Non linear recursive relation....

In summary, Sylvia is looking for a closed form solution to a nonlinear recurrence relation. There are general methods for solving particular types of nonlinear recurrence relations, but this one is nonhomogeneous and nonlinear.
  • #1
chisigma
Gold Member
MHB
1,628
0
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

$\chi$ $\sigma$
 
Physics news on Phys.org
  • #2
My ideas

if a=1 or 0 or -1
it will diverge
[tex]a_1 = 1 -1 = 0 [/tex]
[tex]a_2 = 0 -1 = -1 [/tex]
[tex]a_3 = 1 - 1= 0 [/tex]

if [tex]\mid a \mid > 2 [/tex]
diverge, i choose a=2
[tex]a_1 = 4-1 = 3 [/tex]
[tex]a_2 = 9 -1 = 8 [/tex]
I tried to see if it is increasing or decreasing

[tex]\frac{a_{n+1}}{a_n} = 1 [/tex]
[tex]\frac{a_n ^2 - 1 }{a_n } = 1 [/tex]
[tex]a_n^2 - a_n -1 = 0 [/tex] two zeros

[tex]a_n = \frac{1\mp \sqrt{1 +4}}{2} [/tex]

decreasing between the two zeros, and increasing outside
 
  • #3
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



$\chi$ $\sigma$
 

Attachments

  • MHB10.PNG
    MHB10.PNG
    980 bytes · Views: 53
Last edited:
  • #4
wow, did i get 50% of the answer ?
 
  • #5


Hello Sylvia and welcome to the forum! Nonlinear recurrence relations can be challenging to solve, especially if they are nonhomogeneous. In general, there are no specific methods for solving particular types of nonlinear recurrence relations, but there are some general techniques that can be used.

One approach is to try to find a pattern in the terms of the sequence and see if it can be expressed in a closed form. In your example, it seems that the even terms of the sequence are related to the odd terms, so you could try to express $a_{2n}$ in terms of $a_{2n-1}$.

Another approach is to use generating functions, which can be helpful in finding closed forms for nonlinear recurrence relations. This involves converting the recurrence relation into a power series and then using algebraic manipulations to solve for the coefficients.

If there is no closed form solution to a nonlinear recurrence relation, it is still possible to find some information about the solution. For example, you could analyze the behavior of the sequence and determine if it converges or diverges, and if it converges, what is its limit. This can be done by looking at the behavior of the terms as $n$ approaches infinity and using techniques such as the squeeze theorem or the ratio test.

I hope this helps and good luck with your research on nonlinear recurrence relations!
 

Related to Non linear recursive relation....

1. What is a non-linear recursive relation?

A non-linear recursive relation is a mathematical equation or formula in which the current term or value is dependent on previous terms or values, and the relationship between these terms is not linear (i.e. cannot be expressed as a straight line).

2. How is a non-linear recursive relation different from a linear recursive relation?

A linear recursive relation has a constant rate of change between terms, while a non-linear recursive relation does not have a constant rate of change. In other words, the relationship between terms in a linear recursive relation can be represented by a straight line, while the relationship in a non-linear recursive relation cannot.

3. What are some real-world applications of non-linear recursive relations?

Non-linear recursive relations are often used to model complex systems and processes in areas such as physics, biology, economics, and computer science. For example, they can be used to model population growth, compound interest, and chaotic systems.

4. How do you solve a non-linear recursive relation?

The process for solving a non-linear recursive relation depends on the specific equation or formula. In general, you can use algebraic techniques to rearrange the equation and isolate the current term on one side, and then use substitution or iteration to find its value. In some cases, a closed-form solution may not exist and numerical methods may be used instead.

5. What are some challenges of working with non-linear recursive relations?

One of the main challenges of working with non-linear recursive relations is that they can be difficult to solve, especially if a closed-form solution does not exist. Additionally, small changes in initial values or parameters can lead to significant changes in the outcome, making it challenging to accurately predict the behavior of the system or process being modeled.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
18
Views
2K
  • Atomic and Condensed Matter
Replies
0
Views
772
Replies
8
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
28
Views
5K
  • General Math
Replies
2
Views
875
  • Calculus and Beyond Homework Help
Replies
3
Views
630
  • General Math
Replies
11
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
Back
Top