Exercise 5.7.19* (Sequence of points on the cubic $axy = (x+1)(y+1)(x+y+b)$)

Proof.

(a)

Let $b,x_0,x_1$ be given positive real numbers. We can choose $a=(x_0+1)(x_1+1)(x_0+x_1+b)∕(x_0{x}_1)$ such that the point $(x_0,x_1)$ lies on the curve ${𝒞}_f(ℝ)$, where $$f(x,y)=\mathit{axy}-(x+1)(y+1)(x+y+b).$$

Consider the sequence of real numbers ${(x_n)}_{n\in \mathbb{N}}$ defined by $x_0,x_1$ and

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in {\mathbb{N}}^{\text{∗}},x_{n+1}=\frac{x_n+b}{x_{n-1}}.$$

More precisely,

$$(x_{n+1},x_{n+2})=\left(x_{n+1},\frac{x_{n+1}+b}{x_n}\right)=g(x_n,x_{n+1}),\text{where }g(x,y)=\left(y,\frac{y+b}{x}\right).$$

Since ${ℝ}_{\text{+}}^{\text{∗}}\times {ℝ}_{\text{+}}^{\text{∗}}$ is steady for $g$ (because $b>0$) and $(x_0,x_1)\in {ℝ}_{\text{+}}^{\text{∗}}\times {ℝ}_{\text{+}}^{\text{∗}}$, we know that the sequence ${(x_n)}_{n\in \mathbb{N}}$ is well defined, and that for all $n\in \mathbb{N},x_n>0$.

Consider the property

$$𝒫(n):(x_n,x_{n+1})\in {𝒞}_f(ℝ).$$

By our choice of $a$, we know that $𝒫(0)$ is true. Suppose now that $𝒫(n-1)$ is true for some $n\in {\mathbb{N}}^{\text{∗}}$, so that $(x_{n-1},x_n)\in 𝒞(ℝ)$, that is

$$\begin{array}{lll}\hfill a{x}_{n-1}{x}_n-(x_{n-1}+1)(x_n+1)(x_{n-1}+x_n+b)=0.& & \hfill \text{(1)}\end{array}$$

Substituting $x_{n-1}=\frac{x_n+b}{x_{n+1}}$ in (1), we obtain

$$\begin{array}{lll}\hfill a\frac{x_n+b}{x_{n+1}}{x}_n-\left(\frac{x_n+b}{x_{n+1}}+1\right)\left(x_n+1\right)\left(\frac{x_n+b}{x_{n+1}}+x_n+b\right)=0.& & \hfill \end{array}$$

Multiplying by $x_{n+1}^2$, this gives

$$\begin{array}{llll}\hfill 0& =a(x_n+b)x_n{x}_{n+1}-(x_n+b+x_{n+1})(x_n+1)(x_n+b+x_{n+1}(x_n+b))& \hfill & \\ \hfill & =(x_n+b)\left[a{x}_n{x}_{n+1}-(x_n+b+x_{n+1})(x_n+1)(1+x_{n+1})\right]& \hfill & \\ \hfill & =(x_n+b)f(x_n,x_{n+1}).& \hfill & \end{array}$$

Since $x_n+b>0$, we obtain $f(x_n,x_{n+1})=0$, that is $𝒫(n)$. This induction shows that

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N},(x_n,x_{n+1})\in {𝒞}_f(ℝ).$$

(b)

We suppose in this part that $b=1$ (and as usual $x_0>0,x_1>0$).

Then

$$\begin{array}{llll}\hfill x_2& =\frac{x_1+1}{x_0},& \hfill & \\ \hfill x_3& =\frac{x_2+1}{x_1}=\frac{\frac{x_1+1}{x_0}+1}{x_1}=\frac{x_0+x_1+1}{x_0{x}_1},& \hfill & \\ \hfill x_4& =\frac{x_3+1}{x_2}=\frac{\frac{x_0+x_1+1}{x_0{x}_1}+1}{\frac{x_1+1}{x_0}}=\frac{x_0+x_1+1+x_0{x}_1}{x_1(x_1+1)}=\frac{1+x_0}{x_1},& \hfill & \\ \hfill x_5& =\frac{x_4+1}{x_3}=\frac{\frac{1+x_0}{x_1}+1}{\frac{x_0+x_1+1}{x_0{x}_1}}=x_0,& \hfill & \\ \hfill x_6& =\frac{x_5+1}{x_4}=\frac{x_0+1}{\frac{1+x_0}{x_1}}=x_1.& \hfill & \end{array}$$

Since $(x_5,x_6)=(x_0,x_1)$, and $(x_{n+1},x_{n+2})=g(x_n,x_{n+1})$ for all $n$, where $g$ is defined in part (a), we obtain by induction that for all $n$, $(x_{n+5},x_{n+6})=(x_n,x_{n+1})$. In conclusion,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N},x_{n+5}=x_n.$$

The sequence ${(x_n)}_{n\in \mathbb{N}}$ has period $5$.

(c)

By hypothesis, $b>0,x_0>0,x_1>0$. By part (a), we know that for all $n\in \mathbb{N}$, $x_n>0$, $x_{n+1}>0$ and $(x_n,x_{n+1})\in {𝒞}_f(ℝ)$. Put $x=x_n,y=y_n$. Then $x>0,y>0$ and using $x+1>x>0,y+1>y>0,x+y+b>x+y>0$, we obtain $$\begin{array}{llll}\hfill \mathit{axy}=(x+1)(y+1)(x+y+b)& >\mathit{xy}(x+y),& \hfill & \end{array}$$

therefore

$$a>x+y>0.$$

So $0<x_n<a$ for all indices $n$. The sequence ${(x_n)}_{n\in \mathbb{N}}$ is bounded.