Exercise 15.1.5

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}}

Shows that (15.7) reduces to $(x^2+y^2)^2 = x^2 -y^2$ when $a=b = 1/\sqrt{2}$.

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}
 If we take $a=b=1/\sqrt{2}$ in the formula of the ovals of Cassini
 $$((x-a)^2 + y^2)((x+a)^2 + y^2) = b^4,$$
 we obtain
 \begin{align*}
 \frac{1}{4} &= \left[ \left(x - \frac{1}{\sqrt{2}} ight)^2 + y^2ight ] \left[ \left(x + \frac{1}{\sqrt{2}} ight)^2 + y^2ight ]  \
 &=\left( x^2+y^2 +\frac{1}{2}  - \sqrt{2}\, x ight)\left( x^2+y^2 +\frac{1}{2}  + \sqrt{2}\, x ight)\
 &=\left((x^2 + y^2 + \frac{1}{2}ight)^2 - 2 x^2\
 &=(x^2+y^2)^2  + (x^2+y^2) - 2 x^2\
 &= (x^2+y^2)^2 + y^2 -x^2 + \frac{1}{4}.
 \end{align*}
 Therefore, for $a=b =1/\sqrt{2}$, the equation $((x-a)^2 + y^2)((x+a)^2 + y^2) = b^4$ reduces to
 $$(x^2 + y^2)^2 = x^2 - y^2,$$
 which is the equation of the Lemniscate.
\end{proof}

Exercise 15.1.5

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Exercise 15.1.5

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