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Math Help - 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 a_n>0 for all 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 a_{n+1}= \sqrt{a_{n}} , a_{1}=1 (1)

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

    If the 'recursive relation' is...

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

    ... then is...

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

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

    \chi \sigma
    Last edited by chisigma; March 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 a_{n+1}= \frac{a_n}{2}!

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

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

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