Bourbaki

Proposition 6

Tag 08I2, permanent. Cite it as [Bourbaki, Tag 08I2].

Functions of a Real Variable, chapter VI, § 1. GENERALIZED TAYLOR EXPANSIONS

Let U=DpV=u(D)U = D^p V = u(D) be a composition operator of order pp on K[X]K[X], u(S)u(S) being a formal power series of order pp in K[[S]]K[[S]]. Then

UX(exp⁡(XS))=u(S)exp⁡(XS)(15) U_X \left( \exp(XS) \right) = u(S) \exp(XS) \tag{15}
Spu(S)exp⁡(XS)=∑n=0∞1n!un(X)Sn(16) \frac{S^p}{u(S)} \exp(XS) = \sum_{n=0}^{\infty} \frac{1}{n!} u_n(X) S^n \tag{16}
n being the Appell polynomial of index n attached to U.

By the scholium to th. 1 (VI, p. 271), to establish (15) it is enough to show that for every polynomial f(Y)∈K[Y]f(Y) \in K[Y] one has

UX(exp⁡(XDY)(f(Y)))=u(DY)(exp⁡(XDY)(f(Y))). U_X \left( \exp(XD_Y)(f(Y)) \right) = u(D_Y)(\exp(XD_Y)(f(Y))).

Now the first term in (17) is UX(f(X+Y))U_X(f(X + Y)), and, since U=u(D)U = u(D), the second term in (17) is UY(f(X+Y))U_Y(f(X + Y)), so the identity (17) reduces to (1) (VI, p. 270).

It now suffices to apply (15) to the composition operator V−1=Dp/u(D)V^{-1} = D^p / u(D) to obtain (16), since, by definition, one has

V−1(exp⁡(XS))=∑n=0∞1n!un(X)Sn. V^{-1}(\exp(XS)) = \sum_{n=0}^{\infty} \frac{1}{n!} u_n(X) S^n.

Note that (16) can also be obtained by multiplying the formal series Sp/u(S)S^p / u(S) and exp⁡(XS)\exp(XS), taking account of (6).

One says that the formal series (16) is the generating series of the Appell polynomials attached to UU.

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

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Cites

Transcribed from the printed volume. Not translated.