Exercise 1.3.15

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Use a calculator and Theorem 1.3.3 to compute the roots of the cubic equation $y^3-7y+3 = 0$ to eight decimal places of accuracy.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We use the Vi\`ete's method to
$$y^3 - 7y +3 : \qquad p=-7,q=3.$$

By theorem 1.3.3, the roots are given by
$$y_k =2\sqrt{\frac{-p}{q}} \cos \left(	heta + \frac{2k\pi}{3}ight)= 2\sqrt{\frac{7}{3}} \cos\left (	heta + \frac{2k\pi}{3}ight), k=0,1,2,$$
where
$$	heta = \frac{1}{3}\arccos\left(  \frac{3\sqrt{3}q}{2p\sqrt{-p}}ight)= \frac{1}{3}\arccos\left(  -\frac{9\sqrt{3}}{14\sqrt{7}}ight).$$

$x =  -\frac{9\sqrt{3}}{14\sqrt{7}} \simeq -0.420848788312271,$

$	heta = \frac{1}{3}\arccos(x) \simeq 0.668392376195833,$
\begin{align*}
y_0 &\simeq 2.39766154089,\
y_1 &\simeq-2.83846925239,\
y_2&\simeq 0.44080771150.
\end{align*}
\end{proof}

Exercise 1.3.15

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Exercise 1.3.15

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