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 $(a_i)_{i=1,..,n}$ 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 !

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 $(a_i)_{i=1,..,n}$ 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

3. ## Re: A zero problem.

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

4. ## Re: A zero problem.

Originally Posted by mushroom98
Unfortunatelly, the formula you have written is only true for $\alpha=1$.
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 $(a_i)_{i=1,..,n}$ 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

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