# Thread: convergent sequence with parameter

1. ## convergent sequence with parameter

I have the following sequence: $(x_{n})_{n\geq 0}$, $x_{0}=a$, $x_{n+1}=x^{2}_{n}-4x_{n}+6$

I need to find all $a$ values such that $(x_{n})_{n\geq 0}$ to be convergent.

I took $f(x)=x^{2}-4x+6$ and I found 2 fixed points, $2$ and $3$ , solutions of $f(x)=x$.I know that if $(x_{n})_{n\geq 0}$ is convergent then it converges to one of the points I found.I don't know how to continue.How to approach an exercise like this?

2. ## Re: convergent sequence with parameter

Originally Posted by Vali
I have the following sequence: $(x_{n})_{n\geq 0}$, $x_{0}=a$, $x_{n+1}=x^{2}_{n}-4x_{n}+6$
I need to find all $a$ values such that $(x_{n})_{n\geq 0}$ to be convergent.

3. ## Re: convergent sequence with parameter

I didn;t completely understand the method with induction.
From $x_{n+1}$ I got $]x_{n+1}-2=(x_{n}-2)^{2}$
Then I note $x_{n+1}=a_{n+1}$ and $(x_{n}-2)=a_{n}$ Why it must be noted like this ?
Then I got $a_{n+1}=a_{n}^{2}$ Where $-2$ disappeared ?
Then I got $x_{n}=2+(a-2)^{2n}$
So the limit is finit if the second limit is a number, right ? so $a$ is [1,3]
I understood howI got $x_{n}$ but I didn;t understood the notations and why $-2$ disappeared.

4. ## Re: convergent sequence with parameter

I finally understood.Thank you!