Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/30

22

MÉCANIQUE ANALYTIQUE.

de $\theta$ désignée par $\operatorname {F} (\theta ),$ on fera

$\operatorname {F} '(\theta )={\frac {d\operatorname {F} (\theta )}{d\theta }},$

et l’on aura

$\operatorname {F} (\theta )=\operatorname {F} (u)+f(u)\operatorname {F} '(u){\frac {d\left[f(u)^{2}\operatorname {F} '(u)\right]}{2du}}+{\frac {d^{2}\left[f(u)^{3}\operatorname {F} '(u)\right]}{2.3du^{2}}}+\ldots .$

21. Pour appliquer cette formule à l’équation de l’article 16, on fera

$f(\theta )=e\sin \theta \quad {\text{et}}\quad u=(t-c){\sqrt {\frac {\mathrm {g} }{a^{3}}}}\,;$

on aura immédiatement

$\theta =u+e\sin u+e^{2}{\frac {d\sin ^{2}u}{2du}}+e^{3}{\frac {d^{2}\sin ^{3}u}{2.3du^{2}}}+\ldots ,$

où il n’y aura plus qu’à exécuter les différentiations indiquées ; mais, pour avoir des expressions plus simples, il conviendra de développer auparavant les puissances des sinus en sinus et cosinus d’angles multiples de $u.$

On aura de même

${\begin{aligned}\sin \theta \ \ =&\sin u+e\sin u\cos u+e^{2}{\frac {d\sin ^{2}u\cos u}{2du}}+e^{3}{\frac {d^{2}\sin ^{3}u\cos u}{2.3du^{2}}}+\ldots ,\\\cos \theta \ \ =&\cos u-e\sin ^{2}u-e^{2}{\frac {d\sin ^{3}u}{2du}}-e^{3}{\frac {d^{2}\sin ^{4}u}{2.3du^{2}}}-\ldots ,\\\operatorname {tang} \theta =&\operatorname {tang} u+e{\frac {\sin u}{\cos ^{2}u}}+e^{2}{\frac {d{\dfrac {\sin ^{2}u}{\cos ^{2}u}}}{2du}}+e^{3}{\frac {d^{2}{\dfrac {\sin ^{3}u}{\cos ^{2}u}}}{2.3du^{2}}}+\ldots .\end{aligned}}$

On aura ainsi, par les formules des articles 16 et 17,

${\begin{aligned}r\ =&a\left(1-e\cos u+e^{2}\sin ^{2}u+e^{3}{\frac {d\sin ^{3}u}{2du}}+e^{4}{\frac {d^{2}\sin ^{4}u}{2.3du^{2}}}+\ldots \right),\\r^{n}=&a^{n}\left[(1-e\cos u)^{n}+ne^{2}\sin ^{2}u(1-e\cos u)^{n-1}+{\frac {ne^{3}d\sin ^{3}u(1-e\cos u)^{n-1}}{2du}}+\ldots \right],\\\mathrm {X} =&a\left[\cos u-e\left(1+\sin ^{2}u\right)-e^{2}{\frac {d\sin ^{3}u}{2du}}-e^{3}{\frac {d^{2}\sin ^{4}u}{2.3du^{2}}}-\ldots \right],\\\mathrm {Y} =&a{\sqrt {1-e^{2}}}\left[\sin u+e\sin u\cos u+e^{2}{\frac {d\sin ^{2}u\cos u}{2du}}+e^{3}{\frac {d^{2}\sin ^{3}u\cos u}{2.3du^{2}}}+\ldots \right],\\\operatorname {tang} {\frac {\Phi }{2}}=&{\sqrt {\frac {1+e}{1-e}}}\left[\operatorname {tang} {\frac {u}{2}}+e{\frac {\sin u}{1+\cos u}}+e^{2}{\frac {d{\dfrac {\sin ^{2}u}{1+\cos u}}}{2du}}+e^{3}{\frac {d^{2}{\dfrac {\sin ^{3}u}{1+\cos u}}}{2.3du^{2}}}+\ldots \right].\end{aligned}}$