Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/266

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FRACTIONS

${\begin{aligned}&{\frac {n+1}{1}}.{\frac {n+2}{2}}.{\frac {n+3}{3}}\ldots {\frac {n+m-1}{m-1}}=V^{m},\\&{\frac {n+1}{1}}.{\frac {n+2}{2}}.{\frac {n+3}{3}}\ldots {\frac {n+m-2}{m-2}}=V^{m-1},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}$

et le terme général demandé sera

${\begin{aligned}\theta ^{(n)}&={\frac {p^{n}}{a^{m}}}\left\{V^{m}-{\frac {m}{1}}{\frac {b}{a}}V^{m-1}+{\frac {m}{1}}{\frac {m+1}{2}}{\frac {b^{2}}{a^{2}}}V^{m-2}-\ldots \right\}\\\\&+{\frac {p'^{n}}{a^{m}}}\left\{V^{m}-{\frac {m}{1}}{\frac {b'}{a'}}V^{m-1}+{\frac {m}{1}}{\frac {m+1}{2}}{\frac {b'^{2}}{a'^{2}}}V^{m-2}-\ldots \right\}\end{aligned}}$

4. Si les facteurs de $1-\alpha x+\beta x^{2}$ sont imaginaires, on les représentera par $1-{\mathcal {f}}\left(\operatorname {Cos} .\phi +{\sqrt {-1}}\operatorname {Sin} .\phi \right)x$ et $1-{\mathcal {f}}\left(\operatorname {Cos} .\phi -{\sqrt {-1}}\operatorname {Sin} .\phi \right)x\,;$ et, comme on peut prendre ${\mathcal {f}}=1$ sans que le calcuL en soit moins général, on aura

${\begin{aligned}p&=\operatorname {Cos} .\phi +{\sqrt {-1}}\operatorname {Sin} .\phi ,\\p'&=\operatorname {Cos} .\phi -{\sqrt {-1}}\operatorname {Sin} .\phi \,;\end{aligned}}$

mais, alors $\alpha =2\operatorname {Cos} .\phi ,\ \beta =1\,;$ donc

${\begin{aligned}&a=2\operatorname {Sin} .\phi {\sqrt {-1}}.\left(\operatorname {Cos} .\phi -{\sqrt {-1}}\operatorname {Sin} .\phi \right),\\&b=\operatorname {Cos} .2\phi -{\sqrt {-1}}\operatorname {Sin} .2\phi ,\end{aligned}}$

${\frac {1}{a}}={\frac {\operatorname {Cos} .\phi +{\sqrt {-1}}\operatorname {Sin} .\phi }{2\operatorname {Sin} .\phi {\sqrt {-1}}}}={\frac {\operatorname {Cos} .\left(90^{\circ }-\phi \right)-{\sqrt {-1}}\operatorname {Sin} .\left(90^{\circ }-\phi \right)}{2\operatorname {Sin} .\phi }}.$

On aura semblablement $a',b'$ en changeant le signe de ${\sqrt {-1}}$