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Mathematical background

Let $f=a_n x^n+\ldots+a_0$ be a polynomial and let $f_p=
\vert a_n\vert x^n+\ldots+\vert a_{p+1}x^{p+1}\vert-\vert a_p\vert x^p-\ldots-\vert a_0\vert$ be the associated polynomial. If $f_p$ has two positive roots $\beta>\alpha$, then $f$ has exactly $p$ roots in the disk $\vert x\vert<\alpha$ and no roots in the domain $\alpha <\vert x\vert<\beta$.

Now assume that we have calculated the coefficients of the polynomial $f(x+a+R)=Q(X)$ where $a, R$ are known quantities. If we determine a $Q_p$ polynomial that has 2 positive roots $a_1,a_2$ in the range $[0,R]$, then $Q(x)$ has roots in the disk $\vert X\vert<a_1<R$. Hence the absolute value of the real part of the roots is bounded by $a_1$. As $x=X-a-R$ we get that the real part $R_p$ of the root satisfies $-a_1+a+R\le
R_p<a_1+a+R$. As $R>a_1$ this shows that $f$ has roots whose real part is greater than $a$.

Jean-Pierre Merlet 2012-12-20