§2. Series expansions
371
The proof is strictly the same; the only difference is that now x = z can
be complex, so that in (7) the number h = z/n tends to 0 through complex
and not real values, which explains the necessity of giving the concept of
derivative the meaning which enabled us to show in Chap. II, nO 19 that a
power series is always differentiable.
We again emphasise the fact that, in contrast to the characterisation
of exponential functions given in nO 6, Theorem 2, Corollary 1 applies to
functions with complex values; the usefulness of this generalisation will appear
clearly in the following nO d propos trigonometric functions.
Corollary 2, for its part, yields a fourth proof of Newton's binomial formula in the general case. Since we know that Ns+t(z) = Ns(z)Nt(z) for s, t
and z complex, we have to apply the corollary to the function
(13.9)
e(s) = 1 + sz + s(s - l)z2/2! + ... = Ns(z)
where z is given, of course with Izl < 1. It reduces to proving that the function
(9) has a derivative in the complex sense at s = o. Now
(13.10) e(s) - e(O) = z + (s _ l)z2/2! + (s - 1)(s - 2)z3/3! + ... ;
s
for z given, Izl < 1, the right hand side, considered as a function of s, is a
normally convergent series in all the disc lsi < R as we have seen in nO 11
(second proof) for the binomial series itself. Its terms are continuous functions
of s. So, similarly, is its sum, so that, when SEC tends to 0, the left hand
side tends to the value of the right hand side at s = 0, namely
(13.11)
We can therefore apply Corollary 2, so obtaining Theorem 8 bis again:
(13.12)
exp [s(z - z2/2 + z3/3 - ... )] =
= Ns(z) = L s(s - 1) ... (s - n + l)zn In!.
It is not surprising that, two hundred years after Newton, one of the
more famous of Conan Doyle's characters should be considered as an eminent mathematician by virtue of his profound works on "Newton's binomial".
Let us return to Corollary 2 and scrutinize its hypothesis. Consider, in a
general manner, a function e on C satisfying e(s + t) = e(s)e(t) for all s, t.
For z = x + iy with x and y real, one then has e(z) = e(x)e(iy). It is clear
that each offunctions x >--+ e(x) and y >--+ e(iy) satisfies the eternal functional
equation on JR. If we then assume e continuous, the least we can do, then
Corollary 1 shows that e(x + iy) = exp(ax)exp(by) = exp(ax + by) with
complex constants a and b, and it is clear that every function of this type
satisfies the functional equation on C.
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