258
FRACTIONS
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae93524e6f320d3fd74de88a597243616e3ef028)
et le terme général demandé sera
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99213ab3559b6e5bee1b59bc706fa5f427f9962d)
4. Si les facteurs de
sont imaginaires, on les représentera par
et
et, comme on peut prendre
sans que le calcuL en soit moins
général, on aura
![{\displaystyle {\begin{aligned}p&=\operatorname {Cos} .\phi +{\sqrt {-1}}\operatorname {Sin} .\phi ,\\p'&=\operatorname {Cos} .\phi -{\sqrt {-1}}\operatorname {Sin} .\phi \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/664e4f1a76b4f5fdfdf3a7f3eee783de3302ad2d)
mais, alors
donc
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d7c6b7a91feea19f5dc53e3431774e45df1a9cd)
![{\displaystyle {\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 }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43052c42a7e1bbdfeda1ef259cdc9b8c236f5a25)
On aura semblablement
en changeant le signe de ![{\displaystyle {\sqrt {-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ea1ea9ac61e6e1e84ac39130f78143c18865719)