Annales de mathématiques pures et appliquées/Tome 10/Analise transcendante, article 5

Recherches d’analise relatives au développement des fonctions ; par M. Sarrus.

Annales de mathématiques pures et appliquées, 10, 1820 (p. 245-254).

ANALISE TRANSCENDANTE.

Recherches d’analise, relatives au développement
des fonctions ;

Par M. Frédéric Sarrus, professeur de mathématiques
au collége de Pezenas ;

≈≈≈≈≈≈≈≈≈

Soit l’équation

$x=\operatorname {f} (a,b,c),$

et soient $U,V$ des fonctions entièrement arbitraires de $x$ sans $a,b\,;$ on aura

$\left.{\begin{aligned}&{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} U}{\operatorname {d} b}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} b}}\,;\end{aligned}}\right\}{\text{(1}})\qquad \left.{\begin{aligned}&{\frac {\operatorname {d} V}{\operatorname {d} a}}={\frac {\operatorname {d} V}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} V}{\operatorname {d} b}}={\frac {\operatorname {d} V}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} b}}\,;\end{aligned}}\right\}{\text{(2}})$

les équations des deux couples donnent, par division,

${\frac {\frac {\operatorname {d} x}{\operatorname {d} a}}{\frac {\operatorname {d} x}{\operatorname {d} b}}}={\frac {\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} U}{\operatorname {d} b}}},\qquad {\frac {\frac {\operatorname {d} x}{\operatorname {d} a}}{\frac {\operatorname {d} x}{\operatorname {d} b}}}={\frac {\frac {\operatorname {d} V}{\operatorname {d} a}}{\frac {\operatorname {d} V}{\operatorname {d} b}}},$

et en égalant ces deux valeurs

${\frac {\operatorname {d} U}{\operatorname {d} a}}{\frac {\operatorname {d} V}{\operatorname {d} b}}={\frac {\operatorname {d} U}{\operatorname {d} b}}{\frac {\operatorname {d} V}{\operatorname {d} a}}.\qquad (3)$

L’on a encore

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} b\operatorname {d} a}}+{\frac {\operatorname {d} V}{\operatorname {d} b}}{\frac {\operatorname {d} U}{\operatorname {d} a}},$

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b}}}{\operatorname {d} a}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} b}}+{\frac {\operatorname {d} V}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} b}}\,;$

mais, en vertu de l’équation (3),

$V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} b}}=V{\frac {\operatorname {d} ^{2}U}{\operatorname {d} b\operatorname {d} a}},\quad$et$\quad {\frac {\operatorname {d} V}{\operatorname {d} b}}{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} V}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} b}}\,;$

donc finalement

${\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b}}}{\operatorname {d} a}}\,;\qquad$(4)

et l’on aurait semblablement

${\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} b}}}{\operatorname {d} a}}.$

Si l’on avait

$x=\operatorname {f} (a,b,x,x'),\qquad x_{1}=\operatorname {f_{1}} (a_{1},b_{1},x,x_{1})\,;$

et que $U,V$ fussent des fonctions quelconques de $x,x_{1}$, sans $a,b,a_{1},b_{1},$ on trouverait, comme dans ce qui précède,

$\left.{\begin{aligned}&{\frac {\operatorname {d} U}{\operatorname {d} a}}{\frac {\operatorname {d} V}{\operatorname {d} b}}={\frac {\operatorname {d} U}{\operatorname {d} b}}{\frac {\operatorname {d} V}{\operatorname {d} a}}\\\\&{\frac {\operatorname {d} U}{\operatorname {d} a'}}{\frac {\operatorname {d} V}{\operatorname {d} b'}}={\frac {\operatorname {d} U}{\operatorname {d} b'}}{\frac {\operatorname {d} V}{\operatorname {d} a'}}\end{aligned}}\right\}\quad$(5)

d’où on déduirait

$\left.{\begin{aligned}&{\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b}}}{\operatorname {d} a}}\\\\&{\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} a'}}}{\operatorname {d} b'}}={\frac {\operatorname {d} .V{\frac {\operatorname {d} U}{\operatorname {d} b'}}}{\operatorname {d} a'}}\end{aligned}}\right\}\quad$(6)

et aussi

${\begin{aligned}&{\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} a}}}{\operatorname {d} b}}={\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} b}}}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} a'}}}{\operatorname {d} b'}}={\frac {\operatorname {d} .U{\frac {\operatorname {d} V}{\operatorname {d} b'}}}{\operatorname {d} a'}}.\end{aligned}}$

Les équations (4, 6) peuvent être utilement employées au développement de certaines fonctions. Soit, par exemple, $U$ à développer suivant les puissances de $b,$ lorsque $x$ est donnée par l’équation

$x=\operatorname {f} (a+bz)\,;\qquad$(7)

$z$ désignant une fonction quelconque de $x$ sans $a$ ni $b.$ Nous déduirons d’abord de l’équation (7)

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

$\operatorname {f} '$ désignant ici la fonction prime de $\operatorname {f} ,$ cela donnera

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

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,

${\frac {\operatorname {d} ^{3}U}{\operatorname {d} b^{3}}}={\frac {\operatorname {d} ^{2}.z^{2}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a\operatorname {d} b}}={\frac {\operatorname {d} ^{2}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}}}\,;$

et de celle-là

${\frac {\operatorname {d} ^{4}U}{\operatorname {d} b^{4}}}={\frac {\operatorname {d} ^{3}.z^{3}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{2}\operatorname {d} b}}={\frac {\operatorname {d} ^{3}.z^{4}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{3}}}\,;$

et ainsi de suite ; de manière qu’on aura, en général

${\frac {\operatorname {d} ^{m}U}{\operatorname {d} b^{m}}}={\frac {\operatorname {d} ^{m-1}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{m-1}}}.\qquad$(11)

Une fois parvenu à la formule (11), le développement de $U,$ suivant les puissances de $b$ ne présente plus de difficulté.

Soit encore

$x=\operatorname {f} (a+bz)\,;\qquad x_{1}=\operatorname {f_{1}} (a_{1}+b_{1}z_{1})\,;$

$z,z_{1}$ étant des fonctions quelconques de $x,x_{1}$ sans $a,b,a_{1},b_{1}.$ On trouvera, par la différentiation,

${\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} a}}=\left(1+b{\frac {\operatorname {d} z}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b{\frac {\operatorname {d} z}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f} '(a+bz),\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=\left(b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}\right)\operatorname {f_{1}} '(a_{1}+b_{1}z_{1}).\end{aligned}}$

En posant, pour abréger,

$k={\frac {b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}{1-b_{1}{\frac {\operatorname {d} z_{1}}{\operatorname {d} x_{1}}}\operatorname {f_{1}} '(a_{1}+b_{1}z_{1})}},$

la dernière de ces équations donne

${\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=k{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;$

d’où on conclut, au moyen de la première,

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

On trouvera semblablement

${\frac {\operatorname {d} x_{1}}{\operatorname {d} b}}=k{\frac {\operatorname {d} x}{\operatorname {d} b}},$

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

Ces équations, comparées aux précédentes, donnent

$\left.{\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} b}}=z{\frac {\operatorname {d} x}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} b}}=z{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}.\end{aligned}}\right\}\quad$(12)

En suivant la marche qui a conduit à celles-ci, nous trouverons semblablement

$\left.{\begin{aligned}&{\frac {\operatorname {d} x}{\operatorname {d} b_{1}}}=z_{1}{\frac {\operatorname {d} x}{\operatorname {d} a_{1}}},\\\\&{\frac {\operatorname {d} x_{1}}{\operatorname {d} b_{1}}}=z_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a_{1}}}.\end{aligned}}\right\}\quad$(13)

En observant donc que

${\begin{aligned}&{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a}}+{\frac {\operatorname {d} U}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} U}{\operatorname {d} b}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} b}}+{\frac {\operatorname {d} U}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} b}},\\\\&{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} a_{1}}}+{\frac {\operatorname {d} U}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} a_{1}}},\\\\&{\frac {\operatorname {d} U}{\operatorname {d} b_{1}}}={\frac {\operatorname {d} U}{\operatorname {d} x}}{\frac {\operatorname {d} x}{\operatorname {d} b_{1}}}+{\frac {\operatorname {d} U}{\operatorname {d} x_{1}}}{\frac {\operatorname {d} x_{1}}{\operatorname {d} b_{1}}},\end{aligned}}$

nous aurons

$\left.{\begin{aligned}&{\frac {\operatorname {d} U}{\operatorname {d} b}}=z{\frac {\operatorname {d} U}{\operatorname {d} a}},\\\\&{\frac {\operatorname {d} U}{\operatorname {d} b_{1}}}=z_{1}{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}}.\end{aligned}}\right\}\quad$(14)

Enfin, en suivant la marche qui nous a conduit des équations (4, 8) à la formule (11), nous parviendrons aux suivantes,

$\left.{\begin{aligned}&{\frac {\operatorname {d} ^{m}U}{\operatorname {d} b^{m}}}={\frac {\operatorname {d} ^{m-1}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{m-1}}},\\\\&{\frac {\operatorname {d} ^{m_{1}}U}{\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m_{1}-1}.z_{1}^{m_{1}}{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}}}{\operatorname {d} a_{1}^{m_{1}-1}}}.\end{aligned}}\right\}\quad$(15)

Pour compléter cette recherche, il faut parvenir à l’expression de

${\frac {\operatorname {d} ^{m+m_{1}}U}{\operatorname {d} b^{m}\operatorname {d} b_{1}^{m_{1}}}}.$

Pour cela, soit

$z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} .pU}{\operatorname {d} b}}+{\frac {\operatorname {d} .qU}{\operatorname {d} a}}\,;$

$p,q$ étant des fonctions de $x,x_{1}$, satisfaisant, pour $U,$ aux équations (15), du moins en ce qui concerne $p.$ Avec cette restriction, l’on trouvera, en développant

$z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}=U\left(z{\frac {\operatorname {d} p}{\operatorname {d} a}}+{\frac {\operatorname {d} q}{\operatorname {d} a}}\right)+{\frac {\operatorname {d} U}{\operatorname {d} a}}(zp+q)\,;$

et, comme il suffit de satisfaire à cette equation, nous poserons

$z{\frac {\operatorname {d} p}{\operatorname {d} a}}+{\frac {\operatorname {d} q}{\operatorname {d} a}}=0,$

$zp+q=z^{m}\,;$

différentiant la seconde par rapport à $a,$ elle se réduira, en vertu de la première, à

$p{\frac {\operatorname {d} z}{\operatorname {d} a}}=mz^{m-1}{\frac {\operatorname {d} z}{\operatorname {d} a}},\quad$d’où$\quad p=mz^{m-1}\,;$

mettant cette valeur de $p$ dans la seconde des équations précédentes ; on trouvera pour $q$

$q=-(m-1)z^{m}\,;$

et l’équation (16) deviendra

$z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}=m{\frac {\operatorname {d} .z^{m-1}U}{\operatorname {d} b}}-(m-1){\frac {\operatorname {d} .z^{m}U}{\operatorname {d} a}}\,;$

d’où on conclura

${\frac {\operatorname {d} _{1}^{m}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b_{1}m_{1}}}=m\operatorname {d} .{\frac {\frac {\operatorname {d} _{1}^{m}.z^{m-1}U}{\operatorname {d} b_{1}^{m_{1}}}}{\operatorname {d} b}}-(m-1)\operatorname {d} .{\frac {\frac {\operatorname {d} _{1}^{m}.z^{m}U}{\operatorname {d} b_{1}^{m_{1}}}}{\operatorname {d} a}}\,;$

mais, en vertu des équations (15), on a

${\frac {\operatorname {d} _{1}^{m}.z^{m-1}U}{\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m_{1}-1}.z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m-1}U}{\operatorname {d} a_{1}}}}{\operatorname {d} a_{1}^{m_{1}-1}}},$

et encore

${\frac {\operatorname {d} _{1}^{m}.z^{m}U}{\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m_{1}-1}.z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m}U}{\operatorname {d} a_{1}}}}{\operatorname {d} a_{1}^{m_{1}-1}}}\,;$

ainsi l’on aura

${\frac {\operatorname {d} _{1}^{m}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m_{1}-1}\left(m{\frac {\operatorname {d} .z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m-1}U}{\operatorname {d} a_{1}}}}{\operatorname {d} b}}-(m-1){\frac {\operatorname {d} .z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m}U}{\operatorname {d} a_{1}}}}{\operatorname {d} a}}\right)}{\operatorname {d} a_{1}^{m_{1}-1}}}\,;$

en observant que la quantité dans les parenthèses peut être mise sous cette forme

$z^{m}z_{1}^{m_{1}}{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} a_{1}}}+z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m}}{\operatorname {d} a_{1}}}{\frac {\operatorname {d} U}{\operatorname {d} a}}+z^{m}{\frac {\operatorname {d} .z_{1}^{m_{1}}}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}},$

il viendra finalement

${\frac {\operatorname {d} _{1}^{m}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m_{1}-1}\left(z^{m}z_{1}^{m_{1}}{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} a_{1}}}+z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m}}{\operatorname {d} a_{1}}}{\frac {\operatorname {d} U}{\operatorname {d} a}}+z^{m}{\frac {\operatorname {d} .z_{1}^{m_{1}}}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}}\right)}{\operatorname {d} a_{1}^{m_{1}}}}\,;$

mais on a

${\frac {\operatorname {d} ^{m+m_{1}}U}{\operatorname {d} b^{m}\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m+m_{1}-1}.z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}}{\operatorname {d} a^{m-1}\operatorname {d} b_{1}^{m_{1}}}}\,;$

partant

${\frac {\operatorname {d} ^{m+m_{1}}U}{\operatorname {d} b^{m}\operatorname {d} b_{1}^{m_{1}}}}={\frac {\operatorname {d} ^{m+m_{1}-2}.\left(z^{m}z_{1}^{m_{1}}{\frac {\operatorname {d} ^{2}U}{\operatorname {d} a\operatorname {d} a_{1}}}+z_{1}^{m_{1}}{\frac {\operatorname {d} .z^{m}}{\operatorname {d} a_{1}}}{\frac {\operatorname {d} U}{\operatorname {d} a}}+z^{m}{\frac {\operatorname {d} .z_{1}^{m_{1}}}{\operatorname {d} a}}{\frac {\operatorname {d} U}{\operatorname {d} a_{1}}}\right)}{\operatorname {d} a^{m-1}\operatorname {d} a_{1}^{m_{1}-1}}}.$(17)

Une fois parvenus à la formule (17), le développement de $U,$ suivant les puissances et produits des puissances de $b,b_{1}$ ne saurait plus offrir aucune difficulté.

Il est aisé de voir, d’après ce qui précède, ce qu’on aurait à faire, si, ayant les trois équations

${\begin{aligned}&x\ =\operatorname {f} \ \,(a\ \,+b\ \,z\ \,),\\&x_{1}=\operatorname {f_{1}} (a_{1}+b_{1}z_{1}),\\&x_{2}=\operatorname {f_{2}} (a_{2}+b_{2}z_{2})\,;\end{aligned}}$

où $z,z_{1},z_{2}$ sont supposés des fonctions quelconques de $x,x_{1},x_{2}$ sans $a,b,a_{1},b_{1},a_{2},b_{2},$ il s’agissait de développer $U,$ suivant les puissances et produits de puissances de $b,b_{1},b_{2}\,;$ $U$ étant lui-même une fonction de $x,x_{1},x_{2}$ sans $a,b,a_{1},b_{1},a_{2},b_{2}\,;$ et il en serait de même pour un plus grand nombre d’équations entre un pareil nombre de fonctions.

Nous terminerons par observer que M. Laplace était depuis longtemps parvenu à la formule (17) et à ses analogues ; mais seulement dans la supposition où $b,b_{1}$ devaient, après les différentiations, être faits égaux à zéro ; ce n’est même qu’en admettant cette hypothèse, que M. Lacroix est parvenu aux équations (14) ; voyez son Traité de calcul différentiel et de calcul intégral, deuxième édition, tom. I, pag. 281.