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Math Help - Look at the recursive

  1. #1
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    Look at the recursive

    I´m having a hard time getting through these sum understanding exercises
    This one in particular, If you have some idéa, I value some explaining more then solution so please no: Hey stupid, what the hell, just to the "inversetriangular root extraction recursive absolute value differential" method and you got it!! That just makes me even more confused when I dont understand what I am suppose to do.
    The troublesome exercise is:
    Look at the series of:

    [a_n]_{n=1}^{\infty} of

    a_{n+1}=\frac{5a_n+3}{a_n+3} where 1\leq {n}
    with a_1>0

    and [b_n]_{n=1}^{\infty} Show that b_n is a geometric series with \frac{b_n}{b_{n+1}} = Constant when b_n =:
    b_n=\frac{a_n-3}{a_n+1} where 1\leq{n}
    Here is my current confused work:
    Well I figured lets try some different values on a_1 and se what happens, then clearly a_{n+1} goes towards 3. Well that info could be usefull (I suppose a little sometimes). graph looks like \frac{1}{x} graph shiften left 3 units and up 5 units.
    Now lets get a_n on one side and everything else on another and se what happens:
    so:
    1) a_{n+1}= \frac{5a_n+3}{a_n+3}

    2) a_{n+1}(a_n+3) = 5a_n+3

    3) a_{n+1}*a_n+3a_{n+1} = 5a_n +3

    4) 5a_n-a_{n+1}*a_n = 3a_{n+1}-3

    5) a_n(5-a_{n+1}) = 3a_{n+1}-3

    6) a_n= \frac{3a_{n+1}-3}{5-a_{n+1}}
    if I set a_{n+1} = 3 I get a_n =3
    but if I set a_n = 3 into the b_n formula it goes
    \frac{a_n-3}{a_n+1}= b_n = \frac{3-3}{3+1} = \frac{0}{4}
    And here I am not sure what I´ve done or what I shouldn´t have done...
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  2. #2
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    If you solve for a_n you get

     <br />
a_n = - \frac{b_n + 3}{b_n - 1}<br />

    now substitute this into

     <br />
a_{n+1} = \frac{5a_n+3}{a_n + 3}<br />

    and simplify and see what happens.
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  3. #3
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    Quote Originally Posted by Danny View Post
    If you solve for a_n you get

     <br />
a_n = - \frac{b_n + 3}{b_n - 1}<br />

    now substitute this into

     <br />
a_{n+1} = \frac{5a_n+3}{a_n + 3}<br />

    and simplify and see what happens.
    Yes I get: a_{n+1} =  \frac{b_n+9}{3-b_n}
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  4. #4
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    Quote Originally Posted by Henryt999 View Post
    Yes I get: a_{n+1} = \frac{b_n+9}{3-b_n}
    Yes, but also from

    <br />
a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}

    you get

    <br />
a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}

    which you use.
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  5. #5
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    Quote Originally Posted by Danny View Post
    Yes, but also from

    <br />
a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}

    you get

    <br />
a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}

    which you use.
    Im sorry How do you get that?
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  6. #6
    Member Haven's Avatar
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    He subbed in the original formula for  a_{n+1} and solved for  a_n

    Also, He just incremented the index by 1.
    Note: that if  a_n = b_n then  a_{n+1} = b_{n+1}
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  7. #7
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    Quote Originally Posted by Haven View Post
    He subbed in the original formula for  a_{n+1} and solved for  a_n

    Also, He just incremented the index by 1.
    Note: that if  a_n = b_n for all n \in \mathbb{N} then  a_{n+1} = b_{n+1}
    Small fix!
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