MHB Non linear recursive relation....

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The discussion focuses on the nonlinear recurrence relation a_{n+1} = a_n^2 - 1, with an initial condition a_0 = a. It highlights the challenge of finding closed-form solutions and explores the behavior of the sequence based on the initial value a. The analysis reveals that for |a| > 2, the sequence diverges, while specific values lead to periodic solutions. The existence of attractive and repulsive fixed points is noted, with most solutions diverging rather than converging. Ultimately, the thread suggests that a closed-form solution likely does not exist, and periodic solutions are of interest for further exploration.
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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

$\chi$ $\sigma$
 
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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
 
Let's proceed as explained in...
http://www.mathhelpboards.com/showthread.php?426-Difference-equation-tutorial-draft-of-part-I

... 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...

https://www.physicsforums.com/attachments/125._xfImport

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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wow, did i get 50% of the answer ?
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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