Exercise 9.43

Question

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Find the local maxima and minima of $x^3 - 3px -Ap$ and show that each of the intervals $(-2\sqrt{p}, - \sqrt{p}), (-\sqrt{p},\sqrt{p}), (\sqrt{p},2 \sqrt{p})$ contains exactly one of the values $2 \mathrm{Re}(\omega^k g(\chi_\pi)),\ k=0,1,2$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}} \def\dcro{\mbox{]\hspace{-.15em}]}} 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{ e}{\,\mathrm{Re}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{ }{\mathrm{N}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Write $\chi = \chi_\pi$, and for $k \in \{0,1,2\}$, $$G_k = 2 \mathrm{Re}(\omega^k g(\chi)) = \omega^k g(\chi) + \overline{\omega}^k \overline{g(\chi)},$$ so $G = G_0$. As in section 12, since $g(\chi)^3 = p \pi$, and $|g(\chi)|^2 = p$, \begin{align*} G_k^3 &= g(\chi)^3 + \overline{g(\chi)}^3 + 3 \omega^{2k} g(\chi)^2 \overline{\omega}^k \overline{g(\chi)} + 3 \omega^k g(\chi) \overline{\omega}^{2k} \overline{g(\chi)}^2\ &= p \pi + p \overline{\pi} + 3 g(\chi) \overline{g(\chi)} (\omega^k g(\chi) + \overline{\omega}^k \overline{g(\chi)})\ &= 3 p G_k + p(2a -b)\ &= 3p G_k + pA \end{align*} So $G_0, G_1,G_2$ are the three roots of $f(x) = x^3 - 3p x - Ap$. $f'(x) = 3(x^2-p)<0$ iff $-\sqrt{p} < x < \sqrt{p}$. $f$ is decreasing on $[-\sqrt{p},\sqrt{p}]$, and increasing on $]-\infty, -\sqrt{p}[$, and on $[\sqrt{p}, + \infty[$. Since $4 p = A^2 + 27 B^2$, $|A| < 2 \sqrt{p}$, therefore \begin{align*} f(\sqrt{p}) &= p \sqrt{p} - 3 p \sqrt{p} - Ap\ &=-p(2\sqrt{p} + A)<0, \end{align*} and \begin{align*} f(-\sqrt{p}) &= - p \sqrt{p} + 3 p \sqrt{p} - Ap\ &= p(2\sqrt{p} - A)>0. \end{align*} \end{proof} Since $\lim\limits_{x o - \infty} f(x)= -\infty$ and $\lim\limits_{x o +\infty} f(x)= +\infty$, the intermediate value theorem shows that $f$ has a unique root in each of the intervals $]-\infty, -\sqrt{p}[,]-\sqrt{p},\sqrt{p}[, [\sqrt{p}, + \infty[$. Moreover \begin{align*} f(2\sqrt{p}) &= 8p\sqrt{p} - 6p\sqrt{p} - Ap = p(2\sqrt{p} - A) >0,\ f(-2\sqrt{p}) &= -8p\sqrt{p} + 6 p \sqrt{p} - Ap = p(-2\sqrt{p} - A) <0, \end{align*} therefore $f$ has a unique root in each of the intervals $]-2 \sqrt{p}, -\sqrt{p}[,]-\sqrt{p},\sqrt{p}[, [\sqrt{p}, 2 \sqrt{p}[$.

Exercise 9.43

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Exercise 9.43

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