def f(x):

    return 1.0/x

def brouncker(f,a,b,n):

    s = min( f(a),f(b) )*(b-a)

    H = [0,s]

    for i in range(2,n+1):

        for j in range(1,2**(i-1),2):

            pas = 1.0*(b-a)/2**(i-1)

            h = f(a+j*pas)

            H.append(h)

            H.sort(key = float)

            k = 0

            while h-H[k] > 0:

                k = k+1

            hmin = H[k-1]

            s = s + pas*(h-hmin)

    return s

​

print(brouncker(f,1,2,3))

​