2-1-3-a-1-b-1-c-1-d-a-b-c-d-Z-What-s-b-

Question Number 70885 by naka3546 last updated on 09/Oct/19

\sqrt[3]{2} = a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \dots}}}

a, b, c, d \in Z^{+}

{W h a t}^{'} s b ?

Answered by MJS last updated on 09/Oct/19

the continued fraction of \sqrt[3]{2} is non + periodic

\Rightarrow just calculate it

a = 1

b = ⌊ \frac{1}{\sqrt[3]{2} - 1} ⌋ = ⌊ \sqrt[3]{4} + \sqrt[3]{2} + 1 ⌋ = 3

Facebook Tweet Pin