Exercise 15.2.6

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}}

Show that the subtraction law
$$\varphi(x-y) = \frac{\varphi(x) \varphi'(y) - \varphi(y) \varphi'(x)}{1 + \varphi^2(x) \varphi^2(y)}.
$$
follows from the addition law together with (15.9) and Exercise 3.

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} Starting from the Addition Law for $\varphi$
$$\varphi(x+y) = \frac{\varphi(x) \varphi'(y) + \varphi(y) \varphi'(x)}{1 + \varphi^2(x) \varphi^2(y)}, \qquad x,y \in \R,$$
we obtain for all $x,y \in \R$, substituting $-y$ to $y$,
\begin{align*}
&\varphi(x-y) = \frac{\varphi(x) \varphi'(-y) + \varphi(-y) \varphi'(x)}{1 + \varphi^2(x) \varphi^2(-y)}.
\end{align*}
Since $\varphi$ is odd, and $\varphi'$ even (see 15.9 and Exercise 3), we obtain
$$\varphi(x-y) = \frac{\varphi(x) \varphi'(y) - \varphi(y) \varphi'(x)}{1 + \varphi^2(x) \varphi^2(y)}.
$$
\end{proof}

Exercise 15.2.6

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Exercise 15.2.6

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