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 \(ab = 1\).

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 \(ab = 1\).

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## Comments

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TopNewestMultiply 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 \(c=ab=1\). – Xuming Liang · 2 years, 4 months ago

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