TITLE: Mental $n-$th root of $N$
QUESTION [11 upvotes]: It has been a while since I started thinking about this problem: a fast method to evaluate (in an approximate way) mentally the $n-$th root of a number $N$. I'm talking about great numbers, because otherwise one could handle with the first terms of a Taylor series.
I found time ago a really cute approximation for the square root, which runs like this: you have $N$, and you always can write $N$ as
$$N = q^2 + s$$
Where $q^2$ is the nearest $N$ perfect square, and $s$ is the remainder. Trivial example: $50 = 49 + 1$.
Thanks to that, one can approximate
$$\sqrt{N} = \sqrt{q^2 \pm s} \approx q \pm \frac{s}{2q}$$
Fast example
$$\sqrt{43} = \sqrt{36 + 7} \approx 6 + \frac{7}{12} = 6 + 0.58333 = 6.58333(..)$$
Where actually
$$\sqrt{43} = 6.55743(...)$$
So somehow one may compute it mentally in a quite easy and fast way.
The $n-$th problem
I started then to think about if such a method could be generalized for some $n-$th square root. With intuition (and a lot of naiveness) I thought about a sort of "mathematical symmetry", something like
$$\sqrt[n]{N} = \sqrt[n]{q^n \pm s} \approx q \pm \frac{s}{nq^{n-1}}$$
But I don't know if that may work in general. For example: 
$$\sqrt[4]{600} = \sqrt[4]{625 - 25} = \sqrt[4]{5^4 - 25} \approx 5 - \frac{25}{4\cdot 5^{4-1}} = 5 - 0.05 = 4.95$$
And surprisingly 
$$4.95^4 = 600.3725(...)$$
BUT if we mind with the plus sign...
$$\sqrt[4]{600} = \sqrt[4]{256 + 344} = \sqrt[4]{4^4 + 344} \approx 4 + \frac{344}{4\cdot 4^3} = 1.34375 = 5.34375$$
Where
$$5.34375^4 = 815.4259(...)$$
So it seems like it works if we pick the correct sign namely when the remainder $s$ is quite small.
What I'm asking for is for the existence of a pretty simple form to approximate roots, namely something like
$$\sqrt[n]{N} = \sqrt[n]{q^n + s} \approx q + f(s, q, n)$$
In which $f(s, q, n)$ is some good simple function (simple = not a sum of terms, just one).
Or my "intuition" is good enough to work? (Always with picking the correct decision about $\pm$ et cetera)
Thank you all!

REPLY [2 votes]: There is a pretty simple form to approximate roots, called Newton-Raphson method:
Function (input value, input degree, input num_of_iterations, output root):
    Set root = value
    Repeat num_of_iterations:
        Set temp = root^(degree-1)
        Set root = root-(root*temp-value)/(degree*temp)

Python implementation:
def Function(value,degree,num_of_iterations):
    root = float(value)
    for n in range(0,num_of_iterations):
        temp = root**(degree-1)
        root = root-(root*temp-value)/(degree*temp)
    return root

$3$ or $4$ iterations are typically sufficient for a good approximation.