Talk:Nth root algorithm

I've been playing around with finding (integer) nth roots for large n. Unfortunately, the following implementation of Newton's method (in Python) is ridiculously slow:

def nthroot(y, n):
    x, xp = 1, -1
    while abs(x - xp) > 1:
        xp, x = x, x - x/n + y/(n * x**(n-1))
    while x**n > y:
        x -= 1
    return x

For example, nthroot(12345678901234567890123456789012345**100,100) takes several minutes to finish.

Using binary search, the solution can be found in seconds.

Am I just doing something wrong, or is Newton's method inherently inefficient for such large n? - Fredrik | talk 23:22, 24 May 2005 (UTC)[reply]