5

Let $a_{0}=a>0,a_{1}=b>0$,and such $$a_{n+1}a_{n-1}=\max\{(a_{n},1)\},\forall n\in N^{+}$$ show that $$a_{n+5}=a_{n}$$

Even $a,b$ with the 1 uncertainty,so we can't $$a_{2}=\dfrac{\max{(a_{1},1})}{a_{0}}=\begin{cases}\dfrac{a_{1}}{a_{0}}&a_{1}\ge 1\\ \dfrac{1}{a_{0}},&a_{1}<1 \end{cases}$$, But This sequnece always is period $5$. it looks very interesting.following when I determine$a_{3}$,I can't.Thanks so much for any suggestion.

  • What have you tried? I would start by making a spreadsheet and validating the claim for various $a_0, a_1$ pairs, making sure to include all four possibilities comparing them with $1$. I have done this and it passes. I would then think on how the pattern of which option you choose in the max depends on the starting values, which I have not done. – Ross Millikan Oct 04 '15 at 04:39
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    It's not elegant but why don't you do the four cases: a and b greater or less than 1? – fleablood Oct 04 '15 at 04:44
  • Hmmm... haven't looked at this for more than a minute or so, but would it help that this can be written as $$a_{n+1}a_{n-1}=\frac{a_n+1 + \sqrt{(a_n-1)^2}}{2}$$ Just a thought, I could be totally off :/ – Brevan Ellefsen Oct 04 '15 at 04:46
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    Here's an example sequence: $2,3,\dfrac32,\dfrac12,\dfrac23,2,3, \dfrac32, \dfrac12,\dfrac23,\dotsb$. – Akiva Weinberger Oct 04 '15 at 04:46
  • Having solved this, I do wonder about $a_{n+1} a_n = \max(a_{n-1},1)$ and the general products $$a_{n-1}\prod_{k=1}^l a_{n+k} = \max(a_n,1)\quad \prod_{k=0}^{\infty} a_{n+k} = \max(a_{n-1},1)$$ – Brevan Ellefsen Oct 04 '15 at 05:44

2 Answers2

1

Do 5 different cases,

  1. $b\geq a\geq 1$

  2. $b\geq 1> a$

  3. $a\geq b \geq1$

  4. $b<1,a<1$

  5. $b<1, a\geq 1$

Case 1. $a_2=b/a, a_3=1/a, a_4=a/b, a_5=a, a_6=b, ...$

Case 2. $a_2=b/a, a_3=1/a, a_4=1/b, a_5=a, a_6=b, ...$

Case 3. $a_2=b/a, a_3=1/b, a_4=a/b, a_5=a, a_6=b, ...$

Case 4. $a_2=1/a, a_3=1/ab, a_4=1/b, a_5=a, a_6=b, ...$

Case 5. $a_2=1/a, a_3=1/b, a_4=a/b, a_5=a, a_6=b, ...$

1

I know frankooo has already posted an answer, but I started my post beforehand and I want to go through the whole process of solving one case (as frankooo's answer doesn't show a full proof). I hope this helps in formulating the remaining proofs! Lets start by assuming $a_1 \geq a_0 \geq 1$. We then plug it into the definition $$a_n = \frac{\max(a_{n-1},1)}{a_{n-2}}\qquad$$ To get $$a_2 = \frac{\max(a_1,1)}{a_0} = \frac{a_1}{a_0}$$ Since $a_1\geq a_0$ the quantity $a_2 \geq 1$ and $a_2 \geq a_1$ $$a_3 = \frac{\max(a_2,1)}{a_1} = \frac{a_2}{a_1} = \frac{1}{a_0}$$ Note that we get the last equality above by substituting in the definition of $a_2$. Since we assumed $a_0 \geq 1$ we have that $a_3 \leq 1$ $$a_4 = \frac{\max(a_3,1)}{a_2} = \frac{1}{a_2} = \frac{a_0}{a_1}$$ Note that since $a_1 \geq a_0$ the quantity $a_4 \leq 1$ $$a_5 = \frac{\max(a_4,1)}{a_3} = \frac{1}{a_3} = a_0$$ Finally, since $a_0 \geq 1$ we get $$a_6 = \frac{\max(a_5,1)}{a_4} = \frac{a_5}{a_4} = \frac{a_0}{a_4} = a_1$$ From this we can see that the cycle would continue with the desired cyclic behavior. To preserve my sanity I will leave the rest of the proofs up to you!

Note: there may be a faster way to prove this, but I haven't done any number theory prior to this... I made this up as I went along using Avika's comment to check my answers

Brevan Ellefsen
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