i need help to solve this exercise

To prove the continued fraction of (d^2 +1)^(1/2) is [d;2d,2d,2d,….]

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- Oct 3rd 2009, 11:01 AMuserContinued fraction
i need help to solve this exercise

To prove the continued fraction of (d^2 +1)^(1/2) is [d;2d,2d,2d,….] - Oct 4th 2009, 07:19 AMSoroban
Hello, user!

Quote:

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

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

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

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

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

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

- Oct 5th 2009, 09:16 AMuseranswer
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 - Nov 5th 2009, 11:37 AMuser
Hi friends. How can i prove that the neutral element to the sum is unique.

Thank you

Sincrely, user