A quick doubt, lets study the recurrence sequence: A_{n+1}=(4A_n+2)//A_n+3; A_0<-3.

patylomy7 2022-08-16 Answered
Some questons about recurrence sequences (using a problem).
A quick doubt, lets study the recurrence sequence:
A n + 1 = ( 4 A n + 2 ) / A n + 3
A 0 < 3
First of all i do:
A n + 1 A n < 0
If this is true i can say that A n decrease. This is true for those A n values:
3 > A n > 1
A n > 2
And false (so A n increase) for those A n values:
A n < 3
1 < A n < 2
the case A n < 3 interests me.
The limit L can be -1 or -2 but i cannot say it exists for sure because A n is not limited and monotone for all the A n possible values. For example, the sequence can go from A n > 2 then decrease and go in A n < 3 then again increase and fall in A n > 2 etc...
Another doubt comes from this fact:
It's ok to remove -1 from the possible values of L because in this case the sequence still growing?
Anyways: It happens so many times that i know the sequence increase or decrease in an interval but i don't know if doing it it will fall in another interval where it starts decreasing or increasing and in this scenario i don't know how to demonstrate if it goes on some limit or just starts to "ping-pong" on different intervals.
Or in other words i don't know how many it decrease/increase so i cannot say if it will go out from the interval i'm considering.
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Answers (2)

Dereon Parker
Answered 2022-08-17 Author has 11 answers
Step 1
If you suspect oscillating behavior, then apply the recursion twice, i.e., examine how A n + 2 depends on A n .
In the case of fractions of linear terms one can fully linearize the problem by writing A n = P n / Q n , P 0 = A 0 , Q 0 = 1 and using the one degree of freedom to extract from
P n + 1 Q n + 1 = A n + 1 = 4 A n + 2 A n + 3 = 4 P n + 2 Q n P n + 3 Q n
the linear system [ P n + 1 Q n + 1 ] = [ 4 2 1 3 ] [ P n Q n ]
As the characteristic equation is
0 = z 2 7 z + 10 = ( z 2 ) ( z 5 )
the roots are z 1 = 5 and z 2 = 2 with eigenvectors v 1 = ( 2 1 ) and v 2 = ( 1 1 ) .
Step 2
The solution will thus have the form P n = 2 c 1 5 n + c 2 2 n , Q n = c 1 5 n c 2 2 n , Q 0 = 1 c 2 = c 1 1, P 0 = A 0 c 1 = 1 3 ( A 0 + 1 ) and thus
A n = 2 ( A 0 + 1 ) 5 n + ( A 0 2 ) 2 n ( A 0 + 1 ) 5 n ( A 0 2 ) 2 n = 2 ( A 0 + 1 ) + ( A 0 2 ) ( 2 5 ) n ( A 0 + 1 ) ( A 0 2 ) ( 2 5 ) n
Which will in all but one exceptional case (where c 1 = 0, i.e., A 0 = 1) converge to 2.

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Flambergru
Answered 2022-08-18 Author has 4 answers
Step 1
Assume for the moment that the recursive sequence converges. Then the limit would satisfy the quadratic equation L ( L + 3 ) = 4 L + 2 0 = L 2 L 2 = ( L 2 ) ( L + 1 ).
Take the negative fixed point L = 1 and consider the sequence B n = A n + 1 or A n = 1 + B n . Then
B n + 1 = 1 + 4 + 4 B n + 2 1 + B n + 3 = 5 B n B n + 2 which tells that the iteration will move away from that fixed point with speed around 5 2 except in the case where B 0 = 0, A 0 = 1, which is excluded.
Step 2
For the second fixed point L = 2 a similar consideration with A n = 2 + C n gives
C n + 1 = 2 + 4 ( 2 + C n ) + 2 C n + 2 + 3 = 2 C n C n + 5
Now if A 0 < 3 then C 0 < 5 and C 1 > 0 so that 0 < C n + 1 < 2 5 C n for n > 1. Thus the sequence ( C n ) n converges to 0, in consequence ( A n ) n to 2.

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