Mental $n-$th root of $N$

Let $f(x) = x^{1 \over n}$.

Then $f(N+h) \approx f(N) + hf'(N)$ becomes

$(N+h)^{1 \over n} \approx N^{1 \over n} + \dfrac{hN^{1 \over n}}{nN}$

So, if you want to approximate $600^{1 \over 4}$:

• $4^4 = 256 \lt 600 \lt 625 = 5^4$
• Choose $N = 5^4 = 625$ and $h = 600-625 = -25$.
• $600^{1 \over 4} \approx 5 - \dfrac{25 \cdot 5}{4 \cdot 600} = 5.00 - 0.052 = 4.948$
• To the nearest thousandth, $600^{1 \over 4} = 4.949$

Of course, the smaller $h$ is, relative to $N$, the better the approximation will be.

Approximation of error

\begin{align} (N+x)^{1 \over n} &= N^{1 \over n} \left( 1 + \dfrac hN \right)^{1 \over n} \\ &= N^{1 \over n} \sum_{k=0}^\infty \binom{1 \over n}{k}\left( \dfrac hN \right)^k \\ &\approx N^{1 \over n} \left( 1 + \left( \dfrac 1n\right)\left( \dfrac hN\right) - \left( \dfrac{n-1}{2n^2}\right)\left( \dfrac{h^2}{N^2}\right) \right)\\ &\approx N^{1 \over n} \left( 1 + \dfrac{h}{nN} - \dfrac{(n-1)h^2}{2n^2 N^2} \right) \end{align}

Which implies that the relative error is roughly

$100\dfrac{(n-1)h^2}{2n^2 N^2} \approx \dfrac{50}n \left( \dfrac hN \right)^2\%$

For the above example, this is $\dfrac{50}4 \left( \dfrac{25}{625} \right)^2\% = 0.02\%$