Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/128

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d’où l’on conclura facilement

{\begin{array}{r|r|r}{\overset {1}{y}}_{x}+b_{q}&{\overset {1}{y}}_{x-1}+^{1}\!b_{q}&{\overset {1}{y}}_{x-2}+\ldots \\+\mathrm {A} &+\mathrm {A} b_{q}&\\&+^{1}\!\mathrm {A} &\end{array}}

{\begin{aligned}&=\left(a_{q}\mathrm {B} +c_{q}\right)\left({\overset {q+1}{y}}_{x}+\mathrm {A} {\overset {q+1}{y}}_{x-1}+\ldots \right)\\&+\left(^{1}\!a_{q}\mathrm {B} +a_{q}\,^{1}\!\mathrm {B} +^{1}\!c_{q}+\mathrm {A} c_{q}\right)\left({\overset {q+1}{y}}_{x-1}+\mathrm {A} {\overset {q+1}{y}}_{x-2}+\ldots \right)\\&+\left(^{2}\!a_{q}\mathrm {B} +^{1}\!a_{q}\,^{1}\!\mathrm {B} +a_{q}\,^{2}\!\mathrm {B} +^{2}\!c_{q}+\mathrm {A} \,^{1}\!c_{q}+^{1}\!\mathrm {A} c_{q}\right)\left({\overset {q+1}{y}}_{x-2}+\mathrm {A} {\overset {q+1}{y}}_{x-3}+\ldots \right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\&+a_{q}\mathrm {C} \left({\overset {q+2}{y}}_{x}+\mathrm {A} {\overset {q+2}{y}}_{x-1}+\ldots \right)\\&+\left(^{1}\!a_{q}\mathrm {C} +a_{q}\,^{1}\!\mathrm {C} \right)\left({\overset {q+2}{y}}_{x-1}+\mathrm {A} {\overset {q+2}{y}}_{x-2}+\ldots \right)\\&+\left(^{2}\!a_{q}\mathrm {C} +^{1}\!a_{q}\,^{1}\!\mathrm {C} +a_{q}\,^{2}\!\mathrm {C} \right)\left({\overset {q+2}{y}}_{x-2}+\mathrm {A} {\overset {q+2}{y}}_{x-3}+\ldots \right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\&+{\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)+\ldots \,;\end{aligned}}

or on a

{\begin{aligned}{\overset {1}{y}}_{x}+b_{q+1}{\overset {1}{y}}_{x-1}+\ldots =&a_{q+1}\left({\overset {q+1}{y}}_{x}+\mathrm {A} {\overset {q+1}{y}}_{x-1}+\ldots \right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \\&+c_{q+1}\left({\overset {q+2}{y}}_{x}+\mathrm {A} {\overset {q+2}{y}}_{x-1}+\ldots \right)\\&+\ldots \ldots \ldots \ldots \ldots \ldots \\&+{\overset {q+1}{u}}_{x}\,;\end{aligned}}

d’où l’on aura, en comparant,

{\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 \,;\end{aligned}}

ainsi l’on déterminera $b_{q},^{1}\!b_{q},\ldots \,;$ ensuite

a_{q+1}=a_{q}\mathrm {B} +c_{q}\quad

et

\quad c_{q+1}=a_{q}\mathrm {C} ,\quad

partant

\quad a_{q+1}=a_{q}\mathrm {B} +a_{q-1}\mathrm {C} \,;