ANALISE TRANSCENDANTE.
Recherches d’analise, relatives au développement
des fonctions ;
Par
M. Frédéric Sarrus, professeur de mathématiques
au collége de Pezenas ;
≈≈≈≈≈≈≈≈≈
Soit l’équation
![{\displaystyle x=\operatorname {f} (a,b,c),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/237b46782e44c20d8ecc295331b85176a9a0d45f)
et soient
des fonctions entièrement arbitraires de
sans
on aura
![{\displaystyle \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}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcfe5930296bd8963c879d5d07739b1e2283e3ec)
les équations des deux couples donnent, par division,
![{\displaystyle {\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}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af94ac82a4cd8ca058569d63e21f32924b2533d8)
et en égalant ces deux valeurs
![{\displaystyle {\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a6cef8b09156751dfbbdf6950d5cc437bb7e7a0)
L’on a encore
![{\displaystyle {\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}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f872069ce86dc12f508ac171fa6da4ba0d9cfa6c)
![{\displaystyle {\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}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04c04cc92d9d3b9254056a1061a1d8c2bcee6968)
mais, en vertu de l’équation (3),
![{\displaystyle 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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad17a4807cfdaa0e009751ca28edfee718b0a629)
et
![{\displaystyle \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}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/befe6f0d743b62452b5877a6cac8cd22ff0dd480)
donc finalement
![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/693a05eac86c84b3bc3346daa1a1323fffa73ee1)
(4)
et l’on aurait semblablement
![{\displaystyle {\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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eff4aea4c49b62b5c42bf415597c32d7dfd2150f)
Si l’on avait
![{\displaystyle x=\operatorname {f} (a,b,x,x'),\qquad x_{1}=\operatorname {f_{1}} (a_{1},b_{1},x,x_{1})\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3551a8c93ad7c0a26164c03719a3f80303540e3d)
et que
fussent des fonctions quelconques de
, sans
on trouverait, comme dans ce qui précède,
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a54ad6f94e3e2ea1094a7924c6ea9cd822fd5734)
(5)
d’où on déduirait
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f89994d8ed79e43e11a145de120dbd44c9542981)
(6)
et aussi
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f500a306fc45117d70c767c33a8fd1893d3a8746)
Les équations (4, 6) peuvent être utilement employées au développement de certaines fonctions. Soit, par exemple,
à développer suivant les puissances de
lorsque
est donnée par l’équation
![{\displaystyle x=\operatorname {f} (a+bz)\,;\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3df3275d6feb6af2fba3a992db356259992e5cf1)
(7)
désignant une fonction quelconque de
sans
ni
Nous déduirons d’abord de l’équation (7)
![{\displaystyle {\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)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7a0025733eb6091a1dc9c5fe3aabf350107fd2a)
désignant ici la fonction prime de
cela donnera
![{\displaystyle {\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)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e00db89d61f0ccb5d1522b239e46018408bcf5b3)
on trouverait de la même manière
![{\displaystyle {\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)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8507ad70240862f376ed7dc33565b810d7dd47)
La comparaison de ces deux valeurs donne
![{\displaystyle {\frac {\operatorname {d} x}{\operatorname {d} b}}=z{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16e142035dbea7bc0677f3445fb1839d7b75e791)
comparant ensuite ce résultat avec les équations (1), il viendra
![{\displaystyle {\frac {\operatorname {d} U}{\operatorname {d} b}}=z{\frac {\operatorname {d} U}{\operatorname {d} a}}\,;\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/01046aea6c4e98b58a7f1ea0f2573c0d06e1a2c1)
(8)
ce qui fera devenir l’équation (4)
![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e748f664e26bb775bc110c5a196fc1fe97cd205)
(9)
changeant ensuite
en
on aura, pour ce cas particulier,
![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/04ac08aaebf51d3e16a168b4811d4ea3616af969)
(10)
L’équation (8) donnera donc, en vertu de cette dernière,
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/786ce60c523670d2e7c38be81a726a26d1ce83ef)
d’où, on conclura, de la même manière,
![{\displaystyle {\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}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/184113df869bdf9219ac1592484048dc1e2a4de4)
et de celle-là
![{\displaystyle {\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}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d17ca1742a0c012bb30c97e66d2484b0c4f5abb6)
et ainsi de suite ; de manière qu’on aura, en général
![{\displaystyle {\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cd88b16147819596b99a9c819d03e76203f9eb0)
(11)
Une fois parvenu à la formule (11), le développement de
suivant les puissances de
ne présente plus de difficulté.
Soit encore
![{\displaystyle x=\operatorname {f} (a+bz)\,;\qquad x_{1}=\operatorname {f_{1}} (a_{1}+b_{1}z_{1})\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/017f9aae670c60360f8b3d728ef14dcfd07572a3)
étant des fonctions quelconques de
sans
On trouvera, par la différentiation,
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/674d08c35d5e0e3043d71226fcc85c78717b7403)
En posant, pour abréger,
![{\displaystyle 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})}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5c872f1c21358af8ab6eb7e41e1b192c0d20eb2)
la dernière de ces équations donne
![{\displaystyle {\frac {\operatorname {d} x_{1}}{\operatorname {d} a}}=k{\frac {\operatorname {d} x}{\operatorname {d} a}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25151f7a18af9ef19a42ecda9e195da8d503cbd6)
d’où on conclut, au moyen de la première,
![{\displaystyle {\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)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13485bb58ef760fd30c26376376434cea9b04da1)
On trouvera semblablement
![{\displaystyle {\frac {\operatorname {d} x_{1}}{\operatorname {d} b}}=k{\frac {\operatorname {d} x}{\operatorname {d} b}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45f6eb8c5ae90013a219079910b346bd0d0719c3)
![{\displaystyle {\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)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/149304c049efc9949d8a9e51c145a5e628f816ce)
Ces équations, comparées aux précédentes, donnent
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2ee71ad70a1f2abf44565e7e39e361f0493479c)
(12)
En suivant la marche qui a conduit à celles-ci, nous trouverons semblablement
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b994ac48269a2be6d352ab3258295717c0bfe3aa)
(13)
En observant donc que
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a35193b744c51f0bf40f0e62b3e011ab39acd55d)
nous aurons
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/708fa7e6236c222f8f38d39b1cb6c4a214fc7948)
(14)
Enfin, en suivant la marche qui nous a conduit des équations (4, 8) à la formule (11), nous parviendrons aux suivantes,
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f91b0d3e3bf32d923a96f6f1e37bb72aa6970ba)
(15)
Pour compléter cette recherche, il faut parvenir à l’expression de
![{\displaystyle {\frac {\operatorname {d} ^{m+m_{1}}U}{\operatorname {d} b^{m}\operatorname {d} b_{1}^{m_{1}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6c05d6087d43d0b725858615f327190c92ac8ff)
Pour cela, soit
![{\displaystyle z^{m}{\frac {\operatorname {d} U}{\operatorname {d} a}}={\frac {\operatorname {d} .pU}{\operatorname {d} b}}+{\frac {\operatorname {d} .qU}{\operatorname {d} a}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f5e2efdb028711664a99b92e52db002a574ec41)
étant des fonctions de
, satisfaisant, pour
aux équations (15), du moins en ce qui concerne
Avec cette restriction, l’on trouvera, en développant
![{\displaystyle 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)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/029d00818ba929cd5b17489a050b91fccd13a5fb)
et, comme il suffit de satisfaire à cette equation, nous poserons
![{\displaystyle z{\frac {\operatorname {d} p}{\operatorname {d} a}}+{\frac {\operatorname {d} q}{\operatorname {d} a}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2767837dbf3e285655980eb0349dcf72ccab8a8b)
![{\displaystyle zp+q=z^{m}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95f822da48df8a48cfcdc425ec923e18207fe4a3)
différentiant la seconde par rapport à
elle se réduira, en vertu de la première, à
![{\displaystyle p{\frac {\operatorname {d} z}{\operatorname {d} a}}=mz^{m-1}{\frac {\operatorname {d} z}{\operatorname {d} a}},\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/098c5d6555970be6a5b4d146cabba250d4029800)
d’où
![{\displaystyle \quad p=mz^{m-1}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31bc5dd454637eafc42561fa97407396bc385082)
mettant cette valeur de
dans la seconde des équations précédentes ; on trouvera pour ![{\displaystyle q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06809d64fa7c817ffc7e323f85997f783dbdf71d)
![{\displaystyle q=-(m-1)z^{m}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25a60425c2fe5cb7440885b8d78d53c51086a15b)
et l’équation (16) deviendra
![{\displaystyle 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}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0667542ca481bd828ebcac3b4468c5f21f5e45a4)
d’où on conclura
![{\displaystyle {\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}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4030d874113a10072d03082c8bcd8e75dc02c16b)
mais, en vertu des équations (15), on a
![{\displaystyle {\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}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1af45dde7bc8ccf4d59c83b230ba0185728b7fa)
et encore
![{\displaystyle {\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}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ef63f6f9a62b5e5cb6a66bd67f58889cdd67a65)
ainsi l’on aura
![{\displaystyle {\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}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f86032daef409fb0ea0c56edb6d816a4a40d480)
en observant que la quantité dans les parenthèses peut être mise sous cette forme
![{\displaystyle 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}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64e57409b4de1173efb6a4607dea3c296a9dcf39)
il viendra finalement
![{\displaystyle {\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}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a4ae20f4857eda88c2e74d4003d964a7e76cb0a)
mais on a
![{\displaystyle {\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}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc4d4e9545a72153aa175ae5b2e535b2fd148ec5)
partant
![{\displaystyle {\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}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a029e589340c27f87ec667e84221b66115c8f03e)
(17)
Une fois parvenus à la formule (17), le développement de
suivant les puissances et produits des puissances de
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
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c010ac8cdd05a0cfc048f075d77e9a4037d77206)
où
sont supposés des fonctions quelconques de
sans
il s’agissait de développer
suivant les puissances et produits de puissances de
étant lui-même une fonction de
sans
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
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.