Bilinear Recursion - 1

We have,

$x_{n+1} = f(x_n) = \dfrac{ 6 x_n + 35 }{ 4 x_n + 10 }$

Setting $x_{n+1} = x_{n} = x$ , results in the quadratic equation $x (4x + 10) = 6x + 35$ , which after re-arranging, becomes,

$4 x^2 + 4 x - 35 = 0$

which factors into $(2 x - 5 )(2 x + 7 ) = 0$ , so the possible limits are $\dfrac{5}{2}$ and $- \dfrac{7}{2}$ . To determine which is the limit, we find the derivative of the recursion function $f (x)$ ,

$f'(x) = \dfrac{ -80}{(4 x + 10)^2 }$

It is easy to check that $| f(\dfrac{5}{2}) | \lt 1$ , and that $| f(\dfrac{-7}{2} ) | \gt 1$ , therefore, the limit is $\dfrac{5}{2}$ .

The neatest way to answer this question is @Hosam Hajjir 's solution. However, we can actually solve the recurrence with some substitutions.

Anticipating the result, let $y_n=x_n-\frac52$ . The recurrence then becomes $y_{n+1}=\frac{5}{5+y_n}-1,\;\;\;y_0=\frac{15}{2}$ Now let $z_n=\frac{1}{y_n}$ , and substitute again to get $z_{n+1}=-1-5z_n,\;\;\;z_0=\frac{2}{15}$

This is just a linear recurrence, with solution $z_n=(-5)^n \left(z_0+\frac16 \right) - \frac16$

Clearly this tends to infinity as $n$ tends to infinity; so $y_n \to 0$ and $x_n \to \boxed{\frac52}$ .

An explicit form for $x_n$ is $x_n= \frac52 - \frac{6}{9\cdot (-5)^{n-1} + 1}$

which again shows that $x_n$ converges to $\frac52$ .

The approach used by @Hosam Hajjir , is assuming that $x_n$ converges when $n \to \infty$ . Strictly, we have to prove that $x_n$ converges before we can use the approach. The proof comes from the part 2 of this problem. It is given in the part 2 that $x_n$ can be expressed as:

$x_n = \frac {a+b\lambda^n}{c\lambda^n-d}$

where $a$ , $b$ , $c$ , and $d$ are positive integers and $\lambda$ a negative integer. Then we have

$\begin{aligned} \lim_{n \to \infty} x_n & = \lim_{n \to \infty} \frac {a+b \lambda^n}{c\lambda^n - d} \\ & = \frac {\frac a{\lambda^n}+b}{c-\frac d{\lambda^n}} & \small \blue{\text{for }|\lambda|>1} \\ & = \frac bc \end{aligned}$

So $x_n$ converges if $|\lambda| > 1$ . It is founded that $\lambda = -5$ , therefore $x_n$ as $n \to \infty$ . Let the limit be $L$ , then

$\begin{aligned} x_n & = \frac {4x_n+35}{4x_n+10} \\ \lim_{n \to \infty} x_n & = \lim_{n \to \infty} \frac {4x_n+35}{4x_n+10} \\ L & = \frac {6L+35}{4L+10} \\ 4L^2 + 10L & = 6L + 35 \\ 4L^2 + 4L - 35 & = 0 \\ (2L-5)(2L+7) & = 0 & \small \blue{\text{Since }x_n > 0 \ \forall n \ge 0, \ L >0} \\ \implies L & = \boxed{\frac 52} \end{aligned}$