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DÉVELOPPEMENT EN SÉRIES
Soit
une fonction de
qu’il soit question de développer suivant
les puissances de
étant donnée par l’équation
![{\displaystyle y=a+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8096a0ea7bbebc3e65ec8db136d598bf1e2a92cc)
On aura d’abord cette suite d’équations
![{\displaystyle {\frac {\operatorname {d} u}{\operatorname {d} x}}={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} x}}={\frac {\operatorname {d} u}{\operatorname {d} y}}\left(a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+\ldots \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5f327a659fb7b8b4dc340dac40eaf6c71022be0)
![{\displaystyle \left.{\begin{array}{lll}{\frac {\operatorname {d} u}{\operatorname {d} a}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a}}&=\quad {\frac {\operatorname {d} u}{\operatorname {d} y}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{1}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{1}}}&=x\ \ {\frac {\operatorname {d} u}{\operatorname {d} y}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{2}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{2}}}&=x^{2}{\frac {\operatorname {d} u}{\operatorname {d} y}},\\\ldots &\ldots \ldots &\ldots \ldots \\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&={\frac {\operatorname {d} u}{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} a_{n}}}&=x^{n}{\frac {\operatorname {d} u}{\operatorname {d} y}}\,;\end{array}}\right\}(1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0d3274421a6e0bed6549aa1a2e57ef0ee3131d8)
dans lesquelles
sont traitées comme indépendantes.
Éliminant
entre chacune des équations (1) et la dernière, on trouve
![{\displaystyle \left.{\begin{array}{rl}{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n}\quad {\frac {\operatorname {d} u}{\operatorname {d} a}},\\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n-1}{\frac {\operatorname {d} u}{\operatorname {d} a_{1}}},\\\ldots &\ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} u}{\operatorname {d} a_{n}}}&=x^{n-k}{\frac {\operatorname {d} u}{\operatorname {d} a_{k}}}.\end{array}}\right\}(2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51eae57771c5384b2acd27a4440e41d5ece841cf)