1. ## 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 !

2. ## Re: A zero problem.

Originally Posted by mushroom98
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

3. ## Re: A zero problem.

Unfortunatelly, the formula you have written is only true for $\displaystyle \alpha=1$.

4. ## Re: A zero problem.

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

CB

5. ## Re: A zero problem.

Originally Posted by mushroom98
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

6. ## 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 $\displaystyle \xi + \zeta$.