Proposition 5

Tag 069X, permanent. Cite it as [Bourbaki, Tag 069X].

— For any $u \in \mathbf{A}[\mathbf{X}]$ and $\nu \in \mathbf{N}^{(I)}$ we have

D^\nu u = \nu! \; \Delta^\nu u

Suppose first that $\nu$ has length 1 ; then there exists an element $i$ of I such that $\nu = \varepsilon_i$ , that is, $\nu_i = 1$ and $\nu_j = 0$ for all $j \neq i$ in I. Formula (12) shows that $A''$ is a derivation of the A-algebra $\mathbf{A}[\mathbf{X}]$ which clearly has the value zero on $X_j$ for $j \neq i$ and the value 1 on $X_i$ . We thus have $\Delta^{\varepsilon_i} = D_i$ for each $i \in I$ .

By (13) we have

(14)

(\rho! \; \Delta^\rho) \cdot (\sigma! \; \Delta^\sigma) = (\rho + \sigma)! \; \Delta^{\rho + \sigma}

in the endomorphism algebra of the A-module $\mathbf{A}[\mathbf{X}]$ . It now follows by induction on the length of $\nu$ that $\nu! \; A'' = D^\nu$ .

If $\mathbf{A}$ is a Q-algebra, the formulae (9), (10), (11) may thus be written

(15)

u(\mathbf{X} + \mathbf{Y}) = \sum_{\nu} \frac{1}{\nu!} (D^\nu u)(\mathbf{X}) \mathbf{Y}^\nu

(16)

u(\mathbf{X}) = \sum_{\nu} \frac{1}{\nu!} (D^\nu u)(a)(\mathbf{X} - a)^\nu

(17)

u(\mathbf{X}) = \sum_{\nu} \frac{1}{\nu!} (D^\nu u)(0) \; X^\nu

The formulae (15), (16), (17) are all three called « Taylor's formula ».

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Lemma 2, § 1

Cites

Nothing.