Exercise 1.3.10

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\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}} The goal of this exercise is to prove Theorem 1.3.3. Let $y^3+py+q=0$ be a cubic equation with positive discriminant. Consider the substitution $y = \lambda t$, which transforms the given equation into $\lambda^3 t^3 +\lambda pt+q = 0$. \begin{enumerate} \item[(a)] Show that Exercises 2 and 3 imply that $p<0$. \item[(b)] The equation $\lambda^3 t^3 + \lambda pt + q = 0$ can be written as $$4t^3 - \left(\frac{-4p}{\lambda^2} ight) t - \left(\frac{-4q}{\lambda^3} ight) = 0.$$ Show that this coincides with $4t^3 - 3t -\cos(3 heta) = 0$ if and only if $$\lambda = 2 \sqrt{\frac{-p}{3}}\quad \mathrm{and}\quad \cos(3 heta) = \frac{3\sqrt{3}q}{2p\sqrt{-p}}.$$ Note that $\sqrt{-p}$ is real and nonzero by part (a). \item[(c)] Use $\Delta = -(4p^3+27q^3) >0$ to prove that $$\left | \frac{3\sqrt{3}q}{2p\sqrt{-p}} ight | < 1.$$ \item[(d)] Explain how part (c) implies that the second equation of part (b) can be solved for $ heta$. Also show that $ heta>0$ implies that $\cos(3 heta) eq \pm 1$. \item[(e)] By (1.25), $t_1 = \cos heta, t_2 = \cos\left( heta + \frac{2\pi}{3} ight)$, and $t_3 = \cos\left( heta + \frac{4\pi}{3} ight)$ are the three roots of $\lambda^3 t^3 + \lambda pt + q = 0$. Then show that the theorem follows by transforming this back to $y=\lambda t$ via part (b). \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} The discriminant of $f(y) = y^3+py+q$ is positive by hypothesis : $$\Delta = -4p^3-27q^2>0.$$ Consequently, $f$ has three distinct real roots. \begin{enumerate} \item[(a)] If $p\geq 0$, then $\Delta =-4p^3-27q^2\leq 0$, which is false by hypothesis, so $$p<0.$$ \item[(b)] $g(t) = f(\lambda t) = \lambda^3t^3+\lambda pt + q, \quad \lambda eq 0$. $g(t) = 0 \iff t^3 + \frac{p}{\lambda^2} t + \frac{q}{\lambda^3}=0 \iff 4t^3 -\left(-\frac{4p}{\lambda^2} ight) t - \left(-\frac{4q}{\lambda^3} ight)=0$. Let $$h(t) = \frac{4}{\lambda^3} f(\lambda t) = 4t^3 -\left(-\frac{4p}{\lambda^2} ight) t - \left(-\frac{4q}{\lambda^3} ight)$$ Then $h(t)$ has the same roots as $g(t) = f(\lambda t)$. Moreover $$-\frac{4p}{\lambda^2} = 3 \iff \lambda^2 = -\frac{4}{3} p.$$ If we take $\lambda = 2\sqrt{\frac{-p}{3}}$, then $-\frac{4q}{\lambda^3} = \frac{-4q}{-\frac{8p}{3}\sqrt{-\frac{p}{3}}} = \frac{3\sqrt{3}q}{2p\sqrt{-p}}$, and $$h(t) = 4t^3 -3t-\frac{3\sqrt{3}q}{2p\sqrt{-p}}.$$ So $h(t) = 4t^3 -3t - \cos(3 heta)$ if and only if $$\lambda = 2 \sqrt{\frac{-p}{3}}\quad \mathrm{and}\quad \cos(3 heta) = \frac{3\sqrt{3}q}{2p\sqrt{-p}}.$$ \item[(c)] Since $p<0$, \begin{align*} \left \vert \frac{3\sqrt{3}q}{2p\sqrt{-p}} ight \vert <1 &\iff -\frac{27}{4} \frac{q^2}{p^3} < 1\ &\iff -27q^2 > 4p^3 \ &\iff \Delta = -4p^3-27q^2>0. \end{align*} \item[(d)] As $\Delta > 0$ by hypothesis, $ \left \vert \frac{3\sqrt{3}q}{2p\sqrt{-p}} ight \vert <1$ : so there exists $\beta \in ]0,\pi[$ such that $ \frac{3\sqrt{3}q}{2p\sqrt{-p}} = \cos \beta$. Let $ heta = \beta/3$, then $ heta \in ]0,\pi/3[$ and $$h(t) = 4t^3 - 3t -\cos(3 heta), $$ where $$ heta = \frac{1}{3}\arccos\left(\frac{3\sqrt{3}q}{2p\sqrt{-p}} ight).$$ Since $3 heta \in ]0,\pi[, \cos(3 heta) eq\pm1$. \item[(e)] We will solve the equation $h(t) = 4t^3 - 3t -\cos(3 heta)=0$. Let $u(t) = 4t^3-3t$, where $t \in \R$. $$u'(t) <0 \iff 12t^2-3 = 3(4t^2-1)<0 \iff -\frac{1}{2} \leq t < \frac{1}{2} : $$ $u$ is decreasing on $[-1/2,1/2]$, increasing on $]-\infty,-\frac{1}{2}]$ and on $[\frac{1}{2},+\infty[$. $u(-1/2) = 1,u(0) = 0, u(1/2)=-1,u(1)=1,u(-1) = -1$. Thus $|t|>1 \Rightarrow |u(t)|>1$ and so $$\forall t \in \R,\quad u(t) \in [-1,1] \Rightarrow t \in [-1,1].$$ Let $t$ a root of $h(t)$, then $u(t) = 4t^3 - 3 t = \cos(3 heta) \in [-1,1]$. By the properties of the function $u$ on $[-1,1]$, $t \in [-1,1]$, so there exists $\alpha \in \mathbb{R}$ such that $\cos(\alpha) = t$. \begin{align*} h(t) = 0 &\iff 4 t^3 - 3t = \cos(3 heta)\ &\iff 4 \cos^3(\alpha) - 3 \cos(\alpha) = \cos(3 heta)\ &\iff \cos(3\alpha) = \cos(3 heta)\ &\iff 3 \alpha = 3 heta + 2k \pi \ \mathrm{or} \ 3 \alpha =- 3 heta + 2k \pi, \quad k\in \mathbb{Z}\ &\iff \alpha = \pm heta + k \frac{2\pi}{3}, \quad k \in \Z\ &\iff t = \cos\left ( heta + k \frac{2\pi}{3} ight),\quad k = 0,1,2.\ \end{align*} Therefore the polynomial $$h(t) = 4t^3 - 3t -\cos(3 heta)$$ has three roots : $\cos( heta), \cos( heta+ \frac{2\pi}{3}),\cos( heta+ \frac{4\pi}{3})$. These roots are real, and distinct, since $\Delta>0$. As $h(t) = \frac{4}{\lambda^3} f(\lambda t) $, $f(t)=0 \iff h(t/\lambda)=0$. The roots of $f$ are so $\lambda \cos( heta), \lambda \cos( heta+ \frac{2\pi}{3}),\lambda \cos( heta+ \frac{4\pi}{3})$, where $ \lambda = 2\sqrt{\frac{-p}{3}}$. The Theorem 1.3.3 is proven: {\it Let $y^3+py+q$ be a polynomial with real coefficients and positive discriminant. Then $p<0$, and the roots of the equation are $$y_1 = 2\sqrt{\frac{-p}{3}} \cos( heta),2\sqrt{\frac{-p}{3}} \cos\left( heta+\frac{2\pi}{3} ight),2\sqrt{\frac{-p}{3}} \cos\left( heta+\frac{4\pi}{3} ight),$$ where $$ heta = \frac{1}{3}\arccos\left(\frac{3\sqrt{3}q}{2p\sqrt{-p}} ight).$$ } \end{enumerate} \end{proof}

Exercise 1.3.10

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Exercise 1.3.10

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