22
MÉCANIQUE ANALYTIQUE.
de
désignée par
on fera
![{\displaystyle \operatorname {F} '(\theta )={\frac {d\operatorname {F} (\theta )}{d\theta }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5a1c2b4a059f7a1a3458d6957a3676d14ce8f47)
et l’on aura
![{\displaystyle \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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d08b3dfa8e2d8877890af425d8b86bb6a54494a0)
21. Pour appliquer cette formule à l’équation de l’article 16, on fera
![{\displaystyle f(\theta )=e\sin \theta \quad {\text{et}}\quad u=(t-c){\sqrt {\frac {\mathrm {g} }{a^{3}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4ba5b4b686ddc033e6d1a780ae54dbf1faf6869)
on aura immédiatement
![{\displaystyle \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 ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db8e88bd2e47f17499f88ac549704a251b376249)
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 ![{\displaystyle u.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edd5636410da69bac33da075162221527401793c)
On aura de même
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37a1ff533d7a5e8c53630c4c4cca46b3c299ab41)
On aura ainsi, par les formules des articles 16 et 17,
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b75753ce23b2792484d1f9bd58ccb690b355da5e)