# continued fraction expansion help, / proof

• Oct 22nd 2013, 02:22 AM
Tweety
continued fraction expansion help, / proof
I have attached the question I need help, question 5, and really not sure how to go about,

any help / tips appreciated

thank you
• Oct 22nd 2013, 06:24 AM
SlipEternal
Re: continued fraction expansion help, / proof
Begin with writing out a little bit of the continued fraction:
$\displaystyle y = s + \cfrac{1}{s+ \cfrac{1}{s + \cfrac{1}{s+\cfrac{1}{\ddots}}}}$

Now, you are asked to show that $\displaystyle y = s+\dfrac{1}{y}$. This is obvious. Just replace everything under the top 1 of the continued fraction by y (since it is equal to y).

So, how can you solve the rest of it? Well, solve for $\displaystyle y$. If you multiply everything by $\displaystyle y$, you get $\displaystyle y^2 = sy+1$. Treat $\displaystyle s$ as a constant and you have a quadratic of one variable. Solve for $\displaystyle y$. Then $\displaystyle x = 1 + \dfrac{1}{y}$. Just plug in whatever you get as your answer.
• Oct 22nd 2013, 11:29 AM
Tweety
Re: continued fraction expansion help, / proof
thank you, but do you mean replace the first top 1 by y = 1 + 1/y?
• Oct 22nd 2013, 11:35 AM
SlipEternal
Re: continued fraction expansion help, / proof
Quote:

Originally Posted by Tweety
thank you, but do you mean replace the first top 1 by y = 1 + 1/y?

Huh?

I mean
$\displaystyle y = s + \cfrac{1}{s+ \cfrac{1}{s + \cfrac{1}{s+\cfrac{1}{\ddots}}}} = s + \cfrac{1}{\left(s+ \cfrac{1}{s + \cfrac{1}{s+\cfrac{1}{\ddots}}}\right)}$

According to the first equality, the part that is in parentheses in the rightmost equation is equal to $\displaystyle y$.