Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/368

358

QUESTIONS

Or on a, (pag. 114 du présent volume)

${\frac {\operatorname {Cos} .x}{1}}-{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}-{\frac {\operatorname {Cos} .4x}{4}}+\ldots =\operatorname {Log} .2+\operatorname {Log} .\operatorname {Cos} .{\frac {1}{2}}x,$

c’est-à-dire

${\frac {\operatorname {Cos} .x}{1}}-{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}-{\frac {\operatorname {Cos} .4x}{4}}+\ldots =\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}x\,;$

en changeant donc, tour à tour $x$ en $x+y$ et $x-y,$ il viendra

${\frac {\operatorname {Cos} .(x+y)}{1}}-{\frac {\operatorname {Cos} .2(x+y)}{2}}+{\frac {\operatorname {Cos} .3(x+y)}{3}}-{\frac {\operatorname {Cos} .4(x+y)}{4}}+\ldots$

$=\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x+y)$

${\frac {\operatorname {Cos} .(x-y)}{1}}-{\frac {\operatorname {Cos} .2(x-y)}{2}}+{\frac {\operatorname {Cos} .3(x-y)}{3}}-{\frac {\operatorname {Cos} .4(x-y)}{4}}+\ldots$

$=\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x-y)$

donc, en prenant la demi-somme,

${\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}$

$-\ldots ={\frac {1}{2}}\left\{\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x-y)+\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x+y)\right\}$

ou encore

${\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-\ldots$

$={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)$

mais

$2\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)=\operatorname {Cos} .x+\operatorname {Cos} .y$

d’où

$4\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)=2\left(\operatorname {Cos} .x+\operatorname {Cos} .y\right)$

donc enfin, la somme finie de la troisième des suites infinies proposées est