Faisant les substitutions et les réductions convenables (Article
), on trouvera, pour la valeur de l’anomalie vraie
une expression de cette forme
![{\displaystyle {\begin{aligned}u=t&-{\frac {n\mathrm {K} '}{\sqrt {-1}}}\left(e^{t{\sqrt {-1}}}-e^{-t{\sqrt {-1}}}\right)+{\frac {n^{2}\mathrm {K} ''}{\sqrt {-1}}}\left(e^{2t{\sqrt {-1}}}-e^{-2t{\sqrt {-1}}}\right)\\&-{\frac {n^{3}\mathrm {K} '''}{\sqrt {-1}}}\left(e^{3t{\sqrt {-1}}}-e^{-3t{\sqrt {-1}}}\right)+\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b427bf37da33302d515b2ada6b401168664b750)
c’est-à-dire,
![{\displaystyle u=t-2n\mathrm {K} '\sin t+2n^{2}\mathrm {K} ''\sin 2t-2n^{3}\mathrm {K} '''\sin 3t+2n^{4}\mathrm {K} ^{\scriptscriptstyle {\text{IV}}}\sin 4t-\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43b6086bb5c1110bfa559b8e3ef1dfc7459851da)
dans laquelle les coefficients
seront tels, que si l’on fait, en général,
![{\displaystyle {\begin{aligned}\mathrm {P} =&{\frac {1}{\rho }}-\rho \left({\frac {n}{2}}\right)^{2}+{\frac {\rho ^{3}}{2^{2}}}\left({\frac {n}{2}}\right)^{4}-{\frac {\rho ^{5}}{2^{2}.3^{2}}}\left({\frac {n}{2}}\right)^{6}+\ldots ,\\\mathrm {Q} =&1\ \ -{\frac {\rho ^{2}}{2}}\left({\frac {n}{2}}\right)^{2}+{\frac {\rho ^{4}}{2^{2}.3}}\left({\frac {n}{2}}\right)^{4}-{\frac {\rho ^{6}}{2^{2}.3^{2}.4}}\left({\frac {n}{2}}\right)^{6}+\ldots ,\\\mathrm {R} =&\rho \ \ -{\frac {\rho ^{3}}{2.3}}\left({\frac {n}{2}}\right)^{2}+{\frac {\rho ^{5}}{2^{2}.3.4}}\left({\frac {n}{2}}\right)^{4}-{\frac {\rho ^{7}}{2^{2}.3^{2}.4.5}}\left({\frac {n}{2}}\right)^{6}+\ldots ,\\\mathrm {S} =&\rho ^{2}-{\frac {\rho ^{4}}{2.3.4}}\left({\frac {n}{2}}\right)^{2}-{\frac {\rho ^{6}}{2^{2}.3.4.5}}\left({\frac {n}{2}}\right)^{4}-{\frac {\rho ^{8}}{2^{2}.3^{2}.4.5.6}}\left({\frac {n}{2}}\right)^{6}+\ldots ,\\\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/101473a0b6f77dc87e27da74e54f044f4fb9c48e)
et qu’on dénote par
les valeurs de
qui répondentà
on aura
![{\displaystyle {\begin{aligned}\mathrm {K} '\ =&{\frac {\mathrm {P} '}{1+m}}+\left[1-{\frac {n^{2}}{(1+m)^{2}}}\right]{\frac {\mathrm {Q} '}{2}}+n^{2}\left[1+{\frac {n^{2}}{(1+m)^{2}}}\right]{\frac {\mathrm {R} '}{2^{2}(1+m)}}\\&+n^{4}\left[1-{\frac {n^{2}}{(1+m)^{2}}}\right]{\frac {\mathrm {S} '}{2^{3}(1+m)^{2}}}+n^{6}\left[1+{\frac {n^{2}}{(1+m)^{2}}}\right]{\frac {\mathrm {T} '}{2^{4}(1+m)^{2}}}+\ldots ,\\\mathrm {K} ''=&{\frac {\mathrm {P} ''}{(1+m)^{2}}}+\left[1-{\frac {n^{2}}{(1+m)^{2}}}\right]{\frac {\mathrm {Q} ''}{2(1+m)}}+\left[1+{\frac {n^{4}}{(1+m)^{4}}}\right]{\frac {\mathrm {R} ''}{2^{2}}}\\&+n^{2}\left[1-{\frac {n^{4}}{(1+m)^{4}}}\right]{\frac {\mathrm {S} ''}{2^{3}(1+m)}}+n^{4}\left[1+{\frac {n^{4}}{(1+m)^{4}}}\right]{\frac {\mathrm {T} ''}{2^{4}(1+m)^{2}}}\\&+n^{6}\left[1-{\frac {n^{4}}{(1+m)^{4}}}\right]{\frac {\mathrm {V} ''}{2^{5}(1+m)^{3}}}+\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e85d31d7d9e854fd1086142ea51d4444494eef71)