262
III - Convergence: Continuous variables
Corollary 1. (i) Let f be a complex-valued function defined and differentiable on an interval I; if f' (x) = 0 for all x E I, then f is constant on I.
(ii) Let f and 9 be functions with complex values defined and differentiable
on an interval I; if f'(x) = g'(x) for all x E I, then f(x) = g(x) + C where
C is a constant.
(i) is obvious from (1) in the case of a real-valued function; the complex
case reduces to it by considering the real and imaginary parts of f. (ii) follows
from applying (i) to f - g.
Example 1. From (14.15) or Theorem 3 of Chap. II, n° 10 we know that
log' x = l/x, so that, on putting f(x) = 10g(1 + x),
(16.2)
f'(x) = 1/(1 + x) = 1 - x + x 2 - x 3 + ...
if Ixl < 1. Consider its primitive series (Chap. II, nO 19)
(16.3)
which converges in the same disc. By Chap. II, nO 19, we know that 9 is
differentiable and that g'(x) is the derived series of (3), i.e. precisely (2).
Thus g'(x) = 10g'(1 + x), whence 10g(1 + x) = g(x) since the two sides are
obviously zero for x = O. So we obtain the formula
(16.4)
10g(1 + x) = x - x 2 /2 + x 3 /3 - x4 /4 + ... ,
valid for -1 < x ::; 1 (see nO·S, example 4). This proof depends on a manifest
subterfuge: confusing the derived series of a power series, in the purely formal
sense introduced in Chap. II, nO 19, with the derived function of its sum,
defined by passing to the limit. But we showed in Chap. II that, in the
case of convergent power series, these two senses of the word "derived" were
identical in the interior of the disc of convergence and, in particular, of the
interval of convergence in lR.
Example 2. Let us seek the functions f (x) defined and differentiable on an
interval I and proportional to their derivative:
f'(x) = ef(x)
where e E C is a given constant. The series exp x = L: x[n] trivially satisfies
the relation exp' x = exp x, so that the function exp( ex) satisfies the condition
imposed on f. Since the exponential function never vanishes the function
g(x) = f(x)/ exp(ex) = f(x) exp( -ex) is differentiable on I. We have
g'(x) = f'(x)exp(-ex) - ef(x)exp(-ex) = O.
In consequence 9 is constant, from which we see that the only solutions of
the problem are the constant multiples of exp(ex).
III - Convergence: Continuous variables
Corollary 1. (i) Let f be a complex-valued function defined and differentiable on an interval I; if f' (x) = 0 for all x E I, then f is constant on I.
(ii) Let f and 9 be functions with complex values defined and differentiable
on an interval I; if f'(x) = g'(x) for all x E I, then f(x) = g(x) + C where
C is a constant.
(i) is obvious from (1) in the case of a real-valued function; the complex
case reduces to it by considering the real and imaginary parts of f. (ii) follows
from applying (i) to f - g.
Example 1. From (14.15) or Theorem 3 of Chap. II, n° 10 we know that
log' x = l/x, so that, on putting f(x) = 10g(1 + x),
(16.2)
f'(x) = 1/(1 + x) = 1 - x + x 2 - x 3 + ...
if Ixl < 1. Consider its primitive series (Chap. II, nO 19)
(16.3)
which converges in the same disc. By Chap. II, nO 19, we know that 9 is
differentiable and that g'(x) is the derived series of (3), i.e. precisely (2).
Thus g'(x) = 10g'(1 + x), whence 10g(1 + x) = g(x) since the two sides are
obviously zero for x = O. So we obtain the formula
(16.4)
10g(1 + x) = x - x 2 /2 + x 3 /3 - x4 /4 + ... ,
valid for -1 < x ::; 1 (see nO·S, example 4). This proof depends on a manifest
subterfuge: confusing the derived series of a power series, in the purely formal
sense introduced in Chap. II, nO 19, with the derived function of its sum,
defined by passing to the limit. But we showed in Chap. II that, in the
case of convergent power series, these two senses of the word "derived" were
identical in the interior of the disc of convergence and, in particular, of the
interval of convergence in lR.
Example 2. Let us seek the functions f (x) defined and differentiable on an
interval I and proportional to their derivative:
f'(x) = ef(x)
where e E C is a given constant. The series exp x = L: x[n] trivially satisfies
the relation exp' x = exp x, so that the function exp( ex) satisfies the condition
imposed on f. Since the exponential function never vanishes the function
g(x) = f(x)/ exp(ex) = f(x) exp( -ex) is differentiable on I. We have
g'(x) = f'(x)exp(-ex) - ef(x)exp(-ex) = O.
In consequence 9 is constant, from which we see that the only solutions of
the problem are the constant multiples of exp(ex).