372
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
For the function exp( ax + by) to be of the form exp( cz), it is necessary
and sufficient that b = ia. But if one differentiates the function exp(ax + by)
with respect to x (resp. y), one finds it again, multiplied by a (resp. b);
the relation b = ia is thus, in this case, just Cauchy's relation f~ = if~ of
Chap. II, nO 19. As in the case of the functions exp(cz), we have to impose
the same condition on e(z). But, if e(z) is differentiable in the complex sense
at z = 0, the formula e(z+h) = e(z)e(h) shows that it is so everywhere, with
e'(z) = e'(O)e(z), which proves furthermore that e' is continuous. In other
words, Corollary 2 characterises the holomorphic solutions of our functional
equation.
14 - Imaginary exponentials and trigonometric functions
Let us return to the function exp(z) for z E C. We said in Chap. II, nO 14,
example 2, that the resemblance between the exponential series and those
which represent the functions sin x and cos x led Euler to note that
(14.1')
(14.1")
exp(ix)
exp(-ix)
cos x + isinx
cos x - isinx
for x E JR. The proof of these formulae takes a few lines but presupposes
the power series for the trigonometric functions and even the definition of
these latter; but no considerations of elementary geometry will provide us
with such recondite formulae. Try for example to understand why
This is obvious if you know that the left hand side represents 1 - cos 7r, but
what if you only have sketches on a sheet of paper?
Corollary 1 of Theorem 11 provides a more economical "proof" of (1) because it relies on much more elementary properties of trigonometric functions
than their mysterious series expansions, which will follow from them.
Theorem 12. Let c(x) and s(x) be two real valued functions defined on JR.,
possessing the following properties: (i) they satisfy the addition formulae
of the functions cosx and sinx; (ii) they are differentiable at x = 0, with
c'(0) = 0 and s'(O) = 1. Then
(14.2)
and consequently
(14.3')
(14.3")
c(x)
s(x)
c(x) + is(x) = exp(ix)
1 - X[2] + x[4] - ••• ,
x - x[3] + X[5] - ...
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
For the function exp( ax + by) to be of the form exp( cz), it is necessary
and sufficient that b = ia. But if one differentiates the function exp(ax + by)
with respect to x (resp. y), one finds it again, multiplied by a (resp. b);
the relation b = ia is thus, in this case, just Cauchy's relation f~ = if~ of
Chap. II, nO 19. As in the case of the functions exp(cz), we have to impose
the same condition on e(z). But, if e(z) is differentiable in the complex sense
at z = 0, the formula e(z+h) = e(z)e(h) shows that it is so everywhere, with
e'(z) = e'(O)e(z), which proves furthermore that e' is continuous. In other
words, Corollary 2 characterises the holomorphic solutions of our functional
equation.
14 - Imaginary exponentials and trigonometric functions
Let us return to the function exp(z) for z E C. We said in Chap. II, nO 14,
example 2, that the resemblance between the exponential series and those
which represent the functions sin x and cos x led Euler to note that
(14.1')
(14.1")
exp(ix)
exp(-ix)
cos x + isinx
cos x - isinx
for x E JR. The proof of these formulae takes a few lines but presupposes
the power series for the trigonometric functions and even the definition of
these latter; but no considerations of elementary geometry will provide us
with such recondite formulae. Try for example to understand why
This is obvious if you know that the left hand side represents 1 - cos 7r, but
what if you only have sketches on a sheet of paper?
Corollary 1 of Theorem 11 provides a more economical "proof" of (1) because it relies on much more elementary properties of trigonometric functions
than their mysterious series expansions, which will follow from them.
Theorem 12. Let c(x) and s(x) be two real valued functions defined on JR.,
possessing the following properties: (i) they satisfy the addition formulae
of the functions cosx and sinx; (ii) they are differentiable at x = 0, with
c'(0) = 0 and s'(O) = 1. Then
(14.2)
and consequently
(14.3')
(14.3")
c(x)
s(x)
c(x) + is(x) = exp(ix)
1 - X[2] + x[4] - ••• ,
x - x[3] + X[5] - ...