Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/281

Le texte de cette page a été corrigé et est conforme au fac-similé.

273

DE M. DE STAINVILLE.

Les corrections sont expliquées en page de discussion

{\begin{array}{r|rr|r}\operatorname {f} _{n-1}(a).\operatorname {f} _{0}(b)&{\frac {x^{n-1}}{(n-1)!}}+&\operatorname {f} _{n}(a).\operatorname {f} _{0}(b)&{\frac {x^{n}}{n!}}+\ldots \,;\\\\+{\frac {n-1}{1}}\operatorname {f} _{n-2}(a).\operatorname {f} _{1}(b)&+&{\frac {n}{1}}\operatorname {f} _{n-1}(a).\operatorname {f} _{1}(b)&\\\\+{\frac {n-1}{1}}.{\frac {n-2}{2}}\operatorname {f} _{n-3}(a).\operatorname {f} _{2}(b)&+&{\frac {n}{1}}{\frac {n-1}{2}}\operatorname {f} _{n-2}(a).\operatorname {f} _{2}(b)&\\+\ldots \ldots \ldots \ldots &+&\ldots \ldots \ldots \ldots &\\+{\frac {n-1}{1}}.{\frac {n-2}{2}}\operatorname {f} _{2}(a).\operatorname {f} _{n-3}(b)&+&{\frac {n}{1}}{\frac {n-1}{2}}\operatorname {f} _{2}(a).\operatorname {f} _{n-2}(b)&\\\\+{\frac {n-1}{1}}\operatorname {f} _{1}(a).\operatorname {f} _{n-2}(b)&+&{\frac {n}{1}}\operatorname {f} _{1}(a).\operatorname {f} _{n-1}(b)&\\\\+\operatorname {f} _{0}(a).\operatorname {f} _{n-1}(b)&+&\operatorname {f} _{0}(a).\operatorname {f} _{n}(b)&\end{array}}

opérant sur le premier de ces deux termes comme nous l’avons fait ci-dessus, nous aurons, pour résultat

a\left\{{\begin{array}{lr}&\operatorname {f} _{n-1}(a+k).\operatorname {f} _{0}(b)\\+&{\frac {n-1}{1}}.\operatorname {f} _{n-2}(a+k).\operatorname {f} _{1}(b)\\\\+&{\frac {n-1}{1}}.{\frac {n-2}{2}}.\operatorname {f} _{n-3}(a+k).\operatorname {f} _{2}(b)\\+&\ldots \ldots \ldots \ldots \\+&{\frac {n-1}{1}}.{\frac {n-2}{2}}\operatorname {f} _{2}(a+k).\operatorname {f} _{n-3}(b)\\\\+&{\frac {n-1}{1}}.\operatorname {f} _{1}(a+k).\operatorname {f} _{n-2}(b)\\\\+&\operatorname {f} _{0}(a+k).\operatorname {f} _{n-1}(b)\end{array}}\right\}+b\left\{{\begin{array}{lr}&\operatorname {f} _{n-1}(a).\operatorname {f} _{0}(b+k)\\+&{\frac {n-1}{1}}.\operatorname {f} _{n-2}(a).\operatorname {f} _{1}(b+k)\\\\+&{\frac {n-1}{1}}.{\frac {n-2}{2}}.\operatorname {f} _{n-3}(a).\operatorname {f} _{2}(b+k)\\+&\ldots \ldots \ldots \ldots \\+&{\frac {n-1}{1}}.{\frac {n-2}{2}}\operatorname {f} _{2}(a).\operatorname {f} _{n-3}(b+k)\\\\+&{\frac {n-1}{1}}.\operatorname {f} _{1}(a).\operatorname {f} _{n-2}(b+k)\\\\+&\operatorname {f} _{0}(a).\operatorname {f} _{n-1}(b+k)\end{array}}\right\}\,;