91影视

A polynomial with a root at$\mathrm{x}={\mathrm{x}}_0$ and a sequence ${\mathrm{c}}_1{\mathrm{x}}^{\mathrm{n}}+{\mathrm{c}}_2{\mathrm{x}}^{\mathrm{n}-1}+路路路+{\mathrm{c}}_{\mathrm{n}}\mathrm{x}+{\mathrm{c}}_{\mathrm{n}+1}$.and here${\mathrm{c}}_{\max }$ be the largest absolute value of a ${\mathrm{c}}_{\mathrm{i}}$is always gives $\left|{\mathrm{x}}_0\right|<(\mathrm{n}+1)\frac{{\mathrm{c}}_{\max }}{{\mathrm{c}}_1}$.

Making the polynomial equal zero (in this case, $\mathrm{x}={\mathrm{x}}_0):$

${\mathrm{c}}_1{{\mathrm{x}}_0}^{\mathrm{n}}+{\mathrm{c}}_2{\mathrm{x}}^{\mathrm{n}}-10+\mathrm{...............}+{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0+{\mathrm{c}}_{\mathrm{n}+1}=0$

Rearranging the terms:

${\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}=-({\mathrm{c}}_2{\mathrm{x}}_0^{\mathrm{n}-1}+\mathrm{............}+{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0+{\mathrm{c}}_{\mathrm{n}+1})$

Taking the absolute value of both sides:

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|鈮�\text{鈥娾赌娾赌�}\left|{\mathrm{c}}_2{\mathrm{x}}_0^{\mathrm{n}-1}\right|+\mathrm{........}+{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0+{\mathrm{c}}_{\mathrm{n}+1}|$

Applying triangle inequality:

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|鈮�\text{鈥娾赌娾赌�}\left|{\mathrm{c}}_2{\mathrm{x}}_0^{\mathrm{n}-1}\right|+\mathrm{........}+\left|{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0\right|+\left|{\mathrm{c}}_{\mathrm{n}+1}\right|$

The inequality above still holds if we substitute ${\mathrm{c}}_{\max }$for all coefficients:

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|鈮�\left|{\mathrm{c}}_{\max }\right|(1+\left|{\mathrm{x}}_0\right|+\mathrm{..........}+|{\mathrm{x}}_0^{\mathrm{n}-1}\left|\right)$

The inequality also holds if we substitute,

${\mathrm{nx}}_0^{\mathrm{n}鈭�1}\text{鈥娾赌�}\mathrm{for}\text{鈥娾赌�}(1+\left|{\mathrm{x}}_0\right|+\mathrm{..........}+|{\mathrm{x}}_0^{\mathrm{n}-1}\left|\right)$

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|鈮�\left|{\mathrm{c}}_{\max }\right|\mathrm{n}\left|{\mathrm{x}}_{\mathrm{o}}^{\mathrm{n}鈭�1}\right|$

The above result is very close to the desired result, except that it should be,

$\left|{\mathrm{x}}_0\right|鈮�\mathrm{n}\frac{{\mathrm{c}}_{\max }}{{\mathrm{c}}_1}$

From the above result, it is true that:

$\left|{\mathrm{x}}_0\right|<(\mathrm{n}+1)\frac{{\mathrm{c}}_{\max }}{{\mathrm{c}}_1}$, hence proved.