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Math Help - A zero problem.

  1. #1
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    A zero problem.

    Dear all,

    I am stuck for two weeks on the next problem. All the numerical test I did show that it is true, but I do not find any proof.

    let (a_i)_{i=1,..,n} and (b_i)_{i=1,..,n-1}, reals numbers in (0,1),
    such that :

    \sum_{i=1}^n a_i +\sum_{i=1}^{n-1} b_i =1.

    show that the function :

    \phi(\alpha)=\sum_{i=1}^{n-1} b_i^\alpha + 1 -\sum_{i=1}^n  a_i^\alpha

    as at most one root in (0,1).

    Any idea is welcome ! Thank you !
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  2. #2
    Grand Panjandrum
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    Re: A zero problem.

    Quote Originally Posted by mushroom98 View Post
    Dear all,

    I am stuck for two weeks on the next problem. All the numerical test I did show that it is true, but I do not find any proof.

    let (a_i)_{i=1,..,n} and (b_i)_{i=1,..,n-1}, reals numbers in (0,1),
    such that :

    \sum_{i=1}^n a_i +\sum_{i=1}^{n-1} b_i =1.

    show that the function :

    \phi(\alpha)=\sum_{i=1}^{n-1} b_i^\alpha + 1 -\sum_{i=1}^n  a_i^\alpha

    as at most one root in (0,1).

    Any idea is welcome ! Thank you !
    Maybe I missing something, but:

    \phi(\alpha)=\sum_{i=1}^{n-1} b_i^\alpha + 1 -\sum_{i=1}^n  a_i^\alpha=2 \sum_{i=1}^{n-1} b_i^\alpha

    And since the b_i 's are all positive and presumably these real powers of positive numbers are to be taken positive we have \phi(\alpha)>0.

    That is there is no root of \phi(\alpha) with b_i \in (0,1),\ i=1,...,n-1 and \alpha \in (0,1)

    CB
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  3. #3
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    Re: A zero problem.

    Unfortunatelly, the formula you have written is only true for \alpha=1.
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  4. #4
    Grand Panjandrum
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    Re: A zero problem.

    Quote Originally Posted by mushroom98 View Post
    Unfortunatelly, the formula you have written is only true for \alpha=1.
    Sorry, So I see

    CB
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  5. #5
    Grand Panjandrum
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    Re: A zero problem.

    Quote Originally Posted by mushroom98 View Post
    Dear all,

    I am stuck for two weeks on the next problem. All the numerical test I did show that it is true, but I do not find any proof.

    let (a_i)_{i=1,..,n} and (b_i)_{i=1,..,n-1}, reals numbers in (0,1),
    such that :

    \sum_{i=1}^n a_i +\sum_{i=1}^{n-1} b_i =1.

    show that the function :

    \phi(\alpha)=\sum_{i=1}^{n-1} b_i^\alpha + 1 -\sum_{i=1}^n  a_i^\alpha

    as at most one root in (0,1).

    Any idea is welcome ! Thank you !
    Under the assumptions \xi(\alpha)=\sum_{i=1}^{n-1} b_i^\alpha + 1 is a decreasing function and \zeta(\alpha)=-\sum_{i=1}^n  a_i^\alpha is an increasing function on (0,1), which tells us nothing ... .

    CB
    Last edited by CaptainBlack; October 9th 2011 at 06:54 PM.
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  6. #6
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    Re: A zero problem.

    I do agree that with your definitions, \alpha\mapsto\xi(\alpha) is a decreasing function, and that \alpha\mapsto\zeta(\alpha) is an increasing function on (0,1).

    But I don't consider the roots of \xi - \zeta (which don't exist, for \xi>0 and \zeta<0) but the roots of \xi + \zeta.
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