# How does recurrence $a_n=\frac12(a_{n-1}+\frac\beta{a_{n-1}})$ become $b_n=\frac12\frac{b_{n-1}^2+\beta-\alpha^2}{b_{n-1}+\alpha}$ when $b_n=a_n-a$?

$$a_n = \frac{1}{2}(a_{n-1} + \frac{\beta}{a_{n-1}})$$ 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?

• Before we can answer that, we would have to know what "a" is! Oct 8, 2020 at 17:52

We have $$a_{n}=\frac{1}{2}\big(a_{n-1}+\frac{\beta}{a_{n-1}}\big)$$ $$=\frac{1}{2}\big(\frac{a_{n-1}^2+\beta}{a_{n-1}}\big)$$
and $$b_{n}=a_{n}-\alpha$$, then $$a_{n}=b_{n}+\alpha$$. Thus substituting gives $$b_{n}+\alpha=\frac{1}{2}\big(\frac{(b_{n-1}+\alpha)^2+\beta}{b_{n-1}+\alpha}\big)$$ $$b_{n}=\frac{1}{2}\frac{b_{n-1}^{2}+2b_{n-1}\alpha+\alpha^2+\beta}{b_{n-1}+\alpha}-\alpha$$ $$b_{n}=\frac{1}{2}\frac{b_{n-1}^{2}+2b_{n-1}\alpha+\alpha^2+\beta}{b_{n-1}+\alpha}-\frac{1}{2}\frac{2\alpha(b_{n-1}+\alpha)}{b_{n-1}+\alpha}$$ $$=\frac{1}{2}\big[\frac{b_{n-1}^2+2b_{n-1}\alpha+\alpha^2+\beta-2\alpha b_{n-1}-2\alpha^2}{b_{n-1}+\alpha}\big]$$ $$=\frac{1}{2}\frac{b_{n-1}^2+\beta-\alpha^2}{b_{n-1}+\alpha}$$
$$b_n = a_n - a$$, $$b_{n-1} = a_{n-1} -a$$. Hence
\begin{align}b_n = a_n - a &= \frac12(a_{n-1} + \frac \beta{a_{n-1}})-a \\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a})-a \\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a}-2a) \\&=\frac12(b_{n-1}-a + \frac \beta{b_{n-1}+a}) \\&=\frac1{2(b_{n-1}+a)}((b_{n-1}+a)(b_{n-1}-a)+\beta) \\&=\frac{b_{n-1}^2-a^2+\beta}{2(b_{n-1}+a)} \\&\left(=\frac12\frac{b_{n-1}^2+\beta-a^2}{b_{n-1}+a}\right) \end{align}