Corollary (Leibniz's formula)

Tag 066I, permanent. Cite it as [Bourbaki, Tag 066I].

Let $d_i$ ( $1 \leq i \leq n$ ) be $n$ derivations of $(A, A', A'')$ which are permutable with one another. For all $\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbf{N}^n$ , we write

d^\alpha = d_1^{\alpha_1} d_2^{\alpha_2} \ldots d_n^{\alpha_n}.

Then, for $x \in \mathbf{A}$ and $y \in \mathbf{A}'$ ,

d^\alpha(xy) = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) d^\beta(x)d^\gamma(y)

where we have written (in the notation introduced at the beginning of the chapter)

((\beta, \gamma)) = (\beta + \gamma)!/(\beta! \gamma!).

This follows immediately from the multinomial formula (I, § 8, no. 2)

(T + T')^\alpha = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) T^\beta {T'}^\gamma

and Proposition 2.

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

Cited by

Nothing.

Cites

I, § 8, no. 2
Proposition 2