Page:Mémoires de l’Académie des sciences, Tome 22.djvu/272

Par conséquent, si l’on nomme ${\mathfrak {f}}(x)$ la valeur connue de $u$ entre les limites $x=0,$ $x=a,$ on trouvera

${\begin{array}{ll}f(x)={\mathfrak {f}}(x)\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\text{ entre les limites}}&\left\{{\begin{aligned}x=&0\\x=&a\ \ \end{aligned}}\right\},\\f(x)=-\mathrm {\cfrac {B-\alpha }{B+\alpha }} {\mathfrak {f}}(2a-x)\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots &\left\{{\begin{aligned}x=&a\\x=&2a\end{aligned}}\right\},\\f(x)=\mathrm {{\cfrac {A+\alpha }{A-\alpha }}{\cfrac {B-\alpha }{B+\alpha }}} {\mathfrak {f}}(x-2a)\ldots \ldots \ldots \ldots \ldots \ldots &\left\{{\begin{aligned}x=&2a\\x=&3a\end{aligned}}\right\},\\f(x)=-\mathrm {\left({\cfrac {A+\alpha }{A-\alpha }}{\cfrac {B-\alpha }{B+\alpha }}\right){\cfrac {B-\alpha }{B+\alpha }}} {\mathfrak {f}}(4a-x)\ldots \ldots &\left\{{\begin{aligned}x=&3a\\x=&4a\end{aligned}}\right\},\\f(x)=\mathrm {\left({\cfrac {A+\alpha }{A-\alpha }}{\cfrac {B-\alpha }{B+\alpha }}\right)^{2}} {\mathfrak {f}}(x-4a)..\ldots \ldots \ldots \ldots &\left\{{\begin{aligned}x=&4a\\x=&5a\end{aligned}}\right\},\\f(x)=-\mathrm {\left({\cfrac {A+\alpha }{A-\alpha }}{\cfrac {B-\alpha }{B+\alpha }}\right)^{2}{\cfrac {B-\alpha }{B+\alpha }}} {\mathfrak {f}}(6a-x)\ldots \ldots &\left\{{\begin{aligned}x=&5a\\x=&6a\end{aligned}}\right\},\\{\text{etc}}\ldots &\end{array}}$

On aura au contraire, en vertu de l’équation (69)

${\begin{array}{ll}f(x)=-\mathrm {\cfrac {A-\alpha }{A+\alpha }} {\mathfrak {f}}(-x)\ldots \ldots \ldots \ldots \ldots {\text{ entre les limites}}&\left\{{\begin{aligned}x=&0\\x=&-a\ \ \end{aligned}}\right\},\\f(x)=\mathrm {\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} {\mathfrak {f}}(2a+x).\ldots \ldots \ldots \ldots \ldots &\left\{{\begin{aligned}x=&-a\\x=&-2a\end{aligned}}\right\},\\f(x)=-\mathrm {{\cfrac {A-\alpha }{A+\alpha }}{\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}}} {\mathfrak {f}}(-x-2a)\ldots \ldots &\left\{{\begin{aligned}x=&-2a\\x=&-3a\end{aligned}}\right\},\\f(x)=\left[\mathrm {\cfrac {(A-\alpha )(B+\alpha )}{(A+\alpha )(B-\alpha )}} \right]^{2}{\mathfrak {f}}(4a+x)\ldots \ldots \ldots \ldots .&\left\{{\begin{aligned}x=&-3a\\x=&-4a\end{aligned}}\right\},\\f(x)=-\mathrm {{\cfrac {A-\alpha }{A+\alpha }}\left[{\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}}\right]} {\mathfrak {f}}(-x-4a)\ldots &\left\{{\begin{aligned}x=&-4a\\x=&-5a\end{aligned}}\right\},\\f(x)=\left[\mathrm {\cfrac {(A+\alpha )(B-\alpha )}{(A-\alpha )(B+\alpha )}} \right]^{3}{\mathfrak {f}}(6a+x).\ldots \ldots \ldots \ldots &\left\{{\begin{aligned}x=&-5a\\x=&-6a\end{aligned}}\right\},\\{\text{etc}}\ldots &\end{array}}$

On aura donc généralement, $n$ désignant un nombre entier quelconque,