How would we go about determining the continued fraction expansion of: $\sqrt{5}$?

Question

anonymous · Answer

1+1/(5+1/(5+1/(5+1/5+...)))

anonymous · Answer

I'm wrong.... :-P

amistre64 · Answer

yeah, the result is (2;4,4,4,...)

amistre64 · Answer

all irrational algebraic numbers have a periodic continued fraction expansion.  Which leaves us with transcendentals having an infinite expansion

amistre64 · Answer

this is a slow site today ...

zarkon · Answer

$$\lfloor\sqrt{5}\rfloor=2$$
$$\left\lfloor\frac{1}{\sqrt{5}-2}\right\rfloor=4$$
$$\left\lfloor\frac{1}{\frac{1}{\sqrt{5}-2}-4}\right\rfloor=4$$
...

anonymous · Answer

try newton's method.

zarkon · Answer

continue this way will give the [2,4,4,4...]

anonymous · Answer

amistre, i found this if it helps; it is a special case

anonymous · Answer

http://en.wikipedia.org/wiki/Generalized_continued_fraction#Roots_of_positive_numbers