 # 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:
${A}_{n+1}=\left(4{A}_{n}+2\right)/{A}_{n}+3$
${A}_{0}<-3$
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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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
$\frac{{P}_{n+1}}{{Q}_{n+1}}={A}_{n+1}=\frac{4{A}_{n}+2}{{A}_{n}+3}=\frac{4{P}_{n}+2{Q}_{n}}{{P}_{n}+3{Q}_{n}}$
the linear system $\left[\begin{array}{c}{P}_{n+1}\\ {Q}_{n+1}\end{array}\right]=\left[\begin{array}{cc}4& 2\\ 1& 3\end{array}\right]\left[\begin{array}{c}{P}_{n}\\ {Q}_{n}\end{array}\right]$
As the characteristic equation is
$0={z}^{2}-7z+10=\left(z-2\right)\left(z-5\right)$
the roots are ${z}_{1}=5$ and ${z}_{2}=2$ with eigenvectors ${v}_{1}=\left(\genfrac{}{}{0}{}{2}{1}\right)$ and ${v}_{2}=\left(\genfrac{}{}{0}{}{1}{-1}\right)$.
Step 2
The solution will thus have the form ${P}_{n}=2{c}_{1}{5}^{n}+{c}_{2}{2}^{n}$, ${Q}_{n}={c}_{1}{5}^{n}-{c}_{2}{2}^{n}$, ${Q}_{0}=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{c}_{2}={c}_{1}-1$, ${P}_{0}={A}_{0}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{c}_{1}=\frac{1}{3}\left({A}_{0}+1\right)$ and thus
${A}_{n}=\frac{2\left({A}_{0}+1\right){5}^{n}+\left({A}_{0}-2\right){2}^{n}}{\left({A}_{0}+1\right){5}^{n}-\left({A}_{0}-2\right){2}^{n}}=\frac{2\left({A}_{0}+1\right)+\left({A}_{0}-2\right)\left(\frac{2}{5}{\right)}^{n}}{\left({A}_{0}+1\right)-\left({A}_{0}-2\right)\left(\frac{2}{5}{\right)}^{n}}$
Which will in all but one exceptional case (where ${c}_{1}=0$, i.e., ${A}_{0}=-1$) converge to 2.

We have step-by-step solutions for your answer! Flambergru
Step 1
Assume for the moment that the recursive sequence converges. Then the limit would satisfy the quadratic equation $L\left(L+3\right)=4L+2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}0={L}^{2}-L-2=\left(L-2\right)\left(L+1\right)$.
Take the negative fixed point $L=-1$ and consider the sequence ${B}_{n}={A}_{n}+1$ or ${A}_{n}=-1+{B}_{n}$. Then
${B}_{n+1}=1+\frac{-4+4{B}_{n}+2}{-1+{B}_{n}+3}=\frac{5{B}_{n}}{{B}_{n}+2}$ which tells that the iteration will move away from that fixed point with speed around $\frac{5}{2}$ 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
${C}_{n+1}=-2+\frac{4\left(2+{C}_{n}\right)+2}{{C}_{n}+2+3}=\frac{2{C}_{n}}{{C}_{n}+5}$
Now if ${A}_{0}<-3$ then ${C}_{0}<-5$ and ${C}_{1}>0$ so that $0<{C}_{n+1}<\frac{2}{5}{C}_{n}$ for $n>1$. Thus the sequence $\left({C}_{n}{\right)}_{n}$ converges to 0, in consequence $\left({A}_{n}{\right)}_{n}$ to 2.

We have step-by-step solutions for your answer!