§ 3. Derivatives of higher order

1. Derivatives of order $n$

Let $f$ be a vector function of a real variable, defined, continuous and differentiable on an interval $I$ . If the derivative $f'$ exists on a neighbourhood (with respect to $I$ ) of a point $x_0 \in I$ , and is differentiable at the point $x_0$ , then its derivative is called the second derivative of $f$ at the point $x_0$ , and is denoted by $f''(x_0)$ or $D^2f(x_0)$ . If this second derivative exists at every point of $I$ (which implies that $f'$ exists and is continuous on $I$ ), then $x \mapsto f''(x)$ is a vector function which one denotes by $f''$ or $D^2f$ . We define, in the same way, recursively, the $n^{th}$ derivative (or derivative of order $n$ ) of $f$ , and denote it by $f^{(n)}$ or $D^n f$ ; by definition, its value at the point $x_0 \in I$ is the derivative of the function $f^{(n-1)}$ at the point $x_0$ : this definition presupposes the existence of all the derivatives $f^{(k)}$ of order $k \leq n - 1$ on a neighbourhood of $x_0$ relative to $I$ , and the differentiability of $f^{(n-1)}$ at the point $x_0$ .

We will say that $f$ is n times differentiable at the point $x_0$ (resp. in an interval) if it admits an $n^{th}$ derivative at this point (resp. in this interval). One says that $f$ is indefinitely differentiable on $I$ if for each integer $n > 0$ it admits a derivative of order $n$ on $I$ .

By induction on $m$ one sees that

D^m(D^n f) = D^{m+n} f.

More precisely, when one of the two terms in (1) is defined, then so is the other, and is equal to it.

Proposition 1

The set of vector functions defined on an interval* $I \subset \mathbf{R}$ , taking values in a given topological vector space $E$ , and having an $n^{th}$ derivative on $I$ , is a vector space over $\mathbf{R}$ , and $f \mapsto D^n f$ is a linear mapping of this space into the vector space of linear mappings from $I$ into $E$ .

One proves the formulae

D^n(f + g) = D^n f + D^n g \tag{2}

D^n(fa) = D^n f . a \tag{3}

by induction on $n$ when $f$ and $g$ have an $n^{th}$ derivative on $I$ ( $a$ being constant).

Proposition 2 ("Leibniz' formula")

Let* $E, F, G$ be three topological vector spaces over $\mathbf{R}$ , and $(x, y) \mapsto [x.y]$ a continuous bilinear mapping of $E \times F$ into $G$ . If $f$ (resp. $g$ ) is defined on an interval $I \subset \mathbf{R}$ , takes its values in $E$ (resp. $F$ ) and has an $n^{th}$ derivative on $I$ , then $[f.g]$ has an $n^{th}$ derivative on $I$ , *given by the formula

D^n[f.g] = [f^{(n)}.g] + \binom{n}{1}[f^{(n-1)}.g'] + \cdots + \binom{n}{p}[f^{(n-p)}.g^{(p)}] + \cdots + [f.g^{(n)}]. \tag{4}

Formula (4) is proved by induction on $n$ (using the relation $\binom{n}{p} = \binom{n-1}{p} + \binom{n-1}{p-1}$ for the binomial coefficients).

In the same way one can verify the following formula (where the hypotheses are the same as in prop. 2):

[f^{(n)} . g] + (-1)^{n-1} [f . g^{(n)}] = D([f^{(n-1)} . g] - [f^{(n-2)} . g'] + \cdots + (-1)^{n-1} [f . g^{(n-1)}]).

The preceding propositions have been stated for functions that are $n$ times differentiable on an interval; we leave it to the reader to formulate the analogous propositions for functions that are $n$ times differentiable at a point.

2. Taylor's formula

Let $f$ be a vector function defined on an interval $I \subset \mathbf{R}$ , with values in a normed space $E$ over $\mathbf{R}$ ; to say that $f$ has a derivative at a point $a \in I$ signifies that

\lim_{x \to a,\ x \in I,\ x \neq a} \frac{f(x) - f(a) - f'(a)(x-a)}{x-a} = 0;

or, otherwise, that $f$ is "approximately equal" to the linear function $f(a) + f'(a)(x-a)$ on a neighbourhood of $a$ (cf. chap. V, where this concept is developed in a general manner). We shall see that the existence of the $n^{th}$ order derivative of $f$ at the point $a$ entails in the same way that $f$ is "approximately equal" to a polynomial of degree $n$ in $x$ , with coefficients in $E$ (Gen. Top., X, p. 315) on a neighbourhood of $a$ . To be precise:

Theorem 1

If the function $f$ has an $n^{th}$ derivative at the point $a$ then

\lim_{x \to a,\ x \in I,\ x \neq a} \frac{f(x) - f(a) - f'(a)\frac{(x-a)}{1!} - \cdots - f^{(n)}(a)\frac{(x-a)^n}{n!}}{(x-a)^n} = 0.

We proceed by induction on $n$ . The theorem holds for $n = 1$ . For arbitrary $n$ one can, by the induction hypothesis, apply it to the derivative $f'$ of $f$ : for any $\varepsilon > 0$ there is an $h > 0$ such that, if one puts

g(x) = f(x) - f(a) - f'(a)\frac{(x-a)}{1!} - f''(a)\frac{(x-a)^2}{2!} - \cdots - f^{(n)}(a)\frac{(x-a)^n}{n!}

one has, for $|y-a| \leq h$ and $y \in I$ ,

\|g'(y)\| = \left\| f'(y) - f'(a) - f''(a)\frac{(y-a)}{1!} - \cdots - f^{(n)}(a)\frac{(y-a)^{n-1}}{(n-1)!} \right\|

\leq \varepsilon\ |y-a|^{n-1}.

We apply the mean value theorem (I, p. 15, th. 2) on the interval with endpoints $a,\ x$ (with $|x-a| \leq h$ ) to the vector function $g$ and to the real increasing function equal to $\varepsilon |y - a|^n / n$ if $x > a$ , and to $-\varepsilon |y - a|^n / n$ if $x < a$ ; it follows that $\|g(x)\| \leq \varepsilon |x - a|^n / n$ , which proves the theorem.

We thus can write

f(x) = f(a) + f'(a) \frac{(x - a)}{1!} + f''(a) \frac{(x - a)^2}{2!} + \cdots \\ + f^{(n)}(a) \frac{(x - a)^n}{n!} + u(x) \frac{(x - a)^n}{n!}

where

u(x)

approaches 0 as

x

approaches

a

while remaining in I; this formula is called Taylor’s formula of order

n

at the point

a

, and the right-hand side of (8) is called the Taylor expansion of order

n

of the function

f

at the point

a

. The last term

r_n(x) = u(x)(x - a)^n / n!

is called the remainder in the Taylor formula of order

n

When $f$ has a derivative of order $n + 1$ on I, one can estimate $\|r_n(x)\|$ in terms of this $n + 1^{th}$ derivative, on all of I, and not just on an unspecified neighbourhood of $a$ :

Proposition 3

If $\|f^{(n+1)}(x)\| \leq M$ on I, then we have*

\|r_n(x)\| \leq M \frac{|x - a|^{n+1}}{(n + 1)!}

on I.

Indeed, the formula holds for $n = 0$ , by I, p. 15, th. 2. Let us prove it by induction on $n$ : by the induction hypothesis applied to $f'$ , one has

\|r'_n(y)\| \leq M \frac{|y - a|^n}{n!}

from which the formula (9) follows by the mean value theorem (I, p. 23, th. 2).

Corollary

If $f$ is a finite real function with a derivative of order* $n + 1$ on I, and if $m \leq f^{(n+1)}(x) \leq M$ on I, then for all $x \geq a$ in I one has

m \frac{(x - a)^{n+1}}{(n + 1)!} \leq r_n(x) \leq M \frac{(x - a)^{n+1}}{(n + 1)!}

and the second term cannot be equal to the first (resp. to the third) unless

f^{(n+1)}

is constant and equal to

m

(resp.

M

) on the interval

[a, x]

The proof proceeds in the same way, but applying th. 1 of I, p. 14.

Remark 1

We have already noticed in the proof of th. 1 that if $f$ has a derivative of order $n$ on I, and if

f(x) = a_0 + a_1(x - a) + a_2(x - a)^2 + \cdots + a_n(x - a)^n + r_n(x)

is its Taylor expansion of order

n

at the point

a

, then the Taylor expansion of order

n - 1

for

f'

at the point

a

f'(x) = a_1 + 2a_2(x - a) + \cdots + na_n(x - a)^{n-1} + r'_n(x).

We say that it is obtained from the expansion (11) of $f$ by differentiating term-by-term.

Remark 2

With the same hypotheses, the coefficients $a_i$ in (11) are determined recursively by the relations

a_0 = f(a)

a_1 = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

a_2 = \lim_{x \to a} \frac{f(x) - f(a) - a_1(x - a)}{(x - a)^2}

\ldots

a_n = \lim_{x \to a} \frac{f(x) - f(a) - a_1(x - a) - \cdots - a_{n-1}(x - a)^{n-1}}{(x - a)^n}.

In the case $a = 0$ one concludes, in particular, that if $f(x^p)$ ( $p$ an integer > 0) has a derivative of order $pn$ on a neighbourhood of 0 then the Taylor expansion of order $pn$ of this function is simply

f(x^p) = a_0 + a_1 x^p + a_2 x^{2p} + \cdots + a_n x^{np} + r_n(x^p)

where $r_n(x^p)$ is the remainder in the expansion (cf. V, p. 222).

Remark 3

The definition of the derivative of order $n$ and the preceding results generalize immediately to functions of a complex variable; we shall not pursue this topic further here; it will be treated in detail in a later Book in this Series.

Exercises

See the exercises for § 3.