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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def mete_upscale_iterative_alt(S, N, doublings):
 """ This function is used to upscale from the anchor area. Parameters S : int or float Number of species at anchor scale N : int or float Number of individuals at anchor scale doublings : int Number of doublings of A. Result vector will be length doublings + 1. Returns ------- result : ndarray 1D array of number of species at each doubling """


    # Arrays to store N and S at all doublings
    n_arr = np.empty(doublings+1)
    s_arr = np.empty(doublings+1)

    # Loop through all scales
    for i in xrange(doublings+1):

        # If this is first step (doubling 0), N and S are initial values
        if i == 0:
            n_arr[i] = N
            s_arr[i] = S

        # If not first step
        else:

            # Get previous S
            SA = s_arr[i-1]

            # N is double previous N
            n_arr[i] = 2 * n_arr[i-1]
            N2A = n_arr[i]

            # Eq 8 from Harte 2009, setup to return S2A given input of x
            # x is exp(-lam_phi, 2A)
            def S2A_calc(x, SA, N2A):
                return ((SA +
                        N2A *
                        (1-x)/(x-x**(N2A+1)) *
                        (1 - (x**N2A)/(N2A+1))) /
                        x**-1)

            # Eq 9 from Harte 2009, setup to equal to zero, used to solve x
            # Note that two summations are replaced by known formulas for sum
            # of geometric and logarithmic series.
            # Note "offset" of 1e-23, which is needed because f(a) and f(b) do
            # not have the same sign in solver otherwise. This introduces no
            # more than a 1e-23 error in the calculation of x, which should not
            # cause a significant problem.
            def x_calc(x, SA, N2A):
                return (S2A_calc(x,SA,N2A) /
                        N2A *
                        x*(x**N2A-1)/(x-1) -
                        (x**N2A * (-lerchphi(x,1,N2A+1))-np.log(1-x)) ) - 1e-23

            # Solve for x
            x = (optimize.brentq(x_calc, 1e-24, 1-1e-16, args=(SA,N2A),
                    xtol=1e-16, maxiter=1000, disp=True) + 1e-23)

            # Given x, calculate S2A
            s_arr[i] = S2A_calc(x,SA,N2A)

    return s_arr