Page:Annales de mathématiques pures et appliquées, 1819-1820, Tome 10.djvu/256

248

DÉVELOPPEMENT

${\frac {\operatorname {d} x}{\operatorname {d} a}}={\frac {\operatorname {f} '(a+bz)}{1-b{\frac {\operatorname {d} z}{\operatorname {d} x}}\operatorname {f} '(a+bz)}}\,;$

on trouverait de la même manière

${\frac {\operatorname {d} x}{\operatorname {d} b}}={\frac {z\operatorname {f} '(a+bz)}{1-b{\frac {\operatorname {d} z}{\operatorname {d} x}}\operatorname {f} '(a+bz)}}.$

La comparaison de ces deux valeurs donne

${\frac {\operatorname {d} x}{\operatorname {d} b}}=z{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;$

comparant ensuite ce résultat avec les équations (1), il viendra

${\frac {\operatorname {d} U}{\operatorname {d} b}}=z{\frac {\operatorname {d} U}{\operatorname {d} a}}\,;\qquad$(8)

ce qui fera devenir l’équation (4)

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .zV{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}\,;\qquad$(9)

changeant ensuite $V$ en $z'',$ on aura, pour ce cas particulier,

${\frac {\operatorname {d} .z^{n}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .z^{n+1}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}.\qquad$(10)

L’équation (8) donnera donc, en vertu de cette dernière,

${\frac {\operatorname {d} ^{2}U}{\operatorname {d} b^{2}}}={\frac {\operatorname {d} .z{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .z^{2}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a}}$

d’où, on conclura, de la même manière,