# 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
for?

My attempt:

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

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

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

Step: $$a_n
$$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?

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