164
PROBLÈMES
![{\displaystyle {\frac {\operatorname {d} ^{n+1}\phi }{\operatorname {d} \lambda ^{n+1}}}={\frac {\operatorname {Sin} .\phi }{\operatorname {Cos} .^{n+1}\lambda }}\left\{z\left(nx-2y+xy^{2}\right)+P\left(1-x^{2}\right)+Q\left(2-xy-2y^{2}+xy^{3}\right)\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c51c7ced8de53fe3145564771b2d9c88788c9cb)
Ainsi, pour trouver ces coefficiens, il faudra effectuer les multiplications ; c’est la seule difficulté qu’il restera à surmonter.
7. D’après cela, pour passer du premier
au second
on aura
d’où il résulte
![{\displaystyle {\frac {\operatorname {d} ^{2}\phi }{\operatorname {d} \lambda ^{2}}}={\frac {\operatorname {Sin} .\phi }{\operatorname {Cos} .^{2}\lambda }}\left(5y-x^{2}y-6xy^{2}+2x^{2}y^{3}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5a0e77ec4734b61eb50cf260fa066757194af9e)
On en tire
![{\displaystyle {\begin{aligned}n=2\ ;\ z&=5y-x^{2}y-6xy^{2}+2x^{2}y^{3},\\P&=-2xy-6y^{2}+4xy^{3},\\Q&=5-x^{2}-12xy+6x^{2}y^{2},\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f472a4b2709931a8a187109be9bb1c30a835462a)
donc
![{\displaystyle {\frac {\operatorname {d} ^{3}\phi }{\operatorname {d} \lambda ^{3}}}={\frac {\operatorname {Sin} .\phi }{\operatorname {Cos} .^{3}\lambda }}\left\{{\begin{aligned}&10-2x^{2}+21xy-26y^{2}+x^{3}y\\+&22x^{2}y^{2}+50xy^{3}-8x^{3}y^{3}-34x^{2}y^{4}+8x^{3}y^{5}\\\end{aligned}}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2d306c2f3f9095dad05c22ba795664e7a1e306d)
Faisant ensuite
et
![{\displaystyle {\begin{aligned}z&=10-2x2+21xy-26y^{2}+x^{3}y+22x^{2}y^{2}+50xy^{3}-8x^{3}y^{3}-34x^{2}y^{4}+8x^{3}y^{5},\\P&=-4x-21y+3x^{2}y+44xy^{2}+50y^{3}-24x^{2}y^{3}-68xy^{4}+24x^{2}y^{5},\\Q&=-21x-52y+x^{3}+44x^{2}y+150xy^{2}-24x^{3}y^{2}-136x^{2}y^{3}+40x^{3}y^{4}\,;\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bd0dde9793a8fac7f08b818f689ef67bef1643e)
d’où il résulte
![{\displaystyle {\tfrac {\operatorname {d} ^{4}\phi }{\operatorname {d} \lambda ^{4}}}={\tfrac {\operatorname {Sin} .\phi }{\operatorname {Cos} .^{4}\lambda }}\left\{{\begin{aligned}&-16x-145y+74x^{2}y+412xy^{2}+206y^{3}\\&-x^{4}y-76x^{3}y^{2}-520x^{2}y^{3}-546xy^{4}+26x^{4}y^{3}\\&+288x^{3}y^{4}+564x^{2}y^{5}-72x^{4}y^{5}-266x^{3}y^{6}+48x^{4}y^{7}\\\end{aligned}}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca3a1f1608af6aba79a19bbe60100ed38d736792)
Et ainsi des autres.
8. Il ne reste donc qu’à faire, dans tous ces rapports différentiels,
et par conséquent
On aura
![{\displaystyle B=-2\operatorname {Sin} .A,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6fd9b8b712263280dbf1c92713be8e944619461)