Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/282

274

THÉORÈME

 Les corrections sont expliquées en page de discussion

mais, en vertu de nos définitions (2 et 4), on a

${\begin{array}{rlrl}a\operatorname {f} _{n-1}(a+k)&=\operatorname {f} _{n}(a),&&b\operatorname {f} _{0}(b+k)&=\operatorname {f} _{1}(b),\\a\operatorname {f} _{n-2}(a+k)&=\operatorname {f} _{n-1}(a),&&b\operatorname {f} _{1}(b+k)&=\operatorname {f} _{2}(b),\\a\operatorname {f} _{n-3}(a+k)&=\operatorname {f} _{n-2}(a),&&b\operatorname {f} _{2}(b+k)&=\operatorname {f} _{3}(b),\\\ldots &\ldots \ldots &&\ldots \ldots \ldots &\ldots \\a\operatorname {f} _{2}(a+k)&=\operatorname {f} _{3}(a),&&b\operatorname {f} _{n-3}(b+k)&=\operatorname {f} _{n-2}(b),\\a\operatorname {f} _{1}(a+k)&=\operatorname {f} _{2}(a),&&b\operatorname {f} _{n-2}(b+k)&=\operatorname {f} _{n-1}(b),\\a\operatorname {f} _{0}(a+k)&=\operatorname {f} _{1}(a),&&b\operatorname {f} _{n-1}(b+k)&=\operatorname {f} _{n}(b)\,;\end{array}}$

substituant donc, il viendra

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

observant alors que

$1+{\frac {n-1}{1}}={\frac {n}{1}}$

${\frac {n-1}{1}}+{\frac {n-1}{1}}.{\frac {n-2}{2}}={\frac {n-1}{2}},$