on trouvera, dis-je,
![{\displaystyle {\begin{alignedat}{2}\Gamma (a_{1},a_{3})=&{\frac {\mathrm {A} }{a_{3}^{3}}},\quad &\Gamma _{1}(a_{1},a_{3})=&{\frac {\mathrm {B} }{a_{3}^{3}}},\ldots ,\\\Gamma (a_{1},a_{4})=&{\frac {\mathrm {A} }{a_{4}^{3}}},&\Gamma _{1}(a_{1},a_{4})=&{\frac {\mathrm {B} }{a_{4}^{3}}},\ldots ,\\\Gamma (a_{2},a_{3})=&{\frac {\mathrm {A} }{a_{3}^{3}}},&\Gamma _{1}(a_{2},a_{3})=&{\frac {\mathrm {B} }{a_{3}^{3}}},\ldots \,;\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50e24a687c72d45dcb08c55c9d01cb4b8ad4dc2a)
et ainsi de suite.
À l’égard des quantités
il est évident qu’elles doivent être égales à leurs réciproques
car les fonctions
![{\displaystyle a_{1}^{2}-2a_{1}a_{2}\cos(\varphi _{2}-\varphi _{1})+a_{2}^{2},\quad a_{1}^{2}-2a_{1}a_{3}\cos(\varphi _{3}-\varphi _{1})+a_{3}^{2},\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b6c3b21098ad0a006d44da4ed8478d85555318)
demeurent les mêmes, en changeant
en
et
en
ou bien
en
et
en ![{\displaystyle a_{1},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13f4f666a440687d166ee142e7247a2eaef52d91)
XLIV.
De là on trouvera (Article XXIV),
étant égal à
![{\displaystyle {\begin{aligned}&{\breve {\Gamma }}\ (a_{1},a_{2})={\frac {q^{2}\mathrm {B} -2q^{3}\mathrm {A} }{2}},\\&{\breve {\Gamma }}_{1}(a_{1},a_{2})={\frac {q^{2}\mathrm {C} -2q^{3}\mathrm {B} +2q^{2}\mathrm {A} }{2}}-q^{2},\\&{\breve {\Gamma }}_{2}(a_{1},a_{2})={\frac {q^{2}\mathrm {D} -2q^{3}\mathrm {C} +q^{2}\mathrm {B} }{2}},\\&{\breve {\Gamma }}_{3}(a_{1},a_{2})={\frac {q^{2}\mathrm {E} -2q^{3}\mathrm {D} +q^{2}\mathrm {C} }{2}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbc36db1db1d67a83d9d9dd7d1c3ddc2bfa4ee93)
Et, par l’Article XXV on aura
![{\displaystyle {\begin{aligned}&{\widehat {\Gamma }}_{1}(a_{1},a_{2})={\frac {q^{2}\mathrm {C} -2q^{2}\mathrm {A} }{2}}+q^{2},\\&{\widehat {\Gamma }}_{2}(a_{1},a_{2})={\frac {q^{2}\mathrm {D} -q^{2}\mathrm {B} }{2}},\\&{\widehat {\Gamma }}_{3}(a_{1},a_{2})={\frac {q^{2}\mathrm {E} -q^{2}\mathrm {C} }{2}},\\&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24f2c211ee03658218fcc4275aa8f2624da5ce28)