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Math Help - convergence of recursive series mmn11 2d

  1. #1
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    convergence of recursive series mmn11 2d

    \sum_{k=1}^{infinity}a_{k} where a_{k+1}=cos(a_{k}) for k>=1
    is it converges absolutly ,or with condition(dont know the english term),
    or is it diverges?
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  2. #2
    Ted
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    Are you given the value of \displaystyle a_0 ?
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  3. #3
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    no
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    \sum_{k=1}^{infinity}a_{k} where a_{k+1}=cos(a_{k}) for k>=1
    is it converges absolutly ,or with condition(dont know the english term),
    or is it diverges?
    The recursive relation can be written as...

    \displaystyle \Delta_{k}= a_{k+1} - a_{k} = \cos a_{k} - a_{k} = f(a_{k}) (1)

    The function f(x) is represented here...



    There is only one 'attractive fixed point' at x_{0}= .739085133215... and because \forall x is |f(x)|<|2\ (x_{0}-x)| [red line...] any possible sequence converges at x_{0}. Because is |f(x)|>|x_{0}-x| [blue line...] however the convergence will be not 'monotonic' but 'oscillating'. Of course is \displaystyle \lim_{k \rightarrow \infty} a_{k} \ne 0 the series \displaystyle \sum_{k=1}^{\infty} a_{k} diverges...

    Kind regards

    \chi \sigma
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  5. #5
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    its a recursive series by what test should i test the absolute value?
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  6. #6
    MHF Contributor chisigma's Avatar
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    The general term donsn't tends to 0 if k tends to infinity, so that the series diverges... it is interesting to valuate the series \sum_{k=1}^{\infty} a_{k} where the a_{k} satisfy the recursive relation...

    \displaystyle a_{k+1}= \sin a_{k} (1)

    The (1) can be written as...

    \displaystyle \Delta_{k}= a_{k+1} - a_{k} = \sin a_{k}- a_{k}= f(a_{k}) (2)

    The function f(x) is represented here...



    There is only one 'attractive fixed point' at x_{0}=0 and it is clear that is...

    \displaystyle \lim_{k \rightarrow \infty} a_{k}=0 (3)

    ... no matter which is a_{0} and the convergence is 'monotonic', so that the series has only positive or negative terms. The (3) tells us that the series may converge, but the question is: does it converge or not?... the question isn't trivial... for example the ratio test fails because is...

    \displaystyle \lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} = \lim_{k \rightarrow \infty} \frac{\sin a_{k}}{a_{k}} = 1 (4)

    Does someone have some 'good idea'?...

    Kind regards

    \chi \sigma
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  7. #7
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    i have to foolow the schematics
    first make test on absolute value then go to the original
    here it doesnt work
    ??
    Last edited by transgalactic; March 4th 2011 at 03:18 AM.
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  8. #8
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    ohhh sorry didnt notices that a1=1

    does it help?
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  9. #9
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    It is still true that a_k does NOT go to 0 so the series does NOT converge, conditionally or absolutely.
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  10. #10
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    but we dont have expression for a_n
    its recursive
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  11. #11
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    So what is your point? We don't need an expression for a_n. cosine of any number is between -1 and 1 and cosine of any number between -1 and 1 is larger than .5. The terms do NOT go to 0 so the series cannot converge.
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  12. #12
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    but its a recursive expresion we dont have an expression of a_n with "n" only variable
    we cant say to what a_n goes

    because it is recursive
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  13. #13
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    You can keep saying that as much as you like, but it is not relevant to the question. You don't need to know an explicite formula for a_n to see that this series does not converge.
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  14. #14
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    but i need to prove it mathmatickly
    what kind of limit should i write?
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  15. #15
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    Perhaps I am misunderstanding what you are trying to prove. I thought you were trying to determine whether or not the series \sum a_n converged and prove it.

    One of the very first theorems about infinite series is that if the series \sum a_n converges, then the sequence \{a_n\} must go to 0. The contrapositive of that is that if the sequence does not go to 0, then the series does not converge.

    Now, what exactly are you trying to prove?
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