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

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

    show that the function :

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

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

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

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

    show that the function :

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

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

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

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

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

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

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

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

    show that the function :

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

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

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

    CB
    Last edited by CaptainBlack; Oct 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, $\displaystyle \alpha\mapsto\xi(\alpha)$ is a decreasing function, and that $\displaystyle \alpha\mapsto\zeta(\alpha)$ is an increasing function on $\displaystyle (0,1)$.

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