Continued fraction

user · #1 $Old$ October 3rd, 2009, 01:01 PM

i need help to solve this exercise
To prove the continued fraction of (d^2 +1)^(1/2) is [d;2d,2d,2d,….]

Soroban · #2 $Old$ October 4th, 2009, 09:19 AM

Hello, user!

Quote:

Prove that the continued fraction of . $\sqrt{d^2 +1}$ . is . $\bigg[d,2d,2d,2d, \hdots\bigg]$

$\text{Let: }\;x \;=\;d + \frac{1}{2d + \dfrac{1}{2d + \hdots}}$

$\text{Add }d\text{ to both sides: }\;x + d \;=\;2d + \frac{1}{\left\{2d + \dfrac{1}{2d + \hdots}\right\}} \;\begin{array}{c} \\ \\ \Leftarrow\text{ This is }(x+d) \end{array}$

$\text{Then we have: }\;x + d \;=\;2d + \frac{1}{x+d}$

$\text{Multiply by }(x+d)\!:\;\;(x+d)^2 \;=\;2d(x+d) + 1 \quad\Rightarrow\quad x^2 + 2dx + d^2 \;=\;2dx + 2d^2 + 1$

$\text{Therefore: }\;x^2 \;=\;d^2+1 \quad\Rightarrow\quad x \;=\;\sqrt{d^2+1}$

user · #3 $Old$ October 5th, 2009, 11:16 AM

Hi friend. Thanks for your answer.
I had another solution finding
alpha zero, a0 to obtain that a 3 this is the integer part of alpha 3 is [d; 2d,2d,2d,... ] but this solution is a little more larger than your solution
Bye see you soon

user · #4 $Old$ November 5th, 2009, 01:37 PM

Hi friends. How can i prove that the neutral element to the sum is unique.
Thank you
Sincrely, user