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Thread: prooving monotonicity mmn 11

  1. #1
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    prooving monotonicity mmn 11

    a1=1
    a_n+1=a_n*1/2


    how to prove that
    a_n+1 -a_n<0
    ?
    i tried by induction
    a1>0
    suppose a_n>0

    a_n+1 -a_n>0
    -a_n*1/2<0
    so its monotonicly desendant correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Although it is very easy, you also need to prove that $\displaystyle a_n>0$ for all $\displaystyle n$ .
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    a1=1
    a_n+1=a_n*1/2


    how to prove that
    a_n+1 -a_n<0
    ?
    i tried by induction
    a1>0
    suppose a_n>0

    a_n+1 -a_n>0
    -a_n*1/2<0
    so its monotonicly desendant correct?
    It is a little 'surpising' the fact that You have a total of 973 posts and none of them is written in Latex!... the question is not minor because what You have written can cause misundestanding!...

    If the 'recursive relation' is...

    $\displaystyle \displaystyle a_{n+1}= \sqrt{a_{n}}$ , $\displaystyle a_{1}=1$ (1)

    ... then is $\displaystyle \forall n\ \ a_{n}=1$ and the sequence is neither increasing nor decreasing...

    If the 'recursive relation' is...

    $\displaystyle \displaystyle a_{n}+1= \sqrt{a_{n}}$ , $\displaystyle a_{1}=1$ (2)

    ... then is...

    $\displaystyle \displaystyle \Delta_{n} = a_{n-1}-a_{n} = \sqrt{a_{n}} - a_{n}- 1$ (3)

    ... and $\displaystyle \forall a_{n}>0$ the (3) is negative so that the sequence is decreasing and it finishes at the n for which is $\displaystyle a_{n}<0$...

    There are some other 'possible interpretations' but in any case the 'advatages' of the use of Latex are fully evident!...


    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Mar 4th 2011 at 02:08 AM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by chisigma View Post
    It is a little 'surpising' the fact that You have a total of 973 posts and none of them is written in Latex!... the question is not minor because what You have written can cause misundestanding!...

    I agree.
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  5. #5
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    I would have interpreted that as $\displaystyle a_{n+1}= \frac{a_n}{2}$!

    To prove, by induction that this is a strictly decreasing sequence, that is, that $\displaystyle a_{n+1}< a_n$ for all n, note that $\displaystyle a_1= 1$ and that $\displaystyle a_2= \frac{1}{2}$ so the statement is true for n=1.

    Assume that $\displaystyle a_{k+1}< a_k$ for some k. Then $\displaystyle a_{k+2}= \frac{a_{k+1}}{2}$ but since $\displaystyle a_{k+1}< a_k$, $\displaystyle \frac{a_{k+1}}{2}< \frac{a_k}{2}= a_{k+1}$. That is, $\displaystyle a_{k+2}< a_{k+1}$ and we are done.

    If it is, in fact, $\displaystyle a_{n+1}= \sqrt{a_n}$, then the sequence is neither increasing nor decreasing- $\displaystyle a_n= 1$ for all n!
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