Fast approx to ln

Question

link to problem

I need to write a fast approx of ln in Python and use the 2.4 algorithm. I know I can get the first a_i numbers with:

 def f1(x,i): a0=(x+1)/2 g0=np.sqrt(x) a=[] for j in range(i): a0=(a0+g0)/2 g0=np.sqrt((a0)*g0) a.append(a0) return a

Answer 1

If you use a0 and g0 inside the for loop, you make a mistake. You need to initialise arrays with those values.

def f1(x, n):
    a = [(1 + x) / 2]
    g = [np.sqrt(x)]
    for i in range(n + 2):
        a.append((a[i] + g[i]) / 2)
        g.append(np.sqrt(a[i+1] * g[i]))
    return a[n]

In the function above I also keep the notation consistent with the paper. Let me know if this helps, and please flag my comment as an answer to the question if it does.

Answer 2

Yea but im thinking it could be usefull to have it in array so i can calculate the next d(n,k) in the next function. But i have no idea even what to start with. Im also going to plot how bit the error is with n number of iterations.

Answer 3

this is how im thinking but i know its wrong

 def fast_approx(x,n): a = [(1 + x) / 2] def f1(x, n): a = [(1 + x) / 2] g = [np.sqrt(x)] for i in range(n + 2): a.append((a[i] + g[i]) / 2) g.append(np.sqrt(a[i+1] * g[i])) return a[n] def the_big_D(n,k): dnull=a[n] d=[] for i in range(k): dnull=(a[n-1] - 4**(-k) * a[n-2]) / (1 - 4**(-k)) d.append(dnull) return d[n]