This was nagging me on my drive home - so here is a hand calculation using Newtons method for ##
\sqrt[5]{2}##
. If ## s_n## is the ## n^{th}## estimate of the root, then I get the following for the ##p^{th}## root of ##x##:
[tex]
s_{n+1} = \frac{1}{p}\left[(p-1) s_n + \frac{x}{s_n^{p-1}} \right][/tex]
For us p=5, x=2, so,
[tex]
s_{n+1} = \frac{1}{5}\left[4 s_n + \frac{2}{s_n^{4}} \right][/tex]
if I use ## s_0 = 1##, then ##s_1 = 1.2##. For ##s_2## I get
[tex]
s_{2} = \frac{1}{5}\left[ 4.8 + \frac{2}{2.0736} \right] = \frac{1}{5}\left[ 4.8 + \frac{1}{1.0368} \right] \approx \frac{1}{5}\left[ 4.8 + 1 - 0.0368 \right] <br />
= \frac{5.7632}{5} \approx 1.15[/tex]
So I get ##2^{1.2} \approx 2.30 ##. Better than 1% accuracy in this case...
jason