# Non linear recursive relation...

#### chisigma

##### Well-known member
From mathhelpforum.com...

Hi. This is my first post here so I hope I've posted in the right place. My question concerns finding closed forms of nonlinear recurrence relations such as the following...

$\displaystyle a_{n+1}= a^{2}_{n}-1\ ;\ a_{0}=a$ (1)

This one is both nonlinear and nonhomogeneous. The even terms do form a homogeneous recurrence relation, which is nonetheless still nonlinear. Are there general methods for solving particular types of nonlinear recurrence relations? I've tried googling but the results aren't very helpful...

Sylvia A. Anderson

How to aid Sylvia?... there is a closed form solution to (1)?... if not, there is the way to find some informations of the solution, like the convergence-divergence and the limit in case of convergence?...

Kind regards

$\chi$ $\sigma$

#### Amer

##### Active member
My ideas

if a=1 or 0 or -1
it will diverge
$$a_1 = 1 -1 = 0$$
$$a_2 = 0 -1 = -1$$
$$a_3 = 1 - 1= 0$$

if $$\mid a \mid > 2$$
diverge, i choose a=2
$$a_1 = 4-1 = 3$$
$$a_2 = 9 -1 = 8$$

I tried to see if it is increasing or decreasing

$$\frac{a_{n+1}}{a_n} = 1$$
$$\frac{a_n ^2 - 1 }{a_n } = 1$$
$$a_n^2 - a_n -1 = 0$$ two zeros

$$a_n = \frac{1\mp \sqrt{1 +4}}{2}$$

decreasing between the two zeros, and increasing outside

#### chisigma

##### Well-known member
Let's proceed as explained in...

... so that the first step is to write the deifference equation in the alternative form...

$\displaystyle \Delta_{n}=a_{n+1}-a_{n}= a^{2}_{n}-a_{n}-1= f(a_{n})\ ;\ a_{0}=a$ (1)

The function f(*) is represented in the figure...

There are one 'attractive fixed point' [ a point where is f(x)=0 and f(x) crosses the x axis with negative slope...] at $\displaystyle x_{-}= \frac{1-\sqrt{5}}{2}$ and one 'repulsive fixed point' [a point where is f(x)=0 and f(x) crosses the x axis with positive slope...] at $\displaystyle x_{+}= \frac{1+\sqrt{5}}{2}$. The fact that there is an interval around $\displaystyle x_{-}$ where is $\displaystyle |f(x)|< |2\ (x_{-}-x)|$ however means that in general doesn't exist a solution which tends to $x_{-}$ [see theorems 4.1 and 4.2 of the tutorial post...] and 'almost all' the solutions diverge. As explained in the tutorial post a closed form solution of the (1) 'probably' doesn't exist and what we can do is the search of periodical solution. The solution of periodicity one are of course $a_{n}=x_{-}$ and $a_{n}=x_{+}$. The solution of periodicity two are generated for the values of a satisfying the equation...

$\displaystyle a^{4}-2\ a^{2}-a= a\ (a+1)\ (a^{2}-a-1)=0$ (2)

... that are $a=x_{-}$, $a+x_{+}$, $a=0$ and $a=-1$. The conclusion is the following...

a) for $\displaystyle |a|>x_{+}$ the solution diverges to $+ \infty$...

b) for $\displaystyle |a|<x_{+}\ , a \ne x_{-}$ the solution diverges but tends to the solution with periodicity two '0 -1 0 -1...'

c) for $a=x_{-}$ we have the solution with periodicity one $x_{-}$ and for $a=x_{+}$ we have the solution with periodicity one $x_{+}$...

Solutions with periodicity greater than two, if they exist, have to be found...

Kind regards

$\chi$ $\sigma$

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#### Amer

##### Active member
wow, did i get 50% of the answer ?