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Math Help - Sequence

  1. #1
    Junior Member
    Joined
    Mar 2010
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    40

    Sequence

    Hi,

    (\forall n \in \mathbb{N}) U_{n+1}=\sqrt{12+U_n} and  \ U_0=0

    I showed that (\forall n \in \mathbb{N}) U_n<4

    But i don't know how to show that 4-U_{n+1}\le \frac{1}{4}(4-U_n)

    Can you help me please????
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  2. #2
    Senior Member
    Joined
    Mar 2010
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    4-U(n+1)=4-sqrt(12+U(n)),
    using a=4-U(n)
    4-sqrt(16-a)=
    =4-4*sqrt(1-a/16) (1)
    >=4-4(1-(1/2)a/16)
    4-U(n+1)>=(1/8)(4-U(n)) - aditional rule.
    Using Taulor set of sqrt:
    sqrt(1-x)>=1-x and multiplying by -1, we get
    -sqrt(1-x)<=-(1-x)
    from (1) we get
    4-U(n+1)<=(1/4)(4-U(n)).
    Last edited by zzzoak; March 28th 2010 at 04:37 PM.
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