# Find a closed form solution for $B_n = \frac 3{B_{n-1}}$

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?

• B(1) = 2. Sorry forgot to include that. – DiabeticPie Apr 27 '16 at 19:24
• Have you tried to write out the first few terms? – lulu Apr 27 '16 at 19:24
• 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. – DiabeticPie Apr 27 '16 at 19:30
• If you've figured out that it's repeating then you're about $99\%$ done solving the problem. $\qquad$ – Michael Hardy Apr 27 '16 at 19:37
• "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. – mathguy Apr 27 '16 at 19:38

Hint:

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.

• The odd terms are all equal, right? So how would I make a closed form solution? Wouldn't I need two? – DiabeticPie Apr 27 '16 at 19:27
• @DiabeticPie Yes, they are. Well, I would call SiongthyeGoh's answer a closed form solution. What's your definition? – YoTengoUnLCD Apr 27 '16 at 19:29
• You've got "even" and "odd" switched around. It should be $B_1$ when $n$ is odd, not $B_1$ when $n$ is even. $\qquad$ – Michael Hardy Apr 27 '16 at 19:39
• @MichaelHardy Whoops, in my head the sequence started at $0$ :-P. Thanks. – 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, $B_n=\frac{3}{B_1}.$

• Is it okay to have two closed form solutions? How wold I go about proving this using induction? – DiabeticPie Apr 27 '16 at 19:28
• 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. – Siong Thye Goh Apr 27 '16 at 19:34
• or we can write it as $B_n=1.75-0.25(-1)^n$. – Siong Thye Goh Apr 27 '16 at 19:37
• How would I go about proving this by induction? – DiabeticPie Apr 27 '16 at 20:04
• \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*} – 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: $$\frac{B_1 + (3/B_1)} 2 + (-1)^n \frac {(3/B_1)-B_1} 2.$$