et de là
![{\displaystyle \mathrm {Z} ^{r-1}{\frac {d\mathrm {Z} }{dz}}={\frac {1}{2^{r}}}\left[z^{r-1}+(r+2)\left({\frac {n}{2}}\right)^{2}z^{r+1}+{\frac {(r+3)(r+4)}{2}}\left({\frac {n}{2}}\right)^{4}z^{r+3}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3a268f36dd0666847c2e732d530694353d7c228)
![{\displaystyle \left.+{\frac {r(r+4)(r+5)(r+6)}{2.3}}\left({\frac {n}{2}}\right)^{6}z^{r+5}+\ldots \right]\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0728b8faa078ed042b13645e7da2383600dce6a9)
de sorte qu’en faisant successivement
on aura
![{\displaystyle {\begin{aligned}{\frac {1}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}}=&\quad {\frac {1}{z}}\quad \ +2\left({\frac {n}{2}}\right)^{2}z\,\ +{\frac {3.4}{2}}\left({\frac {n}{2}}\right)^{4}z^{3}+{\frac {4.5.6}{2.3}}\left({\frac {n}{2}}\right)^{6}z^{5}+\ldots \\{\frac {d\mathrm {Z} }{dz}}=&{\frac {1}{2}}\ \ \left[1\ \ +3\left({\frac {n}{2}}\right)^{2}z^{2}+{\frac {4.5}{2}}\left({\frac {n}{2}}\right)^{4}z^{4}+{\frac {5.6.7}{2.3}}\left({\frac {n}{2}}\right)^{6}z^{6}+\ldots \right],\\\mathrm {Z} {\frac {d\mathrm {Z} }{dz}}=&{\frac {1}{2^{2}}}\left[z\ \ +4\left({\frac {n}{2}}\right)^{2}z^{3}+{\frac {5.6}{2}}\left({\frac {n}{2}}\right)^{4}z^{5}+{\frac {6.7.8}{2.3}}\left({\frac {n}{2}}\right)^{6}z^{7}+\ldots \right],\\\mathrm {Z} ^{2}{\frac {d\mathrm {Z} }{dz}}=&{\frac {1}{2^{3}}}\left[z^{2}+5\left({\frac {n}{2}}\right)^{2}z^{4}+{\frac {6.7}{2}}\left({\frac {n}{2}}\right)^{4}z^{6}+{\frac {7.8.9}{2.3}}\left({\frac {n}{2}}\right)^{6}z^{8}+\ldots \right],\\\ldots \ldots &\ldots \ldots \ldots \ldots \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/210dfc15597c35163a5dfad4dc5e703c3c02cce0)
X.
Reprenons le cas de l’Article III où l’on demande l’anomalie de l’excentrique
par l’anomalie moyenne
On fera donc
et
ce qui donnera la fraction
![{\displaystyle {\frac {1}{z(1+nz\sin t)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3eb36ebd5a7298c24c170016647dafe00d5618e)
qui peut se développer en une série de cette forme
![{\displaystyle {\begin{aligned}\mathrm {M} &+\,\ \mathrm {M} 'e^{t{\sqrt {-1}}}+\,\ \mathrm {M} ''e^{2t{\sqrt {-1}}}+\ \mathrm {M} '''e^{3t{\sqrt {-1}}}+\ldots \\&+\mathrm {N} 'e^{-t{\sqrt {-1}}}+\mathrm {N} ''e^{-2t{\sqrt {-1}}}+\mathrm {N} '''e^{-3t{\sqrt {-1}}}+\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f61c5a1df8d6ec570db59d141c831dae401b4e3)
où l’on aura (Article précédent)
![{\displaystyle {\begin{alignedat}{3}\mathrm {M} =&{\frac {1}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}},\quad &\mathrm {M} '=&-{\frac {n}{\sqrt {-1}}}{\frac {d\mathrm {Z} }{dz}},\quad &\mathrm {M} ''=&{\frac {n^{2}}{\left({\sqrt {-1}}\right)^{2}}}\mathrm {Z} {\frac {d\mathrm {Z} }{dz}},\ldots ,\\&&\mathrm {N} '=&{\frac {n}{\sqrt {-1}}}{\frac {d\mathrm {Z} }{dz}},&\mathrm {N} ''=&{\frac {n^{2}}{\left({\sqrt {-1}}\right)^{2}}}\mathrm {Z} {\frac {d\mathrm {Z} }{dz}},\ldots .\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/132451d77a11372d83c9890b7f1c46b553697afc)