Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/393

381

DES SUITES.

$=\pm {\frac {1}{2^{n-1}}}\left\{\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .s+a^{2}}}-\Sigma \operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(s-2t)+a^{2}}}\right.$

$\left.+\Sigma \operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(s-2t-2u)+a^{2}}}-\ldots \right\}.$

par le Théorème III, pour $n$ impair,

${\frac {a\operatorname {Sin} .t\operatorname {Sin} .u\operatorname {Sin} .v\ldots }{1}}-{\frac {a^{2}\operatorname {Sin} .2t\operatorname {Sin} .2u\operatorname {Sin} .2v\ldots }{2}}+{\frac {a^{3}\operatorname {Sin} .3t\operatorname {Sin} .3u\operatorname {Sin} .3v\ldots }{3}}-\ldots$

$=\pm {\frac {1}{2^{n-1}}}\left\{\operatorname {Arc} .\left[\operatorname {Tang} .={\frac {a\operatorname {Sin} .s}{1+a\operatorname {Cos} .s}}\right]-\Sigma \operatorname {Arc} .\left[\operatorname {Tang} .={\frac {a\operatorname {Sin} .(s-2t)}{1+a\operatorname {Cos} .(s-2t)}}\right]\right.$

$\left.+\Sigma \operatorname {Arc} .\left[\operatorname {Tang} .={\frac {a\operatorname {Sin} .(s-2t-2u)}{1+a\operatorname {Cos} .(s-2t-2u)}}\right]-\ldots \right\}.$

par le Corollaire I, quel que soit $n,$

${\frac {a\operatorname {Cos} .^{n}t}{1}}-{\frac {a^{2}\operatorname {Cos} .^{n}2t}{2}}+{\frac {a^{3}\operatorname {Cos} .^{n}3t}{3}}+{\frac {a^{4}\operatorname {Cos} .^{n}4t}{4}}+\ldots$

$=+{\frac {1}{2^{n-1}}}\left\{\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .nt+a^{2}}}+{\frac {n}{1}}\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(n-2)t+a^{2}}}\right.$

$\left.+{\frac {n}{1}}.{\frac {n-1}{2}}\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(n-4)t+a^{2}}}+\ldots \right\}.$

par le Corollaire II, pour $n$ pair,

${\frac {a\operatorname {Sin} .^{n}t}{1}}-{\frac {a^{2}\operatorname {Sin} .^{n}2t}{2}}+{\frac {a^{3}\operatorname {Sin} .^{n}3t}{3}}-{\frac {a^{4}\operatorname {Sin} .^{n}4t}{4}}+\ldots$

$=\pm {\frac {1}{2^{n}}}\left\{\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .nt+a^{2}}}-{\frac {n}{1}}\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(n-2)t+a^{2}}}\right.$

$\left.+{\frac {n}{1}}.{\frac {n-1}{2}}\operatorname {Log} .{\sqrt {1+2a\operatorname {Cos} .(n-4)t+a^{2}}}-\ldots \right\}.$

par le Corollaire III, pour $n$ impair,