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Let $n \in \mathbb{Z}_{\geq 1}$ be some strictly positive integer and let $\alpha \in \left(0,\frac{1}{2}n\right)$ be some real number between $0$ and $\frac{1}{2}n$. Define the polynomial $$f(x) = \sum_{i = 0}^{n} \left(i-\alpha\right)\cdot x^i.$$ Is there some elegant way to show that this function has a unique positive real root? It is not that hard to see that $f(0) = - \alpha < 0$ and $f(1) = \frac{1}{2}(n+1)(n-2\alpha)>0$, so we find that there must be a root in the interval $(0,1)$. With some work we can also show that $f(x) > 0$ for all $x\geq 1$. I am having some trouble with proving that there cannot be more than one root in the interval $(0,1)$. I do believe it to be true. Can someone find an elegant argument?

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Remark: In this proof, $n\geq 2$ is assumed. The case $n=1$ is a trivial exercise (and for which we only need the condition $\alpha\in (0,1)=(0,n)$ for the unique root of this linear polynomial to be positive).

Let $p(x):=x^{n}+x^{n-1}+\ldots+1$ for all $x\in\mathbb{R}$. Take $g(x):=x^{-\alpha}\,p(x)$ for every $x>0$. Then, we have $$x^{\alpha+1}\,g'(x)=x\,p'(x)-\alpha\,p(x)=\sum_{i=0}^n\,(i-\alpha)\,x^i=f(x)$$ for all $x> 0$. We claim that $g''(x)>0$ for every $x>0$. (This will show that $g$ has a unique local optimum in the positive reals, and this local optimum is the global minimum of $g$.)

To prove this claim, observe that $$g''(x)=\sum_{i=0}^n\,(i-\alpha)(i-\alpha-1)\,x^{i-\alpha-2}\,.$$ Thus, $g''(x)$ has at most one term with negative coefficient, namely, the term $x^{j-\alpha-2}$, where $j:=\lceil\alpha\rceil$. Now, we have $$g''(x)\geq \small (\alpha+1-j)(\alpha+2-j)\,x^{j-\alpha-3}+(j+1-\alpha)(j-\alpha)\,x^{j-\alpha-1}-(j-\alpha)(\alpha+1-j)\,x^{j-\alpha-2}$$ for all $x>0$. If $\alpha$ is an integer, then we have $j=\alpha$ and $$g''(x)\geq 2\,x^{-3}>0\text{ for every }x>0\,.$$ Now, we assume that $\alpha \notin \mathbb{Z}$, so that $\alpha<j<\alpha+1$. Using the AM-GM Inequality, we get $$\begin{align} \small (\alpha+1-j)(\alpha+2-j)\,x^{j-\alpha-3}&+(j+1-\alpha)(j-\alpha)\,x^{j-\alpha-1} \\&\geq\small2\sqrt{(\alpha+1-j)(\alpha+2-j)}\,\sqrt{(j+1-\alpha)(j-\alpha)}\,x^{j-\alpha-2} \\&>(\alpha+1-j)(j-\alpha)\,x^{j-\alpha-2}\text{ for each }x>0\,. \end{align}$$ That is, $g''(x)>0$ for each $x>0$.

From the result above, we conclude that $g'(x)$ can have at most one zero. Due to the OP's observation that $x^{\alpha+1}\,g'(x)=f(x)$ has at least one zero in$(0,1)$, we conclude that $g'(x)$ must have exactly one zero in the positive reals, which must lie between $0$ and $1$. The proof is now complete.

P.S. I think this proof works for any $\alpha \in (0,n-1]$. The only change is that, for $\alpha\in\left[\frac{n}{2},n-1\right]$, the unique zero of $g'(x)$ is greater than or equal to $1$. In fact, using Descartes's Rule of Signs, as suggested by Bumblebee in his/her deleted answer, we can also show that, when $n-1<\alpha<n$, $f(x)$ still has a unique root in $\mathbb{R}_{>0}$. Unfortunately, my proof cannot be extended to cover the case where $\alpha\in(n-1,n)$.

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  • $\begingroup$ Great answer. Why do you not consider the case when $x<0$, though. $\endgroup$
    – Allawonder
    Commented Jun 26, 2018 at 17:45
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    $\begingroup$ @Allawonder That is because $x^\alpha$ may not be in $\mathbb{R}$ for $x<0$. $\endgroup$ Commented Jun 26, 2018 at 17:46
  • $\begingroup$ @Allawonder If $n$ is even, then the number of real roots of $P(x)$ is even, and I think it is easy to argue that $P(x)$ cannot have a double root.... So very likely there are negative roots in that case... $\endgroup$
    – N. S.
    Commented Jun 26, 2018 at 18:01

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