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Mathematical background

Let $P(x)$ be an univariate polynomial of degree $n$:

\begin{displaymath}
P(x)= a_0 x^n + a_{1} x^{n-1}+.....a_n=0
\end{displaymath}

Assume $a_0>0$ and let $a_k$ ($k\ge 1$) be the first negative coefficients of $P(x)$ (if $P(x)$ has no negative coefficients then there is no positive real root).

The upper bound $M$ of the value of the positive real root is:

\begin{displaymath}
M=1+ \sqrt[k]{\frac{B}{a_0}}
\end{displaymath}

where $B$ is the greatest absolute value of the negative coefficients of $P(x)$ ,[3],[13].

If we define:

\begin{displaymath}
P(y)=a_n y^n+ a_{n-1}y^{n-1}+.....+a_0
\end{displaymath}

Then the upper bound of the positive real roots of $P(y)$ is the lower bound $m$ of the positive real root of $P(x)$. Consequently if $k$ and $B$ are computed for the polynomial $P(y)$ then

\begin{displaymath}
m=\frac{1}{1+\sqrt[n-k]{\frac{B}{a_n}}}
\end{displaymath}



Jean-Pierre Merlet 2012-12-20