Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/364

354

QUESTIONS

$A_{0}+A_{1}a\operatorname {Cos} .x+A_{2}a^{2}\operatorname {Cos} .2x+A_{3}a^{3}\operatorname {Cos} .3x+A_{4}a^{4}\operatorname {Cos} .4x+\ldots$

$A_{0}+A_{1}a\operatorname {Sin} .x\ +A_{2}a^{2}\operatorname {Sin} .2x\ +A_{3}a^{3}\operatorname {Sin} .3x\ +A_{4}a^{4}\operatorname {Sin} .4x\ +\ldots$

seront respectivement

${\frac {\operatorname {f} \left[a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]+\operatorname {f} \left[a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2}},$

et

${\frac {\operatorname {f} \left[a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]-\operatorname {f} \left[a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2{\sqrt {-1}}}}.$

Cela posé, on a

${\frac {a}{1}}-{\frac {a^{3}}{3}}+{\frac {a^{5}}{5}}-{\frac {a^{7}}{7}}+{\frac {a^{9}}{9}}-\ldots =\operatorname {Arc} .(\operatorname {Tang} .=a),$

donc 1.^o la somme de la série

${\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-{\frac {a^{7}\operatorname {Cos} .7x}{7}}+\ldots$

sera

${\frac {\operatorname {Arc} .\left[\operatorname {Tang} .=a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]+\operatorname {Arc} .\left[\operatorname {Tang} .=a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2}}$

Soient $M$ le premier de ces arcs et $N$ le second, on aura pour la somme de la série ${\frac {M+N}{2}},$ et de plus

$\operatorname {Tang} .M=a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right),$

$\operatorname {Tang} .N=a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\,;$

d’où