leurs valeurs déterminées dans le no 50, on aura
![{\displaystyle {\begin{aligned}(0,\ 1)&=-{\frac {m'n}{2}}\left(a^{2}{\frac {\partial {\rm {A}}^{(0)}}{\partial a}}+{\frac {1}{2}}a^{3}{\frac {\partial ^{2}{\rm {A}}^{(0)}}{\partial a^{2}}}\right),\\\\{\begin{array}{|c|}\hline 0,\ 1\\\hline \end{array}}&={\frac {m'n}{2}}\left(a{\rm {A}}^{(1)}-a^{2}{\frac {\partial {\rm {A}}^{(1)}}{\partial a}}-{\frac {1}{2}}a^{3}{\frac {\partial ^{2}{\rm {A}}^{(1)}}{\partial a^{2}}}\right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23ef5aadf36f3c84ad1e1e07459c0944ed1a16d8)
On a, par le no 49,
![{\displaystyle a^{2}{\frac {\partial {\rm {A}}^{(0)}}{\partial a}}+{\frac {1}{2}}a^{3}{\frac {\partial ^{2}{\rm {A}}^{(0)}}{\partial a^{2}}}=-\alpha ^{2}{\frac {db_{\frac {1}{2}}^{(0)}}{d\alpha }}-{\frac {1}{2}}\alpha ^{3}{\frac {d^{2}b_{\frac {1}{2}}^{(0)}}{d\alpha ^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c255d37dbfc35db0b640261eff2218f47977c81c)
on obtiendra facilement, par le même numéro,
et
en fonctions de
et de
et ces quantités sont données en fonctions linéaires de
et de
on trouvera, cela posé,
![{\displaystyle a^{2}{\frac {\partial {\rm {A}}^{(0)}}{\partial a}}+{\frac {1}{2}}a^{3}{\frac {\partial ^{2}{\rm {A}}^{(0)}}{\partial a^{2}}}={\frac {3\alpha ^{2}b_{-{\frac {1}{2}}}^{(1)}}{2\left(1-\alpha ^{2}\right)^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/962a9f8d8a64f1ff0ac2365a2a7c482d62a8047c)
partant
![{\displaystyle (0,\ 1)=-{\frac {3m'n\alpha ^{2}b_{-{\frac {1}{2}}}^{(1)}}{4\left(1-\alpha ^{2}\right)^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb5b750788adbe45be76d491affe632b2be8a1ea)
Soit
![{\displaystyle \left(a^{2}-2aa'\cos \theta +a'^{2}\right)^{\frac {1}{2}}=(a,a')+(a,a')'\cos \theta +(a,a')''\cos 2\theta +\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef02ace69a6694f898e17c8255f30d165b17b6a0)
on aura, par le no 49,
![{\displaystyle (a,a')={\frac {1}{2}}a'b_{-{\frac {1}{2}}}^{(0)},\qquad (a,a')'=b_{-{\frac {1}{2}}}^{(1)},\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a015f2d083d177701739e222af8480d23b0ba89)
on aura donc
![{\displaystyle (0,\ 1)=-{\frac {3m'na^{2}a'(a,a')'}{4\left(a'^{2}-a^{2}\right)^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/887dc266038da063e2fafd1030823eb06d7697a6)
On a ensuite, par le no 49,
![{\displaystyle a{\rm {A}}^{(1)}-a^{2}{\frac {\partial {\rm {A}}^{(1)}}{\partial a}}-{\frac {1}{2}}a^{3}{\frac {\partial ^{2}{\rm {A}}^{(1)}}{\partial a^{2}}}=-\alpha \left(b_{\frac {1}{2}}^{(1)}-\alpha {\frac {db_{\frac {1}{2}}^{(1)}}{d\alpha }}-{\frac {1}{2}}\alpha ^{2}{\frac {d^{2}b_{\frac {1}{2}}^{(1)}}{d\alpha ^{2}}}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2ba7a5883f52606d1a77fe94d53e5235885d7c2)