I am trying to find a closed form solution for this recurrence relation, but it is a repeating one and I can't seem to figure it out. Any ideas?

  • $\begingroup$ B(1) = 2. Sorry forgot to include that. $\endgroup$ – DiabeticPie Apr 27 '16 at 19:24
  • $\begingroup$ Have you tried to write out the first few terms? $\endgroup$ – lulu Apr 27 '16 at 19:24
  • $\begingroup$ Yes, they alternate between 3/2 and 2. Odds being 2 and evens being 3/2. I just don't know how to get one closed form solution out of it. $\endgroup$ – DiabeticPie Apr 27 '16 at 19:30
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    $\begingroup$ If you've figured out that it's repeating then you're about $99\%$ done solving the problem. $\qquad$ $\endgroup$ – Michael Hardy Apr 27 '16 at 19:37
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    $\begingroup$ "case" definitions of functions (of $n$ in this case) are, indeed, "closed form". You can "close" the form even more, for example you can write $B_n = 1.75 - 0.25(-1)^n$, but this is just a gimmick, it obscures more than it reveals the true nature of the sequence. $\endgroup$ – mathguy Apr 27 '16 at 19:38


Let $n$ be even, then

$$ B_n=\frac {3}{B_{n-1}}=\frac {3}{\frac{3}{B_{n-2}}}=B_{n-2} $$

So the sequence is constant in the even terms. What happens with the odd terms?

As I said in the comments, I'd call

$$ B_n=\begin{cases}B_1, & n\text{ odd.}\\ \frac 3 {B_1}, & n \text{ even.} \end{cases} $$

A closed form solution (there's no recurrence relation). However, I'm not sure what your definition of closed form solution is.

  • $\begingroup$ The odd terms are all equal, right? So how would I make a closed form solution? Wouldn't I need two? $\endgroup$ – DiabeticPie Apr 27 '16 at 19:27
  • $\begingroup$ @DiabeticPie Yes, they are. Well, I would call SiongthyeGoh's answer a closed form solution. What's your definition? $\endgroup$ – YoTengoUnLCD Apr 27 '16 at 19:29
  • $\begingroup$ You've got "even" and "odd" switched around. It should be $B_1$ when $n$ is odd, not $B_1$ when $n$ is even. $\qquad$ $\endgroup$ – Michael Hardy Apr 27 '16 at 19:39
  • $\begingroup$ @MichaelHardy Whoops, in my head the sequence started at $0$ :-P. Thanks. $\endgroup$ – YoTengoUnLCD Apr 27 '16 at 19:40

Let $B_1$ be the first term.

If $n$ is odd, $B_n=B_1$.

If $n$ is even, $B_n=\frac{3}{B_1}.$

  • $\begingroup$ Is it okay to have two closed form solutions? How wold I go about proving this using induction? $\endgroup$ – DiabeticPie Apr 27 '16 at 19:28
  • $\begingroup$ Well, to prove it by induction, YoTengoUnLCD's working has shown us the induction step. We can always group multiple solutions as a single equation. For example, in this case, $B_n=2-0.5\mathbb{1}_{n \text{ is odd }}$ where $\mathbb{1}$ is the indicator function that takes value 1 when $n$ is odd and $0$ otherwise. $\endgroup$ – Siong Thye Goh Apr 27 '16 at 19:34
  • $\begingroup$ or we can write it as $B_n=1.75-0.25(-1)^n$. $\endgroup$ – Siong Thye Goh Apr 27 '16 at 19:37
  • $\begingroup$ How would I go about proving this by induction? $\endgroup$ – DiabeticPie Apr 27 '16 at 20:04
  • $\begingroup$ \begin{align*} B_{n+1}&=\frac{3}{B_n}\\ &=\frac{3}{1.75-0.25(-1)^n}\frac{1.75+0.25(-1)^n}{1.75+0.25(-1)^n}\\ &=\frac{3}{1.75^2-0.25^2(-1)^{2n}}(1.75+0.25(-1)^n)\\ &=\frac{3}{1.75^2-0.25^2}(1.75-0.25(-1)^{n+1})\\ &=\frac{3}{(1.75-0.25)(1.75+0.25)}(1.75-0.25(-1)^{n+1})\\ &=\frac{3}{(1.5)(2)}(1.75-0.25(-1)^{n+1})\\ &=1.75-0.25(-1)^{n+1} \end{align*} $\endgroup$ – Siong Thye Goh Apr 27 '16 at 21:03

$$ B_1 \longmapsto \frac 3 {B_1} \longmapsto \left( \frac 3 {\left( \frac 3 {B_1} \right)} = B_1 \right) \longmapsto \frac 3 {B_1} \mapsto B_1 \longmapsto \frac 3 {B_1} \longmapsto B_1 \longmapsto \cdots $$ That's the solution.

You could say it's $\begin{cases} B_1 & \text{when $n$ is odd,} \\[2pt] 3/B_1 & \text{when $n$ is even.} \end{cases}$

However, maybe the expression "closed form" is what you're worried about. That's a bit vaguely defined but people seem to know what it is when they see it. How about something like this: $$ \frac{B_1 + (3/B_1)} 2 + (-1)^n \frac {(3/B_1)-B_1} 2. $$


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