Recursive sequence depending on the parameter

Mathematics Asked by Invisible on December 1, 2020

For the given parameter $mathbb Rni tgeq 1$, the sequence is
defined recursively: $$a_1=t,;;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
Prove the sequence $(a_n)$ converges and find its limit.

$(b)$ Which parameters $tgeq 1$ is the sequence $(a_n)$ increasing

My attempt:

Bolzano-Weierstrass:A sequence converges if it is monotonous and

$$a_{n+1}a_n=3a_n-2implies a_{n+1}=3-frac{2}{a_n}$$

First few terms: $a_1=4,a_2=frac{5}{2},a_3=frac{11}{5}$

Assumption: the sequence is decreasing

Proof by induction:
the basis (n=1) is trivial: $frac{5}{2}<4$

Assumption: $a_n<a_{n-1},;forall ninmathbb N$

Step: $$a_n<a_{n-1}impliesfrac{1}{a_n}geqfrac{1}{a_{n-1}}Bigg/cdot(-2)$$
$$iff-frac{2}{a_n}leq-frac{2}{a_{n-1}}iff underbrace{3-frac{2}{a_n}}_{a_{n+1}}lequnderbrace{3-frac{2}{a_{n-1}}}_{a_n}$$
The limit: $$L=3-frac{2}{L}implies L^2-3L+2=0$$
I take into account only $2$ because the parabola is convex and $$a_nto L^-.$$
Then I have to prove: $a_ngeq 2;forall ninmathbb N$ after the formal computing: $a_{n+1}geq 3-frac{2}{2}=2$
$underset{implies}{text{Bolzano-Weierstrass theorem}}(a_n)to 2$

$(b)$ Since the sequence doesn’t have to be convergent, only increasing:
$$a_2=3-frac{2}{t}geq timplies tin[1,2]$$
Then, it should follow inductively,analogously to $(a)$, this time it is increasing.
Is this correct?

2 Answers

Let us find the general form for the sequence given by the recursion $$ a_{n+1}= underbrace{ begin{bmatrix}3&-2\ 1&0 end{bmatrix} }_{}cdot a_n , $$ where we use the Möbius action of matrices $2times 2$ on scalars, given in general by $$ begin{bmatrix}a&b\ c&d end{bmatrix}cdot x := frac{ax+b}{cx+d} , $$ see also Möbius transformation, wiki page.

The special matrix $A$ used in the problem can be diagonalized, $$ A= underbrace{ begin{bmatrix}1&1\ 1/2&1 end{bmatrix}}_{T} underbrace{ begin{bmatrix}2&\ &1 end{bmatrix}}_{D} underbrace{ begin{bmatrix}2&-2\-1&2 end{bmatrix}}_{T^{-1}} $$ and because $A^n=TD^nT^{-1}$ we get the general form for $a_n=A^ncdotbegin{bmatrix}4\1end{bmatrix}$, and then passing to the element in the projective space, by taking the quotient, it is: $$ a_n=frac{6cdot 2^n-2}{3cdot 2^n-2} . $$ It converges to $6/3=2$.

For the part (b) a similar study can be started, the general term being $$ a_n(t)= TD^nT^{-1} begin{bmatrix}t\ 1 end{bmatrix}_{Bbb P^1} = begin{bmatrix} 2cdot 2^n-1 & -2cdot 2^n+2\ 2^n-1 & -2cdot 2^n+2 end{bmatrix} begin{bmatrix} t\ 1 end{bmatrix} text{ considered in } {Bbb P^1} . $$ I am stopping here...

Correct answer by dan_fulea on December 1, 2020

If you have already proven $a_n > 2, forall n$ , then

Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - frac{2}{a_n} Rightarrow b_{n+1} + 2=3-frac{2}{b_n+2} Rightarrow b_{n+1} = frac{b_n}{2+b_n} < frac{b_n}{2}$.

Therefore as $nto infty, b_n to 0, a_n to 2.blacksquare$

The next is pure hindsight based on dan_fulea's solution but I believe it can be useful when the two fixed points are distinct.

$a_{n+1} - 1 = 2-frac{2}{a_n} = frac{2(a_n - 1)}{a_n}$

$a_{n+1} - 2 = 1-frac{2}{a_n} = frac{a_n - 2}{a_n}$

Therefore $frac{a_{n+1}-1}{a_{n+1}-2} = 2cdot frac{a_n-1}{a_n-2}$

$frac{a_n-1}{a_n-2}$ is a geometric sequence with initial value $frac{3}{2}$,

so $frac{a_n-1}{a_n-2} = 2^{n-1} frac{3}{2} = 1+frac{1}{a_n-2} Rightarrow a_n = 2+ frac{1}{2^{n-1}frac{3}{2}-1} = frac{6cdot 2^{n-1}-2}{3cdot 2^{n-1}-2}.blacksquare$

In general if there are two distinct fixed points $r$ and $s$ then the ratio $frac{a_n-r}{a_n-s}$ is a geometric sequence.

Answered by Neat Math on December 1, 2020

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