Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/186

$p$ et $\sideset {^{1}}{}p$ étant les deux racines de l’équation

$f^{2}-{\frac {1}{4}}f={\frac {1}{4}},$

c’est-à-dire $p$ étant ${\frac {1+{\sqrt {17}}}{8}}$ et $\sideset {^{1}}{}p$ étant ${\frac {1-{\sqrt {17}}}{8}}.$

Il faut présentement déterminer les constantes arbitraires $\mathrm {A} _{n},\mathrm {N} _{n},\ldots .$ Or, si l’on substitue dans l’équation (1), au lieu de $\sideset {_{n}}{_{x}}y,\sideset {_{n}}{_{x-1}}y,\sideset {_{n-1}}{_{x-1}}y,\ldots$ leurs valeurs tirées de l’expression de $\sideset {_{n}}{_{x}}y,$ on aura

${\frac {\mathrm {A} _{n}}{2^{x}}}$

${\begin{aligned}+\ p^{x}&\left[\ \ \mathrm {N} _{n}{\frac {x(x-2)\ldots (x-n+1)}{1.2.3\ldots (n-2)}}+\left(2\mathrm {N} _{n}+\mathrm {M} _{n}\right){\frac {(x-2)\ldots (x-n+2)}{1.2.3\ldots (n-3)}}\right.\\&+\left(\mathrm {N} _{n}+2\mathrm {M} _{n}+\mathrm {L} _{n}\right){\frac {(x-2)\ldots (x-n+3)}{1.2.3\ldots (n-4)}}\\&+\left(\mathrm {M} _{n}+2\mathrm {L} _{n}+\mathrm {K} _{n}\right){\frac {x(x-2)\ldots (x-n+4)}{1.2.3\ldots (n-5)}}\\&+\left.\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots +\mathrm {C} _{n}{\begin{aligned}\\\\\end{aligned}}\right]\\+\sideset {^{1}}{^{x}}p&\left[\ \ \sideset {^{1}}{_{n}}{\mathrm {N} }_{n}{\frac {(x-2)\ldots (x-n+1)}{1.2.3\ldots (n-2)}}+\ldots \right]\\&={\frac {1}{4}}p^{x}\left\{\mathrm {N} _{n}\left({\frac {1}{p}}+{\frac {1}{p^{2}}}\right){\frac {(x-2)\ldots (x-n+1)}{1.2.3\ldots (n-2)}}+\ldots \right.\\&\qquad \qquad +\left[{\frac {\mathrm {N} _{n}}{p}}+\mathrm {M} _{n}\left({\frac {1}{p}}+{\frac {1}{p^{2}}}\right)+{\frac {\mathrm {N} _{n-1}}{p}}\right]{\frac {(x-2)\ldots (x-n+2)}{1.2.3\ldots (n-3)}}\\&\qquad \qquad +\left[{\frac {\mathrm {M} _{n}}{p}}+\mathrm {L} _{n}\left({\frac {1}{p}}+{\frac {1}{p^{2}}}\right)+{\frac {\mathrm {M} _{n-1}}{p}}\right.\\&\qquad \qquad \qquad \qquad +\left.{\frac {\mathrm {N} _{n-1}}{p}}+{\frac {\mathrm {N} _{n-2}}{p^{2}}}\right]{\frac {(x-2)\ldots (x-n+3)}{1.2.3\ldots (n-4)}}\\&\qquad \qquad +\left[{\frac {\mathrm {L} _{n}}{p}}+\mathrm {K} _{n}\left({\frac {1}{p}}+{\frac {1}{p^{2}}}\right)+{\frac {\mathrm {L} _{n-1}}{p}}\right.\\&\qquad \qquad \qquad \qquad +\left.\left.{\frac {\mathrm {M} _{n-1}}{p}}+{\frac {\mathrm {M} _{n-2}}{p^{2}}}\right]{\frac {(x-2)\ldots (x-n+4)}{1.2.3\ldots (n-5)}}+\ldots \right\}\\&={\frac {1}{4}}\sideset {^{1}}{^{x}}p\left\{\sideset {^{1}}{_{n}}{\mathrm {N} }\left({\frac {1}{\sideset {^{1}}{}p}}+{\frac {1}{\sideset {^{1}}{^{2}}p^{2}}}\right){\frac {(x-2)\ldots (x-n+1)}{1.2.3\ldots (x-2)}}+\ldots \right\}\\&\quad \qquad +{\frac {1}{4}}{\frac {\mathrm {A} _{n}}{2^{x-1}}}+{\frac {1}{4}}{\frac {\mathrm {A} _{n}}{2^{x-2}}}\\&\quad \qquad +{\frac {1}{4}}{\frac {\mathrm {A} _{n-1}}{2^{x-1}}}+{\frac {1}{4}}{\frac {\mathrm {A} _{n-2}}{2^{x-2}}}.\end{aligned}}$