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I have looked on the Wiki page for recurrence relation, and there is a section titled "rational difference equations", which fits my problem of solving $$a_{n+1} = \frac{a_{n}}{a_{n}-1}.$$ But the method on that page does not seem to work. Can anyone point me in the right direction?

Edit: I just realized that this is not useful to what I was trying to do! However it is still interesting to me, and I got some great answers. Thank you. I have posted the real question I am trying to ask in a different thread.

Freddie
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  • what do you mean by solving? Are you looking for an explicit formula for $a_n$? Also the answer will depend on the starting value $a_0$. – Tashi Walde Feb 21 '18 at 01:02
  • for the $a$ variable? –  Feb 21 '18 at 01:03
  • Start with say $a_0=3$ and work out the first few values... I think you will soon be able to guess the answer, and I think you will then easily be able to prove it. – David Feb 21 '18 at 01:06
  • I'm voting to close this question as off-topic because it is not useful to the underlying problem I am trying to solve. – Freddie Feb 21 '18 at 01:09

2 Answers2

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Hint (assuming $a_0 \ne 0\,$): write it as $\,\dfrac{1}{a_{n+1}} = 1 - \dfrac{1}{a_n}\,$ then it becomes a linear recurrence in $\,\dfrac{1}{a_n}\,$.

dxiv
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neither fixpoint is stable. If you have exactly $a_0 = 0,$ it stays there forever. If $a_0 = 2,$ it stays there forever. Otherwise it wanders all over. Let me put in a picture and then look for 2-cycles and 3-cycles...

CORRECTION: it has index 2. You ALWAYS HAVE $$ a_{n+2} = a_n. $$ If $a_0 \neq 0,1,2,$ then your system bounces back and forth between a pair of values forever.

enter image description here

Will Jagy
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