Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/39

35

DES FONCTIONS POLYNOMES.

Mettant ensuite dans ${\frac {\operatorname {d} u}{\operatorname {d} x}}$ les valeurs de ${\frac {\operatorname {d} u}{\operatorname {d} y}},x{\frac {\operatorname {d} u}{\operatorname {d} y}},x^{2}{\frac {\operatorname {d} u}{\operatorname {d} y}},\ldots$ données par les équations (1), on trouve

${\frac {\operatorname {d} u}{\operatorname {d} x}}=a_{1}{\frac {\operatorname {d} u}{\operatorname {d} a}}+2a_{2}{\frac {\operatorname {d} u}{\operatorname {d} a_{1}}}+3a_{3}{\frac {\operatorname {d} u}{\operatorname {d} a_{2}}}+4a_{4}{\frac {\operatorname {d} u}{\operatorname {d} a_{3}}}+\ldots \quad$(3)

Posons, pour abréger,

$u=A,\ {\frac {\operatorname {d} u}{\operatorname {d} x}}=A_{1},\ {\frac {1}{2!}}{\frac {\operatorname {d} ^{2}u}{\operatorname {d} x^{2}}}=A_{2},\ {\frac {1}{3!}}{\frac {\operatorname {d} ^{3}u}{\operatorname {d} x^{3}}}=A_{3},\ldots {\frac {1}{n!}}{\frac {\operatorname {d} ^{n}u}{\operatorname {d} x^{n}}}=A_{n}.\ldots$

La première des équations (2) donne

${\frac {\operatorname {d} .{\frac {\operatorname {d} ^{n-\alpha }u}{\operatorname {d} x^{n-\alpha }}}}{\operatorname {d} a_{n}}}={\frac {\operatorname {d} ^{n-\alpha }.\left(x^{n}{\frac {\operatorname {d} u}{\operatorname {d} a}}\right)}{\operatorname {d} x^{n-\alpha }}}=0\,;$

pourvu qu’après la différentiation on suppose $x=0.$ Partant

${\frac {\operatorname {d} A_{n-\alpha }}{\operatorname {d} a_{n}}}=0.\qquad$(4)

On tirerait de même de la dernière équation (2)

${\frac {\operatorname {d} A_{n+p}}{\operatorname {d} a_{n}}}={\frac {\operatorname {d} A_{k+p}}{\operatorname {d} a_{k}}}\,;\qquad$(5)

et enfin de l’équation (3)

$nA_{n}=a_{1}{\frac {\operatorname {d} A_{n-1}}{\operatorname {d} a}}+2a_{2}{\frac {\operatorname {d} A_{n-1}}{\operatorname {d} a_{1}}}+3a_{3}{\frac {\operatorname {d} A_{n-1}}{\operatorname {d} a_{2}}}+\ldots \qquad$(6)

équation qu’en vertu de l’équation de condition (5) qu’on peut changer en celle-ci,