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Math Help - Prove a sequence is monotone and bounded, and find the limit.

  1. #1
    Junior Member mremwo's Avatar
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    Prove a sequence is monotone and bounded, and find the limit.

    One last thing. So I have this sequence:

    x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N

    I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

    I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either. Help?


    Thanks so much.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mremwo View Post
    x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

    Use induction.

    I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either.

    Take limits in both sides of x_{n+1}=x_n/2+2 .

    P.S. You need also prove that the sequence is lower bounded (choose for example 0 as a lower bound) .



    Fernando Revilla
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mremwo View Post
    One last thing. So I have this sequence:

    x_1:= 8 \ \mbox{and} \ x_{n+1} =\frac{1}{2}x_n + 2 \ \ \mbox{for all} \ n\in N

    I know that it is strictly decreasing, which makes it monotone, but how would I prove that?

    I also think that the limit is 4 just by looking at the first few terms, which means that since it converges that it is bounded, but I don't know the proper way to prove that either. Help?


    Thanks so much.
    The difference equation can be written as...

    \displaystyle \Delta_{n}= x_{n+1} - x_{n} = 2 - \frac{x_{n}}{2} = f(x_{n}) (1)

    The f(*) is linear and is represented here [black line]...



    There is only one 'attractive fixed point' at x_{0}=4 and because for all x is |f(x)|\le |x_{0}-x| , all the possible x_{1} will produce a sequence monotonically convergent in x_{0}. Of course if x_{1}< x_{0} the sequence will be increasing, if x_{1}>x_{0} decreasing...

    Kind regards

    \chi \sigma
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  4. #4
    Member SENTINEL4's Avatar
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    To show that it is monotone :
    x_1=8 and x_2=6 so x_1>x_2 . Suppose that x_n>x_{n+1} then : \frac{1}{2}x_n>\frac{1}{2}x_{n+1}\Rightarrow\frac{  1}{2}x_n+2>\frac{1}{2}x_{n+1}+2 \Rightarrow x_{n+1}>x_{n+2} so it's decreasing.
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