Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/116

112

QUESTIONS

${\frac {a^{n}}{1.2.3\ldots n}}-{\frac {2a^{n}\operatorname {Sin} .^{2}nx}{1.2.3\ldots n}},\qquad {\frac {2a^{n}\operatorname {Cos} .^{2}nx}{1.2.3\ldots n}}-{\frac {a^{n}}{1.2.3\ldots n}},$

il viendra donc, en faisant successivement les deux substitution,

$e^{a\operatorname {Cos} .2x}.\operatorname {Cos} .(a\operatorname {Sin} .2x)=\left(1+{\frac {a}{1}}+{\frac {a^{2}}{1.2}}+{\frac {a^{3}}{1.2.3}}+\ldots \right)$

$-2\left({\frac {a\operatorname {Sin} .^{2}x}{1}}+{\frac {a^{2}\operatorname {Sin} .^{2}2x}{1.2}}+{\frac {a^{3}\operatorname {Sin} .^{2}3x}{1.2.3}}+\ldots \right)$

$e^{a\operatorname {Cos} .2x}.\operatorname {Cos} .(a\operatorname {Sin} .2x)=$

$2\left({\frac {a\operatorname {Cos} .^{2}x}{1}}+{\frac {a^{2}\operatorname {Cos} .^{2}2x}{1.2}}+{\frac {a^{3}\operatorname {Cos} .^{2}3x}{1.2.3}}+\ldots \right)$

$+2-\left(1+{\frac {a}{1}}+{\frac {a^{2}}{1.2}}+{\frac {a^{3}}{1.2.3}}+\ldots \right)\,;$

d’où on tirera, en transposant et remplaçant l’une des séries par sa valeur $e^{a},$

${\frac {a\operatorname {Sin} .^{2}x}{1}}+{\frac {a^{2}\operatorname {Sin} .^{2}2x}{1.2}}+{\frac {a^{3}\operatorname {Sin} .^{2}3x}{1.2.3}}+\ldots =$

${\frac {e^{a}-e^{a\operatorname {Cos} .2x}.\operatorname {Cos} .(a\operatorname {Sin} .2x)}{2}},$

$1+{\frac {a\operatorname {Cos} .^{2}x}{1}}+{\frac {a^{2}\operatorname {Cos} .^{2}2x}{1.2}}+{\frac {a^{3}\operatorname {Cos} .^{2}3x}{1.2.3}}+\ldots =$

${\frac {e^{a}+e^{a\operatorname {Cos} .3x}.\operatorname {Cos} .(a\operatorname {Sin} .2x)}{2}},$

M. Stein remarque encore que, par des moyens semblables à ceux qui ont été appliqués à la série proposée, on parviendrait aussi à sommer la série

${\frac {a\operatorname {Sin} .x}{1}}+{\frac {a^{2}\operatorname {Sin} .2x}{2}}+{\frac {a^{3}\operatorname {Sin} .3x}{3}}+{\frac {a^{4}\operatorname {Sin} .4x}{4}}+\ldots ,$