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

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  • $\begingroup$ Before we can answer that, we would have to know what "a" is! $\endgroup$
    – user247327
    Oct 8, 2020 at 17:52

2 Answers 2

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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 $$b_{n}+\alpha=\frac{1}{2}\big(\frac{(b_{n-1}+\alpha)^2+\beta}{b_{n-1}+\alpha}\big)$$ $$b_{n}=\frac{1}{2}\frac{b_{n-1}^{2}+2b_{n-1}\alpha+\alpha^2+\beta}{b_{n-1}+\alpha}-\alpha$$ $$b_{n}=\frac{1}{2}\frac{b_{n-1}^{2}+2b_{n-1}\alpha+\alpha^2+\beta}{b_{n-1}+\alpha}-\frac{1}{2}\frac{2\alpha(b_{n-1}+\alpha)}{b_{n-1}+\alpha}$$ $$=\frac{1}{2}\big[\frac{b_{n-1}^2+2b_{n-1}\alpha+\alpha^2+\beta-2\alpha b_{n-1}-2\alpha^2}{b_{n-1}+\alpha}\big]$$ $$=\frac{1}{2}\frac{b_{n-1}^2+\beta-\alpha^2}{b_{n-1}+\alpha}$$

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$b_n = a_n - a$, $b_{n-1} = a_{n-1} -a$. Hence

$$\begin{align}b_n = a_n - a &= \frac12(a_{n-1} + \frac \beta{a_{n-1}})-a \\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a})-a \\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a}-2a) \\&=\frac12(b_{n-1}-a + \frac \beta{b_{n-1}+a}) \\&=\frac1{2(b_{n-1}+a)}((b_{n-1}+a)(b_{n-1}-a)+\beta) \\&=\frac{b_{n-1}^2-a^2+\beta}{2(b_{n-1}+a)} \\&\left(=\frac12\frac{b_{n-1}^2+\beta-a^2}{b_{n-1}+a}\right) \end{align}$$

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    $\begingroup$ Here. I added one more step. $\endgroup$
    – player3236
    Oct 8, 2020 at 17:37

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