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Thread: 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:

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

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

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

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

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

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

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

    6)$\displaystyle a_n= \frac{3a_{n+1}-3}{5-a_{n+1}}$
    if I set $\displaystyle a_{n+1} = 3$ I get $\displaystyle a_n =3$
    but if I set $\displaystyle a_n = 3$ into the $\displaystyle b_n$ formula it goes
    $\displaystyle \frac{a_n-3}{a_n+1}= b_n$ = $\displaystyle \frac{3-3}{3+1} = \frac{0}{4}$
    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 $\displaystyle a_n$ you get

    $\displaystyle
    a_n = - \frac{b_n + 3}{b_n - 1}
    $

    now substitute this into

    $\displaystyle
    a_{n+1} = \frac{5a_n+3}{a_n + 3}
    $

    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 $\displaystyle a_n$ you get

    $\displaystyle
    a_n = - \frac{b_n + 3}{b_n - 1}
    $

    now substitute this into

    $\displaystyle
    a_{n+1} = \frac{5a_n+3}{a_n + 3}
    $

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

    $\displaystyle
    a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}$

    you get

    $\displaystyle
    a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}$

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

    $\displaystyle
    a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}$

    you get

    $\displaystyle
    a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}$

    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 $\displaystyle a_{n+1} $ and solved for $\displaystyle a_n $

    Also, He just incremented the index by 1.
    Note: that if $\displaystyle a_n = b_n $ then $\displaystyle 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 $\displaystyle a_{n+1} $ and solved for $\displaystyle a_n $

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