Question

Let ${\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{n}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{+}{\mathbf{c}}_{\mathbf{n}}\mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$be a polynomial with a root at$\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$ . Let role="math" localid="1659797796589" ${\mathbf{c}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$be the largest absolute value of a . Show that

$\mathbf{\left|}{\mathbf{x}}_{\mathbf{0}}\mathbf{\right|}\mathbf{<}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\frac{{\mathbf{c}}_{\mathbf{max}}}{{\mathbf{c}}_{\mathbf{1}}}$

Short answer

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

Step-by-step solution

Algorithm for Polynomial.

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

Step 2: Decidable languages is not closed under homomorphism. a).

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:

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|\le \left|{\mathrm{c}}_2{\mathrm{x}}_0^{\mathrm{n}-1}\right|+\mathrm{........}+{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0+{\mathrm{c}}_{\mathrm{n}+1}|$

Applying triangle inequality:

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|\le \left|{\mathrm{c}}_2{\mathrm{x}}_0^{\mathrm{n}-1}\right|+\mathrm{........}+\left|{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}_0\right|+\left|{\mathrm{c}}_{\mathrm{n}+1}\right|$

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

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

The inequality also holds if we substitute,

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

$\left|{\mathrm{c}}_1{\mathrm{x}}_0^{\mathrm{n}}\right|\le \left|{\mathrm{c}}_{\max }\right|\mathrm{n}\left|{\mathrm{x}}_{\mathrm{o}}^{\mathrm{n}-1}\right|$

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

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

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.