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Thread: convergent sequence with parameter

  1. #1
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    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?
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  2. #2
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    Re: convergent sequence with parameter

    Quote Originally Posted by Vali View Post
    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.
    Does this help you?
    Thanks from topsquark
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  3. #3
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    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.
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  4. #4
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    Re: convergent sequence with parameter

    I finally understood.Thank you!
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