> 
> $a\_n = \frac{1}{2}(a\_{n-1} + \frac{\beta}{a\_{n-1}})$ for $ n > 0$ with $a\_0 = 1$. Changing variables in this recurrence, and letting $b\_n = a\_n - a$, we find by simple algebra that $b\_n = \frac{1}{2}\frac{b\_{n-1} ^2 + \beta - \alpha^2}{b\_{n-1} + \alpha}$.
> 
> 
> 

I've tried making the substitutions and doing algebra but can't come up with it. Would somebody please help show how to get from the expression for $a\_n$ to the one for $b\_n$ using simple algebra?