d’où l’on a, en comparant,
![{\displaystyle {\begin{aligned}b_{q+1}=&\ b_{q}+\mathrm {A} ,\\^{1}\!b_{q+1}=&^{1}\!b_{q}+\mathrm {A} b_{q}+^{1}\!\mathrm {A} ,\\\ldots \ldots &\ldots \ldots \ldots \ldots \,;a_{q+1}=&\ a_{q}+\mathrm {B} ,\\^{1}\!a_{q+1}=&^{1}\!a_{q}\mathrm {B} +a_{q}\,^{1}\!\mathrm {B} ,\\\ldots \ldots &\ldots \ldots \ldots \ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0102019d3dde3231d845038de1fbdfa99ecadaa4)
![{\displaystyle (\Lambda )\left\{{\begin{aligned}&{\overset {q+1}{u}}_{x}={\overset {q}{u}}_{x}+\mathrm {A} {\overset {q}{u}}_{x-1}+^{1}\!\mathrm {A} \,{\overset {q}{u}}_{x-2}+\ldots \\&+\mathrm {X} _{x}a_{q}+\mathrm {X} _{x-1}\left(^{1}\!a_{q}+\mathrm {A} a_{q}\right)+\mathrm {X} _{x-2}\left(^{2}\!a_{q}+\mathrm {A} \,^{1}\!a_{q}+^{1}\!\mathrm {A} a_{q}\right)+\ldots .\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddccfe1c6590ad8cc35dbb0443bb19a2eb504670)
Au moyen de ces équations, on déterminera facilement ![{\displaystyle a_{q},^{1}\!a_{q},\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5814bd7b64bbbb25f2c94985acd5258986338c49)
pour déterminer
j’observe que l’on a
![{\displaystyle {\overset {q}{u}}_{x}=f_{q}\mathrm {X} _{x}+^{1}\!f_{q}\mathrm {X} _{x-1}+^{2}\!f_{q}\mathrm {X} _{x-2}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a6c30a934c781d8dd7d734a5813f778e65aaf73)
je substitue cette valeur dans l’équation
ce qui donne
![{\displaystyle {\begin{aligned}{\overset {q+1}{u}}_{x}=&\mathrm {X} _{x}\left(f_{q}+a_{q}\right)+\mathrm {X} _{x-1}\left(^{1}\!f_{q}+^{1}\!a_{q}+\mathrm {A} a_{q}+\mathrm {A} f_{q}\right)\\&+\mathrm {X} _{x-2}\left(^{2}\!f_{q}+^{2}\!a_{q}+\mathrm {A} \,^{1}\!a_{q}+^{1}\!\mathrm {A} f_{q}+\mathrm {A} \,^{1}\!f_{q}\right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ccbf7ed3b5382dab8ac865e19d6b1fc1f0fd77d)
mais on a
![{\displaystyle {\overset {q+1}{u}}_{x}=f_{q+1}\mathrm {X} _{x}+^{1}\!f_{q+1}\mathrm {X} _{x-1}+^{2}\!f_{q+1}\mathrm {X} _{x-2}+\ldots \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/301460561abdd6c9c05141c0f87c93da27ccbcce)
donc
![{\displaystyle {\begin{aligned}f_{q+1}=&\ f_{q}+a_{q},\\^{1}\!f_{q+1}=&^{1}\!f_{q}+^{1}\!a_{q}+\mathrm {A} f_{q},\\\ldots \ldots &\ldots \ldots \ldots \ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dd56fb440181479886608cdf5fea0cc585b0998)
Au moyen de ces équations on déterminera
et partant
Je suppose maintenant
et l’on aura
![{\displaystyle {\begin{aligned}{\overset {1}{y}}_{x}+b_{n}{\overset {1}{y}}_{x-1}+\ldots =&a_{n}\left({\overset {n}{y}}_{x}+\mathrm {A} {\overset {n}{y}}_{x-1}+\ldots \right)\\&+^{1}\!a_{n}\left({\overset {n}{y}}_{x-1}+^{1}\!\mathrm {A} {\overset {n}{y}}_{x-2}+\ldots \right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \\&+{\overset {n}{u}}_{x}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c004bf3048b1c7a1cbb35d8d2fb628cef1cf6cf8)