Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/273

267

D’ASTRONOMIE.

Cet angle devant s’évanouir en cas de syzygie, on aura

$\alpha +{\frac {2\varpi t}{q}}-\phi =n\varpi ,$

ou

$\phi =\alpha -n\varpi +{\frac {2\varpi t}{q}},\qquad$et$\qquad \operatorname {d} \phi ={\frac {2\varpi t}{q}}\,;$

donc

${\frac {2\varpi t}{q}}={\frac {2\varpi (1-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )^{2}}{p\operatorname {Cos} .^{3}\lambda }}\operatorname {d} t-{\frac {\operatorname {Sin} .\phi (2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )}{\operatorname {Cos} .\lambda }}\operatorname {d} \lambda \,;$

en conséquence

$-{\frac {2\varpi \operatorname {d} t}{pq\operatorname {d} \lambda }}={\frac {\operatorname {Cos} .^{2}\lambda \operatorname {Sin} .\phi (2-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )}{p\operatorname {Cos} .^{3}\lambda -q(1-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi )^{2}}}.$

Tel est le rapport différentiel $\operatorname {d} t:\operatorname {d} \lambda ,$ dont il faudra déduire les coefficiens de la série que nous cherchons.

147. Pour donner à nos formules la simplicité que nos développemens exigent, soient

$\operatorname {Sin} .\lambda =x,\qquad \operatorname {Cos} .\phi =y,$

$\operatorname {Cos} .\lambda =u,\qquad \operatorname {Sin} .\phi =v\,;$

donc

${\frac {\operatorname {d} x}{\operatorname {d} \lambda }}=u,\qquad {\frac {\operatorname {d} y}{\operatorname {d} \lambda }}=-{\frac {v\operatorname {d} \phi }{\operatorname {d} \lambda }}=-{\frac {2\varpi v}{q}}.{\frac {\operatorname {d} t}{\operatorname {d} \lambda }},$

${\frac {\operatorname {d} u}{\operatorname {d} \lambda }}=-x,\qquad {\frac {\operatorname {d} v}{\operatorname {d} \lambda }}=+{\frac {y\operatorname {d} \phi }{\operatorname {d} \lambda }}=+{\frac {2\varpi y}{q}}.{\frac {\operatorname {d} t}{\operatorname {d} \lambda }}.$

On aura ainsi

$-{\frac {2\varpi \operatorname {d} t}{q\operatorname {d} \lambda }}={\frac {pu^{2}v(2-xy)}{pu^{3}-q(1-xy)^{2}}}$

et si, pour abréger, on désigne cette fraction par $h,$ on aura

${\frac {\operatorname {d} x}{\operatorname {d} \lambda }}=+u,\qquad {\frac {\operatorname {d} y}{\operatorname {d} \lambda }}=+hv,$

${\frac {\operatorname {d} u}{\operatorname {d} \lambda }}=-x,\qquad {\frac {\operatorname {d} v}{\operatorname {d} \lambda }}=-hy.$