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Math Help - Correction

  1. #1
    Junior Member simplysparklers's Avatar
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    Correction

    Hi guys
    Can anyone put me right on this? I keep coming up with the opposite answer.

    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}
    (n\geq{1})
    Show that a_{n} is monotonic increasing.

    Anyway I do it, I keep finding a_{n} is monotonic decreasing. I've tried doing it as a_{n+1}-a_{n}\geq{0}, and I've gone very simple and just substituted different random values in for a_{n} to see what kind of resuts it would yield, but it always ends up as a_{n+1}\leq{a_{n}}. Can anyone show me where I'm going wrong please??

    Thanks,
    Jo
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Hi guys
    Can anyone put me right on this? I keep coming up with the opposite answer.

    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}
    (n\geq{1})
    Show that a_{n} is monotonic increasing.

    Anyway I do it, I keep finding a_{n} is monotonic decreasing. I've tried doing it as a_{n+1}-a_{n}\geq{0}, and I've gone very simple and just substituted different random values in for a_{n} to see what kind of resuts it would yield, but it always ends up as a_{n+1}\leq{a_{n}}. Can anyone show me where I'm going wrong please??

    Thanks,
    Jo
    a_{n+1} - a_{n} = \frac{3a_n + 1}{a_n + 3} - a_n = \frac{1 - a_n ^2}{3 + a_n}

    In the previous thread we proved that 0 \leq a_n < 1.This implies 1 - a_n ^2 > 0, 3+a_n > 0. These two together imply a_{n+1} - a_{n} > 0
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  3. #3
    Super Member
    earboth's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Hi guys
    Can anyone put me right on this? I keep coming up with the opposite answer.

    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}
    (n\geq{1})
    Show that a_{n} is monotonic increasing.

    ...
    1.
    <br />
a_{n+1}-a_n=\frac{3 \cdot \frac{3a_n+1}{a_n+3}+1}{\frac{3a_n+1}{a_n+3}+3}- \frac{3a_n+1}{a_n+3} = \frac{6a_n+13-3(a_n)^2}{(a_n+3)(3a_n+5)}

    Since a_n<1 for all n this fraction is positiv thus the sequence is monotonic increasing

    2. I've attached a screenshot of a spreadsheet calculating a_n

    3. What are you doing in Bamberg?
    Attached Thumbnails Attached Thumbnails Correction-folgentabelle.gif  
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  4. #4
    Junior Member simplysparklers's Avatar
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    Thanks Earboth!!
    & I'm on my Erasmus year in Bamberg!I study German and Maths, and I have maths work sent over from my home uni to be done!Awful book to work from, and no lecturer to help me, that's why I'm so grateful I found you MHF guys!!
    Whereabouts are you?
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