How to the fixed points of the following recurrence? X_n=2X_(n−1) (2−3X_(n−1))+X_(n−1)

mksfmasterio

mksfmasterio

Answered question

2022-09-12

How to the fixed points of the following recurrence?
X n = 2 X n 1 ( 2 3 X n 1 ) + X n 1
And therein, determining their stability analytically?

Answer & Explanation

soyafh

soyafh

Beginner2022-09-13Added 17 answers

We can write the recurrence as X n = F ( X n 1 ), where F ( X n 1 ) = 2 X n 1 ( 2 3 X n 1 ) + X n 1 . You can think of F as an iterative map, i.e. if you start with an initial iterate X 0 , F will generates X 1 , X 2 , so that X n converges to the fixed point X of F (i.e. that F ( X ) = X ).
In view of above, one simple way to find the fixed points are to simply solve for X = F ( X ) (assuming that X n converges to the fixed points X )). In this case, we get X = 0 , 2 / 3. To determine the stability of the fixed points, look at the Jacobian of the map F.
To find the solution of the system of recurrences, note that you can recast the problem as:
[ a n b n ] = [ 4 0 3 3 ] [ a n 1 b n 1 ] + [ 6 0 ] .
Note that the above iteration won't converge to a unique solution since the spectral radius of the matrix
[ 4 0 3 3 ]
is greater than 1.

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