Question

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

Answer 1

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

Answer 2

I'm wrong.... :-P

Answer 3

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

Answer 4

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

Answer 5

this is a slow site today ...

Answer 6

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

Answer 7

try newton's method.

Answer 8

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

Answer 9

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

Answer 10

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