# A periodic sequence

The sequence given by $$x_0 = a$$, $$x_1 = b$$ and

$x_{n + 1} = \frac {1}{2} (x_{n - 1} + \frac {1}{x_n})$

is periodic. Prove that $$ab = 1$$.

Note by Sharky Kesa
4 years ago

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Multiply both sides by $$x_n$$ for the recursion and get: $$x_{n+1}x_{n}=\frac {1}{2}(x_nx_{n-1}+1)$$

Notice that if we let $$y_n=x_{n}x_{n-1}$$, then the recursion becomes: $$y_{n+1}=\frac {1}{2}(y_n+1)$$.

It's clear that If $${x_n}$$ is periodic, then so is $${y_n}$$.

If we let the initial value for sequence $${y_n}$$ be $$y_1=ab=c$$, then the closed formula for the sequence is: $$y_n=\frac {c-1}{2^{n-1}}+1$$. If $$c$$ isn't 1, then the sequence either monotonically increases(when $$c<1$$)) or decreases (when $$c>1$$) and eventually converges to $$1$$. Hence it must be true for $$c=ab=1$$.

- 3 years, 12 months ago