Exercise 1.3.11

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}}

Consider the equation $4t^3 - 3t - \cos(3	heta) = 0$, where $\cos(3	heta) 
eq \pm 1$. In (1.25), we expresses the roots in terms of trigonometric functions. In this exercise, you will study what happens when we use Cardan's formulas.
\begin{enumerate}
\item[(a)] Show that Cardan's formulas give the root
$$t_1 = \frac{1}{2} \sqrt[3]{\cos(3	heta) + i \sin(3	heta)} + \frac{1}{2} \sqrt[3]{\cos(3	heta) - i \sin(3	heta)}.$$
\item[(b)] Explain why $\frac{1}{2} e^{i	heta} = \frac{1}{2} (\cos 	heta + i \sin 	heta)$ is a value of $\frac{1}{2} \sqrt[3]{\cos(3	heta) + i \sin(3	heta)}$, and use to show that $t_1$ is just $\cos 	heta$.
\item[(c)] Similarly, show that Cardan's formulas also give the roots $t_2$ and $t_3$ as predicted by (1.25).
\end{enumerate}

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}
\begin{enumerate}

\item[(a)] We apply the Cardan's formulas to $4 t^3 - 3 t -\cos(3	heta) = 0$, which is equivalent to
$$t^3 - \frac{3}{4} t -\frac{1}{4}\cos(3	heta) = 0.$$
$\frac{p}{3} = -\frac{1}{4}, \frac{q}{2} = -\frac{1}{8} \cos(3	heta)$, thus

$\left( \frac{q}{2}ight)^2 + \left ( \frac{p}{3} ight)^3 = \frac{1}{64} (\cos^2(3	heta) - 1) =\left( \frac{i \sin(3	heta)}{8} ight)^2 $

We choose $\sqrt{\left( \frac{q}{2}ight)^2 + \left ( \frac{p}{3} ight)^3} =  \frac{i \sin(3	heta)}{8}$,

then $-\frac{q}{2}+\sqrt{\left( \frac{q}{2}ight)^2 + \left ( \frac{p}{3} ight)^3} = \frac{1}{8}( \cos(3	heta) + i \sin(3 	heta)) = \left (\frac{1}{2} e^{i 	heta} ight )^3$.

\item[(b,c)] 
We choose the cube root $z_1 = \frac{1}{2} e^{i 	heta}$. 
Then $z_2 = \overline{z_1}$.

As $\omega =  e^{i 2 \pi/3} = e^{-i4\pi/3}$, the three real roots are given by
\begin{align*}
t_1 &= z_1 + \overline{z_1} = \frac{1}{2} (e^{i 	heta} + e^{-i	heta}) = \cos (	heta),\
t_2&= e^{i 2 \pi/3} z_1 + e^{-i2\pi/3} \overline{z_1}=\frac{1}{2} (e^{i (	heta + 2\pi/3)} + e^{-i(	heta+ 2\pi/3)}) = \cos\left(	heta + \frac{2\pi}{3}ight),\
t_3&= e^{i 4 \pi/3} z_1 + e^{-i4\pi/3} \overline{z_1}=\frac{1}{2} (e^{i (	heta + 4\pi/3)} + e^{-i(	heta+ 4\pi/3)}) = \cos\left(	heta + \frac{4\pi}{3}ight).\
\end{align*}
\end{enumerate}
\end{proof}

Exercise 1.3.11

Answers

Comments

Exercise 1.3.11

Answers

Comments

Add answer