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 $ab = 1$ .

#Algebra #Sharky

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $c=ab=1$ .

Xuming Liang - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$