Theorem 2

Tag 08I4, permanent. Cite it as [Bourbaki, Tag 08I4].

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

Let f be a function admitting a continuous $(n + 1)^{th}$ derivative on I, and belonging, with all its derivatives $f^{(m)}$ for $1 \leq m \leq n$ , to E. If U is a composition operator of order $p \leq n$ on E one has, for $x \geq 0$ and $h \geq 0$

f^{(p)}(x + h) = \sum_{m=0}^{n} \frac{1}{m!} u_m(x) U_h^\xi \left( f^{(m)}(\xi) \right) + R_n(x, h)

(27)

with

R_n(x, h) = -U_h^\xi \left( \int_0^{\xi - x - h} \frac{1}{n!} u_n(x + \eta) f^{(n+1)}(\xi - \eta) d\eta \right)

(28) (generalized Taylor expansion).

Consider the integral $\int_0^{\xi - x - h} \frac{1}{n!} u_n(x + \eta) f^{(n+1)}(\xi - \eta) d\eta$ , defined for all $\xi \in I$ , and apply to it the formula for integration by parts of order $n + 1$ (II, p. 59, formula (11)); taking account of the relations

u_n^{(k)} = n(n-1)\ldots(n-k+1)u_{n-k}

derived from (7) (VI, p. 273) by recursion, one obtains

\int_0^{\xi - x - h} \frac{1}{n!} u_n(x + \eta) f^{(n+1)}(\xi - \eta) d\eta

= \sum_{m=0}^{n} \frac{1}{m!} u_m(x) f^{(m)}(\xi) - \sum_{m=0}^{n} \frac{1}{m!} u_m(\xi - h) f^{(m)}(x + h).

(29)

We apply the operator $U$ to the two sides of (29), considered as functions of $\xi$ , then take the value of the function so obtained for the value $h$ of the variable $\xi$ ; remarking that, by the formulae (26) (VI, p. 277) and (9) (VI, p. 272), one has

U_h^\xi (u_m(\xi - h)) = U_0^\xi (u_m(\xi)) = \begin{cases} 0 & \text{for } m \neq p \\ p! & \text{for } m = p \end{cases}

one finally obtains (27).

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

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Theorem 2

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