3
$\begingroup$

For the form $\sqrt{b}$, I could just apply the recursive quadratic formula: $P_{k+1} = a_kQ_k - P_k$ $Q_{k+1} = \dfrac{d - P^2_{k+1}}{Q_k}$ $\alpha_k = \dfrac{P_k + \sqrt{d}}{Q_k}$ $a_k = \lfloor \alpha_k \rfloor$

In this case, we have a coefficient namely $a$, so what's $d$? Is it still $b$?

Thanks,

  • 11
    I think I figured it out, $a\sqrt{b} = \sqrt{a^2b}$.2011-05-02

1 Answers 1

2

(So you can have something to "accept"...)

$a \sqrt{b} = \sqrt{a^2b}$ - Chan

  • 0
    @Chan: Hah! I answer even when no one is asking!2011-06-19