipy.special.jvp(0,x_k,n=1) #newton's iteration step
        if numpy.abs(f(x_k)) < tolerance:
             break 
                
                
    return x_k, n