§ 1. Generalized Taylor expansions

1. Composition operators on an algebra of polynomials

Let K be a commutative field of characteristic 0, and K[X] the algebra of polynomials in one indeterminate over K (Alg., IV. 1); throughout this section by an operator on K[X] we shall mean a linear map U of the vector space K[X] (over K) into itself; since the monomials $X^n \ (n \geqslant 0)$ form a basis for this space, U is determined by the polynomials $U(X^n)$ ; specifically, if $f(X) = \sum_{k=0}^\infty \lambda_k X^k$ with $\lambda_k \in K$ , then

U(f) = \sum_{k=0}^\infty \lambda_k U(X^k).

If G is a commutative algebra over K, with an identity element, the G-module G[X] is obtained by extending the field of scalars K of the vector space K[X] to G; every operator U on K[X] extends in a unique manner to a linear map of the G-module G[X] to itself, which we again denote by U (Alg., II, p. 278); for each element

g(X) = \sum_{k=0}^\infty \gamma_k X^k

, with

\gamma_k \in G

, one has

U(g) = \sum_{k=0}^\infty \gamma_k U(X^k)

Consider in particular the case where G = K[Y]; then G[X] is the ring K[X,Y] of polynomials in two indeterminates over K; to avoid confusion we denote the extension of U to G[X] by $U_X$ . For any polynomial $g(X, Y) = \sum_{k=0}^\infty \gamma_k(Y) X^k$ where $\gamma_k(Y) \in K[Y]$ one thus has $U_X(g) = \sum_{k=0}^\infty \gamma_k(Y) U(X^k)$ . Since $U_X$ is linear one sees that if one writes $g(X, Y) = \sum_{h=0}^\infty \beta_h(X) Y^h$ then one also has $U_X(g) = \sum_{h=0}^\infty U(\beta_h) Y^h$ .

Under the canonical isomorphism of K[X] onto K[Y] which associates Y with X the operator U transforms to an operator on K[Y], which we will denote $U_Y$ , to avoid confusion, so that $U_Y(f(Y))$ is the polynomial obtained by replacing X by Y in the polynomial $U(f(X)) = U_X(f(X))$ . This operator $U_Y$ can in turn be extended to an operator (again denoted by $U_Y$ ) on K[X,Y]: if $g(X, Y) = \sum_{h=0}^\infty \beta_h(X) Y^h$ then

U_Y(g(X, Y)) = \sum_{h=0}^\infty \beta_h(X) U_Y(Y^h).

As an example of these extensions we cite the derivation operator D on K[X] (Alg., IV. 6), which gives the partial derivation operators D_X and D_Y on K[X,Y].

For any polynomial $f \in K[X]$ we denote by $T_Y(f)$ the polynomial $f(X + Y)$ in K[X,Y]; the map $T_Y$ is a K-linear map from K[X] into K[X,Y], called a translation operator.

Definition 1

One says that an operator U on K[X] is a composition operator if it commutes with the translation operator, that is, if $U_X T_Y = T_Y U$ .

In other words, if $f$ is an arbitrary polynomial in K[X], and if $g = U(f)$ , one must have $g(X + Y) = U_X(f(X + Y))$ .

It follows immediately from this definition that for every polynomial $f(X) \in K[X]$ one has, in the above notation,

U_X(f(X + Y)) = U_Y(f(X + Y)).

(1)

Example 1

For every $\lambda \in K$ the operator which to every polynomial $f(X)$ associates the polynomial $f(X + \lambda)$ is a composition operator.

Example 2

The derivation D on K[X] is a composition operator (cf. prop. 1).

Remark

Since K is an infinite field the operator U on K[X] is a composition operator if and only if for every polynomial $f \in K[X]$ and every element $\alpha \in K$ one has, putting $g = U(f)$ , that $g(X + \alpha) = U(f(X + \alpha))$ . (Alg., IV. 16, cor.).

It is clear that every linear combination of composition operators, with coefficients in K, is a composition operator; the same is true for the composition of two composition operators. In other words, the composition operators form a subalgebra $\Gamma$ of the algebra of endomorphisms of the vector space K[X].

Proposition 1

For an operator U on K[X] to be a composition operator it is necessary and sufficient that it commute with the derivation D on K[X].

Indeed, the Taylor formula shows that for every polynomial $f \in K[X]$ one has

U_X(f(X + Y)) = U_X \left( \sum_{k=0}^{\infty} \frac{1}{k!} Y^k D^k f(X) \right) = \sum_{k=0}^{\infty} \frac{1}{k!} Y^k U(D^k f(X));

if one puts $g = U(f)$ then one has

g(X + Y) = \sum_{k=0}^{\infty} \frac{1}{k!} Y^k D^k g(X) = \sum_{k=0}^{\infty} \frac{1}{k!} Y^k D^k (U(f(X)));

for U to be a composition operator, one must then have $UD^k = D^k U$ for every integer $k \geq 1$ , and in particular $UD = DU$ . Conversely, if this relation holds, it implies $UD^k = D^k U$ for every integer $k \geq 1$ , by induction on $k$ ; the Taylor formula then shows that $g(X + Y) = U_X(f(X + Y))$ .

For every polynomial $f \in K[X, Y]$ we denote by $U_0(f)$ the term independent of X in the polynomial $U_X(f)$ ; in particular, if $f \in K[X]$ , $U_0(f)$ is the constant term in $U(f)$ , and $U_0$ is a linear form on $K[X]$ . For every polynomial $f \in K[X]$ let $g = U(f)$ ; by def. 1 of VI, p. 270

g(X + Y) = U_X(f(X + Y)) = U_X \left( \sum_{k=0}^{\infty} \frac{1}{k!} X^k D^k f(Y) \right) = \sum_{k=0}^{\infty} \frac{1}{k!} U(X^k) D^k f(Y)

and if, in this formula, one replaces X by 0, one obtains

g(Y) = \sum_{k=0}^{\infty} \frac{1}{k!} U_0(X^k) D^k f(Y).

Thus one sees that

U(f(X)) = \sum_{k=0}^{\infty} \frac{1}{k!} \mu_k D^k f(X) \tag{2}

where $\mu_k$ is the constant term in the polynomial $U(X^k)$ .

This formula shows that the $\mu_k$ determine the composition operator $U$ completely; conversely, if $(\mu_n)$ is an arbitrary sequence of elements of K then the formula (2) defines an operator $U$ which clearly commutes with D, so (VI, p. 270, prop. 1) is a composition operator. From now on we shall write (2) in the form

U = \sum_{k=0}^{\infty} \frac{1}{k!} \mu_k D^k. \tag{3}

This formula can be interpreted in topological language as follows: if one considers the discrete topology on $K[X]$ , and the topology of simple convergence on the algebra $\mathrm{End}(K[X])$ of endomorphisms of $K[X]$ (Gen. Top., X, p. 277), the series with general term $\frac{1}{k!} \mu_k D^k$ is commutatively convergent in $\mathrm{End}(K[X])$ and has sum $U$ (Gen. Top., III, p. 269).

The formula (3) shows that to every formal series $u(S) = \sum_{k=0}^{\infty} \alpha_k S^k$ in one indeterminate over K (Alg., IV. 41) one can associate the composition operator $U = \sum_{k=0}^{\infty} \alpha_k D^k$ , which in future we shall denote by $u(D)$ . This remark can be clarified in the following manner:

Theorem 1

The map which to every formal series* $u(S) = \sum_{k=0}^{\infty} \alpha_k S^k$ in one indeterminate over K associates the composition operator $u(D) = \sum_{k=0}^{\infty} \alpha_k D^k$ on $K[X]$ is an isomorphism of the algebra $K[[S]]$ of formal series onto the algebra $\Gamma$ *of composition operators.

One verifies immediately that this map is an homomorphism. It remains to see that it is injective, in other words, that the relation $\sum_{k=0}^{\infty} \alpha_k D^k = 0$ implies $\alpha_k = 0$ for every $k$ ; now $h! \alpha_h$ is the constant term in the polynomial obtained by applying $\sum_{k=0}^{\infty} \alpha_k D^k$ to $X^h$ , whence the theorem.

Corollary

The algebra $\Gamma$ of composition operators in $K[X]$ is commutative.

Example

If $U$ is the operator which to each polynomial $f(X)$ associates $f(X + \lambda)$ (where $\lambda \in K$ ), one has $U_0(X^k) = \lambda^k$ , and so $U = \sum_{k=0}^{\infty} \frac{1}{k!} (\lambda D)^k$ . By analogy with the series expansion of $e^{\lambda}$ (III, p. 105) we write $e^S$ or $\exp(S)$ for the formal series $\sum_{n=0}^{\infty} \frac{1}{n!} S^n$ in the ring $K[[S]]$ ; so one can write $U = e^{\lambda D}$ . Replacing the field $K$ by the field of rational fractions $K(Y)$ in this argument one sees similarly that the translation operator $T_Y$ can be written $e^{YD}$ .

Furthermore, in the ring $K[[S, T]]$ of formal series in two indeterminates over $K$ one has

(\exp S)(\exp T) = \sum_{p,q} \frac{S^p}{p!} \frac{T^q}{q!}

= \sum_{n=0}^{\infty} \frac{1}{n!} \left( \binom{n}{0} S^n + \binom{n}{1} S^{n-1} T + \cdots + \binom{n}{n} T^n \right)

= \exp(S + T)

and in particular

(\exp S)(\exp(-S)) = 1

which justifies the notation we have introduced.

Scholium

The isomorphism of the algebra $K[[S]]$ of formal series and the algebra $\Gamma$ of composition operators on $K[X]$ sometimes allows one to prove propositions on formal series most simply, by proving them for the corresponding composition operators (cf. VI, p. 274, prop. 6).

2. Appell polynomials attached to a composition operator

Given a composition operator $U = \sum_{k=0}^{\infty} \alpha_k D^k \neq 0$ let $p$ be the least of the integers $k$ such that $\alpha_k \neq 0$ ; we shall say that $p$ is the order of the operator $U$ .

Proposition 2

Every composition operator of order 0 is invertible in the algebra $\Gamma$ of composition operators on $\mathbf{K}[X]$ .

Indeed, a formal series $\sum_{k=0}^{\infty} \alpha_k S^k$ such that $\alpha_0 \neq 0$ is invertible in the ring $\mathbf{K}[[S]]$ (Alg., IV. 41); the proposition thus follows from th. 1 of VI, p. 271.

Proposition 3

Let $U$ be a composition operator of order $p$ ; then $U(f) = 0$ for every polynomial $f$ of degree $< p$ ; for every polynomial $f \neq 0$ of degree $n \geq p$ , $U(f)$ is a polynomial $\neq 0$ of degree $n - p$ .

This is an immediate consequence of formula (2) of VI, p. 271 and of the definition of the order of $U$ .

It is clear that every operator $U$ of order $p$ can be written in a unique way as $U = D^p V = V D^p$ , where $V$ is an operator of order 0 (and so invertible).

Definition 2

Let $U = D^p V$ be a composition operator of order $p$ on $\mathbf{K}[X]$ . The polynomial $u_n(X) = V^{-1}(X^n)$ is called the Appell polynomial of index $n$ attached to the operator $U$ .

If $V^{-1} = \sum_{k=0}^{\infty} \frac{1}{k!} \beta_k D^k$ (with $\beta_0 \neq 0$ ) one thus has

u_n(X) = \sum_{k=0}^{n} \binom{n}{k} \beta_k X^{n-k}.

(6)

One verifies that $u_n$ is a polynomial of degree $n$ (prop. 3); further

u_n(0) = \beta_n.

Proposition 4

The Appell polynomials attached to $U$ satisfy the relations

\frac{d u_n}{d X} = n \, u_{n-1}

(7)

u_n(X + Y) = \sum_{k=0}^{n} \binom{n}{k} u_{n-k}(X) Y^k

(8)

U(u_n(X)) = \frac{n!}{(n-p)!} X^{n-p}.

(9)

These formulae are in fact respectively equivalent to the following relations (bearing def. 2 in mind):

DV^{-1} = V^{-1}D \tag{10}

(\exp(YD_X))V_X^{-1} = V_X^{-1}\exp(YD_X) \tag{11}

UV^{-1} = D^p. \tag{12}

Proposition 5

For every polynomial $f \in K[X]$ and every composition operator of order $p$ one has

f^{(p)}(X + Y) = \sum_{k=0}^{\infty} \frac{1}{k!} U \left( f^{(k)}(X) \right) u_k(Y) \tag{13}

(generalized Taylor expansion).

Indeed, if one puts $U = D^p V = V D^p$ one has (VI, p. 270, formula (1))

V_X^{-1}(f(X + Y)) = V_Y^{-1}(f(X + Y)) = \sum_{k=0}^{\infty} \frac{1}{k!} f^{(k)}(X) u_k(Y) \tag{14}

the Taylor formula and by def. 2 of VI, p. 273; it suffices to apply the operator

U

to the first and last terms of formula (14) to obtain (13).

3. Generating series for the appell polynomials

Let E be the ring of formal series in an indeterminate S, with coefficients in the ring of polynomials $K[X]$ (Alg., IV. 41); in other words, the ring of formal series $g(X, S) = \sum_{n=0}^{\infty} \alpha_n(X) S^n$ where the $\alpha_n$ belong to $K[X]$ . For every operator $U$ on $K[X]$ we defines a map $U_X$ of E to itself by putting $U_X(g(X, S)) = \sum_{n=0}^{\infty} U(\alpha_n) S^n$ . It is clear that E is a module over the ring $K[[S]]$ of formal series in S with coefficients in $K$ ; by the linearity of $U$ on $K[X]$ one verifies immediately that for every element $\theta \in K[[S]]$ and every $g \in E$ one has $U_X(\theta g) = \theta U_X(g)$ ; in other words, $U_X$ is a linear map of the module E into itself.

Proposition 6

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

U_X \left( \exp(XS) \right) = u(S) \exp(XS) \tag{15}

\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 $f(Y) \in K[Y]$ one has

U_X \left( \exp(XD_Y)(f(Y)) \right) = u(D_Y)(\exp(XD_Y)(f(Y))).

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

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

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 $S^p / u(S)$ and $\exp(XS)$ , taking account of (6).

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

4. Bernoulli polynomials

Consider the composition operator $U$ defined by

U(f(X)) = f(X + 1) - f(X);

one can write $U = e^D - 1$ (VI, p. 270, Example 1); this is an operator of order 1, and if one puts $U = DV$ one has $V^{-1} = \frac{D}{e^D - 1}$ . The Appell polynomial of degree $n$ corresponding to the operator $U$ is called the Bernoulli polynomial of degree $n$ and is denoted by $B_n(X)$ ; if one puts $b_n = B_n(0)$ one has the formulae

B_n(X) = \sum_{k=0}^{n} \binom{n}{k} b_{n-k} X^k

(18)

\frac{S \, e^{XS}}{e^S - 1} = \sum_{n=0}^{\infty} \frac{1}{n!} B_n(X) S^n

(19)

and in particular

\frac{S}{e^S - 1} = \sum_{n=0}^{\infty} \frac{1}{n!} b_n S^n

(20)

The formulae (7) and (9) of VI, p. 273, give, for the Bernoulli polynomials, the relations

\frac{dB_n}{dX} = n B_{n-1}(X)

(21)

B_n(X + 1) - B_n(X) = n X^{n-1}.

(22)

In particular, one has $B_n(1) - B_n(0) = 0$ for $n > 1$ , which, taking account of (18), gives the induction relation

\sum_{m=0}^{n-1} \binom{n}{m} b_m = 0 \quad \text{(for } n > 1)

which enables one to calculate the $b_n$ step-by-step. These numbers are clearly rational; since one can write

\frac{S}{e^S - 1} = -\frac{S}{2} + \frac{S}{2} \frac{e^S + 1}{e^S - 1}

and since (VI, p. 272, formula (5))

\frac{e^{-S} + 1}{e^{-S} - 1} = -\frac{e^S + 1}{e^S - 1}

one sees that in the formal series for $\frac{S}{2} \frac{e^S + 1}{e^S - 1}$ all the terms of odd degree have coefficient zero; so one has

b_0 = 1, \qquad b_1 = -\frac{1}{2}, \qquad b_{2n-1} = 0 \quad \text{for } n > 1.

The rational numbers $b_{2n}$ ( $n \geq 1$ ) are called the Bernoulli numbers; we shall see (VI, p. 288) that $b_{2n}$ has the sign of $(-1)^{n-1}$ . The formula (23) gives, for the first values of $n$ ,

\begin{align*} b_2 &= \frac{1}{6}, & b_4 &= -\frac{1}{30}, & b_6 &= \frac{1}{42}, & b_8 &= -\frac{1}{30}, \\ b_{10} &= \frac{5}{66}, & b_{12} &= -\frac{691}{2730}, & b_{14} &= \frac{7}{6}, \\ b_{16} &= -\frac{3617}{510}, & b_{18} &= \frac{43867}{798}, & b_{20} &= -\frac{174611}{330}, & b_{22} &= \frac{854513}{138}, \\ b_{24} &= -\frac{236364091}{2730}, & b_{26} &= \frac{8553103}{6}, & b_{28} &= -\frac{23749461029}{870}. \end{align*}

Note that the numerators 691, 3617, 43867 are prime; the prime factorisations of the others are

\begin{align*} 174611 &= 283 \times 617 \\ 854513 &= 11 \times 131 \times 593 \\ 236364091 &= 103 \times 2294797 \\ 8553103 &= 13 \times 657931 \\ 23749461029 &= 7 \times 9349 \times 362903. \end{align*}

From these we deduce expressions for the first Bernoulli polynomials

\begin{align*} B_0(X) &= 1, & B_1(X) &= X - \frac{1}{2}, & B_2(X) &= X^2 - X + \frac{1}{6}, \\ B_3(X) &= X^3 - \frac{3}{2}X^2 + \frac{1}{2}X, & B_4(X) &= X^4 - 2X^3 + X^2 - \frac{1}{30}. \end{align*}

5. Composition operators on functions of a real variable

Let I be an interval in $\mathbf{R}$ containing the interval $\mathbf{R}_+ = [0, +\infty[$ ; let E be a vector space over the field $\mathbf{C}$ , formed of functions of a real variable with complex values, defined on I. We suppose that for every $a \geqslant 0$ and every function $f \in E$ the function $x \mapsto f(x + a)$ belongs to E; further, we assume that E contains the restrictions to I of the polynomials with complex coefficients and the exponentials $e^{\lambda x}$ , where $\lambda$ is an arbitrary complex number. Any linear map $U$ of E into the space of all maps from I into the field $\mathbf{C}$ of complex numbers will be called an operator on E; if $f \in E$ and $g = U(f)$ it will be convenient to use the notation

g(x) = U_{\xi}^{\xi}(f(\xi))

where $\xi$ is a dummy variable in the functional symbolism of the right-hand term (cf. II, p. 58). For every $a \geqslant 0$ the operator which to any function $f \in E$ associates the restriction to I of the function $x \mapsto f(x + a)$ is called the translation operator by a.

Definition 3

One says that an operator U on E is a composition operator if, for every $a \geqslant 0$ , it is permutable with the translation operator by a*.

In the notation introduced above, this definition becomes the identity

U_{x+a}^{\xi}(f(\xi)) = U_x^{\xi}(f(\xi + a))

in $x$ and $a$ ( $x \in I, a \geqslant 0$ ). One can exchange the rôles of $x$ and $a$ in this identity if $x \geqslant 0$ , then put $a = 0$ ; one thus obtains for $x \geqslant 0$

U_x^{\xi}(f(\xi)) = U_0^{\xi}(f(\xi + x))

where $U_0$ is the linear form on E which to each function $f \in E$ associates the value $g(0)$ of $g = U(f)$ .

If $f$ is a polynomial, one has $f(\xi + x) = \sum_{k=0}^{\infty} \frac{1}{k!} f^{(k)}(\xi) x^k$ , and formula (26) shows that $U(f)$ is also a polynomial; restricted to the set of polynomials in $x$ , with coefficients in $\mathbf{C}$ , (a set which one can identify with the algebra $\mathbf{C}[X]$ ), the operator $U$ is then a composition operator in the sense of def. I of VI, p. 270, and all the results of the preceding sections can be applied to it.

We again write $u_n$ for the Appell polynomials attached to the operator $U$ . To the generalized Taylor expansion of a polynomial (VI, p. 274, formula (13)) there corresponds, for more general functions, the following result:

Theorem 2

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).

6. Indicatrix of a composition operator

With the same hypotheses as in n° 5, the formula (26) of (VI, p. 277) applied to the function $e^{\lambda x}$ , gives

U_\lambda^\xi (e^{\lambda \xi}) = U_0^\xi (e^{\lambda x} e^{\lambda \xi}) = e^{\lambda x} U_0^\xi (e^{\lambda \xi}) = u(\lambda) e^{\lambda x}

(30) on putting

u(\lambda) = U_0^\xi (e^{\lambda \xi})

. One says that the function

u(\lambda)

, defined on

\mathbf{C}

, with complex values, is the indicatrix of the composition operator

U

. Note that if the restriction of

U

to the ring

\mathbf{C}[X]

of polynomials is equal to the series

D^p \sum_{n=0}^{\infty} \alpha_n D^n

(VI, p. 271, th. 1) (which we have denoted

u(D)

in VI, p. 271), the series of complex terms with general term

\alpha_n \lambda^{n+p}

is not necessarily convergent for

\lambda \neq 0

, and that even if it converges for certain values of

\lambda

, the sum is not necessarily equal to the indicatrix

u(\lambda)

of U (VI, p. 291, exerc. 2). We shall say that the composition operator

U

is regular if there exists a neighbourhood of 0 in

\mathbf{C}

such that the series with general term

\alpha_n \lambda^{n+p}

is absolutely convergent and has sum equal to the indicatrix

u(\lambda)

on this neighbourhood

^1

. Let us apply formula (27) of VI, p. 278, to the function

e^{\lambda x}

, making

h = 0

; since

D^m(e^{\lambda x}) = \lambda^m e^{\lambda x}

one has

U_0^\xi \left( D^m(e^{\lambda x}) \right) = \lambda^m u(\lambda)

; it then follows that for every complex

\lambda

such that

u(\lambda) \neq 0

\frac{\lambda^p e^{\lambda x}}{u(\lambda)} = \sum_{m=0}^n u_m(x) \frac{\lambda^m}{m!} - \frac{\lambda^{n+1}}{u(\lambda)} U_0^\xi \left( \int_0^{\xi-x} \frac{1}{n!} u_n(x+\eta) e^{\lambda(\xi-\eta)} d\eta \right)

and in particular, for $x = 0$

\frac{\lambda^p}{u(\lambda)} = \sum_{m=0}^n \beta_m \frac{\lambda^m}{m!} - \frac{\lambda^{n+1}}{u(\lambda)} U_0^\xi \left( \int_0^{\xi} \frac{1}{n!} u_n(\eta) e^{\lambda(\xi-\eta)} d\eta \right)

with $\beta_m = u_m(0)$ .

If $U$ is a regular operator, for all $\lambda \in \mathbf{C}$ such that the entire series $u(\lambda) = \sum_{n=0}^\infty \alpha_n \lambda^{n+p}$ and $\sum_{n=0}^\infty \beta_n \frac{\lambda^n}{n!}$ are absolutely convergent $^2$ , it follows from formula (16) and the formula for the product of two absolutely convergent series (Gen. Top., VIII, p. 115, prop. 1) that one has

\frac{\lambda^p}{u(\lambda)} = \sum_{n=0}^\infty \beta_n \frac{\lambda^n}{n!}.

Similarly, since the Taylor expansion of $e^{\lambda x}$ is absolutely convergent for every $\lambda \in \mathbf{C}$ and every $x \in \mathbf{C}$ (III, p. 106) one also has (formulae (6) (VI, p. 273) and (16) (VI, p. 274)), for all the values considered and for all $x \in \mathbf{C}$

\frac{\lambda^n e^{\lambda x}}{u(\lambda)} = \sum_{n=0}^\infty u_n(x) \frac{\lambda^n}{n!}.

Remark

One can use formula (33) (resp. (34)) to calculate the $\beta_n$ (resp. the $u_n(x)$ ) by using the following lemma on entire series:

(1) Later we shall study series whose general terms are of the form $c_n z^n$ ( $c_n \in \mathbf{C}, z \in \mathbf{C}$ ), which one calls entire series; in particular we shall see that when such a series is absolutely convergent for $z = z_0$ it is normally convergent for $|z| \leq |z_0|$ . (2) It follows from the theory of entire series that when one of these series is absolutely convergent in a neighbourhood $V$ of 0, the other is absolutely convergent in a neighbourhood $W \subset V$ of 0.

emma. If two entire series $\sum_{n=0}^{\infty} c_n \lambda^n,\ \sum_{n=0}^{\infty} d_n \lambda^n$ are absolutely convergent for all $\lambda$ in a neighbourhood of $0$ , and if $\sum_{n=0}^{\infty} c_n \lambda^n = \sum_{n=0}^{\infty} d_n \lambda^n$ for these values of $\lambda$ , then $c_n = d_n$ for all integers $n \geqslant 0$ .

If, by any procedure, one can find an entire series that converges to $\lambda^p / u(\lambda)$ on a neighbourhood of $0$ , then the coefficients of this series are necessarily equal to the $\beta_n / n!$ . Is this procedure that we shall apply in the examples that follow.

Example 1

If $U$ is the identity map one has $u(\lambda) = 1$ and the operator $U$ is clearly singular; since $u_n(x) = x^n$ the formula (27) of VI, p. 278, can be written, putting $\xi = \xi - \eta$

f(x + h) = \sum_{n=0}^{\infty} \frac{1}{m!} f^{(m)}(h) x^m + \int_h^{x+h} f^{(n+1)}(t) \frac{(x + h - t)^n}{n!} dt

that is, it reduces to the Taylor formula (II, p. 62).

Example 2

Let us take for $U$ the composition operator which, to every function $f$ defined on $\mathbf{R}_+$ , associates the function $x \mapsto f(x + 1) - f(x)$ ; then

U_{\lambda}^{\xi} (f(\xi)) = f(x + 1) - f(x);

we have seen (VI, p. 275) that the restriction of $U$ to $\mathbf{C}[X]$ is equal to $e^D - 1$ . Since, on the other hand, $u(\lambda) = e^{\lambda} - 1$ , the operator $U$ is regular; we shall see (VI, p. 288) how to determine the Bernoulli numbers $b_n$ by calculating an expansion of $\frac{\lambda}{e^{\lambda} - 1}$ as a convergent entire series. On applying formula (27) of VI, p. 278 to a primitive of a function $f$ admitting a continuous $n^{th}$ derivative on $\mathbf{R}_+$ , we obtain

f(x + h) = \int_h^{h+1} f(t) dt

+ \sum_{m=1}^n \frac{1}{m!} B_m(x) \left( f^{(m-1)}(h+1) - f^{(m-1)}(h) \right) + R_n(x, h)

with

This lemma is a particular case of a general result that we will prove later; here is the proof. If an entire series $\sum_{n=0}^{\infty} c_n \lambda^n$ is absolutely convergent for $\lambda = \lambda_0$ then for every integer $k \geqslant 0$ the series $\sum_{n=0}^{\infty} c_{n+k} \lambda^n$ is normally convergent for $|\lambda| \leqslant |\lambda_0|$ , so is continuous on this disc (Gen. Top., X, p. 283); one concludes that $\sum_{n=k+1}^{\infty} c_n \lambda^n = o(\lambda^k)$ on a neighbourhood of $0$ . The lemma then follows from the uniqueness of the coefficients of an asymptotic expansion of a function in terms of the $\lambda^n$ (V, p. 223).

R_n(x, h) = - \int_0^{1-x} \frac{B_n(x + \eta)}{n!} f^{(n)}(h + 1 - \eta) d\eta \\ + \int_0^{-x} \frac{B_n(x + \eta)}{n!} f^{(n)}(h - \eta) d\eta.

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Example 3

Let E be the vector space of functions $f$ defined and continuous on $\mathbf{R}$ , and such that the integral $\int_{-\infty}^{+\infty} f(x + \xi) e^{-\xi^2/2} d\xi$ converges for all $x \geq 0$ . The operator U defined by

U_\lambda^\xi \left( f(\xi) \right) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\xi^2/2} f(x + \xi) d\xi

is then defined on E and is clearly a composition operator. The space E contains all the exponentials $e^{\lambda x}$ ( $\lambda$ arbitrary complex), and one has

u(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-(\xi^2/2) + \lambda \xi} d\xi \\ = \frac{1}{\sqrt{2\pi}} e^{\lambda^2/2} \int_{-\infty}^{+\infty} e^{-(\xi - \lambda)^2/2} d\xi = e^{\lambda^2/2}

(cf. III, p. 120, exerc. 24, and VII, p. 313, formula (22)). One has $n! \alpha_n = U_0^\xi (\xi^n) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\xi^2/2} \xi^n d\xi$ . For every integer $n$ one can write

\sum_{k=0}^n \int_{-\infty}^{+\infty} \frac{|\lambda \xi|^k}{k!} e^{-\xi^2/2} d\xi \leq 2 \int_0^{+\infty} e^{-(\xi^2/2) + |\lambda| \xi} d\xi.

The series $\sum_{n=0}^{\infty} e^{-\xi^2/2} \frac{(\lambda \xi)^n}{n!}$ can therefore be integrated term-by-term over $\mathbf{R}$ (II, p. 72, cor. 1), which proves that the series $\sum_{n=0}^{\infty} \alpha_n \lambda^n$ converges absolutely for every $\lambda \in \mathbf{C}$ , and has a sum equal to $u(\lambda) = e^{\lambda^2/2} = \sum_{n=0}^{\infty} \frac{\lambda^{2n}}{2^n n!}$ ; thus the operator $U$ is regular. Applying the lemma mentioned above shows that $\alpha_{2n} = 1/2^n n!,\ \alpha_{2n+1} = 0$ for every $n \geq 0$ ; the operator $U$ is thus of order 0. One has

\frac{1}{u(\lambda)} = e^{-\lambda^2/2} = \sum_{n=0}^{\infty} \frac{(-1)^n \lambda^{2n}}{2^n n!},

the series being absolutely convergent for every $\lambda \in \mathbf{C}$ ; another application of the lemma shows that $\beta_{2n} = \frac{(-1)^n (2n)!}{2^n n!},\ \beta_{2n+1} = 0$ ; further, the series $\sum_{n=0}^{\infty} \frac{\lambda^n}{n!} u_n(x)$ is absolutely convergent for every $\lambda \in \mathbf{C}$ and every $x \in \mathbf{R}$ , and one has

\sum_{n=0}^{\infty} \frac{\lambda^n}{n!} u_n(x) = \exp \left( -\frac{\lambda^2}{2} + \lambda x \right) = \exp \left( \frac{x^2}{2} \right) \exp \left( -\frac{1}{2} (\lambda - x)^2 \right).

On applying the Taylor formula to the function $\exp(-x^2/2)$ one then obtains the following expression for the polynomials $u_n(x)$ :

u_n(x) = (-1)^n e^{\lambda^2/2} \frac{d^n}{dx^n} (e^{-x^2/2}).

This polynomial is called the Hermite polynomial of degree $n$ , and is often denoted by $\mathrm{H}_n(x)$ . The formulae (7), (8) and (9) of VI, p. 273, here give

\frac{d \mathrm{H}_n}{dx} = n \mathrm{H}_{n-1}(x)

\mathrm{H}_n(x+y) = \sum_{k=0}^n \binom{n}{k} \mathrm{H}_{n-k}(x) y^k

\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\xi^2/2} \mathrm{H}_n(x+\xi) d\xi = x^n

and the formula (27) of VI, p. 278, becomes, for $h = 0$

\sqrt{2\pi} f(x) = \sum_{m=0}^n \left( \int_{-\infty}^{+\infty} e^{\xi^2/2} f^{(m)}(\xi) d\xi \right) \frac{\mathrm{H}_m(x)}{m!}

- \int_{-\infty}^{+\infty} d\xi \int_0^\xi \frac{\mathrm{H}_n(x+\eta)}{n!} e^{-(\xi+\eta)^2/2} f^{(n+1)}(x+\xi-\eta) d\eta.

7. The euler-maclaurin summation formula

In the formula (35) of VI, p. 280, let us replace $x$ by 0 and $h$ by $x$ ; since $\mathrm{B}_m(0) = b_m$ it follows from the relations (24) of VI, p. 276, that for each integer $p > 0$ one can write

f(x) = \int_x^{x+1} f(t) dt - \frac{1}{2} (f(x+1) - f(x))

+ \sum_{k=1}^p \frac{b_{2k}}{(2k)!} \left( f^{(2k-1)}(x+1) - f^{(2k-1)}(x) \right) + R_p(x)

with

R_p(x) = - \frac{1}{(2p+1)!} \int_0^1 \mathrm{B}_{2p+1}(t) f^{(2p+1)}(x+1-t) dt.

In this formula let us successively replace

x

x+1, x+2, \ldots, x+n

, and combine the formulae obtained, one by one; we obtain

\begin{cases} f(x) + f(x+1) + \cdots + f(x+n) \\ = \int_x^{x+n+1} f(t) dt - \frac{1}{2} (f(x+n+1) - f(x)) \\ + \sum_{k=1}^p \frac{b_{2k}}{(2k)!} \left( f^{(2k-1)}(x+n+1) - f^{(2k-1)}(x) \right) + T_p(x, n) \end{cases}

with

T_p(x, n) = - \frac{1}{(2p+1)!} \int_0^1 \mathrm{B}_{2p+1}(t) \left( \sum_{k=0}^n f^{(2p+1)}(x+k+1-t) \right) dt.

The remainder $T_p(x, n)$ in this formula can be rewritten in the following way: denote by $\overline{B}_{2p+1}(t)$ the periodic function with period 1 which is equal to $B_{2p+1}(t)$ on the interval $[0, 1[$ . Then

\int_0^1 B_{2p+1}(t) f^{(2p+1)}(x + k + 1 - t) \, dt = \int_k^{k+1} \overline{B}_{2p+1}(1 - s) f^{(2p+1)}(x + s) \, ds

and consequently

T_p(x, n) = - \frac{1}{(2p+1)!} \int_0^{n+1} \overline{B}_{2p+1}(1 - s) f^{(2p+1)}(x + s) ds.

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The formula (39) is called the Euler-Maclaurin summation formula; it is applicable to every complex function having a continuous $(2p + 1)^{th}$ derivative on an interval $[x_0, +\infty[$ , for every $x \geq x_0$ . We shall see (VI, p. 288) how to estimate the remainder $T_p(x, n)$ in this formula.

Exercises

See the exercises for § 1.