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 &lt; − 3First of all i do: A n + 1 − A n &lt; 0If this is true i can say that A n decrease. This is true for those A n values: − 3 &gt; A n &gt; − 1 A n &gt; 2And false (so A n increase) for those A n values: A n &lt; − 3 − 1 &lt; A n &lt; 2the case A n &lt; − 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 &gt; 2 then decrease and go in A n &lt; − 3 then again increase and fall in A n &gt; 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.

Question

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        &amp;lt;  −  3First of all i do:      A          n      +      1        −      A          n        &amp;lt;  0If this is true i can say that       A    n   decrease. This is true for those       A    n   values:  −  3  &amp;gt;      A    n    &amp;gt;  −  1      A    n    &amp;gt;  2And false (so       A    n   increase) for those       A    n   values:      A    n    &amp;lt;  −  3  −  1  &amp;lt;      A    n    &amp;lt;  2the case       A    n    &amp;lt;  −  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    &amp;gt;  2 then decrease and go in       A    n    &amp;lt;  −  3 then again increase and fall in       A    n    &amp;gt;  2 etc...Another doubt comes from this fact:It&#039;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&#039;t know if doing it it will fall in another interval where it starts decreasing or increasing and in this scenario i don&#039;t know how to demonstrate if it goes on some limit or just starts to &quot;ping-pong&quot; on different intervals.Or in other words i don&#039;t know how many it decrease/increase so i cannot say if it will go out from the interval i&#039;m considering.

Dereon Parker · Accepted Answer

Step 1If 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 2The 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.

Flambergru · Accepted Answer

Step 1Assume 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 2For 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    &amp;lt;  −  3 then       C    0    &amp;lt;  −  5 and       C    1    &amp;gt;  0 so that   0  &amp;lt;      C          n      +      1        &amp;lt;            2      5            C    n   for   n  &amp;gt;  1. Thus the sequence   (      C    n        )    n   converges to 0, in consequence   (      A    n        )    n   to 2.

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

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